id,subfield,context,question,solution,final_answer,is_multiple_answer,unit,answer_type,error 800,Mechanics,,"A quarantined physics student decides to perform an experiment to land a small box of mass $m=60 \mathrm{~g}$ onto the center of a target a distance $\Delta d$ away. The student puts the box on a top of a frictionless ramp with height $h_{2}=0.5 \mathrm{~m}$ that is angled $\theta=30^{\circ}$ to the horizontal on a table that is $h_{1}=4 \mathrm{~m}$ above the floor. If the student pushes the spring with spring constant $k=6.5 \mathrm{~N} / \mathrm{m}$ down by $\Delta x=0.3 \mathrm{~m}$ compared to its rest length and lands the box exactly on the target, what is $\Delta d$ ? Answer in meters. You may assume friction is negligible. ![](https://cdn.mathpix.com/cropped/2023_12_21_d6ba160aede69eaaf49ag-1.jpg?height=673&width=1217&top_left_y=585&top_left_x=451)","['By conservation of energy, we have that\n$$\n\\frac{1}{2} k x^{2}=m g x \\sin \\theta+\\frac{1}{2} m v^{2} \\Longrightarrow v=\\sqrt{\\frac{k}{m} x^{2}-2 g x \\sin \\theta}\n$$\n\nBy kinematic formulae for motion with constant acceleration,\n\n$$\n-4.5 \\mathrm{~m}=v \\sin \\theta t-\\frac{1}{2} g t^{2}\n$$\n\nSolving for $t$ through the quadratic, $t=1.099 \\mathrm{~s}$ and the velocity is $v=2.60 \\mathrm{~m} / \\mathrm{s}$. Therefore, the distance, $\\Delta d=v \\cos \\theta t=2.47 \\mathrm{~m}$.']",['2.47'],False,m,Numerical,1e-1 800,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A quarantined physics student decides to perform an experiment to land a small box of mass $m=60 \mathrm{~g}$ onto the center of a target a distance $\Delta d$ away. The student puts the box on a top of a frictionless ramp with height $h_{2}=0.5 \mathrm{~m}$ that is angled $\theta=30^{\circ}$ to the horizontal on a table that is $h_{1}=4 \mathrm{~m}$ above the floor. If the student pushes the spring with spring constant $k=6.5 \mathrm{~N} / \mathrm{m}$ down by $\Delta x=0.3 \mathrm{~m}$ compared to its rest length and lands the box exactly on the target, what is $\Delta d$ ? Answer in meters. You may assume friction is negligible. ","['By conservation of energy, we have that\n$$\n\\frac{1}{2} k x^{2}=m g x \\sin \\theta+\\frac{1}{2} m v^{2} \\Longrightarrow v=\\sqrt{\\frac{k}{m} x^{2}-2 g x \\sin \\theta}\n$$\n\nBy kinematic formulae for motion with constant acceleration,\n\n$$\n-4.5 \\mathrm{~m}=v \\sin \\theta t-\\frac{1}{2} g t^{2}\n$$\n\nSolving for $t$ through the quadratic, $t=1.099 \\mathrm{~s}$ and the velocity is $v=2.60 \\mathrm{~m} / \\mathrm{s}$. Therefore, the distance, $\\Delta d=v \\cos \\theta t=2.47 \\mathrm{~m}$.']",['2.47'],False,m,Numerical,1e-1 801,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A wooden bus of mass $M=20,000 \mathrm{~kg}$ ( $M$ represents the mass excluding the wheels) is on a ramp with angle $30^{\circ}$. Each of the four wheels is composed of a ring of mass $\frac{M}{2}$ and radius $R=1 \mathrm{~m}$ and 6 evenly spaced spokes of mass $\frac{M}{6}$ and length $R$. All components of the truck have a uniform density. Find the acceleration of the bus down the ramp assuming that it rolls without slipping. Answer in $\mathrm{m} / \mathrm{s}^{2}$. ","[""The moment of inertia of each wheel can be thought of as a superposition of the six spokes and a ring. Therefore, we get:\n$$\nI_{\\text {wheel }}=\\frac{M}{2} R^{2}+6\\left(\\frac{1}{3} \\frac{M}{6} R^{2}\\right)=\\frac{5}{6} M R^{2}\n$$\n\nThe moment of inertia of four wheels is:\n\n$$\nI_{\\mathrm{bus}}=4\\left(\\frac{5}{6} M R^{2}\\right)=\\frac{10}{3} M R^{2}\n$$\n\nThe total mass of the bus is\n\n$$\nm=M+4\\left(\\frac{M}{2}+6 \\cdot \\frac{M}{6}\\right)=7 M\n$$\n\nUsing Newton's second law down the ramp,\n\n$$\n7 M a=7 M g \\sin \\theta-4 f\n$$\n\nif $f$ is the friction at each wheel, and the torque balance on each wheel is:\n\n$$\n\\frac{5}{6} M R^{2} \\alpha=f R\n$$\n\nLetting $a=\\alpha r$ for the no slip condition, we can solve for $f$ to be:\n\n$$\nf=\\frac{5}{6} M a\n$$\n\nso our force balance equation becomes:\n\n$$\n7 M a=7 M g \\sin \\theta-\\frac{10}{3} M a \\Longrightarrow a=\\frac{21}{31} g \\sin \\theta=3.32 \\mathrm{~m} / \\mathrm{s}^{2}\n$$""]",['3.32'],False,m,Numerical,1e-1 801,Mechanics,,"A wooden bus of mass $M=20,000 \mathrm{~kg}$ ( $M$ represents the mass excluding the wheels) is on a ramp with angle $30^{\circ}$. Each of the four wheels is composed of a ring of mass $\frac{M}{2}$ and radius $R=1 \mathrm{~m}$ and 6 evenly spaced spokes of mass $\frac{M}{6}$ and length $R$. All components of the truck have a uniform density. Find the acceleration of the bus down the ramp assuming that it rolls without slipping. Answer in $\mathrm{m} / \mathrm{s}^{2}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_d10918c220dd12cc2515g-1.jpg?height=745&width=1008&top_left_y=560&top_left_x=556)","[""The moment of inertia of each wheel can be thought of as a superposition of the six spokes and a ring. Therefore, we get:\n$$\nI_{\\text {wheel }}=\\frac{M}{2} R^{2}+6\\left(\\frac{1}{3} \\frac{M}{6} R^{2}\\right)=\\frac{5}{6} M R^{2}\n$$\n\nThe moment of inertia of four wheels is:\n\n$$\nI_{\\mathrm{bus}}=4\\left(\\frac{5}{6} M R^{2}\\right)=\\frac{10}{3} M R^{2}\n$$\n\nThe total mass of the bus is\n\n$$\nm=M+4\\left(\\frac{M}{2}+6 \\cdot \\frac{M}{6}\\right)=7 M\n$$\n\nUsing Newton's second law down the ramp,\n\n$$\n7 M a=7 M g \\sin \\theta-4 f\n$$\n\nif $f$ is the friction at each wheel, and the torque balance on each wheel is:\n\n$$\n\\frac{5}{6} M R^{2} \\alpha=f R\n$$\n\nLetting $a=\\alpha r$ for the no slip condition, we can solve for $f$ to be:\n\n$$\nf=\\frac{5}{6} M a\n$$\n\nso our force balance equation becomes:\n\n$$\n7 M a=7 M g \\sin \\theta-\\frac{10}{3} M a \\Longrightarrow a=\\frac{21}{31} g \\sin \\theta=3.32 \\mathrm{~m} / \\mathrm{s}^{2}\n$$""]",['3.32'],False,m,Numerical,1e-1 802,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In an old coal factory, a conveyor belt will move at a constant velocity of $20.3 \mathrm{~m} / \mathrm{s}$ and can deliver a maximum power of $15 \mathrm{MW}$. Each wheel in the conveyor belt has a diameter of $2 \mathrm{~m}$. However a changing demand has pushed the coal factory to fill their coal hoppers with a different material with a certain constant specific density. These ""coal"" hoppers have been modified to deliver a constant $18 \mathrm{~m}^{3} \mathrm{~s}^{-1}$ of the new material to the conveyor belt. Assume that the kinetic and static friction are the same and that there is no slippage. What is the maximum density of the material?","['The maximal force the convey belt can provide to a particle is:\n$$\nF=\\frac{P}{v}\n$$\n\nThe conveyor belt must provide an impulse to the particles to have a momentum of $p=m v$, where $m$ is the mass of the particle and $v$ is the velocity.\n\n$$\nF=\\frac{d p}{d t}\n$$\n\nwhere $\\frac{d p}{d t}$ is:\n\n$$\n\\rho \\dot{V} v\n$$\n\nSolving for for the maximum density we get:\n\n$$\n\\begin{gathered}\n\\rho=\\frac{P}{\\dot{V} v^{2}} \\\\\n\\rho=2022.2 \\frac{\\mathrm{kg}}{\\mathrm{m}^{3}}\n\\end{gathered}\n$$']",['$2022.2$'],False,$\frac{\mathrm{kg}}{\mathrm{m}^{3}}$,Numerical,1e-1 803,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Neutrinos are extremely light particles and rarely interact with matter. The Sun emits neutrinos, each with an energy of $8 \times 10^{-14} \mathrm{~J}$ and reaches a flux density of $10^{11}$ neutrinos $/\left(\mathrm{s} \mathrm{cm}^{2}\right)$ at Earth's surface. In the movie 2012, neutrinos have mutated and now are completely absorbed by the Earth's inner core, heating it up. Model the inner core as a sphere of radius $1200 \mathrm{~km}$, density $12.8 \mathrm{~g} / \mathrm{cm}^{3}$, and a specific heat of $0.400 \mathrm{~J} / \mathrm{g} \mathrm{K}$. The time scale, in seconds, that it will take to heat up the inner core by $1^{\circ} \mathrm{C}$ is $t=1 \times 10^{N}$ where $N$ is an integer. What is the value of $N$ ?","['The cross sectional area is $\\pi r^{2}$, so the incoming power generated by the neutrinos is:\n$$\nP=\\pi r^{2} E \\Phi\n$$\n\nwhere $E$ is the energy of each neutrino and $\\Phi$ is the flux density. We want to cause a change in energy of:\n\n$$\n\\Delta Q=m c \\Delta T=\\rho \\frac{4}{3} \\pi r^{3} c \\Delta T\n$$\n\nwhich can be accomplished in a time:\n\n$$\nP t=\\Delta Q \\Longrightarrow t=\\frac{\\rho\\left(4 \\pi r^{3}\\right) c \\Delta T}{3 \\pi r^{2} E \\Phi}=\\frac{4 \\rho r c \\Delta T}{3 E \\Phi}=1 \\times 10^{14} \\mathrm{~s}\n$$']",['$1 \\times 10^{14}$'],False,s,Numerical,1e13 804,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A ball is situated at the midpoint of the bottom of a rectangular ditch with width $1 \mathrm{~m}$. It is shot at a velocity $v=5 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ relative to the horizontal. How many times does the ball collide with the walls of the ditch until it hits the bottom of the ditch again? Assume all collisions to be elastic and that the ball never flies out of the ditch. ",['We use the idea of mirroring the walls of the ditch. We can then draw out the normal path of the projectile and find the number of intersections the projectile makes with the mirror walls. The total time of the projectile to travel a path is given by\n$$\nt=\\frac{2 v_{0} \\sin \\theta}{g}\n$$\n\nThe projectile will cover a horizontal distance\n\n$$\n\\left(N+\\frac{1}{2}\\right) a=v \\cos \\theta t \\Longrightarrow N=\\frac{v}{a} \\cos \\theta t-\\frac{1}{2}\n$$\n\nTaking the ceiling of this gives:\n\n$$\nN=\\left\\lceil\\frac{R-a / 2}{a}\\right\\rceil=2 .\n$$'],['2'],False,,Numerical,2e-1 804,Mechanics,,"A ball is situated at the midpoint of the bottom of a rectangular ditch with width $1 \mathrm{~m}$. It is shot at a velocity $v=5 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ relative to the horizontal. How many times does the ball collide with the walls of the ditch until it hits the bottom of the ditch again? Assume all collisions to be elastic and that the ball never flies out of the ditch. ![](https://cdn.mathpix.com/cropped/2023_12_21_ceba72362d0faa7a8dc8g-1.jpg?height=567&width=813&top_left_y=459&top_left_x=648)",['We use the idea of mirroring the walls of the ditch. We can then draw out the normal path of the projectile and find the number of intersections the projectile makes with the mirror walls. The total time of the projectile to travel a path is given by\n$$\nt=\\frac{2 v_{0} \\sin \\theta}{g}\n$$\n\nThe projectile will cover a horizontal distance\n\n$$\n\\left(N+\\frac{1}{2}\\right) a=v \\cos \\theta t \\Longrightarrow N=\\frac{v}{a} \\cos \\theta t-\\frac{1}{2}\n$$\n\nTaking the ceiling of this gives:\n\n$$\nN=\\left\\lceil\\frac{R-a / 2}{a}\\right\\rceil=2 .\n$$'],['2'],False,,Numerical,2e-1 805,Mechanics,,"A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball (of negligible size) of mass $M=5 \mathrm{~kg}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_4d9f6a3f09e5d7bf56fag-1.jpg?height=357&width=797&top_left_y=407&top_left_x=661) Let $h$ be the minimum height that $m$ was dropped such that $M$ would be able to move all the way around the loop. The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different scenario. Assume that the balls can now collide perfectly inelastically, which means that they stick to each other instantaneously after collision for the rest of the motion. If $m$ was dropped from a height $3 R$, find the minimum value of $\frac{m}{M}$ such that the combined mass can fully move all the way around the loop. Let this minimum value be $k$. Compute $\alpha=\frac{k^{2}}{h^{2}}$. (Note that this question is only asking for $\alpha$ but you need to find $h$ to find $\alpha$ ). Assume the balls are point masses (neglect rotational effects).","['The velocity of $m$ when it gets to the bottom of the track will be given by\n$$\nv_{b}=\\sqrt{2 g h} .\n$$\n\nClaim: The velocity of $M$ after collision will be given by $v_{M}=\\frac{(1+e) m}{m+M} \\sqrt{2 g h}$.\n\nProof: Conservation of momentum before and after the collision is expressed by:\n\n$$\nm v_{b}=m v_{m}+M v_{M}\n$$\n\nBy coefficient of restitution,\n\n$$\nv_{M}-v_{m}=e v_{b}\n$$\n\nThese equations may be solved directly to find $v_{m}, v_{M}$ to give\n\n$$\nv_{M}=\\frac{(1+e) m}{m+M} v_{b}\n$$\n\nNext, when the objects get to the top of the loop, conservation of energy gives the speed of $M$ when it gets to the top as\n\n$$\n\\frac{1}{2} v_{M}^{2}=\\frac{1}{2} v_{t}^{2}+2 g R .\n$$\n\nAt the top of the loop, $M$ must at least have an acceleration $g$ to maintain circular motion, and thus\n\n$$\n\\frac{v_{t}^{2}}{R}=g \\Longrightarrow v_{t}=\\sqrt{g R}\n$$\n\n\n\nSubstituting these results into our conservation equation gives us\n\n$$\n\\begin{aligned}\n\\frac{(1+e)^{2} m^{2}}{(m+M)^{2}}(\\sqrt{2 g h})^{2} & =g R+4 g R \\\\\n\\frac{2(1+e)^{2} m^{2} g h}{(m+M)^{2}} & =5 g R \\\\\nh & =\\frac{5(m+M)^{2} R}{2(1+e)^{2} m^{2}} \\\\\n& =6.805 \\mathrm{~m}\n\\end{aligned}\n$$\n\nIn the second scenario, conservation of momentum before and after the collision gives:\n\n$$\nm v_{b}=(M+m) v_{f} \\Longrightarrow v_{f}=\\frac{m v_{b}}{M+m}\n$$\n\nThe same conservation of energy formula as in the first scenario yields\n\n$$\n\\begin{aligned}\n\\frac{m^{2}}{(M+m)^{2}}(\\sqrt{2 g h})^{2} & =5 g R \\\\\n\\frac{2 m^{2} g h}{(M+m)^{2}} & =5 g R \\\\\n\\frac{6 m^{2} g R}{(M+m)^{2}} & =5 g R \\\\\n\\left(\\frac{k}{k+1}\\right)^{2} & =\\frac{5}{6} \\\\\n\\frac{k}{k+1} & =\\sqrt{\\frac{5}{6}} \\\\\nk & =\\frac{\\sqrt{5}}{\\sqrt{6}-\\sqrt{5}} \\\\\n& =10.477\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\alpha=\\frac{k^{2}}{h^{2}}=2.37 \\mathrm{~m}^{2}\n$$']",['$2.37$'],False,$\mathrm{~m}^{2}$,Numerical,1e-1 805,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball (of negligible size) of mass $M=5 \mathrm{~kg}$. Let $h$ be the minimum height that $m$ was dropped such that $M$ would be able to move all the way around the loop. The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different scenario. Assume that the balls can now collide perfectly inelastically, which means that they stick to each other instantaneously after collision for the rest of the motion. If $m$ was dropped from a height $3 R$, find the minimum value of $\frac{m}{M}$ such that the combined mass can fully move all the way around the loop. Let this minimum value be $k$. Compute $\alpha=\frac{k^{2}}{h^{2}}$. (Note that this question is only asking for $\alpha$ but you need to find $h$ to find $\alpha$ ). Assume the balls are point masses (neglect rotational effects).","['The velocity of $m$ when it gets to the bottom of the track will be given by\n$$\nv_{b}=\\sqrt{2 g h} .\n$$\n\nClaim: The velocity of $M$ after collision will be given by $v_{M}=\\frac{(1+e) m}{m+M} \\sqrt{2 g h}$.\n\nProof: Conservation of momentum before and after the collision is expressed by:\n\n$$\nm v_{b}=m v_{m}+M v_{M}\n$$\n\nBy coefficient of restitution,\n\n$$\nv_{M}-v_{m}=e v_{b}\n$$\n\nThese equations may be solved directly to find $v_{m}, v_{M}$ to give\n\n$$\nv_{M}=\\frac{(1+e) m}{m+M} v_{b}\n$$\n\nNext, when the objects get to the top of the loop, conservation of energy gives the speed of $M$ when it gets to the top as\n\n$$\n\\frac{1}{2} v_{M}^{2}=\\frac{1}{2} v_{t}^{2}+2 g R .\n$$\n\nAt the top of the loop, $M$ must at least have an acceleration $g$ to maintain circular motion, and thus\n\n$$\n\\frac{v_{t}^{2}}{R}=g \\Longrightarrow v_{t}=\\sqrt{g R}\n$$\n\n\n\nSubstituting these results into our conservation equation gives us\n\n$$\n\\begin{aligned}\n\\frac{(1+e)^{2} m^{2}}{(m+M)^{2}}(\\sqrt{2 g h})^{2} & =g R+4 g R \\\\\n\\frac{2(1+e)^{2} m^{2} g h}{(m+M)^{2}} & =5 g R \\\\\nh & =\\frac{5(m+M)^{2} R}{2(1+e)^{2} m^{2}} \\\\\n& =6.805 \\mathrm{~m}\n\\end{aligned}\n$$\n\nIn the second scenario, conservation of momentum before and after the collision gives:\n\n$$\nm v_{b}=(M+m) v_{f} \\Longrightarrow v_{f}=\\frac{m v_{b}}{M+m}\n$$\n\nThe same conservation of energy formula as in the first scenario yields\n\n$$\n\\begin{aligned}\n\\frac{m^{2}}{(M+m)^{2}}(\\sqrt{2 g h})^{2} & =5 g R \\\\\n\\frac{2 m^{2} g h}{(M+m)^{2}} & =5 g R \\\\\n\\frac{6 m^{2} g R}{(M+m)^{2}} & =5 g R \\\\\n\\left(\\frac{k}{k+1}\\right)^{2} & =\\frac{5}{6} \\\\\n\\frac{k}{k+1} & =\\sqrt{\\frac{5}{6}} \\\\\nk & =\\frac{\\sqrt{5}}{\\sqrt{6}-\\sqrt{5}} \\\\\n& =10.477\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\alpha=\\frac{k^{2}}{h^{2}}=2.37 \\mathrm{~m}^{2}\n$$']",['$2.37$'],False,$\mathrm{~m}^{2}$,Numerical,1e-1 806,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Eddie is experimenting with his sister's violin. Allow the ""A"" string of his sister's violin have an ultimate tensile strength $\sigma_{1}$. He tunes a string up to its highest possible frequency $f_{1}$ before it breaks. He then builds an exact copy of the violin, where all lengths have been increased by a factor of $\sqrt{2}$ and tunes the same string again to its highest possible frequency $f_{2}$. What is $f_{2} / f_{1}$ ? The density of the string does not change. Note: The ultimate tensile strength is maximum amount of stress an object can endure without breaking. Stress is defined as $\frac{F}{A}$, or force per unit area.","['We note from a simple dimensional analysis that the angular frequency of the string $\\omega$ will consist\n\n\nof the tension $T$, the length of the string $L$ and the mass of the string $m$.\n\n$$\n\\begin{aligned}\nT & =\\left[M L T^{-2}\\right] \\\\\nL & =[L] \\\\\nm & =[M] \\\\\n\\omega & =\\left[T^{-1}\\right]\n\\end{aligned}\n$$\n\nTherefore, by rearranging, we find that\n\n$$\n\\begin{aligned}\n\\omega & =T^{\\alpha} L^{\\beta} m^{\\gamma} \\\\\n{\\left[T^{-1}\\right] } & =\\left[M L T^{-2}\\right]^{\\alpha}[L]^{\\beta}[M]^{\\gamma}\n\\end{aligned}\n$$\n\nDistributing the exponents, and rearranging gives us\n\n$$\nT^{-1}=M^{\\alpha+\\gamma} L^{\\alpha+\\beta} T^{-2 \\alpha}\n$$\n\nWe now have three equations\n\n$$\n\\begin{aligned}\n\\alpha+\\gamma & =0 \\\\\n\\alpha+\\beta & =0 \\\\\n-2 \\alpha & =-1\n\\end{aligned}\n$$\n\nFrom here, we find that $\\alpha=1 / 2$. Substituting this into the first equation gives us\n\n$$\n1 / 2+\\gamma=0 \\Longrightarrow \\gamma=-1 / 2\n$$\n\nthen substituting $\\alpha$ into the second equation gives us\n\n$$\n1 / 2+\\beta=0 \\Longrightarrow \\beta=-1 / 2\n$$\n\nWe now find that the angular frequency is given by\n\n$$\n\\omega=A \\sqrt{\\frac{T}{L m}}\n$$\n\nwhere $A$ is an arbritary constant. Noting that $\\omega=2 \\pi f$, we find that\n\n$$\nf=\\frac{A}{2 \\pi} \\sqrt{\\frac{T}{L m}}\n$$\n\nFrom this analysis, we can then see that $f_{2} / f_{1}=\\sqrt{2} / 2 \\approx 0.707$.']",['$\\frac{\\sqrt{2}}{2}$'],False,,Numerical,1e-3 807,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A one horsepower propeller powered by a battery and is used to propel a small boat initially at rest. You have two options: 1. Put the propeller on top of the boat and push on the air with an initial force $F_{1}$ 2. Put the propeller underwater and push on the water with an initial force $F_{2}$. The density of water is $997 \mathrm{~kg} / \mathrm{m}^{3}$ while the density of air is $1.23 \mathrm{~kg} / \mathrm{m}^{3}$. Assume that the force is both cases is dependent upon only the density of the medium, the surface area of the propeller, and the power delivered by the battery. What is $F_{2} / F_{1}$ ? You may assume (unrealistically) the efficiency of the propeller does not change. Round to the nearest tenths.","['The force exerted on the fluid is roughly proportional to the change in momentum with respect to time:\n$$\nF=\\frac{d p}{d t}=v \\frac{d m}{d t}=v \\frac{d}{d t}(\\rho A x)=\\rho A v^{2}\n$$\n\nIt is kept at a constant power $P=F v$, which can allow us to solve for the speed $v$ of the propellers.\n\n$$\nP=\\rho A v^{3} \\Longrightarrow v=\\left(\\frac{P}{\\rho A}\\right)^{1 / 3}\n$$\n\nso the force is given by:\n\n$$\nF=\\rho A\\left(\\frac{P}{\\rho A}\\right)^{2 / 3} \\Longrightarrow F \\propto \\rho^{1 / 3}\n$$\n\nTherefore:\n\n$$\nF_{2} / F_{1}=(997 / 1.23)^{1 / 3}=9.26 \\text { times }\n$$']",['9.26'],False,times,Numerical,1e-1 808,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A professional pastry chef is making a sweet which consists of 3 sheets of chocolate. The chef leaves a gap with width $d_{1}=0.1 \mathrm{~m}$ between the top and middle layers and fills it with a chocolate syrup with uniform viscosity $\eta_{1}=10 \mathrm{~Pa} \cdot \mathrm{s}$ and a gap with width $d_{2}=0.2 \mathrm{~m}$ between the middle and bottom sheet and fills it with caramel with uniform viscosity $\eta_{2}=15 \mathrm{~Pa} \cdot \mathrm{s}$. If the chef pulls the top sheet with a velocity $2 \mathrm{~m} / \mathrm{s}$ horizontally, at what speed must he push the bottom sheet horizontally such that the middle sheet remains stationary initially? Ignore the weight of the pastry sheets throughout the problem and the assume the sheets are equally sized. Note: Shear stress is governed by the equation $\tau=\eta \times$ rate of strain.",['The plates are equal sizes so all we have to do is simply balance the shear stresses which act in opposing directions on the middle plate:\n$$\n\\begin{gathered}\n\\tau_{1}=\\tau_{2} \\\\\n\\eta_{1} \\cdot \\frac{v_{1}}{d_{1}}=\\eta_{2} \\cdot \\frac{v_{2}}{d_{2}} \\\\\n10 \\cdot \\frac{2}{0.1}=15 \\frac{v}{0.2} \\\\\nv=2.667 \\mathrm{~m} / \\mathrm{s}\n\\end{gathered}\n$$'],['$2.667$'],False,$\mathrm{~m} / \mathrm{s}$,Numerical,1e-2 809,Electromagnetism,,"The following diagram depicts a single wire that is bent into the shape below. The circuit is placed in a magnetic field pointing out of the page, uniformly increasing at the rate $\frac{d B}{d t}=2.34 \mathrm{~T} / \mathrm{s}$. Calculate the magnitude of induced electromotive force in the wire, in terms of the following labelled areas $\left(\mathrm{m}^{2}\right)$. Note that $B$ is non-inclusive of $C$ and that $A=4.23, B=2.74$, and $C=0.34$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b30a714cc20c5e96a317g-1.jpg?height=374&width=612&top_left_y=2144&top_left_x=754)","['Without loss of generality, let the current around $A$ flow in the counterclockwise direction and let the flux through $A$ be positive. Note that the current will flow in the clockwise direction in $C$ and around $B$, the area enclosed by the loop is $B+C$. The flux will be negative here. Therefore, the total flux is then proportional to:\n$$\n\\Phi \\propto A-B-2 C=0.81\n$$\n\nand the magnitude of the induced electromotive force is:\n\n$$\n\\varepsilon=(A-B-2 C) \\frac{d B}{d t}=1.9 \\mathrm{Tm}^{2} \\mathrm{~s}^{-1} .\n$$']",['$1.9$'],False,$\mathrm{Tm}^{2} \mathrm{~s}^{-1}$,Numerical,1e-1 809,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The following diagram depicts a single wire that is bent into the shape below. The circuit is placed in a magnetic field pointing out of the page, uniformly increasing at the rate $\frac{d B}{d t}=2.34 \mathrm{~T} / \mathrm{s}$. Calculate the magnitude of induced electromotive force in the wire, in terms of the following labelled areas $\left(\mathrm{m}^{2}\right)$. Note that $B$ is non-inclusive of $C$ and that $A=4.23, B=2.74$, and $C=0.34$. ","['Without loss of generality, let the current around $A$ flow in the counterclockwise direction and let the flux through $A$ be positive. Note that the current will flow in the clockwise direction in $C$ and around $B$, the area enclosed by the loop is $B+C$. The flux will be negative here. Therefore, the total flux is then proportional to:\n$$\n\\Phi \\propto A-B-2 C=0.81\n$$\n\nand the magnitude of the induced electromotive force is:\n\n$$\n\\varepsilon=(A-B-2 C) \\frac{d B}{d t}=1.9 \\mathrm{Tm}^{2} \\mathrm{~s}^{-1} .\n$$']",['$1.9$'],False,$\mathrm{Tm}^{2} \mathrm{~s}^{-1}$,Numerical,1e-1 810,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A magnetic field is located within a region enclosed by an elliptical island with semi-minor axis of $a=100 \mathrm{~m}$ and semi-major axis of $b=200 \mathrm{~m}$. A car carrying charge $+Q=1.5 \mathrm{C}$ drives on the boundary of the island at a constant speed of $v=5 \mathrm{~m} / \mathrm{s}$ and has mass $m=2000 \mathrm{~kg}$. Any dimensions of the car can be assumed to be much smaller than the dimensions of the island. Ignore any contributions to the magnetic field from the moving car and assume that the car has enough traction to continue driving in its elliptical path. Let the center of the island be located at the point $(0,0)$ while the semi major and semi minor axes lie on the $x$ and $y$-axes, respectively. On this island, the magnetic field varies as a function of $x$ and $y: B(x, y)=k_{b} e^{c_{b} x y} \hat{z}$ (pointing in the upward direction, perpendicular to the island plane in the positive $z$-direction). The constant $c_{b}=10^{-4} \mathrm{~m}^{-2}$ and the constant $k_{b}=2.1 \mu \mathrm{T}$ At what point on the island is the force from the magnetic field a maximum? Write the distance of this point from the $x$-axis in metres.","['To find a minimum or maximum, the gradient of the constraint function $f(x, y)=\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}-1$ and the gradient of the $B$ field function should be scalar multiples of each other.\n\n$$\n\\begin{aligned}\n& \\frac{2 x}{b^{2}} \\mu=c_{b} y e^{x y} \\\\\n& \\frac{2 y}{a^{2}} \\mu=c_{b} x e^{x y}\n\\end{aligned}\n$$\n\nSolving the two equations, we get that a maximum point $(x, y)$ is of the form $\\left(\\frac{b}{\\sqrt{2}}, \\frac{a}{\\sqrt{2}}\\right)$ or $\\left(-\\frac{b}{\\sqrt{2}},-\\frac{a}{\\sqrt{2}}\\right)$. The distance from the $\\mathrm{y}$-axis is thus $\\frac{a}{\\sqrt{2}}=70.7 \\mathrm{~m}$.']",['70.7'],False,m,Numerical,1e-1 811,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Inside a laboratory at room temperature, a steel tuning fork in the shape of a $\mathrm{U}$ is struck and begins to vibrate at $f=426 \mathrm{~Hz}$. The tuning fork is then brought outside where it is $10^{\circ} \mathrm{C}$ hotter and the experiment is performed again. What is the change in frequency, $\Delta f$ of the tuning fork? (A positive value will indicate an increase in frequency, and a negative value will indicate a decrease.) Note: The linear thermal coefficient of expansion for steel is $\alpha=1.5 \times 10^{-5} \mathrm{~K}^{-1}$ and you may assume the expansion is isotropic and linear. When the steel bends, there is a restoring torque $\tau=-\kappa \theta$ such that $\kappa \equiv G J$ where $G=77 \mathrm{GPa}$ is constant and $J$ depends on the geometry and dimensions of the cross-sectional area.","['Note that $\\kappa$ has units of torque so dimensionally, $J$ must be proportional to $L^{3}$. Therefore, we have:\n$$\n\\beta M L^{2} \\alpha \\propto-L^{3} \\theta \\Longrightarrow f \\propto \\sqrt{L}\n$$\n\nTherefore, we have:\n\n$$\n\\frac{\\Delta f}{f}=\\frac{\\Delta L}{2 L}\n$$\n\nSince $\\frac{\\Delta L}{L}=\\alpha \\Delta T$, this gives us:\n\n$$\n\\Delta f=\\frac{1}{2} f \\alpha \\Delta T=0.0320 \\mathrm{~Hz}\n$$']",['0.0320'],False,Hz,Numerical,1e-3 812,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",A large metal conducting sphere with radius $10 \mathrm{~m}$ at an initial potential of 0 and an infinite supply of smaller conducting spheres of radius $1 \mathrm{~m}$ and potential $10 \mathrm{~V}$ are placed into contact in such a way: the large metal conducting sphere is contacted with each smaller sphere one at a time. You may also assume the spheres are touched using a thin conducting wire that places the two spheres sufficiently far away from each other such that their own spherical charge symmetry is maintained. What is the least number of smaller spheres required to be touched with the larger sphere such that the potential of the larger sphere reaches $9 \mathrm{~V}$ ? Assume that the charges distribute slowly and that the point of contact between the rod and the spheres is not a sharp point.,"['Let each sphere with radius $1 \\mathrm{~m}$ have charge $q$. Note that each time the large metal conducting sphere is contacted with each of the smaller spheres, the potential is equalized between the two objects. The potential on a sphere is proportional to $\\frac{q}{r}$, so the large conducting sphere must retain $\\frac{10}{11}$ of the total charge after it is contacted with a smaller sphere. Furthermore, to reach $9 \\mathrm{~V}$, the required end charge on the sphere of radius $10 \\mathrm{~m}$ is at least $9 q$. Thus, we get a recursion for the charge of the large square $Q$ in terms of the number of small spheres touched $n$.\n$$\nQ(n+1)=(Q(n)+q) \\cdot \\frac{10}{11}\n$$\n\nInductively applying this recursion, we obtain\n\n$$\nQ(n)=q\\left[\\left(\\frac{10}{11}\\right)^{n}+\\cdots+\\frac{10}{11}\\right]\n$$\n\nWe can now sum this geometric series:\n\n$$\nQ(n)=q\\left(\\frac{10}{11} \\cdot \\frac{\\left(\\frac{10}{11}\\right)^{n}-1}{\\frac{10}{11}-1}\\right) .\n$$\n\nThus, using $Q(n) \\geq 9 q$, we find that $\\left(\\frac{10}{11}\\right)^{n} \\leq 0.1$, which provides $n \\geq 24.1588$, or $n=25$ as the answer.']",['25'],False,,Numerical,1e-3 813,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","During high speed motion in a strong electric field, a charged particle can ionize air molecules it collides with. A charged particle of mass $m=0.1 \mathrm{~kg}$ and charge $q=0.5 \mu \mathrm{C}$ is located in the center of a cubical box. Each vertex of the box is fixed in space and has a charge of $Q=-4 \mu \mathrm{C}$. If the side length of the box is $l=1.5 \mathrm{~m}$ what minimum speed (parallel to an edge) should be given to the particle for it to exit the box (even if it's just momentarily)? Let the energy loss from Corona discharge and other radiation effects be $E=0.00250 \mathrm{~J}$.","['Conservation of energy gives:\n$$\nT_{i}+U_{i}=T_{f}+U_{f}+E\n$$\n\nSolving for the initial potential energy gives\n\n$$\nU_{i}=-8 \\frac{k q Q}{l \\sqrt{3} / 2}=-\\frac{16 k q Q}{\\sqrt{3} l}\n$$\n\nAnd since the final kinetic energy is zero, the final potential energy is\n\n$$\nU_{f}=-4 \\frac{k q Q}{l / \\sqrt{2}}-4 \\frac{k q Q}{l \\sqrt{\\frac{3}{2}}}=-\\left(4 \\sqrt{2}+4 \\sqrt{\\frac{2}{3}}\\right) \\frac{k q Q}{l}\n$$\n\nand thus solving for the initial kinetic energy:\n\n$$\n\\frac{1}{2} m v^{2}=\\left(-4 \\sqrt{2}-4 \\sqrt{\\frac{2}{3}}+\\frac{16}{\\sqrt{3}}\\right) \\frac{k q Q}{l}+E\n$$\n\nThe final answer is $v=0.354 \\mathrm{~m} / \\mathrm{s}$.']",['0.354'],False,m/s,Numerical,1e-2 814,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Max finds himself trapped in the center of a mirror walled equilateral triangular room. What minimum beam angle must his flashlight have so that any point of illumination in the room can be traced back to his flashlight with at most 1 bounce? (Answer in degrees.) Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant. The point of illumination refers to any point in the room that is lit.","['Each time light hits a mirror, we can reflect the entire equilateral triangle about that mirror and continue to trace the straight-line path of the light. For a maximum of 1 bounce, we can reflect our triangle about each of its initial sides. In order for the light to hit every part of the triangle (or an image of that part), by symmetry we require a $120^{\\circ}$ angle. In this case, we would just shine the flashlight so that light directly reaches the entirety of one side, and reflections of light will fully reach the other two sides of the triangle.']",['$120$'],False,$^{\circ}$,Numerical,0 815,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","For his art project, Weishaupt cut out $N=20$ wooden equilateral triangular blocks with a side length of $\ell=10 \mathrm{~cm}$ and a thickness of $t=2 \mathrm{~cm}$, each with the same mass and uniform density. He wishes to stack one on top of the other overhanging the edge of his table. In centimeters, what is the maximum overhang? Round to the nearest centimeter. A side view is shown below. Assume that all triangles are parallel to each other. Note: This diagram is not to scale. ","['Let us consider $N=1$ equilateral triangles. From inspection, we need to place the triangle such that the center of mass lies at the edge of the table. The maximum overhang in this case is $(1-f) h$ where $h=\\frac{\\ell \\sqrt{3}}{2}$ is the height of the triangle and $f h=\\frac{h}{3}$ is the location of the center of mass.\nIf we wish to place a second triangle on top, we want to maximize the center of mass to be as far right as possible without the top block toppling. Placing the second block such that its center of mass is at the tip of the first triangle accomplishes this. However, the center of mass of the two triangles combined is now past the edge. Their center of mass is:\n\n$$\nx_{\\mathrm{cm}}=\\frac{f h+h}{2}=\\frac{f+1}{2} h\n$$\n\nThus the maximum overhang of the first block is now:\n\n$$\nh-\\frac{f+1}{2} h=\\frac{1-f}{2} h\n$$\n\nNow, we will place a third block such that it has the maximum overhang with respect to the top block and then shift the entire setup so that the center of mass of the system lies at the edge of the table. Following the same procedures, we find that the maximum overhang of the first block is:\n\n$$\n\\frac{1-f}{3} h\n$$\n\nThe overhang of the top two blocks are $(1-f) h$ and $\\frac{1-f}{2} h$, unchanged from earlier. You can show via induction that the maximum overhang of the $n^{\\text {th }}$ block (counting from the top downwards) is:\n\n$$\n\\frac{1-f}{n} h\n$$\n\nso if there are 20 such blocks, then the total overhang (summing over all the blocks) is:\n\n$$\n\\sum_{k=0}^{20} \\frac{1-f}{k} h=(1-f) h H_{20}=\\frac{2}{3} \\frac{\\ell \\sqrt{3}}{2} H_{20}=\\frac{\\ell \\sqrt{3}}{3} H_{20}=20.77 \\mathrm{~cm} \\approx 21 \\mathrm{~cm}\n$$\n\nwhere $H_{N}$ is the $N^{\\text {th }}$ harmonic number.']",['21'],False,cm,Numerical,5e-1 815,Mechanics,,"For his art project, Weishaupt cut out $N=20$ wooden equilateral triangular blocks with a side length of $\ell=10 \mathrm{~cm}$ and a thickness of $t=2 \mathrm{~cm}$, each with the same mass and uniform density. He wishes to stack one on top of the other overhanging the edge of his table. In centimeters, what is the maximum overhang? Round to the nearest centimeter. A side view is shown below. Assume that all triangles are parallel to each other. Note: This diagram is not to scale. ![](https://cdn.mathpix.com/cropped/2023_12_21_145bbb911bbed723cc96g-1.jpg?height=567&width=940&top_left_y=562&top_left_x=587)","['Let us consider $N=1$ equilateral triangles. From inspection, we need to place the triangle such that the center of mass lies at the edge of the table. The maximum overhang in this case is $(1-f) h$ where $h=\\frac{\\ell \\sqrt{3}}{2}$ is the height of the triangle and $f h=\\frac{h}{3}$ is the location of the center of mass.\nIf we wish to place a second triangle on top, we want to maximize the center of mass to be as far right as possible without the top block toppling. Placing the second block such that its center of mass is at the tip of the first triangle accomplishes this. However, the center of mass of the two triangles combined is now past the edge. Their center of mass is:\n\n$$\nx_{\\mathrm{cm}}=\\frac{f h+h}{2}=\\frac{f+1}{2} h\n$$\n\nThus the maximum overhang of the first block is now:\n\n$$\nh-\\frac{f+1}{2} h=\\frac{1-f}{2} h\n$$\n\nNow, we will place a third block such that it has the maximum overhang with respect to the top block and then shift the entire setup so that the center of mass of the system lies at the edge of the table. Following the same procedures, we find that the maximum overhang of the first block is:\n\n$$\n\\frac{1-f}{3} h\n$$\n\nThe overhang of the top two blocks are $(1-f) h$ and $\\frac{1-f}{2} h$, unchanged from earlier. You can show via induction that the maximum overhang of the $n^{\\text {th }}$ block (counting from the top downwards) is:\n\n$$\n\\frac{1-f}{n} h\n$$\n\nso if there are 20 such blocks, then the total overhang (summing over all the blocks) is:\n\n$$\n\\sum_{k=0}^{20} \\frac{1-f}{k} h=(1-f) h H_{20}=\\frac{2}{3} \\frac{\\ell \\sqrt{3}}{2} H_{20}=\\frac{\\ell \\sqrt{3}}{3} H_{20}=20.77 \\mathrm{~cm} \\approx 21 \\mathrm{~cm}\n$$\n\nwhere $H_{N}$ is the $N^{\\text {th }}$ harmonic number.']",['21'],False,cm,Numerical,5e-1 816,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Kushal finds himself trapped in a large room with mirrors as walls. Being scared of the dark, he has a powerful flashlight to light the room. All references to ""percent"" refer to area. Since the room is large, assume the person is a point does not block light. Visualize the questions in a 2D setup. The floor/ceiling is irrelevant. The point of illumination refers to any point in the room that is lit. What percent of a large circular room can be lit up using a flashlight with a 20 degree beam angle if Kushal stands in the center?","['Each ray emitted follows a straight line, even when it is reflected since it originates from the center of a circle. Thus, the light rays in total trace out two circular sectors with an angle of $\\theta=20^{\\circ}$ each. Thus, the total percent of the room illuminated is:\n\n$$\nf=\\frac{2 \\theta}{360}=11.1 \\% \\text {. }\n$$']",['11.1'],False,%,Numerical,1e-1 817,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two identical neutron stars with mass $m=4 \times 10^{30} \mathrm{~kg}$ and radius $15 \mathrm{~km}$ are orbiting each other a distance $d=700 \mathrm{~km}$ away from each other ( $d$ refers to the initial distance between the cores of the neutron stars). Assume that they orbit as predicted by classical mechanics, except that they generate gravitational waves. The power dissipated through these waves is given by: $$ P=\frac{32 G^{4}}{5}\left(\frac{m}{d c}\right)^{5} $$ How long does it take for the two stars to collide? Answer in seconds. Note: $d$ is the distance between the cores of the stars.","['Due to Virial theorem, we have:\n$$\nK=-\\frac{1}{2} U\n$$\n\nso the total energy is:\n\n$$\nE=U-\\frac{1}{2} U=-\\frac{G m^{2}}{2 R}\n$$\n\nWe know that the power dissipated gives the change in energy, or:\n\n$$\nP=\\frac{32 G^{4}}{5}\\left(\\frac{m}{R c}\\right)^{5}=\\frac{d}{d t} \\frac{G m^{2}}{2 R}\n$$\n\nor:\n\n$$\n\\frac{32 G^{4}}{5}\\left(\\frac{m}{R c}\\right)^{5} d t=-\\frac{G m^{2}}{2 R^{2}} d R \\Longrightarrow \\int_{0}^{t} \\frac{64 G^{3}}{5} \\frac{m^{3}}{c^{5}} d t=\\int_{d}^{2 r}-R^{3} d R\n$$\n\nSolving this leads us to:\n\n$$\n\\frac{64 G^{3} m^{3}}{5 c^{5}} t=\\frac{d^{4}-r^{4}}{4} \\Longrightarrow t=\\frac{5 c^{5}\\left(d^{4}-16 r^{4}\\right)}{256 G^{3} m^{3}}\n$$\n\nPlugging in the numbers gives:\n\n$$\nt=590 \\mathrm{sec}\n$$\n\nNote that we can also assume that $d^{4} \\gg(2 r)^{4}$ which will simplify calculations, and not introduce any noticeable error.']",['590'],False,seconds,Numerical,1e1 818,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The graph provided plots the $y$-component of the velocity against the $x$-component of the velocity of a kiddie roller coaster at an amusement park for a certain duration of time. The ride takes place entirely in a two dimensional plane. Some students made a remark that at one time, the acceleration was perpendicular to the velocity. Using this graph, what is the minimum x-velocity the ride could be travelling at for this to be true? Round to the nearest integer and answer in meters per second. The diagram is drawn to scale, and you may print this page out and make measurements. ","['The solution revolves around the idea that when the acceleration is perpendicular to the velocity, the work done is 0 , and thus, the instantaneous rate of change of the magnitude of velocity $v_{x}^{2}+v_{y}^{2}$ is 0 . Thus, at such points, when the vertical velocity is plotted against the horizontal velocity, the curve will be tangent to a circle centered at the origin because $v_{y}^{2}+v_{x}^{2}$ is nonchanging at that instant.\nThis is equivalent to stating that the line from the origin to the curve is perpendicular to the curve. Drawing such lines to the curve, the first time this occurs is at $v_{x}=1 \\mathrm{~m} / \\mathrm{s}$.']",['1'],False,m/s,Numerical,1e-1 818,Mechanics,,"The graph provided plots the $y$-component of the velocity against the $x$-component of the velocity of a kiddie roller coaster at an amusement park for a certain duration of time. The ride takes place entirely in a two dimensional plane. Some students made a remark that at one time, the acceleration was perpendicular to the velocity. Using this graph, what is the minimum x-velocity the ride could be travelling at for this to be true? Round to the nearest integer and answer in meters per second. The diagram is drawn to scale, and you may print this page out and make measurements. ![](https://cdn.mathpix.com/cropped/2023_12_21_e7e16de1e9199580e484g-1.jpg?height=1900&width=1106&top_left_y=655&top_left_x=496)","['The solution revolves around the idea that when the acceleration is perpendicular to the velocity, the work done is 0 , and thus, the instantaneous rate of change of the magnitude of velocity $v_{x}^{2}+v_{y}^{2}$ is 0 . Thus, at such points, when the vertical velocity is plotted against the horizontal velocity, the curve will be tangent to a circle centered at the origin because $v_{y}^{2}+v_{x}^{2}$ is nonchanging at that instant.\nThis is equivalent to stating that the line from the origin to the curve is perpendicular to the curve. Drawing such lines to the curve, the first time this occurs is at $v_{x}=1 \\mathrm{~m} / \\mathrm{s}$.']",['1'],False,m/s,Numerical,1e-1 819,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In the cosmic galaxy, the Sun is a mainsequence star, generating its energy mainly by nuclear fusion of hydrogen nuclei into helium. In its core, the Sun fuses hydrogen to produce deuterium $(2 \mathrm{H})$ and tritium $(3 \mathrm{H})$, then makes about 600 million metric tons of helium (4He) per second. Of course, there are also some relatively smaller portions of fission reactions in the Sun's core, e.g. a nuclear fission reaction with Uranium-235 (235U). The Fusion reaction: $$ { }^{2} \mathrm{H}+{ }^{3} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+n+\text { Released Energy } $$ The Fission reaction: $$ { }^{235} U+n+(\text { Initial Energy }) \rightarrow{ }^{144} \mathrm{Ba}+{ }^{90} \mathrm{Kr}+2 n+\text { Released Energy } $$ Isotope Mass (at rest) | Isotope Names | Mass (at rest) $(\mathrm{u})$ | | :--- | :--- | | Deuterium $\left({ }^{2} \mathrm{H}\right)$ | 2.0141 | | Tritium $\left({ }^{3} \mathrm{H}\right)$ | 3.0160 | | Helium $\left({ }^{4} \mathrm{He}\right)$ | 4.0026 | | Neutron $(\mathrm{n})$ | 1.0087 | | Uranium-235 $\left({ }^{235} \mathrm{U}\right)$ | 235.1180 | | Barium-144 $\left({ }^{144} \mathrm{Ba}\right)$ | 143.8812 | | Krypton-90 $\left({ }^{90} \mathrm{Kr}\right)$ | 89.9471 | Calculate the kinetic energy (in $\mathrm{MeV}$ ) released by the products in one fusion reaction.","['Let the kinetic energy released be\n\n$$\nK E_{\\text {released }}=-\\Delta m c^{2}\n$$\n\nLet the mass of helium be $m_{h}$, deuterium be $m_{d}$, tritium $m_{t}$, and mass of neutron $m_{n}$ Therefore.\n\n$$\n-\\Delta m=m_{d}+m_{t}-m_{h}-m_{n}=3.0160+2.0141-4.0026-1.0087=0.0188 \\mathrm{u}\n$$\n\nwhich gives\n\n$$\nK E_{\\text {released }}=(0.0188 \\mathrm{u}) \\cdot\\left(931.494 \\frac{\\mathrm{MeV}}{\\mathrm{u}}\\right)=17.51 \\mathrm{MeV}\n$$']",['17.51'],False,MeV,Numerical,1e-1 820,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","While exploring outer space, Darth Vader comes upon a purely reflective spherical planet with radius $R_{p}=$ $40,000 \mathrm{~m}$ and mass $M_{p}=8.128 \times 10^{24} \mathrm{~kg}$. Around the planet is a strange moon of orbital radius $R_{s}=$ $6,400,000 \mathrm{~m}\left(R_{s} \gg R_{p}\right)$ and mass $M_{s}=9.346 \times 10^{19} \mathrm{~kg}\left(M_{s} \ll M_{p}\right)$. The moon can be modelled as a blackbody and absorbs light perfectly. Darth Vader is in the same plane that the planet orbits in. Startled, Darth Vader shoots a laser with constant intensity and power $P_{0}=2 \times 10^{32} \mathrm{~W}$ at the reflective planet and hits the planet a distance of $\frac{R_{p}}{2}$ away from the line from him to the center of the planet. Upon hitting the reflective planet, the light from the laser is plane polarized. The angle of the planet's polarizer is always the same as the angle of reflection. After reflectance, the laser lands a direct hit on the insulator planet. Darth Vader locks the laser in on the planet until it moves right in front of him, when he turns the laser off. Determine the energy absorbed by the satellite. Assume the reflective planet remains stationary and that the reflective planet is a perfect polarizer of light. ","[""The original angle of reflectance can be found with some optical geometry to be $\\theta_{0}=\\sin ^{-1}\\left(\\frac{1}{2}\\right)=$ $\\frac{\\pi}{6}$. By Malus' Law, when the laser hits the planet with an angle $\\theta$, the final power after reflection is $P_{0} \\cos ^{2}(\\theta)$. Also, note that the satellite has to be at an angle $2 \\theta$ due to the law of reflection. Therefore, by Kepler's Third law, the time for the satellite to reach $\\theta=0$ is\n$$\nt=2 \\theta \\sqrt{\\frac{r^{3}}{G M_{p}}} \\Longrightarrow d t=2 \\sqrt{\\frac{r^{3}}{G M_{p}}} d \\theta\n$$\n\nThus, the total power absorbed by the satellite is\n\n$$\n\\int_{0}^{\\theta_{0}} P_{0} \\cos ^{2}(\\theta) d t=2 P_{0} \\sqrt{\\frac{r^{3}}{G M_{p}}} \\int_{0}^{\\theta_{0}} \\cos ^{2}(\\theta) d \\theta=P_{0} \\sqrt{\\frac{r^{3}}{G M_{p}}}\\left(\\theta_{0}+\\sin \\left(\\theta_{0}\\right) \\cos \\left(\\theta_{0}\\right)\\right)=1.33 \\cdot 10^{35} \\mathrm{~J} .\n$$""]",['$1.33 \\cdot 10^{35}$'],False,J,Numerical,1e34 820,Modern Physics,,"While exploring outer space, Darth Vader comes upon a purely reflective spherical planet with radius $R_{p}=$ $40,000 \mathrm{~m}$ and mass $M_{p}=8.128 \times 10^{24} \mathrm{~kg}$. Around the planet is a strange moon of orbital radius $R_{s}=$ $6,400,000 \mathrm{~m}\left(R_{s} \gg R_{p}\right)$ and mass $M_{s}=9.346 \times 10^{19} \mathrm{~kg}\left(M_{s} \ll M_{p}\right)$. The moon can be modelled as a blackbody and absorbs light perfectly. Darth Vader is in the same plane that the planet orbits in. Startled, Darth Vader shoots a laser with constant intensity and power $P_{0}=2 \times 10^{32} \mathrm{~W}$ at the reflective planet and hits the planet a distance of $\frac{R_{p}}{2}$ away from the line from him to the center of the planet. Upon hitting the reflective planet, the light from the laser is plane polarized. The angle of the planet's polarizer is always the same as the angle of reflection. After reflectance, the laser lands a direct hit on the insulator planet. Darth Vader locks the laser in on the planet until it moves right in front of him, when he turns the laser off. Determine the energy absorbed by the satellite. Assume the reflective planet remains stationary and that the reflective planet is a perfect polarizer of light. ![](https://cdn.mathpix.com/cropped/2023_12_21_cc3e4746f35af35f4e0bg-1.jpg?height=776&width=1008&top_left_y=764&top_left_x=556)","[""The original angle of reflectance can be found with some optical geometry to be $\\theta_{0}=\\sin ^{-1}\\left(\\frac{1}{2}\\right)=$ $\\frac{\\pi}{6}$. By Malus' Law, when the laser hits the planet with an angle $\\theta$, the final power after reflection is $P_{0} \\cos ^{2}(\\theta)$. Also, note that the satellite has to be at an angle $2 \\theta$ due to the law of reflection. Therefore, by Kepler's Third law, the time for the satellite to reach $\\theta=0$ is\n$$\nt=2 \\theta \\sqrt{\\frac{r^{3}}{G M_{p}}} \\Longrightarrow d t=2 \\sqrt{\\frac{r^{3}}{G M_{p}}} d \\theta\n$$\n\nThus, the total power absorbed by the satellite is\n\n$$\n\\int_{0}^{\\theta_{0}} P_{0} \\cos ^{2}(\\theta) d t=2 P_{0} \\sqrt{\\frac{r^{3}}{G M_{p}}} \\int_{0}^{\\theta_{0}} \\cos ^{2}(\\theta) d \\theta=P_{0} \\sqrt{\\frac{r^{3}}{G M_{p}}}\\left(\\theta_{0}+\\sin \\left(\\theta_{0}\\right) \\cos \\left(\\theta_{0}\\right)\\right)=1.33 \\cdot 10^{35} \\mathrm{~J} .\n$$""]",['$1.33 \\cdot 10^{35}$'],False,J,Numerical,1e34 821,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",A particle of rest mass $m$ moving at a speed $v=0.7 c$ decomposes into two photons which fly off at a separated angle $\theta$. What is the minimum value of the angle of separation assuming that the two photons have equal wavelength. (Answer in degrees),['Conservation of momentum and energy gives:\n$$\n\\begin{gathered}\np_{m}=2 E_{\\gamma} \\cos (\\theta / 2) \\\\\nE_{m}=2 E_{\\gamma}\n\\end{gathered}\n$$\n\nRelativity demands that:\n\n$$\nE_{m}^{2}=m^{2}+p_{m}^{2}\n$$\n\nSolving this system of three equations gives:\n\n$$\n\\begin{aligned}\n4 E_{\\gamma}^{2} & =m^{2}+p_{m}^{2} \\\\\n\\frac{\\gamma^{2} m^{2} v^{2}}{\\cos ^{2}(\\theta / 2)} & =m^{2}+\\gamma^{2} m^{2} v^{2} \\\\\n\\gamma^{2} v^{2}\\left(-1+\\frac{1}{\\cos ^{2}(\\theta / 2)}\\right) & =1 \\\\\n\\frac{1}{\\cos ^{2}(\\theta / 2)} & =\\frac{1}{\\gamma^{2} v^{2}}+1 \\\\\n\\cos (\\theta / 2) & =v \\\\\n\\theta & =91.1^{\\circ}\n\\end{aligned}\n$$'],['$91.1$'],False,$^{\circ}$,Numerical,1e0 822,Mechanics,,"Mario is racing with Wario on Moo Moo Meadows when a goomba, ready to avenge all of his friends' deaths, came and hijacked Mario's kart. A graph representing the motion of Mario at any instant is shown below. The velocity acquired by Mario is shown on the x-axis, and the net power of his movement is shown on the $\mathrm{y}$-axis. When Mario's velocity is $6 \mathrm{~m} / \mathrm{s}$, he eats a mushroom which gives him a super boost. ![](https://cdn.mathpix.com/cropped/2023_12_21_0f703e5630a441a05081g-1.jpg?height=1309&width=1272&top_left_y=473&top_left_x=421) You may need to make measurements. Feel free to print this picture out as the diagram is drawn to scale. Find the total distance from Mario runs from when his velocity is $0 \mathrm{~m} / \mathrm{s}$ to when his velocity just reaches $9 \mathrm{~m} / \mathrm{s}$ given that Mario's mass is $m=89 \mathrm{~kg}$. Answer in meters and round to one significant digit.","['Our first goal is to find an expression for the power curve $P(v)$. To do this, let us select a few points on the curve. The easiest point to pick is $(0,0)$ since it is fixed at the origin. The next two easiest points to pick are those that are on the lines $x=3$ and $x=6$ which are given as approximately $(3,1.8)$ and $(6,7.2)$. Note that $y$-axis is in units of $100 \\mathrm{~W}$ so in reality these two points are given as $(3,180)$ and $(6,720)$. This curve is resemblant of a quadratic in the form of $y=k_{1} x^{2}$ and upon solving for $k_{1}$ we find that the curve is given as $P=20 v^{2}$. Secondly, the next line remains constant with respect to time as a line $P=900$. Therefore, we can write a piecewise function for power defined by\n$$\nP(v)=\\left\\{\\begin{array}{ll}\n20 v^{2} & \\text { if } v \\geq 0, \\quad \\text { and } \\quad v<6 \\\\\n900 & \\text { if } v \\geq 6\n\\end{array} .\\right.\n$$\n\n\n\nWe need to find the relationship between power, velocity, and displacement of Mario. Consider dividing the displacement into tiny rectangular pieces with width $\\Delta t$ such that\n\n$$\ns=\\sum_{i \\in \\mathbb{N}} \\Delta s_{i}=\\sum_{i} v_{i}(t) \\cdot \\Delta t\n$$\n\nWe want to the displacement to be expressed in terms without $\\Delta t$. This means that we have to find a relationship for $\\Delta t$. Note that\n\n$$\n\\Delta t=\\Delta v \\cdot \\frac{\\Delta t}{\\Delta v}=\\Delta v \\cdot \\frac{1}{\\Delta v / \\Delta t}=\\frac{\\Delta v}{a}\n$$\n\nTherefore, we now know the displacement to be expressed as\n\n$$\ns=\\sum_{i \\rightarrow 0} v_{i}(t) \\frac{\\Delta v}{a}=\\int \\frac{v}{a(v)} d v\n$$\n\nTo find an expression for $a(v)$ in terms of power, we note that\n\n$$\nP(v)=F(v) \\cdot v \\Longrightarrow a(v)=\\frac{P(v)}{v m}\n$$\n\nwhich means that upon substituting,\n\n$$\ns=\\int \\frac{v^{2} m}{P(v)} d v \\Longrightarrow s=\\int_{0}^{6} \\frac{89 v^{2}}{20 v^{2}} d v+\\int_{6}^{9} \\frac{89 v^{2}}{900} d v=43.61 \\approx 40 \\mathrm{~m}\n$$']",['40'],False,m,Numerical,5e0 822,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Mario is racing with Wario on Moo Moo Meadows when a goomba, ready to avenge all of his friends' deaths, came and hijacked Mario's kart. A graph representing the motion of Mario at any instant is shown below. The velocity acquired by Mario is shown on the x-axis, and the net power of his movement is shown on the $\mathrm{y}$-axis. When Mario's velocity is $6 \mathrm{~m} / \mathrm{s}$, he eats a mushroom which gives him a super boost. You may need to make measurements. Feel free to print this picture out as the diagram is drawn to scale. Find the total distance from Mario runs from when his velocity is $0 \mathrm{~m} / \mathrm{s}$ to when his velocity just reaches $9 \mathrm{~m} / \mathrm{s}$ given that Mario's mass is $m=89 \mathrm{~kg}$. Answer in meters and round to one significant digit.","['Our first goal is to find an expression for the power curve $P(v)$. To do this, let us select a few points on the curve. The easiest point to pick is $(0,0)$ since it is fixed at the origin. The next two easiest points to pick are those that are on the lines $x=3$ and $x=6$ which are given as approximately $(3,1.8)$ and $(6,7.2)$. Note that $y$-axis is in units of $100 \\mathrm{~W}$ so in reality these two points are given as $(3,180)$ and $(6,720)$. This curve is resemblant of a quadratic in the form of $y=k_{1} x^{2}$ and upon solving for $k_{1}$ we find that the curve is given as $P=20 v^{2}$. Secondly, the next line remains constant with respect to time as a line $P=900$. Therefore, we can write a piecewise function for power defined by\n$$\nP(v)=\\left\\{\\begin{array}{ll}\n20 v^{2} & \\text { if } v \\geq 0, \\quad \\text { and } \\quad v<6 \\\\\n900 & \\text { if } v \\geq 6\n\\end{array} .\\right.\n$$\n\n\n\nWe need to find the relationship between power, velocity, and displacement of Mario. Consider dividing the displacement into tiny rectangular pieces with width $\\Delta t$ such that\n\n$$\ns=\\sum_{i \\in \\mathbb{N}} \\Delta s_{i}=\\sum_{i} v_{i}(t) \\cdot \\Delta t\n$$\n\nWe want to the displacement to be expressed in terms without $\\Delta t$. This means that we have to find a relationship for $\\Delta t$. Note that\n\n$$\n\\Delta t=\\Delta v \\cdot \\frac{\\Delta t}{\\Delta v}=\\Delta v \\cdot \\frac{1}{\\Delta v / \\Delta t}=\\frac{\\Delta v}{a}\n$$\n\nTherefore, we now know the displacement to be expressed as\n\n$$\ns=\\sum_{i \\rightarrow 0} v_{i}(t) \\frac{\\Delta v}{a}=\\int \\frac{v}{a(v)} d v\n$$\n\nTo find an expression for $a(v)$ in terms of power, we note that\n\n$$\nP(v)=F(v) \\cdot v \\Longrightarrow a(v)=\\frac{P(v)}{v m}\n$$\n\nwhich means that upon substituting,\n\n$$\ns=\\int \\frac{v^{2} m}{P(v)} d v \\Longrightarrow s=\\int_{0}^{6} \\frac{89 v^{2}}{20 v^{2}} d v+\\int_{6}^{9} \\frac{89 v^{2}}{900} d v=43.61 \\approx 40 \\mathrm{~m}\n$$']",['40'],False,m,Numerical,5e0 823,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","At an amusement park, there is a ride with three ""teacups"" that are circular with identical dimensions. Three friends, Ethan, Rishab, and Kushal, all pick a teacup and sit at the edge. Each teacup rotates about its own axis clockwise at an angular speed $\omega=1 \mathrm{rad} / \mathrm{s}$ and can also move linearly at the same time. The teacup Ethan is sitting on (as always) is malfunctional and can only rotate about its own axis. Rishab's teacup is moving linearly at a constant velocity $2 \mathrm{~m} / \mathrm{s}[\mathrm{N}]$ and Kushal's teacup is also moving linearly at a constant velocity of $4 \mathrm{~m} / \mathrm{s}\left[\mathrm{N} 60^{\circ} \mathrm{E}\right]$. All three teacups are rotating as described above. Interestingly, they observe that at some point, all three of them are moving at the same velocity. What is the radius of each teacup? Note: $\left[\mathrm{N} 60^{\circ} \mathrm{E}\right]$ means $60^{\circ}$ clockwise from north e.g. $60^{\circ}$ east of north.","['We can plot the motion on a $v_{y}-v_{z}$ graph instead of carrying out calculations. We have three points at locations $(0,0),(0,2)$, and $(2 \\sqrt{3}, 2)$ which represent the velocity of the center of mass of the teacups. The velocity that they are moving at can be traced as a circle with radius $r \\omega$, centered at these points.\nThe problem now becomes, at what value $r$ will the three circles intersect. Drawing a diagram, or carrying out trigonometric calculations gives $r=2 \\mathrm{~m}$.']",['2'],False,m,Numerical,1e-1 824,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","An engineer has access to a tetrahedron building block with side length $\ell=10 \mathrm{~cm}$. The body is made of a thermal insulator but the edges are wrapped with a thin copper wiring with cross sectional area $S=2 \mathrm{~cm}^{2}$. The thermal conductivity of copper is $385.0 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. He stacks these tetrahedrons (all facing the same direction) to form a large lattice such that the copper wires are all in contact. In the diagram, only the front row of a small section is coloured. Assume that the lattice formed is infinitely large. At some location in the tetrahedral building block, the temperature difference between two adjacent points is $1^{\circ} \mathrm{C}$. What is the heat flow across these two points? Answer in Watts. Note: Two adjacent points refer to two adjacent points on the tetrahedron. ","[""There are many ways to solve this problem. We first identify that this is exactly the same as an infinite lattice resistor problem. To solve these, we can imagine injecting a current at a node and seeing how this current spreads out. However, a faster approach is by applying Foster's Theorem on this lattice.\nThe resistance of a single wire is:\n\n$$\nR=\\frac{\\ell}{k S}=1.299 \\mathrm{~W} / \\mathrm{K}\n$$\n\nFoster's theorem tells us that\n\n$$\nE R=V-1\n$$\n\n\n\nwhere $V$ is number of vertices and $E$ is edges. Taking the limit as $E, V \\rightarrow \\infty$, we get: $E=6 \\mathrm{~V}$ (since each vertex is connected to 12 edges, but each edge is shared by two vertices). Therefore:\n\n$$\nR_{\\mathrm{eff}}=\\frac{1}{6} R=0.2165 \\mathrm{~W} / \\mathrm{K}\n$$\n\nFrom Fourier's Law, we have:\n\n$$\n\\dot{Q}=\\frac{\\Delta T}{R_{\\mathrm{eff}}}=4.62 \\mathrm{~W}\n$$""]",['4.62'],False,N,Numerical,1e-1 824,Thermodynamics,,"An engineer has access to a tetrahedron building block with side length $\ell=10 \mathrm{~cm}$. The body is made of a thermal insulator but the edges are wrapped with a thin copper wiring with cross sectional area $S=2 \mathrm{~cm}^{2}$. The thermal conductivity of copper is $385.0 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. He stacks these tetrahedrons (all facing the same direction) to form a large lattice such that the copper wires are all in contact. In the diagram, only the front row of a small section is coloured. Assume that the lattice formed is infinitely large. At some location in the tetrahedral building block, the temperature difference between two adjacent points is $1^{\circ} \mathrm{C}$. What is the heat flow across these two points? Answer in Watts. Note: Two adjacent points refer to two adjacent points on the tetrahedron. ![](https://cdn.mathpix.com/cropped/2023_12_21_fa0c803e66fc3d4839bfg-1.jpg?height=1285&width=1507&top_left_y=756&top_left_x=301)","[""There are many ways to solve this problem. We first identify that this is exactly the same as an infinite lattice resistor problem. To solve these, we can imagine injecting a current at a node and seeing how this current spreads out. However, a faster approach is by applying Foster's Theorem on this lattice.\nThe resistance of a single wire is:\n\n$$\nR=\\frac{\\ell}{k S}=1.299 \\mathrm{~W} / \\mathrm{K}\n$$\n\nFoster's theorem tells us that\n\n$$\nE R=V-1\n$$\n\n\n\nwhere $V$ is number of vertices and $E$ is edges. Taking the limit as $E, V \\rightarrow \\infty$, we get: $E=6 \\mathrm{~V}$ (since each vertex is connected to 12 edges, but each edge is shared by two vertices). Therefore:\n\n$$\nR_{\\mathrm{eff}}=\\frac{1}{6} R=0.2165 \\mathrm{~W} / \\mathrm{K}\n$$\n\nFrom Fourier's Law, we have:\n\n$$\n\\dot{Q}=\\frac{\\Delta T}{R_{\\mathrm{eff}}}=4.62 \\mathrm{~W}\n$$""]",['4.62'],False,N,Numerical,1e-1 825,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Three unit circles, each with radius 1 meter, lie in the same plane such that the center of each circle is one intersection point between the two other circles, as shown below. Mass is uniformly distributed among all area enclosed by at least one circle. The mass of the region enclosed by the triangle shown above is $1 \mathrm{~kg}$. Let $x$ be the moment of inertia of the area enclosed by all three circles (intersection, not union) about the axis perpendicular to the page and through the center of mass of the triangle. Then, $x$ can be expressed as $\frac{a \pi-b \sqrt{c}}{d \sqrt{e}}$ $\mathrm{kg} \mathrm{m}^{2}$, where $a, b, c, d, e$ are integers such that $\operatorname{gcd}(a, b, d)=1$ and both $c$ and $e$ are squarefree. Compute $a+b+c+d+e$. ","['Define point $O$ as the point in the plane that the axis of rotation passes through. Since moments of inertia simply add about a given axis, we can calculate the moments of inertia of the three ""sectors"" whose union forms the given area and subtract twice the moment of inertia of the triangle, so our answer will be $3 I_{s, O}-2 I_{t, O}$.\nClaim: The center of mass of a sector is $\\frac{2}{\\pi}$ away from the vertex of the sector along its axis of symmetry. Proof: We can divide the sector into arbitrarily small sectors that can be approximated as isosceles triangles. It\'s well known that the center of mass of one such isosceles triangle is $\\frac{2}{3}$ of the way from the central vertex to the base. Therefore, the center of mass of the sector is the center of mass of the arc with central angle $\\frac{\\pi}{3}$ and same center with radius $\\frac{2}{3}$ contained within the sector. Since the center of mass has to lie on the axis of symmetry, we set that as the $\\mathrm{x}$ axis with the vertex of the sector being $x=0$. Then, the $x$-coordinate of a point on the arc whose corresponding radius makes an angle of $\\theta$ with the axis of symmetry is $\\frac{2}{3} \\cos (\\theta)$. We can integrate this over all possible angles $\\left(-\\frac{\\pi}{6} \\leq \\theta \\leq \\frac{\\pi}{6}\\right)$ and then divide by the range $\\left(\\frac{\\pi}{3}\\right)$ to get the average $x$-coordinate, or the center of mass.\n\n$$\n\\begin{gathered}\n\\frac{\\int_{-\\frac{\\pi}{6}}^{\\frac{\\pi}{6}} \\frac{2}{3} \\cos (\\theta) \\mathrm{d} \\theta}{\\frac{\\pi}{3}} \\\\\n\\frac{2}{\\pi} \\int_{-\\frac{\\pi}{6}}^{\\frac{\\pi}{6}} \\cos (\\theta) \\mathrm{d} \\theta \\\\\n\\frac{2}{\\pi}\\left(\\sin \\left(\\frac{\\pi}{6}\\right)-\\sin \\left(-\\frac{\\pi}{6}\\right)\\right)\n\\end{gathered}\n$$\n\n\n\nThis concludes the proof.\n\nNow define point $X$ as the vertex of a sector and point $M$ as the center of mass of that sector. According to the parallel axis theorem,\n\n$$\nI_{s, X}=I_{s, M}+m_{s}\\left(\\frac{2}{\\pi}\\right)^{2}\n$$\n\n. It\'s well known that $I_{s, X}=\\frac{1}{2} m_{s} r^{2}=\\frac{m_{s}}{2}$, and so\n\n$$\nI_{s, M}=\\frac{m_{s}}{2}-\\frac{4 m_{s}}{\\pi^{2}}=m_{s}\\left(\\frac{\\pi^{2}-8}{2 \\pi^{2}}\\right)\n$$\n\nIt\'s also well known that $O$ is on the line of symmetry and a distance of $\\frac{1}{\\sqrt{3}}$ away from $X$, and so $M X=$ $\\frac{2}{\\pi}-\\frac{1}{\\sqrt{3}}$. Therefore,\n\n$$\nI_{s, O}=I_{s, M}+m_{s}\\left(\\frac{2}{\\pi}-\\frac{1}{\\sqrt{3}}\\right)^{2}=m_{s}\\left(\\frac{5 \\pi-8 \\sqrt{3}}{6 \\pi}\\right)\n$$\n\nIt\'s well known that, since $O$ is the center of mass of the triangle,\n\n$$\nI_{t, O}=\\frac{1}{12}\n$$\n\nNow we just need to calculate $m_{s}$. Since the mass of the triangle is $1 \\mathrm{~kg}$, this is equivalent to finding the ratio of the area of a sector to the area of a triangle. Through geometry, this is found to be $\\frac{2 \\pi}{3 \\sqrt{3}}$. Finally, we get our answer to be\n\n$$\n\\left(\\frac{2 \\pi}{\\sqrt{3}}\\right)\\left(\\frac{5 \\pi-8 \\sqrt{3}}{6 \\pi}\\right)-\\frac{1}{6}=\\left(\\frac{10 \\pi-17 \\sqrt{3}}{6 \\sqrt{3}}\\right)\n$$\n\nand $a+b+c+d+e=10+17+3+6+3=039$']",['39'],False,,Numerical,1e0 825,Mechanics,,"Three unit circles, each with radius 1 meter, lie in the same plane such that the center of each circle is one intersection point between the two other circles, as shown below. Mass is uniformly distributed among all area enclosed by at least one circle. The mass of the region enclosed by the triangle shown above is $1 \mathrm{~kg}$. Let $x$ be the moment of inertia of the area enclosed by all three circles (intersection, not union) about the axis perpendicular to the page and through the center of mass of the triangle. Then, $x$ can be expressed as $\frac{a \pi-b \sqrt{c}}{d \sqrt{e}}$ $\mathrm{kg} \mathrm{m}^{2}$, where $a, b, c, d, e$ are integers such that $\operatorname{gcd}(a, b, d)=1$ and both $c$ and $e$ are squarefree. Compute $a+b+c+d+e$. ![](https://cdn.mathpix.com/cropped/2023_12_21_73d8111634dc54a6ce52g-1.jpg?height=491&width=523&top_left_y=1034&top_left_x=801)","['Define point $O$ as the point in the plane that the axis of rotation passes through. Since moments of inertia simply add about a given axis, we can calculate the moments of inertia of the three ""sectors"" whose union forms the given area and subtract twice the moment of inertia of the triangle, so our answer will be $3 I_{s, O}-2 I_{t, O}$.\nClaim: The center of mass of a sector is $\\frac{2}{\\pi}$ away from the vertex of the sector along its axis of symmetry. Proof: We can divide the sector into arbitrarily small sectors that can be approximated as isosceles triangles. It\'s well known that the center of mass of one such isosceles triangle is $\\frac{2}{3}$ of the way from the central vertex to the base. Therefore, the center of mass of the sector is the center of mass of the arc with central angle $\\frac{\\pi}{3}$ and same center with radius $\\frac{2}{3}$ contained within the sector. Since the center of mass has to lie on the axis of symmetry, we set that as the $\\mathrm{x}$ axis with the vertex of the sector being $x=0$. Then, the $x$-coordinate of a point on the arc whose corresponding radius makes an angle of $\\theta$ with the axis of symmetry is $\\frac{2}{3} \\cos (\\theta)$. We can integrate this over all possible angles $\\left(-\\frac{\\pi}{6} \\leq \\theta \\leq \\frac{\\pi}{6}\\right)$ and then divide by the range $\\left(\\frac{\\pi}{3}\\right)$ to get the average $x$-coordinate, or the center of mass.\n\n$$\n\\begin{gathered}\n\\frac{\\int_{-\\frac{\\pi}{6}}^{\\frac{\\pi}{6}} \\frac{2}{3} \\cos (\\theta) \\mathrm{d} \\theta}{\\frac{\\pi}{3}} \\\\\n\\frac{2}{\\pi} \\int_{-\\frac{\\pi}{6}}^{\\frac{\\pi}{6}} \\cos (\\theta) \\mathrm{d} \\theta \\\\\n\\frac{2}{\\pi}\\left(\\sin \\left(\\frac{\\pi}{6}\\right)-\\sin \\left(-\\frac{\\pi}{6}\\right)\\right)\n\\end{gathered}\n$$\n\n\n\nThis concludes the proof.\n\nNow define point $X$ as the vertex of a sector and point $M$ as the center of mass of that sector. According to the parallel axis theorem,\n\n$$\nI_{s, X}=I_{s, M}+m_{s}\\left(\\frac{2}{\\pi}\\right)^{2}\n$$\n\n. It\'s well known that $I_{s, X}=\\frac{1}{2} m_{s} r^{2}=\\frac{m_{s}}{2}$, and so\n\n$$\nI_{s, M}=\\frac{m_{s}}{2}-\\frac{4 m_{s}}{\\pi^{2}}=m_{s}\\left(\\frac{\\pi^{2}-8}{2 \\pi^{2}}\\right)\n$$\n\nIt\'s also well known that $O$ is on the line of symmetry and a distance of $\\frac{1}{\\sqrt{3}}$ away from $X$, and so $M X=$ $\\frac{2}{\\pi}-\\frac{1}{\\sqrt{3}}$. Therefore,\n\n$$\nI_{s, O}=I_{s, M}+m_{s}\\left(\\frac{2}{\\pi}-\\frac{1}{\\sqrt{3}}\\right)^{2}=m_{s}\\left(\\frac{5 \\pi-8 \\sqrt{3}}{6 \\pi}\\right)\n$$\n\nIt\'s well known that, since $O$ is the center of mass of the triangle,\n\n$$\nI_{t, O}=\\frac{1}{12}\n$$\n\nNow we just need to calculate $m_{s}$. Since the mass of the triangle is $1 \\mathrm{~kg}$, this is equivalent to finding the ratio of the area of a sector to the area of a triangle. Through geometry, this is found to be $\\frac{2 \\pi}{3 \\sqrt{3}}$. Finally, we get our answer to be\n\n$$\n\\left(\\frac{2 \\pi}{\\sqrt{3}}\\right)\\left(\\frac{5 \\pi-8 \\sqrt{3}}{6 \\pi}\\right)-\\frac{1}{6}=\\left(\\frac{10 \\pi-17 \\sqrt{3}}{6 \\sqrt{3}}\\right)\n$$\n\nand $a+b+c+d+e=10+17+3+6+3=039$']",['39'],False,,Numerical,1e0 826,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Life on Earth would not exist as we know it without the atmosphere. There are many reasons for this, but one of which is temperature. Let's explore how the atmosphere affects the temperature on Earth. Assume that all thermal energy striking the earth uniformly and ideally distributes itself across the Earth's surface. - Assume that the Earth is a perfect black body with no atmospheric effects. Let the equilibrium temperature of Earth be $T_{0}$. (The sun outputs around $3.846 \times 10^{2^{6}} \mathrm{~W}$, and is $1.496 \times 10^{8} \mathrm{~km}$ away.) - Now assume the Earth's atmosphere is isothermal. The short wavelengths from the sun are nearly unaffected and pass straight through the atmosphere. However, they mostly convert into heat when they strike the ground. This generates longer wavelengths that do interact with the atmosphere. Assume that the albedo of the ground is 0.3 and $e$, the emissivity and absorptivity of the atmosphere, is 0.8 . Let the equilibrium average temperature of the planet now be $T_{1}$. What is the percentage increase from $T_{0}$ to $T_{1}$ ? Note: The emissivity is the degree to which an object can emit longer wavelengths (infrared) and the absorptivity is the degree to which an object can absorb energy. Specifically, the emissivity is the ratio between the energy emitted by an object and the energy emitted by a perfect black body at the same temperature. On the other hand, the absorptivity is the ratio of the amount of energy absorbed to the amount of incident energy.","['Let us solve this problem in the case of the Earth being a graybody first and then substitute values for when it is a blackbody. The portion of energy that reaches the Earth is given by the ratio between the cross-sectional area of the satellite and the area of an imaginary sphere centered around the sun with a radius of $L$. Thus, the incoming radiation is multiplied by a factor of $\\gamma=(R / 2 L)^{2}$. The energy from the sun that the surface absorbs is $\\gamma(1-\\alpha) E$, where $E$ is the energy output of the sun. Here $\\gamma=1 / 4$ as the sphere encompassing will be 4 times the area of its intercept.\nWe can now write two systems of equations at the atmosphere and the ground of the Earth. At the top of the atmosphere, we require equilibrium meaning that zero net radiation leaves the atmosphere or:\n\n$$\n-\\frac{1}{4} S_{0}(1-\\alpha)+\\varepsilon \\sigma T_{a}^{4}+(1-\\varepsilon) \\sigma T_{s}^{4}=0\n$$\n\nSimilarly, at the ground, we write another equilibrium equation of:\n\n$$\n\\frac{1}{4} S_{0}(1-\\alpha)+\\varepsilon \\sigma T_{a}^{4}-\\sigma T_{s}^{4}=0\n$$\n\nThus, solving the ground equilibrium equation yields us $T_{a}=2^{-1 / 4} T_{s}$ and plugging back into the atmosphere equilibrium equation tells us:\n\n$$\n\\frac{1}{4} S_{0}(1-\\alpha)=\\left(1-\\frac{\\varepsilon}{2}\\right) \\sigma T_{s}^{4} \\Longrightarrow T_{s}=289.601 \\mathrm{~K}\n$$']",['289.601'],False,K,Numerical,1e0 827,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two infinitely long current carrying wires carry constant current $i_{1}=2 \mathrm{~A}$ and $i_{2}=3 \mathrm{~A}$ as shown in the diagram. The equations of the wire curvatures are $y^{2}-8 x-6 y+25=0$ and $x=0$. Find the magnitude of force (in Newtons) acting on one of the wires due to the other. Note: The current-carrying wires are rigidly fixed. The units for distances on the graph should be taken in metres.",['The magnetic field from the wire is given by $B=\\frac{\\mu_{0} i_{1}}{2 \\pi x}$. Let $\\theta$ be the direction of a component of force from the vertical. It is then seen that\n$$\nd F=B i_{2} d \\ell \\Longrightarrow d F_{x}=B i_{2} d \\ell \\sin \\theta=B i_{2} d y\n$$\n\nWe only consider the force in the $x$-direction which means that\n\n$$\nF_{x}=\\int_{\\infty}^{\\infty} d F_{x}=\\frac{\\mu_{0} i_{1} i_{2}}{2 \\pi} \\int_{-\\infty}^{\\infty} \\frac{d y}{x}\n$$\n\nSolving the equation in terms of $x$ and then plugging in gives us\n\n$$\nF_{x}=\\frac{8 \\mu_{0} i_{1} i_{2}}{2 \\pi} \\int_{-\\infty}^{\\infty} \\frac{d y}{y^{2}-6 y+25}=\\frac{8 \\mu_{0} i_{1} i_{2}}{2 \\pi} \\cdot \\frac{\\pi}{4}=\\mu_{0} i_{1} i_{2}=7.5398 \\cdot 10^{-6} \\mathrm{~N} .\n$$'],['$7.5398 \\cdot 10^{-6}$'],False,N,Numerical,1e-7 827,Electromagnetism,,"Two infinitely long current carrying wires carry constant current $i_{1}=2 \mathrm{~A}$ and $i_{2}=3 \mathrm{~A}$ as shown in the diagram. The equations of the wire curvatures are $y^{2}-8 x-6 y+25=0$ and $x=0$. Find the magnitude of force (in Newtons) acting on one of the wires due to the other. ![](https://cdn.mathpix.com/cropped/2023_12_21_e2bb42caca393489220dg-1.jpg?height=1041&width=1507&top_left_y=1339&top_left_x=301) Note: The current-carrying wires are rigidly fixed. The units for distances on the graph should be taken in metres.",['The magnetic field from the wire is given by $B=\\frac{\\mu_{0} i_{1}}{2 \\pi x}$. Let $\\theta$ be the direction of a component of force from the vertical. It is then seen that\n$$\nd F=B i_{2} d \\ell \\Longrightarrow d F_{x}=B i_{2} d \\ell \\sin \\theta=B i_{2} d y\n$$\n\nWe only consider the force in the $x$-direction which means that\n\n$$\nF_{x}=\\int_{\\infty}^{\\infty} d F_{x}=\\frac{\\mu_{0} i_{1} i_{2}}{2 \\pi} \\int_{-\\infty}^{\\infty} \\frac{d y}{x}\n$$\n\nSolving the equation in terms of $x$ and then plugging in gives us\n\n$$\nF_{x}=\\frac{8 \\mu_{0} i_{1} i_{2}}{2 \\pi} \\int_{-\\infty}^{\\infty} \\frac{d y}{y^{2}-6 y+25}=\\frac{8 \\mu_{0} i_{1} i_{2}}{2 \\pi} \\cdot \\frac{\\pi}{4}=\\mu_{0} i_{1} i_{2}=7.5398 \\cdot 10^{-6} \\mathrm{~N} .\n$$'],['$7.5398 \\cdot 10^{-6}$'],False,N,Numerical,1e-7 828,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Mountains have two sides: windward and leeward. The windward side faces the wind and typically receives warm, moist air, often from an ocean. As wind hits a mountain, it is forced upward and begins to move towards the leeward side. During social distancing, Rishab decides to cross a mountain from the windward side to the leeward side of the mountain. What he finds is that the air around him has warmed when he is on the leeward side of the mountain. Let us investigate this effect. Consider the warm, moist air mass colliding with the mountain and moving upwards on the mountain. Disregard heat exchange with the air mass and the mountain. Let the humidity of the air on the windward side correspond to a partial vapor pressure $0.5 \mathrm{kPa}$ at $100.2 \mathrm{kPa}$ and have a molar mass of $\mu_{a}=28 \mathrm{~g} / \mathrm{mole}$. The air predominantly consists of diatomic molecules of oxygen and nitrogen. Assume the mountain to be very high which means that at the very top of the mountain, all of the moisture in the air condenses and falls as precipitation. Let the precipitation have a heat of vaporization $L=2.4 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}$ and molar mass $\mu_{p}=18.01 \mathrm{~g} / \mathrm{mole}$. Calculate the total change in temperature from the windward side to the leeward side in degrees Celsius.","['We use the first law of thermodynamics to solve this problem. For diatomic molecules, the internal energy per mole is given by $\\frac{5}{2} R T$. If the molar mass of the air is $\\mu_{a}$, then we have that the change in internal energy of the air is given by\n$$\n\\Delta U=\\frac{5}{2} \\frac{M}{\\mu_{a}} R \\Delta T\n$$\n\nWe also note that the total work performed by the gas is\n\n$$\nW=P_{2} V_{2}-P_{1} V_{1}\n$$\n\nsince the process is adiabatic, we can use the ideal gas equation $P V=\\nu R T=(M / \\mu) R T$ to express the total work as\n\n$$\nW=\\frac{M}{\\mu_{a}} R \\Delta T\n$$\n\nThe heat that is taken away during condensation at the top of the mountain is given by $Q=L \\Delta m$ where $\\Delta m$ is the total mass of the precipitation. According to the ideal gas law, we have that\n\n$$\nP V_{1}=\\frac{\\Delta m}{\\mu_{p}} R T_{1}, \\quad P_{1} V_{1}=\\frac{M}{\\mu_{a}} R T_{1}\n$$\n\nrecombining these equations and equating them gives us\n\n$$\n\\begin{aligned}\n\\frac{\\Delta m}{\\mu_{p} P} R T_{1} & =\\frac{M}{\\mu_{a} P_{1}} R T_{1} \\\\\n\\Delta m & =M \\frac{\\mu_{p} P}{\\mu_{a} P_{1}}\n\\end{aligned}\n$$\n\n\n\nTherefore,\n\n$$\nQ=L M \\frac{\\mu_{a} P}{\\mu_{p} P_{1}}\n$$\n\nWe finally can now use the first law of thermodynamics\n\n$$\nQ=\\Delta U+W \\Longrightarrow L M \\frac{\\mu_{p} P}{\\mu_{a} P_{1}}=\\frac{5}{2} \\frac{M}{\\mu_{a}} R \\Delta T+\\frac{M}{\\mu_{a}} R \\Delta T\n$$\n\nWe then simplify this equation to get\n\n$$\n\\begin{aligned}\nL M \\frac{\\mu_{p} P}{\\mu_{a} P_{1}} & =\\frac{7}{2} \\frac{M}{\\mu_{a}} R \\Delta T \\\\\n\\Delta T & =\\frac{2}{7} \\frac{L \\mu_{p} P}{R P_{1}}=7.41 \\mathrm{~K}\n\\end{aligned}\n$$\n\nA simpler approach could be to assume that the number of moles of water vapour in the atmosphere is equal to number of moles of water condensed. Then, the mass of precipitated water is $\\mu_{p} n \\frac{P}{P_{1}}$, where $n$ is the number of moles of air. Thus,\n\n$$\n\\mu_{p} n \\frac{P}{P_{1}} L=\\frac{7}{2} n R \\Delta T\n$$']",['7.41'],False,K,Numerical,1e-1 829,Electromagnetism,,"Two electrons are in a uniform electric field $\mathbf{E}=E_{0} \hat{\mathbf{z}}$ where $E_{0}=10^{-11} \mathrm{~N} / \mathrm{C}$. One electron is at the origin, and another is $10 \mathrm{~m}$ above the first electron. The electron at the origin is moving at $u=10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ from the line connecting the electrons at $t=0$, while the other electron is at rest at $t=0$. Find the minimum distance between the electrons. You may neglect relativistic effects. ![](https://cdn.mathpix.com/cropped/2023_12_21_a93d620a5fbe0da0bef9g-1.jpg?height=605&width=1049&top_left_y=1351&top_left_x=538)","['Let $\\ell=10 \\mathrm{~m}$. First, switch into the reference frame accelerating at $-\\frac{E q}{m} \\hat{z}$. In this frame, the electrons are not affected by the electric field. Now, switch into the center of mass reference frame from here. In this frame, we have both conservation of angular momentum and conservation of energy. Both electrons in this frame are moving at $\\frac{u}{2}$ initially at an angle of $\\theta=30^{\\circ}$. At the smallest distance, both electrons will be moving perpendicular to the line connecting them. Suppose that they both move with speed $v$ and are a distance $r$ from the center of mass. By conservation of angular momentum,\n$$\n\\begin{gathered}\n2 m \\cdot \\frac{u}{2} \\cdot \\frac{\\ell}{2} \\sin \\theta=2 m v r \\\\\nv r=\\frac{u \\ell}{4} \\sin \\theta\n\\end{gathered}\n$$\n\n\n\nNow, by conservation of energy,\n\n$$\nm v^{2}+\\frac{k e^{2}}{2 r}=\\frac{1}{4} m u^{2}+\\frac{k e^{2}}{\\ell}\n$$\n\nNow, we just solve this system of equations to determine the value of $r$. Substituting $v=\\frac{u l}{4 r} \\sin \\theta$ into the conservation of energy equation, we can solve the ensuing quadratic to find:\n\n$$\nr=\\frac{\\frac{k e^{2}}{2}+\\sqrt{\\left(\\frac{k e^{2}}{2}\\right)^{2}+\\left(m u^{2}+\\frac{4 k e^{2}}{\\ell}\\right)\\left(\\frac{m u^{2} \\ell^{2}}{16} \\sin ^{2}(\\theta)\\right)}}{\\frac{1}{2} m u^{2}+\\frac{2 k e^{2}}{\\ell}}\n$$\n\nFinally, remembering that the distance between the electrons is actually $2 r$, we obtain $2 r=6.84 \\mathrm{~m}$ as the final answer.']",['6.84'],False,m,Numerical,1e-1 829,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two electrons are in a uniform electric field $\mathbf{E}=E_{0} \hat{\mathbf{z}}$ where $E_{0}=10^{-11} \mathrm{~N} / \mathrm{C}$. One electron is at the origin, and another is $10 \mathrm{~m}$ above the first electron. The electron at the origin is moving at $u=10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ from the line connecting the electrons at $t=0$, while the other electron is at rest at $t=0$. Find the minimum distance between the electrons. You may neglect relativistic effects. ","['Let $\\ell=10 \\mathrm{~m}$. First, switch into the reference frame accelerating at $-\\frac{E q}{m} \\hat{z}$. In this frame, the electrons are not affected by the electric field. Now, switch into the center of mass reference frame from here. In this frame, we have both conservation of angular momentum and conservation of energy. Both electrons in this frame are moving at $\\frac{u}{2}$ initially at an angle of $\\theta=30^{\\circ}$. At the smallest distance, both electrons will be moving perpendicular to the line connecting them. Suppose that they both move with speed $v$ and are a distance $r$ from the center of mass. By conservation of angular momentum,\n$$\n\\begin{gathered}\n2 m \\cdot \\frac{u}{2} \\cdot \\frac{\\ell}{2} \\sin \\theta=2 m v r \\\\\nv r=\\frac{u \\ell}{4} \\sin \\theta\n\\end{gathered}\n$$\n\n\n\nNow, by conservation of energy,\n\n$$\nm v^{2}+\\frac{k e^{2}}{2 r}=\\frac{1}{4} m u^{2}+\\frac{k e^{2}}{\\ell}\n$$\n\nNow, we just solve this system of equations to determine the value of $r$. Substituting $v=\\frac{u l}{4 r} \\sin \\theta$ into the conservation of energy equation, we can solve the ensuing quadratic to find:\n\n$$\nr=\\frac{\\frac{k e^{2}}{2}+\\sqrt{\\left(\\frac{k e^{2}}{2}\\right)^{2}+\\left(m u^{2}+\\frac{4 k e^{2}}{\\ell}\\right)\\left(\\frac{m u^{2} \\ell^{2}}{16} \\sin ^{2}(\\theta)\\right)}}{\\frac{1}{2} m u^{2}+\\frac{2 k e^{2}}{\\ell}}\n$$\n\nFinally, remembering that the distance between the electrons is actually $2 r$, we obtain $2 r=6.84 \\mathrm{~m}$ as the final answer.']",['6.84'],False,m,Numerical,1e-1 830,Electromagnetism,,"Consider a long uniform conducting cylinder. First, we divide the cylinder into thirds and remove the middle third. Then, we perform the same steps on the remaining two cylinders. Again, we perform the same steps on the remaining four cylinders and continuing until there are 2048 cylinders. We then connect the terminals of the cylinder to a battery and measure the effective capacitance to be $C_{1}$. If we continue to remove cylinders, the capacitance will reach an asymptotic value of $C_{0}$. What is $C_{1} / C_{0}$ ? You may assume each cylindrical disk to be wide enough to be considered as an infinite plate, such that the radius $R$ of the cylinders is much larger than the $d$ between any successive cylinders. ![](https://cdn.mathpix.com/cropped/2023_12_21_09b501d99d90cb579ed8g-1.jpg?height=192&width=1006&top_left_y=1272&top_left_x=556) Note: The diagram is not to scale.","[""The capacitance is proportional to $C \\propto \\frac{1}{d}$, where $d$ is the distance between successive parallel plates. When we add capacitor plates in series, their effective capacitance will be:\n$$\nC \\propto\\left(\\frac{1}{1 / d_{1}}+\\frac{1}{1 / d_{2}}+\\cdots\\right)^{-1}=\\frac{1}{d_{1}+d_{2}+\\cdots} \\Longrightarrow C \\propto \\frac{1}{d_{\\text {total }}}\n$$\n\nTherefore, this essentially becomes a math problem: What is the total length of the spacing in between? Between successive 'cuts', the length of each cylinder is cut down by $1 / 3$, but the number of gaps double. Therefore, the spacing grows by a factor of $2 / 3$ each time. For $n=2^{1}$, the spacing starts off as $1 / 3$. For $n=2^{10}$, the spacing becomes:\n\n$$\n\\frac{1}{C_{\\text {eff }}} \\propto d=\\frac{1}{3}\\left(\\frac{1-(2 / 3)^{10}}{1-2 / 3}\\right) L=0.983 L\n$$\n\nfor $n \\rightarrow \\infty$, it is clear the total spacing will converge to $L$. Therefore:\n\n$$\nC_{1} / C_{0}=1.017\n$$""]",['1.017'],False,,Numerical,1e-1 830,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a long uniform conducting cylinder. First, we divide the cylinder into thirds and remove the middle third. Then, we perform the same steps on the remaining two cylinders. Again, we perform the same steps on the remaining four cylinders and continuing until there are 2048 cylinders. We then connect the terminals of the cylinder to a battery and measure the effective capacitance to be $C_{1}$. If we continue to remove cylinders, the capacitance will reach an asymptotic value of $C_{0}$. What is $C_{1} / C_{0}$ ? You may assume each cylindrical disk to be wide enough to be considered as an infinite plate, such that the radius $R$ of the cylinders is much larger than the $d$ between any successive cylinders. Note: The diagram is not to scale.","[""The capacitance is proportional to $C \\propto \\frac{1}{d}$, where $d$ is the distance between successive parallel plates. When we add capacitor plates in series, their effective capacitance will be:\n$$\nC \\propto\\left(\\frac{1}{1 / d_{1}}+\\frac{1}{1 / d_{2}}+\\cdots\\right)^{-1}=\\frac{1}{d_{1}+d_{2}+\\cdots} \\Longrightarrow C \\propto \\frac{1}{d_{\\text {total }}}\n$$\n\nTherefore, this essentially becomes a math problem: What is the total length of the spacing in between? Between successive 'cuts', the length of each cylinder is cut down by $1 / 3$, but the number of gaps double. Therefore, the spacing grows by a factor of $2 / 3$ each time. For $n=2^{1}$, the spacing starts off as $1 / 3$. For $n=2^{10}$, the spacing becomes:\n\n$$\n\\frac{1}{C_{\\text {eff }}} \\propto d=\\frac{1}{3}\\left(\\frac{1-(2 / 3)^{10}}{1-2 / 3}\\right) L=0.983 L\n$$\n\nfor $n \\rightarrow \\infty$, it is clear the total spacing will converge to $L$. Therefore:\n\n$$\nC_{1} / C_{0}=1.017\n$$""]",['1.017'],False,,Numerical,1e-1 831,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A square based pyramid (that is symmetrical) is standing on top of a cube with side length $\ell=10 \mathrm{~cm}$ such that their square faces perfectly line up. The cube is initially standing still on flat ground and both objects have the same uniform density. The coefficient of friction between every surface is the same value of $\mu=0.3$. The cube is then given an initial speed $v$ in some direction parallel to the floor. What is the maximum possible value of $v$ such that the base of the pyramid will always remain parallel to the top of the cube? Answer in meters per second. ","['Let $x$ be the relative displacement of the two objects. Then:\n$$\nv_{i}^{2}=2 a x\n$$\n\nwhere the relative acceleration is:\n\n$$\na=\\frac{2 h}{3 \\ell} g \\mu\n$$\n\nwork The acceleration of pyramid is $g \\mu$ and the acceleration of cube is:\n\n$$\n\\rho \\ell^{3} a=\\rho\\left(\\ell^{3}+\\frac{\\ell^{2} h}{3}\\right) g \\mu+\\rho \\frac{\\ell^{2} h}{3} g \\mu \\Longrightarrow a=\\frac{3 \\ell+2 h}{3 \\ell} g \\mu\n$$\n\nTherefore, the relative acceleration is:\n\n$$\na=\\frac{6 \\ell+2 h}{3 \\ell} g \\mu\n$$\n\nWe want to direct the motion of the cube diagonally (such that horizontal sides of the cube form a 45 degree angle with the displacement). Initially, we may think that we need to let $x=\\frac{\\sqrt{2}}{2} \\ell$ but it can start tipping before that. Moving into a non-inertial reference frame for the pyramid, we see that the effective gravity needs to point towards the back corner of the cube, so it needs to satisfy the criteria\n\n$$\n\\frac{h_{\\mathrm{cm}}}{\\frac{\\sqrt{2}}{2} \\ell-x}=\\frac{m g}{m g \\mu} \\Longrightarrow x=\\frac{\\sqrt{2}}{2} \\ell-\\mu \\frac{1}{4} h\n$$\n\n\n\nHere I used the fact that the center of mass was $1 / 4$ of the way up. Substituting, we get:\n\n$$\nv=\\sqrt{2\\left(\\frac{6 \\ell+2 h}{3 \\ell} g \\mu\\right)\\left(\\frac{\\sqrt{2}}{2} l-\\frac{\\mu}{4} h\\right)}=\\sqrt{2 g \\mu\\left(2+\\frac{2 h}{3 \\ell}\\right)\\left(\\frac{\\sqrt{2}}{2} \\ell-\\frac{\\mu}{4} h\\right)}\n$$\n\nThe maximum $v$ occurs at\n\n$$\nh / \\ell=\\frac{-3+\\frac{2 \\sqrt{2}}{\\mu}}{2}\n$$\n\ngiving a maximum $v$ of $v_{\\max }=1.07 \\mathrm{~m} / \\mathrm{s}$.']",['1.07'],False,m/s,Numerical,1e-1 831,Mechanics,,"A square based pyramid (that is symmetrical) is standing on top of a cube with side length $\ell=10 \mathrm{~cm}$ such that their square faces perfectly line up. The cube is initially standing still on flat ground and both objects have the same uniform density. The coefficient of friction between every surface is the same value of $\mu=0.3$. The cube is then given an initial speed $v$ in some direction parallel to the floor. What is the maximum possible value of $v$ such that the base of the pyramid will always remain parallel to the top of the cube? Answer in meters per second. ![](https://cdn.mathpix.com/cropped/2023_12_21_19adaf49165bed09665eg-1.jpg?height=1017&width=772&top_left_y=554&top_left_x=671)","['Let $x$ be the relative displacement of the two objects. Then:\n$$\nv_{i}^{2}=2 a x\n$$\n\nwhere the relative acceleration is:\n\n$$\na=\\frac{2 h}{3 \\ell} g \\mu\n$$\n\nwork The acceleration of pyramid is $g \\mu$ and the acceleration of cube is:\n\n$$\n\\rho \\ell^{3} a=\\rho\\left(\\ell^{3}+\\frac{\\ell^{2} h}{3}\\right) g \\mu+\\rho \\frac{\\ell^{2} h}{3} g \\mu \\Longrightarrow a=\\frac{3 \\ell+2 h}{3 \\ell} g \\mu\n$$\n\nTherefore, the relative acceleration is:\n\n$$\na=\\frac{6 \\ell+2 h}{3 \\ell} g \\mu\n$$\n\nWe want to direct the motion of the cube diagonally (such that horizontal sides of the cube form a 45 degree angle with the displacement). Initially, we may think that we need to let $x=\\frac{\\sqrt{2}}{2} \\ell$ but it can start tipping before that. Moving into a non-inertial reference frame for the pyramid, we see that the effective gravity needs to point towards the back corner of the cube, so it needs to satisfy the criteria\n\n$$\n\\frac{h_{\\mathrm{cm}}}{\\frac{\\sqrt{2}}{2} \\ell-x}=\\frac{m g}{m g \\mu} \\Longrightarrow x=\\frac{\\sqrt{2}}{2} \\ell-\\mu \\frac{1}{4} h\n$$\n\n\n\nHere I used the fact that the center of mass was $1 / 4$ of the way up. Substituting, we get:\n\n$$\nv=\\sqrt{2\\left(\\frac{6 \\ell+2 h}{3 \\ell} g \\mu\\right)\\left(\\frac{\\sqrt{2}}{2} l-\\frac{\\mu}{4} h\\right)}=\\sqrt{2 g \\mu\\left(2+\\frac{2 h}{3 \\ell}\\right)\\left(\\frac{\\sqrt{2}}{2} \\ell-\\frac{\\mu}{4} h\\right)}\n$$\n\nThe maximum $v$ occurs at\n\n$$\nh / \\ell=\\frac{-3+\\frac{2 \\sqrt{2}}{\\mu}}{2}\n$$\n\ngiving a maximum $v$ of $v_{\\max }=1.07 \\mathrm{~m} / \\mathrm{s}$.']",['1.07'],False,m/s,Numerical,1e-1 832,Mechanics,,"During quarantine, the FBI has been monitoring a young physicists suspicious activities. After compiling weeks worth of evidence, the FBI finally has had enough and searches his room. The room's door is opened with a high angular velocity about its hinge. Over a very short period of time, its angular velocity increases to $\omega=8.56 \mathrm{rad} / \mathrm{s}$ due to the force applied at the end opposite from the hinge. For simplicity, treat the door as a uniform thin rod of length $L=1.00 \mathrm{~m}$ and mass $M=9.50 \mathrm{~kg}$. The hinge (pivot) is located at one end of the rod. Ignore gravity. At what distance from the hinge of the door is the door most likely to break? Assume that the door will break at where the bending moment is largest. (Answer in metres.) ![](https://cdn.mathpix.com/cropped/2023_12_21_949eb8cca07484578c58g-1.jpg?height=212&width=605&top_left_y=1225&top_left_x=760)","[""Let $N$ be the force from the pivot and $F$ be the applied force at the end. Let $\\alpha$ be the angular acceleration. Writing the torque equation and Newton's 2nd law for the whole door, we get:\n$$\n\\begin{aligned}\n& F \\cdot L=\\frac{1}{3} M L^{2} \\alpha \\\\\n& N+F=\\frac{1}{2} M L \\alpha\n\\end{aligned}\n$$\n\nSolving, we get $F=\\frac{1}{3} M L \\alpha$ and $N=\\frac{1}{6} M L \\alpha$. Now, we consider the part of the door with length $x$ attached to the pivot. The rest of the door applies a torque $\\tau$ and shear force $f$ on our system. (There is also tension force). Let $\\lambda=\\frac{M}{L}$. We can write the torque equation and Newton's 2 nd law for our system:\n\n$$\n\\begin{aligned}\n& \\tau+f x=\\frac{1}{3} \\lambda x^{3} \\alpha \\\\\n& N+f=\\lambda x \\cdot \\frac{x}{2} \\alpha\n\\end{aligned}\n$$\n\nSolving, we get\n\n$$\n\\tau=\\frac{1}{6} \\lambda x \\alpha\\left(L^{2}-x^{2}\\right)\n$$\n\nand\n\n$$\nf=\\frac{1}{6} \\lambda \\alpha\\left(3 x^{2}-L^{2}\\right)\n$$\n\nWe maximize $\\tau$ (which is equivalent to maximizing bending moment) to get $x=\\frac{L}{\\sqrt{3}}=0.577 \\mathrm{~m}$""]",['$\\frac{\\sqrt{3}}{3}$'],False,m,Numerical,1e-2 832,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","During quarantine, the FBI has been monitoring a young physicists suspicious activities. After compiling weeks worth of evidence, the FBI finally has had enough and searches his room. The room's door is opened with a high angular velocity about its hinge. Over a very short period of time, its angular velocity increases to $\omega=8.56 \mathrm{rad} / \mathrm{s}$ due to the force applied at the end opposite from the hinge. For simplicity, treat the door as a uniform thin rod of length $L=1.00 \mathrm{~m}$ and mass $M=9.50 \mathrm{~kg}$. The hinge (pivot) is located at one end of the rod. Ignore gravity. At what distance from the hinge of the door is the door most likely to break? Assume that the door will break at where the bending moment is largest. (Answer in metres.) ","[""Let $N$ be the force from the pivot and $F$ be the applied force at the end. Let $\\alpha$ be the angular acceleration. Writing the torque equation and Newton's 2nd law for the whole door, we get:\n$$\n\\begin{aligned}\n& F \\cdot L=\\frac{1}{3} M L^{2} \\alpha \\\\\n& N+F=\\frac{1}{2} M L \\alpha\n\\end{aligned}\n$$\n\nSolving, we get $F=\\frac{1}{3} M L \\alpha$ and $N=\\frac{1}{6} M L \\alpha$. Now, we consider the part of the door with length $x$ attached to the pivot. The rest of the door applies a torque $\\tau$ and shear force $f$ on our system. (There is also tension force). Let $\\lambda=\\frac{M}{L}$. We can write the torque equation and Newton's 2 nd law for our system:\n\n$$\n\\begin{aligned}\n& \\tau+f x=\\frac{1}{3} \\lambda x^{3} \\alpha \\\\\n& N+f=\\lambda x \\cdot \\frac{x}{2} \\alpha\n\\end{aligned}\n$$\n\nSolving, we get\n\n$$\n\\tau=\\frac{1}{6} \\lambda x \\alpha\\left(L^{2}-x^{2}\\right)\n$$\n\nand\n\n$$\nf=\\frac{1}{6} \\lambda \\alpha\\left(3 x^{2}-L^{2}\\right)\n$$\n\nWe maximize $\\tau$ (which is equivalent to maximizing bending moment) to get $x=\\frac{L}{\\sqrt{3}}=0.577 \\mathrm{~m}$""]",['$\\frac{\\sqrt{3}}{3}$'],False,m,Numerical,1e-2 833,Mechanics,,"A solid half-disc of mass $m=1 \mathrm{~kg}$ in the shape of a semi-circle of radius $R=1 \mathrm{~m}$ is kept at rest on a smooth horizontal table. QiLin starts applying a constant force of magnitude $F=10 \mathrm{~N}$ at point A as shown, parallel to its straight edge. What is the initial linear acceleration of point B? (Answer in $\mathrm{m} / \mathrm{s}^{2}$ ) ![](https://cdn.mathpix.com/cropped/2023_12_21_325b4a9088b5394ee833g-1.jpg?height=409&width=1004&top_left_y=438&top_left_x=558) Note: the diagram above is a top down view.","[""Let C denote the location of the centre of mass of the disc. It is well known that $O C=\\frac{4 R}{3 \\pi}$. Note that the initial angular velocity of the disc about its centre of mass is 0 , and the linear acceleration is simply\n$$\na_{\\mathrm{CM}}=\\frac{F}{m} \\hat{i}\n$$\n\nNow we use the $\\tau=\\vec{r} \\times \\vec{F}=I_{\\mathrm{CM}} \\alpha_{\\mathrm{CM}}$ about the centre of mass of the disc\n\n$$\n\\left(\\frac{4 R}{3 \\pi}\\right) F=I_{\\mathrm{CM}} \\alpha_{\\mathrm{CM}}\n$$\n\nTo compute the moment of inertia of the disc about its centre of mass, we use Steiner's theorem:\n\n$$\nI_{\\mathrm{CM}}=\\frac{M R^{2}}{2}-M\\left(\\frac{4 R}{3 \\pi}\\right)^{2}\n$$\n\nso the $\\tau=I \\alpha$ equation becomes\n\n$$\n\\left(\\frac{4 R}{3 \\pi}\\right) F=\\left[\\frac{M R^{2}}{2}-M\\left(\\frac{4 R}{3 \\pi}\\right)^{2}\\right] \\alpha_{\\mathrm{CM}} \\Rightarrow \\alpha_{\\mathrm{CM}}=\\frac{\\frac{4 F R}{3 \\pi}}{M R^{2}\\left(\\frac{1}{2}-\\frac{16}{9 \\pi^{2}}\\right)}\n$$\n\nUsing kinematics equation for rotational motion, we have\n\n$$\n\\vec{a}_{B}=\\vec{a}_{\\mathrm{CM}}+\\vec{\\alpha}_{\\mathrm{CM}} \\times \\overrightarrow{\\mathrm{CB}}\n$$\n\nSubstituting the values of $\\overrightarrow{\\mathrm{CB}}=R \\hat{i}-\\frac{4 R}{3 \\pi} \\hat{j}$ and $\\alpha_{\\mathrm{CM}}=\\frac{\\frac{4 F R}{3 \\pi}}{M R^{2}\\left(\\frac{1}{2}-\\frac{16}{9 \\pi^{2}}\\right)} \\hat{k}$ we get\n\n$$\na_{B}=\\frac{F}{m}\\left[\\left(1+\\frac{1}{-1+\\frac{9 \\pi^{2}}{32}}\\right) \\hat{i}+\\frac{2}{3 \\pi\\left(\\frac{1}{2}+\\frac{16}{9 \\pi^{2}}\\right)} \\hat{j}\\right]=15.9395 \\mathrm{~m} / \\mathrm{s}^{2}\n$$\n\nand we are done.""]",['$15.9395$'],False,$\mathrm{m} / \mathrm{s}^{2}$,Numerical,1e-1 833,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A solid half-disc of mass $m=1 \mathrm{~kg}$ in the shape of a semi-circle of radius $R=1 \mathrm{~m}$ is kept at rest on a smooth horizontal table. QiLin starts applying a constant force of magnitude $F=10 \mathrm{~N}$ at point A as shown, parallel to its straight edge. What is the initial linear acceleration of point B? (Answer in $\mathrm{m} / \mathrm{s}^{2}$ ) Note: the diagram above is a top down view.","[""Let C denote the location of the centre of mass of the disc. It is well known that $O C=\\frac{4 R}{3 \\pi}$. Note that the initial angular velocity of the disc about its centre of mass is 0 , and the linear acceleration is simply\n$$\na_{\\mathrm{CM}}=\\frac{F}{m} \\hat{i}\n$$\n\nNow we use the $\\tau=\\vec{r} \\times \\vec{F}=I_{\\mathrm{CM}} \\alpha_{\\mathrm{CM}}$ about the centre of mass of the disc\n\n$$\n\\left(\\frac{4 R}{3 \\pi}\\right) F=I_{\\mathrm{CM}} \\alpha_{\\mathrm{CM}}\n$$\n\nTo compute the moment of inertia of the disc about its centre of mass, we use Steiner's theorem:\n\n$$\nI_{\\mathrm{CM}}=\\frac{M R^{2}}{2}-M\\left(\\frac{4 R}{3 \\pi}\\right)^{2}\n$$\n\nso the $\\tau=I \\alpha$ equation becomes\n\n$$\n\\left(\\frac{4 R}{3 \\pi}\\right) F=\\left[\\frac{M R^{2}}{2}-M\\left(\\frac{4 R}{3 \\pi}\\right)^{2}\\right] \\alpha_{\\mathrm{CM}} \\Rightarrow \\alpha_{\\mathrm{CM}}=\\frac{\\frac{4 F R}{3 \\pi}}{M R^{2}\\left(\\frac{1}{2}-\\frac{16}{9 \\pi^{2}}\\right)}\n$$\n\nUsing kinematics equation for rotational motion, we have\n\n$$\n\\vec{a}_{B}=\\vec{a}_{\\mathrm{CM}}+\\vec{\\alpha}_{\\mathrm{CM}} \\times \\overrightarrow{\\mathrm{CB}}\n$$\n\nSubstituting the values of $\\overrightarrow{\\mathrm{CB}}=R \\hat{i}-\\frac{4 R}{3 \\pi} \\hat{j}$ and $\\alpha_{\\mathrm{CM}}=\\frac{\\frac{4 F R}{3 \\pi}}{M R^{2}\\left(\\frac{1}{2}-\\frac{16}{9 \\pi^{2}}\\right)} \\hat{k}$ we get\n\n$$\na_{B}=\\frac{F}{m}\\left[\\left(1+\\frac{1}{-1+\\frac{9 \\pi^{2}}{32}}\\right) \\hat{i}+\\frac{2}{3 \\pi\\left(\\frac{1}{2}+\\frac{16}{9 \\pi^{2}}\\right)} \\hat{j}\\right]=15.9395 \\mathrm{~m} / \\mathrm{s}^{2}\n$$\n\nand we are done.""]",['$15.9395$'],False,$\mathrm{m} / \mathrm{s}^{2}$,Numerical,1e-1 834,Mechanics,,"A regular tetrahedron of mass $m=1 \mathrm{~g}$ and unknown side length is balancing on top of a hemisphere of mass $M=100 \mathrm{~kg}$ and radius $R=100 \mathrm{~m}$. The hemisphere is placed on a flat surface such that it is at its lowest potential. For a certain value of the length of the regular tetrahedron, the oscillations become unstable. What is this side length of the tetrahedron? ![](https://cdn.mathpix.com/cropped/2023_12_21_ac3dcf6b8fd2f77ccf4bg-1.jpg?height=382&width=403&top_left_y=481&top_left_x=861)","['For stable equilibrium the height resultant from a slight displacement must be greater than the original height of the center of mass of the cube. A small displacement from the original position can be modeled as an $\\mathrm{x}$ displacement of\n$$\nR d \\theta\n$$\n\nwhich raises the height by\n\n$$\nR d \\theta \\sin (d \\theta)\n$$\n\nadded to the height\n\n$$\ns \\cos (d \\theta)\n$$\n\nThis must be greater than $s$, where $s$ is the distance from the center of mass to the point of contact to the sphere. Approximating to the second degree of $\\theta$ using Taylor series, we solve the inequality and get that $s=R$. The altitude of the tetrahedron is therefore $4 s=400$\n\n$$\nl^{2}-(l \\sqrt{3} / 3)^{2}=400^{2}\n$$\n\nTherefore, $l=\\sqrt{400^{2} \\cdot 3 / 2}=490 \\mathrm{~m}$']",['$200\\sqrt{6}$'],False,m,Numerical,5e0 834,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A regular tetrahedron of mass $m=1 \mathrm{~g}$ and unknown side length is balancing on top of a hemisphere of mass $M=100 \mathrm{~kg}$ and radius $R=100 \mathrm{~m}$. The hemisphere is placed on a flat surface such that it is at its lowest potential. For a certain value of the length of the regular tetrahedron, the oscillations become unstable. What is this side length of the tetrahedron? ","['For stable equilibrium the height resultant from a slight displacement must be greater than the original height of the center of mass of the cube. A small displacement from the original position can be modeled as an $\\mathrm{x}$ displacement of\n$$\nR d \\theta\n$$\n\nwhich raises the height by\n\n$$\nR d \\theta \\sin (d \\theta)\n$$\n\nadded to the height\n\n$$\ns \\cos (d \\theta)\n$$\n\nThis must be greater than $s$, where $s$ is the distance from the center of mass to the point of contact to the sphere. Approximating to the second degree of $\\theta$ using Taylor series, we solve the inequality and get that $s=R$. The altitude of the tetrahedron is therefore $4 s=400$\n\n$$\nl^{2}-(l \\sqrt{3} / 3)^{2}=400^{2}\n$$\n\nTherefore, $l=\\sqrt{400^{2} \\cdot 3 / 2}=490 \\mathrm{~m}$']",['$200\\sqrt{6}$'],False,m,Numerical,5e0 835,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A planet has a radius of $10 \mathrm{~km}$ and a uniform density of $5 \mathrm{~g} / \mathrm{cm}^{3}$. A powerful bomb detonates at the center of the planet, releasing $8.93 \times 10^{17} \mathrm{~J}$ of energy, causing the planet to separate into three large sections each with equal masses. You may model each section as a perfect sphere of radius $r^{\prime}$. The initial and final distances between the centers of any two given sections is $2 r^{\prime}$. How long does it take for the three sections to collide again?","['Due to conservation of momentum, the three masses must form an equilateral triangle at all times. Let us determine the force as a function of $r$, the distance between each mass and the center.\n\n\n\n\nThe vector sum of the net force on any individual mass is\n\n$$\nF=\\frac{G m^{2}}{d^{2}} \\sqrt{2-2 \\cos 120^{\\circ}}=\\frac{\\sqrt{3} G m^{2}}{d^{2}}\n$$\n\nwhere $d$ is the distance between the mass and the center.\n\n$$\nd^{2}=3 r^{2}\n$$\n\nThe net force is thus\n\n$$\nF=\\frac{G m^{2}}{\\sqrt{3} r^{2}}\n$$\n\nThe system behaves as if there was a stationary mass $m^{\\prime}=m / \\sqrt{3}$ at the center, simplifying the problem greatly into a restricted two body system. Next, we need to figure out the height of the apoapsis. This can be done via conservation of energy.\n\n$$\nE_{\\text {binding,initial }}+E=3 E_{\\text {binding,final }}-\\frac{G m^{2}}{\\sqrt{3} \\ell} \\Longrightarrow \\ell=-\\frac{G m^{2}}{\\sqrt{3}\\left(-\\frac{3 G M^{2}}{5 R}+8.93 \\cdot 10^{17}-\\left(-3 \\frac{3 G m^{2}}{5 r_{f}}\\right)\\right)}=101,000 \\mathrm{~m}\n$$\n\nIf you have a stationary mass $M$ at the center. The time it takes for an object to fall into it is:\n\n$$\nT=\\pi \\sqrt{\\frac{\\ell^{3}}{8 G M}}\n$$\n\nour time will be double this, and the mass in the center will be $M=m / \\sqrt{3}$. So plugging in numbers gives:\n\n$$\nt=2 \\pi \\sqrt{\\frac{l^{3}}{8 G\\left(\\frac{m}{\\sqrt{3}}\\right)}}=138,000\n$$\n\n\n\nIf we take into account a nonzero radius so final separation is $10 / \\cos \\left(30^{\\circ}\\right) \\mathrm{km}$, then the answer should be:\n\n$$\n-2 \\int_{l}^{\\frac{10000}{\\cos \\left(\\frac{\\pi}{6}\\right)}}\\left(\\sqrt{\\frac{x l}{2 G\\left(\\frac{m}{\\sqrt{3}}\\right)(l-x)}}\\right) d x=136,000 \\mathrm{~s}\n$$\n\nNote if you set the upper bound to zero, you get the same answer as before. Both these answers will be accepted.']",['136000'],False,s,Numerical,1e3 836,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A point charge $+q$ is placed a distance $a$ away from an infinitely large conducting plate. The force of the electrostatic interaction is $F_{0}$. Then, an identical conducting plate is placed a distance $3 a$ from the charge, parallel to the first one such that the charge is ""sandwiched in."" The new electrostatic force the particle feels is $F^{\prime}$. What is $F^{\prime} / F_{0}$ ? Round to the nearest hundredths.","[""We solve this via the method of image charges. Let us first reflect the charge $+q$ across the closest wall. The force of this interaction will be:\n$$\nF_{0}=\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}} \\frac{1}{2^{2}}\n$$\n\nand this will be the only image charge we need to place if there were only one conducting plane. Since there is another conducting plane, another charge will be reflected to a distance $a+4 a$ past the other conducting plane, and thus will be $a+4 a+3 a=8 a$ away from the original charge. All these reflections cause a force that points in the same direction, which we will label as the positive + direction. Therefore:\n\n$$\nF_{+}=\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}}\\left(\\frac{1}{2^{2}}+\\frac{1}{8^{2}}+\\frac{1}{10^{2}}+\\frac{1}{16^{2}} \\frac{1}{18^{2}}+\\frac{1}{24^{2}}+\\cdots\\right)\n$$\n\nNow let us look at what happens if we originally reflect the charge $+q$ across the other wall. Repeating the steps above, we see that through subsequent reflections, each force will point in the negative - direction. Therefore:\n\n$$\nF_{-}=\\frac{-q^{2}}{4 \\pi \\epsilon_{0} a^{2}}\\left(\\frac{1}{6^{2}}+\\frac{1}{8^{2}}+\\frac{1}{14^{2}}+\\frac{1}{16^{2}}+\\frac{1}{22^{2}}+\\frac{1}{24^{2}}+\\cdots\\right)\n$$\n\nThe net force is a result of the superposition of these two forces, giving us:\n\n$$\n\\begin{aligned}\nF^{\\prime}=\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}} & \\left(\\frac{1}{2^{2}}+\\frac{1}{8^{2}}+\\frac{1}{10^{2}}+\\frac{1}{16^{2}} \\frac{1}{18^{2}}+\\frac{1}{24^{2}}+\\cdots\\right. \\\\\n& \\left.-\\frac{1}{6^{2}}-\\frac{1}{8^{2}}-\\frac{1}{14^{2}}-\\frac{1}{16^{2}}-\\frac{1}{22^{2}}-\\frac{1}{24^{2}}-\\cdots\\right)\n\\end{aligned}\n$$\n\nEven terms can be cancelled out to give:\n\n$$\n\\begin{aligned}\nF^{\\prime} & =\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}}\\left(\\frac{1}{2^{2}}-\\frac{1}{6^{2}}+\\frac{1}{10^{2}}-\\frac{1}{14^{2}}+\\frac{1}{18^{2}}-\\frac{1}{22^{2}}+\\cdots\\right) \\\\\n& =\\frac{q^{2}}{4 \\pi \\epsilon_{0} a^{2}} \\frac{1}{4}\\left(\\frac{1}{1^{2}}-\\frac{1}{3^{2}}+\\frac{1}{5^{2}}-\\frac{1}{7^{2}}+\\frac{1}{9^{2}}-\\frac{1}{11^{2}}+\\cdots\\right)\n\\end{aligned}\n$$\n\nYou may recognize the infinite series inside the parentheses to be Catalan's constant $G \\approx 0.916$. Alternatively, you can use a calculator and evaluate the first seven terms to get a rough answer (but will still be correct since we asked for it to be rounded). Therefore:\n\n$$\nF^{\\prime} / F=\\frac{1}{1^{2}}-\\frac{1}{3^{2}}+\\frac{1}{5^{2}}-\\frac{1}{7^{2}}+\\frac{1}{9^{2}}-\\frac{1}{11^{2}}+\\cdots=G \\approx 0.916\n$$""]",['0.916'],False,,Numerical,5e-2 837,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Jerry spots a truckload of his favourite golden yellow Swiss cheese being transported on a cart moving at a constant velocity $v_{0}=5 \mathrm{~m} / \mathrm{s} \hat{i}$ along the x-axis, which is initially placed at $(0,0)$. Jerry, driven by desire immediately starts pursuing the cheese-truck in such a way that his velocity vector always points towards the cheese-truck; however, Jerry is smart and knows that he must maintain a constant distance $\ell=10 \mathrm{~m}$ from the truck to avoid being caught by anyone, no matter what. Note that Jerry starts at coordinates $(0, \ell)$. Let the magnitude of velocity (in $\mathrm{m} / \mathrm{s}$ ) and acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) of Jerry at the moment when the (acute) angle between the two velocity vectors is $\theta=60^{\circ}$ be $\alpha$ and $\beta$ respectively. Compute $\alpha^{2}+\beta^{2}$.","[""If the distance between Jerry and the cheese truck is constant, then Jerry moves in circle of radius $\\ell$ in the reference frame of the cheese truck. There is no radial component of Jerry's velocity in this reference frame, so we must have $\\alpha=v_{0} \\cos \\theta=\\frac{5}{2}$. In this case, the tangential velocity is $v_{0} \\sin \\theta$. Furthermore, the radial acceleration in this frame is given by the centripetal acceleration which is $\\frac{\\left(v_{0} \\sin \\theta\\right)^{2}}{\\ell}=\\frac{v_{0}^{2} \\sin ^{2} \\theta}{\\ell}$. The tangential acceleration is\n\n$$\n\\frac{d}{d t}\\left(v_{0} \\sin \\theta\\right)=v_{0} \\cos \\theta \\cdot \\frac{d \\theta}{d t}=v_{0} \\cos \\theta \\cdot \\frac{-v_{0} \\sin \\theta}{\\ell}=-\\frac{v_{0}^{2} \\sin \\theta \\cos \\theta}{\\ell}\n$$\n\nThe vector sum of these accelerations has magnitude\n\n$$\n\\beta=\\sqrt{\\left(\\frac{v_{0}^{2} \\sin ^{2} \\theta}{\\ell}\\right)^{2}+\\left(\\frac{v_{0}^{2} \\sin \\theta \\cos \\theta}{\\ell}\\right)^{2}}=\\frac{v_{0}^{2} \\sin \\theta}{\\ell}=\\frac{5 \\sqrt{3}}{4} .\n$$\n\nThe final answer is $\\alpha^{2}+\\beta^{2}=10.9375$.""]",['10.9375'],False,,Numerical,1e-1 838,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball of mass $M=5 \mathrm{~kg}$. The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different alternative. Now let the circular loop have a uniform coefficient of friction $\mu=0.6$, while the rest of the path is still frictionless. Assume that the balls can once again collide with a restitution coefficient of $e=\frac{1}{2}$. Considering the balls to be point masses, find the minimum value of $h$ such that the ball of mass $M$ would be able to move all the way around the loop. Both balls can be considered as point masses.","['Let the angle formed by $M$ at any moment of time be angle $\\theta$ with the negative y-axis. The normal force experienced by $M$ is just\n$$\nN=M g \\cos \\theta+M \\frac{v(\\theta)^{2}}{R}\n$$\n\nby balancing the radial forces at this moment. Now, applying the work energy theorem, we have\n\n$$\n\\begin{gathered}\n\\int-\\mu\\left[M g \\cos \\theta+M \\frac{v(\\theta)^{2}}{R}\\right] R \\mathrm{~d} \\theta=\\frac{1}{2} M v(\\theta)^{2}-\\frac{1}{2} M v_{0}^{2}+M g R(1-\\cos \\theta) \\\\\n\\Rightarrow-\\mu\\left[M g \\cos \\theta+M \\frac{v(\\theta)^{2}}{R}\\right] R=\\frac{M}{2} \\frac{\\mathrm{d}\\left(v(\\theta)^{2}\\right)}{\\mathrm{d} \\theta}+M g R \\sin \\theta\n\\end{gathered}\n$$\n\nRearranging, we have\n\n$$\n\\frac{\\left.\\mathrm{d}\\left(v(\\theta)^{2}\\right)\\right)}{\\mathrm{d} \\theta}+2 \\mu v(\\theta)^{2}=-2 g R(\\sin \\theta+\\mu \\cos \\theta)\n$$\n\nLet $v^{2}(\\theta)=y$. Thus we have a first order linear ODE of the form\n\n$$\n\\frac{\\mathrm{d} y}{\\mathrm{~d} \\theta}+P(\\theta) y=Q(\\theta)\n$$\n\nThis is easily solvable using the integrating factor $e^{\\int P(\\theta) \\mathrm{d} \\theta}$. Here the integrating factor is\n\n$$\ne^{\\int 2 \\mu \\mathrm{d} \\theta}=e^{2 \\mu \\theta}\n$$\n\nSo multiplying by the integrating factor, we get\n\n$$\n\\begin{gathered}\n\\int \\mathrm{d}\\left(e^{2 \\mu \\theta} y\\right)=\\int-2 g R(\\sin \\theta+\\mu \\cos \\theta) e^{2 \\mu \\theta} \\mathrm{d} \\theta \\\\\n\\Rightarrow y=\\frac{\\int-2 g R(\\sin \\theta+\\mu \\cos \\theta) e^{2 \\mu \\theta} \\mathrm{d} \\theta}{e^{2 \\mu \\theta}}\n\\end{gathered}\n$$\n\nNow we use the well known integrals\n\n$$\n\\begin{aligned}\n& \\int e^{a x} \\sin x \\mathrm{~d} x=\\frac{e^{a x}}{1+a^{2}}(a \\sin x-\\cos x) \\\\\n& \\int e^{a x} \\cos x \\mathrm{~d} x=\\frac{e^{a x}}{1+a^{2}}(a \\cos x+\\sin x)\n\\end{aligned}\n$$\n\n(These integrals can be computed using integration by parts.) Thus, plugging and chugging these integration formulas into our expression for $y$ and integrating from $\\theta=0$ to $\\theta=\\phi$, we have upon solving\n\n$$\nv^{2}(\\phi)-v_{0}^{2}=\\frac{-2 g R}{1+4 \\mu^{2}}\\left[\\left(3 \\mu \\sin \\phi+\\left(2 \\mu^{2}-1\\right) \\cos \\phi-\\left(2 \\mu^{2}-1\\right) e^{-2 \\mu \\phi}\\right]\\right.\n$$\n\nwhere $v_{0}$ is the velocity at $\\phi=0$. Solving gives us the velocity as a function of angle covered\n\n$$\nv(\\phi)=\\sqrt{v_{0}^{2}-\\frac{2 g R}{1+4 \\mu^{2}}\\left[\\left(3 \\mu \\sin \\phi+\\left(2 \\mu^{2}-1\\right) \\cos \\phi-\\left(2 \\mu^{2}-1\\right) e^{-2 \\mu \\phi}\\right]\\right.}\n$$\n\nBut to cover a complete circle, at the top most point\n\n$$\nN=m g-\\frac{m v^{2}(\\pi)}{R} \\geq 0 \\Rightarrow v(\\pi) \\leq \\sqrt{g R}\n$$\n\nThus\n\n$$\nv_{0} \\leq \\sqrt{g R\\left[1+\\frac{2\\left(1-2 \\mu^{2}\\right)}{1+4 \\mu^{2}}\\left(1+e^{-2 \\mu \\pi}\\right)\\right]}\n$$\n\n\n\nFrom the previous expression,\n\n$$\nv_{0}=\\frac{m(1+e) \\sqrt{2 g h}}{M+m} \\geq \\sqrt{g R\\left[1+\\frac{2\\left(1-2 \\mu^{2}\\right)}{1+4 \\mu^{2}}\\left(1+e^{-2 \\mu \\pi}\\right)\\right]}\n$$\n\nHence\n\n$$\nh \\geq \\frac{R(M+m)^{2}}{2 m^{2}(1+e)^{2}}\\left[1+\\frac{2\\left(1-2 \\mu^{2}\\right)}{1+4 \\mu^{2}}\\left(1+e^{-2 \\mu \\pi}\\right)\\right]\n$$\n\nWe get $h \\geq 72.902 \\mathrm{~m}$ and we are done.']",['$72.902$'],False,m,Numerical,1e-1 838,Mechanics,,"A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball of mass $M=5 \mathrm{~kg}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_e1013618489f33954a52g-1.jpg?height=361&width=797&top_left_y=1725&top_left_x=661) The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different alternative. Now let the circular loop have a uniform coefficient of friction $\mu=0.6$, while the rest of the path is still frictionless. Assume that the balls can once again collide with a restitution coefficient of $e=\frac{1}{2}$. Considering the balls to be point masses, find the minimum value of $h$ such that the ball of mass $M$ would be able to move all the way around the loop. Both balls can be considered as point masses.","['Let the angle formed by $M$ at any moment of time be angle $\\theta$ with the negative y-axis. The normal force experienced by $M$ is just\n$$\nN=M g \\cos \\theta+M \\frac{v(\\theta)^{2}}{R}\n$$\n\nby balancing the radial forces at this moment. Now, applying the work energy theorem, we have\n\n$$\n\\begin{gathered}\n\\int-\\mu\\left[M g \\cos \\theta+M \\frac{v(\\theta)^{2}}{R}\\right] R \\mathrm{~d} \\theta=\\frac{1}{2} M v(\\theta)^{2}-\\frac{1}{2} M v_{0}^{2}+M g R(1-\\cos \\theta) \\\\\n\\Rightarrow-\\mu\\left[M g \\cos \\theta+M \\frac{v(\\theta)^{2}}{R}\\right] R=\\frac{M}{2} \\frac{\\mathrm{d}\\left(v(\\theta)^{2}\\right)}{\\mathrm{d} \\theta}+M g R \\sin \\theta\n\\end{gathered}\n$$\n\nRearranging, we have\n\n$$\n\\frac{\\left.\\mathrm{d}\\left(v(\\theta)^{2}\\right)\\right)}{\\mathrm{d} \\theta}+2 \\mu v(\\theta)^{2}=-2 g R(\\sin \\theta+\\mu \\cos \\theta)\n$$\n\nLet $v^{2}(\\theta)=y$. Thus we have a first order linear ODE of the form\n\n$$\n\\frac{\\mathrm{d} y}{\\mathrm{~d} \\theta}+P(\\theta) y=Q(\\theta)\n$$\n\nThis is easily solvable using the integrating factor $e^{\\int P(\\theta) \\mathrm{d} \\theta}$. Here the integrating factor is\n\n$$\ne^{\\int 2 \\mu \\mathrm{d} \\theta}=e^{2 \\mu \\theta}\n$$\n\nSo multiplying by the integrating factor, we get\n\n$$\n\\begin{gathered}\n\\int \\mathrm{d}\\left(e^{2 \\mu \\theta} y\\right)=\\int-2 g R(\\sin \\theta+\\mu \\cos \\theta) e^{2 \\mu \\theta} \\mathrm{d} \\theta \\\\\n\\Rightarrow y=\\frac{\\int-2 g R(\\sin \\theta+\\mu \\cos \\theta) e^{2 \\mu \\theta} \\mathrm{d} \\theta}{e^{2 \\mu \\theta}}\n\\end{gathered}\n$$\n\nNow we use the well known integrals\n\n$$\n\\begin{aligned}\n& \\int e^{a x} \\sin x \\mathrm{~d} x=\\frac{e^{a x}}{1+a^{2}}(a \\sin x-\\cos x) \\\\\n& \\int e^{a x} \\cos x \\mathrm{~d} x=\\frac{e^{a x}}{1+a^{2}}(a \\cos x+\\sin x)\n\\end{aligned}\n$$\n\n(These integrals can be computed using integration by parts.) Thus, plugging and chugging these integration formulas into our expression for $y$ and integrating from $\\theta=0$ to $\\theta=\\phi$, we have upon solving\n\n$$\nv^{2}(\\phi)-v_{0}^{2}=\\frac{-2 g R}{1+4 \\mu^{2}}\\left[\\left(3 \\mu \\sin \\phi+\\left(2 \\mu^{2}-1\\right) \\cos \\phi-\\left(2 \\mu^{2}-1\\right) e^{-2 \\mu \\phi}\\right]\\right.\n$$\n\nwhere $v_{0}$ is the velocity at $\\phi=0$. Solving gives us the velocity as a function of angle covered\n\n$$\nv(\\phi)=\\sqrt{v_{0}^{2}-\\frac{2 g R}{1+4 \\mu^{2}}\\left[\\left(3 \\mu \\sin \\phi+\\left(2 \\mu^{2}-1\\right) \\cos \\phi-\\left(2 \\mu^{2}-1\\right) e^{-2 \\mu \\phi}\\right]\\right.}\n$$\n\nBut to cover a complete circle, at the top most point\n\n$$\nN=m g-\\frac{m v^{2}(\\pi)}{R} \\geq 0 \\Rightarrow v(\\pi) \\leq \\sqrt{g R}\n$$\n\nThus\n\n$$\nv_{0} \\leq \\sqrt{g R\\left[1+\\frac{2\\left(1-2 \\mu^{2}\\right)}{1+4 \\mu^{2}}\\left(1+e^{-2 \\mu \\pi}\\right)\\right]}\n$$\n\n\n\nFrom the previous expression,\n\n$$\nv_{0}=\\frac{m(1+e) \\sqrt{2 g h}}{M+m} \\geq \\sqrt{g R\\left[1+\\frac{2\\left(1-2 \\mu^{2}\\right)}{1+4 \\mu^{2}}\\left(1+e^{-2 \\mu \\pi}\\right)\\right]}\n$$\n\nHence\n\n$$\nh \\geq \\frac{R(M+m)^{2}}{2 m^{2}(1+e)^{2}}\\left[1+\\frac{2\\left(1-2 \\mu^{2}\\right)}{1+4 \\mu^{2}}\\left(1+e^{-2 \\mu \\pi}\\right)\\right]\n$$\n\nWe get $h \\geq 72.902 \\mathrm{~m}$ and we are done.']",['$72.902$'],False,m,Numerical,1e-1 839,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the ""ground"" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless. Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree. Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$. ","['First, we must have $\\omega^{2} R=g$ and $k=m \\omega^{2}$. Now, we step into the frame of point $A$, rotating around with the spaceship. We will thus have three fictitious forces: translational, centrifugal, and coriolis. Note that because centrifugal is $m \\omega^{2} r$, pointing away from $A$, it cancels with the spring force. Thus, the only forces left to consider are translational and coriolis.\nThe translational force points ""down"" with a constant magnitude of $m g$, like gravity. The coriolis force\n\n\n\npoints perpendicular to the velocity with magnitude $2 m \\omega v$. We recognize this setup is analogous to that of a charged particle moving in an E field and B field. It is well known that the mass will follow a cycloid shape. Writing the equation of the cycloid, and finding where the cycloid hits the circle (spaceship), we can find $\\theta$. Note that we have to use numerical methods.\n\nSpecifically, the cycloid can be parametrized as $\\frac{R}{4}(\\alpha-\\sin \\alpha, 1-\\cos \\alpha)$, and we need to find where this intersects the circle $x^{2}+y^{2}=R^{2}$, so $(\\alpha-\\sin \\alpha)^{2}+(1-\\cos \\alpha)^{2}=16$. Solving gives $\\alpha=3.307$, and since $\\cot \\theta=\\frac{1-\\cos \\alpha}{\\alpha-\\sin \\alpha}$, we have $\\theta=60.2^{\\circ}$.']",['$60.2$'],False,$^{\circ}$,Numerical,2e-1 839,Mechanics,,"Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the ""ground"" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless. Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree. Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$. ![](https://cdn.mathpix.com/cropped/2023_12_21_60ae367721568530fa75g-1.jpg?height=1206&width=1217&top_left_y=964&top_left_x=454)","['First, we must have $\\omega^{2} R=g$ and $k=m \\omega^{2}$. Now, we step into the frame of point $A$, rotating around with the spaceship. We will thus have three fictitious forces: translational, centrifugal, and coriolis. Note that because centrifugal is $m \\omega^{2} r$, pointing away from $A$, it cancels with the spring force. Thus, the only forces left to consider are translational and coriolis.\nThe translational force points ""down"" with a constant magnitude of $m g$, like gravity. The coriolis force\n\n\n\npoints perpendicular to the velocity with magnitude $2 m \\omega v$. We recognize this setup is analogous to that of a charged particle moving in an E field and B field. It is well known that the mass will follow a cycloid shape. Writing the equation of the cycloid, and finding where the cycloid hits the circle (spaceship), we can find $\\theta$. Note that we have to use numerical methods.\n\nSpecifically, the cycloid can be parametrized as $\\frac{R}{4}(\\alpha-\\sin \\alpha, 1-\\cos \\alpha)$, and we need to find where this intersects the circle $x^{2}+y^{2}=R^{2}$, so $(\\alpha-\\sin \\alpha)^{2}+(1-\\cos \\alpha)^{2}=16$. Solving gives $\\alpha=3.307$, and since $\\cot \\theta=\\frac{1-\\cos \\alpha}{\\alpha-\\sin \\alpha}$, we have $\\theta=60.2^{\\circ}$.']",['$60.2$'],False,$^{\circ}$,Numerical,2e-1 840,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",Consider an LC circuit with one inductor and one capacitor. The amplitude of the charge on the plates of the capacitor is $Q=10 \mathrm{C}$ and the two plates are initially at a distance $d=1 \mathrm{~cm}$ away from each other. The plates are then slowly pushed together to a distance $0.5 \mathrm{~cm}$ from each other. Find the resultant amplitude of charge on the parallel plates of the capacitor after this process is completed. Note that the initial current in the circuit is zero and assume that the plates are grounded.,"[""In slow steady periodic processes (when the time for the change in parameters $\\tau$ is much less than the total systems frequency $f$ ), a quantity called the adiabatic invariant $I$ is conserved ${ }^{a}$. The adiabatic invariant corresponds to the area of a closed contour in phase space (a graph with momentum $p$ and position $x$ as its axes). Note the we can electrostatically map this problem to a mechanics one as the charge corresponds to position, while the momentum would correspond to $L I$ where $I$ is the current and $L$ is the inductance. Thus, in phase space, we have an elliptical contour corresponding to the equation: $\\frac{Q^{2}}{2 C}+\\frac{(L I)^{2}}{2 L}=C$ where $C$ is a constant in the system. As the area under the curve is conserved, then it can be written that $\\pi Q_{0} L I_{0}=\\pi Q_{f} L I_{f}$. It is also easy to conserve energy such that $L I^{2}=\\frac{Q^{2}}{C}$ which tells us $I=\\frac{Q}{\\sqrt{L C}}$. As $C \\propto 1 / x$, we then can write the adiabatic invariant as $x q^{4}$ which tells us $Q_{f}=\\sqrt[4]{2} Q$.\nWe can also solve this regularly by looking at the changes analytically. From Gauss's law, the electric field between the plates of the capacitators initially can be estimated as\n\n$$\nE=\\frac{Q}{2 \\varepsilon_{0} A}\n$$\n\nwhere $A$ is the area of the plate. The plates of the capacitator is attracted to the other one with a force of\n\n$$\nF=Q E=\\frac{Q^{2}}{2 \\varepsilon_{0} A}\n$$\n\nThe charges of the plates as a function of time can be approximated as\n\n$$\nQ_{c}= \\pm Q \\sin (\\omega t+\\phi)\n$$\n\nwhere $\\omega=\\frac{1}{\\sqrt{L C}}$. Using this equation, we estimate the average force $\\langle F\\rangle$ applied on the plate after a period of oscillations to be\n\n$$\n\\langle F\\rangle=\\frac{\\left\\langle Q^{2}\\right\\rangle}{2 \\varepsilon_{0} A}=\\frac{Q^{2}}{2 \\varepsilon_{0} A}\\left\\langle\\sin ^{2}(\\omega t+\\phi)\\right\\rangle=\\frac{Q^{2}}{2 \\varepsilon_{0} A} \\cdot\\left(\\frac{1}{2 \\pi} \\int_{0}^{2 \\pi} \\sin ^{2}(x) d x\\right)=\\frac{Q^{2}}{4 \\varepsilon_{0} A}\n$$\n\nthis means that after one period, the amount of work done to push the plates closer together is given by\n\n$$\nW_{F}=\\langle F\\rangle d x=\\frac{Q^{2}}{4 \\varepsilon_{0} A} d x\n$$\n\nIn this cycle, the amount of incremental work done by the $\\mathrm{LC}$ circuit will be given by\n\n$$\nd W_{\\mathrm{LC}}=\\Delta(F x)=\\Delta\\left(\\frac{Q^{2} x}{2 \\varepsilon_{0} A}\\right)=\\frac{Q x}{\\varepsilon_{0} A} d Q+\\frac{Q^{2}}{2 \\varepsilon_{0} A} d x\n$$\n\n\n\nFrom conservation of energy, $W_{F}=W_{L C}$. Or in other words,\n\n$$\n\\frac{Q^{2}}{4 \\varepsilon_{0} A} d x=\\frac{Q x}{\\varepsilon_{0} A} d Q+\\frac{Q^{2}}{2 \\varepsilon_{0} A} d x\n$$\n\nsimplifying gives us\n\n$$\n\\begin{aligned}\n\\frac{Q x}{\\varepsilon_{0} A} d Q & =-\\frac{Q^{2}}{4 \\varepsilon_{0} A} d x \\\\\n\\frac{1}{4} \\int \\frac{d x}{x} & =-\\int \\frac{d Q}{Q} \\\\\n\\frac{1}{4} \\ln x+\\ln Q & =\\text { const. }\n\\end{aligned}\n$$\n\nWe now find our adiabatic invariant to be\n\n$$\nx Q^{4}=\\text { const. }\n$$\n\nSubstituting values into our equation, we find that\n\n$$\nd Q_{i}^{4}=\\frac{d}{2} Q_{f}^{4} \\Longrightarrow Q_{f}=\\sqrt[4]{2} Q=11.892 \\mathrm{C}\n$$[^0]""]",['$11.892$'],False,C,Numerical,2e-1 841,Mechanics,,"A bicycle wheel of mass $M=2.8 \mathrm{~kg}$ and radius $R=0.3 \mathrm{~m}$ is spinning with angular velocity $\omega=5 \mathrm{rad} / \mathrm{s}$ around its axis in outer space, and its center is motionless. Assume that it has all of its mass uniformly concentrated on the rim. A long, massless axle is attached to its center, extending out along its axis. A ball of mass $m=1.0 \mathrm{~kg}$ moves at velocity $v=2 \mathrm{~m} / \mathrm{s}$ parallel to the plane of the wheel and hits the axle at a distance $h=0.5 \mathrm{~m}$ from the center of the wheel. Assume that the collision is elastic and instantaneous, and that the ball's trajectory (before and after the collision) lies on a straight line. ![](https://cdn.mathpix.com/cropped/2023_12_21_ace597fee09961022231g-1.jpg?height=1207&width=1130&top_left_y=562&top_left_x=492) Find the time it takes for the axle to return to its original orientation. Answer in seconds and round to three significant figures.","[""After the collision, let the wheel have speed $v_{1}$ and the ball have speed $v_{2}$. Conserving momentum, energy, and angular momentum gives:\n$$\n\\begin{gathered}\nm v=M v_{1}+m v_{2} \\\\\n\\frac{1}{2} m v^{2}+\\frac{1}{2} M R^{2} \\omega^{2}=\\frac{1}{2} M v_{1}^{2}+\\frac{1}{2} m v_{2}^{2}+\\frac{1}{2} M R^{2} \\omega^{2}+\\frac{1}{2} \\cdot \\frac{1}{2} M R^{2} \\omega_{1}^{2} \\\\\nm\\left(v-v_{2}\\right) h=\\frac{1}{2} M R^{2} \\omega_{1}\n\\end{gathered}\n$$\nwhere $\\omega_{1}$ is the angular velocity (after collision) of the wheel in the direction perp. to the axis and the velocity of the ball.\n\n\n\nSolving for $\\omega_{1}$, we get\n$$\n\\omega_{1}=\\frac{4 h m v}{m\\left(R^{2}+2 h^{2}\\right)+M R^{2}}\n$$\nNow, we realize that the angular momentum of the wheel is given by $I_{x} \\omega \\hat{x}+I_{y} \\omega_{1} \\hat{y}$ where the wheel's axis is the $\\mathrm{x}$-axis and the $\\mathrm{y}$-axis is in the direction of $\\omega_{1}$. Since angular momentum is conserved, the wheel must precess about its angular momentum vector. Let $\\hat{L}$ represent the direction of the angular momentum vector. To find the rate of precession, we can decompose the angular velocity vector $\\omega \\hat{x}+\\omega_{1} \\hat{y}$ into a $\\hat{L}$ component and a $\\hat{x}$ component. Since $I_{x}=2 I_{y}$, the $\\hat{L}$ component is $\\sqrt{(2 \\omega)^{2}+\\omega_{1}^{2}}$, resulting in a precession period of\n$$\nT=\\frac{\\pi}{\\sqrt{\\omega^{2}+\\frac{\\omega_{1}^{2}}{4}}}=0.568 s\n$$""]",['0.568'],False,s,Numerical,5e-2 841,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A bicycle wheel of mass $M=2.8 \mathrm{~kg}$ and radius $R=0.3 \mathrm{~m}$ is spinning with angular velocity $\omega=5 \mathrm{rad} / \mathrm{s}$ around its axis in outer space, and its center is motionless. Assume that it has all of its mass uniformly concentrated on the rim. A long, massless axle is attached to its center, extending out along its axis. A ball of mass $m=1.0 \mathrm{~kg}$ moves at velocity $v=2 \mathrm{~m} / \mathrm{s}$ parallel to the plane of the wheel and hits the axle at a distance $h=0.5 \mathrm{~m}$ from the center of the wheel. Assume that the collision is elastic and instantaneous, and that the ball's trajectory (before and after the collision) lies on a straight line. Find the time it takes for the axle to return to its original orientation. Answer in seconds and round to three significant figures.","[""After the collision, let the wheel have speed $v_{1}$ and the ball have speed $v_{2}$. Conserving momentum, energy, and angular momentum gives:\n$$\n\\begin{gathered}\nm v=M v_{1}+m v_{2} \\\\\n\\frac{1}{2} m v^{2}+\\frac{1}{2} M R^{2} \\omega^{2}=\\frac{1}{2} M v_{1}^{2}+\\frac{1}{2} m v_{2}^{2}+\\frac{1}{2} M R^{2} \\omega^{2}+\\frac{1}{2} \\cdot \\frac{1}{2} M R^{2} \\omega_{1}^{2} \\\\\nm\\left(v-v_{2}\\right) h=\\frac{1}{2} M R^{2} \\omega_{1}\n\\end{gathered}\n$$\nwhere $\\omega_{1}$ is the angular velocity (after collision) of the wheel in the direction perp. to the axis and the velocity of the ball.\n\n\n\nSolving for $\\omega_{1}$, we get\n$$\n\\omega_{1}=\\frac{4 h m v}{m\\left(R^{2}+2 h^{2}\\right)+M R^{2}}\n$$\nNow, we realize that the angular momentum of the wheel is given by $I_{x} \\omega \\hat{x}+I_{y} \\omega_{1} \\hat{y}$ where the wheel's axis is the $\\mathrm{x}$-axis and the $\\mathrm{y}$-axis is in the direction of $\\omega_{1}$. Since angular momentum is conserved, the wheel must precess about its angular momentum vector. Let $\\hat{L}$ represent the direction of the angular momentum vector. To find the rate of precession, we can decompose the angular velocity vector $\\omega \\hat{x}+\\omega_{1} \\hat{y}$ into a $\\hat{L}$ component and a $\\hat{x}$ component. Since $I_{x}=2 I_{y}$, the $\\hat{L}$ component is $\\sqrt{(2 \\omega)^{2}+\\omega_{1}^{2}}$, resulting in a precession period of\n$$\nT=\\frac{\\pi}{\\sqrt{\\omega^{2}+\\frac{\\omega_{1}^{2}}{4}}}=0.568 s\n$$""]",['0.568'],False,s,Numerical,5e-2 842,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A child attaches a small rock of mass $M=0.800 \mathrm{~kg}$ to one end of a uniform elastic string of mass $m=0.100 \mathrm{~kg}$ and natural length $L=0.650 \mathrm{~m}$. He grabs the other end and swings the rock in uniform circular motion around his hand, with angular velocity $\omega=6.30 \mathrm{rad} / \mathrm{s}$. Assume his hand is stationary, and that the elastic string behaves like a spring with spring constant $k=40.0 \mathrm{~N} / \mathrm{m}$. After that, at time $t=0$, a small longitudinal perturbation starts from the child's hand, traveling towards the rock. At time $t=T_{0}$, the perturbation reaches the rock. How far was the perturbation from the child's hand at time $t=\frac{T_{0}}{2}$ ? Ignore gravity.","[""Let $x$ be the distance from a point on the unstretched elastic string to the center of rotation (child's hand). Note that $x$ varies from 0 to $L$. However, the string stretches, so let $r$ be the distance from a point on the stretched string (in steady state) to the center of rotation. Let $T$ be the tension in the string as a function of position. Let $\\lambda=\\frac{m}{L}$. Consider a portion of the string $d x$. We know that the portion as spring constant $k \\frac{L}{d x}$ and it is stretched by $d r-d x$, so by Hooke's Law, we have $T=k \\frac{L}{d x}(d r-d x)=k L\\left(\\frac{d r}{d x}-1\\right)$. Also, by applying Newton's Second Law on the portion, we get $d T=-\\lambda d x \\cdot \\omega^{2} r$, which implies $\\frac{d T}{d x}=-\\lambda \\omega^{2} r$. Combining the two equations, we obtain\n$$\nT=-k L\\left(\\frac{1}{\\lambda \\omega^{2}} T^{\\prime \\prime}+1\\right)\n$$\nWe know that\n$$\nT^{\\prime}(x=0)=0,\n$$\nsince $r=0$ when $x=0$. The general solution is\n$$\nT=A \\cos \\left(\\frac{\\omega}{L} \\sqrt{\\frac{m}{k}} x\\right)-k L\n$$\nfor some constant $A$. Thus, we have\n$$\nr=-\\frac{1}{\\lambda \\omega^{2}} T^{\\prime}=\\frac{A}{\\omega \\sqrt{k m}} \\sin \\left(\\frac{\\omega}{L} \\sqrt{\\frac{m}{k}} x\\right)\n$$\nAlso, we have that\n$$\nT(x=L)=M \\omega^{2} \\int_{0}^{L} r x d x=M \\omega^{2} \\int_{0}^{L}\\left(\\frac{T}{k L}+1\\right) x .\n$$\nPlugging in our general solution, we can get $A \\cos \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)-k L=M \\omega^{2} \\cdot \\frac{A}{k L} \\frac{L}{\\omega} \\sqrt{\\frac{k}{m}} \\sin \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)$. Solving for $A$, we obtain\n$$\nA=\\frac{k L}{\\cos \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)-\\frac{M \\omega}{\\sqrt{k m}} \\sin \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)}\n$$\nWe now introduce a claim:\n\n\n\nClaim. The speed of a longitudinal wave on a spring with spring constant $k$, length $L$, and mass $m$ is given by $v=L \\sqrt{\\frac{k}{m}}$\n\nProof. Let a spring with spring constant $k$ and mass $m$ be stretched to length $L$. The spring constant of a small portion $d x$ of the spring is $k \\frac{L}{d x}$, and the excess tension is $\\delta T=k \\frac{L}{d x} d s=k L \\frac{d s}{d x}$, where $s$ is the displacement from equilibrium. By Newton's second law on the portion, we get $d T=\\frac{m}{L} d x \\cdot \\frac{d^{2} s}{d t^{2}}$, or $\\frac{d T}{d x}=\\frac{m}{L} \\frac{d^{2} s}{d t^{2}}$. Thus, $L^{2} \\frac{k}{m} \\frac{d^{2} s}{d x^{2}}=\\frac{d^{2} s}{d t^{2}}$, which we recognize as the wave equation with speed $v=L \\sqrt{\\frac{k}{m}}$ and the time it takes to traverse the spring is $\\sqrt{\\frac{m}{k}}$.\n\nThus, we have\n$$\nt=\\int d t=\\int_{0}^{x} \\sqrt{\\frac{\\lambda d x}{k \\cdot \\frac{L}{d x}}}=\\int_{0}^{x} \\sqrt{\\frac{\\lambda}{k L}} d x=\\sqrt{\\frac{m}{k}} \\frac{x}{L}\n$$\nSince we know $x=L$ when $t=T_{0}$, we have $x=\\frac{L}{2}$ when $t=\\frac{T_{0}}{2}$. Therefore, our answer is\n$$\nr=\\frac{A}{\\omega \\sqrt{k m}} \\sin \\left(\\frac{\\omega}{L} \\sqrt{\\frac{m}{k}} x\\right)=\\frac{1}{\\omega} \\sqrt{\\frac{k}{m}} \\frac{L \\sin \\left(\\frac{1}{2} \\omega \\sqrt{\\frac{m}{k}}\\right)}{\\cos \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)-\\frac{M \\omega}{\\sqrt{k m}} \\sin \\left(\\omega \\sqrt{\\frac{m}{k}}\\right)}=1.903 \\mathrm{~m}\n$$""]",['1.903'],False,m,Numerical,1e-1 843,Mechanics,,"A small toy car rolls down three ramps with the same height and horizontal length, but different shapes, starting from rest. The car stays in contact with the ramp at all times and no energy is lost. Order the ramps from the fastest to slowest time it takes for the toy car to drop the full $1 \mathrm{~m}$. For example, if ramp 1 is the fastest and ramp 3 is the slowest, then enter 123 as your answer choice. Figure 1 ![](https://cdn.mathpix.com/cropped/2023_12_21_45699e63605c39917edcg-1.jpg?height=439&width=480&top_left_y=645&top_left_x=275) Figure 2 ![](https://cdn.mathpix.com/cropped/2023_12_21_45699e63605c39917edcg-1.jpg?height=442&width=471&top_left_y=641&top_left_x=835) Figure 3 ![](https://cdn.mathpix.com/cropped/2023_12_21_45699e63605c39917edcg-1.jpg?height=439&width=477&top_left_y=645&top_left_x=1385)","['One can solve this by finding the time it takes for each ramp. For ramp 1:\n$$\n\\begin{aligned}\n\\sqrt{2} & =\\frac{1}{2} g \\sin \\left(45^{\\circ}\\right) t^{2} \\\\\n\\Longrightarrow t & =0.639 \\mathrm{~s}\n\\end{aligned}\n$$\n\nFor ramps 2, let the length of the dashed region be $x$. Then:\n\n$$\nx+x / \\tan \\left(30^{\\circ}\\right)=1 \\Longrightarrow x=0.366 \\mathrm{~m}\n\\tag{1}\n$$\n\nDue to symmetry, both the steep and shallow regions of both ramps 2 and 3 have a length of $x / \\cos \\left(60^{\\circ}\\right)=0.732 \\mathrm{~m}$. This results in a time for ramp 2 as:\n\n$$\n\\begin{aligned}\n& 0.732=\\frac{1}{2} g \\sin \\left(60^{\\circ}\\right) t_{1}^{2} \\Longrightarrow t_{1}=0.415 \\mathrm{~s} \\\\\n& 0.732=\\sqrt{2 g(1-x)} t_{2}+\\frac{1}{2} g \\sin \\left(30^{\\circ}\\right) t_{2}^{2} \\Longrightarrow t_{2}=0.184 \\mathrm{~s}\n\\end{aligned}\n$$\n\nfor a total time of $t_{1}+t_{2}=0.599 \\mathrm{~s}$. For ramp 3 ,\n\n$$\n\\begin{aligned}\n& 0.732=\\frac{1}{2} g \\sin \\left(30^{\\circ}\\right) t_{1}^{2} \\Longrightarrow t_{1}=0.546 \\mathrm{~s} \\\\\n& 0.732=\\sqrt{2 g(1-x)} t_{2}+\\frac{1}{2} g \\sin \\left(60^{\\circ}\\right) t_{2}^{2} \\Longrightarrow t_{2}=0.206 \\mathrm{~s}\n\\end{aligned}\n$$\n\nwhich gives a total time of $t_{1}+t_{2}=0.752 \\mathrm{~s}$. From fastest to slowest, the answer becomes 213 . Note that this answer is easily guessable via intuition.']",['213'],False,,Numerical,0 843,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A small toy car rolls down three ramps with the same height and horizontal length, but different shapes, starting from rest. The car stays in contact with the ramp at all times and no energy is lost. Order the ramps from the fastest to slowest time it takes for the toy car to drop the full $1 \mathrm{~m}$. For example, if ramp 1 is the fastest and ramp 3 is the slowest, then enter 123 as your answer choice. Figure 1 Figure 2 Figure 3 ","['One can solve this by finding the time it takes for each ramp. For ramp 1:\n$$\n\\begin{aligned}\n\\sqrt{2} & =\\frac{1}{2} g \\sin \\left(45^{\\circ}\\right) t^{2} \\\\\n\\Longrightarrow t & =0.639 \\mathrm{~s}\n\\end{aligned}\n$$\n\nFor ramps 2, let the length of the dashed region be $x$. Then:\n\n$$\nx+x / \\tan \\left(30^{\\circ}\\right)=1 \\Longrightarrow x=0.366 \\mathrm{~m}\n\\tag{1}\n$$\n\nDue to symmetry, both the steep and shallow regions of both ramps 2 and 3 have a length of $x / \\cos \\left(60^{\\circ}\\right)=0.732 \\mathrm{~m}$. This results in a time for ramp 2 as:\n\n$$\n\\begin{aligned}\n& 0.732=\\frac{1}{2} g \\sin \\left(60^{\\circ}\\right) t_{1}^{2} \\Longrightarrow t_{1}=0.415 \\mathrm{~s} \\\\\n& 0.732=\\sqrt{2 g(1-x)} t_{2}+\\frac{1}{2} g \\sin \\left(30^{\\circ}\\right) t_{2}^{2} \\Longrightarrow t_{2}=0.184 \\mathrm{~s}\n\\end{aligned}\n$$\n\nfor a total time of $t_{1}+t_{2}=0.599 \\mathrm{~s}$. For ramp 3 ,\n\n$$\n\\begin{aligned}\n& 0.732=\\frac{1}{2} g \\sin \\left(30^{\\circ}\\right) t_{1}^{2} \\Longrightarrow t_{1}=0.546 \\mathrm{~s} \\\\\n& 0.732=\\sqrt{2 g(1-x)} t_{2}+\\frac{1}{2} g \\sin \\left(60^{\\circ}\\right) t_{2}^{2} \\Longrightarrow t_{2}=0.206 \\mathrm{~s}\n\\end{aligned}\n$$\n\nwhich gives a total time of $t_{1}+t_{2}=0.752 \\mathrm{~s}$. From fastest to slowest, the answer becomes 213 . Note that this answer is easily guessable via intuition.']",['213'],False,,Numerical,0 844,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","What is the smallest number of $1 \Omega$ resistors needed such that when arranged in a certain arrangement involving only series and parallel connections, that the equivalent resistance is $\frac{7}{6} \Omega$ ?","['We can write:\n$$\n\\frac{7}{6}=\\frac{1}{2}+\\frac{2}{3}\n\\tag{2}\n$$\n\nIt takes two resistors (connected in parallel) to create a $\\frac{1}{2} \\Omega$ resistor. If we write $\\frac{2}{3}=\\frac{2 \\cdot 1}{2+1}$, then it takes three resistors to create a $\\frac{2}{3} \\Omega$ (this is accomplished by connecting a 2 resistor in parallel with a 1 resistor.\n\nCombining the $\\frac{1}{2} \\Omega$ element in series with the $\\frac{2}{3} \\Omega$ element gives us our desired amount. To prove this is the minimum, we can easily check all possible combinations using 4 or fewer resistors.']",['5'],False,,Numerical,0 845,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In a typical derby race, cars start at the top of a ramp, accelerate downwards, and race on a flat track, and are always set-up in the configuration shown below. A common technique is to change the location of the center of mass of the car to gain an advantage. Alice ensures the center of mass of her car is at the rear and Bob puts the center of mass of his car at the very front. Otherwise, their cars are exactly the same. Each car's time is defined as the time from when the car is placed on the top of the ramp to when the front of the car reaches the end of the flat track. At the competition, Alice's car beat Bob's. What is the ratio of Bob's car's time and Alice's car's time? Assume that the wheels are small and light compared to the car body, neglect air resistance, and the height of the cars are small compared to the height of the ramp. In addition, neglect all energy losses during the race and the time it takes to turn onto the horizontal surface from the ramp. Express your answer as a decimal greater than 1 .","[""The speed at which Alice's car hits the ground is given by $v_{A}=\\sqrt{2 g(1)}=4.43 \\mathrm{~m} / \\mathrm{s}$ from energy conservation. The speed at which Bob's car hits the ground is given by $v_{B}=$ $\\sqrt{2 g\\left(1-0.1 \\sin \\left(45^{\\circ}\\right)\\right.}=4.27 \\mathrm{~m} / \\mathrm{s}$ since its center of mass is at the very front of the car.\n\n\nThe time in which Alice and Bob's cars are going down the ramp (and wheels are both on the ramp) is given by:\n\n$$\n\\sqrt{2}-0.1=\\frac{1}{2} g \\sin \\left(45^{\\circ}\\right) t^{2} \\Longrightarrow t=0.616 \\mathrm{~s}\n\\tag{3}\n$$\n\nAnd the time that Alice's car takes on the flat path (and wheels are both on the floor) is given by $t_{A}=\\frac{20}{v_{A}}=4.51$, and for Bob, $t_{B}=\\frac{20}{v_{B}}=4.68$.\n\nWe ignore the time it takes for the car to transition. The justification behind this is that this time would be on the order of magnitude of $\\frac{0.1}{v_{A}}=0.02 \\mathrm{~s} \\ll t_{A}$. In fact, it is smaller than the $1 \\%$ margin allowed (so teams who did not read the clarification would still have gotten the correct answer). The ratio asked for was therefore:\n\n$$\n\\frac{0.616+t_{B}}{0.616+t_{A}}\n\\tag{4}\n$$""]",['1.03'],False,,Numerical,1e-1 845,Mechanics,,"In a typical derby race, cars start at the top of a ramp, accelerate downwards, and race on a flat track, and are always set-up in the configuration shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_72dfd40d081d9f2832ecg-1.jpg?height=634&width=1461&top_left_y=1090&top_left_x=337) A common technique is to change the location of the center of mass of the car to gain an advantage. Alice ensures the center of mass of her car is at the rear and Bob puts the center of mass of his car at the very front. Otherwise, their cars are exactly the same. Each car's time is defined as the time from when the car is placed on the top of the ramp to when the front of the car reaches the end of the flat track. At the competition, Alice's car beat Bob's. What is the ratio of Bob's car's time and Alice's car's time? Assume that the wheels are small and light compared to the car body, neglect air resistance, and the height of the cars are small compared to the height of the ramp. In addition, neglect all energy losses during the race and the time it takes to turn onto the horizontal surface from the ramp. Express your answer as a decimal greater than 1 .","[""The speed at which Alice's car hits the ground is given by $v_{A}=\\sqrt{2 g(1)}=4.43 \\mathrm{~m} / \\mathrm{s}$ from energy conservation. The speed at which Bob's car hits the ground is given by $v_{B}=$ $\\sqrt{2 g\\left(1-0.1 \\sin \\left(45^{\\circ}\\right)\\right.}=4.27 \\mathrm{~m} / \\mathrm{s}$ since its center of mass is at the very front of the car.\n\n\nThe time in which Alice and Bob's cars are going down the ramp (and wheels are both on the ramp) is given by:\n\n$$\n\\sqrt{2}-0.1=\\frac{1}{2} g \\sin \\left(45^{\\circ}\\right) t^{2} \\Longrightarrow t=0.616 \\mathrm{~s}\n\\tag{3}\n$$\n\nAnd the time that Alice's car takes on the flat path (and wheels are both on the floor) is given by $t_{A}=\\frac{20}{v_{A}}=4.51$, and for Bob, $t_{B}=\\frac{20}{v_{B}}=4.68$.\n\nWe ignore the time it takes for the car to transition. The justification behind this is that this time would be on the order of magnitude of $\\frac{0.1}{v_{A}}=0.02 \\mathrm{~s} \\ll t_{A}$. In fact, it is smaller than the $1 \\%$ margin allowed (so teams who did not read the clarification would still have gotten the correct answer). The ratio asked for was therefore:\n\n$$\n\\frac{0.616+t_{B}}{0.616+t_{A}}\n\\tag{4}\n$$""]",['1.03'],False,,Numerical,1e-1 846,Mechanics,,"A simple crane is shown in the below diagram, consisted of light rods with length $1 \mathrm{~m}$ and $\sqrt{2} \mathrm{~m}$. The end of the crane is supporting a $5 \mathrm{kN}$ object. Point $B$ is known as a ""pin."" It is attached to the main body and can exert both a vertical and horizontal force. Point $A$ is known as a ""roller"" and can only exert vertical forces. Rods can only be in pure compression or pure tension. ![](https://cdn.mathpix.com/cropped/2023_12_21_ed61957af961179eb7adg-1.jpg?height=794&width=591&top_left_y=1186&top_left_x=756) In $\mathrm{kN}$, what is the force experienced by the rod $C D$ ? Express a positive number if the member is in tension and a negative number if it is in compression.","['One naive method (though perfectly valid) is to solve for each member individually, starting from the two rods that connect to the $5 \\mathrm{kN}$ weight. At each joint, we can write out force equilibrium equations in the vertical and horizontal directions, and solve a system of linaer equations to get the force in $C D$.\nInstead, we can solve for this force in one line. Consider a horizontal slice right above point $D$.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_ce7d6832dfafbf2eb586g-1.jpg?height=488&width=593&top_left_y=222&top_left_x=755)\n\nSince the net force of this sub-element is still zero, we can do a force balance. The only external forces acting on this system is $E F, E C, C D$, and the $5 \\mathrm{kN}$ weight. If we do a torque balance about $E$, we get:\n\n$$\n5(2 L)=C D(L)\n\\tag{5}\n$$\n\nwhere $L$ is the length of the rod. This immediately gives $C D=10 \\mathrm{kN}$.']",['10'],False,kN,Numerical,1e-1 846,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A simple crane is shown in the below diagram, consisted of light rods with length $1 \mathrm{~m}$ and $\sqrt{2} \mathrm{~m}$. The end of the crane is supporting a $5 \mathrm{kN}$ object. Point $B$ is known as a ""pin."" It is attached to the main body and can exert both a vertical and horizontal force. Point $A$ is known as a ""roller"" and can only exert vertical forces. Rods can only be in pure compression or pure tension. In $\mathrm{kN}$, what is the force experienced by the rod $C D$ ? Express a positive number if the member is in tension and a negative number if it is in compression.","['One naive method (though perfectly valid) is to solve for each member individually, starting from the two rods that connect to the $5 \\mathrm{kN}$ weight. At each joint, we can write out force equilibrium equations in the vertical and horizontal directions, and solve a system of linaer equations to get the force in $C D$.\nInstead, we can solve for this force in one line. Consider a horizontal slice right above point $D$.\n\n\n\n\n\nSince the net force of this sub-element is still zero, we can do a force balance. The only external forces acting on this system is $E F, E C, C D$, and the $5 \\mathrm{kN}$ weight. If we do a torque balance about $E$, we get:\n\n$$\n5(2 L)=C D(L)\n\\tag{5}\n$$\n\nwhere $L$ is the length of the rod. This immediately gives $C D=10 \\mathrm{kN}$.']",['10'],False,kN,Numerical,1e-1 847,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A coaxial cable is cylindrically symmetric and consists of a solid inner cylinder of radius $a=2 \mathrm{~cm}$ and an outer cylindrical shell of inner radius $b=5 \mathrm{~cm}$ and outer radius $c=7 \mathrm{~cm}$. A uniformly distributed current of total magnitude $I=5 \mathrm{~A}$ is flowing in the inner cylinder and a uniformly distributed current of the same magnitude but opposite direction flows in the outer shell. Find the magnitude $B(r)$ of the magnetic field $B$ as a function of distance $r$ from the axis of the cable. As the final result, submit $\int_{0}^{\infty} B(r) \mathrm{d} r$. In case this is infinite, submit 42 .","[""Ampere's law $\\int B \\cdot \\mathrm{d} l=\\mu_{0} I$ is all we need. For every point on the wire, we can write the magnetic field as a function of the distance from its center $r$. Thus,\n$$\nB(r)= \\begin{cases}\\frac{5 \\mu_{0} r}{8 \\pi} & r \\leq 2 \\\\ \\frac{5 \\mu_{0}}{2 \\pi r} & 27\\end{cases}\n$$\n\nNow we just sum each integral from each interval, or in othe words\n\n$$\n\\int_{0}^{\\infty} B(r) \\mathrm{d} r=\\int_{0}^{2} B(r) \\mathrm{d} r+\\int_{2}^{5} B(r) \\mathrm{d} r+\\int_{5}^{7} B(r) \\mathrm{d} r\n$$\n\nThis is now straightforward integration.""]",['$1.6 \\times 10^{-8}$'],False,,Numerical,1e-9 848,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A small block of mass $m$ and charge $Q$ is placed at rest on an inclined plane with a slope $\alpha=40^{\circ}$. The coefficient of friction between them is $\mu=0.3$. A homogenous magnetic field of magnitude $B_{0}$ is applied perpendicular to the slope. The speed of the block after a very long time is given by $v=\beta \frac{m g}{Q B_{0}}$. Determine $\beta$. Do not neglect the effects of gravity. ","['Create a free body diagram. The direction of the magnetic field (into or out of the page) does not matter as we only need to know the magnitude of the terminal velocity. In dynamic equilibrium, we have three forces along the plane: the component of gravity along the plane, friction, and the magnetic force. The component of gravity along the plane is $m g \\sin \\alpha$. Friction has the magnitude $f=\\mu m g \\cos \\alpha$. The magnetic force has magnitude $F_{B}=Q v B_{0}$. At the terminal velocity, these forces are balanced. To finish, note that friction and the magnetic force are perpendicular so the magnitude of their vector sum is equal to the component of gravity along the plane. By the Pythagorean Theorem,\n$$\nF_{B}^{2}+f^{2}=(m g \\sin \\alpha)^{2} \\Longrightarrow v=\\frac{m g}{Q B_{0}} \\sqrt{\\sin ^{2} \\alpha-\\mu^{2} \\cos ^{2} \\alpha}\n$$']",['0.6'],False,,Numerical,5e-2 848,Electromagnetism,,"A small block of mass $m$ and charge $Q$ is placed at rest on an inclined plane with a slope $\alpha=40^{\circ}$. The coefficient of friction between them is $\mu=0.3$. A homogenous magnetic field of magnitude $B_{0}$ is applied perpendicular to the slope. The speed of the block after a very long time is given by $v=\beta \frac{m g}{Q B_{0}}$. Determine $\beta$. Do not neglect the effects of gravity. ![](https://cdn.mathpix.com/cropped/2023_12_21_dafd88496563c6b3fc11g-1.jpg?height=580&width=813&top_left_y=477&top_left_x=661)","['Create a free body diagram. The direction of the magnetic field (into or out of the page) does not matter as we only need to know the magnitude of the terminal velocity. In dynamic equilibrium, we have three forces along the plane: the component of gravity along the plane, friction, and the magnetic force. The component of gravity along the plane is $m g \\sin \\alpha$. Friction has the magnitude $f=\\mu m g \\cos \\alpha$. The magnetic force has magnitude $F_{B}=Q v B_{0}$. At the terminal velocity, these forces are balanced. To finish, note that friction and the magnetic force are perpendicular so the magnitude of their vector sum is equal to the component of gravity along the plane. By the Pythagorean Theorem,\n$$\nF_{B}^{2}+f^{2}=(m g \\sin \\alpha)^{2} \\Longrightarrow v=\\frac{m g}{Q B_{0}} \\sqrt{\\sin ^{2} \\alpha-\\mu^{2} \\cos ^{2} \\alpha}\n$$']",['0.6'],False,,Numerical,5e-2 849,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A train of length $100 \mathrm{~m}$ and mass $10^{5} \mathrm{~kg}$ is travelling at $20 \mathrm{~m} / \mathrm{s}$ along a straight track. The driver engages the brakes and the train starts deccelerating at a constant rate, coming to a stop after travelling a distance $d=2000 \mathrm{~m}$. As the train decelerates, energy released as heat from the brakes goes into the tracks, which have a linear heat capacity of $5000 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$. Assume the rate of heat generation and transfer is uniform across the length of the train at any given moment. If the tracks start at an ambient temperature of $20^{\circ} \mathrm{C}$, there is a function $T(x)$ that describes the temperature (in Celsius) of the tracks at each point $x$, where the rear of where the train starts is at $x=0$. Assume (unrealistically) that $100 \%$ of the original kinetic energy of the train is transferred to the tracks (the train does not absorb any energy), that there is no conduction of heat along the tracks, and that heat transfer between the tracks and the surroundings is negligible. Compute $T(20)+T(500)+T(2021)$ in degrees celsius.","[""Consider a small element of the tracks at position $x$ with width $\\mathrm{d} x$. Since the rate of heat generation is uniform along the length of the train $L$, we have that the rate of heat given to the track element is mav $\\frac{\\mathrm{d} x}{L}$, where $m$ is the train's mass, $a$ is the train's deceleration, and $v$ is the trains speed. Integrating over time gives the total heat given to the track element: $\\mathrm{d} Q=m a \\frac{\\mathrm{d} x}{L} \\Delta x$, where $\\Delta x$ is the total distance the train slips on the track element. Combining with $\\mathrm{d} Q=c \\mathrm{~d} x \\cdot \\Delta T$, we get $T(x)=T_{0}+\\frac{m a}{c L} \\Delta x$, where $c$ is the linear heat capacity. Now we split into 3 cases:\n- $0","[""Method One: Let us look at only the major axis $x$. The wind provides a constant acceleration along $x$, which makes this problem equivalent to throwing a ball up and down a ramp tilted at an angle of $\\beta=\\tan \\frac{a}{g}$. The furthest point along the $x$ axis represents the maximum distance you can throw an object up and down this ramp. This distance can be derived to be:\n\n$$\nd_{\\mathrm{optimal}}=\\frac{v^{2} / g}{1+\\sin \\beta}\n\\tag{6}\n$$\n\n\n\nwhere $\\beta$ is negative if it's tilting downwards. We get the systems of equations:\n\n$$\n\n\\frac{v^{2} / g_{\\mathrm{eff}}}{1+\\sin \\beta}=0.25+0.25 \\cdot \\frac{13}{20}\n\\tag{7}\n$$\n$$\n\\frac{v^{2} / g_{\\mathrm{eff}}}{1-\\sin \\beta}=1\n\n\\tag{8}\n$$\n\nwhere:\n\n$$\ng_{\\mathrm{eff}}=g / \\cos \\beta\n\\tag{9}\n$$\n\nand solving gives $v=2.51 \\mathrm{~m} \\mathrm{~s}^{-1}$.\n\nMethod Two: Looking at the minor axis is much easier. The wind doesn't contribute in this direction, so we know the furthest point must have a vertical displacement of $d_{\\max }=\\frac{v^{2}}{g}$. We measure $d_{\\max }=0.50+0.25 \\cdot \\frac{11}{20}$ and solving for $v$ gives $d_{\\max }=2.50 \\mathrm{~m}$.\n\nNote that the answers aren't exactly the same due to measurement inaccuracies, so we asked for two significant digits. An earlier version of the question had the same diagram, but the labels were accidentally shifted. This did not affect the final answer.""]",['2.5'],False,m/s,Numerical,1e-1 850,Mechanics,,"A sprinkler fountain is in the shape of a semi-sphere that spews out water from all angles at a uniform speed $v$ such that without the presence of wind, the wetted region around the fountain forms a circle in the $X Y$ plane with the fountain centered on it. Now suppose there is a constant wind blowing in a direction parallel to the ground such that the force acting on each water molecule is proportional to their weight. The wetted region forms the shape below where the fountain is placed at $(0,0)$. Determine the exit speed of water $v$ in meters per second. Round to two significant digits. All dimensions are in meters. ![](https://cdn.mathpix.com/cropped/2023_12_21_bc101ee6c7ad2225c8a6g-1.jpg?height=1258&width=1567&top_left_y=661&top_left_x=279)","[""Method One: Let us look at only the major axis $x$. The wind provides a constant acceleration along $x$, which makes this problem equivalent to throwing a ball up and down a ramp tilted at an angle of $\\beta=\\tan \\frac{a}{g}$. The furthest point along the $x$ axis represents the maximum distance you can throw an object up and down this ramp. This distance can be derived to be:\n\n$$\nd_{\\mathrm{optimal}}=\\frac{v^{2} / g}{1+\\sin \\beta}\n\\tag{6}\n$$\n\n\n\nwhere $\\beta$ is negative if it's tilting downwards. We get the systems of equations:\n\n$$\n\n\\frac{v^{2} / g_{\\mathrm{eff}}}{1+\\sin \\beta}=0.25+0.25 \\cdot \\frac{13}{20}\n\\tag{7}\n$$\n$$\n\\frac{v^{2} / g_{\\mathrm{eff}}}{1-\\sin \\beta}=1\n\n\\tag{8}\n$$\n\nwhere:\n\n$$\ng_{\\mathrm{eff}}=g / \\cos \\beta\n\\tag{9}\n$$\n\nand solving gives $v=2.51 \\mathrm{~m} \\mathrm{~s}^{-1}$.\n\nMethod Two: Looking at the minor axis is much easier. The wind doesn't contribute in this direction, so we know the furthest point must have a vertical displacement of $d_{\\max }=\\frac{v^{2}}{g}$. We measure $d_{\\max }=0.50+0.25 \\cdot \\frac{11}{20}$ and solving for $v$ gives $d_{\\max }=2.50 \\mathrm{~m}$.\n\nNote that the answers aren't exactly the same due to measurement inaccuracies, so we asked for two significant digits. An earlier version of the question had the same diagram, but the labels were accidentally shifted. This did not affect the final answer.""]",['2.5'],False,m/s,Numerical,1e-1 851,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a gas of mysterious particles called nieons that all travel at the same speed, $v$. They are enclosed in a cubical box, and there are $\rho$ nieons per unit volume. A very small hole of area $A$ is punched in the side of the box. The number of nieons that escape the box per unit time is given by $$ \alpha v^{\beta} A^{\gamma} \rho^{\delta} \tag{10} $$ where $\alpha, \beta, \gamma$, and $\delta$ are all dimensionless constants. Calculate $\alpha+\beta+\gamma+\delta$.","['The main idea is this: if a nieon is to escape in time $\\Delta t$, then it must be traveling towards the hole and be within a hemisphere of radius $v \\Delta t$, centered on the hole.\n(Note that we will ignore all collisions between nieons. As with many calculations in these kinds of problems, collisions will not affect our answer. Even if they did, we can assume that the hole is so small that we can take $v \\Delta t \\rightarrow 0$, making the radius of the hemisphere much much smaller than the mean free path.)\n\nFirst, we will break up this hemisphere into volume elements. We will calculate how many nieons are inside each volume element. Then we will find out how many of these nieons are traveling in the direction of the hole.\n\nWe define a spherical coordinate system $(r, \\theta, \\phi)$. The volume of each differential element is $r^{2} \\sin \\theta d r d \\theta d \\phi$, which means that there are $\\rho r^{2} \\sin \\theta d r d \\theta d \\phi$ nieons inside this volume element.\n\nWithin this volume element, how many of these nieons are going towards the hole? We must think about what each nieon ""sees."" Visualize a sphere of radius $r$ around the volume element. The surface of the sphere will pass through the small hole in the wall. Each nieon is equally likely to go in any direction and is equally likely to end up going through any patch of this spherical surface. Thus, each nieon within this volume element has a probability $A^{\\prime} / 4 \\pi r^{2}$ of going through the hole, where $A^{\\prime}$ is the perceived area of the hole from where the nieon is situated.\n\n$A^{\\prime}$ can be calculated as follows. $\\hat{\\mathbf{j}}$ is the unit vector pointing perpendicular to the hole. Let $\\hat{\\mathbf{r}}$ be the\n\n\n\nunit vector pointing from the hole to the volume element. Then it can be verified that $A^{\\prime}=A \\cos \\psi$, where $\\psi$ is the angle between $\\hat{\\mathbf{j}}$ and $\\hat{\\mathbf{r}}$.\n\nExpressing $\\hat{\\mathbf{r}}$ in terms of the other unit vectors, it can be shown that $A^{\\prime}=A \\sin \\theta \\sin \\phi$. Hence, the fraction of nieons inside the volume element that are going towards the hole is\n\n$$\n\\frac{A \\sin \\theta \\sin \\phi}{4 \\pi r^{2}}\n$$\n\nIntegrating over the entire hemisphere, the number of nieons escaping through the hole in a time $\\Delta t$ is\n\n$$\n\\int_{0}^{v \\Delta t} d r \\int_{0}^{\\pi} d \\phi \\int_{0}^{\\pi} d \\theta\\left(\\frac{A}{4 \\pi r^{2}} \\sin \\theta \\sin \\phi\\right)\\left(\\rho r^{2} \\sin \\theta\\right)=\\frac{1}{4} v A \\rho \\cdot \\Delta t\n$$']",['3.25'],False,,Numerical,1e-1 852,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Poncho is a very good player of the legendary carnival game known as Pico-Pico. Its setup consists of a steel ball, represented by a point mass, of negligible radius and a frictionless vertical track. The goal of Pico-Pico is to flick the ball from the beginning of the track (point $A$ ) such that it is able to traverse through the track while never leaving the track, successfully reaching the end (point $B)$. The most famous track design is one of parabolic shape; specifically, the giant track is of the shape $h(x)=5-2 x^{2}$ in meters. The starting and ending points of the tracks are where the two points where the track intersects $y=0$. If $\left(v_{a}, v_{b}\right]$ is the range of the ball's initial velocity $v_{0}$ that satisfies the winning condition of Pico-Pico, help Poncho find $v_{b}-v_{a}$. This part is depicted below: ","['Using conservation of energy, the minimum initial velocity of the ball needed to pass the top of the track is $v_{a}=\\sqrt{2 g h}=9.9045 \\frac{\\mathrm{m}}{\\mathrm{s}}$. To find $v_{b}$, the centripetal force at all points on the track must be determined given the initial velocity.\n$$\n\nF_{c} =\\frac{m v^{2}}{R}\n\\tag{11}\n$$\n$$\n=\\frac{m\\left(v_{b}^{2}-2 g h\\right)}{\\frac{\\left|1+\\left(\\frac{d}{d x} h(x)\\right)^{2}\\right|}{\\frac{d^{2}}{d x^{2}} h(x)}}\n\\tag{12}\n$$\n$$\n=\\frac{m\\left(v_{b}^{2}-2 g h\\right)}{\\frac{\\left|1+16 x^{2}\\right|^{\\frac{3}{2}}}{4}}\n\\tag{13}\n$$\n\nFor the boundary condition, the ball leaves if the normal force from the track on the ball $N=m g \\cos \\theta-F_{c}$ becomes 0.\n\n$$\n\\begin{aligned}\nm g \\cos \\theta-F_{c} & =0 \\\\\nm g \\cos \\arctan (-4 x) & =\\frac{4 m\\left(v_{b}^{2}-2 g h\\right)}{\\left|1+16 x^{2}\\right|^{\\frac{3}{2}}} \\\\\n\\frac{g}{\\left|1+16 x^{2}\\right|^{\\frac{1}{2}}} & =\\frac{4\\left(v_{b}^{2}-2 g h\\right)}{\\left|1+16 x^{2}\\right|^{\\frac{3}{2}}} \\\\\ng & =\\frac{4\\left(v_{b}^{2}-2 g h\\right)}{1+16 x^{2}} \\\\\nv_{b \\max } & =\\sqrt{\\frac{g+16 g x^{2}}{4}+2 g h}\n\\end{aligned}\n$$\n\nFrom the derivation, $v_{b \\max }$ is the lowest at $x=0$. Thus,\n\n$$\n\\begin{aligned}\nv_{b \\max } & =\\sqrt{\\frac{g}{4}+2 g h} \\\\\n& =10.0276 \\frac{\\mathrm{m}}{\\mathrm{s}}\n\\end{aligned}\n$$\n\nwhich is our desired $v_{b}$. The final answer, $v_{b}-v_{a}$, can be calculated.']",['0.1231'],False,m/s,Numerical,1e-2 852,Mechanics,,"Poncho is a very good player of the legendary carnival game known as Pico-Pico. Its setup consists of a steel ball, represented by a point mass, of negligible radius and a frictionless vertical track. The goal of Pico-Pico is to flick the ball from the beginning of the track (point $A$ ) such that it is able to traverse through the track while never leaving the track, successfully reaching the end (point $B)$. The most famous track design is one of parabolic shape; specifically, the giant track is of the shape $h(x)=5-2 x^{2}$ in meters. The starting and ending points of the tracks are where the two points where the track intersects $y=0$. If $\left(v_{a}, v_{b}\right]$ is the range of the ball's initial velocity $v_{0}$ that satisfies the winning condition of Pico-Pico, help Poncho find $v_{b}-v_{a}$. This part is depicted below: ![](https://cdn.mathpix.com/cropped/2023_12_21_c2af4dafb1688ec8aa90g-1.jpg?height=501&width=398&top_left_y=1343&top_left_x=858)","['Using conservation of energy, the minimum initial velocity of the ball needed to pass the top of the track is $v_{a}=\\sqrt{2 g h}=9.9045 \\frac{\\mathrm{m}}{\\mathrm{s}}$. To find $v_{b}$, the centripetal force at all points on the track must be determined given the initial velocity.\n$$\n\nF_{c} =\\frac{m v^{2}}{R}\n\\tag{11}\n$$\n$$\n=\\frac{m\\left(v_{b}^{2}-2 g h\\right)}{\\frac{\\left|1+\\left(\\frac{d}{d x} h(x)\\right)^{2}\\right|}{\\frac{d^{2}}{d x^{2}} h(x)}}\n\\tag{12}\n$$\n$$\n=\\frac{m\\left(v_{b}^{2}-2 g h\\right)}{\\frac{\\left|1+16 x^{2}\\right|^{\\frac{3}{2}}}{4}}\n\\tag{13}\n$$\n\nFor the boundary condition, the ball leaves if the normal force from the track on the ball $N=m g \\cos \\theta-F_{c}$ becomes 0.\n\n$$\n\\begin{aligned}\nm g \\cos \\theta-F_{c} & =0 \\\\\nm g \\cos \\arctan (-4 x) & =\\frac{4 m\\left(v_{b}^{2}-2 g h\\right)}{\\left|1+16 x^{2}\\right|^{\\frac{3}{2}}} \\\\\n\\frac{g}{\\left|1+16 x^{2}\\right|^{\\frac{1}{2}}} & =\\frac{4\\left(v_{b}^{2}-2 g h\\right)}{\\left|1+16 x^{2}\\right|^{\\frac{3}{2}}} \\\\\ng & =\\frac{4\\left(v_{b}^{2}-2 g h\\right)}{1+16 x^{2}} \\\\\nv_{b \\max } & =\\sqrt{\\frac{g+16 g x^{2}}{4}+2 g h}\n\\end{aligned}\n$$\n\nFrom the derivation, $v_{b \\max }$ is the lowest at $x=0$. Thus,\n\n$$\n\\begin{aligned}\nv_{b \\max } & =\\sqrt{\\frac{g}{4}+2 g h} \\\\\n& =10.0276 \\frac{\\mathrm{m}}{\\mathrm{s}}\n\\end{aligned}\n$$\n\nwhich is our desired $v_{b}$. The final answer, $v_{b}-v_{a}$, can be calculated.']",['0.1231'],False,m/s,Numerical,1e-2 853,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Now, Poncho has encountered a different Pico-Pico game that uses the same shaped frictionless track, but lays it horizontally on a table with friction and coefficient of friction $\mu=0.8$. In addition, the ball, which can once again be considered a point mass, is placed on the other side of the track as the ball in part 1. Finally, a buzzer on the other side of the track requires the mass to hit with at least velocity $v_{f}=2 \mathrm{~m} / \mathrm{s}$ in order to trigger the buzzer and win the game. Find the minimum velocity $v_{0}$ required for the ball to reach the end of the track with a velocity of at least $v_{f}$. The initial velocity must be directed along the track.","['We simply need to find the work done by friction and we can finish with conservation of energy. To get the work done by friction, since the force is constant, we just need to find the arc length of the track. That is\n$$\n\\ell=\\int_{-\\sqrt{5 / 2}}^{\\sqrt{5 / 2}} \\sqrt{1+h^{\\prime}(x)^{2}} \\mathrm{~d} x=\\int_{-\\sqrt{5 / 2}}^{\\sqrt{5 / 2}} \\sqrt{1+16 x^{2}} \\mathrm{~d} x=10.76 \\mathrm{~m}\n$$\n\nThen by conservation of energy,\n\n$$\n\\frac{1}{2} v_{0}^{2}-\\mu g \\ell=\\frac{1}{2} v_{f}^{2} \\Longrightarrow v_{0}=\\sqrt{v_{f}^{2}+2 \\mu g \\ell}\n$$\n\nWe simply plug in the numbers to finish.']",['$13.1$'],False,m/s,Numerical,1e-1 854,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Anyone who's had an apple may know that pieces of an apple stick together, when picking up one piece a second piece may also come with the first piece. The same idea is tried on a golden apple. Consider two uniform hemispheres with radius $r=4 \mathrm{~cm}$ made of gold of density $\rho_{g}=19300 \mathrm{~kg} \mathrm{~m}^{-3}$. The top half is nailed to a support and the space between is filled with water. Given that the surface tension of water is $\gamma=0.072 \mathrm{~N} \mathrm{~m}^{-1}$ and that the contact angle between gold and water is $\theta=10^{\circ}$, what is the maximum distance between the two hemispheres so that the bottom half doesn't fall? Answer in millimeters.","['Let $h$ be the difference in height. There are 3 forces on the bottom hemisphere. The force from gravity, which has magnitude $\\frac{2}{3} \\pi \\rho_{g} g r^{3}$, the force from the surface tension, and the force from the pressure difference at the top and bottom. The pressure difference is given by the young-laplace equation,\n$$\n\\Delta P=\\gamma\\left(-\\frac{1}{r}+\\frac{2 \\cos \\theta}{h}\\right) \\approx \\frac{2 \\gamma \\cos \\theta}{h}\n$$\n\nThe radii are found by some simple geometry. It is likely that $r$ will be much larger than the height, so we can neglect the $1 / r$ term. Now the force from surface tension is $2 \\pi r \\gamma \\sin \\theta$, since we take the vertical component. So we can now set the net force to 0 ,\n\n$$\n\\frac{2}{3} \\pi \\rho_{g} g r^{3}=\\pi r^{2} \\Delta P+2 \\pi r \\gamma \\sin \\theta \\Longrightarrow \\frac{2}{3} \\rho_{g} g r^{2}=(2 r \\gamma \\cos \\theta) \\frac{1}{h}+2 \\gamma \\sin \\theta\n$$\n\nThis is simple to solve,\n\n$$\nh=\\frac{2 r \\gamma \\cos \\theta}{2 \\rho_{g} g r^{2} / 3-2 \\gamma \\sin \\theta}=2.81 \\times 10^{-5} \\mathrm{~m}=0.0281 \\mathrm{~mm}\n$$\n\nThis is very small so our approximation from earlier is justified.']",['0.0281'],False,mm,Numerical,2e-3 854,Mechanics,,"Anyone who's had an apple may know that pieces of an apple stick together, when picking up one piece a second piece may also come with the first piece. The same idea is tried on a golden apple. Consider two uniform hemispheres with radius $r=4 \mathrm{~cm}$ made of gold of density $\rho_{g}=19300 \mathrm{~kg} \mathrm{~m}^{-3}$. The top half is nailed to a support and the space between is filled with water. ![](https://cdn.mathpix.com/cropped/2023_12_21_8e722241e5b95bed0c25g-1.jpg?height=298&width=317&top_left_y=458&top_left_x=904) Given that the surface tension of water is $\gamma=0.072 \mathrm{~N} \mathrm{~m}^{-1}$ and that the contact angle between gold and water is $\theta=10^{\circ}$, what is the maximum distance between the two hemispheres so that the bottom half doesn't fall? Answer in millimeters.","['Let $h$ be the difference in height. There are 3 forces on the bottom hemisphere. The force from gravity, which has magnitude $\\frac{2}{3} \\pi \\rho_{g} g r^{3}$, the force from the surface tension, and the force from the pressure difference at the top and bottom. The pressure difference is given by the young-laplace equation,\n$$\n\\Delta P=\\gamma\\left(-\\frac{1}{r}+\\frac{2 \\cos \\theta}{h}\\right) \\approx \\frac{2 \\gamma \\cos \\theta}{h}\n$$\n\nThe radii are found by some simple geometry. It is likely that $r$ will be much larger than the height, so we can neglect the $1 / r$ term. Now the force from surface tension is $2 \\pi r \\gamma \\sin \\theta$, since we take the vertical component. So we can now set the net force to 0 ,\n\n$$\n\\frac{2}{3} \\pi \\rho_{g} g r^{3}=\\pi r^{2} \\Delta P+2 \\pi r \\gamma \\sin \\theta \\Longrightarrow \\frac{2}{3} \\rho_{g} g r^{2}=(2 r \\gamma \\cos \\theta) \\frac{1}{h}+2 \\gamma \\sin \\theta\n$$\n\nThis is simple to solve,\n\n$$\nh=\\frac{2 r \\gamma \\cos \\theta}{2 \\rho_{g} g r^{2} / 3-2 \\gamma \\sin \\theta}=2.81 \\times 10^{-5} \\mathrm{~m}=0.0281 \\mathrm{~mm}\n$$\n\nThis is very small so our approximation from earlier is justified.']",['0.0281'],False,mm,Numerical,2e-3 855,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In the following two problems we will look at shooting a basketball. Model the basketball as an elastic hollow sphere with radius 0.1 meters. Model the net and basket as shown below, dimensions marked. Neglect friction between the backboard and basketball, and assume all collisions are perfectly elastic. For this problem, you launch the basketball from the point that is 2 meters above the ground and 4 meters from the backboard as shown. You attempt to make a shot by hitting the basketball off the backboard as depicted above. What is the minimum initial speed required for the ball to make this shot? Note: For this problem, you may assume that the size of the ball is negligible.","[""For this part, we regard the basketball as a point mass. Now to account for the backboard bounce, we reflect the hoop over the backboard. The problem is now equivalent to making it into the hoop that is 4.15 meters away now. So now we need to find the smallest velocity to reach this point. One way to do it is to use the safety parabola, which makes it so minimal calculations are needed, so we present that method here. However, this problem is also readily solvable with only basic kinematics.\nThe point that the ball must hit (4.15 meters in front and 1 meter above) lies on the safety parabola when it is thrown at minimum speed. Since the focus of the safety parabola is the point of launch,\n\n\n\nthe directrix is\n\n$$\n\\sqrt{1^{2}+4.15^{2}}=4.27 \\mathrm{~m}\n$$\n\naway from the top of the hoop. The distance from the focus to the vertex of the parabola is then $(4.27+1) / 2=2.63 \\mathrm{~m}$. So the minimum velocity to go this height is given by\n\n$$\n\\frac{v^{2}}{2 g}=2.63 \\mathrm{~m}\n$$\n\nand that gives $v=7.19 \\mathrm{~m} / \\mathrm{s}$. There is one last bit we need to check to complete this problem. The basketball's trajectory has to actually be possible (i.e. it doesn't go underneath the rim). It is clearly less optimal intuitively. If it passes through the underside, it means that its apex of the trajectory is very near its target.""]",['7.19'],False,m/s,Numerical,1e-2 855,Mechanics,,"In the following two problems we will look at shooting a basketball. Model the basketball as an elastic hollow sphere with radius 0.1 meters. Model the net and basket as shown below, dimensions marked. Neglect friction between the backboard and basketball, and assume all collisions are perfectly elastic. ![](https://cdn.mathpix.com/cropped/2023_12_21_a1e75b3588890e36fda2g-1.jpg?height=1225&width=1634&top_left_y=426&top_left_x=191) For this problem, you launch the basketball from the point that is 2 meters above the ground and 4 meters from the backboard as shown. You attempt to make a shot by hitting the basketball off the backboard as depicted above. What is the minimum initial speed required for the ball to make this shot? Note: For this problem, you may assume that the size of the ball is negligible.","[""For this part, we regard the basketball as a point mass. Now to account for the backboard bounce, we reflect the hoop over the backboard. The problem is now equivalent to making it into the hoop that is 4.15 meters away now. So now we need to find the smallest velocity to reach this point. One way to do it is to use the safety parabola, which makes it so minimal calculations are needed, so we present that method here. However, this problem is also readily solvable with only basic kinematics.\nThe point that the ball must hit (4.15 meters in front and 1 meter above) lies on the safety parabola when it is thrown at minimum speed. Since the focus of the safety parabola is the point of launch,\n\n\n\nthe directrix is\n\n$$\n\\sqrt{1^{2}+4.15^{2}}=4.27 \\mathrm{~m}\n$$\n\naway from the top of the hoop. The distance from the focus to the vertex of the parabola is then $(4.27+1) / 2=2.63 \\mathrm{~m}$. So the minimum velocity to go this height is given by\n\n$$\n\\frac{v^{2}}{2 g}=2.63 \\mathrm{~m}\n$$\n\nand that gives $v=7.19 \\mathrm{~m} / \\mathrm{s}$. There is one last bit we need to check to complete this problem. The basketball's trajectory has to actually be possible (i.e. it doesn't go underneath the rim). It is clearly less optimal intuitively. If it passes through the underside, it means that its apex of the trajectory is very near its target.""]",['7.19'],False,m/s,Numerical,1e-2 856,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Let $A B C$ be a solid right triangle $(A B=5 s, A C=12 s$, and $B C=13 s$ ) with uniform charge density $\sigma$. Let $D$ be the midpoint of $B C$. We denote the electric potential of a point $P$ by $\phi(P)$. The electric potential at infinity is 0 . If $\phi(B)+\phi(C)+\phi(D)=\frac{k \sigma s}{\epsilon_{0}}$ where $k$ is a dimensionless constant, determine $k$.","['If we put two of these right triangles together, we can form a rectangle with side lengths $5 s$ and $12 s$. Let $V$ be the potential at the center of this rectangle. By superposition, $\\phi(D)=\\frac{V}{2}$. Consider the potential at the corner. It can be decomposed into the potential from each of the right triangles, which is precisely $\\phi(B)+\\phi(C)$. Now, also note that if we put 4 of these rectangles side by side, to make a larger rectangle with side lengths $10 s$ and $24 s$, the potential is scaled by a factor of 2 due to dimensional analysis arguments. Thus, the potential at the center of this larger rectangle is $2 \\mathrm{~V}$, but this can be decomposed into the sum of the potentials at the\n\n\ncorners of the 4 smaller rectangles. Thus, the potential in the corner of the smaller rectangle is $\\frac{V}{2}=\\phi(B)+\\phi(C)$. Thus, we obtain $\\phi(B)+\\phi(C)+\\phi(D)=V$.\n\nNow, we will find the potential at the center of the rectangle. Note that it suffices to find the potential at the vertex of an isosceles triangle because we can connect the center of the rectangle to the 4 corners to create 4 isosceles triangles. Suppose we have an isosceles triangle with base $2 x$ and height $y$. The potential at the vertex is\n\n$$\n\\iint \\frac{\\sigma}{4 \\pi \\epsilon_{0} r}(r d r d \\theta)=\\frac{\\sigma}{4 \\pi \\epsilon_{0}} \\int_{-\\tan ^{-1}\\left(\\frac{x}{y}\\right)}^{\\tan ^{-1}\\left(\\frac{x}{y}\\right)} \\int_{0}^{\\frac{y}{\\cos \\theta}} d r d \\theta=\\frac{\\sigma y}{2 \\pi \\epsilon_{0}} \\log \\left(\\frac{x+\\sqrt{x^{2}+y^{2}}}{y}\\right)\n$$\n\nIf the sides of the rectangle are $a$ and $b$, we then obtain the potential at the center is\n\n$$\nV=\\frac{\\sigma}{2 \\pi \\epsilon_{0}}\\left(a \\log \\left(\\frac{b+\\sqrt{a^{2}+b^{2}}}{a}\\right)+b \\log \\left(\\frac{a+\\sqrt{a^{2}+b^{2}}}{b}\\right)\\right)\n$$\n\nIn this case,\n\n$$\nk=\\frac{5}{2 \\pi} \\ln 5+\\frac{6}{\\pi} \\ln \\frac{3}{2}\n$$']",['2.055'],False,,Numerical,6e-2 857,Mechanics,,"Consider a toilet paper roll with some length of it hanging off as shown. The toilet paper roll rests on a cylindrical pole of radius $r=1 \mathrm{~cm}$ and the coefficient of static friction between the role and the pole is $\mu=0.3$. ![](https://cdn.mathpix.com/cropped/2023_12_21_20d264993198809dd70dg-1.jpg?height=612&width=480&top_left_y=1464&top_left_x=817) The length of the paper hanging off has length $\ell=30 \mathrm{~cm}$ and the inner radius of the roll is $R_{i}=2 \mathrm{~cm}$. The toilet paper has thickness $s=0.1 \mathrm{~mm}$ and mass per unit length $\lambda=5 \mathrm{~g} / \mathrm{m}$. What is the minimum outer radius $R_{o}$ such that the toilet paper roll remains static? Answer in centimeters.","[""Due to the length of toilet paper hanging off, the toilet paper will be slightly tilted, in order for torques to balance. The tilt isn't shown in the diagram, since it is meant to be found. So the tilted normal force has to be compensated by the frictional force, and just before slippage,\n\n\nit will look something like:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9bf394e5a782e8f11601g-1.jpg?height=618&width=553&top_left_y=304&top_left_x=778)\n\nLet $\\theta=\\arctan \\mu$. So now balancing torques about the contact point, if $m$ is the mass of the toilet paper roll,\n\n$$\nm g R_{i} \\sin \\theta=(\\ell \\lambda) g\\left(R_{o}-R_{i} \\sin \\theta\\right) .\n$$\n\nAnd now we need $m$. The mass per unit area is $\\sigma=\\lambda / s$, so $m=\\pi\\left(R_{o}^{2}-R_{i}^{2}\\right) \\lambda / s$. So substituting this in,\n\n$$\n\\frac{\\pi\\left(R_{o}^{2}-R_{i}^{2}\\right) \\lambda}{s} g R_{i} \\sin \\theta=\\ell \\lambda g\\left(R_{o}-R_{i} \\sin \\theta\\right)\n$$\n\nSimplifying a bit,\n\n$$\n\\pi\\left(R_{o}^{2}-R_{i}^{2}\\right) R_{i} \\sin \\theta=s \\ell\\left(R_{o}-R_{i} \\sin \\theta\\right) .\n$$\n\nThis is a quadratic and we can just plug in the numbers and use the quadratic formula or use a graphing calculator to finish, yielding $R_{o}=2.061 \\mathrm{~cm}$.""]",['2.061'],False,cm,Numerical,1e-1 857,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a toilet paper roll with some length of it hanging off as shown. The toilet paper roll rests on a cylindrical pole of radius $r=1 \mathrm{~cm}$ and the coefficient of static friction between the role and the pole is $\mu=0.3$. The length of the paper hanging off has length $\ell=30 \mathrm{~cm}$ and the inner radius of the roll is $R_{i}=2 \mathrm{~cm}$. The toilet paper has thickness $s=0.1 \mathrm{~mm}$ and mass per unit length $\lambda=5 \mathrm{~g} / \mathrm{m}$. What is the minimum outer radius $R_{o}$ such that the toilet paper roll remains static? Answer in centimeters.","[""Due to the length of toilet paper hanging off, the toilet paper will be slightly tilted, in order for torques to balance. The tilt isn't shown in the diagram, since it is meant to be found. So the tilted normal force has to be compensated by the frictional force, and just before slippage,\n\n\nit will look something like:\n\n\n\nLet $\\theta=\\arctan \\mu$. So now balancing torques about the contact point, if $m$ is the mass of the toilet paper roll,\n\n$$\nm g R_{i} \\sin \\theta=(\\ell \\lambda) g\\left(R_{o}-R_{i} \\sin \\theta\\right) .\n$$\n\nAnd now we need $m$. The mass per unit area is $\\sigma=\\lambda / s$, so $m=\\pi\\left(R_{o}^{2}-R_{i}^{2}\\right) \\lambda / s$. So substituting this in,\n\n$$\n\\frac{\\pi\\left(R_{o}^{2}-R_{i}^{2}\\right) \\lambda}{s} g R_{i} \\sin \\theta=\\ell \\lambda g\\left(R_{o}-R_{i} \\sin \\theta\\right)\n$$\n\nSimplifying a bit,\n\n$$\n\\pi\\left(R_{o}^{2}-R_{i}^{2}\\right) R_{i} \\sin \\theta=s \\ell\\left(R_{o}-R_{i} \\sin \\theta\\right) .\n$$\n\nThis is a quadratic and we can just plug in the numbers and use the quadratic formula or use a graphing calculator to finish, yielding $R_{o}=2.061 \\mathrm{~cm}$.""]",['2.061'],False,cm,Numerical,1e-1 858,Electromagnetism,,"In the circuit shown below, a capacitor $C=4 \mathrm{~F}$, inductor $L=5 \mathrm{H}$, and resistors $R_{1}=3 \Omega$ and $R_{2}=2 \Omega$ are placed in a diamond shape and are then fed an alternating current with peak voltage $V_{0}=1 \mathrm{~V}$ of unknown frequency. Determine the magnitude of the maximum instantaneous output voltage shown in the diagram. ![](https://cdn.mathpix.com/cropped/2023_12_21_29b1038a6a3b0fabe55eg-1.jpg?height=469&width=616&top_left_y=443&top_left_x=749)","['We use the method of phasors. Consider the following phasor diagram:\n![](https://cdn.mathpix.com/cropped/2023_12_21_29b1038a6a3b0fabe55eg-1.jpg?height=387&width=707&top_left_y=1064&top_left_x=709)\n\nWe define angles as $\\angle A P B=\\angle A Q B=\\gamma, \\angle A O B=2 \\gamma, \\angle B P Q=\\alpha, \\angle A Q P=\\beta$. Note that $\\overrightarrow{P A}=I_{2} \\omega L, \\overrightarrow{A Q}=I_{2} R, \\overrightarrow{P B}=I_{1} R_{1}, \\overrightarrow{B Q}=1 / I_{1} \\omega C$. We seek to maximize the length of $\\mathrm{AB}$. We can write via law of cosines that\n\n$$\n\\mathrm{AB}=\\sqrt{r^{2}+r^{2}-2 r \\cos (2 \\gamma)}=2 r \\sin (\\gamma)\n$$\n\nwhere $r=V_{0}$ is the radius of the circle of which phasors are inscribed in. Since $\\gamma=\\alpha+\\beta$, we can then rewrite the length of $\\mathrm{AB}$ to be\n\n$$\n\\mathrm{AB}=r(\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta)=V_{0}\\left(\\frac{R_{1} R_{2}}{V_{0}^{2}}-\\frac{L}{C V_{0}^{2}}\\right) I_{1} I_{2}=V_{0}\\left(R_{1} R_{2}-\\frac{L}{C}\\right)\\left(\\frac{1}{\\left|Z_{1}\\right|\\left|Z_{2}\\right|}\\right)\n$$\n\nwhere the product of both complex exponentials is simply\n\n$$\n\\left|Z_{1}\\right|\\left|Z_{2}\\right|=\\sqrt{\\omega^{2} L^{2}+R_{2}^{2}} \\sqrt{\\frac{1}{\\omega^{2} C^{2}}+R_{1}^{2}}\n$$\n\nwhich implies the best frequency of the circuit is $\\omega=\\sqrt{\\frac{R_{2}^{2}}{L^{2} R_{1} C}}$. This means with algebra, the lowest value of $\\left|Z_{1}\\right|\\left|Z_{2}\\right|=\\frac{L}{C}-R_{1} R_{2}$. Hence, the maximum value of $\\mathrm{AB}$ is simply\n\n$$\n\\mathrm{AB}=V_{0}\\left(\\frac{R_{1} R_{2}-\\frac{L}{C}}{R_{1} R_{2}+\\frac{L}{C}}\\right)\n$$']",['0.65'],False,V,Numerical,5e-2 858,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In the circuit shown below, a capacitor $C=4 \mathrm{~F}$, inductor $L=5 \mathrm{H}$, and resistors $R_{1}=3 \Omega$ and $R_{2}=2 \Omega$ are placed in a diamond shape and are then fed an alternating current with peak voltage $V_{0}=1 \mathrm{~V}$ of unknown frequency. Determine the magnitude of the maximum instantaneous output voltage shown in the diagram. ","['We use the method of phasors. Consider the following phasor diagram:\n\n\nWe define angles as $\\angle A P B=\\angle A Q B=\\gamma, \\angle A O B=2 \\gamma, \\angle B P Q=\\alpha, \\angle A Q P=\\beta$. Note that $\\overrightarrow{P A}=I_{2} \\omega L, \\overrightarrow{A Q}=I_{2} R, \\overrightarrow{P B}=I_{1} R_{1}, \\overrightarrow{B Q}=1 / I_{1} \\omega C$. We seek to maximize the length of $\\mathrm{AB}$. We can write via law of cosines that\n\n$$\n\\mathrm{AB}=\\sqrt{r^{2}+r^{2}-2 r \\cos (2 \\gamma)}=2 r \\sin (\\gamma)\n$$\n\nwhere $r=V_{0}$ is the radius of the circle of which phasors are inscribed in. Since $\\gamma=\\alpha+\\beta$, we can then rewrite the length of $\\mathrm{AB}$ to be\n\n$$\n\\mathrm{AB}=r(\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta)=V_{0}\\left(\\frac{R_{1} R_{2}}{V_{0}^{2}}-\\frac{L}{C V_{0}^{2}}\\right) I_{1} I_{2}=V_{0}\\left(R_{1} R_{2}-\\frac{L}{C}\\right)\\left(\\frac{1}{\\left|Z_{1}\\right|\\left|Z_{2}\\right|}\\right)\n$$\n\nwhere the product of both complex exponentials is simply\n\n$$\n\\left|Z_{1}\\right|\\left|Z_{2}\\right|=\\sqrt{\\omega^{2} L^{2}+R_{2}^{2}} \\sqrt{\\frac{1}{\\omega^{2} C^{2}}+R_{1}^{2}}\n$$\n\nwhich implies the best frequency of the circuit is $\\omega=\\sqrt{\\frac{R_{2}^{2}}{L^{2} R_{1} C}}$. This means with algebra, the lowest value of $\\left|Z_{1}\\right|\\left|Z_{2}\\right|=\\frac{L}{C}-R_{1} R_{2}$. Hence, the maximum value of $\\mathrm{AB}$ is simply\n\n$$\n\\mathrm{AB}=V_{0}\\left(\\frac{R_{1} R_{2}-\\frac{L}{C}}{R_{1} R_{2}+\\frac{L}{C}}\\right)\n$$']",['0.65'],False,V,Numerical,5e-2 859,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A uniform bar of length $l$ and mass $m$ is connected to a very long thread of negligible mass suspended from a ceiling. It is then rotated such that it is vertically upside down and then released. Initially, the rod is in unstable equilibrium. As it falls down, the minimum tension acting on the thread over the rod's entire motion is given by $\alpha m g$. Determine $\alpha$. ","['First note that the thread is given to be very long. Therefore, only vertical tension and gravitational forces act on the rod allowing for its center of mass to move in a straight vertical line. Using this fact, we can now write the acceleration of the rod in terms of angular velocity $\\omega$ and angular acceleration $\\varepsilon$. Let the angle of the rod at any moment to the vertical be $\\varphi$. For simplicity, we define the length of the rod to be $2 l$. Then, by defining the coordinate $y$ to be the change in vertical length of the rod where $y=l \\cos \\varphi$, one can write for varying $\\varphi \\in[\\varphi, \\varphi+\\mathrm{d} \\varphi)$ that $l \\cos (\\varphi+\\mathrm{d} \\varphi)=y+\\mathrm{d} y \\Longrightarrow \\mathrm{d} y=l \\sin \\varphi \\mathrm{d} \\varphi$. Thus, simple differentiation proves that\n$$\nv=l \\sin \\varphi \\omega \\Longrightarrow a=l \\varepsilon \\sin \\varphi+l \\omega^{2} \\cos \\varphi\n$$\n\nWe can now also write conservation of energy to get another relationship between velocity and angular velocity. At any given moment, it can be written that $m g l=\\frac{1}{2} m v^{2}+\\frac{1}{2} I \\omega^{2}-m g l \\cos \\varphi$. Since $I=\\frac{1}{2} m(2 l)^{2}=\\frac{1}{3} m l^{2}$, then\n\n$$\nm g l(1+\\cos \\varphi)=m l^{2} \\omega^{2}\\left(\\frac{1}{2} \\sin ^{2} \\varphi+\\frac{1}{6}\\right)\n$$\n\nOur third equation comes from the fact that the tension on the rod at any moment can be written as $T=m g-m a \\Longrightarrow m a=m g-T$. We can finally get a fourth equation by equating torques such that $\\frac{1}{3} m l^{2} \\varepsilon=T l \\sin \\varphi$. With these four equations, simple algebra yields the tension at any point is\n\n$$\nT(\\varphi)=\\frac{1+(3 \\cos \\varphi-1)^{2}}{\\left(1+3 \\sin ^{2} \\varphi\\right)^{2}} m g \\Longrightarrow T_{\\min } \\approx 0.165 m g\n$$']",['0.165'],False,mg,Numerical,5e-3 859,Mechanics,,"A uniform bar of length $l$ and mass $m$ is connected to a very long thread of negligible mass suspended from a ceiling. It is then rotated such that it is vertically upside down and then released. Initially, the rod is in unstable equilibrium. As it falls down, the minimum tension acting on the thread over the rod's entire motion is given by $\alpha m g$. Determine $\alpha$. ![](https://cdn.mathpix.com/cropped/2023_12_21_8778e2463d0600bda16eg-1.jpg?height=512&width=160&top_left_y=457&top_left_x=972)","['First note that the thread is given to be very long. Therefore, only vertical tension and gravitational forces act on the rod allowing for its center of mass to move in a straight vertical line. Using this fact, we can now write the acceleration of the rod in terms of angular velocity $\\omega$ and angular acceleration $\\varepsilon$. Let the angle of the rod at any moment to the vertical be $\\varphi$. For simplicity, we define the length of the rod to be $2 l$. Then, by defining the coordinate $y$ to be the change in vertical length of the rod where $y=l \\cos \\varphi$, one can write for varying $\\varphi \\in[\\varphi, \\varphi+\\mathrm{d} \\varphi)$ that $l \\cos (\\varphi+\\mathrm{d} \\varphi)=y+\\mathrm{d} y \\Longrightarrow \\mathrm{d} y=l \\sin \\varphi \\mathrm{d} \\varphi$. Thus, simple differentiation proves that\n$$\nv=l \\sin \\varphi \\omega \\Longrightarrow a=l \\varepsilon \\sin \\varphi+l \\omega^{2} \\cos \\varphi\n$$\n\nWe can now also write conservation of energy to get another relationship between velocity and angular velocity. At any given moment, it can be written that $m g l=\\frac{1}{2} m v^{2}+\\frac{1}{2} I \\omega^{2}-m g l \\cos \\varphi$. Since $I=\\frac{1}{2} m(2 l)^{2}=\\frac{1}{3} m l^{2}$, then\n\n$$\nm g l(1+\\cos \\varphi)=m l^{2} \\omega^{2}\\left(\\frac{1}{2} \\sin ^{2} \\varphi+\\frac{1}{6}\\right)\n$$\n\nOur third equation comes from the fact that the tension on the rod at any moment can be written as $T=m g-m a \\Longrightarrow m a=m g-T$. We can finally get a fourth equation by equating torques such that $\\frac{1}{3} m l^{2} \\varepsilon=T l \\sin \\varphi$. With these four equations, simple algebra yields the tension at any point is\n\n$$\nT(\\varphi)=\\frac{1+(3 \\cos \\varphi-1)^{2}}{\\left(1+3 \\sin ^{2} \\varphi\\right)^{2}} m g \\Longrightarrow T_{\\min } \\approx 0.165 m g\n$$']",['0.165'],False,mg,Numerical,5e-3 860,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",A straight ladder $A B$ of mass $m=1 \mathrm{~kg}$ is positioned almost vertically such that point $B$ is in contact with the ground with a coefficient of friction $\mu=0.15$. It is given an infinitesimal kick at the point $A$ so that the ladder begins rotating about point $B$. Find the value $\phi_{m}$ of angle $\phi$ of the ladder with the vertical at which the lower end $B$ starts slipping on the ground.,"['By conservation of energy, we have $\\frac{1}{2} I \\omega_{m}^{2}=m g \\frac{L}{2}\\left(1-\\cos \\phi_{m}\\right)$ where $I=\\frac{1}{3} m L^{2}$. Thus,\n$$\n\\omega_{m}=\\sqrt{\\frac{3 g\\left(1-\\cos \\phi_{m}\\right)}{L}}\n$$\n\nAlso, by torque analysis about B, we have $\\tau=m g \\frac{L}{2} \\sin \\phi_{m}=I \\alpha_{m}$ which means\n\n$$\n\\alpha_{m}=\\frac{3 g}{2 L} \\sin \\phi_{m}\n$$\n\nThus, the centripetal and tangential accelerations of the ladder are $a_{c}=\\omega_{m}^{2} \\frac{L}{2}=\\frac{3}{2} g\\left(1-\\cos \\phi_{m}\\right)$ and $a_{t}=\\alpha_{m} \\frac{L}{2}=\\frac{3}{4} g \\sin \\phi_{m}$ respectively. The normal force is thus $N=m g-m a_{c} \\cos \\phi-m a_{t} \\sin \\phi$, so\n\n$$\n\\frac{N}{m g}=1-\\frac{3}{2} \\cos \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)-\\frac{3}{4} \\sin ^{2} \\phi_{m}\n$$\n\nThe frictional force is thus $f=m a_{t} \\cos \\phi_{m}-m a_{c} \\sin \\phi_{m}$ so\n\n$$\n\\frac{f}{m g}=\\frac{3}{4} \\sin \\phi_{m} \\cos \\phi_{m}-\\frac{3}{2} \\sin \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)\n$$\n\nSetting $\\frac{f}{N}=\\mu$, we have $6 \\sin \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)-3 \\sin \\phi_{m} \\cos \\phi_{m}=-4 \\mu+6 \\mu \\cos \\phi_{m}\\left(1-\\cos \\phi_{m}\\right)+$ $3 \\mu \\sin ^{2} \\phi_{m}$. Simplifying\n\n$$\n6 \\sin \\phi_{m}-9 \\sin \\phi_{m} \\cos \\phi_{m}+9 \\mu \\cos ^{2} \\phi_{m}-6 \\mu \\cos \\phi_{m}=-\\mu\n$$\n\nWe then can solve for $\\phi_{m}$ numerically.']",['$11.5$'],False,$^{\circ}$,Numerical,1e-1 861,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two Ladders Two straight ladders $A B$ and $C D$, each with length $1 \mathrm{~m}$, are symmetrically placed on smooth ground, leaning on each other, such that they are touching with their ends $B$ and $C$, ends $A$ and $D$ are touching the floor. The friction at any two surfaces is negligible. Initially both ladders are almost parallel and vertical. Find the distance $A D$ when the points $B$ and $C$ lose contact.","['The center of mass of both of the ladders moves in a circle, centered at the point on the ground directly beneath $B / C$. So we find when the required normal force between the two ladders is 0 . That is, when the total net force on one of the ladders is when the two ladders lose contact. Let $2 r=\\ell$. Now by conservation of energy,\n$$\n\\frac{1}{2} m v^{2}+\\frac{1}{2} \\frac{m r^{2}}{3} \\frac{v^{2}}{r^{2}}=m g r(1-\\cos \\theta)\n$$\n\nwhere $\\theta$ is defined as the angle the ladder makes witht he vertical. So we have\n\n$$\nv^{2}\\left(1+\\frac{1}{3}\\right)=2 g r(1-\\cos \\theta) \\Longrightarrow v^{2}=\\frac{3}{2} g r(1-\\cos \\theta)\n$$\n\nSo the centripetal acceleration is\n\n$$\na_{c}=\\frac{v^{2}}{r}=\\frac{3}{2} g(1-\\cos \\theta)\n$$\n\n\n\nAnd the tangential acceleration is\n\n$$\n\\frac{d v}{d t}=\\sqrt{\\frac{3 g r}{2}} \\frac{\\sin \\theta}{2 \\sqrt{1-\\cos \\theta}} \\frac{d \\theta}{d t}\n$$\n\nAnd $\\frac{d \\theta}{d t}=\\frac{v}{r}$, so\n\n$$\na_{\\theta}=\\frac{d v}{d t}=\\frac{3 g}{2} \\frac{\\sin \\theta}{2}\n$$\n\nNow for the total acceleration to be vertical, we need\n\n$$\na_{c} \\tan \\theta=a_{\\theta} \\text {, }\n$$\n\nso\n\n$$\n1-\\cos \\theta=\\frac{\\cos \\theta}{2}\n$$\n\nThis simplifies to\n\n$$\n2=3 \\cos \\theta\n$$\n\nSo the distance between the two ends is $4 r \\sin \\theta=4 r \\sqrt{1-\\frac{4}{9}}=\\frac{4 r \\sqrt{5}}{3}$.']",['$\\frac{2\\sqrt{5}}{3}$'],False,m,Numerical,1e-1 862,Electromagnetism,,"Colliding Conducting Slab A thin conducting square slab with side length $s=5 \mathrm{~cm}$, initial charge $q=0.1 \mu \mathrm{C}$, and mass $m=100 \mathrm{~g}$ is given a kick and sent bouncing between two infinite conducting plates separated by a distance $d=0.5 \mathrm{~cm} \ll s$ and with surface charge density $\pm \sigma= \pm 50 \mu \mathrm{C} / \mathrm{m}^{2}$. After a long time it is observed exactly in the middle of the two plates to be traveling with velocity of magnitude $v=3 \mathrm{~m} / \mathrm{s}$ and direction $\theta=30^{\circ}$ with respect to the horizontal line parallel to the plates. How many collisions occur after it has traveled a distance $L=15 \mathrm{~m}$ horizontally from when it was last observed? Assume that all collisions are elastic, and neglect induced charges. Note that the setup is horizontal so gravity does not need to be accounted for. ![](https://cdn.mathpix.com/cropped/2023_12_21_b6b6c3c8a654288caf6bg-1.jpg?height=280&width=710&top_left_y=1676&top_left_x=705)","[""After the first collision, the charge approaches a constant magnitude. Let's look at a collision with the plate at charge $q$ hitting the $-\\sigma$ plate. Since the slab is thin, to keep the conducting surface an equipotential, the charge on the conduting slab has to become $-\\sigma s^{2}$. The initial charge doesn't matter because the plates are infinite. So the charge is $-\\sigma s^{2}$ after leaving the negative plate and $+\\sigma s^{2}$ after leaving the positive plate.\nNow the acceleration always has a constant magnitude, which is directed toward the plate it is traveling to, which we can find using $F=m a$,\n\n$$\n\\left(\\sigma s^{2}\\right) \\frac{\\sigma}{\\epsilon_{0}}=m a \\Longrightarrow a=\\frac{\\sigma^{2} s^{2}}{\\epsilon_{0} m}=7.06 \\mathrm{~m} / \\mathrm{s}^{2}\n$$\n\n\n\nThen here we can use the trick of reflecting over the plates, and now it's a basic kinematics problem. The time to travel $L$ is $t=\\frac{L}{v \\cos \\theta}$, and then the distance it travels vertically is\n\n$$\ny=v \\sin \\theta \\frac{L}{v \\cos \\theta}+\\frac{1}{2} a\\left(\\frac{L}{v \\cos \\theta}\\right)^{2}=126.36 \\mathrm{~m}\n$$\n\nAnd since $25272 d+d / 2 \\approx 87.129$, the number of collisions is 25273 .""]",['25273'],False,,Numerical,1e2 862,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Colliding Conducting Slab A thin conducting square slab with side length $s=5 \mathrm{~cm}$, initial charge $q=0.1 \mu \mathrm{C}$, and mass $m=100 \mathrm{~g}$ is given a kick and sent bouncing between two infinite conducting plates separated by a distance $d=0.5 \mathrm{~cm} \ll s$ and with surface charge density $\pm \sigma= \pm 50 \mu \mathrm{C} / \mathrm{m}^{2}$. After a long time it is observed exactly in the middle of the two plates to be traveling with velocity of magnitude $v=3 \mathrm{~m} / \mathrm{s}$ and direction $\theta=30^{\circ}$ with respect to the horizontal line parallel to the plates. How many collisions occur after it has traveled a distance $L=15 \mathrm{~m}$ horizontally from when it was last observed? Assume that all collisions are elastic, and neglect induced charges. Note that the setup is horizontal so gravity does not need to be accounted for. ","[""After the first collision, the charge approaches a constant magnitude. Let's look at a collision with the plate at charge $q$ hitting the $-\\sigma$ plate. Since the slab is thin, to keep the conducting surface an equipotential, the charge on the conduting slab has to become $-\\sigma s^{2}$. The initial charge doesn't matter because the plates are infinite. So the charge is $-\\sigma s^{2}$ after leaving the negative plate and $+\\sigma s^{2}$ after leaving the positive plate.\nNow the acceleration always has a constant magnitude, which is directed toward the plate it is traveling to, which we can find using $F=m a$,\n\n$$\n\\left(\\sigma s^{2}\\right) \\frac{\\sigma}{\\epsilon_{0}}=m a \\Longrightarrow a=\\frac{\\sigma^{2} s^{2}}{\\epsilon_{0} m}=7.06 \\mathrm{~m} / \\mathrm{s}^{2}\n$$\n\n\n\nThen here we can use the trick of reflecting over the plates, and now it's a basic kinematics problem. The time to travel $L$ is $t=\\frac{L}{v \\cos \\theta}$, and then the distance it travels vertically is\n\n$$\ny=v \\sin \\theta \\frac{L}{v \\cos \\theta}+\\frac{1}{2} a\\left(\\frac{L}{v \\cos \\theta}\\right)^{2}=126.36 \\mathrm{~m}\n$$\n\nAnd since $25272 d+d / 2 \\approx 87.129$, the number of collisions is 25273 .""]",['25273'],False,,Numerical,1e2 863,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",An evil gamma photon of energy $E_{\gamma 1}=200 \mathrm{keV}$ is heading towards a spaceship. The commander's only choice is shooting another photon in the direction of the gamma photon such that they 'collide' head on and produce an electron-positron pair (both have mass $m_{e}$ ). Find the lower bound on the energy $E_{\gamma 2}$ of the photon as imposed by the principles of special relativity such that this occurs. Answer in keV.,"['The key claim is that energy is minimized when both particles are moving at the same velocity after the collision. This can be proved by transforming into the frame where the total momentum is 0 .\nThis idea is sufficient because it implies both the electron and positron have the same momentum and energy after the collision. Let them both have momentum $p$. We then have\n\n$$\n2 p=\\frac{E_{\\gamma_{1}}-E_{\\gamma_{2}}}{c} \\Longrightarrow p c=\\frac{E_{\\gamma_{1}}-E_{\\gamma_{2}}}{2}\n\\tag{21}\n$$\n\nBy energy conservation, both the electron and positron have energy $\\frac{E_{\\gamma_{1}}+E_{\\gamma_{2}}}{2}$. Using the result $E=(p c)^{2}+\\left(m c^{2}\\right)^{2}$, we obtain,\n\n$$\n\\left(\\frac{E_{\\gamma_{1}}+E_{\\gamma_{2}}}{2}\\right)^{2}=(p c)^{2}+\\left(m_{e} c^{2}\\right)^{2}\n\\tag{22}\n$$\n\nCombining equation 21 and equation 22 , we get\n\n$$\n\\left(\\frac{E_{\\gamma_{1}}+E_{\\gamma_{2}}}{2}\\right)^{2}=\\left(\\frac{E_{\\gamma_{1}}-E_{\\gamma_{2}}}{2}\\right)^{2}+\\left(m_{e} c^{2}\\right)^{2} \\Longrightarrow E_{\\gamma_{2}}=\\frac{m_{e}^{2} c^{4}}{E_{\\gamma_{1}}}\n$$']",['1306'],False,keV,Numerical,1e0 864,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Spinning Cylinder Adithya has a solid cylinder of mass $M=10 \mathrm{~kg}$, radius $R=0.08 \mathrm{~m}$, and height $H=0.20 \mathrm{~m}$. He is running a test in a chamber on Earth over a distance of $d=200 \mathrm{~m}$ as shown below. Assume that the physical length of the chamber is much greater than $d$ (i.e. the chamber extends far to the left and right of the testing area). The chamber is filled with an ideal fluid with uniform density $\rho=700 \mathrm{~kg} / \mathrm{m}^{3}$. Adithya's cylinder is launched with linear velocity $v=10 \mathrm{~m} / \mathrm{s}$ and spins counterclockwise with angular velocity $\omega$. Adithya notices that the cylinder continues on a horizontal path until the end of the chamber. Find the angular velocity $\omega$. Do not neglect forces due to fluid pressure differences. Note that the diagram presents a side view of the chamber (i.e. gravity is oriented downwards with respect to the diagram). Assume the following about the setup and the ideal fluid: - fluid flow is steady in the frame of the center of mass of the cylinder - the ideal fluid is incompressible, irrotational, and has zero viscosity - the angular velocity of the cylinder is approximately constant during its subsequent motion Hint: For a uniform cylinder of radius $R$ rotating counterclockwise at angular velocity $\omega$ situated in an ideal fluid with flow velocity $u$ to the right far away from the cylinder, the velocity potential $\Phi$ is given by $$ \Phi(r, \theta)=u r \cos \theta+u \frac{R^{2}}{r} \cos \theta+\frac{\Gamma \theta}{2 \pi} $$ where $(r, \theta)$ is the polar coordinate system with origin at the center of the cylinder. $\Gamma$ is the circulation and is equal to $2 \pi R^{2} \omega$. The fluid velocity is given by $$ \mathbf{v}=\nabla \Phi=\frac{\partial \Phi}{\partial r} \hat{\mathbf{r}}+\frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat{\theta} $$","[""We will work in the reference frame of the center of mass of the cylinder because the fluid flow is steady in this reference frame. The key intuition here is that the magnitude of the fluid velocity above the cylinder will be higher on the top because the tangential velocity of the cylinder is in the same direction as the velocity of the fluid on the top. By Bernoulli's principle, this means that the pressure on the top is lower than the pressure on the bottom, which will create a lift force on the cylinder.\nWith the given theory, we can model this quantitatively. In our chosen reference frame, the water\n\n\n\nmoves with velocity $v$ to the left. The velocity potential around a cylinder with radius $R$ is\n\n$$\n\\Phi(r, \\theta)=-v r \\cos \\theta-v \\frac{R^{2}}{r} \\cos \\theta+R^{2} \\omega \\theta\n$$\n\nTherefore, we find\n\n$$\n\\mathbf{v}=\\nabla \\Phi=\\frac{\\partial \\Phi}{\\partial r} \\hat{\\mathbf{r}}+\\frac{1}{r} \\frac{\\partial \\Phi}{\\partial \\theta} \\hat{\\theta}=-v\\left(1-\\frac{R^{2}}{r^{2}}\\right) \\cos \\theta \\hat{\\mathbf{r}}+\\left(v\\left(1+\\frac{R^{2}}{r^{2}}\\right) \\sin \\theta+R \\omega\\right) \\hat{\\theta}\n$$\n\nAs expected from boundary conditions, the radial velocity vanishes when $r=R$. Furthermore, on the surface of the cylinder, we have the tangential velocity of the fluid is $2 v \\sin \\theta+R \\omega$ in the counterclockwise direction. Consider points on the cylinder at angles $\\theta$ and $-\\theta$. By Bernoulli's principle (ignoring the height difference which will be accounted with the buoyant force),\n\n$$\np_{-\\theta}-p_{\\theta}=\\frac{1}{2} \\rho\\left((2 v \\sin \\theta+R \\omega)^{2}-(-2 v \\sin \\theta+R \\omega)^{2}\\right)=4 \\rho v R \\omega \\sin \\theta\n$$\n\nIf we integrate this result along the surface of the cylinder, we can find the lift force per unit length. Note that only the vertical components of the pressure will matter as the horizontal components cancel due to symmetry. The vertical component of the pressure difference is then $4 \\rho v r_{0} \\omega \\sin ^{2} \\theta$. Thus, the lift force per unit length is\n\n$$\n\\frac{F_{\\text {lift }}}{H}=\\int_{0}^{\\pi} 4 \\rho v R \\omega \\sin ^{2}(\\theta)(R d \\theta)=2 \\pi \\rho \\omega v R^{2}\n$$\n\nThe total left force is\n\n$$\nF_{\\text {lift }}=2 \\pi \\rho R^{2} H \\omega v \\text {. }\n$$\n\nThe gravitational force is $M g$, and the buoyant force is $\\pi R^{2} H \\rho g$. Therefore, we must have\n\n$$\n\\pi R^{2} H \\rho g+2 \\pi \\rho R^{2} H \\omega v=M g\n$$\n\nSolving for $\\omega$, we obtain\n\n$$\n\\omega=\\frac{M g}{2 \\pi R^{2} H \\rho v}-\\frac{g}{2 v}\n$$\n\n$1.25 \\mathrm{~s}^{-1}$""]",['$1.25$'],False,$\mathrm{s}^{-1}$,Numerical,1e-2 864,Mechanics,,"Spinning Cylinder Adithya has a solid cylinder of mass $M=10 \mathrm{~kg}$, radius $R=0.08 \mathrm{~m}$, and height $H=0.20 \mathrm{~m}$. He is running a test in a chamber on Earth over a distance of $d=200 \mathrm{~m}$ as shown below. Assume that the physical length of the chamber is much greater than $d$ (i.e. the chamber extends far to the left and right of the testing area). The chamber is filled with an ideal fluid with uniform density $\rho=700 \mathrm{~kg} / \mathrm{m}^{3}$. Adithya's cylinder is launched with linear velocity $v=10 \mathrm{~m} / \mathrm{s}$ and spins counterclockwise with angular velocity $\omega$. Adithya notices that the cylinder continues on a horizontal path until the end of the chamber. Find the angular velocity $\omega$. Do not neglect forces due to fluid pressure differences. Note that the diagram presents a side view of the chamber (i.e. gravity is oriented downwards with respect to the diagram). ![](https://cdn.mathpix.com/cropped/2023_12_21_e5db709d919ebb702c65g-1.jpg?height=467&width=1521&top_left_y=707&top_left_x=302) Assume the following about the setup and the ideal fluid: - fluid flow is steady in the frame of the center of mass of the cylinder - the ideal fluid is incompressible, irrotational, and has zero viscosity - the angular velocity of the cylinder is approximately constant during its subsequent motion Hint: For a uniform cylinder of radius $R$ rotating counterclockwise at angular velocity $\omega$ situated in an ideal fluid with flow velocity $u$ to the right far away from the cylinder, the velocity potential $\Phi$ is given by $$ \Phi(r, \theta)=u r \cos \theta+u \frac{R^{2}}{r} \cos \theta+\frac{\Gamma \theta}{2 \pi} $$ where $(r, \theta)$ is the polar coordinate system with origin at the center of the cylinder. $\Gamma$ is the circulation and is equal to $2 \pi R^{2} \omega$. The fluid velocity is given by $$ \mathbf{v}=\nabla \Phi=\frac{\partial \Phi}{\partial r} \hat{\mathbf{r}}+\frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat{\theta} $$","[""We will work in the reference frame of the center of mass of the cylinder because the fluid flow is steady in this reference frame. The key intuition here is that the magnitude of the fluid velocity above the cylinder will be higher on the top because the tangential velocity of the cylinder is in the same direction as the velocity of the fluid on the top. By Bernoulli's principle, this means that the pressure on the top is lower than the pressure on the bottom, which will create a lift force on the cylinder.\nWith the given theory, we can model this quantitatively. In our chosen reference frame, the water\n\n\n\nmoves with velocity $v$ to the left. The velocity potential around a cylinder with radius $R$ is\n\n$$\n\\Phi(r, \\theta)=-v r \\cos \\theta-v \\frac{R^{2}}{r} \\cos \\theta+R^{2} \\omega \\theta\n$$\n\nTherefore, we find\n\n$$\n\\mathbf{v}=\\nabla \\Phi=\\frac{\\partial \\Phi}{\\partial r} \\hat{\\mathbf{r}}+\\frac{1}{r} \\frac{\\partial \\Phi}{\\partial \\theta} \\hat{\\theta}=-v\\left(1-\\frac{R^{2}}{r^{2}}\\right) \\cos \\theta \\hat{\\mathbf{r}}+\\left(v\\left(1+\\frac{R^{2}}{r^{2}}\\right) \\sin \\theta+R \\omega\\right) \\hat{\\theta}\n$$\n\nAs expected from boundary conditions, the radial velocity vanishes when $r=R$. Furthermore, on the surface of the cylinder, we have the tangential velocity of the fluid is $2 v \\sin \\theta+R \\omega$ in the counterclockwise direction. Consider points on the cylinder at angles $\\theta$ and $-\\theta$. By Bernoulli's principle (ignoring the height difference which will be accounted with the buoyant force),\n\n$$\np_{-\\theta}-p_{\\theta}=\\frac{1}{2} \\rho\\left((2 v \\sin \\theta+R \\omega)^{2}-(-2 v \\sin \\theta+R \\omega)^{2}\\right)=4 \\rho v R \\omega \\sin \\theta\n$$\n\nIf we integrate this result along the surface of the cylinder, we can find the lift force per unit length. Note that only the vertical components of the pressure will matter as the horizontal components cancel due to symmetry. The vertical component of the pressure difference is then $4 \\rho v r_{0} \\omega \\sin ^{2} \\theta$. Thus, the lift force per unit length is\n\n$$\n\\frac{F_{\\text {lift }}}{H}=\\int_{0}^{\\pi} 4 \\rho v R \\omega \\sin ^{2}(\\theta)(R d \\theta)=2 \\pi \\rho \\omega v R^{2}\n$$\n\nThe total left force is\n\n$$\nF_{\\text {lift }}=2 \\pi \\rho R^{2} H \\omega v \\text {. }\n$$\n\nThe gravitational force is $M g$, and the buoyant force is $\\pi R^{2} H \\rho g$. Therefore, we must have\n\n$$\n\\pi R^{2} H \\rho g+2 \\pi \\rho R^{2} H \\omega v=M g\n$$\n\nSolving for $\\omega$, we obtain\n\n$$\n\\omega=\\frac{M g}{2 \\pi R^{2} H \\rho v}-\\frac{g}{2 v}\n$$\n\n$1.25 \\mathrm{~s}^{-1}$""]",['$1.25$'],False,$\mathrm{s}^{-1}$,Numerical,1e-2 865,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Adithya is launching a package from New York City $\left(40^{\circ} 43^{\prime} \mathrm{N}\right.$ and $\left.73^{\circ} 56^{\prime} \mathrm{W}\right)$ to Guam $\left(13^{\circ} 27^{\prime} \mathrm{N}\right.$ and $\left.144^{\circ} 48^{\prime} \mathrm{E}\right)$. Find the minimal launch velocity $v_{0}$ from New York City to Guam. Ignore the rotation of the earth, effects due to the atmosphere, and the gravitational force from the sun. Additionally, assume the Earth is a perfect sphere with radius $R_{\oplus}=6.37 \times 10^{6} \mathrm{~m}$ and mass $M_{\oplus}=5.97 \times 10^{24} \mathrm{~kg}$.","['We first want to find the angular distance between New York City and Guam. Let this be $\\theta$. Let New York City be point $A$ and Guam be point $B$. Consider the north pole $P$ and the spherical triangle $P A B$. By the spherical law of cosines,\n$$\n\\cos \\theta=\\cos \\left(90^{\\circ}-\\phi_{A}\\right) \\cos \\left(90^{\\circ}-\\phi_{B}\\right)+\\sin \\left(90^{\\circ}-\\phi_{A}\\right) \\sin \\left(90^{\\circ}-\\phi_{B}\\right) \\cos \\left(\\ell_{B}-\\ell_{A}\\right) .\n\\tag{23}\n$$\n\n\n\nEquation 23 simplifies to\n\n$$\n\\cos \\theta=\\sin \\phi_{A} \\sin \\phi_{B}+\\cos \\phi_{A} \\cos \\phi_{B} \\cos \\left(\\ell_{B}-\\ell_{A}\\right)\n\\tag{24}\n$$\n\nfrom which we find $\\theta=115.05^{\\circ}$.\n\nNow that we have determined the angular distance, we will proceed with the orbital mechanics problem. By the vis-viva equation, the speed at the launch point is\n\n$$\nv_{0}=\\sqrt{G M_{\\oplus}\\left(\\frac{2}{R_{\\oplus}}-\\frac{1}{a}\\right)}\n\\tag{25}\n$$\n\nwhere $a$ is the semimajor axis for the orbit. It is clear that in order to minimize $v_{0}$, we must minimize $a$. The orbit is an ellipse with focii $F_{1}$ and $F_{2}$, where $F_{1}$ is the center of the earth. By the definition of an ellipse,\n\n$$\nA F_{1}+A F_{2}=2 a\n\\tag{26}\n$$\n\nSince $A F_{1}=R_{\\oplus}$, is suffices to minimize $A F_{2}$. By symmetry, line $F_{1} F_{2}$ is the perpendicular bisector of $A B$. Since $F_{2}$ is on a fixed line, to minimize $F_{2}$, we place it at the foot of the perpendicular from $A$ to this line. Then, we obtain $A F_{2}=R_{\\oplus} \\sin \\left(\\frac{\\theta}{2}\\right)$. Thus,\n\n$$\na=\\frac{1+\\sin \\left(\\frac{\\theta}{2}\\right)}{2} R_{\\oplus}\n$$\n\nand from Equation 25 we find\n\n$$\nv_{0}=\\sqrt{\\frac{G M_{\\oplus}}{R_{\\oplus}} \\frac{2 \\sin \\frac{\\theta}{2}}{1+\\sin \\frac{\\theta}{2}}} .\n$$']",['$7564$'],False,m/s,Numerical,1e2 866,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a container filled with argon, with molar mass $39.9 \mathrm{~g} \mathrm{~mol}^{-1}$ whose pressure is much smaller than that of atmospheric pressure. Suppose there is a plate of area $A=10 \mathrm{~mm}^{2}$ moving with a speed $v$ perpendicular to its plane. If the gas has density $\rho=4.8 \times 10^{-7} \mathrm{~g} \mathrm{~cm}^{-3}$, and temperature $T=100 \mathrm{~K}$, find an approximate value for the drag force acting on the plate. Suppose that the speed of the plate is $v=100 \mathrm{~m} \mathrm{~s}^{-1}$.","['Let $N$ be the number of particles colliding with the plate in a unit time, and let all the molecules arrive at the plate with the same speed $u$. The force on the plate will be the momentum imparted over a unit time, i.e. $F=2 N m(v-u) / t$ where the coefficient off two reflects on the fact that the average speed of the molecules hitting the wall is the same as the molecules departing. Note that Maxwells distribution dictates that the average speed of particles in all direction $\\langle u\\rangle=0$ which means that the average force acting on the plate is simply $\\langle F\\rangle=2 N m v / t$. During a time period $t$, these molecules arrive a wall that is of thickness $v_{R} t$ where $v_{R}$ is the relative velocity between the molecules and the plate (which is not negligible as the mean free path and velocity of the plate are comparable.) This means the number of collisions found in this layer will be $N=\\frac{1}{2} n V \\approx A v_{R} t$ where the factor of $1 / 2$ reflects on how half the molecules go to the plate and the other half go the other direction. Thus the force acting on the plate will become $\\left\\langle F_{D}\\right\\rangle=2 m v N / t=2 n m v \\cdot v_{R} A$. To find the relative velocity, consider the vector diagram:\n\n\n\n\nBy law of cosines, the magnitude of $v_{R}$ will simply be\n\n$$\nv_{R}^{2}=v^{2}+u^{2}-2 \\boldsymbol{v} \\cdot \\boldsymbol{u}\n$$\n\nThe direction of $\\boldsymbol{u}$ points homogeneously in all directions as the orientation of molecules changes with each individual one. Therefore, $2 \\boldsymbol{v} \\cdot \\boldsymbol{u}$ will point in all directions averaging to 0 . Thus, the magnitude of $v_{R}$ will simply be $\\left\\langle v_{R}\\right\\rangle=\\sqrt{v^{2}+\\langle u\\rangle^{2}}$ where $\\langle u\\rangle^{2}$ is the average thermal velocity of molecules. In terms of density, we can then express the drag force to be $2 \\rho A v \\sqrt{v^{2}+\\langle u\\rangle^{2}}$. $2.41 \\times 10^{-4} \\mathrm{~N}$.']",['$2.41\\times 10^{-4}$'],False,N,Numerical,1e-5 867,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a rectangular loop made of superconducting material with length $\ell=200 \mathrm{~cm}$ and width $w=2 \mathrm{~cm}$. The radius of this particular wire is $r=0.5 \mathrm{~mm}$. This superconducting rectangular loop initially has a current $I_{1}=5 \mathrm{~A}$ in the counterclockwise direction as shown in the figure below. This rectangular loop is situated a distance $d=1 \mathrm{~cm}$ above an infinitely long wire that initially contains no current. Suppose that the current in the infinitely long wire is increased to some current $I_{2}$ such that there is an attractive force $F$ between the rectangular loop and the long wire. Find the maximum possible value of $F$. Write your answer in newtons. Hint: You may neglect the magnetic field produced by the vertical segments in the rectangular loop. ","[""The key idea is that the superconducting loop must have constant flux. If it did not, by Faraday's Law, an emf\n$$\n\\mathcal{E}=-\\frac{d \\Phi}{d t}\n$$\n\nwould be generated in the loop. Since superconducting materials have no resistance, this would imply an infinite current, hence a contradiction.\n\nWe will first compute the flux through the rectangular loop when there is a current $I_{1}$. Since $w \\ll \\ell$, we can assume that the vertical segments produce negligible amounts of magnetic field. We can furthermore approximate the field produced by one of the horizontal wires a distance $r$ away as\n\n\n\n$\\frac{\\mu_{0} I}{2 \\pi r}$ (this is valid for an infinitely long wire, and therefore is also valid in the regime where $w \\ll \\ell$ ).\n\nThus, the total flux through the rectangular loop when there is a current $I_{1}$ is\n\n$$\n\\Phi_{1}=\\int_{r}^{w} B\\left(\\ell d r^{\\prime}\\right)=\\int_{r}^{w-r}\\left(\\frac{\\mu_{0} I_{1}}{2 \\pi r^{\\prime}}+\\frac{\\mu_{0} I_{1}}{2 \\pi\\left(w-r^{\\prime}\\right)}\\right) \\ell d r^{\\prime}=\\frac{\\mu_{0} I_{1} \\ell}{\\pi} \\ln \\left(\\frac{w}{r}\\right)\n$$\n\nNote that the self inductance of the loop is $L=\\frac{\\Phi}{I_{1}}=\\frac{\\mu_{0} \\ell}{\\pi} \\ln \\left(\\frac{w}{r}\\right)$.\n\nNow, we will determine the flux through the rectangular loop due to the long current-carrying wire. This is\n\n$$\n\\Phi_{2}=\\int_{d}^{d+w} \\frac{\\mu_{0} I_{2}}{2 \\pi r}(\\ell d r)=\\frac{\\mu_{0} I_{2} \\ell}{2 \\pi} \\ln \\left(\\frac{d+w}{d}\\right)\n$$\n\nThe mutual inductance is $M=\\frac{\\Phi_{2}}{I_{2}}=\\frac{\\mu_{0} \\ell}{2 \\pi} \\ln \\left(\\frac{d+w}{d}\\right)$. In to maintain the same flux in the loop, the current will change to $I_{3}$ where\n\n$$\nL I_{1}=M I_{2}+L I_{3}\n$$\n\nor\n\n$$\nI_{3}=I_{1}-\\frac{M}{L} I_{2}\n$$\n\nNow, we compute the force between the rectangular loop and the long, current-carrying wire. The forces on the vertical sides cancel out because the current in the loop is in opposite directions on these sides. From the horizontal sides, we have the force is\n\n$$\n\\begin{aligned}\nF=\\sum\\left(I_{3} \\vec{\\ell} \\times \\vec{B}\\right) & =I_{3} \\ell\\left(\\frac{\\mu_{0} I_{2}}{2 \\pi d}-\\frac{\\mu_{0} I_{2}}{2 \\pi(d+w)}\\right) \\\\\n& =\\frac{\\mu_{0} \\ell w}{2 \\pi d(d+w)}\\left[I_{2}\\left(I_{1}-\\frac{M}{L} I_{2}\\right)\\right] .\n\\end{aligned}\n$$\n\nThis quadratic in $I_{2}$ is maximized when $I_{2}=\\frac{L}{2 M} I_{1}$ in which case the force becomes\n\n$$\nF=\\frac{\\mu_{0} \\ell w}{2 \\pi d(d+w)} \\frac{L I_{1}^{2}}{4 M}=\\frac{\\mu_{0} \\ell w I_{1}^{2}}{4 \\pi d(d+w)} \\frac{\\ln \\left(\\frac{w}{r}\\right)}{\\ln \\left(\\frac{d+w}{d}\\right)}\n$$\n\nNote: If the size of the wires is considered when computing flux, a slightly different answer is obtained. In the contest, all answers between $1.11 \\times 10^{-3}$ and $1.18 \\times 10^{-3}$ were accepted.""]",['$1.12 \\times 10^{-3}$'],False,N,Numerical,1e-4 867,Electromagnetism,,"Consider a rectangular loop made of superconducting material with length $\ell=200 \mathrm{~cm}$ and width $w=2 \mathrm{~cm}$. The radius of this particular wire is $r=0.5 \mathrm{~mm}$. This superconducting rectangular loop initially has a current $I_{1}=5 \mathrm{~A}$ in the counterclockwise direction as shown in the figure below. This rectangular loop is situated a distance $d=1 \mathrm{~cm}$ above an infinitely long wire that initially contains no current. Suppose that the current in the infinitely long wire is increased to some current $I_{2}$ such that there is an attractive force $F$ between the rectangular loop and the long wire. Find the maximum possible value of $F$. Write your answer in newtons. Hint: You may neglect the magnetic field produced by the vertical segments in the rectangular loop. ![](https://cdn.mathpix.com/cropped/2023_12_21_125ad4b445367283c5eag-1.jpg?height=319&width=1214&top_left_y=1689&top_left_x=453)","[""The key idea is that the superconducting loop must have constant flux. If it did not, by Faraday's Law, an emf\n$$\n\\mathcal{E}=-\\frac{d \\Phi}{d t}\n$$\n\nwould be generated in the loop. Since superconducting materials have no resistance, this would imply an infinite current, hence a contradiction.\n\nWe will first compute the flux through the rectangular loop when there is a current $I_{1}$. Since $w \\ll \\ell$, we can assume that the vertical segments produce negligible amounts of magnetic field. We can furthermore approximate the field produced by one of the horizontal wires a distance $r$ away as\n\n\n\n$\\frac{\\mu_{0} I}{2 \\pi r}$ (this is valid for an infinitely long wire, and therefore is also valid in the regime where $w \\ll \\ell$ ).\n\nThus, the total flux through the rectangular loop when there is a current $I_{1}$ is\n\n$$\n\\Phi_{1}=\\int_{r}^{w} B\\left(\\ell d r^{\\prime}\\right)=\\int_{r}^{w-r}\\left(\\frac{\\mu_{0} I_{1}}{2 \\pi r^{\\prime}}+\\frac{\\mu_{0} I_{1}}{2 \\pi\\left(w-r^{\\prime}\\right)}\\right) \\ell d r^{\\prime}=\\frac{\\mu_{0} I_{1} \\ell}{\\pi} \\ln \\left(\\frac{w}{r}\\right)\n$$\n\nNote that the self inductance of the loop is $L=\\frac{\\Phi}{I_{1}}=\\frac{\\mu_{0} \\ell}{\\pi} \\ln \\left(\\frac{w}{r}\\right)$.\n\nNow, we will determine the flux through the rectangular loop due to the long current-carrying wire. This is\n\n$$\n\\Phi_{2}=\\int_{d}^{d+w} \\frac{\\mu_{0} I_{2}}{2 \\pi r}(\\ell d r)=\\frac{\\mu_{0} I_{2} \\ell}{2 \\pi} \\ln \\left(\\frac{d+w}{d}\\right)\n$$\n\nThe mutual inductance is $M=\\frac{\\Phi_{2}}{I_{2}}=\\frac{\\mu_{0} \\ell}{2 \\pi} \\ln \\left(\\frac{d+w}{d}\\right)$. In to maintain the same flux in the loop, the current will change to $I_{3}$ where\n\n$$\nL I_{1}=M I_{2}+L I_{3}\n$$\n\nor\n\n$$\nI_{3}=I_{1}-\\frac{M}{L} I_{2}\n$$\n\nNow, we compute the force between the rectangular loop and the long, current-carrying wire. The forces on the vertical sides cancel out because the current in the loop is in opposite directions on these sides. From the horizontal sides, we have the force is\n\n$$\n\\begin{aligned}\nF=\\sum\\left(I_{3} \\vec{\\ell} \\times \\vec{B}\\right) & =I_{3} \\ell\\left(\\frac{\\mu_{0} I_{2}}{2 \\pi d}-\\frac{\\mu_{0} I_{2}}{2 \\pi(d+w)}\\right) \\\\\n& =\\frac{\\mu_{0} \\ell w}{2 \\pi d(d+w)}\\left[I_{2}\\left(I_{1}-\\frac{M}{L} I_{2}\\right)\\right] .\n\\end{aligned}\n$$\n\nThis quadratic in $I_{2}$ is maximized when $I_{2}=\\frac{L}{2 M} I_{1}$ in which case the force becomes\n\n$$\nF=\\frac{\\mu_{0} \\ell w}{2 \\pi d(d+w)} \\frac{L I_{1}^{2}}{4 M}=\\frac{\\mu_{0} \\ell w I_{1}^{2}}{4 \\pi d(d+w)} \\frac{\\ln \\left(\\frac{w}{r}\\right)}{\\ln \\left(\\frac{d+w}{d}\\right)}\n$$\n\nNote: If the size of the wires is considered when computing flux, a slightly different answer is obtained. In the contest, all answers between $1.11 \\times 10^{-3}$ and $1.18 \\times 10^{-3}$ were accepted.""]",['$1.12 \\times 10^{-3}$'],False,N,Numerical,1e-4 868,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a $1 \mathrm{~cm}$ long slit with negligible height. First, we divide the slit into thirds and cover the middle third. Then, we perform the same steps on the two shorter slits. Again, we perform the same steps on the four even shorter slits and continue for a very long time. Then, we shine a monochromatic, coherent light source of wavelength $500 \mathrm{~nm}$ on our slits, which creates an interference pattern on a wall 10 meters away. On the wall, what is the distance between the central maximum and the first side maximum? Assume the distance to the wall is much greater than the width of the slit. Answer in millimeters.","['This problem is essentially an infinite convolution. In detail, consider two separate amplitude functions $f(x)$ and $g(x)$ that correspond to two different interference patterns. One can think of a convolution of $f$ and $g$ as taking $f$ and placing the same function in every position that exists in $g$. To be more specific, the amplitude of one interference pattern can be thought of as the product of the amplitude patterns of two separate amplitude functions. For example, the diffraction pattern of two slits is the same as the product of the amplitudes for one slit and two light sources (this will be given as an exercise to prove). To make more sense of this, let us consider that very example and let $f$ pertain to the amplitude function of a single slit and let $g$ pertain to the amplitude function of two light sources. One can then write in phase space with a generalized coordinate $x^{\\prime}$ that $\\{f * g\\}(x)=\\int_{-\\infty}^{\\infty} f\\left(x-x^{\\prime}\\right) g\\left(x^{\\prime}\\right) \\mathrm{d} x^{\\prime}$ according to the convolution theorem. The integral then runs through all values and places a copy of $f\\left(x-x^{\\prime}\\right)$ at the peaks of $g\\left(x^{\\prime}\\right)$.\nWith this in hand, we can use the convolution theorem to our advantage. Let us designate the amplitude function of the entire Cantor slit to be $F(\\theta)$. Since $\\theta$ is small, we can designate the phase angle as $\\phi=k x \\theta$ where $x$ is a variable moving through all of the slits. If we consider a single slit that is cut into a third, the angle produced will be a third of its original as well, or the new amplitude function will simply be a function of $\\theta / 3$ or $F(\\theta / 3)$. The distance between the midpoints of the two slits will simply be $d=2 / 3 \\mathrm{~cm}$ which then becomes $2 / 9$, then $2 / 27$ and so on. In terms of the original function, we can decompose it into the function of $F(\\theta / 3)$ and the amplitude function of a single light source. In other words, $F(\\theta)=F(\\theta / 3) \\cos (k d \\theta / 2)$ where $k=\\frac{2 \\pi}{\\lambda}$. To the limit of infinity, this can be otherwise written as\n\n$$\nF(\\theta)=\\prod_{i=0}^{\\infty} \\cos \\left(\\frac{k d \\theta}{2 \\cdot 3^{i}}\\right)=\\cos \\left(\\frac{k d \\theta}{2}\\right) \\cos \\left(\\frac{k d \\theta}{6}\\right) \\cos \\left(\\frac{k d \\theta}{18}\\right) \\ldots\n$$\n\nFinding an exact mathematical solution would be difficult, thus it is simple enough to just multiply the first 4 or 5 terms to achieve the final answer.']",['$0.647$'],False,mm,Numerical,1e-2 869,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A certain planet with radius $R=$ $3 \times 10^{4} \mathrm{~km}$ is made of a liquid with constant density $\rho=1.5 \mathrm{~g} / \mathrm{cm}^{3}$ with the exception of a homogeneous solid core of radius $r=10 \mathrm{~km}$ and mass $m=2.4 \times 10^{16} \mathrm{~kg}$. Normally, the core is situated at the geometric center of the planet. However, a small disturbance has moved the center of the core $x=1 \mathrm{~km}$ away from the geometric center of the planet. The core is released from rest, and the fluid is inviscid and incompressible. Calculate the magnitude of the force due to gravity that now acts on the core. Work under the assumption that $R \gg r$.","['We solve by simplifying the configuration into progressively simpler but mathematically equivalent formulations of the same problem.\nBecause the core is spherical and homogeneous, all the gravitational forces on the core are equivalent to the gravitational forces on a point mass located at the center of the core. Thus, the problem is now the force of gravity on a point mass of mass $m$ located $x$ distance away from the center of planet, with a sphere of radius $r$ evacuated around the point mass.\n\nBut if we filled this evacuated sphere, would the force on the point mass be any different? No! If we filled up the sphere, all the added liquid (of the same density as the rest of the planet) would add no additional force on the point mass because it is symmetrically distributed around the point mass. Thus, the problem is now the force of gravity on a point mass of mass $m$ located $x$ distance away from the center of the planet.\n\nBy shell theorem, we can ignore all the fluid that is more than a distance $x$ away from the center of the planet. Thus, the problem is now the force of gravity on a point mass $m$ situated on the surface of a liquid planet of radius $x$. This force is not difficult to calculate at all:\n\n$$\n\\begin{aligned}\nF & =\\frac{G M m}{x^{2}} \\\\\n& =\\frac{G m}{x^{2}}\\left(\\frac{4}{3} \\pi x^{3} \\rho\\right) \\\\\n& =\\frac{4}{3} \\pi G m \\rho x\n\\end{aligned}\n$$\n\n$1.0058 \\times 10^{13} \\mathrm{~N}$']",['$1.0058 \\times 10^{13}$'],False,N,Numerical,1e11 870,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A scientist is doing an experiment with a setup consisting of 2 ideal solenoids that share the same axis. The lengths of the solenoids are both $\ell$, the radii of the solenoids are $r$ and $2 r$, and the smaller solenoid is completely inside the larger one. Suppose that the solenoids share the same (constant) current $I$, but the inner solenoid has $4 N$ loops while the outer one has $N$, and they have opposite polarities (meaning the current is clockwise in one solenoid but counterclockwise in the other). Model the Earth's magnetic field as one produced by a magnetic dipole centered in the Earth's core. Let $F$ be the magnitude of the total magnetic force the whole setup feels due to Earth's magnetic field. Now the scientist replaces the setup with a similar one: the only differences are that the the radii of the solenoids are $2 r$ (inner) and $3 r$ (outer), the length of each solenoid is $7 \ell$, and the number of loops each solenoid is $27 N$ (inner) and $12 N$ (outer). The scientist now drives a constant current $2 I$ through the setup (the solenoids still have opposite polarities), and the whole setup feels a total force of magnitude $F^{\prime}$ due to the Earth's magnetic field. Assuming the new setup was in the same location on Earth and had the same orientation as the old one, find $F^{\prime} / F$. Assume the dimensions of the solenoids are much smaller than the radius of the Earth.","[""We can solve the problem by assuming that the location of the setup is at the North Pole and that the solenoids are oriented so that their axis intersects the Earth's core. Note that if we had some other location or orientation, then both $F$ and $F^{\\prime}$ would be multiplied by the same factor, so their ratio remains the same.\nSuppose the radii of the solenoids are $r$ and $\\alpha r$, where the number of inner and outer loops are $N$ and $\\frac{N}{\\alpha^{2}}$, respectively. To find the force the Earth's dipole exerts on the solenoids, we can calculate the force the solenoids exert on the dipole. To do this, we need to find the gradient of the magnetic field produced by the solenoids at the dipole's location. Let the radius of the Earth be $R$.\n\nConsider the field produced by 2 concentric, coaxial, ring currents, the inner ring with current $I$ radius $r$ and the outer one with current $\\frac{I}{\\alpha^{2}}$ and radius $\\alpha r$. The currents are in opposite directions. At a distance $R$ away from the center of the rings, along their axis, the magnetic field is given by\n\n$$\n\\begin{aligned}\nB & =\\frac{\\mu_{0} I r^{2}}{2\\left(R^{2}+r^{2}\\right)^{\\frac{3}{2}}}-\\frac{\\mu_{0} I r^{2}}{2\\left(R^{2}+(\\alpha r)^{2}\\right)^{\\frac{3}{2}}} \\\\\n& =\\frac{\\mu_{0} I^{2}}{2 R^{3}}\\left(\\left(1+\\frac{r^{2}}{R^{2}}\\right)^{-\\frac{3}{2}}-\\left(1+\\frac{\\alpha^{2} r^{2}}{R^{2}}\\right)^{-\\frac{3}{2}}\\right) \\\\\n& \\approx \\frac{\\mu_{0} I^{2}}{2 R^{3}}\\left(\\frac{3}{2}\\left(\\alpha^{2}-1\\right) \\frac{r^{2}}{R^{2}}\\right) \\\\\n& =\\frac{3 \\mu_{0} I r^{4}}{4 R^{5}}\\left(\\alpha^{2}-1\\right)\n\\end{aligned}\n$$\n\nThus, the gradient of the magnetic field is proportional to $\\operatorname{Ir}^{4}\\left(\\alpha^{2}-1\\right)$. Now we consider the actual setup. The new setup multiplies the effective current $\\frac{27}{4} \\cdot \\frac{2}{1}=\\frac{27}{2}$ times, while multiplying $r$ by 2 . The factor $\\alpha^{2}-1$ changed from 3 to $\\frac{5}{4}$. Combining, we get $\\frac{F^{\\prime}}{F}=\\frac{27}{2} \\cdot 2^{4} \\cdot \\frac{5}{12}=90$.""]",['90'],False,,Numerical,1e-1 871,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Adithya is in a rocket with proper acceleration $a_{0}=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}^{2}$ to the right, and Eddie is in a rocket with proper acceleration $\frac{a_{0}}{2}$ to the left. Let the frame of Adithya's rocket be $S_{1}$, and the frame of Eddie's rocket be $S_{2}$. Initially, both rockets are at rest with respect to each other, they are at the same location, and Adithya's clock and Eddie's clock are both set to 0 . At the moment Adithya's clock reaches $0.75 \mathrm{~s}$ in $S_{2}$, what is the velocity of Adithya's rocket in $S_{2}$ ?","[""Throughout this solution, we will let $c=a_{0}=1$, for simplicity. We will work in the inertial frame that is initially at rest with both rockets at $t_{1}=t_{2}=0$. First, we determine the velocity of Adithya's rocket in this frame as a function of the proper time that has elapsed $t_{1}$. In frame. $S_{1}$, in a time $d t_{1}$, the rocket's velocity increases by $d t_{1}$, so by velocity addition, the new velocity in the inertial frame is\n$$\n\\frac{v_{1}+d t_{1}}{1+v_{1} d t_{1}} \\approx v_{1}+\\left(1-v_{1}^{2}\\right) d t_{1}\n$$\n\nTherefore, we have\n\n$$\n\\frac{d v_{1}}{d t_{1}}=1-v_{1}^{2}\n$$\n\nUpon separating and integrating, we find $v_{1}=\\tanh \\left(t_{1}\\right)$. Similarly, the velocity of Eddie's rocket in the inertial frame is $v_{2}=\\tanh \\left(t_{2} / 2\\right)$. Now, in this inertial frame, let the time between events $A$ and $B$ be $t$, and let the distance between rockets $A$ and $B$ be $x$. By a Lorentz transformation, the time between the events in frame $S_{2}$ is\n\n$$\nt^{\\prime}=\\gamma\\left(t-v_{2} x\\right)=0\n$$\n\nsince the events are simultaneous in $S_{2}$. Therefore, we must have $t=v_{2} x$. Note that $t=t_{A}-t_{B}$ where $t_{A}$ and $t_{B}$ denote the times of events $A$ and $B$ in the inertial frame, respectively. Also, note that $x=x_{A}+x_{B}$ where $x_{A}$ and $x_{B}$ are the respective displacements of the two rockets in the inertial frame. By the effects of time dilation, we have\n\n$$\nt_{A}=\\int \\gamma d t_{1}=\\int \\frac{1}{\\sqrt{1-v_{1}^{2}}} d t_{1}=\\int_{0}^{t_{1}} \\cosh \\left(t_{1}^{\\prime}\\right) d t_{1}^{\\prime}=\\sinh \\left(t_{1}\\right)\n$$\n\nSimilarly, $t_{B}=2 \\sinh \\left(t_{2} / 2\\right)$, and we obtain $t=2 \\sinh \\left(t_{2} / 2\\right)-\\sinh \\left(t_{1}\\right)$. Additionally, from the above result,\n\n$$\nx_{A}=\\int v_{1} d t=\\int v_{1} \\cosh \\left(t_{1}\\right) d t_{1}=\\int_{0}^{t_{1}} \\sinh \\left(t_{1}\\right) d t_{1}^{\\prime}=\\cosh \\left(t_{1}\\right)-1\n$$\n\nSimilarly, $x_{B}=2 \\cosh \\left(t_{2} / 2\\right)-2$, and $x=\\cosh \\left(t_{1}\\right)+2 \\cosh \\left(t_{2} / 2\\right)-3$. Thus, from $t=v_{2} x$,\n\n$$\n\\begin{gathered}\n2 \\sinh \\left(t_{2} / 2\\right)-\\sinh \\left(t_{1}\\right)=\\tanh \\left(t_{2} / 2\\right)\\left(\\cosh \\left(t_{1}\\right)+2 \\cosh \\left(t_{2} / 2\\right)-3\\right) . \\\\\n-\\sinh \\left(t_{1}\\right)=\\tanh \\left(t_{2} / 2\\right)\\left(\\cosh \\left(t_{1}\\right)-3\\right) . \\\\\n\\tanh \\left(t_{2} / 2\\right)=\\frac{\\sinh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)} .\n\\end{gathered}\n$$\n\nNow, this is the velocity of Eddie's rocket as measured from the inertial frame, so by velocity addition, the velocity of Adithya's rocket as seen by Eddie is\n\n$$\nv=\\frac{\\tanh \\left(t_{1}\\right)+\\frac{\\sinh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)}}{1+\\frac{\\sinh \\left(t_{1}\\right) \\tanh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)}}=\\frac{3 \\tanh \\left(t_{1}\\right)}{3-\\cosh \\left(t_{1}\\right)+\\sinh \\left(t_{1}\\right) \\tanh \\left(t_{1}\\right)} .\n$$""]",['$2.564 \\times 10^{8}$'],False,m/s,Numerical,5e6 872,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Suppose a ping pong ball of radius $R$, thickness $t$, made out of a material with density $\rho_{b}$, and Young's modulus $Y$, is hit so that it resonates in mid-air with small amplitude oscillations. Assume $t \ll R$. The density of air around (and inside) the ball is $\rho_{a}$, and the air pressure is $p$, where $\rho_{a} \ll \rho_{b} \frac{t}{R}$ and $p \ll Y \frac{t^{3}}{R^{3}}$. An estimate for the resonance frequency is $\omega \sim R^{a} t^{b} \rho_{b}^{c} Y^{d}$. Find the value of $4 a^{2}+3 b^{2}+2 c^{2}+d^{2}$. Hint: The surface of the ball will oscillate by ""bending"" instead of ""stretching"", since the former takes much less energy than the latter.","['Throughout the problem, we will work to order of magnitude and ignore prefactors.\nThe hint says the surface of the ball will bend instead of stretch, so we need to develop a theory of bending. First, we consider the simplified scenario of a long beam with thickness $t$, width $w$, length $L$, made out of a material with Young\'s modulus $Y$ and density $\\rho$. When the beam bends, the top part of the beam will be in tension and the bottom part will be in compression. Thus, this is how the potential energy is stored in the beam. Suppose the beam is bent with curvature $\\kappa$. Then the top part of the beam will stretch by $L t \\kappa$, and the bottom part will compress by the same amount. Using Hooke\'s law, we can approximate the total potential energy stored in the beam as $U \\sim \\frac{Y t w}{L}(L t \\kappa)^{2} \\sim Y t^{3} w L \\kappa^{2}$. Note that if the relaxed state of the beam was already curved, we simply replace $\\kappa$ with the change in curvature.\n\nTo find the oscillation frequency of the beam, we need to find the kinetic energy in terms of $\\dot{\\kappa}$. Since curvature is on the order of second derivative of displacement, we can multiply $\\kappa$ by $L^{2}$ to get an estimate for displacement. Then $\\dot{\\kappa} L^{2}$ gives an estimate for speed, so the kinetic energy is $K \\sim \\rho t w L\\left(\\dot{\\kappa} L^{2}\\right)^{2} \\sim \\rho t w L^{5} \\dot{\\kappa}^{2}$. Thus, the frequency of oscillations is $\\omega \\sim \\sqrt{\\frac{Y t^{2}}{\\rho L^{4}}}$. Again, if the beam\n\n\n\nwas already curved, we can replace $\\kappa$ everywhere with the change in curvature.\n\nWe can model the ping pong ball as a ""beam"" of length order $R$, width order $R$, and thickness $t$. This is a very crude approximation, but will give a dimensionally correct answer (since we are ignoring prefactors). The angular frequency is thus $\\omega \\sim \\frac{t}{R^{2}} \\sqrt{\\frac{Y}{\\rho_{b}}}$.']",['19.75'],False,,Numerical,3e-1 873,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A player throws two tennis balls on a level ground at $v=20 \mathrm{~m} / \mathrm{s}$ in the same direction, once at an angle of $\alpha=35^{\circ}$ and once at an angle $\beta=55^{\circ}$ to the horizontal. The distance between the landing spots of the two balls is $d$. Find $d$ in meters. Assume the height of the player is negligble and ignore air resistance.","['The range of a projectile is proportional as $R \\propto \\sin 2 \\theta$, or $R \\propto \\cos \\theta \\sin \\theta$. As $\\cos (90-\\theta)=\\sin \\theta$, and $\\alpha+\\beta=90$, the distance travelled by both projectiles are the same.\n$0 \\mathrm{~m}$']",['0'],False,m,Numerical,0 874,Mechanics,,"Consider the following simple model of a bow and arrow. An ideal elastic string has a spring constant $k=10 \mathrm{~N} / \mathrm{m}$ and relaxed length $L=1 \mathrm{~m}$ which is attached to the ends of an inflexible fixed steel rod of the same length $L$ as shown below. A small ball of mass $m=2 \mathrm{~kg}$ and the thread are pulled by its midpoint away from the rod until each individual part of the thread have the same length of the rod, as shown below. What is the speed of the ball in meters per seconds right after it stops accelerating? Assume the whole setup is carried out in zero gravity. ![](https://cdn.mathpix.com/cropped/2023_12_21_78a63d65ef2bfc8b2947g-1.jpg?height=220&width=273&top_left_y=1164&top_left_x=926)","['We can use conservation of energy. The bow string has its potential increased as\n$$\nE_{p}=\\frac{1}{2} k(2 L-L)^{2}=\\frac{1}{2} k L^{2}\n$$\n\nThis all turns into the kinetic energy of the ball $E_{k}=\\frac{1}{2} m v^{2}$, so\n\n$$\nE_{p}=E_{k} \\Longrightarrow \\frac{1}{2} k L^{2}=\\frac{1}{2} m v^{2} \\Longrightarrow v=L \\sqrt{\\frac{k}{m}}\n$$']",['$2.23$'],False,m/s,Numerical,1e-1 874,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider the following simple model of a bow and arrow. An ideal elastic string has a spring constant $k=10 \mathrm{~N} / \mathrm{m}$ and relaxed length $L=1 \mathrm{~m}$ which is attached to the ends of an inflexible fixed steel rod of the same length $L$ as shown below. A small ball of mass $m=2 \mathrm{~kg}$ and the thread are pulled by its midpoint away from the rod until each individual part of the thread have the same length of the rod, as shown below. What is the speed of the ball in meters per seconds right after it stops accelerating? Assume the whole setup is carried out in zero gravity. ","['We can use conservation of energy. The bow string has its potential increased as\n$$\nE_{p}=\\frac{1}{2} k(2 L-L)^{2}=\\frac{1}{2} k L^{2}\n$$\n\nThis all turns into the kinetic energy of the ball $E_{k}=\\frac{1}{2} m v^{2}$, so\n\n$$\nE_{p}=E_{k} \\Longrightarrow \\frac{1}{2} k L^{2}=\\frac{1}{2} m v^{2} \\Longrightarrow v=L \\sqrt{\\frac{k}{m}}\n$$']",['$2.23$'],False,m/s,Numerical,1e-1 875,Mechanics,,"A truck (denoted by $S$ ) is driving at a speed $v=2 \mathrm{~m} / \mathrm{s}$ in the opposite direction of a car driving at a speed $u=3 \mathrm{~m} / \mathrm{s}$, which is equipped with a rear-view mirror. Both $v$ and $u$ are measured from an observer on the ground. Relative to this observer, what is the speed (in $\mathrm{m} / \mathrm{s}$ ) of the truck's image $S^{\prime}$ through the car's mirror? Car's mirror is a plane mirror. ![](https://cdn.mathpix.com/cropped/2023_12_21_b2b497998ee2d73231a3g-1.jpg?height=488&width=615&top_left_y=192&top_left_x=752)","[""In the mirror's frame of reference, the source speed and the image speed is both $u+v$ but in opposite direction. Now, go back the the observer's frame of reference, the image speed becomes $v^{\\prime}$ :\n$$\nv^{\\prime}=(u+v)+u=2 u+v=8 \\mathrm{~m} / \\mathrm{s}\n\\tag{1}\n$$""]",['$8$'],False,m/s,Numerical,1e-1 875,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A truck (denoted by $S$ ) is driving at a speed $v=2 \mathrm{~m} / \mathrm{s}$ in the opposite direction of a car driving at a speed $u=3 \mathrm{~m} / \mathrm{s}$, which is equipped with a rear-view mirror. Both $v$ and $u$ are measured from an observer on the ground. Relative to this observer, what is the speed (in $\mathrm{m} / \mathrm{s}$ ) of the truck's image $S^{\prime}$ through the car's mirror? Car's mirror is a plane mirror. ","[""In the mirror's frame of reference, the source speed and the image speed is both $u+v$ but in opposite direction. Now, go back the the observer's frame of reference, the image speed becomes $v^{\\prime}$ :\n$$\nv^{\\prime}=(u+v)+u=2 u+v=8 \\mathrm{~m} / \\mathrm{s}\n\\tag{1}\n$$""]",['$8$'],False,m/s,Numerical,1e-1 876,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","For this problem, assume the Earth moves in a perfect circle around the sun in the $x y$ plane, with a radius of $r=1.496 \times 10^{11} \mathrm{~m}$, and the Earth has a mass $m=5.972 \times 10^{24} \mathrm{~kg}$. An alien stands far away from our solar system on the $x$ axis such that it appears the Earth is moving along a one dimensional line, as if there was a zero-length spring connecting the Earth and the Sun. For the alien at this location, it is impossible to tell just from the motion if it's 2D motion via gravity or 1D motion via a spring. Let $U_{g}$ be the gravitational potential energy ignoring its self energy if Earth moves via gravity, taking potential energy at infinity to be 0 and $U_{s}$ be the maximum spring potential energy if Earth moves in $1 \mathrm{D}$ via a spring. Compute $U_{g} / U_{s}$.","['One naive idea is to directly compute $U_{g}$ and $U_{s}$, but we can use the fact that their frequencies are the same, or:\n$$\n\\omega^{2}=\\frac{k}{m}=\\frac{G M}{r^{3}} \\Longrightarrow k r^{2}=\\frac{G M}{r}\n$$\n\nThen,\n\n$$\nU_{g}=-\\frac{G M m}{r}=-k r^{2}\n$$\n\nand\n\n$$\nU_{s}=\\frac{1}{2} k r^{2}\n$$\n\nTherefore,\n\n$$\nU_{g} / U_{s}=-2\n$$']",['-2'],False,,Numerical,1e-1 877,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Battle ropes can be used as a full body workout (see photo). It consists of a long piece of thick rope (ranging from $35 \mathrm{~mm}$ to $50 \mathrm{~mm}$ in diameter), wrapped around a stationary pole. The athlete grabs on to both ends, leans back, and moves their arms up and down in order to create waves, as shown in the photo. The athlete wishes to upgrade from using a $35 \mathrm{~mm}$ diameter rope to a $50 \mathrm{~mm}$ diameter rope, while keeping everything else the same (rope material, rope tension, amplitude, and speed at which her arms move back and forth). By doing so, the power she needs to exert changes from $P_{0}$ to $P_{1}$. Compute $P_{1} / P_{0}$.","['The power transmitted by a wave is given by\n$$\nP=\\frac{1}{2} \\mu \\omega^{2} A^{2} v\n$$\n\nwhere $\\mu=\\frac{m}{L}$ is the linear mass density, $A$ is the amplitude, and $v$ is the speed of the wave. The speed of a wave on a rope is given by\n\n$$\nv=\\sqrt{\\frac{T}{\\mu}}\n\\tag{2}\n$$\n\nwhere $T$ is the tension. Note that $\\omega, A, T$ will all remain constant when changing the radius. Thus, $P \\propto \\sqrt{\\mu} \\propto \\sqrt{m}$. As we increase the radius by a factor of $f=\\frac{50}{35}$, we change the mass by $f^{2}$, so the power changes by a factor of $f$, giving us\n\n$$\nP_{1} / P_{0}=f=1.43\n$$']",['1.43'],False,,Numerical,1e-1 878,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Given vertically polarized light, you're given the task of changing it to horizontally polarized light by passing it through a series of $N=5$ linear polarizers. What is the maximum possible efficiency of this process? (Here, efficiency is defined as the ratio between output light intensity and input light intensity.)","['Let $\\theta_{0}=0$ be the original direction of polarization and $\\theta_{5}=\\pi / 2$ the final direction of polarization. The 5 polarizers are directed along $\\theta_{1}, \\theta_{2}, \\ldots, \\theta_{5}$. Let $\\delta_{k}=\\theta_{k}-\\theta_{k-1}$, so that the efficiency is\n\n$$\n\\eta=\\prod_{k=1}^{5} \\cos ^{2} \\delta_{k}\n$$\n\nWe wish to maximize $\\eta$ subject to the constraint that $\\sum_{k} \\delta_{k}=\\pi / 2$. Clearly, the $\\delta_{k}^{\\prime} s$ should be non-negative, implying that $0 \\leq \\delta_{k} \\leq \\pi / 2$ and thus $\\cos \\delta_{k} \\geq 0$ for all $k$.\n\n\n\nWe claim that the maximum is achieved when all $\\delta_{k}$ are equal. If not, let $\\delta_{i} \\neq \\delta_{i+1}$. Then\n\n$$\n\\begin{aligned}\n\\cos \\delta_{i} \\cos \\delta_{i+1} & =\\frac{1}{2}\\left[\\cos \\left(\\delta_{i}+\\delta_{i+1}\\right)+\\cos \\left(\\delta_{i}-\\delta_{i+1}\\right)\\right] \\\\\n& <\\frac{1}{2}\\left[\\cos \\left(\\delta_{i}+\\delta_{i+1}\\right)+1\\right] \\\\\n& =\\frac{1}{2}\\left[\\cos \\left(\\delta_{i}^{\\prime}+\\delta_{i+1}^{\\prime}\\right)+\\cos \\left(\\delta_{i}^{\\prime}-\\delta_{i+1}^{\\prime}\\right)\\right]\n\\end{aligned}\n$$\n\nwhere $\\delta_{i}^{\\prime}=\\delta_{i+1}^{\\prime}=\\frac{\\delta_{i}+\\delta_{i+1}}{2}$. So replacing $\\delta_{i}, \\delta_{i+1}$ with $\\delta_{i}^{\\prime}, \\delta_{i+1}^{\\prime}$ increases $\\eta$.\n\nSo $\\eta$ is maximized when all $\\delta_{k}$ are equal, i.e., $\\delta_{k}^{*}=\\frac{\\pi}{10}$ for all $k$. Then\n\n$$\n\\eta^{*}=\\cos ^{10}\\left(\\frac{\\pi}{10}\\right) \\approx 0.6054\n$$']",['$\\cos ^{10}(\\frac{\\pi}{10})$'],False,,Numerical,1e-8 879,Optics,,"These days, there are so many stylish rectangular home-designs (see figure A). It is possible from the outline of those houses in their picture to estimate with good precision where the camera was. Consider an outline in one photograph of a rectangular house which has height $H=3$ meters (see figure B for square-grid coordinates). Assume that the camera size is negligible, how high above the ground (in meters) was the camera at the moment this picture was taken?![](https://cdn.mathpix.com/cropped/2023_12_21_6ad9a16bf503154c2401g-1.jpg?height=990&width=892&top_left_y=1260&top_left_x=622)","[""The formation of the house's image seen in the picture is due to pinhole principle, and note that the fish-eye effect here is weak (straight-lines stays straight). Define points $A, B, C, A^{\\prime}, B^{\\prime}, C^{\\prime}$ as in the attached Fig., since $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ stays parallel we know that the camera looked horizontally\n\n\n\nat the time this picture is taken.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_479ef03f149db178c6b9g-1.jpg?height=632&width=1480&top_left_y=313&top_left_x=317)\n\nTo determine the height of the camera at the very same moment, we need to know the where is the horizontal plane passing through the camera in the picture which is collapsed into a line. That can be found by finding the intersection $M$ of $A B \\cap A^{\\prime} B^{\\prime}$ and the intersection $N$ of $B C \\cap B^{\\prime} C^{\\prime}$, then $M N$ is the line of interests. $M N$ intersects $B B^{\\prime}$ at $P$, the position of $P$ can be calculated too be $(22,0.9)$, therefore the height of the camera is the length-ratio $P B^{\\prime} / B B^{\\prime}$ times $3 \\mathrm{~m}$, which equals to $0.9 \\mathrm{~m}$.""]",['$0.9$'],False,m,Numerical,1e-2 879,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","These days, there are so many stylish rectangular home-designs (see figure A). It is possible from the outline of those houses in their picture to estimate with good precision where the camera was. Consider an outline in one photograph of a rectangular house which has height $H=3$ meters (see figure B for square-grid coordinates). Assume that the camera size is negligible, how high above the ground (in meters) was the camera at the moment this picture was taken?","[""The formation of the house's image seen in the picture is due to pinhole principle, and note that the fish-eye effect here is weak (straight-lines stays straight). Define points $A, B, C, A^{\\prime}, B^{\\prime}, C^{\\prime}$ as in the attached Fig., since $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ stays parallel we know that the camera looked horizontally\n\n\n\nat the time this picture is taken.\n\n\n\nTo determine the height of the camera at the very same moment, we need to know the where is the horizontal plane passing through the camera in the picture which is collapsed into a line. That can be found by finding the intersection $M$ of $A B \\cap A^{\\prime} B^{\\prime}$ and the intersection $N$ of $B C \\cap B^{\\prime} C^{\\prime}$, then $M N$ is the line of interests. $M N$ intersects $B B^{\\prime}$ at $P$, the position of $P$ can be calculated too be $(22,0.9)$, therefore the height of the camera is the length-ratio $P B^{\\prime} / B B^{\\prime}$ times $3 \\mathrm{~m}$, which equals to $0.9 \\mathrm{~m}$.""]",['$0.9$'],False,m,Numerical,1e-2 880,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a thin rigid wire-frame MNPP'N'M' in which MNN'M' and NPP'N' are two squares of side $L$ with resistance per unit-length $\lambda$ and their planes are perpendicular. The frame is rotated with a constant angular velocity $\omega$ around an axis passing through $\mathrm{NN}$ ' and put in a region with constant magnetic field $B$ pointing perpendicular to $\mathrm{NN}^{\prime}$. What is the total heat released on the frame per revolution (in Joules)? Use $L=1 \mathrm{~m}, \lambda=1 \Omega / \mathrm{m}, \omega=2 \pi \mathrm{rad} / \mathrm{s}$ and $B=1 \mathrm{~T}$. ","[""In this setting, for every orientation during rotation the total magnetic flux passing though MNPP'N'M' is tthe same as through MPP'M', which has area $S=\\sqrt{2} L^{2}$.\nThe magnetic flux is:\n\n$$\n\\Phi(t)=B S \\sin (\\omega t)=\\sqrt{2} B L^{2} \\sin (\\omega t) .\n\\tag{3}\n$$\n\nThe emf running around the wire-frame is:\n\n$$\nE(t)=\\frac{d}{d t} \\Phi(t)=\\sqrt{2} B L^{2} \\omega \\cos (\\omega t)\n\\tag{4}\n$$\n\nThe electrical current running around the wire-frame is:\n\n$$\nI(t)=\\frac{E(t)}{6 \\lambda L}=\\frac{B L \\omega \\cos (\\omega t)}{3 \\sqrt{2} \\lambda}\n\\tag{5}\n$$\n\nThe heat released power is:\n\n$$\n\\frac{d}{d t} Q(t)=I^{2}(t) \\times 6 \\lambda L=\\frac{B^{2} L^{3} \\omega^{2} \\cos ^{2}(\\omega t)}{3 \\lambda}\n\\tag{6}\n$$\n\nThus the total heat released per revolution is:\n\n$$\nQ=\\int_{0}^{2 \\pi / \\omega} d t \\frac{d}{d t} Q(t)=\\frac{B^{2} L^{3} \\omega^{2} \\int_{0}^{2 \\pi / \\omega} d t \\cos ^{2}(\\omega t)}{3 \\lambda}=\\frac{\\pi B^{2} L^{3} \\omega}{3 \\lambda} \\approx 6.58 \\mathrm{~J}\n\\tag{7}\n$$""]",['$6.58$'],False,J,Numerical,2e-2 880,Electromagnetism,,"Consider a thin rigid wire-frame MNPP'N'M' in which MNN'M' and NPP'N' are two squares of side $L$ with resistance per unit-length $\lambda$ and their planes are perpendicular. The frame is rotated with a constant angular velocity $\omega$ around an axis passing through $\mathrm{NN}$ ' and put in a region with constant magnetic field $B$ pointing perpendicular to $\mathrm{NN}^{\prime}$. What is the total heat released on the frame per revolution (in Joules)? Use $L=1 \mathrm{~m}, \lambda=1 \Omega / \mathrm{m}, \omega=2 \pi \mathrm{rad} / \mathrm{s}$ and $B=1 \mathrm{~T}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_479ef03f149db178c6b9g-1.jpg?height=651&width=681&top_left_y=1710&top_left_x=711)","[""In this setting, for every orientation during rotation the total magnetic flux passing though MNPP'N'M' is tthe same as through MPP'M', which has area $S=\\sqrt{2} L^{2}$.\nThe magnetic flux is:\n\n$$\n\\Phi(t)=B S \\sin (\\omega t)=\\sqrt{2} B L^{2} \\sin (\\omega t) .\n\\tag{3}\n$$\n\nThe emf running around the wire-frame is:\n\n$$\nE(t)=\\frac{d}{d t} \\Phi(t)=\\sqrt{2} B L^{2} \\omega \\cos (\\omega t)\n\\tag{4}\n$$\n\nThe electrical current running around the wire-frame is:\n\n$$\nI(t)=\\frac{E(t)}{6 \\lambda L}=\\frac{B L \\omega \\cos (\\omega t)}{3 \\sqrt{2} \\lambda}\n\\tag{5}\n$$\n\nThe heat released power is:\n\n$$\n\\frac{d}{d t} Q(t)=I^{2}(t) \\times 6 \\lambda L=\\frac{B^{2} L^{3} \\omega^{2} \\cos ^{2}(\\omega t)}{3 \\lambda}\n\\tag{6}\n$$\n\nThus the total heat released per revolution is:\n\n$$\nQ=\\int_{0}^{2 \\pi / \\omega} d t \\frac{d}{d t} Q(t)=\\frac{B^{2} L^{3} \\omega^{2} \\int_{0}^{2 \\pi / \\omega} d t \\cos ^{2}(\\omega t)}{3 \\lambda}=\\frac{\\pi B^{2} L^{3} \\omega}{3 \\lambda} \\approx 6.58 \\mathrm{~J}\n\\tag{7}\n$$""]",['$6.58$'],False,J,Numerical,2e-2 881,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In this problem, we explore how fast an iceberg can melt, through the dominant mode of forced convection. For simplicity, consider a very thin iceberg in the form of a square with side lengths $L=100 \mathrm{~m}$ and a height of $1 \mathrm{~m}$, moving in the arctic ocean at a speed of $0.2 \mathrm{~m} / \mathrm{s}$ with one pair of edges parallel to the direction of motion (Other than the height, these numbers are typical of an average iceberg). The temperature of the surrounding water and air is $2^{\circ} \mathrm{C}$, and the temperature of the iceberg is $0^{\circ} \mathrm{C}$. The density of ice is $917 \mathrm{~kg} / \mathrm{m}^{3}$ and the latent heat of melting is $L_{w}=334 \times 10^{3} \mathrm{~J} / \mathrm{kg}$. The heat transfer rate $\dot{Q}$ between a surface and the surrounding fluid is dependent on the heat transfer coefficient $h$, the surface area in contact with the fluid $A$, and the temperature difference between the surface and the fluid $\Delta T$, via $\dot{Q}=h A \Delta T$. In heat transfer, three useful quantities are the Reynold's number, the Nusselt number, and the Prandtl number. Assume they are constant through and given by (assuming laminar flow): $$ \operatorname{Re}=\frac{\rho v_{\infty} L}{\mu}, \quad \mathrm{Nu}=\frac{h L}{k}, \quad \operatorname{Pr}=\frac{c_{p} \mu}{k} $$ where: - $\rho$ : density of the fluid - $v_{\infty}$ : speed of the fluid with respect to the object (at a very far distance) - $L$ : length of the object in the direction of motion - $\mu$ : dynamic viscosity of the fluid - $k$ : thermal conductivity of the fluid - $c_{p}$ : the specific heat capacity of the fluid Through experiments, the relationship between the three dimensionless numbers is, for a flat plate: $$ \mathrm{Nu}=0.664 \operatorname{Re}^{1 / 2} \operatorname{Pr}^{1 / 3} $$ Use the following values for calculations: | | Air | Water | | :--- | :--- | :--- | | $\rho\left(\mathrm{kg} / \mathrm{m}^{3}\right)$ | 1.29 | 1000 | | $\mu(\mathrm{kg} /(\mathrm{m} \cdot \mathrm{s}))$ | $1.729 \times 10^{-5}$ | $1.792 \times 10^{-3}$ | | $c_{p}(\mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}))$ | 1004 | 4220 | | $k(\mathrm{~W} /(\mathrm{m} \cdot \mathrm{K}))$ | 0.025 | 0.556 | The initial rate of heat transfer is $\dot{Q}$. Assuming this rate is constant (this is not true, but will allow us to obtain an estimate), how long (in days) would it take for the ice to melt completely? Assume convection is only happening on the top and bottom faces. Round to the nearest day.",['The heat transfer coefficient for water-ice and air-ice contact can be figured out with the relationship between the three dimensionless numbers\n$$\n\\mathrm{Nu}=0.664 \\operatorname{Re}^{1 / 2} \\operatorname{Pr}^{1 / 3} \\Longrightarrow h=0.664 \\frac{k}{L}\\left(\\frac{\\rho v_{\\infty} L}{\\mu}\\right)^{1 / 2}\\left(\\frac{c_{p} \\mu}{k}\\right)^{1 / 3}\n$$\n\nAs\n\n$$\n\\frac{\\mathrm{d} Q}{\\mathrm{~d} t}=h A \\Delta T\n$$\n\nwe then have\n\n$$\nt\\left(\\dot{Q}_{a}+\\dot{Q}_{w}\\right)=\\left(h_{w} A_{w}+h_{a} A_{a}\\right) \\Delta T \\Longrightarrow t=\\frac{\\rho L^{2} H L_{w}}{\\Delta T} \\frac{1}{h_{w} L^{2}+h_{a} L^{2}}=59.84 \\approx 60 \\text { days. }\n$$'],['60'],False,days,Numerical,1e0 882,Mechanics,,"A scale of uniform mass $M=3 \mathrm{~kg}$ of length $L=4 \mathrm{~m}$ is kept on a rough table (infinite friction) with $l=1 \mathrm{~m}$ hanging out of the table as shown in the figure below. A small ball of mass $m=1 \mathrm{~kg}$ is released from rest from a height of $h=5 \mathrm{~m}$ above the end of the scale. Find the maximum angle (in degrees) that the scale rotates by in the subsequent motion if ball sticks to the scale after collision. Take gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_4f8039b1096042486648g-1.jpg?height=523&width=612&top_left_y=186&top_left_x=754)",['The ball falls with velocity $\\sqrt{2 g h}=10$. Applying conservation of angular momentum about the end point of the table.\n$$\n\\begin{gathered}\nL_{i}=m v x \\\\\n=10 \\\\\nL_{f}=I_{1} \\omega \\\\\nI_{1}=4+3+1=8 \\\\\n\\Longrightarrow \\omega=1.25\n\\end{gathered}\n$$\n\nNow applying energy conservation\n\n$$\n\\begin{gathered}\nE_{i}=\\frac{1}{2} I \\omega^{2}=6.25 \\\\\nE_{f}=M_{\\text {total }} g x_{c o m} \\sin (\\theta)=20 \\sin \\theta \\\\\n\\theta=18.21^{\\circ}\n\\end{gathered}\n$$'],['$18.21$'],False,$^{\circ}$,Numerical,5e-1 882,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A scale of uniform mass $M=3 \mathrm{~kg}$ of length $L=4 \mathrm{~m}$ is kept on a rough table (infinite friction) with $l=1 \mathrm{~m}$ hanging out of the table as shown in the figure below. A small ball of mass $m=1 \mathrm{~kg}$ is released from rest from a height of $h=5 \mathrm{~m}$ above the end of the scale. Find the maximum angle (in degrees) that the scale rotates by in the subsequent motion if ball sticks to the scale after collision. Take gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$. ",['The ball falls with velocity $\\sqrt{2 g h}=10$. Applying conservation of angular momentum about the end point of the table.\n$$\n\\begin{gathered}\nL_{i}=m v x \\\\\n=10 \\\\\nL_{f}=I_{1} \\omega \\\\\nI_{1}=4+3+1=8 \\\\\n\\Longrightarrow \\omega=1.25\n\\end{gathered}\n$$\n\nNow applying energy conservation\n\n$$\n\\begin{gathered}\nE_{i}=\\frac{1}{2} I \\omega^{2}=6.25 \\\\\nE_{f}=M_{\\text {total }} g x_{c o m} \\sin (\\theta)=20 \\sin \\theta \\\\\n\\theta=18.21^{\\circ}\n\\end{gathered}\n$$'],['$18.21$'],False,$^{\circ}$,Numerical,5e-1 883,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In a galaxy far, far away, there is a planet of mass $M=6 \cdot 10^{27} \mathrm{~kg}$ which is a sphere of radius $R$ and charge $Q=10^{3} \mathrm{C}$ uniformly distributed. Aliens on this planet have devised a device for transportation, which is an insulating rectangular plate with mass $m=1 \mathrm{~kg}$ and charge $q=10^{4} \mathrm{C}$. This transportation device moves in a circular orbit at a distance $r=8 \cdot 10^{6} \mathrm{~m}$ from the center of the planet. The aliens have designated this precise elevation for the device, and do not want the device to deviate at all. In order to maintain its orbit, the device contains a relatively small energy supply. Find the power (in Watts) that the energy supply must release in order to sustain this orbit. The velocity of the device can be assumed to be much smaller than the speed of light, so that relativistic effects can be ignored. The device can also be assumed to be small in comparison to the size of the planet.","['The centripetal force is given by\n$$\n\\frac{m v^{2}}{r}=\\frac{G M m}{r^{2}}-\\frac{q Q}{4 \\pi \\epsilon_{0} r^{2}}\n$$\n\n\n\nwhich implies the centripetal acceleration is\n\n$$\na=\\frac{v^{2}}{r}=\\frac{G M}{r^{2}}-\\frac{q Q}{4 \\pi \\epsilon_{0} m r^{2}}\n$$\n\nNow, the device loses energy due to its acceleration, as given by the Larmor formula. The power needed to sustain motion is\n\n$$\nP=\\frac{q^{2} a^{2}}{6 \\pi \\epsilon_{0} c^{3}}=0.522 \\mathrm{~W}\n$$']",['$0.522$'],False,W,Numerical,5e-2 884,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A raindrop of mass $M=0.035 \mathrm{~g}$ is at height $H=2 \mathrm{~km}$ above a large lake. The raindrop then falls down (without initial velocity), mixing and coming to equilibrium with the lake. Assume that the raindrop, lake, air, and surrounding environment are at the same temperature $T=300 \mathrm{~K}$. Determine the magnitude of entropy change associated with this process (in $J / K$ ).","[""The total heat gain is equal to the change in potential energy of the raindrop, which spreads through out the whole environment at thermally equilibrium temperature $T$ (the environment is very large so any change in $T$ is negligible). The entropy gain $\\Delta S$ is thus generated by the dissipation of this potential energy $M g H$ to internal energy $\\Delta U$ in the environment (given that the specific volume of water doesn't change much, $\\Delta U \\approx M g H$ ). Hence the entropy change associated with this process can be estimated by:\n$$\nS=\\frac{\\Delta U}{T} \\approx \\frac{M g H}{T} \\approx 2.29 \\times 10^{-3} \\mathrm{~J} / \\mathrm{K}\n\\tag{8}\n$$""]",['$2.29 \\times 10^{-3}$'],False,J/K,Numerical,1e-4 885,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A rocket with mass of 563.17 (not including the mass of fuel) metric tons sits on the launchpad of the Kennedy Space Center (latitude $28^{\circ} 31^{\prime} 27^{\prime \prime} \mathrm{N}$, longitude $80^{\circ} 39^{\prime} 03^{\prime \prime} \mathrm{W}$ ), pointing directly upwards. Two solid fuel boosters, each with a mass of $68415 \mathrm{~kg}$ and providing $3421 \mathrm{kN}$ of thrust are pointed directly downwards. The rocket also has a liquid fuel engine, that can be throttled to produce different amounts of thrust and gimbaled to point in various directions. What is the minimum amount of thrust, in $\mathrm{kN}$, that this engine needs to provide for the rocket to lift vertically (to accelerate directly upwards) off the launchpad? Assume $G=6.674 \times 10^{-11} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{3}}$, and that the Earth is a perfect sphere of radius $6370 \mathrm{~km}$ and mass $5.972 \times 10^{24} \mathrm{~kg}$ that completes one revolution every $86164 \mathrm{~s}$ and that the rocket is negligibly small compared to the Earth. Ignore buoyancy forces.","[""Note the additional information provided in the problem (latitude, Earth radius, revolution period), which makes it clear that the effect of the rotation of the Earth must also be considered.\nWe first compute local gravitational acceleration:\n\n$$\ng=\\frac{G M}{R^{2}}=9.823 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}\n$$\n\n\n\nAnd also acceleration due to the Earth's rotation:\n\n$$\na=(R \\cos \\theta) \\omega^{2}=0.02976 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}\n$$\n\nThen vertical acceleration is:\n\n$$\ng-a \\cos \\theta=9.7965 \\frac{\\mathrm{m}}{\\mathrm{s}^{2}}\n$$\n\nAnd horizontal acceleration:\n\n$$\na \\sin \\theta=0.0142 \\frac{m}{s^{2}}\n$$\n\nThe total mass of the craft is 700 metric tons, so the needed forces for liftoff are:\n\n$$\n\\left(F_{x}, F_{y}\\right)=(6857527,9948) \\mathrm{N}\n$$\n\nSubtracting out the effect of the solid fuel boosters, the force the liquid fuel engine needs to provide is:\n\n$$\n\\left(F_{x}, F_{y}\\right)=(15527,9948) \\mathrm{N}\n$$""]",['$18.44$'],False,kN,Numerical,1e-1 886,Electromagnetism,,"The following information applies for the next two problems. A circuit has a power source of $\mathcal{E}=5.82 \mathrm{~V}$ connected to three elements in series: an inductor with $L=12.5 \mathrm{mH}$, a capacitor with $C=48.5 \mu \mathrm{F}$, and a diode with threshold voltage $V_{0}=0.65 \mathrm{~V}$. (Of course, the polarity of the diode is aligned with that of the power source.) You close the switch, and after some time, the voltage across the capacitor becomes constant. (Note: An ideal diode with threshold voltage $V_{0}$ is one whose IV characteristic is given by $I=0$ for $V0$.) ![](https://cdn.mathpix.com/cropped/2023_12_21_1c0bc651808d926dacd3g-1.jpg?height=347&width=618&top_left_y=1450&top_left_x=751) How much time (in seconds) has elapsed before the voltage across the capacitor becomes constant?","['When current is flowing clockwise, the circuit is equivalent to an LC circuit with a power source $\\mathcal{E}-V_{0}$. Thus, the voltage $U$ is sinusoidal about its equilibrium voltage $U_{0}=\\mathcal{E}-V_{0}$ with frequency $\\omega=1 / \\sqrt{L C}$.\n\nWhen the switch is closed, $I=0$ and $U=U_{\\min }=0$. Afterwards, $I$ increases to $I_{\\max }$ and decreases back to 0 , completing half a period of a sine wave. However, $I$ can not go negative due to the presence of the diode. Instead, a reverse voltage builds up on the diode (so that the voltage across the inductor becomes 0 ), and $I$ stays at 0 . At this point, $U$ becomes constant as well.\n\nThe time it took until the system became static was half a period of the LC circuit oscillation, i.e.,\n\n$$\n\\frac{\\pi}{\\omega}=\\pi \\sqrt{L C}=2.446 \\times 10^{-3} \\mathrm{~s}\n$$']",['$2.446 \\times 10^{-3}$'],False,s,Numerical,1e-4 886,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The following information applies for the next two problems. A circuit has a power source of $\mathcal{E}=5.82 \mathrm{~V}$ connected to three elements in series: an inductor with $L=12.5 \mathrm{mH}$, a capacitor with $C=48.5 \mu \mathrm{F}$, and a diode with threshold voltage $V_{0}=0.65 \mathrm{~V}$. (Of course, the polarity of the diode is aligned with that of the power source.) You close the switch, and after some time, the voltage across the capacitor becomes constant. (Note: An ideal diode with threshold voltage $V_{0}$ is one whose IV characteristic is given by $I=0$ for $V0$.) How much time (in seconds) has elapsed before the voltage across the capacitor becomes constant?","['When current is flowing clockwise, the circuit is equivalent to an LC circuit with a power source $\\mathcal{E}-V_{0}$. Thus, the voltage $U$ is sinusoidal about its equilibrium voltage $U_{0}=\\mathcal{E}-V_{0}$ with frequency $\\omega=1 / \\sqrt{L C}$.\n\nWhen the switch is closed, $I=0$ and $U=U_{\\min }=0$. Afterwards, $I$ increases to $I_{\\max }$ and decreases back to 0 , completing half a period of a sine wave. However, $I$ can not go negative due to the presence of the diode. Instead, a reverse voltage builds up on the diode (so that the voltage across the inductor becomes 0 ), and $I$ stays at 0 . At this point, $U$ becomes constant as well.\n\nThe time it took until the system became static was half a period of the LC circuit oscillation, i.e.,\n\n$$\n\\frac{\\pi}{\\omega}=\\pi \\sqrt{L C}=2.446 \\times 10^{-3} \\mathrm{~s}\n$$']",['$2.446 \\times 10^{-3}$'],False,s,Numerical,1e-4 887,Electromagnetism,,"At Hanoi-Amsterdam High School in Vietnam, every subject has its own flag (see Figure A, taken by Tung X. Tran). While the flags differ in color, they share the same central figure. Consider a planar conducting frame of that figure rotating at a constant angular velocity in a uniform magnetic field (see Figure B). The frame is made of thin rigid wires with same uniform curvature and same resistance per unit length. What fraction of the total heat released is released by the outermost wires?![](https://cdn.mathpix.com/cropped/2023_12_21_50c30c0db8a001b34ea6g-1.jpg?height=836&width=1002&top_left_y=1193&top_left_x=562)","['We call the current looping in the wires $I_{1}, I_{2}, I_{3}, I_{4}$ as shown in the attached Fig., and define the resistance of every quarter-circular section (radius $R$ ) of the wires to be $\\rho$, then considering the\n\n\n\nEMF on every loop gives:\n\n$$\n\\begin{aligned}\n& \\left(2 I_{1}+\\left(I_{1}-I_{2}\\right)-I_{4}\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times \\frac{\\pi+2}{2} R^{2} \\\\\n& \\left(2 I_{2}-I_{3}-\\left(I_{1}-I_{2}\\right)\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times \\frac{\\pi+2}{2} R^{2} \\\\\n& \\left(2 I_{4}+2\\left(I_{1}-I_{2}-I_{3}+I_{4}\\right)\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times 2 R^{2} \\\\\n& \\left(2 I_{3}-2\\left(I_{1}-I_{2}-I_{3}+I_{4}\\right)\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times 2 R^{2}\n\\end{aligned}\n\\tag{9}\n$$\n\nin which $B_{\\perp}$ is the perpendicular component of the magnetic field.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6bda60f521822c3b3f7eg-1.jpg?height=781&width=786&top_left_y=778&top_left_x=664)\n\nThe set of equations in Eq. (9) can be solved to get the relation between currents:\n\n$$\nI_{1}=I_{2}, I_{3}=I_{4}, \\frac{I_{1}}{I_{3}}=\\frac{\\pi+4}{4}\n\\tag{10}\n$$\n\nThe fraction of heat released on the outermost wires can be calculated:\n\n$$\n\\frac{Q_{\\text {outermost }}}{Q_{\\text {all }}}=\\frac{\\left(8 I_{1}^{2}+8 I_{2}^{2}\\right) \\rho}{\\left(8 I_{1}^{2}+8 I_{2}^{2}+4 I_{3}^{2}+4 I_{4}^{2}\\right) \\rho}=\\frac{(\\pi+4)^{2}}{(\\pi+4)^{2}+8} \\approx 0.864 \\mathrm{~J}\n\\tag{11}\n$$']",['$0.864$'],False,J,Numerical,5e-2 887,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","At Hanoi-Amsterdam High School in Vietnam, every subject has its own flag (see Figure A, taken by Tung X. Tran). While the flags differ in color, they share the same central figure. Consider a planar conducting frame of that figure rotating at a constant angular velocity in a uniform magnetic field (see Figure B). The frame is made of thin rigid wires with same uniform curvature and same resistance per unit length. What fraction of the total heat released is released by the outermost wires?","['We call the current looping in the wires $I_{1}, I_{2}, I_{3}, I_{4}$ as shown in the attached Fig., and define the resistance of every quarter-circular section (radius $R$ ) of the wires to be $\\rho$, then considering the\n\n\n\nEMF on every loop gives:\n\n$$\n\\begin{aligned}\n& \\left(2 I_{1}+\\left(I_{1}-I_{2}\\right)-I_{4}\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times \\frac{\\pi+2}{2} R^{2} \\\\\n& \\left(2 I_{2}-I_{3}-\\left(I_{1}-I_{2}\\right)\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times \\frac{\\pi+2}{2} R^{2} \\\\\n& \\left(2 I_{4}+2\\left(I_{1}-I_{2}-I_{3}+I_{4}\\right)\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times 2 R^{2} \\\\\n& \\left(2 I_{3}-2\\left(I_{1}-I_{2}-I_{3}+I_{4}\\right)\\right) \\rho=\\frac{d B_{\\perp}}{d t} \\times 2 R^{2}\n\\end{aligned}\n\\tag{9}\n$$\n\nin which $B_{\\perp}$ is the perpendicular component of the magnetic field.\n\n\n\nThe set of equations in Eq. (9) can be solved to get the relation between currents:\n\n$$\nI_{1}=I_{2}, I_{3}=I_{4}, \\frac{I_{1}}{I_{3}}=\\frac{\\pi+4}{4}\n\\tag{10}\n$$\n\nThe fraction of heat released on the outermost wires can be calculated:\n\n$$\n\\frac{Q_{\\text {outermost }}}{Q_{\\text {all }}}=\\frac{\\left(8 I_{1}^{2}+8 I_{2}^{2}\\right) \\rho}{\\left(8 I_{1}^{2}+8 I_{2}^{2}+4 I_{3}^{2}+4 I_{4}^{2}\\right) \\rho}=\\frac{(\\pi+4)^{2}}{(\\pi+4)^{2}+8} \\approx 0.864 \\mathrm{~J}\n\\tag{11}\n$$']",['$0.864$'],False,J,Numerical,5e-2 888,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A spacecraft is orbiting in a very low circular orbit at a velocity $v_{0}$ over the equator of a perfectly spherical moon with uniform density. Relative to a stationary frame, the spacecraft completes a revolution of the moon every 90 minutes, while the moon revolves in the same direction once every 24 hours. The pilot of the spacecraft would like to land on the moon using the following process:So the final answer is 1. Start by firing the engine directly against the direction of motion.So the final answer is 2. Orient the engine over time such that the vertical velocity of the craft remains 0 , while the horizontal speed continues to decrease. 3. Once the velocity of the craft relative to the ground is also 0 , turn off the engine. Assume that the engine of the craft can be oriented instantly in any direction, and the craft has a TWR (thrust-to-weight ratio, where weight refers to the weight at the moon's surface) of 2, which remains constant throughout the burn. If the craft starts at $v_{0}=500 \mathrm{~m} / \mathrm{s}$, compute the delta-v expended to land, minus the initial velocity, i.e. $\Delta v-v_{0}$.","[""The trick in this question is to work in dimensionless units. Let $v$ be the ratio of the craft's velocity to orbital velocity. Then, if the craft has horizontal velocity $v$, the acceleration downwards is the following:\n\n$$\na_{V}=\\frac{v_{0}^{2}}{r}-\\frac{\\left(v_{0} v\\right)^{2}}{r}=\\frac{v_{0}^{2}}{r}\\left(1-v^{2}\\right)=g_{m}\\left(1-v^{2}\\right)\n$$\n\nAs the TWR is 2 , the total acceleration the engine provides is $a_{0}=2 g_{m}$, where $g_{m}$ is the surface gravity of the moon. As this total acceleration is the sum of horizontal and vertical components, and the vertical component cancels out the downwards acceleration:\n\n$$\na_{H}=\\sqrt{\\left(2 g_{m}\\right)^{2}-\\left(g_{m}\\left(1-v^{2}\\right)\\right)^{2}}=g_{m} \\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}\n$$\n\nAnd $a_{H}$ is related to $v$ (which is dimensionless!) by the following relation:\n\n$$\n\\begin{gathered}\n\\frac{\\mathrm{d}\\left(v_{0} v\\right)}{\\mathrm{d} t}=-a_{H} \\\\\n\\frac{\\mathrm{d} v}{\\mathrm{~d} t}=-\\frac{g_{m}}{v_{0}} \\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}} \\\\\n\\frac{\\mathrm{d} v}{\\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}}=-\\frac{g_{m}}{v_{0}} \\mathrm{~d} t\n\\end{gathered}\n$$\n\nAt the start, $v=1$. However, at landing, velocity of the craft is 0 relative to the surface, not a stationary frame! Therefore, we use the orbital periods data to determine the final $v$ to be $\\frac{1.5}{24}=\\frac{1}{16}$. Then integrating:\n\n$$\n\\int_{v=1}^{v=1 / 16} \\frac{\\mathrm{d} v}{\\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}}=-\\frac{g_{m}}{v_{0}} t\n$$\n\nAnd as delta-v is related to time by $\\Delta v=a_{0} t$ :\n\n$$\n\\Delta v=a_{0} t=a_{0} \\frac{v_{0}}{g_{m}} \\int_{1 / 16}^{1} \\frac{\\mathrm{d} v}{\\sqrt{2^{2}-\\left(1-v^{2}\\right)^{2}}}=a_{0} \\frac{v_{0}}{a_{0}} 2(0.503)=500 \\cdot 1.006=503.06 \\frac{\\mathrm{m}}{\\mathrm{s}}\n$$\n\nAnd subtracting:\n\n$$\n\\Delta v-v_{0}=3.06 \\frac{\\mathrm{m}}{\\mathrm{s}}\n$$""]",['$3.06$'],False,m/s,Numerical,1e-1 889,Mechanics,,"A tesseract is a 4 dimensional example of cube. It can be drawn in 3 dimensions by drawing two cubes and connecting their vertices together as shown in the picture below: ![](https://cdn.mathpix.com/cropped/2023_12_21_4d38ea6c92285d4052d0g-1.jpg?height=643&width=659&top_left_y=188&top_left_x=733) Now for the 3D equivalent. The lines connecting the vertices are replaced with ideal springs of constant $k=10 \mathrm{~N} / \mathrm{m}$ (in blue in the figure). Now, suppose the setup is placed in zero-gravity and the outer cube is fixed in place with a sidelength of $b=2 \mathrm{~m}$. The geometric center of the inner cube is placed in the geometric center of the outer cube, and the inner cube has a side-length $a=1 \mathrm{~m}$ and mass $m=1.5 \mathrm{~kg}$. The inner cube is slightly displaced from equilibrium. Consider the period of oscillations - $T_{1}$ : when the springs have a relaxed length of 0 ; - $T_{2}$ : when the springs are initially relaxed before the inner cube is displaced. What is $T_{1}+T_{2}$ ?","['First let us prove that there is a net external torque of $\\vec{\\tau}=0$ on the cube for small displacements which means the inner cube behaves like a point mass. Consider a simple case when the cube is pushed to one side.\n![](https://cdn.mathpix.com/cropped/2023_12_21_4d38ea6c92285d4052d0g-1.jpg?height=612&width=615&top_left_y=1724&top_left_x=755)\n\nIf we label the vertices of the cube from 1 to 4 clockwise, where 1 is the top left side, it is apparent that sides 1 and 2 provide a positive torque while sides 3 and 4 provide a negative torque. As the displacement is small, the angles created are small enough such that $\\sin \\theta \\approx \\theta$. As force is\n\n\n\nproportional to the extension of the spring as $F \\propto x$, we can write that\n\n$$\n\\tau \\propto \\theta\\left(r_{+}+r_{-}-r_{+}-r_{-}\\right) \\propto 0\n$$\n\nIf torque is zero when the cube is displaced in the $x$-direction, then by symmetry, the torque is zero when the cube is displaced in the $y$-direction. Superposing both solutions implies that torque as a function of displacements in the $x$ and $y$ directions $\\alpha \\hat{x}+\\beta \\hat{y}$ is\n\n$$\n\\tau(\\alpha x+\\beta y)=\\tau(\\alpha x)+\\tau(\\beta y)=\\alpha \\tau(x)+\\beta \\tau(y)=0\n$$\n\n1.Label the vertices of the outer cube as $1,2, \\ldots, 8$ and the vectors that point to these vertices from the inner cube as $\\vec{r}_{1}, \\vec{r}_{2}, \\ldots, \\vec{r}_{2}$. Consider when the inner cube deviates from equilibrium with a vector $\\vec{r}$. The force as a function of $\\vec{r}$ is\n$$\n\\begin{aligned}\nF(\\vec{r}) & =k\\left[\\left(\\vec{r}_{1}-\\vec{r}\\right)+\\left(\\vec{r}_{2}-\\vec{r}\\right)+\\ldots\\left(\\vec{r}_{8}-\\vec{r}\\right)\\right] \\\\\n& =k\\left(\\sum_{i=1}^{8} \\vec{r}_{i}-8 \\vec{r}\\right) \\\\\n& =-8 k \\vec{r}\n\\end{aligned}\n$$\n\nThis implies the period of oscillations is\n\n$$\nT_{1}=2 \\pi \\sqrt{\\frac{m}{8 k}}\n$$\n\n2.Let the center of the inner cube be $(0,0,0)$. Consider the coordinates $(a / 2, a / 2, a / 2)$ and $(b / 2, b / 2, b / 2)$ which correspond to the vertex of the inner and larger cube respectively. Consider moving the cube in the $x$-direction. From defining $y=b / 2-a / 2$, the compressional/extension of each spring $\\pm \\Delta \\ell$ is then\n$$\n\\begin{aligned}\n\\Delta \\ell & = \\pm \\sqrt{(x+y)^{2}+2 y^{2}}-\\sqrt{3} y \\\\\n& = \\pm \\sqrt{3} y \\sqrt{1+\\frac{2 x}{3 y}+\\mathcal{O}\\left(x^{2}\\right)}-\\sqrt{3} y \\\\\n& \\approx \\pm \\sqrt{3} y \\frac{x}{3 y} \\\\\n& = \\pm \\frac{x}{\\sqrt{3}}\n\\end{aligned}\n$$\n\nThe total energy in all springs together are then\n\n$$\nE=8 \\times \\frac{1}{2} k\\left(\\frac{x}{\\sqrt{3}}\\right)^{2} \\Longrightarrow F=-\\frac{8 k}{3} x \\Longrightarrow T_{2}=2 \\pi \\sqrt{\\frac{3 m}{8 k}}\n$$\n\nHence, our total answer is\n\n$$\nT_{1}+T_{2}=2 \\pi(1+\\sqrt{3}) \\sqrt{\\frac{m}{8 k}}\n$$']",['2.35'],False,s,Numerical,5e-2 889,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A tesseract is a 4 dimensional example of cube. It can be drawn in 3 dimensions by drawing two cubes and connecting their vertices together as shown in the picture below: Now for the 3D equivalent. The lines connecting the vertices are replaced with ideal springs of constant $k=10 \mathrm{~N} / \mathrm{m}$ (in blue in the figure). Now, suppose the setup is placed in zero-gravity and the outer cube is fixed in place with a sidelength of $b=2 \mathrm{~m}$. The geometric center of the inner cube is placed in the geometric center of the outer cube, and the inner cube has a side-length $a=1 \mathrm{~m}$ and mass $m=1.5 \mathrm{~kg}$. The inner cube is slightly displaced from equilibrium. Consider the period of oscillations - $T_{1}$ : when the springs have a relaxed length of 0 ; - $T_{2}$ : when the springs are initially relaxed before the inner cube is displaced. What is $T_{1}+T_{2}$ ?","['First let us prove that there is a net external torque of $\\vec{\\tau}=0$ on the cube for small displacements which means the inner cube behaves like a point mass. Consider a simple case when the cube is pushed to one side.\n\n\nIf we label the vertices of the cube from 1 to 4 clockwise, where 1 is the top left side, it is apparent that sides 1 and 2 provide a positive torque while sides 3 and 4 provide a negative torque. As the displacement is small, the angles created are small enough such that $\\sin \\theta \\approx \\theta$. As force is\n\n\n\nproportional to the extension of the spring as $F \\propto x$, we can write that\n\n$$\n\\tau \\propto \\theta\\left(r_{+}+r_{-}-r_{+}-r_{-}\\right) \\propto 0\n$$\n\nIf torque is zero when the cube is displaced in the $x$-direction, then by symmetry, the torque is zero when the cube is displaced in the $y$-direction. Superposing both solutions implies that torque as a function of displacements in the $x$ and $y$ directions $\\alpha \\hat{x}+\\beta \\hat{y}$ is\n\n$$\n\\tau(\\alpha x+\\beta y)=\\tau(\\alpha x)+\\tau(\\beta y)=\\alpha \\tau(x)+\\beta \\tau(y)=0\n$$\n\n1.Label the vertices of the outer cube as $1,2, \\ldots, 8$ and the vectors that point to these vertices from the inner cube as $\\vec{r}_{1}, \\vec{r}_{2}, \\ldots, \\vec{r}_{2}$. Consider when the inner cube deviates from equilibrium with a vector $\\vec{r}$. The force as a function of $\\vec{r}$ is\n$$\n\\begin{aligned}\nF(\\vec{r}) & =k\\left[\\left(\\vec{r}_{1}-\\vec{r}\\right)+\\left(\\vec{r}_{2}-\\vec{r}\\right)+\\ldots\\left(\\vec{r}_{8}-\\vec{r}\\right)\\right] \\\\\n& =k\\left(\\sum_{i=1}^{8} \\vec{r}_{i}-8 \\vec{r}\\right) \\\\\n& =-8 k \\vec{r}\n\\end{aligned}\n$$\n\nThis implies the period of oscillations is\n\n$$\nT_{1}=2 \\pi \\sqrt{\\frac{m}{8 k}}\n$$\n\n2.Let the center of the inner cube be $(0,0,0)$. Consider the coordinates $(a / 2, a / 2, a / 2)$ and $(b / 2, b / 2, b / 2)$ which correspond to the vertex of the inner and larger cube respectively. Consider moving the cube in the $x$-direction. From defining $y=b / 2-a / 2$, the compressional/extension of each spring $\\pm \\Delta \\ell$ is then\n$$\n\\begin{aligned}\n\\Delta \\ell & = \\pm \\sqrt{(x+y)^{2}+2 y^{2}}-\\sqrt{3} y \\\\\n& = \\pm \\sqrt{3} y \\sqrt{1+\\frac{2 x}{3 y}+\\mathcal{O}\\left(x^{2}\\right)}-\\sqrt{3} y \\\\\n& \\approx \\pm \\sqrt{3} y \\frac{x}{3 y} \\\\\n& = \\pm \\frac{x}{\\sqrt{3}}\n\\end{aligned}\n$$\n\nThe total energy in all springs together are then\n\n$$\nE=8 \\times \\frac{1}{2} k\\left(\\frac{x}{\\sqrt{3}}\\right)^{2} \\Longrightarrow F=-\\frac{8 k}{3} x \\Longrightarrow T_{2}=2 \\pi \\sqrt{\\frac{3 m}{8 k}}\n$$\n\nHence, our total answer is\n\n$$\nT_{1}+T_{2}=2 \\pi(1+\\sqrt{3}) \\sqrt{\\frac{m}{8 k}}\n$$']",['2.35'],False,s,Numerical,5e-2 890,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider two points $S$ and $S^{\prime}$ randomly placed inside a $D$-dimensional hyper-rectangular room with walls that are perfect-reflecting $(D-1)$-dimensional hyper-plane mirrors. How many different light-rays that start from $S$, reflect $N$ times on one of the walls and $N-1$ times on each of the rest, then go to $S^{\prime}$ ? Use $D=7$ and $N=3$.","[""Using the hyper-rectangular room as the fundamental unit-cell of an infinite hyper-grid in space, then find all possible positions of $S^{\\prime}$-images through reflections: we can realize that there is only one position of $S^{\\prime}$-image inside every unit-cell.\n\n(A)\n\n\n\nConsider two opposite hyper-plane mirrors, since the rest of the mirrors are perpendicular to them, the numbers of reflections on them for any light-path traveled from point $S$ and point $S^{\\prime}$ can only different by 1 or less. If the numbers are both equal and non-zero, then the available positions $S^{\\prime}$ image the light-path from $S$ should reach are two unit-cell hyper-rows that parallel to the mirrors. If the numbers are not equal, then the available positions $S^{\\prime}$-image the light-path from $S$ should reach are one unit-cell hyper-rows that that parallel to the mirrors.\n\nSay, without loss of generality, pick one mirror to be reflected $N$ times and the rest to be reflected $N-1$ times each, then the number of light-rays for that pick should equal to the number of unit-cell intersections between all relevant unit-cell hyper-rows, which is half the total number of vertices a hyper-rectangular has, thus $2^{D-1}$. There are $2 D$ walls, thus the total number of light-rays that satisfies the task given is $2 D \\times 2^{D-1}=D 2^{D}$, independent of $N$ for all values $N>1$.\n\n\n\nTo illustrate the above explanation, let's take a look at the simple case of $D=2$ and $N=2>1$. Consider a rectangular room, with random points $S, S^{\\prime}$ and $2 D=4$ walls $W_{1}, W_{2}, W_{3}, W_{4}$ (see Fig. A). Without loss of generality, we want to find light-rays that go from $S$, reflect $N=2$ times on $W_{1}$ and $N-1=1$ times on $W_{2}, W_{3}, W_{4}$ then come to $S^{\\prime}$. Each image of $S^{\\prime}$ is an unique point in a unit-cell generated by the room (see Fig. B). Note that every light-ray from $S$ that reach $S^{\\prime}$-images in the $\\rightarrow W_{2} \\rightarrow W_{4} \\rightarrow$ and $\\rightarrow W_{4} \\rightarrow W_{2} \\rightarrow$ unit-cell green-rows will satisfy the requirement of one reflection on each of $W_{2}$ and $W_{4}$, every light-ray from $S$ that reach $S^{\\prime}$-images in the $\\rightarrow W_{1} \\rightarrow W_{3} \\rightarrow W_{1} \\rightarrow$ unit-cell blue-rows will satisfy the requirement of two reflection on $W_{1}$ and one reflection on $W_{3}$ (see Fig. C). The intersection of these rows are two unit-cells, corresponds two possible images thuss two possible light-rays that satisfies the requirement (see Fig. D).\n\nFor $D=7$ and $N=3>1$, we get 896 light-rays.""]",['895'],False,,Numerical,0 891,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",Two concentric isolated rings of radius $a=1 \mathrm{~m}$ and $b=2 \mathrm{~m}$ of mass $m_{a}=1 \mathrm{~kg}$ and $m_{b}=2 \mathrm{~kg}$ are kept in a gravity free region. A soap film of surface tension $\sigma=0.05 \mathrm{Nm}^{-1}$ with negligible mass is spread over the rings such that it occupies the region between the rings. The smaller ring is pulled slightly along the axis of the rings. Find the time period of small oscillation in seconds.,"['The force on the two rings when they are a distance $L$ apart follows as\n$$\nF=4 \\pi r \\sigma \\sin \\theta\n$$\n\nIn small displacements, the change in $\\theta$ is small. Therefore,\n\n$$\n\\begin{aligned}\nF & =4 \\pi r \\sigma \\theta \\\\\n\\frac{F}{4 \\pi r \\sigma} & =\\frac{\\mathrm{d} y}{\\mathrm{~d} r} \\\\\n\\int_{a}^{b} \\frac{F}{4 \\pi \\sigma r} \\mathrm{~d} r & =\\int_{0}^{L} \\mathrm{~d} y \\\\\n\\frac{F}{4 \\pi \\sigma} \\ln (b / a) & =L\n\\end{aligned}\n$$\n\nNow let $a_{1}$ and $a_{2}$ be acceleration of a and $\\mathrm{b}$ respectively. We have that\n\n$$\n\na_{\\text {net }}=a_{1}+a_{2}\n\\tag{12}\n$$\n$$\na_{\\text {net }}=F\\left(\\frac{m_{1}+m_{2}}{m_{1} m_{2}}\\right)\n\\tag{13}\n$$\n$$\nL \\omega^{2}=\\frac{4 \\pi \\sigma}{\\ln (b / a)} L\\left(\\frac{m_{1}+m_{2}}{m_{1} m_{2}}\\right)\n\\tag{14}\n$$\n$$\nT=2 \\pi \\sqrt{\\frac{\\ln (b / a) m_{1} m_{2}}{4 \\pi \\sigma\\left(m_{1}+m_{2}\\right)}}\n\\tag{15}\n$$\n$$\nT=2 \\pi \\sqrt{\\frac{10 \\ln (2)}{3 \\pi}}=5.388 \\mathrm{~s}\n\\tag{16}\n$$']",['$2 \\pi \\sqrt{\\frac{10 \\ln (2)}{3 \\pi}}$'],False,s,Numerical,1e-2 892,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","An open electrical circuit contains a wire loop in the shape of a semi-circle, that contains a resistor of resistance $R=0.2 \Omega$. The circuit is completed by a conducting pendulum in the form of a uniform rod with length $\ell=0.1 \mathrm{~m}$ and mass $m=0.05 \mathrm{~kg}$, has no resistance, and stays in contact with the other wires at all times. All electrical components are oriented in the $y z$ plane, and gravity acts in the $z$ direction. A constant magnetic field of strength $B=2 \mathrm{~T}$ is applied in the $+x$ direction. Ignoring self inductance and assuming that $\alpha \ll 1$, the general equation of motion is in the form of $\theta(t)=A(t) \cos (\omega t+\varphi)$, where $A(t) \geq 0$. Find $\omega^{2}$.","[""The area enclosed by the wire loop is\n$$\nA=\\frac{1}{2} \\ell^{2} \\alpha+A_{0}\n$$\n\nfor small angles $\\alpha$, and $A_{0}$ is a constant number (which gets ignored since we really care about how this angle is changing). The flux is $\\Phi=B A$ and from Lenz's Law, we have,\n\n$$\n\\varepsilon=-\\frac{d \\Phi}{d t}=-\\frac{1}{2} B \\ell^{2} \\dot{\\alpha}\n$$\n\nOne can verify that if $\\alpha$ is increasing, the current will flow in the clockwise direction, so we set the counterclockwise direction as positive. The current through the wire is thus,\n\n$$\ni=\\frac{\\varepsilon}{R}=-\\frac{B \\ell^{2}}{2 R} \\dot{\\alpha}\n$$\n\nThe magnetic force acting on it is $F_{B}=i B \\ell$ and the resulting torque is\n\n$$\n\\tau_{B}=F_{B} \\frac{\\ell}{2}=-\\frac{B^{2} \\ell^{4}}{4 R} \\dot{\\alpha}\n$$\n\nPlease verify that the sign is correct. The gravitational torque is $\\tau_{g}=-m g \\frac{\\ell}{2} \\alpha$, so the torque equation gives us\n\n$$\n\\begin{aligned}\n0 & =\\frac{1}{3} m \\ell^{2} \\ddot{\\alpha}+\\frac{B^{2} \\ell^{4}}{4 R} \\dot{\\alpha}+m g \\frac{\\ell}{2} \\alpha \\\\\n0 & =\\ddot{\\alpha}+\\frac{3}{4} \\frac{B^{2} \\ell^{2}}{m R} \\dot{\\alpha}+\\frac{3}{2} \\frac{g}{\\ell} \\alpha\n\\end{aligned}\n$$\n\nRecall that for a damped harmonic oscillator in the form of $\\ddot{\\alpha}+\\gamma \\dot{\\alpha}+\\omega_{0}^{2} \\alpha=0$, the frequency of oscillations is $\\omega^{2}=\\omega_{0}^{2}-\\gamma^{2} / 4$, so in our case, we have\n\n$$\n\\omega^{2}=\\frac{3}{2} \\frac{g}{\\ell}-\\frac{9}{64}\\left(\\frac{B^{2} \\ell^{2}}{m R} \\dot{\\alpha}\\right)^{2}=145 \\mathrm{~s}^{-1}\n$$""]",['$145$'],False,$\mathrm{~s}^{-1}$,Numerical,0 892,Electromagnetism,,"An open electrical circuit contains a wire loop in the shape of a semi-circle, that contains a resistor of resistance $R=0.2 \Omega$. The circuit is completed by a conducting pendulum in the form of a uniform rod with length $\ell=0.1 \mathrm{~m}$ and mass $m=0.05 \mathrm{~kg}$, has no resistance, and stays in contact with the other wires at all times. All electrical components are oriented in the $y z$ plane, and gravity acts in the $z$ direction. A constant magnetic field of strength $B=2 \mathrm{~T}$ is applied in the $+x$ direction. ![](https://cdn.mathpix.com/cropped/2023_12_21_311e0a88c943ba95b98cg-1.jpg?height=463&width=1005&top_left_y=495&top_left_x=560) Ignoring self inductance and assuming that $\alpha \ll 1$, the general equation of motion is in the form of $\theta(t)=A(t) \cos (\omega t+\varphi)$, where $A(t) \geq 0$. Find $\omega^{2}$.","[""The area enclosed by the wire loop is\n$$\nA=\\frac{1}{2} \\ell^{2} \\alpha+A_{0}\n$$\n\nfor small angles $\\alpha$, and $A_{0}$ is a constant number (which gets ignored since we really care about how this angle is changing). The flux is $\\Phi=B A$ and from Lenz's Law, we have,\n\n$$\n\\varepsilon=-\\frac{d \\Phi}{d t}=-\\frac{1}{2} B \\ell^{2} \\dot{\\alpha}\n$$\n\nOne can verify that if $\\alpha$ is increasing, the current will flow in the clockwise direction, so we set the counterclockwise direction as positive. The current through the wire is thus,\n\n$$\ni=\\frac{\\varepsilon}{R}=-\\frac{B \\ell^{2}}{2 R} \\dot{\\alpha}\n$$\n\nThe magnetic force acting on it is $F_{B}=i B \\ell$ and the resulting torque is\n\n$$\n\\tau_{B}=F_{B} \\frac{\\ell}{2}=-\\frac{B^{2} \\ell^{4}}{4 R} \\dot{\\alpha}\n$$\n\nPlease verify that the sign is correct. The gravitational torque is $\\tau_{g}=-m g \\frac{\\ell}{2} \\alpha$, so the torque equation gives us\n\n$$\n\\begin{aligned}\n0 & =\\frac{1}{3} m \\ell^{2} \\ddot{\\alpha}+\\frac{B^{2} \\ell^{4}}{4 R} \\dot{\\alpha}+m g \\frac{\\ell}{2} \\alpha \\\\\n0 & =\\ddot{\\alpha}+\\frac{3}{4} \\frac{B^{2} \\ell^{2}}{m R} \\dot{\\alpha}+\\frac{3}{2} \\frac{g}{\\ell} \\alpha\n\\end{aligned}\n$$\n\nRecall that for a damped harmonic oscillator in the form of $\\ddot{\\alpha}+\\gamma \\dot{\\alpha}+\\omega_{0}^{2} \\alpha=0$, the frequency of oscillations is $\\omega^{2}=\\omega_{0}^{2}-\\gamma^{2} / 4$, so in our case, we have\n\n$$\n\\omega^{2}=\\frac{3}{2} \\frac{g}{\\ell}-\\frac{9}{64}\\left(\\frac{B^{2} \\ell^{2}}{m R} \\dot{\\alpha}\\right)^{2}=145 \\mathrm{~s}^{-1}\n$$""]",['$145$'],False,$\mathrm{~s}^{-1}$,Numerical,0 893,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A table of unknown material has a mass $M=100 \mathrm{~kg}$, width $w=4 \mathrm{~m}$, length $\ell=3 \mathrm{~m}$, and 4 legs of length $L=0.5 \mathrm{~m}$ with a Young's modulus of $Y=1.02 \mathrm{MPa}$ at each of the corners. The cross-sectional area of a table leg is approximately $A=1 \mathrm{~cm}^{2}$. The surface of the table has a coefficient of friction of $\mu=0.1$. A point body with the same mass as the table is put at some position from the geometric center of the table. What is the minimum distance the body must be placed from the center such that it slips on the table surface immediately after? Report your answer in centimeters. The table surface and floor are non-deformable.","['This problem requires some 3 dimensional reasoning. Suppose $\\mathbf{s}=\\left(s_{x}, s_{y}\\right)$ is the gradient of the table. We can use this to calculate the additional torque from the displacement of the mass. The forces from each table leg are\n$$\nF_{i}=\\frac{Y A}{L}\\left( \\pm s_{x} \\frac{\\ell}{2} \\pm s_{y} \\frac{w}{2}\\right)\n$$\n\nTaking the cross product as $\\boldsymbol{\\tau}=\\mathbf{r} \\times \\mathbf{F}_{\\mathbf{i}}$ shows that torque is given as\n\n$$\n\\boldsymbol{\\tau}=\\frac{Y A}{L}\\left(\\begin{array}{c}\n-s_{y} w^{2} \\\\\ns_{x} \\ell^{2}\n\\end{array}\\right)\n$$\n\nwhich must balance out the torque $M g d$ from a point mass. Hence, rewriting yields\n\n$$\nd=\\frac{Y A}{M g L} \\sqrt{s_{y}^{2} w^{4}+s_{x}^{2} \\ell^{4}}\n$$\n\nFurthermore, note that the angle required from slipping is given from a force analysis as\n\n$$\nm g \\sin \\theta=\\mu m g \\cos \\theta \\Longrightarrow \\mu=\\tan \\theta=|\\mathbf{s}|=\\sqrt{s_{x}^{2}+s_{y}^{2}}\n$$\n\nWhen $w>\\ell$, we can rewrite\n\n$$\ns_{y}^{2} w^{4}+s_{x}^{2} \\ell^{4}=s_{y}^{2}\\left(w^{4}-\\ell^{4}\\right)+\\ell^{4} \\mu^{2}\n$$\n\nwhich is minimized to $\\ell^{4} \\mu^{2}$ when $s_{y}=0$. Hence, we obtain\n\n$$\nd=\\frac{\\mu \\ell^{2} Y A}{M g L}\n$$']",['$18.71$'],False,m,Numerical,1e-1 894,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Dipole Conductor An (ideal) electric dipole of magnitude $p=1 \times 10^{-6} \mathrm{C} \cdot \mathrm{m}$ is placed at a distance $a=0.05 \mathrm{~m}$ away from the center of an uncharged, isolated spherical conductor of radius $R=0.02 \mathrm{~m}$. Suppose the angle formed by the dipole vector and the radial vector (the vector pointing from the sphere's center to the dipole's position) is $\theta=20^{\circ}$. Find the (electrostatic) interaction energy between the dipole and the charge induced on the spherical conductor.","[""We can use the fact that if a charge $Q$ is placed at a distance $a$ from a grounded, conducting sphere of radius $R$, as far as the field outside the sphere is concerned, there is an image charge of magnitude $-Q \\frac{R}{a}$ at a position $\\frac{R^{2}}{a}$ from the origin, on the line segment connecting the origin and charge $Q$. It is straightforward to check that indeed, the potential on the sphere due to this image charge prescription is 0 . If the point charge is instead a dipole $p$, we can think of this as a superposition of 2 point charges, and use the fact above. In particular, there is one charge $-Q$ at point $(a, 0,0)$ and another charge $Q$ at point $(a+s \\cos \\theta, s \\sin \\theta, 0)$, where $s$ is small and $Q s=p$. Note that the dipole points in the direction $\\theta$ above the $\\mathrm{x}$-axis. Consequently, there will be an image charge at $\\left(\\frac{R^{2}}{a}, 0,0\\right)$ with magnitude $Q \\frac{R}{a}$ and an image charge at $\\left(\\frac{R^{2}}{a+s \\cos \\theta}, \\frac{R^{2} s \\sin \\theta}{a(a+s \\cos \\theta)}, 0\\right)$ with magnitude $-Q \\frac{R}{a+s \\cos \\theta}$. The image charges are close to each other but do not cancel out exactly,\n\n\n\nso they can be represented as a superposition of an image point charge $Q^{\\prime}$ and an image dipole $p^{\\prime}$. The image point charge has magnitude $Q^{\\prime}=-Q R\\left(\\frac{1}{a+s \\cos \\theta}-\\frac{1}{a}\\right)=\\frac{Q R s \\cos \\theta}{a^{2}}$. The image dipole has magnitude $p^{\\prime}=Q \\frac{R}{a} * \\frac{R^{2} s}{a^{2}}=\\frac{Q R^{3} s}{a^{3}}$ and points towards the direction $\\theta$ below the positive x-axis. Finally, since the sphere in the problem is uncharged instead of grounded, to ensure the net charge in the sphere is 0 , we place another image charge $-Q^{\\prime}$ at the origin.\n\nNow we can calculate the desired interaction energy, which is simply the interaction energy between the image charges and the real dipole. Using the dipole-dipole interaction formula, the interaction between the image dipole and the real dipole is given by:\n\n$$\nU_{1}=\\frac{k p p^{\\prime}}{\\left(a-\\frac{R^{2}}{a}\\right)^{3}}\\left(\\cos (2 \\theta)-3 \\cos ^{2} \\theta\\right)\n$$\n\nThe interaction between the image charge at the image dipole's position and the real dipole is given by:\n\n$$\nU_{2}=-\\frac{k p Q^{\\prime} \\cos \\theta}{\\left(a-\\frac{R^{2}}{a}\\right)^{2}}\n$$\n\nThe interaction between the image charge at the center and the real dipole is given by:\n\n$$\nU_{3}=\\frac{k p Q^{\\prime} \\cos \\theta}{a^{2}}\n$$\n\nThe final answer is $U=U_{1}+U_{2}+U_{3}=\\frac{p^{2} R}{4 \\pi \\epsilon_{0}}\\left(\\frac{\\cos ^{2} \\theta}{a^{4}}-\\frac{a^{2} \\cos ^{2} \\theta+R^{2}}{\\left(a^{2}-R^{2}\\right)^{3}}\\right)=-25.22 \\mathrm{~J} .$""]",['$-25.22$'],False,J,Numerical,1e-1 895,Optics,,"Consider an optical system made of many identical ideal (negligible-thickness) halflenses with focal length $f>0$, organized so that they share the same center and are angular-separated equally at density $n$ (number of lenses per unit-radian). Define the length-scale $\lambda=f / n$. A light-ray arrives perpendicular to the first lens at distance $\lambda / 2$ away from the center, then leaves from the last lens at distance $2 \lambda$ away from the center. Estimate the total deflection angle (in rad) of the light-ray by this system in the limit $n \rightarrow \infty$. ![](https://cdn.mathpix.com/cropped/2023_12_21_31037f3a3fcf7f3d1745g-1.jpg?height=556&width=669&top_left_y=1695&top_left_x=728)","[""We define the angles as in Fig. A. The light-path inside the optical system is $r(\\theta)$, and the angle between the first and last lens is $\\Theta$ (which is an unknown but can be uniquely determined from know information).\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_1dc8e68f97c9d182f754g-1.jpg?height=694&width=1484&top_left_y=225&top_left_x=316)\n\nConsider two consecutive lens at angle $\\theta$ and $\\theta+d \\theta$, in which $d \\theta=1 / n \\rightarrow 0$ in the continuum limit $n \\rightarrow \\infty$. From the ideal-lens' equation, using the approximation that $f$ is very large compare to other relevant length-scales in this optical setting:\n\n$$\n\\frac{1}{f}=\\frac{1}{r \\tan \\phi}+\\frac{1}{r \\tan (\\pi-\\phi-\\delta \\phi)} \\approx \\frac{\\delta \\phi}{r \\sin ^{2} \\phi} \\Rightarrow \\delta \\phi \\approx \\frac{r}{f} \\sin ^{2} \\phi\n\\tag{20}\n$$\n\nthe differential equation for the angle of arrival $\\phi$ can be written as:\n\n$$\nd \\phi=\\delta \\phi-d \\theta \\Rightarrow \\frac{d \\phi}{d \\theta}=\\frac{r}{f / n} \\sin ^{2} \\phi-1=\\frac{r}{\\lambda} \\sin ^{2} \\phi-1\n\\tag{21}\n$$\n\nWe also have the differential relation between radial position $r(\\theta)$ of the light-path and the angle of arrival $\\phi$ as followed:\n\n$$\n\\frac{d r}{d \\theta}=r \\cot \\phi\n\\tag{22}\n$$\n\nFrom Eq. (21) and Eq. (22), we arrive at:\n\n$$\n\\frac{d \\phi}{d r}=\\frac{\\frac{r}{\\lambda}-1}{r \\cot \\phi}\n\\tag{23}\n$$\n\nDefine $\\zeta=\\tan \\phi$, then Eq. (23) becomes:\n\n$$\n\\frac{d \\phi}{d r}=\\frac{1}{1+\\zeta^{2}} \\frac{d \\zeta}{d r}=\\frac{\\frac{r}{\\lambda} \\frac{\\zeta^{2}}{1+\\zeta^{2}}-1}{r / \\zeta} \\Rightarrow-\\frac{d \\zeta}{\\zeta^{3} d r}-\\frac{1}{\\zeta^{2} r}=\\frac{1}{r}-\\frac{1}{\\lambda}\n\\tag{24}\n$$\n\nDefine $\\xi=1 / \\zeta^{2}=1 / \\tan ^{2} \\phi$, then Eq. (23) gives:\n\n$$\n\\frac{d \\zeta}{\\zeta^{2} d r}=-\\frac{1}{2} \\frac{d \\xi}{d r} \\Rightarrow \\frac{d \\xi}{d r}-\\frac{2}{r} \\xi=2\\left(\\frac{1}{r}-\\frac{1}{\\lambda}\\right) \\Rightarrow \\frac{d}{d r}\\left(\\frac{\\xi}{r^{2}}\\right)=\\frac{2}{r^{2}}\\left(\\frac{1}{r}-\\frac{1}{\\lambda}\\right)\n\\tag{25}\n$$\n\nIntegrating both sides, then up to a constant value $C$, Eq. (25) gives:\n\n$$\n\\frac{\\xi}{r^{2}}=-\\frac{1}{r^{2}}+\\frac{2}{\\lambda r}+C \\Rightarrow \\xi=-1+2 \\frac{r}{\\lambda}+C \\frac{r^{2}}{\\lambda^{2}}\n\\tag{26}\n$$\n\n\n\nAt $\\theta=0, r=\\lambda / 2$ and $\\phi=\\pi / 2$ (thus $\\xi=0$ ), we can determine $C=0$. Hence:\n\n$$\n\\cot \\phi=\\sqrt{2 \\frac{r}{\\lambda}-1}\n\\tag{27}\n$$\n\nPlug Eq. (27) into Eq. (22):\n\n$$\n\\frac{d r}{d \\theta}=\\frac{r}{\\lambda} \\sqrt{2 \\frac{r}{\\lambda}-1} \\Rightarrow \\theta=2 \\arctan \\sqrt{2 \\frac{r}{\\lambda}-1}\n\\tag{28}\n$$\n\nAt $\\theta=\\Theta, r=2 \\lambda$ therefore we can use Eq. (28) to get:\n\n$$\n\\Theta=2 \\arctan \\sqrt{3}=\\frac{2 \\pi}{3}\n\\tag{29}\n$$\n\nUsing Eq. (27), the deflection angle $\\Delta$ can be calculated to be:\n\n$$\n\\Delta=\\Theta-\\left.\\phi\\right|_{r=\\lambda / 2}+\\left.\\phi\\right|_{r=2 \\lambda}=\\Theta-\\frac{\\pi}{2}+\\operatorname{arccot} \\sqrt{3}=\\frac{2 \\pi}{3}-\\frac{\\pi}{2}+\\frac{\\pi}{6}=\\frac{\\pi}{3} \\approx 1.05 \\mathrm{rad}\n\\tag{30}\n$$\n\nFor the sake of completeness, we provide the simulated light-path inside the optical system where $n=1000$ using MatLab (which is in great agreement with our theoretical analysis).""]",['$1.05$'],False,rad,Numerical,1e-1 895,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider an optical system made of many identical ideal (negligible-thickness) halflenses with focal length $f>0$, organized so that they share the same center and are angular-separated equally at density $n$ (number of lenses per unit-radian). Define the length-scale $\lambda=f / n$. A light-ray arrives perpendicular to the first lens at distance $\lambda / 2$ away from the center, then leaves from the last lens at distance $2 \lambda$ away from the center. Estimate the total deflection angle (in rad) of the light-ray by this system in the limit $n \rightarrow \infty$. ","[""We define the angles as in Fig. A. The light-path inside the optical system is $r(\\theta)$, and the angle between the first and last lens is $\\Theta$ (which is an unknown but can be uniquely determined from know information).\n\n\n\n\nConsider two consecutive lens at angle $\\theta$ and $\\theta+d \\theta$, in which $d \\theta=1 / n \\rightarrow 0$ in the continuum limit $n \\rightarrow \\infty$. From the ideal-lens' equation, using the approximation that $f$ is very large compare to other relevant length-scales in this optical setting:\n\n$$\n\\frac{1}{f}=\\frac{1}{r \\tan \\phi}+\\frac{1}{r \\tan (\\pi-\\phi-\\delta \\phi)} \\approx \\frac{\\delta \\phi}{r \\sin ^{2} \\phi} \\Rightarrow \\delta \\phi \\approx \\frac{r}{f} \\sin ^{2} \\phi\n\\tag{20}\n$$\n\nthe differential equation for the angle of arrival $\\phi$ can be written as:\n\n$$\nd \\phi=\\delta \\phi-d \\theta \\Rightarrow \\frac{d \\phi}{d \\theta}=\\frac{r}{f / n} \\sin ^{2} \\phi-1=\\frac{r}{\\lambda} \\sin ^{2} \\phi-1\n\\tag{21}\n$$\n\nWe also have the differential relation between radial position $r(\\theta)$ of the light-path and the angle of arrival $\\phi$ as followed:\n\n$$\n\\frac{d r}{d \\theta}=r \\cot \\phi\n\\tag{22}\n$$\n\nFrom Eq. (21) and Eq. (22), we arrive at:\n\n$$\n\\frac{d \\phi}{d r}=\\frac{\\frac{r}{\\lambda}-1}{r \\cot \\phi}\n\\tag{23}\n$$\n\nDefine $\\zeta=\\tan \\phi$, then Eq. (23) becomes:\n\n$$\n\\frac{d \\phi}{d r}=\\frac{1}{1+\\zeta^{2}} \\frac{d \\zeta}{d r}=\\frac{\\frac{r}{\\lambda} \\frac{\\zeta^{2}}{1+\\zeta^{2}}-1}{r / \\zeta} \\Rightarrow-\\frac{d \\zeta}{\\zeta^{3} d r}-\\frac{1}{\\zeta^{2} r}=\\frac{1}{r}-\\frac{1}{\\lambda}\n\\tag{24}\n$$\n\nDefine $\\xi=1 / \\zeta^{2}=1 / \\tan ^{2} \\phi$, then Eq. (23) gives:\n\n$$\n\\frac{d \\zeta}{\\zeta^{2} d r}=-\\frac{1}{2} \\frac{d \\xi}{d r} \\Rightarrow \\frac{d \\xi}{d r}-\\frac{2}{r} \\xi=2\\left(\\frac{1}{r}-\\frac{1}{\\lambda}\\right) \\Rightarrow \\frac{d}{d r}\\left(\\frac{\\xi}{r^{2}}\\right)=\\frac{2}{r^{2}}\\left(\\frac{1}{r}-\\frac{1}{\\lambda}\\right)\n\\tag{25}\n$$\n\nIntegrating both sides, then up to a constant value $C$, Eq. (25) gives:\n\n$$\n\\frac{\\xi}{r^{2}}=-\\frac{1}{r^{2}}+\\frac{2}{\\lambda r}+C \\Rightarrow \\xi=-1+2 \\frac{r}{\\lambda}+C \\frac{r^{2}}{\\lambda^{2}}\n\\tag{26}\n$$\n\n\n\nAt $\\theta=0, r=\\lambda / 2$ and $\\phi=\\pi / 2$ (thus $\\xi=0$ ), we can determine $C=0$. Hence:\n\n$$\n\\cot \\phi=\\sqrt{2 \\frac{r}{\\lambda}-1}\n\\tag{27}\n$$\n\nPlug Eq. (27) into Eq. (22):\n\n$$\n\\frac{d r}{d \\theta}=\\frac{r}{\\lambda} \\sqrt{2 \\frac{r}{\\lambda}-1} \\Rightarrow \\theta=2 \\arctan \\sqrt{2 \\frac{r}{\\lambda}-1}\n\\tag{28}\n$$\n\nAt $\\theta=\\Theta, r=2 \\lambda$ therefore we can use Eq. (28) to get:\n\n$$\n\\Theta=2 \\arctan \\sqrt{3}=\\frac{2 \\pi}{3}\n\\tag{29}\n$$\n\nUsing Eq. (27), the deflection angle $\\Delta$ can be calculated to be:\n\n$$\n\\Delta=\\Theta-\\left.\\phi\\right|_{r=\\lambda / 2}+\\left.\\phi\\right|_{r=2 \\lambda}=\\Theta-\\frac{\\pi}{2}+\\operatorname{arccot} \\sqrt{3}=\\frac{2 \\pi}{3}-\\frac{\\pi}{2}+\\frac{\\pi}{6}=\\frac{\\pi}{3} \\approx 1.05 \\mathrm{rad}\n\\tag{30}\n$$\n\nFor the sake of completeness, we provide the simulated light-path inside the optical system where $n=1000$ using MatLab (which is in great agreement with our theoretical analysis).""]",['$1.05$'],False,rad,Numerical,1e-1 896,Thermodynamics,,"For black body radiation, Wien's Displacement Law states that its spectral radiance will peak at $$ \lambda_{\text {peak }}=\frac{b}{T} $$ where $b=2.89777 \times 10^{-3} \mathrm{mK}$, and $T$ is the temperature of the object. When QiLin tried to reproduce this in a lab, by working with a tungsten-filament lightbulb at $2800 \mathrm{~K}$, he computed a different value for $b$ by measuring the peak wavelength using a spectrometer and multiplying it with the temperature. He hypothesizes that this discrepancy is because tungsten is not an ideal black body. The graph below, courtesy of the CRC Handbook of Chemistry and Physics, shows the emissivity of tungsten at various conditions (the units for wavelength is $\mathrm{nm}$ ). ![](https://cdn.mathpix.com/cropped/2023_12_21_d5c8b11775292675c006g-1.jpg?height=881&width=1762&top_left_y=205&top_left_x=173) Assuming QiLin's hypothesis is correct, and assuming there were no other errors in the experiment, how off was his value for $b$ ? Submit $\frac{\left|b_{\text {theory }}-b_{\text {experiment }}\right|}{b_{\text {theory }}}$ as a decimal number, to one significant digit (giving you room to estimate where the points are).","[""Recall Planck's Law, which says the spectral radiance of a black body is given by\n$$\nB_{0}(\\lambda, T)=\\frac{2 h c^{3}}{\\lambda^{5}} \\frac{1}{\\exp \\left(\\frac{h c}{\\lambda k_{B} T}\\right)-1}\n$$\n\nThe regular Wien's Displacement Law can be derived by finding the peak by computing $\\frac{\\partial B_{0}}{\\partial \\lambda}$, to find the wavelength associated with the maximal radiance. For a nonideal body with emissivity $\\epsilon(\\lambda, T)$, we can write the radiance as\n\n$$\nB(\\lambda, T)=B_{0}(\\lambda, T) \\epsilon(\\lambda, T)\n$$\n\nWe can estimate $\\epsilon(\\lambda, T)$ by looking at the given graph. The tungsten is at $2800 \\mathrm{~K}$, so we will use the red line, and assuming it is near a black body, the peak wavelength should be around $1000 \\mathrm{~nm}$. Performing a linear approximation around $1000 \\mathrm{~nm}$, we get\n\n$$\n\\epsilon(\\lambda, T)=-173333\\left(\\lambda-1000 \\cdot 10^{-9}\\right)+0.366\n$$\n\nwhere $\\lambda$ is in meters. Numerically finding the maximum of $B(\\lambda, T)$ with respect to $\\lambda$ (i.e. with a graphing calculator), we get the new peak wavelength to be $\\lambda_{\\text {new }}=949 \\mathrm{~nm}$, while the old peak wavelength (assuming a perfect blackbody) is $\\lambda_{\\text {old }}=1035 \\mathrm{~nm}$, and their percent difference (rounded to 1 significant digit) is 0.08""]",['0.08'],False,,Numerical,5e-3 896,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","For black body radiation, Wien's Displacement Law states that its spectral radiance will peak at $$ \lambda_{\text {peak }}=\frac{b}{T} $$ where $b=2.89777 \times 10^{-3} \mathrm{mK}$, and $T$ is the temperature of the object. When QiLin tried to reproduce this in a lab, by working with a tungsten-filament lightbulb at $2800 \mathrm{~K}$, he computed a different value for $b$ by measuring the peak wavelength using a spectrometer and multiplying it with the temperature. He hypothesizes that this discrepancy is because tungsten is not an ideal black body. The graph below, courtesy of the CRC Handbook of Chemistry and Physics, shows the emissivity of tungsten at various conditions (the units for wavelength is $\mathrm{nm}$ ). Assuming QiLin's hypothesis is correct, and assuming there were no other errors in the experiment, how off was his value for $b$ ? Submit $\frac{\left|b_{\text {theory }}-b_{\text {experiment }}\right|}{b_{\text {theory }}}$ as a decimal number, to one significant digit (giving you room to estimate where the points are).","[""Recall Planck's Law, which says the spectral radiance of a black body is given by\n$$\nB_{0}(\\lambda, T)=\\frac{2 h c^{3}}{\\lambda^{5}} \\frac{1}{\\exp \\left(\\frac{h c}{\\lambda k_{B} T}\\right)-1}\n$$\n\nThe regular Wien's Displacement Law can be derived by finding the peak by computing $\\frac{\\partial B_{0}}{\\partial \\lambda}$, to find the wavelength associated with the maximal radiance. For a nonideal body with emissivity $\\epsilon(\\lambda, T)$, we can write the radiance as\n\n$$\nB(\\lambda, T)=B_{0}(\\lambda, T) \\epsilon(\\lambda, T)\n$$\n\nWe can estimate $\\epsilon(\\lambda, T)$ by looking at the given graph. The tungsten is at $2800 \\mathrm{~K}$, so we will use the red line, and assuming it is near a black body, the peak wavelength should be around $1000 \\mathrm{~nm}$. Performing a linear approximation around $1000 \\mathrm{~nm}$, we get\n\n$$\n\\epsilon(\\lambda, T)=-173333\\left(\\lambda-1000 \\cdot 10^{-9}\\right)+0.366\n$$\n\nwhere $\\lambda$ is in meters. Numerically finding the maximum of $B(\\lambda, T)$ with respect to $\\lambda$ (i.e. with a graphing calculator), we get the new peak wavelength to be $\\lambda_{\\text {new }}=949 \\mathrm{~nm}$, while the old peak wavelength (assuming a perfect blackbody) is $\\lambda_{\\text {old }}=1035 \\mathrm{~nm}$, and their percent difference (rounded to 1 significant digit) is 0.08""]",['0.08'],False,,Numerical,5e-3 897,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A uniform spherical metallic ball of mass $m$, resistivity $\rho$, and radius $R$ is kept on a smooth friction-less horizontal ground. A horizontal uniform, constant magnetic field $B$ exists in the space parallel to the surface of ground. The ball was suddenly given an impulse perpendicular to magnetic field such that ball begin to move with velocity $v$ without losing the contact with ground. Find the time in seconds required to reduce its velocity by half. Numerical Quantities: $m=2 \mathrm{~kg}, 4 \pi \epsilon_{0} R^{3} B^{2}=3 \mathrm{~kg}, \rho=10^{9} \Omega \mathrm{m}, v=\pi \mathrm{m} / \mathrm{s}$.","['WLOG assuming magnetic field to be into the plane (negative z-axis) and velocity of the block along x-axis. Let any time $\\mathrm{t}$ ball has velocity $\\mathrm{v}$ and surface charge $\\sigma \\cos \\theta$, where $\\theta$ is measured from $\\mathrm{Y}$-axis. As the ball is moving it will also generate and electric field $\\mathrm{E}=\\mathrm{vB}$ along positive Y-axis. Which will be opposed by the electric field of ball. Also we know the electric filed generated by the ball is $\\frac{\\sigma}{3 \\epsilon_{0}}$ in negative $\\mathrm{Z}$-axis. As this charge distribution will arise form a vertical electric field and the subsequent induced charges will also produce only a vertical electric field only thus our assumption about charge distribution and net electric field must be true.\nNow at any point on the surface of the sphere rate of increase in surface charge density is given by-\n\n\n\n$$\n\\begin{gathered}\nJ d A \\cos \\theta=\\frac{d(\\sigma \\cos \\theta)}{d t} d A \\\\\nJ=\\frac{d \\sigma}{d t} \\\\\nE-\\frac{\\sigma}{3 \\epsilon_{0}}=\\rho \\frac{d \\sigma}{d t} \\\\\nv B-\\frac{\\sigma}{3 \\epsilon_{0}}=\\rho \\frac{d \\sigma}{d t}\n\\end{gathered}\n\\tag{44}\n$$\n\nAs the magnetic field is uniform to calculate force path of the current will not matter. Hence assuming it to be straight line between two points located at $\\theta$ and $-\\theta$. So the force can be written as -\n\n$$\n\\begin{gathered}\nd F=B \\times d I \\times l \\\\\nd F=B(2 \\pi(R \\sin \\theta) \\times R d \\theta \\times J)(2 R \\cos (\\theta)) \\\\\nF=4 \\pi R^{3} B \\frac{d \\sigma}{d t} \\int_{0}^{\\pi / 2} \\sin \\theta \\cos ^{2} \\theta d \\theta \\\\\nF=\\frac{4}{3} \\pi R^{3} B \\frac{d \\sigma}{d t}\n\\end{gathered}\n\\tag{45}\n$$\n\nNow writing force equation on the sphere we have\n\n$$\n\\begin{gathered}\nF=-m \\frac{d v}{d t} \\\\\n\\frac{4}{3} \\pi R^{3} B \\frac{d \\sigma}{d t}=-m \\frac{d v}{d t} \\\\\n\\frac{4}{3} \\pi R^{3} B \\int_{0}^{\\sigma} d \\sigma=-m \\int_{v_{0}}^{v} d v \\\\\n\\frac{4}{3} \\pi R^{3} B \\sigma=m v_{0}-m v\n\\end{gathered}\n\\tag{46}\n$$\n\nSolving equation 1 and 3 gives us\n\n$$\n\\left(m+4 \\pi R^{3} B^{2} \\epsilon_{0}\\right) v-m v_{0}=-3 m \\rho \\epsilon_{0} \\frac{d v}{d t}\n$$\n\nIntegrating it from $v_{0}$ to $\\frac{v_{0}}{2}$ gives\n\n$$\n\nt=\\frac{3 m \\rho \\epsilon_{0}}{\\left(m+4 \\pi R^{3} B^{2} \\epsilon_{0}\\right)} \\ln \\left(\\frac{8 \\pi R^{3} B^{2} \\epsilon_{0}}{4 \\pi R^{3} B^{2} \\epsilon_{0}-m}\\right)\n\\tag{47}\n$$\n$$\nt=\\frac{6 \\ln (6)}{5} \\rho \\epsilon_{0}=0.019 \\mathrm{~s}\n\\tag{48}\n$$']",['$0.019$'],False,s,Numerical,1e-3 898,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In quantum mechanics, when calculating the interaction between the electron with the proton in a hydrogen atom, it is necessary to compute the following volume integral (over all space): $$ \mathbf{I}=\int \mathbf{B}(\mathbf{r})|\Psi(\mathbf{r})|^{2} d V $$ where $\Psi(\mathbf{r})$ is the spatial wavefunction of the electron as a function of position $\mathbf{r}$ and $\mathbf{B}(\mathbf{r})$ is the (boldface denotes vector) magnetic field produced by the proton at position $\mathbf{r}$. Suppose the proton is located at the origin and it acts like a finite-sized magnetic dipole (but much smaller than $a_{0}$ ) with dipole moment $\mu_{p}=1.41 \times 10^{-26} \mathrm{~J} / \mathrm{T}$. Let the hydrogen atom be in the ground state, meaning $\Psi(\mathbf{r})=\frac{e^{-r / a_{0}}}{\sqrt{\pi a_{0}^{3}}}$, where $a_{0}=5.29 \times 10^{-11} \mathrm{~m}$ is the Bohr radius. Evaluate the magnitude of the integral $|\mathbf{I}|$ (in SI units).","[""First, note that the result of the integral will be a vector in the direction the dipole is pointing, call it the z-direction. Thus we can replace $\\mathbf{B}$ in the integral with $B_{z}$. Note that for any $R>0$, the integral over the space outside the sphere of radius $R$ is 0 . To show this, since $|\\Psi|$ is exponentially decaying, we only need to show that the integral over a spherical shell is 0 . To show this, we can show that the integral of $\\mathbf{B}$ inside a sphere of radius $R$ is independent of $R$. Indeed, this quickly follows from dimensional analysis (the only relevant quantities are $\\mu_{0}, \\mu_{p}$, and $R$, and one can check that $\\mu_{0} \\mu_{p}$ already gives the right dimensions, and there is no dimensionless combination of these 3 quantities. In fact, we will actually compute this integral at the end.)\n\nNow, it suffices to compute the integral of $\\mathbf{B}|\\Psi|^{2}$ inside the sphere. Since $R$ was arbitrary, we can make it very small, much smaller than $a_{0}$. Then we can replace $|\\Psi(\\mathbf{r})|^{2}$ with $|\\Psi(0)|^{2}=\\frac{1}{\\pi a_{0}^{3}}$, a constant that can be factored out. The problem reduces to computing the integral of $\\mathbf{B}$ inside a sphere of radius $R$.\n\nWe can compute this integral by splitting the sphere up into many thin discs, all perpendicular to the $z$ axis. We have to add up the $\\mathbf{B}$ field integrated over the volume of each disc, which is equivalent to the magnetic flux through the disc times the thickness of the disc. The magnetic flux through each disc can be calculated using the mutual inductance reciprocity theorem. Suppose a current $I$ goes around the boundary of the disc (a ring) with radius $r$. Then the mutual inductance $M$ between the ring and the dipole is given by the flux through the dipole divided by $I$ :\n\n$$\nM=\\frac{B * A}{I}\n$$\n\nwhere $B$ is the magnetic field produced by the ring's current at the dipole's position, and $A$ is the area of the dipole. The dipole itself carries current $i=\\frac{\\mu_{p}}{A}$, so the flux through the ring is given by\n\n$$\n\\Phi=M * i=\\frac{B i A}{I}=\\frac{\\mu_{p} B}{I}\n$$\n\nwhere $B=\\frac{\\mu_{0} I}{4 \\pi} * \\frac{2 \\pi r^{2}}{\\left(r^{2}+z^{2}\\right)^{\\frac{3}{2}}}=\\frac{\\mu_{0} I r^{2}}{2\\left(r^{2}+z^{2}\\right)^{\\frac{3}{2}}}$, where $z$ is the $z$ coordinate of the ring. Using $r=R \\sin \\theta$ and $z=R \\cos \\theta$, we obtain\n\n\n\n$$\n\\Phi(\\theta)=\\frac{\\mu_{0} \\mu_{p} r^{2}}{2\\left(r^{2}+z^{2}\\right)^{\\frac{3}{2}}}=\\frac{\\mu_{0} \\mu_{p} \\sin ^{2} \\theta}{2 R}\n$$\n\nFinally, we integrate over the thickness of the $\\operatorname{disc} R \\sin \\theta d \\theta$ to get :\n\n$$\n\\int_{0}^{\\pi} \\Phi(\\theta) R \\sin \\theta d \\theta=\\frac{1}{2} \\mu_{0} \\mu_{p} \\int_{0}^{\\pi} \\sin ^{3} \\theta d \\theta=\\frac{2}{3} \\mu_{0} \\mu_{p}\n$$\n\nThus, $|\\mathbf{I}|=\\frac{2}{3} \\mu_{0} \\mu_{p} * \\frac{1}{\\pi a_{0}^{3}}=\\frac{2 \\mu_{0} \\mu_{p}}{3 \\pi a_{0}^{3}}=0.0254 \\mathrm{~T}$.""]",['$0.0254$'],False,T,Numerical,5e-3 899,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Zed is trying to model the repulsive interaction between 2 objects, $A$ and $B$ (with masses $m_{A}$ and $m_{B}$, respectively), in a relativistic setting. He knows that in relativity, forces cannot act at a distance, so he models the repulsive force with a small particle of mass $m$ that bounces elastically between $A$ and $B$. Throughout this problem, assume everything moves on the x-axis. Suppose that initially, $A$ and $B$ have positions and velocities $x_{A}, v_{A}$ and $x_{B}, v_{B}$, respectively, where $x_{A}v_{B}$. The particle has an initial (relativistic) speed $v$. For simplicity, assume that the system has no total momentum. You may also assume that $v_{A}, v_{B} \ll v$, and that $p_{m} \ll p_{A}, p_{B}$, where $p_{m}, p_{A}, p_{B}$ are the momenta of the particle, $A$, and $B$, respectively. Do NOT assume $v \ll c$, where $c$ is the speed of light. Find the position (in $\mathrm{m}$ ) of $A$ when its velocity is 0 , given that $m_{A}=1 \mathrm{~kg}, m_{B}=2 \mathrm{~kg}, v_{A}=0.001 c$, $m=1 \times 10^{-6} \mathrm{~kg}, v=0.6 c, x_{A}=0 \mathrm{~m}, x_{B}=1000 \mathrm{~m}$. Note: Answers will be tolerated within $0.5 \%$, unlike other problems.","[""Since total momentum is 0 , we have $m_{A} v_{A}+m_{B} v_{B}=0$, so $v_{B}=-0.0005 c$ By conservation of energy:\n\n$$\n\\frac{1}{2} m_{A} v_{A}^{2}+\\frac{1}{2} m_{B} v_{B}^{2}+\\gamma_{0} m c^{2}=\\gamma m c^{2}\n$$\n\nwhere we define $\\gamma=\\frac{1}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}$ to correspond to the final state of the particle and $\\gamma_{0}$ to the initial state. This equation allows us to solve for $\\gamma$. Since the speed of the particle is much larger than the speed of the masses $A$ and $B$, we can imagine the particle moving in a infinite well potential where the walls are slowly moving. Applying the adiabatic theorem, we get that the adiabatic invariant $p x$ is conserved, where $p$ is the particle's momentum, and $x$ is the distance between $A$ and $B$. Thus, $\\gamma v x$ is conserved, so\n\n$$\n\\gamma v x=\\gamma_{0} v_{0}\\left(x_{B}-x_{A}\\right)\n$$\n\nWe can solve for $x$, since we know $\\gamma$ and $v$. Finally, we realize that the center of mass stays at $x_{c} m=\\frac{m_{A} x_{A}+m_{B} x_{B}}{m_{A}+m_{B}}$, so the final position of $A$ is simply\n\n$$\nx_{c} m-\\frac{m_{B}}{m_{A}+m_{B}} x=378 \\mathrm{~m}\n$$""]",['378'],False,m,Numerical,2e0 900,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider an optical system consisting of two thin lenses sharing the same optical axis. When a cuboid with a side parallel to the optical axis is placed to the left of the left lens, its final image formed by the optical system is also a cuboid but with 500 times the original volume. Assume the two lenses are $10 \mathrm{~cm}$ apart and such a cuboid of volume $1 \mathrm{~cm}^{3}$ is placed such that its right face is $2 \mathrm{~cm}$ to the left of the left lens. What's the maximum possible volume of the intermediate image (i.e., image formed by just the left lens) of the cuboid? Answer in $\mathrm{cm}^{3}$.","[""First, note that the two lenses share a focal point. Here's why. For any cuboid with four edges parallel to the optical axis, consider the four parallel rays of light that these four edges lie on. The intermediate images formed by the left lens of these four edges lie on these same light rays after they've passed through the left lens, and the final images of the edges (images formed by the right lens of the intermediate images) lie on these same light rays after they've also passed through the right lens. Since the initial rays were parallel and the same goes for the final rays, the intermediate rays intersect at a focal point of both the left lens and the right lens at the same time.\n\nNow, let $f, f^{\\prime}$ be the focal lengths of the left and right lenses, respectively. Although they can be any nonzero real number, we will WLOG assume that they are positive. (The following derivation will still hold with either $f$ or $f^{\\prime}$ negative, as long as we have the right sign conventions for all the variables.)\n\nFor a point at a distance $x_{1}$ to the left of $F$, its image is at a distance $x_{2}^{\\prime}=-\\left(\\frac{f^{\\prime}}{f}\\right)^{2} x_{1}$ to the right of $F^{\\prime}$. This follows from applying Newton's formula twice:\n\n$$\nx_{2}^{\\prime}=\\frac{f^{\\prime 2}}{x_{1}^{\\prime}}=-\\frac{f^{\\prime 2}}{x_{2}}=-\\frac{f^{\\prime 2}}{f^{2}} x_{1}\n$$\n\nThus, the optical system magnifies horizontal distances by $\\left(\\frac{f^{\\prime}}{f}\\right)^{2}$.\n\nOn the other hand, for a point at height $h_{1}$ (relative to the optical axis), consider a horizontal light ray through the point. Then the final light ray (after it passes through both lenses) is at a height of\n\n$$\nh_{2}=-\\frac{f^{\\prime}}{f} h_{1}\n$$\n\nwhich is the height of the final image of the point. Hence, the optical system magnifies transverse distances (i.e., distances perpendicular to the optical axis) by $\\frac{f^{\\prime}}{f}$.\n\nThe two results above imply that volumes are magnified by\n\n$$\n\\left(\\frac{f^{\\prime}}{f}\\right)^{2}\\left(\\frac{f^{\\prime}}{f}\\right)^{2}=\\left(\\frac{f^{\\prime}}{f}\\right)^{4}\n$$\n\n(The second factor is squared because there are two transverse dimensions.) Given that volumes are magnified by 500 times, we obtain $\\frac{f^{\\prime}}{f}= \\pm 500^{1 / 4}$.\n\nWe now look at a cuboid with volume $V$ whose right face is at a distance $d$ to the left of $F$. Let it have width $x d$ and transverse cross-sectional area $A=V / x d$. The intermediate image is a frustrum that results from truncating a pyramid with vertex located at $K$.\n\n\n\n\n\nBy Newton's formula, the bases of the frustrum are at distances $\\frac{f^{2}}{d}$ and $\\frac{f^{2}}{d(1+x)}$ to the right of $K$, and they have areas\n\n$$\n\\left(\\frac{\\frac{f^{2}}{d}}{f}\\right)^{2} A=\\frac{V f^{2}}{x d^{3}} \\quad \\text { and } \\quad\\left(\\frac{\\frac{f^{2}}{d(1+x)}}{f}\\right)^{2} A=\\frac{V f^{2}}{x(1+x)^{2} d^{3}}\n$$\n\nThus, the volume of the frustrum is\n\n$$\n\\frac{1}{3}\\left(\\frac{f^{2}}{d} \\frac{V f^{2}}{x d^{3}}-\\frac{f^{2}}{d(1+x)} \\frac{V f^{2}}{x(1+x)^{2} d^{3}}\\right)=\\frac{1}{3} \\frac{V f^{4}}{x d^{4}}\\left(1-\\frac{1}{(1+x)^{3}}\\right) \\leq \\frac{1}{3} \\frac{V f^{4}}{x d^{4}}(1-(1-3 x))=\\frac{V f^{4}}{d^{4}},\n$$\n\nwhere equality is approached as $x \\rightarrow 0$.\n\nSince $f+f^{\\prime}=10 \\mathrm{~cm}$ and $\\frac{f^{\\prime}}{f}= \\pm 500^{1 / 4} \\approx \\pm 4.7287$, either $f=1.7456 \\mathrm{~cm}$ and $d=2 \\mathrm{~cm}-f=$ $0.2544 \\mathrm{~cm}$, which gives $V_{\\max }=\\frac{V f^{4}}{d^{4}}=2216 \\mathrm{~cm}^{3}$, or $f=-2.6819 \\mathrm{~cm}$ and $d=2 \\mathrm{~cm}-f=4.6819 \\mathrm{~cm}$, which gives $V_{\\max }=0.1077 \\mathrm{~cm}^{3}$. The former is larger, so the answer is $2216 \\mathrm{~cm}^{3}$.""]",['$2216$'],False,$\mathrm{~cm}^{3}$,Numerical,5e0 901,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider an infinite square grid of equal resistors where the nodes are exactly the lattice points in the 2D Cartesian plane. A current $I=2.7 \mathrm{~A}$ enters the grid at the origin $(0,0)$. Find the current in Amps through the resistor connecting the nodes $(N, 0)$ and $(N, 1)$, where $N=38$ can be assumed to be much larger than 1.","[""WLOG, let each resistor have unit resistance.\n\nKirchoff's current law says that the total current entering the node $(x, y) \\neq(0,0)$ is zero:\n\n$$\n\\begin{array}{r}\n{[U(x+1, y)-U(x, y)]+[U(x-1, y)-U(x, y)]+[U(x, y+1)-U(x, y)]+[U(x, y-1)-U(x, y)]=0} \\\\\n{[U(x+1, y)-2 U(x, y)+U(x-1, y)]+[U(x, y+1)-2 U(x, y)+U(x, y-1)]=0}\n\\end{array}\n$$\n\nThis is an approximation of the equation\n\n$$\n\\frac{\\partial^{2} U}{\\partial x^{2}}+\\frac{\\partial^{2} U}{\\partial y^{2}}=0\n$$\n\nwhich is Laplace's equation $\\nabla^{2} U=0$ in 2D. Given $U \\rightarrow 0$ as $r \\rightarrow \\infty$, this implies that the potential\n\n\n\nfield approximates that of a point charge at $O$ in 2D. This approximation is valid far from the origin where changes in $U$ over unit length are small.\n\nThe electric field corresponding to this potential field is\n\n$$\n\\mathbf{E}=-\\nabla U=\\left(-\\partial_{x} U,-\\partial_{y} U\\right) \\approx(U(x-1, y)-U(x, y), U(x, y-1)-U(x, y))=\\left(i_{x}(x, y), i_{y}(x, y)\\right)\n$$\n\nfar from the origin, where $i_{x}(x, y), i_{y}(x, y)$ are the horizontal and vertical currents passing through node $(x, y)$. (Note that the current horizontal current is different to the left vs. to the right of the node, and similarly for the vertical current, but the difference is negligible for $N \\gg 1$.)\n\nThe current $i=\\sqrt{i_{x}^{2}+i_{y}^{2}}$ at a distance $r \\gg 1$ away from the origin is given by Gauss's law:\n\n$$\n2 \\pi r i(r) \\approx I\n$$\n\nso at $(N, 0)$ we have\n\n$$\ni_{x}(N, 0)=i(N) \\approx \\frac{I}{2 \\pi N}\n$$\n\nThe difference between the entering and exiting horizontal currents at $(N, 0)$ is approximately\n\n$$\n-\\partial_{x} i_{x}(N, 0)=\\frac{I}{2 \\pi N^{2}}\n$$\n\nThis difference is directed equally into the vertical resistors adjacent to $(N, 0)$, so the final answer is\n\n$$\n\\frac{I}{4 \\pi N^{2}}=1.488 \\times 10^{-4} \\mathrm{~A}\n$$""]",['$1.488 \\times 10^{-4}$'],False,A,Numerical,2e-2 902,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Suppose we have a non-ideal gas, and in a certain volume range and temperature range, it is found to satisfy the state relation $$ p=A V^{\alpha} T^{\beta} $$ where $A$ is a constant, $\alpha=-\frac{4}{5}$ and $\beta=\frac{3}{2}$, and the other variables have their usual meanings. Throughout the problem, we will assume to be always in that volume and temperature range. Assume that $\gamma=\frac{C_{p}}{C_{V}}$ is found to be constant for this gas ( $\gamma$ is independent of the state of the gas), where $C_{p}$ and $C_{v}$ are the heat capacities at constant pressure and volume, respectively. What is the minimum possible value for $\gamma$ ?","['We claim that the conditions given uniquely determine $\\gamma$.\n\nThe fundamental thermodynamic relation gives:\n\n$$\n\\mathrm{d} U=T \\mathrm{~d} S-p \\mathrm{~d} V\n$$\n\nSo\n\n$$\n\\left(\\frac{\\partial U}{\\partial V}\\right)_{T}=T\\left(\\frac{\\partial S}{\\partial V}\\right)_{T}-p=T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}-p\n$$\n\nwhere we have used a Maxwell relation.\n\n\n\n$$\n\\mathrm{d} U=\\left(\\frac{\\partial U}{\\partial T}\\right)_{V} \\mathrm{~d} T+\\left(\\frac{\\partial U}{\\partial V}\\right)_{T} \\mathrm{~d} V=C_{V} \\mathrm{~d} T+\\left(T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}-p\\right) \\mathrm{d} V\n$$\n\nWe have\n\n$$\nC_{p}=\\left(\\frac{\\partial U}{\\partial T}\\right)_{p}+p\\left(\\frac{\\partial V}{\\partial T}\\right)_{p}=C_{V}+T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}\\left(\\frac{\\partial V}{\\partial T}\\right)_{p}\n$$\n\nRearranging, gives\n\n$$\nC_{V}=\\frac{T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}\\left(\\frac{\\partial V}{\\partial T}\\right)_{p}}{\\gamma-1}\n$$\n\nFrom the symmetry of mixed second partial derivatives, we know\n\n$$\n\\left(\\frac{\\partial C_{V}}{\\partial V}\\right)_{T}=\\left(\\frac{\\partial^{2} U}{\\partial T \\partial V}\\right)=\\left(\\frac{\\partial}{\\partial T}\\left(\\frac{\\partial U}{\\partial V}\\right)_{T}\\right)_{V}=\\frac{\\partial}{\\partial T}\\left(T\\left(\\frac{\\partial p}{\\partial T}\\right)_{V}-p\\right)=\\left(\\frac{\\partial^{2} p}{\\partial T^{2}}\\right)_{V}\n$$\n\nPlugging our expression for $C_{V}$ into here, and plugging in the equation of state, we can solve for $\\gamma$ to get\n\n$$\n\\gamma=\\frac{\\alpha+\\beta}{\\alpha(1-\\beta)}=\\frac{7}{4}\n$$']",['$\\frac{7}{4}$'],False,,Numerical,0 903,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The coin flip has long been recognized as a simple and unbiased method to randomly determine the outcome of an event. In the case of an ideal coin, it is well-established that each flip has an equal $50 \%$ chance of landing as either heads or tails. However, coin flips are not entirely random. They appear random to us because we lack sufficient information about the coin's initial conditions. If we possessed this information, we would always be able to predict the outcome without needing to flip the coin. For an intriguing discussion on why this observation is significant, watch this video by Vsauce. Now, consider a scenario where a coin with uniform density and negligible width is tossed directly upward from a height of $h=0.75 \mathrm{~m}$ above the ground. The coin starts with its heads facing upward and is given an initial vertical velocity of $v_{y}=49 \mathrm{~m} / \mathrm{s}$ and a positive angular velocity of $\omega=\pi \mathrm{rad} / \mathrm{s}$. What face does the coin display upon hitting the ground? Submit $\mathbf{0}$ for heads and $\mathbf{1}$ for tails. You only have one attempt for this problem. Assume the floor is padded and it absorbs all of the coin's energy upon contact. The radius of the coin is negligible.","['We have the following quadratic:\n$$\n\\begin{aligned}\nx & =x_{0}+v_{0} t+\\frac{1}{2} a t^{2} \\\\\n0 & =0.75+49 t-4.9 t^{2} \\\\\nt & =-0.01,10.01\n\\end{aligned}\n$$\n\nThe first solution is extraneous so $t=10.01$ is correct. Now, $\\theta=\\omega t \\approx 10 \\pi$. As one full rotation is $\\phi=2 \\pi$, then the coin performs 5 full rotations before landing on the ground. This means the answer is 0 , or heads.']",['0'],False,,Numerical,0 904,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A coin of uniform mass density with a radius of $r=1 \mathrm{~cm}$ is initially at rest and is released from a slight tilt of $\theta=8^{\circ}$ onto a horizontal surface with an infinite coefficient of static friction. The coin has a thicker rim, allowing it to drop and rotate on one point. With every collision, the coin switches pivot points on the rim, and energy is dissipated through heat so that $k=0.9$ of the coin's prior total energy is conserved. How long will it take for the coin to come to a complete stop? A cross-sectional view of the coin before release. The rim can be seen on the edges of the coin.","[""By the parallel axis theorem, the moment of inertia of the coin around the pivot can be expressed as\n$$\nI=I_{x}+m \\ell^{2}=\\frac{1}{4} m r^{2}+m r^{2}=\\frac{5}{4} m r^{2}\n$$\n\n\n\nAs $\\theta$ is small, that means the perpendicular force of gravity is $m g \\cos \\theta \\approx m g$. Hence, Newton's second law to find the angular acceleration as\n\n$$\nI \\alpha=\\tau \\Longrightarrow \\alpha=\\frac{4 g}{5 r} \\cdot(1 \\mathrm{rad})\n$$\n\nUsing rotational kinematics, the time for the first collision will be $\\theta=\\frac{1}{2} \\alpha t^{2} \\Longrightarrow t_{0}=\\sqrt{\\frac{5 r \\theta_{0}}{2 g}}$. By conservation of energy, the angular velocity at the time of collision is $\\frac{1}{2} I \\omega_{0}^{2}=m g r \\sin \\theta \\approx m g r$, meaning that $\\omega_{0}=\\sqrt{\\frac{8 g \\theta_{0}}{5 r}}$. Now consider $k$. This is the ratio of the initial energy $E_{n-1}$ and next energy $E_{n}$. Then, we can say that\n\n$$\nk=\\frac{\\frac{1}{2} I \\omega_{n}^{2}}{\\frac{1}{2} I \\omega_{n-1}^{2}} \\Longrightarrow \\omega_{n}=\\sqrt{k} \\omega_{n-1}\n$$\n\nBy recurrence, $\\omega_{n}=k^{n / 2} \\omega_{0}$. The time of flight for each cycle will be $t_{n}=\\frac{2 \\omega_{n}}{\\alpha}$. Therefore,\n\n$$\n\\begin{aligned}\nT & =t_{0}+\\sum_{n=1}^{\\infty} t_{n} \\\\\n& =\\sqrt{\\frac{5 r \\theta_{0}}{2 g}}+\\sum_{n=1}^{\\infty}\\left(2 k^{n / 2} \\sqrt{\\frac{8 g \\theta_{0}}{5 r}} \\cdot \\frac{5 r}{4 g}\\right) \\\\\n& =\\sqrt{\\frac{5 r \\theta_{0}}{2 g}}+\\sqrt{\\frac{10 r \\theta_{0}}{g}} \\frac{k^{1 / 2}}{1-k^{1 / 2}}\n\\end{aligned}\n$$\n\nConverting $8^{\\circ}$ to $\\frac{2 \\pi}{45}$ radians, $k=0.9, g=9.8 \\mathrm{~m} / \\mathrm{s}^{2}, r=0.01 \\mathrm{~m}$, we find that $T=0.716 \\mathrm{~s}$ which is a reasonable estimate.""]",['0.716'],False,s,Numerical,2e-2 904,Mechanics,,"A coin of uniform mass density with a radius of $r=1 \mathrm{~cm}$ is initially at rest and is released from a slight tilt of $\theta=8^{\circ}$ onto a horizontal surface with an infinite coefficient of static friction. The coin has a thicker rim, allowing it to drop and rotate on one point. With every collision, the coin switches pivot points on the rim, and energy is dissipated through heat so that $k=0.9$ of the coin's prior total energy is conserved. How long will it take for the coin to come to a complete stop? ![](https://cdn.mathpix.com/cropped/2023_12_21_21db419ae935527412b4g-1.jpg?height=135&width=469&top_left_y=2014&top_left_x=752) A cross-sectional view of the coin before release. The rim can be seen on the edges of the coin.","[""By the parallel axis theorem, the moment of inertia of the coin around the pivot can be expressed as\n$$\nI=I_{x}+m \\ell^{2}=\\frac{1}{4} m r^{2}+m r^{2}=\\frac{5}{4} m r^{2}\n$$\n\n\n\nAs $\\theta$ is small, that means the perpendicular force of gravity is $m g \\cos \\theta \\approx m g$. Hence, Newton's second law to find the angular acceleration as\n\n$$\nI \\alpha=\\tau \\Longrightarrow \\alpha=\\frac{4 g}{5 r} \\cdot(1 \\mathrm{rad})\n$$\n\nUsing rotational kinematics, the time for the first collision will be $\\theta=\\frac{1}{2} \\alpha t^{2} \\Longrightarrow t_{0}=\\sqrt{\\frac{5 r \\theta_{0}}{2 g}}$. By conservation of energy, the angular velocity at the time of collision is $\\frac{1}{2} I \\omega_{0}^{2}=m g r \\sin \\theta \\approx m g r$, meaning that $\\omega_{0}=\\sqrt{\\frac{8 g \\theta_{0}}{5 r}}$. Now consider $k$. This is the ratio of the initial energy $E_{n-1}$ and next energy $E_{n}$. Then, we can say that\n\n$$\nk=\\frac{\\frac{1}{2} I \\omega_{n}^{2}}{\\frac{1}{2} I \\omega_{n-1}^{2}} \\Longrightarrow \\omega_{n}=\\sqrt{k} \\omega_{n-1}\n$$\n\nBy recurrence, $\\omega_{n}=k^{n / 2} \\omega_{0}$. The time of flight for each cycle will be $t_{n}=\\frac{2 \\omega_{n}}{\\alpha}$. Therefore,\n\n$$\n\\begin{aligned}\nT & =t_{0}+\\sum_{n=1}^{\\infty} t_{n} \\\\\n& =\\sqrt{\\frac{5 r \\theta_{0}}{2 g}}+\\sum_{n=1}^{\\infty}\\left(2 k^{n / 2} \\sqrt{\\frac{8 g \\theta_{0}}{5 r}} \\cdot \\frac{5 r}{4 g}\\right) \\\\\n& =\\sqrt{\\frac{5 r \\theta_{0}}{2 g}}+\\sqrt{\\frac{10 r \\theta_{0}}{g}} \\frac{k^{1 / 2}}{1-k^{1 / 2}}\n\\end{aligned}\n$$\n\nConverting $8^{\\circ}$ to $\\frac{2 \\pi}{45}$ radians, $k=0.9, g=9.8 \\mathrm{~m} / \\mathrm{s}^{2}, r=0.01 \\mathrm{~m}$, we find that $T=0.716 \\mathrm{~s}$ which is a reasonable estimate.""]",['0.716'],False,s,Numerical,2e-2 905,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Suppose all cars on a (single-lane) highway are identical. Their length is $l=4 \mathrm{~m}$, their wheels have coefficients of friction $\mu=0.7$, and they all travel at speed $v_{0}$. Find the $v_{0}$ which maximizes the flow rate of cars (i.e. how many cars travel across an imaginary line per minute). Assume that they need to be able to stop in time if the car in front instantaneously stops. Disregard reaction time.","['Suppose the maximum speed is $v^{\\prime}$. Notice that $a=\\mu g$ so it takes a time $t^{\\prime}=\\frac{v^{\\prime}}{\\mu g}$ amount of time to stop. This means the car will travel a distance of\n$$\nd=\\frac{1}{2} \\mu g\\left(\\frac{v^{\\prime}}{\\mu g}\\right)^{2}=\\frac{v^{\\prime 2}}{2 \\mu g}\n$$\n\nThus for every distance $d+l=l+\\frac{v^{\\prime 2}}{2 \\mu g}$ there is a car. This means in unit time $t$, there will be $N=\\frac{d+l}{v}$ cars that passes through the line. To maximize the flow rate, we need to maximize\n\n$$\n\\begin{gathered}\nf\\left(v^{\\prime}\\right)=\\frac{l}{v^{\\prime}}+\\frac{v^{\\prime}}{2 \\mu g} \\\\\nf^{\\prime}=\\frac{-l}{v^{\\prime 2}}+\\frac{1}{2 \\mu g}=0 \\Rightarrow v^{\\prime}=\\sqrt{2 \\mu g l}=7.41 \\mathrm{~m} / \\mathrm{s}\n\\end{gathered}\n$$']",['$7.41$'],False,m/s,Numerical,1e-2 906,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Here is a Physoly round button badge, in which the logo is printed on the flat and rigid surface of this badge. Toss it in the air and track the motions of three points (indicated by cyan circles in the figure) separated a straight-line distance of $L=5 \mathrm{~mm}$ apart. At a particular moment, we find that these all have the same speed $V=4 \mathrm{~cm} / \mathrm{s}$ but are heading to different directions which form an angle of $\theta=30^{\circ}$ between each pair. Determine the then angular velocity of the badge (in $\left.\mathrm{rad} / \mathrm{s}\right)$. ","['Call the three tracking points on the Physoly badge A, B, C, and their geometrical center $\\mathrm{O}$. The distance from $\\mathrm{O}$ to these three points are the same and equal to $L / \\sqrt{3}$.\n\n\nDue to symmetry, the velocity vector of $\\mathrm{O}$ has to be perpendicular to the $\\mathrm{ABC}$ plane. In the reference frame of $\\mathrm{O}$, the points $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$ both have the same speed $2 V \\sin (\\theta / 2) / \\sqrt{3}$ but are heading to different directions which form an angle of $120^{\\circ}$ between each pair. Also due to symmetry, the axis of rotation has to be perpendicular to the $\\mathrm{ABC}$ plane, thus the velocity vectors of points $\\mathrm{A}$, $\\mathrm{B}, \\mathrm{C}$ in $\\mathrm{O}$ reference frame looks like described in the attached figure. For $L=5 \\mathrm{~mm}, V=4 \\mathrm{~cm} / \\mathrm{s}$, $\\theta=30^{\\circ}=\\pi / 12$, the angular velocity of the badge can be calculated as:\n\n$$\n\\Omega=\\frac{2 V \\sin (\\theta / 2) / \\sqrt{3}}{L / \\sqrt{3}}=\\left(\\frac{\\sqrt{3}-1}{\\sqrt{2}}\\right) \\frac{V}{L} \\approx 4.1411 \\mathrm{rad} / \\mathrm{s}\n$$']",['$4.1411$'],False,$\mathrm{rad} / \mathrm{s}$,Numerical,2e-1 906,Mechanics,,"Here is a Physoly round button badge, in which the logo is printed on the flat and rigid surface of this badge. Toss it in the air and track the motions of three points (indicated by cyan circles in the figure) separated a straight-line distance of $L=5 \mathrm{~mm}$ apart. At a particular moment, we find that these all have the same speed $V=4 \mathrm{~cm} / \mathrm{s}$ but are heading to different directions which form an angle of $\theta=30^{\circ}$ between each pair. Determine the then angular velocity of the badge (in $\left.\mathrm{rad} / \mathrm{s}\right)$. ![](https://cdn.mathpix.com/cropped/2023_12_21_86ad3cbd2a33fe5168c3g-1.jpg?height=518&width=529&top_left_y=473&top_left_x=798)","['Call the three tracking points on the Physoly badge A, B, C, and their geometrical center $\\mathrm{O}$. The distance from $\\mathrm{O}$ to these three points are the same and equal to $L / \\sqrt{3}$.\n![](https://cdn.mathpix.com/cropped/2023_12_21_86ad3cbd2a33fe5168c3g-1.jpg?height=767&width=1033&top_left_y=1178&top_left_x=554)\n\nDue to symmetry, the velocity vector of $\\mathrm{O}$ has to be perpendicular to the $\\mathrm{ABC}$ plane. In the reference frame of $\\mathrm{O}$, the points $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$ both have the same speed $2 V \\sin (\\theta / 2) / \\sqrt{3}$ but are heading to different directions which form an angle of $120^{\\circ}$ between each pair. Also due to symmetry, the axis of rotation has to be perpendicular to the $\\mathrm{ABC}$ plane, thus the velocity vectors of points $\\mathrm{A}$, $\\mathrm{B}, \\mathrm{C}$ in $\\mathrm{O}$ reference frame looks like described in the attached figure. For $L=5 \\mathrm{~mm}, V=4 \\mathrm{~cm} / \\mathrm{s}$, $\\theta=30^{\\circ}=\\pi / 12$, the angular velocity of the badge can be calculated as:\n\n$$\n\\Omega=\\frac{2 V \\sin (\\theta / 2) / \\sqrt{3}}{L / \\sqrt{3}}=\\left(\\frac{\\sqrt{3}-1}{\\sqrt{2}}\\right) \\frac{V}{L} \\approx 4.1411 \\mathrm{rad} / \\mathrm{s}\n$$']",['$4.1411$'],False,$\mathrm{rad} / \mathrm{s}$,Numerical,2e-1 907,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In a resource-limited ecological system, a population of organisms cannot keep growing forever (such as lab bacteria growing inside culture tube). The effective growth rate $g$ (including contributions from births and deaths) depends on the instantaneous abundance of resource $R(t)$, which in this problem we will consider the simple case of linear-dependency: $$ \frac{\mathrm{d}}{\mathrm{d} t} N=g(R) N=\alpha R N $$ where $N(t)$ is the population size at time $t$. The resources is consumed at a constant rate $\beta$ by each organism: $$ \frac{\mathrm{d}}{\mathrm{d} t} R=-\beta N $$ Initially, the total amount of resources is $R_{0}$ and the population size is $N_{0}$. Given that $\alpha=10^{-9}$ resourceunit $^{-1} \mathrm{~s}^{-1}, \beta=1$ resource-unit/s, $R_{0}=10^{6}$ resource-units and $N_{0}=1$ cell, find the total time it takes from the beginning to when all resources are depleted (in hours).","['We can find the analytical solution for the following set of two ODEs describing the populationresource dynamics:\n\n$$\n\\frac{d N}{d t}=\\alpha R N,\n\\tag{1}\n$$\n$$\n\\frac{d R}{d t}=-\\beta N .\n\\tag{2}\n$$\n\nDivide Eq.(1) by Eq.(2) on both sides, we get a direct relation between the population size $N(t)$ and the amount of resource $R(t)$ :\n\n$$\n\\frac{d N}{d R}=-\\frac{\\alpha}{\\beta} R \\Longrightarrow N=N_{0}+\\frac{\\alpha}{2 \\beta}\\left(R_{0}^{2}-R^{2}\\right)\n$$\n\nPlug this in Eq.(2), we obtain the total time $T$ it takes from the beginning to when the resource is depleted:\n\n$$\n\\begin{aligned}\n\\frac{d R}{d t}=-\\frac{\\alpha}{2}\\left[\\left(\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}\\right)-R^{2}\\right] \\Longrightarrow & \\Longrightarrow \\\\\n\\left.\\right|_{R=0} & =\\frac{2}{\\alpha} \\int_{0}^{R_{0}} d R\\left[\\left(\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}\\right)-R^{2}\\right]^{-1} \\\\\n& =\\frac{2}{\\alpha \\sqrt{\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}}} \\operatorname{arctanh}\\left(\\frac{R_{0}}{\\sqrt{\\frac{2 \\beta}{\\alpha} N_{0}+R_{0}^{2}}}\\right) .\n\\end{aligned}\n$$\n\nUse the given numerical values, we arrive at $\\left.t\\right|_{R=0} \\approx 7594.3 \\mathrm{~s} \\approx 2.1095 \\mathrm{hrs}$.']",['2.1095'],False,h,Numerical,2e-1 908,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","An incandescent lightbulb is connected to a circuit which delivers a maximum power of 10 Watts. The filament of the lightbulb is made of Tungsten and conducts electricity to produce light. The specific heat of Tungsten is $c=235 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{kg})$. If the circuit is alternating such that the temperature inside the lightbulb fluctuates between $T_{0}=3000^{\circ} \mathrm{C}$ and $T_{1}=3200^{\circ} \mathrm{C}$ at a frequency of $\omega=0.02 \mathrm{~s}^{-1}$, estimate the mass of the filament.","['This problem was voided from the test because we gave no method for energy dissipation. Therefore, there was ambiguity and energy would be constantly fed to the lightbulb making it hotter and hotter.\nHowever, the problem can still be solved. Here is a solution written by one of our contestants\n\n\n\nGuangyuan Chen. The circuit continuously delivers sinusoidal power to the lightbulb. In order to maintain a quasi-equilibrium state, there must be some form of heat loss present in the system. We will assume that the main source of heat loss is through radiation, though we will see that the precise form of heat loss is largely inconsequential.\n\nLet the power delivered by the circuit be in the form $P_{d}=P_{0} \\sin ^{2} \\Omega t$. Let the power loss due to the radiation be $P_{l}=-k T^{4}$ where $k$ is some constant and $T$ is the temperature of the filament. We can write:\n\n$$\nm c \\frac{\\mathrm{d} T}{\\mathrm{~d} t}=P_{0} \\sin ^{2} \\Omega t-k T^{4}\n$$\n\nThis is a differential equation that cannot be solved by hand. However, since the deviation in temperature from the equilibrium of roughly $100^{\\circ} \\mathrm{C}$ is much smaller than the equilibrium temperature of roughly $3100^{\\circ} \\mathrm{C}$, we can do a first order approximation of the differential equation. The temperature thus varies sinusoidally with equilibrium temperature $T_{\\text {equi }}=3100^{\\circ} \\mathrm{C}$ and amplitude $T_{a}=100^{\\circ} \\mathrm{C}$. Since the average power delivered is $\\frac{1}{2} P_{0}$, at equilibrium temperature, we have:\n\n$$\n\\begin{gathered}\n\\frac{1}{2} P_{0}=k T_{\\text {equi }}^{4} \\\\\nk=\\frac{P_{0}}{2 T_{e q u i}^{4}}\n\\end{gathered}\n$$\n\nWe rewrite equation 1 with the approximation $T=T_{\\text {equi }}+\\Delta T$ and with the substitution from equation 3.\n\n$$\n\\begin{gathered}\nm c \\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} t}=P_{0} \\sin ^{2} \\Omega t-\\frac{P_{0}}{2 T_{\\text {equi }}^{4}}\\left(T_{\\text {equi }}+\\Delta T\\right)^{4} \\\\\nm c \\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} t} \\approx P_{0} \\sin ^{2} \\Omega t-\\frac{P_{0}}{2}\\left(1+4 \\frac{\\Delta T}{T_{\\text {equi }}}\\right) \\\\\nm c \\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} t} \\approx-\\frac{P_{0}}{2}\\left(\\cos 2 \\Omega t+4 \\frac{\\Delta T}{T_{\\text {equi }}}\\right)\n\\end{gathered}\n$$\n\nAt this point we can see that regardless of the physical mechanism of heat loss, a linear approximation can always be made for sufficiently small variations in temperature which results in an equation of the same form as equation 6 . To solve this equation, we can guess the following solution:\n\n$$\n\\Delta T=A \\sin 2 \\Omega t+B \\cos 2 \\Omega t\n$$\n\nStrictly speaking, we also need to include a term $C e^{-\\lambda t}$. However, for sufficiently long times approaching quasi-equilibrium, this term will go to zero, so we need not include it here. Substituting equation 7 into equation 6 :\n\n$$\n2 m c \\Omega(A \\cos 2 \\Omega t-B \\sin 2 \\Omega t)=-\\frac{P_{0}}{2}\\left(\\cos 2 \\Omega t+\\frac{4}{T_{\\text {equi }}}(A \\sin 2 \\Omega t+B \\cos 2 \\Omega t)\\right)\n$$\n\nEquating the coefficients of the sin and cos:\n\n$$\n\\begin{gathered}\n2 m c \\Omega A=-\\frac{P_{0}}{2}\\left(1+\\frac{4 B}{T_{e q u i}}\\right) \\\\\n-2 m c \\Omega B=-\\frac{2 P_{0} A}{T_{\\text {equi }}}\n\\end{gathered}\n$$\n\n\n\nEquations 9 and 10 can be solved simultaneously to give\n\n$$\n\\begin{aligned}\nA & =-\\frac{m c \\Omega T_{e q u i} P_{0}}{4\\left(\\left(m c \\Omega T_{e q u i}\\right)^{2}+P_{0}^{2}\\right)} T_{e q u i} \\\\\nB & =-\\frac{P_{0}^{2}}{4\\left(\\left(m c \\Omega T_{e q u i}\\right)^{2}+P_{0}^{2}\\right)} T_{e q u i}\n\\end{aligned}\n$$\n\nEquation 7 can be written in the form $T_{a} \\sin (2 \\Omega t+\\phi)$. To find $T_{a}$, we simply have:\n\n$$\nT_{a}=\\sqrt{A^{2}+B^{2}}=\\frac{P_{0}}{4 \\sqrt{\\left(m c \\Omega T_{e q u i}\\right)^{2}+P_{0}^{2}}} T_{e q u i}\n$$\n\nRearranging for $m$, we have:\n\n$$\nm=\\frac{P_{0}}{c \\Omega T_{e q u i}} \\sqrt{\\left(\\frac{T_{e q u i}}{4 T_{a}}\\right)^{2}-1}\n$$\n\nThe final step is to relate $\\Omega$ and $\\omega$. Since $T$ oscillates with angular frequency $2 \\Omega$, its period of oscillation is $\\frac{2 \\pi}{2 \\Omega}$. This is equal to $\\frac{1}{\\omega}$, hence we have:\n\n$$\n\\Omega=\\pi \\omega\n$$\n\nSubstituting into equation 14, we get our final numerical answer.\n\n$$\nm=\\frac{P_{0}}{\\pi c \\omega T_{e q u i}} \\sqrt{\\left(\\frac{T_{e q u i}}{4 T_{a}}\\right)^{2}-1}=1.68 \\times 10^{-3} \\mathrm{~kg}\n$$']",['$1.68 \\times 10^{-3}$'],False,kg,Numerical,2e-5 909,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In hyperdrive, Spaceship-0 is relativistically moving at the velocity $\frac{1}{3} c$ with respect to reference frame $R_{1}$, as measured by Spaceship-1. Spaceship-1 is moving at $\frac{1}{2} c$ with respect to reference frame $R_{2}$, as measured by Spaceship-2. Spaceship- $k$ is moving at speed $v_{k}=\frac{k+1}{k+3} c$ with respect to reference frame $R_{k+1}$. The speed of Spaceship-0 with respect to reference frame $R_{20}$ can be expressed as a decimal fraction of the speed of light which has only $x$ number of 9 s following the decimal point (i.e., in the form of $0 . \underbrace{99 \ldots 9}_{x \text { times }} c)$. Find the value of $x$.","[""Let us define the rapidity as\n$$\n\\tanh \\phi \\equiv \\beta=\\frac{v}{c}\n$$\n\nwhere tanh is the hyperbolic tangent function. Let $\\beta_{1}=\\tanh \\phi_{1}$ and $\\beta_{2}=\\tanh \\phi_{2}$. If we add $\\beta_{1}$ and $\\beta_{2}$ using the relativistic velocity addition formula, we find that\n\n$$\n\\beta=\\frac{\\beta_{1}+\\beta_{2}}{1-\\beta_{1} \\beta_{2}}=\\frac{\\tanh \\phi_{1}+\\tanh \\phi_{2}}{1+\\tanh \\phi_{1} \\tanh \\phi_{2}}=\\tanh \\left(\\phi_{1}+\\phi_{2}\\right)\n$$\n\nWe can then rewrite the problem as\n\n$$\nv_{f}=\\tanh \\left(\\operatorname{arctanh} \\frac{1}{3}+\\operatorname{arctanh} \\frac{2}{4}+\\cdots+\\operatorname{arctanh} \\frac{20}{22}\\right)\n$$\n\n\n\nUsing the fact that $\\operatorname{arctanh}(\\phi)=\\frac{1}{2} \\ln \\left(\\frac{1+\\phi}{1-\\phi}\\right)$, we can find that\n\n$$\n\\begin{aligned}\nv_{f} & =\\tanh \\left(\\frac{1}{2} \\sum_{k=0}^{19} \\ln \\left(\\frac{1+\\frac{k+1}{k+3}}{1-\\frac{k+1}{k+3}}\\right)\\right)=\\tanh \\left(\\frac{1}{2} \\sum_{k=0}^{19} \\ln (k+2)\\right) \\\\\n& =\\tanh (\\ln \\sqrt{2 \\cdot 3 \\cdot 4 \\cdots 21})=\\tanh (\\ln \\sqrt{21 !})\n\\end{aligned}\n$$\n\nAs $\\tanh \\phi=\\left(e^{\\phi}-e^{-\\phi}\\right) /\\left(e^{\\phi}+e^{-\\phi}\\right)$, then\n\n$$\n\\tanh (\\ln (\\phi))=\\frac{\\phi-\\frac{1}{\\phi}}{\\phi+\\frac{1}{\\phi}}=1-\\frac{2}{\\phi^{2}+1} \\Longrightarrow v_{f}=1-\\frac{2}{21 !+1}\n$$\n\nThis implies 19 zeros, but you can also use Stirlings approximation to further approximate the factorial.\n\nAlternate 1: Define $u_{k}$ as Spaceship 0's velocity in frame $R_{k}$, given $c=1$. The recurrence relation from the relativistic velocity addition formula is: $u_{k+1}=\\frac{u_{k}(k+3)+(k+1)}{u_{k}(k+1)+(k+3)}$, starting with $u_{1}=\\frac{1}{3}$. The relation can be simplified as: $u_{k+1}=1+\\frac{2\\left(u_{k}-1\\right)}{u_{k}(k+1)+(k+3)}$. Introducing $v_{k}=u_{k}-1$, $v_{k+1}=v_{k} \\frac{2}{v_{k}(k+1)+(2 k+4)}$. Further simplifying with $w_{k}=\\frac{1}{v_{k}}$, we get $w_{k+1}=w_{k}(k+2)+\\frac{k+1}{2}$. By setting $x_{k}=w_{k}+c$, we find $c=-\\frac{1}{2}$ simplifies to $x_{k+1}=x_{k}(k+2)$. This gives $x_{k}=$ $(k+1)(k)(k-1) \\ldots(4)(3) x_{1}$ and using the initial condition $x_{1}=\\frac{1}{u_{1}-1}+\\frac{1}{2}=1$, we obtain $x_{k}=\\frac{(k+1) !}{2}$. Consequently, $u_{k}=\\frac{(k+1) !-1}{(k+1) !+1}$ and substituting $k=20, u_{20} \\approx 1-3.9 \\times 10^{-20}$, yielding 19 significant digits.\n\nAlternate 2: Let $l=k+2$, then $u_{l}=\\frac{l-1}{l+1}$. Let $u_{m}=\\frac{m-1}{m+1}$. Then you can find that the velocity addition of any $l, m$ will be $\\frac{m l-1}{m l+1}$. Using this identity, we can use recurrence to find that $u_{k}=\\frac{(k+1) !-1}{(k+1) !+1}$.""]",['19'],False,,Numerical,0 910,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The path of an asteroid that comes close to the Earth can be modeled as follows: neglect gravitational effects due to other bodies, and assume the asteroid comes in from far away with some speed $v$ and lever arm distance $r$ to Earth's center. On January 26, 2023, a small asteroid called 2023 BU came to a close distance of $3541 \mathrm{~km}$ to Earth's surface with a speed of $9300 \mathrm{~m} / \mathrm{s}$. Although BU had a very small mass estimated to be about $300,000 \mathrm{~kg}$, if it was much more massive, it could have hit the Earth. How massive would BU have had to have been to make contact with the Earth? Express your answer in scientific notation with 3 significant digits. Use $6357 \mathrm{~km}$ as the radius of the Earth. The parameters that remain constant when the asteroid mass changes are $v$ and $r$, where $v$ is the speed at infinity and $r$ is the impact parameter.","[""Let $v_{1}=9300 \\mathrm{~m} / \\mathrm{s}, d=3541 \\mathrm{~km}$, and $m=300,000 \\mathrm{~kg}$, and let $M$ and $R$ be the Earth's mass and radius.\n\nFirst we find $v$ and $r$. We use the reference frame of the Earth, where the asteroid has reduced mass $\\mu=\\frac{M m}{M+m}$ and the Earth has mass $M+m$. Then by energy and angular momentum conservation, we have\n\n$$\n\\mu v r=\\mu v_{1}(R+d)\n$$\n\n\n\nand\n\n$$\n\\frac{1}{2} \\mu v^{2}=\\frac{1}{2} \\mu v_{1}^{2}-\\frac{G M m}{R+d}\n$$\n\nWe solve for\n\n$$\nv=\\sqrt{2 G(M+m) \\cdot \\frac{R+d}{r^{2}-(R+d)^{2}}}\n$$\n\nso\n\n$$\nv_{1}=\\sqrt{\\frac{2 G(M+m)}{R+d} \\cdot \\frac{r^{2}}{r^{2}-(R+d)^{2}}},\n$$\n\nand we compute $r=37047 \\mathrm{~km}$ and $v=2490 \\mathrm{~m} / \\mathrm{s}$.\n\nNow we consider when the asteroid is massive enough to touch the Earth. We let $m^{\\prime}$ and $\\mu^{\\prime}$ be the mass of the asteroid and its reduced mass, and using a similar method to above, we arrive at\n\n$$\nv=\\sqrt{2 G\\left(M+m^{\\prime}\\right) \\cdot \\frac{R}{r^{2}-R^{2}}}\n$$\n\nso we can solve for $m^{\\prime}=3.74 \\times 10^{24} \\mathrm{~kg}$.""]",['$3.74 \\times 10^{24}$'],False,kg,Numerical,1e23 911,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Pegasi and Betelgeuse are two star systems that can undergo a supernova. Betelgeuse is 548 light-years away from Earth and IK Pegasi is 154 light-years away from Earth. Assume that the two star systems are 500 light-years away from each other. Astronomers on Earth observe that the two star systems undergo a supernova explosion 300 years apart. A spaceship, the OPhO Galaxia Explorer which left Earth in an unknown direction before the first supernova observes both explosions occur simultaneously. Assume that this spaceship travels in a straight line at a constant speed $v$. How far are the two star systems according to the OPhO Galaxia Explorer at the moment of the simultaneous supernovae? Answer in light-years. Note: Like standard relativity problems, we are assuming intelligent observers that know the finite speed of light and correct for it.","[""For any inertial observer, define the 4-distance between two events as $\\Delta s^{\\mu}=(c \\Delta t, \\Delta \\mathbf{x})$, where $\\Delta t$ and $\\Delta \\mathrm{x}$ are the temporal and spatial intervals measured by the observer. By the properties of 4 -vectors, the following quantity is Lorentz-invariant:\n$$\n\\Delta s^{\\mu} \\Delta s_{\\mu}=c^{2} \\Delta t^{2}-\\|\\Delta \\mathbf{x}\\|^{2}\n$$\n\nIn the spaceship's frame, this is equal to $\\Delta s^{\\mu} \\Delta s_{\\mu}=-\\left\\|\\Delta \\mathbf{x}^{\\prime}\\right\\|^{2}$, since the two supernovas are simultaneous; hence\n\n$$\n\\left\\|\\Delta \\mathbf{x}^{\\prime}\\right\\|=\\sqrt{\\|\\Delta \\mathbf{x}\\|^{2}-c^{2} \\Delta t^{2}}\n$$\n\nSince the observers have already taken into account the delay due to the nonzero speed of light, $c \\Delta t=300 \\mathrm{ly},\\|\\Delta \\mathbf{x}\\|=500 \\mathrm{ly}$, and\n\n$$\n\\left\\|\\Delta \\mathbf{x}^{\\prime}\\right\\|=\\sqrt{\\|\\Delta \\mathbf{x}\\|^{2}-c^{2} \\Delta t^{2}}=400 l y\n$$""]",['400'],False,light-years,Numerical,0 912,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$",A ball of mass $1 \mathrm{~kg}$ is thrown vertically upwards and it faces a quadratic drag with a terminal velocity of $20 \mathrm{~m} / \mathrm{s}$. It reaches a maximum height of $30 \mathrm{~m}$ and falls back to the ground. Calculate the energy dissipated until the point of impact (in J).,"[""If we suppose that the magnitude of quadratic drag is $F_{\\mathrm{drag}}=c v^{2}$ for some constant $c$, then when the ball is falling downwards at terminal velocity $v_{t}$ this upwards drag force must cancel gravity to provide zero net force:\n$$\nm g=c v_{t}^{2} \\Longrightarrow c=\\frac{m g}{v_{t}^{2}}\n$$\n\nso we can rewrite our equations of motion in terms of terminal velocity. In the ascending portion of the ball's trajectory, suppose that the ball's speed changes from $v$ to $v+d v$ after traveling an infinitesimal distance from $h$ to $h+d h$. Then:\n\n$d($ kinetic energy $)=-d($ potential energy and energy lost to drag $) \\Longrightarrow m v d v=-\\frac{m g v^{2}}{v_{t}^{2}} d h-m g d h$\n\nwhere we used chain rule and $d W=F \\cdot d x$. This is a separable differential equation for velocity in terms of height (!), which we can solve:\n\n$$\n\\int \\frac{v d v}{1+\\frac{v^{2}}{v_{t}^{2}}}=\\int-g d h+\\text { constant }=\\text { constant }-g h\n$$\n\nWe u-sub the entire denominator $\\left(u=1+v^{2} / v_{t}^{2} \\Longrightarrow d u=2 v / v_{t}^{2} d v\\right)$, which nicely cancels the numerator. Our left-hand integral is\n\n$$\n\\frac{v_{t}^{2}}{2} \\int \\frac{d u}{u}=\\frac{v_{t}^{2}}{2} \\ln \\left|1+\\frac{v^{2}}{v_{t}^{2}}\\right|=\\text { constant }-g h\n$$\n\nNow to find the constant! At our maximum height $h_{0}=30 \\mathrm{~m}$, the speed and left-hand side are zero, so the constant must be $g h_{0}$. Then we can find the initial speed of the projectile by substituting $h=0$, from which:\n\n$$\nv_{\\text {initial }}=v_{t} \\sqrt{e^{\\frac{2 g h_{0}}{v_{t}^{2}}}-1}\n$$\n\nWe can use a similar argument for the downwards trajectory, albeit with drag pointing upwards. We should get that the final speed immediately before impact is\n\n$$\nv_{\\text {final }}=v_{t} \\sqrt{1-e^{\\frac{-2 g h_{0}}{v_{t}^{2}}}}\n$$\n\nfrom which the dissipated energy is the difference in final and initial kinetic energies, or\n\n$$\n\\frac{1}{2} m\\left(v_{\\text {final }}^{2}-v_{\\text {initial }}^{2}\\right) \\approx 515.83 \\text { joules }\n$$""]",['515.83'],False,joules,Numerical,2e-1 913,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In general, we can describe the quadratic drag on an object by the following force law: $$ F_{D}=\frac{1}{2} C_{D} \rho A v^{2} $$ where $A$ is the cross-sectional area of the object exposed to the airflow, $v$ is the speed of the object in a fluid, and $C_{D}$ is the drag coefficient, a dimensionless quantity that varies based on shape. Another useful quantity to know is the Reynold's number, a dimensionless quantity that helps predict fluid flow patterns. It is given by the formula: $$ \operatorname{Re}=\frac{\rho v L}{\mu} $$ where $\rho$ is the density of the surrounding fluid, $\mu$ is the dynamic viscosity of the fluid, and $L$ is a reference length parameter that varies based on each object. For a smooth ${ }^{1}$ sphere traveling in a fluid, its diameter serves as the reference length parameter. A logarithmic graph of $C_{D}$ vs Re of a sphere from the NASA Glenn Research Center. The relationship between the drag coefficient and the Reynold's number holds significant importance. Due to the complexity of fluid dynamics, empirical data is commonly used, as depicted in the figure provided above. Notably, the figure indicates a significant decrease in the drag coefficient around $\operatorname{Re} \approx 4 \times 10^{5}$. This phenomenon, known as the drag crisis, occurs when a sphere transitions from laminar to turbulent flow, resulting in a broad wake and high drag. Let's consider a smooth ball with a radius of $0.2 \mathrm{~m}$ and a mass of $0.1 \mathrm{~kg}$ dropped in air with a constant density of $\rho=1.255 \mathrm{~kg} / \mathrm{m}^{3}$. It is found that at velocity $5 \mathrm{~m} / \mathrm{s}$, the Reynold's number of the ball is $3.41 \cdot 10^{5}$. If the ball is dropped from rest, it approaches a stable terminal velocity $v_{1}$. If the ball is thrown downwards with enough velocity, it will experience turbulence, and approach a stable terminal velocity $v_{2}$. Find $\Delta v=v_{2}-v_{1}$. Ignore any terminal velocities found for Reynold numbers less than an order of magnitude $10^{-1}$. Note: This problem is highly idealized as it assumes the atmosphere has air of constant density and temperature. In reality, this is not true!","['Terminal velocity exists when the net force is 0 . Using $v=\\frac{\\mu \\cdot \\operatorname{Re}}{2 \\rho r}$ where $L=2 r$, we find that\n$$\n\\frac{1}{2} \\rho_{a} C_{D}\\left(\\pi r^{2}\\right)\\left(\\frac{\\mu \\cdot \\mathrm{Re}}{2 \\rho r}\\right)^{2}=m g-\\rho_{a} g\\left(\\frac{4}{3} \\pi r^{3}\\right)\n$$\n\n\nSince $\\rho=\\frac{m}{4 \\pi r^{3} / 3}=2.98 \\mathrm{~kg} / \\mathrm{m}^{3}$ is on the same order as $\\rho_{a}=1.255 \\mathrm{~kg} / \\mathrm{m}^{3}$, the buoyant force must be accounted for and is non-negligible. We can rearrange to find that\n$$\nC_{D} \\operatorname{Re}^{2}=\\frac{8 \\rho_{a}}{\\pi \\mu^{2}}\\left(m g-\\frac{4}{3} \\rho_{a} g \\pi r^{3}\\right)\n$$\nUsing $x$ as $C_{D}$ and $y$ as Re, we can plot an equation $x y^{2}=$ const on the $C_{D}$ vs Re graph. There, we can find three intersections.\n\n\n\nThe intersection in the middle is not stable. So we find the intersections of the other two to be $\\operatorname{Re}_{1} \\approx 2.6 \\times 10^{5}$ and $\\operatorname{Re}_{2}=6 \\times 10^{5}$. Hence, $v_{1}=3.81 \\mathrm{~m} / \\mathrm{s}$ and $v_{2}=8.79 \\mathrm{~m} / \\mathrm{s}$, meaning $\\Delta v=4.98 \\mathrm{~m} / \\mathrm{s}$.']",['$4.98$'],False,m/s,Numerical,1e-1 913,Modern Physics,,"In general, we can describe the quadratic drag on an object by the following force law: $$ F_{D}=\frac{1}{2} C_{D} \rho A v^{2} $$ where $A$ is the cross-sectional area of the object exposed to the airflow, $v$ is the speed of the object in a fluid, and $C_{D}$ is the drag coefficient, a dimensionless quantity that varies based on shape. Another useful quantity to know is the Reynold's number, a dimensionless quantity that helps predict fluid flow patterns. It is given by the formula: $$ \operatorname{Re}=\frac{\rho v L}{\mu} $$ where $\rho$ is the density of the surrounding fluid, $\mu$ is the dynamic viscosity of the fluid, and $L$ is a reference length parameter that varies based on each object. For a smooth ${ }^{1}$ sphere traveling in a fluid, its diameter serves as the reference length parameter. ![](https://cdn.mathpix.com/cropped/2023_12_21_a44161aecfd4a6e63235g-1.jpg?height=574&width=894&top_left_y=556&top_left_x=599) A logarithmic graph of $C_{D}$ vs Re of a sphere from the NASA Glenn Research Center. The relationship between the drag coefficient and the Reynold's number holds significant importance. Due to the complexity of fluid dynamics, empirical data is commonly used, as depicted in the figure provided above. Notably, the figure indicates a significant decrease in the drag coefficient around $\operatorname{Re} \approx 4 \times 10^{5}$. This phenomenon, known as the drag crisis, occurs when a sphere transitions from laminar to turbulent flow, resulting in a broad wake and high drag. Let's consider a smooth ball with a radius of $0.2 \mathrm{~m}$ and a mass of $0.1 \mathrm{~kg}$ dropped in air with a constant density of $\rho=1.255 \mathrm{~kg} / \mathrm{m}^{3}$. It is found that at velocity $5 \mathrm{~m} / \mathrm{s}$, the Reynold's number of the ball is $3.41 \cdot 10^{5}$. If the ball is dropped from rest, it approaches a stable terminal velocity $v_{1}$. If the ball is thrown downwards with enough velocity, it will experience turbulence, and approach a stable terminal velocity $v_{2}$. Find $\Delta v=v_{2}-v_{1}$. Ignore any terminal velocities found for Reynold numbers less than an order of magnitude $10^{-1}$. Note: This problem is highly idealized as it assumes the atmosphere has air of constant density and temperature. In reality, this is not true!","['Terminal velocity exists when the net force is 0 . Using $v=\\frac{\\mu \\cdot \\operatorname{Re}}{2 \\rho r}$ where $L=2 r$, we find that\n$$\n\\frac{1}{2} \\rho_{a} C_{D}\\left(\\pi r^{2}\\right)\\left(\\frac{\\mu \\cdot \\mathrm{Re}}{2 \\rho r}\\right)^{2}=m g-\\rho_{a} g\\left(\\frac{4}{3} \\pi r^{3}\\right)\n$$\n\n\nSince $\\rho=\\frac{m}{4 \\pi r^{3} / 3}=2.98 \\mathrm{~kg} / \\mathrm{m}^{3}$ is on the same order as $\\rho_{a}=1.255 \\mathrm{~kg} / \\mathrm{m}^{3}$, the buoyant force must be accounted for and is non-negligible. We can rearrange to find that\n$$\nC_{D} \\operatorname{Re}^{2}=\\frac{8 \\rho_{a}}{\\pi \\mu^{2}}\\left(m g-\\frac{4}{3} \\rho_{a} g \\pi r^{3}\\right)\n$$\nUsing $x$ as $C_{D}$ and $y$ as Re, we can plot an equation $x y^{2}=$ const on the $C_{D}$ vs Re graph. There, we can find three intersections.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_aa674442fff889231bc7g-1.jpg?height=721&width=827&top_left_y=596&top_left_x=649)\n\nThe intersection in the middle is not stable. So we find the intersections of the other two to be $\\operatorname{Re}_{1} \\approx 2.6 \\times 10^{5}$ and $\\operatorname{Re}_{2}=6 \\times 10^{5}$. Hence, $v_{1}=3.81 \\mathrm{~m} / \\mathrm{s}$ and $v_{2}=8.79 \\mathrm{~m} / \\mathrm{s}$, meaning $\\Delta v=4.98 \\mathrm{~m} / \\mathrm{s}$.']",['$4.98$'],False,m/s,Numerical,1e-1 914,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The following information applies for the next two problems. Pictured is a wheel from a 4wheeled car of weight $1200 \mathrm{~kg}$. The absolute pressure inside the tire is $3.0 \times 10^{5} \mathrm{~Pa}$. Atmospheric pressure is $1.0 \times 10^{5} \mathrm{~Pa}$. Assume the rubber has negligible ""stiffness"" (i.e. a negligibly low Sheer modulus compared to its Young's modulus). The rubber on the bottom of the wheel is completely unstretched. The rubber has a thickness of $7 \mathrm{~mm}$. Based on this information, find the Young's Modulus of the rubber. Is this answer reasonable?","[""Let $P=2.0 \\times 10^{5}$ Pa be the gauge pressure of the tire, $w=20 \\mathrm{~cm}$ be the width of the tire, $r=20 \\mathrm{~cm}$ be the radius of the tire, $t=7 \\mathrm{~mm}$ be the thickness of the rubber, $M=1200 \\mathrm{~kg}$ be the mass of the car, $Y$ be the Young's modulus, and $\\theta$ be the angle subtended by the portion of the tire in contact with the ground. We have\n$$\n\\begin{gathered}\n\\frac{M g}{4}=2 r w P \\sin \\left(\\frac{\\theta}{2}\\right) \\\\\n\\theta=2 \\arcsin \\left(\\frac{M g}{8 R w P}\\right)=0.3696\n\\end{gathered}\n$$\nLet $T$ be the tension in the stretched portion of the rubber. Balance forces on a small section of the tire subtending an angle $d \\theta$.\n$$\n\\operatorname{Pr} w d \\theta=2 T \\sin \\left(\\frac{d \\theta}{2}\\right)\n$$\nUsing the approximation $\\sin \\theta=\\theta$,\n$$\nT=\\operatorname{Pr} w\n$$\nThe stress on the rubber is\n$$\n\\sigma=\\frac{T}{w t}=\\frac{P r}{t}\n$$\nThe strain is\n$$\n\\epsilon=\\frac{\\theta}{2 \\sin \\left(\\frac{\\theta}{2}\\right)}-1\n$$\nThe Young's modulus is\n$$\nY=\\frac{\\sigma}{\\epsilon}=\\frac{\\frac{P r}{t}}{\\frac{b}{2 \\sin \\left(\\frac{b}{2}\\right)}-1}=1.00 \\times 10^{9} \\mathrm{~Pa}\n$$""]",['$1.00 \\times 10^{9}$'],False,Pa,Numerical,1e8 914,Thermodynamics,,"The following information applies for the next two problems. Pictured is a wheel from a 4wheeled car of weight $1200 \mathrm{~kg}$. The absolute pressure inside the tire is $3.0 \times 10^{5} \mathrm{~Pa}$. Atmospheric pressure is $1.0 \times 10^{5} \mathrm{~Pa}$. Assume the rubber has negligible ""stiffness"" (i.e. a negligibly low Sheer modulus compared to its Young's modulus). ![](https://cdn.mathpix.com/cropped/2023_12_21_aa674442fff889231bc7g-1.jpg?height=436&width=653&top_left_y=1820&top_left_x=736) The rubber on the bottom of the wheel is completely unstretched. The rubber has a thickness of $7 \mathrm{~mm}$. Based on this information, find the Young's Modulus of the rubber. Is this answer reasonable?","[""Let $P=2.0 \\times 10^{5}$ Pa be the gauge pressure of the tire, $w=20 \\mathrm{~cm}$ be the width of the tire, $r=20 \\mathrm{~cm}$ be the radius of the tire, $t=7 \\mathrm{~mm}$ be the thickness of the rubber, $M=1200 \\mathrm{~kg}$ be the mass of the car, $Y$ be the Young's modulus, and $\\theta$ be the angle subtended by the portion of the tire in contact with the ground. We have\n$$\n\\begin{gathered}\n\\frac{M g}{4}=2 r w P \\sin \\left(\\frac{\\theta}{2}\\right) \\\\\n\\theta=2 \\arcsin \\left(\\frac{M g}{8 R w P}\\right)=0.3696\n\\end{gathered}\n$$\nLet $T$ be the tension in the stretched portion of the rubber. Balance forces on a small section of the tire subtending an angle $d \\theta$.\n$$\n\\operatorname{Pr} w d \\theta=2 T \\sin \\left(\\frac{d \\theta}{2}\\right)\n$$\nUsing the approximation $\\sin \\theta=\\theta$,\n$$\nT=\\operatorname{Pr} w\n$$\nThe stress on the rubber is\n$$\n\\sigma=\\frac{T}{w t}=\\frac{P r}{t}\n$$\nThe strain is\n$$\n\\epsilon=\\frac{\\theta}{2 \\sin \\left(\\frac{\\theta}{2}\\right)}-1\n$$\nThe Young's modulus is\n$$\nY=\\frac{\\sigma}{\\epsilon}=\\frac{\\frac{P r}{t}}{\\frac{b}{2 \\sin \\left(\\frac{b}{2}\\right)}-1}=1.00 \\times 10^{9} \\mathrm{~Pa}\n$$""]",['$1.00 \\times 10^{9}$'],False,Pa,Numerical,1e8 915,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two parallel square plates of side length $1 \mathrm{~m}$ are placed a distance $30 \mathrm{~cm}$ apart whose centers are at $(-15 \mathrm{~cm}, 0,0)$ and $(15 \mathrm{~cm}, 0,0)$ have uniform charge densities $-10^{-6} \mathrm{C} / \mathrm{m}^{2}$ and $10^{-6} \mathrm{C} / \mathrm{m}^{2}$ respectively. Find the magnitude of the component of the electric field perpendicular to axis passing through the centers of the two plates at $(10 \mathrm{~cm}, 1 \mathrm{~mm}, 0)$.","['By symmetry, the electric field due to the portion of the plates between $y=50 \\mathrm{~cm}$ and $y=-49.8 \\mathrm{~cm}$ will cancel out. We only need to consider the electric field from two $2 \\mathrm{~mm}$ wide strips of charge, which are small enough to be approximated as wires with charge per unit length $\\lambda=\\sigma w= \\pm 2 \\times 10^{-9} \\mathrm{C} / \\mathrm{m}^{2}$ at a distance $y=50 \\mathrm{~cm}$ away. The y-component of the electric field from these wires is then\n$$\n\\begin{aligned}\n& E_{y}=\\frac{0.5 \\lambda}{4 \\pi \\epsilon_{0}} \\int_{-0.5}^{0.5}\\left(\\frac{1}{\\left(z^{2}+0.5^{2}+0.05^{2}\\right)^{\\frac{3}{2}}}-\\frac{1}{\\left(z^{2}+0.5^{2}+0.25^{2}\\right)^{\\frac{3}{2}}}\\right) \\mathrm{d} z \\\\\n& E_{y}=\\frac{0.5 \\lambda}{4 \\pi \\epsilon_{0}}\\left(\\frac{z}{\\left(0.5^{2}+0.05^{2}\\right) \\sqrt{z^{2}+0.5^{2}+0.05^{2}}}-\\frac{z}{\\left(0.5^{2}+0.25^{2}\\right) \\sqrt{z^{2}+0.5^{2}+0.25^{2}}}\\right) \\\\\n& E_{y}=\\frac{0.5 \\lambda}{4 \\pi \\epsilon_{0}}\\left(\\frac{1}{\\left(0.5^{2}+0.05^{2}\\right) \\sqrt{0.5^{2}+0.5^{2}+0.05^{2}}}-\\frac{1}{\\left(0.5^{2}+0.25^{2}\\right) \\sqrt{0.5^{2}+0.5^{2}+0.25^{2}}}\\right)=11.9 \\mathrm{~N} / \\mathrm{C}\n\\end{aligned}\n$$']",['$11.9$'],False,N/C,Numerical,1e-1 916,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A hollow sphere of mass $M$ and radius $R$ is placed under a plank of mass $3 M$ and length $2 R$. The plank is hinged to the floor, and it initially makes an angle $\theta=\frac{\pi}{3}$ rad to the horizontal. Under the weight of the plank, the sphere starts rolling without slipping across the floor. What is the sphere's initial translational acceleration? Assume the plank is frictionless. A not-to-scale picture of the sphere-plank setup.",['Let $N$ be the normal force acting from the ball on the plank. We can use torque at a distance $x=R \\cot \\frac{\\theta}{2}$ from the hinge (by geometry) to write (using $m=3 M$ )\n$$\nm g \\frac{l}{2} \\cos \\theta-N R \\cot \\frac{\\theta}{2}=\\frac{m l^{2}}{3} \\alpha\n$$\nYou can also find $N$ from\n$$\n\\begin{aligned}\nN \\sin \\theta-f & =M a_{0} \\\\\nf R & =\\frac{2}{3} M R^{2} \\alpha \\\\\na_{0} & =R \\alpha\n\\end{aligned}\n$$\nYou can relate the acceleration at the point of contact $a_{p}=\\vec{a}_{0}+\\vec{a}_{0} \\times \\vec{r}=\\alpha_{0}(1-\\cos \\theta)$. Hence combining all equations gives $a=\\frac{9}{32} g \\approx 2.76 \\mathrm{~m} / \\mathrm{s}^{2}$.'],['$2.76$'],False,$\mathrm{~m} / \mathrm{s}^{2}$,Numerical,5e-2 916,Mechanics,,"A hollow sphere of mass $M$ and radius $R$ is placed under a plank of mass $3 M$ and length $2 R$. The plank is hinged to the floor, and it initially makes an angle $\theta=\frac{\pi}{3}$ rad to the horizontal. Under the weight of the plank, the sphere starts rolling without slipping across the floor. What is the sphere's initial translational acceleration? Assume the plank is frictionless. ![](https://cdn.mathpix.com/cropped/2023_12_21_7fcf501c85b9e74489f8g-1.jpg?height=263&width=616&top_left_y=188&top_left_x=749) A not-to-scale picture of the sphere-plank setup.",['Let $N$ be the normal force acting from the ball on the plank. We can use torque at a distance $x=R \\cot \\frac{\\theta}{2}$ from the hinge (by geometry) to write (using $m=3 M$ )\n$$\nm g \\frac{l}{2} \\cos \\theta-N R \\cot \\frac{\\theta}{2}=\\frac{m l^{2}}{3} \\alpha\n$$\nYou can also find $N$ from\n$$\n\\begin{aligned}\nN \\sin \\theta-f & =M a_{0} \\\\\nf R & =\\frac{2}{3} M R^{2} \\alpha \\\\\na_{0} & =R \\alpha\n\\end{aligned}\n$$\nYou can relate the acceleration at the point of contact $a_{p}=\\vec{a}_{0}+\\vec{a}_{0} \\times \\vec{r}=\\alpha_{0}(1-\\cos \\theta)$. Hence combining all equations gives $a=\\frac{9}{32} g \\approx 2.76 \\mathrm{~m} / \\mathrm{s}^{2}$.'],['$2.76$'],False,$\mathrm{~m} / \mathrm{s}^{2}$,Numerical,5e-2 917,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A space elevator consists of a heavy counterweight placed near geostationary orbit, a thread that connects it to the ground (assume this is massless), and elevators that run on the threads (also massless). The mass of the counterweight is $10^{7} \mathrm{~kg}$. Mass is continuously delivered to the counterweight at a rate of $0.001 \mathrm{~kg} / \mathrm{s}$. The elevators move upwards at a rate of $20 \mathrm{~m} / \mathrm{s}$. Assume there are many elevators, so their discreteness can be neglected. The elevators are massless. The counterweight orbits the Earth. Find the minimum possible displacement radially of the counterweight. Specify the sign.","['As the orbit is geostationary, we can balance forces to write that\n$$\n\\frac{G M_{E}}{R_{\\mathrm{GS}}^{2}}=\\omega^{2} R_{\\mathrm{GS}} \\Longrightarrow R_{\\mathrm{GS}}=\\left(\\frac{G M_{E} T^{2}}{4 \\pi}\\right)^{1 / 3}\n$$\nThe total mass of masses $=1790 \\mathrm{~kg} \\ll 10^{6} \\mathrm{~kg}$ which means the displacements are small. The total gravity of masses is\n$$\n\\int_{R_{E}}^{R_{E s}} \\frac{G M_{E}}{x^{2}}(\\lambda d x)=\\operatorname{GME} \\lambda\\left(\\frac{1}{R_{E}}-\\frac{1}{R_{\\cos }}\\right)=2650 \\mathrm{~N}\n$$\nwhere $\\lambda=5 \\cdot 10^{-5} \\mathrm{~kg} / \\mathrm{m}$. The total centrifugal of masses is\n$$\n\\int_{R_{E}}^{R_{E}} \\frac{4 \\pi^{2}}{T^{2}} \\times(\\lambda d x)=\\frac{2 \\pi^{2}}{T^{2}} \\lambda\\left(R_{E s}^{2}-R_{E}^{2}\\right)=230 \\mathrm{~V}\n$$\nHence, the total force is $2420 \\mathrm{~N}$. The outwards force is just\n$$\nF_{\\text {out }}=\\omega^{2} a M-\\frac{\\mathrm{d}}{\\mathrm{d} x}\\left(\\frac{G M_{E}}{x^{2}}\\right) a M\n$$\nwhere $a$ is the horizontal displacement which is much less than $R_{E}, R_{\\mathrm{GS}}$. Hence, this can be rewritten as\n$$\nF_{\\text {out }}=\\left(\\frac{4 \\pi^{2}}{T^{2}}-2 \\frac{G M_{E}}{R_{\\mathrm{GS}}^{2}}\\right) a M=2420 \\mathrm{~N}\n$$\nThis implies that $a=15.21 \\mathrm{~km}$.']",['$15.21$'],False,km,Numerical,1e-2 918,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a spherical shell of thickness $\delta=0.5 \mathrm{~cm}$ and radius $R=5 \mathrm{~cm}$ made of an Ohmic material with resistivity $\rho=10^{-7} \Omega \mathrm{m}$. A spherical laser source with a tuned frequency of $f_{0}$ and intensity $I_{0}=10^{5} \mathrm{~W} / \mathrm{m}^{2}$ is placed at the center of the shell and is turned on. Working in the limit $\delta \ll \frac{c}{f_{0}} \ll R$, approximate the initial average power dissipated by the shell. Neglect inductance.","[""Note that the proper treatment of this problem requires more advanced electromagnetism. The following might seem sketchy, but this will do. We first consider the case of a thin plane made of an ideal conductor placed at $z=0$. A monochromatic electromagnetic wave with an $E$ field: $\\vec{E}_{i}(t, z)=E_{0} \\cos (k z-\\omega t) \\hat{x}$ is incident on the plane. The corresponding $B$ field is of course $\\vec{B}_{i}(t, z)=\\frac{E_{0}}{c} \\cos (k z-\\omega t) \\hat{y}$. We expect a current density $\\hat{K}(t)=K(t) \\hat{x}$ to form in the direction of the $E$ field, as the electric field accelerates the charges on the sheet $-K(t)$ will oscillate with the $E$ field. We know that perfect conductors completely reflect electromagnetic waves, so we expect no EM field in the $z>0$ region. The total EM field is a combination of the incident wave and the wave generated by the oscillating surface charges, so we are to believe that the latter provides just the right EM wave to cancel out $\\vec{B}_{i}$ and $\\vec{E}_{i}$ in the $z>0$ region. Consider an instance where the EM waves have an orientation shown in the figure below. It's easy to verify that the magnetic field generated by a uniform sheet of current pointing out of the page is given as shown. Since the electric wave generated by $\\vec{K}(t)$ must cancel the $\\vec{E}_{i}$ pointing out of page in $z>0$, we get a poynting vector (for the wave generated by $\\vec{K}$ ) in the $\\hat{z}$ direction in $z>0$. By symmetry, this means that the poynting vector points in the $-\\hat{z}$ direction in $z<0$. Such simple symmetry arguments fully determine the wave generated by an oscillating sheet of charge: $\\left|\\vec{E}_{k}(t, z)\\right|=\\left|\\vec{E}_{i}(t, z)\\right|$, $\\left|\\vec{B}_{k}(t, z)\\right|=\\left|\\vec{B}_{i}(t, z)\\right|$ with the orientations shown below.\n\n\nWe therefore get that the total magnetic field in the $z<0$ region is $\\vec{B}=2 \\vec{B}_{i}(t, z)$. Let us now take an Ampere loop as shown in the figure below:\n\n\n\n\n\nWe have:\n$$\nT_{\\epsilon} \\equiv \\oint_{\\Gamma_{\\epsilon}} \\vec{B}(t,-\\epsilon / 2) \\cdot d \\vec{l}=\\mu_{0} I_{e n c}+\\frac{1}{c^{2}} \\frac{\\partial}{\\partial t} \\int_{S_{\\epsilon}} \\vec{E}(t, z) \\cdot d \\vec{A}\n$$\nwhere $S_{\\epsilon}$ is the surface bounded by $\\Gamma_{\\epsilon}$. Since $\\vec{E}=0$ everywhere and $I_{e n c}=K(t) l$,\n$$\n\\lim _{\\epsilon \\rightarrow 0} T_{\\epsilon}=\\frac{2 E_{0} l}{c} \\cos (\\omega t)=\\mu_{0} K(t) l\n$$\nHence, we have $\\vec{K}(t)=\\frac{2 E_{0}}{\\mu_{0} c} \\cos (\\omega t) \\hat{x}$. The key find in this long introduction is that an oscillating sheet of current $\\vec{K}(t)=K_{0} \\cos (\\omega t) \\hat{x}$ generates an EM wave with an electric field $\\vec{E}_{K}(t, 0)=-\\frac{\\mu_{0} c}{2} \\vec{K}(t)$ near the sheet (at $z=0$ ). We finally return to the original problem.\n\nAll laser beams have poynting vectors pointing radially outwards, so the $E_{i}$ field must point in the tangential direction to the sphere at all normal incidence. Say the $E_{i}$ field points in the $\\hat{\\theta}$ direction, so that the current in the shell flows from the north pole to the south pole. The condition $\\delta \\ll \\frac{c}{f_{0}} \\ll R$ allows us to neglect attenuation due to skin effects - essentially, the electric field, thus the surface current, is approximately uniform in the shell. The condition $\\frac{c}{f_{0}} \\ll R$ allows us to treat the incidence of the laser beams as an EM wave hitting an Ohmic plane of thickness $\\delta$, so we consider that problem first. Denote the total electric field inside the the plane at time $t$ as $\\vec{E}(t)$. The current density is given by Ohm's law: $\\vec{J}(t)=\\vec{E}(t) / \\rho$. The corresponding surface current is of course $\\vec{K}(t)=\\delta \\vec{E} / \\rho(t)$. Carefully note the distinction between $\\vec{E}$ and $\\vec{E}_{i}$. Since the source wave $\\vec{E}_{i}(t)$ is sinusoidal, we expect $\\vec{E}(t)$ to be sinusoidal as well so that $\\vec{E}_{K}(t)=-\\frac{\\mu_{0} c}{2} \\vec{K}(t)$ applies. The total electric field inside the sheet is a sum of $\\vec{E}_{K}$ and $\\vec{E}_{i}$, thus:\n$$\n\\vec{E}(t)=\\vec{E}_{i}(t)-\\frac{\\mu_{0} c \\delta}{2 \\rho} \\vec{E}(t)\n$$\nRearranging for $\\vec{E}(t)$, we get:\n$$\n\\vec{E}(t)=\\frac{\\vec{E}_{i}(t)}{1+\\mu_{0} c \\delta / 2 \\rho}\n$$\nOn the sphere, $\\vec{E}(t)$ points in the $\\hat{\\theta}$ direction and has the same magnitude everywhere on the sphere. The relevant cross-section will be the lateral surface of a truncated cone, as shown below:\n\n\n\n\n\nThe current through this surface is therefore:\n$$\nI(\\theta, t)=\\frac{\\pi \\delta(2 R+\\delta)\\left|\\vec{E}_{i}(t)\\right|}{\\rho+\\mu_{0} c \\delta / 2} \\sin \\theta\n$$\nThe power dissipated by a volume generated by $\\theta \\sim \\theta+d \\theta$ is\n$$\nd P(t)=I^{2}(\\theta, t) \\frac{\\rho(R+\\delta / 2) d \\theta}{\\pi \\delta(2 R+\\delta) \\sin \\theta}\n$$\nWe integrate this expression from $\\theta=0$ to $\\theta=\\pi$ :\n$$\nP(t)=\\frac{\\pi \\delta \\rho(2 R+\\delta)^{2}}{\\left(\\rho+\\mu_{0} c \\delta / 2\\right)^{2}}\\left|\\vec{E}_{i}(t)\\right|^{2}\n$$\nNow, the incident electric field has the form $\\left|\\vec{E}_{i}(t)\\right|^{2}=\\frac{2 I_{0}}{\\epsilon_{0} c} \\cos ^{2}\\left(2 \\pi f_{0} t\\right)$. The average of this function from $t=0$ to $t=\\frac{1}{f_{0}}$ is $1 / 2$, so our final answer is:\n$$\n\\bar{P}=\\frac{\\pi \\delta \\rho(2 R+\\delta)^{2}}{\\left(\\rho+\\mu_{0} c \\delta / 2\\right)^{2}} \\frac{I_{0}}{\\epsilon_{0} c}\n$$\nwhich turns out to be $2.39078 \\times 10^{-15} \\mathrm{~W}$""]",['$2.39078 \\times 10^{-15}$'],False,W,Numerical,1e-17 919,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A stable star of radius $R$ has a mass density profile $\rho(r)=\alpha(1-r / R)$. Here, ""stable"" means that the star doesn't collapse under its own gravity. If the internal pressure at the core is provided solely by the radiation of photons, calculate the temperature at the core. Assume the star is a perfect black body and treat photons as a classical ideal gas. Use $R=7 \times 10^{5} \mathrm{~km}$ and $\alpha=3 \mathrm{~g} / \mathrm{cm}^{3}$. Round your answer to the nearest kilokelvin. We treat photons as a classical gas here to neglect any relativistic effects.","[""For a star that doesn't collapse in on itself, there must be some source of pressure $p(r)$ that balances out the pressure due to gravity. Define the function:\n$$\nm(r)=\\int_{0}^{r} 4 \\pi r^{2} \\rho(r) d r\n$$\ninside the star. The gravitational pressure on an infinitesimal shell of radius $r$ and thickness $d r$ is given by:\n$$\nd p_{g}=-\\frac{G m(r) \\rho(r) 4 \\pi r^{2} d r}{r^{2} 4 \\pi r^{2}}=-\\frac{G m(r) \\rho(r)}{r^{2}} d r\n$$\nHence, the pressure source must provide a pressure gradient $\\frac{d p}{d r}$ given by:\n$$\n\\frac{d p}{d r}=\\frac{G m(r) \\rho(r)}{r^{2}}\n$$\nIf $\\rho(r)=\\alpha(1-r / R)$, then $m(r)=\\frac{1}{3} \\pi \\alpha r^{3}(3 r / R-4)$. We need the pressure gradient:\n$$\n\\frac{d p}{d r}=\\frac{1}{3} \\pi G \\alpha^{2} r(3 r / R-4)(r / R-1)\n$$\nWe integrate this equation from $r=0$ to $R$ with the obvious boundary condition $p(R)=0$. We find that $p(0)=\\frac{5}{36} \\pi G \\alpha^{2} R^{2}$. From the Stefan-Boltzmann law, along with elementary kinetic theory, the pressure due to local radiation is given by $\\frac{4 \\sigma}{3 c} T_{c}^{4}$. Plugging in values gives $T_{c}=26718 \\mathrm{kK}$""]",['$26718$'],False,kK,Numerical,1e2 920,Mechanics,,"On a flat playground, choose a Cartesian Oxy coordinate system (in unit of meters). A child running at a constant velocity $V=1 \mathrm{~m} / \mathrm{s}$ around a heart-shaped path satisfies the following order- 6 algebraic equation: $$ \left(x^{2}+y^{2}-L^{2}\right)^{3}-L x^{2} y^{3}=0, L=10 \text {. } $$ When the child is at the position $(x, y)=(L, 0)$, what is the magnitude of their acceleration? ![](https://cdn.mathpix.com/cropped/2023_12_21_23e8f9f667608506c947g-1.jpg?height=688&width=1282&top_left_y=526&top_left_x=408)","['The acceleration can be found from the local geometry of the curves, thus let us study small deviations around the position of interests $(x, y)=(L, 0)$ :\n$$\nx=L+\\delta_{x} \\quad, \\quad y=0+\\delta_{y} \\quad, \\quad\\left|\\delta_{x}\\right|,\\left|\\delta_{y}\\right| \\ll L\n$$\nConsider the 2nd-order approximation in $\\delta_{x}$ of $\\delta_{y}$ with quadratic coefficients $\\alpha$ and $\\beta$ :\n$$\n\\delta_{y} \\approx \\alpha \\delta_{x}+\\frac{\\beta}{L} \\delta_{x}^{2} \\sim \\delta_{x}\n$$\nTo find these coefficients, we look at the algebraic equation of our heart-shape path up to the two lowest-orders of expansions (which are the 3rd and 4th):\n$$\n\\begin{aligned}\n0 & =\\left(x^{2}+y^{2}-L^{2}\\right)^{3}-L x^{2} y^{3} \\approx L^{2}\\left[8 L \\delta_{x}^{3}+12 \\delta_{x}^{4}+12 \\delta_{x}^{2} \\delta_{y}^{2}-2 \\delta_{x} \\delta_{y}^{3}-L \\delta_{y}^{3}+\\mathcal{O}\\left(\\delta_{x}^{5}\\right)\\right] \\\\\n& \\approx L^{2}\\left[8 L \\delta_{x}^{3}+12 \\delta_{x}^{4}+12 \\alpha^{2} \\delta_{x}^{4}-2 \\alpha^{3} \\delta_{x}^{4}-\\left(\\alpha^{3} L \\delta_{x}^{3}+3 \\alpha^{2} \\beta \\delta_{x}^{4}\\right)+\\mathcal{O}\\left(\\delta_{x}^{5}\\right)\\right] \\\\\n& \\propto\\left(8-\\alpha^{3}\\right) L \\delta_{x}^{3}+\\left(12+12 \\alpha^{2}-2 \\alpha^{3}-3 \\alpha^{2} \\beta\\right) \\delta_{x}^{4}+\\mathcal{O}\\left(\\delta_{x}^{5}\\right)\n\\end{aligned}\n$$\nThus, $\\alpha$ and $\\beta$ can be found by solving:\n$$\n8-\\alpha^{3}=0 \\quad, \\quad 12+12 \\alpha^{2}-2 \\alpha^{3}-3 \\alpha^{2} \\beta=0 \\Longrightarrow \\alpha=2, \\beta=\\frac{11}{3} .\n\\tag{3}\n$$\nWe can find the relations between velocities $\\left.(\\dot{x}, \\dot{y})=\\dot{\\delta}_{x}, \\dot{\\delta}_{y}\\right)$ and accelerations $(\\ddot{x}, \\ddot{y})=\\ddot{\\delta}_{x}, \\ddot{\\delta}_{y}$ ) evaluated at the position $(x, y)=(1,0) \\rightarrow\\left(\\delta_{x}, \\delta_{y}\\right)=(0,0)$ by taking the time-derivatives:\n$$\n\\dot{\\delta}_{y}=\\alpha \\dot{\\delta}_{x}+2 \\frac{\\beta}{L} \\delta_{x} \\dot{\\delta}_{x}=\\left(\\alpha+2 \\frac{\\beta}{L} \\delta_{x}\\right) \\dot{\\delta}_{x}=\\alpha \\dot{\\delta}_{x}\n\\tag{4}\n$$\n\n$$\n\\ddot{\\delta}_{y}=\\alpha \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}+2 \\frac{\\beta}{L} \\delta_{x} \\ddot{\\delta}_{x}=\\left(\\alpha+2 \\frac{\\beta}{L} \\delta_{x}\\right) \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}=\\alpha \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}\n\\tag{5}\n$$\nFor a constant running speed $V$, we get:\n$$\nV=\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)^{1 / 2} \\Longrightarrow \\quad \\dot{\\delta}_{x}=\\left(1+\\alpha^{2}\\right)^{-1 / 2} V, \\dot{\\delta}_{y}=\\alpha\\left(1+\\alpha^{2}\\right)^{-1 / 2} V\n$$\nwhich we obtain by applying Eq. (4). Also, the temporal-constraint of constant speed means that the acceleration vector (if non-zero) should be perpendicular to the velocity vector:\n$$\n\\frac{d}{d t} V=0=\\frac{d}{d t}\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)^{1 / 2} \\propto \\dot{\\delta}_{x} \\ddot{\\delta}_{x}+\\dot{\\delta}_{y} \\ddot{\\delta}_{y}=0 \\quad \\Longrightarrow \\quad \\ddot{\\delta}_{x}+\\alpha \\ddot{\\delta}_{y}=0\n$$\nUsing Eq. (5), we can arrive at:\n$$\n\\begin{aligned}\n\\ddot{\\delta}_{x}+\\alpha\\left(\\alpha \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}\\right)=0 \\Longrightarrow \\ddot{\\delta}_{x} & =-2 \\frac{\\beta}{L} \\alpha\\left(1+\\alpha^{2}\\right)^{-1} \\dot{\\delta}_{x}^{2}=-2 \\beta \\alpha\\left(1+\\alpha^{2}\\right)^{-2} \\frac{V^{2}}{L}, \\\\\n\\ddot{\\delta}_{y} & =-\\alpha^{-1} \\dot{\\delta}_{x}=2 \\beta\\left(1+\\alpha^{2}\\right)^{-2} \\frac{V^{2}}{L} .\n\\end{aligned}\n$$\nThe quadratic coefficients are found in Eq. (3), and given that $V=1 \\mathrm{~m} / \\mathrm{s}, L=10 \\mathrm{~m}$, the magnitude of the total acceleration can be calculated:\n$$\na=\\left(\\ddot{\\delta}_{x}^{2}+\\ddot{\\delta}_{y}^{2}\\right)^{1 / 2}=2 \\beta\\left(1+\\alpha^{2}\\right)^{3 / 2} \\frac{V^{2}}{L}=\\frac{22}{15 \\sqrt{5}} \\frac{V^{2}}{L} \\approx 0.066591 \\mathrm{~m} / \\mathrm{s}^{2} .\n$$']",['$0.066591$'],False,$\mathrm{~m} / \mathrm{s}^{2}$,Numerical,5e-3 920,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","On a flat playground, choose a Cartesian Oxy coordinate system (in unit of meters). A child running at a constant velocity $V=1 \mathrm{~m} / \mathrm{s}$ around a heart-shaped path satisfies the following order- 6 algebraic equation: $$ \left(x^{2}+y^{2}-L^{2}\right)^{3}-L x^{2} y^{3}=0, L=10 \text {. } $$ When the child is at the position $(x, y)=(L, 0)$, what is the magnitude of their acceleration? ","['The acceleration can be found from the local geometry of the curves, thus let us study small deviations around the position of interests $(x, y)=(L, 0)$ :\n$$\nx=L+\\delta_{x} \\quad, \\quad y=0+\\delta_{y} \\quad, \\quad\\left|\\delta_{x}\\right|,\\left|\\delta_{y}\\right| \\ll L\n$$\nConsider the 2nd-order approximation in $\\delta_{x}$ of $\\delta_{y}$ with quadratic coefficients $\\alpha$ and $\\beta$ :\n$$\n\\delta_{y} \\approx \\alpha \\delta_{x}+\\frac{\\beta}{L} \\delta_{x}^{2} \\sim \\delta_{x}\n$$\nTo find these coefficients, we look at the algebraic equation of our heart-shape path up to the two lowest-orders of expansions (which are the 3rd and 4th):\n$$\n\\begin{aligned}\n0 & =\\left(x^{2}+y^{2}-L^{2}\\right)^{3}-L x^{2} y^{3} \\approx L^{2}\\left[8 L \\delta_{x}^{3}+12 \\delta_{x}^{4}+12 \\delta_{x}^{2} \\delta_{y}^{2}-2 \\delta_{x} \\delta_{y}^{3}-L \\delta_{y}^{3}+\\mathcal{O}\\left(\\delta_{x}^{5}\\right)\\right] \\\\\n& \\approx L^{2}\\left[8 L \\delta_{x}^{3}+12 \\delta_{x}^{4}+12 \\alpha^{2} \\delta_{x}^{4}-2 \\alpha^{3} \\delta_{x}^{4}-\\left(\\alpha^{3} L \\delta_{x}^{3}+3 \\alpha^{2} \\beta \\delta_{x}^{4}\\right)+\\mathcal{O}\\left(\\delta_{x}^{5}\\right)\\right] \\\\\n& \\propto\\left(8-\\alpha^{3}\\right) L \\delta_{x}^{3}+\\left(12+12 \\alpha^{2}-2 \\alpha^{3}-3 \\alpha^{2} \\beta\\right) \\delta_{x}^{4}+\\mathcal{O}\\left(\\delta_{x}^{5}\\right)\n\\end{aligned}\n$$\nThus, $\\alpha$ and $\\beta$ can be found by solving:\n$$\n8-\\alpha^{3}=0 \\quad, \\quad 12+12 \\alpha^{2}-2 \\alpha^{3}-3 \\alpha^{2} \\beta=0 \\Longrightarrow \\alpha=2, \\beta=\\frac{11}{3} .\n\\tag{3}\n$$\nWe can find the relations between velocities $\\left.(\\dot{x}, \\dot{y})=\\dot{\\delta}_{x}, \\dot{\\delta}_{y}\\right)$ and accelerations $(\\ddot{x}, \\ddot{y})=\\ddot{\\delta}_{x}, \\ddot{\\delta}_{y}$ ) evaluated at the position $(x, y)=(1,0) \\rightarrow\\left(\\delta_{x}, \\delta_{y}\\right)=(0,0)$ by taking the time-derivatives:\n$$\n\\dot{\\delta}_{y}=\\alpha \\dot{\\delta}_{x}+2 \\frac{\\beta}{L} \\delta_{x} \\dot{\\delta}_{x}=\\left(\\alpha+2 \\frac{\\beta}{L} \\delta_{x}\\right) \\dot{\\delta}_{x}=\\alpha \\dot{\\delta}_{x}\n\\tag{4}\n$$\n\n$$\n\\ddot{\\delta}_{y}=\\alpha \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}+2 \\frac{\\beta}{L} \\delta_{x} \\ddot{\\delta}_{x}=\\left(\\alpha+2 \\frac{\\beta}{L} \\delta_{x}\\right) \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}=\\alpha \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}\n\\tag{5}\n$$\nFor a constant running speed $V$, we get:\n$$\nV=\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)^{1 / 2} \\Longrightarrow \\quad \\dot{\\delta}_{x}=\\left(1+\\alpha^{2}\\right)^{-1 / 2} V, \\dot{\\delta}_{y}=\\alpha\\left(1+\\alpha^{2}\\right)^{-1 / 2} V\n$$\nwhich we obtain by applying Eq. (4). Also, the temporal-constraint of constant speed means that the acceleration vector (if non-zero) should be perpendicular to the velocity vector:\n$$\n\\frac{d}{d t} V=0=\\frac{d}{d t}\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)^{1 / 2} \\propto \\dot{\\delta}_{x} \\ddot{\\delta}_{x}+\\dot{\\delta}_{y} \\ddot{\\delta}_{y}=0 \\quad \\Longrightarrow \\quad \\ddot{\\delta}_{x}+\\alpha \\ddot{\\delta}_{y}=0\n$$\nUsing Eq. (5), we can arrive at:\n$$\n\\begin{aligned}\n\\ddot{\\delta}_{x}+\\alpha\\left(\\alpha \\ddot{\\delta}_{x}+2 \\frac{\\beta}{L} \\dot{\\delta}_{x}^{2}\\right)=0 \\Longrightarrow \\ddot{\\delta}_{x} & =-2 \\frac{\\beta}{L} \\alpha\\left(1+\\alpha^{2}\\right)^{-1} \\dot{\\delta}_{x}^{2}=-2 \\beta \\alpha\\left(1+\\alpha^{2}\\right)^{-2} \\frac{V^{2}}{L}, \\\\\n\\ddot{\\delta}_{y} & =-\\alpha^{-1} \\dot{\\delta}_{x}=2 \\beta\\left(1+\\alpha^{2}\\right)^{-2} \\frac{V^{2}}{L} .\n\\end{aligned}\n$$\nThe quadratic coefficients are found in Eq. (3), and given that $V=1 \\mathrm{~m} / \\mathrm{s}, L=10 \\mathrm{~m}$, the magnitude of the total acceleration can be calculated:\n$$\na=\\left(\\ddot{\\delta}_{x}^{2}+\\ddot{\\delta}_{y}^{2}\\right)^{1 / 2}=2 \\beta\\left(1+\\alpha^{2}\\right)^{3 / 2} \\frac{V^{2}}{L}=\\frac{22}{15 \\sqrt{5}} \\frac{V^{2}}{L} \\approx 0.066591 \\mathrm{~m} / \\mathrm{s}^{2} .\n$$']",['$0.066591$'],False,$\mathrm{~m} / \mathrm{s}^{2}$,Numerical,5e-3 921,Mechanics,,"A boy is riding a tricycle across along a sidewalk that is parallel to the $x$-axis. This tricycle contains three identical wheels with radius $0.5 \mathrm{~m}$. The front wheel is free to rotate while the last two wheels are parallel to each other and to the main body of the tricycle. See the diagram. ![](https://cdn.mathpix.com/cropped/2023_12_21_aed9cb2b56be78867ca6g-1.jpg?height=407&width=610&top_left_y=1561&top_left_x=752) The front wheel is rotating at a constant angular speed of $\omega=3 \mathrm{rad} / \mathrm{s}$. The child is controlling the tricycle such that the front wheel is making an angle of $\theta(t)=0.15 \sin ((0.1 \mathrm{rad} / \mathrm{s}) t)$ with the main body of the tricycle. Determine the maximum lateral acceleration in $\mathrm{m} / \mathrm{s}^{2}$. Assume a massless frame. The marked plus sign implies CoM. The degree is in radians.","['Assuming the wheels roll without slipping, the cart will instantaneously rotate about a fixed point. This fixed point can be constructed by drawing perpendicular lines from all the wheels and seeing where they intersect. Consider the below diagram,\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_16605de808d6d47cbc7bg-1.jpg?height=580&width=716&top_left_y=222&top_left_x=691)\n\nWe have:\n$$\n\\sin \\theta_{s}=\\frac{L}{R}, \\quad \\quad \\sin \\theta_{s}^{\\prime}=\\frac{L}{2 R^{\\prime}}\n\\tag{6}\n$$\nThe angular frequency of every point on the robot is the same $\\omega_{n}=\\frac{V_{0}}{R}$. Therefore,\n$$\n\\begin{aligned}\nV_{C M} & =\\omega_{n} R^{\\prime} \\\\\n& =\\frac{V_{0}}{R} \\frac{L}{2 \\sin \\theta_{s}^{\\prime}} \\\\\n& =\\frac{V \\sin \\theta_{s}}{2 \\sin \\theta_{s}^{\\prime}} .\n\\end{aligned}\n$$\nThe naive idea here is to say the lateral velocity is $V_{C M} \\sin \\theta_{s}^{\\prime}$, but note that the robot could already be rotated. Let the angle of rotation with respect to the horizontal line be $\\alpha$, so we have:\n$$\nV_{C M, l a t}=V_{C M} \\sin \\left(\\theta_{s}^{\\prime}+\\alpha\\right)\n$$\nUsing the sine addition formula, we can simplify this to\n$$\n\\begin{aligned}\nV_{C M, l a t} & =V_{C M} \\sin \\theta_{s}^{\\prime} \\cos \\alpha+V_{C M} \\cos \\theta_{s}^{\\prime} \\sin \\alpha \\\\\n& =\\frac{V_{0}}{2} \\sin \\theta_{s} \\cos \\alpha+\\frac{V_{0} \\sin \\theta_{s}}{2 \\tan \\theta_{s}^{\\prime}} \\sin \\alpha\n\\end{aligned}\n$$\nTo simplify this, note that we can write:\n$$\n\\frac{\\sin \\theta_{s}}{\\tan \\theta_{s}^{\\prime}}=2\n$$\nwhere we used the fact that\n$$\n\\sin \\theta_{s} \\approx \\tan \\theta_{s}=2 \\tan \\theta_{s}^{\\prime} \\approx 2 \\sin \\theta_{s}^{\\prime}\n$$\nThen,\n$$\n\\dot{y}_{C M}=\\frac{V_{0}}{2} \\theta_{s}+V_{0} \\alpha\n$$\nAt the center of the two back wheels (marked with an X), the speed is\n$$\nV_{b a c k}=R \\cos \\theta_{s} \\omega_{n}=V_{0} \\cos \\theta_{s} \\approx V_{0}\n$$\n\nso its lateral velocity is\n$$\n\\dot{y}_{b a c k}=V_{0} \\sin \\alpha \\approx V_{0} \\alpha .\n$$\nNote that the angle $\\alpha$ can be written as\n$$\n\\alpha \\approx \\sin \\alpha=\\frac{y_{C M}-y_{b a c k}}{(L / 2)} \\Longrightarrow \\dot{\\alpha}=\\frac{2}{L}\\left(\\dot{y}_{C M}-\\dot{y}_{b a c k}\\right)=\\frac{V_{0}}{L} \\theta_{s}\n$$\nTaking higher derivatives of $\\dot{y}_{C M}$, we have\n$$\n\\begin{aligned}\n\\ddot{y}_{C M} & =\\frac{V_{0}}{2} \\dot{\\theta}_{s}+V_{0} \\dot{\\alpha} \\\\\n& =\\frac{V_{0}}{2} \\dot{\\theta}_{s}+\\frac{V_{0}^{2}}{L} \\theta_{s} .\n\\end{aligned}\n$$\nPlugging in $\\theta(t)=0.15 \\sin (0.1 t)$ and $V_{0}=1.5 \\mathrm{~m} / \\mathrm{s}$, we can optimize $\\ddot{y}_{\\mathrm{CM}}$ to get $0.16912 \\mathrm{~m} / \\mathrm{s}$.']",['$0.16912$'],False,$\mathrm{~m} / \mathrm{s^2}$,Numerical,1e-3 921,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","A boy is riding a tricycle across along a sidewalk that is parallel to the $x$-axis. This tricycle contains three identical wheels with radius $0.5 \mathrm{~m}$. The front wheel is free to rotate while the last two wheels are parallel to each other and to the main body of the tricycle. See the diagram. The front wheel is rotating at a constant angular speed of $\omega=3 \mathrm{rad} / \mathrm{s}$. The child is controlling the tricycle such that the front wheel is making an angle of $\theta(t)=0.15 \sin ((0.1 \mathrm{rad} / \mathrm{s}) t)$ with the main body of the tricycle. Determine the maximum lateral acceleration in $\mathrm{m} / \mathrm{s}^{2}$. Assume a massless frame. The marked plus sign implies CoM. The degree is in radians.","['Assuming the wheels roll without slipping, the cart will instantaneously rotate about a fixed point. This fixed point can be constructed by drawing perpendicular lines from all the wheels and seeing where they intersect. Consider the below diagram,\n\n\n\n\nWe have:\n$$\n\\sin \\theta_{s}=\\frac{L}{R}, \\quad \\quad \\sin \\theta_{s}^{\\prime}=\\frac{L}{2 R^{\\prime}}\n\\tag{6}\n$$\nThe angular frequency of every point on the robot is the same $\\omega_{n}=\\frac{V_{0}}{R}$. Therefore,\n$$\n\\begin{aligned}\nV_{C M} & =\\omega_{n} R^{\\prime} \\\\\n& =\\frac{V_{0}}{R} \\frac{L}{2 \\sin \\theta_{s}^{\\prime}} \\\\\n& =\\frac{V \\sin \\theta_{s}}{2 \\sin \\theta_{s}^{\\prime}} .\n\\end{aligned}\n$$\nThe naive idea here is to say the lateral velocity is $V_{C M} \\sin \\theta_{s}^{\\prime}$, but note that the robot could already be rotated. Let the angle of rotation with respect to the horizontal line be $\\alpha$, so we have:\n$$\nV_{C M, l a t}=V_{C M} \\sin \\left(\\theta_{s}^{\\prime}+\\alpha\\right)\n$$\nUsing the sine addition formula, we can simplify this to\n$$\n\\begin{aligned}\nV_{C M, l a t} & =V_{C M} \\sin \\theta_{s}^{\\prime} \\cos \\alpha+V_{C M} \\cos \\theta_{s}^{\\prime} \\sin \\alpha \\\\\n& =\\frac{V_{0}}{2} \\sin \\theta_{s} \\cos \\alpha+\\frac{V_{0} \\sin \\theta_{s}}{2 \\tan \\theta_{s}^{\\prime}} \\sin \\alpha\n\\end{aligned}\n$$\nTo simplify this, note that we can write:\n$$\n\\frac{\\sin \\theta_{s}}{\\tan \\theta_{s}^{\\prime}}=2\n$$\nwhere we used the fact that\n$$\n\\sin \\theta_{s} \\approx \\tan \\theta_{s}=2 \\tan \\theta_{s}^{\\prime} \\approx 2 \\sin \\theta_{s}^{\\prime}\n$$\nThen,\n$$\n\\dot{y}_{C M}=\\frac{V_{0}}{2} \\theta_{s}+V_{0} \\alpha\n$$\nAt the center of the two back wheels (marked with an X), the speed is\n$$\nV_{b a c k}=R \\cos \\theta_{s} \\omega_{n}=V_{0} \\cos \\theta_{s} \\approx V_{0}\n$$\n\nso its lateral velocity is\n$$\n\\dot{y}_{b a c k}=V_{0} \\sin \\alpha \\approx V_{0} \\alpha .\n$$\nNote that the angle $\\alpha$ can be written as\n$$\n\\alpha \\approx \\sin \\alpha=\\frac{y_{C M}-y_{b a c k}}{(L / 2)} \\Longrightarrow \\dot{\\alpha}=\\frac{2}{L}\\left(\\dot{y}_{C M}-\\dot{y}_{b a c k}\\right)=\\frac{V_{0}}{L} \\theta_{s}\n$$\nTaking higher derivatives of $\\dot{y}_{C M}$, we have\n$$\n\\begin{aligned}\n\\ddot{y}_{C M} & =\\frac{V_{0}}{2} \\dot{\\theta}_{s}+V_{0} \\dot{\\alpha} \\\\\n& =\\frac{V_{0}}{2} \\dot{\\theta}_{s}+\\frac{V_{0}^{2}}{L} \\theta_{s} .\n\\end{aligned}\n$$\nPlugging in $\\theta(t)=0.15 \\sin (0.1 t)$ and $V_{0}=1.5 \\mathrm{~m} / \\mathrm{s}$, we can optimize $\\ddot{y}_{\\mathrm{CM}}$ to get $0.16912 \\mathrm{~m} / \\mathrm{s}$.']",['$0.16912$'],False,$\mathrm{~m} / \mathrm{s^2}$,Numerical,1e-3 922,Thermodynamics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In this problem, we consider a simple model for a thermoacoustic device. The device uses heavily amplified sound to provide work for a pump that can then extract heat. Sound waves form standing waves in a tube of radius $0.25 \mathrm{~mm}$ that is closed on both sides, and a two-plate stack is inserted in the tube. A temperature gradient forms between the plates of the stack, and the parcel of gas trapped between the plates oscillates sinusoidally between a maximum pressure of $1.03 \mathrm{MPa}$ and a minimum of $0.97 \mathrm{MPa}$. The gas is argon, with density $1.78 \mathrm{~kg} / \mathrm{m}^{3}$ and adiabatic constant $5 / 3$. The speed of sound is $323 \mathrm{~m} / \mathrm{s}$. The heat pump itself operates as follows: The parcel of gas starts at minimum pressure. The stack plates adiabatically compress the parcel of gas to its maximum pressure, heating the gas to a temperature higher than that of the hotter stack plate. Then, the gas is allowed to isobarically cool to the temperature of the hotter stack plate. Next, the plates adiabatically expand the gas back to its minimum pressure, cooling it to a temperature lower than that of the colder plate. Finally, the gas is allowed to isobarically heat up to the temperature of the colder stack plate. Find the power at which the thermoacoustic heat pump emits heat.","['The efficiency of the heat engine is $\\epsilon=1-\\left(\\frac{P_{2}}{P_{1}}\\right)^{\\frac{\\gamma-1}{\\gamma}}=0.0237$. The parcel oscillates between pressures $P_{1}$ and $P_{2}$ sinusoidally with amplitude $P_{0}=\\frac{P_{1}-P_{2}}{2}$. For a sound wave, the pressure amplitude is $\\rho s_{0} \\omega v$, where $s_{0}$ is the position amplitude and $v$ is the speed of sound.\n\nThen the average power with which the sound wave does work on the plates is\n$$\n\\langle P\\rangle=\\frac{1}{2} \\rho \\omega^{2} s_{0}^{2} A v=\\frac{P_{0}^{2}}{2 \\rho v} A\n$$\nwhere $A$ is the area of each plate. From this, the heat power generated by the engine is $\\langle P\\rangle / \\epsilon=$ $6.47 \\mathrm{~W}$.']",['$6.47$'],False,W,Numerical,3e-2 923,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The following information applies for the next two problems. For your mass spectroscopy practical you are using an apparatus consisting of a solenoid enclosed by a uniformly charged hollow cylinder of charge density $\sigma=50 \mu \mathrm{C} / \mathrm{m}^{2}$ and radius $r_{0}=7 \mathrm{~cm}$. There exists an infinitesimal slit of insular material between the cylinder and solenoid to stop any charge transfer. Also, assume that there is no interaction between the solenoid and the cylinder, and that the magnetic field produced by the solenoid can be easily controlled to a value of $B_{0}$. An electron is released from rest at a distance of $R=10 \mathrm{~cm}$ from the axis. Assume that it is small enough to pass through the cylinder in both directions without exchanging charge. It is observed that the electron reaches a distance $R$ at different points from the axis 7 times before returning to the original position. Calculate $B_{0}$ under the assumption that the path of the electron does not self-intersect with itself.","['By Gauss\' law, we have\n$$\n\\varepsilon_{0} \\oint \\vec{E} \\cdot \\mathrm{d} \\vec{A}=q_{\\mathrm{enc}} \\Longrightarrow \\varepsilon_{0}(2 \\pi r h) E(r)=\\left(2 \\pi r_{0} h\\right) \\sigma \\Longrightarrow E(r)=\\frac{\\sigma r_{0}}{\\varepsilon_{0} r} \\hat{r}\n$$\nWe find the work done on the electron from a distance $r$ to $R$ is\n$$\nW=-\\int_{R}^{r} q E(r) \\mathrm{d} r=\\int_{r}^{R} \\frac{q \\sigma r_{0}}{\\varepsilon_{0} r} \\mathrm{~d} r=\\frac{q \\sigma r_{0}}{\\varepsilon_{0}} \\ln \\frac{R}{r}\n$$\nTherefore, by conservation of energy:\n$$\n\\frac{1}{2} m v(r)^{2}=\\frac{q \\sigma r_{0}}{\\varepsilon_{0}} \\ln \\frac{R}{r} \\Longrightarrow v(r)=\\sqrt{\\frac{2 q \\sigma r_{0}}{m} \\ln \\frac{R}{r}}\n$$\nSince the electron is moving in a magnetic field, the radius of the electron follows\n$$\n\\frac{m v^{2}}{a}=q v B \\Longrightarrow a=\\frac{m v}{q B}=\\sqrt{\\frac{2 q \\sigma r_{0}}{q B_{0}^{2}} \\ln \\frac{R}{r_{0}}}\n$$\nThis means that $B=\\sqrt{\\frac{2 \\sigma m}{q \\varepsilon_{0} r_{0}} \\ln \\frac{R}{r_{0}}}$. The trajectory of the electron can follow the following paths\n\n\n\nAs the path is specified to be non-intersecting, we analyze the first path. We can say that the radius of each ""circle"" the electron travels through is $s=r \\tan \\frac{\\pi}{8}=\\frac{m v}{q B}$. Hence,\n$$\nB=\\cot \\frac{\\pi}{8} \\sqrt{\\frac{2 \\sigma m}{q \\varepsilon_{0} r_{0}} \\ln \\frac{R}{r_{0}}}\n$$']",['$\\cot \\frac{\\pi}{8} \\sqrt{\\frac{2 \\sigma m}{q \\varepsilon_{0} r_{0}} \\ln \\frac{R}{r_{0}}}$'],False,,Expression, 924,Electromagnetism,,"For any circuit network made of batteries and resistors, if we know the voltages of all the batteries and the resistance values of all the resistors, we can calculate all the electrical currents. However, if we know the voltages of all the batteries and all the currents, it is still not enough to uniquely determine the resistance values of all the resistors. Consider a sail-shape circuit network, in which we connect points $\mathrm{H}$ and $\mathrm{N}$ with a $\mathcal{E}_{\mathrm{HN}}=10 \mathrm{~V}$ battery, points $\mathrm{A}$ and $\mathrm{M}$ with a $\mathcal{E}_{\mathrm{AM}}=20 \mathrm{~V}$ battery. The electrical currents in this network have directions and magnitudes (in $\mathrm{mA}$ ) as shown the figure. The possible resistance values of resistors $R_{\alpha}, R_{\beta}, R_{\gamma}$ is not a single point (corresponds to an unique solution) but a confined region in the three-dimensional $\left(R_{\alpha}, R_{\beta}, R_{\gamma}\right)$-space. Determine the volume of this region (in $\Omega^{3}$ ). ![](https://cdn.mathpix.com/cropped/2023_12_21_1f4c482b1f1adbb769dbg-1.jpg?height=653&width=867&top_left_y=194&top_left_x=626)","[""Let us call the nodes on this sail-shape circuit network as in figure A. Since the electrical current always flow toward lower potential, we have the diagram in figure $\\mathrm{B}$ for the relation between the electrical potential of the nodes (the arrow indicate high-to-low). Use the notation $\\mathcal{V}(\\bullet)$ to indicate the electrical potential of the point\n\nThe trick for this problem is noticing that the Ohm's laws in our circuit network can always be trivially satisfied for every possible solution following the comparison relationship in figure $\\mathrm{B}$, the voltage differences $\\mathcal{V}(\\mathrm{H})-\\mathcal{V}(\\mathrm{N})=10 \\mathrm{~V}, \\mathcal{V}(\\mathrm{A})-\\mathcal{V}(\\mathrm{M})=20 \\mathrm{~V}$ and the equality of in-out currents (in $\\mathrm{H} 5+7$ equals out $\\mathrm{N}-4+16$, in $\\mathrm{A} 8+15$ equals out $\\mathrm{M} 6+17$ ) of the batteries. That indeed responsible to why the resistance values of the resistors in this network are not an unique point but rather a region in the 14-dimensional space.\n\nDefine:\n\n$X=\\mathcal{V}(\\mathrm{P})-\\mathcal{V}(\\mathrm{M})>0, \\quad Y=\\mathcal{V}(\\mathrm{Q})-\\mathcal{V}(\\mathrm{U})>0, Z=\\mathcal{V}(\\mathrm{A})-\\mathcal{V}(\\mathrm{S})>0, a=\\mathcal{V}(\\mathrm{U})-\\mathcal{V}(\\mathrm{M})>0$\n\nWe want to find the volume in the 3 -dimensional subspace of possible $\\left(R_{\\alpha}, R_{\\beta}, R_{\\gamma}\\right)$ :\n$$\n\\int d R_{\\alpha} d R_{\\beta} d R_{\\gamma}=\\frac{\\int d X d Y d Z}{I_{\\alpha} I_{\\beta} I_{\\gamma}}\n$$\nFrom figure $B$ and the fixed voltage differences $\\mathcal{V}(\\mathrm{H})-\\mathcal{V}(\\mathrm{N})=10 \\mathrm{~V}, \\mathcal{V}(\\mathrm{A})-\\mathcal{V}(\\mathrm{M})=20 \\mathrm{~V}$, we can write down the inequality equations to specify the boundary of $\\{(X, Y, Z)\\}$ :\n$$\na<20, \\quad Y<30-a, \\quad a","[""Let us call the nodes on this sail-shape circuit network as in figure A. Since the electrical current always flow toward lower potential, we have the diagram in figure $\\mathrm{B}$ for the relation between the electrical potential of the nodes (the arrow indicate high-to-low). Use the notation $\\mathcal{V}(\\bullet)$ to indicate the electrical potential of the point\n\nThe trick for this problem is noticing that the Ohm's laws in our circuit network can always be trivially satisfied for every possible solution following the comparison relationship in figure $\\mathrm{B}$, the voltage differences $\\mathcal{V}(\\mathrm{H})-\\mathcal{V}(\\mathrm{N})=10 \\mathrm{~V}, \\mathcal{V}(\\mathrm{A})-\\mathcal{V}(\\mathrm{M})=20 \\mathrm{~V}$ and the equality of in-out currents (in $\\mathrm{H} 5+7$ equals out $\\mathrm{N}-4+16$, in $\\mathrm{A} 8+15$ equals out $\\mathrm{M} 6+17$ ) of the batteries. That indeed responsible to why the resistance values of the resistors in this network are not an unique point but rather a region in the 14-dimensional space.\n\nDefine:\n\n$X=\\mathcal{V}(\\mathrm{P})-\\mathcal{V}(\\mathrm{M})>0, \\quad Y=\\mathcal{V}(\\mathrm{Q})-\\mathcal{V}(\\mathrm{U})>0, Z=\\mathcal{V}(\\mathrm{A})-\\mathcal{V}(\\mathrm{S})>0, a=\\mathcal{V}(\\mathrm{U})-\\mathcal{V}(\\mathrm{M})>0$\n\nWe want to find the volume in the 3 -dimensional subspace of possible $\\left(R_{\\alpha}, R_{\\beta}, R_{\\gamma}\\right)$ :\n$$\n\\int d R_{\\alpha} d R_{\\beta} d R_{\\gamma}=\\frac{\\int d X d Y d Z}{I_{\\alpha} I_{\\beta} I_{\\gamma}}\n$$\nFrom figure $B$ and the fixed voltage differences $\\mathcal{V}(\\mathrm{H})-\\mathcal{V}(\\mathrm{N})=10 \\mathrm{~V}, \\mathcal{V}(\\mathrm{A})-\\mathcal{V}(\\mathrm{M})=20 \\mathrm{~V}$, we can write down the inequality equations to specify the boundary of $\\{(X, Y, Z)\\}$ :\n$$\na<20, \\quad Y<30-a, \\quad a\n\n(B)\n\n\n\n(C)\n\n""]",['$4.0741 \\times 10^{10}$'],False,$\Omega^{3}$,Numerical,1e9 925,Electromagnetism,,"Two carts, each with a mass of $300 \mathrm{~g}$, are fixed to move on a horizontal track. As shown in the figure, the first cart has a strong, tiny permanent magnet of dipole moment $0.5 \mathrm{~A} \cdot \mathrm{m}^{2}$ attached to it, which is aligned along the axis of the track pointing toward the other cart. On the second cart, a copper tube of radius $7 \mathrm{~mm}$, thickness $0.5 \mathrm{~mm}$, resistivity $1.73 \cdot 10^{-8} \Omega$, and length $30 \mathrm{~cm}$ is attached. The masses of the magnet and coil are negligible compared to the mass of the carts. At the moment its magnet enters through the right end of the copper tube, the velocity of the first cart is $0.3 \mathrm{~m} / \mathrm{s}$ and the distance between the two ends of each cart is $50 \mathrm{~cm}$, find the minimum distance achieved between the two ends of the carts in centimeters. While on the track, the carts experience an effective coefficient of static friction (i.e., what it would be as if they did not have wheels) of 0.01 . Neglect the self-inductance of the copper tube. ![](https://cdn.mathpix.com/cropped/2023_12_21_f5a0ed653cebecd7d9deg-1.jpg?height=111&width=984&top_left_y=1576&top_left_x=560) A picture of the two cart setup. The black rectangle represents the magnet while the gold rectangle represents the copper tube. Hint 1: The magnetic field due to a dipole of moment $\vec{\mu}$, at a position $\vec{r}$ away from the dipole can be written as $$ \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{(3 \hat{r}(\vec{r} \cdot \vec{\mu})-\vec{\mu})}{r^{3}} \hat{r} $$ where $\hat{r}$ is the unit vector in the direction of $\vec{r}$. Hint 2: The following mathematical identity may be useful: $$ \int_{-\infty}^{\infty} \frac{u^{2} d u}{\left(1+u^{2}\right)^{5}}=\frac{5 \pi}{128} $$","[""We analyze the thin $\\mathrm{dz}$ of copper which center is at distance $\\mathrm{z}$ away from the magnet. Notice that\n\n\n\nthe thickness of the copper tube $(w=0.5 \\mathrm{~mm})$ is much smaller than its radius. Thus, we may model the copper tube as an infinitesimally thin current loop of radius $a_{0}=a+(w / 2)$ (note that $\\frac{w}{2 a}$ is significant!) with resistance $\\rho \\frac{2 \\pi a_{0}}{w d z}$ where $a=7 \\mathrm{~mm}$. Using the formula for the magnetic field due to the dipole moment, it can be found that the B-field along the current loop is:\n$$\nB=\\frac{\\mu_{0} \\mu}{4 \\pi} \\frac{\\left(2 z^{2}-r^{2}\\right) \\hat{z}+3 r z \\hat{r}}{\\left(r^{2}+z^{2}\\right)^{5 / 2}}\n$$\nwhere $\\hat{z}$ is the unit vector parallel to the direction of motion of the magnet and $\\hat{r}$ is the unit vector pointing radially outward from the center of the current loop. The flux through the loop is then:\n$$\n\\Phi_{B}=\\int_{0}^{a_{0}} B_{z}(2 \\pi r) d r=\\frac{1}{2} \\mu_{0} \\mu \\frac{a_{0}^{2}}{\\left(z^{2}+a_{0}^{2}\\right)^{3 / 2}}\n$$\nin which Lenz law can be used to get the current of the loop, using the fact $v=-\\frac{d z}{d t}$, where $v$ is the relative velocity of the right cart to the left.\n$$\nI=-\\frac{d \\Phi_{B}}{d t}\\left(\\rho \\frac{2 \\pi a_{0}}{w d z}\\right)^{-1}=\\frac{3 \\mu_{0} \\mu}{4 \\pi} \\frac{a_{0} w v z d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5 / 2}}\n$$\nThe lorentz force on the current loop due to the magnet then obeys:\n$$\nF=I\\left(2 \\pi a_{0}\\right) B_{r} \\hat{z}=\\frac{9}{8 \\pi} \\mu_{0}^{2} \\mu^{2} \\frac{a_{0}^{3} w v z^{2} d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5}} \\hat{z}\n$$\nwhich is repulsive as expected. By Newton's third law, the force on the magnet due to the current in the copper tube is:\n$$\nF=-\\frac{9}{8 \\pi \\rho} \\mu_{0}^{2} \\mu^{2} a_{0}^{3} w v \\int_{-x}^{L-x} \\frac{z^{2} d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5}} \\hat{z}\n$$\nwhere $x$ is the depth of the magnet into the copper tube. However, when $x, L-x>>a_{0}$,\n$$\n\\int_{-x}^{L-x} \\frac{z^{2} d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5}} \\hat{z} \\approx \\frac{a_{0}^{3}}{a_{0}^{10}} \\int_{-\\infty}^{\\infty} \\frac{u^{2} d u}{\\left(1+u^{2}\\right)^{5}}=\\frac{5 \\pi}{128 a_{0}^{7}}\n$$\nTherefore, for as long as the magnet is sufficiently within the tube, the force is instead:\n$$\nF=-\\frac{45}{1024 \\rho a_{0}^{4}} \\mu_{0}^{2} \\mu^{2} w v \\hat{z}=-\\gamma v \\hat{z}\n$$\nwhere $\\gamma=0.1815 \\mathrm{~kg} / \\mathrm{s}$.\n\nThe velocity of the second cart starts moving when $F>\\mu_{s} M g$ with $\\mu_{s}=0.01$ and $M=0.3 \\mathrm{~kg}$. It turns out that $\\frac{0.01 \\mathrm{Mg}}{\\gamma v_{0}} \\approx 0.5$, so this static friction is overcome very quickly. Once both carts are in motion, and since there is no kinetic friction, momentum is in fact conserved! As we'll see soon, this results in the carts performing an (infinitely-long) perfectly inelastic collision. In addition, it allows us to use a reduced mass of $M_{\\text {red }}=M / 2$ to calculate the closest distance achieved between the two carts. While $x>>a$ is obeyed:\n$$\n-\\gamma v_{r e l}=\\frac{1}{2} M \\frac{d v_{r e l}}{d t}\n$$\ngiving:\n$$\nv_{r e l}=v_{0} e^{-\\frac{2 \\gamma}{M} t}\n$$\n\nin which $v_{r e l}$ goes to 0 as $t$ goes to $\\infty$, resulting in a perfectly inelastic collision. Since $v_{r e l}=-\\frac{d x}{d t}$\n$$\nx_{\\min }=x_{0}-\\int_{0}^{\\infty} v_{r e l} d t=x_{0}-\\frac{m v_{0}}{2 \\gamma}=0.252 m\n$$""]",['25.2'],False,cm,Numerical,2e-1 925,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Two carts, each with a mass of $300 \mathrm{~g}$, are fixed to move on a horizontal track. As shown in the figure, the first cart has a strong, tiny permanent magnet of dipole moment $0.5 \mathrm{~A} \cdot \mathrm{m}^{2}$ attached to it, which is aligned along the axis of the track pointing toward the other cart. On the second cart, a copper tube of radius $7 \mathrm{~mm}$, thickness $0.5 \mathrm{~mm}$, resistivity $1.73 \cdot 10^{-8} \Omega$, and length $30 \mathrm{~cm}$ is attached. The masses of the magnet and coil are negligible compared to the mass of the carts. At the moment its magnet enters through the right end of the copper tube, the velocity of the first cart is $0.3 \mathrm{~m} / \mathrm{s}$ and the distance between the two ends of each cart is $50 \mathrm{~cm}$, find the minimum distance achieved between the two ends of the carts in centimeters. While on the track, the carts experience an effective coefficient of static friction (i.e., what it would be as if they did not have wheels) of 0.01 . Neglect the self-inductance of the copper tube. A picture of the two cart setup. The black rectangle represents the magnet while the gold rectangle represents the copper tube. Hint 1: The magnetic field due to a dipole of moment $\vec{\mu}$, at a position $\vec{r}$ away from the dipole can be written as $$ \vec{B}=\frac{\mu_{0}}{4 \pi} \frac{(3 \hat{r}(\vec{r} \cdot \vec{\mu})-\vec{\mu})}{r^{3}} \hat{r} $$ where $\hat{r}$ is the unit vector in the direction of $\vec{r}$. Hint 2: The following mathematical identity may be useful: $$ \int_{-\infty}^{\infty} \frac{u^{2} d u}{\left(1+u^{2}\right)^{5}}=\frac{5 \pi}{128} $$","[""We analyze the thin $\\mathrm{dz}$ of copper which center is at distance $\\mathrm{z}$ away from the magnet. Notice that\n\n\n\nthe thickness of the copper tube $(w=0.5 \\mathrm{~mm})$ is much smaller than its radius. Thus, we may model the copper tube as an infinitesimally thin current loop of radius $a_{0}=a+(w / 2)$ (note that $\\frac{w}{2 a}$ is significant!) with resistance $\\rho \\frac{2 \\pi a_{0}}{w d z}$ where $a=7 \\mathrm{~mm}$. Using the formula for the magnetic field due to the dipole moment, it can be found that the B-field along the current loop is:\n$$\nB=\\frac{\\mu_{0} \\mu}{4 \\pi} \\frac{\\left(2 z^{2}-r^{2}\\right) \\hat{z}+3 r z \\hat{r}}{\\left(r^{2}+z^{2}\\right)^{5 / 2}}\n$$\nwhere $\\hat{z}$ is the unit vector parallel to the direction of motion of the magnet and $\\hat{r}$ is the unit vector pointing radially outward from the center of the current loop. The flux through the loop is then:\n$$\n\\Phi_{B}=\\int_{0}^{a_{0}} B_{z}(2 \\pi r) d r=\\frac{1}{2} \\mu_{0} \\mu \\frac{a_{0}^{2}}{\\left(z^{2}+a_{0}^{2}\\right)^{3 / 2}}\n$$\nin which Lenz law can be used to get the current of the loop, using the fact $v=-\\frac{d z}{d t}$, where $v$ is the relative velocity of the right cart to the left.\n$$\nI=-\\frac{d \\Phi_{B}}{d t}\\left(\\rho \\frac{2 \\pi a_{0}}{w d z}\\right)^{-1}=\\frac{3 \\mu_{0} \\mu}{4 \\pi} \\frac{a_{0} w v z d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5 / 2}}\n$$\nThe lorentz force on the current loop due to the magnet then obeys:\n$$\nF=I\\left(2 \\pi a_{0}\\right) B_{r} \\hat{z}=\\frac{9}{8 \\pi} \\mu_{0}^{2} \\mu^{2} \\frac{a_{0}^{3} w v z^{2} d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5}} \\hat{z}\n$$\nwhich is repulsive as expected. By Newton's third law, the force on the magnet due to the current in the copper tube is:\n$$\nF=-\\frac{9}{8 \\pi \\rho} \\mu_{0}^{2} \\mu^{2} a_{0}^{3} w v \\int_{-x}^{L-x} \\frac{z^{2} d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5}} \\hat{z}\n$$\nwhere $x$ is the depth of the magnet into the copper tube. However, when $x, L-x>>a_{0}$,\n$$\n\\int_{-x}^{L-x} \\frac{z^{2} d z}{\\rho\\left(a_{0}^{2}+z^{2}\\right)^{5}} \\hat{z} \\approx \\frac{a_{0}^{3}}{a_{0}^{10}} \\int_{-\\infty}^{\\infty} \\frac{u^{2} d u}{\\left(1+u^{2}\\right)^{5}}=\\frac{5 \\pi}{128 a_{0}^{7}}\n$$\nTherefore, for as long as the magnet is sufficiently within the tube, the force is instead:\n$$\nF=-\\frac{45}{1024 \\rho a_{0}^{4}} \\mu_{0}^{2} \\mu^{2} w v \\hat{z}=-\\gamma v \\hat{z}\n$$\nwhere $\\gamma=0.1815 \\mathrm{~kg} / \\mathrm{s}$.\n\nThe velocity of the second cart starts moving when $F>\\mu_{s} M g$ with $\\mu_{s}=0.01$ and $M=0.3 \\mathrm{~kg}$. It turns out that $\\frac{0.01 \\mathrm{Mg}}{\\gamma v_{0}} \\approx 0.5$, so this static friction is overcome very quickly. Once both carts are in motion, and since there is no kinetic friction, momentum is in fact conserved! As we'll see soon, this results in the carts performing an (infinitely-long) perfectly inelastic collision. In addition, it allows us to use a reduced mass of $M_{\\text {red }}=M / 2$ to calculate the closest distance achieved between the two carts. While $x>>a$ is obeyed:\n$$\n-\\gamma v_{r e l}=\\frac{1}{2} M \\frac{d v_{r e l}}{d t}\n$$\ngiving:\n$$\nv_{r e l}=v_{0} e^{-\\frac{2 \\gamma}{M} t}\n$$\n\nin which $v_{r e l}$ goes to 0 as $t$ goes to $\\infty$, resulting in a perfectly inelastic collision. Since $v_{r e l}=-\\frac{d x}{d t}$\n$$\nx_{\\min }=x_{0}-\\int_{0}^{\\infty} v_{r e l} d t=x_{0}-\\frac{m v_{0}}{2 \\gamma}=0.252 m\n$$""]",['25.2'],False,cm,Numerical,2e-1 926,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","In the far future, the Earth received an enormous amount of charge as a result of Mad Scientist ecilA's nefarious experiments. Specifically, the total charge on Earth is $Q=1.0 \times 10^{11} \mathrm{C}$. (compare this with the current $5 \times 10^{5} \mathrm{C}$ ). Estimate the maximum height of a ""mountain"" on Earth that has a circular base with diameter $w=1.0$ $\mathrm{km}$, if it has the shape of a spherical sector. You may assume that $h_{\max } \ll w$. The tensile strength of rock is $10 \mathrm{MPa}$.","['You can approximate the mound as a conducting sphere of radius $r$ connecting by a wire to the Earth of radius $R$. We can therefore say that\n$$\n\\sigma_{\\text {Mound }} \\sim \\frac{R}{r} \\sigma_{\\text {Earth }}\n$$\nas voltages are equal and proportional to $Q / R$. The electrostatic pressure can be given as $P=\\frac{\\sigma E}{2}$, where $E / 2$ comes from the fact that the section does not contribute a force on itself. This can be rewritten as\n$$\nP=\\left(\\frac{R}{r} \\frac{Q}{4 \\pi R^{2}}\\right) \\cdot \\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Q}{R^{2}}=\\frac{Q^{2}}{32 \\pi \\varepsilon_{0} R^{3} r}\n$$\nBy using Pythagorean theorem, we can find that $r^{2}-(r-h)^{2}=w^{2}$ which means that $h=\\frac{w^{2}}{2 r}$. We can hence write the tensile strength as\n$$\nY=\\frac{Q^{2} h}{16 \\pi \\varepsilon_{0} R^{3} w^{2}} \\Longrightarrow h=\\frac{16 \\pi \\varepsilon_{0} R^{3} w^{2} Y}{G^{2}}=115 \\mathrm{~m}\n$$']",['115'],False,m,Numerical,1e-1 927,Mechanics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","The logo of OPhO describes two objects travelling around their center-of-mass, following the same oval-shape trajectory. For simplicity, we assume these objects are point-like, have identical mass, and interacts via an interacting potential $U(d)$ depends on the distance $d$ between them. Choose the polar coordinates $(r, \theta)$ as shown in the figure, where the origin is located at the center of the logo, then the shared trajectory obeys the equation: $$ r(\theta)=\frac{L}{2}[1-\epsilon \cos (2 \theta)]^{-(1+\gamma)}, $$ in which we consider $\epsilon=0.12$ and $\gamma=0.05$. Here the smallest and largest separation between the objects are $d_{\min }$ and $d_{\max }$. Since the interacting potential $U(d)$ is defined up to a constant, let us pick $U(L)=0$. Find the ratio $U\left(d_{\min }\right) / U\left(d_{\max }\right)$. ","['Define the dimensionless inverse-radius $u=\\frac{L}{d}$ and the angular-derivative ${ }^{\\prime}=\\frac{d}{d \\theta}$, then for any central interacting potential we have Binet equation (can be derived directly from the equations of motion in polar coordinates):\n$$\n\\frac{d}{d u} U(u) \\propto-\\left(u^{\\prime \\prime}+u\\right)\n\\tag{26}\n$$\nFor the given oval-shape trajectory:\n$$\nd=L[1-\\epsilon \\cos (2 \\theta)]^{-(1+\\gamma)} \\Longrightarrow u=(1-\\epsilon \\cos \\phi)^{(1+\\gamma)}, \\cos \\phi=\\epsilon^{-1}\\left[1-u^{\\left(\\frac{1}{1+\\gamma}\\right)}\\right]\n\\tag{27}\n$$\nwhere $\\phi=2 \\theta$. The angular 2nd-derivative of the dimensionless inverse-radius $u$ can be calculated:\n$$\n\\begin{aligned}\nu^{\\prime \\prime} & =4 \\frac{d^{2}}{d \\phi^{2}} u=4 \\epsilon(1+\\gamma)\\left[\\frac{\\gamma \\epsilon\\left(1-\\cos ^{2} \\phi\\right)}{1-\\epsilon \\cos \\phi}+\\cos \\phi\\right](1-\\epsilon \\cos \\phi)^{\\gamma} \\\\\n& =4 \\epsilon(1+\\gamma)\\left\\{\\gamma \\epsilon\\left(1-\\epsilon^{-2}\\left[1-u^{\\left(\\frac{1}{1+\\gamma}\\right)}\\right]^{2}\\right) u^{-\\left(\\frac{1}{1+\\gamma}\\right)}+\\epsilon^{-1}\\left[1-u^{\\left(\\frac{1}{1+\\gamma}\\right)}\\right]\\right\\} u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)} \\\\\n& =4(1+\\gamma)\\left\\{\\gamma \\epsilon^{2} u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}-\\gamma u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}\\left[1-2 u^{\\left(\\frac{1}{1+\\gamma}\\right)}+u^{\\left(\\frac{2}{1+\\gamma}\\right)}\\right]+\\left[u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)}-u\\right]\\right\\} \\\\\n& =-4 \\gamma(1+\\gamma)\\left(1-\\epsilon^{2}\\right) u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}+4(1+\\gamma)(1+2 \\gamma) u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)}-4(1+\\gamma)^{2} u .\n\\end{aligned}\n$$\nPlug this back into Eq. (26), we can arrive at:\n$$\n\\begin{aligned}\n\\frac{d}{d u} U(u) & \\propto\\left[4(1+\\gamma)^{2}-1\\right] u+4 \\gamma(1+\\gamma)\\left(1-\\epsilon^{2}\\right) u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}-4(1+\\gamma)(1+2 \\gamma) u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)} \\\\\n\\Longrightarrow U(u) & \\propto\\left(2-\\frac{1}{2(1+\\gamma)^{2}}\\right) u^{2}+2\\left(1-\\epsilon^{2}\\right) u^{\\left(\\frac{2 \\gamma}{1+\\gamma}\\right)}-4 u^{\\left(\\frac{1+2 \\gamma}{1+\\gamma}\\right)}+W\n\\end{aligned}\n\\tag{28}\n$$\nwhere $W$ is a constant so that we can assign $U(u=1)=0$. For $\\epsilon=0.12$ and $\\gamma=0.05$, we get $W=0.48231$. Following that, we use Eq. (27), the maximum distance corresponds to $\\theta=0$ and thus $u_{\\triangleright}=0.87439$, the minimum distance corresponds to $\\theta=\\pi / 2$ and thus $u_{\\triangleleft}=1.1264$. The ratio of interests, therefore, can be found directly:\n$$\n\\frac{U\\left(u=u_{\\triangleleft}\\right)}{U\\left(u=u_{\\triangleright}\\right)} \\approx-0.6864\n$$\nWe plot the interacting potential, up to a pre-factor as given in Eq. (28), in the figure below.\n\n\n\n']",['-0.6864'],False,,Numerical,2e-2 927,Mechanics,,"The logo of OPhO describes two objects travelling around their center-of-mass, following the same oval-shape trajectory. For simplicity, we assume these objects are point-like, have identical mass, and interacts via an interacting potential $U(d)$ depends on the distance $d$ between them. Choose the polar coordinates $(r, \theta)$ as shown in the figure, where the origin is located at the center of the logo, then the shared trajectory obeys the equation: $$ r(\theta)=\frac{L}{2}[1-\epsilon \cos (2 \theta)]^{-(1+\gamma)}, $$ in which we consider $\epsilon=0.12$ and $\gamma=0.05$. Here the smallest and largest separation between the objects are $d_{\min }$ and $d_{\max }$. Since the interacting potential $U(d)$ is defined up to a constant, let us pick $U(L)=0$. Find the ratio $U\left(d_{\min }\right) / U\left(d_{\max }\right)$. ![](https://cdn.mathpix.com/cropped/2023_12_21_7cb4a6b8f7c2770434e1g-1.jpg?height=364&width=1332&top_left_y=392&top_left_x=381)","['Define the dimensionless inverse-radius $u=\\frac{L}{d}$ and the angular-derivative ${ }^{\\prime}=\\frac{d}{d \\theta}$, then for any central interacting potential we have Binet equation (can be derived directly from the equations of motion in polar coordinates):\n$$\n\\frac{d}{d u} U(u) \\propto-\\left(u^{\\prime \\prime}+u\\right)\n\\tag{26}\n$$\nFor the given oval-shape trajectory:\n$$\nd=L[1-\\epsilon \\cos (2 \\theta)]^{-(1+\\gamma)} \\Longrightarrow u=(1-\\epsilon \\cos \\phi)^{(1+\\gamma)}, \\cos \\phi=\\epsilon^{-1}\\left[1-u^{\\left(\\frac{1}{1+\\gamma}\\right)}\\right]\n\\tag{27}\n$$\nwhere $\\phi=2 \\theta$. The angular 2nd-derivative of the dimensionless inverse-radius $u$ can be calculated:\n$$\n\\begin{aligned}\nu^{\\prime \\prime} & =4 \\frac{d^{2}}{d \\phi^{2}} u=4 \\epsilon(1+\\gamma)\\left[\\frac{\\gamma \\epsilon\\left(1-\\cos ^{2} \\phi\\right)}{1-\\epsilon \\cos \\phi}+\\cos \\phi\\right](1-\\epsilon \\cos \\phi)^{\\gamma} \\\\\n& =4 \\epsilon(1+\\gamma)\\left\\{\\gamma \\epsilon\\left(1-\\epsilon^{-2}\\left[1-u^{\\left(\\frac{1}{1+\\gamma}\\right)}\\right]^{2}\\right) u^{-\\left(\\frac{1}{1+\\gamma}\\right)}+\\epsilon^{-1}\\left[1-u^{\\left(\\frac{1}{1+\\gamma}\\right)}\\right]\\right\\} u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)} \\\\\n& =4(1+\\gamma)\\left\\{\\gamma \\epsilon^{2} u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}-\\gamma u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}\\left[1-2 u^{\\left(\\frac{1}{1+\\gamma}\\right)}+u^{\\left(\\frac{2}{1+\\gamma}\\right)}\\right]+\\left[u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)}-u\\right]\\right\\} \\\\\n& =-4 \\gamma(1+\\gamma)\\left(1-\\epsilon^{2}\\right) u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}+4(1+\\gamma)(1+2 \\gamma) u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)}-4(1+\\gamma)^{2} u .\n\\end{aligned}\n$$\nPlug this back into Eq. (26), we can arrive at:\n$$\n\\begin{aligned}\n\\frac{d}{d u} U(u) & \\propto\\left[4(1+\\gamma)^{2}-1\\right] u+4 \\gamma(1+\\gamma)\\left(1-\\epsilon^{2}\\right) u^{-\\left(\\frac{1-\\gamma}{1+\\gamma}\\right)}-4(1+\\gamma)(1+2 \\gamma) u^{\\left(\\frac{\\gamma}{1+\\gamma}\\right)} \\\\\n\\Longrightarrow U(u) & \\propto\\left(2-\\frac{1}{2(1+\\gamma)^{2}}\\right) u^{2}+2\\left(1-\\epsilon^{2}\\right) u^{\\left(\\frac{2 \\gamma}{1+\\gamma}\\right)}-4 u^{\\left(\\frac{1+2 \\gamma}{1+\\gamma}\\right)}+W\n\\end{aligned}\n\\tag{28}\n$$\nwhere $W$ is a constant so that we can assign $U(u=1)=0$. For $\\epsilon=0.12$ and $\\gamma=0.05$, we get $W=0.48231$. Following that, we use Eq. (27), the maximum distance corresponds to $\\theta=0$ and thus $u_{\\triangleright}=0.87439$, the minimum distance corresponds to $\\theta=\\pi / 2$ and thus $u_{\\triangleleft}=1.1264$. The ratio of interests, therefore, can be found directly:\n$$\n\\frac{U\\left(u=u_{\\triangleleft}\\right)}{U\\left(u=u_{\\triangleright}\\right)} \\approx-0.6864\n$$\nWe plot the interacting potential, up to a pre-factor as given in Eq. (28), in the figure below.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_1b9e28ea40fa6fb3d058g-1.jpg?height=828&width=1568&top_left_y=236&top_left_x=274)']",['-0.6864'],False,,Numerical,2e-2 928,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Follin is investigating the electrostatic pendulum. His apparatus consists of an insulating Styrofoam ball with a mass of $14 \mathrm{mg}$ and radius $r=0.5 \mathrm{~cm}$ suspended on a uniform electrically-insulating string of length $1 \mathrm{~m}$ and mass per unit length density of $1.1 \cdot 10^{-5} \mathrm{~kg} / \mathrm{m}$ between two large metal plates separated by a distance $17 \mathrm{~cm}$ with a voltage drop of $10 \mathrm{kV}$ between them, such that when the ball is in equilibrium, its center of mass is exactly equidistant to the two plates. Neglect the possibility of electrical discharge throughout the next two problems. Follin then gives the ball a charge $0.15 \mathrm{nC}$. Assuming that the charge is distributed evenly across the surface of the ball, find the subsequent horizontal deflection of the pendulum bob's center of mass from its hanging point at equilibrium.","[""The force on the Styrofoam ball due to the electric field from the two plates is $F=\\frac{Q V}{d}$. Since the plates are conducting, the ball also experiences an attraction toward the closer plate. This force can be neglected because it is much smaller than $\\frac{Q V}{d}$ (an assumption that will be justified at the end of the solution). The mass of the string, however, needs to be considered. Consider the forces on an infinitesimal segment of the string. The horizontal component of the tension must balance out the electric force on the ball and the vertical component must balance out the weight of everything below it (string and ball). This gives us\n$$\n\\frac{d x}{d h}=\\frac{F}{m g+\\lambda g h}\n$$\nwhere $h$ is the height above the ball, $x$ is the horizontal displacement from equilibrium, $F$ is the electrostatic force, $\\lambda$ is the mass density of the string, and $m$ is the mass of the ball. Separating\n\n\n\nvariables, we have\n$$\nx=F \\int_{0}^{L} \\frac{d h}{m g+\\lambda g h}\n$$\nwhere $L$ is the length of the string. Integrating, we get\n$$\nx=\\frac{F}{\\lambda g} \\ln \\left(\\frac{m+\\lambda L}{m}\\right)=0.0475 \\mathrm{~m}\n$$\nNow, let's justify our assumption that the attraction of the ball towards the plates is negligible. Using the method of image charges, the force due to the closer plate is equivalent to a charge $-Q$ a distance of $d-2 x$ away. Infinitely many image charges exist due to the other plate but they have a smaller effect so they will not be considered. The force due to the image charge is\n$$\nF^{\\prime}=\\frac{Q^{2}}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}=3.59 \\times 10^{-8} \\mathrm{~N}\n$$\nwhich is indeed much less than $F=\\frac{Q V}{d}=8.82 \\times 10^{-6} \\mathrm{~N}$""]",['$0.0475$'],False,m,Numerical,3e-3 929,Electromagnetism,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Hoping to get a larger deflection, Follin replaces the insulating Styrofoam ball with a conducting pith ball of mass $250 \mathrm{mg}$ and $2 \mathrm{~cm}$ and daisy chains 4 additional $10 \mathrm{kV}$ High Voltage Power Supplies to increase the voltage drop across the plates to $50 \mathrm{kV}$. Leaving the plate separation and the string unchanged, he repeats the same experiment as before, but forgets to measure the charge on the ball. Nonetheless, once the ball reaches equilibrium, he measures the deflection from the hanging point to be $5.6 \mathrm{~cm}$. Find the charge on the ball.","['When a conducting ball is placed in a uniform electric field, the charges separate so that the sphere becomes a dipole with electric dipole moment\n$$\np=4 \\pi \\epsilon_{0} r^{3} E_{0}\n$$\nLet $Q$ be the charge on the ball. The ball can be approximated as a point dipole with dipole moment $p$ and a point charge of magnitude $Q$ at the center of the ball. Since the plates are conducting and must be equipotential, the charge and dipole will experience an attractive force to the plates. Using the method of image charges, the force caused by the closer plate is equivalent to that of a point charge $-Q$ and a dipole moment $p$ at a distance $d-2 x$ from the center of the ball, where $d=17 \\mathrm{~cm}$ is the separation between the plates and $x=5.6 \\mathrm{~cm}$ is the deflection of the ball from the hanging point. The forces from the farther plate are negligible compared to the force due to the electric field, $Q E_{0}$. The force between a point charge $Q$ and a point dipole moment $p$ a distance $r$ apart is\n$$\nF=\\frac{p Q}{2 \\pi \\epsilon_{0} r^{3}}\n$$\nThe force between two dipoles with dipole moments of magnitude $p$ in the same direction as the vector from one dipole to the other is\n$$\nF=\\frac{3 p^{2}}{2 \\pi \\epsilon_{0} r^{4}}\n$$\nAdding up the charge-image charge, charge-image dipole, dipole-image charge, and dipole-image dipole interactions, we get that the total electrostatic force on the ball is\n$$\nF=Q E_{0}+\\frac{Q^{2}}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}+\\frac{p Q}{\\pi \\epsilon_{0}(d-2 x)^{3}}+\\frac{3 p^{2}}{2 \\pi \\epsilon_{0}(d-2 x)^{4}}\n$$\nFrom the previous problem, we know that the displacement $x$ of the ball from equilibrium as a function of the electrostatic force $F$ is\n$$\nx=\\frac{F}{l g} \\ln \\left(\\frac{m+l}{m}\\right)\n$$\n\nRearranging this as a quadratic in $Q$ and solving, we have\n$$\n\\begin{gathered}\n\\frac{1}{4 \\pi \\epsilon_{0}(d-2 x)^{2}} Q^{2}+\\left(\\frac{p}{\\pi \\epsilon_{0}(d-2 x)^{3}}+E_{0}\\right) Q+\\left(\\frac{3 p^{2}}{2 \\pi \\epsilon_{0}(d-2 x)^{4}}-\\frac{\\lg x}{\\ln \\left(\\frac{m+l}{m}\\right)}\\right)=0 \\\\\nQ=\\frac{-\\left(\\frac{p}{\\pi \\epsilon_{0}(d-2 x)^{3}}+E_{0}\\right)+\\sqrt{\\left(\\frac{p}{\\pi \\epsilon_{0}(d-2 x)^{3}}+E_{0}\\right)^{2}-4\\left(\\frac{1}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}\\right)\\left(\\frac{3 p^{2}}{2 \\pi \\epsilon_{0}(d-2 x)^{4}}-\\frac{\\lg x}{\\ln \\left(\\frac{m+l}{m}\\right)}\\right)}}{2 \\frac{1}{4 \\pi \\epsilon_{0}(d-2 x)^{2}}} .\n\\end{gathered}\n$$\nPlugging gives $Q=4.48 \\times 10^{-10} \\mathrm{C}$.']",['$4.48 \\times 10^{-10}$'],False,C,Numerical,1e-11 930,Optics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Consider a uniform isosceles triangle prism $\mathrm{ABC}$, with the apex angle $\theta=110^{\circ}$ at vertex $\mathrm{A}$. One of the sides, $\mathrm{AC}$, is coated with silver, allowing it to function as a mirror. When a monochrome light-ray of wavelength $\lambda$ approaches side $\mathrm{AB}$ at an angle of incidence $\alpha$, it first refracts, then reaches side AC, reflects, and continues to base BC. After another refraction, the ray eventually exits the prism at the angle of emergence which is also equal to the angle of incidence (see Fig. A). What is the relative refractive index of the prism for that particular wavelength $\lambda$ with respect to the outside environment, given that $\alpha=70^{\circ}$.","[""The light-path refracts on side $A B$ at point $M$, reflects on side $A C$ at point $N$ and refracts on base $\\mathrm{BC}$ at point $\\mathrm{P}$ (see Fig. A). Define the angle of refraction inside the prism to be $\\beta$, then from Snell's law:\n$$\n\\sin \\alpha=n \\sin \\beta .\n\\tag{29}\n$$\n\nFrom the law of reflection and the $180^{\\circ}$-sum of three interior angles inside any triangles:\n$$\n\\begin{aligned}\n\\widehat{\\mathrm{MNA}} & =180^{\\circ}-\\widehat{\\mathrm{NAM}}-\\widehat{\\mathrm{AMN}}=180^{\\circ}-\\theta-\\left(90^{\\circ}-\\beta\\right) \\\\\n& =\\widehat{\\mathrm{PNC}}=180^{\\circ}-\\widehat{\\mathrm{NCP}}-\\widehat{\\mathrm{CPN}}=180^{\\circ}-\\left(\\frac{180^{\\circ}-\\theta}{2}\\right)-\\left(90^{\\circ}+\\beta\\right),\n\\end{aligned}\n$$\nwe obtain the refraction angle $\\beta$ to be:\n$$\n\\beta=\\frac{3 \\theta-180^{\\circ}}{4}\n$$\nPlug this finding into Eq. (29), we get the relative refraction index of the prism with respect to the outside environment:\n$$\nn=\\frac{\\sin \\alpha}{\\sin \\beta}=\\left.\\frac{\\sin \\alpha}{\\sin \\left(\\frac{3 \\theta-180^{\\circ}}{4}\\right)}\\right|_{\\alpha=70^{\\circ}, \\theta=110^{\\circ}} \\approx 1.5436\n$$""]",['1.5436'],False,,Numerical,5e-2 930,Optics,,"Consider a uniform isosceles triangle prism $\mathrm{ABC}$, with the apex angle $\theta=110^{\circ}$ at vertex $\mathrm{A}$. One of the sides, $\mathrm{AC}$, is coated with silver, allowing it to function as a mirror. When a monochrome light-ray of wavelength $\lambda$ approaches side $\mathrm{AB}$ at an angle of incidence $\alpha$, it first refracts, then reaches side AC, reflects, and continues to base BC. After another refraction, the ray eventually exits the prism at the angle of emergence which is also equal to the angle of incidence (see Fig. A). ![](https://cdn.mathpix.com/cropped/2023_12_21_da5107ae3cf532a9d068g-1.jpg?height=520&width=1732&top_left_y=1152&top_left_x=196) What is the relative refractive index of the prism for that particular wavelength $\lambda$ with respect to the outside environment, given that $\alpha=70^{\circ}$.","[""The light-path refracts on side $A B$ at point $M$, reflects on side $A C$ at point $N$ and refracts on base $\\mathrm{BC}$ at point $\\mathrm{P}$ (see Fig. A). Define the angle of refraction inside the prism to be $\\beta$, then from Snell's law:\n$$\n\\sin \\alpha=n \\sin \\beta .\n\\tag{29}\n$$\n\nFrom the law of reflection and the $180^{\\circ}$-sum of three interior angles inside any triangles:\n$$\n\\begin{aligned}\n\\widehat{\\mathrm{MNA}} & =180^{\\circ}-\\widehat{\\mathrm{NAM}}-\\widehat{\\mathrm{AMN}}=180^{\\circ}-\\theta-\\left(90^{\\circ}-\\beta\\right) \\\\\n& =\\widehat{\\mathrm{PNC}}=180^{\\circ}-\\widehat{\\mathrm{NCP}}-\\widehat{\\mathrm{CPN}}=180^{\\circ}-\\left(\\frac{180^{\\circ}-\\theta}{2}\\right)-\\left(90^{\\circ}+\\beta\\right),\n\\end{aligned}\n$$\nwe obtain the refraction angle $\\beta$ to be:\n$$\n\\beta=\\frac{3 \\theta-180^{\\circ}}{4}\n$$\nPlug this finding into Eq. (29), we get the relative refraction index of the prism with respect to the outside environment:\n$$\nn=\\frac{\\sin \\alpha}{\\sin \\beta}=\\left.\\frac{\\sin \\alpha}{\\sin \\left(\\frac{3 \\theta-180^{\\circ}}{4}\\right)}\\right|_{\\alpha=70^{\\circ}, \\theta=110^{\\circ}} \\approx 1.5436\n$$""]",['1.5436'],False,,Numerical,5e-2 931,Modern Physics,"- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$ - Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$ - Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$ - Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$ - Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$ - Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$ - 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$ - Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$ - Universal Gravitational constant, $$ G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2} $$ - Solar Mass $$ M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg} $$ - Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ - 1 unified atomic mass unit, $$ 1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2} $$ - Planck's constant, $$ h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s} $$ - Permittivity of free space, $$ \epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right) $$ - Coulomb's law constant, $$ k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2} $$ - Permeability of free space, $$ \mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} $$ - Magnetic constant, $$ \frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A} $$ - 1 atmospheric pressure, $$ 1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa} $$ - Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$ - Stefan-Boltzmann constant, $$ \sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4} $$","Field-drive is a locomotion mechanism that is analogous to general relativistic warp-drive. In this mechanism, an active particle continuously climbs up the field-gradient generated by its own influence on the environment so that the particle can bootstrap itself into a constant non-zero velocity motion. Consider a field-drive in one-dimensional (the $\mathrm{O} x$ axis) environment, where the position of the particle at time $t$ is given by $X(t)$ and its instantaneous velocity follows from: $$ \frac{\mathrm{d}}{\mathrm{d} t} X(t)=\left.\kappa \frac{\partial}{\partial x} R(x, t)\right|_{x=X(t)} $$ in which $\kappa$ is called the guiding coefficient and $R(x, t)$ is the field-value in this space. Note that, the operation ... $\left.\right|_{x=X(t)}$ means you have to calculate the part in ... first, then replace $x$ with $X(t)$. For a biological example, the active particle can be a cell, the field can be the nutrient concentration, and the strategy of climbing up the gradient can be chemotaxis. The cell consumes the nutrient and also responses to the local nutrient concentration, biasing its movement toward the direction where the concentration increases the most. If the nutrient is not diffusive and always recovers locally (e.g. a surface secretion) to the value which we defined to be 0 , then its dynamics can usually be approximated by: $$ \frac{\partial}{\partial t} R(x, t)=-\frac{1}{\tau} R(x, t)-\gamma \exp \left\{-\frac{[x-X(t)]^{2}}{2 \lambda^{2}}\right\} $$ where $\tau$ is the timescale of recovery, $\gamma$ is the consumption, and $\lambda$ is the characteristic radius of influence. Before we inoculate the cell into the environment, $R=0$ everywhere at any time. What is the smallest guiding coefficient $\kappa$ (in $\mu \mathrm{m}^{2} / \mathrm{s}$ ) for field-drive to emerge, if the parameters are $\tau=50 \mathrm{~s}, \gamma=1 \mathrm{~s}^{-1}$, and $\lambda=10 \mu \mathrm{m}$. ","['Assume that we inoculate the cell into the environment at position $x=0$ and $t=0$. The field dynamics at $t>0$ can be rewritten as:\n$$\n\\begin{array}{r}\n\\frac{\\partial}{\\partial t} R(x, t)+\\frac{1}{\\tau} R(x, t)=\\exp \\left(-\\frac{t}{\\tau}\\right) \\partial_{t}\\left[\\exp \\left(+\\frac{t}{\\tau}\\right) R(x, t)\\right]=-\\gamma \\exp \\left\\{-\\frac{[x-X(t)]^{2}}{2 \\lambda^{2}}\\right\\} \\\\\n\\Longrightarrow \\exp \\left(+\\frac{t}{\\tau}\\right) R(x, t)=\\int_{0}^{t} d t^{\\prime} \\exp \\left(+\\frac{t^{\\prime}}{\\tau}\\right)\\left(-\\gamma \\exp \\left\\{-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\}\\right) \\\\\n\\Longrightarrow R(x, t)=-\\gamma \\int_{0}^{t} d t^{\\prime} \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\} .\n\\end{array}\n\\tag{39}\n$$\nIf the cell can field-drive at a constant velocity $W>0$, then after a very long time $t \\rightarrow+\\infty$ we expect the cell will be in a steady-state, moving at this velocity. For consistency, this field-drive velocity $W$ should related to the field gradient evaluated at $x=X(t)$ such that:\n$$\nW=\\left.\\kappa \\partial_{x} R(x, t)\\right|_{x=X(t)}\n\\tag{40}\n$$\nFrom Eq. (39) we obtain:\n$$\n\\begin{aligned}\nW & =\\left.\\kappa \\partial_{x}\\left(-\\gamma \\int_{0}^{t} d t^{\\prime} \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\}\\right)\\right|_{x=X(t)} \\\\\n& =\\left.\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{t} d t^{\\prime}\\left[x-X\\left(t^{\\prime}\\right)\\right] \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\}\\right|_{x=X(t)} \\\\\n& =\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{t} d t^{\\prime}\\left[X(t)-X\\left(t^{\\prime}\\right)\\right] \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[X(t)-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\} .\n\\end{aligned}\n$$\nWe then use the steady field-drive condition $X(t)-X\\left(t^{\\prime}\\right)=W\\left(t-t^{\\prime}\\right)$ at $t \\rightarrow+\\infty$ and define $t^{\\prime \\prime}=t-t^{\\prime}$, so that the temporal integration $\\int d t^{\\prime \\prime}$ will run from 0 to $+\\infty$ :\n$$\n\\begin{aligned}\nW & =\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{t} d t^{\\prime}\\left[W\\left(t-t^{\\prime}\\right)\\right] \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[W\\left(t-t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\} \\\\\n& =\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{+\\infty} d t^{\\prime \\prime}\\left(W t^{\\prime \\prime}\\right) \\exp \\left[-\\frac{t^{\\prime \\prime}}{\\tau}-\\frac{\\left(W t^{\\prime \\prime}\\right)^{2}}{2 \\lambda^{2}}\\right] .\n\\end{aligned}\n\\tag{41}\n$$\n\nFor the set of parameter values $(\\kappa, \\tau, \\gamma, \\lambda)$ when the field-drive mechanism start to emerge, we can treat the field-drive velocity as infinitesimal small $W=0^{+}$. Thus, divide both sides of Eq.(41) by $W$, we can arrive at:\n$$\n1=\\left.\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{+\\infty} d t^{\\prime} t^{\\prime \\prime} \\exp \\left[-\\frac{t^{\\prime \\prime}}{\\tau}-\\frac{\\left(W t^{\\prime \\prime}\\right)^{2}}{2 \\lambda^{2}}\\right]\\right|_{W=0^{+}}=\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{+\\infty} d t^{\\prime} t^{\\prime \\prime} \\exp \\left(-\\frac{t^{\\prime \\prime}}{\\tau}\\right)=\\frac{\\kappa \\gamma \\tau^{2}}{\\lambda^{2}}\n$$\nHence, the smallest guiding coefficient that give us field-drive, for $\\tau=50 \\mathrm{~s}, \\gamma=1 \\mathrm{~s}^{-1}, \\lambda=10 \\mu \\mathrm{m}$ :\n$$\n\\kappa=\\frac{\\lambda^{2}}{\\gamma \\tau^{2}}=4 \\times 10^{-2} \\mu \\mathrm{m} / \\mathrm{s}\n$$']",['$4 \\times 10^{-2}$'],False,$\mu \mathrm{m} / \mathrm{s}$,Numerical,1e-3 931,Modern Physics,,"Field-drive is a locomotion mechanism that is analogous to general relativistic warp-drive. In this mechanism, an active particle continuously climbs up the field-gradient generated by its own influence on the environment so that the particle can bootstrap itself into a constant non-zero velocity motion. Consider a field-drive in one-dimensional (the $\mathrm{O} x$ axis) environment, where the position of the particle at time $t$ is given by $X(t)$ and its instantaneous velocity follows from: $$ \frac{\mathrm{d}}{\mathrm{d} t} X(t)=\left.\kappa \frac{\partial}{\partial x} R(x, t)\right|_{x=X(t)} $$ in which $\kappa$ is called the guiding coefficient and $R(x, t)$ is the field-value in this space. Note that, the operation ... $\left.\right|_{x=X(t)}$ means you have to calculate the part in ... first, then replace $x$ with $X(t)$. For a biological example, the active particle can be a cell, the field can be the nutrient concentration, and the strategy of climbing up the gradient can be chemotaxis. The cell consumes the nutrient and also responses to the local nutrient concentration, biasing its movement toward the direction where the concentration increases the most. If the nutrient is not diffusive and always recovers locally (e.g. a surface secretion) to the value which we defined to be 0 , then its dynamics can usually be approximated by: $$ \frac{\partial}{\partial t} R(x, t)=-\frac{1}{\tau} R(x, t)-\gamma \exp \left\{-\frac{[x-X(t)]^{2}}{2 \lambda^{2}}\right\} $$ where $\tau$ is the timescale of recovery, $\gamma$ is the consumption, and $\lambda$ is the characteristic radius of influence. Before we inoculate the cell into the environment, $R=0$ everywhere at any time. What is the smallest guiding coefficient $\kappa$ (in $\mu \mathrm{m}^{2} / \mathrm{s}$ ) for field-drive to emerge, if the parameters are $\tau=50 \mathrm{~s}, \gamma=1 \mathrm{~s}^{-1}$, and $\lambda=10 \mu \mathrm{m}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b3c7c885ee36b4cdd03dg-1.jpg?height=437&width=1014&top_left_y=188&top_left_x=553)","['Assume that we inoculate the cell into the environment at position $x=0$ and $t=0$. The field dynamics at $t>0$ can be rewritten as:\n$$\n\\begin{array}{r}\n\\frac{\\partial}{\\partial t} R(x, t)+\\frac{1}{\\tau} R(x, t)=\\exp \\left(-\\frac{t}{\\tau}\\right) \\partial_{t}\\left[\\exp \\left(+\\frac{t}{\\tau}\\right) R(x, t)\\right]=-\\gamma \\exp \\left\\{-\\frac{[x-X(t)]^{2}}{2 \\lambda^{2}}\\right\\} \\\\\n\\Longrightarrow \\exp \\left(+\\frac{t}{\\tau}\\right) R(x, t)=\\int_{0}^{t} d t^{\\prime} \\exp \\left(+\\frac{t^{\\prime}}{\\tau}\\right)\\left(-\\gamma \\exp \\left\\{-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\}\\right) \\\\\n\\Longrightarrow R(x, t)=-\\gamma \\int_{0}^{t} d t^{\\prime} \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\} .\n\\end{array}\n\\tag{39}\n$$\nIf the cell can field-drive at a constant velocity $W>0$, then after a very long time $t \\rightarrow+\\infty$ we expect the cell will be in a steady-state, moving at this velocity. For consistency, this field-drive velocity $W$ should related to the field gradient evaluated at $x=X(t)$ such that:\n$$\nW=\\left.\\kappa \\partial_{x} R(x, t)\\right|_{x=X(t)}\n\\tag{40}\n$$\nFrom Eq. (39) we obtain:\n$$\n\\begin{aligned}\nW & =\\left.\\kappa \\partial_{x}\\left(-\\gamma \\int_{0}^{t} d t^{\\prime} \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\}\\right)\\right|_{x=X(t)} \\\\\n& =\\left.\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{t} d t^{\\prime}\\left[x-X\\left(t^{\\prime}\\right)\\right] \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[x-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\}\\right|_{x=X(t)} \\\\\n& =\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{t} d t^{\\prime}\\left[X(t)-X\\left(t^{\\prime}\\right)\\right] \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[X(t)-X\\left(t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\} .\n\\end{aligned}\n$$\nWe then use the steady field-drive condition $X(t)-X\\left(t^{\\prime}\\right)=W\\left(t-t^{\\prime}\\right)$ at $t \\rightarrow+\\infty$ and define $t^{\\prime \\prime}=t-t^{\\prime}$, so that the temporal integration $\\int d t^{\\prime \\prime}$ will run from 0 to $+\\infty$ :\n$$\n\\begin{aligned}\nW & =\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{t} d t^{\\prime}\\left[W\\left(t-t^{\\prime}\\right)\\right] \\exp \\left\\{-\\frac{t-t^{\\prime}}{\\tau}-\\frac{\\left[W\\left(t-t^{\\prime}\\right)\\right]^{2}}{2 \\lambda^{2}}\\right\\} \\\\\n& =\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{+\\infty} d t^{\\prime \\prime}\\left(W t^{\\prime \\prime}\\right) \\exp \\left[-\\frac{t^{\\prime \\prime}}{\\tau}-\\frac{\\left(W t^{\\prime \\prime}\\right)^{2}}{2 \\lambda^{2}}\\right] .\n\\end{aligned}\n\\tag{41}\n$$\n\nFor the set of parameter values $(\\kappa, \\tau, \\gamma, \\lambda)$ when the field-drive mechanism start to emerge, we can treat the field-drive velocity as infinitesimal small $W=0^{+}$. Thus, divide both sides of Eq.(41) by $W$, we can arrive at:\n$$\n1=\\left.\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{+\\infty} d t^{\\prime} t^{\\prime \\prime} \\exp \\left[-\\frac{t^{\\prime \\prime}}{\\tau}-\\frac{\\left(W t^{\\prime \\prime}\\right)^{2}}{2 \\lambda^{2}}\\right]\\right|_{W=0^{+}}=\\frac{\\kappa \\gamma}{\\lambda^{2}} \\int_{0}^{+\\infty} d t^{\\prime} t^{\\prime \\prime} \\exp \\left(-\\frac{t^{\\prime \\prime}}{\\tau}\\right)=\\frac{\\kappa \\gamma \\tau^{2}}{\\lambda^{2}}\n$$\nHence, the smallest guiding coefficient that give us field-drive, for $\\tau=50 \\mathrm{~s}, \\gamma=1 \\mathrm{~s}^{-1}, \\lambda=10 \\mu \\mathrm{m}$ :\n$$\n\\kappa=\\frac{\\lambda^{2}}{\\gamma \\tau^{2}}=4 \\times 10^{-2} \\mu \\mathrm{m} / \\mathrm{s}\n$$']",['$4 \\times 10^{-2}$'],False,$\mu \mathrm{m} / \mathrm{s}$,Numerical,1e-3 932,Mechanics,,"1. Restore equilibrium A wooden plank of length $1 \mathrm{~m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height $0.5 \mathrm{~m}$. The specific gravity of the plank (or ratio of plank density to water density) is 0.5 . Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position (exclude the case $\theta=0^{\circ}$ ) ","['\n\nOne could find the net buoyant force by integrating about the point of reference. Or one can directly use the fact this upthrust force acts at the center of portion submerged in water.\n\nA sample solution:\n\nThree forces will act on the plank-\n\n1. Weight which will act at centre of plank.\n\n\n2. Upthrust which will act at centre of submerged portion.\n3. Force from the hinge at $\\mathrm{O}$.\n\nNotation used: Submerged length $=0.5 \\sec \\theta, \\mathrm{F}=$ Upthrust, $w=$ Weight\n\nTaking moments of all three forces about point O. Moment of hinge force will be zero.\n\n$$\n(A \\lg \\rho) \\frac{l}{2} \\sin \\theta=A(0.5 \\sec \\theta)\\left(\\rho_{w}\\right)(g)\\left(\\frac{0.5 \\sec \\theta}{2}\\right) \\sin \\theta\n\\tag{1}\n$$\n\nSolve to get $\\theta=45^{\\circ}$']",['$\\theta=45$'],False,$^{\circ}$,Numerical,0 932,Mechanics,,"1. Restore equilibrium A wooden plank of length $1 \mathrm{~m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height $0.5 \mathrm{~m}$. The specific gravity of the plank (or ratio of plank density to water density) is 0.5 . Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position (exclude the case $\theta=0^{\circ}$ ) ![](https://cdn.mathpix.com/cropped/2023_12_21_97ebfc9f8f7a4705c9ebg-1.jpg?height=570&width=799&top_left_y=623&top_left_x=671)","['![](https://cdn.mathpix.com/cropped/2023_12_21_0c8d119676610ce41c3ag-1.jpg?height=534&width=808&top_left_y=1430&top_left_x=664)\n\nOne could find the net buoyant force by integrating about the point of reference. Or one can directly use the fact this upthrust force acts at the center of portion submerged in water.\n\nA sample solution:\n\nThree forces will act on the plank-\n\n1. Weight which will act at centre of plank.\n\n\n2. Upthrust which will act at centre of submerged portion.\n3. Force from the hinge at $\\mathrm{O}$.\n\nNotation used: Submerged length $=0.5 \\sec \\theta, \\mathrm{F}=$ Upthrust, $w=$ Weight\n\nTaking moments of all three forces about point O. Moment of hinge force will be zero.\n\n$$\n(A \\lg \\rho) \\frac{l}{2} \\sin \\theta=A(0.5 \\sec \\theta)\\left(\\rho_{w}\\right)(g)\\left(\\frac{0.5 \\sec \\theta}{2}\\right) \\sin \\theta\n\\tag{1}\n$$\n\nSolve to get $\\theta=45^{\\circ}$']",['$\\theta=45$'],False,$^{\circ}$,Numerical,0 933,Mechanics,"2. State secrets and the stars In 1949, the physicist G. I. Taylor used publicly released images of the Trinity nuclear test to estimate the yield of the world's first atomic bomb (i.e., the energy released by the detonation). His estimate was stunningly accurate, despite the fact that the number itself was still classified by the government at the time. In this problem, we will follow a simplified version of his estimate, and use the procedure to estimate the energy released by the supernovae which end the lives of massive stars. Figure 1: A snapshot of the Trinity blast wave 15 milliseconds after the explosion, reproduced from Taylor's original paper. A helpful scale bar is included.","(a) After a very short moment, the shock front caused by an explosion sweeps up the material around it, increasing the mass which is blasted outwards. At early times, the energy $E \sim M v^{2} / 2$ can be thought of as roughly conserved. Let $R$ be the radius of the explosion at some time $t$. Estimate the speed of the shock front as $v \sim R / t$ and the swept up mass $M \sim \rho R^{3}$, where $\rho$ is the density of the surrounding material (don't bother to include factors like $4 \pi / 3$ ). Estimate the radius of the explosion $R(t)$ as a function of the energy yield $E$, elapsed time $t$, and the density of the surrounding medium $\rho$.","['Full points should be awarded to a solution as below up to order unity factors.\n\nThe energy of the explosion is $E=M v^{2} / 2 \\sim M v^{2}$. Using $v \\sim R / t$ and $M \\sim \\rho R^{3}$, the total energy of the explosion is\n\n$$\nE \\sim \\frac{\\rho R^{5}}{t^{2}}\n\\tag{2}\n$$\n\nWe can then rearrange this expression as\n\n$$\nR(t) \\sim\\left(\\frac{E}{\\rho}\\right)^{1 / 5} t^{2 / 5}\n\\tag{3}\n$$']",['$\\left(\\frac{E}{\\rho}\\right)^{1 / 5} t^{2 / 5}$'],False,,Expression, 934,Mechanics,"2. State secrets and the stars In 1949, the physicist G. I. Taylor used publicly released images of the Trinity nuclear test to estimate the yield of the world's first atomic bomb (i.e., the energy released by the detonation). His estimate was stunningly accurate, despite the fact that the number itself was still classified by the government at the time. In this problem, we will follow a simplified version of his estimate, and use the procedure to estimate the energy released by the supernovae which end the lives of massive stars. Figure 1: A snapshot of the Trinity blast wave 15 milliseconds after the explosion, reproduced from Taylor's original paper. A helpful scale bar is included. Context question: (a) After a very short moment, the shock front caused by an explosion sweeps up the material around it, increasing the mass which is blasted outwards. At early times, the energy $E \sim M v^{2} / 2$ can be thought of as roughly conserved. Let $R$ be the radius of the explosion at some time $t$. Estimate the speed of the shock front as $v \sim R / t$ and the swept up mass $M \sim \rho R^{3}$, where $\rho$ is the density of the surrounding material (don't bother to include factors like $4 \pi / 3$ ). Estimate the radius of the explosion $R(t)$ as a function of the energy yield $E$, elapsed time $t$, and the density of the surrounding medium $\rho$. Context answer: \boxed{$\left(\frac{E}{\rho}\right)^{1 / 5} t^{2 / 5}$} ","(b) Air has a density $\rho \simeq 1 \mathrm{~kg} \mathrm{~m}^{-3}$. Using the provided image, roughly estimate the yield of the Trinity explosion in kilotons (of TNT). Note that 1 kiloton is equal to approximately $4.2 \times 10^{12} \mathrm{~J}$.","['From the included figure, the explosion looks to be approximately $\\sim 100 \\mathrm{~m}$ in radius at a time $\\sim 15 \\mathrm{~ms}$. We can use Equation 2 with the given values to obtain\n\n$$\nE \\sim 11 \\text { kilotons of TNT }\n\\tag{4}\n$$\n\nNote that this value may vary significantly depending on the estimated radius from the figure (since $E \\propto R^{5}$ depends strongly on $R$ ), but should be in the neighborhood of tens of kilotons of TNT. The official number released by the government at the time was 21 kilotons of TNT, although there has been some uncertainty in this number.']",['$11$'],False,kilotons of TNT,Numerical,5e-5 935,Mechanics,"2. State secrets and the stars In 1949, the physicist G. I. Taylor used publicly released images of the Trinity nuclear test to estimate the yield of the world's first atomic bomb (i.e., the energy released by the detonation). His estimate was stunningly accurate, despite the fact that the number itself was still classified by the government at the time. In this problem, we will follow a simplified version of his estimate, and use the procedure to estimate the energy released by the supernovae which end the lives of massive stars. Figure 1: A snapshot of the Trinity blast wave 15 milliseconds after the explosion, reproduced from Taylor's original paper. A helpful scale bar is included. Context question: (a) After a very short moment, the shock front caused by an explosion sweeps up the material around it, increasing the mass which is blasted outwards. At early times, the energy $E \sim M v^{2} / 2$ can be thought of as roughly conserved. Let $R$ be the radius of the explosion at some time $t$. Estimate the speed of the shock front as $v \sim R / t$ and the swept up mass $M \sim \rho R^{3}$, where $\rho$ is the density of the surrounding material (don't bother to include factors like $4 \pi / 3$ ). Estimate the radius of the explosion $R(t)$ as a function of the energy yield $E$, elapsed time $t$, and the density of the surrounding medium $\rho$. Context answer: \boxed{$\left(\frac{E}{\rho}\right)^{1 / 5} t^{2 / 5}$} Context question: (b) Air has a density $\rho \simeq 1 \mathrm{~kg} \mathrm{~m}^{-3}$. Using the provided image, roughly estimate the yield of the Trinity explosion in kilotons (of TNT). Note that 1 kiloton is equal to approximately $4.2 \times 10^{12} \mathrm{~J}$. Context answer: \boxed{$11$} ","(c) In 1054, Chinese astronomers observed and documented a supernova which was bright enough to be visible during the day for around a month. The rubble left behind is a rapidly expanding supernova remnant called the Crab Nebula, which is intensely studied today. Figure 2: Photographs of the Crab Nebula taken by the Hale Telescope in 1950 (top left) and the Hubble Space Telescope in 2000 (bottom right). The surrounding interstellar medium contains about 1 proton per cubic meter-protons have a mass $m_{p} \approx 1.7 \times 10^{-27} \mathrm{~kg}$. While not perfectly spherical, Crab Nebula has a radius of roughly $1.7 \mathrm{pc}\left(1 \mathrm{pc}=3.1 \times 10^{16} \mathrm{~m}\right)$. From this information and the given images of the Crab Nebula from 1950 and 2000, estimate the energy released by the initial supernova. Your answer should be close to the typical energy scales of supernovae, which briefly outshine their host galaxies and can be seen from across the universe.","['The density of the interstellar medium is\n\n$$\n\\rho \\sim \\frac{m_{p}}{\\left(1 \\mathrm{~cm}^{3}\\right)} \\sim 1.7 \\times 10^{-21} \\mathrm{~kg} \\mathrm{~cm}^{-3}\n\\tag{5}\n$$\n\nWe are given that the Crab Nebula has a radius of approximately 1.7 pc. We see that the explosion seems to expand by $\\sim 25 \\%$ or so during the fifty years of observation along the long axis, which we take to be a very rough measure of its velocity. In particular,\n\n$$\nv \\sim \\frac{25 \\% \\times 1.7 \\mathrm{pc}}{50 \\mathrm{yr}} \\sim 8.3 \\times 10^{6} \\mathrm{~m} \\mathrm{~s}^{-1}\n\\tag{6}\n$$\n\nWe can use $E \\sim M v^{2}$ and $M \\sim \\rho R^{3}$ to write\n\n$$\nE \\sim \\rho R^{3} v^{2} \\sim 1.7 \\times 10^{43} \\mathrm{~J}\n\\tag{7}\n$$\n\nwhere we have used $t \\sim 1000 \\mathrm{yr}$ (the age of the supernova remnant). This is $E \\sim$ $1.7 \\times 10^{50} \\mathrm{erg}$, which is in the neighborhood of true supernovae which are typically at $\\sim 10^{51} \\mathrm{erg}$ (a quantity of energy which has been affectionately named the ""f.o.e."" by astrophysicists because of this fact).']",['$1.7 \\times 10^{43}$'],False,J,Numerical,5e42 935,Mechanics,,"(c) In 1054, Chinese astronomers observed and documented a supernova which was bright enough to be visible during the day for around a month. The rubble left behind is a rapidly expanding supernova remnant called the Crab Nebula, which is intensely studied today. ![](https://cdn.mathpix.com/cropped/2023_12_21_f9686d2b2f582fd8d3cfg-1.jpg?height=837&width=898&top_left_y=232&top_left_x=608) Figure 2: Photographs of the Crab Nebula taken by the Hale Telescope in 1950 (top left) and the Hubble Space Telescope in 2000 (bottom right). The surrounding interstellar medium contains about 1 proton per cubic meter-protons have a mass $m_{p} \approx 1.7 \times 10^{-27} \mathrm{~kg}$. While not perfectly spherical, Crab Nebula has a radius of roughly $1.7 \mathrm{pc}\left(1 \mathrm{pc}=3.1 \times 10^{16} \mathrm{~m}\right)$. From this information and the given images of the Crab Nebula from 1950 and 2000, estimate the energy released by the initial supernova. Your answer should be close to the typical energy scales of supernovae, which briefly outshine their host galaxies and can be seen from across the universe.","['The density of the interstellar medium is\n\n$$\n\\rho \\sim \\frac{m_{p}}{\\left(1 \\mathrm{~cm}^{3}\\right)} \\sim 1.7 \\times 10^{-21} \\mathrm{~kg} \\mathrm{~cm}^{-3}\n\\tag{5}\n$$\n\nWe are given that the Crab Nebula has a radius of approximately 1.7 pc. We see that the explosion seems to expand by $\\sim 25 \\%$ or so during the fifty years of observation along the long axis, which we take to be a very rough measure of its velocity. In particular,\n\n$$\nv \\sim \\frac{25 \\% \\times 1.7 \\mathrm{pc}}{50 \\mathrm{yr}} \\sim 8.3 \\times 10^{6} \\mathrm{~m} \\mathrm{~s}^{-1}\n\\tag{6}\n$$\n\nWe can use $E \\sim M v^{2}$ and $M \\sim \\rho R^{3}$ to write\n\n$$\nE \\sim \\rho R^{3} v^{2} \\sim 1.7 \\times 10^{43} \\mathrm{~J}\n\\tag{7}\n$$\n\nwhere we have used $t \\sim 1000 \\mathrm{yr}$ (the age of the supernova remnant). This is $E \\sim$ $1.7 \\times 10^{50} \\mathrm{erg}$, which is in the neighborhood of true supernovae which are typically at $\\sim 10^{51} \\mathrm{erg}$ (a quantity of energy which has been affectionately named the ""f.o.e."" by astrophysicists because of this fact).']",['$1.7 \\times 10^{43}$'],False,J,Numerical,5e42 936,Optics,"3. Stick a pin there The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature $20 \mathrm{~cm}$. The concave surface has a radius of curvature $60 \mathrm{~cm}$. The convex side is silvered and placed on a horizontal surface.",(a) Where should a pin be placed on the optical axis such that its image is formed at the same place?,"['Image of object will coincide with it if ray of light after refraction from the concave surface fall normally on concave mirror so formed by silvering the convex surface. Or image after refraction from concave surface should form at centre of curvature of concave mirror or at a distance of $20 \\mathrm{~cm}$ on same side of the combination. Let $x$ be the distance of pin from the given optical system.\n\nUsing,\n\n$$\n\\frac{\\mu_{2}}{\\nu}-\\frac{\\mu_{1}}{u}=\\frac{\\mu_{2}-\\mu_{1}}{R}\n\\tag{8}\n$$\n\nWith proper signs,\n\n$$\n\\frac{1.5}{-20}-\\frac{1}{-x}=\\frac{1.5-1}{-60}\n\\tag{9}\n$$\n\nSolve to get $x=15 \\mathrm{~cm}$']",['15'],False,cm,Numerical,5e-1 937,Optics,"3. Stick a pin there The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature $20 \mathrm{~cm}$. The concave surface has a radius of curvature $60 \mathrm{~cm}$. The convex side is silvered and placed on a horizontal surface. Context question: (a) Where should a pin be placed on the optical axis such that its image is formed at the same place? Context answer: \boxed{15} ","(b) If the concave part is filled with water of refractive index $4 / 3$, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin. ","['There could be different approaches to solve this. A sample solution:\n\nNow, before striking with the concave surface, the ray is first refracted from a plane surface. So, let $\\mathrm{x}$ be the distance of pin, then the plane surface will form its image at a distance $\\frac{4}{3} x\\left(h_{a p p}=\\mu h\\right)$ from it.\n\nUsing,\n\n$$\n\\frac{\\mu_{2}}{\\nu}-\\frac{\\mu_{1}}{u}=\\frac{\\mu_{2}-\\mu_{1}}{R}\n\\tag{10}\n$$\n\nwith proper signs,\n\n$$\n\\frac{1.5}{-20}-\\frac{4 / 3}{-4 x / 3}=\\frac{1.5-4 / 3}{-60}\n\\tag{11}\n$$\n\nSolve to get $x=13.84 \\mathrm{~cm}$.\n\nTherefore $\\Delta x=x_{1}-x_{2}=15 \\mathrm{~cm}-13.84 \\mathrm{~cm}=1.16 \\mathrm{~cm}$ (Downwards)']",['1.16'],False,cm,Numerical,2e-1 937,Optics,,"(b) If the concave part is filled with water of refractive index $4 / 3$, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin. ![](https://cdn.mathpix.com/cropped/2023_12_21_c91fdc17266671ace758g-1.jpg?height=341&width=629&top_left_y=949&top_left_x=821)","['There could be different approaches to solve this. A sample solution:\n\nNow, before striking with the concave surface, the ray is first refracted from a plane surface. So, let $\\mathrm{x}$ be the distance of pin, then the plane surface will form its image at a distance $\\frac{4}{3} x\\left(h_{a p p}=\\mu h\\right)$ from it.\n\nUsing,\n\n$$\n\\frac{\\mu_{2}}{\\nu}-\\frac{\\mu_{1}}{u}=\\frac{\\mu_{2}-\\mu_{1}}{R}\n\\tag{10}\n$$\n\nwith proper signs,\n\n$$\n\\frac{1.5}{-20}-\\frac{4 / 3}{-4 x / 3}=\\frac{1.5-4 / 3}{-60}\n\\tag{11}\n$$\n\nSolve to get $x=13.84 \\mathrm{~cm}$.\n\nTherefore $\\Delta x=x_{1}-x_{2}=15 \\mathrm{~cm}-13.84 \\mathrm{~cm}=1.16 \\mathrm{~cm}$ (Downwards)']",['1.16'],False,cm,Numerical,2e-1 938,Mechanics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$","(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$","['We can write down the equations of motion, multiplying the second one by $i$ :\n\n$$\n\\begin{aligned}\n& 0=\\ddot{x}+2 \\Omega \\dot{y}-\\Omega^{2} x \\\\\n& 0=\\ddot{y}-2 \\Omega i \\dot{x}-\\Omega^{2} i y\n\\end{aligned}\n\\tag{19}\n$$\n\nWe can add these equations together to obtain\n\n$$\n0=(\\ddot{x}+i \\ddot{y})+2 \\Omega(\\dot{y}-i \\dot{x})-\\Omega^{2}(x+i y)=\\ddot{\\eta}+2 \\Omega(\\dot{y}-i \\dot{x})-\\Omega^{2} \\eta\n\\tag{20}\n$$\n\nWe note that\n\n$$\n\\dot{y}-i \\dot{x}=-i(\\dot{x}+i \\dot{y})=-i \\eta\n\\tag{21}\n$$\n\nThen\n\n$$\n0=\\ddot{\\eta}-2 i \\Omega \\dot{\\eta}-\\Omega^{2} \\eta\n\\tag{22}\n$$']",,False,,, 939,Mechanics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} ","(b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be?",['We can plug in $\\eta=\\alpha e^{\\lambda t}$ :\n\n$$\n0=\\lambda^{2} \\alpha e^{\\lambda t}-2 i \\lambda \\Omega \\alpha e^{\\lambda t}-\\Omega^{2} \\alpha e^{\\lambda t}\n\\tag{23}\n$$\n\nThen we can cancel common factors and find that\n\n$$\n0=\\lambda^{2}-2 i \\lambda \\Omega-\\Omega^{2}=(\\lambda-i \\Omega)^{2}\n\\tag{24}\n$$\n\nWe see that $\\lambda=i \\Omega$.'],['$\\lambda=i \\Omega$'],False,,Expression, 940,Mechanics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} Context question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be? Context answer: \boxed{$\lambda=i \Omega$} ","(c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$. This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution.","['Using our answer to part (b) and $\\alpha=A e^{i \\phi}$, we have\n\n$$\n\\eta(t)=A e^{i(\\Omega t+\\phi)}\n\\tag{25}\n$$\n\nUsing the Euler identity, we have\n\n$$\nx(t)+i y(t)=A \\cos (\\Omega t+\\phi)+i A \\sin (\\Omega t+\\phi)\n\\tag{26}\n$$\n\nThe real and imaginary parts become\n\n$$\n\\begin{aligned}\n& x(t)=A \\cos (\\Omega t+\\phi) \\\\\n& y(t)=A \\sin (\\Omega t+\\phi)\n\\end{aligned}\n\\tag{27}\n$$']","['$x(t)=A \\cos (\\Omega t+\\phi)$ , $y(t)=A \\sin (\\Omega t+\\phi)$']",True,,Expression, 941,Mechanics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} Context question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be? Context answer: \boxed{$\lambda=i \Omega$} Context question: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$. This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution. Context answer: \boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$} ","(d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by $$ \frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}} \tag{5} $$ A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation.""","['Consider the given differential equation:\n\n$$\n\\frac{\\partial \\psi}{\\partial t}=a \\frac{\\partial^{2} \\psi}{\\partial x^{2}}\n\\tag{28}\n$$\n\nWe can plug in $\\psi(x, t)=A e^{i k x-i \\omega t}$ to find\n\n$$\n-i \\omega A e^{i k x-i \\omega t}=-k^{2} a A e^{i k x-i \\omega t}\n\\tag{29}\n$$\n\nso that\n\n$$\n\\omega=-i k^{2} a\n\\tag{30}\n$$']",['$\\omega=-i k^{2} a$'],False,,Expression, 942,Mechanics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} Context question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be? Context answer: \boxed{$\lambda=i \Omega$} Context question: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$. This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution. Context answer: \boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$} Context question: (d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by $$ \frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}} \tag{5} $$ A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation."" Context answer: \boxed{$\omega=-i k^{2} a$} ","(e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by $$ i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}} \tag{6} $$ Using your answer to part (d), what is the dispersion relation of the Schrödinger equation?","['We see that the free Schrödinger equation takes the form of Equation 28, but with\n\n$$\na=\\frac{i \\hbar}{2 m}\n\\tag{31}\n$$\n\nThen, using our answer from part (d), we have\n\n$$\n\\omega=\\frac{\\hbar k^{2}}{2 m}\n\\tag{32}\n$$']",['$\\omega=\\frac{\\hbar k^{2}}{2 m}$'],False,,Expression, 943,Mechanics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} Context question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be? Context answer: \boxed{$\lambda=i \Omega$} Context question: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$. This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution. Context answer: \boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$} Context question: (d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by $$ \frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}} \tag{5} $$ A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation."" Context answer: \boxed{$\omega=-i k^{2} a$} Context question: (e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by $$ i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}} \tag{6} $$ Using your answer to part (d), what is the dispersion relation of the Schrödinger equation? Context answer: \boxed{$\omega=\frac{\hbar k^{2}}{2 m}$} ","(f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$.","['We can multiply both sides of the answer to part (e) by $\\hbar$ and use $E=\\hbar \\omega$ and $p=\\hbar k$, we have\n\n$$\nE=\\frac{p^{2}}{2 m}\n\\tag{33}\n$$\n\nA classical momentum has $p=m v$, so this gives the classical energy for a free particle (i.e., one without a potential):\n\n$$\nE=\\frac{1}{2} m v^{2}\n\\tag{34}\n$$']",,False,,, 944,Modern Physics,"4. A complex dance In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers. Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary ""growth"" and ""decay"" can be translated into oscillations by the Euler identity: $$ e^{i \theta}=\cos \theta+i \sin \theta \tag{1} $$ Context question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form $$ \vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r})) \tag{2} $$ For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations, $$ \begin{aligned} & 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\ & 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y \end{aligned} \tag{3} $$ where dots represent time derivatives. Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation: $$ 0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta \tag{4} $$ Context answer: \boxed{证明题} Context question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$. Plugging in this guess, what must $\lambda$ be? Context answer: \boxed{$\lambda=i \Omega$} Context question: (c) Using your answer to part (b), and defining $\alpha=A e^{i \phi}$ where $A$ and $\phi$ are real, find $\mathbf{x}(\mathbf{t})$ and $\mathbf{y}(\mathbf{t})$. This is the trajectory for a particle which is stationary with respect to the symmetry axis. While not required for this problem, an additional guess would reveal that $\eta=\beta t e^{\lambda t}$ is also a solution. Context answer: \boxed{$x(t)=A \cos (\Omega t+\phi)$ , $y(t)=A \sin (\Omega t+\phi)$} Context question: (d) The one-dimensional diffusion equation (also called the ""heat equation"") is given (for a free particle) by $$ \frac{\partial \psi}{\partial t}=a \frac{\partial^{2} \psi}{\partial x^{2}} \tag{5} $$ A spatial wave can be written as $\sim e^{i k x}$ (larger $k$ 's correspond to waves oscillating on smaller length scales). Guessing a solution $\psi(x, t)=A e^{i k x-i \omega t}$, find $\omega$ in terms of k. A relationship of this time is called a ""dispersion relation."" Context answer: \boxed{$\omega=-i k^{2} a$} Context question: (e) The most important equation of non-relativistic quantum mechanics is the Schrödinger equation, which is given by $$ i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \psi}{\partial x^{2}} \tag{6} $$ Using your answer to part (d), what is the dispersion relation of the Schrödinger equation? Context answer: \boxed{$\omega=\frac{\hbar k^{2}}{2 m}$} Context question: (f) If the energy of a wave is $E=\hbar \omega$ and the momentum is $p=\hbar k$, show that the dispersion relation found in part (e) resembles the classical expectation for the kinetic energy of a particle, $\mathrm{E}=\mathrm{mv}^{2} / \mathbf{2}$. Context answer: \boxed{证明题} ","(g) The theory of relativity instead posits that the energy of a particle is given by $E=\sqrt{p^{2} c^{2}+m^{2} c^{4}}$. In accordance with this, we can try to guess a relativistic version of the Schrödinger equation: $$ \frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{m^{2} c^{2}}{\hbar^{2}} \phi=0 \tag{7} $$ This is called the Klein-Gordon equation. Using the same guess as before, find $\omega$ in terms of $\mathrm{k}$. Hint: If you are careful, you should find that there is an infinite continuum of energy states extending down to negative infinity. This apparently mathematical issue hints at the existence of antimatter, and ultimately demonstrates to us that we must formulate quantum field theory to properly describe relativistic quantum physics.","['We consider the Klein-Gordon equation:\n\n$$\n\\frac{1}{c^{2}} \\frac{\\partial^{2} \\phi}{\\partial t^{2}}-\\frac{\\partial^{2} \\phi}{\\partial x^{2}}+\\frac{m^{2} c^{2}}{\\hbar^{2}} \\phi=0\n\\tag{35}\n$$\n\nWe can, as before, plug in a guess $\\phi(x, t)=A e^{i k x-i \\omega t}$. This yields\n\n$$\n-\\frac{\\omega^{2}}{c^{2}} A e^{i k x-i \\omega t}+k^{2} A e^{i k x-i \\omega t}+\\frac{m^{2} c^{2}}{\\hbar^{2}} A e^{i k x-i \\omega t}=0\n\\tag{36}\n$$\n\nCancelling out common terms, we see that\n\n$$\n\\omega^{2}=k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}\n\\tag{37}\n$$\n\nor\n\n$$\n\\omega= \\pm \\sqrt{k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}}\n\\tag{38}\n$$\n\nNote that both positive and negative $\\omega$ solve the Klein-Gordon equation.']","['$\\sqrt{k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}}$, $-\\sqrt{k^{2} c^{2}+\\frac{m^{2} c^{4}}{\\hbar^{2}}}$']",True,,Expression, 945,Electromagnetism,,"(a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body.","['In conductors, outer electron of each atom are free to move around the body. So if there is a non-zero electric field inside, this would move those free charges to the surface, at which point they cannot escape further, and they would accumulate there. This process would stop when the electric field created by the polarized charges exactly counter the external electric field inside. Consequently, any external electric field would cause a polarization of the body that results in a net zero field inside. Since $\\nabla \\phi=-\\mathbf{E}=\\mathbf{0}$, we have constant $\\phi$ inside. From the continuity, it is also constant and equal to $\\phi_{\\text {inside }}$ throughout the surface.']",['开放性回答'],False,,Need_human_evaluate, 946,Electromagnetism,"5. Polarization and Oscillation In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations. Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system. Context question: (a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body. Context answer: 开放性回答 ","(b) Laplace's equation is a second order differential equation $$ \nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0 \tag{8} $$ Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary. Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$. Hint: Consider $\phi=\phi_{1}-\phi_{2}$.","[""Suppose that we have two potentials satisfying the Laplace's equation and the boundary conditions. $\\nabla^{2} \\phi_{1}=0, \\nabla^{2} \\phi_{2}=0$ and $\\phi_{1}=\\phi_{2}$ on the boundary of $V$. Define $\\phi_{3}=\\phi_{1}-\\phi_{2}$. By the linearity of Laplace's equation, $\\nabla^{2} \\phi_{2}=0$ and $\\phi_{3}=0$ on the surface. By the maximum principle, all extrema occurs at the surface. Thus, both the minimum and the maximum of $\\phi_{3}$ is 0 . Therefore, $\\phi_{3}=0$. This concludes the proof that $\\phi_{1}=\\phi_{2}$ identically.""]",,False,,, 947,Electromagnetism,"5. Polarization and Oscillation In this problem, we will understand the polarization of metallic bodies and the method of images that simplifies the math in certain geometrical configurations. Throughout the problem, suppose that metals are excellent conductors and they polarize significantly faster than the classical relaxation of the system. Context question: (a) Explain in words why there can't be a non-zero electric field in a metallic body, and why this leads to constant electric potential throughout the body. Context answer: 开放性回答 Context question: (b) Laplace's equation is a second order differential equation $$ \nabla^{2} \phi=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}+\frac{\partial^{2} \phi}{\partial z^{2}}=0 \tag{8} $$ Solutions to this equation are called harmonic functions. One of the most important properties satisfied by these functions is the maximum principle. It states that a harmonic function attains extremes on the boundary. Using this, prove the uniqueness theorem: Solution to Laplace's equation in a volume $V$ is uniquely determined if its solution on the boundary is specified. That is, if $\nabla^{2} \phi_{1}=0$, $\nabla^{2} \phi_{2}=0$ and $\phi_{1}=\phi_{2}$ on the boundary of $V$, then $\phi_{1}=\phi_{2}$ in $V$. Hint: Consider $\phi=\phi_{1}-\phi_{2}$. Context answer: \boxed{证明题} ","(c) The uniqueness theorem allows us to use ""image"" charges in certain settings to describe the system. Consider one such example: There is a point-like charge $q$ at a distance $L$ from a metallic sphere of radius $R$ attached to the ground. As you argued in part (a), sphere will be polarized to make sure the electric potential is constant throughout its body. Since it is attached to the ground, the constant potential will be zero. Place an image charge inside the sphere to counter the non-uniform potential of the outer charge $q$ on the surface. Where should this charge be placed, and what is its value?","['Take the origin of our coordinate system at the center of the sphere and align the axes so that the coordinates of our charge is $(x, y)=(L, 0)$. Now suppose we place an image charge $q_{0}$ at $(x, 0)$ where $0R+a$. ![](https://cdn.mathpix.com/cropped/2023_12_21_78a48c63f6de68a072e3g-1.jpg?height=553&width=634&top_left_y=580&top_left_x=775)","['Force is attractive. Therefore, any deviation from the equilibrium $(\\theta=0)$ induces an attractive torque towards $\\theta=0$.\n\nForce at the equilibrium would be the same as in part (e) with $L$ replaced by $L-a$. Since we are looking for small oscillations, we need to calculate the torque in the first order approximation of $\\theta$. Denote $\\beta$ as the angle between the line segment from the center of the sphere to the particle and from the particle to the wall. Then the equations of motion become\n\n$$\n\\begin{aligned}\n\\tau=-F(\\theta) a \\sin (\\beta) & \\approx-F(0) a \\frac{L}{L-a} \\theta \\\\\n& =I \\frac{d^{2} \\theta}{d t^{2}} \\\\\n& =\\left(m+\\frac{M}{3}\\right) a^{2} \\frac{d^{2} \\theta}{d t^{2}}\n\\end{aligned}\n$$\n\nThus it follows that\n\n$$\n\\omega=\\frac{q}{(L-a)^{2}-R^{2}} \\sqrt{\\frac{R L}{4 \\pi \\epsilon_{0} a\\left(m+\\frac{M}{3}\\right)}}\n$$\n\nIn the limit $a<R+a$. ","['Force is attractive. Therefore, any deviation from the equilibrium $(\\theta=0)$ induces an attractive torque towards $\\theta=0$.\n\nForce at the equilibrium would be the same as in part (e) with $L$ replaced by $L-a$. Since we are looking for small oscillations, we need to calculate the torque in the first order approximation of $\\theta$. Denote $\\beta$ as the angle between the line segment from the center of the sphere to the particle and from the particle to the wall. Then the equations of motion become\n\n$$\n\\begin{aligned}\n\\tau=-F(\\theta) a \\sin (\\beta) & \\approx-F(0) a \\frac{L}{L-a} \\theta \\\\\n& =I \\frac{d^{2} \\theta}{d t^{2}} \\\\\n& =\\left(m+\\frac{M}{3}\\right) a^{2} \\frac{d^{2} \\theta}{d t^{2}}\n\\end{aligned}\n$$\n\nThus it follows that\n\n$$\n\\omega=\\frac{q}{(L-a)^{2}-R^{2}} \\sqrt{\\frac{R L}{4 \\pi \\epsilon_{0} a\\left(m+\\frac{M}{3}\\right)}}\n$$\n\nIn the limit $a< Figure 1: Cylindrical capacitor held above the fluid. 1. Dielectric Fluid Consider a perfectly conducting open-faced cylindrical capacitor with height $l$, inner radius $a$ and outer radius $b$. A constant voltage $V$ is applied continuously across the capacitor by a battery, and one of the open faces is immersed slightly into a fluid with dielectric constant $\kappa$ and mass density $\rho$. Energy considerations cause the fluid to rise up into the capacitor.",(a) What is the total energy stored in the capacitor before any fluid rises as a function of its height $l$ ?,"[""Gauss' law gives $E=\\frac{\\lambda}{2 \\pi \\epsilon_{0} r}$ inside the capacitor. The voltage is then determined by $V=-\\int_{b}^{a} E d r$ which yields $V=\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\log (b / a)$. The capacitance per length is determined by $\\frac{\\lambda}{V}=\\frac{2 \\pi \\epsilon_{0}}{\\log (b / a)}$, and the energy is found to be\n\n$$\nU=\\frac{\\pi \\epsilon_{0} l}{\\log (b / a)} V^{2}\n$$""]",['$U=\\frac{\\pi \\epsilon_{0} l}{\\log (b / a)} V^{2}$'],False,,Expression, 952,Electromagnetism," Figure 1: Cylindrical capacitor held above the fluid. 1. Dielectric Fluid Consider a perfectly conducting open-faced cylindrical capacitor with height $l$, inner radius $a$ and outer radius $b$. A constant voltage $V$ is applied continuously across the capacitor by a battery, and one of the open faces is immersed slightly into a fluid with dielectric constant $\kappa$ and mass density $\rho$. Energy considerations cause the fluid to rise up into the capacitor. Context question: (a) What is the total energy stored in the capacitor before any fluid rises as a function of its height $l$ ? Context answer: \boxed{$U=\frac{\pi \epsilon_{0} l}{\log (b / a)} V^{2}$} ",(b) How high $h$ does the dielectric fluid rise against the force of gravity given by acceleration $g$ ? (Note: it is very easy to get the correct answer while describing the problem incorrectly. Take care for full credit.),"[""The key assumption is that $V$ is held constant, and therefore the capacitor system's energy balance is not isolated, and in general you should take into account the work done by the battery. The total energy is given by $U=U_{\\text {bat }}+U_{\\mathrm{EM}}+U_{\\text {grav }} \\cdot U_{\\text {bat }}=-\\int V d q=-V Q$ at constant $V, U_{\\mathrm{EM}}=\\frac{1}{2} Q V$ is the energy of the capacitor, and\n\n$$\nd U_{\\text {grav }}=g y \\cdot d m \\Longrightarrow U_{\\text {grav }}=\\frac{\\rho g y^{2} \\pi\\left(b^{2}-a^{2}\\right)}{2}\n$$\n\n\n\nthus,\n\n$$\nU=-Q V+\\frac{1}{2} Q V+\\frac{\\rho g y^{2} \\pi\\left(b^{2}-a^{2}\\right)}{2}\n$$\n\nEquilibrium is found by minimizing this function, so\n\n$$\n0=-\\frac{1}{2} \\frac{d Q}{d y} V+\\rho g y \\pi\\left(b^{2}-a^{2}\\right)\n$$\n\nNow we must determine $\\frac{d Q}{d y}$ as follows. From Gauss' law, $\\int D \\cdot d A=Q$. It's clear from the symmetry that the $E, D$ fields could only point radially, and that this automatically satisfies the boundary condition $D_{1}^{\\perp}-D_{2}^{\\perp}=0$ at the air-dielectric boundary. This symmetry allows us to break up the surface integral\n\n$$\nQ=\\int_{\\text {top part }} \\epsilon_{0} E \\cdot d A+\\int_{\\text {bottom part }} \\kappa \\epsilon_{0} E \\cdot d A\n$$\n\nto conclude\n\n$$\nE=\\frac{Q}{2 \\pi \\epsilon_{0} r[(\\kappa-1) y+l]}\n$$\n\nwhich satisfies the remaining boundary condition. Then using $V=-\\int E d r$,\n\n$$\nQ=\\frac{2 \\pi \\epsilon_{0}[(\\kappa-1) y+l]}{\\log (b / a)} V\n$$\n\nand therefore\n\n$$\n\\frac{d Q}{d y}=\\frac{2 \\pi \\epsilon_{0}(\\kappa-1)}{\\log (b / a)} V\n$$\n\nand we conclude\n\n$$\n0=\\frac{\\pi \\epsilon_{0}(1-\\kappa)}{\\log (b / a)} V^{2}+\\rho g y \\pi\\left(b^{2}-a^{2}\\right)\n$$\n\nwhich is solved to yield\n\n$$\nh=\\frac{\\epsilon_{0}(\\kappa-1) V^{2}}{\\log (b / a) \\rho g\\left(b^{2}-a^{2}\\right)}\n$$""]",['$h=\\frac{\\epsilon_{0}(\\kappa-1) V^{2}}{\\log (b / a) \\rho g\\left(b^{2}-a^{2}\\right)}$'],False,,Expression, 953,Electromagnetism," Figure 1: Cylindrical capacitor held above the fluid. 1. Dielectric Fluid Consider a perfectly conducting open-faced cylindrical capacitor with height $l$, inner radius $a$ and outer radius $b$. A constant voltage $V$ is applied continuously across the capacitor by a battery, and one of the open faces is immersed slightly into a fluid with dielectric constant $\kappa$ and mass density $\rho$. Energy considerations cause the fluid to rise up into the capacitor. Context question: (a) What is the total energy stored in the capacitor before any fluid rises as a function of its height $l$ ? Context answer: \boxed{$U=\frac{\pi \epsilon_{0} l}{\log (b / a)} V^{2}$} Context question: (b) How high $h$ does the dielectric fluid rise against the force of gravity given by acceleration $g$ ? (Note: it is very easy to get the correct answer while describing the problem incorrectly. Take care for full credit.) Context answer: \boxed{$h=\frac{\epsilon_{0}(\kappa-1) V^{2}}{\log (b / a) \rho g\left(b^{2}-a^{2}\right)}$} ","(c) Calculate the pressure difference $P$ above atmospheric pressure needed to suck the fluid to the top of the cylinder, assuming this is possible. Assume $l \gg h$ and neglect fluid dynamical and thermodynamic effects. You will find that $P$ consists of a term depending on $l$ and a term depending on $\kappa-1$. Provide a physical interpretation of these terms.","['Consider energy of the overall system. Define $A \\equiv \\frac{\\pi \\epsilon_{0} V^{2}}{\\log (b / a)}$,\n\n$$\n\\begin{aligned}\nU_{i} & =\\frac{\\rho g h^{2} \\pi\\left(b^{2}-a^{2}\\right)}{2}-A[(\\kappa-1) h+l] \\\\\nU_{f} & =\\frac{\\rho g l^{2} \\pi\\left(b^{2}-a^{2}\\right)}{2}-A[(\\kappa-1) l+l]\n\\end{aligned}\n$$\n\nso that\n\n$$\nW=\\frac{\\rho g\\left(l^{2}-h^{2}\\right) \\pi\\left(b^{2}-a^{2}\\right)}{2}-A[(\\kappa-1) l-(\\kappa-1) h]\n$$\n\nThis corresponds to work done by gas expanding on the liquid outside the capacitor with $W=\\int P d V$ where $P$ is the gauge pressure. Then, assuming the pressure is constant with volume and neglecting terms of $O(h / l)$,\n\n$$\nP=\\frac{\\rho g l}{2}-\\frac{\\epsilon_{0} V^{2}(\\kappa-1)}{\\left(b^{2}-a^{2}\\right) \\log (b / a)}\n$$']",['$P=\\frac{\\rho g l}{2}-\\frac{\\epsilon_{0} V^{2}(\\kappa-1)}{\\left(b^{2}-a^{2}\\right) \\log (b / a)}$'],False,,Expression, 954,Mechanics,"2. Trajectory of a point mass A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible,",(a) What value of $\theta$ maximizes the range?,"['From the equation of motion $\\frac{d \\mathbf{v}}{d t}=\\mathbf{g}$ we find that $\\mathbf{v}=\\mathbf{v}_{0}+\\mathbf{g} t$. Then the position of the mass after time $t$ is given by $\\mathbf{r}=\\int_{0}^{t} \\mathbf{v} d t=\\int_{0}^{t}\\left(\\mathbf{v}_{0}+\\mathbf{g} t\\right) d t=\\mathbf{v}_{0} t+\\frac{1}{2} \\mathbf{g} t^{2}$. Denoting the final time as $T$ and the position as $\\mathbf{R}$, we find that\n\n$$\n\\mathbf{R e}_{\\mathbf{y}}=0=v_{0} \\sin \\theta T-\\frac{1}{2} g T^{2}\n\\tag{1}\n$$\n$$\n\\mathbf{R e}_{\\mathbf{x}}=L=v_{0} \\cos \\theta T\n\\tag{2}\n$$\n\nwhere $L$ is the range of the mass. From equations (1) and (2), we get\n\n$$\nL=\\frac{2 v_{0}^{2}}{g} \\sin \\theta \\cos \\theta=\\frac{v_{0}^{2}}{g} \\sin (2 \\theta)\n$$\n\nMaximum $L$ is achieved when $\\sin (2 \\theta)=1$. Therefore, $\\theta=\\frac{\\pi}{4}$']",['$\\frac{\\pi}{4}$'],False,,Numerical,1e-8 955,Mechanics,"2. Trajectory of a point mass A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible, Context question: (a) What value of $\theta$ maximizes the range? Context answer: \boxed{$\frac{\pi}{4}$} ",(b) What value of $\theta$ maximizes the surface area under the trajectory curve?,"['The surface area covered by the trajectory can be found by\n\n$$\n\\mathbf{A}=\\frac{1}{2} \\int \\mathbf{r} \\times \\mathbf{d} \\mathbf{r}=\\frac{1}{2} \\int(\\mathbf{r} \\times \\mathbf{v}) d t\n$$\n\nSince the motion occurs in a plane, the surface area is always orthogonal to the plane. Therefore, the area covered by the trajectory can be found by calculating the magnitude of the vector $\\mathbf{A}$. Now, inserting the equations for the position vector and the velocity as found in (1), we get\n\n$$\n\\begin{aligned}\n\\mathbf{A}=\\frac{1}{2} \\int(\\mathbf{r} \\times \\mathbf{v}) d t & =\\frac{1}{4} \\int\\left(\\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}\\right) t^{2} d t \\\\\n& =\\frac{1}{4}\\left(\\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}\\right) \\int t^{2} d t \\\\\n& \\left.=\\frac{1}{12} \\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}\\right) T^{3}\n\\end{aligned}\n$$\n\n\n\nAs found in (1), $T$ is proportional to $\\sin \\theta$, and $\\mathbf{v}_{\\mathbf{0}} \\times \\mathbf{g}$ is proportional to $\\cos \\theta$. Therefore, the angle that maximizes the area maximizes the function\n\n$$\nf(\\theta)=\\cos \\theta \\sin ^{3} \\theta\n$$\n\nTaking derivative of $f(\\theta)$, we get the equation\n\n$$\n\\frac{d f}{d \\theta}=\\sin ^{2} \\theta\\left(3 \\cos ^{2} \\theta-\\sin ^{2} \\theta\\right)\n$$\n\nwhich gives zero at $\\theta=0$ or $\\theta=\\arctan \\sqrt{3}=\\pi / 3$. Since $\\theta=0$ corresponds to the minimal area (no trajectory, $\\mathrm{A}=0$ ), the answer is $\\theta=\\pi / 3$']",['$\\pi / 3$'],False,rads,Numerical,1e-8 956,Mechanics,"2. Trajectory of a point mass A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible, Context question: (a) What value of $\theta$ maximizes the range? Context answer: \boxed{$\frac{\pi}{4}$} Context question: (b) What value of $\theta$ maximizes the surface area under the trajectory curve? Context answer: \boxed{$\pi / 3$} ","(c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ? Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows $$ \frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v} $$","['If the initial position of the mass is $\\mathbf{r}_{\\mathbf{0}}=h \\mathbf{e}_{\\mathbf{y}}$, equation (1) and (2) becomes\n\n$$\n\\mathbf{R e}_{\\mathbf{y}}=-h=v_{0} \\sin \\theta T-\\frac{1}{2} g T^{2}\n\\tag{3}\n$$\n$$\n\\mathbf{R e}_{\\mathbf{x}}=L=v_{0} \\cos \\theta T\n\\tag{4}\n$$\n\nMaximum $L$ occurs when $\\frac{d L}{d \\theta}=0$. Therefore, from (4), we obtain\n\n$$\n\\frac{d L}{d \\theta}=-v_{0} \\sin \\theta+v_{0} \\cos \\theta \\frac{d T}{d \\theta}=0\n\\tag{5}\n$$\n\nTaking derivative of both sides of (3) to find $\\frac{d T}{d \\theta}$ and inserting it into (5), we get\n\n$$\n\\begin{aligned}\n0 & =v_{0} \\cos \\theta T+\\frac{v_{0} \\sin \\theta-g T}{\\cos \\theta} \\sin \\theta T \\\\\n& \\Longrightarrow v_{0}=g T \\sin \\theta\n\\end{aligned}\n$$\n\nInserting this equation to (3), we obtain\n\n$$\n-h=\\frac{v_{0}^{2}}{g}-\\frac{1}{2} \\frac{v_{0}^{2}}{g \\sin \\theta}\n$$\n\nTherefore, the answer is $\\theta=\\arcsin \\left(\\frac{1}{\\sqrt{2}} \\frac{v_{0}}{\\sqrt{v_{0}^{2}+g h}}\\right)$']",['$\\arcsin \\left(\\frac{1}{\\sqrt{2}} \\frac{v_{0}}{\\sqrt{v_{0}^{2}+g h}}\\right)$'],False,,Expression, 957,Mechanics,"2. Trajectory of a point mass A point mass on the ground is thrown with initial velocity $\vec{v}_{0}$ that makes an angle $\theta$ with the horizontal. Assuming that air friction is negligible, Context question: (a) What value of $\theta$ maximizes the range? Context answer: \boxed{$\frac{\pi}{4}$} Context question: (b) What value of $\theta$ maximizes the surface area under the trajectory curve? Context answer: \boxed{$\pi / 3$} Context question: (c) What is the answer for (a), if the point mass is thrown from an apartment of height $h$ ? Now assume that we have a frictional force that is proportional to the velocity vector, such that the equation of motion is as follows $$ \frac{d \vec{v}}{d t}=\vec{g}-\beta \vec{v} $$ Context answer: \boxed{$\arcsin \left(\frac{1}{\sqrt{2}} \frac{v_{0}}{\sqrt{v_{0}^{2}+g h}}\right)$} ","(d) Supposing that $\beta<<\frac{g}{v_{0}}$, find the duration of the motion $T$.","['Under the presence of friction, equation of motion for $y$ component becomes\n\n$$\n\\frac{d v_{y}}{d t}=-g-\\beta v_{y}\n\\tag{6}\n$$\n\nSolving (6), we get\n\n$$\nv_{y}=v_{0 y}-\\left(v_{0 y}+\\frac{g}{\\beta}\\right)\\left(1-e^{-\\beta t}\\right)\n\\tag{7}\n$$\n\n\n\n$y$ component of the final position vector can be found by integrating $v_{y}$,\n\n$$\ny=0=-\\frac{g T}{\\beta}+\\frac{\\left(v_{0 y}+\\frac{g}{\\beta}\\right)}{\\beta}\\left(1-e^{-\\beta T}\\right)\n\\tag{8}\n$$\n\nSince $\\beta$ is small compared to the dimensions of the system, we can use Taylor expansion for (8) to find an analytic equation for $T$.\n\n$$\n\\begin{aligned}\n& g T=\\left(v_{0 y}+\\frac{g}{\\beta}\\right)\\left(1-e^{-\\beta T}\\right) \\approx\\left(v_{0 y}+\\frac{g}{\\beta}\\right)\\left(\\beta T-\\frac{1}{2} \\beta^{2} T^{2}\\right) \\\\\n& \\Longrightarrow T=\\frac{2 v_{0 y}}{\\beta v_{0 y}+g}\n\\end{aligned}\n$$\n\nTherefore, the answer is\n\n$$\nT=\\frac{2 v_{0} \\sin \\theta}{g} \\frac{1}{1+\\frac{\\beta v_{0} \\sin \\theta}{g}}\n$$\n\nwhich is less than the value found for frictionless trajectory, as expected.']",['$T=\\frac{2 v_{0} \\sin \\theta}{g} \\frac{1}{1+\\frac{\\beta v_{0} \\sin \\theta}{g}}$'],False,,Expression, 958,Electromagnetism,"3. Block on an Incline Plane Consider a block of mass, $\mathrm{M}$, and charge, $Q>0$, sliding down an incline plane at an angle $\alpha$ with the horizontal with initial position $\left(x_{0}, y_{0}\right)$. The system is exposed to an upward electric field given by $E(y)=\beta\left(y_{0}-y\right)$, with $\beta>0$. Figure 2: An incline plane submerged in a medium.","(a) Find an expression for the $y_{e}$, the position where the block experiences net zero vertical force.","['The block experiences net vertical force when the upward force from the electric field equals the downward force from gravity,\n\n$$\nE\\left(y_{e}\\right) Q=M g\n$$\n\nUsing our function for $\\mathrm{E}$ and solving for $y_{e}$ we find, $y_{e}=y_{0}-\\frac{g M}{\\beta Q}$.']",['$y_{e}=y_{0}-\\frac{g M}{\\beta Q}$'],False,,Expression, 959,Electromagnetism,"3. Block on an Incline Plane Consider a block of mass, $\mathrm{M}$, and charge, $Q>0$, sliding down an incline plane at an angle $\alpha$ with the horizontal with initial position $\left(x_{0}, y_{0}\right)$. The system is exposed to an upward electric field given by $E(y)=\beta\left(y_{0}-y\right)$, with $\beta>0$. Figure 2: An incline plane submerged in a medium. Context question: (a) Find an expression for the $y_{e}$, the position where the block experiences net zero vertical force. Context answer: \boxed{$y_{e}=y_{0}-\frac{g M}{\beta Q}$} ",(b) What is the force acting on the block when it is in contact with the incline plane?,['The force acting on the block while on the plane will be the sum of the electromagnetic force (which is dependent on y) and gravitational force projected onto the plane.\n\n$$\nF(y)=\\left(M g-Q \\beta\\left(y_{0}-y\\right)\\right) \\sin (\\alpha)\n$$\n\nThis force is directed parallel to the plane to the right.'],['$F(y)=\\left(M g-Q \\beta\\left(y_{0}-y\\right)\\right) \\sin (\\alpha)$'],False,,Expression, 960,Electromagnetism,,"(c) While sliding down the incline plane, at what point will it lose contact with the plane?","['The block will lose contact with the surface at $y_{e}$. A qualitative answer is sufficient. This can be seen due to the fact that up until $y_{e}$ the block experiences a force parallel to the plane and thus stays in contact with the plane. Suddenly, as it passes $y_{e}$, it begins to experience and upward force due to the electric field surpassing gravity in its asserted force. At this point the block begins to peel away from the incline plane.']",['开放性回答'],False,,Need_human_evaluate, 961,Electromagnetism,,(d) What is the velocity vector of the block when it loses contact with the plane?,"['To find the velocity vector we find the work done by the force vector on the block up until $y_{e}$. At $y_{e}$ we realize that the system has zero potential energy since the net force on the block is zero. We can then treat the total work as kinetic energy at that moment and calculate the magnitude of the velocity.\n\n$$\nW=\\int_{0}^{y_{0}-y_{e}}\\left(M g-Q \\beta\\left(y_{0}-y\\right)\\right) d y=M g\\left(y_{0}-y_{e}\\right)-\\frac{Q \\beta}{2}\\left(y_{0}^{2}-y_{e}^{2}\\right)\n$$\n\nVelocity is then found by, $W=K=\\frac{1}{2} M v^{2}$,\n\n$$\nv=\\sqrt{2 g\\left(y_{0}-y_{e}\\right)-\\frac{Q \\beta}{M}\\left(y_{0}^{2}-y_{e}^{2}\\right)}\n$$\n\nThe velocity vector is then given by,\n\n$$\n\\vec{v}=\\sqrt{2 g\\left(y_{0}-y_{e}\\right)-\\frac{Q \\beta}{M}\\left(y_{0}^{2}-y_{e}^{2}\\right)} \\cdot\\langle\\cos (\\alpha),-\\sin (\\alpha)\\rangle\n$$']","['$\\vec{v}=\\sqrt{2 g\\left(y_{0}-y_{e}\\right)-\\frac{Q \\beta}{M}\\left(y_{0}^{2}-y_{e}^{2}\\right)} \\cdot\\langle\\cos (\\alpha),-\\sin (\\alpha)\\rangle$']",True,,Need_human_evaluate, 962,Electromagnetism,"3. Block on an Incline Plane Consider a block of mass, $\mathrm{M}$, and charge, $Q>0$, sliding down an incline plane at an angle $\alpha$ with the horizontal with initial position $\left(x_{0}, y_{0}\right)$. The system is exposed to an upward electric field given by $E(y)=\beta\left(y_{0}-y\right)$, with $\beta>0$. Figure 2: An incline plane submerged in a medium. Context question: (a) Find an expression for the $y_{e}$, the position where the block experiences net zero vertical force. Context answer: \boxed{$y_{e}=y_{0}-\frac{g M}{\beta Q}$} Context question: (b) What is the force acting on the block when it is in contact with the incline plane? Context answer: \boxed{$F(y)=\left(M g-Q \beta\left(y_{0}-y\right)\right) \sin (\alpha)$} Context question: (c) While sliding down the incline plane, at what point will it lose contact with the plane? Context answer: 开放性回答 Context question: (d) What is the velocity vector of the block when it loses contact with the plane? Context answer: $\vec{v}=\sqrt{2 g\left(y_{0}-y_{e}\right)-\frac{Q \beta}{M}\left(y_{0}^{2}-y_{e}^{2}\right)} \cdot\langle\cos (\alpha),-\sin (\alpha)\rangle$ ",(e) Once the block has lost contact with the plane it begins to oscillate in the field. What frequency does it oscillate at?,"[""After the block has left the incline, consider a displacement around $y_{e}, \\Delta y=y_{e}-$ $y$. We can write the force experienced by the block as a function of $\\Delta y, F(\\Delta y)=$ $\\left(M g-Q \\beta\\left(y_{0}-y\\right)\\right)=\\left(M g-Q \\beta\\left(y_{0}-y_{e}+\\Delta y\\right)\\right)$. Using our value for $y_{e}$, we have, $F(\\Delta y)=$ $\\left(M g-Q \\beta\\left(\\frac{g M}{\\beta Q}+\\Delta y\\right)\\right)$. Thus we have $F(\\Delta y)=-(\\beta Q) \\Delta y$. Noting that $(\\beta Q)$ is constant, we see that this is an example of Hookes' Law and know that the frequency must be given by,\n\n$$\nf=\\frac{1}{2 \\pi} \\sqrt{\\frac{\\beta Q}{m}}\n$$""]",['$f=\\frac{1}{2 \\pi} \\sqrt{\\frac{\\beta Q}{m}}$'],False,,Expression, 963,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost.","(a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.)","[""The homogeneity of space implies the following:\n\n$$\nX\\left(x_{2}+h, t, v\\right)-X\\left(x_{2}, t, v\\right)=X\\left(x_{1}+h, t, v\\right)-X\\left(x_{1}, t, v\\right)\n$$\n\n\n\n\n\nFigure 3: The shaded area is the ruler's trajectory in spacetime. Any measurement of its length must intersect the shaded area.\n\n\n\nnow dividing by $h$ and sending $h \\rightarrow 0$ is the partial derivative, therefore\n\n$$\n\\left.\\frac{\\partial X}{\\partial x}\\right|_{x_{2}}=\\left.\\frac{\\partial X}{\\partial x}\\right|_{x_{1}}\n$$\n\nThe same method is repeated for the other variables.""]",,False,,, 964,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates.","(b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.)","['The isotropy of space implies\n\n$$\n\\begin{aligned}\nX(-x, t,-v) & =-x^{\\prime}=-X(x, t, v) \\\\\nT(-x, t,-v) & =T(x, t, v)\n\\end{aligned}\n$$\n\nand then plugging into (9) we see that\n\n$$\n\\begin{aligned}\n& A(-v)=A(v) \\\\\n& B(-v)=-B(v) \\\\\n& C(-v)=-C(v) \\\\\n& D(-v)=D(v)\n\\end{aligned}\n$$']",,False,,, 965,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} ","(c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$.","['The principle of relativity implies that the coordinate transformation can be inverted such that\n\n$$\n\\begin{aligned}\nX\\left(x^{\\prime}, t^{\\prime},-v\\right) & =x \\\\\nT\\left(x^{\\prime}, t^{\\prime},-v\\right) & =t\n\\end{aligned}\n$$\n\ntherefore, because $X, T$ are linear, we can write the generalized boost in matrix form, and then using the even/oddness of $A, B, C, D$ derived previously\n\n$$\n\\left(\\begin{array}{ll}\nA(v) & B(v) \\\\\nC(v) & D(v)\n\\end{array}\\right)\\left(\\begin{array}{cc}\nA(v) & -B(v) \\\\\n-C(v) & D(v)\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n1 & 0 \\\\\n0 & 1\n\\end{array}\\right)\n$$\n\nwhich is precisely the system of equations listed.']",,False,,, 966,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} ","(d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$","[""The $F^{\\prime}$ frame is defined by $x^{\\prime}=0$, therefore $A(v) x+B(v) t=A(v) v t+B(v) t$ plugging in the equation $x=v t$. This implies that $A v+B=0$. This let's all the undetermined functions $A, B, C, D$ to be solved in terms of $A$.""]",,False,,, 967,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} ","(e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$.","['Consider a frame $F^{\\prime \\prime}$ related to $F^{\\prime}$ by a boost in the $x^{\\prime}$-direction with relative velocity $u$. Therefore, the composition of these two boosts results in a boost $F \\rightarrow F^{\\prime \\prime}$ given by\n\n$$\n\\Lambda_{F \\rightarrow F^{\\prime \\prime}}=\\left(\\begin{array}{cc}\nA(u) & -u A(u) \\\\\n-\\frac{A(u)^{2}-1}{u A(u)} & A(u)\n\\end{array}\\right)\\left(\\begin{array}{cc}\nA(v) & -v A(v) \\\\\n-\\frac{A(v)^{2}-1}{v A(v)} & A(v)\n\\end{array}\\right)\n$$\n\nMultiplying out these matrices and noting that we have shown that the diagonal terms must be equal, we find that\n\n$$\n\\frac{A(v)^{2}-1}{v^{2} A(v)^{2}}=\\frac{A(u)^{2}-1}{u^{2} A(u)^{2}}\n$$\n\nas the left hand side is a function of $v$ only and the right hand side is a function of $u$ only, they must both be equal to some constant $\\kappa$.']",,False,,, 968,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$. Context answer: \boxed{证明题} ","(f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} $$","['Solve for $A$,\n\n$$\nA(v)=\\frac{1}{\\sqrt{1-\\kappa v^{2}}}\n$$\n\nand the dimensions of $\\kappa$ are determined by the restriction that $\\kappa v^{2}$ is being added to a dimensionless number. Substituting this form of $A(v)$ into\n\n$$\n\\begin{aligned}\nx^{\\prime} & =A(v) x-v A(v) t \\\\\nt^{\\prime} & =-\\left(\\frac{A(v)^{2}-1}{v A(v)}\\right) x+A(v) t\n\\end{aligned}\n$$\n\nyields the answer.']",,False,,, 969,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$. Context answer: \boxed{证明题} Context question: (f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} $$ Context answer: \boxed{证明题} ","(g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast.","['If $v$ is unbounded and $\\kappa \\neq 0$, it may be large enough so that the square root gives an imaginary number, and as $x^{\\prime}, t^{\\prime}$ cannot be imaginary, it must be that $\\kappa=0$. There is nothing stopping a particle from traveling arbitrarily fast under a Galilean structure of spacetime, as given any particle we may Galilean boost to an inertial frame moving at $v$ arbitrarily fast, in which the particle is then moving at $-v$. By the principle of relativity, there is nothing wrong with doing physics in this frame, so it must be that it is acceptable to have particles move arbitrarily fast']",,False,,, 970,Modern Physics,"4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: $$ \begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} $$ In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} $$ In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal ""speed limit"" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal ""speed limit,"" which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames $F$ and $F^{\prime}$ be such that the frame $F^{\prime}$ moves at relative velocity $v$ in the positive- $x$ direction compared to frame $F$. Denote the coordinates of $F$ as $(x, t)$ and the coordinates of $F^{\prime}$ as $\left(x^{\prime}, t^{\prime}\right)$. The most general coordinate transformations between $F$ and $F^{\prime}$ are given by functions $X, T$, $$ \begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} $$ which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates $x$ and $t$. Use this fact to show that $\frac{\partial X}{\partial x}$ is independent of the position $x$ and $\frac{\partial T}{\partial t}$ is independent of the time $t$. (Hint: Recall the definition of the partial derivative.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Analogously, we can conclude additionally that $\frac{\partial X}{\partial x}$ is independent of both $x$ and $t$ and $\frac{\partial T}{\partial t}$ is independent of $x$ and $t$. It can be shown that $X, T$ may be given in the form $$ \begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} $$ where $A, B, C, D$ are functions of $v$. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations $X, T$ after setting $x \rightarrow-x$ and $x^{\prime} \rightarrow-x^{\prime}$ and conclude that $A(v), D(v)$ are even functions of $v$ and $B(v), C(v)$ are odd functions of $v$. (Hint: the relative velocity $v$ is a number which is measured by the $F$ frame using $v=\frac{d x}{d t}$.) Context answer: \boxed{证明题} Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations $X, T$ are invertible and their inverses have the same functional form as $X, T$ after setting $v \rightarrow-v$. Use this fact to show the following system of equations hold: $$ \begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} $$ (Hint: It's convenient to write $X, T$ as matrices and recall the definition of matrix inverses.) Physically, we must have that $B(v)$ and $C(v)$ are not both identically zero for nonzero $v$. So, we can conclude from the above that $D(v)=A(v)$ and $C(v)=\frac{A(v)^{2}-1}{B(v)}$. Context answer: \boxed{证明题} Context question: (d) Use the previous results and the fact that the location of the $F^{\prime}$ frame may be given by $x=v t$ in the $F$ frame to conclude that the coordinate transformations have the following form: $$ \begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} $$ Context answer: \boxed{证明题} Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that $$ \frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . $$ for an arbitrary constant $\kappa$. Context answer: \boxed{证明题} Context question: (f) (1 point) Show that $\kappa$ has dimensions of (velocity $)^{-2}$, and show that the generalized boost now has the form $$ \begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} $$ Context answer: \boxed{证明题} Context question: (g) Assume that $v$ may be infinite. Argue that $\kappa=0$ and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast. Context answer: \boxed{证明题} ","(h) Assume that $v$ must be smaller than a finite value. Show that $1 / \sqrt{\kappa}$ is the maximum allowable speed, and that this speed is frame invariant, i.e., $\frac{d x^{\prime}}{d t^{\prime}}=\frac{d x}{d t}$ for something moving at speed $1 / \sqrt{\kappa}$. Experiment has shown that this speed is $c$, the speed of light. Setting $\kappa=1 / c^{2}$, show that you recover the Lorentz boost.","['Again by requiring that $x^{\\prime}, t^{\\prime}$ are real, we can find the desired bound on $v$ from $1-\\kappa v^{2}>0$. One way to show that the speed is frame invariant is by deriving the relativistic velocity addition formula as follows\n\n$$\n\\begin{aligned}\nd x^{\\prime} & =\\gamma(d x-v d t) \\\\\nd t^{\\prime} & =\\gamma\\left(d t-v / c^{2} d x\\right)\n\\end{aligned}\n$$\n\nand dividing to yield\n\n$$\n\\frac{d x^{\\prime}}{d t^{\\prime}}=\\frac{\\frac{d x}{d t}-v}{1-\\frac{v}{c^{2}} \\frac{d x}{d t}}\n$$\n\nLet $w=d x^{\\prime} / d t^{\\prime}, u=d x / d t$, and, as often makes relativity problems easier to deal with, set $c=1$ (this is equivalent to choosing a new system of units). Then, we have\n\n$$\nw=\\frac{u-v}{1-v u}\n$$\n\nnow if $w=c=1$, we can solve the above equation to show that $u=1$, in other words frame $F$ and $F^{\\prime}$ both agree on what velocities move at $c$.']",,False,,, 971,Modern Physics,,"5. Identifying particle processes Note: all numerical values in this problem are given in natural units: $c=1$, mass measured in $\mathrm{GeV}=1$, so no unit conversions are necessary, even when dimensions don't seem to align (this allows us to do things like sum momentum and energy without worrying about units). The 4-momentum of a relativistic particle is the 4-dimensional vector $$ \vec{P}=\left(E, p_{x}, p_{y}, p_{z}\right), $$ where $E$ is the particle's energy and $\left(p_{x}, p_{y}, p_{z}\right)$ is the standard 3-momentum (excluding the time component). The total (summed) 4-momentum of a system is always conserved. The components of 4-momentum can be related to the rest mass $m$ of the particle by $$ E^{2}=m^{2}+p_{x}^{2}+p_{y}^{2}+p_{z}^{2} . $$ A particle accelerator records the 4-momenta of 2 muon-antimuon pairs produced by some particle processes. Other particles produced by the processes are not measured by the detector. Below are two possible processes that involve muon-antimuon production and down quark-antiquark production. We know that out of the two measured 4-momentum pairs, one corresponds to Process A and the other to Process B. Using the muon-antimuon 4-momenta given below, determine which corresponds to each process. The numbers will be truncated for simplicity and slightly fudged from precise values to account for some small error in the measurement, so don't expect precision for small values (this should not affect the problem). The points will be given for an explanation of how you reached your conclusion; simply guessing the process correctly will result in at most one point. Also note that both processes are valid, you do not need to verify that they could actually happen. A table of (mean) particle rest masses may prove useful: | Particle | Rest Mass | | :---: | :---: | | Higgs | $\sim 125$ | | Z | $\sim 91$ | | Top | $\sim 172$ | | Bottom | $\sim 4.2$ | | Up | $\sim 0.0024$ | | Muon | $\sim 0.1$ | | Neutrino | $\sim 0$ | Process A: The collision of an up quark-antiquark pair results in the production of one Higgs boson and one $\mathrm{Z}$ boson. The $\mathrm{Z}$ boson then decays into a muon-antimuon pair Process B: The collision of an up quark-antiquark pair results in the production of a top quark-antiquark pair. The top quark decays into a $\mathrm{W}+$ boson and a bottom quark, while the anti-top decays into a W- boson and an anti-bottom quark. The W- then decays into a muon and an anti-neutrino, while the $\mathrm{W}+$ decays into an anti-muon and neutrino. Detected muon-antimuon 4-momenta: х) $(149.0,12.6,-8.8,148.2) ;(163.7,-38.8,67.3,144.1)$ у) $(319.5,-41.3,-24.2,315.8) ;(305.3,-60.1,25.0,298.1)$","['The sum of the muon-antimuon momenta are:\n\nх) $(312.7,-26.2,58.5,292.3)$\n\nу) $(624.8,101.4,0.8,613.2)$\n\nFor the one that corresponds to Process A, this 4-momentum would correspond to the 4momentum of the $\\mathrm{Z}$ boson. This summed 4-momentum from Process B should not correspond to any particular particle, as the muon-antimuon pair decays off of two different particles. Applying the relation between the 4-momentum and the rest mass, we find that each 4-momentum would correspond to a $\\mathrm{Z}$ with rest mass of:\n\nx) 90.74\n\nу) 63.86\n\nThe mass in $\\mathrm{x}$ ) is much closer to the true mean $\\mathrm{Z}$ mass, so that should be process $\\mathrm{A}$, and $\\mathrm{y}$ ) is process B.']",['x for A and y for B'],False,,Need_human_evaluate, 972,Electromagnetism,"Computing electric fields Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials. Uniform charge distributions Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element.",(a) Compute the electric potential inside and outside the sphere.,"[""From Gauss's theorem:\n\n$$\n\\frac{Q}{\\varepsilon_{0}}=4 \\pi R^{2} E\n$$\n\nSo:\n\n$$\nE=\\frac{Q}{4 \\pi \\varepsilon_{0}}\n$$\n\nBut:\n\n$$\n\\mathbf{E}=-\\nabla \\Phi\n$$\n\nIn this case $E=-\\frac{\\partial \\Phi}{\\partial r}$. To obtain a zero potential at infinity we get:\n\n$$\n\\Phi_{+}=\\frac{Q}{4 \\pi \\varepsilon_{0} r}\n$$\n\nThe electric field inside the sphere is 0 (no charges), so the potential is constant. Because it is continuous on the surface of the sphere (no jumps):\n\n$$\n\\Phi_{-}=\\frac{Q}{4 \\pi \\varepsilon_{0} R}\n$$""]","['$\\Phi_{-}=\\frac{Q}{4 \\pi \\varepsilon_{0} R}$,$\\Phi_{+}=\\frac{Q}{4 \\pi \\varepsilon_{0} r}$']",True,,Expression, 973,Electromagnetism,"Computing electric fields Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials. Uniform charge distributions Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element. Context question: (a) Compute the electric potential inside and outside the sphere. Context answer: \boxed{$\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}$,$\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}$} Extra Supplementary Reading Materials: Legendre polynomials and non-uniform charge distributions Legendre polynomials are a type of orthogonal polynomials essential in mathematical physics. One of their applications is in computing electric potentials for more complicated charge configurations. We will denote the $n$-th Legendre polynomial (having degree $n$ ) as $P_{n}$. Legendre polynomials are defined on $[-1,1]$ and we can express their scalar product as $$ \left\langle P_{m}(x), P_{n}(x)\right\rangle=\int_{-1}^{1} P_{m}(x) P_{n}(x) d x \tag{1} $$ The first two Legendre polynomials are $P_{0}(x)=1$ and $P_{1}(x)=x$.","(a) Knowing that Legendre polynomials are orthogonal $\left(\left\langle P_{m}(x), P_{n}(x)\right\rangle=0\right.$ if $m \neq n)$ and $\operatorname{deg} P_{n}(x)=n$, obtain $P_{2}(x)$ and $P_{3}(x)$. For reaching the usual and most convenient form of these polynomials, divide your results by the norm: $\left\|P_{n}(x)\right\|=\frac{2}{2 n+1}$. Let us now consider a sphere of radius $R$ centered at the origin. Suppose a point charge $q$ is put at the origin and that this is the only charge inside or outside the sphere. Furthermore, the potential is $\Phi=V_{0} \cos \theta$ on the surface of the sphere. We know that we can write the potential induced by the charge on the sphere (without taking into account $q$ ) in the following way: $$ \begin{array}{ll} \Phi_{-}=\sum_{n=0}^{\infty} A_{n} r^{n} P_{n}(\cos \theta), & rR \end{array} $$","['$$\nP_{2}(x)=C_{2}\\left(x^{2}+\\lambda_{1} x+\\lambda_{0}\\right)\n$$\n\nBecause $\\left\\langle P_{2}(x), P_{1}(x)\\right\\rangle=\\int_{-1}^{1} P_{2}(x) P_{1}(x) d x=0$ we get $\\lambda_{1}=0$. Because $\\int_{-1}^{1} x^{2} d x=2 / 3$ and $\\int_{-1}^{1} d x=$ 2 , we get $\\lambda_{0}=-1 / 3$. The condition for the norm implies $P_{2}(x)=\\frac{1}{2}\\left(3 x^{2}-1\\right)$. Similarly:\n\n$$\nP_{3}(x)=C_{3}\\left(x^{3}+\\lambda_{2} x^{2}+\\lambda_{1} x+\\lambda_{0}\\right)\n$$\n\nThrough an identical reasoning:\n\n$$\nP_{3}=\\frac{1}{2}\\left(5 x^{3}-3 x\\right)\n$$']","['$P_{2}(x)=C_{2}\\left(x^{2}+\\lambda_{1} x+\\lambda_{0}\\right)$ , $P_{3}=\\frac{1}{2}\\left(5 x^{3}-3 x\\right)$']",True,,Expression, 974,Electromagnetism,"Computing electric fields Electrostatics relies on multiple methods for computing electric fields and potentials. In this problem, we will explore two of them, Gauss's Law and Legendre polynomials. Uniform charge distributions Let us consider a hollow conducting sphere of radius $R$ charged with the electric charge $Q$, uniformly distributed on its surface. In order to calculate its potential, we can use Gauss's Law, which states that the flux of the electric field $d F=\mathbf{E} \cdot \mathbf{d} \mathbf{A}$ across a closed surface is proportional to the charge enclosed by that surface: $F=Q / \varepsilon_{0}$. We have denoted $\mathbf{d A}=d A \mathbf{n}$ the elementary oriented (towards the exterior) surface element. Context question: (a) Compute the electric potential inside and outside the sphere. Context answer: \boxed{$\Phi_{-}=\frac{Q}{4 \pi \varepsilon_{0} R}$,$\Phi_{+}=\frac{Q}{4 \pi \varepsilon_{0} r}$} Extra Supplementary Reading Materials: Legendre polynomials and non-uniform charge distributions Legendre polynomials are a type of orthogonal polynomials essential in mathematical physics. One of their applications is in computing electric potentials for more complicated charge configurations. We will denote the $n$-th Legendre polynomial (having degree $n$ ) as $P_{n}$. Legendre polynomials are defined on $[-1,1]$ and we can express their scalar product as $$ \left\langle P_{m}(x), P_{n}(x)\right\rangle=\int_{-1}^{1} P_{m}(x) P_{n}(x) d x \tag{1} $$ The first two Legendre polynomials are $P_{0}(x)=1$ and $P_{1}(x)=x$. Context question: (a) Knowing that Legendre polynomials are orthogonal $\left(\left\langle P_{m}(x), P_{n}(x)\right\rangle=0\right.$ if $m \neq n)$ and $\operatorname{deg} P_{n}(x)=n$, obtain $P_{2}(x)$ and $P_{3}(x)$. For reaching the usual and most convenient form of these polynomials, divide your results by the norm: $\left\|P_{n}(x)\right\|=\frac{2}{2 n+1}$. Let us now consider a sphere of radius $R$ centered at the origin. Suppose a point charge $q$ is put at the origin and that this is the only charge inside or outside the sphere. Furthermore, the potential is $\Phi=V_{0} \cos \theta$ on the surface of the sphere. We know that we can write the potential induced by the charge on the sphere (without taking into account $q$ ) in the following way: $$ \begin{array}{ll} \Phi_{-}=\sum_{n=0}^{\infty} A_{n} r^{n} P_{n}(\cos \theta), & rR \end{array} $$ Context answer: \boxed{$P_{2}(x)=C_{2}\left(x^{2}+\lambda_{1} x+\lambda_{0}\right)$ , $P_{3}=\frac{1}{2}\left(5 x^{3}-3 x\right)$} ",(b) Compute the electric potential both inside and outside the sphere.,"[""The potential is given by either Poisson's or Laplace's equation:\n\n$$\n\\begin{array}{ll}\n\\nabla^{2} \\Phi_{-}=-\\frac{q}{\\varepsilon_{0}} \\delta(r), & rR\n\\end{array}\n$$\n\n\n\nThe general solutions finite in the respective regions, taking account of the symmetry, are\n\n$$\n\\begin{aligned}\n& \\Phi_{-}=\\frac{q}{4 \\pi \\varepsilon_{0} r}+\\sum_{n=0}^{\\infty} A_{n} r^{n} P_{n}(\\cos \\theta), \\quad rR\n\\end{aligned}\n$$\n\nThen from the condition $\\Phi_{-}=\\Phi_{+}=V_{0} \\cos \\theta$ at $r=R$ we obtain $A_{0}=-\\frac{q}{4 \\pi \\varepsilon_{0} R}, A_{1}=\\frac{V_{0}}{R}, B_{1}=$ $V_{0} R^{2}, B_{0}=0, A_{n}=B_{n}=0$ for $n \\neq 0,1$, and hence\n\n$$\n\\begin{aligned}\n& \\Phi_{-}=\\frac{q}{4 \\pi \\varepsilon_{0} r}-\\frac{q}{4 \\pi \\varepsilon_{0} R}+\\frac{V_{0} \\cos \\theta}{R} r, \\quad rR .\n\\end{aligned}\n$$""]","['$\\Phi_{-}=\\frac{q}{4 \\pi \\varepsilon_{0} r}-\\frac{q}{4 \\pi \\varepsilon_{0} R}+\\frac{V_{0} \\cos \\theta}{R} r$,$\\Phi_{+}=\\frac{V_{0} R^{2}}{r^{2}} \\cos \\theta$']",True,,Expression, 975,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.","(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.","['A periodic wave on interval $[0, L]$ must fit an integer number of wavelengths $\\lambda$ into the length $L$,\n\n$$\nn \\lambda=L\n\\tag{1}\n$$\n\nTherefore $k_{n}=\\frac{2 \\pi}{\\lambda_{n}}=\\frac{2 \\pi n}{L}$.\n\nFor the next part, use $c^{\\prime}=\\frac{\\omega}{k}$. Therefore,\n\n$$\n\\omega_{n}=c^{\\prime} k_{n}=\\frac{2 \\pi c^{\\prime} n}{L} \\Rightarrow d \\omega_{n}=\\frac{2 \\pi c^{\\prime}}{L} d n \\Rightarrow \\frac{d n}{d \\omega_{n}}=\\frac{L}{2 \\pi c^{\\prime}}\n\\tag{2}\n$$']",,False,,, 976,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} ","(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.","['Use the given Taylor expansion,\n\n$$\n\\left\\langle N_{n}\\right\\rangle=\\frac{1}{\\exp \\frac{\\hbar \\omega_{n}}{k T}-1} \\approx \\frac{1}{\\left(1+\\frac{\\hbar \\omega_{n}}{k T}\\right)-1}=\\frac{k T}{\\hbar \\omega_{n}}\n\\tag{3}\n$$']",,False,,, 977,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} ","(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.","['The electromagnetic modes are excitations of the electromagnetic field like photons, so they are also massless. The energy of a photon is $E=h f=\\hbar \\omega$.\n\nAlternatively, you can start from $E^{2}=\\left(m c^{2}\\right)^{2}+(p c)^{2}$ and use $m=0, p=h / \\lambda$.']",,False,,, 978,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题} ","(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)","['Power equals energy per time. The average energy delivered by one boson to the resistor is $\\hbar \\omega_{n}$.\n\nFor each state $n$, the number of bosons which either enter or exit the resistor per time is equal to their population divided by the time taken to travel the length of the resistor, $t=L / c^{\\prime}$ :\n\n$$\n\\frac{d\\left\\langle N_{n}\\right\\rangle}{d t}=\\frac{\\left\\langle N_{n}\\right\\rangle}{L / c^{\\prime}}=\\frac{k T c^{\\prime}}{\\hbar \\omega_{n} L}\n\\tag{4}\n$$\n\n\n\nFor a frequency interval $d \\omega_{n}$, the number of states is $d n=\\frac{L}{2 \\pi c^{\\prime}} d \\omega_{n}$ (from part (a)). Therefore, the energy delivered per time, per frequency interval is\n\n$$\nd P\\left[\\omega_{n}, \\omega_{n}+d \\omega_{n}\\right]=\\left(\\hbar \\omega_{n}\\right) \\times\\left(\\frac{k T c^{\\prime}}{\\hbar \\omega_{n} L}\\right) \\times\\left(\\frac{L}{2 \\pi c^{\\prime}} d \\omega_{n}\\right)=k T \\frac{d \\omega_{n}}{2 \\pi}\n\\tag{5}\n$$\n\nUsing $f=\\frac{\\omega}{2 \\pi}$,\n\n$$\nd P[f, f+d f]=k T d f\n\\tag{6}\n$$']",,False,,, 979,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题} Context question: (d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Nyquist equivalent noisy voltage source The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.","(a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is $$ P_{\max }=\frac{V^{2}}{4 R} . \tag{7} $$ Give the optimal value of $r$ in terms of $R$ and $V$.","['The total resistance in this circuit is $R+r$. The power dissipated in $r$ is therefore\n\n$$\nP=r I^{2}=r\\left(\\frac{V}{R+r}\\right)^{2}\n\\tag{7}\n$$\n\nThe maximization of $\\frac{r}{(R+r)^{2}}$ is equivalent to the minimization of $\\phi(r)=\\frac{(R+r)^{2}}{r}=\\frac{R^{2}}{r}+2 R+r$. Setting $\\frac{d \\phi}{d r}=0$, for example, gives the solution $r=R$. Substituting $r=R$ into 7 yields\n\n$$\nP=\\frac{V^{2}}{4 R}\n\\tag{8}\n$$']",,False,,, 980,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题} Context question: (d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Nyquist equivalent noisy voltage source The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$. Context question: (a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is $$ P_{\max }=\frac{V^{2}}{4 R} . \tag{7} $$ Give the optimal value of $r$ in terms of $R$ and $V$. Context answer: 证明题 ","(b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$.","['Differentiate the result of the previous part, and apply a time-expectation $\\langle\\rangle: d\\left\\langle V^{2}\\right\\rangle=4 R d\\langle P\\rangle=$ $k T d f$. Therefore $\\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=4 k T R$.']",,False,,, 981,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题} Context question: (d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Nyquist equivalent noisy voltage source The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$. Context question: (a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is $$ P_{\max }=\frac{V^{2}}{4 R} . \tag{7} $$ Give the optimal value of $r$ in terms of $R$ and $V$. Context answer: 证明题 Context question: (b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Other circuit elements We derived the Johnson-Nyquist noise due to a resistor, $R$. It turns out the equation $\frac{d\left\langle V^{2}\right\rangle}{d f}=$ $4 k T R$ is not generalizable to inductors or capacitors.",(a) Explain why no Johnson-Nyquist noise is produced by ideal inductors or capacitors. There are multiple explanations; any explanation will be accepted. (Hint: the impedance of an ideal inductor or capacitor is purely imaginary.),"[""For example, the impedance of an ideal inductor or capacitor is purely imaginary. Replacing $R \\rightarrow i \\omega L, \\frac{1}{i \\omega C}$ in the formula $\\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=4 k T R$ would give an imaginary squared-voltage, which doesn't make sense.\n\nMore physically, the current and voltage in a pure inductor or capacitor are always orthogonal (out-of-phase), so no power is dissipated.""]",,False,,, 982,Electromagnetism,"2. Johnson-Nyquist noise In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ : $$ \frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R \tag{2} $$ Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise. Electromagnetic modes in a resistor We first establish the properties of thermally excited electromagnetic modes $$ V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right) \tag{3} $$ in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$. Context question: (a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$. Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$. Context answer: \boxed{证明题} Context question: (b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is $$ \left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1} \tag{4} $$ In the low-energy limit $\hbar \omega_{n} \ll k T$, show that $$ \left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}} \tag{5} $$ You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$. Context answer: \boxed{证明题} Context question: (c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$. Context answer: \boxed{证明题} Context question: (d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is $$ P[f, f+d f] \approx k T d f . \tag{6} $$ Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.) Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Nyquist equivalent noisy voltage source The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$. Context question: (a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is $$ P_{\max }=\frac{V^{2}}{4 R} . \tag{7} $$ Give the optimal value of $r$ in terms of $R$ and $V$. Context answer: 证明题 Context question: (b) If the average power per frequency interval delivered to the resistor $r$ is $\frac{d\left\langle P_{\max }\right\rangle}{d f}=$ $\frac{d E}{d f}=k T$, show that $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Other circuit elements We derived the Johnson-Nyquist noise due to a resistor, $R$. It turns out the equation $\frac{d\left\langle V^{2}\right\rangle}{d f}=$ $4 k T R$ is not generalizable to inductors or capacitors. Context question: (a) Explain why no Johnson-Nyquist noise is produced by ideal inductors or capacitors. There are multiple explanations; any explanation will be accepted. (Hint: the impedance of an ideal inductor or capacitor is purely imaginary.) Context answer: \boxed{证明题} ","(b) Any real inductor has undesired, or parasitic, resistance. We can model the real inductor as an ideal inductor $L$ in series with a parasitic resistance $R$. Due to the thermal noise $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ of its parasitic resistance, this (real) inductor will support a nonzero per-frequency mean-squared current, $\frac{d\left\langle I^{2}\right\rangle}{d f}$, even when both sides of the inductor are grounded. Compute $\frac{d\left\langle I^{2}\right\rangle}{d f}$ as a function of $f, L, T$ and $R$.","['The total impedance of the circuit is\n\n$$\nZ=R+i \\omega L=R+2 \\pi i f L \\Rightarrow|Z|=\\sqrt{R^{2}+4 \\pi^{2} f^{2} L^{2}}\n\\tag{9}\n$$\n\nTherefore, the circuit equation $V=Z I \\Rightarrow\\left\\langle V^{2}\\right\\rangle=\\left|Z^{2}\\right|\\left\\langle I^{2}\\right\\rangle$ implies\n\n$$\n\\frac{d\\left\\langle I^{2}\\right\\rangle}{d f}=\\frac{1}{\\left|Z^{2}\\right|} \\frac{d\\left\\langle V^{2}\\right\\rangle}{d f}=\\frac{4 k T R}{R^{2}+4 \\pi^{2} f^{2} L^{2}}\n\\tag{10}\n$$']",['$\\frac{4 k T R}{R^{2}+4 \\pi^{2} f^{2} L^{2}}$'],False,,Expression, 983,Mechanics,,"(a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively.","['$$\n\\begin{aligned}\n& M_{1} \\frac{d^{2}}{d t^{2}} \\overrightarrow{r_{1}}=\\frac{G M_{1} M_{2}}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}\\left(\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right) \\\\\n& M_{2} \\frac{d^{2}}{d t^{2}} \\overrightarrow{r_{2}}=\\frac{G M_{1} M_{2}}{\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|^{3}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)\n\\end{aligned}\n$$\n\nHence,\n\n$$\n\\begin{aligned}\n& \\frac{d^{2}}{d t^{2}} \\overrightarrow{r_{1}}=\\frac{G M_{2}}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}\\left(\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right) \\\\\n& \\frac{d^{2}}{d t^{2}} \\overrightarrow{r_{2}}=\\frac{G M_{1}}{\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|^{3}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)\n\\end{aligned}\n$$']",['$$\n\\begin{aligned}\n& \\frac{d^{2}}{d t^{2}} \\overrightarrow{r_{1}}=\\frac{G M_{2}}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}\\left(\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right) \\\\\n& \\frac{d^{2}}{d t^{2}} \\overrightarrow{r_{2}}=\\frac{G M_{1}}{\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|^{3}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)\n\\end{aligned}\n$$'],False,,Need_human_evaluate, 984,Mechanics,"3. The circular restricted three-body problem In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion. Two-body problem Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system. Context question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively. Context answer: $$ \begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} $$ ",(b) Find the period $T$ and angular frequency $\omega$ of the orbital motion.,"['$$\n\\frac{d^{2}}{d t^{2}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)=-\\frac{G\\left(M_{1}+M_{2}\\right)}{\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|^{3}}\\left(\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right)\n$$\n\nUsing polar coordinate, the time derivative of a position vector $\\vec{r}$ with constant magnitude,\n\n$$\n\\begin{aligned}\n& \\frac{d}{d t} \\vec{r}=\\frac{d}{d t} r \\hat{r}=r \\frac{d}{d t} \\hat{r}=\\vec{\\omega} \\times \\vec{r} \\\\\n& \\frac{d 2}{d t^{2}} \\vec{r}=\\vec{\\omega} \\times(\\vec{\\omega} \\times \\vec{r})=-\\omega^{2} \\vec{r}\n\\end{aligned}\n$$\n\nSince $\\left|\\overrightarrow{r_{1}}-\\overrightarrow{r_{2}}\\right|$ is constant,\n\n$$\n\\begin{gathered}\n\\omega=\\sqrt{\\frac{G\\left(M_{1}+M_{2}\\right)}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}} \\\\\nT=2 \\pi / \\omega=2 \\pi \\sqrt{\\frac{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}{G\\left(M_{1}+M_{2}\\right)}}\n\\end{gathered}\n$$']","['$\\omega=\\sqrt{\\frac{G\\left(M_{1}+M_{2}\\right)}{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}}$ , $T=2 \\pi \\sqrt{\\frac{\\left|\\overrightarrow{r_{2}}-\\overrightarrow{r_{1}}\\right|^{3}}{G\\left(M_{1}+M_{2}\\right)}}$']",True,,Expression, 985,Mechanics,"3. The circular restricted three-body problem In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion. Two-body problem Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system. Context question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively. Context answer: $$ \begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} $$ Context question: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion. Context answer: \boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$} Extra Supplementary Reading Materials: Circular restricted three-body problem Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$. The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$.",(a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$.,['$$\n\\begin{gathered}\nM_{1}(-\\alpha R)+M_{2}((1-\\alpha) R)=0 \\\\\n\\alpha=\\frac{M_{2}}{M_{1}+M_{2}}\n\\end{gathered}\n$$'],['$\\alpha=\\frac{M_{2}}{M_{1}+M_{2}}$'],False,,Expression, 986,Mechanics,"3. The circular restricted three-body problem In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion. Two-body problem Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system. Context question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively. Context answer: $$ \begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} $$ Context question: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion. Context answer: \boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$} Extra Supplementary Reading Materials: Circular restricted three-body problem Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$. The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$. Context question: (a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$. Context answer: \boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$} ",(b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given.,['$$\n\\begin{gathered}\n\\rho_{1}(t)=\\sqrt{(x(t)+\\alpha R)^{2}+(y(t))^{2}} \\\\\n\\rho_{2}(t)=\\sqrt{(x(t)-(1-\\alpha) R)^{2}+(y(t))^{2}}\n\\end{gathered}\n$$'],"['$\\rho_{1}(t)=\\sqrt{(x(t)+\\alpha R)^{2}+(y(t))^{2}}$ , $\\rho_{2}(t)=\\sqrt{(x(t)-(1-\\alpha) R)^{2}+(y(t))^{2}}$']",True,,Expression, 987,Mechanics,"3. The circular restricted three-body problem In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion. Two-body problem Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system. Context question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively. Context answer: $$ \begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} $$ Context question: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion. Context answer: \boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$} Extra Supplementary Reading Materials: Circular restricted three-body problem Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$. The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$. Context question: (a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$. Context answer: \boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$} Context question: (b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given. Context answer: \boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$} ","(c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$.",['$$\n\\begin{gathered}\nm \\frac{d^{2}}{d t^{2}} \\vec{r}=-G m M_{1} \\frac{\\vec{r}-\\overrightarrow{r_{1}}}{\\rho_{1}^{3}}-G m M_{2} \\frac{\\vec{r}-\\overrightarrow{r_{2}}}{\\rho_{2}^{3}}+m \\omega^{2} \\vec{r}-2 m \\omega \\times \\vec{r} \\\\\n\\frac{d^{2}}{d t^{2}} \\vec{r}=-G M_{1} \\frac{\\vec{r}-\\overrightarrow{r_{1}}}{\\rho_{1}^{3}}-G M_{2} \\frac{\\vec{r}-\\overrightarrow{r_{2}}}{\\rho_{2}^{3}}+\\omega^{2} \\vec{r}-2 \\omega \\times \\vec{r}\n\\end{gathered}\n$$'],['$\\frac{d^{2}}{d t^{2}} \\vec{r}=-G M_{1} \\frac{\\vec{r}-\\vec{r_{1}}}{\\rho_{1}^{3}}-G M_{2} \\frac{\\vec{r}-\\vec{r_{2}}}{\\rho_{2}^{3}}+\\omega^{2} \\vec{r}-2 \\omega \\times \\vec{r}$'],False,,Expression, 988,Mechanics,"3. The circular restricted three-body problem In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion. Two-body problem Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system. Context question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively. Context answer: $$ \begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} $$ Context question: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion. Context answer: \boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$} Extra Supplementary Reading Materials: Circular restricted three-body problem Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$. The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$. Context question: (a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$. Context answer: \boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$} Context question: (b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given. Context answer: \boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$} Context question: (c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$. Context answer: \boxed{$\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}$} ","(d) Express $\frac{d^{2} x}{d t^{2}}$ and $\frac{d^{2} y}{d t^{2}}$ in terms of $U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right)$.","['$$\n\\begin{gathered}\n\\ddot{x}=-\\frac{G M_{1}(x+\\alpha R)}{\\rho_{1}^{3}}-\\frac{G M_{2}(x-(1-\\alpha) R)}{\\rho_{2}^{3}}+\\omega^{2} x+2 \\omega \\dot{y} \\\\\n\\ddot{y}=-\\frac{G M_{1} y}{\\rho_{1}^{3}}-\\frac{G M_{2} y}{\\rho_{2}^{3}}+\\omega^{2} y-2 \\omega \\dot{x}\n\\end{gathered}\n$$\n\nThen,\n\n$$\n\\begin{aligned}\n\\ddot{x} & =2 \\omega \\dot{y}-\\frac{\\partial U}{\\partial x} \\\\\n\\ddot{y} & =-2 \\omega \\dot{x}-\\frac{\\partial U}{\\partial y}\n\\end{aligned}\n$$']","['$\\ddot{x} =2 \\omega \\dot{y}-\\frac{\\partial U}{\\partial x}$ , $\\ddot{y} =-2 \\omega \\dot{x}-\\frac{\\partial U}{\\partial y}$']",True,,Expression, 989,Mechanics,"3. The circular restricted three-body problem In general, there is no exact solution of the three-body problem, in which three masses move under their mutual gravitational attraction. However, it is possible to make some progress by adding some constraints to the motion. Two-body problem Let's start with the motion of two masses, $M_{1}$ and $M_{2}$. Assume both masses move in circular orbits about their center of mass. Consider the inertial frame whose origin coincides with the center of mass of the system. Context question: (a) Express the equations of motion of $M_{1}$ and $M_{2}$ in terms of the gravitational constant $G$ and the position vectors $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$ which point from the origin to $M_{1}$ and $M_{2}$, respectively. Context answer: $$ \begin{aligned} & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{1}}=\frac{G M_{2}}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}\left(\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right) \\ & \frac{d^{2}}{d t^{2}} \overrightarrow{r_{2}}=\frac{G M_{1}}{\left|\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right|^{3}}\left(\overrightarrow{r_{1}}-\overrightarrow{r_{2}}\right) \end{aligned} $$ Context question: (b) Find the period $T$ and angular frequency $\omega$ of the orbital motion. Context answer: \boxed{$\omega=\sqrt{\frac{G\left(M_{1}+M_{2}\right)}{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}}$ , $T=2 \pi \sqrt{\frac{\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|^{3}}{G\left(M_{1}+M_{2}\right)}}$} Extra Supplementary Reading Materials: Circular restricted three-body problem Let us transform to a non-inertial frame rotating with angular velocity $\vec{\omega}=(0,0, \omega)$ about an axis normal to the orbital plane of masses $M_{1}$ and $M_{2}$, with the origin at their center of mass. In this frame, $M_{1}$ and $M_{2}$ are stationary at the Cartesian coordinates $(-\alpha R, 0,0)$ and $((1-\alpha) R, 0,0)$ respectively. The third mass $m$ is not stationary in this frame; in this non-inertial frame its position is $\vec{r}(t)=(x(t), y(t), 0)$. The masses satisfy $M_{1}, M_{2} \gg m$. Consider $m$ to be so small that it does not affect the motion of $M_{1}$ or $M_{2}$. Context question: (a) Express $\alpha$ in terms of $M_{1}$ and $M_{2}$. Context answer: \boxed{$\alpha=\frac{M_{2}}{M_{1}+M_{2}}$} Context question: (b) Let $\rho_{1}(t)$ and $\rho_{2}(t)$ be the distances from $m$ to $M_{1}$ and $M_{2}$ respectively. Express $\rho_{1}(t)$ and $\rho_{2}(t)$ in terms of the coordinates and constants given. Context answer: \boxed{$\rho_{1}(t)=\sqrt{(x(t)+\alpha R)^{2}+(y(t))^{2}}$ , $\rho_{2}(t)=\sqrt{(x(t)-(1-\alpha) R)^{2}+(y(t))^{2}}$} Context question: (c) By considering the centrifugal acceleration $\omega^{2} \vec{r}$ and Coriolis acceleration $-2 \omega \times$ $\vec{v}$, find the acceleration $\frac{d^{2}}{d t^{2}} \vec{r}$ of the third mass in terms of the coordinates and constants given, including $\rho_{1}$ and $\rho_{2}$. Context answer: \boxed{$\frac{d^{2}}{d t^{2}} \vec{r}=-G M_{1} \frac{\vec{r}-\vec{r_{1}}}{\rho_{1}^{3}}-G M_{2} \frac{\vec{r}-\vec{r_{2}}}{\rho_{2}^{3}}+\omega^{2} \vec{r}-2 \omega \times \vec{r}$} Context question: (d) Express $\frac{d^{2} x}{d t^{2}}$ and $\frac{d^{2} y}{d t^{2}}$ in terms of $U$, where $U=-\frac{G M_{1}}{\rho_{1}}-\frac{G M_{2}}{\rho_{2}}-\frac{\omega^{2}}{2}\left(x^{2}+y^{2}\right)$. Context answer: \boxed{$\ddot{x} =2 \omega \dot{y}-\frac{\partial U}{\partial x}$ , $\ddot{y} =-2 \omega \dot{x}-\frac{\partial U}{\partial y}$} ","(e) Hence, write down an expression of the motion of $m$ which is a constant.","['From d),\n\n$$\n\\begin{gathered}\n\\ddot{x} \\dot{x}=2 \\omega \\dot{x} \\dot{y}-\\dot{x} \\frac{\\partial U}{\\partial x} \\\\\n\\ddot{y} \\dot{y}=-2 \\omega \\dot{x} \\dot{y}-\\dot{y} \\frac{\\partial U}{\\partial y} \\\\\n\\ddot{x} \\dot{x}+\\ddot{y} \\dot{y}=-\\dot{x} \\frac{\\partial U}{\\partial x}-\\dot{y} \\frac{\\partial U}{\\partial y} \\\\\n\\frac{d}{d t}\\left[\\frac{1}{2}\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)\\right]=-\\left(\\dot{x} \\frac{\\partial U}{\\partial x}+\\dot{y} \\frac{\\partial U}{\\partial y}\\right)=-\\frac{d U}{d t} \\\\\n\\frac{d}{d t}\\left[\\frac{1}{2}\\left(\\dot{x}^{2}+\\dot{y}^{2}\\right)+U\\right]=0 \\\\\n\\text { Constant }=-2 U-v^{2}\n\\end{gathered}\n$$']",['$-2 U-v^{2}$'],False,,Expression, 990,Mechanics,,"(a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) ![](https://cdn.mathpix.com/cropped/2023_12_21_2ce10c2f629ebbdfd870g-1.jpg?height=166&width=876&top_left_y=1708&top_left_x=622) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?",['We can always look the infinitesimal process in the comoving frame such that the rocket is at rest in it. Then from conservation of momentum:\n\n$$\n\\mathbf{u} \\Delta m+m \\Delta \\mathbf{v}=0\n\\tag{11}\n$$\n\nDivided by the infinitesimal time $\\Delta t$ we get\n\n$$\n\\frac{\\Delta \\mathbf{v}}{\\Delta t}=-\\frac{\\mathbf{u}}{m} \\frac{\\Delta m}{\\Delta t}\n\\tag{12}\n$$\n\nAs a result\n\n$$\n\\mathbf{a}=\\frac{d \\mathbf{v}}{d t}=-\\frac{\\mu}{m} \\mathbf{u}\n\\tag{13}\n$$'],"['$\\mathbf{u} \\Delta m+m \\Delta \\mathbf{v}=0$ , $\\mathbf{a}=-\\frac{\\mu}{m} \\mathbf{u}$']",True,,"Equation,Expression", 990,Mechanics,"4. The fundamental rocket equation In this problem, we will investigate the acceleration of rockets. Rocket repulsion In empty space, an accelerating rocket must ""throw"" something backward to gain speed from repulsion. Assume there is zero gravity. The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out. Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$. (Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.)","(a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?",['We can always look the infinitesimal process in the comoving frame such that the rocket is at rest in it. Then from conservation of momentum:\n\n$$\n\\mathbf{u} \\Delta m+m \\Delta \\mathbf{v}=0\n\\tag{11}\n$$\n\nDivided by the infinitesimal time $\\Delta t$ we get\n\n$$\n\\frac{\\Delta \\mathbf{v}}{\\Delta t}=-\\frac{\\mathbf{u}}{m} \\frac{\\Delta m}{\\Delta t}\n\\tag{12}\n$$\n\nAs a result\n\n$$\n\\mathbf{a}=\\frac{d \\mathbf{v}}{d t}=-\\frac{\\mu}{m} \\mathbf{u}\n\\tag{13}\n$$'],"['$\\mathbf{u} \\Delta m+m \\Delta \\mathbf{v}=0$ , $\\mathbf{a}=-\\frac{\\mu}{m} \\mathbf{u}$']",True,,"Equation,Expression", 991,Mechanics,"4. The fundamental rocket equation In this problem, we will investigate the acceleration of rockets. Rocket repulsion In empty space, an accelerating rocket must ""throw"" something backward to gain speed from repulsion. Assume there is zero gravity. The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out. Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$. (Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.) Context question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ? Context answer: \boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$}","(b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ? (Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.)","['From part a) we can get a differential equation:\n\n$$\nd v=-u \\frac{d m}{m}\n\\tag{14}\n$$\n\n\n\nNotice that we already get rid of all the vectors and express everything in scalar. It might be tricky to get the sign correct. By integrating it, we find\n\n$$\n\\int_{0}^{v_{f}} d v=-u \\int_{10 m_{0}}^{m_{0}} \\frac{d m}{m}\n\\tag{15}\n$$\n\nAs a result, the final speed of the rocket is\n\n$$\nv_{f}=u \\ln \\frac{10 m_{0}}{m_{0}}=u \\ln 10 \\approx 2.302 u\n\\tag{16}\n$$\n\n']",['$u \\ln 10$'],False,,Expression, 992,Mechanics,"4. The fundamental rocket equation In this problem, we will investigate the acceleration of rockets. Rocket repulsion In empty space, an accelerating rocket must ""throw"" something backward to gain speed from repulsion. Assume there is zero gravity. The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out. Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$. (Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.) Context question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ? Context answer: \boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$} Context question: (b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ? (Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.) Context answer: $u \ln 10$ ",(c) Discuss the factors that limit the final speed of the rocket.,['The ratio of the mass of the rocket it self (the spaceship and the fuel container) to the mass of the fuel it can carry.\n\nThe fuel eject speed $u$.'],,False,,, 993,Mechanics,"4. The fundamental rocket equation In this problem, we will investigate the acceleration of rockets. Rocket repulsion In empty space, an accelerating rocket must ""throw"" something backward to gain speed from repulsion. Assume there is zero gravity. The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out. Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$. (Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.) Context question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ? Context answer: \boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$} Context question: (b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ? (Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.) Context answer: $u \ln 10$ Context question: (c) Discuss the factors that limit the final speed of the rocket. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Chemical rockets Nearly all rockets get energy from the chemical reaction of the fuel (burning with the oxidant) they carry. This is called a combustion reaction. For example, the combustion reaction for hydrogen is $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \tag{8} $$","(a) Suppose that the product of combustion is just one species of molecule $\left(\mathrm{H}_{2} \mathrm{O}\right.$, for example) with mass $m_{p}$ and average kinetic energy $E$. What is the upper limit of exhaust speed $u$ when these molecules are ejected from the rocket?",['$$\nE=\\frac{1}{2} m_{p} u^{2} \\Longrightarrow u=\\sqrt{\\frac{2 E}{m_{p}}}\n\\tag{17}\n$$'],['$\\sqrt{\\frac{2 E}{m_{p}}}$'],False,,Expression, 994,Mechanics,,"(b) To manufacture fast rockets, would you recommend using hydrogen as the fuel? Assume that we only use oxygen as oxidant. If so, please give your reasoning. If not, what fuel would you suggest and why? (There are no numerical values here, so please give the best reasoning based on your own knowledge. Any reasonable answer is acceptable.)","['Yes, I would recommend using (liquid) hydrogen. The reason is that the energy $E$ produced by any chemical reaction is about the same magnitude. Therefore, to get larger ejection speed of the fuel, we hope the product of the reaction to be as light a possible. Since we fix the oxidant to be oxygen, the lightest product we can get is water $\\mathrm{H}_{2} \\mathrm{O}$ from the burning of hydrogen.']",['开放性回答'],False,,Need_human_evaluate, 995,Mechanics,"4. The fundamental rocket equation In this problem, we will investigate the acceleration of rockets. Rocket repulsion In empty space, an accelerating rocket must ""throw"" something backward to gain speed from repulsion. Assume there is zero gravity. The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out. Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$. (Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.) Context question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ? Context answer: \boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$} Context question: (b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ? (Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.) Context answer: $u \ln 10$ Context question: (c) Discuss the factors that limit the final speed of the rocket. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Chemical rockets Nearly all rockets get energy from the chemical reaction of the fuel (burning with the oxidant) they carry. This is called a combustion reaction. For example, the combustion reaction for hydrogen is $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \tag{8} $$ Context question: (a) Suppose that the product of combustion is just one species of molecule $\left(\mathrm{H}_{2} \mathrm{O}\right.$, for example) with mass $m_{p}$ and average kinetic energy $E$. What is the upper limit of exhaust speed $u$ when these molecules are ejected from the rocket? Context answer: \boxed{$\sqrt{\frac{2 E}{m_{p}}}$} Context question: (b) To manufacture fast rockets, would you recommend using hydrogen as the fuel? Assume that we only use oxygen as oxidant. If so, please give your reasoning. If not, what fuel would you suggest and why? (There are no numerical values here, so please give the best reasoning based on your own knowledge. Any reasonable answer is acceptable.) Context answer: 开放性回答 ","(c) The enthalpy of combustion of the reaction $2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}$ is approximately $15.76 \times 10^{6} \mathrm{~J} / \mathrm{kg}$. That is, for every kilogram of water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ produced, the energy released by the reaction is $15.76 \times 10^{6} \mathrm{~J}$. Assuming that all the energy released can be converted to the kinetic energy of the water molecules, what is their exhaust speed $u$ ? (Hint: $\sqrt{15.76 \times 10^{6}} \approx 3970$ )",['Plug in the numbers:\n\n$$\nu=\\sqrt{\\frac{2 E}{m_{p}}} \\approx \\sqrt{2} \\times \\sqrt{15.76 \\times 10^{6} \\mathrm{~J} / \\mathrm{kg}}=\\sqrt{2} \\times 3970 \\mathrm{~m} / \\mathrm{s} \\approx 5614 \\mathrm{~m} / \\mathrm{s}\n\\tag{18}\n$$'],['$\\sqrt{2} \\times 3970$'],False,m/s,Numerical,1e1 996,Mechanics,"4. The fundamental rocket equation In this problem, we will investigate the acceleration of rockets. Rocket repulsion In empty space, an accelerating rocket must ""throw"" something backward to gain speed from repulsion. Assume there is zero gravity. The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out. Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$. (Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.) Context question: (a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.) Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u. Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ? Context answer: \boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$} Context question: (b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ? (Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.) Context answer: $u \ln 10$ Context question: (c) Discuss the factors that limit the final speed of the rocket. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Chemical rockets Nearly all rockets get energy from the chemical reaction of the fuel (burning with the oxidant) they carry. This is called a combustion reaction. For example, the combustion reaction for hydrogen is $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \tag{8} $$ Context question: (a) Suppose that the product of combustion is just one species of molecule $\left(\mathrm{H}_{2} \mathrm{O}\right.$, for example) with mass $m_{p}$ and average kinetic energy $E$. What is the upper limit of exhaust speed $u$ when these molecules are ejected from the rocket? Context answer: \boxed{$\sqrt{\frac{2 E}{m_{p}}}$} Context question: (b) To manufacture fast rockets, would you recommend using hydrogen as the fuel? Assume that we only use oxygen as oxidant. If so, please give your reasoning. If not, what fuel would you suggest and why? (There are no numerical values here, so please give the best reasoning based on your own knowledge. Any reasonable answer is acceptable.) Context answer: 开放性回答 Context question: (c) The enthalpy of combustion of the reaction $2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}$ is approximately $15.76 \times 10^{6} \mathrm{~J} / \mathrm{kg}$. That is, for every kilogram of water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ produced, the energy released by the reaction is $15.76 \times 10^{6} \mathrm{~J}$. Assuming that all the energy released can be converted to the kinetic energy of the water molecules, what is their exhaust speed $u$ ? (Hint: $\sqrt{15.76 \times 10^{6}} \approx 3970$ ) Context answer: \boxed{$\sqrt{2} \times 3970$}","(d) Now, consider the effects of gravity. Given a rocket of mass $m_{0}$ and mass ejection rate $\mu$, what ejection speed $u$ would be required to launch satellites in the Earth's gravitational field, of strength $g$ ? (An estimate, ignoring numerical factors, is acceptable.) What about launching space probes to other planets or out of the solar system? Can you guess why most rockets are multistage?","['There are multiple ways of thinking about this problem.\n\n- (Easiest possible answer) To launch the rocket from earth we need at least the force from repulsion to be stronger than the gravity. That is $a=\\mu u / m_{0}>g$. (We get rid of numerical factors on the total mass.)\n\n\n\n- A nicer perspective: The rocket needs to get to gain enough mechanical energy (kinetic plus potential energy) from the repulsion process. It is known (from multiple ways) that the energy of an object with mass $m_{0}$ orbiting a planet in a orbit whose length of the semimajor axis is $a$ is $-G M m_{0} / 2 a=-g R^{2} m_{0} / 2 a$, where $R$ is the radius of the planet ( $G M$ the standard gravitational parameter). Therefore, the minimum energy is at $a=R$ ( $a$ cannot be lower), which is $-g R m_{0} / 2$. Besides that, we know that before launching, when the rocket is on the ground, its potential energy (kinetic energy being zero) is $-G M m_{0} / R=-g R m_{0}$.\n\nTherefore, the minimum mechanical energy needed is $g R m_{0} / 2$. This situation corresponds to ""launching"" the rocket horizontally. The rocket does not go high into the sky but flies right above the ground. As a result, as a not so bad approximation, the launching process is not affected much by the gravity. The kinetic energy of the rocket after its fuel used up, $m_{0} v_{f}^{2} / 2$ should be greater than $g R m_{0} / 2$. Therefore, we get a result:\n\n$$\nv_{f}^{2}>g R\n\\tag{19}\n$$\n\nAny answer that gets rid of numerical factors like $u^{2} \\geq g R$ is acceptable. If a student remember the value of the first cosmic speed of the earth: $v_{o}=\\sqrt{g R}=7.9 \\mathrm{~km} / \\mathrm{s}$, and states that $v_{f}$, or a multiple of $u$ (like $2.302 u$ we got above), should be greater than $v_{o}$ is also great.\n\n- For launching our rocket to other planets, we need to escape from the gravitational field of the earth. That is, the final mechanical energy of the rocket should be greater than zero:\n\n$$\nv_{f}^{2}>2 g R\n\\tag{20}\n$$\n\nIt is also great if a student remember that the escape speed of the each is $v_{e}=\\sqrt{2 g R}=11.2$ and states that a multiple of $u$ should be larger than $v_{e}$.\n\n- The reason why rockets are multistage is that We see from previous parts that the ejection speed of the lightest fuel with $100 \\%$ energy conversion rate is only $u=5614 \\mathrm{~m} / \\mathrm{s}$, and the final speed $v_{f}$ of a very light rocket (mass ratio between fuel and rocket itself being 9 ) is just $\\approx 2.3 u=12.9 \\mathrm{~km} / \\mathrm{s}$, which is only at the same magnitude of the first cosmic speed and the escape speed.\n\nIn real world, rockets cannot be so ideal. Therefore, a single stage rocket can hardly be launched to orbit around the earth, not to mention escaping. Multistage rockets drop part of its mass to further increase the mass ratio between the remaining fuel and itself in order to make itself to the sky.\n\n']","['$\\sqrt{g R}$ , $\\sqrt{2 g R}$']",True,,Expression, 997,Modern Physics,,a - Determine the positions and widths of the principal maximum on the screen. The width is defined as the distance between the minima on either side of the maxima.,"['a - Consider first the $\\mathrm{x}$-direction. If waves coming from neighbouring slits (with separation $d_{1}$ ) traverse paths of lengths that differ by:\n\n$$\n\\Delta_{1}=n_{1} \\cdot \\lambda\n$$\n\nwhere $\\mathrm{n}_{1}$ is an integer, then a principal maximum occurs. The position on the screen (in the $\\mathrm{x}$-direction) is:\n\n$$\nx_{n_{1}}=\\frac{n_{1} \\cdot \\lambda \\cdot L}{d_{1}}\n$$\n\nsince $\\mathrm{d}_{1}<<\\mathrm{d}_{2}$.\n\nThe path difference between the middle slit and one of the slits at the edge is then:\n\n$$\n\\Delta_{\\left(\\frac{N_{1}}{2}\\right)}=\\frac{N_{1}}{2} \\cdot n_{1} \\cdot \\lambda\n$$\n\nIf on the other hand this path difference is:\n\n$$\n\\Delta_{\\left(\\frac{N_{1}}{2}\\right)}=\\frac{N_{1}}{2} \\cdot n_{1} \\cdot \\lambda+\\frac{\\lambda}{2}\n$$\n\nthen the first minimum, next to the principal maximum, occurs. The position of this minimum on the screen is given by:\n\n$$\n\\begin{gathered}\nx_{n_{1}}+\\Delta x=\\frac{\\left(\\frac{N_{1}}{2} \\cdot n_{1} \\cdot \\lambda+\\frac{\\lambda}{2}\\right) \\cdot L}{\\frac{N_{1}}{2} \\cdot d_{1}}=\\frac{n_{1} \\cdot \\lambda \\cdot L}{d_{1}}+\\frac{\\lambda \\cdot L}{N_{1} \\cdot d_{1}} \\\\\n\\rightarrow \\quad \\Delta x=\\frac{\\lambda \\cdot L}{N_{1} \\cdot d_{1}}\n\\end{gathered}\n$$\n\nThe width of the principal maximum is accordingly:\n\n$$\n2 . \\Delta x=2 \\cdot \\frac{\\lambda . L}{N_{1} \\cdot d_{1}}\n$$\n\nA similar treatment can be made for the $y$-direction, in which there are $\\mathrm{N}_{2}$ slits with separation $\\mathrm{d}_{2}$. The positions and widths of the principal maximal are:\n\n$$\n\\begin{gathered}\n\\left(x_{n_{1},} y_{n_{2}}\\right)=\\left(\\frac{n_{1} \\cdot \\lambda \\cdot L}{d_{1}}, \\frac{n_{2} \\cdot \\lambda \\cdot L}{d_{2}}\\right) \\\\\n2 \\cdot \\Delta x=2 \\cdot \\frac{\\lambda \\cdot L}{N_{1} \\cdot d_{1}} ; \\quad 2 \\cdot \\Delta y=2 \\cdot \\frac{\\lambda \\cdot L}{N_{2} \\cdot d_{2}}\n\\end{gathered}\n$$']","['$(x_{n_{1},} y_{n_{2}})=(\\frac{n_{1} \\cdot \\lambda \\cdot L}{d_{1}}, \\frac{n_{2} \\cdot \\lambda \\cdot L}{d_{2}})$;\n$2 \\cdot \\Delta x=2 \\cdot \\frac{\\lambda \\cdot L}{N_{1} \\cdot d_{1}} ; \\quad 2 \\cdot \\Delta y=2 \\cdot \\frac{\\lambda \\cdot L}{N_{2} \\cdot d_{2}}$']",False,,Need_human_evaluate, 998,Modern Physics,,b - Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta)$.,"[""b - In the x-direction the beam 'sees' a grid with spacing a, so that in this direction we have:\n\n$$\nx_{n_{1}}=\\frac{n_{1} \\cdot \\lambda \\cdot L}{a} \\quad \\Delta x=2 \\cdot \\frac{\\lambda \\cdot L}{N_{0} \\cdot a}\n$$\n\n\n\nIn the y-direction, the beam 'sees' a grid with effective spacing a. $\\cos (\\Theta)$.\n\nAnalogously, we obtain:\n\n$$\ny_{n_{2}}=\\frac{n_{2} \\cdot \\lambda \\cdot L}{a \\cdot \\cos (\\theta)} \\quad \\Delta y=2 \\cdot \\frac{\\lambda \\cdot L}{N_{0} \\cdot a \\cdot \\cos (\\theta)}\n$$\n\nIn the $z$-direction, the beam 'sees' a grid with effective spacing a.sin $(\\Theta)$. This gives rise to principal maxima with position and width:\n\n$$\ny_{n_{3}}^{\\prime}=\\frac{n_{3} \\cdot \\lambda \\cdot L}{a \\cdot \\sin (\\theta)} \\quad \\Delta y^{\\prime}=2 \\cdot \\frac{\\lambda \\cdot L}{N_{1} \\cdot a \\cdot \\sin (\\theta)}\n$$\n\nThis pattern is superimposed on the previous one. Since $\\sin (\\Theta)$ is very small, only the zeroth-order pattern will be seen, and it is very broad, since $\\mathrm{N}_{1} \\cdot \\sin (\\Theta)<<\\mathrm{N}_{0}$. The diffraction pattern from a plane wave falling on a thin plate of a cubic crystal, at a small angle of incidence to the normal, will be almost identical to that from a two-dimensional grid.""]","['position is $(x_{n_{1}},y_{n_{2}},y_{n_{3}}^{\\prime})=(\\frac{n_{1} \\lambda L}{a},\\frac{n_{2} \\lambda L}{a \\cos (\\theta)},\\frac{n_{3} \\lambda L}{a \\sin (\\theta)})$, and the width $\\Delta x=2 \\frac{\\lambda L}{N_{0} a}$, $\\Delta y=2 \\frac{\\lambda L}{N_{0} a \\cos (\\theta)}$, $\\Delta y^{\\prime}=2 \\frac{\\lambda L}{N_{1} a \\sin (\\theta)}$']",True,,Need_human_evaluate, 999,Modern Physics,"X-ray Diffraction from a crystal. We wish to study X-ray diffraction by a cubic crystal lattice. To do this we start with the diffraction of a plane, monochromatic wave that falls perpendicularly on a 2-dimensional grid that consists of $\mathrm{N}_{1} \times \mathrm{N}_{2}$ slits with separations $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$. The diffraction pattern is observed on a screen at a distance $L$ from the grid. The screen is parallel to the grid and $\mathrm{L}$ is much larger than $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$. Context question: a - Determine the positions and widths of the principal maximum on the screen. The width is defined as the distance between the minima on either side of the maxima. Context answer: $(x_{n_{1},} y_{n_{2}})=(\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}, \frac{n_{2} \cdot \lambda \cdot L}{d_{2}})$; $2 \cdot \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot d_{1}} ; \quad 2 \cdot \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{2} \cdot d_{2}}$ Extra Supplementary Reading Materials: We consider now a cubic crystal, with lattice spacing a and size $\mathrm{N}_{0} \cdot a \times \mathrm{N}_{0} \cdot a \times \mathrm{N}_{1} \cdot a \cdot \mathrm{N}_{1}$ is much smaller than $\mathrm{N}_{0}$. The crystal is placed in a parallel X-ray beam along the z-axis at an angle $\Theta$ (see Fig. 1). The diffraction pattern is again observed on a screen at a great distance from the crystal. Figure 1 Diffraction of a parallel X-ray beam along the z-axis. The angle between the crystal and the $y$-axis is $\Theta$. Context question: b - Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta)$. Context answer: position is $(x_{n_{1}},y_{n_{2}},y_{n_{3}}^{\prime})=(\frac{n_{1} \lambda L}{a},\frac{n_{2} \lambda L}{a \cos (\theta)},\frac{n_{3} \lambda L}{a \sin (\theta)})$, and the width $\Delta x=2 \frac{\lambda L}{N_{0} a}$, $\Delta y=2 \frac{\lambda L}{N_{0} a \cos (\theta)}$, $\Delta y^{\prime}=2 \frac{\lambda L}{N_{1} a \sin (\theta)}$ Extra Supplementary Reading Materials: The diffraction pattern can also be derived by means of Bragg's theory, in which it is assumed that the X-rays are reflected from atomic planes in the lattice. The diffraction pattern then arises from interference of these reflected rays with each other.",c - Show that this so-called Bragg reflection yields the same conditions for the maxima as those that you found in $b$.,"['c - In Bragg reflection, the path difference for constructive interference between neighbouring planes:\n\n$$\n\\Delta=2 \\cdot a \\cdot \\sin (\\phi) \\approx 2 \\cdot a \\cdot \\phi=n \\cdot \\lambda \\quad \\rightarrow \\quad \\frac{x}{L} \\approx 2 \\cdot \\phi \\approx \\frac{n \\cdot \\lambda}{a} \\quad \\rightarrow \\quad x \\approx \\frac{n \\cdot \\lambda \\cdot L}{a}\n$$\n\nHere $\\phi$ is the angle of diffraction.\n\nThis is the same condition for a maximum as in section $b$.']",,False,,, 1000,Modern Physics,"X-ray Diffraction from a crystal. We wish to study X-ray diffraction by a cubic crystal lattice. To do this we start with the diffraction of a plane, monochromatic wave that falls perpendicularly on a 2-dimensional grid that consists of $\mathrm{N}_{1} \times \mathrm{N}_{2}$ slits with separations $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$. The diffraction pattern is observed on a screen at a distance $L$ from the grid. The screen is parallel to the grid and $\mathrm{L}$ is much larger than $\mathrm{d}_{1}$ and $\mathrm{d}_{2}$. Context question: a - Determine the positions and widths of the principal maximum on the screen. The width is defined as the distance between the minima on either side of the maxima. Context answer: $(x_{n_{1},} y_{n_{2}})=(\frac{n_{1} \cdot \lambda \cdot L}{d_{1}}, \frac{n_{2} \cdot \lambda \cdot L}{d_{2}})$; $2 \cdot \Delta x=2 \cdot \frac{\lambda \cdot L}{N_{1} \cdot d_{1}} ; \quad 2 \cdot \Delta y=2 \cdot \frac{\lambda \cdot L}{N_{2} \cdot d_{2}}$ Extra Supplementary Reading Materials: We consider now a cubic crystal, with lattice spacing a and size $\mathrm{N}_{0} \cdot a \times \mathrm{N}_{0} \cdot a \times \mathrm{N}_{1} \cdot a \cdot \mathrm{N}_{1}$ is much smaller than $\mathrm{N}_{0}$. The crystal is placed in a parallel X-ray beam along the z-axis at an angle $\Theta$ (see Fig. 1). The diffraction pattern is again observed on a screen at a great distance from the crystal. Figure 1 Diffraction of a parallel X-ray beam along the z-axis. The angle between the crystal and the $y$-axis is $\Theta$. Context question: b - Calculate the position and width of the maxima as a function of the angle $\Theta$ (for small $\Theta)$. Context answer: position is $(x_{n_{1}},y_{n_{2}},y_{n_{3}}^{\prime})=(\frac{n_{1} \lambda L}{a},\frac{n_{2} \lambda L}{a \cos (\theta)},\frac{n_{3} \lambda L}{a \sin (\theta)})$, and the width $\Delta x=2 \frac{\lambda L}{N_{0} a}$, $\Delta y=2 \frac{\lambda L}{N_{0} a \cos (\theta)}$, $\Delta y^{\prime}=2 \frac{\lambda L}{N_{1} a \sin (\theta)}$ Extra Supplementary Reading Materials: The diffraction pattern can also be derived by means of Bragg's theory, in which it is assumed that the X-rays are reflected from atomic planes in the lattice. The diffraction pattern then arises from interference of these reflected rays with each other. Context question: c - Show that this so-called Bragg reflection yields the same conditions for the maxima as those that you found in $b$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: In some measurements the so-called powder method is employed. A beam of X-rays is scattered by a powder of very many, small crystals. (Of course the sizes of the crystals are much larger than the lattice spacing, a). Scattering of X-rays of wavelength $0.15 \mathrm{~nm}$ by Potassium Chloride [KCl] (which has a cubic lattice, see Fig.2) results in the production of concentric dark circles on a photographic plate. The distance between the crystals and the plate is $0.10 \mathrm{~m}$, and the radius of the smallest circle is $0.053 \mathrm{~m}$ (see Fig. 3). $\mathrm{K}^{+}$and $\mathrm{Cl}^{-}$ions have almost the same size, and they may be treated as identical scattering centres. Figure 2. The cubic latice of Potassium Chloride in which the $\mathrm{K}^{+}$and $\mathrm{Cl}^{-}$wons have almost the same size. Figure 3. Scattering of X-rays by a powder of $\mathrm{KCl}$ crystals results in the production of concentric dark circles on a photographic plate.",d - Calculate the distance between two neighbouring $\mathrm{K}$ ions in the crystal.,"['d - For the distance, $\\sqrt{ } 2$.a, between neighbouring $\\mathrm{K}$ ions we have:\n\n$$\n\\begin{gathered}\n\\operatorname{tg}(2 \\phi)=\\frac{x}{L}=\\frac{0,053}{0,1} \\approx 0,53 \\rightarrow a=\\frac{\\lambda}{2 \\cdot \\sin (\\phi)} \\approx \\frac{0,15 \\cdot 10^{-9}}{2 \\cdot 0,24} \\approx 0,31 \\mathrm{~nm} \\\\\nK-K \\approx \\sqrt{2} \\cdot 0,31 \\approx 0,44 \\mathrm{~nm}\n\\end{gathered}\n$$']",['0.44'],False,nm,Numerical,1e-2 1001,Electromagnetism,"Electric experiments in the magnetosphere of the earth. In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane. At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1). S also lies in the equatorial plane. Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit. Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane. The magnetic field (B) is perpendicular to the diagram and is directed towards the reader.","$a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant?","['$a_{1}$ - Since $m_{2}<\n\nFor $\\mathrm{m}_{2}$ we have:\n\n$$\nm_{2} \\cdot L \\cdot \\ddot{\\alpha}=-\\left(F_{g}-F_{c}\\right) \\cdot \\sin (\\alpha)=-\\left(\\frac{G \\cdot m_{2} \\cdot m_{a}}{(R-L \\cdot \\cos (\\alpha))^{2}}-m_{2} \\cdot \\Omega^{2} \\cdot(R-L \\cdot \\cos (\\alpha))\\right) \\cdot \\sin (\\alpha)\n$$\n\nUsing the approximation:\n\n$$\n\\frac{1}{(R-L \\cdot \\cos (\\alpha))^{2}} \\approx \\frac{1}{R^{2}}+\\frac{2 \\cdot L \\cdot \\cos (\\alpha)}{R^{3}}\n$$\n\nand equation (1), one finds:\n\n$$\nL \\cdot \\ddot{\\alpha}=-\\left(\\frac{G \\cdot m_{a}}{R^{2}}+\\frac{2 \\cdot G \\cdot m_{a}}{R^{3}} \\cdot L \\cdot \\cos (\\alpha)-\\frac{G \\cdot m_{a}}{R^{3}} \\cdot R+\\frac{G \\cdot m_{a}}{R^{3}} \\cdot L \\cdot \\cos (\\alpha)\\right) \\cdot \\sin (\\alpha)\n$$\n\nso:\n\n$$\n\\ddot{\\alpha}+3 \\cdot \\Omega^{2} \\cdot \\sin (\\alpha) \\cdot \\cos (\\alpha)=0\n\\tag{2}\n$$\n\nIf $\\alpha$ is constant: $\\ddot{\\alpha}=0 \\quad->\\quad \\sin (\\alpha)=0 \\quad->\\quad \\alpha=0 ; \\quad \\alpha=\\pi$\n\n$$\n->\\quad \\cos (\\alpha)=0 \\quad->\\quad \\alpha=\\pi / 2 ; \\quad \\alpha=3 \\pi / 2\n$$']","['$0,\\pi,\\frac{\\pi}{2},\\frac{3\\pi}{2}$']",True,,Numerical,1e-8 1002,Electromagnetism,"Electric experiments in the magnetosphere of the earth. In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane. At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1). S also lies in the equatorial plane. Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit. Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane. The magnetic field (B) is perpendicular to the diagram and is directed towards the reader. Context question: $a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant? Context answer: \boxed{$0,\pi,\frac{\pi}{2},\frac{3\pi}{2}$} ",$a_{2}$ - Discuss the stability of the equilibrium for each case.,"['$\\mathrm{a}_{2}$ - The situation is stable if the moment $M=m_{2} \\cdot L \\cdot \\ddot{\\alpha} \\cdot L=m_{2} \\cdot L^{2} \\cdot \\ddot{\\alpha}$ changes sign in a manner opposed to that in which the sign of $\\alpha-\\alpha_{0}$ changes:\n\n$\\operatorname{sign}\\left(\\alpha-\\alpha_{0}\\right) \\quad-\\quad+\\quad-\\quad+\\quad-\\quad+\\quad-\\quad+\\quad-\\quad+$\n\n| $\\alpha$ | 0 | $\\pi / 2$ | $\\pi$ | $3 \\pi / 2$ | | $2 \\pi$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\operatorname{sign}(\\mathrm{M})$ | | + | + | $-\\quad+$ | + | |\n| $\\alpha$ | 0 | $\\pi / 2$ | $\\pi$ | $3 \\pi / 2$ | | $2 \\pi$ |\n\nThe equilibrium about the angles 0 en $\\pi$ is thus stable, whereas that around $\\pi / 2$ and $3 \\pi / 2$ is unstable.']",,False,,, 1003,Electromagnetism,"Electric experiments in the magnetosphere of the earth. In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane. At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1). S also lies in the equatorial plane. Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit. Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane. The magnetic field (B) is perpendicular to the diagram and is directed towards the reader. Context question: $a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant? Context answer: \boxed{$0,\pi,\frac{\pi}{2},\frac{3\pi}{2}$} Context question: $a_{2}$ - Discuss the stability of the equilibrium for each case. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Suppose that, at a given moment, the rod deviates from the stable configuration by a small angle. The system will begin to swing like a pendulum.",$\mathrm{b}$ - Express the period of the swinging in terms of the period of revolution of the system around the earth.,"['$\\mathrm{b}$ - For small values of $\\alpha$ equation (2) becomes:\n\n$$\n\\ddot{\\alpha}+3 \\cdot \\Omega^{2} \\cdot \\alpha=0\n$$\n\nThis is the equation of a simple harmonic motion.\n\nThe square of the angular frequency is:\n\n$$\n\\omega^{2}=3 . \\Omega^{2}\n$$\n\nso:\n\n$$\n\\omega=\\Omega \\cdot \\sqrt{3} \\quad \\rightarrow \\quad T_{1}=\\frac{2 \\pi}{\\omega}=\\frac{1}{3} \\sqrt{3} \\cdot\\left(\\frac{2 \\pi}{\\Omega}\\right) \\approx 0,58 \\cdot T_{0}\n$$']",,False,,, 1005,Electromagnetism,"Electric experiments in the magnetosphere of the earth. In May 1991 the spaceship Atlantis will be placed in orbit around the earth. We shall assume that this orbit will be circular and that it lies in the earth's equatorial plane. At some predetermined moment the spaceship will release a satellite $S$, which is attached to a conducting rod of length $\mathrm{L}$. We suppose that the rod is rigid, has negligible mass, and is covered by an electrical insulator. We also neglect all friction. Let $\alpha$ be the angle that the rod makes to the line between the Atlantis and the centre of the earth. (see Fig. 1). S also lies in the equatorial plane. Assume that the mass of the satellite is much smaller than that of the Atlantis, and that $\mathrm{L}$ is much smaller than the radius of the orbit. Figure 1 The spaceship Atlantis (A) with a satellite (S) in an orbit around the earth. The orbit lies in the earth's equatorial plane. The magnetic field (B) is perpendicular to the diagram and is directed towards the reader. Context question: $a_{1}$ - Deduce for which value(s) of $\alpha$ the configuration of the spaceship and satellite remain unchanged (with respect to the earth)? In other words, for which value(s) of $\alpha$ is $\alpha$ constant? Context answer: \boxed{$0,\pi,\frac{\pi}{2},\frac{3\pi}{2}$} Context question: $a_{2}$ - Discuss the stability of the equilibrium for each case. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Suppose that, at a given moment, the rod deviates from the stable configuration by a small angle. The system will begin to swing like a pendulum. Context question: $\mathrm{b}$ - Express the period of the swinging in terms of the period of revolution of the system around the earth. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: In Fig. 1 the magnetic field of the earth is perpendicular to the diagram and is directed towards the reader. Due to the orbital velocity of the rod, a potential difference arises between its ends. The environment (the magnetosphere) is a rarefied, ionised gas with a very good electrical conductivity. Contact with the ionised gas is made by means of electrodes in $\mathrm{A}$ (the Atlantis) and $\mathrm{S}$ (the satellite). As a consequence of the motion, a current, I, flows through the rod. Context question: $\mathrm{c}_{1}$ - In which direction does the current flow through the rod? (Take $\alpha=0$ ) Context answer: \boxed{from the satellite (S) towards the shuttle (A)} Extra Supplementary Reading Materials: Data: - the period of the orbit $$ \begin{aligned} & T=5,4 \cdot 10^{3} \mathrm{~s} \\ & L=2,0 \cdot 10^{4} \mathrm{~m} \\ & B =5,0 \cdot 10^{-5} \mathrm{~Wb} \cdot \mathrm{m}^{-2} \\ & m=1,0 \cdot 10^{5} \mathrm{~kg} \end{aligned} $$ $$ \begin{aligned} & \text { - length of the rod } \\ & \text { - magnetic field strength of the eart at the height } \\ & \text { of the satellite } \\ & \text { - the mass of the shuttle Atlantis } \end{aligned} $$ Next, a current source inside the shuttle is included in the circuit, which maintaines a net direct current of $0.1 \mathrm{~A}$ in the opposite direction.","$\mathrm{c}_{2}$ - How long must this current be maintained to change the altitude of the orbit by 10 m. Assume that $\alpha$ remains zero. Ignore all contributions from currents in the magnetosphere.","['$\\mathrm{c}_{2}$ - For the total energy of the system we have:\n\n$$\nU=U_{k i n}+U_{p o t}=\\frac{1}{2} \\cdot m \\cdot \\Omega^{2} \\cdot R^{2}-\\frac{G \\cdot m \\cdot m_{a}}{R}=-\\frac{1}{2} \\cdot \\frac{G \\cdot m \\cdot m_{a}}{R}\n$$\n\nA small change in the radius of the orbit corresponds to a change in the energy of:\n\n$$\n\\Delta U=\\frac{1}{2} \\cdot \\frac{G \\cdot m \\cdot m_{a}}{R^{2}} \\cdot \\Delta R=\\frac{1}{2} \\cdot m \\cdot \\Omega^{2} \\cdot R \\cdot \\Delta R\n$$\n\nIn the situation under $\\mathrm{c}_{1}$ energy is absorbed from the system as a consequence of which the radius of the orbit will decrease.\n\nIs a current source inside the shuttle included in the circuit, which maintains a net current in the opposite direction, energy is absorbed by the system as a consequence of which the radius of the orbit will increase.\n\nFrom the assumptions in $\\mathrm{c}_{2}$ we have:\n\n$$\n\\Delta U=F_{l} \\cdot v . t=\\text { B.I.L.S.R.t }=\\frac{1}{2} \\cdot m \\cdot \\Omega^{2} \\cdot R \\cdot \\Delta R \\quad \\rightarrow \\quad t=\\frac{1}{2} \\cdot \\frac{m \\cdot \\Omega \\cdot \\Delta R}{\\text { B.I.L }}\n$$\n\nNumerical application gives for the time: $\\mathrm{t} \\approx 5.8\\times 10^{3} \\mathrm{~s}$; which is about the period of the system.']",['$t=\\frac{1}{2} \\cdot \\frac{m \\cdot \\Omega \\cdot \\Delta R}{BIL}$'],False,,Expression, 1006,Mechanics,"The rotating neutron star. A 'millisecond pulsar' is a source of radiation in the universe that emits very short pulses with a period of one to several milliseconds. This radiation is in the radio range of wavelengths; and a suitable radio receiver can be used to detect the separate pulses and thereby to measure the period with great accuracy. These radio pulses originate from the surface of a particular sort of star, the so-called neutron star. These stars are very compact: they have a mass of the same order of magnitude as that of the sun, but their radius is only a few tens of kilometers. They spin very quickly. Because of the fast rotation, a neutron star is slightly flattened (oblate). Assume the axial cross-section of the surface to be an ellipse with almost equal axes. Let $r_{p}$ be the polar and $r_{e}$ the equatorial radii; and let us define the flattening factor by: $$ \epsilon=\frac{\left(r_{e}-r_{p}\right)}{r_{p}} $$ | a mass of | $2.0 \cdot 10^{30} \mathrm{~kg}$, | | :--- | :--- | | an average radius of | $1.0 \cdot 10^{4} \mathrm{~m}$, | | and a rotation period of | $2.0 \cdot 10^{-2} \mathrm{~s}$. |","a - Calculate the flattening factor, given that the gravitational constant is $6.67 \times 10^{-11}$ N.m $\cdot \mathrm{kg}^{-2}$.","['a - 1st method\n\nFor equilibrium we have $\\mathrm{F}_{\\mathrm{c}}=\\mathrm{F}_{\\mathrm{g}}+\\mathrm{N}$ where $\\mathrm{N}$ is normal to the surface.\n\nResolving into horizontal and vertical components, we find:\n\n\n\n$$\n\\begin{aligned}\n& F_{g} \\cdot \\cos (\\phi)=F_{c}+N \\cdot \\sin (\\alpha) \\\\\n& \\quad F_{g} \\cdot \\sin (\\phi)=N \\cdot \\cos (\\alpha)\n\\end{aligned} \\rightarrow F_{g} \\cdot \\cos (\\phi)=F_{c}+F_{g} \\cdot \\sin (\\phi) \\cdot \\operatorname{tg}(\\alpha)\n$$\n\nFrom:\n\n$$\nF_{g}=\\frac{G \\cdot M}{r^{2}}, \\quad F_{c}=\\omega^{2} \\cdot r, x=r \\cdot \\cos (\\phi), y=r \\cdot \\sin (\\phi) \\text { en } \\operatorname{tg}(\\alpha)=\\frac{d y}{d x}\n$$\n\nwe find:\n\n$$\ny \\cdot d y+\\left(1-\\frac{\\omega^{2} \\cdot r^{3}}{G \\cdot M}\\right) \\cdot x \\cdot d x=0\n$$\n\nwhere:\n\n$$\n\\frac{\\omega^{2} \\cdot r^{3}}{G \\cdot M} \\approx 7 \\cdot 10^{-4}\n$$\n\nThis means that, although $r$ depends on $x$ and $y$, the change in the factor in front of $x d x$ is so slight that we can take it to be constant. The solution of Eq. (1) is then an ellipse:\n\n\n\n$$\n\\frac{x^{2}}{r_{e}^{2}}+\\frac{y^{2}}{r_{p}^{2}}=1 \\rightarrow \\frac{r_{p}}{r_{e}}=\\sqrt{1-\\frac{\\omega^{2} \\cdot r^{3}}{G \\cdot M}} \\approx 1-\\frac{\\omega^{2} \\cdot r^{3}}{2 \\cdot G \\cdot M}\n$$\n\nand from this it follows that:\n\n$$\n\\epsilon=\\frac{r_{e}-r_{p}}{r_{e}}=\\frac{\\omega^{2} \\cdot r^{3}}{2 \\cdot G \\cdot M} \\approx 3,7 \\cdot 10^{-4}\n$$', '2nd method \n\nFor a point mass of $1 \\mathrm{~kg}$ on the surface,\n\n$$\nU_{p o t}=-\\frac{G \\cdot M}{r} \\quad U_{k i n}=\\frac{1}{2} \\cdot \\omega^{2} \\cdot r^{2} \\cdot \\cos ^{2}(\\phi)\n$$\n\nThe form of the surface is such that $U_{\\text {pot }}-U_{\\text {kin }}=$ constant. For the equator $(\\Phi=0$, $\\left.r=r_{e}\\right)$ and for the pole $\\left(\\Phi=\\pi / 2, r=r_{p}\\right)$ we have:\n\n$$\n\\frac{G \\cdot M}{r_{p}}=\\frac{G \\cdot M}{r_{e}}+\\frac{1}{2} \\cdot \\omega^{2} \\cdot r_{e}^{2} \\rightarrow \\frac{r_{e}}{r_{p}}=1+\\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M}\n$$\n\nThus:\n\n$$\n\\epsilon=\\frac{r_{e}-r_{p}}{r_{e}}=\\frac{1+\\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M}-1}{1+\\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M}} \\approx \\frac{\\omega^{2} \\cdot r_{e}^{3}}{2 \\cdot G \\cdot M} \\approx 3,7 \\cdot 10^{-4}\n$$']",['$3.7 \\times 10^{-4}$'],False,,Numerical,1e-5 1007,Mechanics,"The rotating neutron star. A 'millisecond pulsar' is a source of radiation in the universe that emits very short pulses with a period of one to several milliseconds. This radiation is in the radio range of wavelengths; and a suitable radio receiver can be used to detect the separate pulses and thereby to measure the period with great accuracy. These radio pulses originate from the surface of a particular sort of star, the so-called neutron star. These stars are very compact: they have a mass of the same order of magnitude as that of the sun, but their radius is only a few tens of kilometers. They spin very quickly. Because of the fast rotation, a neutron star is slightly flattened (oblate). Assume the axial cross-section of the surface to be an ellipse with almost equal axes. Let $r_{p}$ be the polar and $r_{e}$ the equatorial radii; and let us define the flattening factor by: $$ \epsilon=\frac{\left(r_{e}-r_{p}\right)}{r_{p}} $$ | a mass of | $2.0 \cdot 10^{30} \mathrm{~kg}$, | | :--- | :--- | | an average radius of | $1.0 \cdot 10^{4} \mathrm{~m}$, | | and a rotation period of | $2.0 \cdot 10^{-2} \mathrm{~s}$. | Context question: a - Calculate the flattening factor, given that the gravitational constant is $6.67 \times 10^{-11}$ N.m $\cdot \mathrm{kg}^{-2}$. Context answer: \boxed{$3.7 \times 10^{-4}$} Extra Supplementary Reading Materials: In the long run (over many years) the rotation of the star slows down, due to energy loss, and this leads to a decrease in the flattening. The star has however a solid crust that floats on a liquid interior. The solid crust resists a continuous adjustment to equilibrium shape. Instead, starquakes occur with sudden changes in the shape of the crust towards equilibrium. During and after such a star-quake the angular velocity is observed to change according to figure 1. A sudden change in the shape of the crust of a neutron star results in a sudden change of the angular velocity.","$\mathrm{b}$ - Calculate the average radius of the liquid interior, using the data of Fig. 1. Make the approximation that the densities of the crust and the interior are the same. (Ignore the change in shape of the interior).","['b - As a consequence of the star-quake, the moment of inertia of the crust $\\mathrm{I}_{\\mathrm{m}}$ decreases by $\\Delta I_{m}$.\n\nFrom the conservation of angular momentum, we have:\n\n\n\n$$\nI_{m} \\cdot \\omega_{0}=\\left(I_{m}-\\Delta I_{m}\\right) \\cdot \\omega_{1} \\rightarrow \\Delta I_{m}=I_{m} \\cdot \\frac{\\omega_{1}-\\omega_{0}}{\\omega_{1}}\n$$\n\nAfter the internal friction has equalized the angular velocities of the crust and the core, we have:\n\n\n\n$$\n\\begin{gathered}\n\\left(I_{m}+I_{c}\\right) \\cdot \\omega_{0}=\\left(I_{m}+I_{c}-\\Delta I_{m}\\right) \\cdot \\omega_{2} \\rightarrow \\Delta I_{m}=\\left(I_{m}+I_{c}\\right) \\cdot \\frac{\\omega_{2}-\\omega_{0}}{\\omega_{2}} \\\\\n\\frac{I_{m}}{I_{m}+I_{c}}=\\frac{\\left(\\omega_{2}-\\omega_{0}\\right) \\cdot \\omega_{1}}{\\left(\\omega_{1}-\\omega_{0}\\right) \\cdot \\omega_{2}} \\rightarrow 1-\\frac{I_{c}}{I_{m}+I_{c}}=\\frac{\\left(\\omega_{2}-\\omega_{0}\\right) \\cdot \\omega_{1}}{\\left(\\omega_{1}-\\omega_{0}\\right) \\cdot \\omega_{2}} \\\\\nI(:) R^{2} \\\\\n\\rightarrow \\frac{I_{c}}{I_{m}+I_{c}}=\\frac{r_{c}^{2}}{r^{2}} \\rightarrow \\frac{r_{c}}{r}=\\sqrt{1-\\frac{\\left(\\omega_{2}-\\omega_{0}\\right) \\cdot \\omega_{1}}{\\left(\\omega_{1}-\\omega_{0}\\right) \\cdot \\omega_{2}} } \\approx 0.95\n\\end{gathered}\n$$']",['$0.95$'],False,,Numerical,2e-2 1008,Mechanics,"A Three-body Problem and LISA FIGURE 1 Coplanar orbits of three bodies.","1.1 Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_{0}$ of the line joining $M$ and $m$ in terms of $R, r, M, m$ and the universal gravitational constant $G$.","['Let $\\mathrm{O}$ be their centre of mass. Hence\n\n$$\nM R-m r=0\n\\tag{1}\n$$\n$$\n\\begin{aligned}\n& m \\omega_{0}^{2} r=\\frac{G M m}{(R+r)^{2}}\\\\\n& M \\omega_{0}^{2} R=\\frac{G M m}{(R+r)^{2}}\n\\end{aligned}\n\\tag{2}\n$$\n\nFrom Eq. (2), or using reduced mass, $\\omega_{0}^{2}=\\frac{G(M+m)}{(R+r)^{3}}$\n\nHence, \n$$\n\\omega_{0}^{2}=\\frac{G(M+m)}{(R+r)^{3}}=\\frac{G M}{r(R+r)^{2}}=\\frac{G m}{R(R+r)^{2}}\n\\tag{3}\n$$']",['$\\omega_{0}=\\sqrt{\\frac{G m}{R(R+r)^{2}}}$'],False,,Expression, 1009,Mechanics,"A Three-body Problem and LISA FIGURE 1 Coplanar orbits of three bodies. Context question: 1.1 Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_{0}$ of the line joining $M$ and $m$ in terms of $R, r, M, m$ and the universal gravitational constant $G$. Context answer: \boxed{$\omega_{0}=\sqrt{\frac{G m}{R(R+r)^{2}}}$}","1.2 A third body of infinitesimal mass $\mu$ is placed in a coplanar circular orbit about the same centre of mass so that $\mu$ remains stationary relative to both $M$ and $m$ as shown in Figure 1. Assume that the infinitesimal mass is not collinear with $M$ and $m$. Find the values of the following parameters in terms of $R$ and $r$ : 1.2.1 distance from $\mu$ to $M$. 1.2.2 distance from $\mu$ to $m$. 1.2.3 distance from $\mu$ to the centre of mass.","['Since $\\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$. For $\\mu$ to remain stationary relative to both $M$ and $m$ we must have:\n\n$$\n\\frac{G M \\mu}{r_{1}^{2}} \\cos \\theta_{1}+\\frac{G m \\mu}{r_{2}^{2}} \\cos \\theta_{2} =\\mu \\omega_{0}^{2} \\rho=\\frac{G(M+m) \\mu}{(R+r)^{3}} \\rho\n\\tag{4}\n$$\n$$\n\\frac{G M \\mu}{r_{1}^{2}} \\sin \\theta_{1} =\\frac{G m \\mu}{r_{2}^{2}} \\sin \\theta_{2}\n\\tag{5}\n$$\n\nSubstituting $\\frac{G M}{r_{1}^{2}}$ from Eq. (5) into Eq. (4), and using the identity\n\n$\\sin \\theta_{1} \\cos \\theta_{2}+\\cos \\theta_{1} \\sin \\theta_{2}=\\sin \\left(\\theta_{1}+\\theta_{2}\\right)$, we get\n\n$$\nm \\frac{\\sin \\left(\\theta_{1}+\\theta_{2}\\right)}{r_{2}^{2}}=\\frac{(M+m)}{(R+r)^{3}} \\rho \\sin \\theta_{1}\n\\tag{6}\n$$\n\nThe distances $r_{2}$ and $\\rho$, the angles $\\theta_{1}$ and $\\theta_{2}$ are related by two Sine Rule equations\n\n$$\n\\begin{aligned}\n& \\frac{\\sin \\psi_{1}}{\\rho}=\\frac{\\sin \\theta_{1}}{R} \\\\\n& \\frac{\\sin \\psi_{1}}{r_{2}}=\\frac{\\sin \\left(\\theta_{1}+\\theta_{2}\\right)}{R+r}\n\\end{aligned}\n\\tag{7}\n$$\n\nSubstitute (7) into (6)\n\n$$\n\\frac{1}{r_{2}^{3}}=\\frac{R}{(R+r)^{4}} \\frac{(M+m)}{m}\n\\tag{10}\n$$\n\nSince $\\frac{m}{M+m}=\\frac{R}{R+r}$,Eq. (10) gives\n\n$$\nr_{2}=R+r\n\\tag{11}\n$$\n\nBy substituting $\\frac{G m}{r_{2}^{2}}$ from Eq. (5) into Eq. (4), and repeat a similar procedure, we get\n\n$$\nr_{1}=R+r\n\\tag{12}\n$$\n\nAlternatively,\n\n$$\n\\begin{aligned}\n& \\frac{r_{1}}{\\sin \\left(180^{\\circ}-\\phi\\right)}=\\frac{R}{\\sin \\theta_{1}} \\text { and } \\frac{r_{2}}{\\sin \\phi}=\\frac{r}{\\sin \\theta_{2}} \\\\\n& \\frac{\\sin \\theta_{1}}{\\sin \\theta_{2}}=\\frac{R}{r} \\times \\frac{r_{2}}{r_{1}}=\\frac{m}{M} \\times \\frac{r_{2}}{r_{1}}\n\\end{aligned}\n$$\n\nCombining with Eq. (5) gives $r_{1}=r_{2}$\n\n\n\nHence, it is an equilateral triangle with\n\n$$\n\\begin{aligned}\n& \\psi_{1}=60^{\\circ} \\\\\n& \\psi_{2}=60^{\\circ}\n\\end{aligned}\n\\tag{13}\n$$\n\nThe distance $\\rho$ is calculated from the Cosine Rule.\n\n$$\n\\begin{aligned}\n& \\rho^{2}=r^{2}+(R+r)^{2}-2 r(R+r) \\cos 60^{\\circ} \\\\\n& \\rho=\\sqrt{r^{2}+r R+R^{2}}\n\\end{aligned}\n\\tag{14}\n$$', 'Since $\\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$.For $\\mu$ to remain stationary relative to both $M$ and $m$ we must have:\n\n$$\n\\frac{G M \\mu}{r_{1}^{2}} \\cos \\theta_{1}+\\frac{G m \\mu}{r_{2}^{2}} \\cos \\theta_{2}=\\mu \\omega^{2} \\rho=\\frac{G(M+m) \\mu}{(R+r)^{3}} \\rho\n\\tag{4}\n$$\n$$\n\\frac{G M \\mu}{r_{1}^{2}} \\sin \\theta_{1}=\\frac{G m \\mu}{r_{2}^{2}} \\sin \\theta_{2}\n\\tag{5}\n$$\n\nNote that\n\n$$\n\\begin{aligned}\n& \\frac{r_{1}}{\\sin \\left(180^{\\circ}-\\phi\\right)}=\\frac{R}{\\sin \\theta_{1}} \\\\\n& \\frac{r_{2}}{\\sin \\phi}=\\frac{r}{\\sin \\theta_{2}} \\quad \\text { (see figure) }\n\\end{aligned}\n$$\n$$\n\\frac{\\sin \\theta_{1}}{\\sin \\theta_{2}}=\\frac{R}{r} \\times \\frac{r_{2}}{r_{1}}=\\frac{m}{M} \\times \\frac{r_{2}}{r_{1}}\n\\tag{6}\n$$\n\nEquations (5) and (6):\n\n$$\nr_{1}=r_{2}\n\\tag{7}\n$$\n$$\n\\frac{\\sin \\theta_{1}}{\\sin \\theta_{2}}=\\frac{m}{M}\n\\tag{8}\n$$\n$$\n\\psi_{1}=\\psi_{2}\n\\tag{9}\n$$\n\nThe equation (4) then becomes:\n\n$$\nM \\cos \\theta_{1}+m \\cos \\theta_{2}=\\frac{(M+m)}{(R+r)^{3}} r_{1}^{2} \\rho\n\\tag{10}\n$$\n\nEquations (8) and (10): \n$$\n\\sin \\left(\\theta_{1}+\\theta_{2}\\right)=\\frac{M+m}{M} \\frac{r_{1}^{2} \\rho}{(R+r)^{3}} \\sin \\theta_{2}\n\\tag{11}\n$$\n\nNote that from figure, \n$$\n\\quad \\frac{\\rho}{\\sin \\psi_{2}}=\\frac{r}{\\sin \\theta_{2}}\n\\tag{12}\n$$\n\n\n\nEquations (11) and (12): \n$$\n\\sin \\left(\\theta_{1}+\\theta_{2}\\right)=\\frac{M+m}{M} \\frac{r_{1}^{2} r}{(R+r)^{3}} \\sin \\psi_{2}\n\\tag{13}\n$$\n\nAlso from figure,\n\n$$\n(R+r)^{2}=r_{2}^{2}-2 r_{1} r_{2} \\cos \\left(\\theta_{1}+\\theta_{2}\\right)+r_{1}^{2}=2 r_{1}^{2}\\left[1-\\cos \\left(\\theta_{1}+\\theta_{2}\\right)\\right]\n\\tag{14}\n$$\n\nEquations (13) and (14): \n$$\n\\sin \\left(\\theta_{1}+\\theta_{2}\\right)=\\frac{\\sin \\psi_{2}}{2\\left[1-\\cos \\left(\\theta_{1}+\\theta_{2}\\right)\\right]}\n\\tag{15}\n$$\n\n$$\n\\begin{aligned}\n& \\theta_{1}+\\theta_{2}=180^{\\circ}-\\psi_{1}-\\psi_{2}=180^{\\circ}-2 \\psi_{2} \\quad \\text { (see figure) } \\\\\n\\therefore \\quad & \\cos \\psi_{2}=\\frac{1}{2}, \\psi_{2}=60^{\\circ}, \\psi_{1}=60^{\\circ}\n\\end{aligned}\n$$\n\nHence $M$ and $m$ from an equilateral triangle of sides $(R+r)$\n\nDistance $\\mu$ to $M$ is $R+r$\n\nDistance $\\mu$ to $m$ is $R+r$\n\nDistance $\\mu$ to $\\mathrm{O}$ is $\\rho=\\sqrt{\\left(\\frac{R+r}{2}-R\\right)^{2}+\\left\\{(R+r) \\frac{\\sqrt{3}}{2}\\right\\}^{2}}=\\sqrt{R^{2}+R r+r^{2}}$']","['$R+r$, $R+r$, $\\sqrt{r^{2}+r R+R^{2}}$']",True,,Expression, 1010,Mechanics,"A Three-body Problem and LISA FIGURE 1 Coplanar orbits of three bodies. Context question: 1.1 Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_{0}$ of the line joining $M$ and $m$ in terms of $R, r, M, m$ and the universal gravitational constant $G$. Context answer: \boxed{$\omega_{0}=\sqrt{\frac{G m}{R(R+r)^{2}}}$} Context question: 1.2 A third body of infinitesimal mass $\mu$ is placed in a coplanar circular orbit about the same centre of mass so that $\mu$ remains stationary relative to both $M$ and $m$ as shown in Figure 1. Assume that the infinitesimal mass is not collinear with $M$ and $m$. Find the values of the following parameters in terms of $R$ and $r$ : 1.2.1 distance from $\mu$ to $M$. 1.2.2 distance from $\mu$ to $m$. 1.2.3 distance from $\mu$ to the centre of mass. Context answer: distance from $\mu$ to $M$ is $R+r$ distance from $\mu$ to $m$ is $R+r$ distance from $\mu$ to the centre of mass is $\sqrt{r^{2}+r R+R^{2}}$ ","1.3 Consider the case $M=m$. If $\mu$ is now given a small radial perturbation (along $\mathrm{O} \mu$ ), what is the angular frequency of oscillation of $\mu$ about the unperturbed position in terms of $\omega_{0}$ ? Assume that the angular momentum of $\mu$ is conserved.","['The energy of the mass $\\mu$ is given by\n\n$$\nE=-\\frac{G M \\mu}{r_{1}}-\\frac{G m \\mu}{r_{2}}+\\frac{1}{2} \\mu\\left(\\left(\\frac{d \\rho}{d t}\\right)^{2}+\\rho^{2} \\omega^{2}\\right)\n\\tag{15}\n$$\n\nSince the perturbation is in the radial direction, angular momentum is conserved $\\left(r_{1}=r_{2}=\\Re\\right.$ and $m=M$ ),\n\n$$\nE=-\\frac{2 G M \\mu}{\\Re}+\\frac{1}{2} \\mu\\left(\\left(\\frac{d \\rho}{d t}\\right)^{2}+\\frac{\\rho_{0}{ }^{4} \\omega_{0}{ }^{2}}{\\rho^{2}}\\right)\n\\tag{16}\n$$\n\nSince the energy is conserved,\n\n$$\n\\frac{d E}{d t}=0\n$$\n\n$$\n\\frac{d E}{d t}=\\frac{2 G M \\mu}{\\mathfrak{R}^{2}} \\frac{d \\Re}{d t}+\\mu \\frac{d \\rho}{d t} \\frac{d^{2} \\rho}{d t^{2}}-\\mu \\frac{\\rho_{0}{ }^{4} \\omega_{0}{ }^{2}}{\\rho^{3}} \\frac{d \\rho}{d t}=0\n\\tag{17}\n$$\n\n$$\n\\frac{d \\Re}{d t}=\\frac{d \\Re}{d \\rho} \\frac{d \\rho}{d t}=\\frac{d \\rho}{d t} \\frac{\\rho}{\\mathfrak{R}}\n\\tag{18}\n$$\n\n$$\n\\frac{d E}{d t}=\\frac{2 G M \\mu}{\\mathfrak{R}^{3}} \\rho \\frac{d \\rho}{d t}+\\mu \\frac{d \\rho}{d t} \\frac{d^{2} \\rho}{d t^{2}}-\\mu \\frac{\\rho_{0}^{4} \\omega_{0}{ }^{2}}{\\rho^{3}} \\frac{d \\rho}{d t}=0\n\\tag{19}\n$$\n\n\n\n\n\nSince $\\frac{d \\rho}{d t} \\neq 0$, we have\n\n$\\frac{2 G M}{\\mathfrak{R}^{3}} \\rho+\\frac{d^{2} \\rho}{d t^{2}}-\\frac{\\rho_{0}{ }^{4} \\omega_{0}{ }^{2}}{\\rho^{3}}=0$ or\n\n$$\n\\frac{d^{2} \\rho}{d t^{2}}=-\\frac{2 G M}{\\mathfrak{R}^{3}} \\rho+\\frac{\\rho_{0}{ }^{4} \\omega_{0}{ }^{2}}{\\rho^{3}}\n\\tag{20}\n$$\n\nThe perturbation from $\\mathfrak{R}_{0}$ and $\\rho_{0}$ gives $\\mathfrak{R}=\\mathfrak{R}_{0}\\left(1+\\frac{\\Delta \\mathfrak{R}}{\\mathfrak{R}_{0}}\\right)$ and $\\rho=\\rho_{0}\\left(1+\\frac{\\Delta \\rho}{\\rho_{0}}\\right)$.\n\nThen\n\n$$\n\\frac{d^{2} \\rho}{d t^{2}}=\\frac{d^{2}}{d t^{2}}\\left(\\rho_{0}+\\Delta \\rho\\right)=-\\frac{2 G M}{\\mathfrak{R}_{0}^{3}\\left(1+\\frac{\\Delta \\mathfrak{R}}{\\mathfrak{R}_{0}}\\right)^{3}} \\rho_{0}\\left(1+\\frac{\\Delta \\rho}{\\rho_{0}}\\right)+\\frac{\\rho_{0}{ }^{4} \\omega_{0}{ }^{2}}{\\rho_{0}^{3}\\left(1+\\frac{\\Delta \\rho}{\\rho_{0}}\\right)^{3}}\n\\tag{21}\n$$\n\nUsing binomial expansion $(1+\\varepsilon)^{n} \\approx 1+n \\varepsilon$,\n\n$$\n\\frac{d^{2} \\Delta \\rho}{d t^{2}}=-\\frac{2 G M}{\\mathfrak{R}_{0}^{3}} \\rho_{0}\\left(1+\\frac{\\Delta \\rho}{\\rho_{0}}\\right)\\left(1-\\frac{3 \\Delta \\mathfrak{R}}{\\mathfrak{R}_{0}}\\right)+\\rho_{0} \\omega_{0}^{2}\\left(1-\\frac{3 \\Delta \\rho}{\\rho_{0}}\\right)\n\\tag{22}\n$$\n\nUsing $\\Delta \\rho=\\frac{\\Re}{\\rho} \\Delta \\mathfrak{R}$,\n\n$$\n\\frac{d^{2} \\Delta \\rho}{d t^{2}}=-\\frac{2 G M}{\\mathfrak{R}_{0}^{3}} \\rho_{0}\\left(1+\\frac{\\Delta \\rho}{\\rho_{0}}-\\frac{3 \\rho_{0} \\Delta \\rho}{\\mathfrak{R}_{0}^{2}}\\right)+\\rho_{0} \\omega_{0}^{2}\\left(1-\\frac{3 \\Delta \\rho}{\\rho_{0}}\\right)\n\\tag{23}\n$$\n\nSince $\\omega_{0}^{2}=\\frac{2 G M}{\\Re_{0}^{3}}$,\n\n$$\n\\frac{d^{2} \\Delta \\rho}{d t^{2}}=-\\omega_{0}^{2} \\rho_{0}\\left(1+\\frac{\\Delta \\rho}{\\rho_{0}}-\\frac{3 \\rho_{0} \\Delta \\rho}{\\mathfrak{R}_{0}^{2}}\\right)+\\omega_{0}^{2} \\rho_{0}\\left(1-\\frac{3 \\Delta \\rho}{\\rho_{0}}\\right)\n\\tag{24}\n$$\n\n$$\n\\frac{d^{2} \\Delta \\rho}{d t^{2}}=-\\omega_{0}^{2} \\rho_{0}\\left(\\frac{4 \\Delta \\rho}{\\rho_{0}}-\\frac{3 \\rho_{0} \\Delta \\rho}{\\mathfrak{R}_{0}^{2}}\\right)\n\\tag{25}\n$$\n\n$$\n\\frac{d^{2} \\Delta \\rho}{d t^{2}}=-\\omega_{0}^{2} \\Delta \\rho\\left(4-\\frac{3 \\rho_{0}^{2}}{\\mathfrak{R}_{0}^{2}}\\right)\n\\tag{26}\n$$\n\nFrom the figure, $\\rho_{0}=\\mathfrak{R}_{0} \\cos 30^{\\circ}$ or $\\frac{\\rho_{0}{ }^{2}}{\\mathfrak{R}_{0}{ }^{2}}=\\frac{3}{4}$,\n\n$$\n\\frac{d^{2} \\Delta \\rho}{d t^{2}}=-\\omega_{0}^{2} \\Delta \\rho\\left(4-\\frac{9}{4}\\right)=-\\frac{7}{4} \\omega_{0}^{2} \\Delta \\rho\n\\tag{27}\n$$\n\n\n\nAngular frequency of oscillation is $\\frac{\\sqrt{7}}{2} \\omega_{0}$.', ""$M=m$ gives $R=r$ and $\\omega_{0}{ }^{2}=\\frac{G(M+M)}{(R+R)^{3}}=\\frac{G M}{4 R^{3}}$. The unperturbed radial distance of $\\mu$ is $\\sqrt{3} R$, so the perturbed radial distance can be represented by $\\sqrt{3} R+\\zeta$ where $\\zeta<<\\sqrt{3} R$ as shown in the following figure.\n\nUsing Newton's $2^{\\text {nd }}$ law, $-\\frac{2 G M \\mu}{\\left\\{R^{2}+(\\sqrt{3} R+\\zeta)^{2}\\right\\}^{3 / 2}}(\\sqrt{3} R+\\zeta)=\\mu \\frac{d^{2}}{d t^{2}}(\\sqrt{3} R+\\zeta)-\\mu \\omega^{2}(\\sqrt{3} R+\\zeta)$.\n\n(1)\n\nThe conservation of angular momentum gives $\\mu \\omega_{0}(\\sqrt{3} R)^{2}=\\mu \\omega(\\sqrt{3} R+\\zeta)^{2}$.\n\n(2)\n\nManipulate (1) and (2) algebraically, applying $\\zeta^{2} \\approx 0$ and binomial approximation.\n\n$-\\frac{2 G M}{\\left\\{R^{2}+(\\sqrt{3} R+\\zeta)^{2}\\right\\}^{3 / 2}}(\\sqrt{3} R+\\zeta)=\\frac{d^{2} \\zeta}{d t^{2}}-\\frac{\\omega_{0}^{2} \\sqrt{3} R}{(1+\\zeta / \\sqrt{3} R)^{3}}$\n\n$-\\frac{2 G M}{\\left\\{4 R^{2}+2 \\sqrt{3} \\zeta R\\right\\}^{3 / 2}}(\\sqrt{3} R+\\zeta) \\approx \\frac{d^{2} \\zeta}{d t^{2}}-\\frac{\\omega_{0}^{2} \\sqrt{3} R}{(1+\\zeta / \\sqrt{3} R)^{3}}$\n\n$-\\frac{G M}{4 R^{3}} \\sqrt{3} R \\frac{(1+\\zeta / \\sqrt{3} R)}{(1+\\sqrt{3} \\zeta / 2 R)^{3 / 2}}=\\frac{d^{2} \\zeta}{d t^{2}}-\\frac{\\omega_{0}^{2} \\sqrt{3} R}{(1+\\zeta / \\sqrt{3} R)^{3}}$\n\n$-\\omega_{0}^{2} \\sqrt{3} R\\left(1-\\frac{3 \\sqrt{3} \\zeta}{4 R}\\right)\\left(1+\\frac{\\zeta}{\\sqrt{3} R}\\right) \\approx \\frac{d^{2} \\zeta}{d t^{2}}-\\omega_{0}^{2} \\sqrt{3} R\\left(1-\\frac{3 \\zeta}{\\sqrt{3} R}\\right)$\n\n$\\frac{d^{2}}{d t^{2}} \\zeta=-\\left(\\frac{7}{4} \\omega_{0}^{2}\\right) \\zeta$""]",['$\\frac{\\sqrt{7}}{2} \\omega_{0}$'],False,,Expression, 1011,Mechanics,"A Three-body Problem and LISA FIGURE 1 Coplanar orbits of three bodies. Context question: 1.1 Two gravitating masses $M$ and $m$ are moving in circular orbits of radii $R$ and $r$, respectively, about their common centre of mass. Find the angular velocity $\omega_{0}$ of the line joining $M$ and $m$ in terms of $R, r, M, m$ and the universal gravitational constant $G$. Context answer: \boxed{$\omega_{0}=\sqrt{\frac{G m}{R(R+r)^{2}}}$} Context question: 1.2 A third body of infinitesimal mass $\mu$ is placed in a coplanar circular orbit about the same centre of mass so that $\mu$ remains stationary relative to both $M$ and $m$ as shown in Figure 1. Assume that the infinitesimal mass is not collinear with $M$ and $m$. Find the values of the following parameters in terms of $R$ and $r$ : 1.2.1 distance from $\mu$ to $M$. 1.2.2 distance from $\mu$ to $m$. 1.2.3 distance from $\mu$ to the centre of mass. Context answer: distance from $\mu$ to $M$ is $R+r$ distance from $\mu$ to $m$ is $R+r$ distance from $\mu$ to the centre of mass is $\sqrt{r^{2}+r R+R^{2}}$ Context question: 1.3 Consider the case $M=m$. If $\mu$ is now given a small radial perturbation (along $\mathrm{O} \mu$ ), what is the angular frequency of oscillation of $\mu$ about the unperturbed position in terms of $\omega_{0}$ ? Assume that the angular momentum of $\mu$ is conserved. Context answer: \boxed{$\frac{\sqrt{7}}{2} \omega_{0}$} Extra Supplementary Reading Materials: The Laser Interferometry Space Antenna (LISA) is a group of three identical spacecrafts for detecting low frequency gravitational waves. Each of the spacecrafts is placed at the corners of an equilateral triangle as shown in Figure 2 and Figure 3. The sides (or 'arms') are about 5.0 million kilometres long. The LISA constellation is in an earth-like orbit around the Sun trailing the Earth by $20^{\circ}$. Each of them moves on a slightly inclined individual orbit around the Sun. Effectively, the three spacecrafts appear to roll about their common centre one revolution per year. They are continuously transmitting and receiving laser signals between each other. Overall, they detect the gravitational waves by measuring tiny changes in the arm lengths using interferometric means. A collision of massive objects, such as blackholes, in nearby galaxies is an example of the sources of gravitational waves. FIGURE 2 Illustration of the LISA orbit. The three spacecraft roll about their centre of mass with a period of 1 year. Initially, they trail the Earth by $20^{\circ}$. (Picture from D.A. Shaddock, ""An Overview of the Laser Interferometer Space Antenna"", Publications of the Astronomical Society of Australia, 2009, 26, pp.128-132.). FIGURE 3 Enlarged view of the three spacecrafts trailing the Earth. A, B and $\mathrm{C}$ are the three spacecrafts at the corners of the equilateral triangle.","1.4 In the plane containing the three spacecrafts, what is the relative speed of one spacecraft with respect to another?","['Let $v=$ speed of each spacecraft as it moves in circle around the centre $\\mathrm{O}$.\n\nThe relative velocities are denoted by the subscripts A, B and C.\n\nFor example, $v_{\\mathrm{BA}}$ is the velocity of $\\mathrm{B}$ as observed by $\\mathrm{A}$.\n\nThe period of circular motion is 1 year $T=365 \\times 24 \\times 60 \\times 60 \\mathrm{~s}$.\n\nThe angular frequency $\\omega=\\frac{2 \\pi}{T}$\n\nThe speed $v=\\omega \\frac{L}{2 \\cos 30^{\\circ}}=575 \\mathrm{~m} / \\mathrm{s}$\n\n\n\nThe speed is much less than the speed light $\\rightarrow$ Galilean transformation.\n\nIn Cartesian coordinates, the velocities of B and C (as observed by O) are\n\n\nFor B, $\\vec{v}_{B}=v \\cos 60^{\\circ} \\hat{\\mathbf{i}}-v \\sin 60^{\\circ} \\hat{\\mathbf{j}}$\n\nFor C, $\\vec{v}_{C}=v \\cos 60^{\\circ} \\hat{\\mathbf{i}}+v \\sin 60^{\\circ} \\hat{\\mathbf{j}}$\n\nHence $\\vec{v}_{\\mathrm{BC}}=-2 v \\sin 60^{\\circ} \\hat{\\mathbf{j}}=-\\sqrt{3} v \\hat{\\mathbf{j}}$\n\nThe speed of $B$ as observed by $C$ is $\\sqrt{3} v \\approx 996 \\mathrm{~m} / \\mathrm{s}$\n\nNotice that the relative velocities for each pair are anti-parallel.', 'One can obtain $v_{\\mathrm{BC}}$ by considering the rotation about the axis at one of the spacecrafts.\n\n$v_{\\mathrm{BC}}=\\omega L=\\frac{2 \\pi}{365 \\times 24 \\times 60 \\times 60 \\mathrm{~s}}\\left(5 \\times 10^{6} \\mathrm{~km}\\right) \\approx 996 \\mathrm{~m} / \\mathrm{s}$']",['$996 $'],False,$\mathrm{~m} / \mathrm{s}$,Numerical,1e0 1012,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film.","2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.","['The bubble is surrounded by air.\n\n\n\n$P_{a}, T_{a}, \\rho_{a}$\n\nCutting the sphere in half and using the projected area to balance the forces give\n\n$$\n\\begin{aligned}\nP_{i} \\pi R_{0}^{2} & =P_{a} \\pi R_{0}^{2}+2\\left(2 \\pi R_{0} \\gamma\\right) \\\\\nP_{i} & =P_{a}+\\frac{4 \\gamma}{R_{0}}\n\\end{aligned}\n$$\n\nThe pressure and density are related by the ideal gas law:\n\n$P V=n R T$ or $P=\\frac{\\rho R T}{M}$, where $M=$ the molar mass of air.\n\nApply the ideal gas law to the air inside and outside the bubble, we get\n\n$$\n\\begin{aligned}\n& \\rho_{i} T_{i}=P_{i} \\frac{M}{R} \\\\\n& \\rho_{a} T_{a}=P_{a} \\frac{M}{R}, \\\\\n& \\frac{\\rho_{i} T_{i}}{\\rho_{a} T_{a}}=\\frac{P_{i}}{P_{a}}=\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right]\n\\end{aligned}\n$$']",['$1+\\frac{4 \\gamma}{R_{0} P_{a}}$'],False,,Expression, 1013,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film. Context question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$. Context answer: \boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$} ","2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.","['Using $\\gamma=0.025 \\mathrm{Nm}^{-1}, R_{0}=1.0 \\mathrm{~cm}$ and $P_{a}=1.013 \\times 10^{5} \\mathrm{Nm}^{-2}$, the numerical value of the ratio is\n\n$$\n\\frac{\\rho_{i} T_{i}}{\\rho_{a} T_{a}}=1+\\frac{4 \\gamma}{R_{0} P_{a}}=1+0.0001\n$$\n\n(The effect of the surface tension is very small.)']",['1.0001'],False,,Numerical,5e-5 1014,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film. Context question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$. Context answer: \boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$} Context question: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$. Context answer: \boxed{1.0001} ","2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.","['Let $W=$ total weight of the bubble, $F=$ buoyant force due to air around the bubble\n\n$$\n\\begin{aligned}\nW & =(\\text { mass of film }+ \\text { mass of air }) g \\\\\n& =\\left(4 \\pi R_{0}^{2} \\rho_{s} t+\\frac{4}{3} \\pi R_{0}^{3} \\rho_{i}\\right) g \\\\\n& =4 \\pi R_{0}^{2} \\rho_{s} t g+\\frac{4}{3} \\pi R_{0}^{3} \\frac{\\rho_{a} T_{a}}{T_{i}}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right] g\n\\end{aligned}\n$$\n\nThe buoyant force due to air around the bubble is\n\n$$\nB=\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g\n$$\n\nIf the bubble floats in still air,\n\n$$\n\\begin{aligned}\nB & \\geq W \\\\\n\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g & \\geq 4 \\pi R_{0}^{2} \\rho_{s} \\operatorname{tg}+\\frac{4}{3} \\pi R_{0}^{3} \\frac{\\rho_{a} T_{a}}{T_{i}}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right] g\n\\end{aligned}\n$$\n\nRearranging to give\n\n$$\n\\begin{aligned}\nT_{i} & \\geq \\frac{R_{0} \\rho_{a} T_{a}}{R_{0} \\rho_{a}-3 \\rho_{s} t}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right] \\\\\n& \\geq 307.1 \\mathrm{~K}\n\\end{aligned}\n$$\n\nThe air inside must be about $7.1^{\\circ} \\mathrm{C}$ warmer.']",['307.1'],False,K,Numerical,1e-1 1015,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film. Context question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$. Context answer: \boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$} Context question: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$. Context answer: \boxed{1.0001} Context question: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$. Context answer: \boxed{307.1} Extra Supplementary Reading Materials: After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground.","2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.","[""Ignore the radius change $\\rightarrow$ Radius remains $R_{0}=1.0 \\mathrm{~cm}$\n\n(The radius actually decreases by $0.8 \\%$ when the temperature decreases from $307.1 \\mathrm{~K}$ to $300 \\mathrm{~K}$. The film itself also becomes slightly thicker.)\n\nThe drag force from Stokes' Law is $F=6 \\pi \\eta R_{0} u$\n\nIf the bubble floats in the updraught,\n\n$F \\geq W-B$\n\n$6 \\pi \\eta R_{0} u \\geq\\left(4 \\pi R_{0}^{2} \\rho_{s} t+\\frac{4}{3} \\pi R_{0}^{3} \\rho_{i}\\right) g-\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g$\n\nWhen the bubble is in thermal equilibrium $T_{i}=T_{a}$.\n\n$6 \\pi \\eta R_{0} u \\geq\\left(4 \\pi R_{0}^{2} \\rho_{s} t+\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a}\\left[1+\\frac{4 \\gamma}{R_{0} P_{a}}\\right]\\right) g-\\frac{4}{3} \\pi R_{0}^{3} \\rho_{a} g$\n\nRearranging to give\n\n$u \\geq \\frac{4 R_{0} \\rho_{s} t g}{6 \\eta}+\\frac{\\frac{4}{3} R_{0}^{2} \\rho_{a} g\\left(\\frac{4 \\gamma}{R_{0} P_{a}}\\right)}{6 \\eta}$""]",['$\\frac{4 R_{0} \\rho_{s} t g}{6 \\eta}+\\frac{\\frac{4}{3} R_{0}^{2} \\rho_{a} g\\left(\\frac{4 \\gamma}{R_{0} P_{a}}\\right)}{6 \\eta}$'],False,,Expression, 1016,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film. Context question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$. Context answer: \boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$} Context question: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$. Context answer: \boxed{1.0001} Context question: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$. Context answer: \boxed{307.1} Extra Supplementary Reading Materials: After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground. Context question: 2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$. Context answer: \boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$} ",2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$.,['The numerical value is $u \\geq 0.36 \\mathrm{~m} / \\mathrm{s}$.'],['$0.36$'],False,$ \mathrm{~m} / \mathrm{s}$,Numerical,1e-2 1017,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film. Context question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$. Context answer: \boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$} Context question: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$. Context answer: \boxed{1.0001} Context question: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$. Context answer: \boxed{307.1} Extra Supplementary Reading Materials: After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground. Context question: 2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$. Context answer: \boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$} Context question: 2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$. Context answer: \boxed{$0.36$} Extra Supplementary Reading Materials: The above calculations suggest that the terms involving the surface tension $\gamma$ add very little to the accuracy of the result. In all of the questions below, you can neglect the surface tension terms.","2.6 If this spherical bubble is now electrified uniformly with a total charge $q$, find an equation describing the new radius $R_{1}$ in terms of $R_{0}, P_{a}, q$ and the permittivity of free space $\varepsilon_{0}$.","[""When the bubble is electrified, the electrical repulsion will cause the bubble to expand in size and thereby raise the buoyant force.\n\nThe force/area is (e-field on the surface $\\times$ charge/area)\n\n\nConsider a very thin pill box on the soap surface.\n\n\n\n$E=$ electric field on the film surface that results from all other parts of the soap film, excluding the surface inside the pill box itself.\n\n$E_{q}=$ total field just outside the pill box $=\\frac{q}{4 \\pi \\varepsilon_{0} R_{1}^{2}}=\\frac{\\sigma}{\\varepsilon_{0}}$\n\n$=E+$ electric field from surface charge $\\sigma$\n\n$=E+E_{\\sigma}$\n\nUsing Gauss's Law on the pill box, we have $E_{\\sigma}=\\frac{\\sigma}{2 \\varepsilon_{0}}$ perpendicular to the film as a result of symmetry.\n\nTherefore, $E=E_{q}-E_{\\sigma}=\\frac{\\sigma}{\\varepsilon_{0}}-\\frac{\\sigma}{2 \\varepsilon_{0}}=\\frac{\\sigma}{2 \\varepsilon_{0}}=\\frac{1}{2 \\varepsilon_{0}} \\frac{q}{4 \\pi R_{1}^{2}}$\n\n\nThe repulsive force per unit area of the surface of bubble is\n\n$$\n\\left(\\frac{q}{4 \\pi R_{1}^{2}}\\right) E=\\frac{\\left(q / 4 \\pi R_{1}^{2}\\right)^{2}}{2 \\varepsilon_{0}}\n$$\n\nLet $P_{i}^{\\prime}$ and $\\rho_{i}^{\\prime}$ be the new pressure and density when the bubble is electrified.\n\nThis electric repulsive force will augment the gaseous pressure $P_{i}^{\\prime}$.\n\n$P_{i}^{\\prime}$ is related to the original $P_{i}$ through the gas law.\n\n$P_{i}^{\\prime} \\frac{4}{3} \\pi R_{1}^{3}=P_{i} \\frac{4}{3} \\pi R_{0}^{3}$\n\n$P_{i}^{\\prime}=\\left(\\frac{R_{0}}{R_{1}}\\right)^{3} P_{i}=\\left(\\frac{R_{0}}{R_{1}}\\right)^{3} P_{a}$\n\nIn the last equation, the surface tension term has been ignored.\n\nFrom balancing the forces on the half-sphere projected area, we have (again ignoring the surface tension term)\n\n$$\n\\begin{aligned}\n& P_{i}^{\\prime}+\\frac{\\left(q / 4 \\pi R_{1}^{2}\\right)^{2}}{2 \\varepsilon_{0}}=P_{a} \\\\\n& P_{a}\\left(\\frac{R_{0}}{R_{1}}\\right)^{3}+\\frac{\\left(q / 4 \\pi R_{1}^{2}\\right)^{2}}{2 \\varepsilon_{0}}=P_{a}\n\\end{aligned}\n$$\n\n\n\nRearranging to get\n\n$$\n\\left(\\frac{R_{1}}{R_{0}}\\right)^{4}-\\left(\\frac{R_{1}}{R_{0}}\\right)-\\frac{q^{2}}{32 \\pi^{2} \\varepsilon_{0} R_{0}^{4} P_{a}}=0\n\\tag{17}\n$$\n\nNote that (17) yields $\\frac{R_{1}}{R_{0}}=1$ when $q=0$, as expected.""]",['$\\left(\\frac{R_{1}}{R_{0}}\\right)^{4}-\\left(\\frac{R_{1}}{R_{0}}\\right)-\\frac{q^{2}}{32 \\pi^{2} \\varepsilon_{0} R_{0}^{4} P_{a}}=0$'],False,,Equation, 1018,Thermodynamics,"An Electrified Soap Bubble A spherical soap bubble with internal air density $\rho_{i}$, temperature $T_{i}$ and radius $R_{0}$ is surrounded by air with density $\rho_{a}$, atmospheric pressure $P_{a}$ and temperature $T_{a}$. The soap film has surface tension $\gamma$, density $\rho_{s}$ and thickness $t$. The mass and the surface tension of the soap do not change with the temperature. Assume that $R_{0} \gg t$. The increase in energy, $d E$, that is needed to increase the surface area of a soap-air interface by $d A$, is given by $d E=\gamma d A$ where $\gamma$ is the surface tension of the film. Context question: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$. Context answer: \boxed{$1+\frac{4 \gamma}{R_{0} P_{a}}$} Context question: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$. Context answer: \boxed{1.0001} Context question: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$. Context answer: \boxed{307.1} Extra Supplementary Reading Materials: After the bubble is formed for a while, it will be in thermal equilibrium with the surrounding. This bubble in still air will naturally fall towards the ground. Context question: 2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$. Context answer: \boxed{$\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$} Context question: 2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$. Context answer: \boxed{$0.36$} Extra Supplementary Reading Materials: The above calculations suggest that the terms involving the surface tension $\gamma$ add very little to the accuracy of the result. In all of the questions below, you can neglect the surface tension terms. Context question: 2.6 If this spherical bubble is now electrified uniformly with a total charge $q$, find an equation describing the new radius $R_{1}$ in terms of $R_{0}, P_{a}, q$ and the permittivity of free space $\varepsilon_{0}$. Context answer: \boxed{$\left(\frac{R_{1}}{R_{0}}\right)^{4}-\left(\frac{R_{1}}{R_{0}}\right)-\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}=0$} ","2.7 Assume that the total charge is not too large (i.e. $\frac{q^{2}}{\varepsilon_{0} R_{0}^{4}}< An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1. The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem.","3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2. $p=2 a q, \quad r \gg a$ FIGURE 2","[""Using Coulomb's Law, we write the electric field at a distance $r$ is given by\n\n$$\n\\begin{aligned}\n& E_{p}=\\frac{q}{4 \\pi \\varepsilon_{0}(r-a)^{2}}-\\frac{q}{4 \\pi \\varepsilon_{0}(r+a)^{2}} \\\\\n& E_{p}=\\frac{q}{4 \\pi \\varepsilon_{0} r^{2}}\\left(\\frac{1}{\\left(1-\\frac{a}{r}\\right)^{2}}-\\frac{1}{\\left(1+\\frac{a}{r}\\right)^{2}}\\right)\n\\end{aligned}\n\\tag{1}\n$$\n\nUsing binomial expansion for small $a$,\n\n$$\n\\begin{aligned}\nE_{p} & =\\frac{q}{4 \\pi \\varepsilon_{0} r^{2}}\\left(1+\\frac{2 a}{r}-1+\\frac{2 a}{r}\\right) \\\\\n& =+\\frac{4 q a}{4 \\pi \\varepsilon_{0} r^{3}}=+\\frac{q a}{\\pi \\varepsilon_{0} r^{3}} \\\\\n& =\\frac{2 p}{4 \\pi \\varepsilon_{0} r^{3}}\n\\end{aligned}\n\\tag{2}\n$$""]",['$E_{p}=\\frac{2 p}{4 \\pi \\varepsilon_{0} r^{3}}$'],False,,Expression, 1021,Modern Physics,"3. To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1. The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem. Context question: 3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2. $p=2 a q, \quad r \gg a$ FIGURE 2 Context answer: \boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$} ",3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion.,"['The electric field seen by the atom from the ion is\n\n$$\n\\vec{E}_{\\text {ion }}=-\\frac{Q}{4 \\pi \\varepsilon_{0} r^{2}} \\hat{r}\n$$\n\nThe induced dipole moment is then simply\n\n$$\n\\vec{p}=\\alpha \\vec{E}_{\\text {ion }}=-\\frac{\\alpha Q}{4 \\pi \\varepsilon_{0} r^{2}} \\hat{r}\n$$\n\nFrom eq. (2)\n\n$$\n\\vec{E}_{p}=\\frac{2 p}{4 \\pi \\varepsilon_{0} r^{3}} \\hat{r}\n$$\n\nThe electric field intensity $\\vec{E}_{p}$ at the position of an ion at that instant is, using eq. (4),\n\n$$\n\\vec{E}_{p}=\\frac{1}{4 \\pi \\varepsilon_{0} r^{3}}\\left[-\\frac{2 \\alpha Q}{4 \\pi \\varepsilon_{0} r^{2}} \\hat{r}\\right]=-\\frac{\\alpha Q}{8 \\pi^{2} \\varepsilon_{0}^{2} r^{5}} \\hat{r}\n$$\n\nThe force acting on the ion is\n\n$$\n\\vec{f}=Q \\vec{E}_{p}=-\\frac{\\alpha Q^{2}}{8 \\pi^{2} \\varepsilon_{0}^{2} r^{5}} \\hat{r}\n$$\n\nThe ""-"" sign implies that this force is attractive and $Q^{2}$ implies that the force is attractive regardless of the sign of $Q$.']",['$\\vec{f}=-\\frac{\\alpha Q^{2}}{8 \\pi^{2} \\varepsilon_{0}^{2} r^{5}} \\hat{r}$'],False,,Expression, 1022,Modern Physics,"3. To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1. The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem. Context question: 3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2. $p=2 a q, \quad r \gg a$ FIGURE 2 Context answer: \boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$} Context question: 3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion. Context answer: \boxed{$\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$} ","3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ?","['The potential energy of the ion-atom is given by $U=\\int_{r}^{\\infty} \\vec{f} \\cdot d \\vec{r}$\n\nUsing this, $U=\\int_{r}^{\\infty} \\vec{f} \\cdot d \\vec{r}=-\\frac{\\alpha Q^{2}}{32 \\pi^{2} \\varepsilon_{0}^{2} r^{4}}$.']",['$-\\frac{\\alpha Q^{2}}{32 \\pi^{2} \\varepsilon_{0}^{2} r^{4}}$'],False,,Expression, 1023,Modern Physics,"3. To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1. The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem. Context question: 3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2. $p=2 a q, \quad r \gg a$ FIGURE 2 Context answer: \boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$} Context question: 3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion. Context answer: \boxed{$\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$} Context question: 3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ? Context answer: \boxed{$-\frac{\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}$} ","3.4 Find the expression for $r_{\min }$, the distance of the closest approach, as shown in Figure 1.","['At the position $r_{\\min }$ we have, according to the Principle of Conservation of Angular Momentum,\n\n$$\n\\begin{aligned}\nm v_{\\max } r_{\\min } & =m v_{0} b \\\\\nv_{\\max } & =v_{0} \\frac{b}{r_{\\min }}\n\\end{aligned}\n$$\n\nAnd according to the Principle of Conservation of Energy:\n\n$$\n\\frac{1}{2} m v_{\\max }^{2}+\\frac{-\\alpha Q^{2}}{32 \\pi^{2} \\varepsilon_{0}^{2} r^{4}}=\\frac{1}{2} m v_{0}^{2}\n$$\n\nEqs.(12) \\& (13):\n\n$$\n\\begin{array}{r}\n\\left(\\frac{b}{r_{\\text {min }}}\\right)^{2}-\\frac{\\alpha Q^{2} / \\frac{1}{2} m v_{0}^{2}}{32 \\pi^{2} \\varepsilon_{0}^{2} b^{4}}\\left(\\frac{b}{r_{\\text {min }}}\\right)^{4}=1 \\\\\n\\left(\\frac{r_{\\min }}{b}\\right)^{4}-\\left(\\frac{r_{\\min }}{b}\\right)^{2}+\\frac{\\alpha Q^{2}}{16 \\pi^{2} \\varepsilon_{0}^{2} m v_{0}^{2} b^{4}}=0\n\\end{array}\n$$\n\nThe roots of eq. (14) are:\n\n$$\nr_{\\min }=\\frac{b}{\\sqrt{2}}\\left[1 \\pm \\sqrt{1-\\frac{\\alpha Q^{2}}{4 \\pi^{2} \\varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\\right]^{\\frac{1}{2}}\n$$\n\n[Note that the equation (14) implies that $r_{\\min }$ cannot be zero, unless $b$ is itself zero.]\n\nSince the expression has to be valid at $Q=0$, which gives\n\n$$\nr_{\\min }=\\frac{b}{\\sqrt{2}}[1 \\pm 1]^{\\frac{1}{2}}\n$$\n\nWe have to choose ""+"" sign to make $r_{\\min }=b$\n\nHence,\n\n$$\nr_{\\min }=\\frac{b}{\\sqrt{2}}\\left[1+\\sqrt{1-\\frac{\\alpha Q^{2}}{4 \\pi^{2} \\varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\\right]^{\\frac{1}{2}}\n$$']",['$r_{\\min }=\\frac{b}{\\sqrt{2}}\\left[1+\\sqrt{1-\\frac{\\alpha Q^{2}}{4 \\pi^{2} \\varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\\right]^{\\frac{1}{2}}$'],False,,Expression, 1024,Modern Physics,"3. To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1. The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem. Context question: 3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2. $p=2 a q, \quad r \gg a$ FIGURE 2 Context answer: \boxed{$E_{p}=\frac{2 p}{4 \pi \varepsilon_{0} r^{3}}$} Context question: 3.2 Find the expression for the force $\vec{f}$ acting on the ion due to the polarised atom. Show that this force is attractive regardless of the sign of the charge of the ion. Context answer: \boxed{$\vec{f}=-\frac{\alpha Q^{2}}{8 \pi^{2} \varepsilon_{0}^{2} r^{5}} \hat{r}$} Context question: 3.3 What is the electric potential energy of the ion-atom interaction in terms of $\alpha, Q$ and $r$ ? Context answer: \boxed{$-\frac{\alpha Q^{2}}{32 \pi^{2} \varepsilon_{0}^{2} r^{4}}$} Context question: 3.4 Find the expression for $r_{\min }$, the distance of the closest approach, as shown in Figure 1. Context answer: \boxed{$r_{\min }=\frac{b}{\sqrt{2}}\left[1+\sqrt{1-\frac{\alpha Q^{2}}{4 \pi^{2} \varepsilon_{0}^{2} m v_{0}^{2} b^{4}}}\right]^{\frac{1}{2}}$} ","3.5 If the impact parameter $b$ is less than a critical value $b_{0}$, the ion will descend along a spiral to the atom. In such a case, the ion will be neutralized, and the atom is, in turn, charged. This process is known as the ""charge exchange"" interaction. What is the cross sectional area $A=\pi b_{0}^{2}$ of this ""charge exchange"" collision of the atom as seen by the ion?","['A spiral trajectory occurs when (16) is imaginary (because there is no minimum distance of approach). $r_{\\min }$ is real under the condition:\n\n$$\n\\begin{aligned}\n& 1 \\geq \\frac{\\alpha Q^{2}}{4 \\pi^{2} \\varepsilon_{0}^{2} m v_{0}^{2} b^{4}} \\\\\n& b \\geq b_{0}=\\left(\\frac{\\alpha Q^{2}}{4 \\pi^{2} \\varepsilon_{0}^{2} m v_{0}^{2}}\\right)^{\\frac{1}{4}}\n\\end{aligned}\n$$\n\nFor $b",['a) \n\n\n\n\nFrom the figure we get\n\n$n_{A} \\sin \\alpha=n_{1} \\sin \\alpha_{1}=n_{2} \\sin \\alpha_{2}=\\ldots=n_{B} \\sin \\beta$'],,False,,, 1025,Optics,,"a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure. ![](https://cdn.mathpix.com/cropped/2023_12_21_cef941636e625861824cg-1.jpg?height=605&width=937&top_left_y=917&top_left_x=545)",['a) \n\n![](https://cdn.mathpix.com/cropped/2023_12_21_cef941636e625861824cg-1.jpg?height=454&width=534&top_left_y=2217&top_left_x=1155)\n\n\nFrom the figure we get\n\n$n_{A} \\sin \\alpha=n_{1} \\sin \\alpha_{1}=n_{2} \\sin \\alpha_{2}=\\ldots=n_{B} \\sin \\beta$'],['证明题'],False,,Need_human_evaluate, 1026,Optics," Context question: a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure. Context answer: \boxed{证明题} ","b) Assume that you are standing in a large flat desert. At some distance you see what appears to be a water surface. When you approach the ""water"" is seems to move away such that the distance to the ""water"" is always constant. Explain the phenomenon.",['b) The phenomenon is due to total reflexion in a warm layer of air when $\\beta=90^{\\circ}$. This gives\n\n\n\n$$\nn_{A} \\sin \\alpha=n_{B}\n$$'],,False,,, 1027,Thermodynamics," Context question: a) Consider a plane-parallel transparent plate, where the refractive index, $n$, varies with distance, $z$, from the lower surface (see figure). Show that $n_{A} \sin \alpha=n_{B} \sin \beta$. The notation is that of the figure. Context answer: \boxed{证明题} Context question: b) Assume that you are standing in a large flat desert. At some distance you see what appears to be a water surface. When you approach the ""water"" is seems to move away such that the distance to the ""water"" is always constant. Explain the phenomenon. Context answer: \boxed{证明题} ","c) Compute the temperature of the air close to the ground in b) assuming that your eyes are located $1.60 \mathrm{~m}$ above the ground and that the distance to the ""water"" is $250 \mathrm{~m}$. The refractive index of the air at $15{ }^{\circ} \mathrm{C}$ and at normal air pressure $(101.3 \mathrm{kPa})$ is 1.000276 . The temperature of the air more than $1 \mathrm{~m}$ above the ground is assumed to be constant and equal to $30{ }^{\circ} \mathrm{C}$. The atmospheric pressure is assumed to be normal. The refractive index, $n$, is such that $n-1$ is proportional to the density of the air. Discuss the accuracy of your result.","['c) As the density, $\\rho$, of the air is inversely proportional to the absolute temperature, $T$, for fixed pressure we have\n\n$$\nn(T)=1+k \\cdot \\rho=1+k / T\n$$\n\nThe value given at $15{ }^{\\circ} \\mathrm{C}$ determines the value of $k=0.0795$.\n\nIn order to have total reflexion we have $n_{30} \\sin \\alpha=n_{T}$ or\n\n$$\n\\left(1+\\frac{k}{303}\\right) \\cdot \\frac{L}{\\sqrt{h^{2}+L^{2}}}=\\left(1+\\frac{k}{T}\\right) \\text { with } h=1.6 \\mathrm{~m} \\text { and } L=250 \\mathrm{~m}\n$$\n\nAs $h \\ll L$ we can use a power expansion in $h L$ :\n\n$$\nT=\\frac{303}{\\left(\\frac{303}{k}+1\\right) \\frac{1}{\\sqrt{1+h^{2} / L^{2}}}-\\frac{303}{k}} \\approx 303\\left(1+\\frac{303 h^{2}}{2 k L^{2}}\\right)=328 \\mathrm{~K}\n$$']",['328'],False,K,Numerical,1e0 1028,Optics,," In certain lakes there is a strange phenomenon called ""seiching"" which is an oscillation of the water. Lakes in which you can see this phenomenon are normally long compared with the depth and also narrow. It is natural to see waves in a lake but not something like the seiching, where the entire water volume oscillates, like the coffee in a cup that you carry to a waiting guest. In order to create a model of the seiching we look at water in a rectangular container. The length of the container is $L$ and the depth of the water is $h$. Assume that the surface of the water to begin with makes a small angle with the horizontal. The seiching will then start, and we assume that the water surface continues to be plane but oscillates around an axis in the horizontal plane and located in the middle of the container. Create a model of the movement of the water and derive a formula for the oscillation period $T$. The starting conditions are given in figure above. Assume that $\xi< The water surface level in Bastudalen (northern end of lake Vättern) and Jönköping (southern end).","['In the coordinate system of the figure, we have for the centre of mass coordinates of the two triangular parts of the water\n\n$$\n\\left(x_{1}, y_{1}\\right)=(L / 3, h / 2+\\xi / 3) \\quad\\left(x_{2}, y_{2}\\right)=(-L / 3, h / 2-\\xi / 3) .\n$$\n\nFor the entire water mass the centre of mass coordinates will then be\n\n$$\n\\left(x_{C O M}, y_{C O M}\\right)=\\left(\\frac{\\xi L}{6 h}, \\frac{\\xi^{2}}{6 h}\\right)\n$$\n\nDue to that the $y$ component is quadratic in $\\xi$ will be much much smaller than the $x$ component.\n\nThe velocities of the water mass are\n\n$$\n\\left(v_{x}, v_{y}\\right)=\\left(\\frac{g_{L}}{6 h}, \\frac{g_{\\xi}}{3 h}\\right)\n$$\n\nand again the vertical component is much smaller the the horizontal one.\n\nWe now in our model neglect the vertical components. The total energy (kinetic + potential) will then be\n\n$$\nW=W_{K}+W_{P}=\\frac{1}{2} M \\frac{\\xi^{2} L^{2}}{36 h^{2}}+M g \\frac{\\xi^{2}}{6 h^{2}}\n$$\n\nFor a harmonic oscillator we have\n\n$$\nW=W_{K}+W_{P}=\\frac{1}{2} m x^{2}+\\frac{1}{2} m \\omega^{2} x^{2}\n$$\n\nIdentifying gives\n\n$$\n\\omega=\\sqrt{\\frac{12 g h}{L}} \\text { or } T_{\\text {model }}=\\frac{\\pi L}{\\sqrt{3 h}} \\text {. }\n$$\n\nComparing with the experimental data we find $T_{\\text {experiment }} \\approx 1.1 \\cdot T_{\\text {model }}$ our model gives a slight underestimation of the oscillation period.\n\nApplying our corrected model on the Vättern data we have that the oscillation period of the seiching is about 3 hours.']",['$\\frac{\\pi L}{\\sqrt{3 h}}$'],False,,Expression, 1028,Optics,,"![](https://cdn.mathpix.com/cropped/2023_12_21_9085b7bab6de7bb5d3cdg-1.jpg?height=342&width=594&top_left_y=343&top_left_x=1162) In certain lakes there is a strange phenomenon called ""seiching"" which is an oscillation of the water. Lakes in which you can see this phenomenon are normally long compared with the depth and also narrow. It is natural to see waves in a lake but not something like the seiching, where the entire water volume oscillates, like the coffee in a cup that you carry to a waiting guest. In order to create a model of the seiching we look at water in a rectangular container. The length of the container is $L$ and the depth of the water is $h$. Assume that the surface of the water to begin with makes a small angle with the horizontal. The seiching will then start, and we assume that the water surface continues to be plane but oscillates around an axis in the horizontal plane and located in the middle of the container. Create a model of the movement of the water and derive a formula for the oscillation period $T$. The starting conditions are given in figure above. Assume that $\xi< A: without oil B: with oil Figure 1. Parallel plates in water. Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$",A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)?,"['A.1\n\n\n\nFigure 1\n\nLet $h^{\\prime}$ be the height of the column of oil (see Fig. 1). Then pressure at depth $h$ below the water surface must be $p_{h}=\\rho_{\\text {oil }} g h=\\rho_{\\text {oil }} g h^{\\prime}$, from where $h^{\\prime}=\\frac{\\rho_{0}}{\\rho_{\\text {oil }}} h$. Horizontal force on the plate $F_{x}=F_{1}-F_{0}$, where the force due to new fluid is $F_{1}=\\frac{\\rho_{\\mathrm{oil}} h^{\\prime}}{2} \\cdot h^{\\prime} w$ and the force due to water is $F_{0}=\\frac{\\rho_{0} g h}{2} \\cdot h w$.\n\nCombining all the equation above, we get\n\n$$\nF_{x}=\\left(\\frac{\\rho_{0}}{\\rho_{\\mathrm{oil}}}-1\\right) \\frac{\\rho_{0} g h^{2} w}{2}\n$$\n\n$F_{x}=\\left(\\frac{\\rho_{0}}{\\rho_{\\mathrm{oil}}}-1\\right) \\frac{\\rho_{0} g h^{2} w}{2}$.']",['$F_{x}=\\left(\\frac{\\rho_{0}}{\\rho_{\\mathrm{oil}}}-1\\right) \\frac{\\rho_{0} g h^{2} w}{2}$'],False,,Expression, 1030,Thermodynamics,"Mid-ocean ridge Consider a large vessel of water that is situated in a uniform gravitational field with free-fall acceleration $g$. Two vertical rectangular plates parallel to each other are fitted into the vessel so that the vertical edges of the plates are in a tight gap-less contact with the vertical walls of the vessel. Length $h$ of each plate is immersed in water (Fig. 1). The width of the plates along the $y$-axis is $w$, water density is $\rho_{0}$. A: without oil B: with oil Figure 1. Parallel plates in water. Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$ Context question: A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)? Context answer: \boxed{$F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$} Extra Supplementary Reading Materials: Fig. 2 shows a cross-section of a mid-ocean ridge. It consists of overlaying layers of mantle, crust and ocean water. The mantle is composed of rocks that we assume can flow in geological timescales and therefore, in this problem will be treated as a fluid. The thickness of the crust is much smaller than the characteristic length scale in the $x$-direction, hence, the crust behaves as a freely bendable plate. To high accuracy, such a ridge can be modeled as a two-dimensional system, without any variation of variables along the $y$-axis, which is perpendicular to the plane of Fig. 2. Assume that the ridge length $L$ along the $y$-axis is much larger than any other length introduced in this problem. At the centre of the ridge the thickness of the crust is assumed to be zero. As the horizontal distance $x$ from the centre increases, the crust gets thicker and approaches a constant thickness $D$ as $x \rightarrow \infty$. Correspondingly, the ocean floor subsides by a vertical height $h$ below the top of the ridge $\mathrm{O}$, which we define as the origin of our coordinate system (see Fig. 2). Water density $\rho_{0}$ and temperature $T_{0}$ can be assumed to be constant in space and time. The same can be assumed for mantle density $\rho_{1}$ and its temperature $T_{1}$. The temperature of the crust $T$ is also constant in time but can depend on position. It is known that, to high accuracy, the crustal material expands linearly with temperature $T$. Since water and mantle temperatures are assumed to be constant, it is convenient to use a rescaled version of the thermal expansion coefficient. Then $l(T)=l_{1}\left[1-k_{l}\left(T_{1}-T\right) /\left(T_{1}-T_{0}\right)\right]$, where $l$ is the length of a piece of crustal material, $l_{1}$ is its length at temperature $T_{1}$, and $k_{l}$ is the rescaled thermal expansion coefficient, which can be assumed to be constant. Figure 2. Mid-ocean ridge. Note that the $z$-axis is pointing downwards.","A.2 Assuming that the crust is isotropic, find how its density $\rho$ depends on its temperature $T$. Assuming that $\left|k_{l}\right| \ll 1$, write your answer in the approximate form $$ \rho(T) \approx \rho_{1}\left[1+k \frac{T_{1}-T}{T_{1}-T_{0}}\right] $$ where terms of order $k_{l}^{2}$ and higher are neglected. Then, identify constant $k$.","['Consider a rectangular mass element of the crust. Since relation $l(T)=l_{1}\\left[1-k_{l}\\left(T_{1}-T\\right) /\\left(T_{1}-T_{0}\\right)\\right]$ holds for all three dimensions of the solid, its volume $V$ satisfies\n\n$$\nV=V_{1}\\left(1-k_{l} \\frac{T_{1}-T}{T_{1}-T_{0}}\\right)^{3}\n$$\n\nwhere $V_{1}$ is the volume at $T=T_{1}$. If the mass of the element is $m$, density is then\n\n$$\n\\rho(T)=\\frac{m}{V}=\\frac{m}{V_{1}}\\left(1-k_{l} \\frac{T_{1}-T}{T_{1}-T_{0}}\\right)^{-3}=\\rho_{1}\\left(1-k_{l} \\frac{T_{1}-T}{T_{1}-T_{0}}\\right)^{-3}\n$$\n\n\n\nSince $k_{l} \\ll 1$, this can be approximated as\n\n$$\n\\rho(T) \\approx \\rho_{1}\\left(1+3 k_{l} \\frac{T_{1}-T}{T_{1}-T_{0}}\\right)\n$$\n\nso that $k=3 k_{l}$.\nA.2\n\n$\\rho(T) \\approx \\rho_{1}\\left(1+3 k_{l} \\frac{T_{1}-T}{T_{1}-T_{0}}\\right)$.\n\n$k=3 k_{l}$.']","['$\\rho_{1}\\left(1+3 k_{l} \\frac{T_{1}-T}{T_{1}-T_{0}}\\right)$ , $k=3 k_{l}$']",True,,Expression, 1031,Thermodynamics,"Mid-ocean ridge Consider a large vessel of water that is situated in a uniform gravitational field with free-fall acceleration $g$. Two vertical rectangular plates parallel to each other are fitted into the vessel so that the vertical edges of the plates are in a tight gap-less contact with the vertical walls of the vessel. Length $h$ of each plate is immersed in water (Fig. 1). The width of the plates along the $y$-axis is $w$, water density is $\rho_{0}$. A: without oil B: with oil Figure 1. Parallel plates in water. Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$ Context question: A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)? Context answer: \boxed{$F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$} Extra Supplementary Reading Materials: Fig. 2 shows a cross-section of a mid-ocean ridge. It consists of overlaying layers of mantle, crust and ocean water. The mantle is composed of rocks that we assume can flow in geological timescales and therefore, in this problem will be treated as a fluid. The thickness of the crust is much smaller than the characteristic length scale in the $x$-direction, hence, the crust behaves as a freely bendable plate. To high accuracy, such a ridge can be modeled as a two-dimensional system, without any variation of variables along the $y$-axis, which is perpendicular to the plane of Fig. 2. Assume that the ridge length $L$ along the $y$-axis is much larger than any other length introduced in this problem. At the centre of the ridge the thickness of the crust is assumed to be zero. As the horizontal distance $x$ from the centre increases, the crust gets thicker and approaches a constant thickness $D$ as $x \rightarrow \infty$. Correspondingly, the ocean floor subsides by a vertical height $h$ below the top of the ridge $\mathrm{O}$, which we define as the origin of our coordinate system (see Fig. 2). Water density $\rho_{0}$ and temperature $T_{0}$ can be assumed to be constant in space and time. The same can be assumed for mantle density $\rho_{1}$ and its temperature $T_{1}$. The temperature of the crust $T$ is also constant in time but can depend on position. It is known that, to high accuracy, the crustal material expands linearly with temperature $T$. Since water and mantle temperatures are assumed to be constant, it is convenient to use a rescaled version of the thermal expansion coefficient. Then $l(T)=l_{1}\left[1-k_{l}\left(T_{1}-T\right) /\left(T_{1}-T_{0}\right)\right]$, where $l$ is the length of a piece of crustal material, $l_{1}$ is its length at temperature $T_{1}$, and $k_{l}$ is the rescaled thermal expansion coefficient, which can be assumed to be constant. Figure 2. Mid-ocean ridge. Note that the $z$-axis is pointing downwards. Context question: A.2 Assuming that the crust is isotropic, find how its density $\rho$ depends on its temperature $T$. Assuming that $\left|k_{l}\right| \ll 1$, write your answer in the approximate form $$ \rho(T) \approx \rho_{1}\left[1+k \frac{T_{1}-T}{T_{1}-T_{0}}\right] $$ where terms of order $k_{l}^{2}$ and higher are neglected. Then, identify constant $k$. Context answer: \boxed{$\rho_{1}\left(1+3 k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)$ , $k=3 k_{l}$} Extra Supplementary Reading Materials: It is known that $k>0$. Also, thermal conductivity of the crust $\kappa$ can be assumed to be constant. As a consequence, very far away from the ridge axis the temperature of the crust depends linearly with depth.","A.3 By assuming that mantle and water each behave like an incompressible fluid at hydrostatic equilibrium, express the far-distance crust thickness $D$ in terms of $h, \rho_{0}, \rho_{1}$, and $k$. Any motion of the material can be neglected.","['A.3\n\nSince mantle behaves like a fluid in hydrostatic equilibrium, pressure $p(x, z)$ at $z=h+D$ must be the same for all $x$. Therefore,\n\n$$\np(0, h+D)=p(\\infty, h+D) .\n$$\n\nSimilarly, we must have\n\n$$\np(0,0)=p(\\infty, 0) \\text {. }\n$$\n\nHence, the change in pressure between $z=0$ and $z=\\infty$ must be the same at both $x=0$ and $x=\\infty$. At the ridge axis\n\n$$\np(0, h+D)-p(0,0)=\\rho_{1} g(h+D)\n$$\n\nwhile far away\n\n$$\np(\\infty, h+D)-p(\\infty, 0)=\\rho_{0} g h+\\int_{h}^{h+D} \\rho(T(\\infty, z)) g \\mathrm{~d} z\n$$\n\nFar away from the ridge axis the two surfaces of the crust are effectively horizontal, meaning that the law of heat conduction can be written as\n\n$$\n\\frac{\\mathrm{d} T}{\\mathrm{~d} z}=\\text { const. }\n$$\n\nHence, after applying the relevant temperature boundary conditions,\n\n$$\nT(\\infty, z)=T_{0}+\\left(T_{1}-T_{0}\\right) \\frac{z-h}{D}\n$$\n\nFrom all the equations above and by using the density formula given in the problem text,\n\n$$\n\\rho_{1} g(h+D)=\\rho_{0} g h+\\int_{h}^{h+D} \\rho_{1}\\left(1+k \\frac{T_{1}-T_{0}-\\left(T_{1}-T_{0}\\right) \\frac{z-h}{D}}{T_{1}-T_{0}}\\right) g \\mathrm{~d} z\n$$\n\nfrom where we straightforwardly obtain\n\n$$\nD=\\frac{2}{k}\\left(1-\\frac{\\rho_{0}}{\\rho_{1}}\\right) h\n$$\n\n\n\n$D=\\frac{2}{k}\\left(1-\\frac{\\rho_{0}}{\\rho_{1}}\\right) h$.']",['$D=\\frac{2}{k}(1-\\frac{\\rho_{0}}{\\rho_{1}}) h$'],False,,Expression, 1032,Thermodynamics,"Mid-ocean ridge Consider a large vessel of water that is situated in a uniform gravitational field with free-fall acceleration $g$. Two vertical rectangular plates parallel to each other are fitted into the vessel so that the vertical edges of the plates are in a tight gap-less contact with the vertical walls of the vessel. Length $h$ of each plate is immersed in water (Fig. 1). The width of the plates along the $y$-axis is $w$, water density is $\rho_{0}$. A: without oil B: with oil Figure 1. Parallel plates in water. Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$ Context question: A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)? Context answer: \boxed{$F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$} Extra Supplementary Reading Materials: Fig. 2 shows a cross-section of a mid-ocean ridge. It consists of overlaying layers of mantle, crust and ocean water. The mantle is composed of rocks that we assume can flow in geological timescales and therefore, in this problem will be treated as a fluid. The thickness of the crust is much smaller than the characteristic length scale in the $x$-direction, hence, the crust behaves as a freely bendable plate. To high accuracy, such a ridge can be modeled as a two-dimensional system, without any variation of variables along the $y$-axis, which is perpendicular to the plane of Fig. 2. Assume that the ridge length $L$ along the $y$-axis is much larger than any other length introduced in this problem. At the centre of the ridge the thickness of the crust is assumed to be zero. As the horizontal distance $x$ from the centre increases, the crust gets thicker and approaches a constant thickness $D$ as $x \rightarrow \infty$. Correspondingly, the ocean floor subsides by a vertical height $h$ below the top of the ridge $\mathrm{O}$, which we define as the origin of our coordinate system (see Fig. 2). Water density $\rho_{0}$ and temperature $T_{0}$ can be assumed to be constant in space and time. The same can be assumed for mantle density $\rho_{1}$ and its temperature $T_{1}$. The temperature of the crust $T$ is also constant in time but can depend on position. It is known that, to high accuracy, the crustal material expands linearly with temperature $T$. Since water and mantle temperatures are assumed to be constant, it is convenient to use a rescaled version of the thermal expansion coefficient. Then $l(T)=l_{1}\left[1-k_{l}\left(T_{1}-T\right) /\left(T_{1}-T_{0}\right)\right]$, where $l$ is the length of a piece of crustal material, $l_{1}$ is its length at temperature $T_{1}$, and $k_{l}$ is the rescaled thermal expansion coefficient, which can be assumed to be constant. Figure 2. Mid-ocean ridge. Note that the $z$-axis is pointing downwards. Context question: A.2 Assuming that the crust is isotropic, find how its density $\rho$ depends on its temperature $T$. Assuming that $\left|k_{l}\right| \ll 1$, write your answer in the approximate form $$ \rho(T) \approx \rho_{1}\left[1+k \frac{T_{1}-T}{T_{1}-T_{0}}\right] $$ where terms of order $k_{l}^{2}$ and higher are neglected. Then, identify constant $k$. Context answer: \boxed{$\rho_{1}\left(1+3 k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)$ , $k=3 k_{l}$} Extra Supplementary Reading Materials: It is known that $k>0$. Also, thermal conductivity of the crust $\kappa$ can be assumed to be constant. As a consequence, very far away from the ridge axis the temperature of the crust depends linearly with depth. Context question: A.3 By assuming that mantle and water each behave like an incompressible fluid at hydrostatic equilibrium, express the far-distance crust thickness $D$ in terms of $h, \rho_{0}, \rho_{1}$, and $k$. Any motion of the material can be neglected. Context answer: \boxed{$D=\frac{2}{k}(1-\frac{\rho_{0}}{\rho_{1}}) h$} ","A.4 Find, to the leading order in $k$, the net horizontal force $F$ acting on the right half $(x>0)$ of the crust in terms of $\rho_{0}, \rho_{1}, h, L, k$ and $g$.","['A.4\nThe net horizontal force on the half of the ridge is the difference between the pressure forces acting at $x=0$ and $x=\\infty$ :\n\n$$\nF=L \\int_{0}^{h+D}(p(0, z)-p(\\infty, z)) \\mathrm{d} z\n$$\n\nFrom considerations of the previous question, pressure at $x=0$ is\n\n$$\np(0, z)=p(0,0)+\\rho_{1} g z\n$$\n\nwhile very far away\n\n$$\np(\\infty, z)= \\begin{cases}p(\\infty, 0)+\\rho_{0} g z & \\text { if } 0 \\leq z \\leq h \\\\ p(\\infty, 0)+\\rho_{0} g h+\\int_{h}^{z} \\rho_{1}\\left(1+k \\frac{T_{1}-T_{0}-\\left(T_{1}-T_{0}\\right) \\frac{z^{\\prime}-h}{D}}{T_{1}-T_{0}}\\right) g \\mathrm{~d} z^{\\prime} & \\text { if } h \\leq z \\leq h+D\\end{cases}\n$$\n\nThe equations above can be combined into\n\n$$\n\\begin{aligned}\n& F=L \\int_{0}^{h+D}\\left(p(0,0)+\\rho_{1} g z\\right) \\mathrm{d} z-L \\int_{0}^{h}\\left(p(\\infty, 0)+\\rho_{0} g z\\right) \\mathrm{d} z- \\\\\n& \\quad-L \\int_{h}^{h+D}\\left(p(\\infty, 0)+\\rho_{0} g h\\right) \\mathrm{d} z-L \\int_{h}^{h+D}\\left[\\int_{h}^{z} \\rho_{1}\\left(1+k\\left(1-\\frac{z^{\\prime}-h}{D}\\right)\\right) g \\mathrm{~d} z^{\\prime}\\right] \\mathrm{d} z .\n\\end{aligned}\n$$\n\nThe double integral can be easily found either directly or by using a substitution $u=z-h, u^{\\prime}=z^{\\prime}-h$ :\n\n$$\n\\int_{h}^{h+D}\\left[\\int_{h}^{z} \\rho_{1}\\left(1+k\\left(1-\\frac{z^{\\prime}-h}{D}\\right)\\right) g \\mathrm{~d} z^{\\prime}\\right] \\mathrm{d} z=\\int_{0}^{D}\\left[\\int_{0}^{u} \\rho_{1}\\left(1+k\\left(1-\\frac{u}{D}\\right)\\right) g \\mathrm{~d} u^{\\prime}\\right] \\mathrm{d} u\n$$\n\nAfter a straightforward integration and using $p(0,0)=p(\\infty, 0)$ as well as the result of the previous question,\n\n$$\nF=g L\\left[\\rho_{1}\\left(\\frac{h^{2}}{2}+h D-\\frac{k D^{2}}{3}\\right)-\\rho_{0}\\left(\\frac{h^{2}}{2}+h D\\right)\\right]=g L h^{2}\\left(\\rho_{1}-\\rho_{0}\\right)\\left(\\frac{1}{2}+\\frac{2}{3 k}\\left(1-\\frac{\\rho_{0}}{\\rho_{1}}\\right)\\right) .\n$$\n\nSince $k \\ll 1$, the term with $\\frac{1}{k}$ is of the leading order, hence, the required answer is\n\n$$\nF \\approx \\frac{2 g L h^{2}}{3 k} \\frac{\\left(\\rho_{1}-\\rho_{0}\\right)^{2}}{\\rho_{1}}\n$$\n\n\n\n\n$F \\approx \\frac{2 g L h^{2}}{3 k} \\frac{\\left(\\rho_{1}-\\rho_{0}\\right)^{2}}{\\rho_{1}}$.']",['$\\frac{2 g L h^{2}}{3 k} \\frac{\\left(\\rho_{1}-\\rho_{0}\\right)^{2}}{\\rho_{1}}$'],False,,Expression, 1033,Thermodynamics,"Mid-ocean ridge Consider a large vessel of water that is situated in a uniform gravitational field with free-fall acceleration $g$. Two vertical rectangular plates parallel to each other are fitted into the vessel so that the vertical edges of the plates are in a tight gap-less contact with the vertical walls of the vessel. Length $h$ of each plate is immersed in water (Fig. 1). The width of the plates along the $y$-axis is $w$, water density is $\rho_{0}$. A: without oil B: with oil Figure 1. Parallel plates in water. Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$ Context question: A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)? Context answer: \boxed{$F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$} Extra Supplementary Reading Materials: Fig. 2 shows a cross-section of a mid-ocean ridge. It consists of overlaying layers of mantle, crust and ocean water. The mantle is composed of rocks that we assume can flow in geological timescales and therefore, in this problem will be treated as a fluid. The thickness of the crust is much smaller than the characteristic length scale in the $x$-direction, hence, the crust behaves as a freely bendable plate. To high accuracy, such a ridge can be modeled as a two-dimensional system, without any variation of variables along the $y$-axis, which is perpendicular to the plane of Fig. 2. Assume that the ridge length $L$ along the $y$-axis is much larger than any other length introduced in this problem. At the centre of the ridge the thickness of the crust is assumed to be zero. As the horizontal distance $x$ from the centre increases, the crust gets thicker and approaches a constant thickness $D$ as $x \rightarrow \infty$. Correspondingly, the ocean floor subsides by a vertical height $h$ below the top of the ridge $\mathrm{O}$, which we define as the origin of our coordinate system (see Fig. 2). Water density $\rho_{0}$ and temperature $T_{0}$ can be assumed to be constant in space and time. The same can be assumed for mantle density $\rho_{1}$ and its temperature $T_{1}$. The temperature of the crust $T$ is also constant in time but can depend on position. It is known that, to high accuracy, the crustal material expands linearly with temperature $T$. Since water and mantle temperatures are assumed to be constant, it is convenient to use a rescaled version of the thermal expansion coefficient. Then $l(T)=l_{1}\left[1-k_{l}\left(T_{1}-T\right) /\left(T_{1}-T_{0}\right)\right]$, where $l$ is the length of a piece of crustal material, $l_{1}$ is its length at temperature $T_{1}$, and $k_{l}$ is the rescaled thermal expansion coefficient, which can be assumed to be constant. Figure 2. Mid-ocean ridge. Note that the $z$-axis is pointing downwards. Context question: A.2 Assuming that the crust is isotropic, find how its density $\rho$ depends on its temperature $T$. Assuming that $\left|k_{l}\right| \ll 1$, write your answer in the approximate form $$ \rho(T) \approx \rho_{1}\left[1+k \frac{T_{1}-T}{T_{1}-T_{0}}\right] $$ where terms of order $k_{l}^{2}$ and higher are neglected. Then, identify constant $k$. Context answer: \boxed{$\rho_{1}\left(1+3 k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)$ , $k=3 k_{l}$} Extra Supplementary Reading Materials: It is known that $k>0$. Also, thermal conductivity of the crust $\kappa$ can be assumed to be constant. As a consequence, very far away from the ridge axis the temperature of the crust depends linearly with depth. Context question: A.3 By assuming that mantle and water each behave like an incompressible fluid at hydrostatic equilibrium, express the far-distance crust thickness $D$ in terms of $h, \rho_{0}, \rho_{1}$, and $k$. Any motion of the material can be neglected. Context answer: \boxed{$D=\frac{2}{k}(1-\frac{\rho_{0}}{\rho_{1}}) h$} Context question: A.4 Find, to the leading order in $k$, the net horizontal force $F$ acting on the right half $(x>0)$ of the crust in terms of $\rho_{0}, \rho_{1}, h, L, k$ and $g$. Context answer: \boxed{$\frac{2 g L h^{2}}{3 k} \frac{\left(\rho_{1}-\rho_{0}\right)^{2}}{\rho_{1}}$} Extra Supplementary Reading Materials: Suppose that crust is thermally isolated from the rest of the Earth. As a result of heat conduction, the temperatures of the upper and lower surfaces of the crust are going to get closer to each other until the crust reaches thermal equilibrium. Specific heat of the crust is $c$ and can be assumed to be constant.","A.5 By using dimensional analysis or order-of-magnitude analysis, estimate the characteristic time $\tau$ in which the difference between the upper and lower surface temperatures of the crust far away from the ridge axis is going to approach zero. You can assume that $\tau$ does not depend on the two initial surface temperatures of the crust.","['A.5\nThe timescale $\\tau$ is expected to depend only on density of the crust $\\rho_{1}$, its specific heat $c$, thermal conductivity $\\kappa$ and thickness $D$. Hence, we can write $\\tau=A \\rho_{1}^{\\alpha} c^{\\beta} \\kappa^{\\gamma} D^{\\delta}$, where $A$ is a dimensionless constant. We will obtain the powers $\\alpha-\\delta$ via dimensional analysis.\n\nDefine the symbols for different dimensions: $L$ for length, $M$ for mass, $T$ for time and $\\Theta$ for temperature. Then $\\tau, \\rho_{1}, c, \\kappa$ and $D$ have dimensions $\\mathrm{T}, \\mathrm{ML}^{-3}, \\mathrm{~L}^{2} \\mathrm{~T}^{-2} \\Theta^{-1}, \\mathrm{MLT}^{-3} \\Theta^{-1}$ and $\\mathrm{L}$, respectively. The resulting set of linear equations to balance the powers of length, mass, time and temperature, respectively, is\n\n$$\n\\left\\{\\begin{array}{l}\n0=-3 \\alpha+2 \\beta+\\gamma+\\delta \\\\\n0=\\alpha+\\gamma \\\\\n1=-2 \\beta-3 \\gamma \\\\\n0=-\\beta-\\gamma\n\\end{array}\\right.\n$$\n\nThis gives $\\alpha=\\beta=1, \\gamma=-1, \\delta=2$. Hence,\n\n$$\n\\tau=A \\frac{c \\rho_{1} D^{2}}{\\kappa}\n$$\n\n\n$\\tau \\approx \\frac{c \\rho_{1} D^{2}}{\\kappa}$.']",['$\\frac{c \\rho_{1} D^{2}}{\\kappa}$'],False,,Expression, 1034,Optics,"Seismic waves in a stratified medium Suppose that a short earthquake happens at the surface of some planet. The seismic waves can be assumed to originate from a line source situated at $z=x=0$, where $x$ is the horizontal coordinate and $z$ is the depth below the surface (Fig. 3). The seismic wave source can be assumed to be much longer than any other length considered in this question. As a result of the earthquake, a uniform flux of the so-called longitudinal $P$ waves is emitted along all the directions in the $x-z$ plane that have positive component along the $z$-axis. Since the wave theory in a solid is generally complicated, in this problem we neglect all the other waves emitted by the earthquake. The crust of the planet is stratified so that the P-wave speed $v$ depends on depth $z$ according to $v=v_{0}\left(1+z / z_{0}\right)$, where $v_{0}$ is the speed at the surface and $z_{0}$ is a known positive constant. Figure 3. Coordinate system used in part B. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$","B.1 Consider a single ray emitted by the earthquake that makes an initial angle $0<\theta_{0}<\pi / 2$ with the $z$-axis and travels in the $x-z$ plane. What is the horizontal coordinate $x_{1}\left(\theta_{0}\right) \neq 0$ at which this ray can be detected at the surface of the planet? It is known that the ray path is an arc of a circle. Write your answer in the form $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$, where $A$ and $b$ are constants to be found.","[""B.1\nSeismic waves in this problem can be treated by using ray theory. Namely, their propagation is described by the Snell's law of refraction\n\n$$\nn(0) \\sin \\theta_{0}=n(z) \\sin \\theta,\n$$\n\nwhere the refractive index is\n\n$$\nn(z)=\\frac{c}{v(z)}=\\frac{c}{v_{0}\\left(1+\\frac{z}{z_{0}}\\right)}\n$$\n\n\n\n\n\nFigure 2\n\nand $c$ denotes the seismic wave speed in a material with refractive index $n=1$. From the two equations above we have\n\n$$\nv_{0}\\left(1+\\frac{z}{z_{0}}\\right) \\sin \\theta_{0}=v_{0} \\sin \\theta\n$$\n\nMethod 1. Since this describes an arc of a circle, we have that at $\\theta=\\frac{\\pi}{2}, z=R-R \\sin \\theta_{0}$ (fig. 2), giving\n\n$$\n\\left(1+\\frac{R-R \\sin \\theta_{0}}{z_{0}}\\right) \\sin \\theta_{0}=1\n$$\n\nfrom where the circle radius $R=\\frac{z_{0}}{\\sin \\theta_{0}}$. From simple geometry we get\n\n$$\nx_{1}\\left(\\theta_{0}\\right)=2 R \\cos \\theta_{0}\n$$\n\ni.e. $A=2 z_{0}$ and $b=1$.\n\nMethod 2. Implicitly differentiating $v_{0}\\left(1+\\frac{z}{z_{0}}\\right) \\sin \\theta_{0}=v_{0} \\sin \\theta$ gives\n\n$$\n\\frac{\\mathrm{d} z}{z_{0}} \\sin \\theta_{0}=\\cos \\theta \\mathrm{d} \\theta\n$$\n\nAn infinitesimal ray path length $\\mathrm{d} l$ is related to the change in the vertical coordinate via\n\n$$\n\\mathrm{d} z=\\mathrm{d} l \\cos \\theta\n$$\n\ngiving\n\n$$\n\\mathrm{d} l=\\frac{z_{0}}{\\sin \\theta_{0}} \\mathrm{~d} \\theta\n$$\n\nThis is an equation of an arc of a circle of radius $R=\\frac{z_{0}}{\\sin \\theta_{0}}$\n\nAlternatively, instead of considering an infinitesimal ray path length $d l$, one can obtain the answer by writing\n\n$$\n\\cot \\theta=\\frac{\\mathrm{d} z}{\\mathrm{~d} x}=\\frac{\\mathrm{d} z}{\\mathrm{~d} \\theta} \\frac{\\mathrm{d} \\theta}{\\mathrm{d} x}\n$$\n\n\n\nThe first derivative can be eliminated via Snell's law, leading to\n\n$$\n\\cot \\theta=\\frac{z_{0} \\cos \\theta}{\\sin \\theta_{0}} \\frac{\\mathrm{d} \\theta}{\\mathrm{d} x}\n$$\n\nwhich can be integrated to get\n\n$$\nx_{1}=-\\frac{z_{0}}{\\sin \\theta_{0}} \\int_{\\text {start }}^{\\text {end }} \\mathrm{d} \\cos \\theta=\\frac{2 z_{0} \\cos \\theta_{0}}{\\sin \\theta_{0}}\n$$\n\nwhere we used Snell's law again to get that the ray has $\\cos \\theta=-\\cos \\theta_{0}$ at the point where it reaches the surface.\n\n$x_{1}\\left(\\theta_{0}\\right)=2 z_{0} \\cot \\theta_{0}$.""]",['$x_{1}(\\theta_{0})=2 z_{0} \\cot \\theta_{0}$'],False,,Expression, 1035,Optics,"Seismic waves in a stratified medium Suppose that a short earthquake happens at the surface of some planet. The seismic waves can be assumed to originate from a line source situated at $z=x=0$, where $x$ is the horizontal coordinate and $z$ is the depth below the surface (Fig. 3). The seismic wave source can be assumed to be much longer than any other length considered in this question. As a result of the earthquake, a uniform flux of the so-called longitudinal $P$ waves is emitted along all the directions in the $x-z$ plane that have positive component along the $z$-axis. Since the wave theory in a solid is generally complicated, in this problem we neglect all the other waves emitted by the earthquake. The crust of the planet is stratified so that the P-wave speed $v$ depends on depth $z$ according to $v=v_{0}\left(1+z / z_{0}\right)$, where $v_{0}$ is the speed at the surface and $z_{0}$ is a known positive constant. Figure 3. Coordinate system used in part B. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$ Context question: B.1 Consider a single ray emitted by the earthquake that makes an initial angle $0<\theta_{0}<\pi / 2$ with the $z$-axis and travels in the $x-z$ plane. What is the horizontal coordinate $x_{1}\left(\theta_{0}\right) \neq 0$ at which this ray can be detected at the surface of the planet? It is known that the ray path is an arc of a circle. Write your answer in the form $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$, where $A$ and $b$ are constants to be found. Context answer: \boxed{$x_{1}(\theta_{0})=2 z_{0} \cot \theta_{0}$} Extra Supplementary Reading Materials: If you were unable to find $A$ and $b$, in the following questions you can use the result $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$ as given. Suppose that total energy per unit length of the source released as $\mathrm{P}$ waves into the crust during the earthquake is $E$. Assume that waves are completely absorbed when they reach the surface of the planet from below.",B.2 Find how the energy density per unit area $\varepsilon(x)$ absorbed by the surface depends on the distance along the surface $x$.,"['B.2\nIn two dimensions, $\\frac{E}{\\pi} \\mathrm{d} \\theta_{0}$ is the energy carried by rays that are emitted within interval $\\left[\\theta_{0}, \\theta_{0}+\\mathrm{d} \\theta_{0}\\right)$. On the other hand, the energy carried by rays that arrive at $[x, x+\\mathrm{d} x)$ is $\\varepsilon \\mathrm{d} x$. Therefore,\n\n$$\n\\varepsilon=\\frac{E}{\\pi}\\left|\\frac{\\mathrm{d} \\theta_{0}}{\\mathrm{~d} x}\\right|\n$$\n\nUsing the result of question B.1,\n\n$$\n\\frac{\\mathrm{d} x}{\\mathrm{~d} \\theta_{0}}=-\\frac{A b}{\\sin ^{2}\\left(b \\theta_{0}\\right)}=-A b\\left(1+\\cot ^{2}\\left(b \\theta_{0}\\right)\\right)=-\\frac{b\\left(A^{2}+x^{2}\\right)}{A} .\n$$\n\nHence,\n\n$$\n\\varepsilon(x)=\\frac{E A}{\\pi b\\left(A^{2}+x^{2}\\right)}=\\frac{2 E z_{0}}{\\pi\\left(4 z_{0}^{2}+x^{2}\\right)}\n$$\n\n\n$\\varepsilon(x)=\\frac{E A}{\\pi b\\left(A^{2}+x^{2}\\right)}=\\frac{2 E z_{0}}{\\pi\\left(4 z_{0}^{2}+x^{2}\\right)}$.']",['$\\varepsilon(x)=\\frac{2 E z_{0}}{\\pi(4 z_{0}^{2}+x^{2})}$'],False,,Expression, 1036,Optics,"Seismic waves in a stratified medium Suppose that a short earthquake happens at the surface of some planet. The seismic waves can be assumed to originate from a line source situated at $z=x=0$, where $x$ is the horizontal coordinate and $z$ is the depth below the surface (Fig. 3). The seismic wave source can be assumed to be much longer than any other length considered in this question. As a result of the earthquake, a uniform flux of the so-called longitudinal $P$ waves is emitted along all the directions in the $x-z$ plane that have positive component along the $z$-axis. Since the wave theory in a solid is generally complicated, in this problem we neglect all the other waves emitted by the earthquake. The crust of the planet is stratified so that the P-wave speed $v$ depends on depth $z$ according to $v=v_{0}\left(1+z / z_{0}\right)$, where $v_{0}$ is the speed at the surface and $z_{0}$ is a known positive constant. Figure 3. Coordinate system used in part B. The effects of the surface curvature of the planets can be neglected. You might need to use the formula $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. } $$ Context question: B.1 Consider a single ray emitted by the earthquake that makes an initial angle $0<\theta_{0}<\pi / 2$ with the $z$-axis and travels in the $x-z$ plane. What is the horizontal coordinate $x_{1}\left(\theta_{0}\right) \neq 0$ at which this ray can be detected at the surface of the planet? It is known that the ray path is an arc of a circle. Write your answer in the form $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$, where $A$ and $b$ are constants to be found. Context answer: \boxed{$x_{1}(\theta_{0})=2 z_{0} \cot \theta_{0}$} Extra Supplementary Reading Materials: If you were unable to find $A$ and $b$, in the following questions you can use the result $x_{1}\left(\theta_{0}\right)=A \cot \left(b \theta_{0}\right)$ as given. Suppose that total energy per unit length of the source released as $\mathrm{P}$ waves into the crust during the earthquake is $E$. Assume that waves are completely absorbed when they reach the surface of the planet from below. Context question: B.2 Find how the energy density per unit area $\varepsilon(x)$ absorbed by the surface depends on the distance along the surface $x$. Context answer: \boxed{$\varepsilon(x)=\frac{2 E z_{0}}{\pi(4 z_{0}^{2}+x^{2})}$} Extra Supplementary Reading Materials: From now on, assume that the waves are instead fuylly reflected when reaching the surface. Imagine a device positioned at $z=x=0$ that has the same geometry as the previously considered earthquake source. The device is capable of emitting $P$ waves in a freely chosen angular distribution. We make the device emit a signal with a narrow range of emission angles. In particular, the initial angle the signal makes with the vertical belongs to the interval $\left[\theta_{0}-\frac{1}{2} \delta \theta_{0}, \theta_{0}+\frac{1}{2} \delta \theta_{0}\right]$, where $0<\theta_{0}<\pi / 2$, $\delta \theta_{0} \ll 1$ and $\delta \theta_{0} \ll \theta_{0}$.","B.3 At what distance $x_{\max }$ along the surface from the source is the furthest point that the signal does not reach? Write your answer in terms of $\theta_{0}, \delta \theta_{0}$ and other constants given above.","['B.3\nDefine $x_{-}=x_{1}\\left(\\theta_{0}-\\frac{\\delta \\theta_{0}}{2}\\right)$ and $x_{+}=x_{1}\\left(\\theta_{0}+\\frac{\\delta \\theta_{0}}{2}\\right)$. To the leading order in $\\delta \\theta_{0}, x_{-} \\approx x_{+} \\approx x_{1}\\left(\\theta_{0}\\right)$. With each reflection of the signal, the horizontal distance between the points where the edges of\n\nthe signal reflect increases by $\\left|x_{+}-x_{-}\\right|=x_{-}-x_{+}$. When moving along the positive $x$-axis, these zones get wider until they overlap. If this happens after $N$ reflections, then\n\n$$\nN \\approx \\frac{x_{1}\\left(\\theta_{0}\\right)}{x_{-}-x_{+}}\n$$\n\nwhere the approximate sign tends to equality as $\\delta \\theta_{0} \\rightarrow 0$.\n\nThe position where the zones start to overlap is at $x_{\\max }=N x_{1}\\left(\\theta_{0}\\right)$. Therefore,\n\n$$\nx_{\\max }=\\frac{x_{1}\\left(\\theta_{0}\\right)^{2}}{x_{1}\\left(\\theta_{0}-\\frac{\\delta \\theta_{0}}{2}\\right)-x_{1}\\left(\\theta_{0}+\\frac{\\delta \\theta_{0}}{2}\\right)}\n$$\n\nSince $\\delta \\theta_{0} \\ll \\theta_{0}$, we can approximate\n\n$$\nx_{1}\\left(\\theta_{0}-\\frac{\\delta \\theta_{0}}{2}\\right)-x_{1}\\left(\\theta_{0}+\\frac{\\delta \\theta_{0}}{2}\\right) \\approx-\\frac{\\mathrm{d} x_{1}\\left(\\theta_{0}\\right)}{\\mathrm{d} \\theta_{0}} \\delta \\theta_{0}=\\frac{A b}{\\sin ^{2}\\left(b \\theta_{0}\\right)} \\delta \\theta_{0}\n$$\n\nCombining the last two equations and substituting the $x_{1}\\left(\\theta_{0}\\right)$ expression gives\n\n$$\nx_{\\max }=\\frac{A b \\cos ^{2}\\left(b \\theta_{0}\\right)}{\\delta \\theta_{0}}=\\frac{2 z_{0} \\cos ^{2} \\theta_{0}}{\\delta \\theta_{0}}\n$$\n\n\n$x_{\\max }=\\frac{A b \\cos ^{2}\\left(b \\theta_{0}\\right)}{\\delta \\theta_{0}}=\\frac{2 z_{0} \\cos ^{2} \\theta_{0}}{\\delta \\theta_{0}}$.']",['$x_{\\max }=\\frac{2 z_{0} \\cos ^{2} \\theta_{0}}{\\delta \\theta_{0}}$'],False,,Expression, 1037,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point)",A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1).,['The linear charge density of the ring is $\\lambda=q /(2 \\pi R)$. All the points of the ring are situated a distance $\\sqrt{R^{2}+z^{2}}$ away from point A. Integrating over the whole ring we readily obtain:\n\n$$\n\\Phi(z)=\\frac{q}{4 \\pi \\varepsilon_{0}} \\frac{1}{\\sqrt{R^{2}+z^{2}}}\n$$\n\n$\\Phi(z)=\\frac{q}{4 \\pi \\varepsilon_{0}} \\frac{1}{\\sqrt{R^{2}+z^{2}}}$.'],['$\\Phi(z)=\\frac{q}{4 \\pi \\varepsilon_{0}} \\frac{1}{\\sqrt{R^{2}+z^{2}}}$.'],False,,Expression, 1038,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} ","A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$.",['Using an expansion in powers of $z$ we obtain:\n\n$$\n\\Phi(z)=\\frac{q}{4 \\pi \\varepsilon_{0}} \\frac{1}{\\sqrt{R^{2}+z^{2}}}=\\frac{q}{4 \\pi \\varepsilon_{0} R} \\frac{1}{\\sqrt{1+\\left(\\frac{z}{R}\\right)^{2}}} \\approx \\frac{q}{4 \\pi \\varepsilon_{0} R}\\left(1-\\frac{z^{2}}{2 R^{2}}\\right) .\n$$'],['$\\Phi(z) \\approx \\frac{q}{4 \\pi \\varepsilon_{0} R}(1-\\frac{z^{2}}{2 R^{2}})$'],False,,Expression, 1039,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$}","A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron?","['The potential energy of the electron is $V(z)=-e \\Phi(z)$. The force acting on the electron is\n\n$$\nF(z)=-\\frac{\\mathrm{d} V(z)}{\\mathrm{d} z}=+e \\frac{\\mathrm{d} \\Phi}{\\mathrm{d} z}=-\\frac{q e}{4 \\pi \\varepsilon_{0} R^{3}} z\n$$\n\nIf this is a restoring force, it should be negative for positive $z$. Thus, $q>0$.\n\n']",['$F(z)=-\\frac{q e}{4 \\pi \\varepsilon_{0} R^{3}} z$'],,,Expression, 1039,Electromagnetism,,"A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring.","['The potential energy of the electron is $V(z)=-e \\Phi(z)$. The force acting on the electron is\n\n$$\nF(z)=-\\frac{\\mathrm{d} V(z)}{\\mathrm{d} z}=+e \\frac{\\mathrm{d} \\Phi}{\\mathrm{d} z}=-\\frac{q e}{4 \\pi \\varepsilon_{0} R^{3}} z\n$$\n\nIf this is a restoring force, it should be negative for positive $z$. Thus, $q>0$.']","['$F(z)=-\\frac{q e}{4 \\pi \\varepsilon_{0} R^{3}} z$, q>0']",True,,Need_human_evaluate, 1040,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 ",A.4 What is the angular frequency $\omega$ of such harmonic oscillations?,['The equation of motion for an electron is\n\n$$\nm \\ddot{z}+\\frac{q e}{4 \\pi \\varepsilon_{0} R^{3}} z=0\n$$\n\n(here dots denote time derivatives). We therefore get\n\n$$\n\\omega=\\sqrt{\\frac{q e}{4 \\pi m \\varepsilon_{0} R^{3}}}\n$$'],['$\\omega=\\sqrt{\\frac{q e}{4 \\pi m \\varepsilon_{0} R^{3}}}$'],False,,Expression, 1041,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$.",B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above.,"[""There are two different ways to solve this problem: (i) using direct integration; (ii) using Gauss's law and the result of part $A$.\n\n\nFigure 1: Calculating electrostatic potential in the plane of the ring through direct integration.\n\n(i) Direct integration. We will follow the notations of Figure 1. Since the potential has cylindrical symmetry, let the point $\\mathrm{B}$, where we calculate the potential, be on the $x$-axis. Let\n\n$$\n|\\mathrm{OB}|=r ;|\\mathrm{OC}|=R\n$$\n\nThus:\n\n$$\n|\\mathrm{BC}|^{2}=R^{2}+r^{2}-2 R r \\cos \\phi .\n$$\n\n\n\nElectrostatic potential created by ring element $\\mathrm{d} \\phi$ at the point $\\mathrm{B}$ :\n\n$$\n\\mathrm{d} \\Phi=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\lambda R \\mathrm{~d} \\phi}{\\sqrt{R^{2}+r^{2}-2 R r \\cos \\phi}}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\lambda \\mathrm{d} \\phi}{\\sqrt{1+\\frac{r^{2}}{R^{2}}-2 \\frac{r}{R} \\cos \\phi}} .\n$$\n\nUsing the expansion given in the formulation of the problem for $\\varepsilon=-1 / 2$ we have:\n\n$$\n\\mathrm{d} \\Phi \\approx \\frac{\\lambda \\mathrm{d} \\phi}{4 \\pi \\varepsilon_{0}}\\left[1-\\frac{1}{2}\\left(\\frac{r^{2}}{R^{2}}-2 \\frac{r}{R} \\cos \\phi\\right)+\\frac{3}{8}\\left(\\frac{r^{2}}{R^{2}}-2 \\frac{r}{R} \\cos \\phi\\right)^{2}\\right]\n$$\n\nIgnoring the terms of the order $r^{3}$ and $r^{4}$ we get:\n\n$$\n\\mathrm{d} \\Phi \\approx \\frac{\\lambda \\mathrm{d} \\phi}{4 \\pi \\varepsilon_{0}}\\left[1+\\frac{r}{R} \\cos \\phi+\\frac{r^{2}}{R^{2}}\\left(\\frac{3}{2} \\cos ^{2} \\phi-\\frac{1}{2}\\right)\\right] .\n$$\n\nIntegrating over all angles we finally obtain:\n\n$$\n\\begin{gathered}\n\\Phi(r)=\\frac{\\lambda}{4 \\pi \\varepsilon_{0}} \\int_{0}^{2 \\pi}\\left[1+\\frac{r}{R} \\cos \\phi+\\frac{r^{2}}{R^{2}}\\left(\\frac{3}{2} \\cos ^{2} \\phi-\\frac{1}{2}\\right)\\right] \\mathrm{d} \\phi . \\\\\n\\Phi(r)=\\frac{q}{4 \\pi \\varepsilon_{0} R}\\left(1+\\frac{r^{2}}{4 R^{2}}\\right) .\n\\end{gathered}\n$$\n\nFrom here, comparing with the expression $\\Phi(r)=q\\left(\\alpha+\\beta r^{2}\\right)$, we obtain\n\n$$\n\\beta=\\frac{1}{16 \\pi \\varepsilon_{0} R^{3}}\n$$\n\n(ii) Gauss's law.\n\n\n\nFigure 2: Calculating electrostatic potential in the plane of the ring via Gauss's law.\n\nLet us analyze a small cylinder of radius $r$. The center of the cylinder coincides with the center of the ring. In part A we analyzed the potential along the $z$-axis, while in this part we analyze the potential along the radius $r$. For any $z \\ll R$ and $r \\ll R$ the potential has an expression:\n\n$$\n\\Phi(z, r)=\\frac{q}{4 \\pi \\varepsilon_{0} R}\\left(1-\\frac{z^{2}}{2 R^{2}}\\right)+q \\beta r^{2}\n$$\n\n\n\nThe lowest order terms are quadratic in $r$ and $z$. Due to reflection symmetry the potential does not contain terms of the type $r z$. This, for example, immediately gives us $\\alpha=1 /\\left(4 \\pi \\varepsilon_{0} R\\right)$. Thus, for small $r$ and $z$ electric fields in the radial and axial directions are:\n\n$$\n\\mathcal{E}_{z}(z, r)=+\\frac{q}{4 \\pi \\varepsilon_{0} R^{3}} z, \\quad \\mathcal{E}_{r}(z, r)=-2 q \\beta r .\n$$\n\nApplying Gauss's law to the cylinder we obtain:\n\n$$\n\\oint \\overrightarrow{\\mathcal{E}} \\cdot \\mathrm{d} \\vec{S}=0 \\quad \\Rightarrow \\quad \\int_{\\text {side }} \\overrightarrow{\\mathcal{E}} \\cdot \\mathrm{d} \\vec{S}+\\int_{\\text {base }} \\overrightarrow{\\mathcal{E}} \\cdot \\mathrm{d} \\vec{S}=0\n$$\n\nThe second integral is:\n\n$$\n\\int_{\\text {base }} \\overrightarrow{\\mathcal{E}} \\cdot \\mathrm{d} \\vec{S}=2 \\pi r^{2} \\mathcal{E}_{z}(z, r)=\\frac{q z r^{2}}{2 \\varepsilon_{0} R^{3}}\n$$\n\nThe first integral is:\n\n$$\n\\int_{\\text {side }} \\overrightarrow{\\mathcal{E}} \\cdot \\mathrm{d} \\vec{S}=4 \\pi r z \\mathcal{E}_{r}(z, r)=-8 \\pi q \\beta r^{2} z\n$$\n\nGauss's theorem thus gives:\n\n$$\n\\frac{q z r^{2}}{2 \\varepsilon_{0} R^{3}}-8 \\pi q \\beta r^{2} z=0\n$$\n\nThis immediately yields\n\n$$\n\\beta=\\frac{1}{16 \\pi \\varepsilon_{0} R^{3}},\n$$\n\nwhich agrees with the result obtained via direct integration.""]",['$\\beta=\\frac{1}{16 \\pi \\varepsilon_{0} R^{3}}$'],False,,Expression, 1042,Electromagnetism,,"B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring.","['The potential of the electron is $V(r)=-e \\Phi(r)$. Force acting on the electron in the $x y$ plane is\n\n$$\nF(r)=-\\frac{\\mathrm{d} V(r)}{\\mathrm{d} r}=+e \\frac{\\mathrm{d} \\Phi(r)}{\\mathrm{d} r}=\\frac{q e}{8 \\pi \\varepsilon_{0} R^{3}} r\n$$\n\nTo have oscilations we need the force to be negative for $r>0$. Thus, $q<0$.']",['$F(r)=+\\frac{q e}{8 \\pi \\varepsilon_{0} R^{3}} r . \\quad q<0$.'],False,,Need_human_evaluate, 1042,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$}","B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron?","['The potential of the electron is $V(r)=-e \\Phi(r)$. Force acting on the electron in the $x y$ plane is\n\n$$\nF(r)=-\\frac{\\mathrm{d} V(r)}{\\mathrm{d} r}=+e \\frac{\\mathrm{d} \\Phi(r)}{\\mathrm{d} r}=\\frac{q e}{8 \\pi \\varepsilon_{0} R^{3}} r\n$$\n\nTo have oscilations we need the force to be negative for $r>0$. Thus, $q<0$.\n\n']",['$F(r)=+\\frac{q e}{8 \\pi \\varepsilon_{0} R^{3}} r $'],False,,Expression, 1043,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$} Context question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$. Extra Supplementary Reading Materials: Part C. The focal length of the idealized electrostatic lens: instantaneous charging One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called ""active region""). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field ""fills the space"" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring. Figure 2. A model of an electrostatic lens.","C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the ""active region"" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing.","['Let us consider an electron with the velocity $v=\\sqrt{2 E / m}$ at a distance $r$ from the ""optical"" axis (Figure 2 of the problem). The electron crosses the ""active region"" of the lens in time\n\n$$\nt=\\frac{d}{v}\n$$\n\nThe equation of motion in the $r$ direction:\n\n$$\nm \\ddot{r}=2 e q \\beta r .\n$$\n\nDuring the time the electron crosses the active region of the lens, the electron acquires radial velocity:\n\n$$\nv_{r}=\\frac{2 e q \\beta r}{m} \\frac{d}{v}<0\n$$\n\nThe lens will be focusing if $q<0$. The time it takes for an electron to reach the ""optical"" axis is:\n\n$$\nt^{\\prime}=\\frac{r}{\\left|v_{r}\\right|}=-\\frac{m v}{2 e q \\beta d}\n$$\n\nDuring this time the electron travels in the $z$-direction a distance\n\n$$\n\\Delta z=t^{\\prime} v=-\\frac{m v^{2}}{2 e q \\beta d}=-\\frac{E}{e q d \\beta}\n$$\n\n$\\Delta z$ does not depend on the radial distance $r$, therefore all electron will cross the ""optical"" axis (will be focused) in the same spot. Thus,\n\n$$\nf=-\\frac{E}{e q d \\beta}\n$$']",['$f=-\\frac{E}{e q d \\beta}$'],False,,Expression, 1044,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$} Context question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$. Extra Supplementary Reading Materials: Part C. The focal length of the idealized electrostatic lens: instantaneous charging One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called ""active region""). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field ""fills the space"" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring. Figure 2. A model of an electrostatic lens. Context question: C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the ""active region"" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing. Context answer: \boxed{$f=-\frac{E}{e q d \beta}$} Extra Supplementary Reading Materials: In reality the electron source is placed on the $z$-axis at a distance $b>f$ from the center of the ring. Consider that electrons are no longer parallel to the $z$-axis before reaching the ""active region"", but are emitted from a point source at a range of different angles $\gamma \ll 1$ rad to the $z$-axis. Electrons are focused in a point situated at a distance $c$ from the center of the ring.",C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities.,"['\n\nFigure 3: Focusing of electrons.\n\nLet us consider an electron emitted an an angle $\\gamma$ to the optical axis (Figure 3). Its initial velocity in the radial direction is:\n\n$$\nv_{r ; 0}=v \\sin \\gamma \\approx v \\gamma \\approx v \\frac{r}{b}\n$$\n\nwhere $r$ is the radial distance of the electron when it reaches the plane of the ring. The velocity in the $z$-direction is\n\n$$\nv_{z}=v \\cos \\gamma \\approx v\n$$\n\nFor small angles $\\gamma$ the additional velocity in the $r$-direction acquired in the ""active region"" is the same as in part C.1. Thus, the radial velocity after crossing the active region is\n\n$$\nv_{r}=v \\frac{r}{b}+\\frac{2 e q \\beta r}{m} \\frac{d}{v}\n$$\n\nwhere the first term is positive and the second term is negative, since $q<0$. If the electrons are focused, then $v_{r}<0$ (this can be verified after obtaining the final result). The electron will reach the optical axis in time\n\n$$\nt^{\\prime}=\\frac{r}{\\left|v_{r}\\right|}=-\\frac{r}{\\frac{2 e q \\beta r}{m} \\frac{d}{v}+v \\frac{r}{b}}=-\\frac{1}{\\frac{2 e q \\beta}{m} \\frac{d}{v}+\\frac{v}{b}}\n$$\n\nDuring this time it will travel a distance\n\n$$\nc=t^{\\prime} v=-\\frac{1}{\\frac{2 e q \\beta}{m} \\frac{d}{v^{2}}+\\frac{1}{b}}=-\\frac{1}{\\frac{e q \\beta d}{E}+\\frac{1}{b}} .\n$$']",['$c=-\\frac{1}{\\frac{e q \\beta d}{E}+\\frac{1}{b}}$.'],False,,Expression, 1045,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$} Context question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$. Extra Supplementary Reading Materials: Part C. The focal length of the idealized electrostatic lens: instantaneous charging One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called ""active region""). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field ""fills the space"" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring. Figure 2. A model of an electrostatic lens. Context question: C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the ""active region"" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing. Context answer: \boxed{$f=-\frac{E}{e q d \beta}$} Extra Supplementary Reading Materials: In reality the electron source is placed on the $z$-axis at a distance $b>f$ from the center of the ring. Consider that electrons are no longer parallel to the $z$-axis before reaching the ""active region"", but are emitted from a point source at a range of different angles $\gamma \ll 1$ rad to the $z$-axis. Electrons are focused in a point situated at a distance $c$ from the center of the ring. Context question: C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities. Context answer: \boxed{$c=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}}$.} ","C.3 Is the equation of a thin optical lens $$ \frac{1}{b}+\frac{1}{c}=\frac{1}{f} $$ fulfilled for the electrostatic lens? Show it by explicitly calculating $1 / b+1 / c$.",['From the previous answer we obtain:\n\n$$\n\\frac{1}{b}+\\frac{1}{c}=-\\frac{e q \\beta d}{E}\n$$\n\nComparing with the answer of C. 1 we immediately obtain\n\n$$\n\\frac{1}{b}+\\frac{1}{c}=\\frac{1}{f}\n$$\n\ni.e. the equation of a thin optical lens is valid for an electrostatic lens as well.'],,False,,, 1046,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$} Context question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$. Extra Supplementary Reading Materials: Part C. The focal length of the idealized electrostatic lens: instantaneous charging One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called ""active region""). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field ""fills the space"" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring. Figure 2. A model of an electrostatic lens. Context question: C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the ""active region"" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing. Context answer: \boxed{$f=-\frac{E}{e q d \beta}$} Extra Supplementary Reading Materials: In reality the electron source is placed on the $z$-axis at a distance $b>f$ from the center of the ring. Consider that electrons are no longer parallel to the $z$-axis before reaching the ""active region"", but are emitted from a point source at a range of different angles $\gamma \ll 1$ rad to the $z$-axis. Electrons are focused in a point situated at a distance $c$ from the center of the ring. Context question: C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities. Context answer: \boxed{$c=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}}$.} Context question: C.3 Is the equation of a thin optical lens $$ \frac{1}{b}+\frac{1}{c}=\frac{1}{f} $$ fulfilled for the electrostatic lens? Show it by explicitly calculating $1 / b+1 / c$. Context answer: 证明题 Extra Supplementary Reading Materials: Part D. The ring as a capacitor The model of considered above was idealized and we assumed that the ring charged instantaneously. In reality charging is non-instantaneous, as the ring is a capacitor with a finite capacitance $C$. In this part we will analyze the properties of this capacitor. You might need the following integrals: $$ \int \frac{\mathrm{d} x}{\sin x}=-\ln \left|\frac{\cos x+1}{\sin x}\right|+\text { const } $$ and $$ \int \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}=\ln \left|x+\sqrt{1+x^{2}}\right|+\text { const. } $$","D.1 Calculate the capacitance $C$ of the ring. Consider that the ring has a finite width $2 a$, but remember that $a \ll R$.","['\n\nFigure 4: Calculation of the capacitance of the ring.\n\nLet us sub-divide the entire ring into two parts: a part corresponding to the angle $2 \\alpha \\ll 1$, and the rest of the ring, as shown in Figure 4. While the angle is small in comparison to 1, let us assume that the length of the first part, $\\alpha R$, is still large compared to $a(\\alpha R \\gg a)$. Let us calculate the electrostatic potential $\\Phi$ at point K. It it a sum of two terms: the first one produced by the cut-out part with an angle $2 \\alpha$ (contribution $\\Phi_{1}$ ) and the second one originating from the rest of the ring (contribution $\\Phi_{2}$ ).\n\nContribution $\\Phi_{1}$. Since $\\alpha \\ll 1$, we can neglect the curvature of the cylinder that is cut out from the ring. The linear charge density on the ring is $\\lambda=\\frac{q}{2 \\pi R}$. The potential at the center of the\n\n\n\ncylinder is then given by an integral:\n\n$$\n\\Phi_{1}=2 \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q}{2 \\pi R} \\int_{0}^{\\alpha R} \\frac{\\mathrm{d} x}{\\sqrt{x^{2}+a^{2}}}=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\int_{0}^{\\alpha R} \\frac{\\mathrm{d}(x / a)}{\\sqrt{1+(x / a)^{2}}}=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\int_{0}^{\\alpha R / a} \\frac{\\mathrm{d} y}{\\sqrt{1+y^{2}}}\n$$\n\nUsing the integral provided in the description of the problem we get:\n\n$$\n\\Phi_{1}=\\left.\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(y+\\sqrt{1+y^{2}}\\right)\\right|_{0} ^{\\alpha R / a}=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(\\frac{\\alpha R}{a}+\\sqrt{1+\\left(\\frac{\\alpha R}{a}\\right)^{2}}\\right)\n$$\n\nAs $\\alpha R \\gg a$,\n\n$$\n\\Phi_{1} \\approx \\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(\\frac{2 \\alpha R}{a}\\right)\n$$\n\n\n\nFigure 5: Calculation of the capacitance of the ring\n\nContribution $\\Phi_{2}$. In this case we can neglect the thickness $a$. Using the cosine theorem we can derive the distance between points $\\mathrm{K}$ and $\\mathrm{L}$ of Figure 5 :\n\n$$\n|\\mathrm{KL}|=2 R \\sin \\frac{\\phi}{2}\n$$\n\nThe contribution $\\Phi_{2}$ can then be written as an integral:\n\n$$\n\\Phi_{2}=2 \\frac{q}{2 \\pi} \\frac{1}{4 \\pi \\varepsilon_{0}} \\int_{\\alpha}^{\\pi} \\frac{\\mathrm{d} \\phi}{2 R \\sin \\frac{\\phi}{2}}=\\frac{q}{8 \\pi^{2} \\varepsilon_{0} R} \\int_{\\alpha}^{\\pi} \\frac{\\mathrm{d} \\phi}{\\sin \\frac{\\phi}{2}}=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\int_{\\alpha}^{\\pi} \\frac{\\mathrm{d}\\left(\\frac{\\phi}{2}\\right)}{\\sin \\frac{\\phi}{2}}=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\int_{\\alpha / 2}^{\\pi / 2} \\frac{\\mathrm{d} \\chi}{\\sin \\chi} .\n$$\n\nUsing the integral from the formulation of the problem, we calculate:\n\n$$\n\\int_{\\alpha / 2}^{\\pi / 2} \\frac{\\mathrm{d} \\chi}{\\sin \\chi}=-\\left.\\ln \\left(\\frac{\\cos \\chi+1}{\\sin \\chi}\\right)\\right|_{\\alpha / 2} ^{\\pi / 2}=\\ln \\left(\\frac{\\cos \\alpha / 2+1}{\\sin \\alpha / 2}\\right) \\approx \\ln \\left(\\frac{4}{\\alpha}\\right)\n$$\n\nfor $\\alpha \\ll 1$. Therefore\n\n$$\n\\Phi_{2} \\approx \\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(\\frac{4}{\\alpha}\\right)\n$$\n\n\n\nThe total potential and capacitance. The total potential is the sum of $\\Phi_{1}$ and $\\Phi_{2}$ :\n\n$$\n\\Phi=\\Phi_{1}+\\Phi_{2}=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(\\frac{2 \\alpha R}{a}\\right)+\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(\\frac{4}{\\alpha}\\right)=\\frac{q}{4 \\pi^{2} \\varepsilon_{0} R} \\ln \\left(\\frac{8 R}{a}\\right) .\n$$\n\n$\\alpha$ drops out from the expression. From here we obtain the capacitance $C=q / \\Phi$ :\n\n$$\nC=\\frac{4 \\pi^{2} \\varepsilon_{0} R}{\\ln \\left(\\frac{8 R}{a}\\right)}\n$$\n\n$C \\rightarrow 0$ as $a \\rightarrow 0$.']",['$C=\\frac{4 \\pi^{2} \\varepsilon_{0} R}{\\ln (\\frac{8 R}{a})}$'],False,,Expression, 1047,Electromagnetism,,"D.2 Determine the dependence of the charge on the ring as a function of time, $q(t)$. $t=0$ corresponds to a time moment when electrons are in the plane of the ring. What is the charge on the ring $q_{0}$ when the absolute value $q(t)$ is maximum? The capacitance of the ring is $C$ (i.e., you do not have to use the actual expression found in D.1). Remark: the drawn polarity in Figure 3 is for indicative purposes only. The sign should be chosen so that the lens is focusing.","['Let $q(t)$ be the charge on the ring at a time $t$. Potential of the disk is thus $q(t) / C$. Voltage drop of the resistor is $R_{0} I(t)=R_{0} \\mathrm{~d} q / \\mathrm{d} t$. Therefore for time $-\\frac{d}{2 v}\\frac{d}{2 v}$, we get:\n\n$$\n\\frac{q(t)}{C}+R_{0} \\frac{\\mathrm{d} q}{\\mathrm{~d} t}=0\n$$\n\nFrom here:\n\n$$\nq(t)=q_{0} \\mathrm{e}^{-\\frac{t}{R_{0} C}+\\frac{d}{2 v R_{0 C}}}=C V_{0}\\left(\\mathrm{e}^{\\frac{d}{2 v R_{0} C}}-\\mathrm{e}^{-\\frac{d}{2 v R_{0} C}}\\right) \\mathrm{e}^{-\\frac{t}{R C}} .\n$$\n\nTherefore, we obtain:\n\n$$\nq(t)= \\begin{cases}0 & \\text { for } t<-\\frac{d}{2 v} ; \\\\ C V_{0}\\left(1-\\mathrm{e}^{-\\frac{d}{2 v R_{0} C}} \\mathrm{e}^{-\\frac{t}{R_{0} C}}\\right) & \\text { for }-\\frac{d}{2 v}\\frac{d}{2 v} .\\end{cases}\n$$\n\nFor a lens to be focusing we require that charge is negative, therefore $V_{0}<0$. The dependence of charge on time is shown in Figure 6.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d0d67f5b5936c04c5e29g-1.jpg?height=582&width=1006&top_left_y=477&top_left_x=525)\n\nFigure 6: Charge on the ring as a function of time.']","['For $-\\frac{d}{2 v}\\frac{d}{2 v}, \\quad q(t)=C V_{0}\\left(\\mathrm{e}^{\\frac{d}{2 v R_{0} C}}-\\mathrm{e}^{-\\frac{d}{2 v R_{0} C}}\\right) \\mathrm{e}^{-\\frac{t}{R_{0} C}}$.\n\n$q_{0}=C V_{0}\\left(1-\\mathrm{e}^{-\\frac{d}{v R_{0} C}}\\right) . \\quad$']",True,,Need_human_evaluate, 1048,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$} Context question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$. Extra Supplementary Reading Materials: Part C. The focal length of the idealized electrostatic lens: instantaneous charging One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called ""active region""). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field ""fills the space"" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring. Figure 2. A model of an electrostatic lens. Context question: C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the ""active region"" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing. Context answer: \boxed{$f=-\frac{E}{e q d \beta}$} Extra Supplementary Reading Materials: In reality the electron source is placed on the $z$-axis at a distance $b>f$ from the center of the ring. Consider that electrons are no longer parallel to the $z$-axis before reaching the ""active region"", but are emitted from a point source at a range of different angles $\gamma \ll 1$ rad to the $z$-axis. Electrons are focused in a point situated at a distance $c$ from the center of the ring. Context question: C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities. Context answer: \boxed{$c=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}}$.} Context question: C.3 Is the equation of a thin optical lens $$ \frac{1}{b}+\frac{1}{c}=\frac{1}{f} $$ fulfilled for the electrostatic lens? Show it by explicitly calculating $1 / b+1 / c$. Context answer: 证明题 Extra Supplementary Reading Materials: Part D. The ring as a capacitor The model of considered above was idealized and we assumed that the ring charged instantaneously. In reality charging is non-instantaneous, as the ring is a capacitor with a finite capacitance $C$. In this part we will analyze the properties of this capacitor. You might need the following integrals: $$ \int \frac{\mathrm{d} x}{\sin x}=-\ln \left|\frac{\cos x+1}{\sin x}\right|+\text { const } $$ and $$ \int \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}=\ln \left|x+\sqrt{1+x^{2}}\right|+\text { const. } $$ Context question: D.1 Calculate the capacitance $C$ of the ring. Consider that the ring has a finite width $2 a$, but remember that $a \ll R$. Context answer: \boxed{$C=\frac{4 \pi^{2} \varepsilon_{0} R}{\ln (\frac{8 R}{a})}$} Extra Supplementary Reading Materials: When electrons reach the ""active region"", the ring is connected to a source of voltage $V_{0}$ (Figure 3). When electrons pass the ""active region"", the ring is connected to the ground. The resistance of contacts is $R_{0}$ and the resistance of the ring itself can be neglected. Figure 3. Charging of the electrostatic lens. Context question: D.2 Determine the dependence of the charge on the ring as a function of time, $q(t)$. $t=0$ corresponds to a time moment when electrons are in the plane of the ring. What is the charge on the ring $q_{0}$ when the absolute value $q(t)$ is maximum? The capacitance of the ring is $C$ (i.e., you do not have to use the actual expression found in D.1). Remark: the drawn polarity in Figure 3 is for indicative purposes only. The sign should be chosen so that the lens is focusing. Context answer: For $-\frac{d}{2 v}\frac{d}{2 v}, \quad q(t)=C V_{0}\left(\mathrm{e}^{\frac{d}{2 v R_{0} C}}-\mathrm{e}^{-\frac{d}{2 v R_{0} C}}\right) \mathrm{e}^{-\frac{t}{R_{0} C}}$. $q_{0}=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{v R_{0} C}}\right) . \quad$ Extra Supplementary Reading Materials: Part E. Focal length of a more realistic lens: non-instantaneous charging In this part of the problem, we will consider the action of this more realistic lens. Here we will again neglect the width of the ring $2 a$ and will assume that electrons travel parallel to the $z$-axis before reaching the ""active region"". However, the charging of the ring is no longer instantaneous.","E.1 Find the focal length $f$ of the lens. Assume that $f / v \gg R_{0} C$, but $d / v$ and $R_{0} C$ are of the same order of magnitude. Express your answer in terms of the constant $\beta$ from part $\mathrm{B}$ and other known quantities.","['Like in part $C$, the radial equation of motion of an electron is:\n\n$$\nm \\ddot{r}=2 e q(t) \\beta r,\n$$\n\nwhere in this case $q(t)$ depends on time. Using the notation $\\eta=2 e \\beta / m$, we obtain:\n\n$$\n\\ddot{r}-\\eta q(t) r=0 .\n$$\n\nAs $f / v \\gg R_{0} C$, then during charging-decharging the electron does not substantially change its radial position $r$, and we can assume $r$ to be constant during the entire charging-decharging process. In this case the acquired vertical velocity is\n\n$$\nv_{r}=\\eta r \\int_{-d /(2 v)}^{\\infty} q(t) \\mathrm{d} t\n$$\n\n\n\nWe can use the derived equations for $q(t)$ and find the integrals. The integral $\\int_{-d /(2 v)}^{d /(2 v)} q(t) \\mathrm{d} t$ is (using the notation $d / v=t_{0}, R_{0} C=\\tau, C V_{0}=Q_{0}$ ):\n\n$$\n\\int_{-t_{0} / 2}^{t_{0} / 2} q(t) \\mathrm{d} t=\\int_{-t_{0} / 2}^{t_{0} / 2} Q_{0}\\left(1-\\mathrm{e}^{-\\frac{t_{0}}{2 \\tau}} \\mathrm{e}^{-\\frac{t}{\\tau}}\\right) \\mathrm{d} t=Q_{0}\\left(t_{0}-\\tau\\left[1-\\mathrm{e}^{-t_{0} / \\tau}\\right]\\right)\n$$\n\nThe integral $\\int_{d /(2 v)}^{\\infty} q(t) \\mathrm{d} t$ is\n\n$$\n\\int_{t_{0} / 2}^{\\infty} Q_{0}\\left(e^{\\frac{t_{0}}{2 \\tau}}-e^{-\\frac{t_{0}}{2 \\tau}}\\right) \\mathrm{e}^{-\\frac{t}{\\tau}} \\mathrm{d} t=Q_{0} \\tau\\left[1-\\mathrm{e}^{-t_{0} / \\tau}\\right]\n$$\n\nAdding the two integrals we obtain for the final integral:\n\n$$\n\\int_{-t_{0} / 2}^{\\infty} q(t) d t=Q_{0} t_{0}\n$$\n\nInterestingly, it does not depend on $\\tau=R_{0} C$. Therefore, the acquired vertical velocity of the electron is\n\n$$\nv_{r}=\\eta r \\frac{C V_{0} d}{v}=\\frac{2 e \\beta C V_{0} d r}{m v}\n$$\n\nFollowing the logic similar to part $\\mathrm{C}$, we derive the focal length\n\n$$\nf=-\\frac{E}{e C V_{0} d \\beta}\n$$']",['$f=-\\frac{E}{e C V_{0} d \\beta}$'],False,,Expression, 1049,Electromagnetism,"Electrostatic lens Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion) $$ (1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. } $$ Figure 1. A charged ring of radius $R$. Part A. Electrostatic potential on the axis of the ring (1 point) Context question: A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1). Context answer: \boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.} Context question: A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$. Context answer: \boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$} Context question: A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0 Context question: A.4 What is the angular frequency $\omega$ of such harmonic oscillations? Context answer: \boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$} Extra Supplementary Reading Materials: Part B. Electrostatic potential in the plane of the ring In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$. Context question: B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above. Context answer: \boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$} Context question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to harmonic oscillations. The moving electron does not influence the charge distribution on the ring. Context answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r . \quad q<0$. Extra Supplementary Reading Materials: Part C. The focal length of the idealized electrostatic lens: instantaneous charging One wants to build a device to focus electrons-an electrostatic lens. Let us consider the following construction. The ring is situated perpendicularly to the $z$-axis, as shown in Figure 2. We have a source that produces on-demand packets of non-relativistic electrons. Kinetic energy of these electrons is $E=m v^{2} / 2$ ( $v$ is velocity) and they leave the source at precisely controlled moments. The system is programmed so that the ring is charge-neutral most of the time, but its charge becomes $q$ when electrons are closer than a distance $d / 2(d \ll R)$ from the plane of the ring (shaded region in Figure 2, called ""active region""). In part $\mathrm{C}$ assume that charging and de-charging processes are instantaneous and the electric field ""fills the space"" instantaneously as well. One can neglect magnetic fields and assume that the velocity of electrons in the $z$-direction is constant. Moving electrons do not perturb the charge distribution on the ring. Figure 2. A model of an electrostatic lens. Context question: C.1 Determine the focal length $f$ of this lens. Assume that $f \gg d$. Express your answer in terms of the constant $\beta$ from question B. 1 and other known quantities. Assume that before reaching the ""active region"" the electron packet is parallel to the $z$-axis and $r \ll R$. The sign of $q$ is such so that the lens is focusing. Context answer: \boxed{$f=-\frac{E}{e q d \beta}$} Extra Supplementary Reading Materials: In reality the electron source is placed on the $z$-axis at a distance $b>f$ from the center of the ring. Consider that electrons are no longer parallel to the $z$-axis before reaching the ""active region"", but are emitted from a point source at a range of different angles $\gamma \ll 1$ rad to the $z$-axis. Electrons are focused in a point situated at a distance $c$ from the center of the ring. Context question: C.2 Find $c$. Express your answer in terms of the constant $\beta$ from question B.1 and other known quantities. Context answer: \boxed{$c=-\frac{1}{\frac{e q \beta d}{E}+\frac{1}{b}}$.} Context question: C.3 Is the equation of a thin optical lens $$ \frac{1}{b}+\frac{1}{c}=\frac{1}{f} $$ fulfilled for the electrostatic lens? Show it by explicitly calculating $1 / b+1 / c$. Context answer: 证明题 Extra Supplementary Reading Materials: Part D. The ring as a capacitor The model of considered above was idealized and we assumed that the ring charged instantaneously. In reality charging is non-instantaneous, as the ring is a capacitor with a finite capacitance $C$. In this part we will analyze the properties of this capacitor. You might need the following integrals: $$ \int \frac{\mathrm{d} x}{\sin x}=-\ln \left|\frac{\cos x+1}{\sin x}\right|+\text { const } $$ and $$ \int \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}=\ln \left|x+\sqrt{1+x^{2}}\right|+\text { const. } $$ Context question: D.1 Calculate the capacitance $C$ of the ring. Consider that the ring has a finite width $2 a$, but remember that $a \ll R$. Context answer: \boxed{$C=\frac{4 \pi^{2} \varepsilon_{0} R}{\ln (\frac{8 R}{a})}$} Extra Supplementary Reading Materials: When electrons reach the ""active region"", the ring is connected to a source of voltage $V_{0}$ (Figure 3). When electrons pass the ""active region"", the ring is connected to the ground. The resistance of contacts is $R_{0}$ and the resistance of the ring itself can be neglected. Figure 3. Charging of the electrostatic lens. Context question: D.2 Determine the dependence of the charge on the ring as a function of time, $q(t)$. $t=0$ corresponds to a time moment when electrons are in the plane of the ring. What is the charge on the ring $q_{0}$ when the absolute value $q(t)$ is maximum? The capacitance of the ring is $C$ (i.e., you do not have to use the actual expression found in D.1). Remark: the drawn polarity in Figure 3 is for indicative purposes only. The sign should be chosen so that the lens is focusing. Context answer: For $-\frac{d}{2 v}\frac{d}{2 v}, \quad q(t)=C V_{0}\left(\mathrm{e}^{\frac{d}{2 v R_{0} C}}-\mathrm{e}^{-\frac{d}{2 v R_{0} C}}\right) \mathrm{e}^{-\frac{t}{R_{0} C}}$. $q_{0}=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{v R_{0} C}}\right) . \quad$ Extra Supplementary Reading Materials: Part E. Focal length of a more realistic lens: non-instantaneous charging In this part of the problem, we will consider the action of this more realistic lens. Here we will again neglect the width of the ring $2 a$ and will assume that electrons travel parallel to the $z$-axis before reaching the ""active region"". However, the charging of the ring is no longer instantaneous. Context question: E.1 Find the focal length $f$ of the lens. Assume that $f / v \gg R_{0} C$, but $d / v$ and $R_{0} C$ are of the same order of magnitude. Express your answer in terms of the constant $\beta$ from part $\mathrm{B}$ and other known quantities. Context answer: \boxed{$f=-\frac{E}{e C V_{0} d \beta}$} ","E.2 You will see, that the result for $f$ is similar to that obtained in part C, whereby the value $q$ is substituted with $q_{\mathrm{eff}}$. Find the expression for $q_{\mathrm{eff}}$ in terms of quantities given in formulation of the problem.",['Comparing $f=-E /\\left(e C V_{0} d \\beta\\right)$ with $f=-E /(e q d \\beta)$ from part $\\mathrm{C}$ we immediataly obtain $q_{\\mathrm{eff}}=$ $C V_{0}$.'],['$q_{\\mathrm{eff}}=C V_{0}$'],False,,Expression, 1050,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.","A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.","[""The width of the potential well $(L)$ should be equal to the half of the wavelength of the de Broglie standing wave $\\lambda_{\\mathrm{dB}}=h / p$, here $h$ is the Planck's constant and $p$ is the momentum of the particle. Thus $p=h / \\lambda_{\\mathrm{dB}}=h /(2 L)$, and the minimal possible energy of the particle is\n\n$$\nE_{\\min }=\\frac{p^{2}}{2 m}=\\frac{h^{2}}{8 m L^{2}}\n$$""]",['$E_{\\min }=\\frac{h^{2}}{8 m L^{2}}$'],False,,Expression, 1051,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.","A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).","[""The potential well should fit an integer number of the de Broglie half-wavelengths: $L=\\frac{1}{2} \\lambda_{\\mathrm{dB}}^{(n)} \\cdot n$, $n=1,2, \\ldots$ Therefore, particle's momentum, corresponding to the de Broglie wavelength $\\lambda_{\\mathrm{dB}}^{(n)}$ is\n\n$$\np_{n}=\\frac{h}{\\lambda_{\\mathrm{dB}}^{(n)}}=\\frac{h n}{2 L}\n$$\n\nand the corresponding energy is\n\n$$\nE_{n}=\\frac{p_{n}^{2}}{2 m}=\\frac{h^{2} n^{2}}{8 m L^{2}}, \\quad n=1,2,3, \\ldots\n\\tag{1}\n$$""]",['$E_{n}=\\frac{h^{2} n^{2}}{8 m L^{2}}$'],False,,Expression, 1052,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state. Context question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ). Context answer: \boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$} ",A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$.,"[""The energy of the emitted photon, $E=h c / \\lambda$ (here $c$ is the speed of light and $\\lambda$ is the photon's wavelength) should be equal to the energy difference $\\Delta E=E_{2}-E_{1}$, therefore\n\n$$\n\\lambda_{21}=\\frac{h c}{E_{2}-E_{1}}=\\frac{8 m c L^{2}}{3 h}\n$$""]",['$\\lambda_{21}=\\frac{8 m c L^{2}}{3 h}$'],False,,Expression, 1053,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state. Context question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ). Context answer: \boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$} Context question: A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$. Context answer: \boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$} Extra Supplementary Reading Materials: Part C. Bose-Einstein condensation This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value. We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas.","C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants.","['At temperature $T$, the average kinetic energy of translational motion is $\\frac{3}{2} k_{\\mathrm{B}} T$. Equating this result to $p^{2} /(2 m)$, we obtain typical momentum $p=\\sqrt{3 m k_{\\mathrm{B}} T}$ and the de Broglie wavelength\n\n$$\n\\lambda_{\\mathrm{dB}}=\\frac{h}{p}=\\frac{h}{\\sqrt{3 m k_{\\mathrm{B}} T}}\n$$']","['$p=\\sqrt{3 m k_{\\mathrm{B}} T}$ , $\\lambda_{\\mathrm{dB}}=\\frac{h}{\\sqrt{3 m k_{\\mathrm{B}} T}}$']",True,,Expression, 1054,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state. Context question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ). Context answer: \boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$} Context question: A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$. Context answer: \boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$} Extra Supplementary Reading Materials: Part C. Bose-Einstein condensation This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value. We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas. Context question: C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants. Context answer: \boxed{$p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$} ","C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants.","['The volume per particle $V / N$ is a good estimate for $\\ell^{3}$. We obtain $\\ell=n^{-1 / 3}$, with $n=N / V$ and equate $\\ell=\\lambda_{\\mathrm{dB}}$ to express $T_{c}=h^{2} n^{2 / 3} /\\left(3 m k_{\\mathrm{B}}\\right)$.']","['$\\ell=n^{-1 / 3}$ , $T_{c}=\\frac{h^{2} n^{2 / 3}}{3 m k_{\\mathrm{B}}}$']",True,,Expression, 1055,Modern Physics,,"C.3 What is the particle density of the Rb gas $n_{c}$ if the transition takes place at such= a temperature? For the sake of comparison, calculate also the 'ordinary' particle density $n_{0}$ of an ideal gas at the standard temperature and pressure (STP), i.e. $T_{0}=300 \mathrm{~K}$ and $p_{0}=10^{5} \mathrm{~Pa}$. How many times is the 'ordinary' gas denser? You may assume that the mass of the atoms is equal to 87 atomic mass units $\left(m_{\mathrm{amu}}\right)$.","['Using the answer to the previous question, we express $n_{c}=\\left(3 m k_{\\mathrm{B}} T_{c}\\right)^{3 / 2} / h^{3}$. Equation of state for the ideal gas gives $n_{0}=p /\\left(k_{\\mathrm{B}} T\\right)$. Numerical estimations yield $n_{c} \\approx 1.59 \\cdot 10^{18} \\mathrm{~m}^{-3}$ and $n_{0} / n_{c} \\approx 1.5 \\cdot 10^{7}$.']",['Expression: $n_{c}=\\frac{\\left(3 \\cdot 87 m_{\\mathrm{amu}} k_{\\mathrm{B}} T_{c}\\right)^{3 / 2}}{h^{3}}$.\n\nNumerical value: $n_{c} \\approx 1.59 \\cdot 10^{18} \\mathrm{~m}^{-3}$.\n\nExpression: $n_{0}=p /\\left(k_{\\mathrm{B}} T\\right)$.\n\nNumerical value: $n_{0} / n_{c} \\approx 1.5 \\cdot 10^{7}$.'],False,,Need_human_evaluate, 1056,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state. Context question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ). Context answer: \boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$} Context question: A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$. Context answer: \boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$} Extra Supplementary Reading Materials: Part C. Bose-Einstein condensation This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value. We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas. Context question: C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants. Context answer: \boxed{$p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$} Context question: C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants. Context answer: \boxed{$\ell=n^{-1 / 3}$ , $T_{c}=\frac{h^{2} n^{2 / 3}}{3 m k_{\mathrm{B}}}$} Extra Supplementary Reading Materials: To realize BEC in the lab, the experimentalists have to cool gases to temperatures as low as $T_{c}=100 \mathrm{nK}$. Context question: C.3 What is the particle density of the Rb gas $n_{c}$ if the transition takes place at such= a temperature? For the sake of comparison, calculate also the 'ordinary' particle density $n_{0}$ of an ideal gas at the standard temperature and pressure (STP), i.e. $T_{0}=300 \mathrm{~K}$ and $p_{0}=10^{5} \mathrm{~Pa}$. How many times is the 'ordinary' gas denser? You may assume that the mass of the atoms is equal to 87 atomic mass units $\left(m_{\mathrm{amu}}\right)$. Context answer: Expression: $n_{c}=\frac{\left(3 \cdot 87 m_{\mathrm{amu}} k_{\mathrm{B}} T_{c}\right)^{3 / 2}}{h^{3}}$. Numerical value: $n_{c} \approx 1.59 \cdot 10^{18} \mathrm{~m}^{-3}$. Expression: $n_{0}=p /\left(k_{\mathrm{B}} T\right)$. Numerical value: $n_{0} / n_{c} \approx 1.5 \cdot 10^{7}$. Extra Supplementary Reading Materials: Part D. Three-beam optical lattices The first Bose-Einstein condensates were produced back in 1995, and since then the experimental work has branched out in diverse directions. In this part, you will investigate one particularly fruitful idea to load the condensate into spatially periodic potentials created by interfering a number of coherent laser beams. Due to the periodic nature of the resulting interference patterns, they are referred to as optical lattices. The potential energy $V(\vec{r})$ of an atom moving in an optical lattice is proportional to the local intensity of the light, and in your calculations you may assume that $$ V(\vec{r})=-\alpha\left\langle|\vec{E}(\vec{r}, t)|^{2}\right\rangle \tag{2} $$ Here, $\alpha$ is a positive constant, and the angle brackets indicate time-averaging which eliminates the rapidly oscillating terms. The electric field produced by the $i$-th laser is described by $$ \vec{E}_{i}=E_{0, i} \vec{\varepsilon}_{i} \cos \left(\vec{k}_{i} \cdot \vec{r}-\omega t\right) \tag{3} $$ with the amplitude $E_{0, i}$, the wave vector $\vec{k}_{i}$, and the unit-length polarization vector $\vec{\varepsilon}_{i}$. (a) (b) (c) Figure 2. (a) Three-beam optical lattice: three plane waves with wave vectors $\vec{k}_{1,2,3}$ intersect and interfere in the area indicated by the grey circle. (b) Symmetries of a regular hexagon: solid and dashed lines show two sets of symmetry axes. (c) Saddle point: a point on a surface where the slopes in orthogonal directions are all zero, but which is not a local extremum of the plotted function. Travelling along the trajectory marked by the full line one encounters an apparent minimum. Additional analysis of the perpendicular direction (dashed line) is needed to distinguish a true minimum from a saddle point (shown). Your task is to study triangular optical lattices that are produced by interfering three coherent laser beams of equal intensity. A typical setup is shown in Fig. 2a. Here, all three beams are polarized in the $z$ direction, propagate in the $x y$ plane and intersect at equal angles of $120^{\circ}$. Choose the direction of the $x$ axis parallel to the wave vector $\vec{k}_{1}$.","D.1 Using Eqs. 2 and 3 obtain the expression for the potential energy $V(\vec{r})$ as a function of $\vec{r}=(x, y)$ in the plane of the beams. Hint: the result can be neatly expressed as a constant term plus a sum of three cosine functions of arguments $\vec{b}_{i} \cdot \vec{r}$. Please write your result in this form and identify the vectors $\vec{b}_{i}$.","['We sum the three electric fields ( $z$ components)\n\n$$\nE(\\vec{r}, t)=E_{0} \\sum_{i=1}^{3} \\cos \\left(\\vec{k}_{i} \\cdot \\vec{r}-\\omega t\\right)\n\\tag{4}\n$$\n\nand square the result\n\n$$\n\\begin{aligned}\nE^{2}(\\vec{r}, t) & =E_{0}^{2} \\sum_{i=1}^{3} \\sum_{j=1}^{3} \\cos \\left(\\vec{k}_{i} \\cdot \\vec{r}-\\omega t\\right) \\cos \\left(\\vec{k}_{j} \\cdot \\vec{r}-\\omega t\\right) \\\\\n& =\\frac{E_{0}^{2}}{2} \\sum_{i=1}^{3} \\sum_{j=1}^{3}\\left\\{\\cos \\left[\\left(\\vec{k}_{i}-\\vec{k}_{j}\\right) \\cdot \\vec{r}\\right]+\\cos \\left[\\left(\\vec{k}_{i}+\\vec{k}_{j}\\right) \\cdot \\vec{r}-2 \\omega t\\right]\\right\\} .\n\\end{aligned}\n\\tag{5}\n$$\n\nTime averaging gives\n\n$$\n\\left\\langle E^{2}(\\vec{r}, t)\\right\\rangle=\\frac{E_{0}^{2}}{2} \\sum_{i=1}^{3} \\sum_{j=1}^{3} \\cos \\left[\\left(\\vec{k}_{i}-\\vec{k}_{j}\\right) \\cdot \\vec{r}\\right]\n\\tag{6}\n$$\n\nwe analyse the 9 terms and simplify to\n\n$$\n\\left\\langle E^{2}(\\vec{r}, t)\\right\\rangle=E_{0}^{2}\\left(\\frac{3}{2}+\\sum_{j=1}^{3} \\cos \\vec{b}_{j} \\cdot \\vec{r}\\right)\n\\tag{7}\n$$\n\nHere $\\vec{b}_{1,2,3}=\\left(\\vec{k}_{2}-\\vec{k}_{3}\\right),\\left(\\vec{k}_{3}-\\vec{k}_{1}\\right),\\left(\\vec{k}_{1}-\\vec{k}_{2}\\right)$ or in terms of the Levi-Civita symbol, $\\vec{b}_{k}=\\varepsilon_{i j k}\\left(\\vec{k}_{i}-\\vec{k}_{j}\\right)$. Incidentally, they are known as the reciprocal lattice vectors.']","['$V(\\vec{r})=-\\alpha E_{0}^{2}\\left(\\frac{3}{2}+\\sum_{j=1}^{3} \\cos \\vec{b}_{j} \\cdot \\vec{r}\\right)$,$\\vec{b}_{1}=\\vec{k}_{2}-\\vec{k}_{3}, \\quad \\vec{b}_{2}=\\vec{k}_{3}-\\vec{k}_{1}, \\quad \\vec{b}_{3}=\\vec{k}_{1}-\\vec{k}_{2}$']",True,,Expression, 1057,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state. Context question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ). Context answer: \boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$} Context question: A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$. Context answer: \boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$} Extra Supplementary Reading Materials: Part C. Bose-Einstein condensation This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value. We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas. Context question: C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants. Context answer: \boxed{$p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$} Context question: C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants. Context answer: \boxed{$\ell=n^{-1 / 3}$ , $T_{c}=\frac{h^{2} n^{2 / 3}}{3 m k_{\mathrm{B}}}$} Extra Supplementary Reading Materials: To realize BEC in the lab, the experimentalists have to cool gases to temperatures as low as $T_{c}=100 \mathrm{nK}$. Context question: C.3 What is the particle density of the Rb gas $n_{c}$ if the transition takes place at such= a temperature? For the sake of comparison, calculate also the 'ordinary' particle density $n_{0}$ of an ideal gas at the standard temperature and pressure (STP), i.e. $T_{0}=300 \mathrm{~K}$ and $p_{0}=10^{5} \mathrm{~Pa}$. How many times is the 'ordinary' gas denser? You may assume that the mass of the atoms is equal to 87 atomic mass units $\left(m_{\mathrm{amu}}\right)$. Context answer: Expression: $n_{c}=\frac{\left(3 \cdot 87 m_{\mathrm{amu}} k_{\mathrm{B}} T_{c}\right)^{3 / 2}}{h^{3}}$. Numerical value: $n_{c} \approx 1.59 \cdot 10^{18} \mathrm{~m}^{-3}$. Expression: $n_{0}=p /\left(k_{\mathrm{B}} T\right)$. Numerical value: $n_{0} / n_{c} \approx 1.5 \cdot 10^{7}$. Extra Supplementary Reading Materials: Part D. Three-beam optical lattices The first Bose-Einstein condensates were produced back in 1995, and since then the experimental work has branched out in diverse directions. In this part, you will investigate one particularly fruitful idea to load the condensate into spatially periodic potentials created by interfering a number of coherent laser beams. Due to the periodic nature of the resulting interference patterns, they are referred to as optical lattices. The potential energy $V(\vec{r})$ of an atom moving in an optical lattice is proportional to the local intensity of the light, and in your calculations you may assume that $$ V(\vec{r})=-\alpha\left\langle|\vec{E}(\vec{r}, t)|^{2}\right\rangle \tag{2} $$ Here, $\alpha$ is a positive constant, and the angle brackets indicate time-averaging which eliminates the rapidly oscillating terms. The electric field produced by the $i$-th laser is described by $$ \vec{E}_{i}=E_{0, i} \vec{\varepsilon}_{i} \cos \left(\vec{k}_{i} \cdot \vec{r}-\omega t\right) \tag{3} $$ with the amplitude $E_{0, i}$, the wave vector $\vec{k}_{i}$, and the unit-length polarization vector $\vec{\varepsilon}_{i}$. (a) (b) (c) Figure 2. (a) Three-beam optical lattice: three plane waves with wave vectors $\vec{k}_{1,2,3}$ intersect and interfere in the area indicated by the grey circle. (b) Symmetries of a regular hexagon: solid and dashed lines show two sets of symmetry axes. (c) Saddle point: a point on a surface where the slopes in orthogonal directions are all zero, but which is not a local extremum of the plotted function. Travelling along the trajectory marked by the full line one encounters an apparent minimum. Additional analysis of the perpendicular direction (dashed line) is needed to distinguish a true minimum from a saddle point (shown). Your task is to study triangular optical lattices that are produced by interfering three coherent laser beams of equal intensity. A typical setup is shown in Fig. 2a. Here, all three beams are polarized in the $z$ direction, propagate in the $x y$ plane and intersect at equal angles of $120^{\circ}$. Choose the direction of the $x$ axis parallel to the wave vector $\vec{k}_{1}$. Context question: D.1 Using Eqs. 2 and 3 obtain the expression for the potential energy $V(\vec{r})$ as a function of $\vec{r}=(x, y)$ in the plane of the beams. Hint: the result can be neatly expressed as a constant term plus a sum of three cosine functions of arguments $\vec{b}_{i} \cdot \vec{r}$. Please write your result in this form and identify the vectors $\vec{b}_{i}$. Context answer: \boxed{$V(\vec{r})=-\alpha E_{0}^{2}\left(\frac{3}{2}+\sum_{j=1}^{3} \cos \vec{b}_{j} \cdot \vec{r}\right)$,$\vec{b}_{1}=\vec{k}_{2}-\vec{k}_{3}, \quad \vec{b}_{2}=\vec{k}_{3}-\vec{k}_{1}, \quad \vec{b}_{3}=\vec{k}_{1}-\vec{k}_{2}$} ","D.2 The resulting potential energy has a sixfold rotational symmetry axis, i.e., the potential distribution is invariant with respect to a rotation by a multiple of $60^{\circ}$ around the origin. Provide a simple argument to prove that this is indeed the case.","[""Argument: Observe that rotation by $60^{\\circ}$ maps the three vectors $\\vec{b}_{1,2,3}$ into the relabelled triplet of $-\\vec{b}$ 's.""]",,False,,, 1058,Modern Physics,,"D.3 Derive the behavior of the potential $V(\vec{r})$ along the coordinate axes, i.e., determine the functions $V_{X}(x) \equiv V(x, 0)$ and $V_{Y}(y) \equiv V(0, y)$. Identify the locations of the extrema of $V_{X}(x)$ and $V_{Y}(y)$ as functions of a single argument. As these functions are periodic, please include in your lists only one representative from each family of periodically repeated minima and maxima.","[""We find\n\n$$\nV(x, y)=-\\alpha E_{0}^{2}\\left\\{\\frac{3}{2}+\\cos (k y \\sqrt{3})+\\cos \\left(\\frac{3 k x}{2}+\\frac{k y \\sqrt{3}}{2}\\right)+\\cos \\left(\\frac{3 k x}{2}-\\frac{k y \\sqrt{3}}{2}\\right)\\right\\}\n\\tag{8}\n$$\n\nand deduce\n\n$$\nV_{X}(x)=-\\alpha E_{0}^{2}\\left\\{\\frac{5}{2}+2 \\cos \\frac{3 k x}{2}\\right\\}\n\\tag{9}\n$$\n\nThe potential has a simple cosine form, and the origin in an obvious minimum. Its replica appear at multiples of $\\Delta x=4 \\pi /(3 k)$. In the midpoint between any two minima, e.g. at $x=\\Delta x / 2=2 \\pi /(3 k)$, the function $V_{X}(x)$ has its maxima.\n\nConcerning the behaviour along the $y$ axis, we have\n\n$$\nV_{Y}(y)=-\\alpha E_{0}^{2}\\left\\{\\frac{3}{2}+\\cos 2 \\varphi+2 \\cos \\varphi\\right\\}, \\quad \\varphi=\\sqrt{3} k y / 2\n\\tag{10}\n$$\n\nLooking for the extrema, we find the equation\n\n$$\n\\sin 2 \\varphi+\\sin \\varphi=0\n\\tag{11}\n$$\n\n$\\circ \\varphi=0$ is the 'deep' minimum - the lattice site;\n\n- $\\varphi=\\pi$ is the 'shallow' minimum (later shown to be a saddle point of $V(x, y)$ );\n- $\\varphi=2 \\pi / 3$ and $\\varphi=4 \\pi / 3$ are maxima.""]","[""$V_{X}(x)=-\\alpha E_{0}^{2}\\left\\{\\frac{3}{2}+2 \\cos \\frac{3 k x}{2}\\right\\}$.\n\n$V_{Y}(y)=-\\alpha E_{0}^{2}\\left\\{\\frac{3}{2}+\\cos 2 \\varphi+2 \\cos \\varphi\\right\\}, \\quad$ here $\\varphi=\\sqrt{3} k y / 2$.\n\nMinimum $(-\\mathrm{a})$ of $V_{X}(x): x=0$.\n\nMaximum (-a) of $V_{X}(x): x=\\frac{2 \\pi}{3 k}$.\n\nMinimum (-a) of $V_{Y}(y): y=0$ ('deep') and $y=\\frac{2 \\pi}{\\sqrt{3} k}$ ('shallow').\n\nMaximum (-a) of $V_{Y}(y): y=\\frac{4 \\pi}{3 \\sqrt{3} k}$ and $y=\\frac{8 \\pi}{3 \\sqrt{3} k}$.""]",True,,Need_human_evaluate, 1059,Modern Physics,"Particles and Waves Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons. Part A. Quantum particle in a box Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by $$ V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases} \tag{1} $$ While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls. Context question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$. Context answer: \boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$} Extra Supplementary Reading Materials: The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state. Context question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ). Context answer: \boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$} Context question: A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$. Context answer: \boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$} Extra Supplementary Reading Materials: Part C. Bose-Einstein condensation This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value. We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas. Context question: C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants. Context answer: \boxed{$p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$} Context question: C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants. Context answer: \boxed{$\ell=n^{-1 / 3}$ , $T_{c}=\frac{h^{2} n^{2 / 3}}{3 m k_{\mathrm{B}}}$} Extra Supplementary Reading Materials: To realize BEC in the lab, the experimentalists have to cool gases to temperatures as low as $T_{c}=100 \mathrm{nK}$. Context question: C.3 What is the particle density of the Rb gas $n_{c}$ if the transition takes place at such= a temperature? For the sake of comparison, calculate also the 'ordinary' particle density $n_{0}$ of an ideal gas at the standard temperature and pressure (STP), i.e. $T_{0}=300 \mathrm{~K}$ and $p_{0}=10^{5} \mathrm{~Pa}$. How many times is the 'ordinary' gas denser? You may assume that the mass of the atoms is equal to 87 atomic mass units $\left(m_{\mathrm{amu}}\right)$. Context answer: Expression: $n_{c}=\frac{\left(3 \cdot 87 m_{\mathrm{amu}} k_{\mathrm{B}} T_{c}\right)^{3 / 2}}{h^{3}}$. Numerical value: $n_{c} \approx 1.59 \cdot 10^{18} \mathrm{~m}^{-3}$. Expression: $n_{0}=p /\left(k_{\mathrm{B}} T\right)$. Numerical value: $n_{0} / n_{c} \approx 1.5 \cdot 10^{7}$. Extra Supplementary Reading Materials: Part D. Three-beam optical lattices The first Bose-Einstein condensates were produced back in 1995, and since then the experimental work has branched out in diverse directions. In this part, you will investigate one particularly fruitful idea to load the condensate into spatially periodic potentials created by interfering a number of coherent laser beams. Due to the periodic nature of the resulting interference patterns, they are referred to as optical lattices. The potential energy $V(\vec{r})$ of an atom moving in an optical lattice is proportional to the local intensity of the light, and in your calculations you may assume that $$ V(\vec{r})=-\alpha\left\langle|\vec{E}(\vec{r}, t)|^{2}\right\rangle \tag{2} $$ Here, $\alpha$ is a positive constant, and the angle brackets indicate time-averaging which eliminates the rapidly oscillating terms. The electric field produced by the $i$-th laser is described by $$ \vec{E}_{i}=E_{0, i} \vec{\varepsilon}_{i} \cos \left(\vec{k}_{i} \cdot \vec{r}-\omega t\right) \tag{3} $$ with the amplitude $E_{0, i}$, the wave vector $\vec{k}_{i}$, and the unit-length polarization vector $\vec{\varepsilon}_{i}$. (a) (b) (c) Figure 2. (a) Three-beam optical lattice: three plane waves with wave vectors $\vec{k}_{1,2,3}$ intersect and interfere in the area indicated by the grey circle. (b) Symmetries of a regular hexagon: solid and dashed lines show two sets of symmetry axes. (c) Saddle point: a point on a surface where the slopes in orthogonal directions are all zero, but which is not a local extremum of the plotted function. Travelling along the trajectory marked by the full line one encounters an apparent minimum. Additional analysis of the perpendicular direction (dashed line) is needed to distinguish a true minimum from a saddle point (shown). Your task is to study triangular optical lattices that are produced by interfering three coherent laser beams of equal intensity. A typical setup is shown in Fig. 2a. Here, all three beams are polarized in the $z$ direction, propagate in the $x y$ plane and intersect at equal angles of $120^{\circ}$. Choose the direction of the $x$ axis parallel to the wave vector $\vec{k}_{1}$. Context question: D.1 Using Eqs. 2 and 3 obtain the expression for the potential energy $V(\vec{r})$ as a function of $\vec{r}=(x, y)$ in the plane of the beams. Hint: the result can be neatly expressed as a constant term plus a sum of three cosine functions of arguments $\vec{b}_{i} \cdot \vec{r}$. Please write your result in this form and identify the vectors $\vec{b}_{i}$. Context answer: \boxed{$V(\vec{r})=-\alpha E_{0}^{2}\left(\frac{3}{2}+\sum_{j=1}^{3} \cos \vec{b}_{j} \cdot \vec{r}\right)$,$\vec{b}_{1}=\vec{k}_{2}-\vec{k}_{3}, \quad \vec{b}_{2}=\vec{k}_{3}-\vec{k}_{1}, \quad \vec{b}_{3}=\vec{k}_{1}-\vec{k}_{2}$} Context question: D.2 The resulting potential energy has a sixfold rotational symmetry axis, i.e., the potential distribution is invariant with respect to a rotation by a multiple of $60^{\circ}$ around the origin. Provide a simple argument to prove that this is indeed the case. Context answer: 证明题 Extra Supplementary Reading Materials: The above observation of symmetry simplifies the analysis of the two-dimensional potential distribution $V(\vec{r})$. As shown in Fig. 2b, a regular hexagon has symmetry lines that, respectively, connect opposite vertices (solid lines) and midpoints of opposite edges (dashed lines). Therefore, in our situation one does not need to produce and study two-dimensional potential plots as many insights can be deduced by focusing on the coordinate axes $x$ and $y$ that run along the symmetry lines. Context question: D.3 Derive the behavior of the potential $V(\vec{r})$ along the coordinate axes, i.e., determine the functions $V_{X}(x) \equiv V(x, 0)$ and $V_{Y}(y) \equiv V(0, y)$. Identify the locations of the extrema of $V_{X}(x)$ and $V_{Y}(y)$ as functions of a single argument. As these functions are periodic, please include in your lists only one representative from each family of periodically repeated minima and maxima. Context answer: $V_{X}(x)=-\alpha E_{0}^{2}\left\{\frac{3}{2}+2 \cos \frac{3 k x}{2}\right\}$. $V_{Y}(y)=-\alpha E_{0}^{2}\left\{\frac{3}{2}+\cos 2 \varphi+2 \cos \varphi\right\}, \quad$ here $\varphi=\sqrt{3} k y / 2$. Minimum $(-\mathrm{a})$ of $V_{X}(x): x=0$. Maximum (-a) of $V_{X}(x): x=\frac{2 \pi}{3 k}$. Minimum (-a) of $V_{Y}(y): y=0$ ('deep') and $y=\frac{2 \pi}{\sqrt{3} k}$ ('shallow'). Maximum (-a) of $V_{Y}(y): y=\frac{4 \pi}{3 \sqrt{3} k}$ and $y=\frac{8 \pi}{3 \sqrt{3} k}$. Extra Supplementary Reading Materials: We are interested in determining the locations of so-called lattice sites, i.e., the minima of the full twodimensional potential $V(\vec{r})$. The obtained minima of single-argument functions $V_{X}$ and $V_{Y}$ identify their suspected positions but still have to be checked to eliminate the saddle points. As shown in Fig. 2c, when studied along a single line, saddle points may disguise as minima but they are not.",D.5 Calculate the value of $n$ that corresponds to the radius of the Rb Rydberg atom comparable to the wavelength of laser light $\lambda_{\text {las }}=380 \mathrm{~nm}$. Give your answer in terms of $\lambda_{\text {las }}$ and physical constants and find its numerical value.,"[""The atom's core electrons (all but the one promoted to to a state with a high principal quantum number $n$ ) shield the electric field of the nucleus so that the effective potential for the outer electron resembles that of a hydrogen atom. The attractive force acting on that electron, $F=e^{2} /\\left(4 \\pi \\varepsilon_{0} r^{2}\\right)$, gives rise to its centripetal acceleration $a=v^{2} / r$. Equating $F=m_{\\mathrm{e}} a$ and using the expression for the angular momentum $m_{\\mathrm{e}} v r=n \\hbar$ to eliminate the velocity, we find the quantum number $n$ corresponding to the orbit with the radius $r=\\lambda_{\\text {las }}$ :\n\n$$\nn=\\frac{e}{\\hbar} \\sqrt{\\frac{m_{\\mathrm{e}} \\lambda}{4 \\pi \\varepsilon_{0}}} \\approx 85\n\\tag{12}\n$$""]","['$n=\\frac{e}{\\hbar} \\sqrt{\\frac{m_{\\mathrm{e}} \\lambda}{4 \\pi \\varepsilon_{0}}}$, $85$']",True,,"Expression,Numerical",",1e0" 1060,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun :","A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.","[""Stefan's law: $L_{\\odot}=\\left(4 \\pi R_{\\odot}^{2}\\right)\\left(\\sigma T_{\\mathrm{s}}^{4}\\right)$\n\n$$\nT_{\\mathrm{s}}=\\left(\\frac{L_{\\odot}}{4 \\pi R_{\\odot}^{2} \\sigma}\\right)^{1 / 4}=5.76 \\times 10^{3} \\mathrm{~K}\n$$""]",['$5.76 \\times 10^{3} $'],False,K,Numerical,1e1 1061,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.","A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$","['$$\nP_{\\text {in }}=\\int_{0}^{\\infty} u(\\nu) d \\nu=\\int_{0}^{\\infty} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi h}{c^{2}} \\nu^{3} \\exp \\left(-h \\nu / k_{\\mathrm{B}} T_{\\mathrm{s}}\\right) d \\nu\n$$\n\nLet $x=\\frac{h \\nu}{k_{\\mathrm{B}} T_{\\mathrm{s}}}$. Then, $\\nu=\\frac{k_{\\mathrm{B}} T_{\\mathrm{s}}}{h} x \\quad d \\nu=\\frac{k_{\\mathrm{B}} T_{\\mathrm{s}}}{h} d x$.\n\n$$\nP_{\\mathrm{in}}=\\frac{2 \\pi h A R_{\\odot}^{2}}{c^{2} d_{\\odot}^{2}} \\frac{\\left(k_{\\mathrm{B}} T_{\\mathrm{s}}\\right)^{4}}{h^{4}} \\int_{0}^{\\infty} x^{3} e^{-x} d x=\\frac{2 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\cdot 6=\\frac{12 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}}\n$$']",['$P_{\\mathrm{in}}=\\frac{12 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}}$'],False,,Expression, 1062,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} ","A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.",['$$\n\\begin{aligned}\nn_{\\gamma}(\\nu) & =\\frac{u(\\nu)}{h \\nu} \\\\\n& =A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}} \\nu^{2} \\exp \\left(-h \\nu / k_{\\mathrm{B}} T_{\\mathrm{s}}\\right)\n\\end{aligned}\n$$'],['$n_{\\gamma}(\\nu)=A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}} \\nu^{2} e^ {\\frac{-h \\nu }{k_{\\mathrm{B}} T_{\\mathrm{s}}}}$'],False,,Expression, 1063,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy).","A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.","['The useful power output is the useful energy quantum per photon, $E_{\\mathrm{g}} \\equiv h \\nu_{\\mathrm{g}}$, multiplied by the number of photons with energy, $E \\geq E_{\\mathrm{g}}$.\n\n$$\n\\begin{aligned}\nP_{\\text {out }} & =h \\nu_{\\mathrm{g}} \\int_{\\nu_{\\mathrm{g}}}^{\\infty} n_{\\gamma}(\\nu) d \\nu \\\\\n& =h \\nu_{\\mathrm{g}} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}} \\int_{\\nu_{\\mathrm{g}}}^{\\infty} \\nu^{2} \\exp \\left(-h \\nu / k_{\\mathrm{B}} T_{\\mathrm{s}}\\right) d \\nu \\\\\n& =k_{\\mathrm{B}} T_{\\mathrm{s}} x_{\\mathrm{g}} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} \\frac{2 \\pi}{c^{2}}\\left(\\frac{k_{\\mathrm{B}} T_{\\mathrm{s}}}{h}\\right)^{3} \\int_{x_{\\mathrm{g}}}^{\\infty} x^{2} e^{-x} d x \\\\\n& =\\frac{2 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} x_{\\mathrm{g}}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}\n\\end{aligned}\n$$']",['$P_{\\text {out }}=\\frac{2 \\pi k_{\\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\\mathrm{s}}^{4} A \\frac{R_{\\odot}^{2}}{d_{\\odot}^{2}} x_{\\mathrm{g}}(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2) e^{-x_{\\mathrm{g}}}$'],False,,Expression, 1064,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} ","A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.",['Efficiency $\\eta=\\frac{P_{\\text {out }}}{P_{\\text {in }}}=\\frac{x_{\\mathrm{g}}}{6}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}$'],['$\\eta=\\frac{x_{\\mathrm{g}}}{6}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}$'],False,,Expression, 1065,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} ",A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?,"['$$\n\\eta=\\frac{1}{6}\\left(x_{\\mathrm{g}}^{3}+2 x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}\\right) e^{-x_{\\mathrm{g}}}\n$$\n\nPut limiting values, $\\eta(0)=0 \\quad \\eta(\\infty)=0$.\n\nSince the polynomial has all positive coefficients, it increases monotonically; the exponential function decreases monotonically. Therefore, $\\eta$ has only one maximum.\n\n$$\n\\begin{aligned}\n& \\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}=\\frac{1}{6}\\left(-x_{\\mathrm{g}}^{3}+x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}} \\\\\n& \\left.\\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}\\right|_{x_{\\mathrm{g}}=0}=\\left.\\frac{1}{3} \\quad \\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}\\right|_{x_{\\mathrm{g}} \\rightarrow \\infty}=0\n\\end{aligned}\n$$\n\n']","['$\\frac{1}{3}$ , 0']",True,,Numerical,0 1066,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} ",A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.,"['The maximum will be for $\\frac{\\mathrm{d} \\eta}{\\mathrm{d} x_{\\mathrm{g}}}=\\frac{1}{6}\\left(-x_{\\mathrm{g}}^{3}+x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}=0$\n\n$$\n\\Rightarrow p\\left(x_{\\mathrm{g}}\\right) \\equiv x_{\\mathrm{g}}^{3}-x_{\\mathrm{g}}^{2}-2 x_{\\mathrm{g}}-2=0\n$$\n\nA Numerical Solution by the Bisection Method:\n\nNow,\n\n$$\n\\begin{aligned}\np(0) & =-2 \\\\\np(1) & =-4 \\\\\np(2) & =-2 \\\\\np(3) & =10 \\quad \\Rightarrow \\\\\np(2.5) & =2.375 \\quad \\Rightarrow \\quad 2 solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} ","A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value.",['$$\n\\begin{gathered}\nx_{\\mathrm{g}}=\\frac{1.11 \\times 1.60 \\times 10^{-19}}{1.38 \\times 10^{-23} \\times 5763}=2.23 \\\\\n\\eta_{\\mathrm{Si}}=\\frac{x_{\\mathrm{g}}}{6}\\left(x_{\\mathrm{g}}^{2}+2 x_{\\mathrm{g}}+2\\right) e^{-x_{\\mathrm{g}}}=0.457\n\\end{gathered}\n$$'],['0.457'],False,,Numerical,1e-3 1068,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} Context question: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value. Context answer: \boxed{0.457} Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction.","A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.","['The total gravitational potential energy of the Sun: $\\Omega=-\\int_{0}^{M_{\\odot}} \\frac{G m \\mathrm{~d} m}{r}$\n\nFor constant density, $\\rho=\\frac{3 M_{\\odot}}{4 \\pi R_{\\odot}^{3}} \\quad m=\\frac{4}{3} \\pi r^{3} \\rho \\quad \\mathrm{d} m=4 \\pi r^{2} \\rho \\mathrm{d} r$\n\n$$\n\\Omega=-\\int_{0}^{R_{\\odot}} G\\left(\\frac{4}{3} \\pi r^{3} \\rho\\right)\\left(4 \\pi r^{2} \\rho\\right) \\frac{\\mathrm{d} r}{r}=-\\frac{16 \\pi^{2} G \\rho^{2}}{3} \\frac{R_{\\odot}^{5}}{5}=-\\frac{3}{5} \\frac{G M_{\\odot}^{2}}{R_{\\odot}}\n$$']",['$\\Omega=-\\frac{3}{5} \\frac{G M_{\\odot}^{2}}{R_{\\odot}}$'],False,,Expression, 1069,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} Context question: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value. Context answer: \boxed{0.457} Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction. Context question: A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$. Context answer: \boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$} ","A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.",['$$\n\\begin{gathered}\n\\tau_{\\mathrm{KH}}=\\frac{-\\Omega}{L_{\\odot}} \\\\\n\\tau_{\\mathrm{KH}}=\\frac{3 G M_{\\odot}^{2}}{5 R_{\\odot} L_{\\odot}}=1.88 \\times 10^{7} \\text { years }\n\\end{gathered}\n$$'],['$1.88 \\times 10^{7}$'],False,years,Numerical,1e5 1070,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} Context question: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value. Context answer: \boxed{0.457} Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction. Context question: A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$. Context answer: \boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$} Context question: A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. Context answer: \boxed{$1.88 \times 10^{7}$} Extra Supplementary Reading Materials: The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational. B Neutrinos from the Sun : In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: $$ 4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}} $$ The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem.","B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction.",['$$\n\\begin{aligned}\n& \\text { Solution: } \\\\\n& \\qquad \\begin{array}{ll}\n4.0 \\times 10^{-12} \\mathrm{~J} \\leftrightarrow 2 \\nu \\\\\n\\Rightarrow \\Phi_{\\nu}=\\frac{L_{\\odot}}{4 \\pi d_{\\odot}^{2} \\delta E} \\times 2=\\frac{3.85 \\times 10^{26}}{4 \\pi \\times\\left(1.50 \\times 10^{11}\\right)^{2} \\times 4.0 \\times 10^{-12}} \\times 2=6.8 \\times 10^{14} \\mathrm{~m}^{-2} \\mathrm{~s}^{-1} .\n\\end{array}\n\\end{aligned}\n$$'],['$6.8 \\times 10^{14}$'],False,$\mathrm{~m}^{-2} \mathrm{~s}^{-1}$,Numerical,1e13 1071,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} Context question: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value. Context answer: \boxed{0.457} Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction. Context question: A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$. Context answer: \boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$} Context question: A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. Context answer: \boxed{$1.88 \times 10^{7}$} Extra Supplementary Reading Materials: The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational. B Neutrinos from the Sun : In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: $$ 4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}} $$ The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem. Context question: B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction. Context answer: \boxed{$6.8 \times 10^{14}$} Extra Supplementary Reading Materials: Travelling from the core of the Sun to the Earth, some of the electron neutrinos, $v_{e}$, are converted to other types of neutrinos, $v_{\mathrm{x}}$. The efficiency of the detector for detecting $v_{\mathrm{x}}$ is $1 / 6$ of its efficiency for detecting $v_{\mathrm{e}}$. If there is no neutrino conversion, we expect to detect an average of $N_{1}$ neutrinos in a year. However, due to the conversion, an average of $N_{2}$ neutrinos ( $v_{\mathrm{e}}$ and $v_{\mathrm{x}}$ combined) are actually detected per year.","B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$.",['$$\n\\begin{aligned}\n& N_{1}=\\epsilon N_{0} \\\\\n& N_{e}=\\epsilon N_{0}(1-f) \\\\\n& N_{x}=\\epsilon N_{0} f / 6 \\\\\n& N_{2}=N_{e}+N_{x}\n\\end{aligned}\n$$\n\nOR\n\n$$\n\\begin{gathered}\n(1-f) N_{1}+\\frac{f}{6} N_{1}=N_{2} \\\\\n\\Rightarrow f=\\frac{6}{5}\\left(1-\\frac{N_{2}}{N_{1}}\\right)\n\\end{gathered}\n$$'],['$f=\\frac{6}{5}(1-\\frac{N_{2}}{N_{1}})$'],False,,Expression, 1072,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} Context question: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value. Context answer: \boxed{0.457} Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction. Context question: A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$. Context answer: \boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$} Context question: A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. Context answer: \boxed{$1.88 \times 10^{7}$} Extra Supplementary Reading Materials: The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational. B Neutrinos from the Sun : In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: $$ 4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}} $$ The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem. Context question: B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction. Context answer: \boxed{$6.8 \times 10^{14}$} Extra Supplementary Reading Materials: Travelling from the core of the Sun to the Earth, some of the electron neutrinos, $v_{e}$, are converted to other types of neutrinos, $v_{\mathrm{x}}$. The efficiency of the detector for detecting $v_{\mathrm{x}}$ is $1 / 6$ of its efficiency for detecting $v_{\mathrm{e}}$. If there is no neutrino conversion, we expect to detect an average of $N_{1}$ neutrinos in a year. However, due to the conversion, an average of $N_{2}$ neutrinos ( $v_{\mathrm{e}}$ and $v_{\mathrm{x}}$ combined) are actually detected per year. Context question: B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$. Context answer: \boxed{$f=\frac{6}{5}(1-\frac{N_{2}}{N_{1}})$} Extra Supplementary Reading Materials: In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, $n$ ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone.","B.3 Assume that an electron knocked out by a neutrino loses energy at a constant rate of $\alpha$ per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, $\Delta t$, determine the energy imparted to this electron ( $E_{\text {imparted }}$ ) by the neutrino, in terms of $\alpha, \Delta t, n, m_{\mathrm{e}}$ and $c$. (Assume the electron to be at rest before its interaction with the neutrino.)","['When the electron stops emitting Cherenkov radiation, its speed has reduced to $v_{\\text {stop }}=c / n$. Its total energy at this time is\n\n$$\nE_{\\text {stop }}=\\frac{m_{\\mathrm{e}} c^{2}}{\\sqrt{1-v_{\\text {stop }}^{2} / c^{2}}}=\\frac{n m_{\\mathrm{e}} c^{2}}{\\sqrt{n^{2}-1}}\n$$\n\nThe energy of the electron when it was knocked out is\n\n$$\nE_{\\text {start }}=\\alpha \\Delta t+\\frac{n m_{\\mathrm{e}} c^{2}}{\\sqrt{n^{2}-1}}\n$$\n\nBefore interacting, the energy of the electron was equal to $m_{\\mathrm{e}} c^{2}$.\n\nThus, the energy imparted by the neutrino is\n\n$$\nE_{\\text {imparted }}=E_{\\text {start }}-m_{\\mathrm{e}} c^{2}=\\alpha \\Delta t+\\left(\\frac{n}{\\sqrt{n^{2}-1}}-1\\right) m_{\\mathrm{e}} c^{2}\n$$']",['$E_{\\text {imparted }}=\\alpha \\Delta t+(\\frac{n}{\\sqrt{n^{2}-1}}-1) m_{\\mathrm{e}} c^{2}$'],False,,Expression, 1073,Modern Physics,"Particles from the Sun ${ }^{1}$ Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions. Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$. Note: (i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant (ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant (iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant A Radiation from the sun : Context question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface. Context answer: \boxed{$5.76 \times 10^{3} $} Extra Supplementary Reading Materials: The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by $$ u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right) $$ where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation. Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays. Context question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the
solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$ Context answer: \boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$} Context question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of
the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$} Extra Supplementary Reading Materials: The semiconducting material of the solar cell has a ""band gap"" of energy, $E_{\mathrm{g}}$. We assume the following model. Every photon of energy $E \geq E_{\mathrm{g}}$ excites an electron across the band gap. This electron contributes an energy, $E_{\mathrm{g}}$, as the useful output energy, and any extra energy is dissipated as heat (not converted to useful energy). Context question: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$,
$R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$. Context answer: \boxed{$P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$} Context question: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$. Context answer: \boxed{$\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$} Context question: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What
is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ? Context answer: \boxed{$\frac{1}{3}$ , 0} Context question: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the
value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$. Context answer: \boxed{$x_0=2.27$ , $\eta(2.27)=0.457$} Context question: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this
value. Context answer: \boxed{0.457} Extra Supplementary Reading Materials: In the late nineteenth century, Kelvin and Helmholtz $(\mathrm{KH})$ proposed a hypothesis to explain how the Sun shines. They postulated that starting as a very large cloud of matter of mass, $M_{\odot}$, and negligible density, the Sun has been shrinking continuously. The shining of the Sun would then be due to the release of gravitational potential energy through this slow contraction. Context question: A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$. Context answer: \boxed{$\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$} Context question: A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period. Context answer: \boxed{$1.88 \times 10^{7}$} Extra Supplementary Reading Materials: The $\tau_{\mathrm{KH}}$ calculated above does not match the age of the solar system estimated from studies of meteorites. This shows that the energy source of the Sun cannot be purely gravitational. B Neutrinos from the Sun : In 1938, Hans Bethe proposed that nuclear fusion of hydrogen into helium in the core of the Sun is the source of its energy. The net nuclear reaction is: $$ 4{ }^{1} \mathrm{H} \rightarrow{ }^{4} \mathrm{He}+2 \mathrm{e}^{+}+2 v_{\mathrm{e}} $$ The ""electron neutrinos"", $v_{\mathrm{e}}$, produced in this reaction may be taken to be massless. They escape the Sun and their detection on the Earth confirms the occurrence of nuclear reactions inside the Sun. Energy carried away by the neutrinos can be neglected in this problem. Context question: B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction. Context answer: \boxed{$6.8 \times 10^{14}$} Extra Supplementary Reading Materials: Travelling from the core of the Sun to the Earth, some of the electron neutrinos, $v_{e}$, are converted to other types of neutrinos, $v_{\mathrm{x}}$. The efficiency of the detector for detecting $v_{\mathrm{x}}$ is $1 / 6$ of its efficiency for detecting $v_{\mathrm{e}}$. If there is no neutrino conversion, we expect to detect an average of $N_{1}$ neutrinos in a year. However, due to the conversion, an average of $N_{2}$ neutrinos ( $v_{\mathrm{e}}$ and $v_{\mathrm{x}}$ combined) are actually detected per year. Context question: B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$. Context answer: \boxed{$f=\frac{6}{5}(1-\frac{N_{2}}{N_{1}})$} Extra Supplementary Reading Materials: In order to detect neutrinos, large detectors filled with water are constructed. Although the interactions of neutrinos with matter are very rare, occasionally they knock out electrons from water molecules in the detector. These energetic electrons move through water at high speeds, emitting electromagnetic radiation in the process. As long as the speed of such an electron is greater than the speed of light in water (refractive index, $n$ ), this radiation, called Cherenkov radiation, is emitted in the shape of a cone. Context question: B.3 Assume that an electron knocked out by a neutrino loses energy at a constant rate of $\alpha$ per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, $\Delta t$, determine the energy imparted to this electron ( $E_{\text {imparted }}$ ) by the neutrino, in terms of $\alpha, \Delta t, n, m_{\mathrm{e}}$ and $c$. (Assume the electron to be at rest before its interaction with the neutrino.) Context answer: \boxed{$E_{\text {imparted }}=\alpha \Delta t+(\frac{n}{\sqrt{n^{2}-1}}-1) m_{\mathrm{e}} c^{2}$} Extra Supplementary Reading Materials: The fusion of $\mathrm{H}$ into He inside the Sun takes place in several steps. Nucleus of ${ }^{7} \mathrm{Be}$ (rest mass, $m_{\mathrm{Be}}$ ) is produced in one of these intermediate steps. Subsequently, it can absorb an electron, producing a ${ }^{7} \mathrm{Li}$ nucleus (rest mass, $m_{\mathrm{Li}} Figure 1","A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.",['From the principle of Conservation of Mechanical Energy\n\n$$\n\\begin{gathered}\n\\frac{1}{2} m v_{1}^{2}=\\frac{1}{2} m v_{2}^{2}+V_{0} \\\\\nv_{2}=\\left(v_{1}^{2}-\\frac{2 V_{0}}{m}\\right)^{1 / 2}\n\\end{gathered}\n$$'],['$v_{2}=(v_{1}^{2}-\\frac{2 V_{0}}{m})^{1 / 2}$'],False,,Expression, 1075,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} ","A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$.",['At the boundary there is an impulsive force $(\\propto d V / d x)$ in the $-x$ direction. Hence only the velocity component in the $x$-direction $v_{1 x}$ suffers change. The component in the $y$-direction remains unchanged. Therefore\n\n$v_{1 y}=v_{2 y}$\n\n$v_{1} \\sin \\theta_{1}=v_{2} \\sin \\theta_{2}$'],['$v_2=\\frac{v_1 \\sin\\theta_1}{\\sin \\theta_2}$'],False,,Expression, 1076,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).","A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates.","['By definition $A(\\alpha)$ from $O$ to $P$ is\n\n$$\nA(\\alpha)=m v_{1} \\sqrt{x_{1}^{2}+\\alpha^{2}}+m v_{2} \\sqrt{\\left(x_{0}-x_{1}\\right)^{2}+\\left(y_{0}-\\alpha\\right)^{2}}\n$$\n\nDifferentiating w.r.t. $\\alpha$ and setting the derivative of $A(\\alpha)$ to zero\n\n$$\n\\begin{gathered}\n\\frac{v_{1} \\alpha}{\\left(x_{1}^{2}+\\alpha^{2}\\right)^{1 / 2}}-\\frac{v_{2}\\left(y_{0}-\\alpha\\right)}{\\left[\\left(x_{0}-x_{1}\\right)^{2}+\\left(y_{0}-\\alpha\\right)^{2}\\right]^{1 / 2}}=0 \\\\\n\\therefore \\frac{v_{1}}{v_{2}}=\\frac{\\left(y_{0}-\\alpha\\right)\\left(x_{1}^{2}+\\alpha^{2}\\right)^{1 / 2}}{\\alpha\\left[\\left(x_{0}-x_{1}\\right)^{2}+\\left(y_{0}-\\alpha\\right)^{2}\\right]^{1 / 2}}\n\\end{gathered}\n$$\n\nNote this is the same as A2, namely $v_{1} \\sin \\theta_{1}=v_{2} \\sin \\theta_{2}$.']",['$\\frac{v_{1}}{v_{2}}=\\frac{(y_{0}-\\alpha)(x_{1}^{2}+\\alpha^{2})^{1 / 2}}{\\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\\alpha)^{2}]^{1 / 2}}$'],False,,Expression, 1077,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2","B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle.","[""The speed of light in medium I is $c / n_{1}$ and in medium II is $c / n_{2}$, where $c$ is the speed of light in vacuum. Let the two media be separated by the fixed line $y=y_{1}$. Then time $T(\\alpha)$ for light to travel from origin $(0,0)$ and $\\left(x_{0}, y_{0}\\right)$ is\n\n$$\nT(\\alpha)=n_{1}\\left(\\sqrt{y_{1}^{2}+\\alpha^{2}}\\right) / c+n_{2}\\left(\\sqrt{\\left(x_{0}-\\alpha\\right)^{2}+\\left(y_{0}-y_{1}\\right)^{2}}\\right) / c\n$$\n\nDifferentiating w.r.t. $\\alpha$ and setting the derivative of $T(\\alpha)$ to zero\n\n$$\n\\begin{gathered}\n\\frac{n_{1} \\alpha}{\\left(y_{1}^{2}+\\alpha^{2}\\right)^{1 / 2}}-\\frac{n_{2}\\left(y_{0}-\\alpha\\right)}{\\left[\\left(x_{0}-\\alpha\\right)^{2}+\\left(y_{0}-y_{1}\\right)^{2}\\right]^{1 / 2}}=0 \\\\\n\\therefore n_{1} \\sin i_{1}=n_{2} \\sin i_{2}\n\\end{gathered}\n$$\n\n[Note: Derivation is similar to A3. This is Snell's law.]""]",['$n_{1} \\sin i_{1}=n_{2} \\sin i_{2}$'],False,,Equation, 1078,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution",B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$.,"[""From Snell's law $n_{0} \\sin i_{0}=n(y) \\sin i$\n\nThen,\n\n$$\n\\begin{array}{r}\n\\frac{d y}{d x}=-\\cot i \\\\\nn_{0} \\sin i_{0}=\\frac{n(y)}{\\sqrt{1+\\left(\\frac{d y}{d x}\\right)^{2}}} \\\\\n\\frac{d y}{d x}=-\\sqrt{\\left(\\frac{n(y)}{n_{0} \\sin i_{0}}\\right)^{2}-1}\n\\end{array}\n$$""]",['$\\frac{d y}{d x}=-\\sqrt{(\\frac{n(y)}{n_{0} \\sin i_{0}})^{2}-1}$'],False,,Expression, 1079,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} ","B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant","['$$\n\\int \\frac{d y}{\\sqrt{\\left(\\frac{n_{0}-k y}{n_{0} \\sin i_{0}}\\right)^{2}-1}}=-\\int d x\n$$\n\nNote $i_{0}=90^{\\circ}$ so $\\sin i_{0}=1$.\n\n\n\nMethod I We employ the substitution\n\n$$\n\\begin{gathered}\n\\xi=\\frac{n_{0}-k y}{n_{0}} \\\\\n\\int \\frac{d \\xi\\left(-\\frac{n_{0}}{k}\\right)}{\\sqrt{\\xi^{2}-1}}=-\\int d x\n\\end{gathered}\n$$\n\nLet $\\xi=\\sec \\theta$. Then\n\n$$\n\\frac{n_{0}}{k} \\ln (\\sec \\theta+\\tan \\theta)=x+c\n$$\n\nOr METHOD II \n\nWe employ the substition\n\n$$\n\\begin{gathered}\n\\xi=\\frac{n_{0}-k y}{n_{0}} \\\\\n\\int \\frac{d \\xi\\left(-\\frac{n_{0}}{k}\\right)}{\\sqrt{\\xi^{2}-1}}=-\\int d x \\\\\n-\\frac{n_{0}}{k} \\ln \\left(\\frac{n_{0}-k y}{n_{0}}+\\sqrt{\\left(\\frac{n_{0}-k y}{n_{0}}\\right)^{2}-1}\\right)=-x+c\n\\end{gathered}\n$$\n\nNow continuing\n\nConsidering the substitutions and boundary condition, $x=0$ for $y=0$ we obtain that the constant $c=0$.\n\nHence we obtain the following trajectory:\n\n$$\nx=\\frac{n_{0}}{k} \\ln \\left(\\frac{n_{0}-k y}{n_{0}}+\\sqrt{\\left(\\frac{n_{0}-k y}{n_{0}}\\right)^{2}-1}\\right)\n$$']",['$x=\\frac{n_{0}}{k} \\ln (\\frac{n_{0}-k y}{n_{0}}+\\sqrt{(\\frac{n_{0}-k y}{n_{0}})^{2}-1})$'],False,,Expression, 1080,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} ","B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$.",['Given $y_{0}=10.0 \\mathrm{~cm} . \\quad n_{0}=1.50 \\quad k=0.050 \\mathrm{~cm}^{-1}$\n\nFrom (B3)\n\n$$\nx_{0}=\\frac{n_{0}}{k} \\ln \\left[\\left(\\frac{n_{0}-k y}{n_{0}}\\right)+\\left(\\left(\\frac{n_{0}-k y}{n_{0}}\\right)^{2}-1\\right)^{1 / 2}\\right]\n$$\n\nHere $y=-y_{0}$\n\n\n\n$$\n\\begin{gathered}\nx_{0}=\\frac{n_{0}}{k} \\ln \\left[\\frac{\\left(n_{0}+k y_{0}\\right)}{n_{0}}+\\left(\\frac{\\left(n_{0}+k y_{0}\\right)^{2}}{n_{0}^{2}}-1\\right)^{1 / 2}\\right] \\\\\n=30 \\ln \\left[\\frac{2}{1.5}+\\left(\\left(\\frac{2}{1.5}\\right)^{2}-1\\right)^{1 / 2}\\right] \\\\\n=30 \\ln \\left[\\frac{4}{3}+\\left(\\frac{7}{9}\\right)^{1 / 2}\\right] \\\\\n=30 \\ln \\left[\\frac{4}{3}+0.88\\right] \\\\\n=24.0 \\mathrm{~cm}\n\\end{gathered}\n$$'],['24.0'],False,cm,Numerical,1e-1 1081,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} Context question: B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$. Context answer: \boxed{24.0} Extra Supplementary Reading Materials: C The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves.","C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase
of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant.",['From the de Broglie hypothesis\n\n$$\n\\lambda \\rightarrow \\lambda_{d B}=h / m v\n$$\n\nwhere $\\lambda$ is the de Broglie wavelength and the other symbols have their usual meaning\n\n$$\n\\begin{gathered}\n\\Delta \\phi=\\frac{2 \\pi}{\\lambda} \\Delta s \\\\\n=\\frac{2 \\pi}{h} m v \\Delta s \\\\\n=\\frac{2 \\pi \\Delta A}{h}\n\\end{gathered}\n$$'],['$\\Delta \\phi=\\frac{2 \\pi \\Delta A}{h}$'],False,,Expression, 1082,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} Context question: B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$. Context answer: \boxed{24.0} Extra Supplementary Reading Materials: C The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. Context question: C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase
of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant. Context answer: \boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$}","C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$. Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order. Figure 4",['\n\nConsider the extreme trajectories $O C P$ and $O D P$ of $(\\mathrm{C} 1)$\n\nThe geometrical path difference is $E D$ in region I and $C F$ in region II.\n\nThis implies (note: $d \\ll\\left(x_{0}-x_{1}\\right)$ and $d \\ll x_{1}$ )\n\n$$\n\\begin{gathered}\n\\Delta \\phi_{C D}=\\frac{2 \\pi d \\sin \\theta_{1}}{\\lambda_{1}}-\\frac{2 \\pi d \\sin \\theta_{2}}{\\lambda_{2}} \\\\\n\\Delta \\phi_{C D}=\\frac{2 \\pi m v_{1} d \\sin \\theta_{1}}{h}-\\frac{2 \\pi m v_{2} d \\sin \\theta_{2}}{h} \\\\\n=2 \\pi \\frac{m d}{h}\\left(v_{1} \\sin \\theta_{1}-v_{2} \\sin \\theta_{2}\\right) \\\\\n=0 \\quad(\\text { from } \\mathrm{A} 2 \\text { or } \\mathrm{B} 1)\n\\end{gathered}\n$$\n\nThus near the clasical path there is invariably constructive interference.\n\n'],['$\\Delta \\phi_{C D}=0$'],False,,Numerical,0 1082,Mechanics,,"C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$. Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order. ![](https://cdn.mathpix.com/cropped/2023_12_21_1329e5e361501978a8a6g-1.jpg?height=337&width=574&top_left_y=340&top_left_x=1249) Figure 4",['![](https://cdn.mathpix.com/cropped/2023_12_21_1329e5e361501978a8a6g-1.jpg?height=677&width=785&top_left_y=875&top_left_x=681)\n\nConsider the extreme trajectories $O C P$ and $O D P$ of $(\\mathrm{C} 1)$\n\nThe geometrical path difference is $E D$ in region I and $C F$ in region II.\n\nThis implies (note: $d \\ll\\left(x_{0}-x_{1}\\right)$ and $d \\ll x_{1}$ )\n\n$$\n\\begin{gathered}\n\\Delta \\phi_{C D}=\\frac{2 \\pi d \\sin \\theta_{1}}{\\lambda_{1}}-\\frac{2 \\pi d \\sin \\theta_{2}}{\\lambda_{2}} \\\\\n\\Delta \\phi_{C D}=\\frac{2 \\pi m v_{1} d \\sin \\theta_{1}}{h}-\\frac{2 \\pi m v_{2} d \\sin \\theta_{2}}{h} \\\\\n=2 \\pi \\frac{m d}{h}\\left(v_{1} \\sin \\theta_{1}-v_{2} \\sin \\theta_{2}\\right) \\\\\n=0 \\quad(\\text { from } \\mathrm{A} 2 \\text { or } \\mathrm{B} 1)\n\\end{gathered}\n$$\n\nThus near the clasical path there is invariably constructive interference.'],['$\\Delta \\phi_{C D}=0$'],False,,Numerical,0 1083,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} Context question: B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$. Context answer: \boxed{24.0} Extra Supplementary Reading Materials: C The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. Context question: C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase
of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant. Context answer: \boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$} Context question: C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$. Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order. Figure 4 Context answer: \boxed{$\Delta \phi_{C D}=0$} Extra Supplementary Reading Materials: D Matter Wave Interference Consider an electron gun at $\mathrm{O}$ which directs a collimated beam of electrons to a narrow slit at $\mathrm{F}$ in the opaque partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at $x=x_{1}$ such that OFP is a straight line. $\mathrm{P}$ is a point on the screen at $x=x_{0}$ (see Fig. 5). The speed in I is $v_{1}=2.0000 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ and $\theta=10.0000^{\circ}$. The potential in II is such that speed $v_{2}=$ $1.9900 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$. The distance $x_{0}-x_{1}$ is $250.00 \mathrm{~mm}$ $\left(1 \mathrm{~mm}=10^{-3} \mathrm{~m}\right)$. Ignore electron-electron interaction. Figure 5","D.1If the electrons at $\mathrm{O}$ have been accelerated from rest, calculate the accelerating potential $U_{1}$.",['$$\n\\begin{gathered}\nq U_{1}=\\frac{1}{2} m v^{2} \\\\\n=\\frac{9.11 \\times 10^{-31} \\times 4 \\times 10^{14}}{2} \\mathrm{~J} \\\\\n=2 \\times 9.11 \\times 10^{-17} \\mathrm{~J} \\\\\n=\\frac{2 \\times 9.11 \\times 10^{-17}}{1.6 \\times 10^{-19}} \\mathrm{eV} \\\\\n=1.139 \\times 10^{3} \\mathrm{eV}(\\simeq 1100 \\mathrm{eV}) \\\\\nU_{1}=1.139 \\times 10^{3} \\mathrm{~V}\n\\end{gathered}\n$$'],['$1.139 \\times 10^{3}$'],False,V,Numerical,1e0 1084,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} Context question: B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$. Context answer: \boxed{24.0} Extra Supplementary Reading Materials: C The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. Context question: C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase
of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant. Context answer: \boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$} Context question: C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$. Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order. Figure 4 Context answer: \boxed{$\Delta \phi_{C D}=0$} Extra Supplementary Reading Materials: D Matter Wave Interference Consider an electron gun at $\mathrm{O}$ which directs a collimated beam of electrons to a narrow slit at $\mathrm{F}$ in the opaque partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at $x=x_{1}$ such that OFP is a straight line. $\mathrm{P}$ is a point on the screen at $x=x_{0}$ (see Fig. 5). The speed in I is $v_{1}=2.0000 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ and $\theta=10.0000^{\circ}$. The potential in II is such that speed $v_{2}=$ $1.9900 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$. The distance $x_{0}-x_{1}$ is $250.00 \mathrm{~mm}$ $\left(1 \mathrm{~mm}=10^{-3} \mathrm{~m}\right)$. Ignore electron-electron interaction. Figure 5 Context question: D.1If the electrons at $\mathrm{O}$ have been accelerated from rest, calculate the accelerating potential $U_{1}$. Context answer: \boxed{$1.139 \times 10^{3}$} ","D.2Another identical slit $\mathrm{G}$ is made in the partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at a distance of $215.00 \mathrm{~nm}\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right)$ below slit
$\mathrm{F}$ (Fig. 5). If the phase difference between de Broglie waves arriving at $\mathrm{P}$ through the slits $\mathrm{F}$ and $\mathrm{G}$ is $2 \pi \beta$, calculate $\beta$.",['$$\n\\begin{aligned}\n& \\Delta \\phi_{P}=\\frac{2 \\pi d \\sin \\theta}{\\lambda_{1}}-\\frac{2 \\pi d \\sin \\theta}{\\lambda_{2}} \\\\\n& =2 \\pi\\left(v_{1}-v_{2}\\right) \\frac{m d}{h} \\sin 10^{\\circ}=2 \\pi \\beta\n\\end{aligned}\n$$\n\n$$\n\\beta=5.13\n$$'],['5.13'],False,,Numerical,1e-2 1085,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} Context question: B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$. Context answer: \boxed{24.0} Extra Supplementary Reading Materials: C The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. Context question: C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase
of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant. Context answer: \boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$} Context question: C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$. Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order. Figure 4 Context answer: \boxed{$\Delta \phi_{C D}=0$} Extra Supplementary Reading Materials: D Matter Wave Interference Consider an electron gun at $\mathrm{O}$ which directs a collimated beam of electrons to a narrow slit at $\mathrm{F}$ in the opaque partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at $x=x_{1}$ such that OFP is a straight line. $\mathrm{P}$ is a point on the screen at $x=x_{0}$ (see Fig. 5). The speed in I is $v_{1}=2.0000 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ and $\theta=10.0000^{\circ}$. The potential in II is such that speed $v_{2}=$ $1.9900 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$. The distance $x_{0}-x_{1}$ is $250.00 \mathrm{~mm}$ $\left(1 \mathrm{~mm}=10^{-3} \mathrm{~m}\right)$. Ignore electron-electron interaction. Figure 5 Context question: D.1If the electrons at $\mathrm{O}$ have been accelerated from rest, calculate the accelerating potential $U_{1}$. Context answer: \boxed{$1.139 \times 10^{3}$} Context question: D.2Another identical slit $\mathrm{G}$ is made in the partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at a distance of $215.00 \mathrm{~nm}\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right)$ below slit
$\mathrm{F}$ (Fig. 5). If the phase difference between de Broglie waves arriving at $\mathrm{P}$ through the slits $\mathrm{F}$ and $\mathrm{G}$ is $2 \pi \beta$, calculate $\beta$. Context answer: \boxed{5.13} ",D.3 What is the smallest distance $\Delta y$ from $\mathrm{P}$ at which null (zero) electron detection maybe expected on the
screen? [Note: you may find the approximation $\sin (\theta+\Delta \theta) \approx \sin \theta+\Delta \theta \cos \theta$ useful],['\n\nFrom previous part for null (zero) electron detection $\\Delta \\phi=5.5 \\times 2 \\pi$\n\n$$\n\\begin{aligned}\n& \\therefore m v_{1} \\frac{d \\sin \\theta}{h}-\\frac{m v_{2} d \\sin (\\theta+\\Delta \\theta)}{h}=5.5 \\\\\n& \\sin (\\theta+\\Delta \\theta)=\\frac{\\frac{m v_{1} d \\sin \\theta}{h}-5.5}{\\frac{m v_{2} d}{h}} \\\\\n& =\\frac{v_{1}}{v_{2}} \\sin \\theta-\\frac{h}{m} \\frac{5.5}{v_{2} d} \\\\\n& =\\frac{2}{1.99} \\sin 10^{\\circ}-\\frac{5.5}{1374.78 \\times 1.99 \\times 10^{7} \\times \\times 2.15 \\times 10^{-7}} \\\\\n& =0.174521-0.000935\n\\end{aligned}\n$$\n\nThis yields $\\Delta \\theta=-0.0036^{\\circ}$\n\nThe closest distance to $\\mathrm{P}$ is\n\n$$\n\\begin{aligned}\n\\Delta y & =\\left(x_{0}-x_{1}\\right)(\\tan (\\theta+\\Delta \\theta)-\\tan \\theta) \\\\\n& =250(\\tan 9.9964-\\tan 10) \\\\\n& =-0.0162 m m \\\\\n& =-16.2 \\mu \\mathrm{m}\n\\end{aligned}\n$$\n\nThe negative sign means that the closest minimum is below $\\mathrm{P}$.\n\nApproximate Calculation for $\\theta$ and $\\Delta y$\n\nUsing the approximation $\\sin (\\theta+\\Delta \\theta) \\approx \\sin \\theta+\\Delta \\theta \\cos \\theta$\n\nThe phase difference of $5.5 \\times 2 \\pi$ gives\n\n$$\nm v_{1} \\frac{d \\sin 10^{\\circ}}{h}-m v_{2} \\frac{d\\left(\\sin 10^{\\circ}+\\Delta \\theta \\cos 10^{\\circ}\\right)}{h}=5.5\n$$\n\nFrom solution of the previous part\n\n$$\nm v_{1} \\frac{d \\sin 10^{\\circ}}{h}-m v_{2} \\frac{d \\sin 10^{\\circ}}{h}=5.13\n$$\n\n\n\nTherefore\n\n$$\nm v_{2} \\frac{d \\Delta \\theta \\cos 10^{\\circ}}{h}=0.3700\n$$\n\nThis yields $\\Delta \\theta \\approx 0.0036^{\\circ}$\n\n$\\Delta y=-0.0162 \\mathrm{~mm}=-16.2 \\mu \\mathrm{m}$ as before'],['$-16.2 $'],False,$\mu \mathrm{m}$,Numerical,1e-1 1086,Mechanics,"A The Extremum Principle in Mechanics Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts). Figure 1 Context question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$. Context answer: \boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$} Context question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$. Context answer: \boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$} Extra Supplementary Reading Materials: We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA). Context question: A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant
potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and
$\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have
coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the
expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates. Context answer: \boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$} Extra Supplementary Reading Materials: B The Extremum Principle in Optics A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time. Figure 2 Context question: B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle. Context answer: \boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$} Extra Supplementary Reading Materials: Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height. Figure 3: Tank of Sugar Solution Context question: B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$. Context answer: \boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$} Context question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0] You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or
$\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant Context answer: \boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$} Context question: B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$,
$n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$. Context answer: \boxed{24.0} Extra Supplementary Reading Materials: C The Extremum Principle and the Wave Nature of Matter We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves. Context question: C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase
of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant. Context answer: \boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$} Context question: C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$. Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order. Figure 4 Context answer: \boxed{$\Delta \phi_{C D}=0$} Extra Supplementary Reading Materials: D Matter Wave Interference Consider an electron gun at $\mathrm{O}$ which directs a collimated beam of electrons to a narrow slit at $\mathrm{F}$ in the opaque partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at $x=x_{1}$ such that OFP is a straight line. $\mathrm{P}$ is a point on the screen at $x=x_{0}$ (see Fig. 5). The speed in I is $v_{1}=2.0000 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$ and $\theta=10.0000^{\circ}$. The potential in II is such that speed $v_{2}=$ $1.9900 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}$. The distance $x_{0}-x_{1}$ is $250.00 \mathrm{~mm}$ $\left(1 \mathrm{~mm}=10^{-3} \mathrm{~m}\right)$. Ignore electron-electron interaction. Figure 5 Context question: D.1If the electrons at $\mathrm{O}$ have been accelerated from rest, calculate the accelerating potential $U_{1}$. Context answer: \boxed{$1.139 \times 10^{3}$} Context question: D.2Another identical slit $\mathrm{G}$ is made in the partition $\mathrm{A}_{1} \mathrm{~B}_{1}$ at a distance of $215.00 \mathrm{~nm}\left(1 \mathrm{~nm}=10^{-9} \mathrm{~m}\right)$ below slit
$\mathrm{F}$ (Fig. 5). If the phase difference between de Broglie waves arriving at $\mathrm{P}$ through the slits $\mathrm{F}$ and $\mathrm{G}$ is $2 \pi \beta$, calculate $\beta$. Context answer: \boxed{5.13} Context question: D.3 What is the smallest distance $\Delta y$ from $\mathrm{P}$ at which null (zero) electron detection maybe expected on the
screen? [Note: you may find the approximation $\sin (\theta+\Delta \theta) \approx \sin \theta+\Delta \theta \cos \theta$ useful] Context answer: \boxed{$-16.2 $}","D.4 The beam has a square cross section of $500 \mathrm{~nm} \times 500 \mathrm{~nm}$ and the setup is $2 \mathrm{~m}$ long. What should be the
minimum flux density $I_{\min }$ (number of electrons per unit normal area per unit time) if, on an average, there
is at least one electron in the setup at a given time?",['The product of the speed of the electrons and number of electron per unit volume on an average yields the intensity.\n\nThus $N=1$ = Intensity $\\times$ Area $\\times$ Length/ Electron Speed\n\n$=I_{\\min } \\times 0.25 \\times 10^{-12} \\times 2 / 2 \\times 10^{7}$\n\nThis gives $I_{\\min }=4 \\times 10^{19} \\mathrm{~m}^{-2} \\mathrm{~s}^{-1}$\n\n'],['$I_{\\min }=4 \\times 10^{19}$'],False,$\mathrm{~m}^{-2} \mathrm{~s}^{-1}$,Numerical,1e18 1087,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ |",A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance.,"['The energy released during the transformation is\n\n$$\n\\Delta E=\\left[m\\left({ }^{235} \\mathrm{U}\\right)+m\\left({ }^{1} \\mathrm{n}\\right)-m\\left({ }^{94} \\mathrm{Zr}\\right)-m\\left({ }^{140} \\mathrm{Ce}\\right)-2 m\\left({ }^{1} \\mathrm{n}\\right)\\right] \\mathrm{c}^{2}\n$$\n\nSince the data is supplied in terms of unified atomic masses (u), we have\n\n$$\n\\Delta E=\\left[m\\left({ }^{235} \\mathrm{U}\\right)-m\\left({ }^{94} \\mathrm{Zr}\\right)-m\\left({ }^{140} \\mathrm{Ce}\\right)-m\\left({ }^{1} \\mathrm{n}\\right)\\right] \\mathrm{c}^{2}\n$$\n\n$$\n=208.684 \\mathrm{MeV} \\text { [Acceptable Range (208.000 to 209.000)] }\n$$\n\nfrom the given data.\n\n']",['$\\Delta E=208.684 $'],False,$\mathrm{MeV}$,Numerical,7e-1 1088,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} ",A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$.,"['The number of $\\mathrm{UO}_{2}$ molecules per $\\mathrm{m}^{3}$ of the fuel $N_{1}$ is given in the terms of its density $\\rho$, the Avogadro number $N_{A}$ and the average molecular weight $M_{w}$ as\n\n$$\n\\begin{aligned}\nN_{1} & =\\frac{\\rho N_{A}}{M_{w}} \\\\\n& =\\frac{10600 \\times 6.022 \\times 10^{23}}{0.270}=2.364 \\times 10^{28} \\mathrm{~m}^{-3}\n\\end{aligned}\n$$\n\nEach molecule of $\\mathrm{UO}_{2}$ contains one uranium atom. Since only $0.72 \\%$ of these are ${ }^{235} \\mathrm{U}$,\n\n$N=0.0072 \\times N_{1}$\n\n$=1.702 \\times 10^{26} \\mathrm{~m}^{-3}$\n\n[Acceptable Range (1.650 to 1.750.)]']",['$=1.702 \\times 10^{26} $'],False,$\mathrm{~m}^{-3}$,Numerical,5e24 1089,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} ","A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$.",['It is given that $80 \\%$ of the fission energy is available as heat thus the heat energy available per fission $E_{f}$ is from a-(i)\n\n$E_{f}=0.8 \\times 208.7 \\mathrm{MeV}$\n\n$=166.96 \\mathrm{MeV}$\n\n$=2.675 \\times 10^{-11} \\mathrm{~J}$\n\nThe total cross-section per unit volume is $N \\times \\sigma_{f}$. Thus the heat produced per unit volume per unit time Q is \n$$\n\\begin{aligned}\nQ & =N\\times \\sigma_f \\times \\phi \\times E_f \\\\\n& =(1.702\\times 10^{26})\\times(5.4\\times 10^{-26})\\times (2\\times 10^{18})\\times (2.675\\times 10^{-11}) \\mathrm{~W}/\\mathrm{m}^3 \\\\\n&=4.917\\times 10^8 \\mathrm{~W}/\\mathrm{m}^3\n\\end{aligned}\n$$\n\nAcceptable Range(4.800 to 5.000)'],['$4.917\\times 10^8$'],False,$ \mathrm{~W}/\mathrm{m}^3$,Numerical,1e7 1090,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} ","A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis.","['The dimensions of $T_{c}-T_{s}$ is temperature. We write this as $T_{c}-T_{s}=[K]$. Once can similarly write down the dimensions of $Q, a$ and $\\lambda$. Equating the temperature to powers of $Q, a$ and $\\lambda$, one could state the following dimensional equation:\n\n$K=Q^{\\alpha} a^{\\beta} \\lambda^{\\gamma}$\n\n$=\\left[M L^{-1} T^{-3}\\right]^{\\alpha}[L]^{\\beta}\\left[M L^{1} T^{-3} K^{-1}\\right]^{\\gamma}$\n\nThis yields the following algebraic equations\n\n$\\gamma=-1$ equating powers of temperature\n\n$\\alpha+\\gamma=0$ equating powers of mass or time. From the previous equation we get $\\alpha=1$\n\nNext $-\\alpha+\\beta+\\gamma=0$ equating powers of length. This yields $\\beta=2$.\n\nThus we obtain $T_{c}-T_{s}=\\frac{Q a^{2}}{4 \\lambda}$ where we insert the dimensionless factor $1 / 4$ as suggested in the problem. No penalty if the factor $1 / 4$ is not written.']",['$T_{c}-T_{s}=\\frac{Q a^{2}}{4 \\lambda}$'],False,,Expression, 1091,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} Context question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis. Context answer: \boxed{$T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$} ",A.5 The desired temperature of the coolant is $5.770 \times 10^{2} \mathrm{~K}$. Estimate the upper limit $a_{u}$ on the radius $a$ of the pin.,"['The melting point of $\\mathrm{UO}_{2}$ is $3138 \\mathrm{~K}$ and the maximum temperature of the coolant is $577 \\mathrm{~K}$. This sets a limit on the maximum permissible temperature $\\left(T_{c}-T_{s}\\right)$ to be less than $(3138-577=2561 \\mathrm{~K})$ to avoid ""meltdown"". Thus one may take a maximum of $\\left(T_{c}-T_{s}\\right)=2561 \\mathrm{~K}$.\n\nNoting that $\\lambda=3.28 \\mathrm{~W} / \\mathrm{m}-\\mathrm{K}$, we have\n\n$$\na_{u}^{2}=\\frac{2561 \\times 4 \\times 3.28}{4.917 \\times 10^{8}}\n$$\n\nWhere we have used the value of $Q$ from A2. This yields $a_{u} \\simeq 8.267 \\times 10^{-3} \\mathrm{~m}$. So $a_{u}=8.267 \\times 10^{-3} \\mathrm{~m}$ constitutes an upper limit on the radius of the fuel pin.']",['$a_{u}=8.267 \\times 10^{-3}$'],False,m,Numerical,1e-5 1092,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} Context question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis. Context answer: \boxed{$T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$} Context question: A.5 The desired temperature of the coolant is $5.770 \times 10^{2} \mathrm{~K}$. Estimate the upper limit $a_{u}$ on the radius $a$ of the pin. Context answer: \boxed{$a_{u}=8.267 \times 10^{-3}$} Extra Supplementary Reading Materials: B The Moderator Consider the two dimensional elastic collision between a neutron of mass $1 \mathrm{u}$ and a moderator atom of mass $A$ u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let $\overrightarrow{v_{b}}$ and $\overrightarrow{v_{a}}$ be the velocities of the neutron before and after collision respectively in the LF. Let $\overrightarrow{v_{m}}$ be the velocity of the center of mass (CM) frame relative to LF and $\theta$ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. Context question: B.1 The collision in LF is shown schematically with $\theta_{L}$ as the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1,2 and 3 in terms of $\overrightarrow{v_{b}}, \overrightarrow{v_{a}}$ and $\overrightarrow{v_{m}}$. Indicate the scattering angle $\theta$. Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision Context answer: ","B.2 Obtain $v$ and $V$, the speeds of the neutron and the moderator atom in the CM frame after the collision, in terms of $A$ and $v_{b}$.","['Before the collision in the CM frame $\\left(v_{b}-v_{m}\\right)$ and $v_{m}$ will be magnitude of the velocities of the neutron and moderator atom respectively. From momentum conservation in the CM frame, $v_{b}-v_{m}=A v_{m}$ gives $v_{m}=\\frac{v_{b}}{A+1}$.\n\nAfter the collision, let $v$ and $V$ be magnitude of the velocities of neutron and moderator atom respectively in the CM frame. From conservation laws,\n\n$$\nv=A V \\quad \\text { and } \\quad \\frac{1}{2}\\left(v_{b}-v_{m}\\right)^{2}+\\frac{1}{2} A v_{m}^{2}=\\frac{1}{2} v^{2}+\\frac{1}{2} A V^{2} \\cdot(\\rightarrow[0.2+0.2])\n$$\n\n\n\nSolving gives $v=\\frac{A v_{b}}{A+1}$ and $V=\\frac{v_{b}}{A+1}$. (OR) From definition of center of mass frame $v_{m}=\\frac{v_{b}}{A+1}$. Before the collision in the CM frame $v_{b}-v_{m}=\\frac{A v_{b}}{A+1}$ and $v_{m}$ will be magnitude of the velocities of the neutron and moderator atom respectively. In elastic collision the particles are scattered in the opposite direction in the CM frame and so the speeds remain same $v=\\frac{A v_{b}}{A+1}$ and $V=\\frac{v_{b}}{A+1}(\\rightarrow[0.2+0.1])$.']","['$v=\\frac{A v_{b}}{A+1}$ , $V=\\frac{v_{b}}{A+1}$']",True,,Expression, 1093,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} Context question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis. Context answer: \boxed{$T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$} Context question: A.5 The desired temperature of the coolant is $5.770 \times 10^{2} \mathrm{~K}$. Estimate the upper limit $a_{u}$ on the radius $a$ of the pin. Context answer: \boxed{$a_{u}=8.267 \times 10^{-3}$} Extra Supplementary Reading Materials: B The Moderator Consider the two dimensional elastic collision between a neutron of mass $1 \mathrm{u}$ and a moderator atom of mass $A$ u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let $\overrightarrow{v_{b}}$ and $\overrightarrow{v_{a}}$ be the velocities of the neutron before and after collision respectively in the LF. Let $\overrightarrow{v_{m}}$ be the velocity of the center of mass (CM) frame relative to LF and $\theta$ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. Context question: B.1 The collision in LF is shown schematically with $\theta_{L}$ as the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1,2 and 3 in terms of $\overrightarrow{v_{b}}, \overrightarrow{v_{a}}$ and $\overrightarrow{v_{m}}$. Indicate the scattering angle $\theta$. Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision Context answer: Context question: B.2 Obtain $v$ and $V$, the speeds of the neutron and the moderator atom in the CM frame after the collision, in terms of $A$ and $v_{b}$. Context answer: \boxed{$v=\frac{A v_{b}}{A+1}$ , $V=\frac{v_{b}}{A+1}$} ","B.3 Derive an expression for $G(\alpha, \theta)=E_{a} / E_{b}$, where $E_{b}$ and $E_{a}$ are the kinetic energies of the neutron, in the LF, before and after the collision respectively, and $\alpha \equiv[(A-1) /(A+1)]^{2}$,","['Since $\\overrightarrow{v_{a}}=\\vec{v}+\\overrightarrow{v_{m}}, v_{a}^{2}=v^{2}+v_{m}^{2}+2 v v_{m} \\cos \\theta(\\rightarrow[0.3])$. Substituting the values of $v$ and $v_{m}, v_{a}^{2}=\\frac{A^{2} v_{b}^{2}}{(A+1)^{2}}+\\frac{v_{b}^{2}}{(A+1)^{2}}+\\frac{2 A v_{b}^{2}}{(A+1)^{2}} \\cos \\theta(\\rightarrow[0.2])$, so\n\n## Alternate form\n\n$$\nG(\\alpha, \\theta)=\\frac{A^{2}+1}{(A+1)^{2}}+\\frac{2 A}{(A+1)^{2}} \\cos \\theta=\\frac{1}{2}[(1+\\alpha)+(1-\\alpha) \\cos \\theta] .\n$$\n\n$$\n=1-\\frac{(1-\\alpha)(1-\\cos \\theta)}{2}\n$$']","['$G(\\alpha, \\theta)=1-\\frac{(1-\\alpha)(1-\\cos \\theta)}{2}$']",False,,Expression, 1094,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} Context question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis. Context answer: \boxed{$T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$} Context question: A.5 The desired temperature of the coolant is $5.770 \times 10^{2} \mathrm{~K}$. Estimate the upper limit $a_{u}$ on the radius $a$ of the pin. Context answer: \boxed{$a_{u}=8.267 \times 10^{-3}$} Extra Supplementary Reading Materials: B The Moderator Consider the two dimensional elastic collision between a neutron of mass $1 \mathrm{u}$ and a moderator atom of mass $A$ u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let $\overrightarrow{v_{b}}$ and $\overrightarrow{v_{a}}$ be the velocities of the neutron before and after collision respectively in the LF. Let $\overrightarrow{v_{m}}$ be the velocity of the center of mass (CM) frame relative to LF and $\theta$ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. Context question: B.1 The collision in LF is shown schematically with $\theta_{L}$ as the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1,2 and 3 in terms of $\overrightarrow{v_{b}}, \overrightarrow{v_{a}}$ and $\overrightarrow{v_{m}}$. Indicate the scattering angle $\theta$. Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision Context answer: Context question: B.2 Obtain $v$ and $V$, the speeds of the neutron and the moderator atom in the CM frame after the collision, in terms of $A$ and $v_{b}$. Context answer: \boxed{$v=\frac{A v_{b}}{A+1}$ , $V=\frac{v_{b}}{A+1}$} Context question: B.3 Derive an expression for $G(\alpha, \theta)=E_{a} / E_{b}$, where $E_{b}$ and $E_{a}$ are the kinetic energies of the neutron, in the LF, before and after the collision respectively, and $\alpha \equiv[(A-1) /(A+1)]^{2}$, Context answer: \boxed{$G(\alpha, \theta)=1-\frac{(1-\alpha)(1-\cos \theta)}{2}$} ",B.4 Assume that the above expression holds for $\mathrm{D}_{2} \mathrm{O}$ molecule. Calculate the maximum possible fractional energy loss $f_{l} \equiv \frac{E_{b}-E_{a}}{E_{b}}$ of the neutron for the $\mathrm{D}_{2} \mathrm{O}(20 \mathrm{u})$ moderator.,"['The maximum energy loss will be when the collision is head on ie., $E_{a}$ will be minimum for the scattering angle $\\theta=\\pi$.\n\nSo $E_{a}=E_{\\min }=\\alpha E_{b}$.\n\nFor $\\mathrm{D}_{2} \\mathrm{O}, \\alpha=0.819$ and maximum fractional $\\operatorname{loss}\\left(\\frac{E_{b}-E_{\\min }}{E_{b}}\\right)=1-\\alpha=0.181$. [A $\\boldsymbol{c}$ ceptable Range (0.170 to 0.190 )]']",['$f_{l}=0.181$'],False,,Numerical,1e-2 1095,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} Context question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis. Context answer: \boxed{$T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$} Context question: A.5 The desired temperature of the coolant is $5.770 \times 10^{2} \mathrm{~K}$. Estimate the upper limit $a_{u}$ on the radius $a$ of the pin. Context answer: \boxed{$a_{u}=8.267 \times 10^{-3}$} Extra Supplementary Reading Materials: B The Moderator Consider the two dimensional elastic collision between a neutron of mass $1 \mathrm{u}$ and a moderator atom of mass $A$ u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let $\overrightarrow{v_{b}}$ and $\overrightarrow{v_{a}}$ be the velocities of the neutron before and after collision respectively in the LF. Let $\overrightarrow{v_{m}}$ be the velocity of the center of mass (CM) frame relative to LF and $\theta$ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. Context question: B.1 The collision in LF is shown schematically with $\theta_{L}$ as the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1,2 and 3 in terms of $\overrightarrow{v_{b}}, \overrightarrow{v_{a}}$ and $\overrightarrow{v_{m}}$. Indicate the scattering angle $\theta$. Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision Context answer: Context question: B.2 Obtain $v$ and $V$, the speeds of the neutron and the moderator atom in the CM frame after the collision, in terms of $A$ and $v_{b}$. Context answer: \boxed{$v=\frac{A v_{b}}{A+1}$ , $V=\frac{v_{b}}{A+1}$} Context question: B.3 Derive an expression for $G(\alpha, \theta)=E_{a} / E_{b}$, where $E_{b}$ and $E_{a}$ are the kinetic energies of the neutron, in the LF, before and after the collision respectively, and $\alpha \equiv[(A-1) /(A+1)]^{2}$, Context answer: \boxed{$G(\alpha, \theta)=1-\frac{(1-\alpha)(1-\cos \theta)}{2}$} Context question: B.4 Assume that the above expression holds for $\mathrm{D}_{2} \mathrm{O}$ molecule. Calculate the maximum possible fractional energy loss $f_{l} \equiv \frac{E_{b}-E_{a}}{E_{b}}$ of the neutron for the $\mathrm{D}_{2} \mathrm{O}(20 \mathrm{u})$ moderator. Context answer: \boxed{$f_{l}=0.181$} Extra Supplementary Reading Materials: C The Nuclear Reactor To operate the NR at any constant neutron flux $\psi$ (steady state), the leakage of neutrons has to be compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical geometry the leakage rate is $k_{1}\left[(2.405 / R)^{2}+(\pi / H)^{2}\right] \psi$ and the excess production rate is $k_{2} \psi$. The constants $k_{1}$ and $k_{2}$ depend on the material properties of the NR.","C.1 Consider a NR with $k_{1}=1.021 \times 10^{-2} \mathrm{~m}$ and $k_{2}=8.787 \times 10^{-3} \mathrm{~m}^{-1}$. Noting that for a fixed volume the leakage rate is to be minimized for efficient fuel utilization, obtain the dimensions of the NR in the steady state. in the reactor and the mass $M$ of $\mathrm{UO}_{2}$ required to operate the $\mathrm{NR}$ in steady state.","['For constant volume $V=\\pi R^{2} H$,\n\n$$\n\\begin{gathered}\n\\frac{d}{d H}\\left[\\left(\\frac{2.405}{R}\\right)^{2}+\\left(\\frac{\\pi}{H}\\right)^{2}\\right]=0 \\\\\n\\frac{d}{d H}\\left[\\frac{2.405^{2} \\pi H}{V}+\\frac{\\pi^{2}}{H^{2}}\\right]=\\frac{2.405^{2} \\pi}{V}-2 \\frac{\\pi^{2}}{H^{3}}=0\n\\end{gathered}\n$$\n\ngives $\\left(\\frac{2.405}{R}\\right)^{2}=2\\left(\\frac{\\pi}{H}\\right)^{2}$.\n\nFor steady state,\n\n$$\n1.021 \\times 10^{-2}\\left[\\left(\\frac{2.405}{R}\\right)^{2}+\\left(\\frac{\\pi}{H}\\right)^{2}\\right] \\Psi=8.787 \\times 10^{-3} \\Psi\n$$\n\nHence $H=5.866 \\mathrm{~m}$ [Acceptable Range (5.870 to 5.890)]\n\n$R=3.175 \\mathrm{~m}$ [Acceptable Range (3.170 to 3.180)].\n\n## Alternative Non-Calculus Method to Optimize\n\nMinimisation of the expression $\\left(\\frac{2.405}{R}\\right)^{2}+\\left(\\frac{\\pi}{H}\\right)^{2}$, for a fixed volume $V=$ $\\pi R^{2} H$ :\n\nSubstituting for $R^{2}$ in terms of $V, H$ we get $\\frac{2.405^{2} \\pi H}{V}+\\frac{\\pi^{2}}{H^{2}}$,\n\nwhich can be written as, $\\frac{2.405^{2} \\pi H}{2 V}+\\frac{2.405^{2} \\pi H}{2 V}+\\frac{\\pi^{2}}{H^{2}}$.\n\nSince all the terms are positive applying AMGM inequality for three positive terms we get\n\n$$\n\\frac{\\frac{2.405^{2} \\pi H}{2 V}+\\frac{2.405^{2} \\pi H}{2 V}+\\frac{\\pi^{2}}{H^{2}}}{3} \\geq \\sqrt[3]{\\frac{2.405^{2} \\pi H}{2 V} \\times \\frac{2.405^{2} \\pi H}{2 V} \\times \\frac{\\pi^{2}}{H^{2}}}=\\sqrt[3]{\\frac{2.405^{4} \\pi^{4}}{4 V^{2}}}\n$$\n\n\n\nThe RHS is a constant. The LHS is always greater or equal to this constant implies that this is the minimum value the LHS can achieve. The minimum is achieved when all the three positive terms are equal, which gives the condition $\\frac{2.405^{2} \\pi H}{2 \\mathrm{~V}}=$ $\\frac{\\pi^{2}}{H^{2}} \\Rightarrow\\left(\\frac{2.405}{R}\\right)^{2}=2\\left(\\frac{\\pi}{H}\\right)^{2}$.\n\nFor steady state,\n\n$$\n1.021 \\times 10^{-2}\\left[\\left(\\frac{2.405}{R}\\right)^{2}+\\left(\\frac{\\pi}{H}\\right)^{2}\\right] \\Psi=8.787 \\times 10^{-3} \\Psi\n$$\n\nHence $H=5.866 \\mathrm{~m}$ [Acceptable Range (5.870 to 5.890)]\n\n$R=3.175 \\mathrm{~m}$ [Acceptable Range (3.170 to 3.180)].\n\nNote: Putting the condition in the RHS gives the minimum as $\\frac{\\pi^{2}}{H^{2}}$. From the condition we get $\\frac{\\pi^{3}}{H^{3}}=\\frac{2.405^{2} \\pi^{2}}{2 V} \\Rightarrow \\frac{\\pi^{2}}{H^{2}}=\\sqrt[3]{\\frac{2.405^{4} \\pi^{4}}{4 V^{2}}}$.\n\nNote: The radius and height of the Tarapur 3 \\& 4 NR in Western India is $3.192 \\mathrm{~m}$ and $5.940 \\mathrm{~m}$ respectively.']","['$R=3.175$ , $H=5.866$']",True,,Numerical,1e-3 1096,Modern Physics,"Uranium occurs in nature as $\mathrm{UO}_{2}$ with only $0.720 \%$ of the uranium atoms being ${ }^{235} \mathrm{U}$. Neutron induced fission occurs readily in ${ }^{235} \mathrm{U}$ with the emission of 2-3 fission neutrons having high kinetic energy. This fission probability will increase if the neutrons inducing fission have low kinetic energy. So by reducing the kinetic energy of the fission neutrons, one can induce a chain of fissions in other ${ }^{235} \mathrm{U}$ nuclei. This forms the basis of the power generating nuclear reactor (NR). A typical NR consists of a cylindrical tank of height $H$ and radius $R$ filled with a material called moderator. Cylindrical tubes, called fuel channels, each containing a cluster of cylindrical fuel pins of natural $\mathrm{UO}_{2}$ in solid form of height $H$, are kept axially in a square array. Fission neutrons, coming outward from a fuel channel, collide with the moderator, losing energy and reach the surrounding fuel channels with low enough energy to cause fission (Figs I-III). Heat generated from fission in the pin is transmitted to a coolant fluid flowing along its length. In the current problem we shall study some of the physics behind the (A) Fuel Pin, (B) Moderator and (C) NR of cylindrical geometry. # Schematic sketch of the Nuclear Reactor (NR) Fig-I: Enlarged view of a fuel channel (1-Fuel Pins) Fig-II: A view of the NR (2-Fuel Channels) Fig-III: Top view of NR (3-Square Arrangement of Fuel Channels and 4-Typical Neutron Paths). Only components relevant to the problem are shown (e.g. control rods and coolant are not shown). ## A Fuel Pin | Data
for $\mathrm{UO}_{2}$ | 1. Molecular weight $M_{w}=0.270 \mathrm{~kg} \mathrm{~mol}^{-1}$ | 2. | Density $\rho=1.060 \times 10^{4} \mathrm{~kg} \mathrm{~m}^{-3}$ | | :---: | :--- | :--- | :--- | | | 3. Melting point $T_{m}=3.138 \times 10^{3} \mathrm{~K}$ | 4. | Thermal conductivity $\lambda=3.280 \mathrm{~W} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ | Context question: A.1 Consider the following fission reaction of a stationary ${ }^{235} \mathrm{U}$ after it absorbs a neutron of negligible kinetic
energy.
$\qquad{ }^{235} \mathrm{U}+{ }^{1} \mathrm{n} \rightarrow{ }^{94} \mathrm{Zr}+{ }^{140} \mathrm{Ce}+2{ }^{1} \mathrm{n}+\Delta E$
$m\left({ }^{94} \mathrm{Zr}\right)=93.9063 \mathrm{u} ; m\left({ }^{140} \mathrm{Ce}\right)=139.905 \mathrm{u} ; m\left({ }^{1} \mathrm{n}\right)=1.00867$ u and $1 \mathrm{u}=931.502 \mathrm{MeV} \mathrm{c}{ }^{-2}$. Ignore charge
imbalance. Context answer: \boxed{$\Delta E=208.684 $} Context question: A.2 Estimate $N$ the number of ${ }^{235} \mathrm{U}$ atoms per unit volume in natural $\mathrm{UO}_{2}$. Context answer: \boxed{$=1.702 \times 10^{26} $} Context question: A.3 Assume that the neutron flux $\phi=2.000 \times 10^{18} \mathrm{~m}^{-2} \mathrm{~s}^{-1}$ on the fuel is uniform. The fission cross-section (effective area of the target nucleus) of a ${ }^{235} \mathrm{U}$ nucleus is $\sigma_{f}=5.400 \times 10^{-26}$ $\mathrm{m}^{2}$. If $80.00 \%$ of the fission energy is available as heat, estimate $Q$ (in $\mathrm{W} \mathrm{m}^{-3}$ ) the rate of heat production in the pin per unit volume. $1 \mathrm{MeV}=1.602 \times 10^{-13} \mathrm{~J}$. Context answer: \boxed{$4.917\times 10^8$} Context question: A.4 The steady-state temperature difference between the center $\left(T_{c}\right)$ and the surface $\left(T_{s}\right)$ of the pin can be expressed as $T_{c}-T_{s}=k F(Q, a, \lambda)$ where $k=1 / 4$ is a dimensionless constant and $a$ is the radius of the pin. Obtain $F(Q, a, \lambda)$ by dimensional analysis. Context answer: \boxed{$T_{c}-T_{s}=\frac{Q a^{2}}{4 \lambda}$} Context question: A.5 The desired temperature of the coolant is $5.770 \times 10^{2} \mathrm{~K}$. Estimate the upper limit $a_{u}$ on the radius $a$ of the pin. Context answer: \boxed{$a_{u}=8.267 \times 10^{-3}$} Extra Supplementary Reading Materials: B The Moderator Consider the two dimensional elastic collision between a neutron of mass $1 \mathrm{u}$ and a moderator atom of mass $A$ u. Before collision all the moderator atoms are considered at rest in the laboratory frame (LF). Let $\overrightarrow{v_{b}}$ and $\overrightarrow{v_{a}}$ be the velocities of the neutron before and after collision respectively in the LF. Let $\overrightarrow{v_{m}}$ be the velocity of the center of mass (CM) frame relative to LF and $\theta$ be the neutron scattering angle in the CM frame. All the particles involved in collisions are moving at nonrelativistic speeds. Context question: B.1 The collision in LF is shown schematically with $\theta_{L}$ as the scattering angle (Fig-IV). Sketch the collision schematically in CM frame. Label the particle velocities for 1,2 and 3 in terms of $\overrightarrow{v_{b}}, \overrightarrow{v_{a}}$ and $\overrightarrow{v_{m}}$. Indicate the scattering angle $\theta$. Collision in the Laboratory Frame 1-Neutron before collision 2-Neutron after collision 3-Moderator Atom before collision 4-Moderator Atom after collision Context answer: Context question: B.2 Obtain $v$ and $V$, the speeds of the neutron and the moderator atom in the CM frame after the collision, in terms of $A$ and $v_{b}$. Context answer: \boxed{$v=\frac{A v_{b}}{A+1}$ , $V=\frac{v_{b}}{A+1}$} Context question: B.3 Derive an expression for $G(\alpha, \theta)=E_{a} / E_{b}$, where $E_{b}$ and $E_{a}$ are the kinetic energies of the neutron, in the LF, before and after the collision respectively, and $\alpha \equiv[(A-1) /(A+1)]^{2}$, Context answer: \boxed{$G(\alpha, \theta)=1-\frac{(1-\alpha)(1-\cos \theta)}{2}$} Context question: B.4 Assume that the above expression holds for $\mathrm{D}_{2} \mathrm{O}$ molecule. Calculate the maximum possible fractional energy loss $f_{l} \equiv \frac{E_{b}-E_{a}}{E_{b}}$ of the neutron for the $\mathrm{D}_{2} \mathrm{O}(20 \mathrm{u})$ moderator. Context answer: \boxed{$f_{l}=0.181$} Extra Supplementary Reading Materials: C The Nuclear Reactor To operate the NR at any constant neutron flux $\psi$ (steady state), the leakage of neutrons has to be compensated by an excess production of neutrons in the reactor. For a reactor in cylindrical geometry the leakage rate is $k_{1}\left[(2.405 / R)^{2}+(\pi / H)^{2}\right] \psi$ and the excess production rate is $k_{2} \psi$. The constants $k_{1}$ and $k_{2}$ depend on the material properties of the NR. Context question: C.1 Consider a NR with $k_{1}=1.021 \times 10^{-2} \mathrm{~m}$ and $k_{2}=8.787 \times 10^{-3} \mathrm{~m}^{-1}$. Noting that for a fixed volume the leakage rate is to be minimized for efficient fuel utilization, obtain the dimensions of the NR in the steady state. in the reactor and the mass $M$ of $\mathrm{UO}_{2}$ required to operate the $\mathrm{NR}$ in steady state. Context answer: \boxed{$R=3.175$ , $H=5.866$}",C.2 The fuel channels are in a square arrangement (Fig-III) with nearest neighbour distance $0.286 \mathrm{~m}$. The effective radius of a fuel channel (if it were solid) is $3.617 \times 10^{-2} \mathrm{~m}$. Estimate the number of fuel channels $F_{n}$ in the reactor and the mass $M$ of $\mathrm{UO}_{2}$ required to operate the NR in steady state.,"['Since the fuel channels are in square pitch of $0.286 \\mathrm{~m}$, the effective area per channel is $0.286^{2} \\mathrm{~m}^{2}=8.180 \\times 10^{-2} \\mathrm{~m}^{2}$.\n\nThe cross-sectional area of the core is $\\pi R^{2}=3.142 \\times(3.175)^{2}=31.67 \\mathrm{~m}^{2}$, so the maximum number of fuel channels that can be accommodated in the cylinder is the integer part of $\\frac{31.67}{0.0818}=387$.\n\nMass of the fuel $=387 \\times$ Volume of the $\\operatorname{rod} \\times$ density\n\n$$\n=387 \\times\\left(\\pi \\times 0.03617^{2} \\times 5.866\\right) \\times 10600=9.892 \\times 10^{4} \\mathrm{~kg}\n$$\n\n$F_{n}=387$ [Acceptable Range (380 to 394)]\n\n$M=9.892 \\times 10^{4} \\mathrm{~kg}$ [Acceptable Range (9.000 to $\\mathbf{1 0 . 0 0 ) ]}$\n\nNote 1: (Not part of grading) The total volume of the fuel is $387 \\times\\left(\\pi \\times 0.03617^{2} \\times\\right.$ $5.866)=9.332 \\mathrm{~m}^{3}$. If the reactor works at $12.5 \\%$ efficieny then using the result of a-(iii) we have that the power output of the reactor is $9.332 \\times 4.917 \\times 10^{8} \\times 0.125=$\n\n\n\n$573 \\mathrm{MW}$.\n\nNote 2: The Tarapur 3 \\& 4 NR in Western India has 392 channels and the mass of the fuel in it is $10.15 \\times 10^{4} \\mathrm{~kg}$. It produces $540 \\mathrm{MW}$ of power.\n\n']","['$F_{n}=387$ , $M=9.892 \\times 10^{4}$']",True,",kg",Numerical,"1e0,1e2" 1097,Mechanics,,"A small puck of mass $m$ is carefully placed onto the inner surface of the thin hollow thin cylinder of mass $M$ and of radius $R$. Initially, the cylinder rests on the horizontal plane and the puck is located at the height $R$ above the plane as shown in the figure on the left. Find the interaction force $F$ between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory. Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is $g$.","['Consider the forces acting on the puck and the cylinder and depicted in the figure on the right. The puck is subject to the gravity force $m g$ and the reaction force from the cylinder $N$. The cylinder is subject to the gravity force $M g$, the reaction force from the plane $N_{1}$, the friction force $F_{f r}$ and the pressure force from the puck $N^{\\prime}=-N$. The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows\n\n$$\nm a_{x}=N \\sin \\alpha,\n\\tag{A.1}\n$$\n\nwhere $a_{x}$ is the horizontal projection of the puck acceleration.\n\nFor the cylinder the equation of motion with the acceleration $w$ is found as\n\n$$\nM w=N \\sin \\alpha-F_{f r} .\n\\tag{A.2}\n$$\n\nSince the cylinder moves along the plane without sliding its\n\n\nangular acceleration is obtained as\n\n$$\n\\varepsilon=w / R\n\\tag{A.3}\n$$\n\nThen the equation of rotational motion around the center of mass of the cylinder takes the form\n\n$$\nI \\varepsilon=F_{f r} R\n\\tag{A.4}\n$$\n\nwhere the inertia moment of the hollow cylinder is given by\n\n$$\nI=M R^{2}\n\\tag{A.5}\n$$\n\nSolving (A.2)-(A.5) yields\n\n$$\n2 M w=N \\sin \\alpha\n\\tag{A.6}\n$$\n\nFrom equations (A.1) and (A.6) it is easily concluded that\n\n$$\nm a_{x}=2 M w .\n\\tag{A.7}\n$$\n\nSince the initial velocities of the puck and of the cylinder are both equal to zero, then, it follows from (A.7) after integrating that\n\n$$\nm u=2 M v .\n\\tag{A.8}\n$$\n\nIt is obvious that the conservation law for the system is written as\n\n$$\nm g R=\\frac{m u^{2}}{2}+\\frac{M v^{2}}{2}+\\frac{I \\omega^{2}}{2},\n\\tag{A.9}\n$$\n\nwhere the angular velocity of the cylinder is found to be\n\n$$\n\\omega=\\frac{v}{R}\n\\tag{A.10}\n$$\n\nsince it does not slide over the plane.\n\nSolving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory written as\n\n$$\nu=2 \\sqrt{\\frac{M g R}{(2 M+m)}}\n\\tag{A.12}\n$$\n$$\nv=\\frac{m}{M} \\sqrt{\\frac{M g R}{(2 M+m)}}\n\\tag{A.13}\n$$\n\nIn the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius $R$ and, at the lowest point of its trajectory, have the velocity\n\nand the acceleration\n\n$$\nv_{r e l}=u+v\n\\tag{A.14}\n$$\n\n$$\na_{\\text {rel }}=\\frac{v_{\\text {rel }}^{2}}{R}\n\\tag{A.15}\n$$\n\nAt the lowest point of the puck trajectory the acceleration of the cylinder axis is equal to zero, therefore, the puck acceleration in the laboratory reference frame is also given by (A.15).\n\n$$\nF-m g=\\frac{m v_{r e l}^{2}}{R}\n\\tag{A.16}\n$$\n\nthen the interaction force between the puck and the cylinder is finally found as\n\n$$\nF=3 m g\\left(1+\\frac{m}{3 M}\\right) \\text {. }\n\\tag{A.17}\n$$']",['$F=3 m g(1+\\frac{m}{3 M})$'],False,,Expression, 1098,Thermodynamics,"A bubble of radius $r=5.00 \mathrm{~cm}$, containing a diatomic ideal gas, has the soap film of thickness $h=$ $10.0 \mu \mathrm{m}$ and is placed in vacuum. The soap film has the surface tension $\sigma=4.00 \cdot 10^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the density $\rho=1.10 \frac{\mathrm{g}}{\mathrm{cm}^{3}} .","1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.","['1) According to the first law of thermodynamics, the amount of heat transmitted $\\delta Q$ to the gas in the bubble is found as\n\n$$\n\\delta Q=v C_{V} d T+p d V\n\\tag{B.1}\n$$\n\nwhere the molar heat capacity at arbitrary process is as follows\n\n$$\nC=\\frac{1}{v} \\frac{\\delta Q}{d T}=C_{V}+\\frac{p}{v} \\frac{d V}{d T}\n\\tag{B.2}\n$$\n\nHere $C_{V}$ stands for the molar heat capacity of the gas at constant volume, $p$ designates its pressure, $v$ is the total amount of moles of gas in the bubble, $V$ and $T$ denote the volume and temperature of the gas, respectively.\n\nEvaluate the derivative standing on the right hand side of (B.2). According to the Laplace formula, the gas pressure inside the bubble is defined by\n\n$$\np=\\frac{4 \\sigma}{r}\n\\tag{B.3}\n$$\n\nthus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form\n\n$$\np^{3} V=\\text { const. }\n\\tag{B.4}\n$$\n\nThe equation of state of an ideal gas has the form\n\n$$\np V=v R T\n\\tag{B.5}\n$$\n\nand hence equation (B.4) can be rewritten as\n\n$$\nT^{3} V^{-2}=\\text { const. }\n\\tag{B.6}\n$$\n\nDifferentiating (B.6) the derivative with respect to temperature sought is found as\n\n$$\n\\frac{d V}{d T}=\\frac{3 V}{2 T}\n\\tag{B.7}\n$$\n\nTaking into account that the molar heat capacity of a diatomic gas at constant volume is\n\n$$\nC_{V}=\\frac{5}{2} R\n\\tag{B.8}\n$$\n\nand using (B.5) it is finally obtained that\n\n$$\nC=C_{V}+\\frac{3}{2} R=4 R=33.2 \\frac{\\mathrm{J}}{\\mathrm{mole} \\cdot \\mathrm{K}}\n\\tag{B.9}\n$$']",['$33.2 $'],False,$\frac{\mathrm{J}}{\mathrm{mole} \cdot \mathrm{K}}$,Numerical,1e-1 1099,Thermodynamics,"A bubble of radius $r=5.00 \mathrm{~cm}$, containing a diatomic ideal gas, has the soap film of thickness $h=$ $10.0 \mu \mathrm{m}$ and is placed in vacuum. The soap film has the surface tension $\sigma=4.00 \cdot 10^{-2} \frac{\mathrm{N}}{\mathrm{m}}$ and the density $\rho=1.10 \frac{\mathrm{g}}{\mathrm{cm}^{3}} . Context question: 1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$. Context answer: \boxed{$33.2 $} ","2 Find formula for the frequency $\omega$ of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations. Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.","[""Since the heat capacity of the gas is much smaller than the heat capacity of the soap film, and there is heat exchange between them, the gas can be considered as isothermal since the soap film plays the role of thermostat. Consider the fragment of soap film, limited by the angle $\\alpha$ as shown in the figure. It's area is found as\n\n$$\nS=\\pi(\\alpha r)^{2}\n\\tag{B.10}\n$$\n\nand the corresponding mass is obtained as\n\n$$\nm=\\rho S h \\text {. }\n\\tag{B.11}\n$$\n\nLet $x$ be an increase in the radius of the bubble, then the Newton second law for the fragment of the soap film mentioned above takes the form\n\n$$\nm \\ddot{x}=p^{\\prime} S^{\\prime}-F_{\\text {surf }},\n\\tag{B.12}\n$$\n\nwhere $F_{\\text {surf }}$ denotes the projection of the resultant surface tension force acting in the radial direction, $p^{\\prime}$ stands for the gas pressure beneath the surface of the soap film and\n\n$$\nS^{\\prime}=S\\left(1+2 \\frac{x}{r}\\right)\n$$\n\n$F_{\\text {surf }}$ is easily found as\n\n$$\nF_{\\text {surf }}=F_{S T} \\alpha=\\sigma \\cdot 2 \\cdot 2 \\pi[(r+x) \\alpha] \\cdot \\alpha .\n\\tag{B.13}\n$$\n\nSince the gaseous process can be considered isothermal, it is\n\n\nwritten that\n\n$$\np^{\\prime} V^{\\prime}=p V \\text {. }\n\\tag{B.14}\n$$\n\nAssuming that the volume increase is quite small, (B.14) yields\n\n$$\np^{\\prime}=p \\frac{1}{\\left(1+\\frac{x}{r}\\right)^{3}} \\approx p \\frac{1}{\\left(1+\\frac{3 x}{r}\\right)} \\approx p\\left(1-\\frac{3 x}{r}\\right)\n\\tag{B.15}\n$$\n\nThus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as\n\n$$\n\\rho h \\ddot{x}=-\\frac{8 \\sigma}{r^{2}} x\n\\tag{B.16}\n$$\n\n\n\nwith the frequency\n\n$$\n\\omega=\\sqrt{\\frac{8 \\sigma}{\\rho h r^{2}}}=108 \\mathrm{~s}^{-1}\n\\tag{B.17}\n$$""]",['$108 $'],False,$\mathrm{~s}^{-1}$,Numerical,1e0 1100,Electromagnetism,,"Initially, a switch $S$ is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance $2 C$ carries the electric charge $q_{0}$, a capacitor of capacitance $C$ is uncharged, and there are no electric currents in both coils of inductance $L$ and $2 L$, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch $S$ is instantly shorted. Find the maximum current $I_{\max }$ through the switch $S$ thereafter. ","['At the moment when the current in the coils is a maximum, the total voltage across the coils is equal to zero, so the capacitor voltages must be equal in magnitude and opposite in polarity. Let $U$ be a voltage on the capacitors at the time moment just mentioned and $I_{0}$ be that maximum current. According to the law of charge conservation\n\nthus,\n\n$$\nq_{0}=2 C U+C U\n\\tag{C1.1}\n$$\n\n$$\nU=\\frac{q_{0}}{3 C}\n\\tag{C1.2}\n$$\n\nThen, from the energy conservation law\n\n$$\n\\frac{q_{0}^{2}}{2 \\cdot 2 C}=\\frac{L I_{0}^{2}}{2}+\\frac{2 L I_{0}^{2}}{2}+\\frac{C U^{2}}{2}+\\frac{2 C U^{2}}{2}\n\\tag{C1.3}\n$$\n\nthe maximum current is found as\n\n$$\nI_{0}=\\frac{q_{0}}{3 \\sqrt{2 L C}}\n\\tag{C1.4}\n$$\n\nAfter the key $K$ is shortened there will be independent oscillations in both circuits with the frequency\n\n$$\n\\omega=\\frac{1}{\\sqrt{2 L C}}\n\\tag{C1.5}\n$$\n\nand their amplitudes are obtained from the corresponding energy conservation laws written as\n\n$$\n\\frac{2 C U^{2}}{2}+\\frac{L I_{0}^{2}}{2}=\\frac{L J_{1}^{2}}{2}\n\\tag{C1.6}\n$$\n$$\n\\frac{C U^{2}}{2}+\\frac{2 L I_{0}^{2}}{2}=\\frac{2 L J_{2}^{2}}{2} .\n\\tag{C1.7}\n$$\n\nHence, the corresponding amplitudes are found as\n\n$$\nJ_{1} =\\sqrt{5} I_{0},\n\\tag{C1.8}\n$$\n$$\nJ_{2} =\\sqrt{2} I_{0} .\n\\tag{C1.9}\n$$\n\nChoose the positive directions of the currents in the circuits as shown in the figure on the right. Then, the current flowing through the key is written as follows\n\n$$\nI=I_{1}-I_{2}\n\\tag{C1.10}\n$$\n\nThe currents depend on time as\n\n$$\nI_{1}(t)=A \\cos \\omega t+B \\sin \\omega t\n\\tag{C1.11}\n$$\n$$\nI_{2}(t)=D \\cos \\omega t+F \\sin \\omega t\n\\tag{C1.12}\n$$\n\n\n\nThe constants $A, B, D, F$ can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations\n\n$$\nI_{1}(0)=A=I_{0},\n\\tag{C1.13}\n$$\n$$\nA^{2}+B^{2}=J_{1}^{2}, \n\\tag{C1.14}\n$$\n$$\nI_{2}(0)=D=I_{0}, \n\\tag{C1.15}\n$$\n$$\nD^{2}+F^{2}=J_{2}^{2}\n\\tag{C1.16}\n$$\n\nSolving (C1.13)-(C1.16) it is found that\n\n$$\nB=2 I_{0}, \n\\tag{C1.17}\n$$\n$$\nF=-I_{0},\n\\tag{C1.18}\n$$\n\nThe sign in $F$ is chosen negative, since at the time moment of the key shortening the current in the coil $2 L$ decreases.\n\nThus, the dependence of the currents on time takes the following form\n\n$$\nI_{1}(t)=I_{0}(\\cos \\omega t+2 \\sin \\omega t),\n\\tag{C1.19}\n$$\n$$\nI_{2}(t)=I_{0}(\\cos \\omega t-\\sin \\omega t) .\n\\tag{C1.20}\n$$\n\nIn accordance with (C1.10), the current in the key is dependent on time according to\n\n$$\nI(t)=I_{1}(t)-I_{2}(t)=3 I_{0} \\sin \\omega t .\n\\tag{C1.21}\n$$\n\nHence, the amplitude of the current in the key is obtained as\n\n$$\nI_{\\max }=3 I_{0}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C1.22}\n$$', 'Instead of determining the coefficients $A, B, D, F$ the vector diagram shown in the figure on the right can be used. The segment $A C$ represents the current sought and its projection on the current axis is zero at the time of the key shortening. The current $I_{1}$ in the coil of inductance $L$ grows at the same time moment because the capacitor $2 C$ continues to discharge, thus, this current is depicted in the figure by the segment $O A$. The current $I_{2}$ in the coil of inductance $2 L$ decreases at the time of the key shortening since it continues to charge the capacitor $2 C$, that is why this current is depicted in the figure by the segment $O C$.\n\nIt is known for above that $O B=I_{0}, O A=\\sqrt{5} I_{0}, O C=\\sqrt{2} I_{0}$. Hence, it is found from the Pythagorean theorem that\n\n$$\nA B=\\sqrt{O A^{2}-O B^{2}}=2 I_{0}\n\\tag{C2.1}\n$$\n$$\nB C=\\sqrt{O C^{2}-O B^{2}}=I_{0}\n\\tag{C2.2}\n$$\n\n\n\nThus, the current sought is found as\n\n$$\nI_{\\max }=A C=A B+B C=3 I_{0}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C2.3}\n$$\n\nMethod 3. Heuristic approach\n\nIt is clear that the current through the key performs harmonic oscillations with the frequency\n\n$$\n\\omega=\\frac{1}{\\sqrt{2 L C}}\n\\tag{C3.1}\n$$\n\nand it is equal to zero at the time of the key shortening, i.e.\n\n$$\nI(t)=I_{\\max } \\sin \\omega t\n\\tag{C3.2}\n$$\n\nSince the current is equal to zero at the time of the key shortening, then the current amplitude is equal to the current derivative at this time moment divided by the oscillation frequency. Let us find that current derivative. Let the capacitor of capacitance $2 C$ have the charge $q_{1}$. Then the charge on the capacitor of capacitance $C$ is found from the charge conservation law as\n\n$$\nq_{2}=q_{0}-q_{1} .\n\\tag{C3.3}\n$$\n\nAfter shortening the key the rate of current change in the coil of inductance $L$ is obtained as\n\n$$\n\\dot{I}_{1}=\\frac{q_{1}}{2 L C}\n\\tag{C3.4}\n$$\n\nwhereas in the coil of inductance $2 L$ it is equal to\n\n$$\n\\dot{I}_{2}=-\\frac{q_{0}-q_{1}}{2 L C}\n\\tag{C3.5}\n$$\n\nSince the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form\n\n$$\n\\dot{I}=\\dot{I}_{1}-\\dot{I}_{2}=\\frac{q_{0}}{2 L C}=\\omega^{2} q_{0}\n\\tag{C3.6}\n$$\n\nNote that this derivative is independent of the time of the key shortening!\n\nHence, the maximum current is found as\n\n$$\nI_{\\max }=\\frac{\\dot{I}}{\\omega}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C3.7}\n$$\n\nand it is independent of the time of the key shortening!']",['$I_{\\max }=\\frac{q_{0}}{\\sqrt{2 L C}}$'],False,,Expression, 1100,Electromagnetism,,"Initially, a switch $S$ is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance $2 C$ carries the electric charge $q_{0}$, a capacitor of capacitance $C$ is uncharged, and there are no electric currents in both coils of inductance $L$ and $2 L$, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch $S$ is instantly shorted. Find the maximum current $I_{\max }$ through the switch $S$ thereafter. ![](https://cdn.mathpix.com/cropped/2023_12_21_e8fb313ee14bfbe99d8fg-1.jpg?height=397&width=408&top_left_y=1706&top_left_x=1552)","['At the moment when the current in the coils is a maximum, the total voltage across the coils is equal to zero, so the capacitor voltages must be equal in magnitude and opposite in polarity. Let $U$ be a voltage on the capacitors at the time moment just mentioned and $I_{0}$ be that maximum current. According to the law of charge conservation\n\nthus,\n\n$$\nq_{0}=2 C U+C U\n\\tag{C1.1}\n$$\n\n$$\nU=\\frac{q_{0}}{3 C}\n\\tag{C1.2}\n$$\n\nThen, from the energy conservation law\n\n$$\n\\frac{q_{0}^{2}}{2 \\cdot 2 C}=\\frac{L I_{0}^{2}}{2}+\\frac{2 L I_{0}^{2}}{2}+\\frac{C U^{2}}{2}+\\frac{2 C U^{2}}{2}\n\\tag{C1.3}\n$$\n\nthe maximum current is found as\n\n$$\nI_{0}=\\frac{q_{0}}{3 \\sqrt{2 L C}}\n\\tag{C1.4}\n$$\n\nAfter the key $K$ is shortened there will be independent oscillations in both circuits with the frequency\n\n$$\n\\omega=\\frac{1}{\\sqrt{2 L C}}\n\\tag{C1.5}\n$$\n\nand their amplitudes are obtained from the corresponding energy conservation laws written as\n\n$$\n\\frac{2 C U^{2}}{2}+\\frac{L I_{0}^{2}}{2}=\\frac{L J_{1}^{2}}{2}\n\\tag{C1.6}\n$$\n$$\n\\frac{C U^{2}}{2}+\\frac{2 L I_{0}^{2}}{2}=\\frac{2 L J_{2}^{2}}{2} .\n\\tag{C1.7}\n$$\n\nHence, the corresponding amplitudes are found as\n\n$$\nJ_{1} =\\sqrt{5} I_{0},\n\\tag{C1.8}\n$$\n$$\nJ_{2} =\\sqrt{2} I_{0} .\n\\tag{C1.9}\n$$\n\nChoose the positive directions of the currents in the circuits as shown in the figure on the right. Then, the current flowing through the key is written as follows\n\n$$\nI=I_{1}-I_{2}\n\\tag{C1.10}\n$$\n\nThe currents depend on time as\n\n$$\nI_{1}(t)=A \\cos \\omega t+B \\sin \\omega t\n\\tag{C1.11}\n$$\n$$\nI_{2}(t)=D \\cos \\omega t+F \\sin \\omega t\n\\tag{C1.12}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_53b2d60d8c97adffaedag-1.jpg?height=368&width=377&top_left_y=1478&top_left_x=1545)\n\nThe constants $A, B, D, F$ can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations\n\n$$\nI_{1}(0)=A=I_{0},\n\\tag{C1.13}\n$$\n$$\nA^{2}+B^{2}=J_{1}^{2}, \n\\tag{C1.14}\n$$\n$$\nI_{2}(0)=D=I_{0}, \n\\tag{C1.15}\n$$\n$$\nD^{2}+F^{2}=J_{2}^{2}\n\\tag{C1.16}\n$$\n\nSolving (C1.13)-(C1.16) it is found that\n\n$$\nB=2 I_{0}, \n\\tag{C1.17}\n$$\n$$\nF=-I_{0},\n\\tag{C1.18}\n$$\n\nThe sign in $F$ is chosen negative, since at the time moment of the key shortening the current in the coil $2 L$ decreases.\n\nThus, the dependence of the currents on time takes the following form\n\n$$\nI_{1}(t)=I_{0}(\\cos \\omega t+2 \\sin \\omega t),\n\\tag{C1.19}\n$$\n$$\nI_{2}(t)=I_{0}(\\cos \\omega t-\\sin \\omega t) .\n\\tag{C1.20}\n$$\n\nIn accordance with (C1.10), the current in the key is dependent on time according to\n\n$$\nI(t)=I_{1}(t)-I_{2}(t)=3 I_{0} \\sin \\omega t .\n\\tag{C1.21}\n$$\n\nHence, the amplitude of the current in the key is obtained as\n\n$$\nI_{\\max }=3 I_{0}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C1.22}\n$$', 'Instead of determining the coefficients $A, B, D, F$ the vector diagram shown in the figure on the right can be used. The segment $A C$ represents the current sought and its projection on the current axis is zero at the time of the key shortening. The current $I_{1}$ in the coil of inductance $L$ grows at the same time moment because the capacitor $2 C$ continues to discharge, thus, this current is depicted in the figure by the segment $O A$. The current $I_{2}$ in the coil of inductance $2 L$ decreases at the time of the key shortening since it continues to charge the capacitor $2 C$, that is why this current is depicted in the figure by the segment $O C$.\n\nIt is known for above that $O B=I_{0}, O A=\\sqrt{5} I_{0}, O C=\\sqrt{2} I_{0}$. Hence, it is found from the Pythagorean theorem that\n\n$$\nA B=\\sqrt{O A^{2}-O B^{2}}=2 I_{0}\n\\tag{C2.1}\n$$\n$$\nB C=\\sqrt{O C^{2}-O B^{2}}=I_{0}\n\\tag{C2.2}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6d58a6702f578e2add1cg-1.jpg?height=568&width=362&top_left_y=296&top_left_x=1575)\n\nThus, the current sought is found as\n\n$$\nI_{\\max }=A C=A B+B C=3 I_{0}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C2.3}\n$$\n\nMethod 3. Heuristic approach\n\nIt is clear that the current through the key performs harmonic oscillations with the frequency\n\n$$\n\\omega=\\frac{1}{\\sqrt{2 L C}}\n\\tag{C3.1}\n$$\n\nand it is equal to zero at the time of the key shortening, i.e.\n\n$$\nI(t)=I_{\\max } \\sin \\omega t\n\\tag{C3.2}\n$$\n\nSince the current is equal to zero at the time of the key shortening, then the current amplitude is equal to the current derivative at this time moment divided by the oscillation frequency. Let us find that current derivative. Let the capacitor of capacitance $2 C$ have the charge $q_{1}$. Then the charge on the capacitor of capacitance $C$ is found from the charge conservation law as\n\n$$\nq_{2}=q_{0}-q_{1} .\n\\tag{C3.3}\n$$\n\nAfter shortening the key the rate of current change in the coil of inductance $L$ is obtained as\n\n$$\n\\dot{I}_{1}=\\frac{q_{1}}{2 L C}\n\\tag{C3.4}\n$$\n\nwhereas in the coil of inductance $2 L$ it is equal to\n\n$$\n\\dot{I}_{2}=-\\frac{q_{0}-q_{1}}{2 L C}\n\\tag{C3.5}\n$$\n\nSince the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form\n\n$$\n\\dot{I}=\\dot{I}_{1}-\\dot{I}_{2}=\\frac{q_{0}}{2 L C}=\\omega^{2} q_{0}\n\\tag{C3.6}\n$$\n\nNote that this derivative is independent of the time of the key shortening!\n\nHence, the maximum current is found as\n\n$$\nI_{\\max }=\\frac{\\dot{I}}{\\omega}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C3.7}\n$$\n\nand it is independent of the time of the key shortening!']",['$I_{\\max }=\\frac{q_{0}}{\\sqrt{2 L C}}$'],False,,Expression, 1101,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume.",A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.,"['If $V=b$ is substituted into the equation of state, then the gas pressure turns infinite. It is obvious that this is the moment when all the molecules are tightly packed. Therefore, the parameter $b$ is approximately equal to the volume of all molecules, i.e.\n\n$$\nb=N_{A} d^{3}\n\\tag{A1.1}\n$$']",['$b=N_{A} d^{3}$'],False,,Expression, 1102,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy.",A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$.,"['In the most general case thevan der Waals equation of state can be rewritten as\n\n$$\nP_{c} V^{3}-\\left(R T_{c}+b P_{c}\\right) V^{2}+a V-a b=0\n\\tag{A2.1}\n$$\n\nSince at the critical values of the gas parameters the straight line disappears, then, the solution of (A2.1) must have one real triple root, i.e. it can be rewritten as follows\n\n$$\nP_{c}\\left(V-V_{c}\\right)^{3}=0\n\\tag{A2.2}\n$$\n\nobtained\n\nComparing the coefficients of expression (A2.1) and (A2.2), the following set of equations is\n\n$$\n\\left\\{\\begin{array}{c}\n3 P_{c} V_{c}=R T_{c}+b P_{c} \\\\\n3 P_{c} V_{c}^{2}=a \\\\\nP_{c} V_{c}^{3}=a b\n\\end{array}\\right.\n\\tag{A2.3}\n$$\n\nSolution to the set (A2.3) is the following formulas for the van der Waals coefficients\n\n$$\na=\\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}\n\\tag{A2.4}\n$$\n$$\nb=\\frac{R T_{c}}{8 P_{c}}\n\\tag{A2.5}\n$$']","['$a=\\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\\frac{R T_{c}}{8 P_{c}}$']",True,,Expression, 1103,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} ",A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water.,['Numericalcalculationsforwaterproduce the following result\n\n$$\na_{w}=0.56 \\frac{\\mathrm{m}^{6} \\cdot \\mathrm{Pa}}{\\mathrm{mole}^{2}}\n\\tag{A3.1}\n$$\n$$\nb_{w}=3.1 \\cdot 10^{-5} \\frac{\\mathrm{m}^{3}}{\\mathrm{~mole}}\n\\tag{A3.2}\n$$'],"['$a_{w}=0.56$ , $b_{w}=3.1 \\cdot 10^{-5} $']",True,"$ \frac{\mathrm{m}^{6} \cdot \mathrm{Pa}}{\mathrm{mole}^{2}}$, $\frac{\mathrm{m}^{3}}{\mathrm{~mole}}$",Numerical,"1e-2,1e-6" 1104,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} ",A4 Estimate the diameter of water molecules $d_{w},['From equations (A1.4) and (A3.2) it is found that\n\n$$\nd_{w}=\\sqrt[3]{\\frac{b}{N_{A}}}=3.7 \\cdot 10^{-10} \\mathrm{~m} \\approx 4 \\cdot 10^{-10} \\mathrm{~m}\n\\tag{A4.1}\n$$'],['$4 \\cdot 10^{-10}$'],False,m,Numerical,1e-11 1105,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state.","B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$.","['Usingtheinequality $V_{G} \\gg b$, the van der Waals equation of state can be written as\n\n$$\n\\left(p_{0}+\\frac{a}{V_{G}^{2}}\\right) V_{G}=R T\n\\tag{B1.1}\n$$\n\nwhich has the following solutions\n\n$$\nV_{G}=\\frac{R T}{2 p_{0}}\\left(1 \\pm \\sqrt{1-\\frac{4 a p_{0}}{R^{2} T^{2}}}\\right)\n\\tag{B1.2}\n$$\n\n\n\nSmaller root in (B1.2) gives the volume in an unstable state on the rising branch of thevan der Waals isotherm. The volume of gas is given by the larger root, since at $a=0$ an expression for the volume of an ideal gasshould be obtained, i.e.\n\n$$\nV_{G}=\\frac{R T}{2 p_{0}}\\left(1+\\sqrt{1-\\frac{4 a p_{0}}{R^{2} T^{2}}}\\right)\n\\tag{B1.3}\n$$\n\nFor given values of the parameters the value $\\frac{a p_{0}}{(R T)^{2}}=5.8 \\cdot 10^{-3}$. It can therefore be assumed that $\\frac{a p_{0}}{(R T)^{2}} \\ll 1$, then (B1.3)takes the form\n\n$$\nV_{G} \\approx \\frac{R T}{p_{0}}\\left(1-\\frac{a p_{0}}{R^{2} T^{2}}\\right)=\\frac{R T}{p_{0}}-\\frac{a}{R T}\n\\tag{B1.4}\n$$']",['$V_{G}=\\frac{R T}{2 p_{0}}(1+\\sqrt{1-\\frac{4 a p_{0}}{R^{2} T^{2}}})$'],False,,Expression, 1106,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law.","B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$.","['For an ideal gas\n\n$$\nV_{G 0}=\\frac{R T}{p_{0}}\n\\tag{B2.1}\n$$\n\nhence,\n\n$$\n\\left(\\frac{\\Delta V_{G}}{V_{G 0}}\\right)=\\frac{V_{G 0}-V_{G}}{V_{G 0}}=\\frac{1}{2}\\left(1-\\sqrt{1-\\frac{4 a p_{0}}{R^{2} T^{2}}}\\right) \\approx \\frac{a p_{0}}{R^{2} T^{2}}=0.58 \\%\n\\tag{B2.2}\n$$']",['0.58 %'],False,,Numerical,1e-2 1107,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0.","B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ?","['Mechanical stability of a thermodynamic system is inpower provided that\n\n$$\n\\left(\\frac{d P}{d V}\\right)_{T}<0\n\\tag{B3.1}\n$$\n\nThe minimum volume, in which the mattercan still exist in the gaseous state, corresponds to a point in which\n\n$$\nV_{G \\min } \\rightarrow\\left(\\frac{d P}{d V}\\right)_{T}=0\n\\tag{B3.2}\n$$\n\nUsing the van der Waals equation of state (B3.2) is written as\n\n$$\n\\left(\\frac{d P}{d V}\\right)_{T}=-\\frac{R T}{(V-b)^{2}}+\\frac{2 a}{V^{3}}=0\n\\tag{B3.3}\n$$\n\nFrom (B3.2) and (B3.3), and with the help of $V_{G \\min } \\gg b$, it is found that\n\n$$\nV_{Gmin}=\\frac{2a}{RT}\n\\tag{B3.4}\n$$\n\nThus,\n\n$$\nV_{G \\min }=\\frac{2 a}{R T}\n\\tag{B3.5}\n$$\n\n$$\n\\frac{V_{G}}{V_{G \\min }}=\\frac{R^{2} T^{2}}{2 a p_{0}}=86\n$$']",['86'],False,,Numerical,1e0 1108,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$.","B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$.","['Usingtheinequality $P \\ll a / V^{2}$, the van der Waals equation of state is written as\n\nwhose solution is\n\n$$\n\\frac{a}{V_{L}^{2}}\\left(V_{L}-b\\right)=R T,\n\\tag{B4.1}\n$$\n\n$$\nV_{L}=\\frac{a}{2 R T}\\left(1 \\pm \\sqrt{1-\\frac{4 b R T}{a}}\\right)\n\\tag{B4.2}\n$$\n\nIn this case, the smaller root shouldbe taken, since at $T \\rightarrow 0$ the liquid volume $V_{L}=b$ must be obtained according to (B4.1), i.e.\n\n$$\nV_{L}=\\frac{a}{2 R T}\\left(1-\\sqrt{1-\\frac{4 b R T}{a}}\\right) \\approx b\\left(1+\\frac{b R T}{a}\\right)\n\\tag{B4.3}\n$$']",['$V_{L}=\\frac{a}{2 R T}(1-\\sqrt{1-\\frac{4 b R T}{a}})$'],False,,Expression, 1109,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$. Context question: B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$. Context answer: \boxed{$V_{L}=\frac{a}{2 R T}(1-\sqrt{1-\frac{4 b R T}{a}})$} Extra Supplementary Reading Materials: Assuming that $b R T \ll a$, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values!","B5 Express the liquid water density $\rho_{L}$ in some of the terms of $\mu, a, b, R$ and evaluate it.",['Since (B4.3) givesthevolumeoftheonemoleofwaterits mass density is easily found as\n\n$$\n\\rho_{L}=\\frac{\\mu}{V_{L}}=\\frac{\\mu}{b\\left(1+\\frac{b R T}{a}\\right)} \\approx \\frac{\\mu}{b}=5.8 \\cdot 10^{2} \\frac{\\mathrm{kg}}{\\mathrm{m}^{3}}\n\\tag{B5.1}\n$$'],"['$\\rho_{L}=\\frac{\\mu}{b\\left(1+\\frac{b R T}{a}\\right)}$ , $5.8 \\cdot 10^{2} $']",True,",$\frac{\mathrm{kg}}{\mathrm{m}^{3}}$","Expression,Numerical",",1e1" 1110,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$. Context question: B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$. Context answer: \boxed{$V_{L}=\frac{a}{2 R T}(1-\sqrt{1-\frac{4 b R T}{a}})$} Extra Supplementary Reading Materials: Assuming that $b R T \ll a$, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values! Context question: B5 Express the liquid water density $\rho_{L}$ in some of the terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$\rho_{L}=\frac{\mu}{b\left(1+\frac{b R T}{a}\right)}$ , $5.8 \cdot 10^{2} $} ","B6 Express the volume thermal expansion coefficient $\alpha=\frac{1}{V_{L}} \frac{\Delta V_{L}}{\Delta T}$ in terms of $a, b, R$, and evaluate it.",['Inaccordancewith (B4.3) the volume thermal expansion coefficient is derived as\n\n$$\n\\alpha=\\frac{1}{V_{L}} \\frac{\\Delta V_{L}}{\\Delta T}=\\frac{b R}{a+b R T} \\approx \\frac{b R}{a}=4.6 \\cdot 10^{-4} \\mathrm{~K}^{-1}\n\\tag{B6.1}\n$$'],"['$\\alpha=\\frac{b R}{a+b R T}$ , $4.6 \\cdot 10^{-4}$']",True,$\mathrm{~K}^{-1}$,"Expression,Numerical",",1e-5" 1111,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$. Context question: B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$. Context answer: \boxed{$V_{L}=\frac{a}{2 R T}(1-\sqrt{1-\frac{4 b R T}{a}})$} Extra Supplementary Reading Materials: Assuming that $b R T \ll a$, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values! Context question: B5 Express the liquid water density $\rho_{L}$ in some of the terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$\rho_{L}=\frac{\mu}{b\left(1+\frac{b R T}{a}\right)}$ , $5.8 \cdot 10^{2} $} Context question: B6 Express the volume thermal expansion coefficient $\alpha=\frac{1}{V_{L}} \frac{\Delta V_{L}}{\Delta T}$ in terms of $a, b, R$, and evaluate it. Context answer: \boxed{$\alpha=\frac{b R}{a+b R T}$ , $4.6 \cdot 10^{-4}$} ","B7 Express the specific heat of water vaporization $L$ in terms of $\mu, a, b, R$ and evaluate it.","['The heat, required to convert the liquid to gas, is used to overcome the intermolecular forces that create negative pressure $a / V^{2}$, therefore,\n\n$$\nE=L \\mu \\approx \\int_{V_{L}}^{V_{G}} \\frac{a}{V^{2}} d V=a\\left(\\frac{1}{V_{L}}-\\frac{1}{V_{G}}\\right)\n\\tag{B7.1}\n$$\n\nand using $V_{G} \\gg V_{L}$, (B7.1) yields\n\n$$\nL=\\frac{a}{\\mu V_{L}}=\\frac{a}{\\mu b\\left(1+\\frac{b R T}{a}\\right)} \\approx \\frac{a}{\\mu b}=1.0 \\cdot 10^{6} \\frac{\\mathrm{J}}{\\mathrm{kg}}\n\\tag{B7.2}\n$$']","['$L=\\frac{a}{\\mu b(1+\\frac{b R T}{a})}$ , $1.0 \\cdot 10^{6}$']",True,$\frac{\mathrm{J}}{\mathrm{kg}}$,"Expression,Numerical",",1e5" 1112,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$. Context question: B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$. Context answer: \boxed{$V_{L}=\frac{a}{2 R T}(1-\sqrt{1-\frac{4 b R T}{a}})$} Extra Supplementary Reading Materials: Assuming that $b R T \ll a$, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values! Context question: B5 Express the liquid water density $\rho_{L}$ in some of the terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$\rho_{L}=\frac{\mu}{b\left(1+\frac{b R T}{a}\right)}$ , $5.8 \cdot 10^{2} $} Context question: B6 Express the volume thermal expansion coefficient $\alpha=\frac{1}{V_{L}} \frac{\Delta V_{L}}{\Delta T}$ in terms of $a, b, R$, and evaluate it. Context answer: \boxed{$\alpha=\frac{b R}{a+b R T}$ , $4.6 \cdot 10^{-4}$} Context question: B7 Express the specific heat of water vaporization $L$ in terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$L=\frac{a}{\mu b(1+\frac{b R T}{a})}$ , $1.0 \cdot 10^{6}$} ","B8 Considering the monomolecular layer of water, estimate the surface tension $\sigma$ of water.","['Consider some water of volumeV. To make a monolayer of thickness $d$ out of it, the following work must be done\n\n$$\nA=2 \\sigma S\n\\tag{B8.1}\n$$\n\nFabrication of the monomolecular layer may be interpreted as the evaporation of an equivalent volume of water which requires the following amount of heat\n\n$$\nQ=L m\n\\tag{B8.2}\n$$\n\nwhere the mass is given by\n\n$$\nm=\\rho S d\n\\tag{B8.3}\n$$\n\nUsing (A4.1a), (B5.1)and(B7.2), one finally gets\n\n$$\n\\sigma=\\frac{a}{2 b^{2}} d_{w}=0.12 \\cdot 10^{-2} \\frac{\\mathrm{N}}{\\mathrm{m}}\n\\tag{B8.4}\n$$']",['$0.12 \\cdot 10^{-2}$'],False,$\frac{\mathrm{N}}{\mathrm{m}}$,Numerical,1e-4 1113,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$. Context question: B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$. Context answer: \boxed{$V_{L}=\frac{a}{2 R T}(1-\sqrt{1-\frac{4 b R T}{a}})$} Extra Supplementary Reading Materials: Assuming that $b R T \ll a$, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values! Context question: B5 Express the liquid water density $\rho_{L}$ in some of the terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$\rho_{L}=\frac{\mu}{b\left(1+\frac{b R T}{a}\right)}$ , $5.8 \cdot 10^{2} $} Context question: B6 Express the volume thermal expansion coefficient $\alpha=\frac{1}{V_{L}} \frac{\Delta V_{L}}{\Delta T}$ in terms of $a, b, R$, and evaluate it. Context answer: \boxed{$\alpha=\frac{b R}{a+b R T}$ , $4.6 \cdot 10^{-4}$} Context question: B7 Express the specific heat of water vaporization $L$ in terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$L=\frac{a}{\mu b(1+\frac{b R T}{a})}$ , $1.0 \cdot 10^{6}$} Context question: B8 Considering the monomolecular layer of water, estimate the surface tension $\sigma$ of water. Context answer: \boxed{$0.12 \cdot 10^{-2}$} Extra Supplementary Reading Materials: Part C. Liquid-gas system From Maxwell's rule (equalities of areas, by applying trivial integration) and the van der Waals' equation of state together with the approximations made in Part B, it can be shown that the saturated vapor pressure $p_{L G}$ depends on temperature $T$ as follows $$ \ln p_{L G}=A+\frac{B}{T} \tag{3} $$ where $A$ and $B$ are some constants, that can be expressed in terms of $a$ and $b$ as $A=\ln \left(\frac{a}{b^{2}}\right)-1 ; B=-\frac{a}{b R}$ W. Thomson showed that the pressure of saturated vapor depends on the curvature of the liquid surface. Consider a liquid that does not wet the material of a capillary (contact angle $180^{\circ}$ ). When the capillary is immersed into the liquid, the liquid in the capillary drops to a certain level because of the surface tension (see Figure 3).","C1 Find a small change in pressure $\Delta p_{T}$ of the saturated vapor over the curved surface of liquid and express it in terms of the vapor density $\rho_{s}$, the liquid density $\rho_{L}$, the surface tension $\sigma$ and the radius of surface curvature $r$.","['At equilibrium, the pressure in the liquid and gas should be equalat all depths. The pressurepin the fluid at the depth $h$ is related to the pressure of saturated vapor above the flat surface by\n\n$$\np=p_{0}+\\rho_{L} g h\n\\tag{C1.1}\n$$\n\nThe surface tension creates additional pressure defined by the Laplace formula as\n\n$$\n\\Delta p_{L}=\\frac{2 \\sigma}{r}\n\\tag{C1.2}\n$$\n\nThe same pressure $p$ inthefluidatthedepth $h$ depends on the vapor pressure $p_{h}$ over the curved liquid surface and its radiusofcurvature as\n\n$$\np=p_{h}+\\frac{2 \\sigma}{r}\n\\tag{C1.3}\n$$\n\nFurthermore, the vapor pressure at different heights are related by\n\n$$\np_{h}=p_{0}+\\rho_{S} g h\n\\tag{C1.4}\n$$\n\nSolving (C1.1)-(C1.4), it is found that\n\n$$\nh=\\frac{2 \\sigma}{\\left(\\rho_{L}-\\rho_{S}\\right) g r}\n\\tag{C1.5}\n$$\n\nHence,the pressure difference sought is obtained as\n\n$$\n\\Delta p_{T}=p_{h}-p_{0}=\\rho_{S} g h=\\frac{2 \\sigma}{r} \\frac{\\rho_{S}}{\\rho_{L}-\\rho_{S}} \\approx \\frac{2 \\sigma}{r} \\frac{\\rho_{S}}{\\rho_{L}} .\n\\tag{C1.6}\n$$\n\nNote that the vapor pressure over the convex surface of the liquid is larger than the pressure above the flat surface.']",['$\\Delta p_{T}=\\frac{2 \\sigma}{r} \\frac{\\rho_{S}}{\\rho_{L}-\\rho_{S}}$'],False,,Expression, 1114,Thermodynamics,"Van der Waals equation of state In a well-known model of an ideal gas, whose equation of state obeys the Clapeyron-Mendeleev law, the following important physical effects are neglected. First, molecules of a real gas have a finite size and, secondly, they interact with one another. In all parts of this problem one mole of water is considered. Part A. Non-ideal gas equation of state Taking into account the finite size of the molecules, the gaseous equation of state takes the form $$ P(V-b)=R T \tag{1} $$ where $P, V, T$ stands for the gas pressure, its volume per mole and temperature, respectively, $R$ denotes the universal gas constant, and $b$ is a specific constant extracting some volume. Context question: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$. Context answer: \boxed{$b=N_{A} d^{3}$} Extra Supplementary Reading Materials: With account of intermolecular attraction forces, van der Waals proposed the following equation of state that neatly describes both the gaseous and liquid states of matter $$ \left(P+\frac{a}{V^{2}}\right)(V-b)=R T \tag{2} $$ where $a$ is another specific constant. At temperatures $T$ below a certain critical value $T_{c}$ the isotherm of equation (2) is well represented by a non-monotonic curve 1 shown in Figure 1 which is then called van der Waals isotherm. In the same figure curve 2 shows the isotherm of an ideal gas at the same temperature. A real isotherm differs from the van der Waals isotherm by a straight segment $\boldsymbol{A B}$ drawn at some constant pressure $P_{L G}$. This straight segment is located between the volumes $V_{L}$ and $V_{G}$, and corresponds to the equilibrium of the liquid phase (indicated by $L$ ) and the gaseous phase (referred to by $G$ ). From the second law of thermodynamics J. Maxwell showed that the pressure $P_{L G}$ must be chosen such that the areas $I$ and $I I$ shown in Figure 1 must be equal. Figure 1. Van der Waals isotherm of gas/liquid (curve 1) and the isotherm of an ideal gas (curve 2). Figure 2. Several isotherms for van der Waals equation of state. With increasing temperature the straight segment $A B$ on the isotherm shrinks to a single point when the temperature and the pressure reaches some values $T_{c}$ and $P_{L G}=P_{c}$, respectively. The parameters $P_{c}$ and $T_{c}$ are called critical and can be measured experimentally with high degree of accuracy. Context question: A2 Express the van der Waals constants $a$ and $b$ in terms of $T_{c}$ and $P_{c}$. Context answer: \boxed{$a=\frac{27 R^{2} T_{c}^{2}}{64 P_{c}}$ , $b=\frac{R T_{c}}{8 P_{c}}$} Context question: A3 For water $T_{c}=647 \mathrm{~K}$ and $P_{c}=2.2 \cdot 10^{7} \mathrm{~Pa}$. Calculate $a_{w}$ and $b_{w}$ for water. Context answer: \boxed{$a_{w}=0.56$ , $b_{w}=3.1 \cdot 10^{-5} $} Context question: A4 Estimate the diameter of water molecules $d_{w} Context answer: \boxed{$4 \cdot 10^{-10}$} Extra Supplementary Reading Materials: Part B. Properties of gas and liquid This part of the problem deals with the properties of water in the gaseous and liquid states at temperature $T=100{ }^{\circ} \mathrm{C}$. The saturated vapor pressure at this temperature is known to be $p_{L G}=p_{0}=1.0$. $10^{5} \mathrm{~Pa}$, and the molar mass of water is $\mu=1.8 \cdot 10^{-2} \frac{\mathrm{kg}}{\mathrm{mole}}$. Gaseous state It is reasonable to assume that the inequality $V_{G} \gg b$ is valid for the description of water properties in a gaseous state. Context question: B1 Derive the formula for the volume $V_{G}$ and express it in terms of $R, T, p_{0}$, and $a$. Context answer: \boxed{$V_{G}=\frac{R T}{2 p_{0}}(1+\sqrt{1-\frac{4 a p_{0}}{R^{2} T^{2}}})$} Extra Supplementary Reading Materials: Almost the same volume $V_{G 0}$ can be approximately evaluated using the ideal gas law. Context question: B2 Evaluate in percentage the relative decrease in the gas volume due to intermolecular forces, $\frac{\Delta V_{G}}{V_{G 0}}=\frac{V_{G 0}-V_{G}}{V_{G 0}}$. Context answer: \boxed{0.58 %} Extra Supplementary Reading Materials: If the system volume is reduced below $V_{G}$, the gas starts to condense. However, thoroughly purified gas can remain in a mechanically metastable state (called supercooled vapor) until its volume reaches a certain value $V_{G \mathrm{~min}}$. The condition of mechanical stability of supercooled gas at constant temperature is written as: $\frac{d P}{d V}<$ 0. Context question: B3 Find and evaluate how many times the volume of water vapor can be reduced and still remains in a metastable state. In other words, what is $V_{G} / V_{G \text { min }}$ ? Context answer: \boxed{86} Extra Supplementary Reading Materials: Liquid state For the van der Waals' description of water in a liquid state it is reasonable to assume that the following inequality holds $P \ll a / V^{2}$. Context question: B4 Express the volume of liquid water $V_{L}$ in terms of $a, b, R$, and $T$. Context answer: \boxed{$V_{L}=\frac{a}{2 R T}(1-\sqrt{1-\frac{4 b R T}{a}})$} Extra Supplementary Reading Materials: Assuming that $b R T \ll a$, find the following characteristics of water. Do not be surprised if some of the data evaluated do not coincide with the well-known tabulated values! Context question: B5 Express the liquid water density $\rho_{L}$ in some of the terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$\rho_{L}=\frac{\mu}{b\left(1+\frac{b R T}{a}\right)}$ , $5.8 \cdot 10^{2} $} Context question: B6 Express the volume thermal expansion coefficient $\alpha=\frac{1}{V_{L}} \frac{\Delta V_{L}}{\Delta T}$ in terms of $a, b, R$, and evaluate it. Context answer: \boxed{$\alpha=\frac{b R}{a+b R T}$ , $4.6 \cdot 10^{-4}$} Context question: B7 Express the specific heat of water vaporization $L$ in terms of $\mu, a, b, R$ and evaluate it. Context answer: \boxed{$L=\frac{a}{\mu b(1+\frac{b R T}{a})}$ , $1.0 \cdot 10^{6}$} Context question: B8 Considering the monomolecular layer of water, estimate the surface tension $\sigma$ of water. Context answer: \boxed{$0.12 \cdot 10^{-2}$} Extra Supplementary Reading Materials: Part C. Liquid-gas system From Maxwell's rule (equalities of areas, by applying trivial integration) and the van der Waals' equation of state together with the approximations made in Part B, it can be shown that the saturated vapor pressure $p_{L G}$ depends on temperature $T$ as follows $$ \ln p_{L G}=A+\frac{B}{T} \tag{3} $$ where $A$ and $B$ are some constants, that can be expressed in terms of $a$ and $b$ as $A=\ln \left(\frac{a}{b^{2}}\right)-1 ; B=-\frac{a}{b R}$ W. Thomson showed that the pressure of saturated vapor depends on the curvature of the liquid surface. Consider a liquid that does not wet the material of a capillary (contact angle $180^{\circ}$ ). When the capillary is immersed into the liquid, the liquid in the capillary drops to a certain level because of the surface tension (see Figure 3). Context question: C1 Find a small change in pressure $\Delta p_{T}$ of the saturated vapor over the curved surface of liquid and express it in terms of the vapor density $\rho_{s}$, the liquid density $\rho_{L}$, the surface tension $\sigma$ and the radius of surface curvature $r$. Context answer: \boxed{$\Delta p_{T}=\frac{2 \sigma}{r} \frac{\rho_{S}}{\rho_{L}-\rho_{S}}$} Extra Supplementary Reading Materials: Metastable states, considered in part B3, are widely used in real experimental setups, such as the cloud chamber designed for registration of elementary particles. They also occur in natural phenomena such as the formation of morning dew. Supercooled vapor is subject to condensation by forming liquid droplets. Very small droplets evaporate quickly but large enough ones can still grow. Figure 3. Capillary immersed in a liquid that does not wet its material","C2 Suppose that at the evening temperature of $t_{e}=20^{\circ} \mathrm{C}$ the water vapor in the air was saturated, but in the morning the ambient temperature has fallen by a small amount of $\Delta t=5.0^{\circ} \mathrm{C}$. Assuming that the vapor pressure has remained unchanged, estimate the minimum radius of droplets that can grow. Use the tabulated value of water surface tension $\sigma=7.3 \cdot 10^{-2} \mathrm{~N} / \mathrm{m}$.","['Let $P_{e}$ be vapor pressure at a temperature $T_{e}$, and $P_{e}-\\Delta P_{e}$ be vapor pressure at a temperature $T_{e}-\\Delta T_{e}$. In accordance with equation (3) from problem statement, whentheambient temperature falls by an amount of $\\Delta T_{e}$ the saturated vapor pressure changes by an amount\n\n$$\n\\Delta P_{e}=P_{e} \\frac{a}{b R T_{e}^{2}} \\Delta T_{e}\n\\tag{C2.1}\n$$\n\nIn accordance with the Thomson formula obtained in part $\\mathbf{C 1}$, the pressure of saturated vapor above the droplet increases by the amountof $\\Delta p_{T}$. While a droplet is small in size, the vapor above its surface remains unsaturated. Whena droplet hasgrownuptoacertainminimumsize, thevaporaboveitssurface turns saturated.\n\nSince the pressure remains unchanged, the following condition must hold\n\n$$\nP_{e}-\\Delta P_{e}+\\Delta p_{T}=P_{e}\n\\tag{C2.2}\n$$\n\nAssuming the vapor is almost ideal gas, its density can be found as\n\n$$\n\\rho_{S}=\\frac{\\mu P_{e}}{R T_{e}} \\ll \\rho_{L}\n\\tag{C2.3}\n$$\n\nFrom equations (C2.1)-(C2.3), (B5.1) and (C1.6) one finds\n\n$$\n\\frac{2 \\sigma}{r} \\frac{\\mu P_{e}}{R T_{e}\\left(\\frac{\\mu}{b}\\right)}=P_{e} \\frac{a \\Delta T_{e}}{b R T_{e}^{2}}\n\\tag{C2.4}\n$$\n\nThus, it is finally obtained that\n\n$$\nr=\\frac{2 \\sigma b^{2} T_{e}}{a \\Delta T_{e}}=1.5 \\cdot 10^{-8} \\mathrm{~m}\n\\tag{C2.5}\n$$']",['$1.5 \\cdot 10^{-8}$'],False,m,Numerical,1e-9 1115,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent.","A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.","['Let us derive an equation describing the change of the electron number density with time. It is determined by the two processes; the generation of ion pairs by external ionizer and the recombination of electrons with ions. At ionization process electrons and ions are generated in pairs, and at recombination processthey disappear in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.\n\n$$\nn(t)=n_{e}(t)=n_{i}(t)\n\\tag{A1.1}\n$$\n\nThen the equation describing the numberdensityevolution of electrons and ions in time can be written as\n\n$$\n\\frac{d n(t)}{d t}=Z_{e x t}-r n(t)^{2}\n\\tag{A1.2}\n$$\n\nIt is easy to show that at $t \\rightarrow 0$ the function $\\tanh b t \\rightarrow 0$, therefore, by virtue of the initial condition $n(0)=0$,one finds\n\n$$\nn_{0}=0\n\\tag{A1.3}\n$$\n\nSubstituting $n_{e}(t)=a \\tanh b t$ in (A1.2) and separating it in the independent functions (hyperbolic, or 1 and $\\left.e^{x}\\right)$, one gets\n\n$$\na=\\sqrt{\\frac{Z_{e x t}}{r}}\n\\tag{A1.4}\n$$\n$$\nb=\\sqrt{r Z_{e x t}}\n\\tag{A1.5}\n$$']","['$n_{0}=0$ , $a=\\sqrt{\\frac{Z_{e x t}}{r}}$ , $b=\\sqrt{r Z_{e x t}}$']",True,,Expression, 1116,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$.",A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.,"['According to equation (A1.4) the number density of electronsat steady-state is expressed in terms of the external ionizer activity as\n\n$$\nn_{e 1}=\\sqrt{\\frac{Z_{e x t 1}}{r}}\n\\tag{A2.1}\n$$\n$$\nn_{e 2}=\\sqrt{\\frac{Z_{e x t 2}}{r}}\n\\tag{A2.2}\n$$\n$$\nn_{e}=\\sqrt{\\frac{Z_{e x t 1}+Z_{e x t 2}}{r}}\n\\tag{A2.3}\n$$\n\nThus,the following analogue of the Pythagorean theorem is obtained as\n\n$$\nn_{e}=\\sqrt{n_{e 1}^{2}+n_{e 2}^{2}}=20.0 \\cdot 10^{10} \\mathrm{~cm}^{-3}\n\\tag{A2.4}\n$$']",['$20.0 \\cdot 10^{10}$'],False,$ \mathrm{~cm}^{-3}$,Numerical,1e9 1117,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility.","A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge.","['In the steady state, the balance equations of electrons and ions in the tube volume take the form\n\n$$\nZ_{\\text {ext }} S L=r n_{e} n_{i} S L+\\frac{I_{e}}{e}\n\\tag{A3.1}\n$$\n$$\nZ_{\\text {ext }} S L=r n_{e} n_{i} S L+\\frac{I_{i}}{e}\n\\tag{A3.2}\n$$\n\nIt follows from equations (A3.1) and (A3.2) that the ion and electron currents are equal, i.e.\n\n$$\nI_{e}=I_{i}\n\\tag{A3.3}\n$$\n\nAt the same time the total current in each tube section is the sum of the electron and ion currents\n\n$$\nI=I_{e}+I_{i}\n\\tag{A3.4}\n$$\n\nBy definition ofthe current density the following relations hold\n\n$$\nI_{e}=\\frac{I}{2}=e n_{e} v S=e \\beta n_{e} E S\n\\tag{A3.5}\n$$\n$$\nI_{i}=\\frac{I}{2}=e n_{i} v S=e \\beta n_{i} E S\n\\tag{A3.6}\n$$\n\nSubstituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic equation for the current is derived\n\n$$\nZ_{e x t} S L=r S L\\left(\\frac{I}{2 e \\beta E S}\\right)^{2}+\\frac{I}{2 e}\n\\tag{A3.7}\n$$\n\nThe electric field strength in the gas is equal to\n\n$$\nE=\\frac{U}{L}\n\\tag{A3.8}\n$$\n\nand solution to the quadratic equation (A3.7) takes the form\n\n$$\nI=\\frac{e \\beta^{2} U^{2} S}{r L^{3}}\\left(-1 \\pm \\sqrt{1+\\frac{4 r Z_{e x t} L^{4}}{\\beta^{2} U^{2}}}\\right)\n\\tag{A3.9}\n$$\n\n\n\nIt is obvious that only positive root does make sense, i.e.\n\n$$\nI=\\frac{e \\beta^{2} U^{2} S}{r L^{3}}\\left(\\sqrt{1+\\frac{4 r Z_{e x t} L^{4}}{\\beta^{2} U^{2}}}-1\\right)\n\\tag{A3.10}\n$$']",['$I=\\frac{e \\beta^{2} U^{2} S}{r L^{3}}(\\sqrt{1+\\frac{4 r Z_{e x t} L^{4}}{\\beta^{2} U^{2}}}-1)$'],False,,Expression, 1118,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} ","A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$",['At low voltages (A3.10) simplifies and gives the following expression\n\n$$\nI=2 U e \\beta \\sqrt{\\frac{Z_{e x t}}{r}} \\frac{S}{L}\n\\tag{A4.1}\n$$\n\nwhich is actually the Ohm law.\n\nUsing the well-known relation\n\n$$\nR=\\frac{U}{I}\n\\tag{A4.2}\n$$\n\ntogether with\n\n$$\nR=\\rho \\frac{L}{S}\n\\tag{A4.3}\n$$\n\none gets\n\n$$\n\\rho=\\frac{1}{2 e \\beta} \\sqrt{\\frac{r}{Z_{e x t}}}\n\\tag{A4.4}\n$$'],['$\\frac{1}{2 e \\beta} \\sqrt{\\frac{r}{Z_{e x t}}}$'],False,,Expression, 1119,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} Context question: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$ Context answer: \boxed{$\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$} Extra Supplementary Reading Materials: PartB. Self-sustained gas discharge In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining. Attention!In the sequel assume that the external ionizer continues to operatewith the same $Z_{\mathrm{ext}}$ rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored. For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons $\dot{N}_{e}$ per unit timeto the number of ions $\dot{N}_{i}$ hitting the cathode per unit time is called the coefficient of the secondary electron emission, $\gamma=\dot{N}_{e} / \dot{N}_{i}$. The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient $\alpha$, which characterizes an increase in the number of electrons $d N_{e}$ due to moving $N_{e}$ electrons that have passed the distancedl, i.e. $$ \frac{d N_{e}}{d l}=\alpha N_{e} $$ The total current $I$ in any cross section of the gas tube consists of the ion $I_{i}(x)$ and the electron $I_{e}(x)$ currents which, in the steady state, depend on the coordinate $x$, shown in the figureabove. The electron current $I_{e}(x)$ varies along the $x$-axis according to the formula where $A_{1}, A_{2}, C_{1}$ are some constants. $$ I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}, $$","B1 Find $A_{1}, A_{2}$ and express them in terms of $Z_{\text {ext }}, \alpha, e, L, S .$","['Consider a gas layer located between $x$ and $x+d x$. The rate of change in the electron number inside the layer due to the electric current is givenfor a small time interval $d t$ by\n\n$$\nd N_{e}^{I}=\\frac{I_{e}(x+d x)-I_{e}(x)}{e} d t=\\frac{1}{e} \\frac{d I_{e}(x)}{d x} d x d t\n\\tag{B1.1}\n$$\n\nThis change is due to the effect of the external ionization and the electron avalanche formation.\n\nThe external ionizer creates the following number of electrons in the volume $S d x$\n\n$$\nd N_{e}^{e x t}=Z_{e x t} S d x d t\n\\tag{B1.2}\n$$\n\nwhereas the electron avalanche produces the number of electrons found as\n\n$$\nd N_{e}^{a}=\\alpha N_{e} d l=n_{e} S d x v d t=\\alpha \\frac{I_{e}(x)}{e} d x d t\n\\tag{B1.3}\n$$\n\nThe balance equationfor the number of electrons is written as\n\n$$\nd N_{e}^{I}=d N_{e}^{e x t}+d N_{e}^{a}\n\\tag{B1.4}\n$$\n\nwhichresults in the following differential equation for the electron current\n\n$$\n\\frac{d I_{e}(x)}{d x}=e Z_{e x t} S+\\alpha I_{e}(x)\n\\tag{B1.5}\n$$\n\nOn substituting $I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}$, one derives\n\n$$\nA_{1}=\\alpha\n\\tag{B1.6}\n$$\n$$\nA_{2}=-\\frac{e Z_{e x t} S}{\\alpha}\n\\tag{B1.7}\n$$']","['$A_{1}=\\alpha$ , $A_{2}=-\\frac{e Z_{e x t} S}{\\alpha}$']",True,,Expression, 1120,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} Context question: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$ Context answer: \boxed{$\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$} Extra Supplementary Reading Materials: PartB. Self-sustained gas discharge In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining. Attention!In the sequel assume that the external ionizer continues to operatewith the same $Z_{\mathrm{ext}}$ rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored. For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons $\dot{N}_{e}$ per unit timeto the number of ions $\dot{N}_{i}$ hitting the cathode per unit time is called the coefficient of the secondary electron emission, $\gamma=\dot{N}_{e} / \dot{N}_{i}$. The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient $\alpha$, which characterizes an increase in the number of electrons $d N_{e}$ due to moving $N_{e}$ electrons that have passed the distancedl, i.e. $$ \frac{d N_{e}}{d l}=\alpha N_{e} $$ The total current $I$ in any cross section of the gas tube consists of the ion $I_{i}(x)$ and the electron $I_{e}(x)$ currents which, in the steady state, depend on the coordinate $x$, shown in the figureabove. The electron current $I_{e}(x)$ varies along the $x$-axis according to the formula where $A_{1}, A_{2}, C_{1}$ are some constants. $$ I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}, $$ Context question: B1 Find $A_{1}, A_{2}$ and express them in terms of $Z_{\text {ext }}, \alpha, e, L, S .$ Context answer: \boxed{$A_{1}=\alpha$ , $A_{2}=-\frac{e Z_{e x t} S}{\alpha}$} Extra Supplementary Reading Materials: The ion current $I_{i}(x)$ varies along the $x$-axis according to the formula $$ I_{i}(x)=C_{2}+B_{1} e^{B_{2} x} $$ where $B_{1}, B_{2}, C_{2}$ are some constants.","B2 Find $B_{1}, B_{2}$ and express them in terms of $Z_{\mathrm{ext}}, \alpha, e, L, S, C_{1}","['Given the fact that the ions flow in the direction opposite to the electron motion,the balance equationfor the number of ionsis written as\n\nwhere\n\n$$\nd N_{i}^{I}=d N_{i}^{e x t}+d N_{i}^{a}\n\\tag{B2.1}\n$$\n\n$$\nd N_{i}^{I}=\\frac{I_{i}(x)-I_{i}(x+d x)}{e} d t=-\\frac{1}{e} \\frac{d I_{i}(x)}{d x} d x d t \n\\tag{B2.2}\n$$\n$$\nd N_{i}^{e x t}=Z_{e x t} S d x d t\n\\tag{B2.3}\n$$\n$$\nd N_{i}^{a}=\\alpha \\frac{I_{e}(x)}{e} d x d t\n\\tag{B2.4}\n$$\n\nHence, the following differential equation for the ion current is obtained\n\n$$\n-\\frac{d I_{i}(x)}{d x}=e Z_{e x t} S+\\alpha I_{e}(x) .\n\\tag{B2.5}\n$$\n\nOnsubstituting the previouslyfound electron current together with the ion current, $I_{i}(x)=C_{2}+$ $B_{1} e^{B_{2} x}$,yields\n\n$$\nB_{1}=-C_{1}\n\\tag{B2.6}\n$$\n$$\nB_{2}=\\alpha\n\\tag{B2.7}\n$$']","['$B_{1}=-C_{1}$ , $B_{2}=\\alpha$']",True,,Expression, 1121,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} Context question: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$ Context answer: \boxed{$\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$} Extra Supplementary Reading Materials: PartB. Self-sustained gas discharge In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining. Attention!In the sequel assume that the external ionizer continues to operatewith the same $Z_{\mathrm{ext}}$ rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored. For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons $\dot{N}_{e}$ per unit timeto the number of ions $\dot{N}_{i}$ hitting the cathode per unit time is called the coefficient of the secondary electron emission, $\gamma=\dot{N}_{e} / \dot{N}_{i}$. The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient $\alpha$, which characterizes an increase in the number of electrons $d N_{e}$ due to moving $N_{e}$ electrons that have passed the distancedl, i.e. $$ \frac{d N_{e}}{d l}=\alpha N_{e} $$ The total current $I$ in any cross section of the gas tube consists of the ion $I_{i}(x)$ and the electron $I_{e}(x)$ currents which, in the steady state, depend on the coordinate $x$, shown in the figureabove. The electron current $I_{e}(x)$ varies along the $x$-axis according to the formula where $A_{1}, A_{2}, C_{1}$ are some constants. $$ I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}, $$ Context question: B1 Find $A_{1}, A_{2}$ and express them in terms of $Z_{\text {ext }}, \alpha, e, L, S .$ Context answer: \boxed{$A_{1}=\alpha$ , $A_{2}=-\frac{e Z_{e x t} S}{\alpha}$} Extra Supplementary Reading Materials: The ion current $I_{i}(x)$ varies along the $x$-axis according to the formula $$ I_{i}(x)=C_{2}+B_{1} e^{B_{2} x} $$ where $B_{1}, B_{2}, C_{2}$ are some constants. Context question: B2 Find $B_{1}, B_{2}$ and express them in terms of $Z_{\mathrm{ext}}, \alpha, e, L, S, C_{1} Context answer: \boxed{$B_{1}=-C_{1}$ , $B_{2}=\alpha$}",B3 Write down the condition for $I_{i}(x)$ at $x=L$,"['Since the ions starts to move from the anode located at $x=L$, the following condition holds\n\n$$\nI_{i}(L)=0\n\\tag{B3.1}\n$$\n\n']",['$I_{i}(L)=0$'],False,,Expression, 1122,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} Context question: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$ Context answer: \boxed{$\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$} Extra Supplementary Reading Materials: PartB. Self-sustained gas discharge In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining. Attention!In the sequel assume that the external ionizer continues to operatewith the same $Z_{\mathrm{ext}}$ rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored. For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons $\dot{N}_{e}$ per unit timeto the number of ions $\dot{N}_{i}$ hitting the cathode per unit time is called the coefficient of the secondary electron emission, $\gamma=\dot{N}_{e} / \dot{N}_{i}$. The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient $\alpha$, which characterizes an increase in the number of electrons $d N_{e}$ due to moving $N_{e}$ electrons that have passed the distancedl, i.e. $$ \frac{d N_{e}}{d l}=\alpha N_{e} $$ The total current $I$ in any cross section of the gas tube consists of the ion $I_{i}(x)$ and the electron $I_{e}(x)$ currents which, in the steady state, depend on the coordinate $x$, shown in the figureabove. The electron current $I_{e}(x)$ varies along the $x$-axis according to the formula where $A_{1}, A_{2}, C_{1}$ are some constants. $$ I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}, $$ Context question: B1 Find $A_{1}, A_{2}$ and express them in terms of $Z_{\text {ext }}, \alpha, e, L, S .$ Context answer: \boxed{$A_{1}=\alpha$ , $A_{2}=-\frac{e Z_{e x t} S}{\alpha}$} Extra Supplementary Reading Materials: The ion current $I_{i}(x)$ varies along the $x$-axis according to the formula $$ I_{i}(x)=C_{2}+B_{1} e^{B_{2} x} $$ where $B_{1}, B_{2}, C_{2}$ are some constants. Context question: B2 Find $B_{1}, B_{2}$ and express them in terms of $Z_{\mathrm{ext}}, \alpha, e, L, S, C_{1} Context answer: \boxed{$B_{1}=-C_{1}$ , $B_{2}=\alpha$} Context question: B3 Write down the condition for $I_{i}(x)$ at $x=L$ Context answer: $I_{i}(L)=0$",B4 Write down the condition for $I_{i}(x)$ and $I_{e}(x)$ at $x=0$ .,['By definition of secondary electron emission coefficient the following condition should be imposed\n\n$$\nI_{e}(0)=\\gamma I_{i}(0)\n\\tag{B4.1}\n$$\n\n'],['$I_{e}(0)=\\gamma I_{i}(0)$'],False,,Equation, 1123,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} Context question: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$ Context answer: \boxed{$\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$} Extra Supplementary Reading Materials: PartB. Self-sustained gas discharge In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining. Attention!In the sequel assume that the external ionizer continues to operatewith the same $Z_{\mathrm{ext}}$ rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored. For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons $\dot{N}_{e}$ per unit timeto the number of ions $\dot{N}_{i}$ hitting the cathode per unit time is called the coefficient of the secondary electron emission, $\gamma=\dot{N}_{e} / \dot{N}_{i}$. The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient $\alpha$, which characterizes an increase in the number of electrons $d N_{e}$ due to moving $N_{e}$ electrons that have passed the distancedl, i.e. $$ \frac{d N_{e}}{d l}=\alpha N_{e} $$ The total current $I$ in any cross section of the gas tube consists of the ion $I_{i}(x)$ and the electron $I_{e}(x)$ currents which, in the steady state, depend on the coordinate $x$, shown in the figureabove. The electron current $I_{e}(x)$ varies along the $x$-axis according to the formula where $A_{1}, A_{2}, C_{1}$ are some constants. $$ I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}, $$ Context question: B1 Find $A_{1}, A_{2}$ and express them in terms of $Z_{\text {ext }}, \alpha, e, L, S .$ Context answer: \boxed{$A_{1}=\alpha$ , $A_{2}=-\frac{e Z_{e x t} S}{\alpha}$} Extra Supplementary Reading Materials: The ion current $I_{i}(x)$ varies along the $x$-axis according to the formula $$ I_{i}(x)=C_{2}+B_{1} e^{B_{2} x} $$ where $B_{1}, B_{2}, C_{2}$ are some constants. Context question: B2 Find $B_{1}, B_{2}$ and express them in terms of $Z_{\mathrm{ext}}, \alpha, e, L, S, C_{1} Context answer: \boxed{$B_{1}=-C_{1}$ , $B_{2}=\alpha$} Context question: B3 Write down the condition for $I_{i}(x)$ at $x=L$ Context answer: $I_{i}(L)=0$ Context question: B4 Write down the condition for $I_{i}(x)$ and $I_{e}(x)$ at $x=0$ . Context answer: $I_{e}(0)=\gamma I_{i}(0)$ ","B5 Find the total current $I$ and express it in terms of $Z_{\mathrm{ext}}, \alpha, \gamma, e, L, S$. Assume that it remains finite",['Total current in each tube section is the sum of the electron and ion currents:\n\n$$\nI=I_{e}+I_{i}=C_{2}-\\frac{e Z_{e x t} S}{\\alpha}\n\\tag{B5.1}\n$$\n\nAftersubstituting the boundary conditions (B3.1) and (B4.1):\n\n$$\nC_{2}-C_{1} e^{\\alpha L}=0\n\\tag{B5.2}\n$$\n\nand\n\n$$\nC_{1}-\\frac{e Z_{e x t} S}{\\alpha}=\\gamma\\left(C_{2}-C_{1}\\right)\n\\tag{B5.3}\n$$\n\nSolving (B5.2) and (B5.3) one can obtain:\n\n$$\nC_{2}=\\frac{e Z_{e x t} S}{\\alpha}\\left(\\frac{1}{e^{-\\alpha L}(1+\\gamma)-\\gamma}\\right)\n\\tag{B5.4}\n$$\n\nSo the total current:\n\n$$\nI=\\frac{e Z_{e x t} S}{\\alpha}\\left(\\frac{1}{e^{-\\alpha L}(1+\\gamma)-\\gamma}-1\\right)\n\\tag{B5.5}\n$$'],['$I=\\frac{e Z_{e x t} S}{\\alpha}(\\frac{1}{e^{-\\alpha L}(1+\\gamma)-\\gamma}-1)$'],False,,Expression, 1124,Electromagnetism,"Problem3. Simplest model of gas discharge An electric current flowing through a gas is called a gas discharge. There are many types of gas discharges including glow discharge in lighting lamps, arc discharge inwelding and the well known spark discharge that occurs between the clouds and the earth in the form of lightning. Part A. Non-self-sustained gas discharge In this part of the problem the so-called non-self-sustained gas discharge is studied. To maintain it a permanent operationan external ionizer is needed, which creates $Z_{\text {ext }}$ pairs of singly ionized ions and free electrons per unit volume and per unit time uniformly in the volume. When an external ionizer is switched on, the number of electrons and ions starts to grow. Unlimited increase in the number densities of electrons and ions in the gas is prevented by the recombination process in which a free electron recombineswith an ion to form a neutral atom. The number of recombining events $Z_{\text {rec }}$ that occurs in the gas per unit volume and per unit time is given by $$ Z_{\mathrm{rec}}=r n_{e} n_{i}, $$ where $r$ is a constant called the recombination coefficient, and $n_{e}, n_{i}$ denotethe electron and ion number densities, respectively. Suppose that at time $t=0$ the external ionizer is switched on and the initial number densities of electrons and ions in the gas are both equal to zero. Then, the electron number density $n_{e}(t)$ depends on timetas follows: $$ n_{e}(t)=n_{0}+a \tanh b t $$ where $n_{0}, a$ and $b$ are some constants, and $\tanh x$ stands for the hyperbolic tangent. Context question: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$. Context answer: \boxed{$n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$} Extra Supplementary Reading Materials: Assume that there are two external ionizers available. When the first one is switched on, the electron number density in the gas reaches its equilibrium value of be $n_{e 1}=12 \cdot 10^{10} \mathrm{~cm}^{-3}$. When the second external ionizer is switched on, the electron number density reaches its equilibrium value of $n_{e 2}=16$. $10^{10} \mathrm{~cm}^{-3}$. Context question: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously. Context answer: \boxed{$20.0 \cdot 10^{10}$} Extra Supplementary Reading Materials: Attention!In what follows it is assumed that the external ionizer is switched on for quite long period of time such that all processes have become stationary and do not depend on time. Completely neglect the electric field due the charge carriers. Assume that the gas fills in the tube between the two parallel conductive plates of area $S$ separated by the distance $L \ll \sqrt{S}$ from each other. The voltage $U$ is applied across the plates to create an electric field between them. Assume that the number densities of both kinds of charge carriers remain almost constant along the tube. Assumethat both the electrons (denoted by thesubscripte) and the ions (denoted by the subscripti) acquire the same ordered speed $v$ due to the electric field strength $E$ found as $$ v=\beta E $$ where bisa constant called charge mobility. Context question: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge. Context answer: \boxed{$I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$} Context question: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$ Context answer: \boxed{$\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$} Extra Supplementary Reading Materials: PartB. Self-sustained gas discharge In this part of the problemthe ignition of the self-sustained gas discharge is consideredto show how the electric current in the tube becomes self-maintaining. Attention!In the sequel assume that the external ionizer continues to operatewith the same $Z_{\mathrm{ext}}$ rate, neglect the electric field due to the charge carriers such that the electric field is uniform along the tube, and the recombination can be completely ignored. For the self-sustained gas discharge there are two important processes not considered above. The first process isasecondary electron emission, and the second one isa formation of electron avalanche. The secondary electron emission occurs when ions hit on the negative electrode, called a cathode, and the electrons are knocked out of it to move towards the positive electrode, called an anode. The ratio of the number of the knocked electrons $\dot{N}_{e}$ per unit timeto the number of ions $\dot{N}_{i}$ hitting the cathode per unit time is called the coefficient of the secondary electron emission, $\gamma=\dot{N}_{e} / \dot{N}_{i}$. The formation of the electron avalanche isexplained as follows. The electric field accelerates free electrons which acquire enough kinetic energy to ionize the atoms in the gas by hitting them. As a result the number of free electrons moving towards the anodesignificantly increases. This process is described by the Townsend coefficient $\alpha$, which characterizes an increase in the number of electrons $d N_{e}$ due to moving $N_{e}$ electrons that have passed the distancedl, i.e. $$ \frac{d N_{e}}{d l}=\alpha N_{e} $$ The total current $I$ in any cross section of the gas tube consists of the ion $I_{i}(x)$ and the electron $I_{e}(x)$ currents which, in the steady state, depend on the coordinate $x$, shown in the figureabove. The electron current $I_{e}(x)$ varies along the $x$-axis according to the formula where $A_{1}, A_{2}, C_{1}$ are some constants. $$ I_{e}(x)=C_{1} e^{A_{1} x}+A_{2}, $$ Context question: B1 Find $A_{1}, A_{2}$ and express them in terms of $Z_{\text {ext }}, \alpha, e, L, S .$ Context answer: \boxed{$A_{1}=\alpha$ , $A_{2}=-\frac{e Z_{e x t} S}{\alpha}$} Extra Supplementary Reading Materials: The ion current $I_{i}(x)$ varies along the $x$-axis according to the formula $$ I_{i}(x)=C_{2}+B_{1} e^{B_{2} x} $$ where $B_{1}, B_{2}, C_{2}$ are some constants. Context question: B2 Find $B_{1}, B_{2}$ and express them in terms of $Z_{\mathrm{ext}}, \alpha, e, L, S, C_{1} Context answer: \boxed{$B_{1}=-C_{1}$ , $B_{2}=\alpha$} Context question: B3 Write down the condition for $I_{i}(x)$ at $x=L$ Context answer: $I_{i}(L)=0$ Context question: B4 Write down the condition for $I_{i}(x)$ and $I_{e}(x)$ at $x=0$ . Context answer: $I_{e}(0)=\gamma I_{i}(0)$ Context question: B5 Find the total current $I$ and express it in terms of $Z_{\mathrm{ext}}, \alpha, \gamma, e, L, S$. Assume that it remains finite Context answer: \boxed{$I=\frac{e Z_{e x t} S}{\alpha}(\frac{1}{e^{-\alpha L}(1+\gamma)-\gamma}-1)$} Extra Supplementary Reading Materials: Let the Townsend coefficient $\alpha$ be constant. When the length of the tube turns out greater than some critical value, i.e. $L>L_{\mathrm{cr}}$, the external ionizer can be turned off and the discharge becomes self-sustained.","B6 Find $L_{c r}$ and express it in terms of $Z_{\text {ext }}, \alpha, \gamma, e, L, S .","['When the discharge gap length is increased, the denominator in formula (B5.1) decreases. At that moment, when it turns zero, the electric current in the gas becomes self-sustaining and external ionizer can be turned off. Thus,\n\n$$\nL_{c r}=\\frac{1}{\\alpha} \\ln \\left(1+\\frac{1}{\\gamma}\\right)\n\\tag{B6.1}\n$$']",['$L_{c r}=\\frac{1}{\\alpha} \\ln (1+\\frac{1}{\\gamma})$'],False,,Expression, 1125,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)].",a) What is the value of the potential on the sphere?,"['As the metallic sphere is grounded, its potential vanishes, $\\mathrm{V}=0$.']",['0'],False,,Numerical,0 1126,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} ","b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$.","[""Let us consider an arbitrary point B on the surface of the sphere as depicted in Fig. 1.\n\n\n\nFig 1. The potential at point $B$ is zero.\n\nThe distance of point $B$ from the charge $\\mathrm{q}^{\\prime}$ is\n\n$$\nr_{1}=\\sqrt{R^{2}+d^{\\prime 2}-2 R d^{\\prime} \\cos \\alpha}\n\\tag{1}\n$$\n\nwhereas the distance of the point $B$ from the charge $q$ is given with the expression\n\n$$\nr_{2}=\\sqrt{R^{2}+d^{2}-2 R d \\cos \\alpha}\n\\tag{2}\n$$\n\nThe electric potential at the point $B$ is\n\n$$\nV=\\frac{1}{4 \\pi \\varepsilon_{0}}\\left(\\frac{q}{r_{2}}+\\frac{q^{\\prime}}{r_{1}}\\right)\n\\tag{3}\n$$\n\nThis potential must vanish,\n\n$$\n\\frac{q}{r_{2}}+\\frac{q^{\\prime}}{r_{1}}=0\n\\tag{4}\n$$\n\ni.e. its numerical value is $0 \\mathrm{~V}$.\n\n\n\nCombining (1), (2) and (3) we obtain\n\n$$\nR^{2}+d^{2}-2 R d \\cos \\alpha=\\left(\\frac{q}{q^{\\prime}}\\right)^{2}\\left(R^{2}+d^{\\prime 2}-2 R d^{\\prime} \\cos \\alpha\\right)\n\\tag{5}\n$$\n\nAs the surface of the sphere must be equipotential, the condition (5) must be satisfied for every angle $\\alpha$ what leads to the following results\n\n$$\nd^{2}+R^{2}=\\left(\\frac{q}{q^{\\prime}}\\right)^{2}\\left(R^{2}+d^{\\prime 2}\\right)\n\\tag{6}\n$$\n\nand\n\n$$\nd R=\\left(\\frac{q}{q^{\\prime}}\\right)^{2}\\left(d^{\\prime} R\\right)\n\\tag{7}\n$$\n\nBy solving of (6) and (7) we obtain the expression for the distance d' of the charge $q^{\\prime}$ from the center of the sphere\n\n$$\nd^{\\prime}=\\frac{R^{2}}{d}\n\\tag{8}\n$$\n\nand the size of the charge $\\mathrm{q}^{\\prime}$\n\n$$\nq^{\\prime}=-q \\frac{R}{d}\n\\tag{9}\n$$""]","['$d^{\\prime}=\\frac{R^{2}}{d}$ , $q^{\\prime}=-q \\frac{R}{d}$']",True,,Expression, 1128,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere.",a) Find the vector of the electric field at point $A$.,['The electric field at the point $A$ amounts to\n\n\n\n$$\n\\vec{E}_{A}=\\left(\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q}{r^{2}}-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q \\frac{R}{d}}{\\left(r-d+\\frac{R^{2}}{d}\\right)^{2}}\\right) \\hat{r}\n\\tag{11}\n$$'],['$\\vec{E}_{A}=\\left(\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q}{r^{2}}-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q \\frac{R}{d}}{\\left(r-d+\\frac{R^{2}}{d}\\right)^{2}}\\right) \\hat{r}$'],False,,Expression, 1129,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} ","b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$.",['For very large distances $r$ we can apply approximate formula $(1+a)^{-2} \\approx 1-2 a$ to the expression (11) what leads us to\n\n$$\n\\vec{E}_{A}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\left(1-\\frac{R}{d}\\right) q}{r^{2}} \\hat{r}-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{2 q \\frac{R}{d}\\left(d-\\frac{R^{2}}{d}\\right)}{r^{3}} \\hat{r}\n$$\n\nIn general a grounded metallic sphere cannot completely screen a point charge $q$ at a distance $d$ (even in the sense that its electric field would decrease with distance faster than $1 / r^{2}$ ) and the dominant dependence of the electric field on the distance $r$ is as in standard Coulomb law.'],['$\\vec{E}_{A}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{\\left(1-\\frac{R}{d}\\right) q}{r^{2}} \\hat{r}-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{2 q \\frac{R}{d}\\left(d-\\frac{R^{2}}{d}\\right)}{r^{3}} \\hat{r}$'],False,,Expression, 1130,Electromagnetism,,"c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero?",['In the limit $d \\rightarrow R$ the electric field at the point $A$ vanishes and the grounded metallic sphere screens the point charge completely.'],['$d \\rightarrow R$'],False,,Need_human_evaluate, 1131,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} Context question: b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$. Context answer: \boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$} Context question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero? Context answer: $d \rightarrow R$ Extra Supplementary Reading Materials: Task 3 - Small oscillations in the electric field of the grounded metallic sphere A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible. Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum.",a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram,['Let us consider a configuration as in Fig. 2.\n\n\n\nFig 2 . The pendulum formed by a charge near a grounded metallic sphere.\n\nThe distance of the charge $q$ from the center of the sphere is\n\n\n\n$$\nd=\\sqrt{l^{2}+L^{2}-2 l L \\cos \\alpha}\n\\tag{13}\n$$\n\nThe magnitude of the electric force acting on the charge $q$ is\n\n$$\nF=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q q^{\\prime}}{\\left(d-d^{\\prime}\\right)^{2}}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R d}{\\left(d^{2}-R^{2}\\right)^{2}}\n\\tag{14}\n$$\n\nFrom which we have\n\n$$\nF=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R \\sqrt{l^{2}+L^{2}-2 l L \\cos \\alpha}}{\\left(l^{2}+L^{2}-2 l L \\cos \\alpha-R^{2}\\right)^{2}}\n\\tag{15}\n$$'],['$F=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R \\sqrt{l^{2}+L^{2}-2 l L \\cos \\alpha}}{\\left(l^{2}+L^{2}-2 l L \\cos \\alpha-R^{2}\\right)^{2}}$'],False,,Expression, 1132,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} Context question: b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$. Context answer: \boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$} Context question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero? Context answer: $d \rightarrow R$ Extra Supplementary Reading Materials: Task 3 - Small oscillations in the electric field of the grounded metallic sphere A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible. Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum. Context question: a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}}$}","b) Determine the component of this force acting in the direction perpendicular to the thread in terms of $l, L, R, q$ and $\alpha$.","['The direction of the vector of the electric force (17) is described in Fig. 3.\n\n\n\nFig 3. The direction of the force $F$.\n\nThe angles $\\alpha$ and $\\beta$ are related as\n\n$$\nL \\sin \\alpha=d \\sin \\beta\n\\tag{16}\n$$\n\nwhereas for the angle $\\gamma$ the relation $\\gamma=\\alpha+\\beta$ is valid. The component of the force perpendicular to the thread is $F \\sin \\gamma$, that is,\n\n\n\n']",['$F_{\\perp}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R \\sqrt{l^{2}+L^{2}-2 l L \\cos \\alpha}}{\\left(l^{2}+L^{2}-2 l L \\cos \\alpha-R^{2}\\right)^{2}} \\sin \\left(\\alpha+\\arcsin (\\frac{L}{\\sqrt{L^{2}+l^{2}-2 L l \\cos \\alpha}} \\sin \\alpha)\\right)$'],False,,Expression, 1133,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} Context question: b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$. Context answer: \boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$} Context question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero? Context answer: $d \rightarrow R$ Extra Supplementary Reading Materials: Task 3 - Small oscillations in the electric field of the grounded metallic sphere A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible. Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum. Context question: a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}}$} Context question: b) Determine the component of this force acting in the direction perpendicular to the thread in terms of $l, L, R, q$ and $\alpha$. Context answer: $F_{\perp}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}} \sin (\alpha+\beta)$ where $\beta=\arcsin \left(\frac{L}{\sqrt{L^{2}+l^{2}-2 L l \cos \alpha}} \sin \alpha\right)$ ",c) Find the frequency for small oscillations of the pendulum.,"['The equation of motion of the mathematical pendulum is\n\n$$\nm L \\ddot{\\alpha}=-F_{\\perp}\n\\tag{18}\n$$\n\nAs we are interested in small oscillations, the angle $\\alpha$ is small, i.e. for its value in radians we have $\\alpha$ much smaller than 1 . For a small value of argument of trigonometric functions we have approximate relations $\\sin \\mathrm{x} \\approx \\mathrm{x}$ and $\\cos \\mathrm{x} \\approx 1-\\mathrm{x}^{2} / 2$. So for small oscillations of the pendulum we have $\\beta \\approx \\alpha L /(l-L)$ and $\\gamma \\approx l \\alpha /(l-L)$.\n\nCombining these relations with (13) we obtain\n\n$$\nm L \\frac{d^{2} \\alpha}{d t^{2}}+\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R d}{\\left(d^{2}-R^{2}\\right)^{2}}\\left(1+\\frac{L}{d}\\right) \\alpha=0\n\\tag{19}\n$$\n\nWhere $d=l-L$ what leads to\n\n$$\n\\begin{aligned}\n& \\omega=\\frac{q}{d^{2}-R^{2}} \\sqrt{\\frac{R d}{4 \\pi \\varepsilon_{0}} \\frac{1}{m L}\\left(1+\\frac{L}{d}\\right)}= \\\\\n& =\\frac{q}{(l-L)^{2}-R^{2}} \\sqrt{\\frac{R l}{4 \\pi \\varepsilon_{0}} \\frac{1}{m L}}\n\\end{aligned}\n\\tag{20}\n$$']",['$\\omega=\\frac{q}{(l-L)^{2}-R^{2}} \\sqrt{\\frac{R l}{4 \\pi \\varepsilon_{0}} \\frac{1}{m L}}$'],False,,Expression, 1134,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} Context question: b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$. Context answer: \boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$} Context question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero? Context answer: $d \rightarrow R$ Extra Supplementary Reading Materials: Task 3 - Small oscillations in the electric field of the grounded metallic sphere A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible. Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum. Context question: a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}}$} Context question: b) Determine the component of this force acting in the direction perpendicular to the thread in terms of $l, L, R, q$ and $\alpha$. Context answer: $F_{\perp}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}} \sin (\alpha+\beta)$ where $\beta=\arcsin \left(\frac{L}{\sqrt{L^{2}+l^{2}-2 L l \cos \alpha}} \sin \alpha\right)$ Context question: c) Find the frequency for small oscillations of the pendulum. Context answer: \boxed{$\omega=\frac{q}{(l-L)^{2}-R^{2}} \sqrt{\frac{R l}{4 \pi \varepsilon_{0}} \frac{1}{m L}}$} Extra Supplementary Reading Materials: Task 4 - The electrostatic energy of the system For a distribution of electric charges it is important to know the electrostatic energy of the system. In our problem (see Fig. 1a), there is an electrostatic interaction between the external charge $q$ and the induced charges on the sphere, and there is an electrostatic interaction among the induced charges on the sphere themselves. In terms of the charge $q$, radius of the sphere $R$ and the distance $d$ determine the following electrostatic energies:",a) the electrostatic energy of the interaction between charge $q$ and the induced charges on the sphere;,"[""The total energy of the system can be separated into the electrostatic energy of interaction of the external charge with the induced charges on the sphere, $\\mathrm{E}_{\\mathrm{el}, 1}$, and the electrostatic energy of mutual interaction of charges on the sphere, $\\mathrm{E}_{\\mathrm{el}, 2}$, i.e.\n\n$$\nE_{e l}=E_{e l, 1}+E_{e l, 2}\n\\tag{21}\n$$\n\nLet there be $N$ charges induced on the sphere. These charges $q_{j}$ are located at points $\\vec{r}_{j}, j=1, \\ldots, N$ on the sphere. We use the definition of the image charge, i.e., the potential on the surface of the sphere from the image charge is identical to the potential arising from the induced charges:\n\n\n\n$$\n\\frac{q^{\\prime}}{\\left|\\vec{r}-\\vec{d}^{\\prime}\\right|}=\\sum_{j=1}^{N} \\frac{q_{j}}{\\left|\\vec{r}_{j}-\\vec{r}\\right|}\n\\tag{22}\n$$\n\nwhere $\\vec{r}$ is a vector on the sphere and $\\vec{d}$ ' denotes the vector position of the image charge. When $\\vec{r}$ coincides with some $\\vec{r}_{i}$, then we just have\n\n| $\\frac{q^{\\prime}}{\\left\\|\\vec{r}_{i}-\\vec{d}^{\\prime}\\right\\|}=\\sum_{\\substack{j=1 \\\\ j \\neq i}}^{N} \\frac{q_{j}}{\\left\\|\\vec{r}_{j}-\\vec{r}_{i}\\right\\|}$. | (23) |\n\nFrom the requirement that the potential on the surface of the sphere vanishes we have\n\n| $\\frac{q^{\\prime}}{\\left\\|\\vec{r}-\\vec{d}^{\\prime}\\right\\|}+\\frac{q}{\\|\\vec{r}-\\vec{d}\\|}=0, \\quad$ (24) |\n\nwhere $\\vec{d}$ denotes the vector position of the charge $\\vec{q}(\\vec{r}$ is on the sphere).\n\nFor the interaction of the external charge with the induced charges on the sphere we have\n\n$$\nE_{e l, 1}=\\frac{q}{4 \\pi \\varepsilon_{0}} \\sum_{i=1}^{N} \\frac{q_{i}}{\\left|\\vec{r}_{i}-\\vec{d}\\right|}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q q^{\\prime}}{\\left|\\vec{d}^{\\prime}-\\vec{d}\\right|}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q q^{\\prime}}{d-d^{\\prime}}=-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}\n\\tag{25} \n$$\n\nHere the first equality is the definition of this energy as the sum of interactions of the charge $q$ with each of the induced charges on the surface of the sphere. The second equality follows from (21).\n\nIn fact, the interaction energy $E_{e l, 1}$ follows directly from the definition of an image charge.""]","['$E_{e l, 1}=-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}$']",True,,Expression, 1135,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} Context question: b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$. Context answer: \boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$} Context question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero? Context answer: $d \rightarrow R$ Extra Supplementary Reading Materials: Task 3 - Small oscillations in the electric field of the grounded metallic sphere A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible. Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum. Context question: a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}}$} Context question: b) Determine the component of this force acting in the direction perpendicular to the thread in terms of $l, L, R, q$ and $\alpha$. Context answer: $F_{\perp}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}} \sin (\alpha+\beta)$ where $\beta=\arcsin \left(\frac{L}{\sqrt{L^{2}+l^{2}-2 L l \cos \alpha}} \sin \alpha\right)$ Context question: c) Find the frequency for small oscillations of the pendulum. Context answer: \boxed{$\omega=\frac{q}{(l-L)^{2}-R^{2}} \sqrt{\frac{R l}{4 \pi \varepsilon_{0}} \frac{1}{m L}}$} Extra Supplementary Reading Materials: Task 4 - The electrostatic energy of the system For a distribution of electric charges it is important to know the electrostatic energy of the system. In our problem (see Fig. 1a), there is an electrostatic interaction between the external charge $q$ and the induced charges on the sphere, and there is an electrostatic interaction among the induced charges on the sphere themselves. In terms of the charge $q$, radius of the sphere $R$ and the distance $d$ determine the following electrostatic energies: Context question: a) the electrostatic energy of the interaction between charge $q$ and the induced charges on the sphere; Context answer: \boxed{$E_{e l, 1}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}$} ",b) the electrostatic energy of the interaction among the induced charges on the sphere;,"['The energy of mutual interactions of induced charges on the surface of the sphere is given with\n\n$$\n\\begin{aligned}\n& E_{e l, 2}=\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\sum_{i=1}^{N} \\sum_{\\substack{j=1 \\\\\nj \\neq i}}^{N} \\frac{q_{i} q_{j}}{\\left|\\vec{r}_{i}-\\vec{r}_{j}\\right|} \\\\\n& =\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\sum_{i=1}^{N} q_{i} \\frac{q^{\\prime}}{\\left|\\vec{r}_{i}-\\vec{d}^{\\prime}\\right|}= \\\\\n& =-\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\sum_{i=1}^{N} q_{i} \\frac{q}{\\left|\\vec{r}_{i}-\\vec{d}\\right|}= \\\\\n& =-\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q q^{\\prime}}{\\left|\\vec{d}^{\\prime}-\\vec{d}\\right|}=-\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q q^{\\prime}}{d-d^{\\prime}}=\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}\n\\end{aligned}\n\\tag{26}\n$$\n\nHere the second line is obtained using (22). From the second line we obtain the third line applying (23), whereas from the third line we obtain the fourth using (22) again.']","['$E_{e l, 2}=\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}$']",True,,Expression, 1136,Electromagnetism,"Image of a charge in a metallic object Introduction - Method of images A point charge $q$ is placed in the vicinity of a grounded metallic sphere of radius $R$ [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge $q^{\prime}$ placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge $q$ ' reproduces the electric field and the potential only outside the sphere (including its surface). Task 1 - The image charge The symmetry of the problem dictates that the charge $q$ ' should be placed on the line connecting the point charge $q$ and the center of the sphere [see Fig. 1(b)]. Context question: a) What is the value of the potential on the sphere? Context answer: \boxed{0} Context question: b) Express $q^{\prime}$ and the distance $d^{\prime}$ of the charge $q^{\prime}$ from the center of the sphere, in terms of $q, d$, and $R$. Context answer: \boxed{$d^{\prime}=\frac{R^{2}}{d}$ , $q^{\prime}=-q \frac{R}{d}$} Context question: c) Find the magnitude of force acting on charge $q$. Is the force repulsive? Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R d}{\left(d^{2}-R^{2}\right)^{2}}$ , attractive} Extra Supplementary Reading Materials: Task 2 - Shielding of an electrostatic field Consider a point charge $q$ placed at a distance $d$ from the center of a grounded metallic sphere of radius $R$. We are interested in how the grounded metallic sphere affects the electric field at point $A$ on the opposite side of the sphere (see Fig. 2). Point $A$ is on the line connecting charge $q$ and the center of the sphere; its distance from the point charge $q$ is $r$. Fig 2 . The electric field at point $A$ is partially screened by the grounded sphere. Context question: a) Find the vector of the electric field at point $A$. Context answer: \boxed{$\vec{E}_{A}=\left(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}-\frac{1}{4 \pi \varepsilon_{0}} \frac{q \frac{R}{d}}{\left(r-d+\frac{R^{2}}{d}\right)^{2}}\right) \hat{r}$} Context question: b) For a very large distance $r>>d$, find the expression for the electric field by using the approximation $(1+a)^{-2} \approx 1-2 a$, where $a<1$. Context answer: \boxed{$\vec{E}_{A}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(1-\frac{R}{d}\right) q}{r^{2}} \hat{r}-\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q \frac{R}{d}\left(d-\frac{R^{2}}{d}\right)}{r^{3}} \hat{r}$} Context question: c) In which limit of $d$ does the grounded metallic sphere screen the field of the charge $q$ completely, such that the electric field at point $A$ is exactly zero? Context answer: $d \rightarrow R$ Extra Supplementary Reading Materials: Task 3 - Small oscillations in the electric field of the grounded metallic sphere A point charge $q$ with mass $m$ is suspended on a thread of length $L$ which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distancel from the center of the sphere. Assume that the effects of gravity are negligible. Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum. Context question: a) Find the magnitude of the electric force acting on the point charge $q$ for a given angle $\alpha$ and indicate the direction in a clear diagram Context answer: \boxed{$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}}$} Context question: b) Determine the component of this force acting in the direction perpendicular to the thread in terms of $l, L, R, q$ and $\alpha$. Context answer: $F_{\perp}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R \sqrt{l^{2}+L^{2}-2 l L \cos \alpha}}{\left(l^{2}+L^{2}-2 l L \cos \alpha-R^{2}\right)^{2}} \sin (\alpha+\beta)$ where $\beta=\arcsin \left(\frac{L}{\sqrt{L^{2}+l^{2}-2 L l \cos \alpha}} \sin \alpha\right)$ Context question: c) Find the frequency for small oscillations of the pendulum. Context answer: \boxed{$\omega=\frac{q}{(l-L)^{2}-R^{2}} \sqrt{\frac{R l}{4 \pi \varepsilon_{0}} \frac{1}{m L}}$} Extra Supplementary Reading Materials: Task 4 - The electrostatic energy of the system For a distribution of electric charges it is important to know the electrostatic energy of the system. In our problem (see Fig. 1a), there is an electrostatic interaction between the external charge $q$ and the induced charges on the sphere, and there is an electrostatic interaction among the induced charges on the sphere themselves. In terms of the charge $q$, radius of the sphere $R$ and the distance $d$ determine the following electrostatic energies: Context question: a) the electrostatic energy of the interaction between charge $q$ and the induced charges on the sphere; Context answer: \boxed{$E_{e l, 1}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}$} Context question: b) the electrostatic energy of the interaction among the induced charges on the sphere; Context answer: \boxed{$E_{e l, 2}=\frac{1}{2} \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2} R}{d^{2}-R^{2}}$} ","c) the total electrostatic energy of the interaction in the system. Hint: There are several ways of solving this problem: (1) In one of them, you can use the following integral, $\int_{d}^{\infty} \frac{x d x}{\left(x^{2}-R^{2}\right)^{2}}=\frac{1}{2} \frac{1}{d^{2}-R^{2}}$. (2) In another one, you can use the fact that for a collection of $N$ charges $q_{i}$ located at points $\vec{r}_{i}, i=1, \ldots, N$, the electrostatic energy is a sum over all pairs of charges: $V=\frac{1}{2} \sum_{i=1}^{N} \sum_{\substack{j=1 \\ i \neq j}}^{N} \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{i} q_{j}}{\left|\vec{r}_{i}-\vec{r}_{j}\right|}$.","['Combining expressions (19) and (20) with the quantitative results for the image charge we finally obtain the total energy of electrostatic interaction\n\n$$\nE_{e l}(d)=-\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}\n\\tag{27}\n$$', '$$\n\\int_{d}^{\\infty} \\frac{x d x}{\\left(x^{2}-R^{2}\\right)^{2}}=\\frac{1}{2} \\frac{1}{d^{2}-R^{2}}\n\\tag{28}\n$$\n\nWe can obtain the total energy in the system by calculating the work needed to bring the charge $q$ from infinity to the distance $d$ from the center of the sphere:\n\n$$\n\\begin{aligned}\n& E_{e l}(d)=-\\int_{\\infty}^{d} F(\\vec{x}) d \\vec{x}=\\int_{d}^{\\infty} F(\\vec{x}) d \\vec{x}= \\\\\n& =\\int_{d}^{\\infty}(-) \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R x}{\\left(x^{2}-R^{2}\\right)^{2}} d x= \\\\\n& =-\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}\n\\end{aligned}\n\\tag{29}\n$$\n\nThis solves Task 4c).\n\nThe electrostatic energy between the charge $q$ and the sphere must be equal to the energy between the charges $q$ and $q^{\\prime}$ according to the definition of the image charge:\n\n\n\n$$\nE_{e l, 1}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q q^{\\prime}}{\\left(d-d^{\\prime}\\right)}=-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}\n\\tag{30}\n$$\n\nThis solves Task 4a).\n\nFrom this we immediately have that the electrostatic energy among the charges on the sphere is:\n\n$$\nE_{e l, 2}=\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}\n\\tag{31}\n$$\n\nThis solves Task 4b).']",['$E_{e l}(d)=-\\frac{1}{2} \\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{q^{2} R}{d^{2}-R^{2}}$'],False,,Expression, 1137,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1","a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney.","['Let $p(z)$ denote the pressure of air at height $z$; then, according to one of the assumptions $p(z)=p(0)-\\rho_{\\text {Air }} g z$, where $p(0)$ is the atmospheric pressure at zero altitude.\n\nThroughout the chimney the Bernoulli law applies, that is, we can write\n\n$$\n\\frac{1}{2} \\rho_{\\text {Smoke }} \\nu(z)^{2}+\\rho_{\\text {Smoke }} g z+p_{\\text {Smoke }}(z)=\\text { const. },\n\\tag{1}\n$$\n\nwhere $p_{\\text {Smoke }}(z)$ is the pressure of smoke at height $z, \\rho_{\\text {Smoke }}$ is its density, and $v(z)$ denotes the velocity of smoke; here we have used the assumption that the density of smoke does not vary throughout the chimney. Now we apply this equation at two points, (i) in the furnace, that is at point $z=-\\varepsilon$, where $\\varepsilon$ is a negligibly small positive number, and (ii) at the top of the chimney where $z=h$ to obtain:\n\n$$\n\\frac{1}{2} \\rho_{\\text {Smoke }} v(h)^{2}+\\rho_{\\text {Smoke }} g h+p_{\\text {Smoke }}(h) \\approx p_{\\text {Smoke }}(-\\varepsilon)\n\\tag{2}\n$$\n\nOn the right hand side we have used the assumption that the velocity of gases in the furnace is negligible (and also $-\\rho_{\\text {Smoke }} g \\varepsilon \\approx 0$ ).\n\nWe are interested in the minimal height at which the chimney will operate. The pressure of smoke at the top of the chimney has to be equal or larger than the pressure of air at altitude $h$; for minimal height of the chimney we have $p_{\\text {Smoke }}(h) \\approx p(h)$. In the furnace we can use $p_{\\text {Smoke }}(-\\varepsilon) \\approx p(0)$. The Bernoulli law applied in the furnace and at the top of the chimney [Eq. (2)] now reads\n\n$$\n\\frac{1}{2} \\rho_{\\text {Smoke }} v(h)^{2}+\\rho_{\\text {Smoke }} g h+p(h) \\approx p(0)\n\\tag{3}\n$$\n\nFrom this we get\n\n$$\nv(h)=\\sqrt{2 g h\\left(\\frac{\\rho_{\\text {Air }}}{\\rho_{\\text {Smoke }}}-1\\right)}\n\\tag{4}\n$$\n\nThe chimney will be efficient if all of its products are released in the atmosphere, i.e.,\n\n\n\n| $v(h) \\geq \\frac{B}{A}$, | $(5)$ |\n| :--- | :--- |\n| from which we have | |\n| $h \\geq \\frac{B^{2}}{A^{2}} \\frac{1}{2 g} \\frac{1}{\\frac{\\rho_{\\text {Air }}}{\\rho_{\\text {Smoke }}}-1}$. | (6) |\n\nWe can treat the smoke in the furnace as an ideal gas (which is at atmospheric pressure $p(0)$ and temperature $\\left.T_{\\text {Smoke }}\\right)$. If the air was at the same temperature and pressure it would have the same density according to our assumptions. We can use this to relate the ratio $\\rho_{\\text {Air }} / \\rho_{\\text {Smoke }}$ to $T_{\\text {Smoke }} / T_{\\text {Air }}$ that is,\n\n| $\\frac{\\rho_{\\text {Air }}}{\\rho_{\\text {Smoke }}}=\\frac{T_{\\text {Smoke }}}{T_{\\text {Air }}}$, and finally | (7) |\n| :--- | :--- |\n| $h \\geq \\frac{B^{2}}{A^{2}} \\frac{1}{2 g} \\frac{T_{\\text {Air }}}{T_{\\text {Smoke }}-T_{\\text {Air }}}=\\frac{B^{2}}{A^{2}} \\frac{1}{2 g} \\frac{T_{\\text {Air }}}{\\Delta T}$. | (8) |\n\nFor minimal height of the chimney we use the equality sign.']",['$\\frac{B^{2}}{A^{2}} \\frac{1}{2 g} \\frac{T_{\\text {Air }}}{\\Delta T}$'],False,,Expression, 1138,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} ","b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney?",['| $\\frac{h(30)}{h(-30)}=\\frac{\\frac{T(30)}{T_{\\text {Smoke }}-T(30)}}{\\frac{T(-30)}{T_{\\text {Smoke }}-T(-30)}} ; h(30)=145 m$. | (9) |\n| :--- | :--- |'],['145'],False,m,Numerical,1e0 1139,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} ",c) How does the velocity of the gases vary along the height of the chimney? Find its expression.,"['The velocity is constant,\n$$\nv=\\sqrt{2gh\\left(\\frac{\\rho_{Air}}{\\rho_{Smoke}}-1 \\right)}=\\sqrt{2gh\\left(\\frac{T_{Smoke}}{T_{Air}}-1 \\right)}=\\sqrt{2gh\\frac{\\Delta T}{T_{Air}}}\n\\tag{10}\n$$\nThis can be seen from the equation of continuity $A v=$ const. ( $\\rho_{\\text {Smoke }}$ is constant). It has a sudden jump from approximately zero velocity to this constant value when the gases enter the chimney from the furnace. In fact, since the chimney operates at minimal height this constant is equal to $B$, that is $v=B / A$.']",['$v=\\sqrt{2gh\\frac{\\Delta T}{T_{Air}}}$'],False,,Expression, 1140,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} Context question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression. Context answer: \boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$} ",d) How does the pressure of the gases vary along the height of the chimney?,"['At some height $z$, from the Bernoulli equation one gets\n$$\np_{\\text {smoke }}(z)=p(0)-\\left(\\rho_{\\text {Air }}-\\rho_{\\text {Smoke }}\\right) g h-\\rho_{\\text {Smoke }} g z\n\\tag{11}\n$$\n\nThus the pressure of smoke suddenly changes as it enters the chimney from the furnace and acquires velocity.']",['$p_{\\text {smoke }}(z)=p(0)-\\left(\\rho_{\\text {Air }}-\\rho_{\\text {Smoke }}\\right) g h-\\rho_{\\text {Smoke }} g z$'],False,,Expression, 1141,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} Context question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression. Context answer: \boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$} Context question: d) How does the pressure of the gases vary along the height of the chimney? Context answer: \boxed{$p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z$} Extra Supplementary Reading Materials: Solar power plant The flow of gases in a chimney can be used to construct a particular kind of solar power plant (solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of area $S$ with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding (thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air through the chimney powers a turbine, resulting in the production of electrical energy. The energy of solar radiation per unit time per unit of horizontal area of the collector is $G$. Assume that all that energy can be used to heat the air in the collector (the mass heat capacity of the air is $c$, and one can neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to its entry into the chimney. Task 2",a) What is the efficiency of the solar chimney power plant?,"['The kinetic energy of the hot air released in a time interval $\\Delta t$ is\n$$\nE_{\\text {kin }}=\\frac{1}{2}\\left(A v \\Delta t \\rho_{\\text {Hot }}\\right) v^{2}=A v \\Delta t \\rho_{\\text {Hot }} g h \\frac{\\Delta T}{T_{\\text {Atm }}}\n\\tag{12}\n$$\n\nWhere the index ""Hot"" refer to the hot air heated by the Sun. If we denote the mass of the air that exits the chimney in unit time with $w=A v \\rho_{\\text {Hot }}$, then the power which corresponds to kinetic energy above is\n$$\nP_{k i n}=w g h \\frac{\\Delta T}{T_{A i r}}\n\\tag{13}\n$$\nThis is the maximal power that can be obtained from the kinetic energy of the gas flow.\n\nThe Sun power used to heat the air is\n\n| $P_{\\text {Sun }}=G S=w c \\Delta T$ | (14) |\n| :---: | :---: |\n| The efficiency is evidently | |\n| $\\eta=\\frac{P_{\\text {kin }}}{P_{\\text {Sun }}}=\\frac{g h}{c T_{A t m}}$ | $(15)$ |']",['$\\eta=\\frac{g h}{c T_{A t m}}$'],False,,Expression, 1143,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} Context question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression. Context answer: \boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$} Context question: d) How does the pressure of the gases vary along the height of the chimney? Context answer: \boxed{$p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z$} Extra Supplementary Reading Materials: Solar power plant The flow of gases in a chimney can be used to construct a particular kind of solar power plant (solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of area $S$ with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding (thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air through the chimney powers a turbine, resulting in the production of electrical energy. The energy of solar radiation per unit time per unit of horizontal area of the collector is $G$. Assume that all that energy can be used to heat the air in the collector (the mass heat capacity of the air is $c$, and one can neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to its entry into the chimney. Task 2 Context question: a) What is the efficiency of the solar chimney power plant? Context answer: \boxed{$\eta=\frac{g h}{c T_{A t m}}$} Context question: b) How the efficiency of the chimney changes with its height, linear or unlinear. Context answer: \boxed{linear} Extra Supplementary Reading Materials: Manzanares prototype The prototype chimney built in Manzanares, Spain, had a height of $195 \mathrm{~m}$, and a radius $5 \mathrm{~m}$. The collector is circular with diameter of $244 \mathrm{~m}$. The specific heat of the air under typical operational conditions of the prototype solar chimney is $1012 \mathrm{~J} / \mathrm{kg} \mathrm{K}$, the density of the hot air is about $0.9 \mathrm{~kg} / \mathrm{m}^{3}$, and the typical temperature of the atmosphere $T_{\text {Air }}=295 \mathrm{~K}$. In Manzanares, the solar power per unit of horizontal surface is typically $150 \mathrm{~W} / \mathrm{m}^{2}$ during a sunny day. Task 3",a) What is the efficiency of the prototype power plant? Write down the numerical estimate.,['The efficiency is\n$$\n\\eta=\\frac{g h}{c T_{A t m}}=0.0064=0.64 \\%\n\\tag{16}\n$$'],['0.64'],False,,Numerical,0 1144,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} Context question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression. Context answer: \boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$} Context question: d) How does the pressure of the gases vary along the height of the chimney? Context answer: \boxed{$p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z$} Extra Supplementary Reading Materials: Solar power plant The flow of gases in a chimney can be used to construct a particular kind of solar power plant (solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of area $S$ with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding (thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air through the chimney powers a turbine, resulting in the production of electrical energy. The energy of solar radiation per unit time per unit of horizontal area of the collector is $G$. Assume that all that energy can be used to heat the air in the collector (the mass heat capacity of the air is $c$, and one can neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to its entry into the chimney. Task 2 Context question: a) What is the efficiency of the solar chimney power plant? Context answer: \boxed{$\eta=\frac{g h}{c T_{A t m}}$} Context question: b) How the efficiency of the chimney changes with its height, linear or unlinear. Context answer: \boxed{linear} Extra Supplementary Reading Materials: Manzanares prototype The prototype chimney built in Manzanares, Spain, had a height of $195 \mathrm{~m}$, and a radius $5 \mathrm{~m}$. The collector is circular with diameter of $244 \mathrm{~m}$. The specific heat of the air under typical operational conditions of the prototype solar chimney is $1012 \mathrm{~J} / \mathrm{kg} \mathrm{K}$, the density of the hot air is about $0.9 \mathrm{~kg} / \mathrm{m}^{3}$, and the typical temperature of the atmosphere $T_{\text {Air }}=295 \mathrm{~K}$. In Manzanares, the solar power per unit of horizontal surface is typically $150 \mathrm{~W} / \mathrm{m}^{2}$ during a sunny day. Task 3 Context question: a) What is the efficiency of the prototype power plant? Write down the numerical estimate. Context answer: \boxed{0.64} ",b) How much power could be produced in the prototype power plant?,['The power is\n$$\nP=G S \\eta=G(D / 2)^{2} \\pi \\eta=45 \\mathrm{~kW}\n\\tag{17}\n$$'],['45'],False,kW,Numerical,1e0 1145,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} Context question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression. Context answer: \boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$} Context question: d) How does the pressure of the gases vary along the height of the chimney? Context answer: \boxed{$p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z$} Extra Supplementary Reading Materials: Solar power plant The flow of gases in a chimney can be used to construct a particular kind of solar power plant (solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of area $S$ with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding (thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air through the chimney powers a turbine, resulting in the production of electrical energy. The energy of solar radiation per unit time per unit of horizontal area of the collector is $G$. Assume that all that energy can be used to heat the air in the collector (the mass heat capacity of the air is $c$, and one can neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to its entry into the chimney. Task 2 Context question: a) What is the efficiency of the solar chimney power plant? Context answer: \boxed{$\eta=\frac{g h}{c T_{A t m}}$} Context question: b) How the efficiency of the chimney changes with its height, linear or unlinear. Context answer: \boxed{linear} Extra Supplementary Reading Materials: Manzanares prototype The prototype chimney built in Manzanares, Spain, had a height of $195 \mathrm{~m}$, and a radius $5 \mathrm{~m}$. The collector is circular with diameter of $244 \mathrm{~m}$. The specific heat of the air under typical operational conditions of the prototype solar chimney is $1012 \mathrm{~J} / \mathrm{kg} \mathrm{K}$, the density of the hot air is about $0.9 \mathrm{~kg} / \mathrm{m}^{3}$, and the typical temperature of the atmosphere $T_{\text {Air }}=295 \mathrm{~K}$. In Manzanares, the solar power per unit of horizontal surface is typically $150 \mathrm{~W} / \mathrm{m}^{2}$ during a sunny day. Task 3 Context question: a) What is the efficiency of the prototype power plant? Write down the numerical estimate. Context answer: \boxed{0.64} Context question: b) How much power could be produced in the prototype power plant? Context answer: \boxed{45} ",c) How much energy could the power plant produce during a typical sunny day?,['If there are 8 sunny hours per day we get $360 \\mathrm{kWh}$.'],['360'],False,kWh,Numerical,1e0 1146,Thermodynamics,"2. Chimney physics Introduction Gaseous products of burning are released into the atmosphere of temperature $T_{\text {Air }}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{\text {smoke }}$. The volume of gases produced per unit time in the furnace is $B$. Assume that: - the velocity of the gases in the furnace is negligibly small - the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal - the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible - the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2} \rho v^{2}(z)+\rho g z+p(z)=$ const, where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height - the change of the density of the gas is negligible throughout the chimney Task 1 Context question: a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B, A, T_{\text {Air, }} g=9.81 \mathrm{~m} / \mathrm{s}^{2}, \Delta T=T_{\text {Smoke }}-T_{\text {Air }}$. Important: in all subsequent tasks assume that this minimal height is the height of the chimney. Context answer: \boxed{$\frac{B^{2}}{A^{2}} \frac{1}{2 g} \frac{T_{\text {Air }}}{\Delta T}$} Context question: b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of $-30{ }^{\circ} \mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of $30^{\circ} \mathrm{C}$. The temperature of the furnace is $400{ }^{\circ} \mathrm{C}$. It was calculated that the height of the chimney designed to work in cold regions is $100 \mathrm{~m}$. How high is the other chimney? Context answer: \boxed{145} Context question: c) How does the velocity of the gases vary along the height of the chimney? Find its expression. Context answer: \boxed{$v=\sqrt{2gh\frac{\Delta T}{T_{Air}}}$} Context question: d) How does the pressure of the gases vary along the height of the chimney? Context answer: \boxed{$p_{\text {smoke }}(z)=p(0)-\left(\rho_{\text {Air }}-\rho_{\text {Smoke }}\right) g h-\rho_{\text {Smoke }} g z$} Extra Supplementary Reading Materials: Solar power plant The flow of gases in a chimney can be used to construct a particular kind of solar power plant (solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of area $S$ with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding (thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air through the chimney powers a turbine, resulting in the production of electrical energy. The energy of solar radiation per unit time per unit of horizontal area of the collector is $G$. Assume that all that energy can be used to heat the air in the collector (the mass heat capacity of the air is $c$, and one can neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to its entry into the chimney. Task 2 Context question: a) What is the efficiency of the solar chimney power plant? Context answer: \boxed{$\eta=\frac{g h}{c T_{A t m}}$} Context question: b) How the efficiency of the chimney changes with its height, linear or unlinear. Context answer: \boxed{linear} Extra Supplementary Reading Materials: Manzanares prototype The prototype chimney built in Manzanares, Spain, had a height of $195 \mathrm{~m}$, and a radius $5 \mathrm{~m}$. The collector is circular with diameter of $244 \mathrm{~m}$. The specific heat of the air under typical operational conditions of the prototype solar chimney is $1012 \mathrm{~J} / \mathrm{kg} \mathrm{K}$, the density of the hot air is about $0.9 \mathrm{~kg} / \mathrm{m}^{3}$, and the typical temperature of the atmosphere $T_{\text {Air }}=295 \mathrm{~K}$. In Manzanares, the solar power per unit of horizontal surface is typically $150 \mathrm{~W} / \mathrm{m}^{2}$ during a sunny day. Task 3 Context question: a) What is the efficiency of the prototype power plant? Write down the numerical estimate. Context answer: \boxed{0.64} Context question: b) How much power could be produced in the prototype power plant? Context answer: \boxed{45} Context question: c) How much energy could the power plant produce during a typical sunny day? Context answer: \boxed{360} Extra Supplementary Reading Materials: Task 4",What is the mass flow rate of air through the system?,['The result can be obtained by expressing the mass flow of air $w$ as\n\n| $w=A v \\rho_{\\text {Hot }}=A \\sqrt{2 g h \\frac{\\Delta T}{T_{\\text {Air }}}} \\rho_{\\text {Hot }}$ | $(18)$ |\n| :--- | :--- |\n| $w=\\frac{G S}{c \\Delta T}$ | $(19)$ |\n\nwhich yields\n$$\n\\Delta T=\\left(\\frac{G^{2} S^{2} T_{A t m}}{A^{2} c^{2} \\rho_{H o t}^{2} 2 g h}\\right)^{1 / 3} \\approx 9.1 \\mathrm{~K} .\n\\tag{20}\n$$\n\nFrom this we get\n$$\nw=760 \\mathrm{~kg} / \\mathrm{s}\n\\tag{21}\n$$'],['$760$'],False,$ \mathrm{~kg} / \mathrm{s}$,Numerical,1e0 1147,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. ","a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)].","['In the SC-system, in each of 8 corners of a given cube there is one unit (atom, nucleon, etc.), but it is shared by 8 neighboring cubes - this gives a total of one nucleon per cube. If nucleons are touching, as we assume in our simplified model, then $a=2 r_{N}$ is the cube edge length $a$. The volume of one nucleon is then\n$$\nV_{N}=\\frac{4}{3} r_{N}^{3} \\pi=\\frac{4}{3}\\left(\\frac{a}{2}\\right)^{3} \\pi=\\frac{4 a^{3}}{3 \\cdot 8} \\pi=\\frac{\\pi}{6} a^{3}\n\\tag{1}\n$$\nfrom which we obtain\n$$\nf=\\frac{V_{N}}{a^{3}}=\\frac{\\pi}{6} \\approx 0.52\n\\tag{2}\n$$']",['$\\frac{\\pi}{6}$'],False,,Numerical,1e-2 1148,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$.","b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$.","['The mass density of the nucleus is:\n$$\n\\rho_{m}=f \\frac{m_{N}}{V_{N}}=0.52 \\cdot \\frac{1.67 \\cdot 10^{-27}}{4 / 3 \\cdot\\left(0.85 \\cdot 10^{-15}\\right)^{3} \\pi} \\approx 3.40 \\cdot 10^{17} \\frac{\\mathrm{kg}}{\\mathrm{m}^{3}}\n\\tag{4}\n$$\nTaking into account the approximation that the number of protons and neutrons is approximately equal, for charge density we get:\n$$\n\\rho_{c}=\\frac{f}{2} \\frac{e}{V_{N}}=\\frac{0.52}{2} \\cdot \\frac{1.6 \\cdot 10^{-19}}{4 / 3 \\cdot\\left(0.85 \\cdot 10^{-15}\\right)^{3} \\pi} \\approx 1.63 \\cdot 10^{25} \\frac{\\mathrm{C}}{\\mathrm{m}^{3}}\n\\tag{5}\n$$\nThe number of nucleons in a given nucleus is $A$. The total volume occupied by the nucleus is:\n$$\nV=\\frac{A V_{N}}{f}\n\\tag{6}\n$$\nwhich gives the following relation between radii of nucleus and the number of nucleons:\n$$\nR=r_{N}\\left(\\frac{A}{f}\\right)^{1 / 3}=\\frac{r_{N}}{f^{1 / 3}} A^{1 / 3}=\\frac{0.85}{0.52^{1 / 3}} A^{1 / 3}=1.06 \\mathrm{fm} \\cdot A^{1 / 3}\n\\tag{7}\n$$\nThe numerical constant $(1.06 \\mathrm{fm})$ in the equation above will be denoted as $r_{0}$ in the sequel.']","['$3.40 \\cdot 10^{17}$ , $1.63 \\cdot 10^{25} $ , $1.06$']",True,"$ \frac{\mathrm{kg}}{\mathrm{m}^{3}}$, $\frac{\mathrm{C}}{\mathrm{m}^{3}}$, $ \mathrm{fm} \cdot A^{1 / 3}$",Numerical,"1e15,1e23,1e-2" 1149,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$. Context question: b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$. Context answer: \boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$} Extra Supplementary Reading Materials: Task 2 - Binding energy of atomic nuclei - volume and surface terms","Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction.",['First one needs to estimate the number of surface nucleons. The surface nucleons are in a spherical shell of width $2 r_{N}$ at the surface. The volume of this shell is\n\n$$\n\\begin{aligned}\nV_{\\text {surface }} & =\\frac{4}{3} R^{3} \\pi-\\frac{4}{3}\\left(R-2 r_{N}\\right)^{3} \\pi= \\\\\n& =\\frac{4}{3} R^{3} \\pi-\\frac{4}{3} R^{3} \\pi+\\frac{4}{3} \\pi 3 R^{2} 2 r_{N}-\\frac{4}{3} \\pi 3 R 4 r_{N}^{2}+\\frac{4}{3} \\pi 8 r_{N}^{3} \\\\\n& =8 \\pi R r_{N}\\left(R-2 r_{N}\\right)+\\frac{4}{3} \\pi 8 r_{N}^{3}= \\\\\n& =8 \\pi\\left(R^{2} r_{N}-2 R r_{N}^{2}+\\frac{4}{3} r_{N}^{3}\\right)\n\\end{aligned}\n\\tag{8}\n$$\nThe number of surface nucleons is:\n$$\n\\begin{aligned}\nA_{\\text {surface }} & =f \\frac{V_{\\text {surface }}}{V_{N}}=f \\frac{8 \\pi\\left(R^{2} r_{N}-2 R r_{N}^{2}+\\frac{4}{3} r_{N}^{3}\\right)}{\\frac{4}{3} r_{N}^{3} \\pi}= \\\\\n& =f 6\\left(\\left(\\frac{R}{r_{N}}\\right)^{2}-2\\left(\\frac{R}{r_{N}}\\right)+\\frac{4}{3}\\right)= \\\\\n& =f 6\\left(\\left(\\frac{A}{f}\\right)^{2 / 3}-2\\left(\\frac{A}{f}\\right)^{1 / 3}+\\frac{4}{3}\\right)= \\\\\n& =6 f^{1 / 3} A^{2 / 3}-12 f^{2 / 3} A^{1 / 3}+8 f= \\\\\n& =6^{2 / 3} \\pi^{1 / 3} A^{2 / 3}-2 \\cdot 6^{1 / 3} \\pi^{2 / 3} A^{1 / 3}+\\frac{4}{3} \\pi \\approx \\\\\n& \\approx 4.84 A^{2 / 3}-7.80 A^{1 / 3}+4.19 .\n\\end{aligned}\n\\tag{9}\n$$\nThe binding energy is now:\n$$\n\\begin{aligned}\n& E_{b}=\\left(A-A_{\\text {surface }}\\right) a_{V}+A_{\\text {surface }} \\frac{a_{V}}{2}= \\\\\n& =A a_{V}-A_{\\text {surface }} \\frac{a_{V}}{2}= \\\\\n& =A a_{V}-\\left(3 f^{1 / 3} A^{2 / 3}-6 f^{2 / 3} A^{1 / 3}+4 f\\right) a_{V}= \\\\\n& =A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}= \\\\\n& =\\left(15.8 A-38.20 A^{2 / 3}+61.58 A^{1 / 3}-33.09\\right) \\mathrm{MeV}\n\\end{aligned}\n\\tag{10}\n$$'],['$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$'],False,,Expression, 1150,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$. Context question: b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$. Context answer: \boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$} Extra Supplementary Reading Materials: Task 2 - Binding energy of atomic nuclei - volume and surface terms Context question: Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$} Extra Supplementary Reading Materials: Task 3 - Electrostatic (Coulomb) effects on the binding energy The electrostatic energy of a homogeneously charged ball (with radius $R$ and total charge $Q_{0}$ ) is $U_{c}=\frac{3 Q_{0}^{2}}{20 \pi \varepsilon_{0} R}$, where $\varepsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} N^{-1} m^{-2}$.","a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing $Z^{2} \rightarrow Z(Z-1)$ in the obtained formula. Use this correction in subsequent tasks.",['Replacing $Q_{0}$ with $Z e$ gives the electrostatic energy of the nucleus as:\n\n| $U_{c}=\\frac{3(Z e)^{2}}{20 \\pi \\varepsilon_{0} R}=\\frac{3 Z^{2} e^{2}}{20 \\pi \\varepsilon_{0} R}$ | (12) |\n| :--- | :--- |\n\nThe fact that each proton is not acting upon itself is taken into account by replacing $Z^{2}$ with $z(z-1)$ :\n\n$$\nU_{c}=\\frac{3 Z(Z-1) e^{2}}{20 \\pi \\varepsilon_{0} R}\n\\tag{13}\n$$'],['$U_{c}=\\frac{3 Z(Z-1) e^{2}}{20 \\pi \\varepsilon_{0} R}$'],False,,Expression, 1151,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$. Context question: b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$. Context answer: \boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$} Extra Supplementary Reading Materials: Task 2 - Binding energy of atomic nuclei - volume and surface terms Context question: Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$} Extra Supplementary Reading Materials: Task 3 - Electrostatic (Coulomb) effects on the binding energy The electrostatic energy of a homogeneously charged ball (with radius $R$ and total charge $Q_{0}$ ) is $U_{c}=\frac{3 Q_{0}^{2}}{20 \pi \varepsilon_{0} R}$, where $\varepsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} N^{-1} m^{-2}$. Context question: a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing $Z^{2} \rightarrow Z(Z-1)$ in the obtained formula. Use this correction in subsequent tasks. Context answer: \boxed{$U_{c}=\frac{3 Z(Z-1) e^{2}}{20 \pi \varepsilon_{0} R}$} ","b) Write down the complete formula for binding energy, including the main (volume) term, the surface correction term and the obtained electrostatic correction.","['In the formula for the electrostatic energy we should replace $R$ with $r_{N} f^{-1 / 3} A^{1 / 3}$ to obtain\n$$\n\\begin{aligned}\n& \\Delta E_{b}=-\\frac{3 e^{2} f^{1 / 3}}{20 \\pi \\varepsilon_{0} r_{N}} \\frac{Z(Z-1)}{A^{1 / 3}}=-\\frac{Z(Z-1)}{A^{1 / 3}} \\cdot 1.31 \\times 10^{-13} \\mathrm{~J} \\\\\n& =-\\frac{Z(Z-1)}{A^{1 / 3}} \\cdot 0.815 \\mathrm{MeV} \\approx-0.204 \\mathrm{~A}^{5 / 3} \\mathrm{MeV}+0.409 \\mathrm{~A}^{2 / 3} \\mathrm{MeV}\n\\end{aligned}\n\\tag{14}\n$$\nwhere $Z \\approx A / 2$ has been used. The Coulomb repulsion reduces the binding energy, hence the negative sign before the first (main) term. The complete formula for binding energy now gives:\n$$\nE_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}-\\frac{3 e^{2} f^{1 / 3}}{20 \\pi \\varepsilon_{0} r_{N}}\\left(\\frac{A^{5 / 3}}{4}-\\frac{A^{2 / 3}}{2}\\right)\n\\tag{15}\n$$']",['$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}-\\frac{3 e^{2} f^{1 / 3}}{20 \\pi \\varepsilon_{0} r_{N}}\\left(\\frac{A^{5 / 3}}{4}-\\frac{A^{2 / 3}}{2}\\right)$'],False,,Expression, 1152,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$. Context question: b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$. Context answer: \boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$} Extra Supplementary Reading Materials: Task 2 - Binding energy of atomic nuclei - volume and surface terms Context question: Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$} Extra Supplementary Reading Materials: Task 3 - Electrostatic (Coulomb) effects on the binding energy The electrostatic energy of a homogeneously charged ball (with radius $R$ and total charge $Q_{0}$ ) is $U_{c}=\frac{3 Q_{0}^{2}}{20 \pi \varepsilon_{0} R}$, where $\varepsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} N^{-1} m^{-2}$. Context question: a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing $Z^{2} \rightarrow Z(Z-1)$ in the obtained formula. Use this correction in subsequent tasks. Context answer: \boxed{$U_{c}=\frac{3 Z(Z-1) e^{2}}{20 \pi \varepsilon_{0} R}$} Context question: b) Write down the complete formula for binding energy, including the main (volume) term, the surface correction term and the obtained electrostatic correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left(\frac{A^{5 / 3}}{4}-\frac{A^{2 / 3}}{2}\right)$} Extra Supplementary Reading Materials: Task 4 - Fission of heavy nuclei Fission is a nuclear process in which a nucleus splits into smaller parts (lighter nuclei). Suppose that a nucleus with $A$ nucleons splits into only two equal parts as depicted in Fig. 2.","a) Calculate the total kinetic energy of the fission products $E_{\text {kin }}$ when the centers of two lighter nuclei are separated by the distance $d \geq 2 R(A / 2)$, where $R(A / 2)$ is their radius. The large nucleus was initially at rest.","['The kinetic energy comes from the difference of binding energies ( 2 small nuclei - the original large one) and the Coulomb energy between two smaller nuclei (with $Z / 2=A / 4$ nucleons each):\n$$\n\\begin{aligned}\n& E_{k i n}(d)=2 E_{b}\\left(\\frac{A}{2}\\right)-E_{b}(A)-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{A^{2} e^{2}}{4 \\cdot 4 \\cdot d}= \\\\\n& =-3 f^{1 / 3} A^{2 / 3} a_{V}\\left(2^{1 / 3}-1\\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\\left(2^{2 / 3}-1\\right) \\\\\n& -4 f a_{V}-\\frac{3 e^{2} f^{1 / 3}}{20 \\pi \\varepsilon_{0} r_{N}}\\left[\\frac{A^{5 / 3}}{4}\\left(2^{-2 / 3}-1\\right)-\\frac{A^{2 / 3}}{2}\\left(2^{1 / 3}-1\\right)\\right] \\\\\n& -\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{A^{2} e^{2}}{16 d}\n\\end{aligned}\n\\tag{16}\n$$\n(notice that the first term, $A a_{v}$, cancels out).']",['$E_{k i n}(d)=-3 f^{1 / 3} A^{2 / 3} a_{V}\\left(2^{1 / 3}-1\\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\\left(2^{2 / 3}-1\\right)-4 f a_{V}-\\frac{3 e^{2} f^{1 / 3}}{20 \\pi \\varepsilon_{0} r_{N}}\\left[\\frac{A^{5 / 3}}{4}\\left(2^{-2 / 3}-1\\right)-\\frac{A^{2 / 3}}{2}\\left(2^{1 / 3}-1\\right)\\right]-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{A^{2} e^{2}}{16 d}$'],False,,Expression, 1153,Modern Physics,,"b) Assume that $d=2 R(A / 2)$ and evaluate the expression for $E_{k i n}$ obtained in part a) for $A=$ 100, 150, 200 and 250 (express the results in units of MeV). Estimate the values of $A$ for which fission is possible in the model described above? ![](https://cdn.mathpix.com/cropped/2023_12_21_75c6e1b0517dbb434e54g-1.jpg?height=591&width=1513&top_left_y=2035&top_left_x=337)","['The kinetic energy when $d=2 R(A / 2)$ is given with:\n$$\n\\begin{aligned}\n& E_{k i n}=2 E_{b}\\left(\\frac{A}{2}\\right)-E_{b}(A)-\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{2^{1 / 3} A^{2} e^{2}}{16 \\cdot 2 r_{N} A^{1 / 3} f^{-1 / 3}}= \\\\\n& =-3 f^{1 / 3} A^{2 / 3} a_{V}\\left(2^{1 / 3}-1\\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\\left(2^{2 / 3}-1\\right) \\\\\n& -4 f a_{V}-\\frac{e^{2} f^{1 / 3}}{\\pi \\varepsilon_{0} r_{N}}\\left[\\frac{3}{80}\\left(2^{-2 / 3}-1\\right)+\\frac{2^{1 / 3}}{128}\\right] A^{5 / 3}-\\frac{e^{2} f^{1 / 3}}{\\pi \\varepsilon_{0} r_{N}}\\left[\\frac{3}{40}\\left(2^{1 / 3}-1\\right)\\right] A^{2 / 3}= \\\\\n& =\\left(0.02203 A^{5 / 3}-10.0365 A^{2 / 3}+36.175 A^{1 / 3}-33.091\\right) \\mathrm{MeV}\n\\end{aligned}\n\\tag{17}\n$$\nNumerically one gets:\n\n$A=100 \\ldots E_{k i n}=-33.95 \\mathrm{MeV}$,\n\n$A=150 \\ldots E_{k i n}=-30.93 \\mathrm{MeV}$,\n\n\n\n$A=200 \\ldots E_{k i n}=-14.10 \\mathrm{MeV}$,\n\n$A=250 \\ldots E_{\\text {kin }}=+15.06 \\mathrm{MeV}$.\n\nIn our model, fission is possible when $E_{k i n}(d=2 R(A / 2)) \\geq 0$. From the numerical evaluations given above, one sees that this happens approximately halfway between $A=200$ and $A=250$ - a rough estimate would be $A \\approx 225$. Precise numerical evaluation of the equation:\n$$\nE_{\\text {kin }}=\\left(0.02203 A^{5 / 3}-10.0365 A^{2 / 3}+36.175 A^{1 / 3}-33.091\\right) \\mathrm{MeV} \\geq 0\n\\tag{18}\n$$\ngives that for $A \\geq 227$ fission is possible.']","['Numerically one gets:\n$A=100 \\ldots E_{k i n}=-33.95 \\mathrm{MeV}$,\n$A=150 \\ldots E_{k i n}=-30.93 \\mathrm{MeV}$,\n$A=200 \\ldots E_{k i n}=-14.10 \\mathrm{MeV}$,\n$A=250 \\ldots E_{\\text {kin }}=+15.06 \\mathrm{MeV}$.\n\nand $A \\geq 227$']",True,,Need_human_evaluate, 1154,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$. Context question: b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$. Context answer: \boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$} Extra Supplementary Reading Materials: Task 2 - Binding energy of atomic nuclei - volume and surface terms Context question: Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$} Extra Supplementary Reading Materials: Task 3 - Electrostatic (Coulomb) effects on the binding energy The electrostatic energy of a homogeneously charged ball (with radius $R$ and total charge $Q_{0}$ ) is $U_{c}=\frac{3 Q_{0}^{2}}{20 \pi \varepsilon_{0} R}$, where $\varepsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} N^{-1} m^{-2}$. Context question: a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing $Z^{2} \rightarrow Z(Z-1)$ in the obtained formula. Use this correction in subsequent tasks. Context answer: \boxed{$U_{c}=\frac{3 Z(Z-1) e^{2}}{20 \pi \varepsilon_{0} R}$} Context question: b) Write down the complete formula for binding energy, including the main (volume) term, the surface correction term and the obtained electrostatic correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left(\frac{A^{5 / 3}}{4}-\frac{A^{2 / 3}}{2}\right)$} Extra Supplementary Reading Materials: Task 4 - Fission of heavy nuclei Fission is a nuclear process in which a nucleus splits into smaller parts (lighter nuclei). Suppose that a nucleus with $A$ nucleons splits into only two equal parts as depicted in Fig. 2. Context question: a) Calculate the total kinetic energy of the fission products $E_{\text {kin }}$ when the centers of two lighter nuclei are separated by the distance $d \geq 2 R(A / 2)$, where $R(A / 2)$ is their radius. The large nucleus was initially at rest. Context answer: \boxed{$E_{k i n}(d)=-3 f^{1 / 3} A^{2 / 3} a_{V}\left(2^{1 / 3}-1\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\left(2^{2 / 3}-1\right)-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left[\frac{A^{5 / 3}}{4}\left(2^{-2 / 3}-1\right)-\frac{A^{2 / 3}}{2}\left(2^{1 / 3}-1\right)\right]-\frac{1}{4 \pi \varepsilon_{0}} \frac{A^{2} e^{2}}{16 d}$} Context question: b) Assume that $d=2 R(A / 2)$ and evaluate the expression for $E_{k i n}$ obtained in part a) for $A=$ 100, 150, 200 and 250 (express the results in units of MeV). Estimate the values of $A$ for which fission is possible in the model described above? Context answer: Numerically one gets: $A=100 \ldots E_{k i n}=-33.95 \mathrm{MeV}$, $A=150 \ldots E_{k i n}=-30.93 \mathrm{MeV}$, $A=200 \ldots E_{k i n}=-14.10 \mathrm{MeV}$, $A=250 \ldots E_{\text {kin }}=+15.06 \mathrm{MeV}$. and $A \geq 227$ Extra Supplementary Reading Materials: Task 5 - Transfer reactions","a) In modern physics, the energetics of nuclei and their reactions is described in terms of masses. For example, if a nucleus (with zero velocity) is in an excited state with energy $E_{\text {exc }}$ above the ground state, its mass is $m=m_{0}+E_{e x c} / c^{2}$, where $m_{0}$ is its mass in the ground state at rest. The nuclear reaction ${ }^{16} \mathrm{O}+{ }^{54} \mathrm{Fe} \rightarrow{ }^{12} \mathrm{C}+{ }^{58} \mathrm{Ni}$ is an example of the so-called ""transfer reactions"", in which a part of one nucleus (""cluster"") is transferred to the other (see Fig. 3). In our example the transferred part is a ${ }^{4} \mathrm{He}-\mathrm{cluster}(\alpha$-particle). The transfer reactions occur with maximum probability if the velocity of the projectile-like reaction product (in our case: ${ }^{12} \mathrm{C}$ ) is equal both in magnitude and direction to the velocity of projectile (in our case: ${ }^{16} \mathrm{O}$ ). The target ${ }^{54} \mathrm{Fe}$ is initially at rest. In the reaction, ${ }^{58} \mathrm{Ni}$ is excited into one of its higher-lying states. Find the excitation energy of that state (and express it units of $\mathrm{MeV}$ ) if the kinetic energy of the projectile ${ }^{16} \mathrm{O}$ is $50 \mathrm{MeV}$. The speed of light is $\mathrm{c}=3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$. | 1. | $\mathrm{M}\left({ }^{16} \mathrm{O}\right)$ | 15.99491 a.m.u. | | ---: | ---: | :--- | | 2. | $\mathrm{M}\left({ }^{54} \mathrm{Fe}\right)$ | 53.93962 a.m.u. | | 3. | $\mathrm{M}\left({ }^{12} \mathrm{C}\right)$ | 12.00000 a.m.u. | | 4. | $\mathrm{M}\left({ }^{58} \mathrm{Ni}\right)$ | 57.93535 a.m.u. | | { Table 1. The rest masses of the reactants in their ground states. 1 a.m.u. $=1.6605 \cdot 10^{-27} \mathrm{~kg}$}. | | |","[""Non-relativistic solution\n\nFirst one has to find the amount of mass transferred to energy in the reaction (or the energy equivalent, so-called Q-value):\n$$\n\\begin{aligned}\n\\Delta m & =(\\text { total mass })_{\\text {after reaction }}-(\\text { total mass })_{\\text {before reaction }}= \\\\\n& =(57.93535+12.00000) \\text { a.m.u. }-(53.93962+15.99491) \\text { a.m.u. }= \\\\\n& =0.00082 \\text { a.m.u. }= \\\\\n& =1.3616 \\cdot 10^{-30} \\mathrm{~kg} .\n\\end{aligned}\n\\tag{19}\n$$\nUsing the Einstein formula for equivalence of mass and energy, we get:\n$$\n\\begin{aligned}\nQ & =(\\text { total kinetic energy })_{\\text {after reaction }}-(\\text { total kinetic energy })_{\\text {before reaction }}= \\\\\n& =-\\Delta m \\cdot c^{2}= \\\\\n& =-1.3616 \\cdot 10^{-30} \\cdot 299792458^{2}=-1.2237 \\cdot 10^{-13} \\mathrm{~J}\n\\end{aligned}\n\\tag{20}\n$$\nTaking into account that $1 \\mathrm{MeV}$ is equal to $1.602 \\cdot 10^{-13} \\mathrm{~J}$, we get:\n$$\nQ=-1.2237 \\cdot 10^{-13} / 1.602 \\cdot 10^{-13}=-0.761 \\mathrm{MeV}\n\\tag{21}\n$$\nThis exercise is now solved using the laws of conservation of energy and momentum. The latter gives (we are interested only for the case when ${ }^{12} \\mathrm{C}$ and ${ }^{16} \\mathrm{O}$ are having the same direction so we don't need to use vectors):\n$$\nm\\left({ }^{16} \\mathrm{O}\\right) v\\left({ }^{16} \\mathrm{O}\\right)=m\\left({ }^{12} \\mathrm{C}\\right) v\\left({ }^{12} \\mathrm{C}\\right)+m\\left({ }^{58} \\mathrm{Ni}\\right) v\\left({ }^{58} \\mathrm{Ni}\\right)\n\\tag{22}\n$$\nwhile the conservation of energy gives:\n$$\nE_{k}\\left({ }^{16} \\mathrm{O}\\right)+Q=E_{k}\\left({ }^{12} \\mathrm{C}\\right)+E_{k}\\left({ }^{58} \\mathrm{Ni}\\right)+E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)\n\\tag{23}\n$$\n\nwhere $E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)$ is the excitation energy of ${ }^{58} \\mathrm{Ni}$, and $Q$ is calculated in the first part of this task. But since ${ }^{12} \\mathrm{C}$ and ${ }^{16} \\mathrm{O}$ have the same velocity, conservation of momentum reduced to:\n$$\n\\left[m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right] v\\left({ }^{16} \\mathrm{O}\\right)=m\\left({ }^{58} \\mathrm{Ni}\\right) v\\left({ }^{58} \\mathrm{Ni}\\right)\n\\tag{24}\n$$\nNow we can easily find the kinetic energy of ${ }^{58} \\mathrm{Ni}$ :\n$$\n\\begin{aligned}\nE_{k}\\left({ }^{58} \\mathrm{Ni}\\right) & =\\frac{m\\left({ }^{58} \\mathrm{Ni}\\right) v^{2}\\left({ }^{58} \\mathrm{Ni}\\right)}{2}=\\frac{\\left[m\\left({ }^{58} \\mathrm{Ni}\\right) v\\left({ }^{58} \\mathrm{Ni}\\right)\\right]^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}= \\\\\n& =\\frac{\\left.\\left[ m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right] v\\left({ }^{(6} \\mathrm{O}\\right)\\right]^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}= \\\\\n& =E_{k}\\left({ }^{16} \\mathrm{O}\\right) \\frac{\\left[m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right]^{2}}{m\\left({ }^{58} \\mathrm{Ni}\\right) m\\left({ }^{16} \\mathrm{O}\\right)}\n\\end{aligned}\n\\tag{25}\n$$\nand finally the excitation energy of ${ }^{58} \\mathrm{Ni}$ :\n$$\n\\begin{aligned}\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right) & =E_{k}\\left({ }^{16} \\mathrm{O}\\right)+Q-E_{k}\\left({ }^{12} \\mathrm{C}\\right)-E_{k}\\left({ }^{58} \\mathrm{Ni}\\right)= \\\\\n& =E_{k}\\left({ }^{16} \\mathrm{O}\\right)+Q-\\frac{m\\left({ }^{12} \\mathrm{C}\\right) v^{2}\\left({ }^{16} \\mathrm{O}\\right)}{2}-E_{k}\\left({ }^{16} \\mathrm{O}\\right) \\frac{\\left[m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right]^{2}}{m\\left({ }^{58} \\mathrm{Ni}\\right) m\\left({ }^{16} \\mathrm{O}\\right)}= \\\\\n& =Q+E_{k}\\left({ }^{16} \\mathrm{O}\\right)-E_{k}\\left({ }^{16} \\mathrm{O}\\right) \\cdot \\frac{m\\left({ }^{12} \\mathrm{C}\\right)}{m\\left({ }^{16} \\mathrm{O}\\right)}-E_{k}\\left({ }^{16} \\mathrm{O}\\right) \\frac{\\left[m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right]^{2}}{m\\left({ }^{58} \\mathrm{Ni}\\right) m\\left({ }^{16} \\mathrm{O}\\right)}= \\\\\n& =Q+E_{k}\\left({ }^{16} \\mathrm{O}\\right)\\left[1-\\frac{m\\left({ }^{12} \\mathrm{C}\\right)}{m\\left({ }^{16} \\mathrm{O}\\right)}-\\frac{\\left[m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right]^{2}}{m\\left({ }^{58} \\mathrm{Ni}\\right) m\\left({ }^{16} \\mathrm{O}\\right)}\\right]= \\\\\n& =Q+E_{k}\\left({ }^{16} \\mathrm{O}\\right) \\frac{\\left[m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right] \\cdot\\left[m\\left({ }^{58} \\mathrm{Ni}\\right)-m\\left({ }^{16} \\mathrm{O}\\right)+m\\left({ }^{12} \\mathrm{C}\\right)\\right]}{m\\left({ }^{58} \\mathrm{Ni}\\right) m\\left({ }^{16} \\mathrm{O}\\right)}\n\\end{aligned}\n\\tag{26}\n$$\nNote that the first bracket in numerator is approximately equal to the mass of transferred particle (the ${ }^{4} \\mathrm{He}$ nucleus), while the second one is approximately equal to the mass of target nucleus ${ }^{54} \\mathrm{Fe}$. Inserting the numbers we get:\n$$\n\\begin{aligned}\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right) & =-0.761+50 \\cdot \\frac{(15.99491-12 .)(57.93535-15.99491+12 .)}{57.93535 \\cdot 15.99491}= \\\\\n& =10.866 \\mathrm{MeV}\n\\end{aligned}\n\\tag{27}\n$$"", 'Relativistic solution \n\nIn the relativistic version, solution is found starting from the following pair of equations (the first one is the law of conservation of energy and the second one the law of conservation of momentum):\n$$\nm\\left({ }^{54} \\mathrm{Fe}\\right) \\cdot c^{2}+\\frac{m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}{\\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}}=\\frac{m\\left({ }^{12} \\mathrm{C}\\right) \\cdot c^{2}}{\\sqrt{1-v^{2}\\left({ }^{12} \\mathrm{C}\\right) / c^{2}}}+\\frac{m^{*}\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot c^{2}}{\\sqrt{1-v^{2}\\left({ }^{58} \\mathrm{Ni}\\right) / c^{2}}}\n\n\\frac{m\\left({ }^{16} \\mathrm{O}\\right) \\cdot v\\left({ }^{(16} \\mathrm{O}\\right)}{\\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}}=\\frac{\\left.m\\left({ }^{12} \\mathrm{C}\\right) \\cdot v^{(12} \\mathrm{C}\\right)}{\\sqrt{1-v^{2}\\left({ }^{12} \\mathrm{C}\\right) / c^{2}}}+\\frac{m^{*}\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot v\\left({ }^{58} \\mathrm{Ni}\\right)}{\\sqrt{1-v^{2}\\left({ }^{58} \\mathrm{Ni}\\right) / c^{2}}} \\quad\n\\tag{28}\n$$\nAll the masses in the equations are the rest masses; the ${ }^{58} \\mathrm{Ni}$ is NOT in its ground-state, but in one of its excited states (having the mass denoted with $m^{*}$ ). Since ${ }^{12} \\mathrm{C}$ and ${ }^{16} \\mathrm{O}$ have the same velocity, this set of equations reduces to:\n$$\n\\begin{aligned}\n& m\\left({ }^{54} \\mathrm{Fe}\\right)+\\frac{m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)}{\\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}}=\\frac{m^{*}\\left({ }^{58} \\mathrm{Ni}\\right)}{\\sqrt{1-v^{2}\\left({ }^{58} \\mathrm{Ni}\\right) / c^{2}}} \\\\\n& \\frac{\\left(m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right) \\cdot v\\left({ }^{16} \\mathrm{O}\\right)}{\\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}}=\\frac{m^{*}\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot v\\left({ }^{58} \\mathrm{Ni}\\right)}{\\sqrt{1-v^{2}\\left({ }^{58} \\mathrm{Ni}\\right) / c^{2}}}\n\\end{aligned}\n\\tag{29}\n$$\nDividing the second equation with the first one gives:\n$$\nv\\left({ }^{58} \\mathrm{Ni}\\right)=\\frac{\\left(m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right) \\cdot v\\left({ }^{16} \\mathrm{O}\\right)}{\\left(m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right)+m\\left({ }^{54} \\mathrm{Fe}\\right) \\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}}\n\\tag{30}\n$$\nThe velocity of projectile can be calculated from its energy:\n$$\n\\begin{aligned}\n& E_{k i n}\\left({ }^{16} \\mathrm{O}\\right)=\\frac{m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}{\\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}}-m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2} \\\\\n& \\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}=\\frac{m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}{E_{k i n}\\left({ }^{16} \\mathrm{O}\\right)+m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}} \\\\\n& v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}=1-\\left(\\frac{m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}{E_{k i n}\\left({ }^{16} \\mathrm{O}\\right)+m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}\\right)^{2} \\\\\n& v\\left({ }^{16} \\mathrm{O}\\right)=\\sqrt{1-\\left(\\frac{m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}{E_{k i n}\\left({ }^{16} \\mathrm{O}\\right)+m\\left({ }^{16} \\mathrm{O}\\right) \\cdot c^{2}}\\right)^{2}} \\cdot c\n\\end{aligned}\n\\tag{31}\n$$\nFor the given numbers we get:\n$$\n\\begin{aligned}v\\left({ }^{16} \\mathrm{O}\\right) & =\\sqrt{1-\\left(\\frac{15.99491 \\cdot 1.6605 \\cdot 10^{-27} \\cdot\\left(2.9979 \\cdot 10^{8}\\right)^{2}}{50 \\cdot 1.602 \\cdot 10^{-13}+15.99491 \\cdot\\left(2.9979 \\cdot 10^{8}\\right)^{2}}\\right)^{2}} \\cdot c= \\\\ & =\\sqrt{1-0.99666^{2} \\cdot c=0.08172 \\cdot c=2.4498 \\cdot 10^{7} \\mathrm{~km} / \\mathrm{s}}\\end{aligned}\n\\tag{32}\n$$\nNow we can calculate:\n$$\nv\\left({ }^{58} \\mathrm{Ni}\\right)=\\frac{(15.99491-12.0) \\cdot 2.4498 \\cdot 10^{7} \\mathrm{~km} / \\mathrm{s}}{(15.99491-12.0)+53.93962 \\sqrt{1-0.08172^{2}}}=1.6946 \\cdot 10^{6} \\mathrm{~km} / \\mathrm{s}\n\\tag{33}\n$$\n\nThe mass of ${ }^{58} \\mathrm{Ni}$ in its excited state is then:\n$$\n\\begin{aligned}\nm^{*}\\left({ }^{58} \\mathrm{Ni}\\right) & =\\left(m\\left({ }^{16} \\mathrm{O}\\right)-m\\left({ }^{12} \\mathrm{C}\\right)\\right) \\frac{\\sqrt{1-v^{2}\\left({ }^{58} \\mathrm{Ni}\\right) / c^{2}}}{\\sqrt{1-v^{2}\\left({ }^{16} \\mathrm{O}\\right) / c^{2}}} \\cdot \\frac{v\\left({ }^{16} \\mathrm{O}\\right)}{v\\left({ }^{58} \\mathrm{Ni}\\right)}= \\\\\n& =(15.99491-12.0) \\frac{\\sqrt{1-\\left(1.6945 \\cdot 10^{6} / 2.9979 \\cdot 10^{8}\\right)^{2}}}{\\sqrt{1-0.08172^{2}}} \\cdot \\frac{2.4498 \\cdot 10^{7}}{1.6945 \\cdot 10^{6}} \\text { a.m.u. }= \\\\\n& =57.9470 \\text { a.m.u. }\n\\end{aligned}\n\\tag{34}\n$$\nThe excitation energy of ${ }^{58} \\mathrm{Ni}$ is then:\n$$\n\\begin{aligned}\nE_{x} & =\\left[m^{*}\\left({ }^{58} \\mathrm{Ni}\\right)-m\\left({ }^{58} \\mathrm{Ni}\\right)\\right] \\cdot c^{2}=(57.9470-57.93535) \\cdot 1.6605 \\cdot 10^{-27}\\left(2.9979 \\cdot 10^{8}\\right)^{2} \\\\\n& =2.00722 \\cdot 10^{-12} / 1.602 \\cdot 10^{-13} \\mathrm{MeV} / \\mathrm{J}=10.8636 \\mathrm{MeV}\n\\end{aligned}\n\\tag{35}\n$$\nThe relativistic and non-relativistic results are equal within $2 \\mathrm{keV}$ so both can be considered as correct -we can conclude that at the given beam energy, relativistic effects are not important.']",['10.8636'],False,MeV,Numerical,1e-3 1155,Modern Physics,,"b) The ${ }^{58} \mathrm{Ni}$ nucleus produced in the excited state discussed in the part a), deexcites into its ground state by emitting a gamma-photon in the direction of its motion. Consider this decay in the frame of reference in which ${ }^{58} \mathrm{Ni}$ is at rest to find the recoil energy of ${ }^{58} \mathrm{Ni}$ (i.e. kinetic energy which ${ }^{58} \mathrm{Ni}$ acquires after the emission of the photon). What is the photon energy in that system? What is the photon energy in the lab system of reference (i.e. what would be the energy of the photon measured in the detector which is positioned in the direction in which the ${ }^{58} \mathrm{Ni}$ nucleus moves)? ![](https://cdn.mathpix.com/cropped/2023_12_21_566d5d2f42cbf0437902g-1.jpg?height=714&width=1459&top_left_y=1813&top_left_x=296)","['For gamma-emission from the static nucleus, laws of conservation of energy and momentum give:\n$$\n\\begin{array}{r}\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=E_{\\gamma}+E_{\\text {recoil }} \\\\\np_{\\gamma}=p_{\\text {recoil }}\n\\end{array}\n\\tag{36}\n$$\nGamma-ray and recoiled nucleus have, of course, opposite directions. For gamma-ray (photon), energy and momentum are related as:\n$$\nE_{\\gamma}=p_{\\gamma} \\cdot c\n\\tag{37}\n$$\nIn part a) we have seen that the nucleus motion in this energy range is not relativistic, so we have:\n$$\nE_{\\text {recoil }}=\\frac{p_{\\text {recii }}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}=\\frac{p_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}=\\frac{E_{\\gamma}^{2}}{2 m\\left({ }^{(58} \\mathrm{Ni}\\right) \\cdot c^{2}}\n\\tag{38}\n$$\nInserting this into law of energy conservation Eq. (36), we get:\n$$\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=E_{\\gamma}+E_{\\text {recoil }}=E_{\\gamma}+\\frac{E_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot c^{2}}\n\\tag{39}\n$$\nThis reduces to the quadratic equation:\n$$\nE^2_{\\gamma}+2m({}^{58}Ni)c^2\\cdot E_{\\gamma}+2m({}^{58}Ni)c^2 E_x({}^{58}Ni)=0\n\\tag{40}\n$$\n\n$$\n\\begin{aligned}\nE_{\\gamma} &= \\frac{-2m({}^{58}Ni)c^2+\\sqrt{4(m({}^{58}Ni)c^2)^2+8m({}^{58}Ni)c^2E_x({}^{58}Ni)}}{2} \\\\\n&=\\sqrt{(m({}^{58}Ni)c^2)^2+2m({}^{58}Ni)c^2E_x({}^{58}Ni)}-m({}^{58}Ni)c^2\n\\end{aligned}\n\\tag{41}\n$$\nInserting numbers gives:\n$$\nE_{\\gamma}=10.8633 \\mathrm{MeV}\n\\tag{42}\n$$\nThe equation (37) can also be reduced to an approximate equation before inserting numbers:\n$$\nE_{\\gamma}=E_{x}\\left(1-\\frac{E_{x}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}}\\right)=10.8633 \\mathrm{MeV}\n\\tag{43}\n$$\nThe recoil energy is now easily found as:\n$$\nE_{\\text {recoil }}=E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)-E_{\\gamma}=1.1 \\mathrm{keV}\n\\tag{44}\n$$\nDue to the fact that nucleus emitting gamma-ray $\\left({ }^{58} \\mathrm{Ni}\\right)$ is moving with the high velocity, the energy of gamma ray will be changed because of the Doppler effect. The relativistic Doppler effect (when source is moving towards observer/detector) is given with this formula:\n$$\nf_{\\text {detector }}=f_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}\n\\tag{45}\n$$\nand since there is a simple relation between photon energy and frequency ( $E=h f)$, we get the similar expression for energy:\n$$\nE_{\\text {detector }}=E_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}\n\\tag{46}\n$$\nwhere $\\beta=v / c$ and $v$ is the velocity of emitter (the ${ }^{58} \\mathrm{Ni}$ nucleus). Taking the calculated value of the ${ }^{58} \\mathrm{Ni}$ velocity (equation 29 ) we get:\n$$\nE_{\\text {detector }}=E_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}=10.863 \\sqrt{\\frac{1+0.00565}{1-0.00565}}=10.925 \\mathrm{MeV}\n\\tag{47}\n$$']","['10.8633, 10.925']",True,MeV,Numerical,1e-3 1155,Modern Physics,"3. Simple model of an atomic nucleus Introduction Although atomic nuclei are quantum objects, a number of phenomenological laws for their basic properties (like radius or binding energy) can be deduced from simple assumptions: (i) nuclei are built from nucleons (i.e. protons and neutrons); (ii) strong nuclear interaction holding these nucleons together has a very short range (it acts only between neighboring nucleons); (iii) the number of protons $(Z)$ in a given nucleus is approximately equal to the number of neutrons $(N)$, i.e. $Z \approx N \approx A / 2$, where $A$ is the total number of nucleons $(A \gg 1)$. Important: Use these assumptions in Tasks 1-4 below. Task 1 - Atomic nucleus as closely packed system of nucleons In a simple model, an atomic nucleus can be thought of as a ball consisting of closely packed nucleons [see Fig. 1(a)], where the nucleons are hard balls of radius $r_{N}=0.85 \mathrm{fm}\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)$. The nuclear force is present only for two nucleons in contact. The volume of the nucleus $V$ is larger than the volume of all nucleons $A V_{N}$, where $V_{N}=\frac{4}{3} r_{N}^{3} \pi$. The ratio $f=A V_{N} / V$ is called the packing factor and gives the percentage of space filled by the nuclear matter. Context question: a) Calculate what would be the packing factor $f$ if nucleons were arranged in a ""simple cubic"" (SC) crystal system, where each nucleon is centered on a lattice point of an infinite cubic lattice [see Fig. 1(b)]. Context answer: \boxed{$\frac{\pi}{6}$} Extra Supplementary Reading Materials: Important: In all subsequent tasks, assume that the actual packing factor for nuclei is equal to the one from Task 1a. If you are not able to calculate it, in subsequent tasks use $f=1 / 2$. Context question: b) Estimate the average mass density $\rho_{m}$, charge density $\rho_{c}$, and the radius $R$ for a nucleus having $A$ nucleons. The average mass of a nucleon is $1.67 \cdot 10^{-27} \mathrm{~kg}$. Context answer: \boxed{$3.40 \cdot 10^{17}$ , $1.63 \cdot 10^{25} $ , $1.06$} Extra Supplementary Reading Materials: Task 2 - Binding energy of atomic nuclei - volume and surface terms Context question: Binding energy of a nucleus is the energy required to disassemble it into separate nucleons and it essentially comes from the attractive nuclear force of each nucleon with its neighbors. If a given nucleon is not on the surface of the nucleus, it contributes to the total binding energy with $a_{V}=15.8$ $\mathrm{MeV}\left(1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right)$. The contribution of one surface nucleon to the binding energy is approximately $a_{V} / 2$. Express the binding energy $E_{b}$ of a nucleus with $A$ nucleons in terms of $A, a_{V}$, and $f$, and by including the surface correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}$} Extra Supplementary Reading Materials: Task 3 - Electrostatic (Coulomb) effects on the binding energy The electrostatic energy of a homogeneously charged ball (with radius $R$ and total charge $Q_{0}$ ) is $U_{c}=\frac{3 Q_{0}^{2}}{20 \pi \varepsilon_{0} R}$, where $\varepsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} N^{-1} m^{-2}$. Context question: a) Apply this formula to get the electrostatic energy of a nucleus. In a nucleus, each proton is not acting upon itself (by Coulomb force), but only upon the rest of the protons. One can take this into account by replacing $Z^{2} \rightarrow Z(Z-1)$ in the obtained formula. Use this correction in subsequent tasks. Context answer: \boxed{$U_{c}=\frac{3 Z(Z-1) e^{2}}{20 \pi \varepsilon_{0} R}$} Context question: b) Write down the complete formula for binding energy, including the main (volume) term, the surface correction term and the obtained electrostatic correction. Context answer: \boxed{$E_{b}=A a_{V}-3 f^{1 / 3} A^{2 / 3} a_{V}+6 f^{2 / 3} A^{1 / 3} a_{V}-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left(\frac{A^{5 / 3}}{4}-\frac{A^{2 / 3}}{2}\right)$} Extra Supplementary Reading Materials: Task 4 - Fission of heavy nuclei Fission is a nuclear process in which a nucleus splits into smaller parts (lighter nuclei). Suppose that a nucleus with $A$ nucleons splits into only two equal parts as depicted in Fig. 2. Context question: a) Calculate the total kinetic energy of the fission products $E_{\text {kin }}$ when the centers of two lighter nuclei are separated by the distance $d \geq 2 R(A / 2)$, where $R(A / 2)$ is their radius. The large nucleus was initially at rest. Context answer: \boxed{$E_{k i n}(d)=-3 f^{1 / 3} A^{2 / 3} a_{V}\left(2^{1 / 3}-1\right)+6 f^{2 / 3} A^{1 / 3} a_{V}\left(2^{2 / 3}-1\right)-4 f a_{V}-\frac{3 e^{2} f^{1 / 3}}{20 \pi \varepsilon_{0} r_{N}}\left[\frac{A^{5 / 3}}{4}\left(2^{-2 / 3}-1\right)-\frac{A^{2 / 3}}{2}\left(2^{1 / 3}-1\right)\right]-\frac{1}{4 \pi \varepsilon_{0}} \frac{A^{2} e^{2}}{16 d}$} Context question: b) Assume that $d=2 R(A / 2)$ and evaluate the expression for $E_{k i n}$ obtained in part a) for $A=$ 100, 150, 200 and 250 (express the results in units of MeV). Estimate the values of $A$ for which fission is possible in the model described above? Context answer: Numerically one gets: $A=100 \ldots E_{k i n}=-33.95 \mathrm{MeV}$, $A=150 \ldots E_{k i n}=-30.93 \mathrm{MeV}$, $A=200 \ldots E_{k i n}=-14.10 \mathrm{MeV}$, $A=250 \ldots E_{\text {kin }}=+15.06 \mathrm{MeV}$. and $A \geq 227$ Extra Supplementary Reading Materials: Task 5 - Transfer reactions Context question: a) In modern physics, the energetics of nuclei and their reactions is described in terms of masses. For example, if a nucleus (with zero velocity) is in an excited state with energy $E_{\text {exc }}$ above the ground state, its mass is $m=m_{0}+E_{e x c} / c^{2}$, where $m_{0}$ is its mass in the ground state at rest. The nuclear reaction ${ }^{16} \mathrm{O}+{ }^{54} \mathrm{Fe} \rightarrow{ }^{12} \mathrm{C}+{ }^{58} \mathrm{Ni}$ is an example of the so-called ""transfer reactions"", in which a part of one nucleus (""cluster"") is transferred to the other (see Fig. 3). In our example the transferred part is a ${ }^{4} \mathrm{He}-\mathrm{cluster}(\alpha$-particle). The transfer reactions occur with maximum probability if the velocity of the projectile-like reaction product (in our case: ${ }^{12} \mathrm{C}$ ) is equal both in magnitude and direction to the velocity of projectile (in our case: ${ }^{16} \mathrm{O}$ ). The target ${ }^{54} \mathrm{Fe}$ is initially at rest. In the reaction, ${ }^{58} \mathrm{Ni}$ is excited into one of its higher-lying states. Find the excitation energy of that state (and express it units of $\mathrm{MeV}$ ) if the kinetic energy of the projectile ${ }^{16} \mathrm{O}$ is $50 \mathrm{MeV}$. The speed of light is $\mathrm{c}=3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$. | 1. | $\mathrm{M}\left({ }^{16} \mathrm{O}\right)$ | 15.99491 a.m.u. | | ---: | ---: | :--- | | 2. | $\mathrm{M}\left({ }^{54} \mathrm{Fe}\right)$ | 53.93962 a.m.u. | | 3. | $\mathrm{M}\left({ }^{12} \mathrm{C}\right)$ | 12.00000 a.m.u. | | 4. | $\mathrm{M}\left({ }^{58} \mathrm{Ni}\right)$ | 57.93535 a.m.u. | | { Table 1. The rest masses of the reactants in their ground states. 1 a.m.u. $=1.6605 \cdot 10^{-27} \mathrm{~kg}$}. | | | Context answer: \boxed{10.8636} ","b) The ${ }^{58} \mathrm{Ni}$ nucleus produced in the excited state discussed in the part a), deexcites into its ground state by emitting a gamma-photon in the direction of its motion. Consider this decay in the frame of reference in which ${ }^{58} \mathrm{Ni}$ is at rest to find the recoil energy of ${ }^{58} \mathrm{Ni}$ (i.e. kinetic energy which ${ }^{58} \mathrm{Ni}$ acquires after the emission of the photon). What is the photon energy in that system? What is the photon energy in the lab system of reference (i.e. what would be the energy of the photon measured in the detector which is positioned in the direction in which the ${ }^{58} \mathrm{Ni}$ nucleus moves)? ","['For gamma-emission from the static nucleus, laws of conservation of energy and momentum give:\n$$\n\\begin{array}{r}\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=E_{\\gamma}+E_{\\text {recoil }} \\\\\np_{\\gamma}=p_{\\text {recoil }}\n\\end{array}\n\\tag{36}\n$$\nGamma-ray and recoiled nucleus have, of course, opposite directions. For gamma-ray (photon), energy and momentum are related as:\n$$\nE_{\\gamma}=p_{\\gamma} \\cdot c\n\\tag{37}\n$$\nIn part a) we have seen that the nucleus motion in this energy range is not relativistic, so we have:\n$$\nE_{\\text {recoil }}=\\frac{p_{\\text {recii }}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}=\\frac{p_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right)}=\\frac{E_{\\gamma}^{2}}{2 m\\left({ }^{(58} \\mathrm{Ni}\\right) \\cdot c^{2}}\n\\tag{38}\n$$\nInserting this into law of energy conservation Eq. (36), we get:\n$$\nE_{x}\\left({ }^{58} \\mathrm{Ni}\\right)=E_{\\gamma}+E_{\\text {recoil }}=E_{\\gamma}+\\frac{E_{\\gamma}^{2}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) \\cdot c^{2}}\n\\tag{39}\n$$\nThis reduces to the quadratic equation:\n$$\nE^2_{\\gamma}+2m({}^{58}Ni)c^2\\cdot E_{\\gamma}+2m({}^{58}Ni)c^2 E_x({}^{58}Ni)=0\n\\tag{40}\n$$\n\n$$\n\\begin{aligned}\nE_{\\gamma} &= \\frac{-2m({}^{58}Ni)c^2+\\sqrt{4(m({}^{58}Ni)c^2)^2+8m({}^{58}Ni)c^2E_x({}^{58}Ni)}}{2} \\\\\n&=\\sqrt{(m({}^{58}Ni)c^2)^2+2m({}^{58}Ni)c^2E_x({}^{58}Ni)}-m({}^{58}Ni)c^2\n\\end{aligned}\n\\tag{41}\n$$\nInserting numbers gives:\n$$\nE_{\\gamma}=10.8633 \\mathrm{MeV}\n\\tag{42}\n$$\nThe equation (37) can also be reduced to an approximate equation before inserting numbers:\n$$\nE_{\\gamma}=E_{x}\\left(1-\\frac{E_{x}}{2 m\\left({ }^{58} \\mathrm{Ni}\\right) c^{2}}\\right)=10.8633 \\mathrm{MeV}\n\\tag{43}\n$$\nThe recoil energy is now easily found as:\n$$\nE_{\\text {recoil }}=E_{x}\\left({ }^{58} \\mathrm{Ni}\\right)-E_{\\gamma}=1.1 \\mathrm{keV}\n\\tag{44}\n$$\nDue to the fact that nucleus emitting gamma-ray $\\left({ }^{58} \\mathrm{Ni}\\right)$ is moving with the high velocity, the energy of gamma ray will be changed because of the Doppler effect. The relativistic Doppler effect (when source is moving towards observer/detector) is given with this formula:\n$$\nf_{\\text {detector }}=f_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}\n\\tag{45}\n$$\nand since there is a simple relation between photon energy and frequency ( $E=h f)$, we get the similar expression for energy:\n$$\nE_{\\text {detector }}=E_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}\n\\tag{46}\n$$\nwhere $\\beta=v / c$ and $v$ is the velocity of emitter (the ${ }^{58} \\mathrm{Ni}$ nucleus). Taking the calculated value of the ${ }^{58} \\mathrm{Ni}$ velocity (equation 29 ) we get:\n$$\nE_{\\text {detector }}=E_{\\gamma, \\text { emitted }} \\sqrt{\\frac{1+\\beta}{1-\\beta}}=10.863 \\sqrt{\\frac{1+0.00565}{1-0.00565}}=10.925 \\mathrm{MeV}\n\\tag{47}\n$$']","['10.8633, 10.925']",True,MeV,Numerical,1e-3 1156,Optics,"Figure 1.1 A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by $$ y=a \cos [2 \pi(f t-x / \lambda)] $$ where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ).","(i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave. Verify geometrically, from the vector diagram, that $$ A=2 a \cos \theta $$ where $$ \beta=\frac{\pi}{\lambda} d \sin \theta $$","['(i) Vector Diagram\n\n\nIf the phase of the light from the first slit is zero, the phase from second slit is\n\n$$\n\\phi=\\frac{2 \\pi}{\\lambda} d \\sin \\theta\n$$\n\nAdding the two waves with phase difference $\\phi$ where $\\xi=2 \\pi\\left(f t-\\frac{x}{\\lambda}\\right)$,\n\n$$\n\\begin{aligned}\n& a \\cos (\\xi+\\phi)+a \\cos (\\xi)=2 a \\cos (\\phi / 2)(\\xi+\\phi / 2) \\\\\n& a \\cos (\\xi+\\phi)+a \\cos (\\xi)=2 a \\cos \\beta\\{\\cos (\\xi+\\beta)\\}\n\\end{aligned}\n$$\n\nThis is a wave of amplitude $A=2 a \\cos \\beta$ and phase $\\beta$. From vector diagram, in isosceles triangle $\\mathrm{OPQ}$,\n\n$$\n\\beta=\\frac{1}{2} \\phi=\\frac{\\pi}{\\lambda} d \\sin \\theta \\quad(N B \\quad \\phi=2 \\beta)\n$$\n\nand\n\n$$\nA=2 a \\cos \\beta\n$$\n\nThus the sum of the two waves can be obtained by the addition of two vectors of amplitude a and angular directions 0 and $\\phi$.']",,False,,, 1157,Optics,"Figure 1.1 A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by $$ y=a \cos [2 \pi(f t-x / \lambda)] $$ where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ). Context question: (i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave. Verify geometrically, from the vector diagram, that $$ A=2 a \cos \theta $$ where $$ \beta=\frac{\pi}{\lambda} d \sin \theta $$ Context answer: \boxed{证明题} ","(ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by $$ R=\frac{a}{2 \sin \beta} $$ Deduce that the resultant amplitude is $$ \frac{a \sin N \beta}{\sin \beta} $$ and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating.","[""(ii) Each slit in diffraction grating produces a wave of amplitude a with phase $2 \\beta$ relative to previous slit wave. The vector diagram consists of a 'regular' polygon with sides of constant length $a$ and with constant angles between adjacent sides.\n\nLet $\\mathrm{O}$ be the centre of circumscribing circle passing through the vertices of the polygon. Then radial lines such as OS have length $\\mathrm{R}$ and bisect the internal angles of the polygon. Figure 1.2.\n\nFigure 1.2\n\n\n\n\n\n\n\n$$\n\\begin{aligned}\n& O\\hat{ S }T=O\\hat{ T }S=\\frac{1}{2}(180-\\phi) \\\\\n& \\text { and } T \\hat{O }S=\\phi\n\\end{aligned}\n$$\nIn the triangle $T O S$, for example\n$$\n\\begin{gathered}\na=2 R \\sin (\\phi / 2)=2 R \\sin \\beta \\text { as }(\\phi=2 \\beta) \\\\\n\\therefore R=\\frac{a}{2 \\sin \\beta} \\quad \\text { (1) }\n\\end{gathered}\n$$\n\nAs the polygon has $N$ faces then:\n\n$$\nT \\hat{O} Z=N(T \\hat{O} Z)=N \\phi=2 N \\beta\n$$\n\nTherefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by\n\n$$\n2 R \\sin N \\beta \\text {. }\n$$\n\nHence form (1) this amplitude is\n\n$$\n\\frac{a \\sin N \\beta}{\\sin \\beta}\n$$\n\nResultant phase is\n\n\n\n$$\n\\begin{aligned}\n& =Z \\hat{T }S \\\\\n& =O \\hat{T} S-O \\hat{T} Z \\\\\n& \\left(90-\\frac{\\phi}{2}\\right)-\\frac{1}{2}(180-N \\phi) \\\\\n& -\\frac{1}{2}(N-1) \\phi \\\\\n& =(N-1) \\beta\n\\end{aligned}\n$$""]",,False,,, 1158,Optics,"Figure 1.1 A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by $$ y=a \cos [2 \pi(f t-x / \lambda)] $$ where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ). Context question: (i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave. Verify geometrically, from the vector diagram, that $$ A=2 a \cos \theta $$ where $$ \beta=\frac{\pi}{\lambda} d \sin \theta $$ Context answer: \boxed{证明题} Context question: (ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by $$ R=\frac{a}{2 \sin \beta} $$ Deduce that the resultant amplitude is $$ \frac{a \sin N \beta}{\sin \beta} $$ and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating. Context answer: \boxed{证明题} ",(iv) Determine the intensities of the principal intensity maxima.,['(iv) For the principle maxima $\\beta=\\pi p \\quad$ where $p=0 \\pm 1 \\pm 2 \\ldots \\ldots$.\n\n$$\nI_{\\max }=a^{2}\\left(\\frac{N \\beta^{\\prime}}{\\beta^{\\prime}}\\right)=N^{2} a^{2} \\quad \\beta^{\\prime}=0 \\text { and } \\beta=\\pi p+\\beta^{\\prime}\n$$'],['$N^{2} a^{2}$'],False,,Expression, 1159,Optics,"Figure 1.1 A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by $$ y=a \cos [2 \pi(f t-x / \lambda)] $$ where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ). Context question: (i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave. Verify geometrically, from the vector diagram, that $$ A=2 a \cos \theta $$ where $$ \beta=\frac{\pi}{\lambda} d \sin \theta $$ Context answer: \boxed{证明题} Context question: (ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by $$ R=\frac{a}{2 \sin \beta} $$ Deduce that the resultant amplitude is $$ \frac{a \sin N \beta}{\sin \beta} $$ and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating. Context answer: \boxed{证明题} Context question: (iv) Determine the intensities of the principal intensity maxima. Context answer: \boxed{$N^{2} a^{2}$} ","(v) Show that the number of principal maxima cannot exceed $$ \left(\frac{2 d}{\lambda}+1\right) $$","['(v) Adjacent max. estimate $I_{l}$ :\n\n$$\n\\sin ^{2} N \\beta=1, \\quad \\beta=2 \\pi p \\mp \\frac{3 \\pi}{2 N} \\text { i.e } \\beta= \\pm \\frac{3 \\pi}{2 N}\n$$\n\n$$\n\\begin{gathered}\n{\\left[\\beta=\\pi p \\pm \\frac{\\pi}{2 N}\\right] \\text { does not give a maximum as can be observed from the graph. }} \\\\\n\\qquad I_{1}=a^{2} \\frac{1}{\\frac{3 \\pi^{2}}{2 n}}=\\frac{a^{2} N^{2}}{23} \\text { for } N \\gg>1\n\\end{gathered}\n$$\n\nAdjacent zero intensity occurs for $\\beta=\\pi \\rho \\pm \\frac{\\pi}{N}$ i.e. $\\delta= \\pm \\frac{\\pi}{N}$\n\nFor phase differences much greater than $\\delta, \\quad \\mathrm{I}=\\mathrm{a}^{2}\\left(\\frac{\\sin N \\beta}{\\sin \\beta}\\right)=a^{2}$']",,False,,, 1160,Optics,"Figure 1.1 A plane monochromatic light wave, wavelength $\lambda$ and frequency $f$, is incident normally on two identical narrow slits, separated by a distance $d$, as indicated in Figure 1.1. The light wave emerging at each slit is given, at a distance $x$ in a direction $\theta$ at time $t$, by $$ y=a \cos [2 \pi(f t-x / \lambda)] $$ where the amplitude $a$ is the same for both waves. (Assume $x$ is much larger than $d$ ). Context question: (i) Show that the two waves observed at an angle $\theta$ to a normal to the slits, have a resultant amplitude A which can be obtained by adding two vectors, each having magnitude $a$, and each with an associated direction determined by the phase of the light wave. Verify geometrically, from the vector diagram, that $$ A=2 a \cos \theta $$ where $$ \beta=\frac{\pi}{\lambda} d \sin \theta $$ Context answer: \boxed{证明题} Context question: (ii) The double slit is replaced by a diffraction grating with $N$ equally spaced slits, adjacent slits being separated by a distance $d$. Use the vector method of adding amplitudes to show that the vector amplitudes, each of magnitude $a$, form a part of a regular polygon with vertices on a circle of radius $R$ given by $$ R=\frac{a}{2 \sin \beta} $$ Deduce that the resultant amplitude is $$ \frac{a \sin N \beta}{\sin \beta} $$ and obtain the resultant phase difference relative to that of the light from the slit at the edge of the grating. Context answer: \boxed{证明题} Context question: (iv) Determine the intensities of the principal intensity maxima. Context answer: \boxed{$N^{2} a^{2}$} Context question: (v) Show that the number of principal maxima cannot exceed $$ \left(\frac{2 d}{\lambda}+1\right) $$ Context answer: \boxed{证明题} ","(vi) Show that two wavelengths $\lambda$ and $\lambda+\delta \lambda$, where $\delta \lambda<<\lambda$, produce principal maxima with an angular separation given by $$ \Delta \theta=\frac{n \Delta \lambda}{d \cos \theta} \quad \text { where } n=0, \pm 1, \pm 2 \ldots . \text { etc } $$ Calculate this angular separation for the sodium $\mathrm{D}$ lines for which $$ \begin{aligned} & \lambda=589.0 \mathrm{~nm}, \quad \lambda+\Delta \lambda=589.6 \mathrm{~nm}, \quad n=2, \text { and } d=1.2 \times 10^{-6} \mathrm{~m} . \\ & {\left[\text { reminder: } \quad \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right]} \end{aligned} $$","['(vi)\n\n$$\n\\begin{aligned}\n& \\beta=n \\pi \\text { for a principle maximum } \\\\\n& \\text { i.e. } \\frac{\\pi}{\\lambda} d \\sin \\theta=n \\pi \\quad n=0, \\pm 1, \\pm 2 \\ldots \\ldots \\ldots . .\n\\end{aligned}\n$$\n\nDifferentiating w.r.t, $\\lambda$\n\n$d \\cos \\theta \\Delta \\theta=n \\Delta \\lambda$\n\n$$\n\\Delta \\theta=\\frac{n \\Delta \\lambda}{d \\cos \\theta}\n$$\n\nSubstituting $\\lambda=589.0 \\mathrm{~nm}, \\lambda+\\Delta \\lambda=589.6 \\mathrm{~nm} . \\mathrm{n}=2$ and $d=1.2 \\times 10^{-6} \\mathrm{~m}$.\n\n$\\Delta \\theta=\\frac{n \\Delta \\lambda}{d \\sqrt{1-\\left(\\frac{n \\lambda}{d}\\right)^{2}}}$ as $\\sin \\theta=\\frac{n \\lambda}{d}$ and $\\cos \\theta=\\sqrt{1-\\left(\\frac{n \\lambda}{d}\\right)^{2}}$\n\n$\\Rightarrow \\Delta \\theta=5.2 \\times 10^{-3} \\mathrm{rads}$ or $0.30^{\\circ}$']",['$0.30$'],False,^{\circ},Numerical,5e-2 1161,Mechanics,,"(i) If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for each mass.","['Equations of motion:\n\n$$\n\\begin{aligned}\n& m \\frac{d^{2} u_{1}}{d t^{2}}=k\\left(u_{2}-u_{1}\\right)+k\\left(u_{3}-u_{1}\\right) \\\\\n& m \\frac{d^{2} u_{2}}{d t^{2}}=k\\left(u_{3}-u_{2}\\right)+k\\left(u_{1}-u_{2}\\right) \\\\\n& m \\frac{d^{2} u_{3}}{d t^{2}}=k\\left(u_{1}-u_{3}\\right)+k\\left(u_{2}-u_{3}\\right)\n\\end{aligned}\n$$\n\nSubstituting $u_{n}(t)=u_{n}(0) \\cos \\omega t$ and $\\omega_{o}^{2}=\\frac{k}{m}$ :\n\n$$\n\\begin{aligned}\n\\left(2 \\omega_{o}^{2}-\\omega^{2}\\right) u_{1}(0)-\\omega_{o}{ }^{2} u_{2}(0)-\\omega_{o}^{2} u_{3}(0) & =0 \\\\\n-\\omega_{o}{ }^{2} u_{1}(0)+\\left(2 \\omega_{o}^{2}-\\omega^{2}\\right) u_{2}(0)-\\omega_{o}{ }^{2} u_{3}(0) & =0 \\\\\n-\\omega_{o}{ }^{2} u_{1}(0)-\\omega_{o}{ }^{2} u_{2}(0)+\\left(2 \\omega_{o}{ }^{2}-\\omega^{2}\\right) u_{3}(0) & =0\n\\end{aligned}\n$$\n\nSolving for $u_{1}(0)$ and $u_{2}(0)$ in terms of $u_{3}(0)$ using (a) and (b) and substituting into (c) gives the equation equivalent to\n\n$$\n\\begin{gathered}\n\\left(3 \\omega_{o}{ }^{2}-\\omega^{2}\\right)^{2} \\omega^{2}=0 \\\\\n\\omega^{2}=3 \\omega_{o}{ }^{2}, 3 \\omega_{o}{ }^{2} \\text { and } 0 \\\\\n\\omega=\\sqrt{3} \\omega_{o}, \\sqrt{3} \\omega_{o} \\text { and } 0\n\\end{gathered}\n$$']",['$$\n\\begin{aligned}\n& m \\frac{d^{2} u_{1}}{d t^{2}}=k\\left(u_{2}-u_{1}\\right)+k\\left(u_{3}-u_{1}\\right) \\\\\n& m \\frac{d^{2} u_{2}}{d t^{2}}=k\\left(u_{3}-u_{2}\\right)+k\\left(u_{1}-u_{2}\\right) \\\\\n& m \\frac{d^{2} u_{3}}{d t^{2}}=k\\left(u_{1}-u_{3}\\right)+k\\left(u_{2}-u_{3}\\right)\n\\end{aligned}\n$$'],False,,Need_human_evaluate, 1162,Mechanics,"Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 Context question: (i) If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and $u_{3}$ respectively, write down the equation of motion for each mass. Context answer: $$ \begin{aligned} & m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right) \\ & m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right) \\ & m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right) \end{aligned} $$ ","(ii) Verify that the system has simple harmonic solutions of the form $$ u_{n}=a_{n} \cos \omega t $$ with accelerations, $\left(-\omega^{2} u_{n}\right)$ where $a_{n}(n=1,2,3)$ are constant amplitudes, and $\omega$, the angular frequency, can have 3 possible values, $$ \omega_{o} \sqrt{3}, \omega_{o} \sqrt{3} \text { and } 0 \text {. where } \omega_{o}^{2}=\frac{k}{m} \text {. } $$","[""(ii) Equation of motion of the n'th particle:\n\n$$\n\\begin{aligned}\n& m \\frac{d^{2} u_{n}}{d t^{2}}=k\\left(u_{1+n}-u_{n}\\right)+k\\left(u_{n-1}-u_{n}\\right) \\\\\n& \\frac{d^{2} u_{n}}{d t^{2}}=k\\left(u_{1+n}-u_{n}\\right)+\\omega_{o}^{2}\\left(u_{n-1}-u_{n}\\right)\n\\end{aligned}\n$$\n\nSubstituting $u_{n}(t)=u_{n}(0) \\sin \\left(2 n s \\frac{\\pi}{N}\\right) \\cos \\omega_{s} t$\n\n$$\n\\begin{aligned}\n& -\\omega_{s}^{2}\\left(\\sin \\left(2 n s \\frac{\\pi}{N}\\right)\\right)=\\omega_{o}^{2}\\left[\\sin \\left(2(n+1) s \\frac{\\pi}{N}\\right)-2 \\sin \\left(2 n s \\frac{\\pi}{N}\\right)+\\sin \\left(2(n-1) s \\frac{\\pi}{N}\\right)\\right] \\\\\n& -\\omega_{s}^{2}\\left(\\sin \\left(2 n s \\frac{\\pi}{N}\\right)\\right)=2 \\omega_{o}^{2}\\left[\\frac{1}{2} \\sin \\left(2(n+1) s \\frac{\\pi}{N}\\right)+\\sin \\left(2 n s \\frac{\\pi}{N}\\right)-\\frac{1}{2} \\sin \\left(2(n-1) s \\frac{\\pi}{N}\\right)\\right] \\\\\n& -\\omega_{s}^{2}\\left(\\sin \\left(2 n s \\frac{\\pi}{N}\\right)\\right)=2 \\omega_{o}^{2}\\left[\\sin \\left(2 n s \\frac{\\pi}{N}\\right) \\cos \\left(2 s \\frac{\\pi}{N}\\right)-\\sin \\left(2 n s \\frac{\\pi}{N}\\right)\\right] \\\\\n& \\therefore \\omega_{s}^{2}=2 \\omega_{o}^{2}\\left[1-\\cos \\left(2 s \\frac{\\pi}{N}\\right)\\right]: \\quad(s=1,2, \\ldots . . N)\n\\end{aligned}\n$$\n\nAs $2 \\sin ^{2} \\theta=1-\\cos 2 \\theta$\n\nThis gives\n\n$$\n\\omega_{s}=2 \\omega_{o} \\sin \\left(\\frac{s \\pi}{N}\\right) \\quad(s=1,2, \\ldots N)\n$$\n\n$\\omega_{s}$ can have values from 0 to $2 \\omega_{o}=2 \\sqrt{\\frac{k}{m}}$ when $N \\rightarrow \\infty$; corresponding to range $s=1$ to $\\frac{N}{2}$.""]",,False,,, 1163,Mechanics,"Problem T1. Focus on sketches Part A. Ballistics A ball, thrown with an initial speed $v_{0}$, moves in a homogeneous gravitational field in the $x-z$ plane, where the $x$-axis is horizontal, and the $z$-axis is vertical and antiparallel to the free fall acceleration $g$. Neglect the effect of air drag.","i. By adjusting the launching angle for a ball thrown with a fixed initial speed $v_{0}$ from the origin, targets can be hit within the region given by $$ z \leq z_{0}-k x^{2} $$ You can use this fact without proving it. Find the constants $z_{0}$ and $k$.","['When the stone is thrown vertically upwards, it can reach the point $x=0, z=v_{0}^{2} / 2 g$ (as it follows from the energy conservation law). Comparing this with the inequality $z \\leq z_{0}-k x^{2}$ we conclude that\n\n$$\nz_{0}=v_{0}^{2} / 2 g\n$$\n\nLet us consider the asymptotics $z \\rightarrow-\\infty$; the trajectory of the stone is a parabola, and at this limit, the horizontal displacement (for the given $z$ ) is very sensitive with respect to the curvature of the parabola: the flatter the parabola, the larger the displacement. The parabola has the flattest shape when the stone is thrown horizontally, $x=v_{0} t$ and $z=-g t^{2} / 2$, i.e. its trajectory is given by $z=-g x^{2} / 2 v_{0}^{2}$. Now, let us recall that $z \\leq z_{0}-k x^{2}$, i.e. $-g x^{2} / 2 v_{0}^{2} \\leq z_{0}-k x^{2} \\Rightarrow k \\leq g / 2 v_{0}^{2}$. Note that $k minimal initial speed $v_{0}$. Bouncing off the roof prior to hitting the target is not allowed. Sketch qualitatively the shape of the optimal trajectory of the ball (use the designated box on the answer sheet). Note that the marks are given only for the sketch. Context answer: ","iii. What is the minimal launching speed $v_{\min }$ needed to hit the topmost point of a spherical building of radius $R$ ? La Geode, Parc de la Villette, Paris. Photo: katchooo/flickr.com","[""The brute force approach would be writing down the condition that the optimal trajectory intersects with the building at two points and touches at one. This would be described by a fourth order algebraic equation and therefore, it is not realistic to accomplish such a solution within a reasonable time frame.\n\nNote that the interior of the building needs to lie inside the region where the targets can be hit with a stone thrown from the top with initial speed $v_{\\min }$. Indeed, if we can throw over the building, we can hit anything inside by lowering the throwing angle. On the other hand, the boundary of the targetable region needs to touch the building. Indeed, if there were a gap, it would be possible to hit a target just above the point where the optimal trajectory touches the building; the trajectory through that target wouldn't touch the building anywhere, hence we arrive at a contradiction.\n\nSo, with $v_{0}$ corresponding to the optimal trajectory, the targetable region touches the building; due to symmetry, overall there are two touching points (for smaller speeds, there would be four, and for larger speeds, there would be none). With the origin at the top of the building, the intersection points are defined by the following system of equations:\n\n$$\nx^{2}+z^{2}+2 z R=0, \\quad z=\\frac{v_{0}^{2}}{2 g}-\\frac{g x^{2}}{2 v_{0}^{2}}\n$$\n\nUpon eliminating $z$, this becomes a biquadratic equation for $x$ :\n\n$$\nx^{4}\\left(\\frac{g}{2 v_{0}^{2}}\\right)^{2}+x^{2}\\left(\\frac{1}{2}-\\frac{g R}{v_{0}^{2}}\\right)+\\left(\\frac{v_{0}^{2}}{4 g}+R\\right) \\frac{v_{0}^{2}}{g}=0\n$$\n\nHence the speed by which the real-valued solutions disappear can be found from the condition that the discriminant vanishes:\n\n$$\n\\left(\\frac{1}{2}-\\frac{g R}{v_{0}^{2}}\\right)^{2}=\\frac{1}{4}+\\frac{g R}{v_{0}^{2}} \\Longrightarrow \\frac{g R}{v_{0}^{2}}=2\n$$\n\nBearing in mind that due to the energy conservation law, at the ground level the squared speed is increased by $4 g R$. Thus we finally obtain\n\n$$\nv_{\\min }=\\sqrt{v_{0}^{2}+4 g R}=3 \\sqrt{\\frac{g R}{2}} .\n$$""]",['$v_{\\min }=3 \\sqrt{\\frac{g R}{2}}$'],False,,Expression, 1164,Mechanics,,"iii. What is the minimal launching speed $v_{\min }$ needed to hit the topmost point of a spherical building of radius $R$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_72b956b534088c26e196g-1.jpg?height=923&width=968&top_left_y=1680&top_left_x=59) La Geode, Parc de la Villette, Paris. Photo: katchooo/flickr.com","[""The brute force approach would be writing down the condition that the optimal trajectory intersects with the building at two points and touches at one. This would be described by a fourth order algebraic equation and therefore, it is not realistic to accomplish such a solution within a reasonable time frame.\n\nNote that the interior of the building needs to lie inside the region where the targets can be hit with a stone thrown from the top with initial speed $v_{\\min }$. Indeed, if we can throw over the building, we can hit anything inside by lowering the throwing angle. On the other hand, the boundary of the targetable region needs to touch the building. Indeed, if there were a gap, it would be possible to hit a target just above the point where the optimal trajectory touches the building; the trajectory through that target wouldn't touch the building anywhere, hence we arrive at a contradiction.\n\nSo, with $v_{0}$ corresponding to the optimal trajectory, the targetable region touches the building; due to symmetry, overall there are two touching points (for smaller speeds, there would be four, and for larger speeds, there would be none). With the origin at the top of the building, the intersection points are defined by the following system of equations:\n\n$$\nx^{2}+z^{2}+2 z R=0, \\quad z=\\frac{v_{0}^{2}}{2 g}-\\frac{g x^{2}}{2 v_{0}^{2}}\n$$\n\nUpon eliminating $z$, this becomes a biquadratic equation for $x$ :\n\n$$\nx^{4}\\left(\\frac{g}{2 v_{0}^{2}}\\right)^{2}+x^{2}\\left(\\frac{1}{2}-\\frac{g R}{v_{0}^{2}}\\right)+\\left(\\frac{v_{0}^{2}}{4 g}+R\\right) \\frac{v_{0}^{2}}{g}=0\n$$\n\nHence the speed by which the real-valued solutions disappear can be found from the condition that the discriminant vanishes:\n\n$$\n\\left(\\frac{1}{2}-\\frac{g R}{v_{0}^{2}}\\right)^{2}=\\frac{1}{4}+\\frac{g R}{v_{0}^{2}} \\Longrightarrow \\frac{g R}{v_{0}^{2}}=2\n$$\n\nBearing in mind that due to the energy conservation law, at the ground level the squared speed is increased by $4 g R$. Thus we finally obtain\n\n$$\nv_{\\min }=\\sqrt{v_{0}^{2}+4 g R}=3 \\sqrt{\\frac{g R}{2}} .\n$$""]",['$v_{\\min }=3 \\sqrt{\\frac{g R}{2}}$'],False,,Expression, 1165,Electromagnetism,"Part C. Magnetic straws Consider a cylindrical tube made of a superconducting material. The length of the tube is $l$ and the inner radius is $r$ with $l \gg r$. The centre of the tube coincides with the origin, and its axis coincides with the $z$-axis. There is a magnetic flux $\Phi$ through the central cross-section of the tube, $z=0, x^{2}+y^{2} ","ii. Find the tension force $T$ along the $z$-axis in the middle of the tube (i.e. the force by which two halves of the tube, $z>0$ and $z<0$, interact with each other).","['Let us consider the change of the magnetic energy when the tube is stretched (virtually) by a small amount $\\Delta l$. Note that the magnetic flux trough the tube is conserved: any change of flux would imply a non-zero electromotive force $\\frac{d \\Phi}{d t}$, and for a zero resistivity, an infinite current. So, the induction $B=\\frac{\\Phi}{\\pi r^{2}}$. The energy density of the magnetic field is $\\frac{B^{2}}{2 \\mu_{0}}$. Thus, the change of the magnetic energy is calculated as\n\n$$\n\\Delta W=\\frac{B^{2}}{2 \\mu_{0}} \\pi r^{2} \\Delta l=\\frac{\\Phi^{2}}{2 \\mu_{0} \\pi r^{2}} \\Delta l\n$$\n\nThis energy increase is achieved owing to the work done by the stretching force, $\\Delta W=T \\Delta l$. Hence, the force\n\n$$\nT=\\frac{\\Phi^{2}}{2 \\mu_{0} \\pi r^{2}}\n$$']",['$T=\\frac{\\Phi^{2}}{2 \\mu_{0} \\pi r^{2}}$'],False,,Expression, 1166,Electromagnetism,"Part C. Magnetic straws Consider a cylindrical tube made of a superconducting material. The length of the tube is $l$ and the inner radius is $r$ with $l \gg r$. The centre of the tube coincides with the origin, and its axis coincides with the $z$-axis. There is a magnetic flux $\Phi$ through the central cross-section of the tube, $z=0, x^{2}+y^{2} Context question: ii. Find the tension force $T$ along the $z$-axis in the middle of the tube (i.e. the force by which two halves of the tube, $z>0$ and $z<0$, interact with each other). Context answer: \boxed{$T=\frac{\Phi^{2}}{2 \mu_{0} \pi r^{2}}$} ","iii. Consider another tube, identical and parallel to the first one. The second tube has the same magnetic field but in the opposite direction and its centre is placed at $y=l, x=z=0$ (so that the tubes form opposite sides of a square). Determine the magnetic interaction force $F$ between the two tubes.","['Let us analyse, what would be the change of the magnetic energy when one of the straws is displaced to a small distance. The magnetic field inside the tubes will remain constant due to the conservation of magnetic flux, but outside, the magnetic field will be changed. The magnetic field outside the straws is defined by the following condition: there is no circulation of $\\vec{B}$ (because there are no currents outside the straws); there are no sources of the field lines, other than the endpoints of the straws; each of the endpoints of the straws is a source of streamlines with a fixed magnetic flux $\\pm \\Phi$. These are exactly the same condition as those which define the electric field of four charges $\\pm Q$. We know that if the distance between charges is much larger than the geometrical size of a charge, the charges can be considered as point charges (the electric field near the charges remains almost constant, so that the respective contribution to the change of the overall electric field energy is negligible). Therefore we can conclude that the endpoints of the straws can be considered as magnetic point charges. In order to calculate the force between two magnetic charges (magnetic monopoles), we need to establish the correspondence between magnetic and electric quantities.\n\nFor two electric charges $Q$ separated by a distance $a$, the force is $F=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Q^{2}}{a^{2}}$, and at the position of one charge, the electric field of the other charge has energy density $w=\\frac{1}{32 \\pi^{2} \\varepsilon_{0}} \\frac{Q^{2}}{a^{4}}$; hence we can write $F=8 \\pi w a^{2}$. This is a universal expression for the force (for the case when the field lines have the same shape as in the case of two opposite and equal by modulus electric charges) relying only on the energy density, and not related to the nature of the field; so we can apply it to the magnetic\n\n\n\nfield. Indeed, the force can be calculated as a derivative of the full field energy with respect to a virtual displacement of a field line source (electric or magnetic charge); if the energy densities of two fields are respectively equal at one point, they are equal everywhere, and so are equal the full field energies. As it follows from the Gauss law, for a point source of a fixed magnetic flux $\\Phi$ at a distance $a$, the induction $B=\\frac{1}{4 \\pi} \\frac{\\Phi}{a^{2}}$. So, the energy density $w=\\frac{B^{2}}{2 \\mu_{0}}=\\frac{1}{32 \\pi^{2} \\mu_{0}} \\frac{\\Phi^{2}}{a^{4}}$, hence\n\n$$\nF=\\frac{1}{4 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{a^{2}}\n$$\n\nFor the two straws, we have four magnetic charges. The longitudinal (along a straw axis) forces cancel out (the diagonally positioned pairs of same-sign-charges push in opposite directions). The normal force is a superposition of the attraction due to the two pairs of opposite charges, $F_{1}=\\frac{1}{4 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}$, and the repulsive forces of diagonal pairs, $F_{2}=\\frac{\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{2 l^{2}}$. The net attractive force will be\n\n$$\nF=2\\left(F_{1}-F_{2}\\right)=\\frac{4-\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}\n$$']",['$F=\\frac{4-\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}$'],False,,Expression, 1166,Electromagnetism,,"iii. Consider another tube, identical and parallel to the first one. ![](https://cdn.mathpix.com/cropped/2023_12_21_c92eb0b13c54d9857f92g-1.jpg?height=418&width=420&top_left_y=1090&top_left_x=1298) The second tube has the same magnetic field but in the opposite direction and its centre is placed at $y=l, x=z=0$ (so that the tubes form opposite sides of a square). Determine the magnetic interaction force $F$ between the two tubes.","['Let us analyse, what would be the change of the magnetic energy when one of the straws is displaced to a small distance. The magnetic field inside the tubes will remain constant due to the conservation of magnetic flux, but outside, the magnetic field will be changed. The magnetic field outside the straws is defined by the following condition: there is no circulation of $\\vec{B}$ (because there are no currents outside the straws); there are no sources of the field lines, other than the endpoints of the straws; each of the endpoints of the straws is a source of streamlines with a fixed magnetic flux $\\pm \\Phi$. These are exactly the same condition as those which define the electric field of four charges $\\pm Q$. We know that if the distance between charges is much larger than the geometrical size of a charge, the charges can be considered as point charges (the electric field near the charges remains almost constant, so that the respective contribution to the change of the overall electric field energy is negligible). Therefore we can conclude that the endpoints of the straws can be considered as magnetic point charges. In order to calculate the force between two magnetic charges (magnetic monopoles), we need to establish the correspondence between magnetic and electric quantities.\n\nFor two electric charges $Q$ separated by a distance $a$, the force is $F=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Q^{2}}{a^{2}}$, and at the position of one charge, the electric field of the other charge has energy density $w=\\frac{1}{32 \\pi^{2} \\varepsilon_{0}} \\frac{Q^{2}}{a^{4}}$; hence we can write $F=8 \\pi w a^{2}$. This is a universal expression for the force (for the case when the field lines have the same shape as in the case of two opposite and equal by modulus electric charges) relying only on the energy density, and not related to the nature of the field; so we can apply it to the magnetic\n\n\n\nfield. Indeed, the force can be calculated as a derivative of the full field energy with respect to a virtual displacement of a field line source (electric or magnetic charge); if the energy densities of two fields are respectively equal at one point, they are equal everywhere, and so are equal the full field energies. As it follows from the Gauss law, for a point source of a fixed magnetic flux $\\Phi$ at a distance $a$, the induction $B=\\frac{1}{4 \\pi} \\frac{\\Phi}{a^{2}}$. So, the energy density $w=\\frac{B^{2}}{2 \\mu_{0}}=\\frac{1}{32 \\pi^{2} \\mu_{0}} \\frac{\\Phi^{2}}{a^{4}}$, hence\n\n$$\nF=\\frac{1}{4 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{a^{2}}\n$$\n\nFor the two straws, we have four magnetic charges. The longitudinal (along a straw axis) forces cancel out (the diagonally positioned pairs of same-sign-charges push in opposite directions). The normal force is a superposition of the attraction due to the two pairs of opposite charges, $F_{1}=\\frac{1}{4 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}$, and the repulsive forces of diagonal pairs, $F_{2}=\\frac{\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{2 l^{2}}$. The net attractive force will be\n\n$$\nF=2\\left(F_{1}-F_{2}\\right)=\\frac{4-\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}\n$$']",['$F=\\frac{4-\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}$'],False,,Expression, 1167,Thermodynamics,"Problem T2. Kelvin water dropper The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments. A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$. Part A. Single pipe",i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle.,"['Let us write the force balance for the droplet. Since $d \\ll r$, we can neglect the force $\\frac{\\pi}{4} \\Delta p d^{2}$ due to the excess pressure $\\Delta p$ inside the tube. So, the gravity force $\\frac{4}{3} \\pi r_{\\max }^{3} \\rho g$ is balanced by the capillary force. When the droplet separates from the tube, the water surface forms in the vicinity of the nozzle a ""neck"", which has vertical tangent. In the horizontal cross-section of that ""neck"", the capillary force is vertical and can be calculated as $\\pi \\sigma d$. So,\n\n$$\nr_{\\max }=\\sqrt[3]{\\frac{3 \\sigma d}{4 \\rho g}}\n$$']",['$r_{\\max }=\\sqrt[3]{\\frac{3 \\sigma d}{4 \\rho g}}$'],False,,Expression, 1168,Thermodynamics,"Problem T2. Kelvin water dropper The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments. A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$. Part A. Single pipe Context question: i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle. Context answer: \boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$} ","ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$.","[""Since $d \\ll r$, we can neglect the change of the droplet's capacitance due to the tube. On the one hand, the droplet's potential is $\\varphi$; on the other hand, it is $\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Q}{r}$. So,\n\n$$\nQ=4 \\pi \\varepsilon_{0} \\varphi r\n$$""]",['$Q=4 \\pi \\varepsilon_{0} \\varphi r$'],False,,Expression, 1169,Thermodynamics,"Problem T2. Kelvin water dropper The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments. A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$. Part A. Single pipe Context question: i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle. Context answer: \boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$} Context question: ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$. Context answer: \boxed{$Q=4 \pi \varepsilon_{0} \varphi r$} ",iii. Consider the situation in which $r$ is kept constant and $\varphi$ is slowly increased. The droplet becomes unstable and breaks into pieces if the hydrostatic pressure inside the droplet becomes smaller than the atmospheric pressure. Find the critical potential $\varphi_{\max }$ at which this will happen.,"[""Excess pressure inside the droplet is caused by the capillary pressure $2 \\sigma / r$ (increases the inside pressure), and by the electrostatic pressure $\\frac{1}{2} \\varepsilon_{0} E^{2}=\\frac{1}{2} \\varepsilon_{0} \\varphi^{2} / r^{2}$ (decreases the pressure). So, the sign of the excess pressure will change, if $\\frac{1}{2} \\varepsilon_{0} \\varphi_{\\max }^{2} / r^{2}=2 \\sigma / r$, hence\n\n$$\n\\varphi_{\\max }=2 \\sqrt{\\sigma r / \\varepsilon_{0}}\n$$\n\nThe expression for the electrostatic pressure used above can be derived as follows. The electrostatic force acting on a surface charge of density $\\sigma$ and surface area $S$ is given by $F=\\sigma S \\cdot \\bar{E}$, where $\\bar{E}$ is the field at the site without the field created by the surface charge element itself. Note that this force is perpendicular to the surface, so $F / S$ can be interpreted as a pressure. The surface charge gives rise to a field drop on the surface equal to $\\Delta E=\\sigma / \\varepsilon_{0}$ (which follows from the Gauss law); inside the droplet, there is no field due to the conductivity of the droplet: $\\bar{E}-\\frac{1}{2} \\Delta E=0 ;$ outside the droplet, there is field $E=\\bar{E}+\\frac{1}{2} \\Delta E$, therefore $\\bar{E}=\\frac{1}{2} E=\\frac{1}{2} \\Delta E$. Bringing everything together, we obtain the expression used above.\n\nNote that alternatively, this expression can be derived by considering a virtual displacement of a capacitor's surface and comparing the pressure work $p \\Delta V$ with the change of the electrostatic field energy $\\frac{1}{2} \\varepsilon_{0} E^{2} \\Delta V$.\n\nFinally, the answer to the question can be also derived from the requirement that the mechanical work $d A$ done for an infinitesimal droplet inflation needs to be zero. From the energy conservation law, $d W+d W_{\\mathrm{el}}=\\sigma d\\left(4 \\pi r^{2}\\right)+\\frac{1}{2} \\varphi_{\\max }^{2} d C_{d}$, where the droplet's capacitance $C_{d}=4 \\pi \\varepsilon_{0} r$; the electrical work $d W_{\\text {el }}=\\varphi_{\\max } d q=4 \\pi \\varepsilon_{0} \\varphi_{\\max }^{2} d r$. Putting $d W=0$ we obtain an equation for $\\varphi_{\\max }$, which recovers the earlier result.""]",['$\\varphi_{\\max }=2 \\sqrt{\\frac{\\sigma r}{ \\varepsilon_{0}}}$'],False,,Expression, 1170,Thermodynamics,"Problem T2. Kelvin water dropper The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments. A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$. Part A. Single pipe Context question: i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle. Context answer: \boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$} Context question: ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$. Context answer: \boxed{$Q=4 \pi \varepsilon_{0} \varphi r$} Context question: iii. Consider the situation in which $r$ is kept constant and $\varphi$ is slowly increased. The droplet becomes unstable and breaks into pieces if the hydrostatic pressure inside the droplet becomes smaller than the atmospheric pressure. Find the critical potential $\varphi_{\max }$ at which this will happen. Context answer: \boxed{$\varphi_{\max }=2 \sqrt{\frac{\sigma r}{ \varepsilon_{0}}}$} Extra Supplementary Reading Materials: Part B. Two pipes An apparatus called the ""Kelvin water dropper"" consists of two pipes, each identical to the one described in Part A, connected via a T-junction, see fig. The ends of both pipes are at the centres of two cylindrical electrodes (with height $L$ and diameter $D$ with $L \gg D \gg r$ ). For both tubes, the dripping rate is $n$ droplets per unit time. Droplets fall from height $H$ into conductive bowls underneath the nozzles, cross-connected to the electrodes as shown in the diagram. The electrodes are connected via a capacitance $C$. There is no net charge on the system of bowls and electrodes. Note that the top water container is earthed as shown. The first droplet to fall will have some microscopic charge which will cause an imbalance between the two sides and a small charge separation across the capacitor. ","i. Express the absolute value of the charge $Q_{0}$ of the drops as they separate from the tubes, and at the instant when the capacitor's charge is $q$. Express $Q_{0}$ in terms of $r_{\max }$ (from Part A-i) and neglect the effect described in Part A-iii.","[""This is basically the same as Part A-ii, except that the surroundings' potential is that of the surrounding electrode, $-U / 2$ (where $U=q / C$ is the capacitor's voltage) and droplet has the ground potential (0). As it is not defined which electrode is the positive one, opposite sign of the potential may be chosen, if done consistently. Note that since the cylindrical electrode is long, it shields effectively the environment's (ground, wall, etc) potential. So, relative to its surroundings, the droplet's potential is $U / 2$. Using the result of Part A we obtain\n\n$$\nQ=2 \\pi \\varepsilon_{0} U r_{\\max }=2 \\pi \\varepsilon_{0} q r_{\\max } / C .\n$$""]",['$Q_{0}=2 \\pi \\varepsilon_{0} q r_{\\max } / C$'],False,,Expression, 1171,Thermodynamics,"Problem T2. Kelvin water dropper The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments. A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$. Part A. Single pipe Context question: i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle. Context answer: \boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$} Context question: ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$. Context answer: \boxed{$Q=4 \pi \varepsilon_{0} \varphi r$} Context question: iii. Consider the situation in which $r$ is kept constant and $\varphi$ is slowly increased. The droplet becomes unstable and breaks into pieces if the hydrostatic pressure inside the droplet becomes smaller than the atmospheric pressure. Find the critical potential $\varphi_{\max }$ at which this will happen. Context answer: \boxed{$\varphi_{\max }=2 \sqrt{\frac{\sigma r}{ \varepsilon_{0}}}$} Extra Supplementary Reading Materials: Part B. Two pipes An apparatus called the ""Kelvin water dropper"" consists of two pipes, each identical to the one described in Part A, connected via a T-junction, see fig. The ends of both pipes are at the centres of two cylindrical electrodes (with height $L$ and diameter $D$ with $L \gg D \gg r$ ). For both tubes, the dripping rate is $n$ droplets per unit time. Droplets fall from height $H$ into conductive bowls underneath the nozzles, cross-connected to the electrodes as shown in the diagram. The electrodes are connected via a capacitance $C$. There is no net charge on the system of bowls and electrodes. Note that the top water container is earthed as shown. The first droplet to fall will have some microscopic charge which will cause an imbalance between the two sides and a small charge separation across the capacitor. Context question: i. Express the absolute value of the charge $Q_{0}$ of the drops as they separate from the tubes, and at the instant when the capacitor's charge is $q$. Express $Q_{0}$ in terms of $r_{\max }$ (from Part A-i) and neglect the effect described in Part A-iii. Context answer: \boxed{$Q_{0}=2 \pi \varepsilon_{0} q r_{\max } / C$} ",ii. Find the dependence of $q$ on time $t$ by approximating it with a continuous function $q(t)$ and assuming that $q(0)=q_{0}$.,"[""The sign of the droplet's charge is the same as that of the capacitor's opposite plate (which is connected to the farther electrode). So, when the droplet falls into the bowl, it will increase the capacitor's charge by $Q$ :\n\n$$\nd q=2 \\pi \\varepsilon_{0} U r_{\\max } d N=2 \\pi \\varepsilon_{0} r_{\\max } n d t \\frac{q}{C}\n$$\n\nwhere $d N=n d t$ is the number of droplets which fall during the time $d t$ This is a simple linear differential equation which is solved easily to obtain\n\n$$\nq=q_{0} e^{\\gamma t}, \\quad \\gamma=\\frac{2 \\pi \\varepsilon_{0} r_{\\max } n}{C}=\\frac{\\pi \\varepsilon_{0} n}{C} \\sqrt[3]{\\frac{6 \\sigma d}{\\rho g}}\n$$""]","['$q(t)=q_{0} e^{\\gamma t}$ , $\\gamma=\\frac{\\pi \\varepsilon_{0} n}{C} \\sqrt[3]{\\frac{6 \\sigma d}{\\rho g}}$']",True,,Expression, 1172,Thermodynamics,"Problem T2. Kelvin water dropper The following facts about the surface tension may turn out to be useful for this problem. For the molecules of a liquid, the positions at the liquid-air interface are less favourable as compared with the positions in the bulk of the liquid. This interface is described by the so-called surface energy, $U=\sigma S$, where $S$ is the surface area of the interface and $\sigma$ is the surface tension coefficient of the liquid. Moreover, two fragments of the liquid surface pull each other with a force $F=\sigma l$, where $l$ is the length of a straight line separating the fragments. A long metallic pipe with internal diameter $d$ is pointing directly downwards. Water is slowly dripping from a nozzle at its lower end, see fig. Water can be considered to be electrically conducting; its surface tension is $\sigma$ and its density is $\rho$. A droplet of radius $r$ hangs below the nozzle. The radius grows slowly in time until the droplet separates from the nozzle due to the free fall acceleration $g$. Always assume that $d \ll r$. Part A. Single pipe Context question: i. Find the radius $r_{\max }$ of a drop just before it separates from the nozzle. Context answer: \boxed{$r_{\max }=\sqrt[3]{\frac{3 \sigma d}{4 \rho g}}$} Context question: ii. Relative to the far-away surroundings, the pipe's electrostatic potential is $\varphi$. Find the charge $Q$ of a drop when its radius is $r$. Context answer: \boxed{$Q=4 \pi \varepsilon_{0} \varphi r$} Context question: iii. Consider the situation in which $r$ is kept constant and $\varphi$ is slowly increased. The droplet becomes unstable and breaks into pieces if the hydrostatic pressure inside the droplet becomes smaller than the atmospheric pressure. Find the critical potential $\varphi_{\max }$ at which this will happen. Context answer: \boxed{$\varphi_{\max }=2 \sqrt{\frac{\sigma r}{ \varepsilon_{0}}}$} Extra Supplementary Reading Materials: Part B. Two pipes An apparatus called the ""Kelvin water dropper"" consists of two pipes, each identical to the one described in Part A, connected via a T-junction, see fig. The ends of both pipes are at the centres of two cylindrical electrodes (with height $L$ and diameter $D$ with $L \gg D \gg r$ ). For both tubes, the dripping rate is $n$ droplets per unit time. Droplets fall from height $H$ into conductive bowls underneath the nozzles, cross-connected to the electrodes as shown in the diagram. The electrodes are connected via a capacitance $C$. There is no net charge on the system of bowls and electrodes. Note that the top water container is earthed as shown. The first droplet to fall will have some microscopic charge which will cause an imbalance between the two sides and a small charge separation across the capacitor. Context question: i. Express the absolute value of the charge $Q_{0}$ of the drops as they separate from the tubes, and at the instant when the capacitor's charge is $q$. Express $Q_{0}$ in terms of $r_{\max }$ (from Part A-i) and neglect the effect described in Part A-iii. Context answer: \boxed{$Q_{0}=2 \pi \varepsilon_{0} q r_{\max } / C$} Context question: ii. Find the dependence of $q$ on time $t$ by approximating it with a continuous function $q(t)$ and assuming that $q(0)=q_{0}$. Context answer: \boxed{$q(t)=q_{0} e^{\gamma t}$ , $\gamma=\frac{\pi \varepsilon_{0} n}{C} \sqrt[3]{\frac{6 \sigma d}{\rho g}}$} ","iii. The dropper's functioning can be hindered by the effect shown in Part A-iii. In addition, a limit $U_{\max }$ to the achievable potential between the electrodes is set by the electrostatic push between a droplet and the bowl beneath it. Find $U_{\text {max }}$.","[""The droplets can reach the bowls if their mechanical energy $m g H$ (where $m$ is the droplet's mass) is large enough to overcome the electrostatic push: The droplet starts at the point where the electric potential is 0 , which is the sum of the potential $U / 2$, due to the electrode, and of its self-generated potential $-U / 2$. Its motion is not affected by the self-generated field, so it needs to fall from the potential $U / 2$ down to the potential $-U / 2$, resulting in the change of the electrostatic energy equal to $U Q \\leq m g H$, where $Q=2 \\pi \\varepsilon_{0} U r_{\\max }$ (see above). So,\n\n$$\n\\begin{gathered}\nU_{\\max }=\\frac{m g H}{2 \\pi \\varepsilon_{0} U_{\\max } r_{\\max }} \\\\\n\\therefore U_{\\max }=\\sqrt{\\frac{H \\sigma d}{2 \\varepsilon_{0} r_{\\max }}}=\\sqrt[6]{\\frac{H^{3} g \\sigma^{2} \\rho d^{2}}{6 \\varepsilon_{0}^{3}}} .\n\\end{gathered}\n$$""]",['$U_{\\max }=\\sqrt[6]{\\frac{H^{3} g \\sigma^{2} \\rho d^{2}}{6 \\varepsilon_{0}^{3}}}$'],False,,Expression, 1173,Modern Physics,"Problem T3. Protostar formation Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant.","i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.",['$$\n\\begin{aligned}\n& T=\\text { const } \\Longrightarrow p V=\\text { const } \\\\\n& V \\propto r^{3} \\\\\n& \\therefore p \\propto r^{-3} \\Longrightarrow \\frac{p\\left(r_{1}\\right)}{p\\left(r_{0}\\right)}=2^{3}=8 .\n\\end{aligned}\n$$'],['$n=8$'],False,,Numerical,0 1174,Modern Physics,"Problem T3. Protostar formation Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant. Context question: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform. Context answer: \boxed{$n=8$} ",ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.,"[""During the period considered the pressure is negligible. Therefore the gas is in free fall. By Gauss' theorem and symmetry, the gravitational field at any point in the ball is equivalent to the one generated when all the mass closer to the center is compressed into the center. Moreover, while the ball has not yet shrunk much, the field strength on its surface does not change much either. The acceleration of the outermost layer stays approximately constant. Thus,\n\n$$\nt \\approx \\sqrt{\\frac{2\\left(r_{0}-r_{2}\\right)}{g}}\n$$\n\nwhere\n\n$$\n\\begin{aligned}\ng & \\approx \\frac{G m}{r_{0}^{2}} \\\\\n\\therefore t & \\approx \\sqrt{\\frac{2 r_{0}^{2}\\left(r_{0}-r_{2}\\right)}{G m}}=\\sqrt{\\frac{0.1 r_{0}^{3}}{G m}} .\n\\end{aligned}\n$$""]",['$\\sqrt{\\frac{0.1 r_{0}^{3}}{G m}}$'],False,,Expression, 1175,Modern Physics,"Problem T3. Protostar formation Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant. Context question: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform. Context answer: \boxed{$n=8$} Context question: ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle. Context answer: \boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$} ","iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.","[""Gravitationally the outer layer of the ball is influenced by the rest just as the rest were compressed into a point mass. Therefore we have Keplerian motion: the fall of any part of the outer layer consists in a halfperiod of an ultraelliptical orbit. The ellipse is degenerate into a line; its foci are at the ends of the line; one focus is at the center of the ball (by Kepler's $1^{\\text {st }}$ law) and the other one is at $r_{0}$, see figure (instead of a degenerate ellipse, a strongly elliptical ellipse is depicted). The period of the orbit is determined by the longer semiaxis of the ellipse (by Kepler's $3^{\\text {rd }}$ law). The longer semiaxis is $r_{0} / 2$ and we are interested in half a period. Thus, the answer is equal to the halfperiod of a circular orbit of radius $r_{0} / 2$ :\n\n$$\n\\left(\\frac{2 \\pi}{2 t_{r \\rightarrow 0}}\\right)^{2} \\frac{r_{0}}{2}=\\frac{G m}{\\left(r_{0} / 2\\right)^{2}} \\Longrightarrow t_{r \\rightarrow 0}=\\pi \\sqrt{\\frac{r_{0}^{3}}{8 G m}}\n$$\n\ncentre of the cloud initial position of a certain parcel\n\nstrongly elliptical orbit of the gas parcel of gas area covered by the radius vector\n\nAlternatively, one may write the energy conservation law $\\frac{\\dot{r}^{2}}{2}-\\frac{G m}{r}=E$ (that in turn is obtainable from Newton's II law $\\left.\\ddot{r}=-\\frac{G m}{r^{2}}\\right)$ with $E=-\\frac{G m}{r_{0}}$, separate the variables $\\left(\\frac{d r}{d t}=-\\sqrt{2 E+\\frac{2 G m}{r}}\\right)$ and write the integral $t=-\\int \\frac{d r}{\\sqrt{2 E+\\frac{2 G m}{r}}}$ This integral is probably not calculable during the limitted time given during the Olympiad, but a possible approach can be sketched as follows. Substituting $\\sqrt{2 E+\\frac{2 G m}{r}}=\\xi$ and $\\sqrt{2 E}=v$, one gets\n\n$$\n\\begin{aligned}\n\\frac{t_{\\infty}}{4 G m} & =\\int_{0}^{\\infty} \\frac{d \\xi}{\\left(v^{2}-\\xi^{2}\\right)^{2}} \\\\\n& =\\frac{1}{4 v^{3}} \\int_{0}^{\\infty}\\left[\\frac{v}{(v-\\xi)^{2}}+\\frac{v}{(v+\\xi)^{2}}+\\frac{1}{v-\\xi}+\\frac{1}{v+\\xi}\\right] d \\xi\n\\end{aligned}\n$$\n\nHere (after shifting the variable) one can use $\\int \\frac{d \\xi}{\\xi}=\\ln \\xi$ and $\\int \\frac{d \\xi}{\\xi^{2}}=-\\frac{1}{\\xi}$, finally getting the same answer as by Kepler's laws.""]",['$t_{r \\rightarrow 0}=\\pi \\sqrt{\\frac{r_{0}^{3}}{8 G m}}$'],False,,Expression, 1176,Modern Physics,"Problem T3. Protostar formation Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant. Context question: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform. Context answer: \boxed{$n=8$} Context question: ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle. Context answer: \boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$} Context question: iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws. Context answer: \boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$} ","iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.","['By Clapeyron-Mendeleyev law,\n\n$$\np=\\frac{m R T_{0}}{\\mu V}\n$$\n\nWork done by gravity to compress the ball is\n\n$$\nW=-\\int p d V=-\\frac{m R T_{0}}{\\mu} \\int_{\\frac{4}{3} \\pi r_{0}^{3}}^{\\frac{4}{3} \\pi r_{3}^{3}} \\frac{d V}{V}=\\frac{3 m R T_{0}}{\\mu} \\ln \\frac{r_{0}}{r_{3}}\n$$\n\nThe temperature stays constant, so the internal energy does not change; hence, according to the $1^{\\text {st }}$ law of thermodynamics, the compression work $W$ is the heat radiated.']",['$Q=\\frac{3 m R T_{0}}{\\mu} \\ln \\frac{r_{0}}{r_{3}}$'],False,,Expression, 1177,Modern Physics,"Problem T3. Protostar formation Let us model the formation of a star as follows. A spherical cloud of sparse interstellar gas, initially at rest, starts to collapse due to its own gravity. The initial radius of the ball is $r_{0}$ and the mass is $m$. The temperature of the surroundings (much sparser than the gas) and the initial temperature of the gas is uniformly $T_{0}$. The gas may be assumed to be ideal. The average molar mass of the gas is $\mu$ and its adiabatic index is $\gamma>\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant. Context question: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform. Context answer: \boxed{$n=8$} Context question: ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle. Context answer: \boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$} Context question: iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws. Context answer: \boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$} Context question: iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$. Context answer: \boxed{$Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$} ",v. For radii smaller than $r_{3}$ you may neglect heat loss due to radiation. Determine how the temperature $T$ of the ball depends on its radius for $r\frac{4}{3}$. Assume that $G \frac{m \mu}{r_{0}} \gg R T_{0}$, where $R$ is the gas constant and $G$ is the gravitational constant. Context question: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform. Context answer: \boxed{$n=8$} Context question: ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle. Context answer: \boxed{$\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$} Context question: iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws. Context answer: \boxed{$t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$} Context question: iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$. Context answer: \boxed{$Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$} Context question: v. For radii smaller than $r_{3}$ you may neglect heat loss due to radiation. Determine how the temperature $T$ of the ball depends on its radius for $r Hints and Data The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$. Specific heat capacity of the atmosphere in the pertaining temperature range: $\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$ Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$ Specific latent heat of vaporisation of the water within the volume of the cloud: $\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$ $\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$",1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms.,['1. Temperature $T_{1}$ where the cloud ceiling forms\n\n$$\n\\mathrm{T}_{1}=\\mathrm{T}_{0} \\cdot\\left(\\frac{\\mathrm{p}_{1}}{\\mathrm{p}_{0}}\\right)^{1-\\frac{1}{x}}=279 \\mathrm{~K}\n\\tag{1}\n$$'],['279'],False,K,Numerical,1e0 1180,Thermodynamics,"Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$. Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain. Hints and Data The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$. Specific heat capacity of the atmosphere in the pertaining temperature range: $\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$ Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$ Specific latent heat of vaporisation of the water within the volume of the cloud: $\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$ $\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$ Context question: 1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms. Context answer: \boxed{279} ",2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density?,"['2. Height $h_{1}$ of the cloud ceiling:\n\n$$\np_{0}-p_{1}=\\frac{\\rho_{0}+\\rho_{1}}{2} \\cdot g \\cdot h_{1}, with \\rho_{1}=\\rho_{0} \\cdot \\frac{p_{1}}{p_{0}} \\cdot \\frac{T_{0}}{T_{1}}\n$$\n\n$$\n\\mathrm{h}_{1}=1410 \\mathrm{~m}\n\\tag{2}\n$$']",['1410'],False,m,Numerical,1e1 1181,Thermodynamics,"Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$. Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain. Hints and Data The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$. Specific heat capacity of the atmosphere in the pertaining temperature range: $\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$ Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$ Specific latent heat of vaporisation of the water within the volume of the cloud: $\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$ $\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$ Context question: 1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms. Context answer: \boxed{279} Context question: 2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density? Context answer: \boxed{1410} ",3. What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range?,"['3. Temperature $\\mathrm{T}_{2}$ at the ridge of the mountain.\n\nThe temperature difference when the air is ascending from the cloud ceiling to the mountain ridge is caused by two processes:\n\n- adiabatic cooling to temperature $\\mathrm{T}_{\\mathrm{x}}$,\n\n\n\n- heating by $\\Delta \\mathrm{T}$ by condensation.\n\n$$\n\\mathrm{T}_{2} =\\mathrm{T}_{\\mathrm{x}}+\\Delta \\mathrm{T}\n\\tag{3}\n$$\n$$\n\\mathrm{T}_{\\mathrm{x}} =\\mathrm{T}_{1} \\cdot\\left(\\frac{\\mathrm{p}_{2}}{\\mathrm{p}_{1}}\\right)^{1-\\frac{1}{x}}=265 \\mathrm{~K}\n\\tag{4}\n$$\n\nFor each $\\mathrm{kg}$ of air the heat produced by condensation is $\\mathrm{L}_{\\mathrm{v}} \\cdot 2.45 \\mathrm{~g}=6.125 \\mathrm{~kJ}$.\n\n$$\n\\Delta \\mathrm{T}=\\frac{6.125}{\\mathrm{c}_{\\mathrm{p}}} \\cdot \\frac{\\mathrm{kJ}}{\\mathrm{kg}}=6.1 \\mathrm{~K}\n\\tag{5}\n$$\n\n$$\n\\mathrm{T}_{2}=271 \\mathrm{~K}\n\\tag{6}\n$$']",['271'],False,K,Numerical,1e0 1182,Thermodynamics,"Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$. Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain. Hints and Data The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$. Specific heat capacity of the atmosphere in the pertaining temperature range: $\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$ Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$ Specific latent heat of vaporisation of the water within the volume of the cloud: $\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$ $\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$ Context question: 1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms. Context answer: \boxed{279} Context question: 2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density? Context answer: \boxed{1410} Context question: 3. What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range? Context answer: \boxed{271} ","4. Determine the height of the water column (precipitation level) precipitated by the air stream in 3 hours, assuming a homogeneous rainfall between points $M_{1}$ and $M_{2}$.",['4. Height of precipitated water column\n\n$$\n\\mathrm{h}=35 \\mathrm{~mm}\n\\tag{7}\n$$'],['35'],False,mm,Numerical,1e0 1183,Thermodynamics,"Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of $100 \mathrm{kPa}$ are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of $70 \mathrm{kPa}$ at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ} \mathrm{C}$. As the air is ascending, cloud formation sets in at $84.5 \mathrm{kPa}$. Consider a quantity of moist air ascending the mountain with a mass of $2000 \mathrm{~kg}$ over each square meter. This moist air reaches the mountain ridge (station $\mathrm{M}_{2}$ ) after 1500 seconds. During that rise an amount of $2.45 \mathrm{~g}$ of water per kilogram of air is precipitated as rain. Hints and Data The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of $1 \mathrm{~K}$, the height of the cloud ceiling to an accuracy of $10 \mathrm{~m}$ and the precipitation level to an accuracy of $1 \mathrm{~mm}$. Specific heat capacity of the atmosphere in the pertaining temperature range: $\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}$ Atmospheric density for $\mathrm{p}_{0}$ and $\mathrm{T}_{0}$ at station $\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3}$ Specific latent heat of vaporisation of the water within the volume of the cloud: $\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}$ $\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4$ and $\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}$ Context question: 1. Determine temperature $T_{1}$ at $M_{1}$ where the cloud ceiling forms. Context answer: \boxed{279} Context question: 2. What is the height $h_{1}\left(\right.$ at $\left.M_{1}\right)$ above station $M_{0}$ of the cloud ceiling assuming a linear decrease of atmospheric density? Context answer: \boxed{1410} Context question: 3. What temperature $\mathrm{T}_{2}$ is measured at the ridge of the mountain range? Context answer: \boxed{271} Context question: 4. Determine the height of the water column (precipitation level) precipitated by the air stream in 3 hours, assuming a homogeneous rainfall between points $M_{1}$ and $M_{2}$. Context answer: \boxed{35} ","5. What temperature $\mathrm{T}_{3}$ is measured in the back of the mountain range at station $\mathrm{M}_{3}$ ? Discuss the state of the atmosphere at station $\mathrm{M}_{3}$ in comparison with that at station $\mathrm{M}_{0}$.",['5. Temperature $\\mathrm{T}_{3}$ behind the mountain\n\n$$\n\\mathrm{T}_{3}=\\mathrm{T}_{2} \\cdot\\left(\\frac{\\mathrm{p}_{3}}{\\mathrm{p}_{2}}\\right)^{1-\\frac{1}{x}}=300 \\mathrm{~K}\n\\tag{8}\n$$\n\nThe air has become warmer and dryer. The temperature gain is caused by condensation of vapour.'],['300'],False,K,Numerical,1e0 1184,Electromagnetism,"A beam of electrons emitted by a point source $\mathrm{P}$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small $\left(2 \cdot \alpha_{0}<<1\right)$. The injection of the electrons occurs on the mean radius $\mathrm{R}$ of the toroid with acceleration voltage $\mathrm{V}_{0}$. Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant. Data: $\frac{\mathrm{e}}{\mathrm{m}}=1.76 \cdot 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1} ; \quad \mathrm{V}_{0}=3 \mathrm{kV} ; \quad \mathrm{R}=50 \mathrm{~mm}$",1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus.,"['1. Determination of $\\mathrm{B}$ :\n\nThe vector of the velocity of any electron is divided into components parallel with and perpendicular to the magnetic field $\\overrightarrow{\\mathrm{B}}$ :\n\n$$\n\\overrightarrow{\\mathrm{v}}=\\overrightarrow{\\mathrm{v}}_{\\|}+\\overrightarrow{\\mathrm{v}}_{\\perp}\n\\tag{1}\n$$\n\nThe Lorentz force $\\overrightarrow{\\mathrm{F}}=-\\mathrm{e} \\cdot(\\overrightarrow{\\mathrm{v}} \\times \\overrightarrow{\\mathrm{B}})$ influences only the perpendicular component, it acts as a radial force:\n\n$$\n\\mathrm{m} \\cdot \\frac{\\mathrm{v}_{\\perp}^{2}}{\\mathrm{r}}=\\mathrm{e} \\cdot \\mathrm{v}_{\\perp} \\cdot \\mathrm{B}\n\\tag{2}\n$$\n\nHence the radius of the circular path that has been travelled is\n\n$$\n\\mathrm{r}=\\frac{\\mathrm{m}}{\\mathrm{e}} \\cdot \\frac{\\mathrm{v}_{\\perp}}{\\mathrm{B}}\n\\tag{3}\n$$\n\n\n\nand the period of rotation which is independent of $\\mathrm{v}_{\\perp}$ is\n\n$$\n\\mathrm{T}=\\frac{2 \\cdot \\pi \\cdot \\mathrm{r}}{\\mathrm{v}_{\\perp}}=\\frac{2 \\cdot \\pi \\cdot \\mathrm{m}}{\\mathrm{B} \\cdot \\mathrm{e}}\n\\tag{4}\n$$\n\nThe parallel component of the velocity does not vary. Because of $\\alpha_{0}<<1$ it is approximately equal for all electrons:\n\n$$\n\\mathrm{V}_{\\| 0}=\\mathrm{V}_{0} \\cdot \\cos \\alpha_{0} \\approx \\mathrm{v}_{0}\n\\tag{5}\n$$\n\nHence the distance $b$ between the focusing points, using eq. (5), is\n\n$$\n\\mathrm{b}=\\mathrm{v}_{\\| 0} \\cdot \\mathrm{T} \\approx \\mathrm{v}_{0} \\cdot \\mathrm{T}\n\\tag{6}\n$$\n\nFrom the law of conservation of energy follows the relation between the acceleration voltage $\\mathrm{V}_{0}$ and the velocity $\\mathrm{v}_{0}$ :\n\n$$\n\\frac{\\mathrm{m}}{2} \\cdot \\mathrm{v}_{0}^{2}=\\mathrm{e} \\cdot \\mathrm{V}_{0}\n\\tag{7}\n$$\n\nUsing eq. (7) and eq. (4) one obtains from eq. (6)\n\n$$\n\\mathrm{b}=\\frac{2 \\cdot \\pi}{\\mathrm{B}} \\cdot \\sqrt{2 \\cdot \\frac{\\mathrm{m}}{\\mathrm{e}} \\cdot \\mathrm{V}_{0}}\n\\tag{8}\n$$\n\nand because of $b=\\frac{2 \\cdot \\pi \\cdot R}{4}$ one obtains\n\n$$\n\\mathrm{B}=\\frac{4}{\\mathrm{R}} \\cdot \\sqrt{2 \\cdot \\frac{\\mathrm{m}}{\\mathrm{e}} \\cdot \\mathrm{V}_{0}}=1.48 \\cdot 10^{-2} \\frac{\\mathrm{Vs}}{\\mathrm{m}^{2}}\n\\tag{9}\n$$']",['$1.48 \\cdot 10^{-2}$'],False,$\frac{\mathrm{Vs}}{\mathrm{m}^{2}}$,Numerical,1e-4 1185,Electromagnetism,"A beam of electrons emitted by a point source $\mathrm{P}$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small $\left(2 \cdot \alpha_{0}<<1\right)$. The injection of the electrons occurs on the mean radius $\mathrm{R}$ of the toroid with acceleration voltage $\mathrm{V}_{0}$. Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant. Data: $\frac{\mathrm{e}}{\mathrm{m}}=1.76 \cdot 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1} ; \quad \mathrm{V}_{0}=3 \mathrm{kV} ; \quad \mathrm{R}=50 \mathrm{~mm}$ Context question: 1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus. Context answer: \boxed{$1.48 \cdot 10^{-2}$} ","2. Determine the value of $\vec{B}$ which gives four focussing points separated by $\pi / 2$ as indicated in the diagram. Note: When considering the electron paths you may disregard the curvature of the magnetic field.",['2. Determination of $\\mathrm{B}_{1}$ :\n\nAnalogous to eq. (2)\n\n$$\n\\mathrm{m} \\cdot \\frac{\\mathrm{v}_{0}^{2}}{\\mathrm{R}}=\\mathrm{e} \\cdot \\mathrm{v}_{0} \\cdot \\mathrm{B}_{1}\n\\tag{10}\n$$\n\nmust hold.\n\nFrom eq. (7) follows\n\n$$\n\\mathrm{B}_{1}=\\frac{1}{\\mathrm{R}} \\cdot \\sqrt{2 \\cdot \\frac{\\mathrm{m}}{\\mathrm{e}} \\cdot \\mathrm{V}_{0}}=0.37 \\cdot 10^{-2} \\frac{\\mathrm{Vs}}{\\mathrm{m}^{2}}\n\\tag{11}\n$$'],['$0.37 \\cdot 10^{-2}$'],False,$\frac{\mathrm{Vs}}{\mathrm{m}^{2}}$,Numerical,1e-4 1186,Electromagnetism,"A beam of electrons emitted by a point source $\mathrm{P}$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small $\left(2 \cdot \alpha_{0}<<1\right)$. The injection of the electrons occurs on the mean radius $\mathrm{R}$ of the toroid with acceleration voltage $\mathrm{V}_{0}$. Neglect any interaction between the electrons. The magnitude of $\vec{B}, B$, is assumed to be constant. Data: $\frac{\mathrm{e}}{\mathrm{m}}=1.76 \cdot 10^{11} \mathrm{C} \cdot \mathrm{kg}^{-1} ; \quad \mathrm{V}_{0}=3 \mathrm{kV} ; \quad \mathrm{R}=50 \mathrm{~mm}$ Context question: 1. To guide the electron in the toroidal field a homogeneous magnetic deflection field $\vec{B}_{1}$ is required. Calculate $\vec{B}_{1}$ for an electron moving on a circular orbit of radius $R$ in the torus. Context answer: \boxed{$1.48 \cdot 10^{-2}$} Context question: 2. Determine the value of $\vec{B}$ which gives four focussing points separated by $\pi / 2$ as indicated in the diagram. Note: When considering the electron paths you may disregard the curvature of the magnetic field. Context answer: \boxed{$0.37 \cdot 10^{-2}$} Extra Supplementary Reading Materials: 3. The electron beam cannot stay in the toroid without a deflection field $\vec{B}_{1}$, but will leave it with a systematic motion (drift) perpendicular to the plane of the toroid. Note: The angle of aperture of the electron beam can be neglected. Use the laws of conservation of energy and of angular momentum.",a) Show that the radial deviation of the electrons from the injection radius is finite.,"['3. Finiteness of $r_{1}$ and direction of the drift velocity\n\nIn the magnetic field the lines of force are circles with their centres on the symmetry axis (z-axis) of the toroid.\n\nIn accordance with the symmetry of the problem, polar coordinates $r$ and $\\varphi$ are introduced into the plane perpendicular to the z-axis (see figure below) and the occurring vector quantities (velocity, magnetic field $\\overrightarrow{\\mathrm{B}}$, Lorentz force) are divided into the corresponding components.\n\n\n\nSince the angle of aperture of the beam can be neglected examine a single electron injected tangentially into the toroid with velocity $\\mathrm{v}_{0}$ on radius $\\mathrm{R}$.\n\nIn a static magnetic field the kinetic energy is conserved, thus\n\n$$\n\\mathrm{E}=\\frac{\\mathrm{m}}{2}\\left(\\mathrm{v}_{\\mathrm{r}}^{2}+\\mathrm{v}_{\\varphi}^{2}+\\mathrm{v}_{\\mathrm{z}}^{2}\\right)=\\frac{\\mathrm{m}}{2} \\mathrm{v}_{0}^{2}\n\\tag{12}\n$$\n\nThe radial points of inversion of the electron are defined by the condition\n\n$$\n\\mathrm{v}_{\\mathrm{r}}=0\n$$\n\nUsing eq. (12) one obtains\n\n$$\n\\mathrm{v}_{0}^{2}=\\mathrm{v}_{\\varphi}^{2}+\\mathrm{v}_{\\mathrm{z}}^{2}\n\\tag{13}\n$$\n\nSuch an inversion point is obviously given by\n\n$$\n\\mathrm{r}=\\mathrm{R} \\cdot\\left(\\mathrm{v}_{\\varphi}=\\mathrm{v}_{0}, \\mathrm{v}_{\\mathrm{r}}=0, \\mathrm{v}_{\\mathrm{z}}=0\\right)\n$$\n\nTo find further inversion points and thus the maximum radial deviation of the electron the components of velocity $\\mathrm{v}_{\\varphi}$ and $\\mathrm{v}_{\\mathrm{Z}}$ in eq. (13) have to be expressed by the radius.\n\n$\\mathrm{V}_{\\varphi}$ will be determined by the law of conservation of angular momentum. The Lorentz force obviously has no component in the $\\varphi$ - direction (parallel to the magnetic field). Therefore it cannot produce a torque around the z-axis. From this follows that the $\\mathrm{z}$-component of the angular momentum is a constant, i.e. $\\mathrm{L}_{\\mathrm{z}}=\\mathrm{m} \\cdot \\mathrm{v}_{\\varphi} \\cdot \\mathrm{r}=\\mathrm{m} \\cdot \\mathrm{v}_{0} \\cdot \\mathrm{R}$ and therefore \n$$\n\\mathrm{v}_{\\varphi}=\\mathrm{v}_{0} \\cdot \\frac{\\mathrm{R}}{\\mathrm{r}}\n\\tag{14}\n$$\n\n$\\mathrm{v}_{\\mathrm{z}}$ will be determined from the equation of motion in the z-direction. The z-component of the Lorentz force is $\\mathrm{F}_{\\mathrm{z}}=-\\mathrm{e} \\cdot \\mathrm{B} \\cdot \\mathrm{v}_{\\mathrm{r}}$. Thus the acceleration in the $\\mathrm{z}$-direction is\n\n$$\n\\mathrm{a}_{\\mathrm{z}}=-\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot \\mathrm{v}_{\\mathrm{r}} .\n\\tag{15}\n$$\n\nThat means, since B is assumed to be constant, a change of $v_{z}$ is related to a change of $r$ as follows:\n\n$$\n\\Delta \\mathrm{v}_{\\mathrm{z}}=-\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot \\Delta \\mathrm{r}\n$$\n\nBecause of $\\Delta \\mathrm{r}=\\mathrm{r}-\\mathrm{R}$ and $\\Delta \\mathrm{v}_{\\mathrm{z}}=\\mathrm{v}_{\\mathrm{z}}$ one finds\n\n$$\n\\mathrm{v}_{\\mathrm{z}}=-\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot(\\mathrm{r}-\\mathrm{R})\n\\tag{16}\n$$\n\nUsing eq. (14) and eq. (15) one obtains for eq. (13)\n\n$$\n1=\\left(\\frac{\\mathrm{R}}{\\mathrm{r}}\\right)^{2}+\\mathrm{A}^{2} \\cdot\\left(\\frac{\\mathrm{r}}{\\mathrm{R}}-1\\right)^{2}\n\\tag{17}\n$$\n\nwhere $\\mathrm{A}=\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot \\frac{\\mathrm{R}}{\\mathrm{v}_{0}}$\n\nDiscussion of the curve of the right side of eq. (17) gives the qualitative result shown in the following diagram:\n\n\n\nHence $\\mathrm{r}_{1}$ is finite. Since $\\mathrm{R} \\leq \\mathrm{r} \\leq \\mathrm{r}_{1}$ eq. (16) yields $\\mathrm{v}_{\\mathrm{z}}<0$. Hence the drift is in the direction of the negative $z$-axis.']",,False,,, 1187,Electromagnetism,,b) Determine the direction of the drift velocity.,"['3. Finiteness of $r_{1}$ and direction of the drift velocity\n\nIn the magnetic field the lines of force are circles with their centres on the symmetry axis (z-axis) of the toroid.\n\nIn accordance with the symmetry of the problem, polar coordinates $r$ and $\\varphi$ are introduced into the plane perpendicular to the z-axis (see figure below) and the occurring vector quantities (velocity, magnetic field $\\overrightarrow{\\mathrm{B}}$, Lorentz force) are divided into the corresponding components.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_85f45440123514ebf734g-1.jpg?height=402&width=734&top_left_y=807&top_left_x=707)\n\nSince the angle of aperture of the beam can be neglected examine a single electron injected tangentially into the toroid with velocity $\\mathrm{v}_{0}$ on radius $\\mathrm{R}$.\n\nIn a static magnetic field the kinetic energy is conserved, thus\n\n$$\n\\mathrm{E}=\\frac{\\mathrm{m}}{2}\\left(\\mathrm{v}_{\\mathrm{r}}^{2}+\\mathrm{v}_{\\varphi}^{2}+\\mathrm{v}_{\\mathrm{z}}^{2}\\right)=\\frac{\\mathrm{m}}{2} \\mathrm{v}_{0}^{2}\n\\tag{12}\n$$\n\nThe radial points of inversion of the electron are defined by the condition\n\n$$\n\\mathrm{v}_{\\mathrm{r}}=0\n$$\n\nUsing eq. (12) one obtains\n\n$$\n\\mathrm{v}_{0}^{2}=\\mathrm{v}_{\\varphi}^{2}+\\mathrm{v}_{\\mathrm{z}}^{2}\n\\tag{13}\n$$\n\nSuch an inversion point is obviously given by\n\n$$\n\\mathrm{r}=\\mathrm{R} \\cdot\\left(\\mathrm{v}_{\\varphi}=\\mathrm{v}_{0}, \\mathrm{v}_{\\mathrm{r}}=0, \\mathrm{v}_{\\mathrm{z}}=0\\right)\n$$\n\nTo find further inversion points and thus the maximum radial deviation of the electron the components of velocity $\\mathrm{v}_{\\varphi}$ and $\\mathrm{v}_{\\mathrm{Z}}$ in eq. (13) have to be expressed by the radius.\n\n$\\mathrm{V}_{\\varphi}$ will be determined by the law of conservation of angular momentum. The Lorentz force obviously has no component in the $\\varphi$ - direction (parallel to the magnetic field). Therefore it cannot produce a torque around the z-axis. From this follows that the $\\mathrm{z}$-component of the angular momentum is a constant, i.e. $\\mathrm{L}_{\\mathrm{z}}=\\mathrm{m} \\cdot \\mathrm{v}_{\\varphi} \\cdot \\mathrm{r}=\\mathrm{m} \\cdot \\mathrm{v}_{0} \\cdot \\mathrm{R}$ and therefore \n$$\n\\mathrm{v}_{\\varphi}=\\mathrm{v}_{0} \\cdot \\frac{\\mathrm{R}}{\\mathrm{r}}\n\\tag{14}\n$$\n\n$\\mathrm{v}_{\\mathrm{z}}$ will be determined from the equation of motion in the z-direction. The z-component of the Lorentz force is $\\mathrm{F}_{\\mathrm{z}}=-\\mathrm{e} \\cdot \\mathrm{B} \\cdot \\mathrm{v}_{\\mathrm{r}}$. Thus the acceleration in the $\\mathrm{z}$-direction is\n\n$$\n\\mathrm{a}_{\\mathrm{z}}=-\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot \\mathrm{v}_{\\mathrm{r}} .\n\\tag{15}\n$$\n\nThat means, since B is assumed to be constant, a change of $v_{z}$ is related to a change of $r$ as follows:\n\n$$\n\\Delta \\mathrm{v}_{\\mathrm{z}}=-\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot \\Delta \\mathrm{r}\n$$\n\nBecause of $\\Delta \\mathrm{r}=\\mathrm{r}-\\mathrm{R}$ and $\\Delta \\mathrm{v}_{\\mathrm{z}}=\\mathrm{v}_{\\mathrm{z}}$ one finds\n\n$$\n\\mathrm{v}_{\\mathrm{z}}=-\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot(\\mathrm{r}-\\mathrm{R})\n\\tag{16}\n$$\n\nUsing eq. (14) and eq. (15) one obtains for eq. (13)\n\n$$\n1=\\left(\\frac{\\mathrm{R}}{\\mathrm{r}}\\right)^{2}+\\mathrm{A}^{2} \\cdot\\left(\\frac{\\mathrm{r}}{\\mathrm{R}}-1\\right)^{2}\n\\tag{17}\n$$\n\nwhere $\\mathrm{A}=\\frac{\\mathrm{e}}{\\mathrm{m}} \\cdot \\mathrm{B} \\cdot \\frac{\\mathrm{R}}{\\mathrm{v}_{0}}$\n\nDiscussion of the curve of the right side of eq. (17) gives the qualitative result shown in the following diagram:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_7b02e42b91aa41787752g-1.jpg?height=334&width=663&top_left_y=1923&top_left_x=885)\n\nHence $\\mathrm{r}_{1}$ is finite. Since $\\mathrm{R} \\leq \\mathrm{r} \\leq \\mathrm{r}_{1}$ eq. (16) yields $\\mathrm{v}_{\\mathrm{z}}<0$. Hence the drift is in the direction of the negative $z$-axis.']",['negative z-axis'],False,,Need_human_evaluate, 1188,Electromagnetism,,"a) Determine how $\Phi$ depends on $\omega, \mathrm{L}$ and $\mathrm{C}$ ( $\omega$ is the angular frequency of the sine wave).","['a)\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4abc4f73a4283ca2dd46g-1.jpg?height=454&width=1100&top_left_y=1937&top_left_x=615)\n\nCurrent law: \n$$\n\\quad \\mathrm{I}_{\\mathrm{L}_{\\mathrm{n}-1}}+\\mathrm{I}_{\\mathrm{C}_{\\mathrm{n}}}-\\mathrm{I}_{\\mathrm{L}_{\\mathrm{n}}}=0\n\\tag{1}\n$$\n\nVoltage law: \n$$\n\\quad \\mathrm{V}_{\\mathrm{C}_{\\mathrm{n}-1}}+\\mathrm{V}_{\\mathrm{L}_{\\mathrm{n}-1}}-\\mathrm{V}_{\\mathrm{C}_{\\mathrm{n}}}=0\n\\tag{2}\n$$\n\n\n\nCapacitive voltage drop: \n$$\n\\mathrm{V}_{\\mathrm{C}_{\\mathrm{n}-1}}=\\frac{1}{\\omega \\cdot \\mathrm{C}} \\cdot \\tilde{\\mathrm{I}}_{\\mathrm{C}_{\\mathrm{n}-1}}\n\\tag{3}\n$$\n\nNote: In eq. (3) $\\tilde{\\mathrm{I}}_{\\mathrm{C}_{\\mathrm{n}-1}}$ is used instead of $\\mathrm{I}_{\\mathrm{C}_{\\mathrm{n}-1}}$ because the current leads the voltage by $90^{\\circ}$.\n\nInductive voltage drop: \n$$\n\\quad \\mathrm{V}_{\\mathrm{L}_{\\mathrm{n}-1}}=\\omega \\cdot \\mathrm{L} \\cdot \\tilde{\\mathrm{I}}_{\\mathrm{L}_{\\mathrm{n}-1}}\n\\tag{4}\n$$\n\nNote: In eq. (4) $\\tilde{\\mathrm{I}}_{\\mathrm{L}_{\\mathrm{n}-1}}$ is used instead of $\\mathrm{I}_{\\mathrm{L}_{\\mathrm{n}-1}}$ because the current lags behind the voltage by $90^{\\circ}$.\n\nThe voltage $\\mathrm{V}_{\\mathrm{C}_{\\mathrm{n}}}$ is given by: \n$$\n\\mathrm{V}_{\\mathrm{C}_{\\mathrm{n}}}=\\mathrm{V}_{0} \\cdot \\sin (\\omega \\cdot \\mathrm{t}+\\mathrm{n} \\cdot \\varphi)\n\\tag{5}\n$$\n\nFormula (5) follows from the problem.\n\nFrom eq. (3) and eq. (5): \n$$\n\\mathrm{I}_{\\mathrm{C}_{\\mathrm{n}}}=\\omega \\cdot \\mathrm{C} \\cdot \\mathrm{V}_{0} \\cdot \\cos (\\omega \\cdot \\mathrm{t}+\\mathrm{n} \\cdot \\varphi)\n\\tag{6}\n$$\n\nFrom eq. (4) and eq. (2) and with eq. (5)\n\n$$\nI_{L_{n-1}}=\\frac{V_{0}}{\\omega \\cdot L} \\cdot\\left[2 \\cdot \\sin \\left(\\omega \\cdot t+\\left(n-\\frac{1}{2}\\right) \\cdot \\varphi\\right) \\cdot \\sin \\frac{\\varphi}{2}\\right]\n\\tag{7}\n$$\n$$\nI_{L_{n}}=\\frac{V_{0}}{\\omega \\cdot L} \\cdot\\left[2 \\cdot \\sin \\left(\\omega \\cdot t+\\left(n+\\frac{1}{2}\\right) \\cdot \\varphi\\right) \\cdot \\sin \\frac{\\varphi}{2}\\right]\n\\tag{8}\n$$\n\nEqs. (6), (7) and (8) must satisfy the current law. This gives the dependence of $\\varphi$ on $\\omega, \\mathrm{L}$ and $\\mathrm{C}$.\n\n$0=\\mathrm{V}_{0} \\cdot \\omega \\cdot \\mathrm{C} \\cdot \\cos (\\omega \\cdot \\mathrm{t}+\\mathrm{n} \\cdot \\varphi)+2 \\cdot \\frac{\\mathrm{V}_{0}}{\\omega \\cdot \\mathrm{L}} \\cdot \\sin \\frac{\\varphi}{2} \\cdot\\left[2 \\cdot \\cos (\\omega \\cdot \\mathrm{t}+\\mathrm{n} \\cdot \\varphi) \\cdot \\sin \\left(-\\frac{\\varphi}{2}\\right)\\right]$\n\nThis condition must be true for any instant of time. Therefore it is possible to divide by $\\mathrm{V}_{0} \\cdot \\cos (\\omega \\cdot \\mathrm{t}+\\mathrm{n} \\cdot \\varphi)$.\n\nHence $\\omega^{2} \\cdot \\mathrm{L} \\cdot \\mathrm{C}=4 \\cdot \\sin ^{2}\\left(\\frac{\\varphi}{2}\\right)$. The result is\n\n$$\n\\varphi=2 \\cdot \\arcsin \\left(\\frac{\\omega \\cdot \\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}{2}\\right) with 0 \\leq \\omega \\leq \\frac{2}{\\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}\n\\tag{9}\n$$']",['$\\varphi=2 \\cdot \\arcsin \\left(\\frac{\\omega \\cdot \\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}{2}\\right)$ with $0 \\leq \\omega \\leq \\frac{2}{\\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}$'],False,,Need_human_evaluate, 1189,Electromagnetism,"When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. Formulae: $\cos \alpha-\cos \beta=-2 \cdot \sin \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$ $\sin \alpha-\sin \beta=2 \cdot \cos \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$ Context question: a) Determine how $\Phi$ depends on $\omega, \mathrm{L}$ and $\mathrm{C}$ ( $\omega$ is the angular frequency of the sine wave). Context answer: $\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)$ with $0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}$ ",b) Determine the velocity of propagation of the waves if the length of each unit is $\ell$.,['b) The distance $\\ell$ is covered in the time $\\Delta \\mathrm{t}$ thus the propagation velocity is\n\n$$\n\\mathrm{v}=\\frac{\\ell}{\\Delta \\mathrm{t}}=\\frac{\\omega \\cdot \\ell}{\\varphi} \\quad \\text { or } \\quad \\mathrm{v}=\\frac{\\omega \\cdot \\ell}{2 \\cdot \\arcsin \\left(\\frac{\\omega \\cdot \\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}{2}\\right)}\n\\tag{10}\n$$'],['$\\frac{\\omega \\cdot \\ell}{2 \\cdot \\arcsin \\left(\\frac{\\omega \\cdot \\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}{2}\\right)}$'],False,,Expression, 1190,Electromagnetism,"When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. Formulae: $\cos \alpha-\cos \beta=-2 \cdot \sin \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$ $\sin \alpha-\sin \beta=2 \cdot \cos \left(\frac{\alpha+\beta}{2}\right) \cdot \sin \left(\frac{\alpha-\beta}{2}\right)$ Context question: a) Determine how $\Phi$ depends on $\omega, \mathrm{L}$ and $\mathrm{C}$ ( $\omega$ is the angular frequency of the sine wave). Context answer: $\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)$ with $0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{L} \cdot \mathrm{C}}}$ Context question: b) Determine the velocity of propagation of the waves if the length of each unit is $\ell$. Context answer: \boxed{$\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right)}$} ",c) State under what conditions the propagation velocity of the waves is almost independent of $\omega$. Determine the velocity in this case.,"['c)\n\n\n\nSlightly dependent means arc sin $\\left(\\frac{\\omega \\cdot \\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}{2}\\right) \\sim \\omega$, since $\\mathrm{v}$ is constant in that case.\n\nThis is true only for small values of $\\omega$. That means $\\frac{\\omega \\cdot \\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}{2} \\ll 1$ and therefore\n\n$$\n\\mathrm{v}_{0}=\\frac{\\ell}{\\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}\n\\tag{11}\n$$']",['$\\frac{\\ell}{\\sqrt{\\mathrm{L} \\cdot \\mathrm{C}}}$'],False,,Expression, 1191,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base.","A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base.",['Geometric solution: use that torque with respect to point of contact is $0 \\Rightarrow$ center of gravity has to be vertically above point of contact.\n\n$$\n\\begin{aligned}\n& \\sin \\phi=\\frac{D}{b} \\\\\n& \\sin \\Theta=\\frac{D}{r_{1}}\n\\end{aligned}\n$$\n\nHere $D$ may be called another name. Solve this:\n\n$$\n\\sin \\phi=\\frac{r_{1}}{b} \\sin \\Theta \\Rightarrow b=\\frac{r_{1} \\sin \\Theta}{\\sin \\phi}\n$$'],['$b=\\frac{r_{1} \\sin \\Theta}{\\sin \\phi}$'],False,,Expression, 1192,Mechanics,,"A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small.","['Kinetic energy: $\\frac{1}{2} I_{S} \\dot{\\varphi}^{2}$ and potential energy: $-b M g \\cos \\varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion.\n\nAngular equation of motion from torque, $\\tau=I_{S} \\ddot{\\varphi}=-M g b \\sin \\varphi$.\n\n\nCorrect equation (either energy conservation or torque equation of motion)\n\n$$\nT=2 \\pi \\sqrt{\\frac{I_{S}}{M g b}} \\Rightarrow I_{S}=\\frac{M g b T^{2}}{4 \\pi^{2}}\n$$\n\n\n\n(Derivation:\n\n$$\n\\Rightarrow \\ddot{\\varphi}=-\\frac{b M g}{I_{S}} \\sin \\varphi \\simeq-\\frac{b g M}{I_{S}} \\varphi\n$$\n\nso that\n\n$$\n\\omega^{2}=\\frac{b g M}{I_{S}}\n$$']","['Kinetic energy: $\\frac{1}{2} I_{S} \\dot{\\varphi}^{2}$ and potential energy: $-b M g \\cos \\varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion.\n\nAngular equation of motion from torque, $\\tau=I_{S} \\ddot{\\varphi}=-M g b \\sin \\varphi$.\n\n\nCorrect equation (either energy conservation or torque equation of motion)\n\n$$\nT=2 \\pi \\sqrt{\\frac{I_{S}}{M g b}} \\Rightarrow I_{S}=\\frac{M g b T^{2}}{4 \\pi^{2}}\n$$\n\n\n\n(Derivation:\n\n$$\n\\Rightarrow \\ddot{\\varphi}=-\\frac{b M g}{I_{S}} \\sin \\varphi \\simeq-\\frac{b g M}{I_{S}} \\varphi\n$$\n\nso that\n\n$$\n\\omega^{2}=\\frac{b g M}{I_{S}}\n$$']",True,,Need_human_evaluate, 1193,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder.","A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5.","['Some version of the center of mass equation, e.g.\n\n$$\nb=\\frac{d M_{2}}{M_{1}+M_{2}}\n$$\n\ncorrect solution:\n\n$$\nd=\\frac{b M}{\\pi h_{2} r_{2}^{2}\\left(\\rho_{2}-\\rho_{1}\\right)}\n$$']",['$d=\\frac{b M}{\\pi h_{2} r_{2}^{2}(\\rho_{2}-\\rho_{1})}$'],False,,Expression, 1194,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} ","A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5.","[""correct answer for moment of inertia of homogeneous disk\n\n$$\nI_{1}=\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}\n$$\n\nMass wrong\n\nFactor $1 / 2$ wrong in formula for moment of inertia of a disk\n\nCorrect answer for moment of inertia of 'excess' disk:\n\n$$\nI_{2}=\\frac{1}{2} \\pi h_{2}\\left(\\rho_{2}-\\rho_{1}\\right) r_{2}^{4}\n$$\n\nUsing Steiner's theorem:\n\n$$\nI_{S}=I_{1}+I_{2}+d^{2} \\pi r_{2}^{2} h_{2}\\left(\\rho_{2}-\\rho_{1}\\right)\n$$\n\ncorrect solution:\n\n$$\nI_{S}=\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}+\\frac{1}{2} \\pi h_{2}\\left(\\rho_{2}-\\rho_{1}\\right) r_{2}^{4}+\\frac{b^{2} M^{2}}{\\pi r_{2}^{2} h_{2}\\left(\\rho_{2}-\\rho_{1}\\right)}\n$$\n\n\n\n$$\nI_{S}=\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}+\\frac{1}{2} \\pi h_{2}\\left(\\rho_{2}-\\rho_{1}\\right) r_{2}^{4}+d^{2} \\pi r_{2}^{2} h_{2}\\left(\\rho_{2}-\\rho_{1}\\right)\n$$""]",['$I_{S}=\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}+\\frac{1}{2} \\pi h_{2}(\\rho_{2}-\\rho_{1}) r_{2}^{4}+d^{2} \\pi r_{2}^{2} h_{2}(\\rho_{2}-\\rho_{1})$'],False,,Expression, 1195,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} ","A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$.",['It is not clear how exactly students will attempt to solve this system of equations. It is likely that they will use the following equation:\n\n$$\nM=\\pi r_{1}^{2} h_{1} \\rho_{1}+\\pi r_{2}^{2} h_{2}\\left(\\rho_{2}-\\rho_{1}\\right)\n$$\n\nsolve $I_{S}$ for $r_{2}^{2}$ :\n\n$$\nr_{2}^{2}=\\frac{2}{M-\\pi r_{1}^{2} h_{1} \\rho_{1}}\\left(I_{S}-\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}-b^{2} \\frac{M^{2}}{M-\\pi r_{1}^{2} h_{1} \\rho_{1}}\\right)\n$$\n\nreplace $I_{S}$ by $T$ :\n\n$$\nI_{S}=\\frac{M g b T^{2}}{4 \\pi^{2}}\n$$\n\nsolve correctly for $r_{2}$ :\n\n$$\nr_{2}=\\sqrt{\\frac{2}{M-\\pi r_{1}^{2} h_{1} \\rho_{1}}\\left(M \\frac{b g T^{2}}{4 \\pi^{2}}-\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}-b^{2} \\frac{M^{2}}{M-\\pi r_{1}^{2} h_{1} \\rho_{1}}\\right)}\n$$\n\nwrite down an equation for $h_{2}$ along the lines of $M=\\pi r_{1}^{2} \\rho_{1} h_{1}+\\pi r_{2}^{2}\\left(\\rho_{2}-\\rho_{1}\\right) h_{2}$ and solve it correctly:\n\n$$\nh_{2}=\\frac{M-\\pi r_{1}^{2} \\rho_{1} h_{1}}{\\pi r_{2}^{2}\\left(\\rho_{2}-\\rho_{1}\\right)}\n$$'],"['$r_{2}=\\sqrt{\\frac{2}{M-\\pi r_{1}^{2} h_{1} \\rho_{1}}(M \\frac{b g T^{2}}{4 \\pi^{2}}-\\frac{1}{2} \\pi h_{1} \\rho_{1} r_{1}^{4}-b^{2} \\frac{M^{2}}{M-\\pi r_{1}^{2} h_{1} \\rho_{1}})}$ , $h_{2}=\\frac{M-\\pi r_{1}^{2} \\rho_{1} h_{1}}{\\pi r_{2}^{2}(\\rho_{2}-\\rho_{1})}$']",True,,Expression, 1196,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored.",B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface?,"['An equation for the centrifugal force along the lines of\n\n$$\nF_{c e}=m \\omega^{2} r\n$$\n\n\n\nBalancing the forces, correct equation\n\n$$\ng_{E}=\\omega_{s s}^{2} R\n$$\n\nCorrect solution\n\n$$\n\\omega_{s s}=\\sqrt{g_{E} / R}\n$$']",['$\\omega_{s s}=\\sqrt{\\frac{g_{E}}{R}}$'],False,,Expression, 1197,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction.","B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure?",['Correct result:\n\n$$\n\\omega_{E}=\\sqrt{k / m}\n$$'],['$\\omega_{E}=\\sqrt{k / m}$'],False,,Expression, 1198,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction. Context question: B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure? Context answer: \boxed{$\omega_{E}=\sqrt{k / m}$} ",B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station?,['some version of the correct equation for force\n\n$$\nF=-k x \\pm m \\omega_{s s}^{2} x\n$$\n\ngetting the sign right\n\n$$\nF=-k x+m \\omega_{s s}^{2} x\n$$\n\nFind correct differential equation\n\n$$\nm \\ddot{x}+\\left(k-m \\omega_{s s}^{2}\\right) x=0\n$$\n\nDerive correct result\n\n$$\n\\omega=\\sqrt{k / m-\\omega_{s s}^{2}}\n$$\n\nUsing $g_{E} / R$ instead of $\\omega_{s s}^{2}$ is also correct.'],"['$\\sqrt{k / m-\\omega_{s s}^{2}}$', '$\\sqrt{k / m-\\frac{g_E}{R}}$']",False,,Expression, 1199,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction. Context question: B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure? Context answer: \boxed{$\omega_{E}=\sqrt{k / m}$} Context question: B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station? Context answer: \boxed{$\sqrt{k / m-\omega_{s s}^{2}}$} Extra Supplementary Reading Materials: Alice is convinced that her experiment proves that they are on a rotating space station. Bob remains sceptical. He claims that when taking into account the change in gravity above the surface of the Earth, one finds a similar effect. In the following tasks we investigate whether Bob is right. Figure 4: Space station",B.4 Derive an expression of the gravity $g_{E}(h)$ for small heights $h$ above the surface of the Earth and compute the oscillation frequency $\tilde{\omega}_{F}$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_{E}$. Neglect the rotation of Earth.,"['$$\ng_{E}(h)=-G M /\\left(R_{E}+h\\right)^{2}\n$$\n\nlinear approximation of gravity:\n\n$$\ng_{E}(h)=-\\frac{G M}{R_{E}^{2}}+2 h \\frac{G M}{R_{E}^{3}}+\\ldots\n$$\n\nRealize that $g_{E}=G M / R_{E}^{2}$ :\n\n$$\ng_{E}(h)=-g_{E}+2 h g_{E} / R_{E}+\\ldots\n$$\n\nOpposite sign is also correct, as long as it is opposite in both terms.\n\nRealize what this means for force, i.e. that the constant term can be eliminated by shifting the equilibrium point:\n\n$$\nF=-k x+2 x m g_{E} / R_{E}\n$$\n\nFind correct differential equation\n\n$$\nm \\ddot{x}+\\left(k-2 m g_{E} / R_{E}\\right) x=0\n$$\n\ncorrect result\n\n$$\n\\tilde{\\omega}_{E}=\\sqrt{k / m-2 g_{E} / R_{E}}\n$$']","['$g_{E}(h)=\\frac{-G M}{(R_{E}+h)^{2}}$ , $\\tilde{\\omega}_{E}=\\sqrt{k / m-2 g_{E} / R_{E}}$']",True,,Expression, 1200,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction. Context question: B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure? Context answer: \boxed{$\omega_{E}=\sqrt{k / m}$} Context question: B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station? Context answer: \boxed{$\sqrt{k / m-\omega_{s s}^{2}}$} Extra Supplementary Reading Materials: Alice is convinced that her experiment proves that they are on a rotating space station. Bob remains sceptical. He claims that when taking into account the change in gravity above the surface of the Earth, one finds a similar effect. In the following tasks we investigate whether Bob is right. Figure 4: Space station Context question: B.4 Derive an expression of the gravity $g_{E}(h)$ for small heights $h$ above the surface of the Earth and compute the oscillation frequency $\tilde{\omega}_{F}$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_{E}$. Neglect the rotation of Earth. Context answer: \boxed{$g_{E}(h)=\frac{-G M}{(R_{E}+h)^{2}}$ , $\tilde{\omega}_{E}=\sqrt{k / m-2 g_{E} / R_{E}}$} Extra Supplementary Reading Materials: Indeed, for this space station, Alice does find that the spring pendulum oscillates with the frequency that Bob predicted.",B.5 For what radius $R$ of the space station does the oscillation frequency $\omega$ match the oscillation frequency $\tilde{\omega}_{E}$ on the Earth? Express your answer in terms of $R_{E}$.,['Write down equation\n\n$$\n\\omega_{s s}^{2}=2 g_{E} / R_{E}\n$$\n\nSolve\n\n$$\nR=R_{E} / 2\n$$'],['$R=R_{E} / 2$'],False,,Expression, 1201,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction. Context question: B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure? Context answer: \boxed{$\omega_{E}=\sqrt{k / m}$} Context question: B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station? Context answer: \boxed{$\sqrt{k / m-\omega_{s s}^{2}}$} Extra Supplementary Reading Materials: Alice is convinced that her experiment proves that they are on a rotating space station. Bob remains sceptical. He claims that when taking into account the change in gravity above the surface of the Earth, one finds a similar effect. In the following tasks we investigate whether Bob is right. Figure 4: Space station Context question: B.4 Derive an expression of the gravity $g_{E}(h)$ for small heights $h$ above the surface of the Earth and compute the oscillation frequency $\tilde{\omega}_{F}$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_{E}$. Neglect the rotation of Earth. Context answer: \boxed{$g_{E}(h)=\frac{-G M}{(R_{E}+h)^{2}}$ , $\tilde{\omega}_{E}=\sqrt{k / m-2 g_{E} / R_{E}}$} Extra Supplementary Reading Materials: Indeed, for this space station, Alice does find that the spring pendulum oscillates with the frequency that Bob predicted. Context question: B.5 For what radius $R$ of the space station does the oscillation frequency $\omega$ match the oscillation frequency $\tilde{\omega}_{E}$ on the Earth? Express your answer in terms of $R_{E}$. Context answer: \boxed{$R=R_{E} / 2$} Extra Supplementary Reading Materials: Exasperated with Bob's stubbornness, Alice comes up with an experiment to prove her point. To this end she climbs on a tower of height $H$ over the floor of the space station and drops a mass. This experiment can be understood in the rotating reference frame as well as in an inertial reference frame. In a uniformly rotating reference frame, the astronauts perceive a fictitious force $\vec{F}_{C}$ called the Coriolis force. The force $\vec{F}_{C}$ acting on an object of mass $m$ moving at velocity $\vec{v}$ in a rotating frame with constant angular frequency $\vec{\omega}_{s s}$ is given by $$ \vec{F}_{C}=2 m \vec{v} \times \vec{\omega}_{s s} . \tag{2} $$ In terms of the scalar quantities you may use $$ F_{C}=2 m v \omega_{s s} \sin \phi, \tag{3} $$ where $\phi$ is the angle between the velocity and the axis of rotation. The force is perpendicular to both the velocity $v$ and the axis of rotation. The sign of the force can be determined from the right-hand rule, but in what follows you may choose it freely.","B.6 Calculate the horizontal velocity $v_{x}$ and the horizontal displacement $d_{x}$ (relative to the base of the tower, in the direction perpendicular to the tower) of the mass at the moment it hits the floor. You may assume that the height $H$ of the tower is small, so that the acceleration as measured by the astronauts is constant during the fall. Also, you may assume that $d_{x} \ll H$.","['- Velocity $v_{x}$\n\nEquation for Coriolis force with correct velocity:\n\n$$\nF_{C}(t)=2 m \\omega_{s s}^{2} R t \\omega_{s s}=2 m \\omega_{s s}^{3} R t\n$$\n\nIntegrate this, or realize that it is like uniform acceleration for the velocity:\n\n$$\nv_{x}(t)=\\omega_{s s}^{3} R t^{2}\n$$\n\nplug in correct value for\n\n$$\nt=\\sqrt{2 H / \\omega_{s s}^{2} R}\n$$\n\noverall correct result\n\n$$\nv_{x}=2 H \\omega_{s s}\n$$\n\n- The displacement $d_{x}$ :\n\nIntegrate $v_{x}(t)$ :\n\n$$\nd_{x}=\\frac{1}{3} R \\omega_{s s}^{3} t^{3}\n$$\n\nPlug in value for $t$\n\n$$\nd_{x}=\\frac{1}{3} R \\omega_{s s}^{3}\\left(2 H / \\omega_{s s}^{2} R\\right)^{3 / 2}=\\frac{1}{3} 2^{3 / 2} H^{3 / 2} R^{-1 / 2}=\\frac{1}{3} \\sqrt{\\frac{8 H^{3}}{R}}\n$$', 'Using inertial frame This solution is similar to the way to solve B7, but needs more complicated approximations than Solution one.\n- $v_{x}$\n\nHere $\\phi$ denotes the angle swept by the mass and $\\alpha$ the angle the astronauts (and tower) has rotated when the mass lands on the floor, see\n\nInitially the velocity of the mass in an inertial frame is $v_{x}=\\omega_{s s}(R-H)$.\n\nWhen the mass lands, the $x$-direction has been rotated by $\\phi$ so the new horizontal velocity component is then\n\n$$\n\\omega_{s s}(R-H) \\cos \\phi\n$$\n\n(Student may also write $\\cos \\alpha$ instead of $\\cos \\phi$, since $d_{x} \\ll H$.)\n\n$$\n\\cos \\phi=\\frac{R-H}{R}=1-\\frac{H}{R}\n$$\n\nTransforming to the rotating reference frame, one needs to subtract $\\omega_{s s} R$.\n\nFinally in the reference frame of the astronauts\n\n$$\nv_{x}=\\omega_{s s} R\\left(1-\\frac{H}{R}\\right)^{2}-\\omega_{s s} R \\approx \\omega_{s s} R\\left(1-2 \\frac{H}{R}\\right)-\\omega_{s s} R=-2 \\omega_{s s} H\n$$\n\nThe sign of the velocity depend on the choice of reference direction, so a positive sign is also correct.\n\n\n\n- $d_{x}$\n\nWith the notation from the calculation of $v_{x}$\n\n$$\n\\begin{gathered}\nd_{x}=(\\alpha-\\phi) R \\\\\n\\phi=\\arccos \\left(1-\\frac{H}{R}\\right) \\\\\n\\alpha=\\omega_{s s} t\n\\end{gathered}\n$$\n\nwhere $t$ is the fall time of the mass, which is given by\n\n$$\nt=\\frac{\\sqrt{R^{2}-(R-H)^{2}}}{\\omega_{s s}(R-H)}\n$$\n\n(see solution to $\\mathrm{B} 7$ )\n\nWriting $\\xi \\equiv H / R$ this means\n\n$$\nd_{x}=\\left[\\frac{\\sqrt{1-(1-\\xi)^{2}}}{1-\\xi}-\\arccos (1-\\xi)\\right] R\n$$\n\nwhich is a valid end answer to the problem. It is possible, but not necessary, to approximate this for small $\\xi$ :\n\n$$\n\\arccos (1-\\xi) \\approx \\sqrt{2 \\xi}\\left(1+\\frac{\\xi}{12}\\right)\n$$\n\nwhich after insertion into the equation for $d_{x}$ and approximation of small $\\xi$ yields the same result as in Solution one:\n\n$$\nd_{x}=\\frac{2}{3} \\sqrt{\\frac{2 H^{3}}{R}}\n$$']","['$2 H \\omega_{s s}$,$\\frac{1}{3} \\sqrt{\\frac{8 H^{3}}{R}}$', '$-2 H \\omega_{s s}$,$\\frac{1}{3} \\sqrt{\\frac{8 H^{3}}{R}}$']",True,,Expression, 1202,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction. Context question: B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure? Context answer: \boxed{$\omega_{E}=\sqrt{k / m}$} Context question: B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station? Context answer: \boxed{$\sqrt{k / m-\omega_{s s}^{2}}$} Extra Supplementary Reading Materials: Alice is convinced that her experiment proves that they are on a rotating space station. Bob remains sceptical. He claims that when taking into account the change in gravity above the surface of the Earth, one finds a similar effect. In the following tasks we investigate whether Bob is right. Figure 4: Space station Context question: B.4 Derive an expression of the gravity $g_{E}(h)$ for small heights $h$ above the surface of the Earth and compute the oscillation frequency $\tilde{\omega}_{F}$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_{E}$. Neglect the rotation of Earth. Context answer: \boxed{$g_{E}(h)=\frac{-G M}{(R_{E}+h)^{2}}$ , $\tilde{\omega}_{E}=\sqrt{k / m-2 g_{E} / R_{E}}$} Extra Supplementary Reading Materials: Indeed, for this space station, Alice does find that the spring pendulum oscillates with the frequency that Bob predicted. Context question: B.5 For what radius $R$ of the space station does the oscillation frequency $\omega$ match the oscillation frequency $\tilde{\omega}_{E}$ on the Earth? Express your answer in terms of $R_{E}$. Context answer: \boxed{$R=R_{E} / 2$} Extra Supplementary Reading Materials: Exasperated with Bob's stubbornness, Alice comes up with an experiment to prove her point. To this end she climbs on a tower of height $H$ over the floor of the space station and drops a mass. This experiment can be understood in the rotating reference frame as well as in an inertial reference frame. In a uniformly rotating reference frame, the astronauts perceive a fictitious force $\vec{F}_{C}$ called the Coriolis force. The force $\vec{F}_{C}$ acting on an object of mass $m$ moving at velocity $\vec{v}$ in a rotating frame with constant angular frequency $\vec{\omega}_{s s}$ is given by $$ \vec{F}_{C}=2 m \vec{v} \times \vec{\omega}_{s s} . \tag{2} $$ In terms of the scalar quantities you may use $$ F_{C}=2 m v \omega_{s s} \sin \phi, \tag{3} $$ where $\phi$ is the angle between the velocity and the axis of rotation. The force is perpendicular to both the velocity $v$ and the axis of rotation. The sign of the force can be determined from the right-hand rule, but in what follows you may choose it freely. Context question: B.6 Calculate the horizontal velocity $v_{x}$ and the horizontal displacement $d_{x}$ (relative to the base of the tower, in the direction perpendicular to the tower) of the mass at the moment it hits the floor. You may assume that the height $H$ of the tower is small, so that the acceleration as measured by the astronauts is constant during the fall. Also, you may assume that $d_{x} \ll H$. Context answer: \boxed{$2 H \omega_{s s}$,$\frac{1}{3} \sqrt{\frac{8 H^{3}}{R}}$} Extra Supplementary Reading Materials: To get a good result, Alice decides to conduct this experiment from a much taller tower than before. To her surprise, the mass hits the floor at the base of the tower, so that $d_{x}=0$.",B.7 Find a lower bound for the height of the tower for which it can happen that $d_{x}=0$.,"['The key is to use a non-rotating frame of reference. If the mass is released close enough\n\n\n\nto the center, its linear velocity will be small enough for the space station to rotate more than $2 \\pi$ before it hits the ground.\n\nThe velocity is given by\n\n$$\nv=\\omega_{s s}(R-H)\n$$\n\ndistance $d$ that the mass flies before hitting the space station\n\n$$\nd^{2}=R^{2}-(R-H)^{2}\n$$\n\nuse non-rotating frame of reference to obtain time $t$ until impact\n\n$$\nt=d / v=\\frac{\\sqrt{R^{2}-(R-H)^{2}}}{\\omega_{s s}(R-H)}\n$$\n\n\n$$\nt=\\frac{R \\sin \\phi}{\\omega_{s s} R \\cos \\phi}\n$$\n\nThis time must match $t=\\phi / \\omega_{s s}$. Obtain the equation\n\n$$\n\\phi=\\tan \\phi\n$$\n\nRealizing that there is an infinite number of solutions.\n\nThis equation has one trivial solution $\\phi=0$, next solution is slightly less than $3 \\pi / 2$ which corresponds to the case $H>R$ (and is thus not correct). The one that gives a lower bound for $H$ is the third solution\n\n$$\n\\phi \\approx 5 \\pi / 2\n$$\n\nThe equation $\\phi=\\tan \\phi$ can be solved graphically or numerically to obtain a close value $(\\phi=7.725 \\mathrm{rad})$ which means\n\n$$\nH / R=(1-\\cos \\phi) \\approx 0.871\n$$', 'The key is to use a non-rotating frame of reference. If the mass is released close enough\n\n\n\nto the center, its linear velocity will be small enough for the space station to rotate more than $2 \\pi$ before it hits the ground.\n\nThe velocity is given by\n\n$$\nv=\\omega_{s s}(R-H)\n$$\n\ndistance $d$ that the mass flies before hitting the space station\n\n$$\nd^{2}=R^{2}-(R-H)^{2}\n$$\n\nuse non-rotating frame of reference to obtain time $t$ until impact\n\n$$\nt=d / v=\\frac{\\sqrt{R^{2}-(R-H)^{2}}}{\\omega_{s s}(R-H)}\n$$\n\n\nrelation between $H$ and rotated angle $\\phi$\n\n$$\n\\frac{R-H}{R}=\\cos \\phi\n$$\n\nobtain equation of the form\n\n$$\n\\frac{H}{R}=1-\\cos \\left(\\frac{\\sqrt{1-(1-H / R)^{2}}}{1-H / R}\\right)\n$$\n\nFigure 3 gives a plot of $f(x)=1-\\cos \\left(\\frac{\\sqrt{1-(1-x)^{2}}}{1-x}\\right)$. The goal is to find an approximate solution for the second intersection. The first intersection is discarded - it is introduced because of $\\cos \\phi=\\cos (-\\phi)$ and corresponds to a situation with $H>R$.\n\nRealizing that there is an infinite number of solutions.\n\n\n\n\n\nFigure 3: Plot of $f(H / R)$ and $H / R$\n\n\n\nFigure 4: Plot of $g(x)$ and $x$\n\n- introduce new variable $x:=1-H / R$, so that the equation becomes\n\n$$\nx=\\cos \\left(\\sqrt{1-x^{2}} / x\\right)=: g(x)\n$$\n\n- $g(x)$ is then smaller than $x$ up to the first solution. In particular it is negative in some region (see figure 4). Finding the third zero thus gives a lower bound for the solution:\n\n$$\n\\frac{\\sqrt{1-x^{2}}}{x}=5 \\pi / 2\n$$\n\n- give lower bound\n\n$$\nx=1 / \\sqrt{25 \\pi^{2} / 4+1} \\Rightarrow H=R\\left(1-1 / \\sqrt{25 \\pi^{2} / 4+1}\\right) \\approx 0.874\n$$\n\nNote: the actual result is $H / R=0.871 \\ldots$\n\nUse the same points for the numerical answer as was mentioned in solution one.\n\nIf the student plots $f$ rather than $g$, find solution to $f=1$ : is equivalent to the solution above. Give same number of points.\n\nIt is also possible to use $\\cos \\left(\\frac{\\sqrt{1-x^{2}}}{x}\\right)=\\sin (1 / x)$.']",['0.871'],False,,Numerical,1e-3 1203,Mechanics,"Two Problems in Mechanics Please read the general instructions in the separate envelope before you start this problem. Part A.The Hidden Disk We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$. In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup. The goal of this task is to determine the size and the position of the metal disk. In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known: $$ r_{1}, h_{1}, \rho_{1}, \rho_{2}, M \tag{1} $$ The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements. а) b) Figure 1: a) side view b) view from above We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure. Figure 2: Cylinder on an inclined base. Context question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base. Context answer: \boxed{$b=\frac{r_{1} \sin \Theta}{\sin \phi}$} Extra Supplementary Reading Materials: From now on, we can assume that the value of $b$ is known. Figure 3: Suspended system. Next we want to measure the moment of inertia $I_{S}$ of the system with respect to the symmetry axis $S$. To this end, we suspend the wooden cylinder at its symmetry axis from a rigid rod. We then turn it away from its equilibrium position by a small angle $\varphi$, and let it go. See figure 3 for the setup. We find that $\varphi$ describes a periodic motion with period $T$. Context question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small. Context answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion. Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$. Correct equation (either energy conservation or torque equation of motion) $$ T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}} $$ (Derivation: $$ \Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi $$ so that $$ \omega^{2}=\frac{b g M}{I_{S}} $$ Extra Supplementary Reading Materials: From the measurements in questions $\mathbf{A} .1$ and $\mathbf{A . 2}$, we now want to determine the geometry and the position of the metal disk inside the wooden cylinder. Context question: A.3 Find an expression for the distance $d$ as a function of $b$ and the quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$d=\frac{b M}{\pi h_{2} r_{2}^{2}(\rho_{2}-\rho_{1})}$} Context question: A.4 Find an expression for the moment of inertia $I_{S}$ in terms of $b$ and the known quantities (1). You may also include $r_{2}$ and $h_{2}$ as variables in your expression, as they will be calculated in subtask A.5. Context answer: \boxed{$I_{S}=\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}+\frac{1}{2} \pi h_{2}(\rho_{2}-\rho_{1}) r_{2}^{4}+d^{2} \pi r_{2}^{2} h_{2}(\rho_{2}-\rho_{1})$} Context question: A.5 Using all the above results, write down an expression for $h_{2}$ and $r_{2}$ in terms of $b, T$ and the known quantities (1). You may express $h_{2}$ as a function of $r_{2}$. Context answer: \boxed{$r_{2}=\sqrt{\frac{2}{M-\pi r_{1}^{2} h_{1} \rho_{1}}(M \frac{b g T^{2}}{4 \pi^{2}}-\frac{1}{2} \pi h_{1} \rho_{1} r_{1}^{4}-b^{2} \frac{M^{2}}{M-\pi r_{1}^{2} h_{1} \rho_{1}})}$ , $h_{2}=\frac{M-\pi r_{1}^{2} \rho_{1} h_{1}}{\pi r_{2}^{2}(\rho_{2}-\rho_{1})}$} Extra Supplementary Reading Materials: Part B.Rotating Space Station Alice is an astronaut living on a space station. The space station is a gigantic wheel of radius $R$ rotating around its axis, thereby providing artificial gravity for the astronauts. The astronauts live on the inner side of the rim of the wheel. The gravitational attraction of the space station and the curvature of the floor can be ignored. Context question: B.1 At what angular frequency $\omega_{s s}$ does the space station rotate so that the astronauts experience the same gravity $g_{E}$ as on the Earth's surface? Context answer: \boxed{$\omega_{s s}=\sqrt{\frac{g_{E}}{R}}$} Extra Supplementary Reading Materials: Alice and her astronaut friend Bob have an argument. Bob does not believe that they are in fact living in a space station and claims that they are on Earth. Alice wants to prove to Bob that they are living on a rotating space station by using physics. To this end, she attaches a mass $m$ to a spring with spring constant $k$ and lets it oscillate. The mass oscillates only in the vertical direction, and cannot move in the horizontal direction. Context question: B.2 Assuming that on Earth gravity is constant with acceleration $g_{E}$, what would be the angular oscillation frequency $\omega_{E}$ that a person on Earth would measure? Context answer: \boxed{$\omega_{E}=\sqrt{k / m}$} Context question: B.3 What angular oscillation frequency $\omega$ does Alice measure on the space station? Context answer: \boxed{$\sqrt{k / m-\omega_{s s}^{2}}$} Extra Supplementary Reading Materials: Alice is convinced that her experiment proves that they are on a rotating space station. Bob remains sceptical. He claims that when taking into account the change in gravity above the surface of the Earth, one finds a similar effect. In the following tasks we investigate whether Bob is right. Figure 4: Space station Context question: B.4 Derive an expression of the gravity $g_{E}(h)$ for small heights $h$ above the surface of the Earth and compute the oscillation frequency $\tilde{\omega}_{F}$ of the oscillating mass (linear approximation is enough). Denote the radius of the Earth by $R_{E}$. Neglect the rotation of Earth. Context answer: \boxed{$g_{E}(h)=\frac{-G M}{(R_{E}+h)^{2}}$ , $\tilde{\omega}_{E}=\sqrt{k / m-2 g_{E} / R_{E}}$} Extra Supplementary Reading Materials: Indeed, for this space station, Alice does find that the spring pendulum oscillates with the frequency that Bob predicted. Context question: B.5 For what radius $R$ of the space station does the oscillation frequency $\omega$ match the oscillation frequency $\tilde{\omega}_{E}$ on the Earth? Express your answer in terms of $R_{E}$. Context answer: \boxed{$R=R_{E} / 2$} Extra Supplementary Reading Materials: Exasperated with Bob's stubbornness, Alice comes up with an experiment to prove her point. To this end she climbs on a tower of height $H$ over the floor of the space station and drops a mass. This experiment can be understood in the rotating reference frame as well as in an inertial reference frame. In a uniformly rotating reference frame, the astronauts perceive a fictitious force $\vec{F}_{C}$ called the Coriolis force. The force $\vec{F}_{C}$ acting on an object of mass $m$ moving at velocity $\vec{v}$ in a rotating frame with constant angular frequency $\vec{\omega}_{s s}$ is given by $$ \vec{F}_{C}=2 m \vec{v} \times \vec{\omega}_{s s} . \tag{2} $$ In terms of the scalar quantities you may use $$ F_{C}=2 m v \omega_{s s} \sin \phi, \tag{3} $$ where $\phi$ is the angle between the velocity and the axis of rotation. The force is perpendicular to both the velocity $v$ and the axis of rotation. The sign of the force can be determined from the right-hand rule, but in what follows you may choose it freely. Context question: B.6 Calculate the horizontal velocity $v_{x}$ and the horizontal displacement $d_{x}$ (relative to the base of the tower, in the direction perpendicular to the tower) of the mass at the moment it hits the floor. You may assume that the height $H$ of the tower is small, so that the acceleration as measured by the astronauts is constant during the fall. Also, you may assume that $d_{x} \ll H$. Context answer: \boxed{$2 H \omega_{s s}$,$\frac{1}{3} \sqrt{\frac{8 H^{3}}{R}}$} Extra Supplementary Reading Materials: To get a good result, Alice decides to conduct this experiment from a much taller tower than before. To her surprise, the mass hits the floor at the base of the tower, so that $d_{x}=0$. Context question: B.7 Find a lower bound for the height of the tower for which it can happen that $d_{x}=0$. Context answer: \boxed{0.871} Extra Supplementary Reading Materials: Alice is willing to make one last attempt at convincing BoB.She wants to use her spring oscillator to show the effect of the Coriolis force. To this end she changes the original setup: She attaches her spring to a ring which can slide freely on a horizontal rod in the $x$ direction without any friction. The spring itself oscillates in the $y$ direction. The rod is parallel to the floor and perpendicular to the axis of rotation of the space station. The $x y$ plane is thus perpendicular to the axis of rotation, with the $y$ direction pointing straight towards the center of rotation of the station. Figure 5: Setup.","B.8 Alice pulls the mass a distance $d$ downwards from the equilibrium point $x=0$, $y=0$, and then lets it go (see figure 5). - Give an algebraic expression of $x(t)$ and $y(t)$. You may assume that $\omega_{s s} d$ is small, and neglect the Coriolis force for motion along the $y$-axis.","[""Note: we did not specify the overall sign of the Coriolis force. Give same amount of points if using opposite convention, but it has to be consistent! Otherwise: subtract $0.1 p t$ for each instance of inconsistency.\n\nStudents are allowed to express everything in terms of $\\omega$, they don't need to write $\\sqrt{k / m-\\omega_{s s}^{2}}$ explicitly. Deduct $0.1 p t$ however if they use $k / m$ instead of $\\omega$..\n\nRealize that $y(t)$ is standard harmonic oscillation:\n\n$$\ny(t)=A \\cos \\omega t+B\n$$\n\n0.1\n\nGive correct constants from initial conditions\n\n$$\ny(t)=-d \\cos \\omega t\n$$\n\nCorrect expression for $v_{y}(t)$ :\n\n$$\nv_{y}(t)=-d \\omega \\sin \\omega t\n$$\n\nCoriolis force in $x$-direction\n\n$$\nF_{x}(t)=2 m \\omega_{s s} v_{y}(t)=-2 m \\omega_{s s} d \\omega \\sin \\omega t\n$$\n\nRealize that this implies that $x(t)$ is also a harmonic oscillation... getting the correct amplitude:\n\n$$\nA=\\frac{2 \\omega_{s s} d}{\\omega}\n$$\n\nCorrect answer with correct initial conditions:\n\n$$\nx(t)=\\frac{2 \\omega_{s s} d}{\\omega} \\sin \\omega t-2 \\omega_{s s} d t\n$$\n\nSketch:\n\n""]","['$y(t)=-d \\cos \\omega t$ , $x(t)=\\frac{2 \\omega_{s s} d}{\\omega} \\sin \\omega t-2 \\omega_{s s} d t$']",True,,Expression, 1204,Electromagnetism,"Nonlinear Dynamics in Electric Circuits Please read the general instructions in the separate envelope before you start this problem. Introduction Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches and generators of electromagnetic oscillations. The primary field of applications of thyristors is controlling alternating currents in power electronics, for instance rectification of AC current to DC at the megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern nonlinear science. Goals To study instabilities and nontrivial dynamics of circuits including elements with non-linear $I-V$ characteristics. To discover possible applications of such circuits in engineering and in modeling of biological systems. Part A.Stationary states and instabilities Fig. 1 shows the so-called S-shaped $I-V$ characteristics of a non-linear element $X$. In the voltage range between $U_{\mathrm{h}}=4.00 \mathrm{~V}$ (the holding voltage) and $U_{\text {th }}=10.0 \mathrm{~V}$ (the threshold voltage) this $I-V$ characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it is extended. This approximation gives a good description of real thyristors. Figure 1: $I-V$ characteristics of the non-linear element $X$.","A.1 Using the graph, determine the resistance $R_{\mathrm{on}}$ of the element $X$ on the upper branch of the $I-V$ characteristics, and $R_{\text {off }}$ on the lower branch, respectively. The middle branch is described by the equation $$ I=I_{0}-\frac{U}{R_{\text {int }}} \tag{1} $$ Find the values of the parameters $I_{0}$ and $R_{\text {int }}$.","['By looking at the $I-V$ graph, we obtain\n\n$$\n\\begin{gathered}\nR_{\\text {off }}=10.0 \\Omega, \\\\\nR_{\\text {on }}=1.00 \\Omega, \\\\\nR_{\\text {int }}=2.00 \\Omega, \\\\\nI_{0}=6.00 \\mathrm{~A} .\n\\end{gathered}\n$$\n\n']","['$R_{int}=2.00 $ , $I_{0}=6.00$']",True,"$\Omega$, A",Numerical,0 1205,Electromagnetism,,A.2 What are the possible numbers of stationary states that the circuit of Fig. 2 may have for a fixed value of $\mathcal{E}$ and for $R=3.00 \Omega$ ? How does the answer change for $R=1.00 \Omega$ ?,"['Kirchoff law for the circuit ( $U$ is the voltage of the bistable element):\n\n$$\n\\mathcal{E}=I R+U\n$$\n\nThis yields\n\n$$\nI=\\frac{\\mathcal{E}-U}{R}\n$$\n\nHence, stationary states of the circuit are intersections of the line defined by this equation and the $I-V$ graph of $X$.\n\nFor $R=3.00 \\Omega$, one always gets exactly one intersection.\n\nFor $R=1.00 \\Omega$, one gets 1,2 or 3 intersections depending on the value of $\\mathcal{E}$.']","['For $R=3.00 \\Omega$, one always gets exactly one intersection.\n\nFor $R=1.00 \\Omega$, one gets 1,2 or 3 intersections depending on the value of $\\mathcal{E}$.']",True,,Need_human_evaluate, 1206,Electromagnetism,"Nonlinear Dynamics in Electric Circuits Please read the general instructions in the separate envelope before you start this problem. Introduction Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches and generators of electromagnetic oscillations. The primary field of applications of thyristors is controlling alternating currents in power electronics, for instance rectification of AC current to DC at the megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern nonlinear science. Goals To study instabilities and nontrivial dynamics of circuits including elements with non-linear $I-V$ characteristics. To discover possible applications of such circuits in engineering and in modeling of biological systems. Part A.Stationary states and instabilities Fig. 1 shows the so-called S-shaped $I-V$ characteristics of a non-linear element $X$. In the voltage range between $U_{\mathrm{h}}=4.00 \mathrm{~V}$ (the holding voltage) and $U_{\text {th }}=10.0 \mathrm{~V}$ (the threshold voltage) this $I-V$ characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it is extended. This approximation gives a good description of real thyristors. Figure 1: $I-V$ characteristics of the non-linear element $X$. Context question: A.1 Using the graph, determine the resistance $R_{\mathrm{on}}$ of the element $X$ on the upper branch of the $I-V$ characteristics, and $R_{\text {off }}$ on the lower branch, respectively. The middle branch is described by the equation $$ I=I_{0}-\frac{U}{R_{\text {int }}} \tag{1} $$ Find the values of the parameters $I_{0}$ and $R_{\text {int }}$. Context answer: \boxed{$R_{int}=2.00 $ , $I_{0}=6.00$} Extra Supplementary Reading Materials: The element $X$ is connected in series (see Fig.2) with a resistor $R$, an inductor $L$ and an ideal voltage source $\mathcal{E}$. One says that the circuit is in a stationary state if the current is constant in time, $I(t)=$ const. Figure 2: Circuit with element $X$, resistor $R$, inductor $L$ and voltage source $\mathcal{E}$. Context question: A.2 What are the possible numbers of stationary states that the circuit of Fig. 2 may have for a fixed value of $\mathcal{E}$ and for $R=3.00 \Omega$ ? How does the answer change for $R=1.00 \Omega$ ? Context answer: For $R=3.00 \Omega$, one always gets exactly one intersection. For $R=1.00 \Omega$, one gets 1,2 or 3 intersections depending on the value of $\mathcal{E}$.","A.3 Let $R=3.00 \Omega, L=1.00 \mu \mathrm{H}$ and $\mathcal{E}=15.0 \mathrm{~V}$ in the circuit shown in Fig.2 Determine the values of the current $I_{\text {stationary }}$ and the voltage $V_{\text {stationary }}$ on the non-linear element $X$ in the stationary state.","['The stationary state is on the intermediate branch, one can thus use the corresponding equation:\n\n\n\n$$\n\\begin{aligned}\nI_{\\text {stationary }} & =\\frac{\\mathcal{E}-R_{\\text {int }} I_{0}}{R-R_{\\text {int }}} \\\\\n& =3.00 \\mathrm{~A} \\\\\nU_{\\text {stationary }} & =R_{\\text {int }}\\left(I_{0}-I\\right) \\\\\n& =6.00 \\mathrm{~V}\n\\end{aligned}\n$$\n\n']","['$I_{\\text {stationary }}=3.00$ , $U_{\\text {stationary }}=6.00$']",True,"A,V",Numerical,0 1208,Electromagnetism,,B.2 Find expressions for the times $t_{1}$ and $t_{2}$ that the system spends on each branch of the $I-V$ graph during the oscillation cycle. Determine their numerical values. Find the numerical value of the oscillation period $T$ assuming that the time needed for jumps between the branches of the $I-V$ graph is negligible.,"['Since the non-linear element is oscillating between the switched on and switched off branches we can put $U_{X}=R_{\\mathrm{on} / \\text { off }} I_{X}$. On either of the branches, the circuit behaves as a standard RC-circuit with conductance $C$ and resistance $R_{\\mathrm{on} / \\mathrm{off}} R /\\left(R_{\\mathrm{on} / \\mathrm{off}}+R\\right)$ (the resistor and the element $X$ being connected in parallel). Another way to express it is to\n\n\n\nwrite the Kirchhoff law for the switched on and switched off branches\n\n$$\nR_{\\text {on } / \\mathrm{off}} R C \\frac{d I_{X}}{d t}=\\mathcal{E}-\\left(R_{\\mathrm{on} / \\mathrm{off}}+R\\right) I_{X}\n$$\n\nThe time constant of the circuit is\n\n$$\n\\frac{R_{\\text {on } / \\text { off }} R}{R_{\\text {on } / \\text { off }}+R} C\n$$\n\nIf the branch in question (switched on or switched off) extended indefinitely, after a long time the system would have landed in a stationary state with the voltage\n\n$$\nU_{\\text {on } / \\text { off }}=\\frac{R_{\\text {on } / \\text { off }}}{R_{\\text {on } / \\text { off }}+R} \\mathcal{E}\n$$\n\nThen, the time dependence of the voltage drop on the non-linear element is a sum of the constant term $U_{\\text {on/off }}$ and of the exponentially decaying term:\n\n$$\nU_{X}(t)=\\frac{R_{\\text {on } / \\text { off }}}{R_{\\text {on } / \\text { off }}+R} \\mathcal{E}+\\left(U_{\\text {on } / \\text { off }}-\\frac{R_{\\text {on } / \\text { off }}}{R_{\\text {on } / \\text { off }}+R} \\mathcal{E}\\right) e^{-\\frac{R_{\\text {on } / \\text { off }}+R}{R_{\\text {on } / \\text { off }}^{R C}} t}\n$$\n\nTime spent by the system on the switched on branch during one cycle:\n\n$$\nt_{\\mathrm{on}}=\\frac{R_{\\mathrm{on}} R}{R_{\\mathrm{on}}+R} C \\log \\left(\\frac{U_{\\mathrm{th}}-U_{\\mathrm{on}}}{U_{\\mathrm{h}}-U_{\\mathrm{on}}}\\right)=2.41 \\cdot 10^{-6} \\mathrm{~s},\n$$\n\n\nTime spent by the system on the switched off branch during one cycle:\n\n$$\nt_{\\mathrm{off}}=\\frac{R_{\\mathrm{off}} R}{R_{\\mathrm{off}}+R} C \\log \\left(\\frac{U_{\\mathrm{off}}-U_{\\mathrm{h}}}{U_{\\mathrm{off}}-U_{\\mathrm{th}}}\\right)=3.71 \\cdot 10^{-6} s\n$$\n\nThe total period of oscillations:\n\n$$\nT=t_{\\mathrm{on}}+t_{\\mathrm{off}}=6.12 \\cdot 10^{-6} \\mathrm{~s}\n$$']","['$t_{\\mathrm{on}}=2.41 \\cdot 10^{-6} \\mathrm{~s}$, $t_{\\mathrm{off}}=3.71 \\cdot 10^{-6} \\mathrm{~s}$,$T=6.12 \\cdot 10^{-6} \\mathrm{~s}$']",True,,Need_human_evaluate, 1209,Electromagnetism,"Nonlinear Dynamics in Electric Circuits Please read the general instructions in the separate envelope before you start this problem. Introduction Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches and generators of electromagnetic oscillations. The primary field of applications of thyristors is controlling alternating currents in power electronics, for instance rectification of AC current to DC at the megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern nonlinear science. Goals To study instabilities and nontrivial dynamics of circuits including elements with non-linear $I-V$ characteristics. To discover possible applications of such circuits in engineering and in modeling of biological systems. Part A.Stationary states and instabilities Fig. 1 shows the so-called S-shaped $I-V$ characteristics of a non-linear element $X$. In the voltage range between $U_{\mathrm{h}}=4.00 \mathrm{~V}$ (the holding voltage) and $U_{\text {th }}=10.0 \mathrm{~V}$ (the threshold voltage) this $I-V$ characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it is extended. This approximation gives a good description of real thyristors. Figure 1: $I-V$ characteristics of the non-linear element $X$. Context question: A.1 Using the graph, determine the resistance $R_{\mathrm{on}}$ of the element $X$ on the upper branch of the $I-V$ characteristics, and $R_{\text {off }}$ on the lower branch, respectively. The middle branch is described by the equation $$ I=I_{0}-\frac{U}{R_{\text {int }}} \tag{1} $$ Find the values of the parameters $I_{0}$ and $R_{\text {int }}$. Context answer: \boxed{$R_{int}=2.00 $ , $I_{0}=6.00$} Extra Supplementary Reading Materials: The element $X$ is connected in series (see Fig.2) with a resistor $R$, an inductor $L$ and an ideal voltage source $\mathcal{E}$. One says that the circuit is in a stationary state if the current is constant in time, $I(t)=$ const. Figure 2: Circuit with element $X$, resistor $R$, inductor $L$ and voltage source $\mathcal{E}$. Context question: A.2 What are the possible numbers of stationary states that the circuit of Fig. 2 may have for a fixed value of $\mathcal{E}$ and for $R=3.00 \Omega$ ? How does the answer change for $R=1.00 \Omega$ ? Context answer: For $R=3.00 \Omega$, one always gets exactly one intersection. For $R=1.00 \Omega$, one gets 1,2 or 3 intersections depending on the value of $\mathcal{E}$. Context question: A.3 Let $R=3.00 \Omega, L=1.00 \mu \mathrm{H}$ and $\mathcal{E}=15.0 \mathrm{~V}$ in the circuit shown in Fig.2 Determine the values of the current $I_{\text {stationary }}$ and the voltage $V_{\text {stationary }}$ on the non-linear element $X$ in the stationary state. Context answer: \boxed{$I_{\text {stationary }}=3.00$ , $U_{\text {stationary }}=6.00$} Extra Supplementary Reading Materials: The circuit in Fig. 2 is in the stationary state with $I(t)=I_{\text {stationary }}$. This stationary state is said to be stable if after a small displacement (increase or decrease in the current), the current returns towards the stationary state. And if the system keeps moving away from the stationary state, it is said to be unstable. Context question: A.4 Use numerical values of the question A.3 and study the stability of the stationary state with $I(t)=I_{\text {stationary }}$. Is it stable or unstable? Context answer: \boxed{stable} Extra Supplementary Reading Materials: Part B.Bistable non-linear elements in physics: radio transmitter We now investigate a new circuit configuration (see Fig. 3). This time, the non-linear element $X$ is connected in parallel to a capacitor of capacitance $C=1.00 \mu \mathrm{F}$. This block is then connected in series to a resistor of resistance $R=3.00 \Omega$ and an ideal constant voltage source of voltage $\mathcal{E}=15.0 \mathrm{~V}$. It turns out that this circuit undergoes oscillations with the non-linear element $X$ jumping from one branch of the $I-V$ characteristics to another over the course of one cycle. Figure 3: Circuit with element $X$, capacitor $C$, resistor $R$ and voltage source $\mathcal{E}$. Context question: B.1 Draw the oscillation cycle on the $I-V$ graph, including its direction (clockwise or anticlockwise). Justify your answer with equations and sketches. Context answer: Context question: B.2 Find expressions for the times $t_{1}$ and $t_{2}$ that the system spends on each branch of the $I-V$ graph during the oscillation cycle. Determine their numerical values. Find the numerical value of the oscillation period $T$ assuming that the time needed for jumps between the branches of the $I-V$ graph is negligible. Context answer: $t_{\mathrm{on}}=2.41 \cdot 10^{-6} \mathrm{~s}$ $t_{\mathrm{off}}=3.71 \cdot 10^{-6} \mathrm{~s}$ $T==6.12 \cdot 10^{-6} \mathrm{~s}$ ",B.3 Estimate the average power $P$ dissipated by the non-linear element over the course of one oscillation. An order of magnitude is sufficient.,"['Neglect the energy consumed on the switched off branch. The energy consumed\n\n\n\non the switched on branch during the cycle is estimated by\n\n$$\nE=\\frac{1}{R_{\\mathrm{on}}}\\left(\\frac{U_{h}+U_{t h}}{2}\\right)^{2} t_{\\mathrm{on}}=1.18 \\cdot 10^{-4} \\mathrm{~J}\n$$\n\nFor the power, this gives an estimate of\n\n$$\nP \\sim \\frac{E}{T}=19.3 \\mathrm{~W} .\n$$']",['19.3'],False,W,Numerical,1e-1 1210,Electromagnetism,"Nonlinear Dynamics in Electric Circuits Please read the general instructions in the separate envelope before you start this problem. Introduction Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches and generators of electromagnetic oscillations. The primary field of applications of thyristors is controlling alternating currents in power electronics, for instance rectification of AC current to DC at the megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern nonlinear science. Goals To study instabilities and nontrivial dynamics of circuits including elements with non-linear $I-V$ characteristics. To discover possible applications of such circuits in engineering and in modeling of biological systems. Part A.Stationary states and instabilities Fig. 1 shows the so-called S-shaped $I-V$ characteristics of a non-linear element $X$. In the voltage range between $U_{\mathrm{h}}=4.00 \mathrm{~V}$ (the holding voltage) and $U_{\text {th }}=10.0 \mathrm{~V}$ (the threshold voltage) this $I-V$ characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it is extended. This approximation gives a good description of real thyristors. Figure 1: $I-V$ characteristics of the non-linear element $X$. Context question: A.1 Using the graph, determine the resistance $R_{\mathrm{on}}$ of the element $X$ on the upper branch of the $I-V$ characteristics, and $R_{\text {off }}$ on the lower branch, respectively. The middle branch is described by the equation $$ I=I_{0}-\frac{U}{R_{\text {int }}} \tag{1} $$ Find the values of the parameters $I_{0}$ and $R_{\text {int }}$. Context answer: \boxed{$R_{int}=2.00 $ , $I_{0}=6.00$} Extra Supplementary Reading Materials: The element $X$ is connected in series (see Fig.2) with a resistor $R$, an inductor $L$ and an ideal voltage source $\mathcal{E}$. One says that the circuit is in a stationary state if the current is constant in time, $I(t)=$ const. Figure 2: Circuit with element $X$, resistor $R$, inductor $L$ and voltage source $\mathcal{E}$. Context question: A.2 What are the possible numbers of stationary states that the circuit of Fig. 2 may have for a fixed value of $\mathcal{E}$ and for $R=3.00 \Omega$ ? How does the answer change for $R=1.00 \Omega$ ? Context answer: For $R=3.00 \Omega$, one always gets exactly one intersection. For $R=1.00 \Omega$, one gets 1,2 or 3 intersections depending on the value of $\mathcal{E}$. Context question: A.3 Let $R=3.00 \Omega, L=1.00 \mu \mathrm{H}$ and $\mathcal{E}=15.0 \mathrm{~V}$ in the circuit shown in Fig.2 Determine the values of the current $I_{\text {stationary }}$ and the voltage $V_{\text {stationary }}$ on the non-linear element $X$ in the stationary state. Context answer: \boxed{$I_{\text {stationary }}=3.00$ , $U_{\text {stationary }}=6.00$} Extra Supplementary Reading Materials: The circuit in Fig. 2 is in the stationary state with $I(t)=I_{\text {stationary }}$. This stationary state is said to be stable if after a small displacement (increase or decrease in the current), the current returns towards the stationary state. And if the system keeps moving away from the stationary state, it is said to be unstable. Context question: A.4 Use numerical values of the question A.3 and study the stability of the stationary state with $I(t)=I_{\text {stationary }}$. Is it stable or unstable? Context answer: \boxed{stable} Extra Supplementary Reading Materials: Part B.Bistable non-linear elements in physics: radio transmitter We now investigate a new circuit configuration (see Fig. 3). This time, the non-linear element $X$ is connected in parallel to a capacitor of capacitance $C=1.00 \mu \mathrm{F}$. This block is then connected in series to a resistor of resistance $R=3.00 \Omega$ and an ideal constant voltage source of voltage $\mathcal{E}=15.0 \mathrm{~V}$. It turns out that this circuit undergoes oscillations with the non-linear element $X$ jumping from one branch of the $I-V$ characteristics to another over the course of one cycle. Figure 3: Circuit with element $X$, capacitor $C$, resistor $R$ and voltage source $\mathcal{E}$. Context question: B.1 Draw the oscillation cycle on the $I-V$ graph, including its direction (clockwise or anticlockwise). Justify your answer with equations and sketches. Context answer: Context question: B.2 Find expressions for the times $t_{1}$ and $t_{2}$ that the system spends on each branch of the $I-V$ graph during the oscillation cycle. Determine their numerical values. Find the numerical value of the oscillation period $T$ assuming that the time needed for jumps between the branches of the $I-V$ graph is negligible. Context answer: $t_{\mathrm{on}}=2.41 \cdot 10^{-6} \mathrm{~s}$ $t_{\mathrm{off}}=3.71 \cdot 10^{-6} \mathrm{~s}$ $T==6.12 \cdot 10^{-6} \mathrm{~s}$ Context question: B.3 Estimate the average power $P$ dissipated by the non-linear element over the course of one oscillation. An order of magnitude is sufficient. Context answer: \boxed{19.3} Extra Supplementary Reading Materials: The circuit in Fig. 3 is used to build a radio transmitter. For this purpose, the element $X$ is attached to one end of a linear antenna (a long straight wire) of length $s$. The other end of the wire is free. In the antenna, an electromagnetic standing wave is formed. The speed of electromagnetic waves along the antenna is the same as in vacuum. The transmitter is using the main harmonic of the system, which has period $T$ of question B.2 .",B.4 What is the optimal value of $s$ assuming that it cannot exceed $1 \mathrm{~km}$ ?,['The only choice which is below $1 \\mathrm{~km}$ is $s=\\lambda / 4=459 \\mathrm{~m}$.'],['459'],False,m,Numerical,1e0 1211,Electromagnetism,"Nonlinear Dynamics in Electric Circuits Please read the general instructions in the separate envelope before you start this problem. Introduction Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches and generators of electromagnetic oscillations. The primary field of applications of thyristors is controlling alternating currents in power electronics, for instance rectification of AC current to DC at the megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern nonlinear science. Goals To study instabilities and nontrivial dynamics of circuits including elements with non-linear $I-V$ characteristics. To discover possible applications of such circuits in engineering and in modeling of biological systems. Part A.Stationary states and instabilities Fig. 1 shows the so-called S-shaped $I-V$ characteristics of a non-linear element $X$. In the voltage range between $U_{\mathrm{h}}=4.00 \mathrm{~V}$ (the holding voltage) and $U_{\text {th }}=10.0 \mathrm{~V}$ (the threshold voltage) this $I-V$ characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it is extended. This approximation gives a good description of real thyristors. Figure 1: $I-V$ characteristics of the non-linear element $X$. Context question: A.1 Using the graph, determine the resistance $R_{\mathrm{on}}$ of the element $X$ on the upper branch of the $I-V$ characteristics, and $R_{\text {off }}$ on the lower branch, respectively. The middle branch is described by the equation $$ I=I_{0}-\frac{U}{R_{\text {int }}} \tag{1} $$ Find the values of the parameters $I_{0}$ and $R_{\text {int }}$. Context answer: \boxed{$R_{int}=2.00 $ , $I_{0}=6.00$} Extra Supplementary Reading Materials: The element $X$ is connected in series (see Fig.2) with a resistor $R$, an inductor $L$ and an ideal voltage source $\mathcal{E}$. One says that the circuit is in a stationary state if the current is constant in time, $I(t)=$ const. Figure 2: Circuit with element $X$, resistor $R$, inductor $L$ and voltage source $\mathcal{E}$. Context question: A.2 What are the possible numbers of stationary states that the circuit of Fig. 2 may have for a fixed value of $\mathcal{E}$ and for $R=3.00 \Omega$ ? How does the answer change for $R=1.00 \Omega$ ? Context answer: For $R=3.00 \Omega$, one always gets exactly one intersection. For $R=1.00 \Omega$, one gets 1,2 or 3 intersections depending on the value of $\mathcal{E}$. Context question: A.3 Let $R=3.00 \Omega, L=1.00 \mu \mathrm{H}$ and $\mathcal{E}=15.0 \mathrm{~V}$ in the circuit shown in Fig.2 Determine the values of the current $I_{\text {stationary }}$ and the voltage $V_{\text {stationary }}$ on the non-linear element $X$ in the stationary state. Context answer: \boxed{$I_{\text {stationary }}=3.00$ , $U_{\text {stationary }}=6.00$} Extra Supplementary Reading Materials: The circuit in Fig. 2 is in the stationary state with $I(t)=I_{\text {stationary }}$. This stationary state is said to be stable if after a small displacement (increase or decrease in the current), the current returns towards the stationary state. And if the system keeps moving away from the stationary state, it is said to be unstable. Context question: A.4 Use numerical values of the question A.3 and study the stability of the stationary state with $I(t)=I_{\text {stationary }}$. Is it stable or unstable? Context answer: \boxed{stable} Extra Supplementary Reading Materials: Part B.Bistable non-linear elements in physics: radio transmitter We now investigate a new circuit configuration (see Fig. 3). This time, the non-linear element $X$ is connected in parallel to a capacitor of capacitance $C=1.00 \mu \mathrm{F}$. This block is then connected in series to a resistor of resistance $R=3.00 \Omega$ and an ideal constant voltage source of voltage $\mathcal{E}=15.0 \mathrm{~V}$. It turns out that this circuit undergoes oscillations with the non-linear element $X$ jumping from one branch of the $I-V$ characteristics to another over the course of one cycle. Figure 3: Circuit with element $X$, capacitor $C$, resistor $R$ and voltage source $\mathcal{E}$. Context question: B.1 Draw the oscillation cycle on the $I-V$ graph, including its direction (clockwise or anticlockwise). Justify your answer with equations and sketches. Context answer: Context question: B.2 Find expressions for the times $t_{1}$ and $t_{2}$ that the system spends on each branch of the $I-V$ graph during the oscillation cycle. Determine their numerical values. Find the numerical value of the oscillation period $T$ assuming that the time needed for jumps between the branches of the $I-V$ graph is negligible. Context answer: $t_{\mathrm{on}}=2.41 \cdot 10^{-6} \mathrm{~s}$ $t_{\mathrm{off}}=3.71 \cdot 10^{-6} \mathrm{~s}$ $T==6.12 \cdot 10^{-6} \mathrm{~s}$ Context question: B.3 Estimate the average power $P$ dissipated by the non-linear element over the course of one oscillation. An order of magnitude is sufficient. Context answer: \boxed{19.3} Extra Supplementary Reading Materials: The circuit in Fig. 3 is used to build a radio transmitter. For this purpose, the element $X$ is attached to one end of a linear antenna (a long straight wire) of length $s$. The other end of the wire is free. In the antenna, an electromagnetic standing wave is formed. The speed of electromagnetic waves along the antenna is the same as in vacuum. The transmitter is using the main harmonic of the system, which has period $T$ of question B.2 . Context question: B.4 What is the optimal value of $s$ assuming that it cannot exceed $1 \mathrm{~km}$ ? Context answer: \boxed{459} Extra Supplementary Reading Materials: Part C. Bistable non-linear elements in biology: neuristor In this part of the problem, we consider an application of bistable non-linear elements to modeling of biological processes. A neuron in a human brain has the following property: when excited by an external signal, it makes one single oscillation and then returns to its initial state. This feature is called excitability. Due to this property, pulses can propagate in the network of coupled neurons constituting the nerve systems. A semiconductor chip designed to mimic excitability and pulse propagation is called a neuristor (from neuron and transistor). We attempt to model a simple neuristor using a circuit that includes the non-linear element $X$ that we investigated previously. To this end, the voltage $\mathcal{E}$ in the circuit of Fig. 3 is decreased to the value $\mathcal{E}^{\prime}=$ $12.0 \mathrm{~V}$. The oscillations stop, and the system reaches its stationary state. Then, the voltage is rapidly increased back to the value $\mathcal{E}=15.0 \mathrm{~V}$, and after a period of time $\tau$ (with $\tau\tau_{\text {crit }}$. Figure 4: Voltage of the voltage source as a function of time. Context question: C.1 Sketch the graphs of the time dependence of the current $I_{X}(t)$ on the non-linear element $X$ for $\tau<\tau_{\text {crit }}$ and for $\tau>\tau_{\text {crit }}$. Context answer: ",C.2 Find the expression and the numerical value of the critical time $\tau_{\text {crit }}$ for which the scenario switches.,['The time needed to reach the threshold voltage is given by\n\n$$\n\\tau_{\\text {crit }}=\\frac{R_{\\text {off }} R}{R_{\\text {off }}+R} C \\log \\left(\\frac{U_{\\text {off }}-\\tilde{U}}{U_{\\text {off }}-U_{\\text {th }}}\\right)=9.36 \\cdot 10^{-7} s .\n$$'],['$\\tau_{\\text {crit }}=9.36 \\cdot 10^{-7}$'],False,s,Numerical,1e-8 1213,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.","A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.",['Conservation of energy:\n\n$$\nm_{p} \\cdot c^{2}+V \\cdot e=m_{p} \\cdot c^{2} \\cdot \\gamma=\\frac{m_{p} \\cdot c^{2}}{\\sqrt{1-v^{2} / c^{2}}}\n$$\n\nPenalties\n\n$$\n\\text { No or incorrect total energy }\n$$\n\nSolve for velocity:\n\n$$\nv=c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e}\\right)^{2}}\n$$'],['$v=c \\cdot \\sqrt{1-(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e})^{2}}$'],False,,Expression, 1214,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.",A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.,['velocity (from previous question):\n\n$$\nv=c \\cdot \\sqrt{1-\\left(\\frac{m_{e} \\cdot c^{2}}{m_{e} \\cdot c^{2}+V \\cdot e}\\right)^{2}} \\text { or } c \\cdot \\sqrt{1-\\left(\\frac{m_{e} \\cdot c^{2}}{V \\cdot e}\\right)^{2}}\n$$\n\nrelative difference:\n\n$$\n\\begin{gathered}\n\\Delta=\\frac{c-v}{c}=1-\\frac{v}{c} \\\\\n\\rightarrow \\Delta \\simeq \\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{m_{e} \\cdot c^{2}+V \\cdot e}\\right)^{2} \\text { or } \\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{V \\cdot e}\\right)^{2}\n\\end{gathered}\n$$\n\nrelative difference\n\n$$\n\\Delta=3.63 \\cdot 10^{-11}\n$$'],"['$\\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{m_{e} \\cdot c^{2}+V \\cdot e}\\right)^{2}$, $\\Delta=3.63 \\cdot 10^{-11}$', '$\\frac{1}{2}\\left(\\frac{m_{e} \\cdot c^{2}}{V \\cdot e}\\right)^{2}$, $\\Delta=3.63 \\cdot 10^{-11}$']",True,,"Expression,Numerical",",1e-12" 1215,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$. Context question: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants. Context answer: \boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$} Extra Supplementary Reading Materials: We now return to the protons in the LHC. Assume that the beam pipe has a circular shape.","A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits. Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.","['Balance of forces:\n\n$$\n\\frac{\\gamma \\cdot m_{p} \\cdot v^{2}}{r}=\\frac{m_{p} \\cdot v^{2}}{r \\cdot \\sqrt{1-\\frac{v^{2}}{c^{2}}}}=e \\cdot v \\cdot B\n$$\n\nIn case of a mistake, partial points can be given for intermediate steps (up to max 0.2). Examples:\n\n$$\n\\begin{array}{ll}\n\\text { Example: } & \\text { Lorentz force } \\\\\n\\text { Example: } & \\frac{\\gamma \\cdot m_{p} \\cdot v^{2}}{r}\n\\end{array}\n$$\n\nEnergy:\n\n$$\nE=(\\gamma-1) \\cdot m_{p} \\cdot c^{2} \\simeq \\gamma \\cdot m_{p} \\cdot c^{2} \\rightarrow \\gamma=\\frac{E}{m_{p} c^{2}}\n$$\n\nTherefore:\n\n$$\n\\frac{E \\cdot v}{c^{2} \\cdot r}=e \\cdot B\n$$\n\nWith\n\n$$\nv \\simeq c \\text { and } r=\\frac{L}{2 \\pi}\n$$\n\nfollows:\n\n$$\n\\rightarrow B=\\frac{2 \\pi \\cdot E}{e \\cdot c \\cdot L}\n$$\n\nSolution:\n\n$$\nB=5.50 \\mathrm{~T}\n$$']","['$B=\\frac{2 \\pi \\cdot E}{e \\cdot c \\cdot L}$ , $B=5.50$']",True,",$\mathrm{~T}$","Expression,Numerical",",1e-1" 1216,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$. Context question: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants. Context answer: \boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$} Extra Supplementary Reading Materials: We now return to the protons in the LHC. Assume that the beam pipe has a circular shape. Context question: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits. Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons. Context answer: \boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$} Extra Supplementary Reading Materials: Radiated Power: An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$.",A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$.,"['Ansatz:\n\n$$\nP_{\\text {rad }}=a^{\\alpha} \\cdot q^{\\beta} \\cdot c^{\\gamma} \\cdot \\epsilon_{0}^{\\delta}\n$$\n\nDimensions: $[\\mathrm{a}]=\\mathrm{ms}^{-2},[\\mathrm{q}]=\\mathrm{C}=\\mathrm{As},[\\mathrm{c}]=\\mathrm{ms}^{-1},\\left[\\epsilon_{0}\\right]=\\mathrm{As}(\\mathrm{Vm})^{-1}=\\mathrm{A}^{2} \\mathrm{~s}^{2}\\left(\\mathrm{Nm}^{2}\\right)^{-1}=\\mathrm{A}^{2} \\mathrm{~s}^{4}\\left(\\mathrm{kgm}^{3}\\right)^{-}$']",['$P_{\\text {rad }}=a^{\\alpha} \\cdot q^{\\beta} \\cdot c^{\\gamma} \\cdot \\epsilon_{0}^{\\delta}$'],False,,Expression, 1217,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$. Context question: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants. Context answer: \boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$} Extra Supplementary Reading Materials: We now return to the protons in the LHC. Assume that the beam pipe has a circular shape. Context question: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits. Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons. Context answer: \boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$} Extra Supplementary Reading Materials: Radiated Power: An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$. Context question: A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$. Context answer: \boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$} Extra Supplementary Reading Materials: The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$.","A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations.",['Radiated Power:\n\n$$\nP_{r a d}=\\frac{\\gamma^{4} \\cdot a^{2} \\cdot e^{2}}{6 \\pi \\cdot c^{3} \\cdot \\epsilon_{0}}\n$$\n\nEnergy:\n\n$$\nE=(\\gamma-1) m_{p} \\cdot c^{2} \\text { or equally valid } E \\simeq \\gamma \\cdot m_{p} \\cdot c^{2}\n$$\n\nAcceleration:\n\n$$\na \\simeq \\frac{c^{2}}{r} \\text { with } \\quad r=\\frac{L}{2 \\pi}\n$$\n\nTherefore:\n\n$$\n\\begin{gathered}\nP_{\\text {rad }}=\\left(\\frac{E}{m_{p} c^{2}}+1\\right)^{4} \\cdot \\frac{e^{2} \\cdot c}{6 \\pi \\epsilon_{0} \\cdot r^{2}} \\text { or }\\left(\\frac{E}{m_{p} c^{2}}\\right)^{4} \\cdot \\frac{e^{2} \\cdot c}{6 \\pi \\epsilon_{0} \\cdot r^{2}} \\\\\n\\left(\\text { not required } P_{\\text {rad }}=7.94 \\cdot 10^{-12} \\mathrm{~W}\\right)\n\\end{gathered}\n$$\n\nTotal radiated power:\n\n$$\nP_{t o t}=2 \\cdot 2808 \\cdot 1.15 \\cdot 10^{11} \\cdot P_{r a d}=5.13 \\mathrm{~kW}\n$$'],['5.13'],False,kW,Numerical,1e-1 1218,Modern Physics,,"A.6 Determine the time T that the protons take to pass through this field ![](https://cdn.mathpix.com/cropped/2023_12_21_b9b9ac37d6714962e913g-1.jpg?height=429&width=1262&top_left_y=762&top_left_x=403) Figure 1: Sketch of an accelerator module.","[""2nd Newton's law\n\n$$\n\\begin{gathered}\nF=\\frac{d p}{d t} \\quad \\text { leads to } \\\\\n\\frac{V \\cdot e}{d}=\\frac{p_{f}-p_{i}}{T} \\text { with } p_{i}=0\n\\end{gathered}\n$$\n\nConservation of energy:\n\n$$\nE_{t o t}=m \\cdot c^{2}+e \\cdot V\n$$\n\nSince\n\n$$\n\\begin{gathered}\nE_{t o t}^{2}=\\left(m \\cdot c^{2}\\right)^{2}+\\left(p_{f} \\cdot c\\right)^{2} \\\\\n\\rightarrow p_{f}=\\frac{1}{c} \\cdot \\sqrt{\\left(m \\cdot c^{2}+e \\cdot V\\right)^{2}-\\left(m \\cdot c^{2}\\right)^{2}}=\\sqrt{2 e \\cdot m \\cdot V+\\left(\\frac{e \\cdot V}{c}\\right)^{2}} \\\\\n\\rightarrow T=\\frac{d \\cdot p_{f}}{V \\cdot e}=\\frac{d}{V \\cdot e} \\sqrt{2 e \\cdot m_{p} \\cdot V+\\left(\\frac{e \\cdot V}{c}\\right)^{2}} \\\\\nT=218 \\mathrm{~ns}\n\\end{gathered}\n$$"", ""2nd Newton's law\n\n$$\n\\frac{V \\cdot e}{d}=\\frac{p_{f}-p_{i}}{T} \\text { with } p_{i}=0\n$$\n\nvelocity from A1 or from conservation of energy\n\n$$\nv=c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e}\\right)^{2}}\n$$\n\nand hence for $\\gamma$\n\n$$\n\\begin{gathered}\n\\gamma=1 / \\sqrt{1-\\frac{v^{2}}{c^{2}}}=1+\\frac{e \\cdot V}{m_{p} \\cdot c^{2}} \\\\\n\\rightarrow p_{f}=\\gamma \\cdot m_{p} \\cdot v=\\left(1+\\frac{e \\cdot V}{m_{p} \\cdot c^{2}}\\right) \\cdot m_{p} \\cdot c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e}\\right)^{2}} \\\\\n\\rightarrow T=\\frac{d \\cdot p_{f}}{V \\cdot e}=\\frac{d \\cdot m_{p} \\cdot c}{V \\cdot e} \\cdot \\sqrt{\\left(\\frac{m_{p} \\cdot c^{2}+e \\cdot V}{m_{p} \\cdot c^{2}}\\right)^{2}-1}=\\frac{d}{V \\cdot e} \\sqrt{2 e \\cdot m_{p} \\cdot V+\\left(\\frac{e \\cdot V}{c}\\right)^{2}} \\\\\nT=218 \\mathrm{~ns}\n\\end{gathered}\n$$"", 'Energy increases linearly with distance $\\mathrm{x}$\n\n$$\n\\begin{gathered}\nE(x)=\\frac{e \\cdot V \\cdot x}{d} \\\\\nt=\\int d t=\\int_{0}^{d} \\frac{d x}{v(x)} \\\\\nv(x)=c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+\\frac{e \\cdot V \\cdot x}{d}}\\right)^{2}}=c \\cdot \\frac{\\sqrt{\\left(m_{p} \\cdot c^{2}+\\frac{e \\cdot V \\cdot x}{d}\\right)^{2}-\\left(m_{p} \\cdot c^{2}\\right)^{2}}}{m_{p} \\cdot c^{2}+\\frac{e \\cdot V \\cdot x}{d}} \\\\\n=c \\cdot \\frac{\\sqrt{\\left(1+\\frac{e \\cdot V \\cdot x}{d \\cdot m_{p} \\cdot c^{2}}\\right)^{2}-1}}{1+\\frac{e \\cdot V \\cdot x}{d \\cdot m_{p} \\cdot c^{2}}}\n\\end{gathered}\n$$\n\nSubstitution : $\\xi=\\frac{e \\cdot V \\cdot x}{d \\cdot m_{p} \\cdot c^{2}} \\quad \\frac{d \\xi}{d x}=\\frac{e \\cdot V}{d \\cdot m_{p} \\cdot c^{2}}$\n\n$$\n\\rightarrow t=\\frac{1}{c} \\int_{0}^{b} \\frac{1+\\xi}{\\sqrt{(1+\\xi)^{2}-1}} \\frac{d \\cdot m_{p} \\cdot c^{2}}{e \\cdot V} d \\xi \\quad b=\\frac{e \\cdot V}{m_{p} \\cdot c^{2}}\n$$\n\n$$\n1+\\xi:=\\cosh (s) \\quad \\frac{d \\xi}{d s}=\\sinh (s)\n$$\n\n0.1\n\n$$\nt=\\frac{m_{p} \\cdot c \\cdot d}{e \\cdot V} \\int \\frac{\\cosh (s) \\cdot \\sinh (s) d s}{\\sqrt{\\cosh ^{2}(s)-1}}=\\frac{m_{p} \\cdot c \\cdot d}{e \\cdot V}[\\sinh (s)]_{b_{1}}^{b_{2}}\n$$\n\n$$\n\\text { with } \\quad b_{1}=\\cosh ^{-1}(1), \\quad b_{2}=\\cosh ^{-1}\\left(1+\\frac{e \\cdot V}{m_{p} \\cdot c^{2}}\\right)\n$$\n\n$$\nT=218 \\mathrm{~ns}\n$$', 'Alternative: differential equation \n\n$$\n\\begin{gathered}\n\\rightarrow \\frac{V \\cdot e}{d}=\\frac{\\mathrm{d}}{\\mathrm{d} t}\\left(\\frac{m \\cdot v}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}\\right)=\\frac{m \\cdot a\\left(1-\\frac{v^{2}}{c^{2}}\\right)+m \\cdot a \\frac{v^{2}}{c^{2}}}{\\left(1-\\frac{v^{2}}{c^{2}}\\right)^{\\frac{3}{2}}}=\\gamma^{3} \\cdot m \\cdot a \\\\\na=\\ddot{s}=\\frac{V \\cdot e}{d \\cdot m}\\left(1-\\frac{\\dot{s}^{2}}{c^{2}}\\right)^{\\frac{3}{2}}\n\\end{gathered}\n$$\n\n$$\n\\begin{gathered}\n\\rightarrow s(t)=\\frac{c}{V \\cdot e}\\left(\\sqrt{e^{2} \\cdot V^{2} \\cdot t^{2}+c^{2} \\cdot m^{2} \\cdot d^{2}}-c \\cdot m \\cdot d\\right) \\\\\ns=d \\rightarrow T=\\frac{d}{V \\cdot e} \\sqrt{\\left(\\frac{V \\cdot e}{c}\\right)^{2}+2 V \\cdot e \\cdot m} \\\\\nT=218 \\mathrm{~ns}\n\\end{gathered}\n$$']",['218'],False,ns,Numerical,2e0 1218,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$. Context question: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants. Context answer: \boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$} Extra Supplementary Reading Materials: We now return to the protons in the LHC. Assume that the beam pipe has a circular shape. Context question: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits. Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons. Context answer: \boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$} Extra Supplementary Reading Materials: Radiated Power: An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$. Context question: A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$. Context answer: \boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$} Extra Supplementary Reading Materials: The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$. Context question: A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations. Context answer: \boxed{5.13} Extra Supplementary Reading Materials: Linear Acceleration: At CERN, protons at rest are accelerated by a linear accelerator of length $d=30.0 \mathrm{~m}$ through a potential difference of $V=500 \mathrm{MV}$. Assume that the electrical field is homogeneous. A linear accelerator consists of two plates as sketched in Figure 1.","A.6 Determine the time T that the protons take to pass through this field Figure 1: Sketch of an accelerator module.","[""2nd Newton's law\n\n$$\n\\begin{gathered}\nF=\\frac{d p}{d t} \\quad \\text { leads to } \\\\\n\\frac{V \\cdot e}{d}=\\frac{p_{f}-p_{i}}{T} \\text { with } p_{i}=0\n\\end{gathered}\n$$\n\nConservation of energy:\n\n$$\nE_{t o t}=m \\cdot c^{2}+e \\cdot V\n$$\n\nSince\n\n$$\n\\begin{gathered}\nE_{t o t}^{2}=\\left(m \\cdot c^{2}\\right)^{2}+\\left(p_{f} \\cdot c\\right)^{2} \\\\\n\\rightarrow p_{f}=\\frac{1}{c} \\cdot \\sqrt{\\left(m \\cdot c^{2}+e \\cdot V\\right)^{2}-\\left(m \\cdot c^{2}\\right)^{2}}=\\sqrt{2 e \\cdot m \\cdot V+\\left(\\frac{e \\cdot V}{c}\\right)^{2}} \\\\\n\\rightarrow T=\\frac{d \\cdot p_{f}}{V \\cdot e}=\\frac{d}{V \\cdot e} \\sqrt{2 e \\cdot m_{p} \\cdot V+\\left(\\frac{e \\cdot V}{c}\\right)^{2}} \\\\\nT=218 \\mathrm{~ns}\n\\end{gathered}\n$$"", ""2nd Newton's law\n\n$$\n\\frac{V \\cdot e}{d}=\\frac{p_{f}-p_{i}}{T} \\text { with } p_{i}=0\n$$\n\nvelocity from A1 or from conservation of energy\n\n$$\nv=c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e}\\right)^{2}}\n$$\n\nand hence for $\\gamma$\n\n$$\n\\begin{gathered}\n\\gamma=1 / \\sqrt{1-\\frac{v^{2}}{c^{2}}}=1+\\frac{e \\cdot V}{m_{p} \\cdot c^{2}} \\\\\n\\rightarrow p_{f}=\\gamma \\cdot m_{p} \\cdot v=\\left(1+\\frac{e \\cdot V}{m_{p} \\cdot c^{2}}\\right) \\cdot m_{p} \\cdot c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+V \\cdot e}\\right)^{2}} \\\\\n\\rightarrow T=\\frac{d \\cdot p_{f}}{V \\cdot e}=\\frac{d \\cdot m_{p} \\cdot c}{V \\cdot e} \\cdot \\sqrt{\\left(\\frac{m_{p} \\cdot c^{2}+e \\cdot V}{m_{p} \\cdot c^{2}}\\right)^{2}-1}=\\frac{d}{V \\cdot e} \\sqrt{2 e \\cdot m_{p} \\cdot V+\\left(\\frac{e \\cdot V}{c}\\right)^{2}} \\\\\nT=218 \\mathrm{~ns}\n\\end{gathered}\n$$"", 'Energy increases linearly with distance $\\mathrm{x}$\n\n$$\n\\begin{gathered}\nE(x)=\\frac{e \\cdot V \\cdot x}{d} \\\\\nt=\\int d t=\\int_{0}^{d} \\frac{d x}{v(x)} \\\\\nv(x)=c \\cdot \\sqrt{1-\\left(\\frac{m_{p} \\cdot c^{2}}{m_{p} \\cdot c^{2}+\\frac{e \\cdot V \\cdot x}{d}}\\right)^{2}}=c \\cdot \\frac{\\sqrt{\\left(m_{p} \\cdot c^{2}+\\frac{e \\cdot V \\cdot x}{d}\\right)^{2}-\\left(m_{p} \\cdot c^{2}\\right)^{2}}}{m_{p} \\cdot c^{2}+\\frac{e \\cdot V \\cdot x}{d}} \\\\\n=c \\cdot \\frac{\\sqrt{\\left(1+\\frac{e \\cdot V \\cdot x}{d \\cdot m_{p} \\cdot c^{2}}\\right)^{2}-1}}{1+\\frac{e \\cdot V \\cdot x}{d \\cdot m_{p} \\cdot c^{2}}}\n\\end{gathered}\n$$\n\nSubstitution : $\\xi=\\frac{e \\cdot V \\cdot x}{d \\cdot m_{p} \\cdot c^{2}} \\quad \\frac{d \\xi}{d x}=\\frac{e \\cdot V}{d \\cdot m_{p} \\cdot c^{2}}$\n\n$$\n\\rightarrow t=\\frac{1}{c} \\int_{0}^{b} \\frac{1+\\xi}{\\sqrt{(1+\\xi)^{2}-1}} \\frac{d \\cdot m_{p} \\cdot c^{2}}{e \\cdot V} d \\xi \\quad b=\\frac{e \\cdot V}{m_{p} \\cdot c^{2}}\n$$\n\n$$\n1+\\xi:=\\cosh (s) \\quad \\frac{d \\xi}{d s}=\\sinh (s)\n$$\n\n0.1\n\n$$\nt=\\frac{m_{p} \\cdot c \\cdot d}{e \\cdot V} \\int \\frac{\\cosh (s) \\cdot \\sinh (s) d s}{\\sqrt{\\cosh ^{2}(s)-1}}=\\frac{m_{p} \\cdot c \\cdot d}{e \\cdot V}[\\sinh (s)]_{b_{1}}^{b_{2}}\n$$\n\n$$\n\\text { with } \\quad b_{1}=\\cosh ^{-1}(1), \\quad b_{2}=\\cosh ^{-1}\\left(1+\\frac{e \\cdot V}{m_{p} \\cdot c^{2}}\\right)\n$$\n\n$$\nT=218 \\mathrm{~ns}\n$$', 'Alternative: differential equation \n\n$$\n\\begin{gathered}\n\\rightarrow \\frac{V \\cdot e}{d}=\\frac{\\mathrm{d}}{\\mathrm{d} t}\\left(\\frac{m \\cdot v}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}\\right)=\\frac{m \\cdot a\\left(1-\\frac{v^{2}}{c^{2}}\\right)+m \\cdot a \\frac{v^{2}}{c^{2}}}{\\left(1-\\frac{v^{2}}{c^{2}}\\right)^{\\frac{3}{2}}}=\\gamma^{3} \\cdot m \\cdot a \\\\\na=\\ddot{s}=\\frac{V \\cdot e}{d \\cdot m}\\left(1-\\frac{\\dot{s}^{2}}{c^{2}}\\right)^{\\frac{3}{2}}\n\\end{gathered}\n$$\n\n$$\n\\begin{gathered}\n\\rightarrow s(t)=\\frac{c}{V \\cdot e}\\left(\\sqrt{e^{2} \\cdot V^{2} \\cdot t^{2}+c^{2} \\cdot m^{2} \\cdot d^{2}}-c \\cdot m \\cdot d\\right) \\\\\ns=d \\rightarrow T=\\frac{d}{V \\cdot e} \\sqrt{\\left(\\frac{V \\cdot e}{c}\\right)^{2}+2 V \\cdot e \\cdot m} \\\\\nT=218 \\mathrm{~ns}\n\\end{gathered}\n$$']",['218'],False,ns,Numerical,2e0 1219,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$. Context question: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants. Context answer: \boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$} Extra Supplementary Reading Materials: We now return to the protons in the LHC. Assume that the beam pipe has a circular shape. Context question: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits. Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons. Context answer: \boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$} Extra Supplementary Reading Materials: Radiated Power: An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$. Context question: A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$. Context answer: \boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$} Extra Supplementary Reading Materials: The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$. Context question: A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations. Context answer: \boxed{5.13} Extra Supplementary Reading Materials: Linear Acceleration: At CERN, protons at rest are accelerated by a linear accelerator of length $d=30.0 \mathrm{~m}$ through a potential difference of $V=500 \mathrm{MV}$. Assume that the electrical field is homogeneous. A linear accelerator consists of two plates as sketched in Figure 1. Context question: A.6 Determine the time T that the protons take to pass through this field Figure 1: Sketch of an accelerator module. Context answer: \boxed{218} Extra Supplementary Reading Materials: Part B.Particle Identification Time of flight: It is important to identify the high energy particles that are generated in the collision in order to interpret the interaction process. A simple method is to measure the time $(t)$ that a particle with known momentum needs to pass a length $l$ in a so-called Time-of-Flight (ToF) detector. Typical particles which are identified in the detector, together with their masses, are listed in table 2. | Particle | Mass $\left[\mathrm{MeV} / \mathrm{c}^{2}\right]$ | | :--- | :--- | | Deuteron | 1876 | | Proton | 938 | | charged Kaon | 494 | | charged Pion | 140 | | Electron | 0.511 | Table 2: Particles and their masses. Figure 2: Schematic view of a time-of-flight detector.","B.1 Express the particle mass $m$ in terms of of the momentum $p$, the flight length $l$ and the flight time $t$, assuming that particles have elementary charge $e$ and travel with velocity close to $c$ on straight tracks in the ToF detector and that they travel perpendicular to the two detection planes (see figure 2).","['with velocity\n\n$$\nv=\\frac{l}{t}\n$$\n\n0.1\n\nrelativistic momentum\n\n$$\np=\\frac{m \\cdot v}{\\sqrt{1-\\frac{v^{2}}{c^{2}}}}\n$$\n\ngets\n\n$$\np=\\frac{m \\cdot l}{t \\cdot \\sqrt{1-\\frac{l^{2}}{t^{2} \\cdot c^{2}}}}\n$$\n\n$\\rightarrow$ mass\n\n$$\nm=\\frac{p \\cdot t}{l} \\cdot \\sqrt{1-\\frac{l^{2}}{t^{2} \\cdot c^{2}}}=\\frac{p}{l \\cdot c} \\cdot \\sqrt{t^{2} \\cdot c^{2}-l^{2}}\n$$', 'with flight distance: $l$, flight time $\\mathrm{t}$ gets:\n\n$$\nt=\\frac{l}{(c \\cdot \\beta)}\n$$\n\nrelativistic momentum\n\n$$\np=\\frac{m \\cdot \\beta \\cdot c}{\\sqrt{1-\\beta^{2}}}\n$$\n\ntherefore the velocity:\n\n$$\n\\beta=\\frac{p}{\\sqrt{m^{2} \\cdot c^{2}+p^{2}}}\n$$\n\ninsert into the expression for $t$ :\n\n$$\nt=l \\frac{\\sqrt{m^{2} \\cdot c^{2}+p^{2}}}{c \\cdot p}\n$$\n\n$\\rightarrow$ mass:\n\n$$\nm=\\sqrt{\\left(\\frac{p \\cdot t}{l}\\right)^{2}-\\left(\\frac{p}{c}\\right)^{2}}=\\frac{p}{l \\cdot c} \\sqrt{(t \\cdot c)^{2}-(l)^{2}}\n$$']",['$m=\\frac{p}{l \\cdot c} \\cdot \\sqrt{t^{2} \\cdot c^{2}-l^{2}}$'],False,,Expression, 1220,Modern Physics,"Large Hadron Collider Please read the general instructions in the separate envelope before you start this problem. In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 . | LHC ring | | | :--- | :--- | | Circumference of ring | $26659 \mathrm{~m}$ | | Number of bunches per proton beam | 2808 | | Number of protons per bunch | $1.15 \times 10^{11}$ | | Proton beams | | | Energy of protons | $7.00 \mathrm{TeV}$ | | Centre of mass energy | $14.0 \mathrm{TeV}$ | Table 1: Typical numerical values of relevant LHC parameters. Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$. The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$. Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN. Part A.LHC accelerator Acceleration: Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles. Context question: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants. Context answer: \boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$} Extra Supplementary Reading Materials: A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$. Context question: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants. Context answer: \boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$} Extra Supplementary Reading Materials: We now return to the protons in the LHC. Assume that the beam pipe has a circular shape. Context question: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits. Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons. Context answer: \boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$} Extra Supplementary Reading Materials: Radiated Power: An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$. Context question: A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$. Context answer: \boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$} Extra Supplementary Reading Materials: The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$. Context question: A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations. Context answer: \boxed{5.13} Extra Supplementary Reading Materials: Linear Acceleration: At CERN, protons at rest are accelerated by a linear accelerator of length $d=30.0 \mathrm{~m}$ through a potential difference of $V=500 \mathrm{MV}$. Assume that the electrical field is homogeneous. A linear accelerator consists of two plates as sketched in Figure 1. Context question: A.6 Determine the time T that the protons take to pass through this field Figure 1: Sketch of an accelerator module. Context answer: \boxed{218} Extra Supplementary Reading Materials: Part B.Particle Identification Time of flight: It is important to identify the high energy particles that are generated in the collision in order to interpret the interaction process. A simple method is to measure the time $(t)$ that a particle with known momentum needs to pass a length $l$ in a so-called Time-of-Flight (ToF) detector. Typical particles which are identified in the detector, together with their masses, are listed in table 2. | Particle | Mass $\left[\mathrm{MeV} / \mathrm{c}^{2}\right]$ | | :--- | :--- | | Deuteron | 1876 | | Proton | 938 | | charged Kaon | 494 | | charged Pion | 140 | | Electron | 0.511 | Table 2: Particles and their masses. Figure 2: Schematic view of a time-of-flight detector. Context question: B.1 Express the particle mass $m$ in terms of of the momentum $p$, the flight length $l$ and the flight time $t$, assuming that particles have elementary charge $e$ and travel with velocity close to $c$ on straight tracks in the ToF detector and that they travel perpendicular to the two detection planes (see figure 2). Context answer: \boxed{$m=\frac{p}{l \cdot c} \cdot \sqrt{t^{2} \cdot c^{2}-l^{2}}$} ","B.2 Calculate the minimal length $l$ of a ToF detector that allows to safely distinguish a charged kaon from a charged pion, given both their momenta are measured to be $1.00 \mathrm{GeV} / \mathrm{c}$. For a good separation it is required that the difference in the time-of-flight is larger than three times the time resolution of the detector. The typical resolution of a ToF detector is $150 \mathrm{ps}\left(1 \mathrm{ps}=10^{-12} \mathrm{~s}\right)$.",['Flight time difference between kaon and pion\n\n$$\n\\Delta t=450 \\mathrm{ps}=450 \\cdot 10^{-12} \\mathrm{~s}\n$$\n\nFlight time difference between kaon and pion\n\n$$\n\\begin{gathered}\n\\Delta t=\\frac{l}{c p}\\left(\\sqrt{m_{\\pi}^{2} \\cdot c^{2}+p^{2}}-\\sqrt{m_{K}^{2} \\cdot c^{2}+p^{2}}\\right)=450 \\mathrm{ps}=450 \\cdot 10^{-12} \\mathrm{~s} \\\\\n\\rightarrow l=\\frac{\\Delta t \\cdot p}{\\sqrt{m_{K}^{2}+p^{2} / c^{2}}-\\sqrt{m_{\\pi}^{2}+p^{2} / c^{2}}} \\\\\n\\sqrt{m_{K}^{2}+p^{2} / c^{2}}=1.115 \\mathrm{GeV} / c^{2} \\text { and } \\sqrt{m_{\\pi}^{2}+p^{2} / c^{2}}=1.010 \\mathrm{GeV} / c^{2} \\\\\nl=450 \\cdot 10^{-12} \\cdot \\frac{1}{1.115-1.010} \\mathrm{~s} \\mathrm{GeV} c^{2} /(\\mathrm{GeV} c) \\\\\nl=4285.710^{-12} \\mathrm{~s} \\cdot c=4285.7 \\cdot 10^{-12} \\cdot 2.998 \\cdot 10^{8} \\mathrm{~m}=1.28 \\mathrm{~m}\n\\end{gathered}\n$$'],['1.28'],False,m,Numerical,1e-1 1221,Modern Physics,,"3.1 Assume that the ion which has just one electron left the shell. $\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state. In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$. Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form.",['3.1\n\n$\\mathrm{r}_{0}^{2}=(\\Delta \\mathrm{x})^{2}+(\\Delta \\mathrm{y})^{2}+(\\Delta \\mathrm{z})^{2}$\n\n$\\mathrm{p}_{0}^{2}=\\left(\\Delta \\mathrm{p}_{\\mathrm{x}}\\right)^{2}+\\left(\\Delta \\mathrm{p}_{\\mathrm{y}}\\right)^{2}+\\left(\\Delta \\mathrm{p}_{\\mathrm{z}}\\right)^{2}$\n\nsince\n\n$\\Delta \\mathrm{p}_{\\mathrm{x}} \\geq \\frac{\\hbar}{2 \\cdot \\Delta \\mathrm{x}} \\quad \\Delta \\mathrm{p}_{\\mathrm{y}} \\geq \\frac{\\hbar}{2 \\cdot \\Delta \\mathrm{y}} \\quad \\Delta \\mathrm{p}_{\\mathrm{z}} \\geq \\frac{\\hbar}{2 \\cdot \\Delta \\mathrm{z}}$\n\ngives\n\n$\\mathrm{p}_{0}^{2} \\geq \\frac{\\hbar^{2}}{4} \\cdot\\left[\\frac{1}{(\\Delta \\mathrm{x})^{2}}+\\frac{1}{(\\Delta \\mathrm{y})^{2}}+\\frac{1}{(\\Delta \\mathrm{z})^{2}}\\right]$\n\nand\n\n$(\\Delta \\mathrm{x})^{2}=(\\Delta \\mathrm{y})^{2}=(\\Delta \\mathrm{z})^{2}=\\frac{\\mathrm{r}_{0}^{2}}{3}$\n\nthus\n\n$$\n\\mathrm{p}_{0}^{2} \\cdot \\mathrm{r}_{0}^{2} \\geq \\frac{9}{4} \\cdot \\hbar^{2}\n$$'],['$\\mathrm{p}_{0}^{2} \\cdot \\mathrm{r}_{0}^{2} \\geq \\frac{9}{4} \\cdot \\hbar^{2}$'],False,,Need_human_evaluate, 1222,Modern Physics,,"3.2 The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon. Write down an equation which is to be used for calculation the frequency of an emitted photon.",['3.2\n\n$\\left|\\vec{v}_{\\mathrm{e}}\\right| \\ldots . .$. speed of the external electron before the capture\n\n$\\left|\\vec{V}_{i}\\right| \\ldots . .$. speed of $A^{(Z-1)+}$ before capturing\n\n$\\left|\\vec{V}_{\\mathrm{f}}\\right| \\ldots . .$. speed of $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ after capturing\n\n$\\mathrm{E}_{\\mathrm{n}}=\\mathrm{h} . v \\quad \\ldots . .$. energy of the emitted photon\n\nconservation of energy:\n\n$\\frac{1}{2} \\cdot m_{e} \\cdot V_{e}^{2}+\\frac{1}{2} \\cdot\\left(M+m_{e}\\right) \\cdot V_{i}^{2}+E\\left[A^{(Z-1)+}\\right]=\\frac{1}{2} \\cdot\\left(M+2 \\cdot m_{e}\\right) \\cdot V_{f}^{2}+E\\left[A^{(Z-2)+}\\right]$\n\nwhere $\\mathrm{E}\\left[\\mathrm{A}^{(\\mathrm{Z}-1)+}\\right)$ and $\\mathrm{E}\\left[\\mathrm{A}^{(\\mathrm{Z}-2)+}\\right]$ denotes the energy of the electron in the outermost shell of ions $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ and $\\mathrm{A}^{(\\mathrm{Z}-2)+}$ respectively.\n\nconservation of momentum:\n\n$m_{e} \\cdot \\vec{v}_{e}+(M+m) \\cdot \\vec{v}_{i}=\\left(M+2 \\cdot m_{e}\\right) \\cdot \\vec{v}_{f}+\\frac{h \\cdot v}{c} \\cdot \\overrightarrow{1}$\n\nwhere $\\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.'],['3.2\n\n$\\left|\\vec{v}_{\\mathrm{e}}\\right| \\ldots . .$. speed of the external electron before the capture\n\n$\\left|\\vec{V}_{i}\\right| \\ldots . .$. speed of $A^{(Z-1)+}$ before capturing\n\n$\\left|\\vec{V}_{\\mathrm{f}}\\right| \\ldots . .$. speed of $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ after capturing\n\n$\\mathrm{E}_{\\mathrm{n}}=\\mathrm{h} . v \\quad \\ldots . .$. energy of the emitted photon\n\nconservation of energy:\n\n$\\frac{1}{2} \\cdot m_{e} \\cdot V_{e}^{2}+\\frac{1}{2} \\cdot\\left(M+m_{e}\\right) \\cdot V_{i}^{2}+E\\left[A^{(Z-1)+}\\right]=\\frac{1}{2} \\cdot\\left(M+2 \\cdot m_{e}\\right) \\cdot V_{f}^{2}+E\\left[A^{(Z-2)+}\\right]$\n\nwhere $\\mathrm{E}\\left[\\mathrm{A}^{(\\mathrm{Z}-1)+}\\right)$ and $\\mathrm{E}\\left[\\mathrm{A}^{(\\mathrm{Z}-2)+}\\right]$ denotes the energy of the electron in the outermost shell of ions $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ and $\\mathrm{A}^{(\\mathrm{Z}-2)+}$ respectively.\n\nconservation of momentum:\n\n$m_{e} \\cdot \\vec{v}_{e}+(M+m) \\cdot \\vec{v}_{i}=\\left(M+2 \\cdot m_{e}\\right) \\cdot \\vec{v}_{f}+\\frac{h \\cdot v}{c} \\cdot \\overrightarrow{1}$\n\nwhere $\\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.'],False,,Need_human_evaluate, 1223,Modern Physics,"Recombination of Positive and Negative Ions in Ionized Gas Introduction A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $\mathrm{A}^{(\mathrm{Z}-1)+}$ Constants: | electric field constant | $\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}$ | | :--- | :--- | | elementary charge | $\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}$ | | | $\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}$ | | Planck's constant | $\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ | | (rest) mass of an electron | $\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}$ | | Bohr's atomic radius | $\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}$ | | Rydberg's energy | $\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}$ | | (rest) mass of a proton | $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}$ | Context question: 3.1 Assume that the ion which has just one electron left the shell. $\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state. In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$. Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form. Context answer: $\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$ Context question: 3.2 The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon. Write down an equation which is to be used for calculation the frequency of an emitted photon. Context answer: 3.2 $\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture $\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing $\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing $\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon conservation of energy: $\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$ where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively. conservation of momentum: $m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$ where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon. ","3.3 Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles: 3.3.A The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$. (ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem). 3.3.B In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$","['3.3 \n\nDetermination of the energy of $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ :\n\npotential energy $=-\\frac{Z \\cdot e^{2}}{4 \\cdot \\pi \\cdot \\varepsilon_{0} \\cdot r_{0}}=-\\frac{Z \\cdot q^{2}}{r_{0}}$\n\nkinetic energy $=\\frac{p^{2}}{2 \\cdot m}$\n\nIf the motion of the electrons is confined within the $x-y$-plane, principles of uncertainty in 3.1 can be written as\n\n$r_{0}^{2}=(\\Delta \\mathrm{x})^{2}+(\\Delta \\mathrm{y})^{2}$\n\n$\\mathrm{p}_{0}^{2}=\\left(\\Delta \\mathrm{p}_{\\mathrm{x}}\\right)^{2}+\\left(\\Delta \\mathrm{p}_{\\mathrm{y}}\\right)^{2}$\n\n$\\mathrm{p}_{0}^{2}=\\frac{\\hbar^{2}}{4} \\cdot\\left[\\frac{1}{(\\Delta \\mathrm{x})^{2}}+\\frac{1}{(\\Delta \\mathrm{y})^{2}}\\right]=\\frac{\\hbar^{2}}{4} \\cdot\\left[\\frac{2}{\\mathrm{r}_{0}^{2}}+\\frac{2}{\\mathrm{r}_{0}^{2}}\\right]=\\frac{\\hbar^{2}}{4} \\cdot \\frac{4}{\\mathrm{r}_{0}^{2}}$\n\nthus\n\n$\\mathrm{p}_{0}^{2} \\cdot \\mathrm{r}_{0}^{2}=\\hbar^{2}$\n\n$E\\left[A^{(z-1)+}\\right]=\\frac{p_{0}^{2}}{2 \\cdot m_{e}}-\\frac{Z \\cdot q^{2}}{r_{0}}=\\frac{\\hbar^{2}}{2 \\cdot m_{e} \\cdot r_{e}}-\\frac{Z \\cdot q^{2}}{r_{0}}$\n\nEnergy minimum exists, when $\\frac{\\mathrm{dE}}{\\mathrm{dr}_{0}}=0$.\n\nHence\n\n$\\frac{\\mathrm{dE}}{\\mathrm{dr}_{0}}=-\\frac{\\hbar^{2}}{\\mathrm{~m}_{\\mathrm{e}} \\cdot \\mathrm{r}_{\\mathrm{e}}^{3}}+\\frac{\\mathrm{Z} \\cdot \\mathrm{q}^{2}}{\\mathrm{r}_{0}^{2}}=0$\n\nthis gives $\\frac{1}{r_{0}}=\\frac{Z \\cdot q^{2} \\cdot m_{e}}{\\hbar^{2}}$\n\nhence\n\n$$\n\\begin{gathered}\nE\\left[A^{(Z-1)+}\\right]=\\frac{\\hbar^{2}}{2 \\cdot m_{e}} \\cdot\\left(\\frac{Z \\cdot q^{2} \\cdot m_{e}}{\\hbar}\\right)^{2}-Z \\cdot q^{2} \\cdot \\frac{Z \\cdot q^{2} \\cdot m_{e}}{\\hbar^{2}}=-\\frac{m_{e}}{2} \\cdot\\left(\\frac{Z \\cdot q^{2}}{\\hbar}\\right)^{2}=-\\frac{q^{2} \\cdot Z^{2}}{2 \\cdot r_{B}}=-E_{R} \\cdot Z^{2} \\\\\nE\\left[A^{(Z-1)+}\\right]=-E_{R} \\cdot Z^{2}\n\\end{gathered}\n$$']",['$-E_{R} \\cdot Z^{2}$'],False,,Expression, 1224,Modern Physics,"Recombination of Positive and Negative Ions in Ionized Gas Introduction A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $\mathrm{A}^{(\mathrm{Z}-1)+}$ Constants: | electric field constant | $\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}$ | | :--- | :--- | | elementary charge | $\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}$ | | | $\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}$ | | Planck's constant | $\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ | | (rest) mass of an electron | $\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}$ | | Bohr's atomic radius | $\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}$ | | Rydberg's energy | $\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}$ | | (rest) mass of a proton | $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}$ | Context question: 3.1 Assume that the ion which has just one electron left the shell. $\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state. In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$. Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form. Context answer: $\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$ Context question: 3.2 The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon. Write down an equation which is to be used for calculation the frequency of an emitted photon. Context answer: 3.2 $\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture $\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing $\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing $\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon conservation of energy: $\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$ where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively. conservation of momentum: $m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$ where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon. Context question: 3.3 Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles: 3.3.A The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$. (ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem). 3.3.B In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$ Context answer: \boxed{$-E_{R} \cdot Z^{2}$} ","3.4 Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$. Given the average distance of each of the two electrons in the outermost shell (same as $r_{0}$ given in 3.3) denoted by $r_{1}$ and $r_{2}$, assume the average distance between the two electrons is given by $r_{1}+r_{2}$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie. $\mathrm{p}_{1}^{2} \cdot \mathrm{r}_{1}^{2} \approx \hbar^{2}$ and $\mathrm{p}_{2}^{2} \cdot \mathrm{r}_{2}^{2} \approx \hbar^{2}$ hint: Make use of the information that in the ground state $r_{1}=r_{2}$",['3.4 \n\nIn the case of $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ ion captures a second electron\n\npotential energy of both electrons $=-2 \\cdot \\frac{Z \\cdot q^{2}}{r_{0}}$\n\nkinetic energy of the two electrons $=2 \\cdot \\frac{p^{2}}{2 \\cdot m}=\\frac{\\hbar^{2}}{m_{e} \\cdot r_{0}^{2}}$\n\npotential energy due to interaction between the two electrons $=\\frac{q^{2}}{\\left|\\vec{r}_{1}-\\vec{r}_{2}\\right|}=\\frac{q^{2}}{2 \\cdot r_{0}}$\n\n$E\\left[A^{(z-2)+}\\right]=\\frac{\\hbar^{2}}{m_{e} \\cdot r_{0}^{2}}-\\frac{2 \\cdot Z \\cdot q^{2}}{r_{0}^{2}}+\\frac{q^{2}}{2 \\cdot r_{0}}$\n\ntotal energy is lowest when $\\frac{\\mathrm{dE}}{\\mathrm{dr}}=0$\n\nhence\n\n$0=-\\frac{2 \\cdot \\hbar^{2}}{m_{e} \\cdot r_{0}^{3}}+\\frac{2 \\cdot Z \\cdot q^{2}}{r_{0}^{3}}-\\frac{q^{2}}{2 \\cdot r_{0}^{2}}$\n\nhence\n\n$\\frac{1}{r_{0}}=\\frac{q^{2} \\cdot m_{e}}{2 \\cdot \\hbar^{2}} \\cdot\\left(2 \\cdot z-\\frac{1}{2}\\right)=\\frac{1}{r_{B}} \\cdot\\left(z-\\frac{1}{4}\\right)$\n\n$E\\left[A^{(Z-2)+}\\right]=\\frac{\\hbar^{2}}{m_{e}} \\cdot\\left(\\frac{q^{2} \\cdot m_{e}}{2 \\cdot \\hbar^{2}}\\right)^{2}-\\frac{q^{2} \\cdot\\left(2 \\cdot Z-\\frac{1}{2}\\right)}{\\hbar} \\cdot \\frac{q^{2} \\cdot m_{e} \\cdot\\left(2 \\cdot Z-\\frac{1}{2}\\right)}{2 \\cdot \\hbar}$\n\n$E\\left[A^{(Z-2)+}\\right]=-\\frac{m_{e}}{4} \\cdot\\left[\\frac{q^{2} \\cdot\\left(2 \\cdot Z-\\frac{1}{2}\\right)}{\\hbar}\\right]^{2}=-\\frac{m_{e} \\cdot\\left[q^{2} \\cdot\\left(Z-\\frac{1}{4}\\right)\\right]^{2}}{\\hbar^{2}}=-\\frac{q^{2} \\cdot\\left(Z-\\frac{1}{4}\\right)^{2}}{\\hbar^{2}}$\n\nthis gives\n\n$$\nE\\left[A^{(z-2)+}\\right]=-2 \\cdot E_{R} \\cdot\\left(z-\\frac{1}{4}\\right)^{2}\n$$'],['$-2 \\cdot E_{R} \\cdot(z-\\frac{1}{4})^{2}$'],False,,Expression, 1225,Modern Physics,"Recombination of Positive and Negative Ions in Ionized Gas Introduction A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $\mathrm{A}^{(\mathrm{Z}-1)+}$ Constants: | electric field constant | $\varepsilon_{\mathrm{0}}=8.85 \cdot 10^{-12} \frac{\mathrm{A} \cdot \mathrm{s}}{\mathrm{V} \cdot \mathrm{m}}$ | | :--- | :--- | | elementary charge | $\mathrm{e}= \pm 1.602 \cdot 10^{-19} \mathrm{~A} \cdot \mathrm{s}$ | | | $\mathrm{q}^{2}=\frac{\mathrm{e}^{2}}{4 \cdot \pi \cdot \varepsilon_{\mathrm{0}}}=2.037 \cdot 10^{-28} \mathrm{~J} \cdot \mathrm{m}$ | | Planck's constant | $\hbar=1.054 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ | | (rest) mass of an electron | $\mathrm{m}_{\mathrm{e}}=9.108 \cdot 10^{-31} \mathrm{~kg}$ | | Bohr's atomic radius | $\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m} \cdot \mathrm{q}^{2}}=5.92 \cdot 10^{-11} \mathrm{~m}$ | | Rydberg's energy | $\mathrm{E}_{\mathrm{R}}=\frac{\mathrm{q}^{2}}{2 \cdot \mathrm{r}_{\mathrm{B}}}=2.180 \cdot 10^{-18} \mathrm{~J}$ | | (rest) mass of a proton | $\mathrm{m}_{\mathrm{P}} \cdot \mathrm{c}^{2}=1.503 \cdot 10^{-10} \mathrm{~J}$ | Context question: 3.1 Assume that the ion which has just one electron left the shell. $\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state. In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$. Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form. Context answer: $\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$ Context question: 3.2 The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon. Write down an equation which is to be used for calculation the frequency of an emitted photon. Context answer: 3.2 $\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture $\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing $\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing $\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon conservation of energy: $\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$ where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively. conservation of momentum: $m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$ where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon. Context question: 3.3 Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles: 3.3.A The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$. (ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem). 3.3.B In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$ Context answer: \boxed{$-E_{R} \cdot Z^{2}$} Context question: 3.4 Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$. Given the average distance of each of the two electrons in the outermost shell (same as $r_{0}$ given in 3.3) denoted by $r_{1}$ and $r_{2}$, assume the average distance between the two electrons is given by $r_{1}+r_{2}$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie. $\mathrm{p}_{1}^{2} \cdot \mathrm{r}_{1}^{2} \approx \hbar^{2}$ and $\mathrm{p}_{2}^{2} \cdot \mathrm{r}_{2}^{2} \approx \hbar^{2}$ hint: Make use of the information that in the ground state $r_{1}=r_{2}$ Context answer: \boxed{$-2 \cdot E_{R} \cdot(z-\frac{1}{4})^{2}$}","3.5 Consider in particular the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ is at rest in the ground state when capturing an additional electron and the captured electron is also at rest prior to the capturing. Determine the numerical value of $Z$, if the frequency of the emitted photon accompanying electron capturing is $2,057\times 10^{17} \mathrm{rad} / \mathrm{s}$. Identify the element which gives rise to the ion.","['3.5 \n\nThe ion $\\mathrm{A}^{(\\mathrm{Z}-1)+}$ is at rest when it captures the second electron also at rest before capturing.\n\nFrom the information provided in the problem, the frequency of the photon emitted is given by\n\n$v=\\frac{\\omega}{2 \\cdot \\pi}=\\frac{2,057 \\cdot 10^{17}}{2 \\cdot \\pi} \\mathrm{Hz}$\n\nThe energy equation can be simplified to $\\quad E\\left[A^{(Z-1)+}\\right]-E\\left[A^{(Z-2)+}\\right]=\\hbar \\cdot \\omega=h \\cdot v$ that is\n\n$-E_{R} \\cdot Z^{2}-\\left[-2 \\cdot E_{R} \\cdot\\left(Z-\\frac{1}{4}\\right)^{2}\\right]=\\hbar \\cdot \\omega$\n\nputting in known numbers follows\n\n$2,180 \\cdot 10^{-18} \\cdot\\left[-Z^{2}+2 \\cdot\\left(Z-\\frac{1}{4}\\right)^{2}\\right]=1,05 \\cdot 10^{-34} \\cdot 2,607 \\cdot 10^{17}$\n\nthis gives\n\n$Z^{2}-Z-12,7=0$\n\nwith the physical sensuous result $\\quad Z=\\frac{1+\\sqrt{1+51}}{2}=4,1$\n\nThis implies $\\mathrm{Z}=4$, and that means Beryllium\n\n']","['Z=4, Beryllium']",True,,Numerical, 1226,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun.","1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit.",['$L_{1}=I_{E} \\omega_{E 1}+I_{M 1} \\omega_{M 1}$'],['$L_{1}=I_{E} \\omega_{E 1}+I_{M 1} \\omega_{M 1}$'],False,,Expression, 1227,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear.","1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon.",['$L_{2}=I_{E} \\omega_{2}+I_{M 2} \\omega_{2}$'],['$L_{2}=I_{E} \\omega_{2}+I_{M 2} \\omega_{2}$'],False,,Expression, 1228,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$}","1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem.",['$I_{E} \\omega_{E 1}+I_{M 1} \\omega_{M 1}=I_{M 2} \\omega_{2}=L_{1}$\n\n'],['$I_{E} \\omega_{E 1}+I_{M 1} \\omega_{M 1}=I_{M 2} \\omega_{2}=L_{1}$'],False,,Equation, 1229,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum.","2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant.",['$\\omega_{2}^{2} D_{2}^{3}=G M_{E}$'],['$\\omega_{2}^{2} D_{2}^{3}=G M_{E}$'],False,,Equation, 1230,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} ","2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$.",['$D_{2}=\\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$'],['$D_{2}=\\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$'],False,,Equation, 1231,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} ","2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$.",['$\\omega_{2}=\\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$'],['$\\omega_{2}=\\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$'],False,,Equation, 1232,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth.","2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3).",['The moment of inertia of the Earth will be the addition of the moment of inertia of a sphere with radius $r_{o}$ and density $\\rho_{o}$ and of a sphere with radius $r_{i}$ and density $\\rho_{i}-\\rho_{o}$ :\n\n$I_{E}=\\frac{2}{5} \\frac{4 \\pi}{3}\\left[r_{o}^{5} \\rho_{o}+r_{i}^{5}\\left(\\rho_{i}-\\rho_{o}\\right)\\right]$.'],['$I_{E}=\\frac{2}{5} \\frac{4 \\pi}{3}\\left[r_{o}^{5} \\rho_{o}+r_{i}^{5}\\left(\\rho_{i}-\\rho_{o}\\right)\\right]$'],False,,Equation, 1233,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits.","2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$.",['$\\quad I_{E}=\\frac{2}{5} \\frac{4 \\pi}{3}\\left[r_{o}^{5} \\rho_{o}+r_{i}^{5}\\left(\\rho_{i}-\\rho_{o}\\right)\\right]=8.0 \\times 10^{37} \\mathrm{~kg} \\mathrm{~m}^{2}$'],['$8.0 \\times 10^{37} $'],False,$\mathrm{~kg} \mathrm{~m}^{2}$,Numerical,1e36 1234,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$.","2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$.",['$\\quad L_{1}=I_{E} \\omega_{E 1}+I_{M 1} \\omega_{M 1}=3.4 \\times 10^{34} \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-1}$'],['$3.4 \\times 10^{34}$'],False,$\mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$,Numerical,1e33 1235,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} ",2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$.,"['$D_{2}=5.4 \\times 10^{8} \\mathrm{~m}$, that is $D_{2}=1.4 D_{1}$']",['$5.4 \\times 10^{8}$'],False,m,Numerical,1e7 1236,Mechanics,,"2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days.","['$\\omega_{2}=1.6 \\times 10^{-6} \\mathrm{~s}^{-1}$, that is, a period of 46 days']","['$\\omega_{2}=1.6 \\times 10^{-6} \\mathrm{~s}^{-1}$, that is, a period of 46 days']",True,,Need_human_evaluate, 1237,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity.",2i Find the ratio of the final angular momentum of the Earth to that of the
Moon.,"['Since $I_{E} \\omega_{2}=1.3 \\times 10^{32} \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-1}$ and $I_{M 2} \\omega_{2}=3.4 \\times 10^{34} \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-1}$, the
approximation is justified since the final angular momentum of the Earth
is $1 / 260$ of that of the Moon.']",['$\\frac{1}{260}$'],False,,Numerical,0 1238,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale.","3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass.","['Using the law of cosines, the magnitude of the force produced by the mass
$m$ closest to the Moon will be:
$F_{c}=\\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \\cos (\\theta)}$']",['$F_{c}=\\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \\cos (\\theta)}$'],False,,Expression, 1239,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} ","3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass.","['Using the law of cosines, the magnitude of the force produced by the mass
$m$ farthest to the Moon will be:
$F_{f}=\\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \\cos (\\theta)}$']",['$F_{f}=\\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \\cos (\\theta)}$'],False,,Expression, 1240,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses.","3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass.","['Using the law of sines, the torque will be $\\tau_{c}=F_{c} \\frac{\\sin (\\theta) r_{0} D_{1}}{\\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \\cos (\\theta)\\right]^{1 / 2}}=\\frac{G m M_{M} \\sin (\\theta) r_{0} D_{1}}{\\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \\cos (\\theta)\\right]^{3 / 2}}$']",['$\\tau_{c}=\\frac{G m M_{M} \\sin (\\theta) r_{0} D_{1}}{\\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \\cos (\\theta)\\right]^{3 / 2}}$'],False,,Expression, 1241,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} ","3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass.","['Using the law of sines, the torque will be
$\\tau_{f}=F_{f} \\frac{\\sin (\\theta) r_{0} D_{1}}{\\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \\cos (\\theta)\\right]^{1 / 2}}=\\frac{G m M_{M} \\sin (\\theta) r_{0} D_{1}}{\\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \\cos (\\theta)\\right]^{3 / 2}}$']",['$\\tau_{f}=\\frac{G m M_{M} \\sin (\\theta) r_{0} D_{1}}{\\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \\cos (\\theta)\\right]^{3 / 2}}$'],False,,Expression, 1242,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} ","3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$.",['$$\n\\begin{aligned}\n & \\tau_{c}-\\tau_{f}=G m M_{M} \\sin (\\theta) r_{0} D_{1}^{-2}\\left(1-\\frac{3 r_{o}^{2}}{2 D_{1}^{2}}+\\frac{3 r_{o} \\cos (\\theta)}{D_{1}}-1+\\frac{3 r_{o}^{2}}{2 D_{1}^{2}}+\\frac{3 r_{o} \\cos (\\theta)}{D_{1}}\\right) \\\\\n& =\\frac{6 G m M_{M} r_{o}^{2} \\sin (\\theta) \\cos (\\theta)}{D_{1}^{3}}\n\\end{aligned}\n$$'],['$\\frac{6 G m M_{M} r_{o}^{2} \\sin (\\theta) \\cos (\\theta)}{D_{1}^{3}}$'],False,,Expression, 1243,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$. Context answer: \boxed{$\frac{6 G m M_{M} r_{o}^{2} \sin (\theta) \cos (\theta)}{D_{1}^{3}}$} ","3f Calculate the numerical value of the total torque $\tau$, taking into account 0.5 that $\theta=3^{\circ}$ and that $m=3.6 \times 10^{16} \mathrm{~kg}$ (note that this mass is of the order of $10^{-8}$ times the mass of the Earth).",['$\\tau=\\frac{6 G m M_{M} r_{o}^{2} \\sin (\\theta) \\cos (\\theta)}{D_{1}^{3}}=4.1 \\times 10^{16} \\mathrm{~N} \\mathrm{~m}$'],['$4.1 \\times 10^{16} $'],False,$\mathrm{~N} \mathrm{~m}$,Numerical,1e15 1244,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$. Context answer: \boxed{$\frac{6 G m M_{M} r_{o}^{2} \sin (\theta) \cos (\theta)}{D_{1}^{3}}$} Context question: 3f Calculate the numerical value of the total torque $\tau$, taking into account 0.5 that $\theta=3^{\circ}$ and that $m=3.6 \times 10^{16} \mathrm{~kg}$ (note that this mass is of the order of $10^{-8}$ times the mass of the Earth). Context answer: \boxed{$4.1 \times 10^{16} $} Extra Supplementary Reading Materials: Since the torque is the rate of change of angular momentum with time, find the increase in the distance Earth-Moon at present, per year. For this step, express the angular momentum of the Moon in terms of $M_{M}, M_{E}, D_{1}$ and $G$ only.","3g Find the increase in the distance Earth-Moon at present, per year.","['Since $\\omega^2_{M1}D^3_1=GM_E$, we have that the angular momentum of the Moon is\n\n$$\nI_{M1}\\omega_{M1}=M_MD^2_1\\left[\\frac{GM_E}{D^3_1} \\right]^{1/2}=M_M[D_1GM_E]^{1/2}\n$$\n\nThe torque will be:\n\n$$\n\\tau=\\frac{M_M[GM_E]^{1/2}\\Delta(D_1^{1/2})}{\\Delta t}=\\frac{M_M[GM_E]^{1/2}\\Delta D_1}{2[D_1]^{1/2}\\Delta t}\n$$\n\nSo, we have that\n\n$$\n\\Delta D_1=\\frac{2\\tau \\Delta t}{M_M}\\left[\\frac{D_1}{GM_E} \\right]^{1/2}\n$$\n\nThat for $\\Delta t=3.1\\times 10^7 \\mathrm{~s}=$ 1 year, gives $\\Delta D_1=0.034m$.\nThis is the yearly increase in the Earth-Moon distance.']",['0.034'],False,m,Numerical,1e-3 1245,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$. Context answer: \boxed{$\frac{6 G m M_{M} r_{o}^{2} \sin (\theta) \cos (\theta)}{D_{1}^{3}}$} Context question: 3f Calculate the numerical value of the total torque $\tau$, taking into account 0.5 that $\theta=3^{\circ}$ and that $m=3.6 \times 10^{16} \mathrm{~kg}$ (note that this mass is of the order of $10^{-8}$ times the mass of the Earth). Context answer: \boxed{$4.1 \times 10^{16} $} Extra Supplementary Reading Materials: Since the torque is the rate of change of angular momentum with time, find the increase in the distance Earth-Moon at present, per year. For this step, express the angular momentum of the Moon in terms of $M_{M}, M_{E}, D_{1}$ and $G$ only. Context question: 3g Find the increase in the distance Earth-Moon at present, per year. Context answer: \boxed{0.034} Extra Supplementary Reading Materials: Finally, estimate how much the length of the day is increasing each year.",3h Find the decrease of $\omega_{E1}$ per year and how much is the length of the day at present increasing each year.,"['We now use that\n\n$$\n\\tau = -\\frac{I_E\\Delta \\omega_{E1}}{\\Delta t}\n$$\n\nfrom where we get\n\n$$\n\\Delta \\omega_{E1}=-\\frac{\\tau \\Delta t}{I_E}\n$$\n\nthat for $\\Delta t=3.1\\times 10^7 s$ = 1 year gives\n$\\Delta \\omega_{E1}=-1.6\\times 10^{-14}s^{-1}$\n\nIf $P_E$ is the period of time considered, we have that:\n\n$\\frac{\\Delta P_E}{P_E}=-\\frac{\\Delta \\omega_{E1}}{\\omega_E}$\n\nsince $P_E=1\\text{ day}=8.64\\times 10^4s$, we get\n$\\Delta P_E=1.9\\times 10^{-5}s$.\n\nThis is the amount of time that the day lengthens in a year.']","['$\\Delta \\omega_{E1}=-\\frac{\\tau \\Delta t}{I_E}$ , $\\Delta P_E=1.9\\times 10^{-5}s$']",True,",s","Expression,Numerical",",1e-6" 1246,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$. Context answer: \boxed{$\frac{6 G m M_{M} r_{o}^{2} \sin (\theta) \cos (\theta)}{D_{1}^{3}}$} Context question: 3f Calculate the numerical value of the total torque $\tau$, taking into account 0.5 that $\theta=3^{\circ}$ and that $m=3.6 \times 10^{16} \mathrm{~kg}$ (note that this mass is of the order of $10^{-8}$ times the mass of the Earth). Context answer: \boxed{$4.1 \times 10^{16} $} Extra Supplementary Reading Materials: Since the torque is the rate of change of angular momentum with time, find the increase in the distance Earth-Moon at present, per year. For this step, express the angular momentum of the Moon in terms of $M_{M}, M_{E}, D_{1}$ and $G$ only. Context question: 3g Find the increase in the distance Earth-Moon at present, per year. Context answer: \boxed{0.034} Extra Supplementary Reading Materials: Finally, estimate how much the length of the day is increasing each year. Context question: 3h Find the decrease of $\omega_{E1}$ per year and how much is the length of the day at present increasing each year. Context answer: \boxed{$\Delta \omega_{E1}=-\frac{\tau \Delta t}{I_E}$ , $\Delta P_E=1.9\times 10^{-5}s$} Extra Supplementary Reading Materials: 4. Where is the energy going? In contrast to the angular momentum, that is conserved, the total (rotational plus gravitational) energy of the system is not. We will look into this in this last section.","4a Write down an equation for the total (rotational plus gravitational) energy of the Earth-Moon system at present, $E$. Put this equation in terms of $I_{E}$, $u_{E 1}, M_{M}, M_{E}, D_{1}$ and $G$ only.","['The present total (rotational plus gravitational) energy of the system is: $E=\\frac{1}{2} I_{E} \\omega_{E 1}^{2}+\\frac{1}{2} I_{M} \\omega_{M 1}^{2}-\\frac{G M_{E} M_{M}}{D_{1}}$.\n\nUsing that\n\n$\\omega_{M 1}^{2} D_{1}^{3}=G M_{E}$, we get\n\n\n\n$E=\\frac{1}{2} I_{E} \\omega_{E 1}^{2}-\\frac{1}{2} \\frac{G M_{E} M_{M}}{D_{1}}$']",['$E=\\frac{1}{2} I_{E} \\omega_{E 1}^{2}-\\frac{1}{2} \\frac{G M_{E} M_{M}}{D_{1}}$'],False,,Equation, 1247,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$. Context answer: \boxed{$\frac{6 G m M_{M} r_{o}^{2} \sin (\theta) \cos (\theta)}{D_{1}^{3}}$} Context question: 3f Calculate the numerical value of the total torque $\tau$, taking into account 0.5 that $\theta=3^{\circ}$ and that $m=3.6 \times 10^{16} \mathrm{~kg}$ (note that this mass is of the order of $10^{-8}$ times the mass of the Earth). Context answer: \boxed{$4.1 \times 10^{16} $} Extra Supplementary Reading Materials: Since the torque is the rate of change of angular momentum with time, find the increase in the distance Earth-Moon at present, per year. For this step, express the angular momentum of the Moon in terms of $M_{M}, M_{E}, D_{1}$ and $G$ only. Context question: 3g Find the increase in the distance Earth-Moon at present, per year. Context answer: \boxed{0.034} Extra Supplementary Reading Materials: Finally, estimate how much the length of the day is increasing each year. Context question: 3h Find the decrease of $\omega_{E1}$ per year and how much is the length of the day at present increasing each year. Context answer: \boxed{$\Delta \omega_{E1}=-\frac{\tau \Delta t}{I_E}$ , $\Delta P_E=1.9\times 10^{-5}s$} Extra Supplementary Reading Materials: 4. Where is the energy going? In contrast to the angular momentum, that is conserved, the total (rotational plus gravitational) energy of the system is not. We will look into this in this last section. Context question: 4a Write down an equation for the total (rotational plus gravitational) energy of the Earth-Moon system at present, $E$. Put this equation in terms of $I_{E}$, $u_{E 1}, M_{M}, M_{E}, D_{1}$ and $G$ only. Context answer: \boxed{$E=\frac{1}{2} I_{E} \omega_{E 1}^{2}-\frac{1}{2} \frac{G M_{E} M_{M}}{D_{1}}$} ","4b Write down an equation for the change in $E, \Delta E$, as a function of the changes in $D_{1}$ and in $\boldsymbol{a}_{E 1}$. Evaluate the numerical value of $\Delta E$ for a year, using the values of changes in $D_{1}$ and in $a_{E 1}$ found in questions $3 \mathrm{~g}$ and $3 h$.","['$\\Delta E=I_{E} \\omega_{E 1} \\Delta \\omega_{E 1}+\\frac{1}{2} \\frac{G M_{E} M_{M}}{D_{1}^{2}} \\Delta D_{1}$, that gives\n\n$\\Delta E=-9.0 \\times 10^{19} \\mathrm{~J}$']","['$\\Delta E=I_{E} \\omega_{E 1} \\Delta \\omega_{E 1}+\\frac{1}{2} \\frac{G M_{E} M_{M}}{D_{1}^{2}} \\Delta D_{1}$ , $-9.0 \\times 10^{19}$']",True,",J","Equation,Numerical",",1e18" 1248,Mechanics,"EVOLUTION OF THE EARTH-MOON SYSTEM Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1). Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon. With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon. Figure 2. The Moon's gravity produces tidal deformations or ""bulges"" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale. 1. Conservation of Angular Momentum. Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun. Context question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit. Context answer: \boxed{$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$} Extra Supplementary Reading Materials: This process of transfer of angular momentum will end when the period of rotation of the Earth and the period of revolution of the Moon around the Earth have the same duration. At this point the tidal bulges produced by the Moon on the Earth will be aligned with the line between the Moon and the Earth and the torque will disappear. Context question: 1b Write down the equation for the final total angular momentum $L_{2}$ of the Earth-Moon system. Make the same assumptions as in Question 1a. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{2}$, the final angular frequency of the Earth's rotation and Moon's translation; and $I_{M 2}$, the final moment of inertia of the Moon. Context answer: \boxed{$L_{2}=I_{E} \omega_{2}+I_{M 2} \omega_{2}$} Context question: 1c Neglecting the contribution of the Earth's rotation to the final total angular momentum, write down the equation that expresses the angular momentum conservation for this problem. Context answer: $I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}=I_{M 2} \omega_{2}=L_{1}$ Extra Supplementary Reading Materials: 2. Final Separation and Final Angular Frequency of the Earth-Moon System. Assume that the gravitational equation for a circular orbit (of the Moon around the Earth) is always valid. Neglect the contribution of the Earth's rotation to the final total angular momentum. Context question: 2a Write down the gravitational equation for the circular orbit of the Moon around the Earth, at the final state, in terms of $M_{E}, a_{2}, G$ and the final separation $D_{2}$ between the Earth and the Moon. $M_{E}$ is the mass of the Earth and $G$ is the gravitational constant. Context answer: \boxed{$\omega_{2}^{2} D_{2}^{3}=G M_{E}$} Context question: 2b Write down the equation for the final separation $D_{2}$ between the Earth and the Moon in terms of the known parameters, $L_{1}$, the total angular momentum of the system, $M_{E}$ and $M_{M}$, the masses of the Earth and Moon, respectively, and $G$. Context answer: \boxed{$D_{2}=\frac{L_{1}^{2}}{G M_{E} M_{M}^{2}}$} Context question: 2c Write down the equation for the final angular frequency $a_{2}$ of the Earth-Moon system in terms of the known parameters $L_{1}, M_{E}, M_{M}$ and $G$. Context answer: \boxed{$\omega_{2}=\frac{G^{2} M_{E}^{2} M_{M}^{3}}{L_{1}^{3}}$} Extra Supplementary Reading Materials: Below you will be asked to find the numerical values of $D_{2}$ and $c_{2}$. For this you need to know the moment of inertia of the Earth. Context question: 2d Write down the equation for the moment of inertia of the Earth $I_{E}$ assuming it is a sphere with inner density $\rho_{i}$ from the center to a radius $r_{i}$, and with outer density $\rho_{o}$ from the radius $r_{i}$ to the surface at a radius $r_{o}$ (see Figure 3). Context answer: \boxed{$I_{E}=\frac{2}{5} \frac{4 \pi}{3}\left[r_{o}^{5} \rho_{o}+r_{i}^{5}\left(\rho_{i}-\rho_{o}\right)\right]$} Extra Supplementary Reading Materials: Figure 3. The Earth as a sphere with two densities, $\rho_{i}$ and $\rho_{o}$. Determine the numerical values requested in this problem always to two significant digits. Context question: 2e Evaluate the moment of inertia of the Earth $I_{E}$, using $\rho_{i}=1.3 \times 0^{4} \mathrm{~kg} \mathrm{~m}^{-3}, \quad 0.2$ $r_{i}=3.5 \times 0^{6} \mathrm{~m}, \rho_{o}=4.0 \times 0^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and $r_{o}=6.4 \times 0^{6} \mathrm{~m}$. Context answer: \boxed{$8.0 \times 10^{37} $} Extra Supplementary Reading Materials: The masses of the Earth and Moon are $M_{E}=6.0 \times 0^{24} \mathrm{~kg}$ and $M_{M}=7.3 \times 0^{22} \mathrm{~kg}$, respectively. The present separation between the Earth and the Moon is $D_{1}=3.8 \times 0^{8} \mathrm{~m}$. The present angular frequency of the Earth's rotation is $\omega_{E 1}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$. The present angular frequency of the Moon's translation around the Earth is $\omega_{M 1}=2.7 \times 0^{-6} \mathrm{~s}^{-1}$, and the gravitational constant is $G=6.7 \times 0^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$. Context question: 2f Evaluate the numerical value of the total angular momentum of the system, $L_{1}$. Context answer: \boxed{$3.4 \times 10^{34}$} Context question: 2g Find the final separation $D_{2}$ in meters and in units of the present separation $D_{1}$. Context answer: \boxed{$5.4 \times 10^{8}$} Context question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days. Context answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days Extra Supplementary Reading Materials: Verify that the assumption of neglecting the contribution of the Earth's rotation to the final total angular momentum is justified by finding the ratio of the final angular momentum of the Earth to that of the Moon. This should be a small quantity. Context question: 2i Find the ratio of the final angular momentum of the Earth to that of the
Moon. Context answer: \boxed{$\frac{1}{260}$} Extra Supplementary Reading Materials: 3. How much is the Moon receding per year? Now, you will find how much the Moon is receding from the Earth each year. For this, you will need to know the equation for the torque acting at present on the Moon. Assume that the tidal bulges can be approximated by two point masses, each of mass $m$, located on the surface of the Earth, see Fig. 4. Let $\theta$ be the angle between the line that goes through the bulges and the line that joins the centers of the Earth and the Moon. Figure 4. Schematic diagram to estimate the torque produced on the Moon by the bulges on the Earth. The drawing is not to scale. Context question: 3a Find $F_{c}$, the magnitude of the force produced on the Moon by the closest point mass. Context answer: \boxed{$F_{c}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)}$} Context question: 3b Find $F_{f}$, the magnitude of the force produced on the Moon by the farthest point mass. Context answer: \boxed{$F_{f}=\frac{G m M_{M}}{D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)}$} Extra Supplementary Reading Materials: You may now evaluate the torques produced by the point masses. Context question: 3c Find the magnitude of $\boldsymbol{\tau}_{c}$, the torque produced by the closest point mass. Context answer: \boxed{$\tau_{c}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}-2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3d Find the magnitude of $\tau_{f}$, the torque produced by the farthest point mass. Context answer: \boxed{$\tau_{f}=\frac{G m M_{M} \sin (\theta) r_{0} D_{1}}{\left[D_{1}^{2}+r_{o}^{2}+2 D_{1} r_{o} \cos (\theta)\right]^{3 / 2}}$} Context question: 3e Find the magnitude of the total torque $\tau$ produced by the two masses. Since $r_{o} \ll D_{1}$ you should approximate your expression to lowest significant order in $r_{o} / D_{1}$. You may use that $(1+x)^{a} \approx 1+a x$, if $x \ll 1$. Context answer: \boxed{$\frac{6 G m M_{M} r_{o}^{2} \sin (\theta) \cos (\theta)}{D_{1}^{3}}$} Context question: 3f Calculate the numerical value of the total torque $\tau$, taking into account 0.5 that $\theta=3^{\circ}$ and that $m=3.6 \times 10^{16} \mathrm{~kg}$ (note that this mass is of the order of $10^{-8}$ times the mass of the Earth). Context answer: \boxed{$4.1 \times 10^{16} $} Extra Supplementary Reading Materials: Since the torque is the rate of change of angular momentum with time, find the increase in the distance Earth-Moon at present, per year. For this step, express the angular momentum of the Moon in terms of $M_{M}, M_{E}, D_{1}$ and $G$ only. Context question: 3g Find the increase in the distance Earth-Moon at present, per year. Context answer: \boxed{0.034} Extra Supplementary Reading Materials: Finally, estimate how much the length of the day is increasing each year. Context question: 3h Find the decrease of $\omega_{E1}$ per year and how much is the length of the day at present increasing each year. Context answer: \boxed{$\Delta \omega_{E1}=-\frac{\tau \Delta t}{I_E}$ , $\Delta P_E=1.9\times 10^{-5}s$} Extra Supplementary Reading Materials: 4. Where is the energy going? In contrast to the angular momentum, that is conserved, the total (rotational plus gravitational) energy of the system is not. We will look into this in this last section. Context question: 4a Write down an equation for the total (rotational plus gravitational) energy of the Earth-Moon system at present, $E$. Put this equation in terms of $I_{E}$, $u_{E 1}, M_{M}, M_{E}, D_{1}$ and $G$ only. Context answer: \boxed{$E=\frac{1}{2} I_{E} \omega_{E 1}^{2}-\frac{1}{2} \frac{G M_{E} M_{M}}{D_{1}}$} Context question: 4b Write down an equation for the change in $E, \Delta E$, as a function of the changes in $D_{1}$ and in $\boldsymbol{a}_{E 1}$. Evaluate the numerical value of $\Delta E$ for a year, using the values of changes in $D_{1}$ and in $a_{E 1}$ found in questions $3 \mathrm{~g}$ and $3 h$. Context answer: \boxed{$\Delta E=I_{E} \omega_{E 1} \Delta \omega_{E 1}+\frac{1}{2} \frac{G M_{E} M_{M}}{D_{1}^{2}} \Delta D_{1}$ , $-9.0 \times 10^{19}$} Extra Supplementary Reading Materials: Verify that this loss of energy is consistent with an estimate for the energy dissipated as heat in the tides produced by the Moon on the Earth. Assume that the tides rise, on the average by $0.5 \mathrm{~m}$, a layer of water $h=0.5 \mathrm{~m}$ deep that covers the surface of the Earth (for simplicity assume that all the surface of the Earth is covered with water). This happens twice a day. Further assume that $10 \%$ of this gravitational energy is dissipated as heat due to viscosity when the water descends. Take the density of water to be $\rho_{\text {water }}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$, and the gravitational acceleration on the surface of the Earth to be $g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$.",4c What is the mass of this surface layer of water?,['$M_{\\text {water }}=4 \\pi r_{o}^{2} \\times h \\times \\rho_{\\text {water }} \\mathrm{kg}=2.6 \\times 10^{17} \\mathrm{~kg}$'],['$2.6 \\times 10^{17} $'],False,$\mathrm{~kg}$,Numerical,1e16 1250,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption.",1a Write down the resonance condition for the absorption of the photon.,['$\\omega_{0} \\approx \\omega_{L}\\left(1+\\frac{v}{C}\\right)$'],['$\\omega_{L}\\left(1+\\frac{v}{C}\\right)$'],False,,Expression, 1251,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} ","1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory.",['$p_{a t}=p-\\hbar q \\approx m v-\\frac{\\hbar \\omega_{L}}{c}$'],['$m v-\\frac{\\hbar \\omega_{L}}{c}$'],False,,Expression, 1252,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} ","1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory.",['$\\varepsilon_{a t}=\\frac{p_{a t}^{2}}{2 m}+\\hbar \\omega_{0} \\approx \\frac{m v^{2}}{2}+\\hbar \\omega_{L}$'],['$\\frac{m v^{2}}{2}+\\hbar \\omega_{L}$'],False,,Expression, 1253,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction.","2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory.","['First, one calculates the energy of the emitted photon, as seen in the lab reference frame. One must be careful to keep the correct order; this is because the velocity of the atom changes after the absorption, however, this is second order correction for the emitted frequency:\n\n$\\omega_{p h} \\approx \\omega_{0}\\left(1-\\frac{v^{\\prime}}{c}\\right) \\quad$ with $\\quad v^{\\prime} \\approx v-\\frac{\\hbar q}{m}$\n\nthus,\n\n$$\n\\begin{aligned}\n\\omega_{p h} & \\approx \\omega_{0}\\left(1-\\frac{v}{c}+\\frac{\\hbar q}{m c}\\right) \\\\\n& \\approx \\omega_{L}\\left(1+\\frac{v}{c}\\right)\\left(1-\\frac{v}{c}+\\frac{\\hbar q}{m c}\\right) \\\\\n& \\approx \\omega_{L}\\left(1+\\frac{\\hbar q}{m c}\\right) \\\\\n& \\approx \\omega_{L}\\left(1+\\left(\\frac{\\hbar q}{m v}\\right)\\left(\\frac{v}{c}\\right)\\right) \\\\\n& \\approx \\omega_{L}\n\\end{aligned}\n$$']",['$\\hbar \\omega_{L}$'],False,,Expression, 1254,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} ","2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory.",[''],['$-\\hbar \\omega_{L} / c$'],False,,Expression, 1255,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} ","2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory.",['Use conservation of momentum (see 1b):\n\n$p_{a t}+p_{p h} \\approx p-\\hbar q$'],['$m v$'],False,,Expression, 1256,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} ","2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory.",['$\\varepsilon_{a t} \\approx \\frac{p^{2}}{2 m}=\\frac{m v^{2}}{2}$'],['$\\frac{m v^{2}}{2}$'],False,,Expression, 1257,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction.","3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory.",['$\\varepsilon_{p h} \\approx \\hbar \\omega_{0}\\left(1+\\frac{v}{c}\\right) \\approx \\hbar \\omega_{L}\\left(1+\\frac{v}{c}\\right)\\left(1+\\frac{v}{c}\\right) \\approx \\hbar \\omega_{L}\\left(1+2 \\frac{v}{c}\\right)$'],['$\\hbar \\omega_{L}\\left(1+2 \\frac{v}{c}\\right)$'],False,,Expression, 1258,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} ","3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory.",['$p_{p h} \\approx \\frac{\\hbar \\omega_{L}}{c}\\left(1+2 \\frac{v}{c}\\right)$'],['$\\frac{\\hbar \\omega_{L}}{c}\\left(1+2 \\frac{v}{c}\\right)$'],False,,Expression, 1259,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} ","3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory.",['$p_{a t}=p-\\hbar q-p_{p h} \\approx p-\\hbar q-\\frac{\\hbar \\omega_{L}}{c}\\left(1+2 \\frac{v}{C}\\right) \\approx m v-2 \\frac{\\hbar \\omega_{L}}{c}$'],['$m v-2 \\frac{\\hbar \\omega_{L}}{c}$'],False,,Expression, 1260,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} ","3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory.",['$\\varepsilon_{a t}=\\frac{p_{a t}^{2}}{2 m} \\approx \\frac{m v^{2}}{2}\\left(1-2 \\frac{\\hbar q}{m v}\\right)$'],['$\\frac{m v^{2}}{2}\\left(1-2 \\frac{\\hbar q}{m v}\\right)$'],False,,Expression, 1261,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions.","4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process.",['$\\varepsilon_{p h}=\\frac{1}{2} \\varepsilon_{p h}^{+}+\\frac{1}{2} \\varepsilon_{p h}^{-} \\approx \\hbar \\omega_{L}\\left(1+\\frac{v}{c}\\right)$'],['$\\hbar \\omega_{L}\\left(1+\\frac{v}{c}\\right)$'],False,,Expression, 1262,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} ","4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process.",['$p_{p h}=\\frac{1}{2} p_{p h}^{+}+\\frac{1}{2} p_{p h}^{-} \\approx \\frac{\\hbar \\omega_{L}}{c} \\frac{v}{c}=m v\\left(\\frac{\\hbar q}{m v} \\frac{v}{c}\\right) \\approx 0 \\quad$ second order'],"['$m v\\left(\\frac{\\hbar q}{m v} \\frac{v}{c}\\right)$', '$0$']",False,,Expression, 1263,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} ","4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process.",['$\\varepsilon_{a t}=\\frac{1}{2} \\varepsilon_{a t}^{+}+\\frac{1}{2} \\varepsilon_{a t}^{-} \\approx \\frac{m v^{2}}{2}\\left(1-\\frac{\\hbar q}{m v}\\right)$'],['$\\frac{m v^{2}}{2}\\left(1-\\frac{\\hbar q}{m v}\\right)$'],False,,Expression, 1264,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} ","4d Write down the average momentum of the atom $p_{a t}$, after the emission process.",['$p_{a t}=\\frac{1}{2} p_{a t}^{+}+\\frac{1}{2} p_{a t}^{-} \\approx p-\\frac{\\hbar \\omega_{L}}{c}$'],['$p-\\frac{\\hbar \\omega_{L}}{c}$'],False,,Expression, 1265,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom.",5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process.,['$\\Delta \\varepsilon=\\varepsilon_{a t}^{a f t e r}-\\varepsilon_{a t}^{\\text {before }} \\approx-\\frac{1}{2} \\hbar q v=-\\frac{1}{2} \\hbar \\omega_{L} \\frac{v}{C}$'],['$-\\frac{1}{2} \\hbar \\omega_{L} \\frac{v}{C}$'],False,,Expression, 1266,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} ",5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process.,['$\\Delta p=p_{\\text {at }}^{\\text {after }}-p_{\\text {at }}^{\\text {before }} \\approx-\\hbar q=-\\frac{\\hbar \\omega_{L}}{c}$'],['$-\\frac{\\hbar \\omega_{L}}{c}$'],False,,Expression, 1267,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions:",6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process.,['$\\Delta \\varepsilon=\\varepsilon_{a t}^{a f t e r}-\\varepsilon_{a t}^{b e f o r e} \\approx+\\frac{1}{2} \\hbar q v=+\\frac{1}{2} \\hbar \\omega_{L}^{\\prime} \\frac{v}{c}$'],['$+\\frac{1}{2} \\hbar \\omega_{L}^{\\prime} \\frac{v}{c}$'],False,,Expression, 1268,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} ",6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process.,['$\\Delta p=p_{a t}^{a f t e r}-p_{a t}^{b e f o r e} \\approx+\\hbar q=+\\frac{\\hbar \\omega_{L}^{\\prime}}{c}$'],['$+\\frac{\\hbar \\omega_{L}^{\\prime}}{c}$'],False,,Expression, 1269,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers.","7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$.","['On the average, the fraction of atoms found in the excited state is given by,\n\n$$\nP_{e x c}=\\frac{N_{e x c}}{N}=\\frac{\\Omega_{R}^{2}}{\\left(\\omega_{0}-\\omega_{L}\\right)^{2}+\\frac{\\Gamma^{2}}{4}+2 \\Omega_{R}^{2}}\n$$\n\nwhere $\\omega_{0}$ is the resonance frequency of the atoms and $\\Omega_{R}$ is the so-called Rabi frequency; $\\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. The lifetime of the excited energy level of the atom is $\\Gamma^{-1}$.\n\nThe force is calculated as the number of absorption-emission cycles, times the momentum exchange in each event, divided by the time of each event. CAREFUL! One must take into account the Doppler shift of each laser, as seen by the atoms:\n\nWith the information found so far, find the force that the lasers exert on the atomic beam. You must assume that $mv>> \\hbar q$.\n\n$F = N\\Delta p^- P^-_{exc}\\Gamma+N\\Delta p^+ P^+_{exc}\\Gamma$\n\n$$\n=\\left(\\frac{\\Omega^2_R}{(\\omega_0 - \\omega_L + \\omega_L \\frac{v}{c})^2+\\frac{\\Gamma^2}{4}+2\\Omega^2_R} - \\frac{\\Omega^2_R}{(\\omega_0 - \\omega_L - \\omega_L \\frac{v}{c})^2+\\frac{\\Gamma^2}{4}+2\\Omega^2_R} \\right) N\\Gamma\\hbar q\n$$']",['$F=\\left(\\frac{\\Omega^2_R}{(\\omega_0 - \\omega_L + \\omega_L \\frac{v}{c})^2+\\frac{\\Gamma^2}{4}+2\\Omega^2_R} - \\frac{\\Omega^2_R}{(\\omega_0 - \\omega_L - \\omega_L \\frac{v}{c})^2+\\frac{\\Gamma^2}{4}+2\\Omega^2_R} \\right) N\\Gamma\\hbar q$'],False,,Expression, 1270,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$.","8a Find an expression for the force found in Question (7a), in this limit.",['$F \\approx-\\frac{4 N \\hbar q^{2} \\Omega_{R}^{2} \\Gamma}{\\left(\\left(\\omega_{0}-\\omega_{L}\\right)^{2}+\\frac{\\Gamma^{2}}{4}+2 \\Omega_{R}^{2}\\right)^{2}}\\left(\\omega_{0}-\\omega_{L}\\right) v$'],['$-\\frac{4 N \\hbar q^{2} \\Omega_{R}^{2} \\Gamma}{\\left(\\left(\\omega_{0}-\\omega_{L}\\right)^{2}+\\frac{\\Gamma^{2}}{4}+2 \\Omega_{R}^{2}\\right)^{2}}\\left(\\omega_{0}-\\omega_{L}\\right) v$'],False,,Expression, 1271,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$. Context question: 8a Find an expression for the force found in Question (7a), in this limit. Context answer: \boxed{$-\frac{4 N \hbar q^{2} \Omega_{R}^{2} \Gamma}{\left(\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}\right)^{2}}\left(\omega_{0}-\omega_{L}\right) v$} Extra Supplementary Reading Materials: Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation.",8b Write down the condition to obtain a positive force (speeding up the atoms).,['$\\omega_{0}<\\omega_{L}$\n\n'],['$\\omega_{0}<\\omega_{L}$'],False,,Expression, 1272,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$. Context question: 8a Find an expression for the force found in Question (7a), in this limit. Context answer: \boxed{$-\frac{4 N \hbar q^{2} \Omega_{R}^{2} \Gamma}{\left(\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}\right)^{2}}\left(\omega_{0}-\omega_{L}\right) v$} Extra Supplementary Reading Materials: Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation. Context question: 8b Write down the condition to obtain a positive force (speeding up the atoms). Context answer: $\omega_{0}<\omega_{L}$ ",8c Write down the condition to obtain a zero force.,['$\\omega_{0}=\\omega_{L}$'],['$\\omega_{0}=\\omega_{L}$'],False,,Equation, 1273,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$. Context question: 8a Find an expression for the force found in Question (7a), in this limit. Context answer: \boxed{$-\frac{4 N \hbar q^{2} \Omega_{R}^{2} \Gamma}{\left(\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}\right)^{2}}\left(\omega_{0}-\omega_{L}\right) v$} Extra Supplementary Reading Materials: Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation. Context question: 8b Write down the condition to obtain a positive force (speeding up the atoms). Context answer: $\omega_{0}<\omega_{L}$ Context question: 8c Write down the condition to obtain a zero force. Context answer: \boxed{$\omega_{0}=\omega_{L}$}",8d Write down the condition to obtain a negative force (slowing down the atoms).,['$\\omega_{0}>\\omega_{L}$\n\n'],['$\\omega_{0}>\\omega_{L}$'],False,,Expression, 1274,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$. Context question: 8a Find an expression for the force found in Question (7a), in this limit. Context answer: \boxed{$-\frac{4 N \hbar q^{2} \Omega_{R}^{2} \Gamma}{\left(\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}\right)^{2}}\left(\omega_{0}-\omega_{L}\right) v$} Extra Supplementary Reading Materials: Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation. Context question: 8b Write down the condition to obtain a positive force (speeding up the atoms). Context answer: $\omega_{0}<\omega_{L}$ Context question: 8c Write down the condition to obtain a zero force. Context answer: \boxed{$\omega_{0}=\omega_{L}$} Context question: 8d Write down the condition to obtain a negative force (slowing down the atoms). Context answer: $\omega_{0}>\omega_{L}$",8e Consider now that the atoms are moving with a velocity $-v$ (in the $-x$ direction). Write down the condition to obtain a slowing down force on the atoms.,['$\\omega_{0}>\\omega_{L}$\n\n'],['$\\omega_{0}>\\omega_{L}$'],False,,Expression, 1275,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$. Context question: 8a Find an expression for the force found in Question (7a), in this limit. Context answer: \boxed{$-\frac{4 N \hbar q^{2} \Omega_{R}^{2} \Gamma}{\left(\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}\right)^{2}}\left(\omega_{0}-\omega_{L}\right) v$} Extra Supplementary Reading Materials: Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation. Context question: 8b Write down the condition to obtain a positive force (speeding up the atoms). Context answer: $\omega_{0}<\omega_{L}$ Context question: 8c Write down the condition to obtain a zero force. Context answer: \boxed{$\omega_{0}=\omega_{L}$} Context question: 8d Write down the condition to obtain a negative force (slowing down the atoms). Context answer: $\omega_{0}>\omega_{L}$ Context question: 8e Consider now that the atoms are moving with a velocity $-v$ (in the $-x$ direction). Write down the condition to obtain a slowing down force on the atoms. Context answer: $\omega_{0}>\omega_{L}$ Extra Supplementary Reading Materials: 9. Optical molasses. In the case of a negative force, one obtains a frictional dissipative force. Assume that initially, at $t=0$, the gas of atoms has velocity $v_{0}$. This model does not allow you to go to arbitrarily low temperatures.","9a In the limit of low velocities, find the velocity of the atoms after the laser beams have been on for a time $\tau$.",['$F=-\\beta v \\Rightarrow m \\frac{d v}{d t} \\approx-\\beta v \\quad \\beta$ can be read from (8a)
$\\Rightarrow v=v_{0} e^{-\\beta t / m}$'],['$v=v_{0} e^{-\\beta t / m}$'],False,,Expression, 1276,Modern Physics,"THEORETICAL PROBLEM 2 DOPPLER LASER COOLING AND OPTICAL MOLASSES The purpose of this problem is to develop a simple theory to understand the so-called ""laser cooling"" and ""optical molasses"" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997. The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called ""optical molasses"" because the dissipative optical force resembles the viscous drag on a body moving through molasses. In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension. PART I: BASICS OF LASER COOLING Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities. internal energy levels of the atom Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$. Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions: 1. Absorption. Context question: 1a Write down the resonance condition for the absorption of the photon. Context answer: \boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$} Context question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory. Context answer: \boxed{$m v-\frac{\hbar \omega_{L}}{c}$} Context question: 1c Write down the total energy $\varepsilon_{a t}$ of the atom after absorption, as seen in the 0.2 laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}+\hbar \omega_{L}$} Extra Supplementary Reading Materials: 2. Spontaneous emission of a photon in the $-x$ direction. At some time after the absorption of the incident photon, the atom may emit a photon in the $-x$ direction. Context question: 2a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}$} Context question: 2b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$-\hbar \omega_{L} / c$} Context question: 2c Write down the momentum of the atom $p_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$m v$} Context question: 2d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $-x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}$} Extra Supplementary Reading Materials: 3. Spontaneous emission of a photon in the $+x$ direction. At some time after the absorption of the incident photon, the atom may instead emit a photon in the $+x$ direction. Context question: 3a Write down the energy of the emitted photon, $\varepsilon_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\hbar \omega_{L}\left(1+2 \frac{v}{c}\right)$} Context question: 3b Write down the momentum of the emitted photon $p_{p h}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{\hbar \omega_{L}}{c}\left(1+2 \frac{v}{c}\right)$} Context question: 3c Write down the momentum of the atom $p_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$m v-2 \frac{\hbar \omega_{L}}{c}$} Context question: 3d Write down the total energy of the atom $\varepsilon_{a t}$, after the emission process in the $+x$ direction, as seen in the laboratory. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-2 \frac{\hbar q}{m v}\right)$} Extra Supplementary Reading Materials: 4. Average emission after the absorption. The spontaneous emission of a photon in the $-x$ or in the $+x$ directions occurs with the same probability. Taking this into account, answer the following questions. Context question: 4a Write down the average energy of an emitted photon, $\varepsilon_{p h}$, after the
emission process. Context answer: \boxed{$\hbar \omega_{L}\left(1+\frac{v}{c}\right)$} Context question: 4b Write down the average momentum of an emitted photon $p_{p h}$, after the
emission process. Context answer: \boxed{$m v\left(\frac{\hbar q}{m v} \frac{v}{c}\right)$} Context question: 4c Write down the average total energy of the atom $\varepsilon_{a t}$, after the emission process. Context answer: \boxed{$\frac{m v^{2}}{2}\left(1-\frac{\hbar q}{m v}\right)$} Context question: 4d Write down the average momentum of the atom $p_{a t}$, after the emission process. Context answer: \boxed{$p-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 5. Energy and momentum transfer. Assuming a complete one-photon absorption-emission process only, as described above, there is a net average momentum and energy transfer between the laser radiation and the atom. Context question: 5a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete
one-photon absorption-emission process. Context answer: \boxed{$-\frac{1}{2} \hbar \omega_{L} \frac{v}{C}$} Context question: 5b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$-\frac{\hbar \omega_{L}}{c}$} Extra Supplementary Reading Materials: 6. Energy and momentum transfer by a laser beam along the $+x$ direction. Consider now that a laser beam of frequency $\omega_{L}$ is incident on the atom along the $+x$ direction, while the atom moves also in the $+x$ direction with velocity $v$. Assuming a resonance condition between the internal transition of the atom and the laser beam, as seen by the atom, answer the following questions: Context question: 6a Write down the average energy change $\Delta \varepsilon$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{1}{2} \hbar \omega_{L}^{\prime} \frac{v}{c}$} Context question: 6b Write down the average momentum change $\Delta p$ of the atom after a complete one-photon absorption-emission process. Context answer: \boxed{$+\frac{\hbar \omega_{L}^{\prime}}{c}$} Extra Supplementary Reading Materials: PART II: DISSIPATION AND THE FUNDAMENTALS OF OPTICAL MOLASSES Nature, however, imposes an inherent uncertainty in quantum processes. Thus, the fact that the atom can spontaneously emit a photon in a finite time after absorption, gives as a result that the resonance condition does not have to be obeyed exactly as in the discussion above. That is, the frequency of the laser beams $a_{L}$ and $\omega_{L}^{\prime}$ may have any value and the absorption-emission process can still occur. These will happen with different (quantum) probabilities and, as one should expect, the maximum probability is found at the exact resonance condition. On the average, the time elapsed between a single process of absorption and emission is called the lifetime of the excited energy level of the atom and it is denoted by $\Gamma^{-1}$. Consider a collection of $N$ atoms at rest in the laboratory frame of reference, and a laser beam of frequency $a_{L}$ incident on them. The atoms absorb and emit continuously such that there is, on average, $N_{\text {exc }}$ atoms in the excited state (and therefore, $N-N_{\text {exc }}$ atoms in the ground state). A quantum mechanical calculation yields the following result: $$ N_{e x c}=N \frac{\Omega_{R}^{2}}{\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}} $$ where $a_{0}$ is the resonance frequency of the atomic transition and $\Omega_{R}$ is the so-called Rabi frequency; $\Omega_{R}^{2}$ is proportional to the intensity of the laser beam. As mentioned above, you can see that this number is different from zero even if the resonance frequency $a_{0}$ is different from the frequency of the laser beam $a_{L}$. An alternative way of expressing the previous result is that the number of absorption-emission processes per unit of time is $N_{\text {exc }} \Gamma$. Consider the physical situation depicted in Figure 2, in which two counter propagating laser beams with the same but arbitrary frequency $a_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. Figure 2. Two counter propagating laser beams with the same but arbitrary frequency $u_{L}$ are incident on a gas of $N$ atoms that move in the $+x$ direction with velocity $v$. 7. Force on the atomic beam by the lasers. Context question: 7a With the information found so far, find the force that the lasers exert on the atomic beam. You should assume that $m v \gg \hbar q$. Context answer: \boxed{$F=\left(\frac{\Omega^2_R}{(\omega_0 - \omega_L + \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} - \frac{\Omega^2_R}{(\omega_0 - \omega_L - \omega_L \frac{v}{c})^2+\frac{\Gamma^2}{4}+2\Omega^2_R} \right) N\Gamma\hbar q$} Extra Supplementary Reading Materials: 8. Low velocity limit. Assume now that the velocity of the atoms is small enough, such that you can expand the force up to first order in $v$. Context question: 8a Find an expression for the force found in Question (7a), in this limit. Context answer: \boxed{$-\frac{4 N \hbar q^{2} \Omega_{R}^{2} \Gamma}{\left(\left(\omega_{0}-\omega_{L}\right)^{2}+\frac{\Gamma^{2}}{4}+2 \Omega_{R}^{2}\right)^{2}}\left(\omega_{0}-\omega_{L}\right) v$} Extra Supplementary Reading Materials: Using this result, you can find the conditions for speeding up, slowing down, or no effect at all on the atoms by the laser radiation. Context question: 8b Write down the condition to obtain a positive force (speeding up the atoms). Context answer: $\omega_{0}<\omega_{L}$ Context question: 8c Write down the condition to obtain a zero force. Context answer: \boxed{$\omega_{0}=\omega_{L}$} Context question: 8d Write down the condition to obtain a negative force (slowing down the atoms). Context answer: $\omega_{0}>\omega_{L}$ Context question: 8e Consider now that the atoms are moving with a velocity $-v$ (in the $-x$ direction). Write down the condition to obtain a slowing down force on the atoms. Context answer: $\omega_{0}>\omega_{L}$ Extra Supplementary Reading Materials: 9. Optical molasses. In the case of a negative force, one obtains a frictional dissipative force. Assume that initially, at $t=0$, the gas of atoms has velocity $v_{0}$. This model does not allow you to go to arbitrarily low temperatures. Context question: 9a In the limit of low velocities, find the velocity of the atoms after the laser beams have been on for a time $\tau$. Context answer: \boxed{$v=v_{0} e^{-\beta t / m}$} ","9b Assume now that the gas of atoms is in thermal equilibrium at a temperature $T_{0}$. Find the temperature $T$ after the laser beams have been on for a time $\tau$.","['Recalling that $\\frac{1}{2} m v^{2}=\\frac{1}{2} k T$ in 1 dimension, and using $v$ as the average
thermal velocity in the equation of (9a), we can write down
$T=T e^{-2 \\beta t / m}$']",['$T=T e^{-2 \\beta t / m}$'],False,,Expression, 1277,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision.","1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures.","['We equate the initial kinetic energy of the two protons to the electric potential energy at the distance of closest approach:\n\n$2\\left(\\frac{1}{2} m_{p} v_{r m s}^{2}\\right)=\\frac{q^{2}}{4 \\pi \\varepsilon_{0} d_{c}} ;$ and since\n\n$\\frac{3}{2} k T_{c}=\\frac{1}{2} m_{p} v_{r m s}^{2}$, we obtain\n\n$T_{c}=\\frac{q^{2}}{12 \\pi \\varepsilon_{0} d_{c} k}=5.5 \\times 10^{9} \\mathrm{~K}$']",['$5.5 \\times 10^{9}$'],False,K,Numerical,1e8 1278,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas.","2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only.","[""Since we have that
$\\frac{\\Delta P}{\\Delta r}=-\\frac{G M_{r} \\rho_{r}}{r^{2}}$, making the assumptions given above, we obtain that:
$P_{c}=\\frac{G M \\rho_{c}}{R}$. Now, the pressure of an ideal gas is
$P_{c}=\\frac{2 \\rho_{c} k T_{c}}{m_{p}}$, where $k$ is Boltzmann's constant, $T_{c}$ is the central
temperature of the star, and $m_{p}$ is the proton mass. The factor of 2 in the
previous equation appears because we have two particles (one proton and
one electron) per proton mass and that both contribute equally to the
pressure. Equating the two previous equations, we finally obtain that:
$T_{c}=\\frac{G M m_{p}}{2 k R}$""]",['$T_{c}=\\frac{G M m_{p}}{2 k R}$'],False,,Expression, 1279,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity:",2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only.,['From section (2a) we have that:\n\n$\\frac{M}{R}=\\frac{2 k T_{c}}{G m_{p}}$'],['$\\frac{M}{R}=\\frac{2 k T_{c}}{G m_{p}}$'],False,,Expression, 1280,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} ",2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star.,"['$\\quad$ From section (2b) we have that, for $T_{c}=5.5 \\times 10^{9} \\mathrm{~K}$ :\n\n$$\n\\frac{M}{R}=\\frac{2 k T_{c}}{G m_{p}}=1.4 \\times 10^{24} \\mathrm{~kg} \\mathrm{~m}^{-1}\n$$']",['$1.4 \\times 10^{24} $'],False,$\mathrm{~kg} \mathrm{~m}^{-1}$,Numerical,1e23 1281,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} ","2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$.","['For the Sun we have that: $\\frac{M(\\text { Sun })}{R(\\text { Sun })}=2.9 \\times 10^{21} \\mathrm{~kg} \\mathrm{~m}^{-1}$, that is, three orders of magnitude smaller.']",['$2.9 \\times 10^{21}$'],False,$ \mathrm{~kg} \mathrm{~m}^{-1}$,Numerical,1e20 1282,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse.","3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only.","['We have that \n\n$\\lambda_{p}=\\frac{h}{m_{p} v_{r m s}}$, and since\n\n$\\frac{3}{2} k T_{c}=\\frac{1}{2} m_{p} v_{r m s}^{2}$, and\n\n$T_{c}=\\frac{q^{2}}{12 \\pi \\varepsilon_{0} d_{c} k}$, we obtain:\n\n$T_{c}=\\frac{q^{4} m_{p}}{24 \\pi^{2} \\varepsilon_{0}^{2} k h^{2}} \\cdot$']",['$T_{c}=\\frac{q^{4} m_{p}}{24 \\pi^{2} \\varepsilon_{0}^{2} k h^{2}}$'],False,,Expression, 1283,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} ",3b Evaluate numerically the value of $T_{c}$ obtained in (3a).,['$T_{c}=\\frac{q^{4} m_{p}}{24 \\pi^{2} \\varepsilon_{0}^{2} k h^{2}}=9.7 \\times 10^{6} \\mathrm{~K}$.'],['$9.7 \\times 10^{6}$'],False,K,Numerical,1e5 1284,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} ","3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed.","['From section (2b) we have that, for $T_{c}=9.7 \\times 10^{6} \\mathrm{~K}$ :\n\n$\\frac{M}{R}=\\frac{2 k T_{c}}{G m_{p}}=2.4 \\times 10^{21} \\mathrm{~kg} \\mathrm{~m}^{-1}$; while for the Sun we have that:\n\n$\\frac{M(\\text { Sun })}{R(\\text { Sun })}=2.9 \\times 10^{21} \\mathrm{~kg} \\mathrm{~m}^{-1}$.']",['$2.4 \\times 10^{21}$'],False,$ \mathrm{~kg} \mathrm{~m}^{-1}$,Numerical,1e20 1285,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} Context question: 3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed. Context answer: \boxed{$2.4 \times 10^{21}$} Extra Supplementary Reading Materials: Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow this ratio for a large range of masses. 4. The mass/radius ratio of the stars. The previous agreement suggests that the quantum mechanical approach for estimating the temperature at the center of the Sun is correct.","4a Use the previous results to demonstrate that for any star fusing hydrogen, the ratio of mass $M$ to radius $R$ is the same and depends only on physical constants. Find the equation for the ratio $M / R$ for stars fusing hydrogen.","['Taking into account that\n\n$\\frac{M}{R}=\\frac{2 k T_{c}}{G m_{p}}$, and that\n\n$T_{c}=\\frac{q^{4} m_{p}}{24 \\pi^{2} \\varepsilon_{0}^{2} k h^{2}}$, we obtain:\n\n$\\frac{M}{R}=\\frac{q^{4}}{12 \\pi^{2} \\varepsilon_{0}^{2} G h^{2}}$.']",['$\\frac{M}{R}=\\frac{q^{4}}{12 \\pi^{2} \\varepsilon_{0}^{2} G h^{2}}$'],False,,Expression, 1286,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} Context question: 3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed. Context answer: \boxed{$2.4 \times 10^{21}$} Extra Supplementary Reading Materials: Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow this ratio for a large range of masses. 4. The mass/radius ratio of the stars. The previous agreement suggests that the quantum mechanical approach for estimating the temperature at the center of the Sun is correct. Context question: 4a Use the previous results to demonstrate that for any star fusing hydrogen, the ratio of mass $M$ to radius $R$ is the same and depends only on physical constants. Find the equation for the ratio $M / R$ for stars fusing hydrogen. Context answer: \boxed{$\frac{M}{R}=\frac{q^{4}}{12 \pi^{2} \varepsilon_{0}^{2} G h^{2}}$} Extra Supplementary Reading Materials: 5. The mass and radius of the smallest star. The result found in (4a) suggests that there could be stars of any mass as long as such a relationship is fulfilled; however, this is not true. The gas inside normal stars fusing hydrogen is known to behave approximately as an ideal gas. This means that $d_{e}$, the typical separation between electrons is on the average larger that $\lambda_{e}$, their typical de Broglie wavelength. If closer, the electrons would be in a so-called degenerate state and the stars would behave differently. Note the distinction in the ways we treat protons and electrons inside the star. For protons, their de Broglie waves should overlap closely as they collide in order to fuse, whereas for electrons their de Broglie waves should not overlap in order to remain as an ideal gas. The density in the stars increases with decreasing radius. Nevertheless, for this order-ofmagnitude estimate assume they are of uniform density. You may further use that $m_{p} \gg m_{e}$.","5a Find an equation for $n_{e}$, the average electron number density inside the
star.",['$n_{e}=\\frac{M}{(4 / 3) \\pi R^{3} m_{p}}$'],['$n_{e}=\\frac{M}{(4 / 3) \\pi R^{3} m_{p}}$'],False,,Expression, 1287,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} Context question: 3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed. Context answer: \boxed{$2.4 \times 10^{21}$} Extra Supplementary Reading Materials: Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow this ratio for a large range of masses. 4. The mass/radius ratio of the stars. The previous agreement suggests that the quantum mechanical approach for estimating the temperature at the center of the Sun is correct. Context question: 4a Use the previous results to demonstrate that for any star fusing hydrogen, the ratio of mass $M$ to radius $R$ is the same and depends only on physical constants. Find the equation for the ratio $M / R$ for stars fusing hydrogen. Context answer: \boxed{$\frac{M}{R}=\frac{q^{4}}{12 \pi^{2} \varepsilon_{0}^{2} G h^{2}}$} Extra Supplementary Reading Materials: 5. The mass and radius of the smallest star. The result found in (4a) suggests that there could be stars of any mass as long as such a relationship is fulfilled; however, this is not true. The gas inside normal stars fusing hydrogen is known to behave approximately as an ideal gas. This means that $d_{e}$, the typical separation between electrons is on the average larger that $\lambda_{e}$, their typical de Broglie wavelength. If closer, the electrons would be in a so-called degenerate state and the stars would behave differently. Note the distinction in the ways we treat protons and electrons inside the star. For protons, their de Broglie waves should overlap closely as they collide in order to fuse, whereas for electrons their de Broglie waves should not overlap in order to remain as an ideal gas. The density in the stars increases with decreasing radius. Nevertheless, for this order-ofmagnitude estimate assume they are of uniform density. You may further use that $m_{p} \gg m_{e}$. Context question: 5a Find an equation for $n_{e}$, the average electron number density inside the
star. Context answer: \boxed{$n_{e}=\frac{M}{(4 / 3) \pi R^{3} m_{p}}$} ","5b Find an equation for $d_{e}$, the typical separation between electrons inside
the star.",['$d_{e}=n_{e}^{-1 / 3}=\\left(\\frac{M}{(4 / 3) \\pi R^{3} m_{p}}\\right)^{-1 / 3}$'],['$d_{e}=\\left(\\frac{M}{(4 / 3) \\pi R^{3} m_{p}}\\right)^{-1 / 3}$'],False,,Expression, 1288,Modern Physics,,5c Use the $d_{e} \geq \frac{\lambda_{e}}{2^{1 / 2}}$ condition to write down an equation for the radius of
the smallest normal star possible. Take the temperature at the center of the
star as typical for all the stellar interior.,"['We assume that
$d_{e} \\geq \\frac{\\lambda_{e}}{2^{1 / 2}} \\cdot$ Since
$\\lambda_{e}=\\frac{h}{m_{e} v_{r m s}(\\text { electron })}$,
$\\frac{3}{2} k T_{c}=\\frac{1}{2} m_{e} v_{r m s}^{2}($ electron $)$,
$T_{c}=\\frac{q^{4} m_{p}}{24 \\pi^{2} \\varepsilon_{0}^{2} k h^{2}}$,
$\\frac{M}{R}=\\frac{q^{4}}{12 \\pi^{2} \\varepsilon_{0}^{2} G h^{2}}$, and
$d_{e}=\\left(\\frac{M}{(4 / 3) \\pi R^{3} m_{p}}\\right)^{-1 / 3}$,
we get that
$R \\geq \\frac{\\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$']",['$R \\geq \\frac{\\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$'],False,,Need_human_evaluate, 1289,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} Context question: 3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed. Context answer: \boxed{$2.4 \times 10^{21}$} Extra Supplementary Reading Materials: Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow this ratio for a large range of masses. 4. The mass/radius ratio of the stars. The previous agreement suggests that the quantum mechanical approach for estimating the temperature at the center of the Sun is correct. Context question: 4a Use the previous results to demonstrate that for any star fusing hydrogen, the ratio of mass $M$ to radius $R$ is the same and depends only on physical constants. Find the equation for the ratio $M / R$ for stars fusing hydrogen. Context answer: \boxed{$\frac{M}{R}=\frac{q^{4}}{12 \pi^{2} \varepsilon_{0}^{2} G h^{2}}$} Extra Supplementary Reading Materials: 5. The mass and radius of the smallest star. The result found in (4a) suggests that there could be stars of any mass as long as such a relationship is fulfilled; however, this is not true. The gas inside normal stars fusing hydrogen is known to behave approximately as an ideal gas. This means that $d_{e}$, the typical separation between electrons is on the average larger that $\lambda_{e}$, their typical de Broglie wavelength. If closer, the electrons would be in a so-called degenerate state and the stars would behave differently. Note the distinction in the ways we treat protons and electrons inside the star. For protons, their de Broglie waves should overlap closely as they collide in order to fuse, whereas for electrons their de Broglie waves should not overlap in order to remain as an ideal gas. The density in the stars increases with decreasing radius. Nevertheless, for this order-ofmagnitude estimate assume they are of uniform density. You may further use that $m_{p} \gg m_{e}$. Context question: 5a Find an equation for $n_{e}$, the average electron number density inside the
star. Context answer: \boxed{$n_{e}=\frac{M}{(4 / 3) \pi R^{3} m_{p}}$} Context question: 5b Find an equation for $d_{e}$, the typical separation between electrons inside
the star. Context answer: \boxed{$d_{e}=\left(\frac{M}{(4 / 3) \pi R^{3} m_{p}}\right)^{-1 / 3}$} Context question: 5c Use the $d_{e} \geq \frac{\lambda_{e}}{2^{1 / 2}}$ condition to write down an equation for the radius of
the smallest normal star possible. Take the temperature at the center of the
star as typical for all the stellar interior. Context answer: $R \geq \frac{\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$ ","5d Find the numerical value of the radius of the smallest normal star possible, in meters.",['$R \\geq \\frac{\\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}=6.9 \\times 10^{7} \\mathrm{~m}=0.10 R($ Sun $)$'],['$6.9 \\times 10^{7}$'],False,m,Numerical,1e6 1290,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} Context question: 3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed. Context answer: \boxed{$2.4 \times 10^{21}$} Extra Supplementary Reading Materials: Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow this ratio for a large range of masses. 4. The mass/radius ratio of the stars. The previous agreement suggests that the quantum mechanical approach for estimating the temperature at the center of the Sun is correct. Context question: 4a Use the previous results to demonstrate that for any star fusing hydrogen, the ratio of mass $M$ to radius $R$ is the same and depends only on physical constants. Find the equation for the ratio $M / R$ for stars fusing hydrogen. Context answer: \boxed{$\frac{M}{R}=\frac{q^{4}}{12 \pi^{2} \varepsilon_{0}^{2} G h^{2}}$} Extra Supplementary Reading Materials: 5. The mass and radius of the smallest star. The result found in (4a) suggests that there could be stars of any mass as long as such a relationship is fulfilled; however, this is not true. The gas inside normal stars fusing hydrogen is known to behave approximately as an ideal gas. This means that $d_{e}$, the typical separation between electrons is on the average larger that $\lambda_{e}$, their typical de Broglie wavelength. If closer, the electrons would be in a so-called degenerate state and the stars would behave differently. Note the distinction in the ways we treat protons and electrons inside the star. For protons, their de Broglie waves should overlap closely as they collide in order to fuse, whereas for electrons their de Broglie waves should not overlap in order to remain as an ideal gas. The density in the stars increases with decreasing radius. Nevertheless, for this order-ofmagnitude estimate assume they are of uniform density. You may further use that $m_{p} \gg m_{e}$. Context question: 5a Find an equation for $n_{e}$, the average electron number density inside the
star. Context answer: \boxed{$n_{e}=\frac{M}{(4 / 3) \pi R^{3} m_{p}}$} Context question: 5b Find an equation for $d_{e}$, the typical separation between electrons inside
the star. Context answer: \boxed{$d_{e}=\left(\frac{M}{(4 / 3) \pi R^{3} m_{p}}\right)^{-1 / 3}$} Context question: 5c Use the $d_{e} \geq \frac{\lambda_{e}}{2^{1 / 2}}$ condition to write down an equation for the radius of
the smallest normal star possible. Take the temperature at the center of the
star as typical for all the stellar interior. Context answer: $R \geq \frac{\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$ Context question: 5d Find the numerical value of the radius of the smallest normal star possible, in meters. Context answer: \boxed{$6.9 \times 10^{7}$} ","5e Find the numerical value of the mass of the smallest normal star possible, in $\mathrm{kg}$.","['The mass to radius ratio is:
$\\frac{M}{R}=\\frac{q^{4}}{12 \\pi^{2} \\varepsilon_{0}^{2} G h^{2}}=2.4 \\times 10^{21} \\mathrm{~kg} \\mathrm{~m}^{-1}$, from where we derive that
$M \\geq 1.7 \\times 10^{29} \\mathrm{~kg}=0.09 M($ Sun $)$']",['$1.7 \\times 10^{29}$'],False,kg,Numerical, 1291,Modern Physics,"WHY ARE STARS SO LARGE? The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen. Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts. $$ \begin{aligned} & \text { USEFUL CONSTANTS } \\ & \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\ & \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\ & \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\ & \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\ & \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\ & \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\ & \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\ & \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\ & \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg} \end{aligned} $$ 1. A classical estimate of the temperature at the center of the stars. Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision. Context question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures. Context answer: \boxed{$5.5 \times 10^{9}$} Extra Supplementary Reading Materials: 2. Finding that the previous temperature estimate is wrong. To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by $$ \frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}} $$ where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab. Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity. An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations: $\Delta P=P_{o}-P_{c}$, where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively. Since $P_{c} \gg P_{o}$, we can assume that $\Delta P=-P_{c}$. Within the same approximation, we can write $\Delta r \approx R$, where $R$ is the total radius of the star, and $M_{r} \approx M_{R}=M$, with $M$ the total mass of the star. The density may be approximated by its value at the center, $\rho_{r}=\rho_{c}$. You can assume that the pressure is that of an ideal gas. Context question: 2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only. Context answer: \boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$} Extra Supplementary Reading Materials: We can use now the following prediction of this model as a criterion for its validity: Context question: 2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only. Context answer: \boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$} Context question: 2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star. Context answer: \boxed{$1.4 \times 10^{24} $} Context question: 2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$. Context answer: \boxed{$2.9 \times 10^{21}$} Extra Supplementary Reading Materials: 3. A quantum mechanical estimate of the temperature at the center of the stars The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse. Context question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton
with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants
only. Context answer: \boxed{$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$} Context question: 3b Evaluate numerically the value of $T_{c}$ obtained in (3a). Context answer: \boxed{$9.7 \times 10^{6}$} Context question: 3c Use the value of $T_{c}$ derived in (3b) to find the numerical value of the 0.5 ratio $M / R$ expected for a star, using the formula derived in (2b). Verify that this value is quite similar to the ratio $M($ Sun $) / R($ Sun $)$ observed. Context answer: \boxed{$2.4 \times 10^{21}$} Extra Supplementary Reading Materials: Indeed, stars in the so-called main sequence (fusing hydrogen) approximately do follow this ratio for a large range of masses. 4. The mass/radius ratio of the stars. The previous agreement suggests that the quantum mechanical approach for estimating the temperature at the center of the Sun is correct. Context question: 4a Use the previous results to demonstrate that for any star fusing hydrogen, the ratio of mass $M$ to radius $R$ is the same and depends only on physical constants. Find the equation for the ratio $M / R$ for stars fusing hydrogen. Context answer: \boxed{$\frac{M}{R}=\frac{q^{4}}{12 \pi^{2} \varepsilon_{0}^{2} G h^{2}}$} Extra Supplementary Reading Materials: 5. The mass and radius of the smallest star. The result found in (4a) suggests that there could be stars of any mass as long as such a relationship is fulfilled; however, this is not true. The gas inside normal stars fusing hydrogen is known to behave approximately as an ideal gas. This means that $d_{e}$, the typical separation between electrons is on the average larger that $\lambda_{e}$, their typical de Broglie wavelength. If closer, the electrons would be in a so-called degenerate state and the stars would behave differently. Note the distinction in the ways we treat protons and electrons inside the star. For protons, their de Broglie waves should overlap closely as they collide in order to fuse, whereas for electrons their de Broglie waves should not overlap in order to remain as an ideal gas. The density in the stars increases with decreasing radius. Nevertheless, for this order-ofmagnitude estimate assume they are of uniform density. You may further use that $m_{p} \gg m_{e}$. Context question: 5a Find an equation for $n_{e}$, the average electron number density inside the
star. Context answer: \boxed{$n_{e}=\frac{M}{(4 / 3) \pi R^{3} m_{p}}$} Context question: 5b Find an equation for $d_{e}$, the typical separation between electrons inside
the star. Context answer: \boxed{$d_{e}=\left(\frac{M}{(4 / 3) \pi R^{3} m_{p}}\right)^{-1 / 3}$} Context question: 5c Use the $d_{e} \geq \frac{\lambda_{e}}{2^{1 / 2}}$ condition to write down an equation for the radius of
the smallest normal star possible. Take the temperature at the center of the
star as typical for all the stellar interior. Context answer: $R \geq \frac{\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$ Context question: 5d Find the numerical value of the radius of the smallest normal star possible, in meters. Context answer: \boxed{$6.9 \times 10^{7}$} Context question: 5e Find the numerical value of the mass of the smallest normal star possible, in $\mathrm{kg}$. Context answer: \boxed{$1.7 \times 10^{29}$} Extra Supplementary Reading Materials: 6. Fusing helium nuclei in older stars. As stars get older they will have fused most of the hydrogen in their cores into helium $(\mathrm{He})$, so they are forced to start fusing helium into heavier elements in order to continue shining. A helium nucleus has two protons and two neutrons, so it has twice the charge and approximately four times the mass of a proton. We saw before that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition for the protons to fuse.","6a Set the equivalent condition for helium nuclei and find $v_{r m s}(H e)$, the rms velocity of the helium nuclei and $T(\mathrm{He})$, the temperature needed for helium fusion.",['For helium we have that
$\\frac{4 q^{2}}{4 \\pi \\varepsilon_{0} m_{\\mathrm{He}} v_{r m s}^{2}(\\mathrm{He})}=\\frac{h}{2^{1 / 2} m_{\\mathrm{He}} v_{r m s}(\\mathrm{He})} ;$ from where we get
$v_{r m s}(\\mathrm{He})=\\frac{2^{1 / 2} q^{2}}{\\pi \\varepsilon_{0} h}=2.0 \\times 10^{6} \\mathrm{~m} \\mathrm{~s}^{-1}$.
We now use:
$T(\\mathrm{He})=\\frac{v_{r m s}^{2}(\\mathrm{He}) m_{\\mathrm{He}}}{3 k}=6.5 \\times 10^{8} \\mathrm{~K}$.
This value is of the order of magnitude of the estimates of stellar models.'],"['$2.0 \\times 10^{6}$ , $6.5 \\times 10^{8}$']",True,"$\mathrm{~m} \mathrm{~s}^{-1}$, K",Numerical,"1e5,1e7" 1292,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics","A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring?","['The force $F$ causes the spring to change its length from $L_{0}$ to $L$. Since equal parts of the spring are extended to equal lengths, we get: $\\frac{\\Delta y}{\\Delta l}=\\frac{L}{L_{0}} \\rightarrow \\Delta y=\\frac{L}{L_{0}} \\Delta l$.\n\nSince $L=\\max \\left\\{\\frac{F}{k}, L_{0}\\right\\}$, we get $\\Delta y=\\max \\left\\{\\frac{F}{k L_{0}} \\Delta l, \\Delta l\\right\\}$. From this result we see that any piece of length $\\Delta l$ the spring behaves as a ZLS with spring constant $k^{*}=k \\frac{L_{0}}{\\Delta l}$.\n\n']","['$\\Delta y=\\max \\left\\{\\frac{F}{k L_{0}} \\Delta l, \\Delta l\\right\\}$ , $k^{*}=k \\frac{L_{0}}{\\Delta l}$']",True,,Expression, 1293,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics Context question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring? Context answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$ ","A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$.","['Let us compute the work of the force. From Task A.1: $d W=F(x) d x=\\frac{k L_{0}}{\\Delta l} x d x$.\n\nHence, $\\Delta W=\\int_{\\Delta l}^{\\Delta y} \\frac{k L_{0}}{\\Delta l} x d x=\\left.\\frac{k L_{0}}{\\Delta l} \\frac{x^{2}}{2}\\right|_{\\Delta l} ^{\\Delta y}=\\frac{k L_{0}}{2 \\Delta l}\\left(\\Delta y^{2}-\\Delta l^{2}\\right)$.']",['$\\Delta W=\\frac{k L_{0}}{2 \\Delta l}\\left(\\Delta y^{2}-\\Delta l^{2}\\right)$'],False,,Expression, 1294,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics Context question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring? Context answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$ Context question: A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$. Context answer: \boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$} Extra Supplementary Reading Materials: Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches.","A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$.","['At every point along the statically hanging spring the weight of the mass below is balanced by the tension from above. This implies that at the bottom of the spring there is a section of length $l_{0}$ whose turns are still touching each other, as their weight is insufficient to exceed the threshold force $k L_{0}$ to pull them apart. The length $l_{0}$ can be derived from the equation:\n\n$\\frac{l_{0}}{L_{0}} M g=k L_{0}$, hence $l_{0}=\\frac{k L_{0}^{2}}{M g}=\\alpha L_{0}$.\n\nFor $l>l_{0}$, a segment of the unstretched spring between $l$ and $l+\\mathrm{d} l$ feels a weight of $\\frac{l}{L_{0}} M g$ from beneath, which causes its length to stretch from $\\mathrm{d} l$ to $d y=\\frac{F}{k L_{0}} d l=\\frac{l}{L_{0}} M g \\frac{d l}{k L_{0}}=$ $\\frac{M g}{k L_{0}^{2}} l d l=\\frac{l}{l_{0}} d l$.\n\nIntegration of the last expression over the stretched region, up to the point $L_{0}$, gives its height when the spring is stretched\n\n$$\nH=l_{0}+\\int_{l_{0}}^{L_{0}} \\frac{l}{l_{0}} d l=l_{0}+\\left.\\frac{l^{2}}{2 l_{0}}\\right|_{l_{0}} ^{L_{0}}=l_{0}+\\frac{1}{2 l_{0}}\\left(L_{0}^{2}-l_{0}^{2}\\right)=\\frac{L_{0}^{2}}{2 l_{0}}+\\frac{l_{0}}{2}=\\frac{L_{0}}{2}\\left(\\alpha+\\frac{1}{\\alpha}\\right)\n$$']",['$H=\\frac{L_{0}}{2}(\\alpha+\\frac{1}{\\alpha})$'],False,,Expression, 1295,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics Context question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring? Context answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$ Context question: A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$. Context answer: \boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$} Extra Supplementary Reading Materials: Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches. Context question: A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$. Context answer: \boxed{$H=\frac{L_{0}}{2}(\alpha+\frac{1}{\alpha})$} Extra Supplementary Reading Materials: Part B: Dynamics Experiments show that when the spring is hung at rest and then released, it gradually contracts from the top, while the lower part remains stationary (see Figure 2). As time advances, the contracting part moves as a solid chunk and accumulates additional turns of the spring, while the stationary part becomes shorter. Every point on the spring begins to move only when the moving part reaches it. The bottom end of the spring starts moving only when the spring is fully collapsed and reaches its unstretched length $L_{0}$. After that, the contracted spring continues falling straight downwards, without tumbling, as a rigid body under the influence of gravity. Figure 2: Left: a sequence of pictures taken during the free fall of slinky. Right: the moving part I and the stationary part II during the free fall of the spring. In the remaining parts of the question, you are asked to base your solution on this described model. You may neglect air resistance, but you are not allowed to neglect $L_{0}$.","B.1 Calculate the time $t_{c}$ it takes from the moment the spring is released, until it fully collapses back to its minimal length $L_{0}$. Express your answer in terms of $L_{0}, g$ and $\alpha$. Compute the numerical value of $t_{c}$ for a spring with $k=1.02 \mathrm{~N} / \mathrm{m}, L_{0}=0.055 \mathrm{~m}$ and $M=0.201 \mathrm{~kg}$, while taking $g$ to be $9.80 \mathrm{~m} / \mathrm{s}^{2}$.","['From Task A. 3 we have $H(l)=\\frac{l^{2}}{2 l_{0}}+\\frac{l_{0}}{2}$. We now calculate the position of the center of mass of the suspended spring. The contribution of the unstretched section of height $l_{0}$ at the bottom, having a mass of $\\frac{l_{0}}{L_{0}} M=\\alpha M$, is $\\alpha M \\frac{l_{0}}{2}$. The position of the center of mass is obtained by summing the contributions of its elements:\n\n$$\n\\begin{array}{r}\nH_{c m}=\\frac{1}{M}\\left[\\frac{l_{0}}{2} \\alpha M+\\int_{l_{0}}^{L_{0}} H(l) d m\\right]=\\frac{1}{M}\\left[\\frac{\\alpha L_{0}}{2} \\alpha M+\\int_{l_{0}}^{L_{0}}\\left(\\frac{l^{2}}{2 l_{0}}+\\frac{l_{0}}{2}\\right) \\frac{M d l}{L_{0}}\\right] \\\\\n=\\frac{\\alpha^{2} L_{0}}{2}+\\frac{1}{L_{0}}\\left[\\frac{l^{3}}{6 l_{0}}+\\frac{l_{0}}{2} l\\right]_{l_{0}}^{L_{0}}=\\frac{\\alpha^{2} L_{0}}{2}+\\frac{1}{L_{0}}\\left[\\frac{L_{0}^{3}-l_{0}^{3}}{6 l_{0}}+\\frac{l_{0}}{2}\\left(L_{0}-l_{0}\\right)\\right]\n\\end{array}\n$$\n\nWhere we have used $d m=\\frac{d l}{L_{0}} M$. Substituting $l_{0}=\\alpha L_{0}$ yields\n\n$$\nH_{c m}=L_{0}\\left[\\frac{1}{6 \\alpha}-\\frac{\\alpha^{2}}{6}+\\frac{\\alpha}{2}\\right]\n$$\n\nWhen the spring is contracted to its free length $L_{0}$, its center of mass is located at $\\frac{L_{0}}{2}$. From the falling of the center of mass at acceleration $g$ we get:\n\n$$\n\\frac{g}{2} t_{c}^{2}=H_{c m}-\\frac{L_{0}}{2}=L_{0}\\left[\\frac{1}{6 \\alpha}-\\frac{\\alpha^{2}}{6}+\\frac{\\alpha}{2}-\\frac{1}{2}\\right]=\\frac{L_{0}}{6 \\alpha}(1-\\alpha)^{3}\n$$\n\nHence, $t_{c}=\\sqrt{\\frac{L_{0}}{3 g \\alpha}(1-\\alpha)^{3}}$.\n\nFor $k=1.02 \\mathrm{~N} / \\mathrm{m}, L_{0}=0.055 \\mathrm{~m}, M=0.201 \\mathrm{~kg}$, and $g=9.80 \\mathrm{~m} / \\mathrm{s}^{2}$, we have $\\alpha=0.0285$, and $t_{c}=0.245 \\mathrm{~s}$.']","['$t_{c}=\\sqrt{\\frac{L_{0}}{3 g \\alpha}(1-\\alpha)^{3}}$ , $t_{c}=0.245$']",True,",s","Expression,Numerical",",1e-2" 1296,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics Context question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring? Context answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$ Context question: A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$. Context answer: \boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$} Extra Supplementary Reading Materials: Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches. Context question: A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$. Context answer: \boxed{$H=\frac{L_{0}}{2}(\alpha+\frac{1}{\alpha})$} Extra Supplementary Reading Materials: Part B: Dynamics Experiments show that when the spring is hung at rest and then released, it gradually contracts from the top, while the lower part remains stationary (see Figure 2). As time advances, the contracting part moves as a solid chunk and accumulates additional turns of the spring, while the stationary part becomes shorter. Every point on the spring begins to move only when the moving part reaches it. The bottom end of the spring starts moving only when the spring is fully collapsed and reaches its unstretched length $L_{0}$. After that, the contracted spring continues falling straight downwards, without tumbling, as a rigid body under the influence of gravity. Figure 2: Left: a sequence of pictures taken during the free fall of slinky. Right: the moving part I and the stationary part II during the free fall of the spring. In the remaining parts of the question, you are asked to base your solution on this described model. You may neglect air resistance, but you are not allowed to neglect $L_{0}$. Context question: B.1 Calculate the time $t_{c}$ it takes from the moment the spring is released, until it fully collapses back to its minimal length $L_{0}$. Express your answer in terms of $L_{0}, g$ and $\alpha$. Compute the numerical value of $t_{c}$ for a spring with $k=1.02 \mathrm{~N} / \mathrm{m}, L_{0}=0.055 \mathrm{~m}$ and $M=0.201 \mathrm{~kg}$, while taking $g$ to be $9.80 \mathrm{~m} / \mathrm{s}^{2}$. Context answer: \boxed{$t_{c}=\sqrt{\frac{L_{0}}{3 g \alpha}(1-\alpha)^{3}}$ , $t_{c}=0.245$} ","B.2 In this task $\ell$ is used to denote the coordinate of the boundary between parts I (in figure 2, the moving part) and II (the stationary part). At a certain moment, while a stationary part still exists its mass is $m(\ell)=\frac{\ell}{L_{0}} M$, and the moving part moves with uniform instantaneous velocity $v_{I}(\ell)$. Show that at this moment (while there exists a stationary part) the velocity of the moving part is $v_{I}(\ell)=$ $\sqrt{A \ell+B}$. Express the constants $A$ and $B$ in terms of $L_{0}, g$ and $\alpha$.","[""The moving top section of the spring is pulled down by its own weight, $m_{t o p} g=M g \\frac{\\left(L_{0}-l\\right)}{L_{0}}$ and also by the tension in the spring below, which is equal to the weight $M g l / L_{0}$ of the stationary section of the spring. Thus, the moving top section experiences a constant force $F=$ $M g$ throughout its whole fall. Another way to see that, is that a total force of $M g$ is exerted on the spring, but only the moving part experiences it. Let's calculate the position of the center of mass at equilibrium of the upper part, i.e., all points with $l^{\\prime}>l$ for some $l>l_{0}$. From part A, the position of a small portion $\\Delta l^{\\prime}$ with coordinate $l^{\\prime}$ is: $H\\left(l^{\\prime}\\right)=\\frac{l^{\\prime 2}}{2 l_{0}}+\\frac{l_{0}}{2}$ and the center of mass of this part is:\n\n$$\n\\begin{aligned}\nH_{\\text {cm-upper }-i} & =\\frac{L_{0}}{M\\left(L_{0}-l\\right)} \\int_{l}^{L_{0}} H\\left(l^{\\prime}\\right) d m=\\frac{L_{0}}{M\\left(L_{0}-l\\right)} \\int_{l}^{L_{0}}\\left(\\frac{l^{\\prime 2}}{2 l_{0}}+\\frac{l_{0}}{2}\\right) d m \\\\\n& =\\frac{L_{0}}{M\\left(L_{0}-l\\right)} \\int_{l}^{L_{0}}\\left(\\frac{l^{\\prime 2}}{2 l_{0}}+\\frac{l_{0}}{2}\\right) \\frac{M d l^{\\prime}}{L_{0}}=\\frac{1}{\\left(L_{0}-l\\right)} \\int_{l}^{L_{0}}\\left(\\frac{l^{\\prime 2}}{2 l_{0}}+\\frac{l_{0}}{2}\\right) d l^{\\prime} \\\\\n& =\\frac{1}{\\left(L_{0}-l\\right)}\\left[\\frac{l^{\\prime 3}}{6 l_{0}}+\\frac{l_{0} l^{\\prime}}{2}\\right]_{l}^{L_{-} 0}=\\frac{L_{0}^{2}+L_{0} l+l^{2}}{6 l_{0}}+\\frac{l_{0}}{2}\n\\end{aligned}\n$$\n\nThe position of the upper part of $\\mathrm{CM}$ when it contracts to a length $L_{0}-l$ is $H_{c m-u p p e r-f}=$ $\\frac{l^{2}}{2 l_{0}}+\\frac{l_{0}}{2}+\\frac{1}{2}\\left(L_{0}-l\\right)$. The change in the CM during the contraction process is: $\\Delta H_{c m-u p p e r}=$ $H_{\\text {cm-upper }-i}-H_{\\text {cm-upper }-f}=\\frac{L_{0}^{2}+L_{0} l-2 l^{2}}{6 l_{0}}-\\frac{1}{2}\\left(L_{0}-l\\right)=\\frac{\\left(L_{0}-l\\right)\\left(L_{0}+2 l\\right)}{6 l_{0}}-\\frac{1}{2}\\left(L_{0}-l\\right)$.\n\nThe acceleration of the $\\mathrm{CM}$ of the upper part is $a_{C M}=\\frac{F L_{0}}{M\\left(L_{0}-l\\right)}=\\frac{g L_{0}}{L_{0}-l}$. From the work energy theorem we get the equation $v_{\\text {upper-f }}^{2}=2 a_{C M} \\Delta H_{\\text {cm-upper }}$, hence\n\n$$\n\\begin{gathered}\nv_{\\text {upper }-f}^{2}=2 \\frac{g L_{0}}{L_{0}-l}\\left[\\frac{\\left(L_{0}-l\\right)\\left(L_{0}+2 l\\right)}{6 \\alpha L_{0}}-\\frac{1}{2}\\left(L_{0}-l\\right)\\right]=2 g\\left[\\frac{L_{0}+2 l}{6 \\alpha}-\\frac{1}{2} L_{0}\\right] \\\\\n=\\frac{2 g}{3 \\alpha} l+\\left(\\frac{1}{3 \\alpha}-1\\right) g L_{0}\n\\end{gathered}\n$$\n\nTherefore, $A=\\frac{2 g}{3 \\alpha}$ and $B=\\left(\\frac{1}{3 \\alpha}-1\\right) g L_{0}$.\n\nNote that for $l=L_{0}$, we have $v_{\\text {upper }-f}^{2}=L_{0} g \\frac{1-\\alpha}{\\alpha}$ and for $l=l_{0}=\\alpha L_{0}$, we get $v_{u p p e r-f}^{2}=$ $L_{0} g \\frac{1-\\alpha}{3 \\alpha}$, hence, the moment we release the spring its velocity is finite (not zero, the meaning is that it accumulate this velocity in time that is much shorter than the contracting time $t_{c}$ ) and it decreases to $\\frac{1}{\\sqrt{3}}$ of the initial value when $l=l_{0}$.""]","['$A=\\frac{2 g}{3 \\alpha}$ , $B=(\\frac{1}{3 \\alpha}-1) g L_{0}$']",True,,Expression, 1297,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics Context question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring? Context answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$ Context question: A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$. Context answer: \boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$} Extra Supplementary Reading Materials: Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches. Context question: A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$. Context answer: \boxed{$H=\frac{L_{0}}{2}(\alpha+\frac{1}{\alpha})$} Extra Supplementary Reading Materials: Part B: Dynamics Experiments show that when the spring is hung at rest and then released, it gradually contracts from the top, while the lower part remains stationary (see Figure 2). As time advances, the contracting part moves as a solid chunk and accumulates additional turns of the spring, while the stationary part becomes shorter. Every point on the spring begins to move only when the moving part reaches it. The bottom end of the spring starts moving only when the spring is fully collapsed and reaches its unstretched length $L_{0}$. After that, the contracted spring continues falling straight downwards, without tumbling, as a rigid body under the influence of gravity. Figure 2: Left: a sequence of pictures taken during the free fall of slinky. Right: the moving part I and the stationary part II during the free fall of the spring. In the remaining parts of the question, you are asked to base your solution on this described model. You may neglect air resistance, but you are not allowed to neglect $L_{0}$. Context question: B.1 Calculate the time $t_{c}$ it takes from the moment the spring is released, until it fully collapses back to its minimal length $L_{0}$. Express your answer in terms of $L_{0}, g$ and $\alpha$. Compute the numerical value of $t_{c}$ for a spring with $k=1.02 \mathrm{~N} / \mathrm{m}, L_{0}=0.055 \mathrm{~m}$ and $M=0.201 \mathrm{~kg}$, while taking $g$ to be $9.80 \mathrm{~m} / \mathrm{s}^{2}$. Context answer: \boxed{$t_{c}=\sqrt{\frac{L_{0}}{3 g \alpha}(1-\alpha)^{3}}$ , $t_{c}=0.245$} Context question: B.2 In this task $\ell$ is used to denote the coordinate of the boundary between parts I (in figure 2, the moving part) and II (the stationary part). At a certain moment, while a stationary part still exists its mass is $m(\ell)=\frac{\ell}{L_{0}} M$, and the moving part moves with uniform instantaneous velocity $v_{I}(\ell)$. Show that at this moment (while there exists a stationary part) the velocity of the moving part is $v_{I}(\ell)=$ $\sqrt{A \ell+B}$. Express the constants $A$ and $B$ in terms of $L_{0}, g$ and $\alpha$. Context answer: \boxed{$A=\frac{2 g}{3 \alpha}$ , $B=(\frac{1}{3 \alpha}-1) g L_{0}$} ","B.3 Based on B.2, find the minimum speed $v_{\min }$ of the moving part of the spring in the course of its motion, after its release and before it hits the ground. Express your answer in terms of $L_{0}, \alpha, A$ and $B$.","['Note that even though the center of mass of the spring accelerates downwards constantly, the moving top section actually decelerates, while the position of the center of mass moves down the spring. The speed of the top section $v(l)$, calculated in Task B2, decreases and approaches the value $\\sqrt{A \\alpha L_{0}+B}$ immediately before it attaches to the bottom section of height $l_{0}=\\alpha L_{0}$, which was unstretched and at rest. Once the moving top section attaches to the resting bottom section, its momentum is shared between both sections, so the speed further decreases just before the whole spring starts accelerating downwards as a single mass. Thus, the minimum speed is that of the whole spring immediately after its full collapse. From momentum conservation, we have\n\n$$\n\\begin{gathered}\nM v_{\\text {min }}=m_{\\text {top }} v\\left(l_{0}\\right)=M\\left(1-\\frac{l_{0}}{L_{0}}\\right) \\sqrt{A \\alpha L_{0}+B} \\\\\nv_{\\text {min }}=(1-\\alpha) \\sqrt{A \\alpha L_{0}+B}\n\\end{gathered}\n$$']",['$v_{min}=(1-\\alpha) \\sqrt{A \\alpha L_{0}+B}$'],False,,Expression, 1298,Mechanics,"Zero-length springs and slinky coils A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$. Figure 1: the relation between the force $F$ and the spring length $L$ A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as ""slinky"" may be (but not necessarily) such a ZLS. Part A: Statics Context question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring? Context answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$ Context question: A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$. Context answer: \boxed{$\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$} Extra Supplementary Reading Materials: Throughout this question, we will denote a point on the spring by its distance $0 \leq \ell \leq L_{0}$ from the bottom of the spring when it is unstretched. In particular, for every point on the spring, $\ell$ remains unchanged as the spring stretches. Context question: A.3 Suppose that we hang the spring by its top end, so that it stretches under its own weight. What is the total length $H$ of the suspended spring in equilibrium? Express your answers in terms of $L_{0}$ and $\alpha$. Context answer: \boxed{$H=\frac{L_{0}}{2}(\alpha+\frac{1}{\alpha})$} Extra Supplementary Reading Materials: Part B: Dynamics Experiments show that when the spring is hung at rest and then released, it gradually contracts from the top, while the lower part remains stationary (see Figure 2). As time advances, the contracting part moves as a solid chunk and accumulates additional turns of the spring, while the stationary part becomes shorter. Every point on the spring begins to move only when the moving part reaches it. The bottom end of the spring starts moving only when the spring is fully collapsed and reaches its unstretched length $L_{0}$. After that, the contracted spring continues falling straight downwards, without tumbling, as a rigid body under the influence of gravity. Figure 2: Left: a sequence of pictures taken during the free fall of slinky. Right: the moving part I and the stationary part II during the free fall of the spring. In the remaining parts of the question, you are asked to base your solution on this described model. You may neglect air resistance, but you are not allowed to neglect $L_{0}$. Context question: B.1 Calculate the time $t_{c}$ it takes from the moment the spring is released, until it fully collapses back to its minimal length $L_{0}$. Express your answer in terms of $L_{0}, g$ and $\alpha$. Compute the numerical value of $t_{c}$ for a spring with $k=1.02 \mathrm{~N} / \mathrm{m}, L_{0}=0.055 \mathrm{~m}$ and $M=0.201 \mathrm{~kg}$, while taking $g$ to be $9.80 \mathrm{~m} / \mathrm{s}^{2}$. Context answer: \boxed{$t_{c}=\sqrt{\frac{L_{0}}{3 g \alpha}(1-\alpha)^{3}}$ , $t_{c}=0.245$} Context question: B.2 In this task $\ell$ is used to denote the coordinate of the boundary between parts I (in figure 2, the moving part) and II (the stationary part). At a certain moment, while a stationary part still exists its mass is $m(\ell)=\frac{\ell}{L_{0}} M$, and the moving part moves with uniform instantaneous velocity $v_{I}(\ell)$. Show that at this moment (while there exists a stationary part) the velocity of the moving part is $v_{I}(\ell)=$ $\sqrt{A \ell+B}$. Express the constants $A$ and $B$ in terms of $L_{0}, g$ and $\alpha$. Context answer: \boxed{$A=\frac{2 g}{3 \alpha}$ , $B=(\frac{1}{3 \alpha}-1) g L_{0}$} Context question: B.3 Based on B.2, find the minimum speed $v_{\min }$ of the moving part of the spring in the course of its motion, after its release and before it hits the ground. Express your answer in terms of $L_{0}, \alpha, A$ and $B$. Context answer: \boxed{$v_{min}=(1-\alpha) \sqrt{A \alpha L_{0}+B}$} Extra Supplementary Reading Materials: Part C: Energetics","C.1 Calculate the amount of mechanical energy $Q$ that was lost by generating heat, from the moment the spring is released until just before the spring hits the ground. Express your answer in terms of $L_{0}, M, g$ and $\alpha$.","['From the moment the spring is released, the acceleration of its center of mass is governed by the external force $M g$ and therefore the gravitational potential energy of the spring is fully converted into the kinetic energy of the center of mass of the spring, which just before hitting the ground is equal to the kinetic energy of the spring.\n\nAll that is left is the elastic energy stored in the spring, which is converted into heat, sound, etc. To calculate it, we consider the elastic energy stored in a segment $d h$ of the stretched spring, which when unstretched lies between $l$ and $l+\\mathrm{d} l$, using the result of Task A.2, $\\Delta W=$ $\\frac{k L_{0}}{2 \\Delta l}\\left(\\Delta l_{2}^{2}-\\Delta l^{2}\\right.$ ), by choosing $\\Delta l=d l$ and $\\Delta l_{2}=d y$, and using $d y=\\frac{l}{l_{0}} d l$ (which was obtained in Task A.3), we get:\n\n$d W=\\frac{k L_{0}}{2}\\left(\\frac{l^{2}}{l_{0}^{2}}-1\\right) d l$. Integrating from $l_{0}$ to $L_{0}$ we find\n\n$$\n\\begin{aligned}\n& W=\\int_{l_{0}}^{L_{0}} \\frac{k L_{0}}{2}\\left(\\frac{l^{2}}{l_{0}^{2}}-1\\right) d l=\\frac{k L_{0}}{2}\\left[\\frac{l^{3}}{3 l_{0}^{2}}-l\\right]_{l_{0}}^{L_{0}}=\\frac{k L_{0}}{2}\\left(\\frac{L_{0}^{3}-l_{0}^{3}}{3 l_{0}^{2}}-\\left(L_{0}-l_{0}\\right)\\right) \\\\\n& =\\frac{k L_{0}^{2}}{2}\\left(\\frac{1-\\alpha^{3}}{3 \\alpha^{2}}-(1-\\alpha)\\right)=\\frac{k L_{0}^{2}}{6 \\alpha^{2}}(1-\\alpha)^{2}(2 \\alpha+1) \\\\\n& =M g L_{0} \\frac{(1-\\alpha)^{2}(2 \\alpha+1)}{6 \\alpha}\n\\end{aligned}\n$$']",['$W=M g L_{0} \\frac{(1-\\alpha)^{2}(2 \\alpha+1)}{6 \\alpha}$'],False,,Expression, 1299,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively.","A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.)","['The frequency of an LC circuit is $f=\\omega / 2 \\pi=1 /(2 \\pi \\sqrt{L C})$. If the total electric current flowing along the boundary of the cavity is $I$, it generates a magnetic field whose magnitude (by the assumptions of the question) is $0.6 \\mu_{0} I / h$, and a total magnetic flux equal to $\\pi R^{2} \\times$ $0.6 \\mu_{0} I / h$, hence the inductance of the resonator is $L=0.6 \\pi \\mu_{0} R^{2} / h$. Approximating the capacitor as a plate capacitor, its capacitance is $C=\\varepsilon_{0} l h / d$. Putting everything together, we find\n\n$$\nf_{e s t}=\\frac{1}{2 \\pi} \\frac{1}{\\sqrt{L C}}=\\frac{1}{2 \\pi} \\sqrt{\\frac{h}{0.6 \\pi R^{2} \\mu_{0}} \\frac{d}{\\varepsilon_{0} l h}}=\\frac{1}{2 \\pi} \\frac{c}{R} \\sqrt{\\frac{d}{0.6 \\pi l}}=\\frac{1}{2 \\pi} \\frac{3 \\cdot 10^{8}}{7 \\cdot 10^{-3}} \\sqrt{\\frac{1}{3.6 \\pi}}=2.0 \\cdot 10^{9}\n$$\n\n$\\mathrm{Hz}$']",['$2.0 \\times 10^{9}$'],False,Hz,Numerical,1e8 1300,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively.","A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$.","['Denoting the electron velocity by $\\vec{u}(t)$, in this case the total force applied on it is\n\n$$\n\\vec{F}=-e\\left(-E_{0} \\hat{y}+\\vec{u}(t) \\times B_{0} \\hat{z}\\right)\n$$\n\nLet us write $\\vec{u}(t)=\\vec{u}_{D}+\\vec{u}^{\\prime}(t)$, with $\\vec{u}_{D}=\\left(-E_{0} / B_{0}\\right) \\hat{x}$ being the drift velocity of a charged particle in the crossed electric and magnetic fields (the velocity at which the electric and magnetic forces cancel each other exactly). Then $\\vec{F}=-e \\vec{u}^{\\prime}(t) \\times B_{0} \\hat{z}$. Thus, in a frame moving at the drift velocity $\\vec{u}_{D}$, the electron trajectory is a circle with constant-magnitude velocity $u^{\\prime}=$ $\\left|\\vec{u}^{\\prime}(t)\\right|$, and radius $r=m u^{\\prime} / e B_{0}$. In the lab frame this circular motion is superimposed upon the drift at the constant velocity $\\vec{u}_{D}$. Hence:\n\n1. For $\\vec{u}(0)=\\left(3 E_{0} / B_{0}\\right) \\hat{x}$ we find $u^{\\prime}=4 E_{0} / B_{0}$ and $r=4 m E_{0} / e B_{0}^{2}$.\n2. For $\\vec{u}(0)=-\\left(3 E_{0} / B_{0}\\right) \\hat{x}$ we find $u^{\\prime}=2 E_{0} / B_{0}$ and $r=2 m E_{0} / e B_{0}^{2}$.']","['$u^{\\prime}=4 E_{0} / B_{0}$,$u^{\\prime}=2 E_{0} / B_{0}$']",True,,Expression, 1301,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively.","A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial.","['The velocity of the electron in a frame of reference where the motion is approximately circular is $u^{\\prime}$. From A.2 we get that $u_{D}+u^{\\prime}=v_{\\max }$ and $u_{D}-u^{\\prime}=v_{\\min }$, hence $u^{\\prime}=\\left(v_{\\max }-v_{\\min }\\right) / 2 Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively. Context question: A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial. Context answer: \boxed{0.3} Extra Supplementary Reading Materials: Figure 2 Context question: A.4 Figure 2 depicts the alternating electric field lines between the anode and the cathode at a given moment in time (the static field is not plotted). Indicate in the Answer Sheet which of the electrons positioned at A,B,C,D and E will drift towards the anode, which will drift towards the cathode and which will drift at a direction perpendicular to the radius at that moment. Context answer: Extra Supplementary Reading Materials: Figure 3 Figure 3 depicts the alternating electric field lines between the anode and the cathode (the static field is not plotted) at a given moment in time. The positions of six electrons at that moment are denoted by A, $B, C, D, E$ and $F$. All electrons are at the same distance from the cathode. Context question: A.5 Consider the situation shown in Figure 3. For each of the six electron pairs $A B$, $A C, B C, D E, D F, E F$, indicate in the Answer Sheet whether their drift will cause the angle between their position vectors (measured from the cathode's center $0)$ to increase or decrease at that moment. Context answer: | points | angle decreases | angle increases | indeterminate | | ---: | ---: | :--- | ---: | | AB | $X$ | | | | BC | $X$ | | | | CA | | $X$ | | | DE | | $X$ | | | EF | | $X$ | | | DF | | | | Extra Supplementary Reading Materials: Figure 4 The pattern you have discovered in Task A. 5 acts as a focusing mechanism, concentrating the electrons in the space between the cathode and anode into spokes. Figure 4 depicts one such spoke, denoted by S. Context question: A.6 Depict in the Answer Sheet the other spokes at that moment. Indicate by arrows their direction of rotation, and calculate their average angular velocity $\omega_{s}$. Context answer: Extra Supplementary Reading Materials: Make the approximation that the total electric field half-way between the cathode and the anode is equal to its average static value along a radial line from the cathode to the anode, and that the spokes are approximately radial in that region. The cathode and anode radii ( $a$ and $b$, respectively) are defined in Figure 4.",A.7 Find an approximate expression for the static voltage $V_{0}$ required for operating the magnetron in the manner described. (The expression you will find gives an approximation for the minimal value required for the magnetron operation; the optimal voltage is somewhat higher.),"['The magnitude of the electric field in the region considered, $r=(b+a) / 2$, is the magnitude of the static field, that is, $E=V_{0} /(b-a)$, giving rise to an azimuthal drift velocity of magnitude $u_{D}=E / B_{0}=V_{0} /\\left[B_{0}(b-a)\\right]$. Equating $u_{D} / r$ with the angular velocity found in the previous task we find $V_{0}=\\pi f B_{0}\\left(b^{2}-a^{2}\\right) / 4$']",['$V_{0}=\\pi f B_{0}\\left(b^{2}-a^{2}\\right) / 4$'],False,,Expression, 1303,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively. Context question: A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial. Context answer: \boxed{0.3} Extra Supplementary Reading Materials: Figure 2 Context question: A.4 Figure 2 depicts the alternating electric field lines between the anode and the cathode at a given moment in time (the static field is not plotted). Indicate in the Answer Sheet which of the electrons positioned at A,B,C,D and E will drift towards the anode, which will drift towards the cathode and which will drift at a direction perpendicular to the radius at that moment. Context answer: Extra Supplementary Reading Materials: Figure 3 Figure 3 depicts the alternating electric field lines between the anode and the cathode (the static field is not plotted) at a given moment in time. The positions of six electrons at that moment are denoted by A, $B, C, D, E$ and $F$. All electrons are at the same distance from the cathode. Context question: A.5 Consider the situation shown in Figure 3. For each of the six electron pairs $A B$, $A C, B C, D E, D F, E F$, indicate in the Answer Sheet whether their drift will cause the angle between their position vectors (measured from the cathode's center $0)$ to increase or decrease at that moment. Context answer: | points | angle decreases | angle increases | indeterminate | | ---: | ---: | :--- | ---: | | AB | $X$ | | | | BC | $X$ | | | | CA | | $X$ | | | DE | | $X$ | | | EF | | $X$ | | | DF | | | | Extra Supplementary Reading Materials: Figure 4 The pattern you have discovered in Task A. 5 acts as a focusing mechanism, concentrating the electrons in the space between the cathode and anode into spokes. Figure 4 depicts one such spoke, denoted by S. Context question: A.6 Depict in the Answer Sheet the other spokes at that moment. Indicate by arrows their direction of rotation, and calculate their average angular velocity $\omega_{s}$. Context answer: Extra Supplementary Reading Materials: Make the approximation that the total electric field half-way between the cathode and the anode is equal to its average static value along a radial line from the cathode to the anode, and that the spokes are approximately radial in that region. The cathode and anode radii ( $a$ and $b$, respectively) are defined in Figure 4. Context question: A.7 Find an approximate expression for the static voltage $V_{0}$ required for operating the magnetron in the manner described. (The expression you will find gives an approximation for the minimal value required for the magnetron operation; the optimal voltage is somewhat higher.) Context answer: \boxed{$V_{0}=\pi f B_{0}\left(b^{2}-a^{2}\right) / 4$} Extra Supplementary Reading Materials: Part B: The interaction of microwave radiation with water molecules This part deals with the usage of microwave radiation (radiated by the magnetron antenna into the food chamber) for cooking, that is, heating up a lossy dielectric material such as water, either pure or salty (which is our model for, say, soup). An electric dipole is a configuration of two equal and opposite electric charges $q$ and $-q$ a small distance $d$ apart. The electric dipole vector points from the negative to the positive charge, and its magnitude is $p=q d$. A time-dependent electric field $\vec{E}(t)=E(t) \hat{x}$ is applied on a single dipole of moment $\vec{p}(t)$ with constant magnitude $p_{0}=|\vec{p}(t)|$. The angle between the dipole and the electric field is $\theta(t)$.","B.1 Write expressions for both the magnitude of the torque $\tau(t)$ applied by the electric field on the dipole and the power $H_{i}(t)$ delivered by the field to the dipole, in terms of $p_{0}, E(t), \theta(t)$ and their derivatives.","['The torque at time $t$ is given by $\\tau(t)=-q d \\sin [\\theta(t)] E(t)=-p_{0} \\sin [\\theta(t)] E(t)$, hence the instantaneous power delivered to the dipole by the electric field is\n\n$$\nH_{i}(t)=\\tau(t) \\dot{\\theta}(t)=-p_{0} E(t) \\sin \\theta(t) \\dot{\\theta}(t)=E(t) \\frac{d}{d t}\\left(p_{0} \\cos \\theta(t)\\right)=E(t) \\frac{d p_{x}(t)}{d t}\n$$']","['$\\tau(t)=-p_{0} \\sin [\\theta(t)] E(t)$ , $H_{i}(t)=-p_{0} E(t) \\sin \\theta(t) \\dot{\\theta}(t)$']",True,,Expression, 1304,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively. Context question: A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial. Context answer: \boxed{0.3} Extra Supplementary Reading Materials: Figure 2 Context question: A.4 Figure 2 depicts the alternating electric field lines between the anode and the cathode at a given moment in time (the static field is not plotted). Indicate in the Answer Sheet which of the electrons positioned at A,B,C,D and E will drift towards the anode, which will drift towards the cathode and which will drift at a direction perpendicular to the radius at that moment. Context answer: Extra Supplementary Reading Materials: Figure 3 Figure 3 depicts the alternating electric field lines between the anode and the cathode (the static field is not plotted) at a given moment in time. The positions of six electrons at that moment are denoted by A, $B, C, D, E$ and $F$. All electrons are at the same distance from the cathode. Context question: A.5 Consider the situation shown in Figure 3. For each of the six electron pairs $A B$, $A C, B C, D E, D F, E F$, indicate in the Answer Sheet whether their drift will cause the angle between their position vectors (measured from the cathode's center $0)$ to increase or decrease at that moment. Context answer: | points | angle decreases | angle increases | indeterminate | | ---: | ---: | :--- | ---: | | AB | $X$ | | | | BC | $X$ | | | | CA | | $X$ | | | DE | | $X$ | | | EF | | $X$ | | | DF | | | | Extra Supplementary Reading Materials: Figure 4 The pattern you have discovered in Task A. 5 acts as a focusing mechanism, concentrating the electrons in the space between the cathode and anode into spokes. Figure 4 depicts one such spoke, denoted by S. Context question: A.6 Depict in the Answer Sheet the other spokes at that moment. Indicate by arrows their direction of rotation, and calculate their average angular velocity $\omega_{s}$. Context answer: Extra Supplementary Reading Materials: Make the approximation that the total electric field half-way between the cathode and the anode is equal to its average static value along a radial line from the cathode to the anode, and that the spokes are approximately radial in that region. The cathode and anode radii ( $a$ and $b$, respectively) are defined in Figure 4. Context question: A.7 Find an approximate expression for the static voltage $V_{0}$ required for operating the magnetron in the manner described. (The expression you will find gives an approximation for the minimal value required for the magnetron operation; the optimal voltage is somewhat higher.) Context answer: \boxed{$V_{0}=\pi f B_{0}\left(b^{2}-a^{2}\right) / 4$} Extra Supplementary Reading Materials: Part B: The interaction of microwave radiation with water molecules This part deals with the usage of microwave radiation (radiated by the magnetron antenna into the food chamber) for cooking, that is, heating up a lossy dielectric material such as water, either pure or salty (which is our model for, say, soup). An electric dipole is a configuration of two equal and opposite electric charges $q$ and $-q$ a small distance $d$ apart. The electric dipole vector points from the negative to the positive charge, and its magnitude is $p=q d$. A time-dependent electric field $\vec{E}(t)=E(t) \hat{x}$ is applied on a single dipole of moment $\vec{p}(t)$ with constant magnitude $p_{0}=|\vec{p}(t)|$. The angle between the dipole and the electric field is $\theta(t)$. Context question: B.1 Write expressions for both the magnitude of the torque $\tau(t)$ applied by the electric field on the dipole and the power $H_{i}(t)$ delivered by the field to the dipole, in terms of $p_{0}, E(t), \theta(t)$ and their derivatives. Context answer: \boxed{$\tau(t)=-p_{0} \sin [\theta(t)] E(t)$ , $H_{i}(t)=-p_{0} E(t) \sin \theta(t) \dot{\theta}(t)$} Extra Supplementary Reading Materials: Water molecules are polar, hence can be treated as electric dipoles. Due to the strong hydrogen bonds between water molecules in liquid water, one cannot treat them as independent dipoles. Rather, one should refer to the polarization vector $\vec{P}(t)$, which is the dipole moment density (average dipole moment per unit volume of an ensemble of water molecules). The polarization $\vec{P}(t)$ is parallel to the local applied alternating electric field (of the microwave radiation), $\vec{E}(t)$, and oscillates in time with an amplitude that is proportional to the amplitude of the local alternating electric field, but with a phase lag $\delta$. The local alternating electric field at a given location inside the water is $\vec{E}(t)=E_{0} \sin (\omega t) \hat{x}$, where $\omega=2 \pi f$, giving rise to polarization $\vec{P}(t)=\beta \varepsilon_{0} E_{0} \sin (\omega t-\delta) \hat{x}$, where the dimensionless constant $\beta$ is a property of water.","B.2 Find an expression for the time-averaged power $\langle H(t)\rangle$ per unit volume absorbed by the water. The time-average for a time dependent periodic variable $f(t)$ over its period $T$ is defined as: $$ \langle f(t)\rangle=\frac{1}{T} \int_{t_{0}}^{t_{0}+T} f(t) \mathrm{d} t \tag{1} $$","['Since the average dipole density (hence the average of each molecular dipole) is parallel to the field, the absorbed power density is (angular brackets, $\\langle\\cdots\\rangle$, denote average over time)\n\n$$\n\\begin{aligned}\n& \\langle H(t)\\rangle=\\left\\langle E_{0} \\sin \\left(\\omega_{f} t\\right) \\frac{d P_{x}}{d t}\\right\\rangle=\\left\\langle E_{0} \\sin \\left(\\omega_{f} t\\right) \\frac{d}{d t}\\left(\\beta \\varepsilon_{0} E_{0} \\sin \\left(\\omega_{f} t-\\delta\\right)\\right)\\right\\rangle= \\\\\n& E_{0}^{2} \\beta \\varepsilon_{0} \\omega_{f}\\left\\langle\\sin \\left(\\omega_{f} t\\right) \\cos \\left(\\omega_{f} t-\\delta\\right)\\right\\rangle=0.5 E_{0}^{2} \\beta \\varepsilon_{0} \\omega_{f}\\left\\langle\\sin \\delta+\\sin \\left(2 \\omega_{f} t-\\delta\\right)\\right\\rangle=0.5 E_{0}^{2} \\beta \\varepsilon_{0} \\omega_{f} \\sin \\delta\n\\end{aligned}\n$$']",['$\\langle H(t)\\rangle=0.5 E_{0}^{2} \\beta \\varepsilon_{0} \\omega_{f} \\sin \\delta$'],False,,Expression, 1305,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively. Context question: A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial. Context answer: \boxed{0.3} Extra Supplementary Reading Materials: Figure 2 Context question: A.4 Figure 2 depicts the alternating electric field lines between the anode and the cathode at a given moment in time (the static field is not plotted). Indicate in the Answer Sheet which of the electrons positioned at A,B,C,D and E will drift towards the anode, which will drift towards the cathode and which will drift at a direction perpendicular to the radius at that moment. Context answer: Extra Supplementary Reading Materials: Figure 3 Figure 3 depicts the alternating electric field lines between the anode and the cathode (the static field is not plotted) at a given moment in time. The positions of six electrons at that moment are denoted by A, $B, C, D, E$ and $F$. All electrons are at the same distance from the cathode. Context question: A.5 Consider the situation shown in Figure 3. For each of the six electron pairs $A B$, $A C, B C, D E, D F, E F$, indicate in the Answer Sheet whether their drift will cause the angle between their position vectors (measured from the cathode's center $0)$ to increase or decrease at that moment. Context answer: | points | angle decreases | angle increases | indeterminate | | ---: | ---: | :--- | ---: | | AB | $X$ | | | | BC | $X$ | | | | CA | | $X$ | | | DE | | $X$ | | | EF | | $X$ | | | DF | | | | Extra Supplementary Reading Materials: Figure 4 The pattern you have discovered in Task A. 5 acts as a focusing mechanism, concentrating the electrons in the space between the cathode and anode into spokes. Figure 4 depicts one such spoke, denoted by S. Context question: A.6 Depict in the Answer Sheet the other spokes at that moment. Indicate by arrows their direction of rotation, and calculate their average angular velocity $\omega_{s}$. Context answer: Extra Supplementary Reading Materials: Make the approximation that the total electric field half-way between the cathode and the anode is equal to its average static value along a radial line from the cathode to the anode, and that the spokes are approximately radial in that region. The cathode and anode radii ( $a$ and $b$, respectively) are defined in Figure 4. Context question: A.7 Find an approximate expression for the static voltage $V_{0}$ required for operating the magnetron in the manner described. (The expression you will find gives an approximation for the minimal value required for the magnetron operation; the optimal voltage is somewhat higher.) Context answer: \boxed{$V_{0}=\pi f B_{0}\left(b^{2}-a^{2}\right) / 4$} Extra Supplementary Reading Materials: Part B: The interaction of microwave radiation with water molecules This part deals with the usage of microwave radiation (radiated by the magnetron antenna into the food chamber) for cooking, that is, heating up a lossy dielectric material such as water, either pure or salty (which is our model for, say, soup). An electric dipole is a configuration of two equal and opposite electric charges $q$ and $-q$ a small distance $d$ apart. The electric dipole vector points from the negative to the positive charge, and its magnitude is $p=q d$. A time-dependent electric field $\vec{E}(t)=E(t) \hat{x}$ is applied on a single dipole of moment $\vec{p}(t)$ with constant magnitude $p_{0}=|\vec{p}(t)|$. The angle between the dipole and the electric field is $\theta(t)$. Context question: B.1 Write expressions for both the magnitude of the torque $\tau(t)$ applied by the electric field on the dipole and the power $H_{i}(t)$ delivered by the field to the dipole, in terms of $p_{0}, E(t), \theta(t)$ and their derivatives. Context answer: \boxed{$\tau(t)=-p_{0} \sin [\theta(t)] E(t)$ , $H_{i}(t)=-p_{0} E(t) \sin \theta(t) \dot{\theta}(t)$} Extra Supplementary Reading Materials: Water molecules are polar, hence can be treated as electric dipoles. Due to the strong hydrogen bonds between water molecules in liquid water, one cannot treat them as independent dipoles. Rather, one should refer to the polarization vector $\vec{P}(t)$, which is the dipole moment density (average dipole moment per unit volume of an ensemble of water molecules). The polarization $\vec{P}(t)$ is parallel to the local applied alternating electric field (of the microwave radiation), $\vec{E}(t)$, and oscillates in time with an amplitude that is proportional to the amplitude of the local alternating electric field, but with a phase lag $\delta$. The local alternating electric field at a given location inside the water is $\vec{E}(t)=E_{0} \sin (\omega t) \hat{x}$, where $\omega=2 \pi f$, giving rise to polarization $\vec{P}(t)=\beta \varepsilon_{0} E_{0} \sin (\omega t-\delta) \hat{x}$, where the dimensionless constant $\beta$ is a property of water. Context question: B.2 Find an expression for the time-averaged power $\langle H(t)\rangle$ per unit volume absorbed by the water. The time-average for a time dependent periodic variable $f(t)$ over its period $T$ is defined as: $$ \langle f(t)\rangle=\frac{1}{T} \int_{t_{0}}^{t_{0}+T} f(t) \mathrm{d} t \tag{1} $$ Context answer: \boxed{$\langle H(t)\rangle=0.5 E_{0}^{2} \beta \varepsilon_{0} \omega_{f} \sin \delta$} Extra Supplementary Reading Materials: Let us now consider the propagation of the radiation through the water. The relative dielectric constant of water (at the electromagnetic field frequency) is $\varepsilon_{r}$, and the corresponding index of refraction of water is $n=\sqrt{\varepsilon_{r}}$. The momentary energy density of the electric field is given by $\frac{1}{2} \varepsilon_{r} \varepsilon_{0} E^{2}$. The time-averaged energy density of the electric and magnetic fields are equal.","B.3 Let us denote the time-averaged radiation energy flux density by $I(z)$ (average radiation power flow per unit area). Here $z$ is the depth of penetration into the water, and the radiation propagates in the $z$ direction. Find an expression for the dependence of the flux density $I(z)$ on $z$. The flux density at the water surface, $I(0)$, may appear in your result.","['The energy density of the electromagnetic field at penetration depth $z$, which is twice the electric energy density, is $2 \\times \\varepsilon_{r} \\varepsilon_{0}\\left\\langle E^{2}(z, t)\\right\\rangle / 2=\\varepsilon_{r} \\varepsilon_{0} E_{0}^{2}(z)\\left\\langle\\sin ^{2}(\\omega t)\\right\\rangle=\\varepsilon_{r} \\varepsilon_{0} E_{0}^{2}(z) / 2$. Therefore, the time-averaged flux density at depth $z$ is:\n\n$$\nI(z)=\\frac{1}{2} \\varepsilon_{r} \\varepsilon_{0} E_{0}^{2}(z) \\times \\frac{c}{n}=\\frac{1}{2} \\sqrt{\\varepsilon_{r}} \\varepsilon_{0} c E_{0}^{2}(z),\n$$\n\nwhere $c$ is the speed of light in vacuum. $I$ decreases with $z$ due to the absorbed power calculated in the previous task we find\n\n$$\n\\frac{d I(z)}{d z}=-\\frac{1}{2} \\beta \\varepsilon_{0} \\omega E_{0}^{2}(z) \\sin \\delta=-\\frac{\\beta \\omega \\sin \\delta}{c \\sqrt{\\varepsilon_{r}}} I(z)\n$$\n\nhence $I(z)=I(0) \\exp \\left[-z \\beta \\omega \\sin \\delta /\\left(c \\sqrt{\\varepsilon_{r}}\\right)\\right]$.']",['$I(z)=I(0) e^{[-z \\beta \\omega \\sin \\delta /(c \\sqrt{\\varepsilon_{r}})]}$'],False,,Expression, 1306,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively. Context question: A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial. Context answer: \boxed{0.3} Extra Supplementary Reading Materials: Figure 2 Context question: A.4 Figure 2 depicts the alternating electric field lines between the anode and the cathode at a given moment in time (the static field is not plotted). Indicate in the Answer Sheet which of the electrons positioned at A,B,C,D and E will drift towards the anode, which will drift towards the cathode and which will drift at a direction perpendicular to the radius at that moment. Context answer: Extra Supplementary Reading Materials: Figure 3 Figure 3 depicts the alternating electric field lines between the anode and the cathode (the static field is not plotted) at a given moment in time. The positions of six electrons at that moment are denoted by A, $B, C, D, E$ and $F$. All electrons are at the same distance from the cathode. Context question: A.5 Consider the situation shown in Figure 3. For each of the six electron pairs $A B$, $A C, B C, D E, D F, E F$, indicate in the Answer Sheet whether their drift will cause the angle between their position vectors (measured from the cathode's center $0)$ to increase or decrease at that moment. Context answer: | points | angle decreases | angle increases | indeterminate | | ---: | ---: | :--- | ---: | | AB | $X$ | | | | BC | $X$ | | | | CA | | $X$ | | | DE | | $X$ | | | EF | | $X$ | | | DF | | | | Extra Supplementary Reading Materials: Figure 4 The pattern you have discovered in Task A. 5 acts as a focusing mechanism, concentrating the electrons in the space between the cathode and anode into spokes. Figure 4 depicts one such spoke, denoted by S. Context question: A.6 Depict in the Answer Sheet the other spokes at that moment. Indicate by arrows their direction of rotation, and calculate their average angular velocity $\omega_{s}$. Context answer: Extra Supplementary Reading Materials: Make the approximation that the total electric field half-way between the cathode and the anode is equal to its average static value along a radial line from the cathode to the anode, and that the spokes are approximately radial in that region. The cathode and anode radii ( $a$ and $b$, respectively) are defined in Figure 4. Context question: A.7 Find an approximate expression for the static voltage $V_{0}$ required for operating the magnetron in the manner described. (The expression you will find gives an approximation for the minimal value required for the magnetron operation; the optimal voltage is somewhat higher.) Context answer: \boxed{$V_{0}=\pi f B_{0}\left(b^{2}-a^{2}\right) / 4$} Extra Supplementary Reading Materials: Part B: The interaction of microwave radiation with water molecules This part deals with the usage of microwave radiation (radiated by the magnetron antenna into the food chamber) for cooking, that is, heating up a lossy dielectric material such as water, either pure or salty (which is our model for, say, soup). An electric dipole is a configuration of two equal and opposite electric charges $q$ and $-q$ a small distance $d$ apart. The electric dipole vector points from the negative to the positive charge, and its magnitude is $p=q d$. A time-dependent electric field $\vec{E}(t)=E(t) \hat{x}$ is applied on a single dipole of moment $\vec{p}(t)$ with constant magnitude $p_{0}=|\vec{p}(t)|$. The angle between the dipole and the electric field is $\theta(t)$. Context question: B.1 Write expressions for both the magnitude of the torque $\tau(t)$ applied by the electric field on the dipole and the power $H_{i}(t)$ delivered by the field to the dipole, in terms of $p_{0}, E(t), \theta(t)$ and their derivatives. Context answer: \boxed{$\tau(t)=-p_{0} \sin [\theta(t)] E(t)$ , $H_{i}(t)=-p_{0} E(t) \sin \theta(t) \dot{\theta}(t)$} Extra Supplementary Reading Materials: Water molecules are polar, hence can be treated as electric dipoles. Due to the strong hydrogen bonds between water molecules in liquid water, one cannot treat them as independent dipoles. Rather, one should refer to the polarization vector $\vec{P}(t)$, which is the dipole moment density (average dipole moment per unit volume of an ensemble of water molecules). The polarization $\vec{P}(t)$ is parallel to the local applied alternating electric field (of the microwave radiation), $\vec{E}(t)$, and oscillates in time with an amplitude that is proportional to the amplitude of the local alternating electric field, but with a phase lag $\delta$. The local alternating electric field at a given location inside the water is $\vec{E}(t)=E_{0} \sin (\omega t) \hat{x}$, where $\omega=2 \pi f$, giving rise to polarization $\vec{P}(t)=\beta \varepsilon_{0} E_{0} \sin (\omega t-\delta) \hat{x}$, where the dimensionless constant $\beta$ is a property of water. Context question: B.2 Find an expression for the time-averaged power $\langle H(t)\rangle$ per unit volume absorbed by the water. The time-average for a time dependent periodic variable $f(t)$ over its period $T$ is defined as: $$ \langle f(t)\rangle=\frac{1}{T} \int_{t_{0}}^{t_{0}+T} f(t) \mathrm{d} t \tag{1} $$ Context answer: \boxed{$\langle H(t)\rangle=0.5 E_{0}^{2} \beta \varepsilon_{0} \omega_{f} \sin \delta$} Extra Supplementary Reading Materials: Let us now consider the propagation of the radiation through the water. The relative dielectric constant of water (at the electromagnetic field frequency) is $\varepsilon_{r}$, and the corresponding index of refraction of water is $n=\sqrt{\varepsilon_{r}}$. The momentary energy density of the electric field is given by $\frac{1}{2} \varepsilon_{r} \varepsilon_{0} E^{2}$. The time-averaged energy density of the electric and magnetic fields are equal. Context question: B.3 Let us denote the time-averaged radiation energy flux density by $I(z)$ (average radiation power flow per unit area). Here $z$ is the depth of penetration into the water, and the radiation propagates in the $z$ direction. Find an expression for the dependence of the flux density $I(z)$ on $z$. The flux density at the water surface, $I(0)$, may appear in your result. Context answer: \boxed{$I(z)=I(0) e^{[-z \beta \omega \sin \delta /(c \sqrt{\varepsilon_{r}})]}$} Extra Supplementary Reading Materials: The phase $\operatorname{lag} \delta$ is the result of the interaction between the water molecules. It depends on the dimensionless dielectric loss coefficient $\varepsilon_{\ell}$ and the relative dielectric constant $\varepsilon_{r}$ (both of which depend on the radiation angular frequency $\omega$ and the temperature) via the relation $\tan \delta=\varepsilon_{\ell} / \varepsilon_{r}$. When $\delta$ is small enough, the electric field at penetration depth $z$ into the water is given by: $$ \vec{E}(z, t)=\vec{E}_{0} e^{-\frac{1}{2} n k_{0} z \tan \delta} \sin \left(n k_{0} z-\omega t\right) $$ where $k_{0}=\omega / c$ and $c=3.0 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}$ is the speed of light in vacuum.",B.4 Employ the approximation $\tan \delta \approx \sin \delta$ and find an expression for the coefficient $\beta$ defined in Task B. 2 in terms of the other parameters.,"['Similarly to the previous task, the energy flux corresponding to the given field is\n\n$$\nI(z)=\\sqrt{\\varepsilon_{r}} \\varepsilon_{0} c\\left\\langle E^{2}(z, t)\\right\\rangle=\\frac{1}{2} \\sqrt{\\varepsilon_{r}} \\varepsilon_{0} c E_{0}^{2} e^{-z \\omega \\sqrt{\\varepsilon_{r}} \\tan \\delta / c} .\n$$\n\nEquating the argument of the exponent in the last expression with the result of the previous task, and using the given approximation $\\tan \\delta \\approx \\sin \\delta$ leads to $\\beta=\\varepsilon_{r}$.']",['$\\beta=\\varepsilon_{r}$'],False,,Expression, 1307,Electromagnetism,"The Physics of a Microwave Oven This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called ""magnetron"". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food. Figure 1 $$ \begin{gathered} d=1 \mathrm{~mm} \\ l=6 \mathrm{~mm} \\ R=7 \mathrm{~mm} \\ h=17.5 \mathrm{~mm} \end{gathered} $$ Part A: The structure and operation of a magnetron A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron. A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as ""resonators"". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$. A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations. The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field. Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively. Context question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.) Context answer: \boxed{$2.0 \times 10^{9}$} Extra Supplementary Reading Materials: Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively. Context question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if: 1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ 2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$. Context answer: \boxed{$u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$} Extra Supplementary Reading Materials: We now resume our discussion of the magnetron. The distance between the cathode and the anode is $15 \mathrm{~mm}$. Assume that, due to the aforementioned energy loss to the alternating fields, the maximal kinetic energy of each electron does not exceed $K_{\max }=800 \mathrm{eV}$. The static magnetic field strength is $B_{0}=0.3 \mathrm{~T}$. The electron mass and charge are $m=9.1 \cdot 10^{-31} \mathrm{~kg}$ and $-e=-1.6 \cdot 10^{-19} \mathrm{C}$, respectively. Context question: A.3 Numerically estimate the maximal radius $r$ of the electron motion trajectory in the reference frame in which this motion is approximately circular, considering this reference frame as approximately inertial. Context answer: \boxed{0.3} Extra Supplementary Reading Materials: Figure 2 Context question: A.4 Figure 2 depicts the alternating electric field lines between the anode and the cathode at a given moment in time (the static field is not plotted). Indicate in the Answer Sheet which of the electrons positioned at A,B,C,D and E will drift towards the anode, which will drift towards the cathode and which will drift at a direction perpendicular to the radius at that moment. Context answer: Extra Supplementary Reading Materials: Figure 3 Figure 3 depicts the alternating electric field lines between the anode and the cathode (the static field is not plotted) at a given moment in time. The positions of six electrons at that moment are denoted by A, $B, C, D, E$ and $F$. All electrons are at the same distance from the cathode. Context question: A.5 Consider the situation shown in Figure 3. For each of the six electron pairs $A B$, $A C, B C, D E, D F, E F$, indicate in the Answer Sheet whether their drift will cause the angle between their position vectors (measured from the cathode's center $0)$ to increase or decrease at that moment. Context answer: | points | angle decreases | angle increases | indeterminate | | ---: | ---: | :--- | ---: | | AB | $X$ | | | | BC | $X$ | | | | CA | | $X$ | | | DE | | $X$ | | | EF | | $X$ | | | DF | | | | Extra Supplementary Reading Materials: Figure 4 The pattern you have discovered in Task A. 5 acts as a focusing mechanism, concentrating the electrons in the space between the cathode and anode into spokes. Figure 4 depicts one such spoke, denoted by S. Context question: A.6 Depict in the Answer Sheet the other spokes at that moment. Indicate by arrows their direction of rotation, and calculate their average angular velocity $\omega_{s}$. Context answer: Extra Supplementary Reading Materials: Make the approximation that the total electric field half-way between the cathode and the anode is equal to its average static value along a radial line from the cathode to the anode, and that the spokes are approximately radial in that region. The cathode and anode radii ( $a$ and $b$, respectively) are defined in Figure 4. Context question: A.7 Find an approximate expression for the static voltage $V_{0}$ required for operating the magnetron in the manner described. (The expression you will find gives an approximation for the minimal value required for the magnetron operation; the optimal voltage is somewhat higher.) Context answer: \boxed{$V_{0}=\pi f B_{0}\left(b^{2}-a^{2}\right) / 4$} Extra Supplementary Reading Materials: Part B: The interaction of microwave radiation with water molecules This part deals with the usage of microwave radiation (radiated by the magnetron antenna into the food chamber) for cooking, that is, heating up a lossy dielectric material such as water, either pure or salty (which is our model for, say, soup). An electric dipole is a configuration of two equal and opposite electric charges $q$ and $-q$ a small distance $d$ apart. The electric dipole vector points from the negative to the positive charge, and its magnitude is $p=q d$. A time-dependent electric field $\vec{E}(t)=E(t) \hat{x}$ is applied on a single dipole of moment $\vec{p}(t)$ with constant magnitude $p_{0}=|\vec{p}(t)|$. The angle between the dipole and the electric field is $\theta(t)$. Context question: B.1 Write expressions for both the magnitude of the torque $\tau(t)$ applied by the electric field on the dipole and the power $H_{i}(t)$ delivered by the field to the dipole, in terms of $p_{0}, E(t), \theta(t)$ and their derivatives. Context answer: \boxed{$\tau(t)=-p_{0} \sin [\theta(t)] E(t)$ , $H_{i}(t)=-p_{0} E(t) \sin \theta(t) \dot{\theta}(t)$} Extra Supplementary Reading Materials: Water molecules are polar, hence can be treated as electric dipoles. Due to the strong hydrogen bonds between water molecules in liquid water, one cannot treat them as independent dipoles. Rather, one should refer to the polarization vector $\vec{P}(t)$, which is the dipole moment density (average dipole moment per unit volume of an ensemble of water molecules). The polarization $\vec{P}(t)$ is parallel to the local applied alternating electric field (of the microwave radiation), $\vec{E}(t)$, and oscillates in time with an amplitude that is proportional to the amplitude of the local alternating electric field, but with a phase lag $\delta$. The local alternating electric field at a given location inside the water is $\vec{E}(t)=E_{0} \sin (\omega t) \hat{x}$, where $\omega=2 \pi f$, giving rise to polarization $\vec{P}(t)=\beta \varepsilon_{0} E_{0} \sin (\omega t-\delta) \hat{x}$, where the dimensionless constant $\beta$ is a property of water. Context question: B.2 Find an expression for the time-averaged power $\langle H(t)\rangle$ per unit volume absorbed by the water. The time-average for a time dependent periodic variable $f(t)$ over its period $T$ is defined as: $$ \langle f(t)\rangle=\frac{1}{T} \int_{t_{0}}^{t_{0}+T} f(t) \mathrm{d} t \tag{1} $$ Context answer: \boxed{$\langle H(t)\rangle=0.5 E_{0}^{2} \beta \varepsilon_{0} \omega_{f} \sin \delta$} Extra Supplementary Reading Materials: Let us now consider the propagation of the radiation through the water. The relative dielectric constant of water (at the electromagnetic field frequency) is $\varepsilon_{r}$, and the corresponding index of refraction of water is $n=\sqrt{\varepsilon_{r}}$. The momentary energy density of the electric field is given by $\frac{1}{2} \varepsilon_{r} \varepsilon_{0} E^{2}$. The time-averaged energy density of the electric and magnetic fields are equal. Context question: B.3 Let us denote the time-averaged radiation energy flux density by $I(z)$ (average radiation power flow per unit area). Here $z$ is the depth of penetration into the water, and the radiation propagates in the $z$ direction. Find an expression for the dependence of the flux density $I(z)$ on $z$. The flux density at the water surface, $I(0)$, may appear in your result. Context answer: \boxed{$I(z)=I(0) e^{[-z \beta \omega \sin \delta /(c \sqrt{\varepsilon_{r}})]}$} Extra Supplementary Reading Materials: The phase $\operatorname{lag} \delta$ is the result of the interaction between the water molecules. It depends on the dimensionless dielectric loss coefficient $\varepsilon_{\ell}$ and the relative dielectric constant $\varepsilon_{r}$ (both of which depend on the radiation angular frequency $\omega$ and the temperature) via the relation $\tan \delta=\varepsilon_{\ell} / \varepsilon_{r}$. When $\delta$ is small enough, the electric field at penetration depth $z$ into the water is given by: $$ \vec{E}(z, t)=\vec{E}_{0} e^{-\frac{1}{2} n k_{0} z \tan \delta} \sin \left(n k_{0} z-\omega t\right) $$ where $k_{0}=\omega / c$ and $c=3.0 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}$ is the speed of light in vacuum. Context question: B.4 Employ the approximation $\tan \delta \approx \sin \delta$ and find an expression for the coefficient $\beta$ defined in Task B. 2 in terms of the other parameters. Context answer: \boxed{$\beta=\varepsilon_{r}$} Extra Supplementary Reading Materials: Figure 5. The arrows indicate the variation with temperature across the curves from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. Figure 5 depicts $\varepsilon_{\ell}$ (blue) and $\varepsilon_{r}$ (red) for both pure water (solid lines) and a dilute solution of salt in water (dashed lines) as functions of wavelength or frequency, at several different temperatures. The angular frequency $\omega=2 \pi \cdot 2.45 \cdot 10^{9} \mathrm{~s}^{-1}$ is indicated by a bold vertical line. Below we will consider microwave radiation at this frequency only.","B.5 Use Figure 5 to address the following question: For water at $20^{\circ} \mathrm{C}$, find the penetration depth $z_{1 / 2}$ at which the power per unit volume is reduced to half of its value at $z=0$.","['Using previous results, the radiation power per unit area is reduced to half of its $z=0$ value at $z_{1 / 2}=c \\ln 2 /\\left(\\omega \\sqrt{\\varepsilon_{r}} \\tan \\delta\\right)=c \\sqrt{\\varepsilon_{r}} \\ln 2 /\\left(\\omega \\varepsilon_{l}\\right)$. From the given graph, at the given frequency $\\varepsilon_{r} \\approx 78$ and $\\varepsilon_{l} \\approx 10$, hence $z_{1 / 2} \\approx 12 \\mathrm{~mm}$.']",['$12$'],False,mm,Numerical,1e0 1308,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1","A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system?","['The boundary conditions are: $u(0, t)=u(L, t)=0$. As a result, $\\sin \\left(\\frac{2 \\pi}{\\lambda} L\\right)=0$, so we get $\\lambda_{\\max }=2 L$.']",['$\\lambda_{\\max }=2 L$'],False,,Expression, 1309,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values.","A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$.","['We get\n\n$$\nV(x, t)=S \\cdot(\\Delta x+u(x+\\Delta x, t)-u(x, t))=S \\Delta x \\cdot\\left(1+u^{\\prime}\\right)=V_{0}+V_{0} u^{\\prime} .\n$$\n\nThus,\n\n$$\nV(x, t)=V_{0}+a k V_{0} \\cos (k x) \\cos (\\omega t) \\quad \\Rightarrow \\quad V_{1}(x)=a k V_{0} \\cos (k x) .\n$$']",['$V_{1}(x)=a k V_{0} \\cos (k x)$'],False,,Expression, 1310,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} ","A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$.","[""We use Newton's Second Law $\\rho_{0} \\ddot{u}=-p^{\\prime}$ to deduce $p^{\\prime}=-\\rho_{0} \\ddot{u}=\\rho_{0} a \\omega^{2} \\sin (k x) \\cos (\\omega t)$, so that\n\n$$\np(x, t)=p_{0}-a \\frac{\\omega^{2}}{k} \\rho_{0} \\cos (k x) \\cos (\\omega t) \\quad \\Rightarrow \\quad p_{1}(x)=a \\frac{\\omega^{2}}{k} \\rho_{0} \\cos (k x) .\n$$""]",['$p_{1}(x)=a \\frac{\\omega^{2}}{k} \\rho_{0} \\cos (k x)$'],False,,Expression, 1311,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant.","A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$.","['Using $a \\ll L$, we obtain $\\frac{p_{1}(x)}{p_{0}}=\\gamma \\frac{V_{1}(x)}{V_{0}}$. As a result, $\\frac{\\rho_{0}}{p_{0}} \\frac{\\omega^{2}}{k}=\\gamma \\cdot k$, and $c=\\sqrt{\\frac{\\gamma p_{0}}{\\rho_{0}}}$.']",['$c=\\sqrt{\\frac{\\gamma p_{0}}{\\rho_{0}}}$'],False,,Expression, 1312,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} ","A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$.","['The relative change in $T(x, t)$ is the sum of the relative changes in $V(x, t)$ and $p(x, t)$. As a result,\n\n$$\nT_{1}(x)=\\frac{T_{0}}{p_{0}} p_{1}(x)-\\frac{T_{0}}{V_{0}} V_{1}(x)=(\\gamma-1) \\frac{T_{0}}{V_{0}} V_{1}(x)=a k(\\gamma-1) T_{0} \\cos (k x) .\n$$']",['$a k(\\gamma-1) T_{0} \\cos (k x)$'],False,,Expression, 1313,Optics,,"A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. ![](https://cdn.mathpix.com/cropped/2023_12_21_2fffc9c901c17aa00e2eg-1.jpg?height=191&width=826&top_left_y=2417&top_left_x=615) Figure 2","['The movement of the gas parcels inside the tube conveys heat along its boundary. To determine the direction of the convection, we combining the result of Task A. 5 and the expression (1) for $u(x, t)$. We see that when $0 Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} ","A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2","['The movement of the gas parcels inside the tube conveys heat along its boundary. To determine the direction of the convection, we combining the result of Task A. 5 and the expression (1) for $u(x, t)$. We see that when $0 Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack.","B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$.","['We get\n\n$$\nT_{\\text {env }}(t)=T_{\\text {plate }}\\left(x_{0}+u\\left(x_{0}, t\\right)\\right)=T_{0}-\\frac{\\tau}{\\ell} \\cdot u\\left(x_{0}, t\\right)\n$$\n\nso that:\n\n$$\nT_{\\mathrm{st}}=\\frac{a \\tau}{\\ell} \\sin \\left(k x_{0}\\right)=\\frac{a \\tau}{\\ell \\sqrt{2}}\n$$']",['$\\frac{a \\tau}{\\ell \\sqrt{2}}$'],False,,Expression, 1315,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack. Context question: B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$. Context answer: \boxed{$\frac{a \tau}{\ell \sqrt{2}}$} ","B.2 Above which critical temperature difference $\tau_{\text {cr }}$ will the gas be conveying heat from the hot reservoir to the cold one? Express $\tau_{\mathrm{cr}}$ in terms of $T_{0}, \gamma, k$ and $\ell$.","['The gas will convey heat from the hot reservoir to the cold one if the parcels are colder than the environment when $u\\left(x_{0}, t\\right)<0$, and hotter when $u\\left(x_{0}, t\\right)>0$. This occurs precisely if\n\n$$\nT_{\\mathrm{st}}>T_{1} .\n$$\n\nPlugging in the results of Tasks A. 5 and B.1, we get\n\n$$\n\\frac{a \\tau_{\\mathrm{cr}}}{\\ell} \\sin \\left(k x_{0}\\right)=a k(\\gamma-1) T_{0} \\cos \\left(k x_{0}\\right) \\quad \\Rightarrow \\quad \\tau_{\\mathrm{cr}}=k \\ell(\\gamma-1) T_{0}\n$$']",['$\\tau_{\\mathrm{cr}}=k \\ell(\\gamma-1) T_{0}$'],False,,Expression, 1316,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack. Context question: B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$. Context answer: \boxed{$\frac{a \tau}{\ell \sqrt{2}}$} Context question: B.2 Above which critical temperature difference $\tau_{\text {cr }}$ will the gas be conveying heat from the hot reservoir to the cold one? Express $\tau_{\mathrm{cr}}$ in terms of $T_{0}, \gamma, k$ and $\ell$. Context answer: \boxed{$\tau_{\mathrm{cr}}=k \ell(\gamma-1) T_{0}$} ","B.3 Obtain the general approximate expression for the heat flow $\frac{d Q}{d t}$ into a small parcel of gas as a linear function of its volume and pressure change rates. Express your answer in terms of the rate of volume change $\frac{d V}{d t}$, the rate of pressure change $\frac{d p}{d t}$, the unperturbed equilibrium values of parcel pressure and volume $p_{0}, V_{0}$ and the adiabatic index $\gamma$. (You may use the expression for the molar heat capacity at constant volume $c_{v}=\frac{R}{\gamma-1}$, where $R$ is the gas constant.)","['Using the first law of thermodynamics, we get\n\n$$\n\\frac{d Q}{d t}=\\frac{d E}{d t}+p \\frac{d V}{d t}\n$$\n\nPlugging in the relation $E=\\frac{1}{\\gamma-1} p V$, we see that:\n\n$$\n\\frac{d Q}{d t}=\\frac{1}{\\gamma-1} \\frac{d}{d t}(p V)+p \\frac{d V}{d t}=\\frac{1}{\\gamma-1} V \\frac{d p}{d t}+\\frac{\\gamma}{\\gamma-1} p \\frac{d V}{d t} \\approx \\frac{1}{\\gamma-1} V_{0} \\frac{d p}{d t}+\\frac{\\gamma}{\\gamma-1} p_{0} \\frac{d V}{d t}\n$$']",['$\\frac{1}{\\gamma-1} V_{0} \\frac{d p}{d t}+\\frac{\\gamma}{\\gamma-1} p_{0} \\frac{d V}{d t}$'],False,,Expression, 1317,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack. Context question: B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$. Context answer: \boxed{$\frac{a \tau}{\ell \sqrt{2}}$} Context question: B.2 Above which critical temperature difference $\tau_{\text {cr }}$ will the gas be conveying heat from the hot reservoir to the cold one? Express $\tau_{\mathrm{cr}}$ in terms of $T_{0}, \gamma, k$ and $\ell$. Context answer: \boxed{$\tau_{\mathrm{cr}}=k \ell(\gamma-1) T_{0}$} Context question: B.3 Obtain the general approximate expression for the heat flow $\frac{d Q}{d t}$ into a small parcel of gas as a linear function of its volume and pressure change rates. Express your answer in terms of the rate of volume change $\frac{d V}{d t}$, the rate of pressure change $\frac{d p}{d t}$, the unperturbed equilibrium values of parcel pressure and volume $p_{0}, V_{0}$ and the adiabatic index $\gamma$. (You may use the expression for the molar heat capacity at constant volume $c_{v}=\frac{R}{\gamma-1}$, where $R$ is the gas constant.) Context answer: \boxed{$\frac{1}{\gamma-1} V_{0} \frac{d p}{d t}+\frac{\gamma}{\gamma-1} p_{0} \frac{d V}{d t}$} Extra Supplementary Reading Materials: The limited heat flow rate between the parcel and the stack causes a phase difference between the pressure and volume oscillations of the parcel. We will see how this generates work. Let the heat flux into the parcel from the stack be proportional to the temperature difference between the parcel and the neighboring element of the stack, given approximately by $\frac{d Q}{d t}=-\beta V_{0}\left(T_{\mathrm{st}}-T_{1}\right) \cos (\omega t)$. Here $T_{1}$ and $T_{\mathrm{st}}$ are the temperature oscillation amplitudes of the gas parcel and the neighbouring stack from Tasks A. 5 and B. 1 , respectively, and $\beta>0$ is a constant. Assume that at the machine's operating frequencies, the change in gas temperature as a result of this heat flow is insignificant compared to both $T_{1}$ and $T_{\mathrm{st}}$.","B.4 In order to calculate work, we will consider a change to the volume of the moving parcel as a result of the thermal contact with the stack. Let us write the pressure and the volume of the parcel under the stack's influence in the form: $$ \begin{gathered} p=p_{0}+p_{a} \sin (\omega t)-p_{b} \cos (\omega t), \\ V=V_{0}+V_{a} \sin (\omega t)+V_{b} \cos (\omega t) . \end{gathered} \tag{6} $$ Given $p_{a}$ and $p_{b}$, find the coefficients $V_{a}$ and $V_{b}$. Express your answer in terms of $p_{a}, p_{b}, p_{0}, V_{0}, \gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a$ and $\ell$.","['We plug the expression for $\\frac{d Q}{d t}$ into the result of Task B.3. This gives:\n\n$$\n\\frac{1}{\\gamma-1} V_{0} \\frac{d p}{d t}+\\frac{\\gamma}{\\gamma-1} p_{0} \\frac{d V}{d t}=\\beta V_{0}\\left(T_{\\mathrm{st}}-T_{1}\\right) \\cdot \\cos (\\omega t)\n$$\n\nWe now plug in the data given in equation (6), and get (by considering terms with $\\cos (\\omega t)$ and $\\sin (\\omega t)$ separately):\n\n$$\n\\begin{gathered}\n\\frac{1}{\\gamma-1} V_{0} p_{a} \\omega+\\frac{\\gamma}{\\gamma-1} p_{0} V_{a} \\omega=\\beta V_{0}\\left(T_{\\mathrm{st}}-T_{1}\\right) \\\\\n\\frac{1}{\\gamma-1} V_{0} p_{b} \\omega-\\frac{\\gamma}{\\gamma-1} p_{0} V_{b} \\omega=0\n\\end{gathered}\n$$\n\nand thus, we can already express $V_{b}$ as\n\n$$\nV_{b}=\\frac{1}{\\gamma} p_{b} \\cdot \\frac{V_{0}}{p_{0}}\n$$\n\nFor $V_{a}$, we plug in the results of Tasks B. 1 and B.2,\n\n$$\nT_{\\mathrm{st}}-T_{1}=\\frac{a}{\\ell \\sqrt{2}}\\left(\\tau-\\tau_{\\mathrm{cr}}\\right)\n$$\n\ngiving:\n\n$$\nV_{a}=\\left(-\\frac{1}{\\gamma} p_{a}-\\frac{\\gamma-1}{\\gamma} \\frac{\\beta}{\\omega} \\frac{a}{\\ell \\sqrt{2}}\\left(\\tau-\\tau_{\\mathrm{cr}}\\right)\\right) \\cdot \\frac{V_{0}}{p_{0}}\n$$']","['$V_{b}=\\frac{1}{\\gamma} p_{b} \\cdot \\frac{V_{0}}{p_{0}}$ , $V_{a}=(-\\frac{1}{\\gamma} p_{a}-\\frac{\\gamma-1}{\\gamma} \\frac{\\beta}{\\omega} \\frac{a}{\\ell \\sqrt{2}}(\\tau-\\tau_{\\mathrm{cr}})) \\cdot \\frac{V_{0}}{p_{0}}$']",True,,Expression, 1318,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack. Context question: B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$. Context answer: \boxed{$\frac{a \tau}{\ell \sqrt{2}}$} Context question: B.2 Above which critical temperature difference $\tau_{\text {cr }}$ will the gas be conveying heat from the hot reservoir to the cold one? Express $\tau_{\mathrm{cr}}$ in terms of $T_{0}, \gamma, k$ and $\ell$. Context answer: \boxed{$\tau_{\mathrm{cr}}=k \ell(\gamma-1) T_{0}$} Context question: B.3 Obtain the general approximate expression for the heat flow $\frac{d Q}{d t}$ into a small parcel of gas as a linear function of its volume and pressure change rates. Express your answer in terms of the rate of volume change $\frac{d V}{d t}$, the rate of pressure change $\frac{d p}{d t}$, the unperturbed equilibrium values of parcel pressure and volume $p_{0}, V_{0}$ and the adiabatic index $\gamma$. (You may use the expression for the molar heat capacity at constant volume $c_{v}=\frac{R}{\gamma-1}$, where $R$ is the gas constant.) Context answer: \boxed{$\frac{1}{\gamma-1} V_{0} \frac{d p}{d t}+\frac{\gamma}{\gamma-1} p_{0} \frac{d V}{d t}$} Extra Supplementary Reading Materials: The limited heat flow rate between the parcel and the stack causes a phase difference between the pressure and volume oscillations of the parcel. We will see how this generates work. Let the heat flux into the parcel from the stack be proportional to the temperature difference between the parcel and the neighboring element of the stack, given approximately by $\frac{d Q}{d t}=-\beta V_{0}\left(T_{\mathrm{st}}-T_{1}\right) \cos (\omega t)$. Here $T_{1}$ and $T_{\mathrm{st}}$ are the temperature oscillation amplitudes of the gas parcel and the neighbouring stack from Tasks A. 5 and B. 1 , respectively, and $\beta>0$ is a constant. Assume that at the machine's operating frequencies, the change in gas temperature as a result of this heat flow is insignificant compared to both $T_{1}$ and $T_{\mathrm{st}}$. Context question: B.4 In order to calculate work, we will consider a change to the volume of the moving parcel as a result of the thermal contact with the stack. Let us write the pressure and the volume of the parcel under the stack's influence in the form: $$ \begin{gathered} p=p_{0}+p_{a} \sin (\omega t)-p_{b} \cos (\omega t), \\ V=V_{0}+V_{a} \sin (\omega t)+V_{b} \cos (\omega t) . \end{gathered} \tag{6} $$ Given $p_{a}$ and $p_{b}$, find the coefficients $V_{a}$ and $V_{b}$. Express your answer in terms of $p_{a}, p_{b}, p_{0}, V_{0}, \gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a$ and $\ell$. Context answer: \boxed{$V_{b}=\frac{1}{\gamma} p_{b} \cdot \frac{V_{0}}{p_{0}}$ , $V_{a}=(-\frac{1}{\gamma} p_{a}-\frac{\gamma-1}{\gamma} \frac{\beta}{\omega} \frac{a}{\ell \sqrt{2}}(\tau-\tau_{\mathrm{cr}})) \cdot \frac{V_{0}}{p_{0}}$} ","B.5 Obtain an approximate expression for the acoustic work per unit volume $w$ produced by the gas parcel over one cycle. Integrate over the volume of the stack to obtain the total work $W_{\text {tot }}$ generated by the gas over one cycle. Express $W_{\text {tot }}$ in terms of $\gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a, k$ and $S$.","['We want to integrate the mechanical work generated, $\\int p d V$, and averaging the result over a long time. To do this, we substitute our expressions (6) for the perturbed $p$ and $V$. Since the average of $\\cos (\\omega t) \\sin (\\omega t)$ is 0 , and that of $\\sin ^{2}(\\omega t)$ and $\\cos ^{2}(\\omega t)$ is $\\frac{1}{2}$, we get:\n\n$$\n\\frac{V_{0}}{S \\ell} W_{\\text {tot }}=-\\pi \\cdot\\left(p_{a} V_{b}+p_{b} V_{a}\\right)\n$$\n\nUsing the result of B.4, we get\n\n$$\n\\frac{V_{0}}{S \\ell} W_{\\text {tot }}=\\frac{\\pi}{\\omega} \\cdot \\frac{\\gamma-1}{\\gamma} \\beta \\frac{a}{\\ell \\sqrt{2}}\\left(\\tau-\\tau_{\\mathrm{cr}}\\right) \\cdot V_{0} \\frac{p_{b}}{p_{0}}\n$$\n\nTo leading order, $p_{b}$ is the unperturbed wave $p_{b} \\approx p_{1}\\left(x_{0}\\right)=a \\frac{\\omega^{2}}{k} \\rho_{0} \\cos \\left(k x_{0}\\right)=a k \\gamma p_{0} \\frac{1}{\\sqrt{2}}$. Simplifying, we get\n\n$$\nW_{\\text {tot }}=\\frac{\\pi}{\\omega} S \\cdot \\frac{\\gamma-1}{\\gamma} \\beta \\frac{a}{\\sqrt{2}}\\left(\\tau-\\tau_{\\mathrm{cr}}\\right) \\cdot \\frac{p_{b}}{p_{0}}=\\frac{\\pi}{2 \\omega}(\\gamma-1) \\beta\\left(\\tau-\\tau_{\\mathrm{cr}}\\right) k a^{2} S .\n$$']",['$\\frac{\\pi}{2 \\omega}(\\gamma-1) \\beta(\\tau-\\tau_{\\mathrm{cr}}) k a^{2} S$'],False,,Expression, 1319,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack. Context question: B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$. Context answer: \boxed{$\frac{a \tau}{\ell \sqrt{2}}$} Context question: B.2 Above which critical temperature difference $\tau_{\text {cr }}$ will the gas be conveying heat from the hot reservoir to the cold one? Express $\tau_{\mathrm{cr}}$ in terms of $T_{0}, \gamma, k$ and $\ell$. Context answer: \boxed{$\tau_{\mathrm{cr}}=k \ell(\gamma-1) T_{0}$} Context question: B.3 Obtain the general approximate expression for the heat flow $\frac{d Q}{d t}$ into a small parcel of gas as a linear function of its volume and pressure change rates. Express your answer in terms of the rate of volume change $\frac{d V}{d t}$, the rate of pressure change $\frac{d p}{d t}$, the unperturbed equilibrium values of parcel pressure and volume $p_{0}, V_{0}$ and the adiabatic index $\gamma$. (You may use the expression for the molar heat capacity at constant volume $c_{v}=\frac{R}{\gamma-1}$, where $R$ is the gas constant.) Context answer: \boxed{$\frac{1}{\gamma-1} V_{0} \frac{d p}{d t}+\frac{\gamma}{\gamma-1} p_{0} \frac{d V}{d t}$} Extra Supplementary Reading Materials: The limited heat flow rate between the parcel and the stack causes a phase difference between the pressure and volume oscillations of the parcel. We will see how this generates work. Let the heat flux into the parcel from the stack be proportional to the temperature difference between the parcel and the neighboring element of the stack, given approximately by $\frac{d Q}{d t}=-\beta V_{0}\left(T_{\mathrm{st}}-T_{1}\right) \cos (\omega t)$. Here $T_{1}$ and $T_{\mathrm{st}}$ are the temperature oscillation amplitudes of the gas parcel and the neighbouring stack from Tasks A. 5 and B. 1 , respectively, and $\beta>0$ is a constant. Assume that at the machine's operating frequencies, the change in gas temperature as a result of this heat flow is insignificant compared to both $T_{1}$ and $T_{\mathrm{st}}$. Context question: B.4 In order to calculate work, we will consider a change to the volume of the moving parcel as a result of the thermal contact with the stack. Let us write the pressure and the volume of the parcel under the stack's influence in the form: $$ \begin{gathered} p=p_{0}+p_{a} \sin (\omega t)-p_{b} \cos (\omega t), \\ V=V_{0}+V_{a} \sin (\omega t)+V_{b} \cos (\omega t) . \end{gathered} \tag{6} $$ Given $p_{a}$ and $p_{b}$, find the coefficients $V_{a}$ and $V_{b}$. Express your answer in terms of $p_{a}, p_{b}, p_{0}, V_{0}, \gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a$ and $\ell$. Context answer: \boxed{$V_{b}=\frac{1}{\gamma} p_{b} \cdot \frac{V_{0}}{p_{0}}$ , $V_{a}=(-\frac{1}{\gamma} p_{a}-\frac{\gamma-1}{\gamma} \frac{\beta}{\omega} \frac{a}{\ell \sqrt{2}}(\tau-\tau_{\mathrm{cr}})) \cdot \frac{V_{0}}{p_{0}}$} Context question: B.5 Obtain an approximate expression for the acoustic work per unit volume $w$ produced by the gas parcel over one cycle. Integrate over the volume of the stack to obtain the total work $W_{\text {tot }}$ generated by the gas over one cycle. Express $W_{\text {tot }}$ in terms of $\gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a, k$ and $S$. Context answer: \boxed{$\frac{\pi}{2 \omega}(\gamma-1) \beta(\tau-\tau_{\mathrm{cr}}) k a^{2} S$} ","B.6 Obtain an approximate expression for the heat $Q_{\text {tot }}$ transported from the left side of the plane $x=x_{0}$ to the right, over a cycle. Express your answer in terms of $\tau, \tau_{\text {cr }}, \beta, \omega, a, S, \ell$","['We want to compute the amount of heat convection over one cycle. This means that we need to take the amount of heat moving in or out of the parcel, and weigh it by the position of the parcel at that time. Thus, the total heat conveyed by the parcel, integrated along a cycle, is:\n\n$$\nQ_{\\text {tot }}=\\frac{1}{\\Delta x} \\int \\frac{d Q}{d t} u \\cdot d t\n$$\n\nThis expression can be computed to leading order using $\\frac{d Q}{d t}=\\beta V_{0}\\left(T_{\\mathrm{st}}-T_{1}\\right) \\cdot \\cos (\\omega t)$ and the unperturbed displacement $u\\left(x_{0}, t\\right)=\\frac{a}{\\sqrt{2}} \\cos (\\omega t)$. This gives\n\n$$\nQ_{\\mathrm{tot}}=\\frac{\\pi}{\\omega} \\beta V_{0}\\left(T_{\\mathrm{st}}-T_{1}\\right) \\frac{a}{\\sqrt{2}}=\\frac{\\pi}{\\omega} \\beta V_{0} \\cdot \\frac{a}{\\ell \\sqrt{2}}\\left(\\tau-\\tau_{\\mathrm{cr}}\\right) \\cdot \\frac{a}{\\sqrt{2}}=\\frac{\\pi}{2 \\omega} \\beta\\left(\\tau-\\tau_{\\mathrm{cr}}\\right) \\frac{a^{2} S}{\\ell} .\n$$']",['$\\frac{\\pi}{2 \\omega} \\beta(\\tau-\\tau_{\\mathrm{cr}}) \\frac{a^{2} S}{\\ell}$'],False,,Expression, 1320,Optics,"Thermoacoustic Engine A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself. The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices. Part A: Sound wave in a closed tube Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions. Figure 1 Context question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by $$ u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t) \tag{1} $$ (please note the $u$ here describes the displacement of a gas element) where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system? Context answer: \boxed{$\lambda_{\max }=2 L$} Extra Supplementary Reading Materials: We will assume throughout the question an oscillation mode of $\lambda=\lambda_{\max }$. Now, consider a narrow parcel of gas, located at rest between $x$ and $x+\Delta x(\Delta x \ll L)$. As a result of the displacement wave of Task A.1, the parcel oscillates along the $x$ axis and undergoes a change in volume and other thermodynamic properties. Throughout the following tasks assume all these changes to the thermodynamic properties to be small compared to the unperturbed values. Context question: A.2 The parcel volume $V(x, t)$ oscillates around the equilibrium value of $V_{0}=S \Delta x$ and has the form $$ V(x, t)=V_{0}+V_{1}(x) \cos (\omega t) . \tag{2} $$ Obtain an expression for $V_{1}(x)$ in terms of $V_{0}, a, k$ and $x$. Context answer: \boxed{$V_{1}(x)=a k V_{0} \cos (k x)$} Context question: A.3 Assume that the total pressure of the gas, as a result of the sound wave, takes the approximate form $$ p(x, t)=p_{0}-p_{1}(x) \cos (\omega t) . \tag{3} $$ Considering the forces acting on the parcel of gas, compute the amplitude $p_{1}(x)$ of the pressure oscillation to leading order, in terms of the position $x$, the equilibrium density $\rho_{0}$, the displacement amplitude $a$ and the wave parameters $k$ and $\omega$. Context answer: \boxed{$p_{1}(x)=a \frac{\omega^{2}}{k} \rho_{0} \cos (k x)$} Extra Supplementary Reading Materials: At acoustic frequencies, the thermal conductivity of the gas can be neglected. We will treat the expansion and contraction of gas parcels as purely adiabatic, satisfying the relation $p V^{\gamma}=$ const, where $\gamma$ is the adiabatic constant. Context question: A.4 Use the relation above and the results of the previous tasks to obtain an expression for the speed of sound waves $c=\omega / k$ in the tube, to first order. Express your answer in terms of $p_{0}, \rho_{0}$ and the adiabatic constant $\gamma$. Context answer: \boxed{$c=\sqrt{\frac{\gamma p_{0}}{\rho_{0}}}$} Context question: A.5 The change in the gas temperature due to the adiabatic expansion and contraction, as a result of the sound wave, takes the form: $$ T(x, t)=T_{0}-T_{1}(x) \cos (\omega t) \tag{4} $$ Compute the amplitude $T_{1}(x)$ of the temperature oscillations in terms of $T_{0}, \gamma$, $a, k$ and $x$. Context answer: \boxed{$a k(\gamma-1) T_{0} \cos (k x)$} Context question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected. For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time. Figure 2 Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part B: Sound wave amplification induced by external thermal contact A stack of thin well-spaced solid plates is placed inside the tube. The plates of the stack are aligned in parallel to the tube axis, so as not to obstruct the flow of gas along the tube. The center of the stack is positioned at $x_{0}=L / 4$, and spans a width of $\ell \ll L$ along the tube axis, filling its entire cross section. The right and left edges of the stack are held at temperature difference $\tau$. The left edge of the stack, at $x_{H}=x_{0}-\ell / 2$, is held by an external thermal reservoir at temperature $T_{H}=T_{0}+\tau / 2$, and at the same time, its right edge, at $x_{C}=x_{0}+\ell / 2$, is held at a temperature $T_{C}=T_{0}-\tau / 2$. The plate stack allows a slight longitudinal heat flow to maintain a constant temperature gradient between its edges, such that $T_{\text {plate }}(x)=T_{0}-\frac{x-x_{0}}{\ell} \tau$. Figure 3. A sketch of the system. (A) and (B) denote the hot and cold heat reservoirs respectively. (D) denotes the stack. To analyze the effect of the thermal contact between the plate stack and the gas on the sound waves in the tube, make the following assumptions: - As in the previous part, all changes to the thermodynamic properties are small compared to the unperturbed values. - The system operates in the fundamental standing-wave mode of the longest possible wavelength. It is only slightly modified by the presence of the plate stack. - The stack is much shorter than the wavelength $\ell \ll \lambda_{\max }$, and can be positioned far enough from both displacement and pressure nodes, so that the displacement $u(x, t) \approx u\left(x_{0}, t\right)$ and the pressure $p(x, t) \approx p\left(x_{0}, t\right)$ may be considered uniform over the entire length of the stack. - We may neglect any edge effects, caused by the parcels moving in and out of the stack. - The temperature difference between the ends of the plate stack, i.e. between the hot and the cold reservoirs, is small compared to the absolute temperature: $\tau \ll T_{0}$. - Heat conduction through the stack, through the gas, and along the tube are all negligible. The only significant sources of heat transfer are convection due to the motion of the gas and conduction between the gas and the stack. Context question: B.1 Consider a specific parcel of gas in the region of the stack, originally at $x_{0}=L / 4 .$ As the parcel moves within the stack, the local temperature of the nearby part of the stack changes as follows: $$ T_{\mathrm{env}}(t)=T_{0}-T_{\mathrm{st}} \cos (\omega t) \tag{5} $$ Express $T_{\mathrm{st}}$ in terms of $a, \tau$ and $\ell$. Context answer: \boxed{$\frac{a \tau}{\ell \sqrt{2}}$} Context question: B.2 Above which critical temperature difference $\tau_{\text {cr }}$ will the gas be conveying heat from the hot reservoir to the cold one? Express $\tau_{\mathrm{cr}}$ in terms of $T_{0}, \gamma, k$ and $\ell$. Context answer: \boxed{$\tau_{\mathrm{cr}}=k \ell(\gamma-1) T_{0}$} Context question: B.3 Obtain the general approximate expression for the heat flow $\frac{d Q}{d t}$ into a small parcel of gas as a linear function of its volume and pressure change rates. Express your answer in terms of the rate of volume change $\frac{d V}{d t}$, the rate of pressure change $\frac{d p}{d t}$, the unperturbed equilibrium values of parcel pressure and volume $p_{0}, V_{0}$ and the adiabatic index $\gamma$. (You may use the expression for the molar heat capacity at constant volume $c_{v}=\frac{R}{\gamma-1}$, where $R$ is the gas constant.) Context answer: \boxed{$\frac{1}{\gamma-1} V_{0} \frac{d p}{d t}+\frac{\gamma}{\gamma-1} p_{0} \frac{d V}{d t}$} Extra Supplementary Reading Materials: The limited heat flow rate between the parcel and the stack causes a phase difference between the pressure and volume oscillations of the parcel. We will see how this generates work. Let the heat flux into the parcel from the stack be proportional to the temperature difference between the parcel and the neighboring element of the stack, given approximately by $\frac{d Q}{d t}=-\beta V_{0}\left(T_{\mathrm{st}}-T_{1}\right) \cos (\omega t)$. Here $T_{1}$ and $T_{\mathrm{st}}$ are the temperature oscillation amplitudes of the gas parcel and the neighbouring stack from Tasks A. 5 and B. 1 , respectively, and $\beta>0$ is a constant. Assume that at the machine's operating frequencies, the change in gas temperature as a result of this heat flow is insignificant compared to both $T_{1}$ and $T_{\mathrm{st}}$. Context question: B.4 In order to calculate work, we will consider a change to the volume of the moving parcel as a result of the thermal contact with the stack. Let us write the pressure and the volume of the parcel under the stack's influence in the form: $$ \begin{gathered} p=p_{0}+p_{a} \sin (\omega t)-p_{b} \cos (\omega t), \\ V=V_{0}+V_{a} \sin (\omega t)+V_{b} \cos (\omega t) . \end{gathered} \tag{6} $$ Given $p_{a}$ and $p_{b}$, find the coefficients $V_{a}$ and $V_{b}$. Express your answer in terms of $p_{a}, p_{b}, p_{0}, V_{0}, \gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a$ and $\ell$. Context answer: \boxed{$V_{b}=\frac{1}{\gamma} p_{b} \cdot \frac{V_{0}}{p_{0}}$ , $V_{a}=(-\frac{1}{\gamma} p_{a}-\frac{\gamma-1}{\gamma} \frac{\beta}{\omega} \frac{a}{\ell \sqrt{2}}(\tau-\tau_{\mathrm{cr}})) \cdot \frac{V_{0}}{p_{0}}$} Context question: B.5 Obtain an approximate expression for the acoustic work per unit volume $w$ produced by the gas parcel over one cycle. Integrate over the volume of the stack to obtain the total work $W_{\text {tot }}$ generated by the gas over one cycle. Express $W_{\text {tot }}$ in terms of $\gamma, \tau, \tau_{\text {cr }}, \beta, \omega, a, k$ and $S$. Context answer: \boxed{$\frac{\pi}{2 \omega}(\gamma-1) \beta(\tau-\tau_{\mathrm{cr}}) k a^{2} S$} Context question: B.6 Obtain an approximate expression for the heat $Q_{\text {tot }}$ transported from the left side of the plane $x=x_{0}$ to the right, over a cycle. Express your answer in terms of $\tau, \tau_{\text {cr }}, \beta, \omega, a, S, \ell$ Context answer: \boxed{$\frac{\pi}{2 \omega} \beta(\tau-\tau_{\mathrm{cr}}) \frac{a^{2} S}{\ell}$} ","B.7 Find the efficiency $\eta$ of the thermoacoustic engine. The efficiency is defined as the ratio of the generated acoustic work to the heat drawn from the hot reservoir. Express your answer in terms of the temperature difference $\tau$ between the hot and the cold reservoir, the critical temperature difference $\tau_{\text {cr }}$ and the Carnot efficiency $\eta_{c}=1-T_{C} / T_{H}$.","['Dividing the results of Tasks B. 5 and B.6, we obtain the expression:\n\n$$\n\\eta=\\frac{W_{\\mathrm{tot}}}{Q_{\\mathrm{tot}}}=(\\gamma-1) k \\ell=\\frac{\\tau_{\\mathrm{cr}}}{T_{0}}=\\frac{\\tau_{\\mathrm{cr}}}{\\tau} \\cdot \\frac{\\tau}{T_{0}}=\\frac{\\tau_{\\mathrm{cr}}}{\\tau} \\cdot \\eta_{c}\n$$']",['$\\frac{\\tau_{\\mathrm{cr}}}{\\tau} \\cdot \\eta_{c}$'],False,,Expression, 1321,Thermodynamics,"Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or $\mathrm{B})$ of their saturated vapors obey, to a good approximation, the formula: $$ \ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B} $$ where $p_{o}$ denotes the normal atmospheric pressure, $T$ - the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}$ ( $i=\mathrm{A}$ or $\left.\mathrm{B}\right)$ - certain constants depending on the liquid. (The symbol $\ln$ denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots)$ The values of the ratio $p_{i} / p_{0}$ for the liquids $\mathrm{A}$ and $\mathrm{B}$ at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1. Table 1.1 | $t\left[{ }^{\circ} \mathrm{C}\right]$ | $i=\mathrm{A}$ | $p_{i} / p_{0}$ | | :---: | :---: | :---: | | | 0.284 | 0.07278 | | 40 | 1.476 | 0.6918 | | 90 | | | The errors of these values are negligible.",A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$.,"['PART A\n\nThe liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid $i(i-\\mathrm{A}$ or $\\mathrm{B}$ ), one should determine such a temperature $T_{b i}$ (or $t_{b i}$ ) for which $p_{i} / p_{0}=1$.\n\nThen $\\ln \\left(p_{i} / p_{0}\\right)=0$, and we have:\n\n$$\nT_{b i}=-\\frac{\\alpha_{i}}{\\beta_{i}}\n$$\n\n\n\nThe coefficients $\\alpha_{i}$ and $\\beta_{i}$ are not given explicitly. However, they can be calculated from the formula given in the text of the problem. For this purpose one should make use of the numerical data given in the Tab. 1.1.\n\nFor the liquid A, we have:\n\n$$\n\\begin{aligned}\n& \\ln 0.284=\\frac{\\alpha_{A}}{(40+273.15) \\mathrm{K}}+\\beta_{A}, \\\\\n& \\ln 1.476=\\frac{\\alpha_{A}}{(90+273.15) \\mathrm{K}}+\\beta_{A} .\n\\end{aligned}\n$$\n\nAfter subtraction of these equations, we get:\n\n$$\n\\begin{gathered}\n\\ln 0.284-\\ln 1.476=\\alpha_{A}\\left(\\frac{1}{40+273.15}-\\frac{1}{90+273.15}\\right) \\mathrm{K}^{-1} . \\\\\n\\alpha_{A}=\\frac{\\ln \\frac{0.284}{1.476}}{\\frac{1}{40+273.15}-\\frac{1}{90+273.15}} \\mathrm{~K} \\approx-3748.49 \\mathrm{~K}\n\\end{gathered}\n$$\n\nHence,\n\n$$\n\\beta_{A}=\\ln 0.284-\\frac{\\alpha_{A}}{(40+273.15) \\mathrm{K}} \\approx 10.711\n$$\n\nThus, the boiling temperature of the liquid $\\mathrm{A}$ is equal to\n\n$$\nT_{b A}=3748.49 \\mathrm{~K} / 10.711 \\approx 349.95 \\mathrm{~K} .\n$$\n\nIn the Celsius scale the boiling temperature of the liquid $\\mathrm{A}$ is\n\n$$\nt_{b A}=(349.95-273.15)^{\\circ} \\mathrm{C}=76.80^{\\circ} \\mathrm{C} \\approx 77^{\\circ} \\mathrm{C} .\n$$\n\nFor the liquid B, in the same way, we obtain:\n\n$$\n\\begin{aligned}\n\\alpha_{B} & \\approx-5121.64 \\mathrm{~K} \\\\\n\\beta_{B} & \\approx 13.735 \\\\\nT_{b B} & \\approx 372-89 \\mathrm{~K} \\\\\nt_{b B} & \\approx 99.74^{\\circ} \\mathrm{C} \\approx 100^{\\circ} \\mathrm{C} .\n\\end{aligned}\n$$']","['$76.8$ , $100$']",True,$^{\circ}\mathrm{C}$,Numerical,1e0 1322,Thermodynamics,,"B. The liquids A and B were poured into a vessel in which the layers shown in Fig. 1.1 were formed. The surface of the liquid B has been covered with a thin layer of a non-volatile liquid $\mathrm{C}$, which is insoluble in the liquids $\mathrm{A}$ and $\mathrm{B}$ and vice versa, thereby preventing any free evaporation from the upper surface of the liquid $B$, The ratio of molecular masses of the liquids A and B (in the gaseous phase) is: $$ \gamma=\mu_{A} / \mu_{B}=8 $$ ![](https://cdn.mathpix.com/cropped/2023_12_21_21a6109b4be0c7e77f4eg-1.jpg?height=440&width=625&top_left_y=685&top_left_x=270) Fig. 1.1 ![](https://cdn.mathpix.com/cropped/2023_12_21_21a6109b4be0c7e77f4eg-1.jpg?height=463&width=596&top_left_y=682&top_left_x=1067) Fig. 1.2 The masses of the liquids A and B were initially the same, each equal to $m=100 \mathrm{~g}$. The heights of the layers of the liquids in the vessel and the densities of the liquids are small enough to make the assumption that the pressure in any point in the vessel is practically equal to the normal atmospheric pressure $p_{0}$. The system of liquids in the vessel is slowly, but continuously and uniformly, heated. It was established that the temperature $t$ of the liquids changed with time $\tau$ as shown schematically in the Fig. 1.2. Determine the temperatures $t_{1}$ and $t_{2}$ corresponding to the horizontal parts of the diagram and the masses of the liquids $\mathrm{A}$ and $\mathrm{B}$ at the time $\tau_{1}$. The temperatures should be rounded to the nearest degree (in ${ }^{\circ} \mathrm{C}$ ) and the masses of the liquids should be determined to one-tenth of gram. REMARK: Assume that the vapors of the liquids, to a good approximation, (1) obey the Dalton law stating that the pressure of a mixture of gases is equal to the sum of the partial pressures of the gases forming the mixture and (2) can be treated as perfect gases up to the pressures corresponding to the saturated vapors","['PART B\n\nAs the liquids are in thermal contact with each other, their temperatures increase in time in the same way.\n\nAt the beginning of the heating, what corresponds to the left sloped part of the diagram, no evaporation can occur. The free evaporation from the upper surface of the liquid B cannot occur - it is impossible due to the layer of the non-volatile liquid $\\mathrm{C}$. The evaporation from the inside of the system is considered below.\n\n\n\nLet us consider a bubble formed in the liquid $\\mathrm{A}$ or in the liquid $\\mathrm{B}$ or on the surface that separates these liquids. Such a bubble can be formed due to fluctuations or for many other reasons, which will not be analyzed here.\n\nThe bubble can get out of the system only when the pressure inside it equals to the external pressure $p_{0}$ (or when it is a little bit higher than $p_{0}$ ). Otherwise, the bubble will collapse.\n\nThe pressure inside the bubble formed in the volume of the liquid A or in the volume of the liquid B equals to the pressure of the saturated vapor of the liquid A or B, respectively. However, the pressure inside the bubble formed on the surface separating the liquids A and B is equal to the sum of the pressures of the saturated vapors of both these liquids, as then the bubble is in a contact with the liquids $\\mathrm{A}$ and $\\mathrm{B}$ at the same time. In the case considered the pressure inside the bubble is greater than the pressures of the saturated vapors of each of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ (at the same temperature).\n\nTherefore, when the system is heated, the pressure $p_{0}$ is reached first in the bubbles that were formed on the surface separating the liquids. Thus, the temperature $t_{1}$ corresponds to a kind of common boiling of both liquids that occurs in the region of their direct contact. The temperature $t_{1}$ is for sure lower than the boiling temperatures of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ as then the pressures of the saturated vapors of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ are less then $p_{0}$ (their sum equals to $p_{0}$ and each of them is greater than zero).\n\nIn order to determine the value of $t_{1}$ with required accuracy, we can calculate the values of the sum of the saturated vapors of the liquids A and B for several values of the temperature $t$ and look when one gets the value $p_{0}$.\n\nFrom the formula given in the text of the problem, we have:\n\n$$\n\\begin{aligned}\n& \\frac{p_{A}}{p_{0}}=e^{\\frac{\\alpha_{A}}{T}+\\beta_{A}}, \\\\\n& \\frac{p_{B}}{p_{0}}=e^{\\frac{\\alpha_{B}}{T}+\\beta_{B}} .\n\\end{aligned}\n$$\n\n$p_{A}+p_{B}$ equals to $p_{0}$ if\n\n$$\n\\frac{p_{A}}{p_{0}}+\\frac{p_{B}}{p_{0}}=1\n$$\n\nThus, we have to calculate the values of the following function:\n\n$$\ny(x)=e^{\\frac{\\alpha_{A}}{t+t_{0}}+\\beta_{A}}+e^{\\frac{\\alpha_{B}}{t+t_{0}}+\\beta_{B}}\n$$\n\n(where $t_{0}=273.15^{\\circ} \\mathrm{C}$ ) and to determine the temperature $t=t_{1}$, at which $y(t)$ equals to 1 . When calculating the values of the function $y(t)$ we can divide the intervals of the temperatures $t$ by 2 (approximately) and look whether the results are greater or less than 1 .\n\nWe have:\n\n\n\nTable 1.2\n\n| $t$ | $y(t)$ |\n| :--- | :--- |\n| $40^{\\circ} \\mathrm{C}$ | $<1($ see Tab. 1.1$)$ |\n| $77^{\\circ} \\mathrm{C}$ | $>1\\left(\\right.$ as $t_{1}$ is less than $\\left.t_{b A}\\right)$ |\n| $59^{\\circ} \\mathrm{C}$ | $0.749<1$ |\n| $70^{\\circ} \\mathrm{C}$ | $1.113>1$ |\n| $66^{\\circ} \\mathrm{C}$ | $0.966<1$ |\n| $67^{\\circ} \\mathrm{C}$ | $1.001>1$ |\n| $66.5^{\\circ} \\mathrm{C}$ | $0.983<1$ |\n\nTherefore, $t_{1} \\approx 67^{\\circ} \\mathrm{C}$ (with required accuracy).\n\nNow we calculate the pressures of the saturated vapors of the liquids $A$ and $B$ at the temperature $t_{1} \\approx 67^{\\circ} \\mathrm{C}$, i.e. the pressures of the saturated vapors of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ in each bubble formed on the surface separating the liquids. From the equations (1) and (2), we get:\n\n$$\n\\begin{aligned}\n& p_{A} \\approx 0.734 p_{0}, \\\\\n& p_{B} \\approx 0.267 p_{0}, \\\\\n& \\left(p_{A}+p_{B}=1.001 p_{0} \\approx p_{0}\\right) .\n\\end{aligned}\n$$\n\nThese pressures depend only on the temperature and, therefore, they remain constant during the motion of the bubbles through the liquid $\\mathrm{B}$. The volume of the bubbles during this motion also cannot be changed without violation of the relation $p_{A}+p_{B}=p_{0}$. It follows from the above remarks that the mass ratio of the saturated vapors of the liquids A and B in each bubble is the same. This conclusion remains valid as long as both liquids are in the system. After total evaporation of one of the liquids the temperature of the system will increase again (second sloped part of the diagram). Then, however, the mass of the system remains constant until the temperature reaches the value $t_{2}$ at which the boiling of the liquid (remained in the vessel) starts. Therefore, the temperature $t_{2}$ (the higher horizontal part of the diagram) corresponds to the boiling of the liquid remained in the vessel.\n\nThe mass ratio $m_{A} / m_{B}$ of the saturated vapors of the liquids $A$ and $B$ in each bubble leaving the system at the temperature $t_{1}$ is equal to the ratio of the densities of these vapors $\\rho_{A} / \\rho_{B}$. According to the assumption 2, stating that the vapors can be treated as ideal gases, the last ratio equals to the ratio of the products of the pressures of the saturated vapors by the molecular masses:\n\n$$\n\\frac{m_{A}}{m_{B}}=\\frac{\\rho_{A}}{\\rho_{B}}=\\frac{p_{A} \\mu_{A}}{p_{B} \\mu_{B}}=\\frac{p_{A}}{p_{B}} \\mu\n$$\n\nThus,\n\n$$\n\\frac{m_{A}}{m_{B}} \\approx 22.0\n$$\n\nWe see that the liquid A evaporates 22 times faster than the liquid B. The evaporation of $100 \\mathrm{~g}$ of the liquid A during the ""surface boiling"" at the temperature $t_{1}$ is associated with the\n\n\n\nevaporation of $100 \\mathrm{~g} / 22 \\approx 4.5 \\mathrm{~g}$ of the liquid B. Thus, at the time $\\tau_{1}$ the vessel contains 95.5 $\\mathrm{g}$ of the liquid $\\mathrm{B}$ (and no liquid $\\mathrm{A}$ ). The temperature $t_{2}$ is equal to the boiling temperature of the liquid $\\mathrm{B}: t_{2}=100^{\\circ} \\mathrm{C}$.']","['$67$ , $100$']",True,$^{\circ}\mathrm{C}$,Numerical,1e0 1322,Thermodynamics,"Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or $\mathrm{B})$ of their saturated vapors obey, to a good approximation, the formula: $$ \ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B} $$ where $p_{o}$ denotes the normal atmospheric pressure, $T$ - the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}$ ( $i=\mathrm{A}$ or $\left.\mathrm{B}\right)$ - certain constants depending on the liquid. (The symbol $\ln$ denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots)$ The values of the ratio $p_{i} / p_{0}$ for the liquids $\mathrm{A}$ and $\mathrm{B}$ at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1. Table 1.1 | $t\left[{ }^{\circ} \mathrm{C}\right]$ | $i=\mathrm{A}$ | $p_{i} / p_{0}$ | | :---: | :---: | :---: | | | 0.284 | 0.07278 | | 40 | 1.476 | 0.6918 | | 90 | | | The errors of these values are negligible. Context question: A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$. Context answer: \boxed{$76.8$ , $100$} ","B. The liquids A and B were poured into a vessel in which the layers shown in Fig. 1.1 were formed. The surface of the liquid B has been covered with a thin layer of a non-volatile liquid $\mathrm{C}$, which is insoluble in the liquids $\mathrm{A}$ and $\mathrm{B}$ and vice versa, thereby preventing any free evaporation from the upper surface of the liquid $B$, The ratio of molecular masses of the liquids A and B (in the gaseous phase) is: $$ \gamma=\mu_{A} / \mu_{B}=8 $$ Fig. 1.1 Fig. 1.2 The masses of the liquids A and B were initially the same, each equal to $m=100 \mathrm{~g}$. The heights of the layers of the liquids in the vessel and the densities of the liquids are small enough to make the assumption that the pressure in any point in the vessel is practically equal to the normal atmospheric pressure $p_{0}$. The system of liquids in the vessel is slowly, but continuously and uniformly, heated. It was established that the temperature $t$ of the liquids changed with time $\tau$ as shown schematically in the Fig. 1.2. Determine the temperatures $t_{1}$ and $t_{2}$ corresponding to the horizontal parts of the diagram and the masses of the liquids $\mathrm{A}$ and $\mathrm{B}$ at the time $\tau_{1}$. The temperatures should be rounded to the nearest degree (in ${ }^{\circ} \mathrm{C}$ ) and the masses of the liquids should be determined to one-tenth of gram. REMARK: Assume that the vapors of the liquids, to a good approximation, (1) obey the Dalton law stating that the pressure of a mixture of gases is equal to the sum of the partial pressures of the gases forming the mixture and (2) can be treated as perfect gases up to the pressures corresponding to the saturated vapors","['PART B\n\nAs the liquids are in thermal contact with each other, their temperatures increase in time in the same way.\n\nAt the beginning of the heating, what corresponds to the left sloped part of the diagram, no evaporation can occur. The free evaporation from the upper surface of the liquid B cannot occur - it is impossible due to the layer of the non-volatile liquid $\\mathrm{C}$. The evaporation from the inside of the system is considered below.\n\n\n\nLet us consider a bubble formed in the liquid $\\mathrm{A}$ or in the liquid $\\mathrm{B}$ or on the surface that separates these liquids. Such a bubble can be formed due to fluctuations or for many other reasons, which will not be analyzed here.\n\nThe bubble can get out of the system only when the pressure inside it equals to the external pressure $p_{0}$ (or when it is a little bit higher than $p_{0}$ ). Otherwise, the bubble will collapse.\n\nThe pressure inside the bubble formed in the volume of the liquid A or in the volume of the liquid B equals to the pressure of the saturated vapor of the liquid A or B, respectively. However, the pressure inside the bubble formed on the surface separating the liquids A and B is equal to the sum of the pressures of the saturated vapors of both these liquids, as then the bubble is in a contact with the liquids $\\mathrm{A}$ and $\\mathrm{B}$ at the same time. In the case considered the pressure inside the bubble is greater than the pressures of the saturated vapors of each of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ (at the same temperature).\n\nTherefore, when the system is heated, the pressure $p_{0}$ is reached first in the bubbles that were formed on the surface separating the liquids. Thus, the temperature $t_{1}$ corresponds to a kind of common boiling of both liquids that occurs in the region of their direct contact. The temperature $t_{1}$ is for sure lower than the boiling temperatures of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ as then the pressures of the saturated vapors of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ are less then $p_{0}$ (their sum equals to $p_{0}$ and each of them is greater than zero).\n\nIn order to determine the value of $t_{1}$ with required accuracy, we can calculate the values of the sum of the saturated vapors of the liquids A and B for several values of the temperature $t$ and look when one gets the value $p_{0}$.\n\nFrom the formula given in the text of the problem, we have:\n\n$$\n\\begin{aligned}\n& \\frac{p_{A}}{p_{0}}=e^{\\frac{\\alpha_{A}}{T}+\\beta_{A}}, \\\\\n& \\frac{p_{B}}{p_{0}}=e^{\\frac{\\alpha_{B}}{T}+\\beta_{B}} .\n\\end{aligned}\n$$\n\n$p_{A}+p_{B}$ equals to $p_{0}$ if\n\n$$\n\\frac{p_{A}}{p_{0}}+\\frac{p_{B}}{p_{0}}=1\n$$\n\nThus, we have to calculate the values of the following function:\n\n$$\ny(x)=e^{\\frac{\\alpha_{A}}{t+t_{0}}+\\beta_{A}}+e^{\\frac{\\alpha_{B}}{t+t_{0}}+\\beta_{B}}\n$$\n\n(where $t_{0}=273.15^{\\circ} \\mathrm{C}$ ) and to determine the temperature $t=t_{1}$, at which $y(t)$ equals to 1 . When calculating the values of the function $y(t)$ we can divide the intervals of the temperatures $t$ by 2 (approximately) and look whether the results are greater or less than 1 .\n\nWe have:\n\n\n\nTable 1.2\n\n| $t$ | $y(t)$ |\n| :--- | :--- |\n| $40^{\\circ} \\mathrm{C}$ | $<1($ see Tab. 1.1$)$ |\n| $77^{\\circ} \\mathrm{C}$ | $>1\\left(\\right.$ as $t_{1}$ is less than $\\left.t_{b A}\\right)$ |\n| $59^{\\circ} \\mathrm{C}$ | $0.749<1$ |\n| $70^{\\circ} \\mathrm{C}$ | $1.113>1$ |\n| $66^{\\circ} \\mathrm{C}$ | $0.966<1$ |\n| $67^{\\circ} \\mathrm{C}$ | $1.001>1$ |\n| $66.5^{\\circ} \\mathrm{C}$ | $0.983<1$ |\n\nTherefore, $t_{1} \\approx 67^{\\circ} \\mathrm{C}$ (with required accuracy).\n\nNow we calculate the pressures of the saturated vapors of the liquids $A$ and $B$ at the temperature $t_{1} \\approx 67^{\\circ} \\mathrm{C}$, i.e. the pressures of the saturated vapors of the liquids $\\mathrm{A}$ and $\\mathrm{B}$ in each bubble formed on the surface separating the liquids. From the equations (1) and (2), we get:\n\n$$\n\\begin{aligned}\n& p_{A} \\approx 0.734 p_{0}, \\\\\n& p_{B} \\approx 0.267 p_{0}, \\\\\n& \\left(p_{A}+p_{B}=1.001 p_{0} \\approx p_{0}\\right) .\n\\end{aligned}\n$$\n\nThese pressures depend only on the temperature and, therefore, they remain constant during the motion of the bubbles through the liquid $\\mathrm{B}$. The volume of the bubbles during this motion also cannot be changed without violation of the relation $p_{A}+p_{B}=p_{0}$. It follows from the above remarks that the mass ratio of the saturated vapors of the liquids A and B in each bubble is the same. This conclusion remains valid as long as both liquids are in the system. After total evaporation of one of the liquids the temperature of the system will increase again (second sloped part of the diagram). Then, however, the mass of the system remains constant until the temperature reaches the value $t_{2}$ at which the boiling of the liquid (remained in the vessel) starts. Therefore, the temperature $t_{2}$ (the higher horizontal part of the diagram) corresponds to the boiling of the liquid remained in the vessel.\n\nThe mass ratio $m_{A} / m_{B}$ of the saturated vapors of the liquids $A$ and $B$ in each bubble leaving the system at the temperature $t_{1}$ is equal to the ratio of the densities of these vapors $\\rho_{A} / \\rho_{B}$. According to the assumption 2, stating that the vapors can be treated as ideal gases, the last ratio equals to the ratio of the products of the pressures of the saturated vapors by the molecular masses:\n\n$$\n\\frac{m_{A}}{m_{B}}=\\frac{\\rho_{A}}{\\rho_{B}}=\\frac{p_{A} \\mu_{A}}{p_{B} \\mu_{B}}=\\frac{p_{A}}{p_{B}} \\mu\n$$\n\nThus,\n\n$$\n\\frac{m_{A}}{m_{B}} \\approx 22.0\n$$\n\nWe see that the liquid A evaporates 22 times faster than the liquid B. The evaporation of $100 \\mathrm{~g}$ of the liquid A during the ""surface boiling"" at the temperature $t_{1}$ is associated with the\n\n\n\nevaporation of $100 \\mathrm{~g} / 22 \\approx 4.5 \\mathrm{~g}$ of the liquid B. Thus, at the time $\\tau_{1}$ the vessel contains 95.5 $\\mathrm{g}$ of the liquid $\\mathrm{B}$ (and no liquid $\\mathrm{A}$ ). The temperature $t_{2}$ is equal to the boiling temperature of the liquid $\\mathrm{B}: t_{2}=100^{\\circ} \\mathrm{C}$.']","['$67$ , $100$']",True,$^{\circ}\mathrm{C}$,Numerical,1e0 1323,Mechanics,,"Three non-collinear points $P_{1}, P_{2}$ and $P_{3}$, with known masses $m_{1}, m_{2}$ and $m_{3}$, interact with one another through their mutual gravitational forces only; they are isolated in free space and do not interact with any other bodies. Let $\sigma$ denote the axis going through the center-ofmass of the three masses, and perpendicular to the triangle $P_{1} P_{2} P_{3}$. What conditions should the angular velocities $\omega$ of the system (around the axis $\sigma$ ) and the distances: $$ P_{1} P_{2}=a_{12}, \quad P_{2} P_{3}=a_{23}, \quad P_{1} P_{3}=a_{13} $$ fulfill to allow the shape and size of the triangle $P_{1} P_{2} P_{3}$ unchanged during the motion of the system, i.e. under what conditions does the system rotate around the axis $\sigma$ as a rigid body?","[""As the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points $\\left(\\mathrm{P}_{1}, \\mathrm{P}_{2}\\right),\\left(\\mathrm{P}_{2}, \\mathrm{P}_{3}\\right)$ and $\\left(\\mathrm{P}_{1}, \\mathrm{P}_{3}\\right)$. It depends only on the distances $a_{12}, a_{23}$ and $a_{23}$ which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis $\\sigma$ depends only on the distances from the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ to the axis $\\sigma$ which, for fixed $a_{12}, a_{23}$ and $a_{23}$ do not depend on time. This means that the moment of inertia $I$ is constant. Therefore, the angular velocity of the system must also be constant:\n\n$$\n\\omega=\\text { const. }\n\\tag{1}\n$$\n\nThis is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performing calculations, it is desirable to specify a convenient coordinates system in which the calculations are expected to be simple.\n\nLet the positions of the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ be given by the vectors $\\mathbf{r}_{1}, \\mathbf{r}_{2}$ and $\\mathbf{r}_{3}$. For simplicity we assume that the origin of the coordinate system is localized at the center of mass of the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ and that all the vectors $\\mathbf{r}_{1}, \\mathbf{r}_{2}$ and $\\mathbf{r}_{3}$ are in the same coordinate plane, e.g. in the plane $(x, y)$. Then the axis $\\sigma$ is the axis $z$.\n\nIn this coordinate system, according to the definition of the center of mass, we have:\n\n$$\nm_{1} \\mathbf{r}_{1}+m_{2} \\mathbf{r}_{2}+m_{3} \\mathbf{r}_{2}=0\n\\tag{2}\n$$\n\n\nConsider the point $\\mathrm{P}_{1}$ with the mass $m_{1}$. The points $\\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ act on it with the forces:\n\n$$\n\\mathbf{F}_{21}=G \\frac{m_{1} m_{2}}{a_{12}^{3}}\\left(\\mathbf{r}_{2}-\\mathbf{r}_{1}\\right), \n\\tag{3}\n$$\n$$\n\\mathbf{F}_{31}=G \\frac{m_{1} m_{3}}{a_{13}^{3}}\\left(\\mathbf{r}_{3}-\\mathbf{r}_{1}\\right) .\n\\tag{4}\n$$\n\nwhere $G$ denotes the gravitational constant.\n\nIn the inertial frame the sum of these forces is the centripetal force\n\n$$\n\\mathbf{F}_{r 1}=-m_{1} \\omega^{2} \\mathbf{r}_{1}\n$$\n\nwhich causes the movement of the point $\\mathrm{P}_{1}$ along a circle with the angular velocity $\\omega$. (The moment of this force with respect to the axis $\\sigma$ is equal to zero.) Thus, we have:\n\n$$\n\\mathbf{F}_{21}+\\mathbf{F}_{31}=\\mathbf{F}_{r 1}\n\\tag{5}\n$$\n\nIn the non-inertial frame, rotating around the axis $\\sigma$ with the angular velocity $\\omega$, the sum of the forces (3), (4) and the centrifugal force\n\n\n\n$$\n\\mathbf{F}_{r 1}^{\\prime}=m_{1} \\omega^{2} \\mathbf{r}_{1}\n$$\n\nshould be equal to zero:\n\n$$\n\\mathbf{F}_{21}+\\mathbf{F}_{31}+\\mathbf{F}_{r 1}^{\\prime}=0\n\\tag{6}\n$$\n\n(The moment of this sum with respect to any axis equals to zero.)\n\nThe conditions (5) and (6) are equivalent. They give the same vector equality:\n\n$$\nG \\frac{m_{1} m_{2}}{a_{12}^{3}}\\left(\\mathbf{r}_{2}-\\mathbf{r}_{1}\\right)+G \\frac{m_{1} m_{3}}{a_{13}^{3}}\\left(\\mathbf{r}_{3}-\\mathbf{r}_{1}\\right)+m_{1} \\omega^{2} \\mathbf{r}_{1}=0,\n\\tag{7'}\n$$\n$$\nG \\frac{m_{1}}{a_{12}^{3}} m_{2} \\mathbf{r}_{2}+G \\frac{m_{1}}{a_{13}^{3}} m_{3} \\mathbf{r}_{3}+m_{1} \\mathbf{r}_{1}\\left(\\omega^{2}-\\frac{G m_{2}}{a_{12}^{3}}-\\frac{G m_{3}}{a_{13}^{3}}\\right)=0\n\\tag{7''}\n$$\n\nFrom the formula (2), we get:\n\n$$\nm_{2} \\mathbf{r}_{2}=-m_{1} \\mathbf{r}_{1}-m_{3} \\mathbf{r}_{3}\n\\tag{8}\n$$\n\nUsing this relation, we write the formula (7) in the following form:\n\n$$\nG \\frac{m_{1}}{a_{12}^{3}}\\left(-m_{1} \\mathbf{r}_{1}-m_{3} \\mathbf{r}_{3}\\right)+G \\frac{m_{1}}{a_{13}^{3}} m_{3} \\mathbf{r}_{3}+m_{1} \\mathbf{r}_{1}\\left(\\omega^{2}-\\frac{G m_{2}}{a_{12}^{3}}-\\frac{G m_{3}}{a_{13}^{3}}\\right)=0,\n$$\n\ni.e.\n\n$$\n\\mathbf{r}_{1} m_{1}\\left(\\omega^{2}-\\frac{G m_{2}}{a_{12}^{3}}-\\frac{G m_{3}}{a_{13}^{3}}-\\frac{G m_{1}}{a_{12}^{3}}\\right)+\\mathbf{r}_{3}\\left(\\frac{1}{a_{13}^{3}}-\\frac{1}{a_{12}^{3}}\\right) G m_{1} m_{3}=0 .\n$$\n\nThe vectors $\\mathbf{r}_{1}$ and $\\mathbf{r}_{3}$ are non-collinear. Therefore, the coefficients in the last formula must be equal to zero:\n\n$$\n\\begin{gathered}\n\\left(\\frac{1}{a_{13}^{3}}-\\frac{1}{a_{12}^{3}}\\right) G m_{1} m_{3}=0 \\\\\nm_{1}\\left(\\omega^{2}-\\frac{G m_{2}}{a_{12}^{3}}-\\frac{G m_{3}}{a_{13}^{3}}-\\frac{G m_{1}}{a_{12}^{3}}\\right)=0 .\n\\end{gathered}\n$$\n\nThe first equality leads to:\n\n$$\n\\frac{1}{a_{13}^{3}}=\\frac{1}{a_{12}^{3}}\n$$\n\nand hence,\n\n$$\na_{13}=a_{12} \\text {. }\n$$\n\nLet $a_{13}=a_{12}=a$. Then the second equality gives:\n\n$$\n\\omega^{2} a^{3}=G M\n\\tag{9}\n$$\n\nwhere\n\n$$\nM=m_{1}+m_{2}+m_{3}\n\\tag{10}\n$$\n\ndenotes the total mass of the system.\n\n\n\nIn the same way, for the points $\\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$, one gets the relations:\n\na) the point $\\mathrm{P}_{2}$ :\n\n$$\na_{23}=a_{12} ; \\quad \\omega^{2} a^{3}=G M\n$$\n\nb) the point $\\mathrm{P}_{3}$ :\n\n$$\na_{13}=a_{23} ; \\quad \\omega^{2} a^{3}=G M\n$$\n\nSummarizing, the system can rotate as a rigid body if all the distances between the masses are equal:\n\n$$\na_{12}=a_{23}=a_{13}=a,\n\\tag{11}\n$$\n\nthe angular velocity $\\omega$ is constant and the relation (9) holds."", 'As the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points $\\left(\\mathrm{P}_{1}, \\mathrm{P}_{2}\\right),\\left(\\mathrm{P}_{2}, \\mathrm{P}_{3}\\right)$ and $\\left(\\mathrm{P}_{1}, \\mathrm{P}_{3}\\right)$. It depends only on the distances $a_{12}, a_{23}$ and $a_{23}$ which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis $\\sigma$ depends only on the distances from the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ to the axis $\\sigma$ which, for fixed $a_{12}, a_{23}$ and $a_{23}$ do not depend on time. This means that the moment of inertia $I$ is constant. Therefore, the angular velocity of the system must also be constant:\n\n$$\n\\omega=\\text { const. }\n\\tag{1}\n$$\n\nThis is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performing calculations, it is desirable to specify a convenient coordinates system in which the calculations are expected to be simple.\n\nLet the positions of the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ be given by the vectors $\\mathbf{r}_{1}, \\mathbf{r}_{2}$ and $\\mathbf{r}_{3}$. For simplicity we assume that the origin of the coordinate system is localized at the center of mass of the points $\\mathrm{P}_{1}, \\mathrm{P}_{2}$ and $\\mathrm{P}_{3}$ with the masses $m_{1}, m_{2}$ and $m_{3}$ and that all the vectors $\\mathbf{r}_{1}, \\mathbf{r}_{2}$ and $\\mathbf{r}_{3}$ are in the same coordinate plane, e.g. in the plane $(x, y)$. Then the axis $\\sigma$ is the axis $z$.\n\nIn this coordinate system, according to the definition of the center of mass, we have:\n\n$$\nm_{1} \\mathbf{r}_{1}+m_{2} \\mathbf{r}_{2}+m_{3} \\mathbf{r}_{2}=0\n\\tag{2}\n$$\n\n\nAt the beginning we find the moment of inertia $I$ of the system with respect to the axis $\\sigma$. Using the relation (2), we can write:\n\n$$\n0=\\left(m_{1} \\mathbf{r}_{1}+m_{2} \\mathbf{r}_{2}+m_{3} \\mathbf{r}_{3}\\right)^{2}=m_{1}^{2} \\mathbf{r}_{1}^{2}+m_{2}^{2} \\mathbf{r}_{2}^{2}+m_{3}^{2} \\mathbf{r}_{3}^{2}+2 m_{1} m_{2} \\mathbf{r}_{1} \\mathbf{r}_{2}+2 m_{1} m_{3} \\mathbf{r}_{1} \\mathbf{r}_{3}+2 m_{3} m_{2} \\mathbf{r}_{3} \\mathbf{r}_{2} .\n$$\n\nOf course,\n\n$$\n\\mathbf{r}_{i}^{2}=r_{i}^{2} \\quad i=1,2,3\n$$\n\nThe quantities $2 \\mathbf{r}_{i} \\mathbf{r}_{j}(i, j=1,2,3)$ can be determined from the following obvious relation:\n\n$$\na_{i j}^{2}=\\left|\\mathbf{r}_{i}-\\mathbf{r}_{j}\\right|^{2}=\\left(\\mathbf{r}_{i}-\\mathbf{r}_{j}\\right)^{2}=\\mathbf{r}_{i}^{2}+\\mathbf{r}_{j}^{2}-2 \\mathbf{r}_{i} \\mathbf{r}_{j}=r_{i}^{2}+r_{j}^{2}-2 \\mathbf{r}_{i} \\mathbf{r}_{j} .\n$$\n\nWe get:\n\n$$\n2 \\mathbf{r}_{i} \\mathbf{r}_{j}=r_{i}^{2}+r_{j}^{2}-a_{i j}^{2}\n$$\n\nWith help of this relation, after simple transformations, we obtain:\n\n$$\n0=\\left(m_{1} \\mathbf{r}_{1}+m_{2} \\mathbf{r}_{2}+m_{3} \\mathbf{r}_{3}\\right)^{2}=\\left(m_{1}+m_{2}+m_{3}\\right)\\left(m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}\\right)-\\sum_{i Figure 1 A water-powered rice-pounding mortar OPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTAR b) Figure 2 a) At the beginning there is no water in the bucket, the pestle rests on the mortar. Water flows into the bucket with a small rate, but for some time the lever remains in the horizontal position. b) At some moment the amount of water is enough to lift the lever up. Due to the tilt, water rushes to the farther side of the bucket, tilting the lever more quickly. Water starts to flow out at $\alpha=\alpha_{1}$. c) As the angle $\alpha$ increases, water starts to flow out. At some particular tilt angle, $\alpha=\beta$, the total torque is zero. d) $\alpha$ continues increasing, water continues to flow out until no water remains in the bucket. e) $\alpha$ keeps increasing because of inertia. Due to the shape of the bucket, water falls into the bucket but immediately flows out. The inertial motion of the lever continues until $\alpha$ reaches the maximal value $\alpha_{0}$. f) With no water in the bucket, the weight of the lever pulls it back to the initial horizontal position. The pestle gives the mortar (with rice inside) a pound and a new cycle begins. C. The problem Consider a water-powered rice-pounding mortar with the following parameters (Figure 3) The mass of the lever (including the pestle but without water) is $M=30 \mathrm{~kg}$, The center of mass of the lever is G. The lever rotates around the axis T (projected onto the point $\mathrm{T}$ on the figure). The moment of inertia of the lever around T is $I=12 \mathrm{~kg} \cdot \mathrm{m}^{2}$. When there is water in the bucket, the mass of water is denoted as $m$, the center of mass of the water body is denoted as $\mathrm{N}$. The tilt angle of the lever with respect to the horizontal axis is $\alpha$. The main length measurements of the mortar and the bucket are as in Figure 3. Neglect friction at the rotation axis and the force due to water falling onto the bucket. In this problem, we make an approximation that the water surface is always horizontal. Figure 3 Design and dimensions of the rice-pounding mortar 1. The structure of the mortar At the beginning, the bucket is empty, and the lever lies horizontally. Then water flows into the bucket until the lever starts rotating. The amount of water in the bucket at this moment is $m=1.0 \mathrm{~kg}$.",1.1. Determine the distance from the center of mass $G$ of the lever to the rotation axis T. It is known that GT is horizontal when the bucket is empty.,"['The volume of water in the bucket is $V=1000 \\mathrm{~cm}^{3}=10^{-3} \\mathrm{~m}^{3}$. The length of the bottom of the bucket is $d=L-h \\tan 60^{\\circ}=\\left(0.74-0.12 \\tan 60^{\\circ}\\right) \\mathrm{m}=0.5322 \\mathrm{~m}$. (as the initial data are given with two significant digits, we shall keep only two significant digits in the final answer, but we keep more digits in the intermediate steps). The height $c$ of the water layer in the bucket is calculated from the formula:\n\n$$\nV=b c d+b \\frac{c}{2} c \\tan 60^{0} \\Rightarrow c=\\frac{\\left(d^{2}+2 \\sqrt{3} V / b\\right)^{1 / 2}-d}{\\sqrt{3}}\n$$\n\nInserting numerical values for $V, b$ and $d$, we find $c=0.01228 \\mathrm{~m}$.\n\nWhen the lever lies horizontally, the distance, on the horizontal axis, between the rotation axis and the center of mass of water $\\mathrm{N}$, is $\\mathrm{TH} \\approx a+\\frac{d}{2}+\\frac{c}{4} \\tan 60^{\\circ}=0.4714 \\mathrm{~m}$, and $\\mathrm{TG}=(m / M) \\mathrm{TH}=0.01571 \\mathrm{~m}($ see the figure below $)$.\n\n']",['0.016'],False,m,Numerical,1e-3 1327,Mechanics,"WATER-POWERED RICE-POUNDING MORTAR A. Introduction Rice is the main staple food of most people in Vietnam. To make white rice from paddy rice, one needs separate of the husk (a process called ""hulling"") and separate the bran layer (""milling""). The hilly parts of northern Vietnam are abundant with water streams, and people living there use water-powered rice-pounding mortar for bran layer separation. Figure 1 shows one of such mortars., Figure 2 shows how it works. B. Design and operation 1. Design. The rice-pounding mortar shown in Figure 1 has the following parts: The mortar, basically a wooden container for rice. The lever, which is a tree trunk with one larger end and one smaller end. It can rotate around a horizontal axis. A pestle is attached perpendicularly to the lever at the smaller end. The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally. The larger end of the lever is carved hollow to form a bucket. The shape of the bucket is crucial for the mortar's operation. 2. Modes of operation The mortar has two modes. Working mode. In this mode, the mortar goes through an operation cycle illustrated in Figure 2. The rice-pounding function comes from the work that is transferred from the pestle to the rice during stage f) of Figure 2. If, for some reason, the pestle never touches the rice, we say that the mortar is not working. Rest mode with the lever lifted up. During stage c) of the operation cycle (Figure 2), as the tilt angle $\alpha$ increases, the amount of water in the bucket decreases. At one particular moment in time, the amount of water is just enough to counterbalance the weight of the lever. Denote the tilting angle at this instant by $\beta$. If the lever is kept at angle $\beta$ and the initial angular velocity is zero, then the lever will remain at this position forever. This is the rest mode with the lever lifted up. The stability of this position depends on the flow rate of water into the bucket, $\Phi$. If $\Phi$ exceeds some value $\Phi_{2}$, then this rest mode is stable, and the mortar cannot be in the working mode. In other words, $\Phi_{2}$ is the minimal flow rate for the mortar not to work. Figure 1 A water-powered rice-pounding mortar OPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTAR b) Figure 2 a) At the beginning there is no water in the bucket, the pestle rests on the mortar. Water flows into the bucket with a small rate, but for some time the lever remains in the horizontal position. b) At some moment the amount of water is enough to lift the lever up. Due to the tilt, water rushes to the farther side of the bucket, tilting the lever more quickly. Water starts to flow out at $\alpha=\alpha_{1}$. c) As the angle $\alpha$ increases, water starts to flow out. At some particular tilt angle, $\alpha=\beta$, the total torque is zero. d) $\alpha$ continues increasing, water continues to flow out until no water remains in the bucket. e) $\alpha$ keeps increasing because of inertia. Due to the shape of the bucket, water falls into the bucket but immediately flows out. The inertial motion of the lever continues until $\alpha$ reaches the maximal value $\alpha_{0}$. f) With no water in the bucket, the weight of the lever pulls it back to the initial horizontal position. The pestle gives the mortar (with rice inside) a pound and a new cycle begins. C. The problem Consider a water-powered rice-pounding mortar with the following parameters (Figure 3) The mass of the lever (including the pestle but without water) is $M=30 \mathrm{~kg}$, The center of mass of the lever is G. The lever rotates around the axis T (projected onto the point $\mathrm{T}$ on the figure). The moment of inertia of the lever around T is $I=12 \mathrm{~kg} \cdot \mathrm{m}^{2}$. When there is water in the bucket, the mass of water is denoted as $m$, the center of mass of the water body is denoted as $\mathrm{N}$. The tilt angle of the lever with respect to the horizontal axis is $\alpha$. The main length measurements of the mortar and the bucket are as in Figure 3. Neglect friction at the rotation axis and the force due to water falling onto the bucket. In this problem, we make an approximation that the water surface is always horizontal. Figure 3 Design and dimensions of the rice-pounding mortar 1. The structure of the mortar At the beginning, the bucket is empty, and the lever lies horizontally. Then water flows into the bucket until the lever starts rotating. The amount of water in the bucket at this moment is $m=1.0 \mathrm{~kg}$. Context question: 1.1. Determine the distance from the center of mass $G$ of the lever to the rotation axis T. It is known that GT is horizontal when the bucket is empty. Context answer: \boxed{0.016} ","1.2. Water starts flowing out of the bucket when the angle between the lever and the horizontal axis reaches $\alpha_{1}$. The bucket is completely empty when this angle is $\alpha_{2}$. Determine $\alpha_{1}$ and $\alpha_{2}$.","['When the lever tilts with angle $\\alpha_{1}$, water level is at the edge of the bucket. At that point the water volume is $10^{-3} \\mathrm{~m}^{3}$. Assume $\\mathrm{PQ}']","['$\\alpha_{1}=20.6^{\\circ}$ , $\\alpha_{2}=30^{\\circ}$']",True,,Numerical,1e-1 1328,Mechanics,"WATER-POWERED RICE-POUNDING MORTAR A. Introduction Rice is the main staple food of most people in Vietnam. To make white rice from paddy rice, one needs separate of the husk (a process called ""hulling"") and separate the bran layer (""milling""). The hilly parts of northern Vietnam are abundant with water streams, and people living there use water-powered rice-pounding mortar for bran layer separation. Figure 1 shows one of such mortars., Figure 2 shows how it works. B. Design and operation 1. Design. The rice-pounding mortar shown in Figure 1 has the following parts: The mortar, basically a wooden container for rice. The lever, which is a tree trunk with one larger end and one smaller end. It can rotate around a horizontal axis. A pestle is attached perpendicularly to the lever at the smaller end. The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally. The larger end of the lever is carved hollow to form a bucket. The shape of the bucket is crucial for the mortar's operation. 2. Modes of operation The mortar has two modes. Working mode. In this mode, the mortar goes through an operation cycle illustrated in Figure 2. The rice-pounding function comes from the work that is transferred from the pestle to the rice during stage f) of Figure 2. If, for some reason, the pestle never touches the rice, we say that the mortar is not working. Rest mode with the lever lifted up. During stage c) of the operation cycle (Figure 2), as the tilt angle $\alpha$ increases, the amount of water in the bucket decreases. At one particular moment in time, the amount of water is just enough to counterbalance the weight of the lever. Denote the tilting angle at this instant by $\beta$. If the lever is kept at angle $\beta$ and the initial angular velocity is zero, then the lever will remain at this position forever. This is the rest mode with the lever lifted up. The stability of this position depends on the flow rate of water into the bucket, $\Phi$. If $\Phi$ exceeds some value $\Phi_{2}$, then this rest mode is stable, and the mortar cannot be in the working mode. In other words, $\Phi_{2}$ is the minimal flow rate for the mortar not to work. Figure 1 A water-powered rice-pounding mortar OPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTAR b) Figure 2 a) At the beginning there is no water in the bucket, the pestle rests on the mortar. Water flows into the bucket with a small rate, but for some time the lever remains in the horizontal position. b) At some moment the amount of water is enough to lift the lever up. Due to the tilt, water rushes to the farther side of the bucket, tilting the lever more quickly. Water starts to flow out at $\alpha=\alpha_{1}$. c) As the angle $\alpha$ increases, water starts to flow out. At some particular tilt angle, $\alpha=\beta$, the total torque is zero. d) $\alpha$ continues increasing, water continues to flow out until no water remains in the bucket. e) $\alpha$ keeps increasing because of inertia. Due to the shape of the bucket, water falls into the bucket but immediately flows out. The inertial motion of the lever continues until $\alpha$ reaches the maximal value $\alpha_{0}$. f) With no water in the bucket, the weight of the lever pulls it back to the initial horizontal position. The pestle gives the mortar (with rice inside) a pound and a new cycle begins. C. The problem Consider a water-powered rice-pounding mortar with the following parameters (Figure 3) The mass of the lever (including the pestle but without water) is $M=30 \mathrm{~kg}$, The center of mass of the lever is G. The lever rotates around the axis T (projected onto the point $\mathrm{T}$ on the figure). The moment of inertia of the lever around T is $I=12 \mathrm{~kg} \cdot \mathrm{m}^{2}$. When there is water in the bucket, the mass of water is denoted as $m$, the center of mass of the water body is denoted as $\mathrm{N}$. The tilt angle of the lever with respect to the horizontal axis is $\alpha$. The main length measurements of the mortar and the bucket are as in Figure 3. Neglect friction at the rotation axis and the force due to water falling onto the bucket. In this problem, we make an approximation that the water surface is always horizontal. Figure 3 Design and dimensions of the rice-pounding mortar 1. The structure of the mortar At the beginning, the bucket is empty, and the lever lies horizontally. Then water flows into the bucket until the lever starts rotating. The amount of water in the bucket at this moment is $m=1.0 \mathrm{~kg}$. Context question: 1.1. Determine the distance from the center of mass $G$ of the lever to the rotation axis T. It is known that GT is horizontal when the bucket is empty. Context answer: \boxed{0.016} Context question: 1.2. Water starts flowing out of the bucket when the angle between the lever and the horizontal axis reaches $\alpha_{1}$. The bucket is completely empty when this angle is $\alpha_{2}$. Determine $\alpha_{1}$ and $\alpha_{2}$. Context answer: \boxed{$\alpha_{1}=20.6^{\circ}$ , $\alpha_{2}=30^{\circ}$} ",1.3. Let $\mu(\alpha)$ be the total torque (relative to the axis $\mathrm{T}$ ) which comes from the weight of the lever and the water in the bucket. $\mu(\alpha)$ is zero when $\alpha=\beta$. Determine $\beta$ and the mass $m_{1}$ of water in the bucket at this instant.,"['Denote $\\mathrm{PQ}=x(\\mathrm{~m})$. The amount of water in the bucket is $m=\\rho_{\\text {water }} \\frac{x h b}{2}=9 x(\\mathrm{~kg})$.\n\n$\\mu=0$ when the torque coming from the water in the bucket cancels out the torque coming from the weight of the lever. The cross section of the water in the bucket is the triangle PQR in the figure. The center of mass $\\mathrm{N}$ of water is located at $2 / 3$ of the meridian RI, therefore NTG lies on a straight line. Then: $m g \\times \\mathrm{TN}=M g \\times \\mathrm{TG}$ or\n\n$$\nm \\times \\mathrm{TN}=M \\times \\mathrm{TG}=30 \\times 0.1571=0.4714\n\\tag{1}\n$$\n\nCalculating TN from $x$ then substitute (1):\n\n$$\n\\mathrm{TN}=L+a-\\frac{2}{3}\\left(h \\sqrt{3}+\\frac{x}{2}\\right)=0.94-0.08 \\sqrt{3}-\\frac{x}{3}=0.8014-\\frac{x}{3}\n$$\n\nwhich implies $m \\times \\mathrm{TN}=9 x(0.8014-x / 3)=-3 x^{2}+7.213 x$ (2)\n\nSo we find an equation for $x$ :\n\n$$\n-3 x^{2}+7.213 x=0.4714\n\\tag{3}\n$$\n\nThe solutions to (3) are $x=2.337$ and $x=0.06723$. Since $x$ has to be smaller than 0.5322 , we have to take $x=x_{0}=0.06723$ and $m=9 x_{0}=0.6051 \\mathrm{~kg}$.\n\n$$\n\\tan \\beta=\\frac{h}{x+h \\sqrt{3}}=0.4362 \\text {, or } \\quad \\beta=23.57^{\\circ} \\text {. }\n$$']","['$m=0.61 $ , $\\beta=23.6$']",True,"$\mathrm{~kg}$, $^{\circ}$",Numerical,"1e-2,5e-1" 1329,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: ",1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$.,"['\n\nFigure 1\n\nLet us consider a plane containing the particle trajectory. At $t=0$, the particle position is at point $\\mathrm{A}$. It reaches point $\\mathrm{B}$ at $t=t_{1}$. According to the Huygens principle, at moment $0\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$",3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light.,"['For the Cherenkov effect to occur it is necessary that $n>\\frac{C}{v}$, that is $n_{\\min }=\\frac{C}{v}$.\n\nPutting $\\zeta=n-1=2.7 \\times 10^{-4} P$, we get\n\n$$\n\\zeta_{\\min }=2.7 \\times 10^{-4} P_{\\min }=\\frac{C}{v}-1=\\frac{1}{\\beta}-1\n\\tag{1}\n$$\n\nBecause\n\n$$\n\\frac{M c^{2}}{p c}=\\frac{M c}{p}=\\frac{M c}{\\frac{M v}{\\sqrt{1-\\beta^{2}}}}=\\frac{\\sqrt{1-\\beta^{2}}}{\\beta}=K\n\\tag{2}\n$$\n\nthen $K=0.094 ; 0.05 ; 0.014$ for proton, kaon and pion, respectively.\n\nFrom (2) we can express $\\beta$ through $K$ as\n\n$$\n\\beta=\\frac{1}{\\sqrt{1+K^{2}}}\n\\tag{3}\n$$\n\nSince $K^{2}<<1$ for all three kinds of particles we can neglect the terms of order higher than 2 in $K$. We get\n\n$$\n1-\\beta=1-\\frac{1}{\\sqrt{1+K^{2}}} \\approx \\frac{1}{2} K^{2}=\\frac{1}{2}\\left(\\frac{M c}{p}\\right)^{2}\n\\tag{3a}\n$$\n$$\n\\frac{1}{\\beta}-1=\\sqrt{1+K^{2}}-1 \\approx \\frac{1}{2} K^{2}=\\frac{1}{2}\\left(\\frac{M c}{p}\\right)^{2}\n\\tag{3b}\n$$\n\nPutting (3b) into (1), we obtain\n\n\n\n$$\nP_{\\min }=\\frac{1}{2.7 \\times 10^{-4}} \\times \\frac{1}{2} K^{2}\n\\tag{4}\n$$\n\nWe get the following numerical values of the minimal pressure:\n\n$$\n\\begin{array}{ll}\nP_{\\min }=16 \\mathrm{~atm} & \\text { for protons, } \\\\\nP_{\\min }=4.6 \\mathrm{~atm} & \\text { for kaons, } \\\\\nP_{\\min }=0.36 \\mathrm{~atm} & \\text { for pions. }\n\\end{array}\n$$\n\n']","['$16, 4.6, 0.36$']",True,\mathrm{~atm},Numerical,"1e-1, 1e-2, 1e-3" 1331,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$","3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure?","['For $\\theta_{\\pi}=2 \\theta_{\\kappa} \\quad$ we have\n\n$$\n\\cos \\theta_{\\pi}=\\cos 2 \\theta_{\\kappa}=2 \\cos ^{2} \\theta_{\\kappa}-1\n\\tag{5}\n$$\n\nWe denote\n\n$$\n\\varepsilon=1-\\beta=1-\\frac{1}{\\sqrt{1+K^{2}}} \\approx \\frac{1}{2} K^{2}\n\\tag{6}\n$$\n\nFrom (5) we obtain\n\n$$\n\\frac{1}{\\beta_{\\pi} n}=\\frac{2}{\\beta_{\\kappa}^{2} n^{2}}-1\n\\tag{7}\n$$\n\nSubstituting $\\beta=1-\\varepsilon$ and $n=1+\\zeta$ into (7), we get approximately:\n\n$$\n\\begin{aligned}\n& \\zeta_{\\frac{1}{2}}=\\frac{4 \\varepsilon_{\\kappa}-\\varepsilon_{\\pi}}{3}=\\frac{1}{6}\\left(4 K_{\\kappa}^{2}-K_{\\pi}^{2}\\right)=\\frac{1}{6}\\left[4 .(0.05)^{2}-(0.014)^{2}\\right], \\\\\n& P_{\\frac{1}{2}}=\\frac{1}{2.7 \\times 10^{-4}} \\zeta_{\\frac{1}{2}}=6 \\mathrm{~atm} .\n\\end{aligned}\n$$\n\nThe corresponding value of refraction index is $n=1.00162$. We get:\n\n$$\n\\theta_{\\kappa}=1.6^{\\circ} ; \\quad \\theta_{\\pi}=2 \\theta_{\\kappa}=3.2^{\\circ} \\text {. }\n$$\n\nWe do not observe the ring image of protons since\n\n$$\nP_{\\frac{1}{2}}=6 \\mathrm{~atm}<16 \\mathrm{~atm}=P_{\\min } \\text { for protons. }\n$$\n\n']","['$P_{\\frac{1}{2}}=6 $ , $\\theta_{\\kappa}=1.6$ , $\\theta_{\\pi}=3.2$']",True,"$\mathrm{~atm}$, $^{\circ}$, $^{\circ}$",Numerical, 1332,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$ Context question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: \boxed{$P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$} Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at $10 \mathrm{GeV} / \mathrm{c}$ having a half width at half height $\Delta p$. This makes the ring image broaden, correspondingly $\theta$ distribution has a half width at half height $\Delta \theta$. The pressure of the radiator is $P_{\frac{1}{2}}$ determined in 3.2.","4.1. Calculate $\frac{\Delta \theta_{\kappa}}{\Delta p}$ and $\frac{\Delta \theta_{\pi}}{\Delta p}$, the values taken by $\frac{\Delta \theta}{\Delta p}$ in the pions and kaons cases.","['Taking logarithmic differentiation of both sides of the equation $\\cos \\theta=\\frac{1}{\\beta n}$, we obtain\n\n$$\n\\frac{\\sin \\theta \\times \\Delta \\theta}{\\cos \\theta}=\\frac{\\Delta \\beta}{\\beta}\n\\tag{8}\n$$\n\nLogarithmically differentiating equation (3a) gives\n\n$$\n\\frac{\\Delta \\beta}{1-\\beta}=2 \\frac{\\Delta p}{p}\n\\tag{9}\n$$\n\nCombining (8) and (9), taking into account (3b) and putting approximately $\\tan \\theta=\\theta$, we derive\n\n$$\n\\frac{\\Delta \\theta}{\\Delta p}=\\frac{2}{\\theta} \\times \\frac{1-\\beta}{p \\beta}=\\frac{K^{2}}{\\theta p}\n\\tag{10}\n$$\n\nWe obtain\n\n-for kaons $K_{\\kappa}=0.05, \\quad \\theta_{\\kappa}=1.6^{\\circ}=1.6 \\frac{\\pi}{180} \\mathrm{rad}$, and so, $\\frac{\\Delta \\theta_{\\kappa}}{\\Delta p}=0.51 \\frac{1^{0}}{\\mathrm{GeV} / \\mathrm{c}}$,\n\n-for\n\npions\n\n$$\nK_{\\pi}=0.014\n$$\n\n$$\n\\theta_{\\pi}=3.2^{\\circ}\n$$\n\n$\\frac{\\Delta \\theta_{\\pi}}{\\Delta p}=0.02 \\frac{1^{0}}{\\mathrm{GeV} / c}$.\n\n']","['$\\frac{\\Delta \\theta_{\\kappa}}{\\Delta p}=0.51 $ , $\\frac{\\Delta \\theta_{\\pi}}{\\Delta p}=0.02$']",True,$\frac{1^{0}}{\mathrm{GeV} / \mathrm{c}}$,Numerical,1e-2 1333,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$ Context question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: \boxed{$P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$} Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at $10 \mathrm{GeV} / \mathrm{c}$ having a half width at half height $\Delta p$. This makes the ring image broaden, correspondingly $\theta$ distribution has a half width at half height $\Delta \theta$. The pressure of the radiator is $P_{\frac{1}{2}}$ determined in 3.2. Context question: 4.1. Calculate $\frac{\Delta \theta_{\kappa}}{\Delta p}$ and $\frac{\Delta \theta_{\pi}}{\Delta p}$, the values taken by $\frac{\Delta \theta}{\Delta p}$ in the pions and kaons cases. Context answer: \boxed{$\frac{\Delta \theta_{\kappa}}{\Delta p}=0.51 $ , $\frac{\Delta \theta_{\pi}}{\Delta p}=0.02$} ","4.2. When the separation between the two ring images, $\theta_{\pi}-\theta_{\kappa}$, is greater than 10 times the half-width sum $\Delta \theta=\Delta \theta_{\kappa}+\Delta \theta_{\pi}$, that is $\theta_{\pi}-\theta_{\kappa}>10 \Delta \theta$, it is possible to distinguish well the two ring images. Calculate the maximal value of $\Delta p$ such that the two ring images can still be well distinguished.",['$\\frac{\\Delta \\theta_{\\kappa}+\\Delta \\theta_{\\pi}}{\\Delta p} \\equiv \\frac{\\Delta \\theta}{\\Delta p}=(0.51+0.02) \\frac{1^{0}}{\\mathrm{GeV} / c}=0.53 \\frac{1^{0}}{\\mathrm{GeV} / c}$.\n\nThe condition for two ring images to be distinguishable is $\\Delta \\theta<0.1\\left(\\theta_{\\pi}-\\theta_{\\kappa}\\right)=0.16^{0}$.\n\nIt follows $\\quad \\Delta p<\\frac{1}{10} \\times \\frac{1.6}{0.53}=0.3 \\mathrm{GeV} / c$.'],['$0.3 $'],False,$\mathrm{GeV} / c$,Numerical,1e-1 1334,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$ Context question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: \boxed{$P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$} Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at $10 \mathrm{GeV} / \mathrm{c}$ having a half width at half height $\Delta p$. This makes the ring image broaden, correspondingly $\theta$ distribution has a half width at half height $\Delta \theta$. The pressure of the radiator is $P_{\frac{1}{2}}$ determined in 3.2. Context question: 4.1. Calculate $\frac{\Delta \theta_{\kappa}}{\Delta p}$ and $\frac{\Delta \theta_{\pi}}{\Delta p}$, the values taken by $\frac{\Delta \theta}{\Delta p}$ in the pions and kaons cases. Context answer: \boxed{$\frac{\Delta \theta_{\kappa}}{\Delta p}=0.51 $ , $\frac{\Delta \theta_{\pi}}{\Delta p}=0.02$} Context question: 4.2. When the separation between the two ring images, $\theta_{\pi}-\theta_{\kappa}$, is greater than 10 times the half-width sum $\Delta \theta=\Delta \theta_{\kappa}+\Delta \theta_{\pi}$, that is $\theta_{\pi}-\theta_{\kappa}>10 \Delta \theta$, it is possible to distinguish well the two ring images. Calculate the maximal value of $\Delta p$ such that the two ring images can still be well distinguished. Context answer: \boxed{$0.3 $} Extra Supplementary Reading Materials: 5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light.","5.1. Find out the minimal kinetic energy $T_{\min }$ of a particle with a rest mass $M$ moving in water, such that it emits Cherenkov light. The index of refraction of water is $n=1.33$.","['The lower limit of $\\beta$ giving rise to Cherenkov effect is\n\n$$\n\\beta=\\frac{1}{n}=\\frac{1}{1.33}\n\\tag{11}\n$$\n\nThe kinetic energy of a particle having rest mass $M$ and energy $E$ is given by the expression\n\n$$\nT=E-M c^{2}=\\frac{M c^{2}}{\\sqrt{1-\\beta^{2}}}-M c^{2}=M c^{2}\\left[\\frac{1}{\\sqrt{1-\\beta^{2}}}-1\\right]\n\\tag{12}\n$$\n\nSubstituting the limiting value (11) of $\\beta$ into (12), we get the minimal kinetic energy of the particle for Cherenkov effect to occur:\n\n$$\nT_{\\min }=M c^{2}\\left[\\frac{1}{\\sqrt{1-\\left(\\frac{1}{1.33}\\right)^{2}}}-1\\right]=0.517 M c^{2}\n\\tag{13}\n$$\n\n']",['$0.517 M c^{2}$'],False,,Expression, 1335,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$ Context question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: \boxed{$P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$} Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at $10 \mathrm{GeV} / \mathrm{c}$ having a half width at half height $\Delta p$. This makes the ring image broaden, correspondingly $\theta$ distribution has a half width at half height $\Delta \theta$. The pressure of the radiator is $P_{\frac{1}{2}}$ determined in 3.2. Context question: 4.1. Calculate $\frac{\Delta \theta_{\kappa}}{\Delta p}$ and $\frac{\Delta \theta_{\pi}}{\Delta p}$, the values taken by $\frac{\Delta \theta}{\Delta p}$ in the pions and kaons cases. Context answer: \boxed{$\frac{\Delta \theta_{\kappa}}{\Delta p}=0.51 $ , $\frac{\Delta \theta_{\pi}}{\Delta p}=0.02$} Context question: 4.2. When the separation between the two ring images, $\theta_{\pi}-\theta_{\kappa}$, is greater than 10 times the half-width sum $\Delta \theta=\Delta \theta_{\kappa}+\Delta \theta_{\pi}$, that is $\theta_{\pi}-\theta_{\kappa}>10 \Delta \theta$, it is possible to distinguish well the two ring images. Calculate the maximal value of $\Delta p$ such that the two ring images can still be well distinguished. Context answer: \boxed{$0.3 $} Extra Supplementary Reading Materials: 5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light. Context question: 5.1. Find out the minimal kinetic energy $T_{\min }$ of a particle with a rest mass $M$ moving in water, such that it emits Cherenkov light. The index of refraction of water is $n=1.33$. Context answer: $0.517 M c^{2}$ ","5.2. The radioactive source used by Cherenkov emits either $\alpha$ particles (i.e. helium nuclei) having a rest mass $M_{\alpha}=3.8 \mathrm{GeV} / c^{2}$ or $\beta$ particles (i.e. electrons) having a rest mass $M_{\mathrm{e}}=0.51 \mathrm{MeV} / c^{2}$. Calculate the numerical values of $T_{\min }$ for $\alpha$ particles and $\beta$ particles. Knowing that the kinetic energy of particles emitted by radioactive sources never exceeds a few $\mathrm{MeV}$, find out which particles give rise to the radiation observed by Cherenkov.","['For $\\alpha$ particles, $T_{\\min }=0.517 \\times 3.8 \\mathrm{GeV}=1.96 \\mathrm{GeV}$.\n\nFor electrons, $\\quad T_{\\text {min }}=0.517 \\times 0.51 \\mathrm{MeV}=0.264 \\mathrm{MeV}$.\n\nSince the kinetic energy of the particles emitted by radioactive source does not exceed a few $\\mathrm{MeV}$, these are electrons which give rise to Cherenkov radiation in the considered experiment.']","['1.96, 0.264']",True,$\mathrm{GeV}$,Numerical,1e-2 1336,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$ Context question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: \boxed{$P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$} Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at $10 \mathrm{GeV} / \mathrm{c}$ having a half width at half height $\Delta p$. This makes the ring image broaden, correspondingly $\theta$ distribution has a half width at half height $\Delta \theta$. The pressure of the radiator is $P_{\frac{1}{2}}$ determined in 3.2. Context question: 4.1. Calculate $\frac{\Delta \theta_{\kappa}}{\Delta p}$ and $\frac{\Delta \theta_{\pi}}{\Delta p}$, the values taken by $\frac{\Delta \theta}{\Delta p}$ in the pions and kaons cases. Context answer: \boxed{$\frac{\Delta \theta_{\kappa}}{\Delta p}=0.51 $ , $\frac{\Delta \theta_{\pi}}{\Delta p}=0.02$} Context question: 4.2. When the separation between the two ring images, $\theta_{\pi}-\theta_{\kappa}$, is greater than 10 times the half-width sum $\Delta \theta=\Delta \theta_{\kappa}+\Delta \theta_{\pi}$, that is $\theta_{\pi}-\theta_{\kappa}>10 \Delta \theta$, it is possible to distinguish well the two ring images. Calculate the maximal value of $\Delta p$ such that the two ring images can still be well distinguished. Context answer: \boxed{$0.3 $} Extra Supplementary Reading Materials: 5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light. Context question: 5.1. Find out the minimal kinetic energy $T_{\min }$ of a particle with a rest mass $M$ moving in water, such that it emits Cherenkov light. The index of refraction of water is $n=1.33$. Context answer: $0.517 M c^{2}$ Context question: 5.2. The radioactive source used by Cherenkov emits either $\alpha$ particles (i.e. helium nuclei) having a rest mass $M_{\alpha}=3.8 \mathrm{GeV} / c^{2}$ or $\beta$ particles (i.e. electrons) having a rest mass $M_{\mathrm{e}}=0.51 \mathrm{MeV} / c^{2}$. Calculate the numerical values of $T_{\min }$ for $\alpha$ particles and $\beta$ particles. Knowing that the kinetic energy of particles emitted by radioactive sources never exceeds a few $\mathrm{MeV}$, find out which particles give rise to the radiation observed by Cherenkov. Context answer: \boxed{1.96, 0.264} Extra Supplementary Reading Materials: 6. In the previous sections of the problem, the dependence of the Cherenkov effect on wavelength $\lambda$ has been ignored. We now take into account the fact that the Cherenkov radiation of a particle has a broad continuous spectrum including the visible range (wavelengths from $0.4 \mu \mathrm{m}$ to $0.8 \mu \mathrm{m}$ ). We know also that the index of refraction $n$ of the radiator decreases linearly by $2 \%$ of $n-1$ when $\lambda$ increases over this range.",6.1. Consider a beam of pions with definite momentum of $10.0 \mathrm{GeV} / \mathrm{c}$ moving in air at pressure $6 \mathrm{~atm}$. Find out the angular difference $\delta \theta$ associated with the two ends of the visible range.,"['For a beam of particles having a definite momentum the dependence of the angle $\\theta$ on the refraction index $n$ of the medium is given by the expression\n\n$$\n\\cos \\theta=\\frac{1}{n \\beta}\n\\tag{14}\n$$\n\nLet $\\delta \\theta$ be the difference of $\\theta$ between two rings corresponding to two wavelengths limiting the visible range, i.e. to wavelengths of $0.4 \\mu \\mathrm{m}$ (violet) and $0.8 \\mu \\mathrm{m}$ (red), respectively. The difference in the refraction indexes at these wavelengths is $n_{\\mathrm{v}}-n_{\\mathrm{r}}=\\delta n=0.02(n-1)$.\n\nLogarithmically differentiating both sides of equation (14) gives\n\n\n\n$$\n\\frac{\\sin \\theta \\times \\delta \\theta}{\\cos \\theta}=\\frac{\\delta n}{n}\n\\tag{15}\n$$\n\nCorresponding to the pressure of the radiator $P=6 \\mathrm{~atm}$ we have from 4.2. the values $\\theta_{\\pi}=3.2^{\\circ}, \\quad n=1.00162$.\n\nPutting approximately $\\tan \\theta=\\theta$ and $n=1$, we get $\\delta \\theta=\\frac{\\delta n}{\\theta}=0.033^{\\circ}$.']",['$0.033$'],False,$^{\circ}$,Numerical,1e-3 1337,Optics,"CHERENKOV LIGHT AND RING IMAGING COUNTER Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle $$ \theta=\arccos \frac{1}{\beta n} \tag{1} $$ where $\beta=\frac{v}{c}$. 1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$. At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time. Context question: 1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle. Context answer: Context question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$. Context answer: \boxed{$\varphi=\arcsin \frac{1}{\beta n}$} Context question: 2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror. Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image. This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator. Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected. Context answer: Extra Supplementary Reading Materials: 3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$. The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$ Context question: 3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light. Context answer: $$ \begin{array}{ll} P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\ P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\ P_{\min }=0.36 \mathrm{~atm} & \text { for pions. } \end{array} $$ Context question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case. Is it possible to observe the ring image of protons under this pressure? Context answer: \boxed{$P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$} Extra Supplementary Reading Materials: 4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at $10 \mathrm{GeV} / \mathrm{c}$ having a half width at half height $\Delta p$. This makes the ring image broaden, correspondingly $\theta$ distribution has a half width at half height $\Delta \theta$. The pressure of the radiator is $P_{\frac{1}{2}}$ determined in 3.2. Context question: 4.1. Calculate $\frac{\Delta \theta_{\kappa}}{\Delta p}$ and $\frac{\Delta \theta_{\pi}}{\Delta p}$, the values taken by $\frac{\Delta \theta}{\Delta p}$ in the pions and kaons cases. Context answer: \boxed{$\frac{\Delta \theta_{\kappa}}{\Delta p}=0.51 $ , $\frac{\Delta \theta_{\pi}}{\Delta p}=0.02$} Context question: 4.2. When the separation between the two ring images, $\theta_{\pi}-\theta_{\kappa}$, is greater than 10 times the half-width sum $\Delta \theta=\Delta \theta_{\kappa}+\Delta \theta_{\pi}$, that is $\theta_{\pi}-\theta_{\kappa}>10 \Delta \theta$, it is possible to distinguish well the two ring images. Calculate the maximal value of $\Delta p$ such that the two ring images can still be well distinguished. Context answer: \boxed{$0.3 $} Extra Supplementary Reading Materials: 5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light. Context question: 5.1. Find out the minimal kinetic energy $T_{\min }$ of a particle with a rest mass $M$ moving in water, such that it emits Cherenkov light. The index of refraction of water is $n=1.33$. Context answer: $0.517 M c^{2}$ Context question: 5.2. The radioactive source used by Cherenkov emits either $\alpha$ particles (i.e. helium nuclei) having a rest mass $M_{\alpha}=3.8 \mathrm{GeV} / c^{2}$ or $\beta$ particles (i.e. electrons) having a rest mass $M_{\mathrm{e}}=0.51 \mathrm{MeV} / c^{2}$. Calculate the numerical values of $T_{\min }$ for $\alpha$ particles and $\beta$ particles. Knowing that the kinetic energy of particles emitted by radioactive sources never exceeds a few $\mathrm{MeV}$, find out which particles give rise to the radiation observed by Cherenkov. Context answer: \boxed{1.96, 0.264} Extra Supplementary Reading Materials: 6. In the previous sections of the problem, the dependence of the Cherenkov effect on wavelength $\lambda$ has been ignored. We now take into account the fact that the Cherenkov radiation of a particle has a broad continuous spectrum including the visible range (wavelengths from $0.4 \mu \mathrm{m}$ to $0.8 \mu \mathrm{m}$ ). We know also that the index of refraction $n$ of the radiator decreases linearly by $2 \%$ of $n-1$ when $\lambda$ increases over this range. Context question: 6.1. Consider a beam of pions with definite momentum of $10.0 \mathrm{GeV} / \mathrm{c}$ moving in air at pressure $6 \mathrm{~atm}$. Find out the angular difference $\delta \theta$ associated with the two ends of the visible range. Context answer: \boxed{$0.033$} Extra Supplementary Reading Materials: 6.2. On this basis, study qualitatively the effect of the dispersion on the ring image of pions with momentum distributed over an interval centered at $p=10 \mathrm{GeV} / c$ and having a half width at half height $\Delta p=0.3 \mathrm{GeV} / c$.",6.2.1. Calculate the broadening due to dispersion (varying refraction index) and that due to achromaticity of the beam (varying momentum).,"['The broadening due to dispersion in terms of half width at half height is, according to (6.1), $\\frac{1}{2} \\delta \\theta=0.017^{\\circ}$.']",['$0.017^{\\circ}$'],False,,Numerical,1e-3 1338,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude.",1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.,"['For an altitude change $d z$, the atmospheric pressure change is :\n\n$$\nd p=-\\rho g d z\n\\tag{1}\n$$\n\nwhere $g$ is the acceleration of gravity, considered constant, $\\rho$ is the specific mass of air, which is considered as an ideal gas:\n\n$$\n\\rho=\\frac{m}{V}=\\frac{p \\mu}{R T}\n$$\n\nPut this expression in (1) :\n\n$$\n\\frac{d p}{p}=-\\frac{\\mu g}{R T} d z\n$$\n\nIf the air temperature is uniform and equals $T_{0}$, then\n\n$$\n\\frac{d p}{p}=-\\frac{\\mu g}{R T_{0}} d z\n$$\n\nAfter integration, we have :\n\n$$\np(z)=p(0) \\mathbf{e}^{-\\frac{\\mu g}{R T_{0}} z}\n\\tag{2}\n$$']",['$p(z)=p(0) e^{-\\frac{\\mu g}{R T_{0}} z}$'],False,,Expression, 1339,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ).",1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.,"['If\n\n$$\nT(z)=T(0)-\\Lambda z\n\\tag{3}\n$$\n\nthen\n\n$$\n\\frac{d p}{p}=-\\frac{\\mu g}{R[T(0)-\\Lambda z]} d z\n\\tag{4}\n$$\n\nKnowing that :\n\n$$\n\\int \\frac{d z}{T(0)-\\Lambda z}=-\\frac{1}{\\Lambda} \\int \\frac{d[T(0)-\\Lambda z]}{T(0)-\\Lambda z}=-\\frac{1}{\\Lambda} \\ln (T(0)-\\Lambda z)\n$$\n\nby integrating both members of (4), we obtain :\n\n$$\n\\ln \\frac{p(z)}{p(0)}=\\frac{\\mu g}{R \\Lambda} \\ln \\frac{T(0)-\\Lambda z}{T(0)}=\\frac{\\mu g}{R \\Lambda} \\ln \\left(1-\\frac{\\Lambda z}{T(0)}\\right)\n$$\n$$\np(z)=p(0)\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\mu g}{R \\Lambda}}\n\\tag{5}\n$$']",['$p(z)=p(0)\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\mu g}{R \\Lambda}}$'],False,,Expression, 1340,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} ",1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?,['The free convection occurs if:\n\n$$\n\\frac{\\rho(z)}{\\rho(0)}>1\n$$\n\nThe ratio of specific masses can be expressed as follows:\n\n$$\n\\frac{\\rho(z)}{\\rho(0)}=\\frac{p(z)}{p(0)} \\frac{T(0)}{T(z)}=\\left(1-\\frac{\\Lambda z}{T(0)}\\right)^{\\frac{\\mu g}{R \\Lambda}-1}\n$$\n\nThe last term is larger than unity if its exponent is negative:\n\n$$\n\\frac{\\mu g}{R \\Lambda}-1<0\n$$\n\nThen :\n\n$$\n\\Lambda>\\frac{\\mu g}{R}=\\frac{0.029 \\times 9.81}{8.31}=0.034 \\frac{\\mathrm{K}}{\\mathrm{m}}\n$$'],['$0.034$'],False,$ \frac{\mathrm{K}}{\mathrm{m}}$,Numerical,1e-3 1341,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$.","2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.","['In vertical motion, the pressure of the parcel always equals that of the surrounding air, the latter depends on the altitude. The parcel temperature $T_{\\text {parcel }}$ depends on the pressure.\n\nWe can write:\n\n$$\n\\frac{d T_{\\text {parcel }}}{d z}=\\frac{d T_{\\text {parcel }}}{d p} \\frac{d p}{d z}\n$$\n\n$p$ is simultaneously the pressure of air in the parcel and that of the surrounding air.\n\nExpression for $\\frac{d T_{\\text {parcel }}}{d p}$\n\nBy using the equation for adiabatic processes $p V^{\\gamma}=$ const and equation of state, we can deduce the equation giving the change of pressure and temperature in a quasi-equilibrium adiabatic process of an air parcel:\n\n$$\nT_{\\text {parcel }} p^{\\frac{1-\\gamma}{\\gamma}}=\\text { const }\n\\tag{6}\n$$\n\n\n\nwhere $\\gamma=\\frac{c_{p}}{c_{V}}$ is the ratio of isobaric and isochoric thermal capacities of air. By logarithmic differentiation of the two members of (6), we have:\n\n$$\n\\frac{d T_{\\text {parcel }}}{T_{\\text {parcel }}}+\\frac{1-\\gamma}{\\gamma} \\frac{d p}{p}=0\n$$\n\nOr\n\n$$\n\\frac{d T_{\\text {parcel }}}{d p}=\\frac{T_{\\text {parcel }}}{p} \\frac{\\gamma-1}{\\gamma}\n\\tag{7}\n$$\n\nNote: we can use the first law of thermodynamic to calculate the heat received by the parcel in an elementary process: $d Q=\\frac{m}{\\mu} c_{V} d T_{\\text {parcel }}+p d V$, this heat equals zero in an adiabatic process. Furthermore, using the equation of state for air in the parcel $p V=\\frac{m}{\\mu} R T_{\\text {parcel }}$ we can derive (6)\n\nExpression for $\\frac{d p}{d z}$\n\nFrom (1) we can deduce:\n\n$$\n\\frac{d p}{d z}=-\\rho g=-\\frac{p g \\mu}{R T}\n$$\n\nwhere $T$ is the temperature of the surrounding air.\n\nOn the basis of these two expressions, we derive the expression for $d T_{\\text {parcel }} / d z$ :\n\n$$\n\\frac{d T_{\\text {parcel }}}{d z}=-\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R} \\frac{T_{\\text {parcel }}}{T}=-G\n\\tag{8}\n$$\n\nIn general, $G$ is not a constant.']",['$\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R} \\frac{T_{\\text {parcel }}}{T}$'],False,,Expression, 1342,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate.",2.2.1. Derive the expression of $\Gamma$,"[""If at any altitude, $T=T_{\\text {parcel }}$, then instead of $G$ in (8), we have :\n\n$$\n\\Gamma=\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R}=\\mathrm{const}\n\\tag{9}\n$$\n\nor\n\n\n\n$$\n\\Gamma=\\frac{\\mu g}{c_{p}}\n\\tag{9'}\n$$""]",['$\\Gamma=\\frac{\\mu g}{c_{p}}$'],False,,Expression, 1343,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} ",2.2.2. Calculate the numerical value of $\Gamma$.,['Numerical value:\n\n$$\n\\Gamma=\\frac{1.4-1}{1.4} \\frac{0.029 \\times 9.81}{8.31}=0.00978 \\frac{\\mathrm{K}}{\\mathrm{m}} \\approx 10^{-2} \\frac{\\mathrm{K}}{\\mathrm{m}}\n$$'],['$0.00978 $'],False,$\frac{\mathrm{K}}{\mathrm{m}}$,Numerical,1e-3 1344,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} ",2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.,"['Thus, the expression for the temperature at the altitude $z$ in this special atmosphere (called adiabatic atmosphere) is :\n\n$$\nT(z)=T(0)-\\Gamma z\n\\tag{10}\n$$']",['$T(z)=T(0)-\\Gamma z$'],False,,Expression, 1345,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} ","2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.","['Search for the expression of $T_{\\text {parcel }}(z)$\n\nSubstitute $T$ in (7) by its expression given in (3), we have:\n\n$$\n\\frac{d T_{\\text {parcel }}}{T_{\\text {parcel }}}=-\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R} \\frac{d z}{T(0)-\\Lambda z}\n$$\n\nIntegration gives:\n\n$$\n\\ln \\frac{T_{\\text {parcel }}(z)}{T_{\\text {parcel }}(0)}=-\\frac{\\gamma-1}{\\gamma} \\frac{\\mu g}{R}\\left(-\\frac{1}{\\Lambda}\\right) \\ln \\frac{T(0)-\\Lambda z}{T(0)}\n$$\n\nFinally, we obtain:\n\n$$\nT_{\\text {parcel }}(z)=T_{\\text {parcel }}(0)\\left(\\frac{T(0)-\\Lambda z}{T(0)}\\right)^{\\frac{\\Gamma}{\\Lambda}}\n\\tag{11}\n$$']",['$T_{\\text {parcel }}(z)=T_{\\text {parcel }}(0)\\left(\\frac{T(0)-\\Lambda z}{T(0)}\\right)^{\\frac{\\Gamma}{\\Lambda}}$'],False,,Expression, 1346,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$. Context answer: \boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$} ",2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<0$ ), $\\quad T_{\\text {parcel }}\\left(z_{0}+d\\right)=T_{\\text {parcel }}\\left(z_{0}\\right)-\\Gamma d$ and $T\\left(z_{0}+d\\right)=T\\left(z_{0}\\right)-\\Lambda d$.\n\n- In the case the atmosphere has temperature lapse rate $\\Lambda>\\Gamma$, we have $T_{\\text {parcel }}\\left(z_{0}+d\\right)>T\\left(z_{0}+d\\right)$, then $\\rho<\\rho^{\\prime}$. The buoyant force is then larger than the air parcel weight, their resultant is oriented upward and tends to push the parcel away from the equilibrium position.\n\nConversely, if the air parcel is lowered to the altitude $z_{0}-d(d>0)$, $T_{\\text {parcel }}\\left(z_{0}-d\\right)\\rho^{\\prime}$.\n\nThe buoyant force is then smaller than the air parcel weight; their resultant is oriented downward and tends to push the parcel away from the equilibrium position (see Figure 1)\n\nSo the equilibrium of the parcel is unstable, and we found that: An atmosphere with a temperature lapse rate $\\Lambda>\\Gamma$ is unstable.\n\n- In an atmosphere with temperature lapse rate $\\Lambda<\\Gamma$, if the air parcel is lifted to a higher position, at altitude $z_{0}+d$ (with $d>0$ ), $\\quad T_{\\text {parcel }}\\left(z_{0}+d\\right)\\rho^{\\prime}$. The buoyant force is then smaller than the air parcel weight, their resultant is oriented downward and tends to push the parcel back to the equilibrium position.\n\nConversely, if the air parcel is lowered to altitude $z_{0}-d(d>0)$, $T_{\\text {parcel }}\\left(\\mathrm{z}_{0}-d\\right)>T\\left(\\mathrm{z}_{0}-d\\right)$ and then $\\rho<\\rho^{\\prime}$. The buoyant force is then larger than the air parcel weight, their resultant is oriented upward and tends to push the parcel also back to the equilibrium position (see Figure 2).\n\nSo the equilibrium of the parcel is stable, and we found that: An atmosphere with a temperature lapse rate $\\Lambda<\\Gamma$ is stable.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_cb6ddd00156af4e446fbg-1.jpg?height=648&width=691&top_left_y=981&top_left_x=425)\n\n$$\n\\begin{aligned}\n& T_{\\text {parcel }}>T \\Rightarrow \\rho_{\\text {parcel }}<\\rho \\quad \\text { up } \\uparrow \\\\\n& T_{\\text {parcel }}\\rho \\quad \\text { down } \\downarrow\n\\end{aligned}\n$$\n\nFigure 1\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_cb6ddd00156af4e446fbg-1.jpg?height=654&width=699&top_left_y=1803&top_left_x=427)\n\n$$\n\\begin{aligned}\n& T_{\\text {parcel }}\\rho \\text { down } \\downarrow \\\\\n& T_{\\text {parcel }}>T \\Rightarrow \\rho_{\\text {parcel }}<\\rho \\quad \\text { up } \\uparrow\n\\end{aligned}\n$$\n\nstable\n\nFigure 2\n\n\n\n- In an atmosphere with lapse rate $\\Lambda=\\Gamma$, if the parcel is brought from equilibrium position and put in any other position, it will stay there, the equilibrium is indifferent. $A n$ atmosphere with a temperature lapse rate $\\Lambda=\\Gamma$ is neutral']",['$\\Lambda>\\Gamma$ is unstable\n$\\Lambda<\\Gamma$ is stable\n$\\Lambda=\\Gamma$ is neutral'],False,,Need_human_evaluate, 1348,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$. Context answer: \boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$} Context question: 2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<\Gamma$ is unstable $\Lambda<\Gamma$ is stable $\Lambda=\Gamma$ is neutral ",3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$.,"['In a stable atmosphere, with $\\Lambda<\\Gamma$, a parcel, which on ground has temperature $T_{\\text {parcel }}(0)>T(0)$ and pressure $p(0)$ equal to that of the atmosphere, can rise and reach a maximal altitude $h$, where $T_{\\text {parcel }}(h)=T(h)$.\n\nIn vertical motion from the ground to the altitude $h$, the air parcel realizes an adiabatic quasi-static process, in which its temperature changes from $T_{\\text {parcel }}(0)$ to $T_{\\text {parcel }}(h)=T(h)$. Using (11), we can write:\n\n$$\n\\begin{aligned}\n& \\left(1-\\frac{\\Lambda h}{T(0)}\\right)^{-\\frac{\\Gamma}{\\Lambda}}=\\frac{T_{\\text {parcel }}(0)}{T(h)}=\\frac{T_{\\text {parcel }}(0)}{T(0)\\left(1-\\frac{\\Lambda h}{T(0)}\\right)} \\\\\n& \\left(1-\\frac{\\Lambda h}{T(0)}\\right)^{1-\\frac{\\Gamma}{\\Lambda}}=T_{\\text {parcel }}(0) \\times T^{-1}(0) \\\\\n& 1-\\frac{\\Lambda h}{T(0)}=T_{\\text {parcel }}^{\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) \\times T^{-\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) \\\\\n& h=\\frac{1}{\\Lambda} T(0)\\left[1-T_{\\text {parcel }}^{\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) \\times T^{-\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0)\\right] \\\\\n& =\\frac{1}{\\Lambda}\\left[T(0)-T_{\\text {parcel }}^{-\\frac{\\Lambda}{\\Lambda-\\Gamma}}(0) T^{\\frac{\\Gamma}{\\Gamma-\\Lambda}}(0)\\right]\n\\end{aligned}\n$$\n\nSo that the maximal altitude $h$ has the following expression:\n\n$$\nh=\\frac{1}{\\Lambda}\\left[T(0)-\\left(\\frac{(T(0))^{\\Gamma}}{\\left(T_{\\text {parcel }}(0)\\right)^{\\Lambda}}\\right)^{\\frac{1}{\\Gamma-\\Lambda}}\\right]\n\\tag{13}\n$$']",['$h=\\frac{1}{\\Lambda}\\left[T(0)-\\left(\\frac{(T(0))^{\\Gamma}}{\\left(T_{\\text {parcel }}(0)\\right)^{\\Lambda}}\\right)^{\\frac{1}{\\Gamma-\\Lambda}}\\right]$'],False,,Expression, 1349,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$. Context answer: \boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$} Context question: 2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<\Gamma$ is unstable $\Lambda<\Gamma$ is stable $\Lambda=\Gamma$ is neutral Context question: 3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$. Context answer: \boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$} Extra Supplementary Reading Materials: 4. The mixing height Table 1 Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. | Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ | | :---: | :---: | | 5 | 21.5 | | 60 | 20.6 | | 64 | 20.5 | | 69 | 20.5 | | 75 | 20.4 | | 81 | 20.3 | | 90 | 20.2 | | 96 | 20.1 | | 102 | 20.1 | | 109 | 20.1 | | 113 | 20.1 | | 119 | 20.1 | | 128 | 20.2 | | 136 | 20.3 | | 145 | 20.4 | | 153 | 20.5 | | 159 | 20.6 | | 168 | 20.8 | | 178 | 21.0 | | 189 | 21.5 | | 202 | 21.8 | | 215 | 22.0 | | 225 | 22.1 | | 234 | 22.2 | | 246 | 22.3 | | 257 | 22.3 |","4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$. Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$.","['Using data from the Table, we obtain the plot of $z$ versus $T$ shown in Figure 3.\n\n\n\nFigure 3\n\nWe can divide the atmosphere under 200m into three layers, corresponding to the following altitudes:\n\n1) $0\Gamma$ is unstable $\Lambda<\Gamma$ is stable $\Lambda=\Gamma$ is neutral Context question: 3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$. Context answer: \boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$} Extra Supplementary Reading Materials: 4. The mixing height Table 1 Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. | Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ | | :---: | :---: | | 5 | 21.5 | | 60 | 20.6 | | 64 | 20.5 | | 69 | 20.5 | | 75 | 20.4 | | 81 | 20.3 | | 90 | 20.2 | | 96 | 20.1 | | 102 | 20.1 | | 109 | 20.1 | | 113 | 20.1 | | 119 | 20.1 | | 128 | 20.2 | | 136 | 20.3 | | 145 | 20.4 | | 153 | 20.5 | | 159 | 20.6 | | 168 | 20.8 | | 178 | 21.0 | | 189 | 21.5 | | 202 | 21.8 | | 215 | 22.0 | | 225 | 22.1 | | 234 | 22.2 | | 246 | 22.3 | | 257 | 22.3 | Context question: 4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$. Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$. Context answer: \boxed{$294.04$, $293.81 $} ","4.2. Determine the maximal elevation $H$ the parcel can reach, and the temperature $T_{\text {parcel }}(H)$ of the parcel. $H$ is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer.","['In the layer 3), starting from $119 \\mathrm{~m}$, by using (13) we find the maximal elevation $h=23 \\mathrm{~m}$, and the corresponding temperature $293.6 \\mathrm{~K}$ (or $20.6{ }^{\\circ} \\mathrm{C}$ ).\n\nFinally, the mixing height is\n\n$$\nH=119+23=142 \\mathrm{~m} .\n$$\n\nAnd\n\n$$\nT_{\\text {parcel }}(142 \\mathrm{~m})=293.6 \\mathrm{~K} \\text { that is } 20.6^{\\circ} \\mathrm{C}\n$$\n\nFrom this relation, we can find $T_{\\text {parcel }}(119 \\mathrm{~m}) \\approx 293.82 \\mathrm{~K}$ and $h=23 \\mathrm{~m}$.\n\nNote: By using approximate expression (12) we can easily find $T_{\\text {parcel }}(z)=294 \\mathrm{~K}$ and 293.8 K at elevations $96 \\mathrm{~m}$ and $119 \\mathrm{~m}$, respectively. At $119 \\mathrm{~m}$ elevation, the difference between parcel and surrounding air temperatures is $0.7 \\mathrm{~K}$ (= 293.8 - 293.1), so that the maximal distance the parcel will travel in the third layer is $0.7 /\\left(\\Gamma-\\Lambda_{3}\\right)=0.7 / 0.03=23 \\mathrm{~m}$.']","['142, 293.6']",True,"m, K",Numerical,"1e-0,1e-1" 1351,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$. Context answer: \boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$} Context question: 2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<\Gamma$ is unstable $\Lambda<\Gamma$ is stable $\Lambda=\Gamma$ is neutral Context question: 3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$. Context answer: \boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$} Extra Supplementary Reading Materials: 4. The mixing height Table 1 Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. | Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ | | :---: | :---: | | 5 | 21.5 | | 60 | 20.6 | | 64 | 20.5 | | 69 | 20.5 | | 75 | 20.4 | | 81 | 20.3 | | 90 | 20.2 | | 96 | 20.1 | | 102 | 20.1 | | 109 | 20.1 | | 113 | 20.1 | | 119 | 20.1 | | 128 | 20.2 | | 136 | 20.3 | | 145 | 20.4 | | 153 | 20.5 | | 159 | 20.6 | | 168 | 20.8 | | 178 | 21.0 | | 189 | 21.5 | | 202 | 21.8 | | 215 | 22.0 | | 225 | 22.1 | | 234 | 22.2 | | 246 | 22.3 | | 257 | 22.3 | Context question: 4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$. Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$. Context answer: \boxed{$294.04$, $293.81 $} Context question: 4.2. Determine the maximal elevation $H$ the parcel can reach, and the temperature $T_{\text {parcel }}(H)$ of the parcel. $H$ is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer. Context answer: \boxed{142, 293.6} Extra Supplementary Reading Materials: 5. Estimation of carbon monoxide (CO) pollution during a morning motorbike rush hour in Hanoi. Hanoi metropolitan area can be approximated by a rectangle with base dimensions $L$ and $W$ as shown in the figure, with one side taken along the south-west bank of the Red River. It is estimated that during the morning rush hour, from 7:00 am to 8:00 am, there are $8 \times 10^{5}$ motorbikes on the road, each running on average $5 \mathrm{~km}$ and emitting $12 \mathrm{~g}$ of CO per kilometer. The amount of CO pollutant is approximately considered as emitted uniformly in time, at a constant rate $M$ during the rush hour. At the same time, the clean north-east wind blows perpendicularly to the Red River (i.e. perpendicularly to the sides $L$ of the rectangle) with velocity $u$, passes the city with the same velocity, and carries a part of the CO-polluted air out of the city atmosphere. Also, we use the following rough approximate model: - The CO spreads quickly throughout the entire volume of the mixing layer above the Hanoi metropolitan area, so that the concentration $C(t)$ of $\mathrm{CO}$ at time $t$ can be assumed to be constant throughout that rectangular box of dimensions $L, W$ and $H$. - The upwind air entering the box is clean and no pollution is assumed to be lost from the box through the sides parallel to the wind. - Before 7:00 am, the $\mathrm{CO}$ concentration in the atmosphere is negligible.",5.1. Derive the differential equation determining the CO pollutant concentration $C(t)$ as a function of time.,"['Consider a volume of atmosphere of Hanoi metropolitan area being a parallelepiped with height $H$, base sides $L$ and $W$. The emission rate of CO gas by motorbikes from 7:00 am to 8:00 am\n\n$$\nM=800000 \\times 5 \\times 12 / 3600=13300 \\mathrm{~g} / \\mathrm{s}\n$$\n\nThe CO concentration in air is uniform at all points in the parallelepiped and denoted by $C(t)$.\n\nAfter an elementary interval of time $d t$, due to the emission of the motorbikes, the mass of CO gas in the box increases by Mdt . The wind blows parallel to the short sides $W$, bringing away an amount of CO gas with mass $L H C(t) u d t$. The remaining part raises the $\\mathrm{CO}$ concentration by a quantity $d C$ in all over the box. Therefore:\n\n$$\nM d t-L H C(t) u d t=L W H d C\n$$\n\nor\n\n\n\n$$\n\\frac{d C}{d t}+\\frac{u}{W} C(t)=\\frac{M}{L W H}\n\\tag{14}\n$$']",['$\\frac{d C}{d t}+\\frac{u}{W} C(t)=\\frac{M}{L W H}$'],False,,Equation, 1352,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$. Context answer: \boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$} Context question: 2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<\Gamma$ is unstable $\Lambda<\Gamma$ is stable $\Lambda=\Gamma$ is neutral Context question: 3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$. Context answer: \boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$} Extra Supplementary Reading Materials: 4. The mixing height Table 1 Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. | Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ | | :---: | :---: | | 5 | 21.5 | | 60 | 20.6 | | 64 | 20.5 | | 69 | 20.5 | | 75 | 20.4 | | 81 | 20.3 | | 90 | 20.2 | | 96 | 20.1 | | 102 | 20.1 | | 109 | 20.1 | | 113 | 20.1 | | 119 | 20.1 | | 128 | 20.2 | | 136 | 20.3 | | 145 | 20.4 | | 153 | 20.5 | | 159 | 20.6 | | 168 | 20.8 | | 178 | 21.0 | | 189 | 21.5 | | 202 | 21.8 | | 215 | 22.0 | | 225 | 22.1 | | 234 | 22.2 | | 246 | 22.3 | | 257 | 22.3 | Context question: 4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$. Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$. Context answer: \boxed{$294.04$, $293.81 $} Context question: 4.2. Determine the maximal elevation $H$ the parcel can reach, and the temperature $T_{\text {parcel }}(H)$ of the parcel. $H$ is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer. Context answer: \boxed{142, 293.6} Extra Supplementary Reading Materials: 5. Estimation of carbon monoxide (CO) pollution during a morning motorbike rush hour in Hanoi. Hanoi metropolitan area can be approximated by a rectangle with base dimensions $L$ and $W$ as shown in the figure, with one side taken along the south-west bank of the Red River. It is estimated that during the morning rush hour, from 7:00 am to 8:00 am, there are $8 \times 10^{5}$ motorbikes on the road, each running on average $5 \mathrm{~km}$ and emitting $12 \mathrm{~g}$ of CO per kilometer. The amount of CO pollutant is approximately considered as emitted uniformly in time, at a constant rate $M$ during the rush hour. At the same time, the clean north-east wind blows perpendicularly to the Red River (i.e. perpendicularly to the sides $L$ of the rectangle) with velocity $u$, passes the city with the same velocity, and carries a part of the CO-polluted air out of the city atmosphere. Also, we use the following rough approximate model: - The CO spreads quickly throughout the entire volume of the mixing layer above the Hanoi metropolitan area, so that the concentration $C(t)$ of $\mathrm{CO}$ at time $t$ can be assumed to be constant throughout that rectangular box of dimensions $L, W$ and $H$. - The upwind air entering the box is clean and no pollution is assumed to be lost from the box through the sides parallel to the wind. - Before 7:00 am, the $\mathrm{CO}$ concentration in the atmosphere is negligible. Context question: 5.1. Derive the differential equation determining the CO pollutant concentration $C(t)$ as a function of time. Context answer: \boxed{$\frac{d C}{d t}+\frac{u}{W} C(t)=\frac{M}{L W H}$} ",5.2. Write down the solution of that equation for $C(t)$.,"['The general solution of (14) is :\n\n$$\nC(t)=K \\exp \\left(-\\frac{u t}{W}\\right)+\\frac{M}{L H u}\n\\tag{15}\n$$\n\nFrom the initial condition $C(0)=0$, we can deduce :\n\n$$\nC(t)=\\frac{M}{L H u}\\left[1-\\exp \\left(-\\frac{u t}{W}\\right)\\right]\n\\tag{16}\n$$']",['$C(t)=\\frac{M}{L H u}\\left[1-\\exp \\left(-\\frac{u t}{W}\\right)\\right]$'],False,,Expression, 1353,Thermodynamics,"CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere. We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$. Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$. Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas. The student may use the following data if necessary: The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K). The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$ The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$ The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air. The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air. Mathematical hints a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$ b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$. c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$ 1. Change of pressure with altitude. Context question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$. Context answer: \boxed{$p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$} Extra Supplementary Reading Materials: 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation $$ T(z)=T(0)-\Lambda z $$ where $\Lambda$ is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is $-\Lambda$ ). Context question: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$. Context answer: \boxed{$p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$} Context question: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur? Context answer: \boxed{$0.034$} Extra Supplementary Reading Materials: 2. Change of the temperature of an air parcel in vertical motion Consider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase. As the size of the parcel is not large, the atmospheric pressure at different points on the parcel boundary can be considered to have the same value $p(z)$, with $z$ - the altitude of the parcel center. The temperature in the parcel is uniform and equals to $T_{\text {parcel }}(z)$, which is generally different from the temperature of the surrounding air $T(z)$. In parts 2.1 and 2.2, we do not make any assumption about the form of $T(z)$. Context question: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$. Context answer: \boxed{$\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$} Extra Supplementary Reading Materials: 2.2. Consider a special atmospheric condition in which at any altitude $z$ the temperature $T$ of the atmosphere equals to that of the parcel $T_{\text {parcel }}, T(z)=T_{\text {parcel }}(z)$. We use $\Gamma$ to denote the value of $G$ when $T=T_{\text {parcel }}$, that is $\Gamma=-\frac{d T_{\text {parcel }}}{d z}$ (with $\left.T=T_{\text {parcel }}\right) . \Gamma$ is called dry adiabatic lapse rate. Context question: 2.2.1. Derive the expression of $\Gamma$ Context answer: \boxed{$\Gamma=\frac{\mu g}{c_{p}}$} Context question: 2.2.2. Calculate the numerical value of $\Gamma$. Context answer: \boxed{$0.00978 $} Context question: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude. Context answer: \boxed{$T(z)=T(0)-\Gamma z$} Context question: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$. Context answer: \boxed{$T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$} Context question: 2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<\Gamma$ is unstable $\Lambda<\Gamma$ is stable $\Lambda=\Gamma$ is neutral Context question: 3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$. Context answer: \boxed{$h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$} Extra Supplementary Reading Materials: 4. The mixing height Table 1 Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. | Altitude, m | Temperature, ${ }^{\circ} \mathrm{C}$ | | :---: | :---: | | 5 | 21.5 | | 60 | 20.6 | | 64 | 20.5 | | 69 | 20.5 | | 75 | 20.4 | | 81 | 20.3 | | 90 | 20.2 | | 96 | 20.1 | | 102 | 20.1 | | 109 | 20.1 | | 113 | 20.1 | | 119 | 20.1 | | 128 | 20.2 | | 136 | 20.3 | | 145 | 20.4 | | 153 | 20.5 | | 159 | 20.6 | | 168 | 20.8 | | 178 | 21.0 | | 189 | 21.5 | | 202 | 21.8 | | 215 | 22.0 | | 225 | 22.1 | | 234 | 22.2 | | 246 | 22.3 | | 257 | 22.3 | Context question: 4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$. Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$. Context answer: \boxed{$294.04$, $293.81 $} Context question: 4.2. Determine the maximal elevation $H$ the parcel can reach, and the temperature $T_{\text {parcel }}(H)$ of the parcel. $H$ is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer. Context answer: \boxed{142, 293.6} Extra Supplementary Reading Materials: 5. Estimation of carbon monoxide (CO) pollution during a morning motorbike rush hour in Hanoi. Hanoi metropolitan area can be approximated by a rectangle with base dimensions $L$ and $W$ as shown in the figure, with one side taken along the south-west bank of the Red River. It is estimated that during the morning rush hour, from 7:00 am to 8:00 am, there are $8 \times 10^{5}$ motorbikes on the road, each running on average $5 \mathrm{~km}$ and emitting $12 \mathrm{~g}$ of CO per kilometer. The amount of CO pollutant is approximately considered as emitted uniformly in time, at a constant rate $M$ during the rush hour. At the same time, the clean north-east wind blows perpendicularly to the Red River (i.e. perpendicularly to the sides $L$ of the rectangle) with velocity $u$, passes the city with the same velocity, and carries a part of the CO-polluted air out of the city atmosphere. Also, we use the following rough approximate model: - The CO spreads quickly throughout the entire volume of the mixing layer above the Hanoi metropolitan area, so that the concentration $C(t)$ of $\mathrm{CO}$ at time $t$ can be assumed to be constant throughout that rectangular box of dimensions $L, W$ and $H$. - The upwind air entering the box is clean and no pollution is assumed to be lost from the box through the sides parallel to the wind. - Before 7:00 am, the $\mathrm{CO}$ concentration in the atmosphere is negligible. Context question: 5.1. Derive the differential equation determining the CO pollutant concentration $C(t)$ as a function of time. Context answer: \boxed{$\frac{d C}{d t}+\frac{u}{W} C(t)=\frac{M}{L W H}$} Context question: 5.2. Write down the solution of that equation for $C(t)$. Context answer: \boxed{$C(t)=\frac{M}{L H u}\left[1-\exp \left(-\frac{u t}{W}\right)\right]$} ","5.3. Calculate the numerical value of the concentration $C(t)$ at 8:00 a.m. Given $L=15 \mathrm{~km}, W=8 \mathrm{~km}, u=1 \mathrm{~m} / \mathrm{s}$.","['Taking as origin of time the moment 7:00 am, then 8:00 am corresponds to $t=3600 \\mathrm{~s}$. Putting the given data in (15), we obtain :\n\n$$\nC(3600 \\mathrm{~s})=6.35 \\times(1-0.64)=2.3 \\mathrm{mg} / \\mathrm{m}^{3}\n$$']",['$2.3 $'],False,$\mathrm{mg} / \mathrm{m}^{3}$,Numerical,1e-1 1354,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits","A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.","[""Apply Newton's law to mass $M_{1}$ :\n\n$$\nM_{1} \\frac{\\mathrm{d}^{2} \\vec{r}_{1}}{\\mathrm{~d} t^{2}}=G \\frac{M_{1} M_{2}}{\\left|\\vec{r}_{2}-\\vec{r}_{1}\\right|^{2}} \\frac{\\vec{r}_{2}-\\vec{r}_{1}}{\\left|\\vec{r}_{2}-\\overrightarrow{r_{1}}\\right|}\n\\tag{1}\n$$\n\nUse, from eq. (1) of the question sheet\n\n$$\n\\vec{r}_{2}=-\\frac{M_{1}}{M_{2}} \\vec{r}_{1}\n\\tag{2}\n$$\n\nin eq. (1) above, to obtain\n\n$$\n\\frac{\\mathrm{d}^{2} \\vec{r}_{1}}{\\mathrm{~d} t^{2}}=-\\frac{G M_{2}^{3}}{\\left(M_{1}+M_{2}\\right)^{2} r_{1}^{2}} \\frac{\\vec{r}_{1}}{r_{1}} .\n\\tag{3}\n$$""]","['$n=3$ , $\\alpha=\\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$']",True,,"Numerical,Expression","0," 1355,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits Context question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$. Context answer: \boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$} ","A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as: $$ E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} $$ where $$ \mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2} \tag{4} $$ are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.","['The total energy of the system is the sum of the two kinetic energies plus the gravitational potential energy. For circular motions, the linear velocity of each of the masses reads\n\n$$\n\\left|\\vec{v}_{1}\\right|=r_{1} \\Omega, \\quad\\left|\\vec{v}_{2}\\right|=r_{2} \\Omega\n\\tag{4}\n$$\n\nThus, the total energy is\n\n$$\nE=\\frac{1}{2}\\left(M_{1} r_{1}^{2}+M_{2} r_{2}^{2}\\right) \\Omega^{2}-\\frac{G M_{1} M_{2}}{L}\n\\tag{5}\n$$\n\nNow,\n\n$$\n\\left(M_{1} r_{1}-M_{2} r_{2}\\right)^{2}=0 \\quad \\Rightarrow \\quad M_{1} r_{1}^{2}+M_{2} r_{2}^{2}=\\mu L^{2}\n\\tag{6}\n$$\n\nThus,\n\n$$\nE=\\frac{1}{2} \\mu L^{2} \\Omega^{2}-G \\frac{M \\mu}{L} .\n\\tag{7}\n$$']","['$A(\\mu, \\Omega, L)=\\frac{1}{2} \\mu L^{2} \\Omega^{2}$']",False,,Expression, 1356,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits Context question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$. Context answer: \boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$} Context question: A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as: $$ E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} $$ where $$ \mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2} \tag{4} $$ are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$. Context answer: \boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$} ",A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.,"[""Energy (3) of the question sheet can be interpreted as describing a system of a mass $\\mu$ in a circular orbit with angular velocity $\\Omega$, radius $L$, around a mass $M$ (at rest). Equating the gravitational acceleration to the centripetal acceleration:\n\n$$\nG \\frac{M}{L^{2}}=\\Omega^{2} L\n\\tag{8}\n$$\n\nThis is indeed Kepler's third law (for circular orbits). Then, from 7,\n\n$$\nE=-\\frac{1}{2} G \\frac{M \\mu}{L}\n\\tag{9}\n$$""]",['$\\beta=-\\frac{1}{2}$'],False,,Numerical,0 1357,Mechanics,,"B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as: $$ Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t . \tag{8} $$ Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$.","['Some simple trigonometry for the $x, y$ motion of the masses (in an appropriate Cartesian system) yields:\n\n$$\n\\left(x_{1}, y_{1}\\right)=r_{1}(\\cos (\\Omega t), \\sin (\\Omega t)), \\quad\\left(x_{2}, y_{2}\\right)=-r_{2}(\\cos (\\Omega t), \\sin (\\Omega t))\n\\tag{10}\n$$\n\nThen,\n\n$$\nQ_{i j}=\\frac{M_{1} r_{1}^{2}+M_{2} r_{2}^{2}}{2}\\left(\\begin{array}{ccc}\n\\frac{4}{3} \\cos ^{2}(\\Omega t)-\\frac{2}{3} \\sin ^{2}(\\Omega t) & 2 \\sin (\\Omega t) \\cos (\\Omega t) & 0 \\\\\n2 \\sin (\\Omega t) \\cos (\\Omega t) & \\frac{4}{3} \\sin ^{2}(\\Omega t)-\\frac{2}{3} \\cos ^{2}(\\Omega t) & 0 \\\\\n0 & 0 & -\\frac{2}{3}\n\\end{array}\\right)\n\\tag{11}\n$$\n\nor, using some simple trigonometry and (6),\n\n$$\nQ_{i j}=\\frac{\\mu L^{2}}{2}\\left(\\begin{array}{ccc}\n\\frac{1}{3}+\\cos 2 \\Omega t & \\sin 2 \\Omega t & 0 \\\\\n\\sin 2 \\Omega t & \\frac{1}{3}-\\cos 2 \\Omega t & 0 \\\\\n0 & 0 & -\\frac{2}{3}\n\\end{array}\\right)\n\\tag{12}\n$$']","['$$\nk=2 \\Omega, \\quad a_{1}=a_{2}=\\frac{1}{3}, a_{3}=-\\frac{2}{3}, \\quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \\stackrel{\\text { otherwise }}{=} 0\n$$']",True,,Need_human_evaluate, 1358,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits Context question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$. Context answer: \boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$} Context question: A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as: $$ E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} $$ where $$ \mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2} \tag{4} $$ are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$. Context answer: \boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$} Context question: A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$. Context answer: \boxed{$\beta=-\frac{1}{2}$} Extra Supplementary Reading Materials: Part B: Introducing relativistic dissipation The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy. Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula, $$ \mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right) \tag{5} $$ Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number) $$ Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right), \tag{6} $$ $$ Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A}, \tag{7} $$ and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame. Context question: B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as: $$ Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t . \tag{8} $$ Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$. Context answer: $$ k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0 $$ ","B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain: $$ \mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6} \tag{9} $$ What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.]",['First take the derivatives:\n\n$$\n\\frac{\\mathrm{d}^{3} Q_{i j}}{\\mathrm{~d} t^{3}}=4 \\Omega^{3} \\mu L^{2}\\left(\\begin{array}{ccc}\n\\sin 2 \\Omega t & -\\cos 2 \\Omega t & 0 \\\\\n-\\cos 2 \\Omega t & -\\sin 2 \\Omega t & 0 \\\\\n0 & 0 & 0\n\\end{array}\\right)\n\\tag{13}\n$$\n\nThen perform the sum:\n\n$$\n\\frac{\\mathrm{d} E}{\\mathrm{~d} t}=\\frac{G}{5 c^{5}}\\left(4 \\Omega^{3} \\mu L^{2}\\right)^{2}\\left[2 \\sin ^{2}(2 \\Omega t)+2 \\cos ^{2}(2 \\Omega t)\\right]=\\frac{32}{5} \\frac{G}{c^{5}} \\mu^{2} L^{4} \\Omega^{6} .\n\\tag{14}\n$$'],['$\\xi=\\frac{32}{5}$'],False,,Numerical,0 1359,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits Context question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$. Context answer: \boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$} Context question: A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as: $$ E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} $$ where $$ \mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2} \tag{4} $$ are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$. Context answer: \boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$} Context question: A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$. Context answer: \boxed{$\beta=-\frac{1}{2}$} Extra Supplementary Reading Materials: Part B: Introducing relativistic dissipation The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy. Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula, $$ \mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right) \tag{5} $$ Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number) $$ Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right), \tag{6} $$ $$ Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A}, \tag{7} $$ and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame. Context question: B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as: $$ Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t . \tag{8} $$ Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$. Context answer: $$ k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0 $$ Context question: B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain: $$ \mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6} \tag{9} $$ What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.] Context answer: \boxed{$\xi=\frac{32}{5}$} ","B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form $$ \left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5} \tag{10} $$ where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name ""chirp"" is inspired by the high pitch sound (increasing frequency) produced by small birds.]","[""Now we assume a sequency of Keplerian orbits, with decreasing energy, which is being taken from the system by the GWs.\n\nFirst, from (9), differentiating with respect to time,\n\n$$\n\\frac{\\mathrm{d} E}{\\mathrm{~d} t}=\\frac{G M \\mu}{2 L^{2}} \\frac{\\mathrm{d} L}{\\mathrm{~d} t}\n\\tag{15}\n$$\n\nSince this loss of energy is due to GWs, we equate it with (minus) the luminosity of GWs, given by (14)\n\n$$\n\\frac{G M \\mu}{2 L^{2}} \\frac{\\mathrm{d} L}{\\mathrm{~d} t}=-\\frac{32}{5} \\frac{G}{c^{5}} \\mu^{2} L^{4} \\Omega^{6}\n\\tag{16}\n$$\n\nWe can eliminate the $L$ and $\\mathrm{d} L / \\mathrm{d} t$ dependence in this equation in terms of $\\Omega$ and $\\mathrm{d} \\Omega / \\mathrm{d} t$, by using Kepler's third law (8), which relates:\n\n$$\nL^{3}=G \\frac{M}{\\Omega^{2}}, \\quad \\frac{\\mathrm{d} L}{\\mathrm{~d} t}=-\\frac{2}{3} \\frac{L}{\\Omega} \\frac{\\mathrm{d} \\Omega}{\\mathrm{d} t}\n\\tag{17}\n$$\n\n\n\nSubstituting in (16), we obtain:\n\n$$\n\\left(\\frac{\\mathrm{d} \\Omega}{\\mathrm{d} t}\\right)^{3}=\\left(\\frac{96}{5}\\right)^{3} \\frac{\\Omega^{11}}{c^{15}} G^{5} \\mu^{3} M^{2} \\equiv\\left(\\frac{96}{5}\\right)^{3} \\frac{\\Omega^{11}}{c^{15}}\\left(G M_{\\mathrm{c}}\\right)^{5}\n\\tag{18}\n$$""]",['$M_{\\mathrm{c}}=(\\mu^{3} M^{2})^{1 / 5}$'],False,,Expression, 1360,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits Context question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$. Context answer: \boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$} Context question: A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as: $$ E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} $$ where $$ \mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2} \tag{4} $$ are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$. Context answer: \boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$} Context question: A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$. Context answer: \boxed{$\beta=-\frac{1}{2}$} Extra Supplementary Reading Materials: Part B: Introducing relativistic dissipation The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy. Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula, $$ \mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right) \tag{5} $$ Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number) $$ Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right), \tag{6} $$ $$ Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A}, \tag{7} $$ and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame. Context question: B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as: $$ Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t . \tag{8} $$ Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$. Context answer: $$ k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0 $$ Context question: B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain: $$ \mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6} \tag{9} $$ What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.] Context answer: \boxed{$\xi=\frac{32}{5}$} Context question: B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form $$ \left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5} \tag{10} $$ where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name ""chirp"" is inspired by the high pitch sound (increasing frequency) produced by small birds.] Context answer: \boxed{$M_{\mathrm{c}}=(\mu^{3} M^{2})^{1 / 5}$} ","B.4 Using the information provided above, relate the orbital angular velocity $\Omega$ with the GW frequency $f_{\mathrm{GW}}$. Knowing that, for any smooth function $F(t)$ and $a \neq 1$, $$ \frac{\mathrm{d} F(t)}{\mathrm{d} t}=\chi F(t)^{a} \quad \Rightarrow \quad F(t)^{1-a}=\chi(1-a)\left(t-t_{0}\right) \tag{11} $$ where $\chi$ is a constant and $t_{0}$ is an integration constant, show that (10) implies that the GW frequency is $$ f_{\mathrm{GW}}^{-8 / 3}=8 \pi^{8 / 3} \xi\left(\frac{G M_{c}}{c^{3}}\right)^{(2 / 3)+p}\left(t_{0}-t\right)^{2-p} \tag{12} $$ and determine the constant $p$.","['Angular and cycle frequencies are related as $\\Omega=2 \\pi f$. From the information provided above: GWs have a frequency which is twice as large as the orbital frequency, we have\n\n$$\n\\frac{\\Omega}{2 \\pi}=\\frac{f_{\\mathrm{GW}}}{2}\n\\tag{19}\n$$\n\nFormula (10) of the question sheet has the form\n\n$$\n\\frac{\\mathrm{d} \\Omega}{\\mathrm{d} t}=\\chi \\Omega^{11 / 3}, \\quad \\chi \\equiv \\frac{96}{5} \\frac{\\left(G M_{\\mathrm{c}}\\right)^{5 / 3}}{c^{5}} .\n\\tag{20}\n$$\n\nThus, from (11) of the question sheet\n\n$$\n\\Omega(t)^{-8 / 3}=\\frac{8}{3} \\chi\\left(t_{0}-t\\right),\n\\tag{21}\n$$\n\nor, using (20) and the definition of $\\chi$ gives\n\n$$\nf_{\\mathrm{GW}}^{-8 / 3}(t)=\\frac{(8 \\pi)^{8 / 3}}{5}\\left(\\frac{G M_{\\mathrm{c}}}{c^{3}}\\right)^{5 / 3}\\left(t_{0}-t\\right)\n\\tag{22}\n$$']",['p=1'],False,,Numerical,0 1361,Mechanics,"LIGO-GW150914 In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal. Part A: Newtonian (conservative) orbits Context question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is, $$ M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 . \tag{!} $$ The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as $$ \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} $$ where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$. Context answer: \boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$} Context question: A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as: $$ E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} $$ where $$ \mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2} \tag{4} $$ are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$. Context answer: \boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$} Context question: A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$. Context answer: \boxed{$\beta=-\frac{1}{2}$} Extra Supplementary Reading Materials: Part B: Introducing relativistic dissipation The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy. Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula, $$ \mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right) \tag{5} $$ Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number) $$ Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right), \tag{6} $$ $$ Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A}, \tag{7} $$ and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame. Context question: B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as: $$ Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t . \tag{8} $$ Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$. Context answer: $$ k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0 $$ Context question: B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain: $$ \mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6} \tag{9} $$ What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.] Context answer: \boxed{$\xi=\frac{32}{5}$} Context question: B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form $$ \left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5} \tag{10} $$ where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name ""chirp"" is inspired by the high pitch sound (increasing frequency) produced by small birds.] Context answer: \boxed{$M_{\mathrm{c}}=(\mu^{3} M^{2})^{1 / 5}$} Context question: B.4 Using the information provided above, relate the orbital angular velocity $\Omega$ with the GW frequency $f_{\mathrm{GW}}$. Knowing that, for any smooth function $F(t)$ and $a \neq 1$, $$ \frac{\mathrm{d} F(t)}{\mathrm{d} t}=\chi F(t)^{a} \quad \Rightarrow \quad F(t)^{1-a}=\chi(1-a)\left(t-t_{0}\right) \tag{11} $$ where $\chi$ is a constant and $t_{0}$ is an integration constant, show that (10) implies that the GW frequency is $$ f_{\mathrm{GW}}^{-8 / 3}=8 \pi^{8 / 3} \xi\left(\frac{G M_{c}}{c^{3}}\right)^{(2 / 3)+p}\left(t_{0}-t\right)^{2-p} \tag{12} $$ and determine the constant $p$. Context answer: \boxed{p=1} Extra Supplementary Reading Materials: On September 14, 2015 GW150914 was registered by the LIGO detectors, consisting of two L-shaped arms, each $4 \mathrm{~km}$ long. These arms changed by a relative length according to Fig. 1. The arms of the detector respond linearly to a passing gravitational wave, and the response pattern mimics the wave. This wave was created by two black holes on quasi-circular orbits; the loss of energy through gravitational radiation caused the orbit to shrink and the black holes to eventually collide. The collision point corresponds, roughly, to the peak of the signal after point D, in Fig. 1. Figure 1. Strain, i.e. relative variation of the size of each arm, at the LIGO detector H1. The horizontal axis is time, and the points A, B, C, D correspond to $t=0.000,0.009,0.034,0.040$ seconds, respectively.","B.5 From the figure, estimate $f_{\mathrm{GW}}(t)$ at $$ t_{\overline{\mathrm{AB}}}=\frac{t_{\mathrm{B}}+t_{\mathrm{A}}}{2} \quad \text { and } \quad t_{\overline{\mathrm{CD}}}=\frac{t_{\mathrm{D}}+t_{\mathrm{C}}}{2} \tag{13} $$ Assuming that (12) is valid all the way until the collision (which strictly speaking is not true) and that the two objects have equal mass, estimate the chirp mass, $M_{c}$, and total mass of the system, in terms of solar masses $M_{\odot} \simeq 2 \times 10^{30} \mathrm{~kg}$.","[""From the figure, we consider the two $\\Delta t$ 's as half periods. Thus, the (cycle) GW frequency is $f_{\\mathrm{GW}}=$ $1 /(2 \\Delta t)$. Then, the four given points allow us to compute the frequency at the mean time of the two intervals as\n\n| | $t_{\\overline{\\mathrm{AB}}}$ | $t_{\\overline{\\mathrm{CD}}}$ |\n| :---: | :---: | :---: |\n| $t(\\mathrm{~s})$ | 0.0045 | 0.037 |\n| $f_{\\mathrm{GW}}(\\mathrm{Hz})$ | $(2 \\times 0.009)^{-1}$ | $(2 \\times 0.006)^{-1}$ |\n\nNow, using (22) we have two pairs of $\\left(f_{\\mathrm{GW}}, t\\right)$ values for two unknowns $\\left(t_{0}, M_{\\mathrm{c}}\\right)$. Expressing (22) for both $t_{\\overline{\\mathrm{AB}}}$ and $t_{\\overline{\\mathrm{CD}}}$ and dividing the two equations we obtain:\n\n$$\nt_{0}=\\frac{A t_{\\overline{\\mathrm{CD}}}-t_{\\overline{\\mathrm{AB}}}}{A-1}, \\quad A \\equiv\\left(\\frac{f_{\\mathrm{GW}}\\left(t_{\\overline{\\mathrm{AB}}}\\right)}{f_{\\mathrm{GW}}\\left(t_{\\overline{\\mathrm{CD}}}\\right)}\\right)^{-8 / 3}\n\\tag{23}\n$$\n\nReplacing by the numerical values, $A \\simeq 2.95$ and $t_{0} \\simeq 0.054 \\mathrm{~s}$. Now we can use (22) for either of the two values $t_{\\overline{\\mathrm{AB}}}$ or $t_{\\overline{\\mathrm{CD}}}$ and determine $M_{\\mathrm{c}}$. One obtains for the chirp mass\n\n$$\nM_{\\mathrm{c}} \\simeq 6 \\times 10^{31} \\mathrm{~kg} \\simeq 30 \\times M_{\\odot}\n\\tag{24}\n$$\n\nThus, the total mass $M$ is\n\n$$\nM=4^{3 / 5} M_{\\mathrm{c}} \\simeq 69 \\times M_{\\odot} .\n\\tag{25}\n$$\n\nThis result is actually remarkably close to the best estimates using the full theory of General Relativity! [Even though the actual objects do not have precisely equal masses and the theory we have just used is not valid very close to the collision.]""]","['$30$ ,$69$']",True,$\times M_{\odot}$,Numerical,1e0 1362,Mechanics,,"B.6 Estimate the minimal orbital separation between the two objects at $t_{\overline{\mathrm{CD}}}$. Hence estimate a maximum size for each object, $R_{\max }$. Obtain $R_{\odot} / R_{\max }$ to compare this size with the radius of our Sun, $R_{\odot} \simeq 7 \times 10^{5} \mathrm{~km}$. Estimate also their orbital linear velocity at the same instant, $v_{\text {col }}$, comparing it with the speed of light, $v_{\text {col }} / c$.","[""From (8), Kepler's law states that $L=\\left(G M / \\Omega^{2}\\right)^{1 / 3}$. The second pair of points highlighted in the plot correspond to the cycle prior to merger. Thus, we use (19) to obtain the orbital angular velocity at $t_{\\overline{\\mathrm{CD}}}$ :\n\n$$\n\\Omega_{t_{\\overline{\\mathrm{CD}}}} \\sim 2.6 \\times 10^{2} \\mathrm{rad} / \\mathrm{s}\n\\tag{26}\n$$\n\nThen, using the total mass (25) we find\n\n$$\nL \\sim 5 \\times 10^{2} \\mathrm{~km}\n\\tag{27}\n$$\n\nThus, these objects have a maximum radius of $R_{\\max } \\sim 250 \\mathrm{~km}$. Hence they have over 30 times more mass and,\n\n$$\n\\frac{R_{\\odot}}{R_{\\max }} \\sim 3 \\times 10^{3}\n\\tag{28}\n$$\n\nthey are 3000 times smaller than the Sun and!\n\nTheir linear velocity is\n\n$$\nv_{\\mathrm{col}}=\\frac{L}{2} \\Omega \\simeq 7 \\times 10^{4} \\mathrm{~km} / \\mathrm{s}\n\\tag{29}\n$$\n\nThey are moving at over $20 \\%$ of the velocity of light!""]","['$$\nL_{\\text {collision }} \\sim 5 \\times 10^{2} \\mathrm{~km}, \\quad \\frac{R_{\\odot}}{R_{\\max }} \\sim 3 \\times 10^{3}, \\quad \\frac{v_{\\text {col }}}{c} \\sim 0.2\n$$']",True,,Need_human_evaluate, 1363,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics","A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.","['The magnetic force is the centripetal force:\n\n$$\nm \\frac{v^{2}}{r}=e v B \\Rightarrow r=\\frac{m v}{e B}\n$$\n\nFirst express the velocity in terms of the kinetic energy,\n\n$$\nK=\\frac{1}{2} m v^{2} \\Rightarrow v=\\sqrt{\\frac{2 K}{m}},\n$$\n\nand then insert it in the expression above for the radius to get']",['$r=\\frac{\\sqrt{2 K m}}{e B}$'],False,,Expression, 1364,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered.","A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.","['The radius of the circular motion of a charged particle in the presence of a uniform magnetic field is given by,\n\n$$\nr=\\frac{m v}{e B}\n$$\n\nThis formula is valid in the relativistic scenario if the mass correction, $m \\rightarrow \\gamma m$ is included:\n\n$$\nr=\\frac{\\gamma m v}{e B}=\\frac{p}{e B} \\Rightarrow p=r e B\n$$\n\nNote that the radius of the circular motion is half the radius of the inner part of the detector. One obtains $\\left[1 \\mathrm{MeV} / c=5.34 \\times 10^{-22} \\mathrm{~m} \\mathrm{~kg} \\mathrm{~s}^{-1}\\right.$ ]']",['$p=330$'],False,$ \mathrm{MeV} / c$,Numerical,1e0 1365,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$.","A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$.","['The acceleration for the particle is $a=\\frac{e v B}{\\gamma m} \\sim \\frac{e c B}{\\gamma m}$, in the ultrarelativistic limit. Then,\n\n$$\nP=\\frac{e^{4} c^{2} \\gamma^{4} B^{2}}{6 \\pi \\epsilon_{0} c^{3} \\gamma^{2} m^{2}}=\\frac{e^{4} \\gamma^{2} c^{4} B^{2}}{6 \\pi \\epsilon_{0} c^{5} m^{2}}\n$$\n\nSince $E=\\gamma m c^{2}$ we can obtain $\\gamma^{2} c^{4}=\\frac{E^{2}}{m^{2}}$ and, finally,\n\n\n\n$$\nP=\\frac{e^{4}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} E^{2} B^{2}\n$$']","['$\\xi=\\frac{1}{6 \\pi}, n=5$ , $k=4$']",True,,Numerical,0 1366,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} ","A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$.","['The power emitted by the particle is given by,\n\n$$\nP=-\\frac{\\mathrm{d} E}{\\mathrm{~d} t}=\\frac{e^{4}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} E^{2} B^{2}\n$$\n\nThe energy of the particle as a function of time can be calculated from\n\n$$\n\\int_{E_{0}}^{E(t)} \\frac{1}{E^{2}} \\mathrm{~d} E=-\\int_{0}^{t} \\frac{e^{4}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} B^{2} \\mathrm{~d} t\n$$\n\nwhere $E(0)=E_{0}$. This leads to,\n\n$$\n\\frac{1}{E(t)}-\\frac{1}{E_{0}}=\\frac{e^{4} B^{2}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} t \\quad \\Rightarrow \\quad E(t)=\\frac{E_{0}}{1+\\alpha E_{0} t},\n$$']",['$\\alpha=\\frac{e^{4} B^{2}}{6 \\pi \\epsilon_{0} m^{4} c^{5}}$'],False,,Expression, 1367,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} ",A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV.,"['If the initial energy of the electron is $100 \\mathrm{GeV}$, the radius of curvature is extremely large ( $r=\\frac{E}{e B c} \\approx 167 \\mathrm{~m}$ ). Therefore, in approximation, one can consider the electron is moving in the inner part of the ATLAS detector along a straight line. The time of flight of the electron is $t=R / c$, where $R=1.1 \\mathrm{~m}$ is the radius of the inner part of the detector. The total energy lost due to synchrotron radiation is,\n\n$$\n\\Delta E=E(R / c)-E_{0}=\\frac{E_{0}}{1+\\alpha E_{0} \\frac{R}{c}}-E_{0} \\approx-\\alpha E_{0}^{2} \\frac{R}{c}\n$$']",['-56'],False,MeV,Numerical,1e0 1368,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} Context question: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV. Context answer: \boxed{-56} ",A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit.,"['In the ultrarelativistic limit, $v \\approx c$ and $E \\approx p c$. The cyclotron frequency is,\n\n$$\n\\omega(t)=\\frac{c}{r(t)}=\\frac{e c B}{p(t)}=\\frac{e c^{2} B}{E(t)}\n$$']",['$\\omega(t)=\\frac{e c^{2} B}{E_{0}}(1+\\frac{e^{4} B^{2}}{6 \\pi \\epsilon_{0} m^{4} c^{5}} E_{0} t)$'],False,,Expression, 1369,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} Context question: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV. Context answer: \boxed{-56} Context question: A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit. Context answer: \boxed{$\omega(t)=\frac{e c^{2} B}{E_{0}}(1+\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} E_{0} t)$} Extra Supplementary Reading Materials: Part B.Finding the neutrino The collision between the two protons shown in Figure 1 leads to the production of a top quark $(t)$ and an anti-top quark $(\bar{t})$, the heaviest elementary particles ever detected. The top quark decays into a $W^{+}$ boson and a bottom quark (b), while the anti-top quark decays into a $W^{-}$boson and an anti-bottom quark $(\bar{b})$. In the case depicted in Figure 1, the $W^{+}$boson decays into a anti-muon $\left(\mu^{+}\right)$and a neutrino $(\nu)$, and the $W^{-}$boson decays into a quark and an anti-quark. The task of this problem is to reconstruct the full momentum of the neutrino using the momenta of some detected particles. For simplicity, all particles and jets in this problem will be considered massless, except for the top quark and $\mathbf{W}^{ \pm}$bosons. The momenta of the top quark decay products can be determined from the experiment (see Table), except for the neutrino momentum component along the ( $z$-axis). The total linear momentum of the final state particles caught by the detector is only zero on the transverse plane ( $x y$ plane), and not along the collision line (z-axis). As such, one can find the transverse momentum of the neutrino from the missing momentum in the transverse plane. On June 4, 2015, the ATLAS experiment at the LHC recorded a proton-proton collision at 00:21:24 GMT+1 like the one represented in Figure 1. Figure 1. Schematic representation of the ATLAS detector coordinate system (left) and proton-proton collision (right). The linear momenta of the three final-state particles coming from the top quark decay, including the neutrino, is presented below for each component. | Particle | $p_{x}(\mathrm{GeV} / c)$ | $p_{y}(\mathrm{GeV} / c)$ | $p_{z}(\mathrm{GeV} / c)$ | | :--- | :---: | :---: | :---: | | anti-muon $\left(\mu^{+}\right)$ | -24.7 | -24.9 | -12.4 | | jet 1 $\left(j_{1}\right)$ | -14.2 | +50.1 | +94.1 | | neutrino $(\nu)$ | -104.1 | +5.3 | --- |","B.1 Find an equation which relates the square of the $W^{+}$boson mass, $m_{\mathrm{W}}^{2}$, with the neutrino and anti-muon momentum components presented in the table above. Express your answer in terms of the neutrino and anti-muon transverse momentum, $$ \vec{p}_{\mathrm{T}}{ }^{(\nu)}=p_{x}{ }^{(\nu)} \hat{\imath}+p_{y}{ }^{(\nu)} \hat{\jmath} \text { and } \vec{p}_{\mathrm{T}}{ }^{(\mu)}=p_{x}{ }^{(\mu)} \hat{\imath}+p_{y}{ }^{(\mu)} \hat{\jmath}, $$ and their $z$-axis momentum components, $p_{z}{ }^{(\mu)}$ and $p_{z}{ }^{(\nu)}$.","['Since the $W^{+}$boson decays into an anti-muon and a neutrino, one can use principles of conservation of energy and linear momentum to calculate the unknown $p_{z}^{(\\nu)}$ of the neutrino. Moreover, the antimuon and the neutrino can be considered massless, which implies that the magnitude of their momenta (times $c$ ) and their energies are the same. Therefore, the conservation of linear momentum can be expressed as\n\n$$\n\\vec{p}^{(W)}=\\vec{p}^{(\\mu)}+\\vec{p}^{(\\nu)},\n$$\n\nand the conservation of energy as,\n\n$$\nE^{(W)}=c p^{(\\mu)}+c p^{(\\nu)} .\n$$\n\nIn addition, one can also relate the energy and the momentum of the $W^{+}$boson through its mass,\n\n$$\nm_{W}^{2}=\\left(E^{(W)}\\right)^{2} / c^{4}-\\left(p^{(W)}\\right)^{2} / c^{2}\n$$\n\nwhich leads to a quadratic equation on $p_{z}{ }^{(\\nu)}$,\n\n$$\n\\begin{aligned}\nm_{W}^{2} & =\\left[\\left(p^{(\\mu)}+p^{(\\nu)}\\right)^{2}-\\left(\\vec{p}^{(\\mu)}+\\vec{p}^{(\\nu)}\\right)^{2}\\right] / c^{2} \\\\\n& =\\left(2 p^{(\\mu)} p^{(\\nu)}-2 \\vec{p}^{(\\mu)} \\cdot \\vec{p}^{(\\nu)}\\right) / c^{2}\n\\end{aligned}\n$$']",['$m_{W}^{2}=\\frac{1}{c^{2}}(2 p^{(\\mu)} \\sqrt{(p_{\\mathrm{T}}^{(\\nu)})^{2}+(p_{z}^{(\\nu)})^{2}}-2 \\vec{p}_{\\mathrm{T}}^{(\\mu)} \\cdot \\vec{p}_{\\mathrm{T}}^{(\\nu)}-2 p_{z}^{(\\mu)} p_{z}^{(\\nu)})$'],False,,Expression, 1370,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} Context question: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV. Context answer: \boxed{-56} Context question: A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit. Context answer: \boxed{$\omega(t)=\frac{e c^{2} B}{E_{0}}(1+\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} E_{0} t)$} Extra Supplementary Reading Materials: Part B.Finding the neutrino The collision between the two protons shown in Figure 1 leads to the production of a top quark $(t)$ and an anti-top quark $(\bar{t})$, the heaviest elementary particles ever detected. The top quark decays into a $W^{+}$ boson and a bottom quark (b), while the anti-top quark decays into a $W^{-}$boson and an anti-bottom quark $(\bar{b})$. In the case depicted in Figure 1, the $W^{+}$boson decays into a anti-muon $\left(\mu^{+}\right)$and a neutrino $(\nu)$, and the $W^{-}$boson decays into a quark and an anti-quark. The task of this problem is to reconstruct the full momentum of the neutrino using the momenta of some detected particles. For simplicity, all particles and jets in this problem will be considered massless, except for the top quark and $\mathbf{W}^{ \pm}$bosons. The momenta of the top quark decay products can be determined from the experiment (see Table), except for the neutrino momentum component along the ( $z$-axis). The total linear momentum of the final state particles caught by the detector is only zero on the transverse plane ( $x y$ plane), and not along the collision line (z-axis). As such, one can find the transverse momentum of the neutrino from the missing momentum in the transverse plane. On June 4, 2015, the ATLAS experiment at the LHC recorded a proton-proton collision at 00:21:24 GMT+1 like the one represented in Figure 1. Figure 1. Schematic representation of the ATLAS detector coordinate system (left) and proton-proton collision (right). The linear momenta of the three final-state particles coming from the top quark decay, including the neutrino, is presented below for each component. | Particle | $p_{x}(\mathrm{GeV} / c)$ | $p_{y}(\mathrm{GeV} / c)$ | $p_{z}(\mathrm{GeV} / c)$ | | :--- | :---: | :---: | :---: | | anti-muon $\left(\mu^{+}\right)$ | -24.7 | -24.9 | -12.4 | | jet 1 $\left(j_{1}\right)$ | -14.2 | +50.1 | +94.1 | | neutrino $(\nu)$ | -104.1 | +5.3 | --- | Context question: B.1 Find an equation which relates the square of the $W^{+}$boson mass, $m_{\mathrm{W}}^{2}$, with the neutrino and anti-muon momentum components presented in the table above. Express your answer in terms of the neutrino and anti-muon transverse momentum, $$ \vec{p}_{\mathrm{T}}{ }^{(\nu)}=p_{x}{ }^{(\nu)} \hat{\imath}+p_{y}{ }^{(\nu)} \hat{\jmath} \text { and } \vec{p}_{\mathrm{T}}{ }^{(\mu)}=p_{x}{ }^{(\mu)} \hat{\imath}+p_{y}{ }^{(\mu)} \hat{\jmath}, $$ and their $z$-axis momentum components, $p_{z}{ }^{(\mu)}$ and $p_{z}{ }^{(\nu)}$. Context answer: \boxed{$m_{W}^{2}=\frac{1}{c^{2}}(2 p^{(\mu)} \sqrt{(p_{\mathrm{T}}^{(\nu)})^{2}+(p_{z}^{(\nu)})^{2}}-2 \vec{p}_{\mathrm{T}}^{(\mu)} \cdot \vec{p}_{\mathrm{T}}^{(\nu)}-2 p_{z}^{(\mu)} p_{z}^{(\nu)})$} ","B.2 Assuming a $W^{+}$boson mass of $m_{\mathrm{W}}=80.4 \mathrm{GeV} / c^{2}$ calculate the two possible solutions for the neutrino momentum along the $z$-axis, $p_{z}{ }^{(\nu)}$. Express your answer in $\mathrm{GeV} / \mathrm{c}$.","['The numerical substitution directly in the answer of B.1, using\n\n$$\n\\begin{gathered}\np^{(\\mu)}=37.2 \\mathrm{GeV} / c \\quad m_{W}^{2} c^{2}=6464.2(\\mathrm{GeV} / c)^{2} \\quad p_{\\mathrm{\\top}}^{(\\nu) 2}=10864.9(\\mathrm{GeV} / c)^{2} \\\\\n\\vec{p}_{\\mathrm{\\top}}^{(\\mu)} \\cdot \\vec{p}_{\\mathrm{\\top}}^{(\\nu)}=2439.3(\\mathrm{GeV} / c)^{2} \\quad p_{z}^{(\\mu)}=-12.4 \\mathrm{GeV} / c,\n\\end{gathered}\n$$\n\nleads to\n\n$$\n6464.2=74.4 \\sqrt{10864.9+p_{z}^{(\\nu) 2}}-4878.6+24.8 p^{(\\nu)} .\n$$\n\nThis is a quadratic equation, equivalent to\n\n$$\n0.88889 p_{z}^{(\\nu) 2}+101.64 p_{z}^{(\\nu)}-12378=0\n$$\n\nwhose solutions are:\n$p_{z}^{(\\nu)}=74.0 \\mathrm{GeV} / c $ and $ p_{z}^{(\\nu)}=-188.3 \\mathrm{GeV} / c$\n\n\nThe general solution of the equation above in B.1 leads to\n\n$$\n\\begin{aligned}\np_{z}^{(\\nu)}= & \\frac{2 \\vec{p}_{\\mathrm{T}}^{(\\mu)} \\cdot \\vec{p}_{\\mathrm{T}}^{(\\nu)} p_{z}^{(\\mu)}+m_{W}^{2} c^{2} p_{z}^{(\\mu)}}{2\\left(p_{\\mathrm{T}}^{(\\mu)}\\right)^{2}} \\\\\n& \\pm \\frac{p^{(\\mu)} \\sqrt{-4\\left(p_{\\mathrm{T}}^{(\\mu)}\\right)^{2}\\left(p_{\\mathrm{T}}^{(\\nu)}\\right)^{2}+4\\left(\\vec{p}_{\\mathrm{T}}^{(\\mu)} \\cdot \\vec{p}_{\\mathrm{T}}^{(\\nu)}\\right)^{2}+4 \\vec{p}_{\\mathrm{T}}^{(\\mu)} \\cdot \\vec{p}_{\\mathrm{T}}^{(\\nu)} m_{W}^{2} c^{2}+m_{W}^{4} c^{4}}}{2\\left(p_{\\mathrm{T}}^{(\\mu)}\\right)^{2}}\n\\end{aligned}\n$$\n\nNumerical substitution leads to the above mentioned values for $p_{z}^{(\\nu)}$.']","['$p_{z}^{(\\nu)}=74.0 $ , $ p_{z}^{(\\nu)}=-188.3$']",True,$\mathrm{GeV} / c$,Numerical,1e0 1371,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} Context question: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV. Context answer: \boxed{-56} Context question: A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit. Context answer: \boxed{$\omega(t)=\frac{e c^{2} B}{E_{0}}(1+\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} E_{0} t)$} Extra Supplementary Reading Materials: Part B.Finding the neutrino The collision between the two protons shown in Figure 1 leads to the production of a top quark $(t)$ and an anti-top quark $(\bar{t})$, the heaviest elementary particles ever detected. The top quark decays into a $W^{+}$ boson and a bottom quark (b), while the anti-top quark decays into a $W^{-}$boson and an anti-bottom quark $(\bar{b})$. In the case depicted in Figure 1, the $W^{+}$boson decays into a anti-muon $\left(\mu^{+}\right)$and a neutrino $(\nu)$, and the $W^{-}$boson decays into a quark and an anti-quark. The task of this problem is to reconstruct the full momentum of the neutrino using the momenta of some detected particles. For simplicity, all particles and jets in this problem will be considered massless, except for the top quark and $\mathbf{W}^{ \pm}$bosons. The momenta of the top quark decay products can be determined from the experiment (see Table), except for the neutrino momentum component along the ( $z$-axis). The total linear momentum of the final state particles caught by the detector is only zero on the transverse plane ( $x y$ plane), and not along the collision line (z-axis). As such, one can find the transverse momentum of the neutrino from the missing momentum in the transverse plane. On June 4, 2015, the ATLAS experiment at the LHC recorded a proton-proton collision at 00:21:24 GMT+1 like the one represented in Figure 1. Figure 1. Schematic representation of the ATLAS detector coordinate system (left) and proton-proton collision (right). The linear momenta of the three final-state particles coming from the top quark decay, including the neutrino, is presented below for each component. | Particle | $p_{x}(\mathrm{GeV} / c)$ | $p_{y}(\mathrm{GeV} / c)$ | $p_{z}(\mathrm{GeV} / c)$ | | :--- | :---: | :---: | :---: | | anti-muon $\left(\mu^{+}\right)$ | -24.7 | -24.9 | -12.4 | | jet 1 $\left(j_{1}\right)$ | -14.2 | +50.1 | +94.1 | | neutrino $(\nu)$ | -104.1 | +5.3 | --- | Context question: B.1 Find an equation which relates the square of the $W^{+}$boson mass, $m_{\mathrm{W}}^{2}$, with the neutrino and anti-muon momentum components presented in the table above. Express your answer in terms of the neutrino and anti-muon transverse momentum, $$ \vec{p}_{\mathrm{T}}{ }^{(\nu)}=p_{x}{ }^{(\nu)} \hat{\imath}+p_{y}{ }^{(\nu)} \hat{\jmath} \text { and } \vec{p}_{\mathrm{T}}{ }^{(\mu)}=p_{x}{ }^{(\mu)} \hat{\imath}+p_{y}{ }^{(\mu)} \hat{\jmath}, $$ and their $z$-axis momentum components, $p_{z}{ }^{(\mu)}$ and $p_{z}{ }^{(\nu)}$. Context answer: \boxed{$m_{W}^{2}=\frac{1}{c^{2}}(2 p^{(\mu)} \sqrt{(p_{\mathrm{T}}^{(\nu)})^{2}+(p_{z}^{(\nu)})^{2}}-2 \vec{p}_{\mathrm{T}}^{(\mu)} \cdot \vec{p}_{\mathrm{T}}^{(\nu)}-2 p_{z}^{(\mu)} p_{z}^{(\nu)})$} Context question: B.2 Assuming a $W^{+}$boson mass of $m_{\mathrm{W}}=80.4 \mathrm{GeV} / c^{2}$ calculate the two possible solutions for the neutrino momentum along the $z$-axis, $p_{z}{ }^{(\nu)}$. Express your answer in $\mathrm{GeV} / \mathrm{c}$. Context answer: \boxed{$p_{z}^{(\nu)}=74.0 $ , $ p_{z}^{(\nu)}=-188.3$} ","B.3 Calculate the top quark mass for each one of the two previous solutions. Express your answer in $\mathrm{GeV} / c^{2}$. IIf you did not obtain the two solutions in B.2, use $$ \left.p_{z}^{(\nu)}=70 \mathrm{GeV} / c \text { and } p_{z}{ }^{(\nu)}=-180 \mathrm{GeV} / c .\right] $$","['The final state particles of the top quark decay are the anti-muon, the neutrino and jet 1 . Since the neutrino is now fully reconstructed the energy and linear momentum of the top quark can be calculated as,\n\n$$\n\\begin{aligned}\nE^{(\\mathrm{t})} & =c p^{(\\mu)}+c p^{(\\nu)}+c p^{\\left(j_{1}\\right)} \\\\\n\\vec{p}^{(\\mathrm{t})} & =\\vec{p}^{(\\mu)}+\\vec{p}^{(\\nu)}+\\vec{p}^{\\left(j_{1}\\right)} .\n\\end{aligned}\n$$\n\nThe top quark mass is,\n\n$$\n\\begin{aligned}\nm_{\\mathrm{t}} & =\\sqrt{\\left(E^{(\\mathrm{t})}\\right)^{2} / c^{4}-\\left(\\vec{p}^{(\\mathrm{t})}\\right)^{2} / c^{2}} \\\\\n& =c^{-1} \\sqrt{\\left(p^{(\\mu)}+p^{(\\nu)}+p^{\\left(j_{1}\\right)}\\right)^{2}-\\left(\\vec{p}^{(\\mu)}+\\vec{p}^{(\\nu)}+\\vec{p}^{\\left(j_{1}\\right)}\\right)^{2}} .\n\\end{aligned}\n$$']",['$m_{\\mathrm{t}}=169.3 $ and $m_{\\mathrm{t}}=311.2$'],False,$\mathrm{GeV} / c^{2}$,Numerical,1e-1 1372,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} Context question: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV. Context answer: \boxed{-56} Context question: A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit. Context answer: \boxed{$\omega(t)=\frac{e c^{2} B}{E_{0}}(1+\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} E_{0} t)$} Extra Supplementary Reading Materials: Part B.Finding the neutrino The collision between the two protons shown in Figure 1 leads to the production of a top quark $(t)$ and an anti-top quark $(\bar{t})$, the heaviest elementary particles ever detected. The top quark decays into a $W^{+}$ boson and a bottom quark (b), while the anti-top quark decays into a $W^{-}$boson and an anti-bottom quark $(\bar{b})$. In the case depicted in Figure 1, the $W^{+}$boson decays into a anti-muon $\left(\mu^{+}\right)$and a neutrino $(\nu)$, and the $W^{-}$boson decays into a quark and an anti-quark. The task of this problem is to reconstruct the full momentum of the neutrino using the momenta of some detected particles. For simplicity, all particles and jets in this problem will be considered massless, except for the top quark and $\mathbf{W}^{ \pm}$bosons. The momenta of the top quark decay products can be determined from the experiment (see Table), except for the neutrino momentum component along the ( $z$-axis). The total linear momentum of the final state particles caught by the detector is only zero on the transverse plane ( $x y$ plane), and not along the collision line (z-axis). As such, one can find the transverse momentum of the neutrino from the missing momentum in the transverse plane. On June 4, 2015, the ATLAS experiment at the LHC recorded a proton-proton collision at 00:21:24 GMT+1 like the one represented in Figure 1. Figure 1. Schematic representation of the ATLAS detector coordinate system (left) and proton-proton collision (right). The linear momenta of the three final-state particles coming from the top quark decay, including the neutrino, is presented below for each component. | Particle | $p_{x}(\mathrm{GeV} / c)$ | $p_{y}(\mathrm{GeV} / c)$ | $p_{z}(\mathrm{GeV} / c)$ | | :--- | :---: | :---: | :---: | | anti-muon $\left(\mu^{+}\right)$ | -24.7 | -24.9 | -12.4 | | jet 1 $\left(j_{1}\right)$ | -14.2 | +50.1 | +94.1 | | neutrino $(\nu)$ | -104.1 | +5.3 | --- | Context question: B.1 Find an equation which relates the square of the $W^{+}$boson mass, $m_{\mathrm{W}}^{2}$, with the neutrino and anti-muon momentum components presented in the table above. Express your answer in terms of the neutrino and anti-muon transverse momentum, $$ \vec{p}_{\mathrm{T}}{ }^{(\nu)}=p_{x}{ }^{(\nu)} \hat{\imath}+p_{y}{ }^{(\nu)} \hat{\jmath} \text { and } \vec{p}_{\mathrm{T}}{ }^{(\mu)}=p_{x}{ }^{(\mu)} \hat{\imath}+p_{y}{ }^{(\mu)} \hat{\jmath}, $$ and their $z$-axis momentum components, $p_{z}{ }^{(\mu)}$ and $p_{z}{ }^{(\nu)}$. Context answer: \boxed{$m_{W}^{2}=\frac{1}{c^{2}}(2 p^{(\mu)} \sqrt{(p_{\mathrm{T}}^{(\nu)})^{2}+(p_{z}^{(\nu)})^{2}}-2 \vec{p}_{\mathrm{T}}^{(\mu)} \cdot \vec{p}_{\mathrm{T}}^{(\nu)}-2 p_{z}^{(\mu)} p_{z}^{(\nu)})$} Context question: B.2 Assuming a $W^{+}$boson mass of $m_{\mathrm{W}}=80.4 \mathrm{GeV} / c^{2}$ calculate the two possible solutions for the neutrino momentum along the $z$-axis, $p_{z}{ }^{(\nu)}$. Express your answer in $\mathrm{GeV} / \mathrm{c}$. Context answer: \boxed{$p_{z}^{(\nu)}=74.0 $ , $ p_{z}^{(\nu)}=-188.3$} Context question: B.3 Calculate the top quark mass for each one of the two previous solutions. Express your answer in $\mathrm{GeV} / c^{2}$. IIf you did not obtain the two solutions in B.2, use $$ \left.p_{z}^{(\nu)}=70 \mathrm{GeV} / c \text { and } p_{z}{ }^{(\nu)}=-180 \mathrm{GeV} / c .\right] $$ Context answer: \boxed{$m_{\mathrm{t}}=169.3 $ and $m_{\mathrm{t}}=311.2$} Extra Supplementary Reading Materials: The normalised number of collision events for the measurement of the top quark mass(as determined from the experiment), has two components: the so-called ""signal"" (corresponding to the decay of top quarks) and ""background"" (corresponding to events from other processes that do not include top quarks). Experimental data include both processes, see Fig. 2. Figure 2. Top quark mass distribution as determined from the experiment, i.e. the normalised number of events plotted against the top quark mass. The dots correspond to the datA.The dashed line corresponds to the ""signal"" and the shade to the ""background"".","B.4 According to the top quark mass distribution, which one of the two previous solutions is more likely to be the right one? Estimate the probability for the most likely solution.","['According to the frequency distribution for signal (dashed line), the probability of the $m_{\\mathrm{t}}=169.3 \\mathrm{GeV} / c^{2}$ solution is roughly 0.1 while the probability of the $m_{\\mathrm{t}}=311.2 \\mathrm{GeV} / \\mathrm{c}^{2}$ solution is below 0.01 .']",['$m_{\\mathrm{t}}=169.3$'],False,$ \mathrm{GeV} / \mathrm{c}^{2}$,Numerical,1e-1 1373,Modern Physics,"Where is the neutrino? When two protons collide with a very high energy at the Large Hadron Collider (LHC), several particles may be produced as a result of that collision, such as electrons, muons, neutrinos, quarks, and their respective anti-particles. Most of these particles can be detected by the particle detector surrounding the collision point. For example, quarks undergo a process called hadronisation in which they become a shower of subatomic particles, called ""jet"". In addition, the high magnetic field present in the detectors allows even very energetic charged particles to curve enough for their momentum to be determined. The ATLAS detector uses a superconducting solenoid system that produces a constant and uniform 2.00 Tesla magnetic field in the inner part of the detector, surrounding the collision point. Charged particles with momenta below a certain value will be curved so strongly that they will loop repeatedly in the field and most likely not be measured. Due to its nature, the neutrino is not detected at all, as it escapes through the detector without interacting. Data: Electron rest mass, $m=9.11 \times 10^{-31} \mathrm{~kg}$; Elementary charge, $e=1.60 \times 10^{-19} \mathrm{C}$; Speed of light, $c=3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$; Vacuum permittivity, $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \mathrm{~m}^{-1}$ Part A.ATLAS Detector physics Context question: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle. Context answer: \boxed{$r=\frac{\sqrt{2 K m}}{e B}$} Extra Supplementary Reading Materials: Electrons produced inside the ATLAS detector must be treated relativistically. However, the formula for the cyclotron radius also holds for relativistic motion when the relativistic momentum is considered. Context question: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$. Context answer: \boxed{$p=330$} Extra Supplementary Reading Materials: When accelerated perpendicularly to the velocity, relativistic particles of charge $e$ and rest mass $m$ emitt electromagnetic radiation, called synchrotron radiation. The emitted power is given by $$ P=\frac{e^{2} a^{2} \gamma^{4}}{6 \pi \epsilon_{0} c^{3}} $$ where $a$ is the acceleration and $\gamma=\left[1-(v / c)^{2}\right]^{-1 / 2}$. Context question: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as: $$ P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2} $$ where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$. Context answer: \boxed{$\xi=\frac{1}{6 \pi}, n=5$ , $k=4$} Context question: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is: $$ E(t)=\frac{E_{0}}{1+\alpha E_{0} t}, $$ where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$. Context answer: \boxed{$\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$} Context question: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV. Context answer: \boxed{-56} Context question: A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit. Context answer: \boxed{$\omega(t)=\frac{e c^{2} B}{E_{0}}(1+\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} E_{0} t)$} Extra Supplementary Reading Materials: Part B.Finding the neutrino The collision between the two protons shown in Figure 1 leads to the production of a top quark $(t)$ and an anti-top quark $(\bar{t})$, the heaviest elementary particles ever detected. The top quark decays into a $W^{+}$ boson and a bottom quark (b), while the anti-top quark decays into a $W^{-}$boson and an anti-bottom quark $(\bar{b})$. In the case depicted in Figure 1, the $W^{+}$boson decays into a anti-muon $\left(\mu^{+}\right)$and a neutrino $(\nu)$, and the $W^{-}$boson decays into a quark and an anti-quark. The task of this problem is to reconstruct the full momentum of the neutrino using the momenta of some detected particles. For simplicity, all particles and jets in this problem will be considered massless, except for the top quark and $\mathbf{W}^{ \pm}$bosons. The momenta of the top quark decay products can be determined from the experiment (see Table), except for the neutrino momentum component along the ( $z$-axis). The total linear momentum of the final state particles caught by the detector is only zero on the transverse plane ( $x y$ plane), and not along the collision line (z-axis). As such, one can find the transverse momentum of the neutrino from the missing momentum in the transverse plane. On June 4, 2015, the ATLAS experiment at the LHC recorded a proton-proton collision at 00:21:24 GMT+1 like the one represented in Figure 1. Figure 1. Schematic representation of the ATLAS detector coordinate system (left) and proton-proton collision (right). The linear momenta of the three final-state particles coming from the top quark decay, including the neutrino, is presented below for each component. | Particle | $p_{x}(\mathrm{GeV} / c)$ | $p_{y}(\mathrm{GeV} / c)$ | $p_{z}(\mathrm{GeV} / c)$ | | :--- | :---: | :---: | :---: | | anti-muon $\left(\mu^{+}\right)$ | -24.7 | -24.9 | -12.4 | | jet 1 $\left(j_{1}\right)$ | -14.2 | +50.1 | +94.1 | | neutrino $(\nu)$ | -104.1 | +5.3 | --- | Context question: B.1 Find an equation which relates the square of the $W^{+}$boson mass, $m_{\mathrm{W}}^{2}$, with the neutrino and anti-muon momentum components presented in the table above. Express your answer in terms of the neutrino and anti-muon transverse momentum, $$ \vec{p}_{\mathrm{T}}{ }^{(\nu)}=p_{x}{ }^{(\nu)} \hat{\imath}+p_{y}{ }^{(\nu)} \hat{\jmath} \text { and } \vec{p}_{\mathrm{T}}{ }^{(\mu)}=p_{x}{ }^{(\mu)} \hat{\imath}+p_{y}{ }^{(\mu)} \hat{\jmath}, $$ and their $z$-axis momentum components, $p_{z}{ }^{(\mu)}$ and $p_{z}{ }^{(\nu)}$. Context answer: \boxed{$m_{W}^{2}=\frac{1}{c^{2}}(2 p^{(\mu)} \sqrt{(p_{\mathrm{T}}^{(\nu)})^{2}+(p_{z}^{(\nu)})^{2}}-2 \vec{p}_{\mathrm{T}}^{(\mu)} \cdot \vec{p}_{\mathrm{T}}^{(\nu)}-2 p_{z}^{(\mu)} p_{z}^{(\nu)})$} Context question: B.2 Assuming a $W^{+}$boson mass of $m_{\mathrm{W}}=80.4 \mathrm{GeV} / c^{2}$ calculate the two possible solutions for the neutrino momentum along the $z$-axis, $p_{z}{ }^{(\nu)}$. Express your answer in $\mathrm{GeV} / \mathrm{c}$. Context answer: \boxed{$p_{z}^{(\nu)}=74.0 $ , $ p_{z}^{(\nu)}=-188.3$} Context question: B.3 Calculate the top quark mass for each one of the two previous solutions. Express your answer in $\mathrm{GeV} / c^{2}$. IIf you did not obtain the two solutions in B.2, use $$ \left.p_{z}^{(\nu)}=70 \mathrm{GeV} / c \text { and } p_{z}{ }^{(\nu)}=-180 \mathrm{GeV} / c .\right] $$ Context answer: \boxed{$m_{\mathrm{t}}=169.3 $ and $m_{\mathrm{t}}=311.2$} Extra Supplementary Reading Materials: The normalised number of collision events for the measurement of the top quark mass(as determined from the experiment), has two components: the so-called ""signal"" (corresponding to the decay of top quarks) and ""background"" (corresponding to events from other processes that do not include top quarks). Experimental data include both processes, see Fig. 2. Figure 2. Top quark mass distribution as determined from the experiment, i.e. the normalised number of events plotted against the top quark mass. The dots correspond to the datA.The dashed line corresponds to the ""signal"" and the shade to the ""background"". Context question: B.4 According to the top quark mass distribution, which one of the two previous solutions is more likely to be the right one? Estimate the probability for the most likely solution. Context answer: \boxed{$m_{\mathrm{t}}=169.3$} ","B.5 Calculate the distance traveled by the top quark before decaying, using the most likely solution. Assume the top quark has a mean lifetime at rest of $5 \times 10^{-25}$ s.",['The top quark energy for the most likely candidate is $E^{(\\mathrm{t})}=c p^{(\\mu)}+c p^{(\\nu)}+c p^{\\left(j_{1}\\right)}=272.6 \\mathrm{GeV}$.\n\n$$\nd=v t=v \\gamma t_{0}=\\frac{p^{(\\mathrm{t})}}{m_{\\mathrm{t}}} t_{0}=c t_{0} \\sqrt{\\frac{E^{(\\mathrm{t})^{2}}}{m_{\\mathrm{t}}^{2} c^{4}}-1}\n$$'],['$d=2 \\times 10^{-16}$'],False,m,Numerical,1e-17 1374,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles.","A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$.","['Since the vessel network is symmetrical, the flow in a vessel of level $i+1$ is half the flow in a vessel of level $i$.\n\nIn this way, we can sum the pressure differences in all levels:\n\n$$\n\\Delta P=\\sum_{i=0}^{N-1} Q_{i} R_{i}=Q_{0} \\sum_{i=0}^{N-1} \\frac{R_{i}}{2^{i}}\n$$\n\nIntroducing the radii dependences yields\n\n$$\n\\Delta P=Q_{0} \\sum_{i=0}^{N-1} \\frac{8 \\ell_{i} \\eta}{2^{i} \\pi r_{i}^{4}}=Q_{0} \\frac{8 \\ell_{0} \\eta}{\\pi r_{0}^{4}} \\sum_{i=0}^{N-1} \\frac{2^{4 i / 3}}{2^{i} 2^{i / 3}}=Q_{0} N \\frac{8 \\ell_{0} \\eta}{\\pi r_{0}^{4}}\n$$\n\nTherefore\n\n$$\nQ_{0}=\\Delta P \\frac{\\pi r_{0}^{4}}{8 N \\ell_{0} \\eta}\n$$']",['$Q_{i}=\\Delta P \\frac{\\pi r_{0}^{4}}{2^{i+3} N \\ell_{0} \\eta}$'],False,,Expression, 1375,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles. Context question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$. Context answer: \boxed{$Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$} ","A.2 Calculate the numerical value of the volumetric flow rate $Q_{0}$ of the arteriole at level 0 , if its radius is $6.0 \times 10^{-5} \mathrm{~m}$ and its length is $2.0 \times 10^{-3} \mathrm{~m}$. Consider that the pressure at the arteriole inlet is $55 \mathrm{mmHg}$ and the vessel network has $N=6$ levels linking this arteriole to the capillary bed at the pressure $30 \mathrm{mmHg}$. Consider that the blood viscosity is $\eta=3.5 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$. Express your result in $\mathrm{ml} / \mathrm{h}$.",['Replace values in the formula and change units appropriately\n\n$$\n\\begin{aligned}\nQ_{0} & =\\frac{\\Delta P \\pi r_{0}^{4}}{8 N \\ell_{0} \\eta}= \\\\\n& =\\frac{(55-30) \\times 1.013 \\times 10^{5} \\times 3.1415 \\times\\left(6.0 \\times 10^{-5}\\right)^{4}}{760 \\times 48 \\times 2.0 \\times 10^{-3} \\times 3.5 \\times 10^{-3}}=4.0 \\times 10^{-10} \\mathrm{~m}^{3} / \\mathrm{s}\n\\end{aligned}\n$$'],['$1.5$'],False,$\mathrm{~m} \ell / \mathrm{h}$,Numerical,1e-1 1376,Thermodynamics,,"A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$.","['The current is given by\n\n$$\nI=\\frac{P_{\\mathrm{in}} \\mathrm{e}^{i \\omega t}}{R+i \\omega L+\\frac{1}{i \\omega C}}\n$$\n\nThe pressure difference in the capacitor is\n\n$$\nP_{\\text {out }} \\mathrm{e}^{i(\\omega t+\\phi)}=\\frac{P_{\\text {in }} \\mathrm{e}^{i \\omega t}}{R+i \\omega L+\\frac{1}{i \\omega C}} \\frac{1}{i \\omega C}=\\frac{P_{\\text {in }} \\mathrm{e}^{i \\omega t}}{i \\omega C R-\\omega^{2} L C+1} .\n$$\n\nThe amplitude is\n\n$$\nP_{\\text {out }}=\\frac{P_{\\text {in }}}{\\sqrt{\\left(1-\\omega^{2} L C\\right)^{2}+\\omega^{2} C^{2} R^{2}}} .\n$$\n\nTo be smaller than $P_{\\text {in }}$, for $\\omega \\rightarrow 0$ :\n\n$$\n\\left(1-\\omega^{2} L C\\right)^{2}+\\omega^{2} C^{2} R^{2}>1 \\Longleftrightarrow-2 C L+C^{2} R^{2}>0\n$$\n\nReplacing the expressions for $L, C$, and $R$ we get: $\\frac{642^{2} \\ell^{2}}{3 E h r^{3} \\rho}>1$.']",['$$\nP_{\\text {out }}=\\frac{P_{\\text {in }}}{\\sqrt{\\left(1-\\omega^{2} L C\\right)^{2}+\\omega^{2} C^{2} R^{2}}} .\n$$\n\nCondition:\n\n$$\n\\frac{64 \\eta^{2} \\ell^{2}}{3 E h r^{3} \\rho}>1\n$$'],False,,Need_human_evaluate, 1377,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles. Context question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$. Context answer: \boxed{$Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$} Context question: A.2 Calculate the numerical value of the volumetric flow rate $Q_{0}$ of the arteriole at level 0 , if its radius is $6.0 \times 10^{-5} \mathrm{~m}$ and its length is $2.0 \times 10^{-3} \mathrm{~m}$. Consider that the pressure at the arteriole inlet is $55 \mathrm{mmHg}$ and the vessel network has $N=6$ levels linking this arteriole to the capillary bed at the pressure $30 \mathrm{mmHg}$. Consider that the blood viscosity is $\eta=3.5 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$. Express your result in $\mathrm{ml} / \mathrm{h}$. Context answer: \boxed{$1.5$} Extra Supplementary Reading Materials: A blood vessel as an LCR circuit The approximation of rigid cylindrical vessels falls short for several reasons. It is particularly important to include the time dependent flow and to take into account the change in vessel diameter that occurs when the pressure varies during a blood pumping cycle done by the heart. Moreover, it is observed that in the larger vessels the blood pressure varies significantly during a cycle, while in the smaller vessels the amplitude of the oscillations in pressure is much smaller, and the flow is almost time independent. When the pressure increases in a single elastic vessel, there will be an increase in its diameter, thus permitting to store more fluid in the vessel, and to deliver it when the pressure drops. For this reason, the elastic behaviour of the vessel can be simulated by adding a capacitor to our initial description. Moreover, when taking into account the time dependent blood flow rate, one has to consider the inertia of the fluid, proportional to its density $\rho=1.05 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. This inertia can be described by an inductance in our model. In Figure 2 we represent the equivalent circuit for a single vessel in this model. The equivalent capacitance and inductance are given by $$ C=\frac{3 \ell \pi r^{3}}{2 E h} \quad \text { and } \quad L=\frac{9 \ell \rho}{4 \pi r^{2}} \tag{2} $$ respectively, where $h$ is the width of the vessel wall and $E$ is the artery Young's modulus, a coefficient that describes the alteration in size of the vessel tissue when a force is applied. The Young's modulus has units of pressure and is on the order of $E=0.06 \mathrm{MPa}$ for arterioles. Figure 2. Equivalent electric circuit for a single vessel. Context question: A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$. Context answer: $$ P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} . $$ Condition: $$ \frac{64 \eta^{2} \ell^{2}}{3 E h r^{3} \rho}>1 $$ ",A.4 For the vessel network in A.2 estimate the maximum arteriole wall thickness $h$ so that the condition established in A.3 is satisfied (consider that $h$ is level independent).,"['The previous condition can also be expressed as\n\n$$\nh<\\frac{64 \\eta^{2} \\ell^{2}}{3 E r^{3} \\rho}\n$$\n\nFor the network referred to in $\\mathbf{A} .2$\n\n$$\nh<\\frac{64 \\eta^{2} \\ell_{0}^{2} \\times 2^{i}}{3 \\times 2^{2 i / 3} E r_{0}^{3} \\rho}=\\frac{64 \\times\\left(3.5 \\times 10^{-3}\\right)^{2} \\times\\left(2.0 \\times 10^{-3}\\right)^{2}}{3 \\times 0.06 \\times 10^{6} \\times\\left(6.0 \\times 10^{-5}\\right)^{3} \\times 1.05 \\times 10^{3}} \\times 2^{i / 3}=7.7 \\times 10^{-5} \\times 2^{i / 3}\n$$\n\n\n\nFor $i=0$, in the worse case scenario,\n\n$$\nh_{\\max }=7.7 \\times 10^{-5} \\times 2^{0}=7.7 \\times 10^{-5} \\mathrm{~m}\n$$\n\nThis value is certainly observed in these vessels since their radius range from $18 \\mu \\mathrm{m}$ to $60 \\mu \\mathrm{m}$. A wall width smaller than $80 \\mu \\mathrm{m}$ is certainly reasonable.']",['$h=8 \\times 10^{-5}$'],False,m,Numerical,1e-6 1378,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles. Context question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$. Context answer: \boxed{$Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$} Context question: A.2 Calculate the numerical value of the volumetric flow rate $Q_{0}$ of the arteriole at level 0 , if its radius is $6.0 \times 10^{-5} \mathrm{~m}$ and its length is $2.0 \times 10^{-3} \mathrm{~m}$. Consider that the pressure at the arteriole inlet is $55 \mathrm{mmHg}$ and the vessel network has $N=6$ levels linking this arteriole to the capillary bed at the pressure $30 \mathrm{mmHg}$. Consider that the blood viscosity is $\eta=3.5 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$. Express your result in $\mathrm{ml} / \mathrm{h}$. Context answer: \boxed{$1.5$} Extra Supplementary Reading Materials: A blood vessel as an LCR circuit The approximation of rigid cylindrical vessels falls short for several reasons. It is particularly important to include the time dependent flow and to take into account the change in vessel diameter that occurs when the pressure varies during a blood pumping cycle done by the heart. Moreover, it is observed that in the larger vessels the blood pressure varies significantly during a cycle, while in the smaller vessels the amplitude of the oscillations in pressure is much smaller, and the flow is almost time independent. When the pressure increases in a single elastic vessel, there will be an increase in its diameter, thus permitting to store more fluid in the vessel, and to deliver it when the pressure drops. For this reason, the elastic behaviour of the vessel can be simulated by adding a capacitor to our initial description. Moreover, when taking into account the time dependent blood flow rate, one has to consider the inertia of the fluid, proportional to its density $\rho=1.05 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. This inertia can be described by an inductance in our model. In Figure 2 we represent the equivalent circuit for a single vessel in this model. The equivalent capacitance and inductance are given by $$ C=\frac{3 \ell \pi r^{3}}{2 E h} \quad \text { and } \quad L=\frac{9 \ell \rho}{4 \pi r^{2}} \tag{2} $$ respectively, where $h$ is the width of the vessel wall and $E$ is the artery Young's modulus, a coefficient that describes the alteration in size of the vessel tissue when a force is applied. The Young's modulus has units of pressure and is on the order of $E=0.06 \mathrm{MPa}$ for arterioles. Figure 2. Equivalent electric circuit for a single vessel. Context question: A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$. Context answer: $$ P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} . $$ Condition: $$ \frac{64 \eta^{2} \ell^{2}}{3 E h r^{3} \rho}>1 $$ Context question: A.4 For the vessel network in A.2 estimate the maximum arteriole wall thickness $h$ so that the condition established in A.3 is satisfied (consider that $h$ is level independent). Context answer: \boxed{$h=8 \times 10^{-5}$} Extra Supplementary Reading Materials: Part B.Tumour growth Tumour growth is a very complex process where biological mechanisms such as cell proliferation and natural selection are intertwined with physics. In this problem we will consider a simplified model of tumour growth that addresses the increase in pressure commonly observed in solid tumors. Consider a group of normal cells forming a tissue surrounded by an inextensible basement membrane, which forces the tissue to maintain always the same form: a sphere of radius $R$ (Figure 3). Figure 3. Simplified tumour. Initially the tissue does not have residual stresses, i.e. the pressure at every point is equal to the atmospheric pressure. At time $t=0$, a tumour starts growing at the centre of this sphere and, as it grows, the pressure inside the tissue increases. Consider that both tissues (normal, N, and tumour, $\mathrm{T}$ ) are compressible such that their densities, $\rho_{\mathrm{N}}$ and $\rho_{\mathrm{T}}$, increase linearly with pressure: $$ \rho_{\mathrm{N}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{N}}}\right), \quad \rho_{\mathrm{T}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{T}}}\right) $$ where $\rho_{0}$ is the rest tissue density, $p$ is the pressure difference to the atmospheric pressure and $K_{\mathrm{N}}, K_{\mathrm{T}}$ are the compressibility moduli (bulk moduli) of the normal and tumour tissues, respectively. In general, tumours are stiffer and so they have a higher bulk modulus.","B.1 The mass of normal cells is not altered while the tumour is growing. Obtain the ratio between the tumour volume and the total tissue volume, $v=V_{\mathrm{T}} / V$, as a function of the ratio between the tumour mass $\left(M_{\mathrm{T}}\right)$ and the normal tissue mass $\left(M_{\mathrm{N}}\right), \mu=M_{\mathrm{T}} / M_{\mathrm{N}}$ and the ratio of the bulk moduli, $\kappa=K_{\mathrm{N}} / K_{\mathrm{T}}$.","['The expressions for the masses of tumour and normal tissue are written as:\n\n$$\n\\left\\{\\begin{array}{l}\nM_{\\mathrm{T}}=V_{\\mathrm{T}} \\rho_{\\mathrm{T}}=V_{\\mathrm{T}} \\rho_{0}\\left(1+\\frac{p}{K_{\\mathrm{T}}}\\right) \\\\\nM_{\\mathrm{N}}=V \\rho_{0}=\\left(V-V_{\\mathrm{T}}\\right) \\rho_{0}\\left(1+\\frac{p}{K_{\\mathrm{N}}}\\right)\n\\end{array}\\right.\n$$\n\nThe pressure, $p$, can be expressed as\n\n$$\np=\\frac{M_{\\mathrm{T}} K_{\\mathrm{T}}}{V_{\\mathrm{T}} \\rho_{0}}-K_{\\mathrm{T}}\n\\tag{3}\n$$\n\nand, then, used in the equation for $M_{\\mathrm{N}}$ :\n\n$$\nM_{\\mathrm{N}}=\\left(V-V_{\\mathrm{T}}\\right) \\frac{M_{\\mathrm{N}}}{V}\\left[\\left(1-\\frac{K_{\\mathrm{T}}}{K_{\\mathrm{N}}}\\right)+\\frac{M_{\\mathrm{T}} V K_{\\mathrm{T}}}{V_{\\mathrm{T}} M_{\\mathrm{N}} K_{\\mathrm{N}}}\\right]\n$$\n\nSimplifying and rearranging the terms, the equation for $v$ becomes\n\n$$\n(1-\\kappa) v^{2}-(1+\\mu) v+\\mu=0,\n$$\n\nfor which the solution is (the other solution of the quadratic equation is not physically relevant since does not lead to $v=0$ for $\\mu=0$ )']",['$v=\\frac{1+\\mu-\\sqrt{(1+\\mu)^{2}-4 \\mu(1-\\kappa)}}{2(1-\\kappa)}$'],False,,Expression, 1379,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles. Context question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$. Context answer: \boxed{$Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$} Context question: A.2 Calculate the numerical value of the volumetric flow rate $Q_{0}$ of the arteriole at level 0 , if its radius is $6.0 \times 10^{-5} \mathrm{~m}$ and its length is $2.0 \times 10^{-3} \mathrm{~m}$. Consider that the pressure at the arteriole inlet is $55 \mathrm{mmHg}$ and the vessel network has $N=6$ levels linking this arteriole to the capillary bed at the pressure $30 \mathrm{mmHg}$. Consider that the blood viscosity is $\eta=3.5 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$. Express your result in $\mathrm{ml} / \mathrm{h}$. Context answer: \boxed{$1.5$} Extra Supplementary Reading Materials: A blood vessel as an LCR circuit The approximation of rigid cylindrical vessels falls short for several reasons. It is particularly important to include the time dependent flow and to take into account the change in vessel diameter that occurs when the pressure varies during a blood pumping cycle done by the heart. Moreover, it is observed that in the larger vessels the blood pressure varies significantly during a cycle, while in the smaller vessels the amplitude of the oscillations in pressure is much smaller, and the flow is almost time independent. When the pressure increases in a single elastic vessel, there will be an increase in its diameter, thus permitting to store more fluid in the vessel, and to deliver it when the pressure drops. For this reason, the elastic behaviour of the vessel can be simulated by adding a capacitor to our initial description. Moreover, when taking into account the time dependent blood flow rate, one has to consider the inertia of the fluid, proportional to its density $\rho=1.05 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. This inertia can be described by an inductance in our model. In Figure 2 we represent the equivalent circuit for a single vessel in this model. The equivalent capacitance and inductance are given by $$ C=\frac{3 \ell \pi r^{3}}{2 E h} \quad \text { and } \quad L=\frac{9 \ell \rho}{4 \pi r^{2}} \tag{2} $$ respectively, where $h$ is the width of the vessel wall and $E$ is the artery Young's modulus, a coefficient that describes the alteration in size of the vessel tissue when a force is applied. The Young's modulus has units of pressure and is on the order of $E=0.06 \mathrm{MPa}$ for arterioles. Figure 2. Equivalent electric circuit for a single vessel. Context question: A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$. Context answer: $$ P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} . $$ Condition: $$ \frac{64 \eta^{2} \ell^{2}}{3 E h r^{3} \rho}>1 $$ Context question: A.4 For the vessel network in A.2 estimate the maximum arteriole wall thickness $h$ so that the condition established in A.3 is satisfied (consider that $h$ is level independent). Context answer: \boxed{$h=8 \times 10^{-5}$} Extra Supplementary Reading Materials: Part B.Tumour growth Tumour growth is a very complex process where biological mechanisms such as cell proliferation and natural selection are intertwined with physics. In this problem we will consider a simplified model of tumour growth that addresses the increase in pressure commonly observed in solid tumors. Consider a group of normal cells forming a tissue surrounded by an inextensible basement membrane, which forces the tissue to maintain always the same form: a sphere of radius $R$ (Figure 3). Figure 3. Simplified tumour. Initially the tissue does not have residual stresses, i.e. the pressure at every point is equal to the atmospheric pressure. At time $t=0$, a tumour starts growing at the centre of this sphere and, as it grows, the pressure inside the tissue increases. Consider that both tissues (normal, N, and tumour, $\mathrm{T}$ ) are compressible such that their densities, $\rho_{\mathrm{N}}$ and $\rho_{\mathrm{T}}$, increase linearly with pressure: $$ \rho_{\mathrm{N}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{N}}}\right), \quad \rho_{\mathrm{T}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{T}}}\right) $$ where $\rho_{0}$ is the rest tissue density, $p$ is the pressure difference to the atmospheric pressure and $K_{\mathrm{N}}, K_{\mathrm{T}}$ are the compressibility moduli (bulk moduli) of the normal and tumour tissues, respectively. In general, tumours are stiffer and so they have a higher bulk modulus. Context question: B.1 The mass of normal cells is not altered while the tumour is growing. Obtain the ratio between the tumour volume and the total tissue volume, $v=V_{\mathrm{T}} / V$, as a function of the ratio between the tumour mass $\left(M_{\mathrm{T}}\right)$ and the normal tissue mass $\left(M_{\mathrm{N}}\right), \mu=M_{\mathrm{T}} / M_{\mathrm{N}}$ and the ratio of the bulk moduli, $\kappa=K_{\mathrm{N}} / K_{\mathrm{T}}$. Context answer: \boxed{$v=\frac{1+\mu-\sqrt{(1+\mu)^{2}-4 \mu(1-\kappa)}}{2(1-\kappa)}$} Extra Supplementary Reading Materials: Hyperthermia is sometimes used together with chemotherapy and radiotherapy in the treatment of cancer. In hyperthermia the cancer cells are selectively heated from the normal human body temperature, $37^{\circ} \mathrm{C}$, to temperatures above $43^{\circ} \mathrm{C}$, inducing their death. Researchers are currently developing carbon nanotubes covered with special proteins capable of binding to tumour cells. When the tissue is irradiated with near-infrared radiation, the nanotubes absorb it in a much greater extent than the surrounding tissues and therefore can be selectively heated as well as the tumour cells to which they are attached. Consider that the tumour, the normal cells and the surrounding tissue have a constant thermal conductivity $k$, i.e. in the geometry of this problem, the energy that crosses a spherical surface of radius $r$ per unit time and per unit area is equal to $k$ times the derivative of the temperature with respect to $r$. The nanotubes are uniformly distributed in the tumour volume and are able to deliver a power $\mathcal{P}$ of thermal energy per unit volume. Assume that the temperature is equal to the normal human body temperature very far away from the tumour.","B.2 Obtain, for the stationary state, the temperature at the centre of the tumour as a function of $\mathcal{P}, k$, the human body temperature and the tumour radius, $R_{\mathrm{T}}$.","['For $rR_{\\mathrm{T}}$, the conservation of energy implies that\n\n$$\n4 \\pi r^{2}(-k) \\frac{\\mathrm{d} T}{\\mathrm{~d} r}=\\mathcal{P}_{\\frac{4}{3}}^{4} \\pi R_{\\mathrm{\\top}}^{3}\n$$\n\nTherefore, the temperature difference to $37^{\\circ} \\mathrm{C}$ is\n\n$$\n\\Delta T(r)=\\frac{\\mathcal{P} R_{\\mathrm{T}}^{3}}{3 k r}\n$$\n\nIn this case there is no constant, since very far away the increase in temperature is zero.\n\nMatching the two solutions at $r=R_{\\mathrm{T}}$ gives\n\n$$\nC=\\frac{\\mathcal{P} R_{\\mathrm{T}}^{2}}{2 k}\n$$\n\nTherefore the temperature at the centre of the tumour, in SI units, is']",['$310.15+\\frac{\\mathcal{P} R_{\\uparrow}^{2}}{2 k}$'],False,,Expression, 1380,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles. Context question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$. Context answer: \boxed{$Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$} Context question: A.2 Calculate the numerical value of the volumetric flow rate $Q_{0}$ of the arteriole at level 0 , if its radius is $6.0 \times 10^{-5} \mathrm{~m}$ and its length is $2.0 \times 10^{-3} \mathrm{~m}$. Consider that the pressure at the arteriole inlet is $55 \mathrm{mmHg}$ and the vessel network has $N=6$ levels linking this arteriole to the capillary bed at the pressure $30 \mathrm{mmHg}$. Consider that the blood viscosity is $\eta=3.5 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$. Express your result in $\mathrm{ml} / \mathrm{h}$. Context answer: \boxed{$1.5$} Extra Supplementary Reading Materials: A blood vessel as an LCR circuit The approximation of rigid cylindrical vessels falls short for several reasons. It is particularly important to include the time dependent flow and to take into account the change in vessel diameter that occurs when the pressure varies during a blood pumping cycle done by the heart. Moreover, it is observed that in the larger vessels the blood pressure varies significantly during a cycle, while in the smaller vessels the amplitude of the oscillations in pressure is much smaller, and the flow is almost time independent. When the pressure increases in a single elastic vessel, there will be an increase in its diameter, thus permitting to store more fluid in the vessel, and to deliver it when the pressure drops. For this reason, the elastic behaviour of the vessel can be simulated by adding a capacitor to our initial description. Moreover, when taking into account the time dependent blood flow rate, one has to consider the inertia of the fluid, proportional to its density $\rho=1.05 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. This inertia can be described by an inductance in our model. In Figure 2 we represent the equivalent circuit for a single vessel in this model. The equivalent capacitance and inductance are given by $$ C=\frac{3 \ell \pi r^{3}}{2 E h} \quad \text { and } \quad L=\frac{9 \ell \rho}{4 \pi r^{2}} \tag{2} $$ respectively, where $h$ is the width of the vessel wall and $E$ is the artery Young's modulus, a coefficient that describes the alteration in size of the vessel tissue when a force is applied. The Young's modulus has units of pressure and is on the order of $E=0.06 \mathrm{MPa}$ for arterioles. Figure 2. Equivalent electric circuit for a single vessel. Context question: A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$. Context answer: $$ P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} . $$ Condition: $$ \frac{64 \eta^{2} \ell^{2}}{3 E h r^{3} \rho}>1 $$ Context question: A.4 For the vessel network in A.2 estimate the maximum arteriole wall thickness $h$ so that the condition established in A.3 is satisfied (consider that $h$ is level independent). Context answer: \boxed{$h=8 \times 10^{-5}$} Extra Supplementary Reading Materials: Part B.Tumour growth Tumour growth is a very complex process where biological mechanisms such as cell proliferation and natural selection are intertwined with physics. In this problem we will consider a simplified model of tumour growth that addresses the increase in pressure commonly observed in solid tumors. Consider a group of normal cells forming a tissue surrounded by an inextensible basement membrane, which forces the tissue to maintain always the same form: a sphere of radius $R$ (Figure 3). Figure 3. Simplified tumour. Initially the tissue does not have residual stresses, i.e. the pressure at every point is equal to the atmospheric pressure. At time $t=0$, a tumour starts growing at the centre of this sphere and, as it grows, the pressure inside the tissue increases. Consider that both tissues (normal, N, and tumour, $\mathrm{T}$ ) are compressible such that their densities, $\rho_{\mathrm{N}}$ and $\rho_{\mathrm{T}}$, increase linearly with pressure: $$ \rho_{\mathrm{N}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{N}}}\right), \quad \rho_{\mathrm{T}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{T}}}\right) $$ where $\rho_{0}$ is the rest tissue density, $p$ is the pressure difference to the atmospheric pressure and $K_{\mathrm{N}}, K_{\mathrm{T}}$ are the compressibility moduli (bulk moduli) of the normal and tumour tissues, respectively. In general, tumours are stiffer and so they have a higher bulk modulus. Context question: B.1 The mass of normal cells is not altered while the tumour is growing. Obtain the ratio between the tumour volume and the total tissue volume, $v=V_{\mathrm{T}} / V$, as a function of the ratio between the tumour mass $\left(M_{\mathrm{T}}\right)$ and the normal tissue mass $\left(M_{\mathrm{N}}\right), \mu=M_{\mathrm{T}} / M_{\mathrm{N}}$ and the ratio of the bulk moduli, $\kappa=K_{\mathrm{N}} / K_{\mathrm{T}}$. Context answer: \boxed{$v=\frac{1+\mu-\sqrt{(1+\mu)^{2}-4 \mu(1-\kappa)}}{2(1-\kappa)}$} Extra Supplementary Reading Materials: Hyperthermia is sometimes used together with chemotherapy and radiotherapy in the treatment of cancer. In hyperthermia the cancer cells are selectively heated from the normal human body temperature, $37^{\circ} \mathrm{C}$, to temperatures above $43^{\circ} \mathrm{C}$, inducing their death. Researchers are currently developing carbon nanotubes covered with special proteins capable of binding to tumour cells. When the tissue is irradiated with near-infrared radiation, the nanotubes absorb it in a much greater extent than the surrounding tissues and therefore can be selectively heated as well as the tumour cells to which they are attached. Consider that the tumour, the normal cells and the surrounding tissue have a constant thermal conductivity $k$, i.e. in the geometry of this problem, the energy that crosses a spherical surface of radius $r$ per unit time and per unit area is equal to $k$ times the derivative of the temperature with respect to $r$. The nanotubes are uniformly distributed in the tumour volume and are able to deliver a power $\mathcal{P}$ of thermal energy per unit volume. Assume that the temperature is equal to the normal human body temperature very far away from the tumour. Context question: B.2 Obtain, for the stationary state, the temperature at the centre of the tumour as a function of $\mathcal{P}, k$, the human body temperature and the tumour radius, $R_{\mathrm{T}}$. Context answer: \boxed{$310.15+\frac{\mathcal{P} R_{\uparrow}^{2}}{2 k}$}","B.3 Obtain the minimum power per unit volume, $\mathcal{P}_{\min }$, needed to heat up all tumour cells in a tumour with $5.0 \mathrm{~cm}$ radius to a temperature larger than $43.0^{\circ} \mathrm{C}$. Take the thermal conductivity of the tissue to be equal to $k=0.60 \mathrm{~W} \mathrm{~K}^{-1} \mathrm{~m}^{-1}$.","['The increase in temperature at the tumour surface (the lower temperature in the tumour) is\n\n$$\n\\Delta T\\left(R_{\\mathrm{T}}\\right)=\\frac{\\mathcal{P} R_{\\mathrm{T}}^{2}}{3 k}\n$$\n\nThis increase should be equal to $6.0 \\mathrm{~K}$. Therefore,\n\n$$\n\\mathcal{P}=\\frac{3 \\Delta T k}{R_{\\top}^{2}}=\\frac{3 \\times 6 \\times 0.6}{0.05^{2}}=4.3 \\mathrm{~kW} / \\mathrm{m}^{3}\n$$\n\n']",['$\\mathcal{P}_{\\min }=4.3$'],False,$ \mathrm{~kW} / \mathrm{m}^{3}$,Numerical,1e-1 1381,Thermodynamics,"Physics of Live Systems Data: Normal atmospheric pressure, $P_{0}=1.013 \times 10^{5} \mathrm{~Pa}=760 \mathrm{mmHg}$ Part A.The physics of blood flow In this part you will analyse two simplified models of blood flow in vessels. Blood vessels are approximately cylindrical in shape, and it is known that for a steady, non turbulent flow of an incompressible fluid in a rigid cylinder, the difference in pressure of the fluid at the two ends of the cylinder is given by $$ \Delta P=\frac{8 \ell \eta}{\pi r^{4}} Q \tag{1} $$ where $\ell$ and $r$ are the length and radius of the cylinder, $\eta$ is the fluid viscosity and $Q$ is the volumetric flow rate, i.e. the fluid volume that passes the cylinder cross section per unit time. This expression is often able to provide the correct order of magnitude for the pressure difference in a vessel, even without taking into account the pulsatile flow, the vessel's compressibility and irregular shape, and the fact that blood is not a simple fluid but a mixture of cells and plasmA.Moreover, this expression has the same form as Ohm's law, with the volumetric flow rate being interpreted as a current, the difference in pressure as a voltage, and the factor $R=\frac{8 \ell \eta}{\pi r^{4}}$ as a resistance. Consider for example the symmetrical network of arterioles (small arteries) depicted in Figure 1 that delivers blood to the capillary bed of a tissue. In this network, at each bifurcation a vessel is divided in two identical vessels. However, the vessels of higher levels are thinner and shorter: consider that the radii and lengths of vessels in two consecutive levels, $i$ and $i+1$, are related by $r_{i+1}=r_{i} / 2^{1 / 3}$ and by $\ell_{i+1}=\ell_{i} / 2^{1 / 3}$. Level $\quad \begin{array}{lllllll}0 & 1 & 2 & 3 & \cdots & N-1\end{array}$ Figure 1. Network of arterioles. Context question: A.1 Obtain an expression for the volumetric flow rate, $Q_{i}$, in a vessel at any level $i$, as a function of the total number of levels $N$, of the viscosity $\eta$, of the radius $r_{0}$ and length $\ell_{0}$ of the first vessel, and of the difference $\Delta P=P_{0}-P_{\text {cap }}$ between the pressure at the arteriole at level $0, P_{0}$, and the pressure at the capillary bed, $P_{\text {cap }}$. Context answer: \boxed{$Q_{i}=\Delta P \frac{\pi r_{0}^{4}}{2^{i+3} N \ell_{0} \eta}$} Context question: A.2 Calculate the numerical value of the volumetric flow rate $Q_{0}$ of the arteriole at level 0 , if its radius is $6.0 \times 10^{-5} \mathrm{~m}$ and its length is $2.0 \times 10^{-3} \mathrm{~m}$. Consider that the pressure at the arteriole inlet is $55 \mathrm{mmHg}$ and the vessel network has $N=6$ levels linking this arteriole to the capillary bed at the pressure $30 \mathrm{mmHg}$. Consider that the blood viscosity is $\eta=3.5 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$. Express your result in $\mathrm{ml} / \mathrm{h}$. Context answer: \boxed{$1.5$} Extra Supplementary Reading Materials: A blood vessel as an LCR circuit The approximation of rigid cylindrical vessels falls short for several reasons. It is particularly important to include the time dependent flow and to take into account the change in vessel diameter that occurs when the pressure varies during a blood pumping cycle done by the heart. Moreover, it is observed that in the larger vessels the blood pressure varies significantly during a cycle, while in the smaller vessels the amplitude of the oscillations in pressure is much smaller, and the flow is almost time independent. When the pressure increases in a single elastic vessel, there will be an increase in its diameter, thus permitting to store more fluid in the vessel, and to deliver it when the pressure drops. For this reason, the elastic behaviour of the vessel can be simulated by adding a capacitor to our initial description. Moreover, when taking into account the time dependent blood flow rate, one has to consider the inertia of the fluid, proportional to its density $\rho=1.05 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. This inertia can be described by an inductance in our model. In Figure 2 we represent the equivalent circuit for a single vessel in this model. The equivalent capacitance and inductance are given by $$ C=\frac{3 \ell \pi r^{3}}{2 E h} \quad \text { and } \quad L=\frac{9 \ell \rho}{4 \pi r^{2}} \tag{2} $$ respectively, where $h$ is the width of the vessel wall and $E$ is the artery Young's modulus, a coefficient that describes the alteration in size of the vessel tissue when a force is applied. The Young's modulus has units of pressure and is on the order of $E=0.06 \mathrm{MPa}$ for arterioles. Figure 2. Equivalent electric circuit for a single vessel. Context question: A.3 Obtain, in the stationary regime, the pressure amplitude at the vessel outlet, $P_{\text {out }}$, as a function of the pressure amplitude at the inlet, $P_{\text {in }}$, the equivalent resistance, $R$, inductance, $L$ and capacitance, $C$, for a flow with angular frequency $\omega$. Establish the condition between $\eta, \rho, E, h, r$ and $\ell$ so that, for low frequencies, the pressure oscillation amplitude at the outlet is smaller than that of $P_{\text {in }}$. Context answer: $$ P_{\text {out }}=\frac{P_{\text {in }}}{\sqrt{\left(1-\omega^{2} L C\right)^{2}+\omega^{2} C^{2} R^{2}}} . $$ Condition: $$ \frac{64 \eta^{2} \ell^{2}}{3 E h r^{3} \rho}>1 $$ Context question: A.4 For the vessel network in A.2 estimate the maximum arteriole wall thickness $h$ so that the condition established in A.3 is satisfied (consider that $h$ is level independent). Context answer: \boxed{$h=8 \times 10^{-5}$} Extra Supplementary Reading Materials: Part B.Tumour growth Tumour growth is a very complex process where biological mechanisms such as cell proliferation and natural selection are intertwined with physics. In this problem we will consider a simplified model of tumour growth that addresses the increase in pressure commonly observed in solid tumors. Consider a group of normal cells forming a tissue surrounded by an inextensible basement membrane, which forces the tissue to maintain always the same form: a sphere of radius $R$ (Figure 3). Figure 3. Simplified tumour. Initially the tissue does not have residual stresses, i.e. the pressure at every point is equal to the atmospheric pressure. At time $t=0$, a tumour starts growing at the centre of this sphere and, as it grows, the pressure inside the tissue increases. Consider that both tissues (normal, N, and tumour, $\mathrm{T}$ ) are compressible such that their densities, $\rho_{\mathrm{N}}$ and $\rho_{\mathrm{T}}$, increase linearly with pressure: $$ \rho_{\mathrm{N}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{N}}}\right), \quad \rho_{\mathrm{T}}=\rho_{0}\left(1+\frac{p}{K_{\mathrm{T}}}\right) $$ where $\rho_{0}$ is the rest tissue density, $p$ is the pressure difference to the atmospheric pressure and $K_{\mathrm{N}}, K_{\mathrm{T}}$ are the compressibility moduli (bulk moduli) of the normal and tumour tissues, respectively. In general, tumours are stiffer and so they have a higher bulk modulus. Context question: B.1 The mass of normal cells is not altered while the tumour is growing. Obtain the ratio between the tumour volume and the total tissue volume, $v=V_{\mathrm{T}} / V$, as a function of the ratio between the tumour mass $\left(M_{\mathrm{T}}\right)$ and the normal tissue mass $\left(M_{\mathrm{N}}\right), \mu=M_{\mathrm{T}} / M_{\mathrm{N}}$ and the ratio of the bulk moduli, $\kappa=K_{\mathrm{N}} / K_{\mathrm{T}}$. Context answer: \boxed{$v=\frac{1+\mu-\sqrt{(1+\mu)^{2}-4 \mu(1-\kappa)}}{2(1-\kappa)}$} Extra Supplementary Reading Materials: Hyperthermia is sometimes used together with chemotherapy and radiotherapy in the treatment of cancer. In hyperthermia the cancer cells are selectively heated from the normal human body temperature, $37^{\circ} \mathrm{C}$, to temperatures above $43^{\circ} \mathrm{C}$, inducing their death. Researchers are currently developing carbon nanotubes covered with special proteins capable of binding to tumour cells. When the tissue is irradiated with near-infrared radiation, the nanotubes absorb it in a much greater extent than the surrounding tissues and therefore can be selectively heated as well as the tumour cells to which they are attached. Consider that the tumour, the normal cells and the surrounding tissue have a constant thermal conductivity $k$, i.e. in the geometry of this problem, the energy that crosses a spherical surface of radius $r$ per unit time and per unit area is equal to $k$ times the derivative of the temperature with respect to $r$. The nanotubes are uniformly distributed in the tumour volume and are able to deliver a power $\mathcal{P}$ of thermal energy per unit volume. Assume that the temperature is equal to the normal human body temperature very far away from the tumour. Context question: B.2 Obtain, for the stationary state, the temperature at the centre of the tumour as a function of $\mathcal{P}, k$, the human body temperature and the tumour radius, $R_{\mathrm{T}}$. Context answer: \boxed{$310.15+\frac{\mathcal{P} R_{\uparrow}^{2}}{2 k}$} Context question: B.3 Obtain the minimum power per unit volume, $\mathcal{P}_{\min }$, needed to heat up all tumour cells in a tumour with $5.0 \mathrm{~cm}$ radius to a temperature larger than $43.0^{\circ} \mathrm{C}$. Take the thermal conductivity of the tissue to be equal to $k=0.60 \mathrm{~W} \mathrm{~K}^{-1} \mathrm{~m}^{-1}$. Context answer: \boxed{$\mathcal{P}_{\min }=4.3$} Extra Supplementary Reading Materials: Consider that the tumour is irrigated by a vessel network with a branched structure like in question A.1. As the tumour grows, when its pressure $p$ becomes larger than the pressure $P_{\mathrm{cap}}$ at the thinnest vessels, the radii of these vessels will decrease by a small amount $\delta r$. If this pressure reaches a critical value $p_{c}$ (which would correspond to a radius decrease of $\delta r_{c}$ ), the thinnest vessels would collapse, compromising seriously the irrigation to the tumour. The pressure and the radius change can be related by the following phenomenological relation: $$ \frac{p}{P_{\text {cap }}}-1=\left(\frac{p_{\mathrm{c}}}{P_{\text {cap }}}-1\right)\left(2-\frac{\delta r}{\delta r_{\mathrm{c}}}\right) \frac{\delta r}{\delta r_{\mathrm{c}}} \tag{4} $$ Consider that just the smallest vessels (of level $N-1$ ) have their radius altered when the tumour increases its pressure.","B.4 In the linear regime (i.e. consider that $p-P_{\text {cap }}$ is very small), express the relative drop in the flow rate, $\frac{\delta Q_{N-1}}{Q_{N-1}}$, in these thinnest vessels, as a function of the tumour volume ratio $v=V_{\mathrm{T}} / V$ and $K_{N}, N, p_{\mathrm{c}}, \delta r_{\mathrm{c}}, r_{N-1}, P_{\text {cap }}$.","['We can relate $\\delta r$ with the pressure in the tumour, using the relation given in the text up to leading order in $p-P_{\\text {cap }}: \\delta r=\\frac{p-P_{\\text {cap }}}{2\\left(p_{\\mathrm{c}}-P_{\\text {cap }}\\right)} \\delta r_{\\mathrm{c}}$. Therefore, if $p-P_{\\text {cap }}$ is very small, also it is $\\delta r$.\n\nThe pressure can be related with the volume. We know that\n\n$$\n\\frac{M_{\\mathrm{N}}}{V_{\\mathrm{N}}}=\\frac{\\rho_{0} V}{V-V_{\\mathrm{T}}}=\\frac{\\rho_{0}}{1-v}=\\rho_{0}\\left(1+\\frac{p}{K_{\\mathrm{N}}}\\right) .\n$$\n\n\n\nAnd so $p=\\frac{K_{N} v}{1-v}$.\n\nWhen the thinner vessels are narrower, the flow rate in the main vessel is altered:\n\n$$\n\\begin{gathered}\n\\Delta P=\\left(Q_{0}+\\delta Q_{0}\\right) \\sum_{i=0}^{N-1} \\frac{8 \\ell_{i} \\eta}{2^{i} \\pi r_{i}^{4}}=\\left(Q_{0}+\\delta Q_{0}\\right) \\frac{8 \\ell_{0} \\eta}{\\pi r_{0}^{4}}\\left(\\sum_{i=0}^{N-2} \\frac{2^{4 i / 3}}{2^{i} 2^{i / 3}}+\\frac{2^{4(N-1) / 3}}{2^{N-1} 2^{(N-1) / 3}\\left(1-\\frac{\\delta r}{r_{0} / 2^{(N-1) / 3}}\\right)^{4}}\\right) \\\\\n\\Longrightarrow \\Delta P \\simeq\\left(Q_{0}+\\delta Q_{0}\\right) \\frac{\\Delta P}{N Q_{0}}\\left(N-1+1+\\frac{4 \\delta r}{r_{N-1}}\\right)\n\\end{gathered}\n$$\n\nNoting that $\\frac{\\delta Q_{N-1}}{Q_{N-1}}=\\frac{\\delta Q_{0}}{Q_{0}}$, we obtain\n\n$$\n1+\\frac{\\delta Q_{N-1}}{Q_{N-1}}=\\frac{1}{1+\\frac{4 \\delta r}{N r_{N-1}}} \\simeq 1-\\frac{4 \\delta r}{N r_{N-1}}\n$$\n\nAnd so:\n\n$$\n\\frac{\\delta Q_{N-1}}{Q_{N-1}} \\simeq-\\frac{4}{N} \\frac{\\delta r}{r_{N-1}}\n$$\n\nPutting all together']",['$-\\frac{2}{N} \\frac{K_{\\mathrm{N}} v-(1-v) P_{\\text {cap }}}{(1-v)\\left(p_{\\mathrm{c}}-P_{\\mathrm{cap}}\\right)} \\frac{\\delta r_{\\mathrm{c}}}{r_{N-1}}$'],False,,Expression, 1382,Mechanics,"## Toffee Pudding A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over.",a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.,"['This is when the spring force equals the maximal static friction, $k s=\\mu_{0} m g$, so $s=\\mu_{0} m g / k$.']",['$s=\\mu_{0} m g / k$'],False,,Expression, 1383,Mechanics,"## Toffee Pudding A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over. Context question: a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving. Context answer: \boxed{$s=\mu_{0} m g / k$} ","b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.","['The net stretching of the spring is $s+v_{0} t-x$, leading to a rightward force $k s$. When the box is moving, it is always moving to the right, so the kinetic friction force $\\mu m g$ is always in the leftward direction, which means\n\n\n\n$$\n\nm a=k\\left(s+v_{0} t-x\\right)-\\mu m g\n\n$$\n\n\n\nwhich means\n\n\n\n$$\n\na=\\frac{k}{m}\\left(s+v_{0} t-x\\right)-\\mu g .\n\n$$']",['$a=\\frac{k}{m}\\left(s+v_{0} t-x\\right)-\\mu g$'],False,,Expression, 1384,Mechanics,"## Toffee Pudding A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over. Context question: a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving. Context answer: \boxed{$s=\mu_{0} m g / k$} Context question: b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving. Context answer: \boxed{$a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$} Extra Supplementary Reading Materials: The position of the box as a function of time $t$ as defined in part (b) is $$ x(t)=\frac{v_{0}}{\omega}(\omega t-\sin \omega t)+(1-r) s(1-\cos \omega t), $$ where $\omega=\sqrt{k / m}$ and $r=\mu / \mu_{0}$. This expression applies as long as the box is still moving, and you can use it in the parts below. Express all your answers in terms of $v_{0}, \omega, s$, and $r$.",c. Find the time $t_{0}$ when the box stops for the first time.,"['Taking the derivative, the velocity of the box is\n\n\n\n$$\n\nv=v_{0}(1-\\cos \\omega t)+(1-r) s \\omega \\sin \\omega t .\n\n$$\n\n\n\nThe box stops when this is equal to zero for the first time. There are several ways to\n\n\n\n\n\n\n\nevaluate this condition. First, we can use half-angle identities to find\n\n\n\n$$\n\n0=2 v_{0} \\sin ^{2} \\frac{\\omega t}{2}+2(1-r) s \\omega \\sin \\frac{\\omega t}{2} \\cos \\frac{\\omega t}{2}\n\n$$\n\n\n\nAs a result, the box stops when\n\n\n\n$$\n\n\\tan \\frac{\\omega t}{2}=-\\frac{(1-r) s \\omega}{v_{0}}\n\n$$\n\n\n\nUsing a basic property of the tangent function,\n\n\n\n$$\n\n\\tan \\left(\\pi-\\frac{\\omega t}{2}\\right)=\\frac{(1-r) s \\omega}{v_{0}}\n\n$$\n\n\n\nSolving for $t$, we conclude that\n\n\n\n$$\n\nt_{0}=\\frac{2 \\pi-2 \\alpha}{\\omega}, \\quad \\alpha=\\tan ^{-1} \\frac{(1-r) s \\omega}{v_{0}}\n\n$$\n\n\n\nNote that we cancelled a factor of $\\sin (\\omega t / 2)$, which has a zero at $t=2 \\pi / \\omega$. However, this is a larger time than the one we just found, so it is irrelevant.\n\n\n\nAnother way to arrive at the answer is to rewrite the original condition as\n\n\n\n$$\n\n1=\\cos \\omega t-\\tan \\alpha \\sin \\omega t\n\n$$\n\n\n\nSquaring both sides and using some trigonometric identities gives\n\n\n\n$$\n\n\\tan \\omega t=-\\frac{2 \\tan \\alpha}{1-\\tan ^{2} \\alpha}\n\n$$\n\n\n\nThis can then be further simplified using the tangent half-angle identity, upon which we recover the same result as above.']","['$t_{0}=\\frac{2 \\pi-2 \\alpha}{\\omega}, \\quad \\alpha=\\tan ^{-1} \\frac{(1-r) s \\omega}{v_{0}}$']",True,,Expression, 1385,Mechanics,,d. For what values of $r$ will the spring always be at least as long as its rest length?,"['The spring is stretched by $\\Delta \\ell=s+v_{0} t-x$. Inserting the solution for $x$, we have\n\n\n\n$$\n\n\\Delta \\ell=r s+\\frac{v_{0}}{\\omega} \\sin \\omega t+(1-r) s \\cos \\omega t .\n\n$$\n\n\n\nThe most convenient way to write this is to use the sine addition formula in reverse, getting\n\n\n\n$$\n\n\\Delta \\ell=r s+\\frac{v_{0}}{\\omega \\cos \\alpha} \\sin (\\omega t+\\alpha)\n\n$$\n\n\n\nThe minimum stretch thus occurs when\n\n\n\n$$\n\n\\omega t+\\alpha=\\frac{3 \\pi}{2}\n\n$$\n\n\n\n\n\n\n\nOf course, we should check that this time is before the box stops; comparing with the answer to part (c) shows that it is. For the spring to always be as long as its rest length, we need the stretch at this time to be nonnegative,\n\n\n\n$$\n\nr s-\\frac{v_{0}}{\\omega \\cos \\alpha} \\geq 0\n\n$$\n\n\n\nSolving the triangle, we have\n\n\n\n$$\n\n\\cos \\alpha=\\frac{v_{0}}{\\sqrt{v_{0}^{2}+((1-r) s \\omega)^{2}}}\n\n$$\n\n\n\nPlugging this in and simplifying gives the answer,\n\n\n\n$$\n\nr \\geq \\frac{1}{2}\\left(1+\\left(\\frac{v_{0}}{s \\omega}\\right)^{2}\\right)\n\n$$\n\n\n\nNote that if $v_{0} / s \\omega$ is too large, then it is impossible to satisfy this condition, since we need to have $r<1$.']",['$r \\geq \\frac{1}{2}\\left(1+\\left(\\frac{v_{0}}{s \\omega}\\right)^{2}\\right)$'],False,,Need_human_evaluate, 1386,Mechanics,"## Toffee Pudding A box of mass $m$ is at rest on a horizontal floor. The coefficients of static and kinetic friction between the box and the floor are $\mu_{0}$ and $\mu$ (less than $\mu_{0}$ ), respectively. One end of a spring with spring constant $k$ is attached to the right side of the box, and the spring is initially held at its relaxed length. The other end of the spring is pulled horizontally to the right with constant velocity $v_{0}$. As a result, the box will move in fits and starts. Assume the box does not tip over. Context question: a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving. Context answer: \boxed{$s=\mu_{0} m g / k$} Context question: b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving. Context answer: \boxed{$a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$} Extra Supplementary Reading Materials: The position of the box as a function of time $t$ as defined in part (b) is $$ x(t)=\frac{v_{0}}{\omega}(\omega t-\sin \omega t)+(1-r) s(1-\cos \omega t), $$ where $\omega=\sqrt{k / m}$ and $r=\mu / \mu_{0}$. This expression applies as long as the box is still moving, and you can use it in the parts below. Express all your answers in terms of $v_{0}, \omega, s$, and $r$. Context question: c. Find the time $t_{0}$ when the box stops for the first time. Context answer: \boxed{$t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}$} Context question: d. For what values of $r$ will the spring always be at least as long as its rest length? Context answer: $r \geq \frac{1}{2}\left(1+\left(\frac{v_{0}}{s \omega}\right)^{2}\right)$ ","e. After the box stops, how long will it stay at rest before starting to move again?","['Using a result we found in part (d), the stretch is\n\n\n\n$$\n\n\\Delta \\ell=r s+\\frac{v_{0}}{\\omega \\cos \\alpha} \\sin \\left(\\omega t_{0}+\\alpha\\right)\n\n$$\n\n\n\nwhen the box stops. Plugging in the value of $t_{0}$ found in part (c),\n\n\n\n$$\n\n\\Delta \\ell=r s+\\frac{v_{0}}{\\omega \\cos \\alpha} \\sin (2 \\pi-\\alpha)=r s-\\frac{v_{0}}{\\omega} \\tan \\alpha=(2 r-1) s\n\n$$\n\n\n\nThe box starts to move again when the stretch becomes $s$, so the time is\n\n\n\n$$\n\n\\frac{s-(2 r-1) s}{v_{0}}=\\frac{2(1-r) s}{v_{0}}\n\n$$\n\n\n\nThe pattern of motion investigated in this problem is known as ""stick-slip"" and occurs in many practical contexts.']",['$\\frac{2(1-r) s}{v_{0}}$'],False,,Expression, 1387,Modern Physics,"## Flashlight Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame. a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.",i. What is the horizontal velocity of the muons in Bob's reference frame?,"[""The muons were fired straight down in Alice's frame, so in Bob's frame their horizontal velocity is the same as Alice's, $v=\\beta c$.\n\n\n\nOne way to see this is to imagine Alice was carrying a vertical pole with her. In her reference frame, the muons travel along the length of the pole. This must remain true in any frame, so in Bob's frame the muons must have the same horizontal velocity as Alice.""]",['$v=\\beta c$'],False,,Expression, 1388,Modern Physics,"## Flashlight Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame. a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants. Context question: i. What is the horizontal velocity of the muons in Bob's reference frame? Context answer: \boxed{$v=\beta c$} ",ii. What is the vertical velocity of the muons in Bob's reference frame?,"['The muons have speed $c$ and horizontal velocity $v=\\beta c$, so they have vertical velocity $c \\sqrt{1-\\beta^{2}}$ by the Pythagorean theorem.']",['$c \\sqrt{1-\\beta^{2}}$'],False,,Expression, 1389,Modern Physics,"## Flashlight Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame. a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants. Context question: i. What is the horizontal velocity of the muons in Bob's reference frame? Context answer: \boxed{$v=\beta c$} Context question: ii. What is the vertical velocity of the muons in Bob's reference frame? Context answer: \boxed{$c \sqrt{1-\beta^{2}}$} ",iii. How long does it take the muons to reach the ground in Bob's reference frame?,"[""The time is the height divided by the vertical velocity of the muons,\n\n\n\n$$\n\n\\Delta t=\\frac{h}{c \\sqrt{1-\\beta^{2}}}\n\n$$\n\n\n\nAlice's velocity $v$ is directed an angle $\\theta$ away from Bob. For the rest of the problem, you may additionally express your answers in terms of $\\theta$.\n\n\n\n""]",['$\\Delta t=\\frac{h}{c \\sqrt{1-\\beta^{2}}}$'],False,,Expression, 1390,Modern Physics,"## Flashlight Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame. a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants. Context question: i. What is the horizontal velocity of the muons in Bob's reference frame? Context answer: \boxed{$v=\beta c$} Context question: ii. What is the vertical velocity of the muons in Bob's reference frame? Context answer: \boxed{$c \sqrt{1-\beta^{2}}$} Context question: iii. How long does it take the muons to reach the ground in Bob's reference frame? Context answer: \boxed{$\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$} ","b. In Bob's reference frame, how much time is there between when he sees Alice first fire the beam, and when he sees the beam first hit the ground? (Hint: remember to account for the travel time of light to Bob's eyes.)","[""In Bob's frame, during the time the muons take to reach the ground, Alice's car moves a distance $\\Delta r=(v \\Delta t) \\cos \\theta$ away from Bob, which means the light from the muons hitting the ground takes an extra time $(\\Delta r) / c$ to reach Bob. Thus, the time interval Bob sees, with his eyes, is\n\n\n\n$$\n\n\\Delta t+\\frac{v \\cos \\theta}{c} \\Delta t=\\frac{h}{c} \\frac{1+\\beta \\cos \\theta}{\\sqrt{1-\\beta^{2}}}\n\n$$""]",['$\\frac{h}{c} \\frac{1+\\beta \\cos \\theta}{\\sqrt{1-\\beta^{2}}}$'],False,,Expression, 1391,Modern Physics,"## Flashlight Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame. a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants. Context question: i. What is the horizontal velocity of the muons in Bob's reference frame? Context answer: \boxed{$v=\beta c$} Context question: ii. What is the vertical velocity of the muons in Bob's reference frame? Context answer: \boxed{$c \sqrt{1-\beta^{2}}$} Context question: iii. How long does it take the muons to reach the ground in Bob's reference frame? Context answer: \boxed{$\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$} Context question: b. In Bob's reference frame, how much time is there between when he sees Alice first fire the beam, and when he sees the beam first hit the ground? (Hint: remember to account for the travel time of light to Bob's eyes.) Context answer: \boxed{$\frac{h}{c} \frac{1+\beta \cos \theta}{\sqrt{1-\beta^{2}}}$} ","c. In this part, suppose that $\beta=1 / 2$. Does there exist a value of $\theta$ so that the time it takes the muons to hit the ground in Alice's frame is equal to the time taken according to Bob's eyes, in Bob's frame? If so, find the value of $\theta$ in degrees. If not, briefly explain why not.","['This will be true if\n\n\n\n$$\n\n\\frac{h}{c \\sqrt{1-\\beta^{2}}}(1+\\beta \\cos \\theta)=\\frac{h}{c}\n\n$$\n\n\n\nSolving for $\\theta$, we find\n\n\n\n$$\n\n\\theta=\\cos ^{-1}\\left(\\sqrt{\\beta^{-2}-1}-\\beta^{-1}\\right) .\n\n$$\n\n\n\nThis has a solution as long as the argument of the inverse cosine is less than 1 . In the case $\\beta=0.5$, it is, and the result is\n\n\n\n$$\n\n\\theta=105.5^{\\circ} \\text {. }\n\n$$\n\n\n\nThat is, this value of $\\beta$ is low enough so that the motion of Alice towards Bob can make up for the time dilation effect.']",['$105.5$'],False,$^{\circ}$,Numerical,1e0 1392,Modern Physics,"## Flashlight Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame. a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants. Context question: i. What is the horizontal velocity of the muons in Bob's reference frame? Context answer: \boxed{$v=\beta c$} Context question: ii. What is the vertical velocity of the muons in Bob's reference frame? Context answer: \boxed{$c \sqrt{1-\beta^{2}}$} Context question: iii. How long does it take the muons to reach the ground in Bob's reference frame? Context answer: \boxed{$\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$} Context question: b. In Bob's reference frame, how much time is there between when he sees Alice first fire the beam, and when he sees the beam first hit the ground? (Hint: remember to account for the travel time of light to Bob's eyes.) Context answer: \boxed{$\frac{h}{c} \frac{1+\beta \cos \theta}{\sqrt{1-\beta^{2}}}$} Context question: c. In this part, suppose that $\beta=1 / 2$. Does there exist a value of $\theta$ so that the time it takes the muons to hit the ground in Alice's frame is equal to the time taken according to Bob's eyes, in Bob's frame? If so, find the value of $\theta$ in degrees. If not, briefly explain why not. Context answer: \boxed{$105.5$} ",d. Suppose Alice is carrying a radio transmitter set to frequency $f$. To what frequency would Bob have to set his radio receiver in order to receive Alice's transmission?,"[""The key point is that all the logic in part (b) still works, if we replace Bob's eyes with the radio receiver. Since the frequency of a wave is the inverse of the time between maxima,\n\n\n\n$$\n\nf^{\\prime}=\\frac{\\sqrt{1-\\beta^{2}}}{1+\\beta \\cos \\theta} f\n\n$$\n\n\n\n\n\n\n\nThis is the two-dimensional relativistic Doppler shift, and the secret point of this problem was to derive it in a simple way.""]",['$f^{\\prime}=\\frac{\\sqrt{1-\\beta^{2}}}{1+\\beta \\cos \\theta} f$'],False,,Expression, 1393,Electromagnetism,"## Electroneering An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron. Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.","a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral: $$ \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} $$","[""A point on the sphere's equator moves at a speed $\\omega R$, where $\\omega$ is the angular velocity of rotation. Setting $\\omega R=c$ gives $\\omega=c / R$.\n\n\n\nThe spinning sphere can be thought of as a stack of infinitesimal current loops, all of which have a magnetic moment pointing in the same direction. Consider making a thin, circular slice of the sphere's surface, corresponding to polar angles in the range $(\\theta, \\theta+d \\theta)$. This slice has a radius $R \\sin \\theta$, so that the surface area of the slice is\n\n\n\n$$\n\nd s=2 \\pi R \\sin \\theta R d \\theta\n\n$$\n\n\n\nThe charge of the slice is\n\n\n\n$$\n\nd Q=-\\frac{q d s}{4 \\pi R^{2}}=-\\frac{q \\sin \\theta}{2}\n\n$$\n\n\n\nSince the charge $d Q$ moves around the rotation axis one time per period $T=2 \\pi / \\omega$, the corresponding current is\n\n\n\n$$\n\nd I=\\frac{d Q}{T}=-\\frac{\\omega q \\sin \\theta}{4 \\pi}\n\n$$\n\n\n\nThe magnitude of the magnetic moment of this slice is\n\n\n\n$$\n\nd \\mu=\\pi(R \\sin \\theta)^{2}|d I|=\\frac{1}{4} q \\omega R^{2} \\sin ^{3} \\theta d \\theta\n\n$$\n\n\n\nUsing the provided integral, the total magnetic moment is\n\n\n\n$$\n\n\\mu=\\int_{0}^{\\pi} \\frac{1}{4} q \\omega R^{2} \\sin ^{3} \\theta d \\theta=\\frac{1}{3} q c R\n\n$$\n\n\n\nIf you weren't able to do this, you could also have given the answer $\\mu \\sim q c R$, which can be derived by dimensional analysis, for partial credit."", 'Note that for a uniformly charged ring of mass $d m$, charge $d q$, and radius $r$, rotating with angular velocity $\\omega$, the ratio of the magnetic moment and the\n\n\n\n\n\n\n\nangular momentum is\n\n\n\n$$\n\n\\frac{\\mu}{L}=\\frac{\\pi r^{2}(\\omega d q / 2 \\pi)}{\\left(r^{2} d m\\right) \\omega}=\\frac{1}{2} \\frac{d q}{d m}\n\n$$\n\n\n\nThe ratio is independent of $r$ and $\\omega$. Since the sphere can be decomposed into such rings, the total magnetic moment and total angular momentum must have the same ratio,\n\n\n\n$$\n\n\\frac{\\mu}{L}=\\frac{1}{2} \\frac{q}{m}\n\n$$\n\n\n\nFinally, we know that $L=(2 / 3) m R^{2} \\omega$ for a spherical shell. Plugging this in and using $\\omega=c / R$ gives $\\mu=q c R / 3$ as before, but with no integration required.']",['$\\mu=q c R / 3$'],False,,Expression, 1394,Electromagnetism,"## Electroneering An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron. Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center. Context question: a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral: $$ \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} $$ Context answer: \boxed{$\mu=q c R / 3$} ","b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants.","['Since the magnetic moment is fixed, and we want the radius to be small, we want the electron to be spinning as fast as possible. Thus, the magnetic moment has the value found in part (a), and equating this to the known value gives\n\n\n\n$$\n\nR=\\frac{3}{2} \\frac{\\hbar}{m c}\n\n$$\n\n\n\nAgain, you can get $R \\sim \\hbar / m c$ by dimensional analysis.']",['$R=\\frac{3}{2} \\frac{\\hbar}{m c}$'],False,,Expression, 1395,Electromagnetism,"## Electroneering An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron. Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center. Context question: a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral: $$ \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} $$ Context answer: \boxed{$\mu=q c R / 3$} Context question: b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants. Context answer: \boxed{$R=\frac{3}{2} \frac{\hbar}{m c}$} ","c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric field of the electron? Express your answer in terms of $E_{0}=m c^{2}$ and the fine structure constant, $$ \alpha=\frac{q^{2}}{4 \pi \epsilon_{0} \hbar c} \approx \frac{1}{137} $$","[""For a collection of charges, the total energy stored in the electrostatic field is\n\n\n\n$$\n\nU_{E}=\\frac{1}{2} \\sum_{i} q_{i} V_{i}\n\n$$\n\n\n\nwhere $V_{i}$ is the electric potential at $q_{i}$. In this case, the total charge is $q$, and all of the charge is at potential $q / 4 \\pi \\epsilon_{0} R$, so\n\n\n\n$$\n\nU_{E}=\\frac{q^{2}}{8 \\pi \\epsilon_{0} R}\n\n$$\n\n\n\nUsing the result of part (b),\n\n\n\n$$\n\nU_{E}=\\frac{1}{3} \\alpha E_{0}\n\n$$\n\n\n\nNote that you can't get this answer by dimensional analysis alone, since $\\alpha$ is dimensionless. (However, if you found $R$ by dimensional analysis, and additionally reasoned that $U_{E}$ could\ndepend only on $q, \\epsilon_{0}$, and $R$, then you could derive $U_{E} \\sim \\alpha E_{0}$, for partial credit.)""]",['$U_{E}=\\frac{1}{3} \\alpha E_{0}$'],False,,Expression, 1396,Electromagnetism,"## Electroneering An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron. Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center. Context question: a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral: $$ \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} $$ Context answer: \boxed{$\mu=q c R / 3$} Context question: b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants. Context answer: \boxed{$R=\frac{3}{2} \frac{\hbar}{m c}$} Context question: c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric field of the electron? Express your answer in terms of $E_{0}=m c^{2}$ and the fine structure constant, $$ \alpha=\frac{q^{2}}{4 \pi \epsilon_{0} \hbar c} \approx \frac{1}{137} $$ Context answer: \boxed{$U_{E}=\frac{1}{3} \alpha E_{0}$} ","d. Roughly estimate the total energy stored in the magnetic field of the electron, in terms of $E_{0}$ and $\alpha$. (Hint: one way to do this is to suppose the magnetic field has roughly constant magnitude inside the sphere and is negligible outside of it, then estimate the field inside the sphere.)","['Following the hint, we can estimate\n\n\n\n$$\n\nU_{B} \\sim \\frac{B_{0}^{2}}{2 \\mu_{0}}\\left(\\frac{4}{3} \\pi R^{3}\\right)\n\n$$\n\n\n\nwhere $B_{0}$ is the typical magnetic field inside the sphere. Actually finding the value of $B_{0}$ would require doing some complicated integrals. To get a rough estimate, note that if we replaced the sphere with a ring of charge, then at the center of the ring,\n\n\n\n$$\n\nB_{0} \\sim \\frac{\\mu_{0} I}{R} \\sim \\frac{\\mu_{0} q c}{R^{2}}\n\n$$\n\n\n\nThus, we have\n\n\n\n$$\n\nU_{B} \\sim \\frac{1}{\\mu_{0}}\\left(\\frac{\\mu_{0} q c}{R^{2}}\\right)^{2} R^{3} \\sim \\frac{\\mu_{0} q^{2} c^{2}}{R} \\sim \\frac{\\mu_{0} m c^{3} q^{2}}{\\hbar}\n\n$$\n\n\n\nTo get this in terms of the fine structure constant, we use $c^{2}=1 / \\mu_{0} \\epsilon_{0}$, giving\n\n\n\n$$\n\nU_{B} \\sim m c^{2} \\frac{q^{2}}{\\epsilon_{0} \\hbar c} \\sim \\alpha E_{0}\n\n$$\n\n\n\nAn even faster way to get this result is to note that in general, the energy stored in magnetic fields tends to be a factor of order $(v / c)^{2}$ smaller than the energy stored in electric fields, where $v$ is the speed of the charge. In this problem the charge is all moving relativistically, so we must have $U_{B} \\sim U_{E}$.']",['$\\alpha E_{0}$'],False,,Expression, 1397,Electromagnetism,,e. How does your estimate for the total energy in the electric and magnetic fields compare to $E_{0}$ ?,"[""Both $U_{E}$ and $U_{B}$ are much smaller than $E_{0}$, by a factor of $\\alpha \\ll 1$. Thus, this classical model cannot explain the origin of the electron's rest energy.\n\n\n\nThere were many attempts to make classical models of the electron in the early $20^{\\text {th }}$ century, but they all ran into difficulties like this one. For more on this subject, see chapter II-28 of the Feynman lectures.""]","[""Both $U_{E}$ and $U_{B}$ are much smaller than $E_{0}$, by a factor of $\\alpha \\ll 1$. Thus, this classical model cannot explain the origin of the electron's rest energy.""]",True,,Need_human_evaluate, 1398,Mechanics,,"a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$. ![](https://cdn.mathpix.com/cropped/2023_12_21_d0fccf1ebe41035d3cadg-1.jpg?height=474&width=491&top_left_y=907&top_left_x=833) Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ?","['During the first leg of the trip, the disk has angular velocity\n\n\n\n$$\n\n\\omega=-\\frac{2 m v}{M R}\n\n$$\n\n\n\nby conservation of angular momentum. Thus, the disk rotates through an angle\n\n\n\n$$\n\n\\phi_{1}=-\\frac{2 m v}{M R} \\frac{\\theta R}{v}=-\\frac{2 m \\theta}{M}\n\n$$\n\n\n\nto leading order in $m / M$. (Here we have neglected the fact that the disk rotates under the ant as it is walking, somewhat reducing the distance it has to walk; this changes the answer only to higher order in $m / M$. The exact answer is a more complicated function of $m / M$. By going to ""leading order"", we mean we are expanding that exact answer in a series in $m / M$, such as with the binomial theorem, and keeping only the first nonzero term.)\n\n\n\nWhen the ant is moving radially, $\\omega=0$, so no rotation occurs. On the last leg of the trip, the disk has angular velocity\n\n\n\n$$\n\n\\omega=\\frac{2 m v(R-h)}{M R^{2}}\n\n$$\n\n\n\n\n\nwhich means the disk rotates through an angle\n\n\n\n$$\n\n\\phi_{2}=\\frac{2 m v(R-h)}{M R^{2}} \\frac{\\theta(R-h)}{v}=\\frac{2 m \\theta}{M}\\left(1-\\frac{h}{R}\\right)^{2}\n\n$$\n\n\n\nThe net rotation is\n\n\n\n$$\n\n\\phi_{1}+\\phi_{2}=\\frac{2 m \\theta}{M}\\left(\\left(1-\\frac{h}{R}\\right)^{2}-1\\right) \\approx-\\frac{4 m}{M} \\frac{h \\theta}{R}\n\n$$\n\n\n\nThe sign is not important since it is convention-dependent. (Solutions that were not fully approximated were also accepted; however, not approximating early dramatically increases the amount of work you have to do.)\n\n\n\nIncidentally, you might have thought the answer had to be zero, by angular momentum conservation. After all, when a system has zero total linear momentum, its center of mass can\'t move. But this problem shows that systems with zero total angular momentum can perform net rotations, which is the reason, e.g. that a falling cat can always land on its feet. In more advanced physics, this would be described by saying the constraint on the disc\'s motion coming from angular momentum conservation is not holonomic.']","['$\\frac{2 m \\theta}{M}\\left(\\left(1-\\frac{h}{R}\\right)^{2}-1\\right)$', '$-\\frac{4 m}{M} \\frac{h \\theta}{R}$']",False,,Expression, 1398,Mechanics,"## Disk Jockey A disk of uniform mass density, mass $M$, and radius $R$ sits at rest on a frictionless floor. The disk is attached to the floor by a frictionless pivot at its center, which keeps the center of the disk in place, but allows the disk to rotate freely. An ant of mass $m \ll M$ is initially standing on the edge of the disk; you may give your answers to leading order in $m / M$.","a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$. Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ?","['During the first leg of the trip, the disk has angular velocity\n\n\n\n$$\n\n\\omega=-\\frac{2 m v}{M R}\n\n$$\n\n\n\nby conservation of angular momentum. Thus, the disk rotates through an angle\n\n\n\n$$\n\n\\phi_{1}=-\\frac{2 m v}{M R} \\frac{\\theta R}{v}=-\\frac{2 m \\theta}{M}\n\n$$\n\n\n\nto leading order in $m / M$. (Here we have neglected the fact that the disk rotates under the ant as it is walking, somewhat reducing the distance it has to walk; this changes the answer only to higher order in $m / M$. The exact answer is a more complicated function of $m / M$. By going to ""leading order"", we mean we are expanding that exact answer in a series in $m / M$, such as with the binomial theorem, and keeping only the first nonzero term.)\n\n\n\nWhen the ant is moving radially, $\\omega=0$, so no rotation occurs. On the last leg of the trip, the disk has angular velocity\n\n\n\n$$\n\n\\omega=\\frac{2 m v(R-h)}{M R^{2}}\n\n$$\n\n\n\n\n\nwhich means the disk rotates through an angle\n\n\n\n$$\n\n\\phi_{2}=\\frac{2 m v(R-h)}{M R^{2}} \\frac{\\theta(R-h)}{v}=\\frac{2 m \\theta}{M}\\left(1-\\frac{h}{R}\\right)^{2}\n\n$$\n\n\n\nThe net rotation is\n\n\n\n$$\n\n\\phi_{1}+\\phi_{2}=\\frac{2 m \\theta}{M}\\left(\\left(1-\\frac{h}{R}\\right)^{2}-1\\right) \\approx-\\frac{4 m}{M} \\frac{h \\theta}{R}\n\n$$\n\n\n\nThe sign is not important since it is convention-dependent. (Solutions that were not fully approximated were also accepted; however, not approximating early dramatically increases the amount of work you have to do.)\n\n\n\nIncidentally, you might have thought the answer had to be zero, by angular momentum conservation. After all, when a system has zero total linear momentum, its center of mass can\'t move. But this problem shows that systems with zero total angular momentum can perform net rotations, which is the reason, e.g. that a falling cat can always land on its feet. In more advanced physics, this would be described by saying the constraint on the disc\'s motion coming from angular momentum conservation is not holonomic.']","['$\\frac{2 m \\theta}{M}\\left(\\left(1-\\frac{h}{R}\\right)^{2}-1\\right)$', '$-\\frac{4 m}{M} \\frac{h \\theta}{R}$']",False,,Expression, 1399,Mechanics,,"b. Now suppose the ant walks with speed $v$ along a circle of radius $r$, tangent to its starting point. ![](https://cdn.mathpix.com/cropped/2023_12_21_6172b297d6dc286a9e8fg-1.jpg?height=480&width=491&top_left_y=1278&top_left_x=836) Through what net angle does the disk rotate?","[""First solution: The overall rotation angle of the disk is\n\n\n\n$$\n\n\\phi=\\int \\omega d t=\\frac{2}{M R^{2}} \\int L d t=\\frac{2 m}{M R^{2}} \\int \\mathbf{r} \\times \\mathbf{v} d t\n\n$$\n\n\n\nwhere we again work to leading order in $m / M$, and $\\mathbf{r}$ and $\\mathbf{v}$ are the position and velocity of the ant. The coordinates of a point on the circle are given by\n\n\n\n$$\n\n(r \\sin \\theta,(R-r)+r \\cos \\theta)\n\n$$\n\n\n\n\n\n\n\nIf the speed of the ant is $v$, the velocity is given by\n\n\n\n$$\n\n(v \\cos \\theta,-v \\sin \\theta)\n\n$$\n\n\n\nTo evaluate the angular momentum, note that\n\n\n\n$$\n\n|\\mathbf{v} \\times \\mathbf{r}|=v r \\cos ^{2} \\theta+v r \\sin ^{2} \\theta+v(R-r) \\cos \\theta=v r+v(R-r) \\cos \\theta\n\n$$\n\n\n\nThus, we have\n\n\n\n$$\n\nL=m v(r+(R-r) \\cos \\theta)\n\n$$\n\n\n\nPlugging this into the time integral above,\n\n\n\n$$\n\n\\phi=\\int \\frac{2 m v}{M R^{2}}(r+(R-r) \\cos (\\theta(t))) \\mathrm{d} t\n\n$$\n\n\n\nChanging this to an integral over $\\theta$ using $d \\theta=v d t / r$,\n\n\n\n$$\n\n\\phi=\\int_{0}^{2 \\pi} \\frac{2 m r}{M R^{2}}(r+(R-r) \\cos (\\theta(t))) \\mathrm{d} \\theta=\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}\n\n$$\n\n\n\nOf course, the problem could also be solved by parameterizing the ant's path in a different way, such as by using polar coordinates with the origin at the center of the disk. The way we set it up here is the simplest, since it makes the integral easy. (For most students, the hardest part was finding a compact expression for $L$. A common mistake was assuming $L=m v_{x} r$ or a variant thereof.)"", ""The net effect on the disk of one ant going in the circular path is the same as two ants going along the path, each with half the mass, and thus the same as four ants each with a quarter the mass, and so on. By repeating this logic, we can thus replace the ant with a ring of radius $r$ and mass $m$ of uniform density, which rotates around once. Therefore, the rotation angle is\n\n\n\n$$\n\n\\phi=2 \\pi \\frac{I_{\\mathrm{disk}}}{I_{\\text {ring }}}=2 \\pi \\frac{m r^{2}}{M R^{2} / 2}=\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}\n\n$$\n\n\n\nThis is very simple, though it's a trick that only works for a circular trajectory."", ""Starting from the first line of the first solution, we notice that\n\n\n\n$$\n\n\\int \\mathbf{r} \\times \\mathbf{v} d t=\\int \\mathbf{r} \\times d \\mathbf{r}=2 A\n\n$$\n\n\n\nwhere $A$ is the area of the ant's trajectory. Thus, we have\n\n\n\n$$\n\n\\phi=\\frac{4 m}{M} \\frac{A}{R^{2}}=\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}\n\n$$\n\n\n\nThis makes it clear why the answer had to be simple in general: the angle can only depend on a geometric property of the ant's trajectory, namely its area. This kind of phenomenon occurs in many fields of physics, and is generally known as a geometric phase."", 'We can decompose the circle into a stack of thin rectangles. The effect of a single ant going around the circle is the same as the effect of one ant going around each rectangle. But by slightly generalizing your result in part (a), you can show that the net rotation due to each rectangle is $(4 m / M)\\left(d A / R^{2}\\right)$ where $d A$ is the area of that rectangle. Summing the areas gives the answer. Like the third solution, this works for any ant trajectory, and it makes it clear why it was the area of the trajectory that mattered.']",['$\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}$'],False,,Expression, 1399,Mechanics,"## Disk Jockey A disk of uniform mass density, mass $M$, and radius $R$ sits at rest on a frictionless floor. The disk is attached to the floor by a frictionless pivot at its center, which keeps the center of the disk in place, but allows the disk to rotate freely. An ant of mass $m \ll M$ is initially standing on the edge of the disk; you may give your answers to leading order in $m / M$. Context question: a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$. Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ? Context answer: \boxed{$\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right)$} ","b. Now suppose the ant walks with speed $v$ along a circle of radius $r$, tangent to its starting point. Through what net angle does the disk rotate?","[""First solution: The overall rotation angle of the disk is\n\n\n\n$$\n\n\\phi=\\int \\omega d t=\\frac{2}{M R^{2}} \\int L d t=\\frac{2 m}{M R^{2}} \\int \\mathbf{r} \\times \\mathbf{v} d t\n\n$$\n\n\n\nwhere we again work to leading order in $m / M$, and $\\mathbf{r}$ and $\\mathbf{v}$ are the position and velocity of the ant. The coordinates of a point on the circle are given by\n\n\n\n$$\n\n(r \\sin \\theta,(R-r)+r \\cos \\theta)\n\n$$\n\n\n\n\n\n\n\nIf the speed of the ant is $v$, the velocity is given by\n\n\n\n$$\n\n(v \\cos \\theta,-v \\sin \\theta)\n\n$$\n\n\n\nTo evaluate the angular momentum, note that\n\n\n\n$$\n\n|\\mathbf{v} \\times \\mathbf{r}|=v r \\cos ^{2} \\theta+v r \\sin ^{2} \\theta+v(R-r) \\cos \\theta=v r+v(R-r) \\cos \\theta\n\n$$\n\n\n\nThus, we have\n\n\n\n$$\n\nL=m v(r+(R-r) \\cos \\theta)\n\n$$\n\n\n\nPlugging this into the time integral above,\n\n\n\n$$\n\n\\phi=\\int \\frac{2 m v}{M R^{2}}(r+(R-r) \\cos (\\theta(t))) \\mathrm{d} t\n\n$$\n\n\n\nChanging this to an integral over $\\theta$ using $d \\theta=v d t / r$,\n\n\n\n$$\n\n\\phi=\\int_{0}^{2 \\pi} \\frac{2 m r}{M R^{2}}(r+(R-r) \\cos (\\theta(t))) \\mathrm{d} \\theta=\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}\n\n$$\n\n\n\nOf course, the problem could also be solved by parameterizing the ant's path in a different way, such as by using polar coordinates with the origin at the center of the disk. The way we set it up here is the simplest, since it makes the integral easy. (For most students, the hardest part was finding a compact expression for $L$. A common mistake was assuming $L=m v_{x} r$ or a variant thereof.)"", ""The net effect on the disk of one ant going in the circular path is the same as two ants going along the path, each with half the mass, and thus the same as four ants each with a quarter the mass, and so on. By repeating this logic, we can thus replace the ant with a ring of radius $r$ and mass $m$ of uniform density, which rotates around once. Therefore, the rotation angle is\n\n\n\n$$\n\n\\phi=2 \\pi \\frac{I_{\\mathrm{disk}}}{I_{\\text {ring }}}=2 \\pi \\frac{m r^{2}}{M R^{2} / 2}=\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}\n\n$$\n\n\n\nThis is very simple, though it's a trick that only works for a circular trajectory."", ""Starting from the first line of the first solution, we notice that\n\n\n\n$$\n\n\\int \\mathbf{r} \\times \\mathbf{v} d t=\\int \\mathbf{r} \\times d \\mathbf{r}=2 A\n\n$$\n\n\n\nwhere $A$ is the area of the ant's trajectory. Thus, we have\n\n\n\n$$\n\n\\phi=\\frac{4 m}{M} \\frac{A}{R^{2}}=\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}\n\n$$\n\n\n\nThis makes it clear why the answer had to be simple in general: the angle can only depend on a geometric property of the ant's trajectory, namely its area. This kind of phenomenon occurs in many fields of physics, and is generally known as a geometric phase."", 'We can decompose the circle into a stack of thin rectangles. The effect of a single ant going around the circle is the same as the effect of one ant going around each rectangle. But by slightly generalizing your result in part (a), you can show that the net rotation due to each rectangle is $(4 m / M)\\left(d A / R^{2}\\right)$ where $d A$ is the area of that rectangle. Summing the areas gives the answer. Like the third solution, this works for any ant trajectory, and it makes it clear why it was the area of the trajectory that mattered.']",['$\\frac{4 m}{M} \\frac{\\pi r^{2}}{R^{2}}$'],False,,Expression, 1400,Thermodynamics,"## Hot Pocket a. It's winter and you want to keep warm. The temperature is $T_{0}=263 \mathrm{~K}$ outside and $T_{1}=290 \mathrm{~K}$ in your room. You have started a fire, which acts as a hot reservoir at temperature $T_{2}=1800 \mathrm{~K}$. You want to add a small amount of heat $d Q_{1}$ to your room. The simplest method would be to extract heat $-d Q_{2, \mathrm{dump}}=d Q_{1}$ from the fire and directly transfer it to your room. However, it is possible to heat your room more efficiently. Suppose that you can exchange heat between any pair of reservoirs. You cannot use any external source of work, such as the electrical grid, but the work extracted from running heat engines can be stored and used without dissipation.","i. What is the minimum heat extraction $-d Q_{2, \text { min }}$ required by the laws of thermodynamics to heat up the room by $d Q_{1}$ ?","['The second law of thermodynamics implies that, no matter what you do, you must have $\\mathrm{d} S_{\\text {universe }} \\geq 0$, and if your process is to be as efficient as possible, we can assume it is reversible, so\n\n\n\n$$\n\n\\mathrm{d} S_{\\text {universe; reversible }}=0\n\n$$\n\n\n\nIf we do extract any work while allowing heat to transfer between reservoirs, we will later use that work to transfer more heat. So in the entire process, there are only heat transfers, and by conservation of energy,\n\n\n\n$$\n\n\\mathrm{d} Q_{0}+\\mathrm{d} Q_{1}+\\mathrm{d} Q_{2}=0\n\n$$\n\n\n\nThe entropy change associated with each reversible heat transfer is $d S=d Q / T$, so our assumption of zero entropy production becomes\n\n\n\n$$\n\n\\frac{\\mathrm{d} Q_{0}}{T_{0}}+\\frac{\\mathrm{d} Q_{1}}{T_{1}}+\\frac{\\mathrm{d} Q_{2}}{T_{2}}=0\n\n$$\n\n\n\nBy combining these equations, we can eliminate $d Q_{0}$ and solve for $d Q_{2}$, giving\n\n\n\n$$\n\n-\\mathrm{d} Q_{2 ; \\min }=\\frac{T_{2}}{T_{1}} \\frac{T_{1}-T_{0}}{T_{2}-T_{0}} \\mathrm{~d} Q_{1}\n\n$$\n\n\n\nFor the provided numbers, this happens to be about $0.11 d Q_{1}$. That is, a heat pump can be much more efficient than direct heating. This problem was inspired by Jaynes, E. T, ""Note on thermal heating efficiency."", American Journal of Physics 71.2 (2003): 180-182. (You can also solve the problem by considering an explicit procedure using Carnot engines. But since Carnot engines are reversible, all such procedures will just give the same answer.)']",['$\\frac{T_{2}}{T_{1}} \\frac{T_{1}-T_{0}}{T_{2}-T_{0}} \\mathrm{~d} Q_{1}$'],False,,Expression, 1401,Thermodynamics,"b. When the air at the bottom of a container is heated, it becomes less dense than the surrounding air and rises. Simultaneously, cooler air falls downward. This process of net upward heat transfer is known as convection. Consider a closed, rectangular box of height $h$ filled with air initially of uniform temperature $T_{0}$. Next, suppose the bottom of the box is heated so that the air there instantly reaches temperature $T_{0}+\Delta T$. The hot parcel of air at the bottom rises upward until it hits the top of the box, where its temperature is instantly reduced to $T_{0}$. You may neglect any heat transfer and friction between the parcel of air and the surrounding air, and assume that the temperature difference is not too large. In addition, you may assume the height $h$ is small enough so that the pressure $P_{0}$ and density $\rho_{0}$ of the surrounding air are very nearly constant throughout the container. More precisely, assume that $\rho_{0} g h / P_{0} \ll \Delta T / T_{0} \ll 1$. Express your answers in terms of $P_{0}, g, h, \Delta T$, and $T_{0}$.","i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion.","['The temperature of the air is higher than its surroundings by a fractional amount of order $\\Delta T / T$. Thus, by the ideal gas law, the density is lower than its surroundings by a fraction of order $\\Delta T / T$, which means the upward acceleration due to the buoyant force is of order $a=g \\Delta T / T$. Since this is roughly uniformly accelerated motion, $v_{0}^{2} \\propto a h$,\n\n\n\n\n\n\n\nwhich implies\n\n\n\n$$\n\nv_{0} \\sim \\sqrt{g h \\frac{\\Delta T}{T_{0}}}\n\n$$\n\n\n\nNote that because $d P / d z=-\\rho g$ in hydrostatic equilibrium, the pressure of the surrounding air varies between the bottom and top of the container, by a fractional amount of order $\\rho_{0} g h / P_{0}$. But since we assumed $\\rho_{0} g h / P_{0} \\ll \\Delta T / T_{0}$, we can neglect this effect.']",['$\\sqrt{g h \\frac{\\Delta T}{T_{0}}}$'],False,,Expression, 1402,Thermodynamics,"b. When the air at the bottom of a container is heated, it becomes less dense than the surrounding air and rises. Simultaneously, cooler air falls downward. This process of net upward heat transfer is known as convection. Consider a closed, rectangular box of height $h$ filled with air initially of uniform temperature $T_{0}$. Next, suppose the bottom of the box is heated so that the air there instantly reaches temperature $T_{0}+\Delta T$. The hot parcel of air at the bottom rises upward until it hits the top of the box, where its temperature is instantly reduced to $T_{0}$. You may neglect any heat transfer and friction between the parcel of air and the surrounding air, and assume that the temperature difference is not too large. In addition, you may assume the height $h$ is small enough so that the pressure $P_{0}$ and density $\rho_{0}$ of the surrounding air are very nearly constant throughout the container. More precisely, assume that $\rho_{0} g h / P_{0} \ll \Delta T / T_{0} \ll 1$. Express your answers in terms of $P_{0}, g, h, \Delta T$, and $T_{0}$. Context question: i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion. Context answer: \boxed{$\sqrt{g h \frac{\Delta T}{T_{0}}}$} ","ii. In the steady state, warm parcels of air are continuously moving upward from the bottom, and cold parcels of air are continuously moving downward from the top. Find a rough estimate for the net rate of upward energy transfer per area.","['The extra energy carried by a parcel of gas is\n\n\n\n$$\n\nn C_{p} \\Delta T \\sim n R \\Delta T \\sim P_{0} V \\frac{\\Delta T}{T_{0}}\n\n$$\n\n\n\nwhere $V$ is the volume of the parcel. The net volume of warm air transported upward per unit time is of order $A v_{0}$, where $A$ is the cross-sectional area of the box. Thus, the average power per area is roughly\n\n\n\n$$\n\nP_{0} v_{0} \\frac{\\Delta T}{T_{0}} \\sim P_{0} \\sqrt{g h}\\left(\\frac{\\Delta T}{T_{0}}\\right)^{3 / 2}\n\n$$\n\n\n\nThis is a simplified version of the mixing length theory of convection, which is essential for modeling the interiors of stars.']","['$P_{0} v_{0} \\frac{\\Delta T}{T_{0}}$', '$ P_{0} \\sqrt{g h}\\left(\\frac{\\Delta T}{T_{0}}\\right)^{3 / 2}$']",False,,Expression, 1403,Mechanics,,"a. The number and nature of the equilibrium points on the hemisphere depends on the value of the spring constant $k$. Consider the semicircular arc shown above as a dashed line, which is parameterized by angles in the range $-\pi / 2 \leq \theta \leq \pi / 2$. Make a table indicating the number of equilibrium points on the arc, and the number that are stable, for each range of $k$ values. A blank table for your reference is given below. (You may need more or fewer rows than shown.) | Range of $k\left(k_{\min }\\theta^{\\star}$, and the mass is again forced toward $\\theta^{\\star}$. Thus, we have three total equilibria (two stable, one unstable).\n\n\n\nFinally, for sufficiently large $k$ the spring force is stronger than the gravitational force even for arbitrarily small $\\theta$, and so we have just one stable equilibrium point at the top of the hemisphere in this third regime.\n\n\n\nWe now compute the two critical points for us to fill out the table. The first critical point\n\n\n\n\n\n\n\nis given by the balancing of the two forces at $\\theta=\\pi / 2$, so we compute\n\n\n\n$$\n\nm g \\sin \\pi / 2=k_{1} R(\\pi / 2) \\Longrightarrow k_{1}=\\frac{2 m g}{\\pi R}\n\n$$\n\n\n\nThe second critical point is given by the balancing of the two forces for $\\theta \\ll 1$, so we compute\n\n\n\n$$\n\nm g \\sin \\theta \\approx m g \\theta=k_{2} R \\theta \\Longrightarrow k_{2}=m g / R\n\n$$\n\n\n\nWe thus get the following table:\n\n\n\n| Range of $k$ values $\\left(k_{\\min } Context question: a. The number and nature of the equilibrium points on the hemisphere depends on the value of the spring constant $k$. Consider the semicircular arc shown above as a dashed line, which is parameterized by angles in the range $-\pi / 2 \leq \theta \leq \pi / 2$. Make a table indicating the number of equilibrium points on the arc, and the number that are stable, for each range of $k$ values. A blank table for your reference is given below. (You may need more or fewer rows than shown.) | Range of $k\left(k_{\min } Context question: i. Indicate which of the following trajectories the mass takes for a short time after $t=0$ and briefly explain your reasoning. The differences between the paths are exaggerated. Context answer: Several explanations would work here. Here are two. - If we go to the rotating frame of reference, there is an outward centrifugal force that the mass experiences, pushing it down the sphere. - For the mass to go in a circle around the sphere, the spring force not only has to compensate for the gravitational force but also must provide centripetal acceleration. Therefore, the spring must get longer. ","ii. What is the total radial force (i.e., normal to the surface of the hemisphere) on the mass at $t=0$ ? Express your answer in terms of $m, v, R, g$, and $\theta_{0}$.","['We draw a free-body diagram. It is helpful to draw the diagram in the noninertial reference frame that revolves around the central axis of the hemisphere with speed $v$ at the location of the mass.\n\n\n\n\n\n\n\nHere, $F_{c}=\\frac{m v^{2}}{R \\sin \\theta_{0}}$ is the centrifugal force, and $F_{r}$ is the radial force from the hemisphere. The forces in the radial direction must balance for the mass to be constrained to the surface of the sphere. Thus,\n\n\n\n$$\n\nF_{r}+F_{c}^{\\perp}=m g \\cos \\theta_{0}\n\n$$\n\n\n\nThe perpendicular part of the centrifugal force is $F_{c} \\sin \\theta_{0}$. so we get\n\n\n\n$$\n\nF_{r}=m g \\cos \\theta_{0}-\\frac{m v^{2}}{R}\n\n$$\n\n\n\nIncidentally, there\'s a simple way to understand why the second term has to be exactly $m v^{2} / R$. Consider decomposing the total force on the mass into radial and tangential parts. The radial part simply keeps the mass on the hemisphere; in the absence of a tangential force, the mass would travel in a great circle of radius $R$. Adding a tangential force deflects the mass away from this great circle trajectory, but doesn\'t change the radial force required, so the net radial force always has to be $m v^{2} / R$ inward.\n\n\n\nNote: The phrase ""total radial force"" could also validly be interpreted as the net radial force. Thus, we accepted both $m g \\cos \\theta_{0}-m v^{2} / R$ and $-m v^{2} / R$.']",['$F_{r}=m g \\cos \\theta_{0}-\\frac{m v^{2}}{R}$'],False,,Expression, 1406,Mechanics,"## The Mad Hatter A frictionless hemisphere of radius $R$ is fixed on top of a flat cylinder. One end of a spring with zero relaxed length and spring constant $k$ (i.e. the force from the spring when stretched to length $\ell$ is $-k \ell$ ) is fixed to the top of the hemisphere. Its other end is attached to a point mass of mass $m$. Context question: a. The number and nature of the equilibrium points on the hemisphere depends on the value of the spring constant $k$. Consider the semicircular arc shown above as a dashed line, which is parameterized by angles in the range $-\pi / 2 \leq \theta \leq \pi / 2$. Make a table indicating the number of equilibrium points on the arc, and the number that are stable, for each range of $k$ values. A blank table for your reference is given below. (You may need more or fewer rows than shown.) | Range of $k\left(k_{\min } Context question: i. Indicate which of the following trajectories the mass takes for a short time after $t=0$ and briefly explain your reasoning. The differences between the paths are exaggerated. Context answer: Several explanations would work here. Here are two. - If we go to the rotating frame of reference, there is an outward centrifugal force that the mass experiences, pushing it down the sphere. - For the mass to go in a circle around the sphere, the spring force not only has to compensate for the gravitational force but also must provide centripetal acceleration. Therefore, the spring must get longer. Context question: ii. What is the total radial force (i.e., normal to the surface of the hemisphere) on the mass at $t=0$ ? Express your answer in terms of $m, v, R, g$, and $\theta_{0}$. Context answer: \boxed{$F_{r}=m g \cos \theta_{0}-\frac{m v^{2}}{R}$} ","c. A cylinder of radius $r \ll R \theta_{0}$ is placed on top of the sphere. Suppose the mass is launched at an angle $\alpha$ away from the direction of the spring's displacement with kinetic energy $K$, as shown. What is the maximum angle $\alpha_{\max }$ at which the mass can be launched such that it can still hit the cylinder? Express your answer in terms of $K, m, g, \theta_{0}, r$, and $R$. You may assume $K$ is large enough for the mass to reach the cylinder for $\alpha=0$. (view from above)","['The initial energy of the system is given by\n\n\n\n$$\n\nm g R \\cos \\theta_{0}+\\frac{1}{2} k R^{2} \\theta_{0}^{2}+K\n\n$$\n\n\n\nSuppose the mass is launched with speed $v$. Then, the speed in the $\\theta$ direction is $v \\cos \\alpha$ and the speed in the $\\phi$ direction is $v \\sin \\alpha$, and therefore, the $z$-component of the angular momentum of the mass is\n\n\n\n$$\n\nL=m v \\sin \\alpha(R \\sin \\theta)\n\n$$\n\n\n\nWe now compute the distance of closest approach. If the distance of closest approach is equal to $r$ (as it does for $\\alpha_{\\max }$, then at $r$, the motion of the mass has no inward component, and the speed of the object at $r$ is given by conservation of angular momentum:\n\n\n\n$$\n\nm u r=m v \\sin \\alpha(R \\sin \\theta) \\Longrightarrow u=\\frac{v \\sin \\alpha(R \\sin \\theta)}{r}\n\n$$\n\n\n\nBecause $r \\ll R$, at the point of closest approach, the energy of the system is roughly\n\n\n\n$$\n\nm g R+\\frac{1}{2} m u^{2} \\approx m g R+\\frac{1}{2} \\frac{m v^{2} \\sin ^{2} \\alpha R^{2} \\sin ^{2} \\theta}{r^{2}} \\approx m g R+K\\left(\\frac{\\alpha^{2} R^{2} \\sin ^{2} \\theta}{r^{2}}\\right)\n\n$$\n\n\n\nEquating with the initial energy gives us\n\n\n\n$$\n\nm g R \\cos \\theta_{0}+\\frac{1}{2} k R^{2} \\theta_{0}^{2}+K=m g R+K\\left(\\frac{\\alpha^{2} R^{2} \\sin ^{2} \\theta}{r^{2}}\\right)\n\n$$\n\n\n\nBefore finishing the calculation, we now compute the required $k$ for the object to be at equilibrium (since our answer cannot contain $k$ ). Setting $k R \\theta_{0}=m g \\sin \\theta_{0}$ gives us $k=$\n\n\n\n\n\n\n\n$\\frac{m g \\sin \\theta_{0}}{R \\theta_{0}}$. Then,\n\n\n\n$$\n\nm g R \\cos \\theta_{0}+\\frac{1}{2} m g R \\theta_{0} \\sin \\theta_{0}+K=m g R+K\\left(\\frac{\\alpha^{2} R^{2} \\sin ^{2} \\theta_{0}}{r^{2}}\\right)\n\n$$\n\n\n\nSolving for $\\alpha$ gives us\n\n\n\n$$\n\n\\alpha=\\frac{r}{R \\sin \\theta_{0}} \\sqrt{1-\\frac{m g R\\left(1-\\cos \\theta_{0}\\right)-(1 / 2) m g R \\theta_{0} \\sin \\theta_{0}}{K}}\n\n$$\n\n\n\nSolving for $\\alpha$ without using the small angle approximation for $\\alpha$ also earned full credit. (The answer $\\alpha=\\pi$, which is technically also correct, earned partial credit.)']",['$\\alpha=\\frac{r}{R \\sin \\theta_{0}} \\sqrt{1-\\frac{m g R\\left(1-\\cos \\theta_{0}\\right)-(1 / 2) m g R \\theta_{0} \\sin \\theta_{0}}{K}}$'],False,,Expression, 1406,Mechanics,,"c. A cylinder of radius $r \ll R \theta_{0}$ is placed on top of the sphere. Suppose the mass is launched at an angle $\alpha$ away from the direction of the spring's displacement with kinetic energy $K$, as shown. What is the maximum angle $\alpha_{\max }$ at which the mass can be launched such that it can still hit the cylinder? Express your answer in terms of $K, m, g, \theta_{0}, r$, and $R$. You may assume $K$ is large enough for the mass to reach the cylinder for $\alpha=0$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ab96511c90603f8e400bg-1.jpg?height=615&width=618&top_left_y=408&top_left_x=775) (view from above)","['The initial energy of the system is given by\n\n\n\n$$\n\nm g R \\cos \\theta_{0}+\\frac{1}{2} k R^{2} \\theta_{0}^{2}+K\n\n$$\n\n\n\nSuppose the mass is launched with speed $v$. Then, the speed in the $\\theta$ direction is $v \\cos \\alpha$ and the speed in the $\\phi$ direction is $v \\sin \\alpha$, and therefore, the $z$-component of the angular momentum of the mass is\n\n\n\n$$\n\nL=m v \\sin \\alpha(R \\sin \\theta)\n\n$$\n\n\n\nWe now compute the distance of closest approach. If the distance of closest approach is equal to $r$ (as it does for $\\alpha_{\\max }$, then at $r$, the motion of the mass has no inward component, and the speed of the object at $r$ is given by conservation of angular momentum:\n\n\n\n$$\n\nm u r=m v \\sin \\alpha(R \\sin \\theta) \\Longrightarrow u=\\frac{v \\sin \\alpha(R \\sin \\theta)}{r}\n\n$$\n\n\n\nBecause $r \\ll R$, at the point of closest approach, the energy of the system is roughly\n\n\n\n$$\n\nm g R+\\frac{1}{2} m u^{2} \\approx m g R+\\frac{1}{2} \\frac{m v^{2} \\sin ^{2} \\alpha R^{2} \\sin ^{2} \\theta}{r^{2}} \\approx m g R+K\\left(\\frac{\\alpha^{2} R^{2} \\sin ^{2} \\theta}{r^{2}}\\right)\n\n$$\n\n\n\nEquating with the initial energy gives us\n\n\n\n$$\n\nm g R \\cos \\theta_{0}+\\frac{1}{2} k R^{2} \\theta_{0}^{2}+K=m g R+K\\left(\\frac{\\alpha^{2} R^{2} \\sin ^{2} \\theta}{r^{2}}\\right)\n\n$$\n\n\n\nBefore finishing the calculation, we now compute the required $k$ for the object to be at equilibrium (since our answer cannot contain $k$ ). Setting $k R \\theta_{0}=m g \\sin \\theta_{0}$ gives us $k=$\n\n\n\n\n\n\n\n$\\frac{m g \\sin \\theta_{0}}{R \\theta_{0}}$. Then,\n\n\n\n$$\n\nm g R \\cos \\theta_{0}+\\frac{1}{2} m g R \\theta_{0} \\sin \\theta_{0}+K=m g R+K\\left(\\frac{\\alpha^{2} R^{2} \\sin ^{2} \\theta_{0}}{r^{2}}\\right)\n\n$$\n\n\n\nSolving for $\\alpha$ gives us\n\n\n\n$$\n\n\\alpha=\\frac{r}{R \\sin \\theta_{0}} \\sqrt{1-\\frac{m g R\\left(1-\\cos \\theta_{0}\\right)-(1 / 2) m g R \\theta_{0} \\sin \\theta_{0}}{K}}\n\n$$\n\n\n\nSolving for $\\alpha$ without using the small angle approximation for $\\alpha$ also earned full credit. (The answer $\\alpha=\\pi$, which is technically also correct, earned partial credit.)']",['$\\alpha=\\frac{r}{R \\sin \\theta_{0}} \\sqrt{1-\\frac{m g R\\left(1-\\cos \\theta_{0}\\right)-(1 / 2) m g R \\theta_{0} \\sin \\theta_{0}}{K}}$'],False,,Expression, 1407,Mechanics,"## Braking up An infinitely long wire with linear charge density $-\lambda$ lies along the $z$-axis. An infinitely long insulating cylindrical shell of radius $a$ is concentric with the wire and can rotate freely about the $z$-axis. The shell has moment of inertia per unit length $I$. Charge is uniformly distributed on the shell, with surface charge density $\frac{\lambda}{2 \pi a}$. The system is immersed in an external magnetic field $B_{0} \hat{\mathbf{z}}$, and is initially at rest. Starting at $t=0$, the external magnetic field is slowly reduced to zero over a time $T \gg a / c$, where $c$ is the speed of light. ",a. Find an expression of the final angular velocity $\omega$ of the cylinder in terms of the symbols given and other constants.,"[""From Faraday's law, you can find the induced electric field inside the cylinder at a distance $r$ from the wire:\n\n\n\n$$\n\n\\begin{aligned}\n\n\\oint \\vec{E}_{\\text {ind }} \\cdot d \\vec{\\ell} & =-\\frac{d \\Phi_{B}}{d t} \\tag{A1-1}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n\nE_{\\text {ind }}(r) & =-\\frac{r}{2} \\frac{d B}{d t} \\tag{A-2}\n\n\\end{aligned}\n\n$$\n\n\n\nThis induced field exerts a torque on the cylinder, causing it to rotate:\n\n\n\n$$\n\n\\begin{aligned}\n\n\\tau & =2 \\pi a \\cdot \\frac{\\lambda}{2 \\pi a} \\cdot E_{\\text {ind }}(a) \\cdot a=I \\cdot \\frac{d \\omega}{d t}\n\\end{aligned} \\tag{A1-3}\n$$\n$$\n\\begin{aligned}\n\n& \\Rightarrow \\frac{d \\omega}{d t}=-\\frac{\\lambda a^{2}}{2 I} \\cdot \\frac{d B}{d t}\n\\tag{A1-4}\n\\end{aligned}\n\n$$\n\n\n\nIntegrate on both sides, and noting that $\\omega(t=0)=0$, we have:\n\n\n\n$$\n\n\\omega(T)=-\\frac{\\lambda a^{2}}{2 I}\\left(B(T)-B_{0}\\right) \\tag{A1-5}\n\n$$\n\n\n\nIt is important to note that $B(T) \\neq 0$. Even though the external field decreases to zero, the now-rotating charged cylinder generates a magnetic field. Using Ampere's Law, you can find at $t=T$, the magnetic field is:\n\n\n\n$$\n\n\\begin{gathered}\n\n\\oint \\vec{B}_{\\mathrm{ind}} \\cdot d \\vec{\\ell}=\\mu_{0} I_{\\mathrm{enc}},\n\\end{gathered} \\tag{A1-6}\n$$\n$$\n\\begin{gathered}\n\n\\text { where } I_{\\mathrm{enc}}=\\frac{\\lambda}{2 \\pi a} \\cdot \\omega(T) \\cdot a \n\\end{gathered} \\tag{A1-7}\n$$\n$$\n\\begin{gathered}\n\n\\Rightarrow B(T)=\\mu_{0} \\frac{\\lambda}{4 \\pi^{2} a} \\omega(T)\n\n\\end{gathered} \\tag{A1-8}\n\n$$\n\n\n\nCombining equations A1-5 and A1-8, we have:\n\n\n\n$$\n\n\\omega(T)=\\frac{\\frac{\\lambda a^{2}}{2 I} B_{0}}{1+\\mu_{0} \\frac{\\lambda^{2} a}{8 \\pi I}}\n\n$$""]",['$\\omega(T)=\\frac{\\frac{\\lambda a^{2}}{2 I} B_{0}}{1+\\mu_{0} \\frac{\\lambda^{2} a}{8 \\pi I}}$'],False,,Expression, 1408,Mechanics,"## Braking up An infinitely long wire with linear charge density $-\lambda$ lies along the $z$-axis. An infinitely long insulating cylindrical shell of radius $a$ is concentric with the wire and can rotate freely about the $z$-axis. The shell has moment of inertia per unit length $I$. Charge is uniformly distributed on the shell, with surface charge density $\frac{\lambda}{2 \pi a}$. The system is immersed in an external magnetic field $B_{0} \hat{\mathbf{z}}$, and is initially at rest. Starting at $t=0$, the external magnetic field is slowly reduced to zero over a time $T \gg a / c$, where $c$ is the speed of light. Context question: a. Find an expression of the final angular velocity $\omega$ of the cylinder in terms of the symbols given and other constants. Context answer: \boxed{$\omega(T)=\frac{\frac{\lambda a^{2}}{2 I} B_{0}}{1+\mu_{0} \frac{\lambda^{2} a}{8 \pi I}}$} ","b. You may be surprised that the expression you find above is not zero! However, the electric and magnetic fields can have angular momentum. Analogous to the ""regular"" angular momentum definition, the EM field angular momentum per unit volume at a displacement $\mathbf{r}$ from the axis of rotation is: $$ \mathcal{L}(\mathbf{r})=\mathbf{r} \times \mathcal{P}(\mathbf{r}) $$ $\mathcal{P}(\mathbf{r})$ is a vector analogous to momentum, given by $$ \mathcal{P}(\mathbf{r})=\alpha \cdot(\mathbf{E}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})) . $$ where $\alpha$ is some proportionality constant. Find an expression for $\alpha$ in terms of given variables and fundamental constants.",['The electric field inside the cylindrical shell is given by $\\mathbf{E}(r)=-\\frac{\\lambda}{2 \\pi \\epsilon_{0} r} \\hat{\\mathbf{r}}$ inward. The magnetic field is given by $B(t) \\hat{\\mathbf{z}}$. Then:\n\n\n\n$$\n\n\\mathcal{P}(\\mathbf{r})=\\alpha \\frac{\\lambda B(t)}{2 \\pi \\epsilon_{0} r} \\hat{\\theta}\n\n$$\n\n\n\nThe angular momentum per unit volume is then:\n\n\n\n$$\n\n\\mathcal{L}(\\mathbf{r})=-\\alpha \\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\hat{\\mathbf{z}}\n\n$$\n\n\n\nThe angular momentum per unit length is then:\n\n\n\n$$\n\n\\mathbf{L}=-\\frac{\\alpha \\lambda B(t) a^{2}}{2} \\hat{\\mathbf{z}}\n\n$$\n\n\n\nComparing this to Equation shows that $\\alpha=\\epsilon_{0}$.'],['$\\alpha=\\epsilon_{0}$'],False,,Expression, 1409,Mechanics,"## Swoosh! In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from "" 12 "" to "" 6 "", later on they would move to the "" $3-9$ "" plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out. Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame. First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$. John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane.","a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$.","[""John's reference frame is inertial and the point of attachment stationary, so this is a freemoving pendulum obeying simple harmonic motion in each axis:\n\n\n\n$$\n\na_{x}+\\omega^{2} x=0 ; a_{y}+\\omega^{2} y=0 \\tag{A2-1}\n\n$$""]","['$a_{x}+\\omega^{2} x=0 , a_{y}+\\omega^{2} y=0$']",True,,Equation, 1410,Mechanics," Swoosh! In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from "" 12 "" to "" 6 "", later on they would move to the "" $3-9$ "" plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out. Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame. First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$. John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane. Context question: a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$. Context answer: \boxed{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$}","b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ?","['The solution to Eq. A2-1) is the familiar simple harmonic motion. In general, if the displacement is $r=A \\cos \\omega t$, then the velocity is $v=-A \\omega \\sin \\omega t$. Using the inital conditions\n\n\n\n\n\n\n\nprovided, we have:\n\n\n\n$$\n\nx(t)=A \\cos (\\omega t) ; y(t)=\\frac{V}{\\omega} \\sin (\\omega t) \\tag{A2-2}\n\n$$\n\n\n\nNote that this corresponds to an ellipse.\n\n']","['$x(t)=A \\cos (\\omega t), y(t)=\\frac{V}{\\omega} \\sin (\\omega t)$']",True,,Expression, 1411,Mechanics,"## Swoosh! In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from "" 12 "" to "" 6 "", later on they would move to the "" $3-9$ "" plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out. Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame. First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$. John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane. Context question: a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$. Context answer: \boxed{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$} Context question: b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ? Context answer: $x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t)$ ","c. Ella, an observer resides at the North Pole, is also looking at the pendulum. What are the coordinates, $\tilde{x}(t)$ and $\tilde{y}(t)$, as observed by Ella? Assume that at time $t=0$, the coordinate systems of John's and Ella's overlap.","['In the rotating frame, we have $\\tilde{x}=x \\cos (\\Omega t)+y \\sin (\\Omega t), \\tilde{y}=-x \\sin (\\Omega t)+y \\cos (\\Omega t)$ (with $2 \\pi / \\Omega=24 \\mathrm{hrs})$. Plugging in the form of $x(t)$ and $y(t)$ we find:\n\n\n\n$$\n\n\\tilde{x}=A \\cos (\\omega t) \\cos (\\Omega t)+\\frac{V}{\\omega} \\sin (\\omega t) \\sin (\\Omega t) \\tag{A2-3}\n\n$$\n\n\n\nand:\n\n\n\n$$\n\n\\tilde{y}=-A \\cos (\\omega t) \\sin (\\Omega t)+\\frac{V}{\\omega} \\sin (\\omega t) \\cos (\\Omega t) \\tag{A2-4}\n\n$$']","['$\\tilde{x}=A \\cos (\\omega t) \\cos (\\Omega t)+\\frac{V}{\\omega} \\sin (\\omega t) \\sin (\\Omega t)$,$\\tilde{y}=-A \\cos (\\omega t) \\sin (\\Omega t)+\\frac{V}{\\omega} \\sin (\\omega t) \\cos (\\Omega t)$']",True,,Expression, 1412,Mechanics," Swoosh! In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from "" 12 "" to "" 6 "", later on they would move to the "" $3-9$ "" plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out. Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame. First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$. John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane. Context question: a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$. Context answer: \boxed{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$} Context question: b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ? Context answer: $x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t)$ Context question: c. Ella, an observer resides at the North Pole, is also looking at the pendulum. What are the coordinates, $\tilde{x}(t)$ and $\tilde{y}(t)$, as observed by Ella? Assume that at time $t=0$, the coordinate systems of John's and Ella's overlap. Context answer: \boxed{$\tilde{x}=A \cos (\omega t) \cos (\Omega t)+\frac{V}{\omega} \sin (\omega t) \sin (\Omega t)$,$\tilde{y}=-A \cos (\omega t) \sin (\Omega t)+\frac{V}{\omega} \sin (\omega t) \cos (\Omega t)$}",d. What is the speed of the pendulum bob observed by Ella at $t=0$ ?,"[""In John's frame, the velocity at this time is $(0, V)$. To get the velocity in Ella's frame, we can either take the derivative of the result of part (c) directly, or transform the velocity obtained in John's frame to Ella's frame, not forgetting to add the term $-\\Omega A$ to the initial velocity in the $y$ axis. This gives\n\n\n\n$$\n\n\\tilde{v}_{x} \\approx(V-\\Omega A) \\sin (\\Omega t) ; \\tilde{v}_{y}=(V-\\Omega A) \\cos (\\Omega t)\n\n$$\n\n\n\nAt $t=0, \\tilde{v}_{x}=0$ and $\\tilde{v}_{y}=V-\\Omega A$.\n\n""]","['$\\tilde{v}_{x}=0$ , $\\tilde{v}_{y}=V-\\Omega A$']",True,,Expression, 1413,Mechanics,,"e. Find the initial conditions for $A, V$, such that as measured in Ella's frame: i. the pendulum passes precisely through its resting position.","[""Considering the motion in John's frame, clearly the pendulum will pass through the resting position if and only if $V=0$.""]","[""Considering the motion in John's frame, clearly the pendulum will pass through the resting position if and only if $V=0$.""]",True,,Need_human_evaluate, 1414,Mechanics," Swoosh! In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from "" 12 "" to "" 6 "", later on they would move to the "" $3-9$ "" plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out. Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame. First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$. John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane. Context question: a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$. Context answer: \boxed{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$} Context question: b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ? Context answer: $x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t)$ Context question: c. Ella, an observer resides at the North Pole, is also looking at the pendulum. What are the coordinates, $\tilde{x}(t)$ and $\tilde{y}(t)$, as observed by Ella? Assume that at time $t=0$, the coordinate systems of John's and Ella's overlap. Context answer: \boxed{$\tilde{x}=A \cos (\omega t) \cos (\Omega t)+\frac{V}{\omega} \sin (\omega t) \sin (\Omega t)$,$\tilde{y}=-A \cos (\omega t) \sin (\Omega t)+\frac{V}{\omega} \sin (\omega t) \cos (\Omega t)$} Context question: d. What is the speed of the pendulum bob observed by Ella at $t=0$ ? Context answer: $\tilde{v}_{x}=0$ , $\tilde{v}_{y}=V-\Omega A$. Context question: e. Find the initial conditions for $A, V$, such that as measured in Ella's frame: i. the pendulum passes precisely through its resting position. Context answer: Considering the motion in John's frame, clearly the pendulum will pass through the resting position if and only if $V=0$.","e. Find the initial conditions for $A, V$, such that as measured in Ella's frame: ii. it has a ""spike"" at the points of maximal amplitude (see figure below) instead of a ""rounded"" trajectory.","['\n\n\n\nFigure 2: Two possible trajectories with ""spike""(left) and more ""rounded"" (right).\n\n\nTo have a spike, we need the velocity to vanish at the extremal points in Ella\'s frame. This gives the condition:\n\n\n\n$$\n\nV=\\Omega A \\tag{A2-6}\n\n$$\n\n\n\nNote that in Ella\'s frame, this implies releasing the pendulum from rest at some amplitude.\n\n']",['$V=\\Omega A$'],False,,Expression, 1415,Mechanics,"## Swoosh! In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from "" 12 "" to "" 6 "", later on they would move to the "" $3-9$ "" plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out. Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame. First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$. John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane. Context question: a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$. Context answer: \boxed{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$} Context question: b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ? Context answer: $x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t)$ Context question: c. Ella, an observer resides at the North Pole, is also looking at the pendulum. What are the coordinates, $\tilde{x}(t)$ and $\tilde{y}(t)$, as observed by Ella? Assume that at time $t=0$, the coordinate systems of John's and Ella's overlap. Context answer: \boxed{$\tilde{x}=A \cos (\omega t) \cos (\Omega t)+\frac{V}{\omega} \sin (\omega t) \sin (\Omega t)$,$\tilde{y}=-A \cos (\omega t) \sin (\Omega t)+\frac{V}{\omega} \sin (\omega t) \cos (\Omega t)$} Context question: d. What is the speed of the pendulum bob observed by Ella at $t=0$ ? Context answer: $\tilde{v}_{x}=0$ , $\tilde{v}_{y}=V-\Omega A$. Context question: e. Find the initial conditions for $A, V$, such that as measured in Ella's frame: i. the pendulum passes precisely through its resting position. Context answer: Considering the motion in John's frame, clearly the pendulum will pass through the resting position if and only if $V=0$. Context question: e. Find the initial conditions for $A, V$, such that as measured in Ella's frame: ii. it has a ""spike"" at the points of maximal amplitude (see figure below) instead of a ""rounded"" trajectory. Context answer: $V=\Omega A$ Extra Supplementary Reading Materials: In a rotating frame, a fictitious force known as the Coriolis force acts on the particles. For Foucault's pendulum, the Coriolis force acts primarily in the horizontal plane, in a direction perpendicular to the velocity of the mass in the Earth's frame with magnitude: $$ F=2 m \Omega v \cdot \sin \theta \tag{2-7} $$ where $m$ and $v$ are the pendulum's mass and its velocity, and $\theta$ the latitude $\left(90^{\circ}\right.$ for the North Pole). Note that when the velocity changes sign, so does the Coriolis force.","f. How long would it take for the plane of oscillation of Foucault's pendulum to return to its initial value in Paris, which has a latitude of about $49^{\circ}$.","['Since the expression for the Coriolis force only depends on the combination $\\Omega \\sin (\\theta)$, and since the solution at the North Pole must be $\\pi / \\Omega=12$ hours, the time at a general latitude must be:\n\n\n\n$$\n\nT=\\frac{\\pi}{\\Omega \\sin (\\theta)} \\tag{A2-8}\n\n$$\n\n\n\nFor Paris, the time is about 16 hours.']",['16'],False,hours,Numerical,5e-1 1416,Modern Physics,"## Spin Cycle Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.",a. How much time elapses on Carla's stopwatch with each revolution?,"[""From time dilation, her clock ticks slower by a factor $\\gamma$. Therefore, each revolution takes\n\n\n\n$$\n\n\\frac{2 \\pi R}{\\gamma v}=\\frac{2 \\pi R \\sqrt{1-v^{2} / c^{2}}}{v}\n\n$$\n\n\n\nwhen measured by Carla's stopwatch.""]",['$\\frac{2 \\pi R \\sqrt{1-v^{2} / c^{2}}}{v}$'],False,,Expression, 1417,Modern Physics,"## Spin Cycle Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light. Context question: a. How much time elapses on Carla's stopwatch with each revolution? Context answer: \boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$} Extra Supplementary Reading Materials: Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training. For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time.","b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?","[""Carla's motion is perpendicular to the displacement between Clock $A$ and Clock $B$ when they are synchronized in CIRF. Therefore, the simultaneous synchronization in CIRF is also simultaneous in the lab frame. Thus, the offset is 0 .\n\n\n\nTo understand why this offset is 0 , you can also imagine placing an lightbulb halfway between the two clocks and having it send a light pulse at some known time. In both Carla's frame and the lab frame, the light pulse reaches the two clocks simultaneously.""]",['0'],False,,Numerical,0 1418,Modern Physics,"## Spin Cycle Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light. Context question: a. How much time elapses on Carla's stopwatch with each revolution? Context answer: \boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$} Extra Supplementary Reading Materials: Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training. For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time. Context question: b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ? Context answer: \boxed{0} ","c. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock A measure in CIRF?","[""By symmetry, the speed at which the center clock ticks according to CIRF cannot change. In one revolution, Carla's stopwatch measures $\\frac{2 \\pi R \\sqrt{1-v^{2} / c^{2}}}{v}$, while the center clock measures $\\frac{2 \\pi R}{v}$. Then,\n\n\n\n\n\n$$\n\nt_{A}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}\n\n$$""]",['$t_{A}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}$'],False,,Expression, 1419,Modern Physics,"## Spin Cycle Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light. Context question: a. How much time elapses on Carla's stopwatch with each revolution? Context answer: \boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$} Extra Supplementary Reading Materials: Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training. For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time. Context question: b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ? Context answer: \boxed{0} Context question: c. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock A measure in CIRF? Context answer: \boxed{$t_{A}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}$} ","d. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock B measure in CIRF?","[""The readings on Clock B and on Clock A are not necessarily identical once Carla moves through the circle (because her motion becomes more parallel with the displacement between the two clocks, and thus simultaneity is lost).\n\n\n\nSuppose Carla is at $(R \\cos \\theta, R \\sin \\theta)$, so her velocity is given by $(-v \\sin \\theta, v \\cos \\theta)$. Suppose we place a light bulb between the two clocks and having it propagate a light pulse. In the lab frame, the light pulse reaches the two clocks simultaneously. In CIRF, the math is a little more complicated.\n\n\n\nWe first rotate our lab coordinates so that $\\hat{\\mathbf{a}}=-\\sin \\theta \\hat{\\mathbf{x}}+\\cos \\theta \\hat{\\mathbf{y}}$, and $\\hat{\\mathbf{b}}=\\cos \\theta \\hat{\\mathbf{x}}+\\sin \\theta \\hat{\\mathbf{y}}$. We now give the coordinates of the clocks and bulb in the rotated lab frame: Clock A, $(a, b)=(0,0)$; Clock B, $(a, b)=(-R \\sin \\theta, R \\cos \\theta)$; bulb, $(a, b)=(-R \\sin \\theta, R \\cos \\theta) / 2$. In the lab frame, a light pulse is emitted at\n\n\n\n$$\n\nt=0, a=-(R / 2) \\sin \\theta, b=(R / 2) \\cos \\theta\n\n$$\n\n\n\nThe light pulse reaches Clock A at\n\n\n\n$$\n\nt=R / 2, a=0, b=0\n\n$$\n\n\n\nand Clock B at\n\n\n\n$$\n\nt=R / 2, a=-R \\sin \\theta, b=R \\cos \\theta .\n\n$$\n\n\n\nUnder a Lorentz tranformation from the lab frame to CIRF, we have that the light pulse reaches Clock $A$ at $t^{\\prime}=\\gamma R / 2$ and Clock $B$ at $t^{\\prime}=\\gamma R / 2+\\gamma v R \\sin \\theta$. Thus, Clock $B$ reads the same time as Clock $A$ with offset $\\gamma v R \\sin \\theta$ in the reference frame moving at $v_{a}=v$, $v_{b}=0$. Note that Clock $A$ ticks slower by a factor of $\\gamma$ in this frame. Therefore, the time on clock $B$ is $v R \\sin \\theta$ behind the time on clock $A$.\n\n\n\nThen,\n\n\n\n$$\n\nt_{B}(\\tau)=t_{A}(\\tau)-v R \\sin \\theta=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}-v R \\sin \\theta\n\n$$\n\n\n\n(This is the answer we expect from the rear clock ahead effect!) Finally, we use that $\\theta=\\omega \\tau$ and $\\omega=\\frac{2 \\pi}{T}$, where $T$ is the period in Carla's frame. Then,\n\n\n\n$$\n\nt_{B}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}-\\frac{v R}{c^{2}} \\sin \\left(\\frac{v \\tau}{R \\sqrt{1-v^{2} / c^{2}}}\\right)\n\n$$""]",['$t_{B}(\\tau)=\\frac{\\tau}{\\sqrt{1-v^{2} / c^{2}}}-\\frac{v R}{c^{2}} \\sin \\left(\\frac{v \\tau}{R \\sqrt{1-v^{2} / c^{2}}}\\right)$'],False,,Expression, 1420,Mechanics,## String Cheese,"a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up. Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout.","['We can use energy conservation to answer this question. For a bit of water with mass $m$, the total energy $E$ is the sum of the kinetic and gravitational potential energies,\n\n\n\n$$\n\nE=\\frac{1}{2} m v^{2}+m g y \\text {. } \\tag{B1-1}\n\n$$\n\n\n\n(With this sign convention, $g \\approx 10 \\mathrm{~m} / \\mathrm{s}^{2}$ is positive. As $y$ decreases, so does the potential energy.)\n\n\n\nAs the bit of water falls, its energy remains constant, and is equal to the initial value of\n\n\n\n$$\n\nE=\\frac{1}{2} m v_{0}^{2}+m g y_{0} \\tag{B1-2}\n\n$$\n\n\n\nEquating eliminating $E$ from equations B1-1 and B1-2, we have\n\n\n\n$$\n\n\\frac{1}{2} m v^{2}+m g y=\\frac{1}{2} m v_{0}^{2}+m g y_{0},\n\n$$\n\n\n\nand solving for $v$, we get\n\n\n\n$$\n\nv=\\sqrt{v_{0}^{2}+2 g\\left(y_{0}-y\\right)}\n\n$$\n\n\n\nThe equation for $v^{2}$ has three terms, but we were asked to choose the zero of $y$ such that there is only one. Evidently, two of the terms must cancel, and these must be the two constant terms, since the final term varies with $y$.\n\n\n\nThat means we need\n\n\n\n$$\n\nv_{0}^{2}+2 g y_{0}=0\n\n$$\n\n\n\nSolving for $y_{0}$, the vertical position of the spout is\n\n\n\n$$\n\ny_{0}=\\frac{-v_{0}^{2}}{2 g}\n\n$$\n\n\n\n\n\n\n\nWith this choice of the zero of $y$, the equation for $v$ simplifies to\n\n\n\n$$\n\nv=\\sqrt{-2 g y} \\tag{B1-3}\n\n$$\n\n\n\nWe note that the result of this equation is real because $y<0$ at the spout, and decreases as the water falls, so this equation shows that $v$ is real and increases as the water falls.']","['$y_{0}=\\frac{-v_{0}^{2}}{2 g}$ ,$v=\\sqrt{-2 g y}$']",True,,Expression, 1421,Mechanics,"## String Cheese Context question: a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up. Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout. Context answer: \boxed{$y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$} ","b. Assume that the stream of water falling from the faucet is cylindrically symmetric about a vertical axis through the center of the stream. Also assume that the volume of water per unit time exiting the spout is constant, and that the shape of the stream of water is constant over time. In this case, the radius $r$ of the stream of water is a function of vertical position $y$. Let the radius at the faucet be $r_{0}$. Using your result from part (a), find $r(y)$. If $r(y)$ is not constant, it implies that the water has some radial velocity during its fall, in contradiction to our assumptions in part (a) that the motion is purely vertical. You may assume throughout the problem that any such radial velocity is negligibly small.","[""The same volume of water must fall through any horizontal cross-section of the stream each second because water doesn't disappear during its fall, and its density if constant. That volume per unit time $Q$ is the cross-sectional area of the stream multiplied by the speed of the water in the vertical direction. As an equation,\n\n\n\n$$\n\nQ=v \\pi r^{2} \\tag{B1-4}\n\n$$\n\n\n\n$Q$ is the same at all $y$, and is equal to its initial value of\n\n\n\n$$\n\nQ=v_{0} \\pi r_{0}^{2} \\tag{B1-5}\n\n$$\n\n\n\nEliminating $Q$ from B1-4 and B1-5 and solving for $r$ gives\n\n\n\n$$\n\nr=r_{0} \\sqrt{\\frac{v_{0}}{v}}\n\n$$\n\n\n\nPlugging in our equation B1-3 for $v$,\n\n\n\n$$\n\nr=r_{0} \\sqrt[4]{\\frac{v_{0}^{2}}{-2 g y}}\n\n$$""]",['$r=r_{0} \\sqrt[4]{\\frac{v_{0}^{2}}{-2 g y}}$'],False,,Expression, 1422,Mechanics,"## String Cheese Context question: a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up. Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout. Context answer: \boxed{$y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$} Context question: b. Assume that the stream of water falling from the faucet is cylindrically symmetric about a vertical axis through the center of the stream. Also assume that the volume of water per unit time exiting the spout is constant, and that the shape of the stream of water is constant over time. In this case, the radius $r$ of the stream of water is a function of vertical position $y$. Let the radius at the faucet be $r_{0}$. Using your result from part (a), find $r(y)$. If $r(y)$ is not constant, it implies that the water has some radial velocity during its fall, in contradiction to our assumptions in part (a) that the motion is purely vertical. You may assume throughout the problem that any such radial velocity is negligibly small. Context answer: \boxed{$r=r_{0} \sqrt[4]{\frac{v_{0}^{2}}{-2 g y}}$} ","c. The water-air interface has some surface tension, $\sigma$. The effect of surface tension is to change the pressure in the stream according to the Young-Laplace equation, $$ \Delta P=\sigma\left(\frac{1}{r}+\frac{1}{R}\right) $$ where $\Delta P$ is the difference in pressure between the stream and the atmosphere and $R$ is the radius of curvature of the vertical profile of the stream, visualized below. ( $R<0$ for the stream of water; the radius of curvature would be positive only if the stream profile curved inwards.) For this part of the problem, we assume that $|R| \gg|r|$, so that the curvature of the vertical profile of the stream can be ignored. Also assume that water is incompressible. Accounting for the pressure in the stream, find a new equation relating for $r(y)$ in terms of $\sigma, r_{0}, v_{0}$, and $\rho$, the density of water. You do not need to solve the equation for $r$.","['Our conservation of energy approach from part (b) needs to be modified to account for the work done against pressure. As we look further down in the stream, the radius is smaller. This means the pressure is higher there, and the water is slowed compared to when we assumed only gravity acted on the water.\n\n\n\nThe result of accounting for changes in pressure in a flow where no energy is dissipated is the Bernoulli equation,\n\n\n\n$$\n\n\\frac{1}{2} \\rho v^{2}+\\rho g y+P=\\frac{1}{2} \\rho v_{0}^{2}+\\rho g y_{0}+P_{0}\n\n$$\n\n\n\nwhere $P_{0}$ is the pressure in the stream at the spout.\n\n\n\nUsing the Young-Laplace equation to replace $P$ and $P_{0}$, we have\n\n\n\n$$\n\n\\frac{1}{2} \\rho v^{2}+\\rho g y+\\frac{\\sigma}{r}=\\frac{1}{2} \\rho v_{0}^{2}+\\rho g y_{0}+\\frac{\\sigma}{r_{0}}\n\n$$\n\n\n\nIf we substitute in $y_{0}=-\\frac{v_{0}^{2}}{2 g}$ and $v=v_{0} \\frac{r_{0}^{2}}{r^{2}}$, this becomes\n\n\n\n$$\n\n\\frac{1}{2} \\rho v_{0}^{2} \\frac{r_{0}^{4}}{r^{4}}+\\rho g y+\\frac{\\sigma}{r}=\\frac{1}{2} \\rho v_{0}^{2}-\\rho g \\frac{v_{0}^{2}}{2 g}+\\frac{\\sigma}{r_{0}}\n\n$$\n\n\n\nThis may be simplified to\n\n\n\n$$\n\n\\frac{1}{2} \\rho v_{0}^{2} \\frac{r_{0}^{4}}{r^{4}}+\\rho g y=\\sigma\\left(\\frac{1}{r_{0}}-\\frac{1}{r}\\right)\n\n$$']",['$\\frac{1}{2} \\rho v_{0}^{2} \\frac{r_{0}^{4}}{r^{4}}+\\rho g y=\\sigma\\left(\\frac{1}{r_{0}}-\\frac{1}{r}\\right)$'],False,,Equation, 1422,Mechanics,,"c. The water-air interface has some surface tension, $\sigma$. The effect of surface tension is to change the pressure in the stream according to the Young-Laplace equation, $$ \Delta P=\sigma\left(\frac{1}{r}+\frac{1}{R}\right) $$ where $\Delta P$ is the difference in pressure between the stream and the atmosphere and $R$ is the radius of curvature of the vertical profile of the stream, visualized below. ( $R<0$ for the stream of water; the radius of curvature would be positive only if the stream profile curved inwards.) ![](https://cdn.mathpix.com/cropped/2023_12_21_d83191acb7c4227786c9g-1.jpg?height=325&width=265&top_left_y=431&top_left_x=930) For this part of the problem, we assume that $|R| \gg|r|$, so that the curvature of the vertical profile of the stream can be ignored. Also assume that water is incompressible. Accounting for the pressure in the stream, find a new equation relating for $r(y)$ in terms of $\sigma, r_{0}, v_{0}$, and $\rho$, the density of water. You do not need to solve the equation for $r$.","['Our conservation of energy approach from part (b) needs to be modified to account for the work done against pressure. As we look further down in the stream, the radius is smaller. This means the pressure is higher there, and the water is slowed compared to when we assumed only gravity acted on the water.\n\n\n\nThe result of accounting for changes in pressure in a flow where no energy is dissipated is the Bernoulli equation,\n\n\n\n$$\n\n\\frac{1}{2} \\rho v^{2}+\\rho g y+P=\\frac{1}{2} \\rho v_{0}^{2}+\\rho g y_{0}+P_{0}\n\n$$\n\n\n\nwhere $P_{0}$ is the pressure in the stream at the spout.\n\n\n\nUsing the Young-Laplace equation to replace $P$ and $P_{0}$, we have\n\n\n\n$$\n\n\\frac{1}{2} \\rho v^{2}+\\rho g y+\\frac{\\sigma}{r}=\\frac{1}{2} \\rho v_{0}^{2}+\\rho g y_{0}+\\frac{\\sigma}{r_{0}}\n\n$$\n\n\n\nIf we substitute in $y_{0}=-\\frac{v_{0}^{2}}{2 g}$ and $v=v_{0} \\frac{r_{0}^{2}}{r^{2}}$, this becomes\n\n\n\n$$\n\n\\frac{1}{2} \\rho v_{0}^{2} \\frac{r_{0}^{4}}{r^{4}}+\\rho g y+\\frac{\\sigma}{r}=\\frac{1}{2} \\rho v_{0}^{2}-\\rho g \\frac{v_{0}^{2}}{2 g}+\\frac{\\sigma}{r_{0}}\n\n$$\n\n\n\nThis may be simplified to\n\n\n\n$$\n\n\\frac{1}{2} \\rho v_{0}^{2} \\frac{r_{0}^{4}}{r^{4}}+\\rho g y=\\sigma\\left(\\frac{1}{r_{0}}-\\frac{1}{r}\\right)\n\n$$']",['$\\frac{1}{2} \\rho v_{0}^{2} \\frac{r_{0}^{4}}{r^{4}}+\\rho g y=\\sigma\\left(\\frac{1}{r_{0}}-\\frac{1}{r}\\right)$'],False,,Equation, 1423,Mechanics,,"d. After falling for some distance, the water stream usually breaks into smaller droplets. This occurs because small random perturbations to the shape of the stream grow over time, eventually breaking the stream into apart. For the rest of this problem we ignore the change in the radius of the stream due to changing speed of the water, as considered earlier. Instead, we examine small random variations in the radius of the stream. Random variations can be broken down into a sum of sinusoidal variations in stream radius, each with a different wavenumber $k$. We can analyze these different sinusoidal variations independently. Consider a stream of water whose radius obeys $$ r(y)=r_{0}+A \cos (k y) $$ where $A \ll r_{0}$ is the perturbation amplitude. To analyze such a stream, it is sufficient to consider only the thickest and thinnest parts of the stream. Accounting for both sources of curvature, find a condition on $r_{0}$ and $k$ such that the size of perturbations increases with time. ![](https://cdn.mathpix.com/cropped/2023_12_21_514efaf567aad8907c0ag-1.jpg?height=479&width=379&top_left_y=1002&top_left_x=884)","['If the size of the perturbation increases with time, water must be flowing from the thin parts of the stream to the thick parts. For that to happen, the pressure needs to be higher in the thin parts of the stream than in the thick parts of the stream so that the pressure gradient will force water towards the thick parts, eventually breaking the stream into droplets.\n\n\n\nWe consider a small patch with side lengths $h$ on the surface of the stream at the thinnest part of the stream. The pressure is\n\n\n\n$$\n\n\\Delta P_{\\text {thin }}=\\sigma\\left(\\frac{1}{r_{\\text {thin }}}+\\frac{1}{R_{\\text {thin }}}\\right)\n\n$$\n\n\n\nAnd at the thickest part of the stream,\n\n\n\n$$\n\n\\Delta P_{\\text {thick }}=\\sigma\\left(\\frac{1}{r_{\\text {thick }}}+\\frac{1}{R_{\\text {thick }}}\\right)\n\n$$\n\n\n\nWe are looking for the wavenumbers such that\n\n\n\n$$\n\n\\Delta P_{\\text {thin }}>\\Delta P_{\\text {thick }}\n\n$$\n\n\n\n\n\n\n\nUsing the Young-Laplace equation, this becomes\n\n\n\n$$\n\n\\sigma\\left(\\frac{1}{r_{\\text {thin }}}+\\frac{1}{R_{\\text {thin }}}\\right)>\\sigma\\left(\\frac{1}{r_{\\text {thick }}}+\\frac{1}{R_{\\text {thick }}}\\right)\n\n$$\n\n\n\nDropping the common factor $\\sigma$,\n\n\n\n$$\n\n\\left(\\frac{1}{r_{\\text {thin }}}+\\frac{1}{R_{\\text {thin }}}\\right)>\\left(\\frac{1}{r_{\\text {thick }}}+\\frac{1}{R_{\\text {thick }}}\\right)\n\n$$\n\n\n\nTo simplify this further, we will need to find $r$ and $R$ in terms of $A$ and $k$, the variables given in the problem statement.\n\n\n\n$r$ is the thickness of the stream, which from the equation given, varies sinusoidally. So\n\n\n\n$$\n\n\\begin{aligned}\n\nr_{\\text {thin }} & =r_{0}-A . \\\\\n\nr_{\\text {thick }} & =r_{0}+A\n\n\\end{aligned}\n\n$$\n\n\n\nWe are going to need $\\frac{1}{r}$ to use in the Young Laplace equation, so we make the approximations\n\n\n\n$$\n\n\\begin{aligned}\n\n& \\frac{1}{r_{\\text {thin }}} \\approx \\frac{1}{r_{0}}+\\frac{A}{r_{0}^{2}} \\\\\n\n& \\frac{1}{r_{\\text {thick }}} \\approx \\frac{1}{r_{0}}-\\frac{A}{r_{0}^{2}}\n\n\\end{aligned}\n\n$$\n\n\n\n(To find these, recall $\\frac{1}{1-\\epsilon} \\approx 1+\\epsilon$ for small $\\epsilon$.)\n\n\n\nThe inequality now becomes\n\n\n\n$$\n\n\\frac{1}{r_{0}}+\\frac{A}{r_{0}^{2}}+\\frac{1}{R_{\\text {thin }}}>\\frac{1}{r_{0}}-\\frac{A}{r_{0}^{2}}+\\frac{1}{R_{\\text {thick }}}\n\n$$\n\n\n\nThis simplifies to\n\n\n\n$$\n\n\\frac{2 A}{r_{0}^{2}}>\\frac{1}{R_{\\text {thick }}}-\\frac{1}{R_{\\text {thin }}}\n\n$$\n\n\n\nNext we need to determine the radius of curvature $R$ of the sinusoidal as a function of $k$ and $A$.\n\n\n\n\n\n\n\nTo do this, we compare the sinusoidal function and a circle at small deviations from the thickest part of the stream.\n\n\n\nRecall that, for small $\\theta$,\n\n\n\n$$\n\n\\cos \\theta \\approx 1-\\frac{1}{2} \\theta^{2}\n\n$$\n\n\n\nwhich means that for small $x$,\n\n\n\n$$\n\ny_{\\text {sinusoidal }}=A \\cos (k x) \\approx A\\left(1-\\frac{1}{2} k^{2} x^{2}\\right)\n\n$$\n\n\n\nNext we consider a circle of radius $R$. If a particle moves along such a circle at speed $v$, its acceleration is $v^{2} / R$. This means that if the particle moves forward for a short time $t$, it moves forward a distance $v t$ and falls a distance $\\frac{1}{2} \\frac{v^{2}}{R} t^{2}$. If we set $v t=x$, then the $y$ position of the particle is given by\n\n\n\n$$\n\ny_{\\text {circle }} \\approx y_{0}-\\frac{1}{2} \\frac{x^{2}}{R} .\n\n$$\n\n\n\nComparing $y_{\\text {circle }}$ and $y_{\\text {sinusoidal }}$, they give the same motion if $A k^{2}=\\frac{1}{R}$.\n\n\n\nThen\n\n\n\n$$\n\n\\begin{aligned}\n\n\\frac{1}{R_{\\text {thin }}} & =-A k^{2} . \\\\\n\n\\frac{1}{R_{\\text {thick }}} & =A k^{2} .\n\n\\end{aligned}\n\n$$\n\n\n\nPutting these into the inequality,\n\n\n\n$$\n\n\\frac{2 A}{r_{0}^{2}}>2 A k^{2}\n\n$$\n\n\n\nThis simplifies to\n\n\n\n$$\n\nk<\\frac{1}{r_{0}}\n\n$$\n\n\n\nSo the perturbations will grow as long as they have a wavenumber greater than one over the radius, or equivalently when the wavelength of the perturbation is longer than the circumference of the stream.\n\n\n\nThis result was discovered experimentally by Plateau and derived theoretically by Rayleigh. The breaking up of a stream into droplets is called the Plateau-Rayleigh instability.']",['$k<\\frac{1}{r_{0}}$'],False,,Need_human_evaluate, 1424,Optics,"## Mirror Mirror on the Wall Consider a square room with side length $L$. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength $\lambda$ is placed a distance $d$ above the center of the mirror, where $\lambda \ll d \ll L$. *Remember that the phase of light reflected by a mirror changes by $180^{\circ}$.","a. On the right wall, an interference pattern emerges. What is the distance $y$ between the bottom corner and the closest bright fringe above it? Hint: you may assume $\lambda \ll y \ll L$ as well.","['This setup is essentially a double-slit experiment with the second slit being the image of the point source on the other side of the mirror, with the additional phase shift from the mirror. The distance between the source and a spot $y$ on the wall is given by $\\sqrt{(d-y)^{2}+(L / 2)^{2}}$ and the distance between the image and a spot $y$ is given by $\\sqrt{(d+y)^{2}+(L / 2)^{2}}$. Subtracting the two distances and adding in the phase shift gives us approximately\n\n\n\n$$\n\nL / 2\\left(\\frac{2(d+y)^{2}}{L^{2}}-\\frac{2(d-y)^{2}}{L^{2}}\\right)+\\lambda / 2\n\n$$\n\n\n\nThis distance must be a multiple of $\\lambda$ for interference to occur. Then,\n\n\n\n$$\n\n\\frac{4 d y}{L}+\\lambda / 2=m \\lambda\n\n$$\n\n\n\nSubstituting $m=1$ gives us $y=\\frac{\\lambda L}{8 d}$.']",['$y=\\frac{\\lambda L}{8 d}$'],False,,Expression, 1425,Optics,"## Mirror Mirror on the Wall Consider a square room with side length $L$. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength $\lambda$ is placed a distance $d$ above the center of the mirror, where $\lambda \ll d \ll L$. *Remember that the phase of light reflected by a mirror changes by $180^{\circ}$. Context question: a. On the right wall, an interference pattern emerges. What is the distance $y$ between the bottom corner and the closest bright fringe above it? Hint: you may assume $\lambda \ll y \ll L$ as well. Context answer: \boxed{$y=\frac{\lambda L}{8 d}$} ","c. Now suppose we place a transparent hemispherical shell of thickness $s$ and index of refraction $n$ over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell. hemispherical shell At what $y$ is the fringe closest to the bottom-most corner now? (You may find it convenient to use $\lfloor x\rfloor$, the largest integer below $x$.)","['Now the optical distance between the source and a spot $y$ on the wall is increased by $(n-1) s$. Then, we need\n\n\n\n$$\n\n\\frac{4 d y}{L}-(n-1) s+\\lambda / 2=m \\lambda\n\n$$\n\n\n\nTo minimize $y$, we take $m$ to be $-\\left\\lfloor\\frac{(n-1) s}{\\lambda}-\\frac{1}{2}\\right\\rfloor$. Then,\n\n\n\n$$\n\ny=\\frac{L}{4 d}\\left((n-1) s-\\lambda\\left\\lfloor\\frac{(n-1) s}{\\lambda}-\\frac{1}{2}\\right\\rfloor-\\frac{\\lambda}{2}\\right)\n\n$$\n\n\n\nBecause $(n-1) s$ is just an offset, the spacing between the fringes does not change, i.e., the spacing is still $\\lambda L /(4 d)$.']",['$y=\\frac{L}{4 d}\\left((n-1) s-\\lambda\\left\\lfloor\\frac{(n-1) s}{\\lambda}-\\frac{1}{2}\\right\\rfloor-\\frac{\\lambda}{2}\\right)$'],False,,Expression, 1425,Optics,,"c. Now suppose we place a transparent hemispherical shell of thickness $s$ and index of refraction $n$ over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell. hemispherical shell ![](https://cdn.mathpix.com/cropped/2023_12_21_051b35decfa15bf56402g-1.jpg?height=187&width=358&top_left_y=1370&top_left_x=927) At what $y$ is the fringe closest to the bottom-most corner now? (You may find it convenient to use $\lfloor x\rfloor$, the largest integer below $x$.)","['Now the optical distance between the source and a spot $y$ on the wall is increased by $(n-1) s$. Then, we need\n\n\n\n$$\n\n\\frac{4 d y}{L}-(n-1) s+\\lambda / 2=m \\lambda\n\n$$\n\n\n\nTo minimize $y$, we take $m$ to be $-\\left\\lfloor\\frac{(n-1) s}{\\lambda}-\\frac{1}{2}\\right\\rfloor$. Then,\n\n\n\n$$\n\ny=\\frac{L}{4 d}\\left((n-1) s-\\lambda\\left\\lfloor\\frac{(n-1) s}{\\lambda}-\\frac{1}{2}\\right\\rfloor-\\frac{\\lambda}{2}\\right)\n\n$$\n\n\n\nBecause $(n-1) s$ is just an offset, the spacing between the fringes does not change, i.e., the spacing is still $\\lambda L /(4 d)$.']",['$y=\\frac{L}{4 d}\\left((n-1) s-\\lambda\\left\\lfloor\\frac{(n-1) s}{\\lambda}-\\frac{1}{2}\\right\\rfloor-\\frac{\\lambda}{2}\\right)$'],False,,Expression, 1426,Optics,"## Mirror Mirror on the Wall Consider a square room with side length $L$. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength $\lambda$ is placed a distance $d$ above the center of the mirror, where $\lambda \ll d \ll L$. *Remember that the phase of light reflected by a mirror changes by $180^{\circ}$. Context question: a. On the right wall, an interference pattern emerges. What is the distance $y$ between the bottom corner and the closest bright fringe above it? Hint: you may assume $\lambda \ll y \ll L$ as well. Context answer: \boxed{$y=\frac{\lambda L}{8 d}$} Context question: c. Now suppose we place a transparent hemispherical shell of thickness $s$ and index of refraction $n$ over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell. hemispherical shell At what $y$ is the fringe closest to the bottom-most corner now? (You may find it convenient to use $\lfloor x\rfloor$, the largest integer below $x$.) Context answer: \boxed{$y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)$} ","d. Now, suppose the hemispherical shell is removed, and we instead observe the interference pattern on the top wall. To the nearest integer, what is the total number of fringes that appear on the top wall? You may assume that $d \ll L$.","['Now, the distance between the source and a spot $x$ on the wall is given by $\\sqrt{(L-d)^{2}+x^{2}}$ and the distance between the image and a spot on the wall is $\\sqrt{(L+d)^{2}+x^{2}}+\\lambda / 2$. We do not assume $x \\ll L$ this time. Subtracting the two distances gives us roughly\n\n\n\n$$\n\n\\sqrt{L^{2}+x^{2}} \\sqrt{1+\\frac{2 d L}{L^{2}+x^{2}}}-\\sqrt{L^{2}+x^{2}} \\sqrt{1-\\frac{2 d L}{L^{2}+x^{2}}}+\\lambda / 2=m \\lambda\n\n$$\n\n\n\nTaylor expanding gives us\n\n\n\n$$\n\n\\frac{2 d L}{\\sqrt{L^{2}+x^{2}}}=(m-1 / 2) \\lambda\n\n$$\n\n\n\nThen,\n\n\n\n$$\n\nx= \\pm L \\sqrt{\\frac{4 d^{2}}{(m-1 / 2)^{2} \\lambda^{2}}-1}\n\n$$\n\n\n\nFor $x$ to be physical, we require that $m-1 / 2 \\leq 2 d / \\lambda$.\n\n\n\nThe maximum allowed $x$ is $L / 2$. Then,\n\n\n\n$$\n\n\\sqrt{\\frac{4 d^{2}}{(m-1 / 2)^{2} \\lambda^{2}}-1} \\leq \\frac{1}{2}\n\n$$\n\n\n\nso\n\n\n\n$$\n\n\\frac{4 d^{2}}{(m-1 / 2)^{2} \\lambda^{2}} \\leq \\frac{5}{4}\n\n$$\n\n\n\nThus, we have that\n\n\n\n$$\n\nm-1 / 2 \\geq \\frac{4 d}{\\sqrt{5} \\lambda}\n\n$$\n\n\n\nThen, the number of fringes is\n\n\n\n$$\n\n2 \\cdot \\frac{2 d}{\\lambda}\\left(1-\\frac{2}{\\sqrt{5}}\\right)\n\n$$\n\n\n\nwhere the extra factor of 2 comes from there being two sides to the interference pattern.']",['$2 \\cdot \\frac{2 d}{\\lambda}\\left(1-\\frac{2}{\\sqrt{5}}\\right)$'],False,,Expression, 1427,Thermodynamics,"## Real Expansion Consider a ""real"" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this ""real"" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.","a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by $$ u(r)= \begin{cases}0 & r","Determine the range of values for $F$ so that the the top wedge does not slip on the bottom wedge. Express your answer(s) in terms of any or all of $m, g, \theta$, and $\mu$.","['Assume the block does not slip. Considering the horizontal forces on the entire system gives\n\n\n\n$$\n\nF-3 \\mu m g=3 m a \\quad \\Rightarrow \\quad a=\\frac{F}{3 m}-\\mu g\n\n$$\n\n\n\nWhen $F$ is small, the top wedge wants to slide downward, so static friction points up the ramp. Considering the horizontal and vertical forces on the block gives\n\n\n\n$$\n\nN \\cos \\theta+f \\sin \\theta=m g, \\quad N \\sin \\theta-f \\cos \\theta=m a\n\n$$\n\n\n\nWhen the minimal force is applied, the friction is maximal, $f=\\mu N$. Eliminating $N$ gives\n\n\n\n$$\n\nm a=m g \\frac{\\sin \\theta-\\mu \\cos \\theta}{\\cos \\theta+\\mu \\sin \\theta}\n\n$$\n\n\n\nand plugging in our first equation gives\n\n\n\n$$\n\nF_{\\min }=3 m g\\left(\\mu+\\frac{\\sin \\theta-\\mu \\cos \\theta}{\\cos \\theta+\\mu \\sin \\theta}\\right)=3 m g \\frac{\\left(1+\\mu^{2}\\right) \\tan \\theta}{1+\\mu \\tan \\theta}\n\n$$\n\n\n\nWhen $F$ is large, the top wedge wants to slide upward, so static friction points down the ramp, and\n\n\n\n$$\n\nN \\cos \\theta-f \\sin \\theta=m g, \\quad N \\sin \\theta+f \\cos \\theta=m a\n\n$$\n\n\n\nNow setting $f=\\mu N$ gives\n\n\n\n$$\n\nF_{\\max }=3 m g\\left(\\mu+\\frac{\\sin \\theta+\\mu \\cos \\theta}{\\cos \\theta-\\mu \\sin \\theta}\\right)=3 m g \\frac{2 \\mu+\\left(1-\\mu^{2}\\right) \\tan \\theta}{1-\\mu \\tan \\theta}\n\n$$\n\n\n\nTherefore, naively the range of forces so that the block will not slip is\n\n\n\n$$\n\nF \\in\\left[F_{\\min }, F_{\\max }\\right]\n\n$$\n\n\n\n\n\n\n\nHowever, to get full credit, students must account for two edge cases. First, when $\\mu>\\tan \\theta$, no force is required at all to keep the block in place, so the minimum force is zero. Second, when $\\mu>\\cot \\theta$, the block will not slip up under any circumstances, so there is no maximal force.', 'Assume the block does not slip. Considering the horizontal forces on the entire system gives\n\n\n\n$$\n\nF-3 \\mu m g=3 m a \\quad \\Rightarrow \\quad a=\\frac{F}{3 m}-\\mu g\n\n$$\n\nThe problem can also be solved geometrically. In general, the no slip condition is\n\n\n\n$$\n\n\\mu>\\tan \\phi\n\n$$\n\n\n\nwhere $\\phi$ is the angle between the vertical and the normal to the plane. Working in the noninertial reference frame of the plane, the fictitious force due to the acceleration is equivalent to a tilting of the gravity vector by an angle\n\n\n\n$$\n\n\\tan \\beta=\\frac{a}{g}\n\n$$\n\n\n\nwhere, as in above,\n\n\n\n$$\n\na=\\frac{F}{3 m}-\\mu g\n\n$$\n\n\n\nThen the top block will not slip as long as $|\\theta-\\beta| \\leq \\phi$. At the minimum acceleration $a_{\\min }, \\beta=\\theta-\\phi$, and taking the tangent of both sides gives\n\n\n\n$$\n\n\\frac{a_{\\min }}{g}=\\frac{\\tan \\theta-\\mu}{1+\\mu \\tan \\theta}\n\n$$\n\n\n\nThis is only meaningful for $\\tan \\theta>\\mu$, otherwise the answer is simply $a_{\\min }=0$. At the maximum acceleration $a_{\\max }, \\beta=\\theta+\\phi$, which gives\n\n\n\n$$\n\n\\frac{a_{\\max }}{g}=\\frac{\\tan \\theta+\\mu}{1-\\mu \\tan \\theta}\n\n$$\n\n\n\nThis is only meaningful for $\\cot \\theta>\\mu$, otherwise the answer is simply $a_{\\max }=\\infty$.']","['$[3 m g \\frac{\\left(1+\\mu^{2}\\right) \\tan \\theta}{1+\\mu \\tan \\theta}, 3 m g \\frac{2 \\mu+\\left(1-\\mu^{2}\\right) \\tan \\theta}{1-\\mu \\tan \\theta}]$']",False,,Interval, 1434,Thermodynamics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine.","a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine.","['Since a Carnot engine is reversible, it produces no entropy,\n\n\n\n$$\n\nd S_{1}+d S_{2}=\\frac{d Q_{1}}{T_{1}}+\\frac{d Q_{2}}{T_{2}}=0\n\n$$\n\n\nBy the definition of heat capacity, $d Q_{i}=C d T_{i}$, so\n\n\n\n$$\n\n\\frac{d T_{1}}{T_{1}}=-\\frac{d T_{2}}{T_{2}}\n\n$$\n\n\n\nIntegrating this equation shows that $T_{1} T_{2}$ is constant, so the final temperature is\n\n\n\n$$\n\nT_{f}=\\sqrt{T_{1} T_{2}}\n\n$$\n\n\n\nThe change in thermal energy of the objects is\n\n\n\n$$\n\nC\\left(T_{f}-T_{1}\\right)+C\\left(T_{f}-T_{2}\\right)=C\\left[2 \\sqrt{T_{1} T_{2}}-T_{1}-T_{2}\\right]\n\n$$\n\n\n\nBy the First Law of Thermodynamics, the missing energy has been used to do work, so\n\n\n\n$$\n\nW=C\\left[T_{1}+T_{2}-2 \\sqrt{T_{1} T_{2}}\\right]\n\n$$']","['$T_{f}=\\sqrt{T_{1} T_{2}}$, $W=C\\left[T_{1}+T_{2}-2 \\sqrt{T_{1} T_{2}}\\right]$']",True,,Expression, 1435,Thermodynamics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine. Context question: a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine. Context answer: \boxed{$T_{f}=\sqrt{T_{1} T_{2}}$, $W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]$} Extra Supplementary Reading Materials: Now consider three objects with equal and constant heat capacity at initial temperatures $T_{1}=100 \mathrm{~K}, T_{2}=300 \mathrm{~K}$, and $T_{3}=300 \mathrm{~K}$. Suppose we wish to raise the temperature of the third object. To do this, we could run a Carnot engine between the first and second objects, extracting work $W$. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object. Assume that all work produced by running engines can be stored and used without dissipation.",b. Find the minimum temperature $T_{L}$ to which the first object can be lowered.,"['By the Second Law of Thermodynamics, we must have $T_{L}=100 \\mathrm{~K}$. Otherwise, we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one.']",['$100$'],False,K,Numerical,1e0 1436,Thermodynamics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine. Context question: a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine. Context answer: \boxed{$T_{f}=\sqrt{T_{1} T_{2}}$, $W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]$} Extra Supplementary Reading Materials: Now consider three objects with equal and constant heat capacity at initial temperatures $T_{1}=100 \mathrm{~K}, T_{2}=300 \mathrm{~K}$, and $T_{3}=300 \mathrm{~K}$. Suppose we wish to raise the temperature of the third object. To do this, we could run a Carnot engine between the first and second objects, extracting work $W$. This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object. Assume that all work produced by running engines can be stored and used without dissipation. Context question: b. Find the minimum temperature $T_{L}$ to which the first object can be lowered. Context answer: \boxed{$100$} ",c. Find the maximum temperature $T_{H}$ to which the third object can be raised.,"['The entropy of an object with constant heat capacity is\n\n\n\n$$\n\nS=\\int \\frac{d Q}{T}=C \\int \\frac{d T}{T}=C \\ln T\n\n$$\n\n\n\n\n\n\n\nSince the total entropy remains constant, $T_{1} T_{2} T_{3}$ is constant; this is a direct generalization of the result for $T_{f}$ found in part (a). Energy is also conserved, as it makes no sense to leave stored energy unused, so $T_{1}+T_{2}+T_{3}$ is constant.\n\n\n\nWhen one object is at temperature $T_{H}$, the other two must be at the same lower temperature $T_{0}$, or else further work could be extracted from their temperature difference, so\n\n\n\n$$\n\nT_{1}+T_{2}+T_{3}=T_{H}+2 T_{0}, \\quad T_{1} T_{2} T_{3}=T_{H} T_{0}^{2}\n\n$$\n\n\n\nPlugging in temperatures with values divided by 100 for convenience, and eliminating $T_{0}$ gives\n\n\n\n$$\n\nT_{H}\\left(7-T_{H}\\right)^{2}=36\n\n$$\n\n\n\nWe know that $T_{H}=1$ is one (spurious) solution, since this is the minimum possible final temperature as found in part (b). The other roots are $T_{H}=4$ and $T_{H}=9$ by the quadratic formula. The solution $T_{H}=9$ is impossible by energy conservation, so\n\n\n\n$$\n\nT_{H}=400 \\mathrm{~K}\n\n$$\n\n\n\nIt is also possible to solve the problem more explicitly. For example, one can run a Carnot cycle between the first two objects until they are at the same temperature, then run a Carnot cycle in reverse between the last two objects using the stored work. At this point, the first two objects will no longer be at the same temperature, so we can repeat the procedure; this yields an infinite series for $T_{H}$. Some students did this, and took only the first term of the series. This yields a fairly good approximation of $T_{H} \\approx 395 \\mathrm{~K}$.\n\n\n\nAnother explicit method is to continuously switch between running one Carnot engine forward and another Carnot engine in reverse; this yields three differential equations for $T_{1}, T_{2}$, and $T_{3}$. Solving the equations and setting $T_{1}=T_{2}$ yields $T_{3}=T_{H}$.']",['395'],False,K,Numerical,5e0 1437,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship. Let the strength of the Earth's magnetic field near the ship be $B_{e}$, and the orientation of the field be horizontal, pointing directly toward true north. The Earth's magnetic field $B_{e}$ will magnetize the ship, which will then create a second magnetic field $B_{s}$ in the vicinity of the ship's compass given by $$ \overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right) $$ where $K_{b}$ and $K_{s}$ are positive constants, $\theta$ is the angle between the heading of the ship and magnetic north, measured clockwise, $\hat{\mathbf{b}}$ and $\hat{\mathbf{s}}$ are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively. Because of the ship's magnetic field, the ship's compass will no longer necessarily point North.","a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$.","[""We add the fields to get the local field. The northward component is\n\n\n\n$$\n\nB_{\\text {north }}=B_{e}-B_{e} K_{b} \\cos \\theta \\cos \\theta-B_{e} K_{s} \\sin \\theta \\sin \\theta\n\n$$\n\n\n\nwhile the eastward component is\n\n\n\n$$\n\nB_{\\text {east }}=-B_{e} K_{b} \\sin \\theta \\cos \\theta+B_{e} K_{s} \\cos \\theta \\sin \\theta\n\n$$\n\n\n\nThe deviation is given by\n\n\n\n$$\n\n\\tan \\delta \\theta=\\left(K_{s}-K_{b}\\right) \\frac{\\sin \\theta \\cos \\theta}{1-K_{b} \\cos ^{2} \\theta-K_{s} \\sin ^{2} \\theta}\n\n$$\n\n\n\nThis form is particularly nice, because as we'll see below, $K_{b}$ and $K_{s}$ are small enough to ignore in the denominator.""]",['$\\delta \\theta=\\arctan \\left(K_{s}-K_{b}\\right) \\frac{\\sin \\theta \\cos \\theta}{1-K_{b} \\cos ^{2} \\theta-K_{s} \\sin ^{2} \\theta}$'],False,,Expression, 1438,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship. Let the strength of the Earth's magnetic field near the ship be $B_{e}$, and the orientation of the field be horizontal, pointing directly toward true north. The Earth's magnetic field $B_{e}$ will magnetize the ship, which will then create a second magnetic field $B_{s}$ in the vicinity of the ship's compass given by $$ \overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right) $$ where $K_{b}$ and $K_{s}$ are positive constants, $\theta$ is the angle between the heading of the ship and magnetic north, measured clockwise, $\hat{\mathbf{b}}$ and $\hat{\mathbf{s}}$ are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively. Because of the ship's magnetic field, the ship's compass will no longer necessarily point North. Context question: a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$. Context answer: \boxed{$\delta \theta=\arctan \left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta}$} ","b. Assuming that $K_{b}$ and $K_{s}$ are both much smaller than one, at what heading(s) $\theta$ will the deviation $\delta \theta$ be largest?","[""By inspection, $\\theta=45^{\\circ}$ will yield the largest deviation. It's also acceptable to list $45^{\\circ}, 135^{\\circ}$, $225^{\\circ}$, and $315^{\\circ}$.""]","['$45$, $135$, $225$, $315$']",True,$^{\circ}$,Numerical,0 1439,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship. Let the strength of the Earth's magnetic field near the ship be $B_{e}$, and the orientation of the field be horizontal, pointing directly toward true north. The Earth's magnetic field $B_{e}$ will magnetize the ship, which will then create a second magnetic field $B_{s}$ in the vicinity of the ship's compass given by $$ \overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right) $$ where $K_{b}$ and $K_{s}$ are positive constants, $\theta$ is the angle between the heading of the ship and magnetic north, measured clockwise, $\hat{\mathbf{b}}$ and $\hat{\mathbf{s}}$ are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively. Because of the ship's magnetic field, the ship's compass will no longer necessarily point North. Context question: a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$. Context answer: \boxed{$\delta \theta=\arctan \left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta}$} Context question: b. Assuming that $K_{b}$ and $K_{s}$ are both much smaller than one, at what heading(s) $\theta$ will the deviation $\delta \theta$ be largest? Context answer: \boxed{$45$, $135$, $225$, $315$} Extra Supplementary Reading Materials: A pair of iron balls placed in the same horizontal plane as the compass but a distance $d$ away can be used to help correct for the error caused by the induced magnetism of the ship. A binnacle, protecting the ship's compass in the center, with two soft iron spheres to help correct for errors in the compass heading. The use of the spheres was suggested by Lord Kelvin. Just like the ship, the iron balls will become magnetic because of the Earth's field $B_{e}$. As spheres, the balls will individually act like dipoles. A dipole can be thought of as the field produced by two magnetic monopoles of strength $\pm m$ at two different points. The magnetic field of a single pole is $$ \overrightarrow{\mathbf{B}}= \pm m \frac{\hat{\mathbf{r}}}{r^{2}} $$ where the positive sign is for a north pole and the negative for a south pole. The dipole magnetic field is the sum of the two fields: a north pole at $y=+a / 2$ and a south pole at $y=-a / 2$, where the $y$ axis is horizontal and pointing north. $a$ is a small distance much smaller than the radius of the iron balls; in general $a=K_{i} B_{e}$ where $K_{i}$ is a constant that depends on the size of the iron sphere. ","c. Derive an expression for the magnetic field $\overrightarrow{\mathbf{B}}_{i}$ from the iron a distance $d \gg a$ from the center of the ball. Note that there will be a component directed radially away from the ball and a component directed tangent to a circle of radius $d$ around the ball, so using polar coordinates is recommended.","['This problem is not nearly as difficult as it looks.\n\n\n\n\n\n\n\nConsider the colored triangle above. The black side has length $a$. The angle between the green and black sides is $\\phi$, so the length of the red side is $a \\sin \\phi$ and the length of the green side is $a \\cos \\phi$.\n\n\n\nThe magnetic field strength from one magnetic pole a distance $d$ away is given by\n\n\n\n$$\n\nB= \\pm m \\frac{1}{d^{2}}\n\n$$\n\n\n\nThe sum of the two fields has two components. The angular component is a measure of the ""opening"" of the triangle formed by the two vectors, and since the two vectors basically have the same length, we can use similar triangles to conclude\n\n\n\n$$\n\n\\frac{a \\sin \\phi}{d} \\approx \\frac{B_{\\phi}}{B} \\quad \\Rightarrow \\quad B_{\\phi}=m \\frac{a}{d^{3}} \\sin \\phi=B_{e} \\frac{m K_{i}}{d^{3}} \\sin \\phi\n\n$$\n\n\n\nAs expected, this component vanishes for $\\phi=0$.\n\n\n\nThe radial component is given by the difference in the lengths of the two field vectors, or\n\n\n\n$$\n\nB_{r}=m\\left(\\frac{1}{d^{2}}-\\frac{1}{(d+x)^{2}}\\right)=\\frac{m}{d^{2}}\\left(1-\\frac{1}{(1+x / d)^{2}}\\right) \\approx \\frac{m}{d^{2}} \\frac{2 x}{d}\n\n$$\n\n\n\nwhere $x=a \\cos \\phi$ is the length of the green side, so\n\n\n\n$$\n\nB_{r}=2 B_{e} \\frac{m K_{i}}{d^{3}} \\cos \\phi\n\n$$\n\n\n\nThat wasn\'t so bad, was it?']",['$B_{r}=2 B_{e} \\frac{m K_{i}}{d^{3}} \\cos \\phi$'],False,,Expression, 1440,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: A ship can be thought of as a symmetric arrangement of soft iron. In the presence of an external magnetic field, the soft iron will become magnetized, creating a second, weaker magnetic field. We want to examine the effect of the ship's field on the ship's compass, which will be located in the middle of the ship. Let the strength of the Earth's magnetic field near the ship be $B_{e}$, and the orientation of the field be horizontal, pointing directly toward true north. The Earth's magnetic field $B_{e}$ will magnetize the ship, which will then create a second magnetic field $B_{s}$ in the vicinity of the ship's compass given by $$ \overrightarrow{\mathbf{B}}_{s}=B_{e}\left(-K_{b} \cos \theta \hat{\mathbf{b}}+K_{s} \sin \theta \hat{\mathbf{s}}\right) $$ where $K_{b}$ and $K_{s}$ are positive constants, $\theta$ is the angle between the heading of the ship and magnetic north, measured clockwise, $\hat{\mathbf{b}}$ and $\hat{\mathbf{s}}$ are unit vectors pointing in the forward direction of the ship (bow) and directly right of the forward direction (starboard), respectively. Because of the ship's magnetic field, the ship's compass will no longer necessarily point North. Context question: a. Derive an expression for the deviation of the compass, $\delta \theta$, from north as a function of $K_{b}$, $K_{s}$, and $\theta$. Context answer: \boxed{$\delta \theta=\arctan \left(K_{s}-K_{b}\right) \frac{\sin \theta \cos \theta}{1-K_{b} \cos ^{2} \theta-K_{s} \sin ^{2} \theta}$} Context question: b. Assuming that $K_{b}$ and $K_{s}$ are both much smaller than one, at what heading(s) $\theta$ will the deviation $\delta \theta$ be largest? Context answer: \boxed{$45$, $135$, $225$, $315$} Extra Supplementary Reading Materials: A pair of iron balls placed in the same horizontal plane as the compass but a distance $d$ away can be used to help correct for the error caused by the induced magnetism of the ship. A binnacle, protecting the ship's compass in the center, with two soft iron spheres to help correct for errors in the compass heading. The use of the spheres was suggested by Lord Kelvin. Just like the ship, the iron balls will become magnetic because of the Earth's field $B_{e}$. As spheres, the balls will individually act like dipoles. A dipole can be thought of as the field produced by two magnetic monopoles of strength $\pm m$ at two different points. The magnetic field of a single pole is $$ \overrightarrow{\mathbf{B}}= \pm m \frac{\hat{\mathbf{r}}}{r^{2}} $$ where the positive sign is for a north pole and the negative for a south pole. The dipole magnetic field is the sum of the two fields: a north pole at $y=+a / 2$ and a south pole at $y=-a / 2$, where the $y$ axis is horizontal and pointing north. $a$ is a small distance much smaller than the radius of the iron balls; in general $a=K_{i} B_{e}$ where $K_{i}$ is a constant that depends on the size of the iron sphere. Context question: c. Derive an expression for the magnetic field $\overrightarrow{\mathbf{B}}_{i}$ from the iron a distance $d \gg a$ from the center of the ball. Note that there will be a component directed radially away from the ball and a component directed tangent to a circle of radius $d$ around the ball, so using polar coordinates is recommended. Context answer: \boxed{$B_{r}=2 B_{e} \frac{m K_{i}}{d^{3}} \cos \phi$} ","d. If placed directly to the right and left of the ship compass, the iron balls can be located at a distance $d$ to cancel out the error in the magnetic heading for any angle(s) where $\delta \theta$ is largest. Assuming that this is done, find the resulting expression for the combined deviation $\delta \theta$ due to the ship and the balls for the magnetic heading for all angles $\theta$.","['Note that the two iron balls create a magnetic field near the compass that behaves like that of the ship as a whole. There is a component directed toward the bow given by\n\n\n\n$$\n\nB_{b}=-2 B_{\\theta} \\propto \\sin \\phi \\propto \\cos \\theta\n\n$$\n\n\n\nand a component directed toward the starboard given by\n\n\n\n$$\n\nB_{s}=2 B_{r} \\propto \\cos \\phi \\propto \\sin \\theta\n\n$$\n\n\n\nwhere the factors of 2 are because there are two balls. Note that $\\theta$ is the ship heading while $\\phi$ is the angle between North and the location of the compass relative to one of the balls. Thus, if the field is corrected for the maximum angles it will necessarily cancel out the induced ship field for all of the angles, so that\n\n\n\n$$\n\n\\delta \\theta=0\n\n$$\n\n\n\nfor all $\\theta$. Effectively, this means placing the balls to make $K_{b}=K_{s}$.']",['0'],False,,Numerical,0 1441,Modern Physics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Relativistic particles obey the mass energy relation $$ E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2} $$ where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light. A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or ""delta"". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.","a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.","[""We can solve the problem exactly, approximating only in the last step. This certainly isn't necessary; we do this to illustrate a useful technique. We set $c=1$ throughout, and transform to an inertial frame where the proton is initially at rest. Before the collision,\n\n\n\n$$\n\nE_{p}=m_{p}, \\quad E_{\\gamma}=\\left|p_{\\gamma}\\right|\n\n$$\n\n\n\nFor the $\\Delta$ particle, we have the usual relativistic relation\n\n\n\n$$\n\nE_{\\Delta}^{2}=p_{\\Delta}^{2}+m_{\\Delta}^{2}\n\n$$\n\n\n\nBy energy-momentum conservation,\n\n\n\n$$\n\nE_{p}+E_{\\gamma}=E_{\\Delta}, \\quad p_{\\gamma}=p_{\\Delta} .\n\n$$\n\n\n\nCombining these results gives\n\n\n\n$$\n\n\\left(m_{p}+E_{\\gamma}\\right)^{2}=E_{\\gamma}^{2}+m_{\\Delta}^{2} \\Rightarrow E_{\\gamma}=\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{2 m_{p}}\n\n$$\n\n\n\nNow we need to transform back to the original frame, where the energy of the photon is $E_{b}$. We can use the Lorentz transformation for this, but it's a little easier to realize that $E=h f$ for photons, and apply the Doppler shift. Then\n\n\n\n$$\n\n\\alpha \\equiv \\frac{E_{b}}{E_{\\gamma}}=\\sqrt{\\frac{1-\\beta}{1+\\beta}}\n\n$$\n\n\n\nwhere $\\beta$ is the velocity parameter of the proton in the inertial frame where the photon has energy $E_{b}$. Solving for $\\beta$ in terms of the energy ratio $\\alpha$,\n\n\n\n$$\n\n\\beta=\\frac{1-\\alpha^{2}}{1+\\alpha^{2}}\n\n$$\n\n\n\nTo calculate the proton energy, we need the Lorentz factor,\n\n\n\n$$\n\n\\gamma=\\frac{1}{\\sqrt{1-\\beta^{2}}}=\\frac{1+\\alpha^{2}}{2 \\alpha}\n\n$$\n\n\n\n\n\n\n\nThen the proton energy in the original frame is\n\n\n\n$$\n\nE_{p}=\\gamma m_{p}=\\frac{m_{p}}{2}\\left(\\alpha+\\frac{1}{\\alpha}\\right)=\\frac{m_{p}}{2}\\left(\\frac{2 m_{p} E_{b}}{m_{\\Delta}^{2}-m_{p}^{2}}+\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{2 m_{p} E_{b}}\\right)\n\n$$\n\n\n\nwhich is the exact answer.\n\n\n\nAt this point, we can approximate. The second term is much larger than the first, so\n\n\n\n$$\n\nE_{p} \\approx \\frac{m_{\\Delta}^{2}-m_{p}^{2}}{4 E_{b}}\n\n$$\n\n\n\nwhich was the required answer for this problem."", ""We now show another method that approximates throughout. We'll do everything in the lab frame, so the symbols in this solution don't mean the same things they did in solution 1. In the lab frame, energy-momentum conservation gives\n\n\n\n$$\n\np_{p}-p_{b}=p_{\\Delta}, \\quad E_{p}+E_{b}=E_{\\Delta}\n\n$$\n\n\n\nSquaring both expressions and dropping $E_{b}^{2}$ terms, since $E_{b}$ is small,\n\n\n\n$$\n\np_{p}^{2}-2 p_{p} p_{b} \\approx p_{\\Delta}^{2}, \\quad E_{p}^{2}+2 E_{p} E_{b} \\approx E_{\\Delta}^{2}\n\n$$\n\n\n\nSubtracting these equations gives\n\n\n\n$$\n\nm_{p}^{2}+2 E_{p} E_{b}+2 p_{p} E_{b}=m_{\\Delta}^{2} \\Rightarrow E_{p}+p_{p}=\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{2 E_{b}}\n\n$$\n\n\n\nSince $E_{b}$ is small, this quantity must be large. But this means the protons are ultrarelativistic, so $E_{p} \\approx p_{p}$, giving\n\n\n\n$$\n\nE_{p} \\approx \\frac{m_{\\Delta}^{2}-m_{p}^{2}}{4 E_{b}}\n\n$$\n\n\n\nas desired.""]",['$\\frac{m_{\\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$'],False,,Expression, 1442,Modern Physics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Relativistic particles obey the mass energy relation $$ E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2} $$ where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light. A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or ""delta"". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum. Context question: a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small. Context answer: \boxed{$\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$} ","b. In this case, the photon energy $E_{b}$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06 \mathrm{~mm}$. Determine the energy of the photons, writing your answer in electron volts.",['Plugging in the numbers gives\n\n\n\n$$\n\nE=\\frac{h c}{\\lambda}=0.00112 \\mathrm{eV}\n\n$$'],['$0.00112$'],False,eV,Numerical,2e-5 1443,Modern Physics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Relativistic particles obey the mass energy relation $$ E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2} $$ where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light. A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or ""delta"". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum. Context question: a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small. Context answer: \boxed{$\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$} Context question: b. In this case, the photon energy $E_{b}$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06 \mathrm{~mm}$. Determine the energy of the photons, writing your answer in electron volts. Context answer: \boxed{$0.00112$} ","c. Assuming this value for $E_{b}$, what is the energy of the proton, in electron volts, that will allow the above reaction? This sets an upper limit on the energy of cosmic rays. The mass of the proton is given by $m_{p} c^{2}=938 \mathrm{MeV}$ and the mass of the $\Delta$ is given by $m_{\Delta} c^{2}=1232 \mathrm{MeV}$.",['Restoring the factors of $c$ and plugging in the numbers gives\n\n\n\n$$\n\nE_{p}=1.4 \\times 10^{20} \\mathrm{eV}\n\n$$\n\n\n\nThis is known as the GZK bound for cosmic rays.'],['$1.4 \\times 10^{20}$'],False,eV,Numerical,5e18 1444,Mechanics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Suppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table. ","a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum $p$. i. At what height $H$ above the table is the impulse $p$ required to topple the domino smallest? ii. What is the minimum value of $p$ to topple the domino?","[""First we'll look for the height $H$ at which we should push to topple the domino with the least momentum. A convenient method is to look at the angular momentum in the domino because this is easy to calculate and describes rotational motion. Because the push is horizontal, the moment arm of the push (about the domino's rotational axis) is purely vertical. That means the angular momentum of the push is $p H$, where $p$ is the momentum imparted and $H$ is the height of the push. There is some minimum angular momentum $L_{\\min }$ to topple the domino, so we set $L_{\\min }=p H$. The bigger $H$, the smaller $p$, so we should choose the largest possible $H$. In other words, we should push at the very top of the domino, $H=h$. While we're on this part, note that if the push weren't constrained to be horizontal, the push could be a little bit smaller since the moment arm could be the entire diagonal of the thin edge of the domino.\n\n\n\nNext we calculate the minimum $p$ using energy. As the domino rotates, it converts kinetic energy to potential energy, so we'll calculate both. The domino's potential energy is greatest when its center of mass is directly over the contact point of the domino and the table. That height is half the diagonal of the domino, so the distance from the contact point to the center of the domino is $\\frac{1}{2} \\sqrt{t^{2}+h^{2}}$. From the push until it reaches this point, the domino's potential energy increases by\n\n\n\n$$\n\n\\Delta U=\\frac{1}{2} m g\\left(\\sqrt{t^{2}+h^{2}}-h\\right)\n\n$$\n\n\n\nThen the domino topples. By conservation of energy, $\\Delta U$ is how much rotational kinetic energy the domino must have begun with.\n\n\n\nNext we find the kinetic energy using the rotational kinetic energy formula, $T=L^{2} / 2 I$. The moment of inertia of the domino about its contact point with the table is $I=\\frac{1}{3} m\\left(h^{2}+t^{2}\\right)$.\n\n\n\n\n\n\n\nYou can find this with an integral, or if you know the moment of inertia about the center $\\left(\\frac{1}{12} m\\left(h^{2}+t^{2}\\right)\\right.$ ), you can use the parallel axis theorem. We know the momentum $L=p h$, so the kinetic energy is\n\n\n\n$$\n\nT=\\frac{1}{2} \\frac{L^{2}}{I}=\\frac{1}{2} \\frac{(p h)^{2}}{(1 / 3) m\\left(h^{2}+t^{2}\\right)}=\\frac{3}{2} \\frac{p^{2} h^{2}}{m\\left(h^{2}+t^{2}\\right)}\n\n$$\n\n\n\nSetting the initial kinetic energy equal to the gain in potential energy and solving for $p$,\n\n\n\n$$\n\np_{\\min }=\\frac{1}{\\sqrt{3}} \\frac{m}{h} \\sqrt{g\\left(\\sqrt{t^{2}+h^{2}}-h\\right)\\left(h^{2}+t^{2}\\right)}\n\n$$""]","['$H=h$ , $p_{\\min }=\\frac{1}{\\sqrt{3}} \\frac{m}{h} \\sqrt{g\\left(\\sqrt{t^{2}+h^{2}}-h\\right)\\left(h^{2}+t^{2}\\right)}$']",True,,Expression, 1445,Mechanics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Suppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table. Context question: a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum $p$. i. At what height $H$ above the table is the impulse $p$ required to topple the domino smallest? ii. What is the minimum value of $p$ to topple the domino? Context answer: \boxed{$H=h$ , $p_{\min }=\frac{1}{\sqrt{3}} \frac{m}{h} \sqrt{g\left(\sqrt{t^{2}+h^{2}}-h\right)\left(h^{2}+t^{2}\right)}$} ","b. Next, imagine a long row of dominoes with equal spacing $l$ between the nearest sides of any pair of adjacent dominos, as shown above. When a domino topples, it collides with the next domino in the row. Imagine this collision to be completely inelastic. What fraction of the total kinetic energy is lost in the collision of the first domino with the second domino?","[""Right after the collision, the dominoes touch at a height $\\sqrt{h^{2}-l^{2}}$ above the table. The second domino is vertical, while the first is rotated so that the angle between its leading edge and the table is $\\theta=\\arccos (l / h)$. Let the dominoes' angular velocities be $\\omega_{1}$ and $\\omega_{2}$ respectively. After the dominoes collide, they stick together. If we take the two parts of the dominoes that are in contact, they must have the same horizontal velocity component in order for the dominoes to stay in contact. For the first domino, this velocity is $\\omega_{1} h \\sin \\theta=\\omega_{1} \\sqrt{h^{2}-l^{2}}$. For the second domino it is $\\omega_{2}$ times the height of the impact point, so $\\omega_{2} \\sqrt{h^{2}-l^{2}}$. Because these velocity components must be equal, $\\omega_{1}=\\omega_{2}$.\n\n\n\nThis means the dominoes have the same angular momentum as each other, measured relative to their respective rotation axes. If we look at the collision, the forces between the dominoes are purely horizontal because the dominoes' faces are frictionless, so they only exert normal forces. The second domino is vertical, so all its normal forces are purely horizontal. These horizontal normal forces exchange angular momentum between the two dominoes. They have the same moment arm (i.e. the height of the collision), so the amount of angular momentum transferred out of the first domino is equal to the amount gained by the second domino, both measured relative to the dominoes' respective rotation axes. Because the dominoes are identical and have the same angular velocity, they have the same angular momentum. This means each domino has half as much angular momentum about its rotation axis as the first domino had just before the collision. (Note that forces from the table cannot change the angular momentum of the dominoes about their respective rotational axes because such forces have zero moment arm, so only the inter-domino forces need to be examined here.)\n\n\n\nKinetic energy scales with the square of angular momentum, so each domino has a quarter as much kinetic energy after the collision as the first domino had before it. That means the total kinetic energy after the collision is half what it was before the collision. The fraction of kinetic energy lost is one half.""]",['50'],False,percent,Numerical,0 1446,Mechanics,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Suppose a domino stands upright on a table. It has height $h$, thickness $t$, width $w$ (as shown below), and mass $m$. The domino is free to rotate about its edges, but will not slide across the table. Context question: a. Suppose we give the domino a sharp, horizontal impulsive push with total momentum $p$. i. At what height $H$ above the table is the impulse $p$ required to topple the domino smallest? ii. What is the minimum value of $p$ to topple the domino? Context answer: \boxed{$H=h$ , $p_{\min }=\frac{1}{\sqrt{3}} \frac{m}{h} \sqrt{g\left(\sqrt{t^{2}+h^{2}}-h\right)\left(h^{2}+t^{2}\right)}$} Context question: b. Next, imagine a long row of dominoes with equal spacing $l$ between the nearest sides of any pair of adjacent dominos, as shown above. When a domino topples, it collides with the next domino in the row. Imagine this collision to be completely inelastic. What fraction of the total kinetic energy is lost in the collision of the first domino with the second domino? Context answer: \boxed{50} ","c. After the collision, the dominoes rotate in such a way so that they always remain in contact. Assume that there is no friction between the dominoes and the first domino was given the smallest possible push such that it toppled. What is the minimum $l$ such that the second domino will topple? You may work to lowest nontrivial order in the angles through which the dominoes have rotated. Equivalently, you may approximate $t, l \ll h$.","[""As in part (a), we will find the highest potential energy of the system and make sure the initial kinetic energy is high enough to get the system to that point.\n\n\n\nAt any time after the collision, let us call the angle that the first domino has rotated past its point of highest potential energy $\\alpha$, and the angle that the second domino has rotated past its point of highest potential energy $\\beta$. Also, let's call the angle a domino rotates from its standing position up to its point of highest potential energy $\\phi$, so the dominoes have rotated $\\phi+\\alpha$ and $\\phi+\\beta$ respectively.\n\n\n\nThe dominoes need to be in contact. The top right corner of the first domino has moved horizontally a distance $h \\sin (\\phi+\\alpha)$ which we will approximate as $h(\\phi+\\alpha)$ using the small angle approximation. The top left corner of the second domino moves horizontally forward by $h \\sin (\\phi+\\beta)+t(1-\\cos (\\phi+\\beta))$. We ignore the cosine term as second order in the rotation angle and approximate this as $h(\\phi+\\beta)$.\n\n\n\nThe $y$-coordinate of the upper right corner of the first domino is $h(1-\\cos (\\phi+\\alpha)) \\approx h$. The $y$-coordinate of the upper left corner of the second domino is $h(1-\\cos (\\phi+\\beta))+t \\sin (\\phi+\\beta) \\approx$ $h+t(\\phi+\\beta)$ There is a first-order difference in $y$-coordinates of the two corners, but this means the difference in $x$ coordinate between the top left corner of the second domino and the top right corner of the first domino is second-order in the rotation angles. We conclude that to first order\n\n\n\n$$\n\nh(\\phi+\\alpha)=l+h(\\phi+\\beta) \\quad \\Rightarrow \\quad \\alpha=\\frac{l}{h}+\\beta\n\n$$\n\n\n\nbecause this condition puts the top right corner of the first domino at the same position as the top left of the second domino.\n\n\n\nThe potential energy of the first domino, setting zero potential energy to be when the domino is upright, is $U_{1}=\\Delta U\\left(1-(\\alpha / \\phi)^{2}\\right)$ to second order in $\\alpha$, and for the second domino, $U_{2}=$ $\\Delta U\\left(1-(\\beta / \\phi)^{2}\\right)$. These figures come from fitting a quadratic whose peak is when the center of mass is above the rotation point and which is zero when the domino is upright. The total potential energy is $U_{1}+U_{2}$, and using the relation between $\\alpha$ and $\\beta$, it is minimized for\n\n\n\n$$\n\n\\beta_{\\max }=-\\frac{l}{2 h}\n\n$$\n\n\n\nIn other words, the second domino is as far away from rotating to the top of arc as the first domino has rotated past the top of its arc. This gives a maximum potential energy\n\n\n\n$$\n\nU_{\\max }=2 \\Delta U\\left(1-\\frac{l^{2}}{4 t^{2}}\\right)\n\n$$\n\n\n\nwhere we have used the approximation $\\phi \\approx t / h$.\n\n\nAt impact, $U_{\\text {impact }} \\approx \\Delta U\\left(-l^{2} / t^{2}+2 l / t\\right)$. Before impact, the kinetic energy is $\\Delta U-U_{\\text {impact }}$ because the maximum potential energy before impact was $\\Delta U$, and the kinetic energy was zero there. The kinetic energy just after the collision is then $\\left(\\Delta U-U_{\\text {impact }}\\right) / 2$. Setting this equal to the potential energy gain as the two dominoes rotate to their highest potential energy $U_{\\max }$,\n\n\n\n$$\n\n\\frac{1}{2}\\left(\\Delta U-U_{\\text {impact }}\\right)=U_{\\max }-U_{\\text {impact }}\n\n$$\n\n\n\nor\n\n\n\n$$\n\n\\frac{1}{2} \\Delta U=U_{\\max }-\\frac{1}{2} U_{\\text {impact }}\n\n$$\n\n\n\nPlugging in the earlier expressions for all these gives\n\n\n\n$$\n\n\\frac{1}{2} \\Delta U=2 \\Delta U\\left(1-\\frac{l^{2}}{4 t^{2}}\\right)-\\Delta U\\left(\\frac{l}{t}-\\frac{l^{2}}{2 t^{2}}\\right)\n\n$$\n\n\n\nand solving this yields\n\n\n\n$$\n\nl=\\frac{3}{2} t\n\n$$\n\n\n\nto first order in $t$.""]",['$l=\\frac{3}{2} t$'],False,,Expression, 1447,Mechanics,,"d. A row of toppling dominoes can be considered to have a propagation speed of the length $l+t$ divided by the time between successive collisions. When the first domino is given a minimal push just large enough to topple and start a chain reaction of toppling dominoes, the speed increases with each domino, but approaches an asymptotic speed $v$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ccfe1ffced15d55565eag-1.jpg?height=246&width=358&top_left_y=1449&top_left_x=905) Suppose there is a row of dominoes on another planet. These dominoes have the same density as the dominoes previously considered, but are twice as tall, wide, and thick, and placed with a spacing of $2 l$ between them. If this row of dominoes topples with the same asymptotic speed $v$ previously found, what is the local gravitational acceleration on this planet?","['This part is independent of the others and requires only dimensional analysis. The speed $v$ can depend on $g, h, w, l$. To get a quantity with dimensions $\\left[L T^{-1}\\right]$, we must take\n\n\n\n$$\n\nv=c \\sqrt{g L}\n\n$$\n\n\n\nwhere $L$ is some length made from $h, w$, and $l$. On the new planet, $v$ is the same and $L$ is twice as much, so $g$ must be half as great on the new planet.']",['So the final answer is $g$ must be half as great on the new planet'],False,,Need_human_evaluate, 1448,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target.","a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome.",['The electric potential is given by\n\n\n\n$$\n\nV=\\frac{q}{4 \\pi \\epsilon_{0} a}\n\n$$\n\n\n\nand the electric field is given by\n\n\n\n$$\n\nE=\\frac{q}{4 \\pi \\epsilon_{0} a^{2}}\n\n$$\n\n\n\n\n\n\n\nso\n\n\n\n$$\n\nE=\\frac{V}{a}=10^{6} \\mathrm{~V} / \\mathrm{m}\n\n$$'],['$10^{6}$'],False,V/m,Numerical,5e5 1449,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} ","b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value.",['The time constant is given by\n\n\n\n$$\n\n\\tau=R C\n\n$$\n\n\n\nwhere\n\n\n\n$$\n\nC=Q / V=4 \\pi \\epsilon_{0} a=5.56 \\times 10^{-11} \\mathrm{~F}\n\n$$\n\n\n\nand\n\n\n\n$$\n\nR=20 r_{0}=10^{10} \\Omega\n\n$$\n\n\n\nso\n\n\n\n$$\n\n\\tau=R C=0.556 \\mathrm{~s} .\n\n$$'],['0.556'],False,s,Numerical,5e-2 1450,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} Context question: b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value. Context answer: \boxed{0.556} ","c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome.","['There are several ways the dome can discharge, two of which are along the resistors and the proton beam. We will ignore any other path.\n\n\n\nAt $500 \\mathrm{kV}$, the current through the resistor chain is $50 \\mu \\mathrm{A}$, from $V=I R$. So the total current $I$ needed to be supplied to the dome is $75 \\mu \\mathrm{A}$. This is sprayed onto the belt, which moves at a rate of\n\n\n\n$$\n\n\\frac{\\delta A}{\\Delta t}=v_{b} w\n\n$$\n\n\n\nso the necessary surface charge density is\n\n\n\n$$\n\n\\sigma=\\frac{I}{v_{b} w}=\\frac{(75 \\mu \\mathrm{C} / \\mathrm{s})}{(20 \\mathrm{~m} / \\mathrm{s})(0.10 \\mathrm{~m})}=37.5 \\mu \\mathrm{C} / \\mathrm{m}^{2}\n\n$$']",['$37.5$'],False,$\mu \mathrm{C} / \mathrm{m}^{2}$,Numerical,5e-1 1451,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} Context question: b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value. Context answer: \boxed{0.556} Context question: c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome. Context answer: \boxed{$37.5$} ","d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength. ","['Start with $F=q v B$ where $F$ is the force on the protons, and $v$ the velocity. The protons are non-relativistic, so\n\n\n\n$$\n\n\\frac{1}{2} m v^{2}=q V\n\n$$\n\n\n\nThey move in a circle of radius $r$ inside the field, given by\n\n\n\n$$\n\n\\frac{m v^{2}}{r}=q v B\n\n$$\n\n\n\nCombining these equations gives\n\n\n\n$$\n\nr=\\frac{m v}{q B}=\\frac{m}{q B} \\sqrt{\\frac{2 q V}{m}}=\\frac{1}{B} \\sqrt{\\frac{2 m V}{q}}\n\n$$\n\n\n\nSolving for the magnetic field strength,\n\n\n\n$$\n\nB=\\frac{1}{r} \\sqrt{\\frac{2 m V}{q}}\n\n$$\n\n\n\nTo relate this to the angle, we need to do some geometry. Sketch two circles, one of radius $b$, the other of radius $r$, that intersect perpendicular to each other, as in the diagram above. Then drawing a triangle gives\n\n\n\n$$\n\n\\tan \\frac{\\theta}{2}=\\frac{b}{r} .\n\n$$\n\n\n\nCombining our results, the answer is\n\n\n\n$$\n\nB=\\frac{\\tan \\theta / 2}{b} \\sqrt{\\frac{2 m V}{q}}=0.0894 \\mathrm{~T}\n\n$$']",['0.0894'],False,T,Numerical,1e-3 1451,Electromagnetism,,"d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength. ![](https://cdn.mathpix.com/cropped/2023_12_21_3059736d5ab3a83051d2g-1.jpg?height=342&width=659&top_left_y=363&top_left_x=836)","['Start with $F=q v B$ where $F$ is the force on the protons, and $v$ the velocity. The protons are non-relativistic, so\n\n\n\n$$\n\n\\frac{1}{2} m v^{2}=q V\n\n$$\n\n\n\nThey move in a circle of radius $r$ inside the field, given by\n\n\n\n$$\n\n\\frac{m v^{2}}{r}=q v B\n\n$$\n\n\n\nCombining these equations gives\n\n\n\n$$\n\nr=\\frac{m v}{q B}=\\frac{m}{q B} \\sqrt{\\frac{2 q V}{m}}=\\frac{1}{B} \\sqrt{\\frac{2 m V}{q}}\n\n$$\n\n\n\nSolving for the magnetic field strength,\n\n\n\n$$\n\nB=\\frac{1}{r} \\sqrt{\\frac{2 m V}{q}}\n\n$$\n\n\n\nTo relate this to the angle, we need to do some geometry. Sketch two circles, one of radius $b$, the other of radius $r$, that intersect perpendicular to each other, as in the diagram above. Then drawing a triangle gives\n\n\n\n$$\n\n\\tan \\frac{\\theta}{2}=\\frac{b}{r} .\n\n$$\n\n\n\nCombining our results, the answer is\n\n\n\n$$\n\nB=\\frac{\\tan \\theta / 2}{b} \\sqrt{\\frac{2 m V}{q}}=0.0894 \\mathrm{~T}\n\n$$']",['0.0894'],False,T,Numerical,1e-3 1452,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} Context question: b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value. Context answer: \boxed{0.556} Context question: c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome. Context answer: \boxed{$37.5$} Context question: d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength. Context answer: \boxed{0.0894} ","e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral. Copyright (c)2017 American Association of Physics Teachers ","['Treat the problem as two dimensional. The area of the spiral is\n\n\n\n$$\n\nA=\\frac{\\pi}{4}\\left(D_{o}^{2}-D_{i}^{2}\\right)\n\n$$\n\n\n\nThe area of the pipe is\n\n\n\n$$\n\nA=L d_{o}\n\n$$\n\n\n\nEquating and solving,\n\n\n\n$$\n\nL=\\frac{\\pi\\left(D_{o}^{2}-D_{i}^{2}\\right)}{4 d_{o}}=33 \\mathrm{~m}\n\n$$']",['33'],False,m,Numerical,5e-1 1452,Electromagnetism,,"e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral. Copyright (c)2017 American Association of Physics Teachers ![](https://cdn.mathpix.com/cropped/2023_12_21_b3dd0e9beea33cd68ffdg-1.jpg?height=379&width=306&top_left_y=220&top_left_x=950)","['Treat the problem as two dimensional. The area of the spiral is\n\n\n\n$$\n\nA=\\frac{\\pi}{4}\\left(D_{o}^{2}-D_{i}^{2}\\right)\n\n$$\n\n\n\nThe area of the pipe is\n\n\n\n$$\n\nA=L d_{o}\n\n$$\n\n\n\nEquating and solving,\n\n\n\n$$\n\nL=\\frac{\\pi\\left(D_{o}^{2}-D_{i}^{2}\\right)}{4 d_{o}}=33 \\mathrm{~m}\n\n$$']",['33'],False,m,Numerical,5e-1 1453,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} Context question: b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value. Context answer: \boxed{0.556} Context question: c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome. Context answer: \boxed{$37.5$} Context question: d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength. Context answer: \boxed{0.0894} Context question: e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral. Copyright (c)2017 American Association of Physics Teachers Context answer: \boxed{33} ","f. Hollow pipe is used instead of solid conductors in order to allow for cooling of the magnet. If the resistivity of copper is $\rho=1.7 \times 10^{-8} \Omega \cdot \mathrm{m}$, determine the electrical resistance of one spiral.","['We have\n\n\n\n$$\n\nr_{s}=\\frac{\\rho L}{A}\n\n$$\n\n\n\nwhere $A$ is the cross sectional area of the pipe, or\n\n\n\n$$\n\nA=\\frac{\\pi}{4}\\left(d_{o}^{2}-d_{i}^{2}\\right)=7.1 \\times 10^{-6} \\mathrm{~m}^{2}\n\n$$\n\n\n\nCombining, we have\n\n\n\n$$\n\nr_{s}=\\frac{\\rho}{d} \\frac{D_{o}^{2}-D_{i}^{2}}{d_{o}^{2}-d_{i}^{2}}=0.079 \\Omega\n\n$$']",['0.079'],False,$\Omega$,Numerical,1e-3 1454,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} Context question: b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value. Context answer: \boxed{0.556} Context question: c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome. Context answer: \boxed{$37.5$} Context question: d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength. Context answer: \boxed{0.0894} Context question: e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral. Copyright (c)2017 American Association of Physics Teachers Context answer: \boxed{33} Context question: f. Hollow pipe is used instead of solid conductors in order to allow for cooling of the magnet. If the resistivity of copper is $\rho=1.7 \times 10^{-8} \Omega \cdot \mathrm{m}$, determine the electrical resistance of one spiral. Context answer: \boxed{0.079} ",g. There are $N=24$ coils stacked on top of each other. Tap water with an initial temperature of $T_{c}=18^{\circ} \mathrm{C}$ enters the spiral through the copper pipe to keep it from over heating; the water exits at a temperature of $T_{h}=31^{\circ} \mathrm{C}$. The copper pipe carries a direct $45 \mathrm{Amp}$ current in order to generate the necessary magnetic field. At what rate must the cooling water flow be provided to the electromagnet? Express your answer in liters per second with only one significant digit. The specific heat capacity of water is $4200 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{kg}$; the density of water is $1000 \mathrm{~kg} / \mathrm{m}^{3}$.,"['The rate of heat generation in the coils is given by\n\n\n\n$$\n\nP=I^{2} R=I^{2} N r_{s}=3850 \\mathrm{~W}\n\n$$\n\n\n\nThis must be dissipated via the increase in water temperature,\n\n\n\n$$\n\nP=C \\Delta T Q\n\n$$\n\n\n\nwhere $C$ is the specific heat capacity in liters, and $Q$ is the flow rate in liters per second. But since one liter of water is one kilogram, we can use either $C$. Combining, we have\n\n\n\n$$\n\nQ=\\frac{I^{2} N r}{C \\Delta T}=0.07 \\mathrm{l} / \\mathrm{s}\n\n$$']",['0.07'],False,l/s,Numerical,5e-3 1455,Electromagnetism,"$g=9.8 \mathrm{~N} / \mathrm{kg}$ $k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$ $c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$ $\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$ $1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$ $m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$ $m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$ $\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$ $$ \begin{aligned} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ & e=1.602 \times 10^{-19} \mathrm{C} \\ & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ & \ln (1+x) \approx x \text { for }|x| \ll 1 \\ & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{aligned} $$ $$ \begin{array}{lrl} \text { velocity parameter } & \beta & =\frac{v}{c} \\ \text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\ \text { relativistic momentum } & p & =\gamma \beta m c \\ \text { relativistic energy } & E & =\gamma m c^{2} \\ \text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}} \end{array} $$ Extra Supplementary Reading Materials: Beloit College has a ""homemade"" $500 \mathrm{kV}$ VanDeGraff proton accelerator, designed and constructed by the students and faculty. Accelerator dome (assume it is a sphere); accelerating column; bending electromagnet The accelerator dome, an aluminum sphere of radius $a=0.50$ meters, is charged by a rubber belt with width $w=10 \mathrm{~cm}$ that moves with speed $v_{b}=20 \mathrm{~m} / \mathrm{s}$. The accelerating column consists of 20 metal rings separated by glass rings; the rings are connected in series with $500 \mathrm{M} \Omega$ resistors. The proton beam has a current of $25 \mu \mathrm{A}$ and is accelerated through $500 \mathrm{kV}$ and then passes through a tuning electromagnet. The electromagnet consists of wound copper pipe as a conductor. The electromagnet effectively creates a uniform field $B$ inside a circular region of radius $b=10 \mathrm{~cm}$ and zero outside that region. Only six of the 20 metals rings and resistors are shown in the figure. The fuzzy grey path is the path taken by the protons as they are accelerated from the dome, through the electromagnet, into the target. Context question: a. Assuming the dome is charged to $500 \mathrm{kV}$, determine the strength of the electric field at the surface of the dome. Context answer: \boxed{$10^{6}$} Context question: b. Assuming the proton beam is off, determine the time constant for the accelerating dome (the time it takes for the charge on the dome to decrease to $1 / e \approx 1 / 3$ of the initial value. Context answer: \boxed{0.556} Context question: c. Assuming the $25 \mu \mathrm{A}$ proton beam is on, determine the surface charge density that must be sprayed onto the charging belt in order to maintain a steady charge of $500 \mathrm{kV}$ on the dome. Context answer: \boxed{$37.5$} Context question: d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength. Context answer: \boxed{0.0894} Context question: e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral. Copyright (c)2017 American Association of Physics Teachers Context answer: \boxed{33} Context question: f. Hollow pipe is used instead of solid conductors in order to allow for cooling of the magnet. If the resistivity of copper is $\rho=1.7 \times 10^{-8} \Omega \cdot \mathrm{m}$, determine the electrical resistance of one spiral. Context answer: \boxed{0.079} Context question: g. There are $N=24$ coils stacked on top of each other. Tap water with an initial temperature of $T_{c}=18^{\circ} \mathrm{C}$ enters the spiral through the copper pipe to keep it from over heating; the water exits at a temperature of $T_{h}=31^{\circ} \mathrm{C}$. The copper pipe carries a direct $45 \mathrm{Amp}$ current in order to generate the necessary magnetic field. At what rate must the cooling water flow be provided to the electromagnet? Express your answer in liters per second with only one significant digit. The specific heat capacity of water is $4200 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{kg}$; the density of water is $1000 \mathrm{~kg} / \mathrm{m}^{3}$. Context answer: \boxed{0.07} ",h. The protons are fired at a target consisting of Fluorine atoms $(Z=9)$. What is the distance of closest approach to the center of the Fluorine nuclei for the protons? You can assume that the Fluorine does not move.,"['Conservation of energy gives\n\n\n\n$$\n\nq V=\\frac{1}{4 \\pi \\epsilon_{0}} \\frac{Z q^{2}}{r}\n\n$$\n\n\n\nwhere $r$ is the radius of closest approach. Then\n\n\n\n$$\n\nr=\\frac{1}{4 \\pi \\epsilon_{0}} \\frac{Z q}{V}=2.59 \\times 10^{-14} \\mathrm{~m}\n\n$$\n\n\n\nSince this is about the size of a Fluorine nucleus, we can potentially get a nuclear reaction. Actually, the important reaction occurs at about $380 \\mathrm{kV}$.']",['$2.59 \\times 10^{-14}$'],False,m,Numerical,1e-16 1456,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Collision Course Two blocks, $A$ and $B$, of the same mass are on a fixed inclined plane, which makes a $30^{\circ}$ angle with the horizontal. At time $t=0, A$ is a distance $\ell=5 \mathrm{~cm}$ along the incline above $B$, and both blocks are at rest. Suppose the coefficients of static and kinetic friction between the blocks and the incline are $$ \mu_{A}=\frac{\sqrt{3}}{6}, \quad \mu_{B}=\frac{\sqrt{3}}{3} $$ and that the blocks collide perfectly elastically. Let $v_{A}(t)$ and $v_{B}(t)$ be the speeds of the blocks down the incline. For this problem, use $g=10 \mathrm{~m} / \mathrm{s}^{2}$, assume both blocks stay on the incline for the entire time, and neglect the sizes of the blocks. Context question: a. Graph the functions $v_{A}(t)$ and $v_{B}(t)$ for $t$ from 0 to 1 second on the provided answer sheet, with a solid and dashed line respectively. Mark the times at which collisions occur. Context answer: ","b. Derive an expression for the total distance block $A$ has moved from its original position right after its $n^{\text {th }}$ collision, in terms of $\ell$ and $n$.","['The easiest way to calculate the total distance after $n$ collisions is through the graph, namely the area enclosed by the blue line: When $n=1$, the distance traveled is $d_{A}(1)=\\ell$, for $n=2, d_{A}(2)=d_{A}(1)+4 \\ell$, and so on, so\n\n\n\n$$\n\nd_{A}(n)=\\ell+4 \\ell+8 \\ell+\\ldots+(n-1) \\times 4 \\ell=\\ell\\left(2 n^{2}-2 n+1\\right) .\n\n$$']",['$d_{A}(n)=\\ell \\left(2 n^{2}-2 n+1\\right)$'],False,,Expression, 1457,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Collision Course Two blocks, $A$ and $B$, of the same mass are on a fixed inclined plane, which makes a $30^{\circ}$ angle with the horizontal. At time $t=0, A$ is a distance $\ell=5 \mathrm{~cm}$ along the incline above $B$, and both blocks are at rest. Suppose the coefficients of static and kinetic friction between the blocks and the incline are $$ \mu_{A}=\frac{\sqrt{3}}{6}, \quad \mu_{B}=\frac{\sqrt{3}}{3} $$ and that the blocks collide perfectly elastically. Let $v_{A}(t)$ and $v_{B}(t)$ be the speeds of the blocks down the incline. For this problem, use $g=10 \mathrm{~m} / \mathrm{s}^{2}$, assume both blocks stay on the incline for the entire time, and neglect the sizes of the blocks. Context question: a. Graph the functions $v_{A}(t)$ and $v_{B}(t)$ for $t$ from 0 to 1 second on the provided answer sheet, with a solid and dashed line respectively. Mark the times at which collisions occur. Context answer: Context question: b. Derive an expression for the total distance block $A$ has moved from its original position right after its $n^{\text {th }}$ collision, in terms of $\ell$ and $n$. Context answer: \boxed{$d_{A}(n)=\ell \left(2 n^{2}-2 n+1\right)$} Extra Supplementary Reading Materials: Now suppose that the coefficient of block $B$ is instead $\mu_{B}=\sqrt{3} / 2$, while $\mu_{A}=\sqrt{3} / 6$ remains the same. Context question: Again, graph the functions $v_{A}(t)$ and $v_{B}(t)$ for $t$ from 0 to 1 second on the provided answer sheet, with a solid and dashed line respectively. Mark the times at which collisions occur. Context answer: ","At time $t=1 \mathrm{~s}$, how far has block $A$ moved from its original position?","['Again, using the graph, it is easy to see that at $t=1 \\mathrm{~s}, A$ just finished the 5 th collision, and the total distance it moved is: $5 \\ell=25 \\mathrm{~cm}$.']",['$25$'],False,cm,Numerical,1e-8 1458,Thermodynamics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Green Revolution ${ }^{1}$ In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies.",a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.),"[""The intensity of solar radiation at the surface of the sun is $\\sigma T_{s}^{4}$, so the intensity at the planet's orbit radius is\n\n\n\n$$\n\nI=\\sigma T_{s}^{4} \\frac{R_{s}^{2}}{R_{0}^{2}}\n\n$$\n\n\n\nThe area subtended by the planet is $\\pi R^{2}$, so\n\n\n\n$$\n\nP=\\pi \\sigma T_{s}^{4} \\frac{R^{2} R_{s}^{2}}{R_{0}^{2}}\n\n$$""]",['$P=\\pi \\sigma T_{s}^{4} \\frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$'],False,,Expression, 1459,Thermodynamics,,"b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced?","[""The minimum value is simply zero; in this case zero heat is transferred to the dark side of the planet. Since no heat is transferred, the heat engine can't run, so $P_{w}=0$. (To show this a bit more carefully, note that the entropy exhausted by the heat engine is $Q_{2} / T_{2} \\propto T_{2}^{3}$ by the Stefan-Boltzmann law. In the limit $T_{2} \\rightarrow 0$, the entropy out goes to zero, so the entropy in and hence the heat intake also goes to zero.)\n\n\n\nThe maximum value is $T_{2} / T_{1}=1$. It cannot be any higher by the second law of thermodynamics. In this case, there is no temperature difference, so the heat engine has zero efficiency and $P_{w}=0$. Power $P / 2$ is simply transferred from the bright side to the dark side as heat.""]","['The minimum value is simply zero, $P_{w}=0$.\nThe maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$.']",True,,Need_human_evaluate, 1460,Thermodynamics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Green Revolution ${ }^{1}$ In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies. Context question: a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.) Context answer: \boxed{$P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$} Extra Supplementary Reading Materials: In order to keep both $T_{1}$ and $T_{2}$ constant, heat must be continually transferred from the bright side to the dark side. By viewing the two hemispheres as the two reservoirs of a reversible heat engine, work can be performed from this temperature difference, which appears in the form of wind power. For simplicity, we assume all of this power is immediately captured and stored by windmills. Context question: b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced? Context answer: The minimum value is simply zero, $P_{w}=0$. The maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$. ",c. Find the wind power $P_{w}$ in terms of $P$ and the temperature ratio $T_{2} / T_{1}$.,"['Let heat be transferred from the bright side at a rate $Q_{1}$ and transferred to the dark side at a rate $Q_{2}$. Then by conservation of energy,\n\n\n\n$$\n\nQ_{1}=P_{w}+Q_{2}\n\n$$\n\n\n\nSince the two hemispheres have constant temperatures, energy balance for each gives\n\n\n\n$$\n\nP=Q_{1}+\\left(2 \\pi R^{2} \\sigma\\right) T_{1}^{4}, \\quad Q_{2}=A T_{2}^{4}\n\n$$\n\n\n\nFinally, since the engine is reversible,\n\n\n\n$$\n\n\\frac{Q_{1}}{T_{1}}=\\frac{Q_{2}}{T_{2}}\n\n$$\n\n\n\nBy combining the first three equations, and defining $x=T_{2} / T_{1}$, we have\n\n\n\n$$\n\nP_{w}=Q_{1}-Q_{2}=P-\\left(2 \\pi R^{2} \\sigma\\right)\\left(T_{1}^{4}+T_{2}^{4}\\right)=P-\\left(2 \\pi R^{2} \\sigma\\right) T_{1}^{4}\\left(1+x^{4}\\right) .\n\n$$\n\n\n\nThis is not yet in terms of $P$ and $x$, so now we use the reversibility condition,\n\n\n\n$$\n\n\\frac{P-A T_{1}^{4}}{T_{1}}=\\frac{A T_{2}^{4}}{T_{2}}\n\n$$\n\n\n\nwhich simplifies to\n\n\n\n$$\n\nP=\\left(2 \\pi R^{2} \\sigma\\right)\\left(T_{1}^{4}+T_{1} T_{2}^{3}\\right)=\\left(2 \\pi R^{2} \\sigma\\right) T_{1}^{4}\\left(1+x^{3}\\right)\n\n$$\n\n\n\nPlugging this in above, we find\n\n\n\n$$\n\nP_{w}=P-\\frac{P}{1+x^{3}}\\left(1+x^{4}\\right)=\\frac{x^{3}(1-x)}{1+x^{3}} P\n\n$$']",['$P_{w}=\\frac{x^{3}(1-x)}{1+x^{3}} P$'],False,,Expression, 1461,Thermodynamics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Green Revolution ${ }^{1}$ In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies. Context question: a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.) Context answer: \boxed{$P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$} Extra Supplementary Reading Materials: In order to keep both $T_{1}$ and $T_{2}$ constant, heat must be continually transferred from the bright side to the dark side. By viewing the two hemispheres as the two reservoirs of a reversible heat engine, work can be performed from this temperature difference, which appears in the form of wind power. For simplicity, we assume all of this power is immediately captured and stored by windmills. Context question: b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced? Context answer: The minimum value is simply zero, $P_{w}=0$. The maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$. Context question: c. Find the wind power $P_{w}$ in terms of $P$ and the temperature ratio $T_{2} / T_{1}$. Context answer: \boxed{$P_{w}=\frac{x^{3}(1-x)}{1+x^{3}} P$}","d. Estimate the maximum possible value of $P_{w}$ as a fraction of $P$, to one significant figure. Briefly explain how you obtained this estimate.","['There are many ways to get the required answer. For example, by sketching the function, one can see that there is a unique maximum at an intermediate value of $x$, and furthermore that this maximum is at $x>0.5$, because of the rapid rise of the $x^{3}$ factor. One could then compute $P_{w} / P$ with a calculator at a few trial values such as $x=0.5,0.7,0.9$, which are already enough to get the desired accuracy.\n\n\n\nThe optimum value is $x=0.69$, at which point\n\n\n\n$$\n\nP_{w}^{\\max }=0.077 P\n\n$$\n\n\n\n\n\n\n\nHence in this model, at most $7.7 \\%$ of solar energy can be converted into wind energy. Any answer within $15 \\%$ of this value was accepted.\n\n']",['$P_{w}^{\\max }=0.077 P$'],False,,Expression, 1462,Electromagnetism,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Electric Slide Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.","a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.","['In the steady state, the current is the same everywhere. Consider the region $(x, x+d x)$. The time it takes for the charge in the second region to leave is $\\frac{\\mathrm{d} x}{v(x)}$. The amount of charge that leaves is $\\rho A \\mathrm{~d} x$. The current is thus given by $\\rho A v$, so $\\rho v$ is constant. Alternatively, one can write this as\n\n\n\n$$\n\nv \\frac{\\mathrm{d} \\rho}{\\mathrm{d} x}+\\rho \\frac{\\mathrm{d} v}{\\mathrm{~d} x}=0\n\n$$\n\n\n\nBoth forms were accepted.\n\n']",['$v \\frac{\\mathrm{d} \\rho}{\\mathrm{d} x}+\\rho \\frac{\\mathrm{d} v}{\\mathrm{~d} x}=0$'],False,,Equation, 1463,Electromagnetism,,"ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.)","[""Let us find an expression for the electric field at position $x$. The position $x$ is effectively in between two uniform sheets of charge density. The sheet on the left has charge density $\\int_{0}^{x} \\rho \\mathrm{d} x+\\sigma_{0}$, where $\\sigma_{0}$ is the charge density on the left plate, and the sheet on the right has charge density $\\int_{x}^{d} \\rho \\mathrm{d} x+\\sigma_{d}$, where $\\sigma_{d}$ is the charge density on the left plate. Then, the electric field is given by\n\n\n\n$$\n\nE=\\sigma_{0} /\\left(2 \\epsilon_{0}\\right)+\\int_{0}^{x} \\rho /\\left(2 \\epsilon_{0}\\right) \\mathrm{d} x-\\int_{x}^{d} \\rho /\\left(2 \\epsilon_{0}\\right) \\mathrm{d} x-\\sigma_{d} /\\left(2 \\epsilon_{0}\\right)\n\n$$\n\n\n\nThen, by the Fundamental Theorem of Calculus\n\n\n\n$$\n\n\\frac{d E}{d x}=\\frac{\\rho}{\\epsilon_{0}}\n\n$$\n\n\n\n\n\n\n\nso\n\n\n\n$$\n\n\\frac{d^{2} V}{d x^{2}}=-\\frac{\\rho}{\\epsilon_{0}}\n\n$$\n\n\n\nThis can also be derived from the differential form of Gauss's Law more easily and is known as Poisson's equation.""]",['$$\n\n\\frac{d^{2} V}{d x^{2}}=-\\frac{\\rho}{\\epsilon_{0}}\n\n$$'],False,,Need_human_evaluate, 1464,Electromagnetism,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Electric Slide Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium. Context question: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives. Context answer: $v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$ Extra Supplementary Reading Materials: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. Context question: ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.) Context answer: $$ \frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}} $$ ","iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.","[""We have that\n\n\n\n$$\n\n\\rho \\frac{d v}{d x}+v \\frac{d \\rho}{d x}=0\n\n$$\n\n\n\nand now that $v=-\\mu \\frac{d V}{d x}$, so substituting in Poisson's equation gives us that\n\n\n\n$$\n\n\\left(\\frac{d^{2} V}{d x^{2}}\\right)^{2}+\\frac{d V}{d x}\\left(\\frac{d^{3} V}{d x^{3}}\\right)=0\n\n$$\n\n\n\nUsing $V(x)=-V_{0}(x / d)^{b}$ gives\n\n\n\n$$\n\nb(b-1) b(b-1)=-b b(b-1)(b-2)\n\n$$\n\n\n\nThe solution with $b=0$ cannot satisfy the boundary conditions, while $b=1$ has zero current. Assuming $b$ is neither of these values, we have $b-1=-(b-2)$, so $b=3 / 2$. Substituting gives\n\n\n\n$$\n\nv=-\\frac{3 V_{0} \\mu x^{1 / 2}}{2 d^{3 / 2}}\n\n$$\n\n\n\nand\n\n\n\n$$\n\n\\rho=-\\frac{3 V_{0} \\epsilon_{0}}{4 d^{3 / 2} x^{1 / 2}}\n\n$$\n\n\n\nso\n\n\n\n$$\n\nI=\\rho A v=\\frac{9 \\epsilon_{0} \\mu A V_{0}^{2}}{8 d^{3}}\n\n$$\n\n\n\nwith the current flowing from left to right.""]",['$I=\\frac{9 \\epsilon_{0} \\mu A V_{0}^{2}}{8 d^{3}}$'],False,,Expression, 1465,Electromagnetism,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Electric Slide Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium. Context question: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives. Context answer: $v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$ Extra Supplementary Reading Materials: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. Context question: ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.) Context answer: $$ \frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}} $$ Context question: iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters. Context answer: \boxed{$I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$} Extra Supplementary Reading Materials: b. For small $V_{0}$, the positive charges move by diffusion. The current due to diffusion is given by Fick's Law, $$ I=-A D \frac{\mathrm{d} \rho}{\mathrm{d} x} $$ Here, $D$ is the diffusion constant, which you can assume to be described by the Einstein relation $$ D=\frac{\mu k_{B} T}{q} $$ where $T$ is the temperature of the system.","i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x) \propto x^{b^{\prime}}$, where $b^{\prime}$ is another exponent you must find. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.","[""We again have that $V(x)=V_{0}(x / d)^{b}$. Note that from Poisson's equation, $\\frac{d \\rho}{d x}=-\\epsilon_{0} \\frac{d^{3} V}{d x^{3}}$, so we need $b=3$ for this expression to be constant. Therefore,\n\n\n\n$$\n\nI=\\frac{6 \\mu k_{B} T A \\epsilon_{0} V_{0}}{q d^{3}}\n\n$$""]",['$I=\\frac{6 \\mu k_{B} T A \\epsilon_{0} V_{0}}{q d^{3}}$'],False,,Expression, 1466,Electromagnetism,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Electric Slide Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium. Context question: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives. Context answer: $v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$ Extra Supplementary Reading Materials: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. Context question: ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.) Context answer: $$ \frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}} $$ Context question: iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters. Context answer: \boxed{$I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$} Extra Supplementary Reading Materials: b. For small $V_{0}$, the positive charges move by diffusion. The current due to diffusion is given by Fick's Law, $$ I=-A D \frac{\mathrm{d} \rho}{\mathrm{d} x} $$ Here, $D$ is the diffusion constant, which you can assume to be described by the Einstein relation $$ D=\frac{\mu k_{B} T}{q} $$ where $T$ is the temperature of the system. Context question: i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x) \propto x^{b^{\prime}}$, where $b^{\prime}$ is another exponent you must find. Derive an expression for the current in terms of $V_{0}$ and the other given parameters. Context answer: \boxed{$I=\frac{6 \mu k_{B} T A \epsilon_{0} V_{0}}{q d^{3}}$} ",ii. At roughly what voltage $V_{0}$ does the system transition from this regime to the high voltage regime of the previous part?,"['We find the crossover voltage by equating our answers in previous parts, to get\n\n\n\n$$\n\n\\frac{6 \\mu k_{B} T A \\epsilon_{0} V_{0}}{q d^{3}}=\\frac{9 \\epsilon_{0} \\mu A V_{0}^{2}}{8 d^{3}}\n\n$$\n\n\n\nor\n\n\n\n$$\n\nV_{0}=\\frac{16 k_{B} T}{3 q}\n\n$$']",['$V_{0}=\\frac{16 k_{B} T}{3 q}$'],False,,Expression, 1467,Modern Physics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Strain in the Membrane ${ }^{2}$ The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature. The neuron has ""ion pumps"" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$.","a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane?","['One charge layer by itself creates an electric field $E_{1}=Q /(2 \\epsilon A)$ in each direction. So the force between the two sides of the membrane $F_{E}=Q E_{1}=Q^{2} /(2 \\epsilon A)$.\n\n\n\nThis electric force is balanced by the spring force $F_{s}=k x$, where $x=d_{0}-d$. Equating these two forces and solving for $d$ gives gives\n\n\n\n$$\n\nd=d_{0}-\\frac{Q^{2}}{2 \\epsilon A k}\n\n$$']",['$d=d_{0}-\\frac{Q^{2}}{2 \\epsilon A k}$'],False,,Expression, 1468,Modern Physics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Strain in the Membrane ${ }^{2}$ The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature. The neuron has ""ion pumps"" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$. Context question: a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane? Context answer: \boxed{$d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$} ",b. Derive an expression for the voltage difference $V$ between the outer and inner surfaces of the membrane in terms of $Q$ and the other parameters given.,"['The electric field inside the membrane (as produced by both the left and right plates) is $E=Q /\\left(\\epsilon_{0} \\kappa A\\right)$. So the voltage between them is\n\n\n\n$$\n\nV=E d=\\frac{Q}{\\epsilon A} d\n\n$$\n\n\n\nInserting the expression for $Q$ from part (a) gives\n\n\n\n$$\n\nV=\\frac{Q}{\\epsilon A}\\left(d_{0}-\\frac{Q^{2}}{2 \\epsilon A k}\\right)\n\n$$\n\n\n\nThis equation implies that as the charge $Q$ is increased, the voltage first increases and then decreases again.']",['$V=\\frac{Q}{\\epsilon A}\\left(d_{0}-\\frac{Q^{2}}{2 \\epsilon A k}\\right)$'],False,,Expression, 1469,Modern Physics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Strain in the Membrane ${ }^{2}$ The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature. The neuron has ""ion pumps"" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$. Context question: a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane? Context answer: \boxed{$d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$} Context question: b. Derive an expression for the voltage difference $V$ between the outer and inner surfaces of the membrane in terms of $Q$ and the other parameters given. Context answer: \boxed{$V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)$} ","c. Suppose that the ion pumps are first turned on in the uncharged state, and the membrane is charged very slowly (quasistatically). The pumps will only turn off when the voltage difference across the membrane becomes larger than a particular value $V_{\text {th }}$. How large must the spring constant $k$ be so that the ion pumps turn off before the membrane collapses?","['The voltage $V$ first increases and then decreases as a function of $Q$, which implies that there is a maximum voltage $V_{\\max }$ to which the membrane can be charged. This voltage can be found by taking the derivative $d V / d Q$ and setting it equal to zero. This procedure gives\n\n\n\n$$\n\nV_{\\max }=\\sqrt{\\frac{k d_{0}^{3}}{\\epsilon A}}\\left(\\frac{2}{3}\\right)^{3 / 2}\n\n$$\n\n\n\nThe corresponding charge at the maximum voltage is given by\n\n\n\n$$\n\nQ_{\\mathrm{V} \\max }^{2}=\\frac{2}{3} \\epsilon A k d_{0}\n\n$$\n\n\n\nFor the ion pumps to turn off, we must have $V_{\\max }>V_{\\mathrm{th}}$. Otherwise the pumps will continue to move charge across the membrane until it collapses. Setting $V_{\\max }>V_{\\text {th }}$ and solving for $k$ gives\n\n\n\n$$\n\nk>\\left(\\frac{3}{2}\\right)^{3} \\frac{V_{\\mathrm{th}}^{2} \\epsilon A}{d_{0}^{3}}\n\n$$']",['$\\left(\\frac{3}{2}\\right)^{3} \\frac{V_{\\mathrm{th}}^{2} \\epsilon A}{d_{0}^{3}}$'],False,,Expression, 1470,Modern Physics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Strain in the Membrane ${ }^{2}$ The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature. The neuron has ""ion pumps"" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$. Context question: a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane? Context answer: \boxed{$d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$} Context question: b. Derive an expression for the voltage difference $V$ between the outer and inner surfaces of the membrane in terms of $Q$ and the other parameters given. Context answer: \boxed{$V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)$} Context question: c. Suppose that the ion pumps are first turned on in the uncharged state, and the membrane is charged very slowly (quasistatically). The pumps will only turn off when the voltage difference across the membrane becomes larger than a particular value $V_{\text {th }}$. How large must the spring constant $k$ be so that the ion pumps turn off before the membrane collapses? Context answer: \boxed{$\left(\frac{3}{2}\right)^{3} \frac{V_{\mathrm{th}}^{2} \epsilon A}{d_{0}^{3}}$} Extra Supplementary Reading Materials: d. How much work is done by the ion pumps in each of the following situations? Express your answers in terms of $k$ and $d_{0}$.",i. $k$ is infinitesimally larger than the value derived in part (c).,"['If $k$ is larger than the value in part (c), then the threshold voltage $V_{\\mathrm{th}}>V_{\\max }$, and the ion pumps turn off before the membrane thickness $d$ reaches zero. The work $W$ done by the ion pumps is equal to the potential energy of the system relative to the uncharged state (with $Q=0$ and $d=0$ ). That is, $W=\\frac{1}{2} k x^{2}+Q^{2} /(2 C)$, where $C=\\epsilon A / d$ is the capacitance of the membrane. Writing this equation in terms of $Q$ gives $W=Q^{2} d_{0} /(2 \\epsilon A)-Q^{4} /\\left(8 \\epsilon^{2} A^{2} k\\right)$.\n\n\n\n$k$ being infinitesimally larger than the critical value means that the ion pumps turn off just as the voltage maximum $V_{\\mathrm{th}}$ is reached. At this point the charge $Q$ approaches\n\n\n\n\n\n\n\n$Q_{\\mathrm{Vmax}}$, derived in the previous answer. Inserting the value into the expression for $W$ gives\n\n\n\n$$\n\nW=\\frac{5}{18} k d_{0}^{2}\n\n$$']",['$W=\\frac{5}{18} k d_{0}^{2}$'],False,,Expression, 1471,Modern Physics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Strain in the Membrane ${ }^{2}$ The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature. The neuron has ""ion pumps"" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$. Context question: a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane? Context answer: \boxed{$d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$} Context question: b. Derive an expression for the voltage difference $V$ between the outer and inner surfaces of the membrane in terms of $Q$ and the other parameters given. Context answer: \boxed{$V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)$} Context question: c. Suppose that the ion pumps are first turned on in the uncharged state, and the membrane is charged very slowly (quasistatically). The pumps will only turn off when the voltage difference across the membrane becomes larger than a particular value $V_{\text {th }}$. How large must the spring constant $k$ be so that the ion pumps turn off before the membrane collapses? Context answer: \boxed{$\left(\frac{3}{2}\right)^{3} \frac{V_{\mathrm{th}}^{2} \epsilon A}{d_{0}^{3}}$} Extra Supplementary Reading Materials: d. How much work is done by the ion pumps in each of the following situations? Express your answers in terms of $k$ and $d_{0}$. Context question: i. $k$ is infinitesimally larger than the value derived in part (c). Context answer: \boxed{$W=\frac{5}{18} k d_{0}^{2}$} ",ii. $k$ is infinitesimally smaller than the value derived in part (c).,"['If $k$ is smaller than the value in part (c), then the threshold voltage is larger than the stopping voltage $V_{\\mathrm{th}}$, and the ion pumps continue to work until the membrane collapses to $d=0$. At this point there is no electrostatic energy in the membrane (the membrane capacitance is infinite), and\n\n\n\n$$\n\nW=\\frac{1}{2} k d_{0}^{2}\n\n$$\n\n\n\nAssume in each case that the membrane thickness $d$ cannot become negative.']",['$W=\\frac{1}{2} k d_{0}^{2}$'],False,,Expression, 1472,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Pitfall A bead is placed on a horizontal rail, along which it can slide frictionlessly. It is attached to the end of a rigid, massless rod of length $R$. A ball is attached at the other end. Both the bead and the ball have mass $M$. The system is initially stationary, with the ball directly above the bead. The ball is then given an infinitesimal push, parallel to the rail. Assume that the rod and ball are designed in such a way (not shown explicitly in the diagram) so that they can pass through the rail without hitting it. In other words, the rail only constrains the motion of the bead. Two subsequent states of the system are shown below. ","a. Derive an expression for the force in the rod when it is horizontal, as shown at left above.","[""The motion of the rod can be decomposed as a superposition of rotation about the center of mass, and translation of the center of mass. We claim that the bead is stationary at this moment. To see this, note that the bead cannot have a vertical velocity component since it is fixed to the rod. Moreover, it cannot have a horizontal velocity component, because since the system experiences no external forces in the horizontal direction, the motion of the center of mass has no horizontal component.\n\n\n\nTherefore, the bead is stationary, and since the released gravitational potential energy is $M g R$, the speed $v$ of the ball obeys\n\n\n\n$$\n\n\\frac{1}{2} M v^{2}=M g R \\quad \\Rightarrow \\quad v=\\sqrt{2 g R}\n\n$$\n\n\n\nThe motion of the system consists of a downward speed $v / 2$ of the center of mass, superposed\n\n\n\n\n\n\n\non a rotation about the center of mass which gives the masses speed $v / 2$.\n\n\n\nNow consider the forces on the ball. The only horizontal force on the ball is the force in the rod, so we need to find the horizontal acceleration of the ball. Part of this is due to the centripetal force associated with the rotational motion,\n\n\n\n$$\n\nF=\\frac{M(v / 2)^{2}}{R / 2}=M g\n\n$$\n\n\n\nAnother part of the ball's acceleration is due to the acceleration of the center of mass, but this is purely vertical and hence irrelevant here. A third part is due to the angular acceleration about the center of mass, but this also is associated with a vertical acceleration of the ball and hence irrelevant. Thus the force in the rod is just a tension\n\n\n\n$$\n\nT=M g\n\n$$\n\n\n\nA common pitfall to think of the motion of the system as pure rotation about the bead, giving the incorrect answer $F=M v^{2} / R=2 M g$. The reason this is incorrect is that, while it correctly describes the instantaneous velocity, to compute the acceleration one must also account for the rate of change of the pivot point. It can be easy to miss this, because the point at which the pivot is currently located (the bead, in this case) always has zero instantaneous velocity. Yet the pivot point itself does have an instantaneous velocity, which must be accounted for. It is possible to solve the problem this way, but it's subtle."", 'Here we present a more complicated, but more straightforward alternative. Let $x$ be the horizontal position of the bead. The position of the ball is\n\n\n\n$$\n\n\\mathbf{r}=(x+R \\sin \\theta, R \\cos \\theta)\n\n$$\n\n\n\nBy conservation of momentum, the center of mass must be at the same horizontal position as it started, so\n\n\n\n$$\n\n2 x+R \\sin \\theta=0\n\n$$\n\n\n\nTherefore, the position of the ball is\n\n\n\n$$\n\n\\mathbf{r}=((R / 2) \\sin \\theta, R \\cos \\theta)\n\n$$\n\n\n\nThat is, it follows an ellipse whose major axis is twice the minor axis.\n\n\n\nDifferentiating this twice, the acceleration of the ball is\n\n\n\n$$\n\na_{x}=\\frac{R}{2}\\left(\\alpha \\cos \\theta-\\omega^{2} \\sin \\theta\\right), \\quad a_{y}=-R\\left(\\alpha \\sin \\theta+\\omega^{2} \\cos \\theta\\right)\n\n$$\n\n\n\nIn this case, $\\theta=90^{\\circ}$ and we are only interested in $a_{x}$, giving\n\n\n\n$$\n\na_{x}=-\\frac{R \\omega^{2}}{2}\n\n$$\n\n\n\nAt this point, one needs to find $\\omega(\\theta)$ using conservation of energy. Plugging in the result from either part (c) below or the first solution above gives $a_{x}=-g$, which indicates a force of $M g$ to the left, and hence a tension in the rod. Incidentally, yet another solution is to use $F=M v^{2} / R$ where $R$ is the instantaneous radius of curvature of the ellipse, which may be computed directly.', 'We can work in the noninertial reference frame of the bead. In this frame, the ball simply rotates around the bead, with the naive centripetal force\n\n\n\n$$\n\nF=\\frac{M v^{2}}{R}=2 M g\n\n$$\n\n\n\nHowever, since the tension accelerates the bead to the right, there is a fictitious force of magnitude $T$ pulling the ball to the left in this frame. The real tension force and the fictitious force add up to the centripetal force, so $T+T=F$, giving $T=M g$.']",['$T=M g$'],False,,Expression, 1473,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Pitfall A bead is placed on a horizontal rail, along which it can slide frictionlessly. It is attached to the end of a rigid, massless rod of length $R$. A ball is attached at the other end. Both the bead and the ball have mass $M$. The system is initially stationary, with the ball directly above the bead. The ball is then given an infinitesimal push, parallel to the rail. Assume that the rod and ball are designed in such a way (not shown explicitly in the diagram) so that they can pass through the rail without hitting it. In other words, the rail only constrains the motion of the bead. Two subsequent states of the system are shown below. Context question: a. Derive an expression for the force in the rod when it is horizontal, as shown at left above. Context answer: \boxed{$T=M g$} ","b. Derive an expression for the force in the rod when the ball is directly below the bead, as shown at right above.","['At this point the released gravitational potential energy is $2 M g R$, and both masses are moving horizontally with speed $v$, where\n\n\n\n$$\n\n2 \\frac{1}{2} M v^{2}=2 M g R \\quad \\Rightarrow \\quad v=\\sqrt{2 g R}\n\n$$\n\n\n\nWork in the frame moving to the right with speed $v$. In this frame the bead is stationary and the ball has velocity $2 v$ and is instantaneously rotating about the bead, so it must be experiencing a centripetal force\n\n\n\n$$\n\n\\frac{M(2 v)^{2}}{R}=8 M g\n\n$$\n\n\n\nUnlike in part (a), there are no additional contribution from the acceleration of the rotation center, because the bead can only ever accelerate horizontally, and the force in the rod at this moment is vertical. Since the ball also experiences a downward force of $M g$ due to gravity, the force in the rod is a tension\n\n\n\n$$\n\nT=9 M g\n\n$$', 'We can also solve the problem using the algebraic approach. Following what we found in the second solution to part (a),\n\n\n\n$$\n\na_{y}=-R\\left(\\alpha \\sin \\theta+\\omega^{2} \\cos \\theta\\right)=R \\omega^{2}\n\n$$\n\n\n\nPlugging in the result for $\\omega^{2}$, one finds\n\n\n\n$$\n\na_{y}=8 g\n\n$$\n\n\n\nwhich, as before, indicates a tension of $T=9 \\mathrm{Mg}$.', 'Work in the original frame, where the masses are instantaneously rotating about the midpoint of the rod. Because of this motion, the ball must be experiencing a\n\n\n\n\n\n\n\ncentripetal force\n\n\n\n$$\n\n\\frac{M v^{2}}{R / 2}=4 M g\n\n$$\n\n\n\nThis makes answering $5 \\mathrm{Mg}$ a tempting pitfall; to get the right answer, we must also account for the vertical acceleration of the center of mass. The height of the center of mass is $y_{\\mathrm{CM}}=(R / 2) \\cos \\theta$, so\n\n\n\n$$\n\na_{\\mathrm{CM}}=\\frac{d}{d t}\\left(-\\frac{R}{2} \\omega \\sin \\theta\\right)=-\\frac{R}{2}\\left(\\omega^{2} \\cos \\theta+\\alpha \\sin \\theta\\right)\n\n$$\n\n\n\nAt this point, $\\sin \\theta=0$ and $\\cos \\theta=-1$, so\n\n\n\n$$\n\na_{\\mathrm{CM}}=\\frac{R}{2} \\omega^{2}=\\frac{v^{2}}{R / 2}=4 g\n\n$$\n\n\n\nwhere we used $v=\\omega(R / 2)$. Therefore, the total upward acceleration of the ball is $8 g$, as we found earlier, so the force in the rod is a tension $T=9 \\mathrm{Mg}$.\n\n\n\nIn summary, there are many valid ways to arrive at the correct answers of $M g$ and $9 M g$, but all of them require careful thought. When this exam was given, only about $2 \\%$ of students successfully found these answers.']",['$T=9 Mg$'],False,,Expression, 1474,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: ## Pitfall A bead is placed on a horizontal rail, along which it can slide frictionlessly. It is attached to the end of a rigid, massless rod of length $R$. A ball is attached at the other end. Both the bead and the ball have mass $M$. The system is initially stationary, with the ball directly above the bead. The ball is then given an infinitesimal push, parallel to the rail. Assume that the rod and ball are designed in such a way (not shown explicitly in the diagram) so that they can pass through the rail without hitting it. In other words, the rail only constrains the motion of the bead. Two subsequent states of the system are shown below. Context question: a. Derive an expression for the force in the rod when it is horizontal, as shown at left above. Context answer: \boxed{$T=M g$} Context question: b. Derive an expression for the force in the rod when the ball is directly below the bead, as shown at right above. Context answer: \boxed{$T=9 Mg$}","c. Let $\theta$ be the angle the rod makes with the vertical, so that the $\operatorname{rod} \operatorname{begins}$ at $\theta=0$. Find the angular velocity $\omega=d \theta / d t$ as a function of $\theta$.","['The height of the ball is $R \\cos \\theta$, so the released gravitational potential energy is\n\n\n\n$$\n\nM g R(1-\\cos \\theta)\n\n$$\n\n\n\nThe kinetic energy may be decomposed into rotation about the center of mass and translation of the center of mass; this translation is purely vertical by conservation of momentum. The vertical speed of the center of mass is $(R / 2) \\omega \\sin \\theta$, so the kinetic energy is\n\n\n\n$$\n\n\\frac{1}{2} I \\omega^{2}+\\frac{1}{2}(2 M) v_{\\mathrm{CM}}^{2}=\\frac{1}{4} M R^{2} \\omega^{2}+\\frac{1}{4} M R^{2} \\omega^{2} \\sin ^{2} \\theta\n\n$$\n\n\n\nEquating these two expressions and simplifying gives\n\n\n\n$$\n\n\\omega^{2}=\\frac{4 g}{R} \\frac{1-\\cos \\theta}{1+\\sin ^{2} \\theta}\n\n$$\n\n\n\nSome students instead treated the motion of the bead and ball separately; this can lead to the correct answer, but it is easy to make a mistake. Another common route was to apply Lagrangian mechanics, solving the Euler-Lagrange equations, or equivalently to solve the $F=m a$ equations. These are quite complicated, and nobody managed to integrate them to get the correct answer.\n\n']",['$\\omega^{2}=\\frac{4 g}{R} \\frac{1-\\cos \\theta}{1+\\sin ^{2} \\theta}$'],False,,Equation, 1475,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$","a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?","[""During the collision, the block receives impulses from the normal force, friction, and gravity. Since the collision is very short, the impulse due to gravity is negligible. Let $p_{N}$ and $p_{F}$ be the magnitudes of the impulses from the normal force and friction force.\n\n\n\nSince the block stays on the ramp after the collision, its final momentum is parallel to the ramp. Then the normal force must completely eliminate the block's initial momentum perpendicular to the ramp, so\n\n\n\n$$\n\np_{N}=m v \\cos \\theta\n\n$$\n\n\n\nSince the block still moves after the collision,\n\n\n\n$$\n\np_{F}=\\mu p_{N}\n\n$$\n\n\n\nThe block's initial momentum parallel to the ramp is $m v \\sin \\theta$, so\n\n\n\n$$\n\nm v \\sin \\theta-p_{F}=m u\n\n$$\n\n\n\nwhere $u$ is the final speed of the block. Solving for $u$ gives\n\n\n\n$$\n\nu=v(\\sin \\theta-\\mu \\cos \\theta)\n\n$$""]",['$u=v(\\sin \\theta-\\mu \\cos \\theta)$'],False,,Expression, 1476,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Context question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: \boxed{$u=v(\sin \theta-\mu \cos \theta)$} ",ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?,"['We set $u=0$ to obtain\n\n\n\n$$\n\n\\mu=\\tan \\theta\n\n$$\n\n\n\nNote that this is simply the no-slip condition for a block resting on an inclined plane! This is because in both cases, equality is achieved when the normal force and maximal friction force sum to a purely vertical force.']",['$\\mu=\\tan \\theta$'],False,,Expression, 1477,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Context question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: \boxed{$u=v(\sin \theta-\mu \cos \theta)$} Context question: ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ? Context answer: \boxed{$\mu=\tan \theta$} ","b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$. i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?","[""If the sphere immediately begins to roll without slipping, we can calculate the frictional impulse independently of the normal impulse. We have\n\n\n\n$$\n\nm v \\sin \\theta-p_{F}=m u\n\n$$\n\n\n\nThe frictional impulse is responsible for the sphere's rotation, so its angular momentum about its center of mass is $L=p_{F} R$. But we also know that\n\n\n\n$$\n\nL=\\beta m R^{2} \\omega=\\beta m R u\n\n$$\n\n\n\nThen\n\n\n\n$$\n\np_{F}=\\beta m u \\text {. }\n\n$$\n\n\n\nSubstituting into the previous expression gives\n\n\n\n$$\n\nm v \\sin \\theta=(1+\\beta) m u \\Rightarrow u=\\frac{v \\sin \\theta}{1+\\beta}\n\n$$""]",['$u=\\frac{v \\sin \\theta}{1+\\beta}$'],False,,Expression, 1478,Mechanics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Context question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: \boxed{$u=v(\sin \theta-\mu \cos \theta)$} Context question: ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ? Context answer: \boxed{$\mu=\tan \theta$} Context question: b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$. i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case? Context answer: \boxed{$u=\frac{v \sin \theta}{1+\beta}$} ",ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision?,"['As in part (a), the normal impulse is $p_{N}=m v \\cos \\theta$ and the maximal frictional impulse is $p_{F}=\\mu p_{N}$. From the previous part, we need\n\n\n\n$$\n\np_{F}=\\frac{\\beta m v \\sin \\theta}{1+\\beta}\n\n$$\n\n\n\nand equating these expressions gives\n\n\n\n$$\n\n\\mu=\\frac{\\beta \\tan \\theta}{1+\\beta}\n\n$$']",['$\\mu=\\frac{\\beta \\tan \\theta}{1+\\beta}$'],False,,Expression, 1479,Thermodynamics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: A vacuum system consists of a chamber of volume $V$ connected to a vacuum pump that is a cylinder with a piston that moves left and right. The minimum volume in the pump cylinder is $V_{0}$, and the maximum volume in the cylinder is $V_{0}+\Delta V$. You should assume that $\Delta V \ll V$. The cylinder has two valves. The inlet valve opens when the pressure inside the cylinder is lower than the pressure in the chamber, but closes when the piston moves to the right. The outlet valve opens when the pressure inside the cylinder is greater than atmospheric pressure $P_{a}$, and closes when the piston moves to the left. A motor drives the piston to move back and forth. The piston moves at such a rate that heat is not conducted in or out of the gas contained in the cylinder during the pumping cycle. One complete cycle takes a time $\Delta t$. You should assume that $\Delta t$ is a very small quantity, but $\Delta V / \Delta t=R$ is finite. The gas in the chamber is ideal monatomic and remains at a fixed temperature of $T_{a}$. Start with assumption that $V_{0}=0$ and there are no leaks in the system.",a. At $t=0$ the pressure inside the chamber is $P_{a}$. Find an equation for the pressure at a later time $t$.,"['During each cycle, the system sucks gas out of the chamber and pushes it into the atmosphere. Since $V_{0}=0$, the inlet valve opens the moment the piston starts moving to the left. When the piston is all the way to the left, a fraction $\\Delta V /(V+\\Delta V)$ of the gas is in the cylinder. As the piston moves to the right, all of this gas is pushed out, so after a single cycle,\n\n\n\n$$\n\nP_{f}=P_{i}\\left(\\frac{V}{V+\\Delta V}\\right)\n\n$$\n\n\n\nand in general,\n\n\n\n$$\n\nP(t)=P_{a}\\left(\\frac{V}{V+\\Delta V}\\right)^{t / \\Delta t}\n\n$$\n\n\n\nWhile this is technically correct, it can be simplified significantly. Write\n\n\n\n$$\n\nP(t)=P_{a}\\left(1+\\frac{\\Delta V}{V}\\right)^{-t / \\Delta t}=P_{a}\\left((1+x)^{1 / x}\\right)^{-R t / V}\n\n$$\n\n\n\nwhere $x=\\Delta V / V \\ll 1$. Then using the definition of $e$,\n\n\n\n$$\n\ne=\\lim _{x \\rightarrow 0}(1+x)^{1 / x}\n\n$$\n\n\n\nwe have\n\n\n\n$$\n\nP(t)=P_{a} e^{-R t / V}\n\n$$']",['$P(t)=P_{a} e^{-R t / V}$'],False,,Expression, 1480,Thermodynamics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: A vacuum system consists of a chamber of volume $V$ connected to a vacuum pump that is a cylinder with a piston that moves left and right. The minimum volume in the pump cylinder is $V_{0}$, and the maximum volume in the cylinder is $V_{0}+\Delta V$. You should assume that $\Delta V \ll V$. The cylinder has two valves. The inlet valve opens when the pressure inside the cylinder is lower than the pressure in the chamber, but closes when the piston moves to the right. The outlet valve opens when the pressure inside the cylinder is greater than atmospheric pressure $P_{a}$, and closes when the piston moves to the left. A motor drives the piston to move back and forth. The piston moves at such a rate that heat is not conducted in or out of the gas contained in the cylinder during the pumping cycle. One complete cycle takes a time $\Delta t$. You should assume that $\Delta t$ is a very small quantity, but $\Delta V / \Delta t=R$ is finite. The gas in the chamber is ideal monatomic and remains at a fixed temperature of $T_{a}$. Start with assumption that $V_{0}=0$ and there are no leaks in the system. Context question: a. At $t=0$ the pressure inside the chamber is $P_{a}$. Find an equation for the pressure at a later time $t$. Context answer: \boxed{$P(t)=P_{a} e^{-R t / V}$} ",b. Find an expression for the temperature of the gas as it is emitted from the pump cylinder into the atmosphere. Your answer may depend on time.,"['When the piston is all the way to the left, the pressure is $P(t)$ and the temperature is $T_{a}$. As the piston moves to the right, the gas is adiabatically compressed until its pressure reaches $P_{a}$ and the outlet valve opens. Since $P V^{\\gamma}$ is constant during adiabatic compression and $P V / T$ is constant by the ideal gas law,\n\n\n\n$$\n\nT_{\\text {out }}(t)=T_{a}\\left(\\frac{P_{a}}{P(t)}\\right)^{1-1 / \\gamma}=T_{a}\\left(\\frac{P_{a}}{P(t)}\\right)^{2 / 5}=T_{a} e^{2 R t / 5 V}\n\n$$\n\n\n\nwhere we used $\\gamma=5 / 3$ for a monatomic ideal gas.']",['$T_{a} e^{2 R t / 5 V}$'],False,,Expression, 1481,Thermodynamics,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: A vacuum system consists of a chamber of volume $V$ connected to a vacuum pump that is a cylinder with a piston that moves left and right. The minimum volume in the pump cylinder is $V_{0}$, and the maximum volume in the cylinder is $V_{0}+\Delta V$. You should assume that $\Delta V \ll V$. The cylinder has two valves. The inlet valve opens when the pressure inside the cylinder is lower than the pressure in the chamber, but closes when the piston moves to the right. The outlet valve opens when the pressure inside the cylinder is greater than atmospheric pressure $P_{a}$, and closes when the piston moves to the left. A motor drives the piston to move back and forth. The piston moves at such a rate that heat is not conducted in or out of the gas contained in the cylinder during the pumping cycle. One complete cycle takes a time $\Delta t$. You should assume that $\Delta t$ is a very small quantity, but $\Delta V / \Delta t=R$ is finite. The gas in the chamber is ideal monatomic and remains at a fixed temperature of $T_{a}$. Start with assumption that $V_{0}=0$ and there are no leaks in the system. Context question: a. At $t=0$ the pressure inside the chamber is $P_{a}$. Find an equation for the pressure at a later time $t$. Context answer: \boxed{$P(t)=P_{a} e^{-R t / V}$} Context question: b. Find an expression for the temperature of the gas as it is emitted from the pump cylinder into the atmosphere. Your answer may depend on time. Context answer: \boxed{$T_{a} e^{2 R t / 5 V}$} Extra Supplementary Reading Materials: For the remainder of this problem $00$, the inlet valve will not open immediately when the piston begins moving to the left; instead it will open once the pressure in the cylinder equals the pressure in the chamber. Since the expansion of the cylinder is adiabatic, $P V^{\\gamma}$ is constant, so\n\n\n\n$$\n\nP_{\\min }=P_{a}\\left(\\frac{V_{0}}{V_{0}+\\Delta V}\\right)^{\\gamma}=P_{a}\\left(1+\\frac{\\Delta V}{V_{0}}\\right)^{-\\gamma}\n\n$$']",['$P_{a}\\left(1+\\frac{\\Delta V}{V_{0}}\\right)^{-\\gamma}$'],False,,Expression, 1482,Electromagnetism,"$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: The electric potential at the center of a cube with uniform charge density $\rho$ and side length $a$ is $$ \Phi \approx \frac{0.1894 \rho a^{2}}{\epsilon_{0}} $$ You do not need to derive this. ${ }^{1}$ For the entirety of this problem, any computed numerical constants should be written to three significant figures.","a. What is the electric potential at a corner of the same cube? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.","['By dimensional analysis, the answer takes the form\n\n\n\n$$\n\n\\Phi_{c}(a, \\rho) \\approx \\frac{C \\rho a^{2}}{\\epsilon_{0}}\n\n$$\n\n\n\nfor a dimensionless constant $C$. Note that a cube of side length $a$ consists of 8 cubes of side length $a / 2$, each with a corner at the center of the larger cube. Then\n\n\n\n$$\n\n\\frac{0.1894 \\rho a^{2}}{\\epsilon_{0}}=8 \\frac{C \\rho(a / 2)^{2}}{\\epsilon_{0}}\n\n$$\n\n\n\nso $C=0.1894 / 2=0.0947$.\n\n']","['$\\Phi_{c}(a, \\rho) \\approx \\frac{C \\rho a^{2}}{\\epsilon_{0}}$, $C=0.0947$']",False,,"Equation,Numerical",",1e-3" 1483,Electromagnetism,,"b. What is the electric potential at the tip of a pyramid with a square base of side length $a$, height $a / 2$, and uniform charge density $\rho$ ? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.","['A cube of side length $a$ consists of 6 such pyramids. Then we simply compute $0.1894 / 6$ for\n\n\n\n$$\n\n\\Phi_{p}(a, \\rho) \\approx \\frac{0.0316 \\rho a^{2}}{\\epsilon_{0}}\n\n$$']","['$\\Phi_{p}(a, \\rho) \\approx \\frac{0.0316 \\rho a^{2}}{\\epsilon_{0}}$']",False,,Need_human_evaluate, 1484,Electromagnetism,,"c. What is the electric potential due to a square plate with side length $a$ of uniform charge density $\sigma$ at a height $a / 2$ above its center? Write your answer in terms of $\sigma, a, \epsilon_{0}$, and any necessary numerical constants.","['Let the potential due to such a square be $\\Phi_{s}(a, \\sigma)$. Note that adding a square plate of infinitesimal thickness $d z$ and side length $a$ to a square pyramid with base side length $a$ and height $a / 2$ yields a square pyramid with base side length $a+2 d z$ and height $a / 2+d z$.\n\n\n\nThe surface charge density of a square plate with thickness $d z$ and volume charge density $\\rho$ is $\\sigma=\\rho d z$. Then by the principle of superposition,\n\n\n\n$$\n\n\\Phi_{s}(a, \\rho d z)=\\Phi_{p}(a+2 d z, \\rho)-\\Phi_{p}(a, \\rho) \\approx \\frac{0.0316 \\rho\\left((a+2 d z)^{2}-a^{2}\\right)}{\\epsilon_{0}}=\\frac{0.126 a \\rho d z}{\\epsilon_{0}}\n\n$$\n\n\n\nso we have\n\n\n\n$$\n\n\\Phi_{s}(a, \\sigma) \\approx \\frac{0.126 a \\sigma}{\\epsilon_{0}}\n\n$$']","['$\\Phi_{s}(a, \\sigma) \\approx \\frac{0.126 a \\sigma}{\\epsilon_{0}}$']",False,,Need_human_evaluate, 1485,Electromagnetism,,"d. Let $E(z)$ be the electric field at a height $z$ above the center of a square with charge density $\sigma$ and side length $a$. If the electric potential at the center of the square is approximately $\frac{0.281 a \sigma}{\epsilon_{0}}$, estimate $E(a / 2)$ by assuming that $E(z)$ is linear in $z$ for $0",a. Find the volume of the lower chamber $V_{0}$ when the gas starts to leak between the chambers.,"['Let $p_{1}, V_{1}, T_{1}$ denote the (time dependent) pressure, volume, and temperature in the upper chamber, and $p_{2}, V_{2}$, $T_{2}$ - those in the lower one. Note that $V_{1} \\equiv V$ does not change.\n\nConsider a parcel of volume $v$ below the diaphragm containing $n$ moles of helium. It is convenient to imagine it bounded by two fictitious free thin massless pistons. During slow perturbations the parcel undergoes an adiabatic process. The pressure and the temperature for the parcel are actually the pressure and the temperature for entire lower chamber $p_{2}$ and $T_{2}$. The energy conservation for the parcel is\n\n$$\n0=p_{2} \\mathrm{~d} v+\\mathrm{d}\\left(\\frac{3}{2} n_{v} R T_{2}\\right)=\\frac{5}{2} p_{2} \\mathrm{~d} v+\\frac{3}{2} v \\mathrm{~d} p_{2}\n$$\n\nThis gives\n\n$$\nv^{5} p_{2}^{3} =\\text { const }\n\\tag{1}\n$$\n\n$$\nT_{2}^{5} p_{2}^{-2} =\\text { const }\n\\tag{2}\n$$\n\nThe leak begins when the pressure below the diaphragm exceeds that in the upper chamber by $\\Delta p \\equiv$ $p_{0}-p=m g H / V=p x$, where\n\n$$\nx=\\frac{m g H}{p V} \\tag{(a., b., c.)}\n$$\n\n\n\na. We may let $v=V_{2}$ before that.\n\n$$\n\\begin{gathered}\nV^{5} p^{3}=V_{0}^{5}(p+\\Delta p)^{3}=V_{0}^{5} p^{3}(1+x)^{3} \\\\\nV_{0}=V(1+x)^{-3 / 5}\n\\end{gathered}\n$$']",['$V(1+x)^{-3 / 5}$'],False,,Expression, 1496,Thermodynamics,"A hollow insulated cylinder of height $2 \mathrm{H}$ and volume $2 \mathrm{~V}$ is closed from below by an insulating piston. The cylinder is divided into two initially identical chambers by an insulating diaphragm of mass $m$. The diaphragm rests on a circular ledge and a gasket between them provides tight contact. Both chambers are filled with gaseous helium at pressure $p$ and temperature $T$. A force is applied to the piston, so that it moves upwards slowly. Context question: a. Find the volume of the lower chamber $V_{0}$ when the gas starts to leak between the chambers. Context answer: \boxed{$V(1+x)^{-3 / 5}$} ",b. Find the temperature $T_{1}$ in the upper chamber when the piston touches the diaphragm.,"['Let $p_{1}, V_{1}, T_{1}$ denote the (time dependent) pressure, volume, and temperature in the upper chamber, and $p_{2}, V_{2}$, $T_{2}$ - those in the lower one. Note that $V_{1} \\equiv V$ does not change.\n\nConsider a parcel of volume $v$ below the diaphragm containing $n$ moles of helium. It is convenient to imagine it bounded by two fictitious free thin massless pistons. During slow perturbations the parcel undergoes an adiabatic process. The pressure and the temperature for the parcel are actually the pressure and the temperature for entire lower chamber $p_{2}$ and $T_{2}$. The energy conservation for the parcel is\n\n$$\n0=p_{2} \\mathrm{~d} v+\\mathrm{d}\\left(\\frac{3}{2} n_{v} R T_{2}\\right)=\\frac{5}{2} p_{2} \\mathrm{~d} v+\\frac{3}{2} v \\mathrm{~d} p_{2}\n$$\n\nThis gives\n\n$$\nv^{5} p_{2}^{3} =\\text { const }\n\\tag{1}\n$$\n\n$$\nT_{2}^{5} p_{2}^{-2} =\\text { const }\n\\tag{2}\n$$\n\nThe leak begins when the pressure below the diaphragm exceeds that in the upper chamber by $\\Delta p \\equiv$ $p_{0}-p=m g H / V=p x$, where\n\n$$\nx=\\frac{m g H}{p V} \\tag{(a., b., c.)}\n$$\n\n\n\na. We may let $v=V_{2}$ before that.\n\n$$\n\\begin{gathered}\nV^{5} p^{3}=V_{0}^{5}(p+\\Delta p)^{3}=V_{0}^{5} p^{3}(1+x)^{3} \\\\\nV_{0}=V(1+x)^{-3 / 5}\n\\end{gathered}\n$$\n\nb. The energy conservation for the whole system:\n\n$$\n\\begin{gathered}\n0=p_{1} \\mathrm{~d} V_{1}+\\mathrm{d}\\left(\\frac{3}{2} n_{1} R T_{1}\\right)+p_{2} \\mathrm{~d} V_{2}+\\mathrm{d}\\left(\\frac{3}{2} n_{2} R T_{2}\\right)= \\\\\n\\frac{5}{2}\\left(p_{1} \\mathrm{~d} V_{1}+p_{2} \\mathrm{~d} V_{2}\\right)+\\frac{3}{2}\\left(V_{1} \\mathrm{~d} p_{1}+V_{2} \\mathrm{~d} p_{2}\\right)= \\\\\n\\frac{5}{2} p_{2} \\mathrm{~d}\\left(V+V_{2}\\right)+\\frac{3}{2}\\left(V+V_{2}\\right) \\mathrm{d} p_{2}\n\\end{gathered}\n$$\n\nsince the pressure above the diaphragm remains lower than that below by the same margin $\\Delta p$ during the later process and $\\mathrm{d} V_{2}=\\mathrm{d}\\left(V+V_{2}\\right)$. Similarly to (1), we get\n\n$$\n\\left(V+V_{2}\\right)^{5} p_{2}^{3}=\\text { const }\n$$\n\nThe pressure $p_{2}^{\\prime}$ in the lower chamber when the piston touches the diaphragm is found from the equation\n\n$$\n\\begin{gathered}\nV^{5} p_{2}^{\\prime 3}=\\left(V+V_{0}\\right)^{5} p_{0}^{3}=V^{5}\\left(1+\\frac{1}{(1+x)^{3 / 5}}\\right)^{5} p^{3}(1+x)^{3} \\\\\np_{2}^{\\prime}=p\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}\n\\end{gathered}\n\\tag{3}\n$$\n\nThe pressure in the upper chamber at this moment is\n\n$$\np_{1}^{\\prime}=p_{2}^{\\prime}-\\Delta p=p\\left(\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}-x\\right)\n$$\n\nThe temperature in the upper chamber is found from the equations of state $p V=n R T$ and $p^{\\prime} V=(2 n) R T^{\\prime}$\n\n$$\nT_{1}^{\\prime}=\\frac{T}{2}\\left(\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}-x\\right)\n$$']",['$\\frac{T}{2}\\left(\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}-x\\right)$'],False,,Expression, 1497,Thermodynamics,"A hollow insulated cylinder of height $2 \mathrm{H}$ and volume $2 \mathrm{~V}$ is closed from below by an insulating piston. The cylinder is divided into two initially identical chambers by an insulating diaphragm of mass $m$. The diaphragm rests on a circular ledge and a gasket between them provides tight contact. Both chambers are filled with gaseous helium at pressure $p$ and temperature $T$. A force is applied to the piston, so that it moves upwards slowly. Context question: a. Find the volume of the lower chamber $V_{0}$ when the gas starts to leak between the chambers. Context answer: \boxed{$V(1+x)^{-3 / 5}$} Context question: b. Find the temperature $T_{1}$ in the upper chamber when the piston touches the diaphragm. Context answer: \boxed{$\frac{T}{2}\left(\left(1+(1+x)^{3 / 5}\right)^{5 / 3}-x\right)$} ",c. Find the temperature $T_{2}$ in the lower chamber immediately before the piston touches the diaphragm.,"['Let $p_{1}, V_{1}, T_{1}$ denote the (time dependent) pressure, volume, and temperature in the upper chamber, and $p_{2}, V_{2}$, $T_{2}$ - those in the lower one. Note that $V_{1} \\equiv V$ does not change.\n\nConsider a parcel of volume $v$ below the diaphragm containing $n$ moles of helium. It is convenient to imagine it bounded by two fictitious free thin massless pistons. During slow perturbations the parcel undergoes an adiabatic process. The pressure and the temperature for the parcel are actually the pressure and the temperature for entire lower chamber $p_{2}$ and $T_{2}$. The energy conservation for the parcel is\n\n$$\n0=p_{2} \\mathrm{~d} v+\\mathrm{d}\\left(\\frac{3}{2} n_{v} R T_{2}\\right)=\\frac{5}{2} p_{2} \\mathrm{~d} v+\\frac{3}{2} v \\mathrm{~d} p_{2}\n$$\n\nThis gives\n\n$$\nv^{5} p_{2}^{3} =\\text { const }\n\\tag{1}\n$$\n\n$$\nT_{2}^{5} p_{2}^{-2} =\\text { const }\n\\tag{2}\n$$\n\nThe leak begins when the pressure below the diaphragm exceeds that in the upper chamber by $\\Delta p \\equiv$ $p_{0}-p=m g H / V=p x$, where\n\n$$\nx=\\frac{m g H}{p V} \\tag{(a., b., c.)}\n$$\n\n\n\na. We may let $v=V_{2}$ before that.\n\n$$\n\\begin{gathered}\nV^{5} p^{3}=V_{0}^{5}(p+\\Delta p)^{3}=V_{0}^{5} p^{3}(1+x)^{3} \\\\\nV_{0}=V(1+x)^{-3 / 5}\n\\end{gathered}\n$$\n\nb. The energy conservation for the whole system:\n\n$$\n\\begin{gathered}\n0=p_{1} \\mathrm{~d} V_{1}+\\mathrm{d}\\left(\\frac{3}{2} n_{1} R T_{1}\\right)+p_{2} \\mathrm{~d} V_{2}+\\mathrm{d}\\left(\\frac{3}{2} n_{2} R T_{2}\\right)= \\\\\n\\frac{5}{2}\\left(p_{1} \\mathrm{~d} V_{1}+p_{2} \\mathrm{~d} V_{2}\\right)+\\frac{3}{2}\\left(V_{1} \\mathrm{~d} p_{1}+V_{2} \\mathrm{~d} p_{2}\\right)= \\\\\n\\frac{5}{2} p_{2} \\mathrm{~d}\\left(V+V_{2}\\right)+\\frac{3}{2}\\left(V+V_{2}\\right) \\mathrm{d} p_{2}\n\\end{gathered}\n$$\n\nsince the pressure above the diaphragm remains lower than that below by the same margin $\\Delta p$ during the later process and $\\mathrm{d} V_{2}=\\mathrm{d}\\left(V+V_{2}\\right)$. Similarly to (1), we get\n\n$$\n\\left(V+V_{2}\\right)^{5} p_{2}^{3}=\\text { const }\n$$\n\nThe pressure $p_{2}^{\\prime}$ in the lower chamber when the piston touches the diaphragm is found from the equation\n\n$$\n\\begin{gathered}\nV^{5} p_{2}^{\\prime 3}=\\left(V+V_{0}\\right)^{5} p_{0}^{3}=V^{5}\\left(1+\\frac{1}{(1+x)^{3 / 5}}\\right)^{5} p^{3}(1+x)^{3} \\\\\np_{2}^{\\prime}=p\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}\n\\end{gathered}\n$$\n\nThe pressure in the upper chamber at this moment is\n\n$$\np_{1}^{\\prime}=p_{2}^{\\prime}-\\Delta p=p\\left(\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}-x\\right)\n$$\n\nThe temperature in the upper chamber is found from the equations of state $p V=n R T$ and $p^{\\prime} V=(2 n) R T^{\\prime}$\n\n$$\nT_{1}^{\\prime}=\\frac{T}{2}\\left(\\left(1+(1+x)^{3 / 5}\\right)^{5 / 3}-x\\right)\n$$\n\n\nThe temperature and the pressure in the lower chamber are related by (2). Substituting (3) we get\n\n$$\nT_{2}^{\\prime}=T\\left(\\frac{p_{2}^{\\prime}}{p}\\right)^{2 / 5}=T\\left(1+(1+x)^{3 / 5}\\right)^{2 / 3}\n$$']",['$T\\left(1+(1+x)^{3 / 5}\\right)^{2 / 3}$'],False,,Expression, 1498,Mechanics,,"T2: Thread around a cylinder One end of a thread is tied into a loop of length $L>2 \pi R$, and a cylinder of radius $R$ is put through the loop. The coefficient of friction between the thread and the cylinder is $\mu$. The free end of the thread is being pulled parallel to the axis of the cylinder (as shown by arrow in the photo) while keeping the cylinder at rest. If the length of the loop is longer than a critical value, $L>L_{0}$, the loop can slide along the cylinder without changing its shape, otherwise the friction ""locks"" it into a place and increasing the pulling force would eventually just break the thread. Find this critical value $L_{0}$. The weight of the thread is to be neglected; the thread will not twist when being pulled. It might be useful to know that $$ 2 \int \sqrt{1+x^{2}} \mathrm{~d} x=x \sqrt{1+x^{2}}+\operatorname{arcsinh} x, $$ where $\operatorname{arcsinh} x \equiv \ln \left(x+\sqrt{1+x^{2}}\right)$.","['Fig. 1 shows the cylinder and the loop from three different angles; $P$ denotes the pulled point of the loop, while $O$ is the top point of the thread.\n\n\nFig. 1\n\nImagine that the side of the cylinder is cut along the generatrix $A B$ passing through point $P$, and then the side (including the loop) is unfolded as shown in Fig. 2. In this figure points $A$ and $A^{\\prime}, B$ and $B^{\\prime}, P$ and $P^{\\prime}$ are\n\n\n\nequivalent, respectively. Let us introduce a Cartesian coordinate system on this unfolded plane so that point $O$ is the origin, axis $z$ is parallel with the axis of cylinder and directed downwards, axis $x$ is perpendicular to $z$ (i.e. horizontal).\n\n\n\nFig. 2\n\nConsider the forces acting on a small piece of the thread (with horizontal projection $\\Delta x$ ) indicated with red line in both figures. These are the tensions at both ends of the small piece exerted by neighbouring parts of the thread, the normal force $\\Delta N$ and the friction force $\\Delta f$ exerted by the cylinder. On the verge of slipping, the direction of $\\Delta f$ is parallel to the $z$-axis. Since the small piece of thread is in equilibrium, the $x$-component of the tension is the same everywhere:\n\n$$\nT_{x}=\\text { const. }\n$$\n\nThe normal force $\\Delta N$ can be determined by looking at the top view of the loop in Fig. 1. The polar angle corresponding to small piece of thread is $\\Delta \\varphi=\\Delta x / R$, so the force balance in the radial direction can be written as\n\n$$\n2 T_{x} \\underbrace{\\sin \\frac{\\Delta \\varphi}{2}}_{\\approx \\Delta \\varphi / 2}-\\Delta N=0 \\quad \\longrightarrow \\quad \\Delta N=T_{x} \\frac{\\Delta x}{R} .\n\\tag{4}\n$$\n\nThe frictional force on the verge of slipping is given by\n\n$$\n\\Delta f=\\mu \\Delta N\n\\tag{5}\n$$\n\nThus, the force balance on the small piece of thread in the $z$ direction (see Fig. 2):\n\n$$\n\\left.T_{x} \\frac{\\mathrm{d} z}{\\mathrm{~d} x}\\right|_{x+\\Delta x}-\\left.T_{x} \\frac{\\mathrm{d} z}{\\mathrm{~d} x}\\right|_{x}-\\Delta f=0\n\\tag{6}\n$$\n\nwhere we expressed the $z$-component of the tension forces with $T_{x}$ and the tangent $\\mathrm{d} z / \\mathrm{d} x$. Using the three equations above and taking the limit $\\Delta x \\rightarrow 0$, we get the differential equation\n\n$$\n\\frac{\\mathrm{d}^{2} z}{\\mathrm{~d} x^{2}}=\\frac{\\mu}{R}\n$$\n\nwhere we used the fact that $T_{x} \\neq 0$. By direct integration and taking into account the boundary conditions $z(0)=0$ and $z^{\\prime}(0)=0$ we get\n\n$$\nz(x)=\\frac{\\mu}{2 R} x^{2}\n$$\n\nso the shape of the thread on the unfolded side surface (Fig. 2) can be described by a parabola.\nThe thread needs to span over the entire cylinder, so its length can be calculated as\n\n$$\nL_{0}=\\int_{-\\pi R}^{\\pi R} \\sqrt{\\mathrm{d} x^{2}+\\mathrm{d} z^{2}}=2 \\int_{0}^{\\pi R} \\sqrt{1+\\left(\\frac{\\mathrm{d} z}{\\mathrm{~d} x}\\right)^{2}} \\mathrm{~d} x\n$$\n\nSubstituting the $z(x)$ function we get:\n\n$$\nL_{0}=2 \\int_{0}^{\\pi R} \\sqrt{1+\\left(\\frac{\\mu x}{R}\\right)^{2}} \\mathrm{~d} x=\\left(\\frac{R}{\\mu}\\right) 2 \\int_{0}^{\\pi \\mu} \\sqrt{1+\\xi^{2}} \\mathrm{~d} \\xi\n$$\n\nwhere we introduced the notation $\\xi=\\mu x / R$. Using the integral given in the text of the problem:\n\n$$\nL_{0}=\\pi R \\sqrt{1+(\\pi \\mu)^{2}}+\\frac{R}{\\mu} \\operatorname{arcsinh}(\\pi \\mu) .\n$$\n\nIf the length of the thread is shorter than the length calculated here, then there is no solution satisfying the thread length constraint, i.e. the thread cannot slip.\n\nNote. In the limit $\\mu \\rightarrow 0$ the loop slips even if $L=2 \\pi R$, which should be reproduced by our final formula. Using the relation for the inverse hyperbolic function given in the problem text, then expanding the logarithm in Taylor series in linear order around 0 we get:\n\n$$\n\\operatorname{arcsinh} x \\equiv \\ln \\left(x+\\sqrt{1+x^{2}}\\right) \\approx \\ln (x+1) \\approx x\n$$\n\nso for small values of $\\mu$ we get\n\n$$\nL_{0} \\approx \\pi R+\\frac{R}{\\mu}(\\pi \\mu)=2 \\pi R\n$$']",['$2 \\pi R$'],False,,Expression, 1498,Mechanics,,"T2: Thread around a cylinder One end of a thread is tied into a loop of length $L>2 \pi R$, and a cylinder of radius $R$ is put through the loop. The coefficient of friction between the thread and the cylinder is $\mu$. The free end of the thread is being pulled parallel to the axis of the cylinder (as shown by arrow in ![](https://cdn.mathpix.com/cropped/2023_12_21_acad72c0ff0a2ad974c9g-1.jpg?height=445&width=466&top_left_y=1482&top_left_x=561) the photo) while keeping the cylinder at rest. If the length of the loop is longer than a critical value, $L>L_{0}$, the loop can slide along the cylinder without changing its shape, otherwise the friction ""locks"" it into a place and increasing the pulling force would eventually just break the thread. Find this critical value $L_{0}$. The weight of the thread is to be neglected; the thread will not twist when being pulled. It might be useful to know that $$ 2 \int \sqrt{1+x^{2}} \mathrm{~d} x=x \sqrt{1+x^{2}}+\operatorname{arcsinh} x, $$ where $\operatorname{arcsinh} x \equiv \ln \left(x+\sqrt{1+x^{2}}\right)$.","['Fig. 1 shows the cylinder and the loop from three different angles; $P$ denotes the pulled point of the loop, while $O$ is the top point of the thread.\n![](https://cdn.mathpix.com/cropped/2023_12_21_68ab93ca6ae203241e21g-1.jpg?height=764&width=636&top_left_y=1802&top_left_x=1188)\n\nFig. 1\n\nImagine that the side of the cylinder is cut along the generatrix $A B$ passing through point $P$, and then the side (including the loop) is unfolded as shown in Fig. 2. In this figure points $A$ and $A^{\\prime}, B$ and $B^{\\prime}, P$ and $P^{\\prime}$ are\n\n\n\nequivalent, respectively. Let us introduce a Cartesian coordinate system on this unfolded plane so that point $O$ is the origin, axis $z$ is parallel with the axis of cylinder and directed downwards, axis $x$ is perpendicular to $z$ (i.e. horizontal).\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6a02edc5e0df952b6ba8g-1.jpg?height=454&width=740&top_left_y=453&top_left_x=178)\n\nFig. 2\n\nConsider the forces acting on a small piece of the thread (with horizontal projection $\\Delta x$ ) indicated with red line in both figures. These are the tensions at both ends of the small piece exerted by neighbouring parts of the thread, the normal force $\\Delta N$ and the friction force $\\Delta f$ exerted by the cylinder. On the verge of slipping, the direction of $\\Delta f$ is parallel to the $z$-axis. Since the small piece of thread is in equilibrium, the $x$-component of the tension is the same everywhere:\n\n$$\nT_{x}=\\text { const. }\n$$\n\nThe normal force $\\Delta N$ can be determined by looking at the top view of the loop in Fig. 1. The polar angle corresponding to small piece of thread is $\\Delta \\varphi=\\Delta x / R$, so the force balance in the radial direction can be written as\n\n$$\n2 T_{x} \\underbrace{\\sin \\frac{\\Delta \\varphi}{2}}_{\\approx \\Delta \\varphi / 2}-\\Delta N=0 \\quad \\longrightarrow \\quad \\Delta N=T_{x} \\frac{\\Delta x}{R} .\n\\tag{4}\n$$\n\nThe frictional force on the verge of slipping is given by\n\n$$\n\\Delta f=\\mu \\Delta N\n\\tag{5}\n$$\n\nThus, the force balance on the small piece of thread in the $z$ direction (see Fig. 2):\n\n$$\n\\left.T_{x} \\frac{\\mathrm{d} z}{\\mathrm{~d} x}\\right|_{x+\\Delta x}-\\left.T_{x} \\frac{\\mathrm{d} z}{\\mathrm{~d} x}\\right|_{x}-\\Delta f=0\n\\tag{6}\n$$\n\nwhere we expressed the $z$-component of the tension forces with $T_{x}$ and the tangent $\\mathrm{d} z / \\mathrm{d} x$. Using the three equations above and taking the limit $\\Delta x \\rightarrow 0$, we get the differential equation\n\n$$\n\\frac{\\mathrm{d}^{2} z}{\\mathrm{~d} x^{2}}=\\frac{\\mu}{R}\n$$\n\nwhere we used the fact that $T_{x} \\neq 0$. By direct integration and taking into account the boundary conditions $z(0)=0$ and $z^{\\prime}(0)=0$ we get\n\n$$\nz(x)=\\frac{\\mu}{2 R} x^{2}\n$$\n\nso the shape of the thread on the unfolded side surface (Fig. 2) can be described by a parabola.\nThe thread needs to span over the entire cylinder, so its length can be calculated as\n\n$$\nL_{0}=\\int_{-\\pi R}^{\\pi R} \\sqrt{\\mathrm{d} x^{2}+\\mathrm{d} z^{2}}=2 \\int_{0}^{\\pi R} \\sqrt{1+\\left(\\frac{\\mathrm{d} z}{\\mathrm{~d} x}\\right)^{2}} \\mathrm{~d} x\n$$\n\nSubstituting the $z(x)$ function we get:\n\n$$\nL_{0}=2 \\int_{0}^{\\pi R} \\sqrt{1+\\left(\\frac{\\mu x}{R}\\right)^{2}} \\mathrm{~d} x=\\left(\\frac{R}{\\mu}\\right) 2 \\int_{0}^{\\pi \\mu} \\sqrt{1+\\xi^{2}} \\mathrm{~d} \\xi\n$$\n\nwhere we introduced the notation $\\xi=\\mu x / R$. Using the integral given in the text of the problem:\n\n$$\nL_{0}=\\pi R \\sqrt{1+(\\pi \\mu)^{2}}+\\frac{R}{\\mu} \\operatorname{arcsinh}(\\pi \\mu) .\n$$\n\nIf the length of the thread is shorter than the length calculated here, then there is no solution satisfying the thread length constraint, i.e. the thread cannot slip.\n\nNote. In the limit $\\mu \\rightarrow 0$ the loop slips even if $L=2 \\pi R$, which should be reproduced by our final formula. Using the relation for the inverse hyperbolic function given in the problem text, then expanding the logarithm in Taylor series in linear order around 0 we get:\n\n$$\n\\operatorname{arcsinh} x \\equiv \\ln \\left(x+\\sqrt{1+x^{2}}\\right) \\approx \\ln (x+1) \\approx x\n$$\n\nso for small values of $\\mu$ we get\n\n$$\nL_{0} \\approx \\pi R+\\frac{R}{\\mu}(\\pi \\mu)=2 \\pi R\n$$']",['$2 \\pi R$'],False,,Expression, 1499,Electromagnetism,"## T1: Solenoid and loop A closed circular loop of radius $r$ consists of an ideal battery of electromotive force $\mathcal{E}$ and a wire of resistance $R$. A long thin air-core solenoid is aligned with the axis of the loop ( $z$-axis). Its length is $\ell \gg r$, cross-sectional area is $A(\sqrt{A} \ll r)$, and the number of turns is $N$. The solenoid is powered by a constant current $I$ provided by an ideal current source. The directions of the currents in the solenoid and in the loop are the same (clockwise in the figure). ",a. Find the force $F_{1}$ acting on the solenoid when its head $O_{1}$ is positioned in the loop centre $O$. What is the force $F_{2}$ acting on the solenoid when its tail $O_{2}$ is located in the centre of the loop?,"[""According to Newton's third law, the force acting on the solenoid is equal in magnitude, but opposite in direction to the force acting on the loop. The latter can be obtained from Lorentz's law by summation of infinitesimal forces $\\vec{F}=J \\Delta \\vec{l} \\times \\vec{B}$ acting on the individual loop elements $\\Delta \\vec{l}$, where $J=\\mathcal{E} / R$ is the current in the loop.\n\nSince the solenoid is long and thin, the magnetic field lines inside it are directed in $+z$ direction, and escape only from the immediate vicinity of its ends. The magnetic field outside the solenoid is vortex-free and sourcefree. The same requirements are satisfied by the electric field in empty space. Hence, the magnetic field outside the solenoid can be well approximated by a field created by two magnetic poles: a North pole residing close to point $O_{1}$ and a South pole located near to $O_{2}$. The flux $\\Phi$ emerging from the North pole (and the flux entering the South pole) is the same as the flux passing through the solenoid's cross-section:\n\n$$\n\\Phi=B_{\\text {in }} A=\\mu_{0} \\frac{N I}{\\ell} A\n$$\n\nWhen the endpoint $O_{1}$ of the solenoid is placed in the loop centre $O$, the magnetic field of the other end $\\left(O_{2}\\right)$ near the loop is negligible. The field created by the North pole located at $O_{1}$ is pointing radially outwards, and its magnitude at the loop circumference is (from spherical symmetry):\n\n$$\nB(r)=\\frac{\\Phi}{4 \\pi r^{2}}=\\frac{\\mu_{0}}{4 \\pi} \\frac{N I A}{\\ell r^{2}}\n$$\n\nNote. it is also correct to assume that the flux from the $B$-field (which carries a factor of $1 / 2$ compared to the field above at the end of the solenoid is uniformly distributed over hemisphere.\n\nThe forces acting on all elements of the loop (and also the net force) point in the $-z$ direction, which can be expected also from the same current directions (i.e. the loop and the solenoid attract each other). Hence, the reaction force acting on the solenoid points in the $+z$ direction, and its magnitude is given by:\n\n$$\nF_{1}=\\frac{\\mathcal{E}}{R} \\cdot 2 \\pi r \\cdot \\frac{\\mu_{0}}{4 \\pi} \\frac{N I A}{\\ell r^{2}}=\\frac{\\mu_{0} N I A \\mathcal{E}}{2 \\ell R r} .\n$$\n\nWhen the endpoint $\\mathrm{O}_{2}$ is located at the centre of the loop, the magnetic field produced by the South pole exerts a force on the loop. Since this field is directed radially inward, the force acting on the solenoid is the same in magnitude, but opposite in direction $(-z)$ as the force calculated above:\n\n$$\nF_2 = -F_1\n$$"", 'In this solution the force acting on the solenoid is calculated, as the force acting on the magnetic pole placed at the centre of the current-carrying loop. For this we need to find an expression for the magnetic pole strength (or magnetic charge) $Q_{\\mathrm{m}}$, which is defined as the ratio of the force and the magnetic field.\n\nThe total dipole moment $m$ of the solenoid is the product of the number of turns and the dipole moment $I A$ of each turn:\n\n$$\nm=N I A .\n$$\n\nThis can be also expressed with the magnetic pole strength and the distance of the poles: $m=Q_{\\mathrm{m}} \\ell$. From this we arrive to the expression\n\n$$\nQ_{\\mathrm{m}}=\\frac{N I A}{\\ell}=\\frac{\\Phi}{\\mu_{0}}\n$$\n\nwhere $\\Phi$ is the total flux emerging from the pole (see solution I).\n\nNote. The same result can be obtained from the analogy between electrostatic and magnetostatic fields. The Coulomb force between two point charges $\\pm Q$ can be derived from the principle of virtual work. The force is the derivative of the interaction part of the field energy with respect to the distance between the charges. The force between two magnetic charges $\\pm Q_{\\mathrm{m}}$ can be also calculated this way. From the expressions of electric and magnetic energy densities we can conclude the formula of the magnetic interaction force:\n\n$$\n\\begin{gathered}\nw_{E}=\\frac{1}{2} \\varepsilon_{0} E^{2} \\quad \\longleftrightarrow w_{B}=\\frac{1}{2 \\mu_{0}} B^{2}, \\\\\nF_{E}=\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Q^{2}}{r^{2}} \\quad \\longleftrightarrow \\quad F_{B}=\\frac{\\mu_{0}}{4 \\pi} \\frac{Q_{\\mathrm{m}}^{2}}{r^{2}} .\n\\end{gathered}\n$$\n\nAs it can be seen, the well-known formulae known in electrostatics can be also used in magnetostatics with the substitutions $\\varepsilon_{0}^{-1} \\longleftrightarrow \\mu_{0}, E \\longleftrightarrow B, Q \\longleftrightarrow Q_{\\mathrm{m}}$. Carrying on this analogy the magnetic pole strength can be figured out:\n\n$$\nQ=\\varepsilon_{0} \\Psi \\quad \\longleftrightarrow \\quad Q_{\\mathrm{m}}=\\frac{\\Phi}{\\mu_{0}}=\\frac{N}{\\ell} I A\n$$\n\nwhere $\\Psi$ and $\\Phi$ are the electric and magnetic flux for a closed surface containing the electric and magnetic charge, respectively.\n\nWhen endpoint $O_{1}$ of the solenoid is located at point $O$, a North pole resides at the centre of the loop. Here the magnetic field created by the loop can be expressed from Biot-Savart-law:\n\n$$\nB_{\\text {loop }}^{(\\text {at center })}=\\frac{\\mu_{0} J}{2 r}=\\frac{\\mu_{0} \\mathcal{E}}{2 R r},\n$$\n\n\n\npointing in the $+z$ direction. So the magnitude of the force acting on this end of the solenoid is:\n\n$$\nF_{1}=Q_{\\mathrm{m}} B_{\\text {loop }}^{(\\text {at center })}=\\frac{\\mu_{0} \\mathcal{E} N I A}{2 \\ell R r}\n$$\n\nand it is directed to $+z$. When the endpoint $\\mathrm{O}_{2}$ is located at the center of the loop, the force acting on the South pole should be calculated, resulting a force of same magnitude, but opposite direction.', 'Some of the magnetic field lines created by the loop enter into the near end $O_{1}$ of the solenoid; this entering flux is given by\n\n$$\n\\Phi_{\\text {in }}=B_{\\text {loop }}^{\\text {(at center) }} A=\\frac{\\mu_{0} J}{2 r} A=\\frac{\\mu_{0} \\mathcal{E} A}{2 R r} .\n$$\n\nSince the end $O_{2}$ is far from the loop, the flux created by the loop escaping there is negligibly small. This means that almost all the flux $\\Phi_{\\text {in }}$ escapes from the solenoid through its side.\n\nDenote the radial component of the magnetic field vector produced by the current-carrying loop at the perimeter of the $i$ th turn of the solenoid by $B_{i}$. Only this component contributes to the net force acting on the solenoid, as the axial component produces a radial force which is cancelled due to rotational symmetry. The axial force acting on the $i$ th turn of the solenoid is given by\n\n$$\nF_{1, i}=2 \\sqrt{A \\pi} I B_{i}\n$$\n\nwhere $2 \\sqrt{A \\pi}$ is the circumference of one turn, and the force points in the $+z$ direction. Summing up both sides gives the net force:\n\n$$\nF_{1}=\\sum_{i} F_{i}=\\sum_{i} 2 \\sqrt{A \\pi} I B_{i}\n$$\n\nTake out the factor $I$ from the summation and insert 1 written in the unusual way $(\\ell / N) \\cdot(N / \\ell)$ :\n\n$$\nF_{1}=I \\frac{N}{\\ell} \\sum_{i} 2 \\sqrt{A \\pi} \\frac{\\ell}{N} B_{i}\n$$\n\nThe sum on the right hand side is the flux escaping through the side of the solenoid, which equals $\\Phi_{i n}$, so the force:\n\n$$\nF_{1}=I \\frac{N}{\\ell} \\Phi_{\\text {in }}=\\frac{\\mu_{0} \\mathcal{E} A N I}{2 R r \\ell}\n$$', 'The force acting on a current loop of magnetic moment $\\vec{m}$ placed in magnetic field $\\vec{B}$ is given by $\\vec{\\nabla}(\\vec{m} \\cdot \\vec{B})$. Divide the solenoid into short circular coils of equal length $\\Delta \\ell$, then the magnetic moment of each short coil is\n\n$$\n\\Delta \\vec{m}=I A \\frac{N \\Delta \\ell}{\\ell} \\vec{e}_{z}\n$$\n\nwhere $\\vec{e}_{z}$ denotes the unit vector in $z$-direction. This magnetic moment is parallel to the field $\\vec{B}_{\\text {loop }}$ created by the large current-carrying loop, so the force acting on each short segment of the solenoid in $z$-direction can be written as:\n\n$$\n\\Delta F_{1}=\\Delta m \\frac{\\mathrm{d} B_{\\text {loop }}}{\\mathrm{d} z}=I A \\frac{N \\Delta \\ell}{\\ell} \\frac{\\mathrm{d} B_{\\text {loop }}}{\\mathrm{d} z}\n$$\n\nThe total force on the solenoid can be determined from integration of the force contributions along the solenoid:\n\n$$\nF_{1}=\\int_{-\\ell}^{0} \\mathrm{~d} \\ell \\frac{\\Delta F_{1}}{\\Delta \\ell}=I A \\frac{N}{\\ell}\\left(B_{\\text {loop }}(0)-B_{\\text {loop }}(-\\ell)\\right) .\n$$\n\nUsing the Biot-Savart-law we can compute the magnetic field $B_{\\text {loop }}(z)$ of the current-carrying loop along the $z$-axis to\n\n$$\nB_{\\text {loop }}(z)=\\frac{\\mu_{0} \\mathcal{E}}{2 R} \\frac{r^{2}}{\\left(z^{2}+r^{2}\\right)^{3 / 2}}\n$$\n\nThis expression for $B_{\\text {loop }}$ yields\n\n$$\nF_{1}=\\frac{\\mu_{0} \\mathcal{E} A N I r^{2}}{2 R \\ell}\\left(\\frac{1}{r^{3}}-\\frac{1}{\\left(\\ell^{2}+r^{2}\\right)^{3 / 2}}\\right) \\stackrel{\\ell}{\\approx} r \\frac{\\mu_{0} \\mathcal{E} A N I}{2 \\ell R r},\n$$\n\nand $F_{1}$ is directed to $+z$. From a similar calculation we get $\\vec{F}_{2}=-\\vec{F}_{1}$.\n\n\n\nNote: Using the idea of dividing the solenoid into small segments other solutions are possible as well (e.g. considering small dipole contributions). One recurring solution of this type is to divide the solenoid into small segments of length $\\mathrm{d} z$, compute the $B$-field in $z$-direction effected by current-loop from Biot-Savart, use Gauss law to relate the radial $B$ flux through the segment to $\\mathrm{d} B / \\mathrm{d} z$. compute the force on the segment using Lorentz forces and integrate the force over the length of the solenoid. In this case the marking scheme is adapted in the second aspect, awarding $0.5 \\mathrm{p}$ for the idea of using Gauss law to find radial $B$-field', ""In this solution we relate the force acting on the solenoid to the change in energy of the system. Investigate the case when point $O_{1}$ is located at $O$ first. Due to the same current directions, the magnetic force $\\vec{F}_{1}$ acting on the solenoid points in direction $+z$. While keeping the solenoid in equilibrium with external force $-\\vec{F}_{1}$, let it move by a small displacement $\\delta z$ in the positive $z$ direction. The work done by the external force is equal to the change in energy of the system:\n\n$$\n-\\vec{F}_{1} \\cdot \\delta \\vec{z}=-F_{1} \\delta z=\\delta E_{\\text {total }}\n$$\n\nWe should be aware of the fact that the system is not closed: there is also a battery and a current source included in the circuits. Hence, $\\delta E_{\\text {total }}$ contains the change in field energy and the change of energy of the power sources:\n\n$$\n\\delta E_{\\text {total }}=\\delta E_{\\text {field }}+\\delta E_{\\text {sources }}\n$$\n\nSince the force does not depend on what kind of power supplies we have, let us replace the battery with a current source providing constant current $J=\\mathcal{E} / R$.\n\nNow we find a relation between $\\delta E_{\\text {field }}$ and $\\delta E_{\\text {sources }}$ The energy stored in the field can be expressed as\n\n$$\nE_{\\text {field }}=\\frac{1}{2} L_{1} I^{2}+\\frac{1}{2} L_{2} J^{2}+L_{12} I J\n$$\n\nwhere $L_{1}$ is the inductance of the solenoid, $L_{2}$ is that of the loop and $L_{12}$ is the mutual inductance of the system. Upon small displacement $\\delta z$, only the last term changes, so\n\n$$\n\\delta E_{\\text {field }}=\\delta L_{12} \\cdot I J\n$$\n\nThe small displacement results in a change of the flux enclosed by the loop and the solenoid. The flux created by the solenoid on the loop is $L_{12} I$, and the flux created by the loop through the solenoid is $L_{21} J=L_{12} J$ (here we used the symmetry property of mutual inductance). During the short time $\\delta t$ of the displacement $\\delta z$, the e.m.f. induced in the loop ( $\\left.V_{\\text {ind }}^{\\text {loop }}\\right)$ and the solenoid $\\left(V_{\\text {ind }}^{\\text {solenoid }}\\right)$ can be expressed with Faraday's law:\n\n$$\nV_{\\text {ind }}^{\\text {loop }}=-\\frac{\\delta L_{12}}{\\delta t} I, \\quad V_{\\text {ind }}^{\\text {solenoid }}=-\\frac{\\delta L_{12}}{\\delta t} J .\n$$\n\nIn order to keep the current in the circuits constant, the current sources need to provide an additional power, so they give away extra energy (in addition to Joule heat). This energy change of the sources is given by:\n\n$$\n\\delta E_{\\text {sources }}=\\left(V_{\\text {ind }}^{\\text {loop }} J+V_{\\text {ind }}^{\\text {solenoid }} I\\right) \\delta t\n$$\n\nUsing the previous results we finally get:\n\n$$\n\\delta E_{\\text {sources }}=-2 \\delta L_{12} I J=-2 \\delta E_{\\text {field }}\n$$\n\nwhich means $\\delta E_{\\text {total }}=-\\delta E_{\\text {field }}$, and hence $F_{1} \\delta z=$ $\\delta E_{\\text {field }}=\\delta L_{12} I J$.\n\nNote 1. Naively, one might think that we get the result $\\delta E_{\\text {total }}=\\delta E_{\\text {field }}$ if we imagine superconducting wires without power supplies. One can show with detailed calculation that in that case the currents in the loop and the coil change, as the total flux enclosed by a superconducting circuit must remain constant. The correct physical justification of the appearing negative sign is an important part of the solution.\n\nNow we calculate the change in mutual inductance $\\delta L_{12}$. The small displacement $\\delta z$ can be imagined as we take a short segment from the tail $\\mathrm{O}_{2}$ of the coil (consisting of $N \\delta z / \\ell$ turns) and move it to the head $O_{1}$. As a result, the flux produced by the loop on the solenoid increases by\n\n$$\n\\delta \\Phi_{12}=\\delta L_{21} J=\\underbrace{\\frac{\\mu_{0} J}{2 r}}_{\\substack{\\text { (at center) } \\\\ \\text { loop }}} A \\frac{N}{\\ell} \\delta z\n$$\n\nFrom this we get:\n\n$$\nF_{1}=\\frac{\\delta L_{12} I J}{\\delta z}=\\frac{\\mu_{0} \\mathcal{E} N A I}{2 \\ell R r}\n$$\n\nIf the tail $\\mathrm{O}_{2}$ of the solenoid is located at point $O$, the coefficient of mutual inductance decreases upon small displacement, which results in $\\vec{F}_{2}=-\\vec{F}_{1}$.\n\nNote 2. The field energy can be also calculated from the energy density integrated for the whole space. Instead of calculating the total field energy, it is easier to find its change using the same idea presented above, i.e. take a segment of length $\\delta z$ from the tail and move it to the head of the solenoid. Assuming $\\ell \\gg \\delta z \\gg \\sqrt{A}$, the field created by the solenoid inside that segment is $\\mu_{0} N I / \\ell$ (because the field differs from this only at distance $\\sim \\sqrt{A}$ from the ends). At the end we get the same result for the change in field energy using the expression:\n\n$$\n\\delta E_{\\text {field }}=\\frac{1}{2 \\mu_{0}}\\left[\\left(B_{\\text {loop }}^{\\text {(at centre) }}+B_{\\text {sol }}\\right)^{2}-\\left(B_{\\text {loop }}^{\\text {(at centre) })}\\right)^{2}-B_{\\text {sol }}^{2}\\right] A \\delta z .\n$$\n\nNote 3. A third possibility is to calculate the potential energy change of the displaced few turns of the solenoid. The magnetic moment of a segment of length $\\delta z$ is $\\vec{m}=\\vec{e}_{z} I A N \\delta z / \\ell$, and its energy in external field is $E_{\\mathrm{pot}}=-\\vec{m} \\vec{B}$. Important to highlight that this potential energy already contains the factor of -1 discussed at the beginning of the solution, so the force acting on the solenoid can be expressed as\n\n$$\nF_{1}=-\\frac{\\delta E_{\\mathrm{pot}}}{\\delta z}\n$$\n\nThe external field is the superposition of the field $\\vec{B}_{\\text {loop }}$ created by the loop and the field $\\vec{B}_{\\text {sol }}$ created by the coil (note that this latter contains a factor of $1 / 2$ compared to the field in the middle of the solenoid). Since $\\vec{B}_{\\text {sol }}$ is the same at the two ends $O_{1}$ and $O_{2}$, the energy change is:\n\n$$\n\\delta E_{\\mathrm{pot}}=-\\vec{m} \\vec{B}_{\\text {loop }}^{\\text {(at center) }}-\\vec{m} \\vec{B}_{\\text {loop }}(z=\\ell) .\n$$\n\nThe second term can be neglected, and we get\n\n$$\n\\delta E_{\\mathrm{pot}}=-\\frac{I A N \\delta z}{\\ell} \\frac{\\mu_{0} \\mathcal{E}}{2 R r}\n$$\n\n\n\nwhich gives the same answer for $F_{1}$ as the other ideas.""]","['$F_{1}=\\frac{\\mu_{0} N I A \\mathcal{E}}{2 \\ell R r}$, $F_2=-F_1$']",True,,"Expression,Equation", 1500,Mechanics,,"## T2: Mechanical accelerator A massless thread makes $N$ turns around statically fixed cylinder, as shown in the figure. Initially, the free (unwound) ends of the thread are parallel to the axis $X$. Then, a heavy point-like object $P$ is attached to one end of the thread while the other end is pulled with a constant velocity $u$ along $X$. Find the maximum velocity attained by the heavy object. ![](https://cdn.mathpix.com/cropped/2023_12_21_ac051a2ec661e03601f7g-1.jpg?height=174&width=737&top_left_y=2209&top_left_x=197) The thread is inextendable and flexible. Suppose that the turns of the thread are wound tightly to one another and are placed practically in the same plane, perpendicular to the cylinder axis. Neglect any friction in the system. Do not consider the force of gravity.","['![](https://cdn.mathpix.com/cropped/2023_12_21_9cc90a1201132700af30g-1.jpg?height=366&width=371&top_left_y=1733&top_left_x=1065)\n\nRotating system of reference\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9cc90a1201132700af30g-1.jpg?height=166&width=468&top_left_y=1779&top_left_x=1502)\n\nFigure 2: Mechanical accelerator\n\nPart 1: Thread in contact with the cylinder. The velocity of the mass $P$ can be decomposed into longitudinal component $v_{l}$ along the thread, and a transverse component $v_{\\perp}$ perpendicular to the thread:\n\n$$\n\\vec{v}=v_{l} \\vec{e}_{1}+v_{\\perp} \\vec{e}_{2}\n$$\n\nwhere the unit vectors $\\vec{e}_{1}$ and $\\vec{e}_{2}$ are parallel and perpendicular to the thread, respectively (see Fig. 2). Since the thread is inextensible, the longitudinal component is constant: $v_{l}=-u$, i.e.\n\n$$\n\\vec{v}=-u \\vec{e}_{1}+v_{\\perp} \\vec{e}_{2}\n$$\n\nThe acceleration of $P$ is, respectively:\n\n$$\n\\vec{a}=\\frac{d \\vec{v}}{d t}=-u \\frac{d \\vec{e}_{1}}{d t}+v_{\\perp} \\frac{d \\vec{e}_{2}}{d t}+\\frac{d v_{\\perp}}{d t} \\vec{e}_{2}\n$$\n\n\n\nVectors $\\vec{e}_{1}$ and $\\vec{e}_{2}$ form a coordinate system, which rotates as a rigid object with an angular velocity:\n\n$$\n\\vec{\\omega}=\\frac{d \\phi}{d t} \\vec{e}_{3}\n$$\n\nwhere $\\vec{e}_{3}=\\vec{e}_{1} \\times \\vec{e}_{2}$ is a unit vector perpendicular to the plane of motion, i.e. along the cylinder axis, and $\\phi$ is the angle between the thread and the X-axis. Therefore, the time derivatives of the basis vectors are:\n\n$$\n\\frac{d \\vec{e}_{1}}{d t}=\\vec{\\omega} \\times \\vec{e}_{1}=\\frac{d \\phi}{d t} \\vec{e}_{2}\n$$\n\nand\n\n$$\n\\frac{d \\vec{e}_{2}}{d t}=\\vec{\\omega} \\times \\vec{e}_{2}=-\\frac{d \\phi}{d t} \\vec{e}_{1}\n$$\n\nIn this way, the acceleration of $P$ can be represented in terms of the angular velocity:\n\n$$\n\\vec{a}=-v_{\\perp} \\frac{d \\phi}{d t} \\vec{e}_{1}+\\left(-u \\frac{d \\phi}{d t}+\\frac{d v_{\\perp}}{d t}\\right) \\vec{e}_{2}\n$$\n\nThe only force, acting on $P$, is the tension of the thread. Therefore, the component of the acceleration perpendicular to the tread, i.e. along $\\vec{e}_{2}$, is null:\n\n$$\n-u \\frac{d \\phi}{d t}+\\frac{d v_{\\perp}}{d t}=0\n$$\n\nAfter integration over time, we obtain a relationship between the transverse velocity, acquired by $P$, and the angle of rotation of the thread:\n\n$$\nv_{\\perp}=u \\phi\n$$\n\nThe end of the tread turns at a total angle of $2 \\pi N$ until the tread detaches from the cylinder completely. Therefore, the transverse component of the velocity of $P$ at the moment of detachment is:\n\n$$\nv_{\\perp}=2 \\pi N u\n$$\n\nand the magnitude of velocity:\n\n$$\nv=\\sqrt{v_{l}^{2}+v_{\\perp}^{2}}=u \\sqrt{(2 \\pi N)^{2}+1}\n$$\n\nPart 2: Thread detached fom the cylinder. This expression, however, still does not represent the maximum velocity attained by $P$. In the frame of reference of the free end of the thread, the mass continues to rotate about the end of the thread. The velocity of $P$, relative to Earth, reaches maximum in the moment when the thread reaches right angle with $\\mathrm{X}$-axis, i.e. the transverse component of velocity of $P$ aligns with $\\vec{u}$ :\n\n$$\nv_{\\max }=u(2 \\pi N+1)\n$$', ""![](https://cdn.mathpix.com/cropped/2023_12_21_9cc90a1201132700af30g-1.jpg?height=366&width=371&top_left_y=1733&top_left_x=1065)\n\nRotating system of reference\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9cc90a1201132700af30g-1.jpg?height=166&width=468&top_left_y=1779&top_left_x=1502)\n\nFigure 2: Mechanical accelerator\n\nPart 1: Thread in contact with the cylinder. Consider a point $Q$ on the end of the thread being pulled that coincided with $P$ at the moment when it touched the cylinder. Consider motion of the thread in a system of reference (SR), which rotates at angular speed $\\omega=u / R$ around the center of the cylinder. In that SR the part of the thread in contact with the cylinder is at rest and the point $Q$ rotates around the cylinder with the angular velocity $\\omega_{Q}=-\\omega$ (see the figure).\nSince the middle part of the thread is at rest, energy of the mass $P$ is conserved. For the same reason, the velocity $v_{P}$ of $P$ is perpendicular to the thread. Therefore the kinetic energy acquired by the mass in the rotating frame is equal to the decrease of its centrifugal potential energy:\n\n$$\n\\frac{1}{2} m v_{P}^{2}=-\\frac{1}{2} m \\omega^{2} R^{2}+\\frac{1}{2} m \\omega^{2} r^{2}=\\frac{1}{2} m \\omega^{2} l_{P}^{2}\n$$\n\nwhere $l_{P}$ is the length of the unwound part of the thread on the side of the mass $P$ (see the figure). Therefore, the mass $P$ rotates around the fixture point of the thread with a velocity:\n\n$$\nv_{P}=\\frac{u}{R} l_{P}\n$$\n\nand a constant angular velocity:\n\n$$\n\\omega_{P}=\\frac{u}{R}=\\omega\n$$\n\nSince $\\omega_{P}=-\\omega_{Q}$, in the rotating SR the two ends of the thread will unwind symmetrically and the lengths of the two straight parts of the string will be equal at any moment of time. Therefore, at the moment of detachment:\n\n$$\nl_{P}=\\frac{1}{2}(2 \\pi R N)=\\pi N R\n$$\n\nand the detachment velocity of $P$ is, respectively:\n\n$$\nv_{P}=\\pi N u\n$$\n\nWhen transforming the velocity of $P$ to the Earth's SR, the velocity $\\vec{v}_{P}$ should be added to the rotational velocity $\\vec{\\omega} \\times \\vec{r}$. It is easy to establish that the result for the transverse component of $P$ is:\n\n$$\nv_{\\perp}=2 v_{P}=2 \\pi N u\n$$\n\nPart 2: Thread detached fom the cylinder. This expression, however, still does not represent the maximum velocity attained by $P$. In the frame of reference of the free end of the thread, the mass continues to rotate about the end of the thread. The velocity of $P$, relative to Earth, reaches maximum in the moment when the thread reaches right angle with $\\mathrm{X}$-axis, i.e. the transverse component of velocity of $P$ aligns with $\\vec{u}$ :\n\n$$\nv_{\\max }=u(2 \\pi N+1)\n$$"", ""![](https://cdn.mathpix.com/cropped/2023_12_21_9cc90a1201132700af30g-1.jpg?height=366&width=371&top_left_y=1733&top_left_x=1065)\n\nRotating system of reference\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9cc90a1201132700af30g-1.jpg?height=166&width=468&top_left_y=1779&top_left_x=1502)\n\nFigure 2: Mechanical accelerator\n\nPart 1: The thread in contact with the cylinder\n\nLike in Solution I we decompose the velocity into longitudinal and transverse components, and come to the conclusion that the longitudinal component is $v_{l}=$ $-u$. Afterwards, the acceleration of $P$ is expressed. In this case, however, we consider the longitudinal (centripetal) component of the acceleration:\n\n$$\na_{l}=-v_{\\perp}^{2} / l \\equiv-v_{\\perp} \\frac{d \\phi}{d t}\n$$\n\nFrom the second Newton's law we obtain the tension $F$ of the thread:\n\n$$\nF=-m v_{\\perp} \\frac{d \\phi}{d t}\n$$\n\nThe rate of change of the kinetic energy of the mass is equal to the power of the tension force:\n\n$$\n\\frac{d E_{k}}{d t}=F v_{l}=+m v_{\\perp} \\frac{d \\phi}{d t} u\n$$\n\nTaking into account that:\n\n$$\nE_{k}=\\frac{1}{2} m\\left(u^{2}+v_{\\perp}^{2}\\right)\n$$\n\n\nand taking the first derivative from that expression, we obtain: $m v_{\\perp} d v_{\\perp} / d t=m v_{\\perp} d \\phi / d t u$, or:\n\n$$\n\\frac{d v_{\\perp}}{d t}=u \\frac{d \\phi}{d t}\n$$\n\nPart 2: Thread detached fom the cylinder. This expression, however, still does not represent the maximum velocity attained by $P$. In the frame of reference of the free end of the thread, the mass continues to rotate about the end of the thread. The velocity of $P$, relative to Earth, reaches maximum in the moment when the thread reaches right angle with $\\mathrm{X}$-axis, i.e. the transverse component of velocity of $P$ aligns with $\\vec{u}$ :\n\n$$\nv_{\\max }=u(2 \\pi N+1)\n$$""]",['$v_{\\max }=u(2 \\pi N+1)$'],False,,Expression, 1500,Mechanics,,"## T2: Mechanical accelerator A massless thread makes $N$ turns around statically fixed cylinder, as shown in the figure. Initially, the free (unwound) ends of the thread are parallel to the axis $X$. Then, a heavy point-like object $P$ is attached to one end of the thread while the other end is pulled with a constant velocity $u$ along $X$. Find the maximum velocity attained by the heavy object. The thread is inextendable and flexible. Suppose that the turns of the thread are wound tightly to one another and are placed practically in the same plane, perpendicular to the cylinder axis. Neglect any friction in the system. Do not consider the force of gravity.","['\n\nRotating system of reference\n\n\n\nFigure 2: Mechanical accelerator\n\nPart 1: Thread in contact with the cylinder. The velocity of the mass $P$ can be decomposed into longitudinal component $v_{l}$ along the thread, and a transverse component $v_{\\perp}$ perpendicular to the thread:\n\n$$\n\\vec{v}=v_{l} \\vec{e}_{1}+v_{\\perp} \\vec{e}_{2}\n$$\n\nwhere the unit vectors $\\vec{e}_{1}$ and $\\vec{e}_{2}$ are parallel and perpendicular to the thread, respectively (see Fig. 2). Since the thread is inextensible, the longitudinal component is constant: $v_{l}=-u$, i.e.\n\n$$\n\\vec{v}=-u \\vec{e}_{1}+v_{\\perp} \\vec{e}_{2}\n$$\n\nThe acceleration of $P$ is, respectively:\n\n$$\n\\vec{a}=\\frac{d \\vec{v}}{d t}=-u \\frac{d \\vec{e}_{1}}{d t}+v_{\\perp} \\frac{d \\vec{e}_{2}}{d t}+\\frac{d v_{\\perp}}{d t} \\vec{e}_{2}\n$$\n\n\n\nVectors $\\vec{e}_{1}$ and $\\vec{e}_{2}$ form a coordinate system, which rotates as a rigid object with an angular velocity:\n\n$$\n\\vec{\\omega}=\\frac{d \\phi}{d t} \\vec{e}_{3}\n$$\n\nwhere $\\vec{e}_{3}=\\vec{e}_{1} \\times \\vec{e}_{2}$ is a unit vector perpendicular to the plane of motion, i.e. along the cylinder axis, and $\\phi$ is the angle between the thread and the X-axis. Therefore, the time derivatives of the basis vectors are:\n\n$$\n\\frac{d \\vec{e}_{1}}{d t}=\\vec{\\omega} \\times \\vec{e}_{1}=\\frac{d \\phi}{d t} \\vec{e}_{2}\n$$\n\nand\n\n$$\n\\frac{d \\vec{e}_{2}}{d t}=\\vec{\\omega} \\times \\vec{e}_{2}=-\\frac{d \\phi}{d t} \\vec{e}_{1}\n$$\n\nIn this way, the acceleration of $P$ can be represented in terms of the angular velocity:\n\n$$\n\\vec{a}=-v_{\\perp} \\frac{d \\phi}{d t} \\vec{e}_{1}+\\left(-u \\frac{d \\phi}{d t}+\\frac{d v_{\\perp}}{d t}\\right) \\vec{e}_{2}\n$$\n\nThe only force, acting on $P$, is the tension of the thread. Therefore, the component of the acceleration perpendicular to the tread, i.e. along $\\vec{e}_{2}$, is null:\n\n$$\n-u \\frac{d \\phi}{d t}+\\frac{d v_{\\perp}}{d t}=0\n$$\n\nAfter integration over time, we obtain a relationship between the transverse velocity, acquired by $P$, and the angle of rotation of the thread:\n\n$$\nv_{\\perp}=u \\phi\n$$\n\nThe end of the tread turns at a total angle of $2 \\pi N$ until the tread detaches from the cylinder completely. Therefore, the transverse component of the velocity of $P$ at the moment of detachment is:\n\n$$\nv_{\\perp}=2 \\pi N u\n$$\n\nand the magnitude of velocity:\n\n$$\nv=\\sqrt{v_{l}^{2}+v_{\\perp}^{2}}=u \\sqrt{(2 \\pi N)^{2}+1}\n$$\n\nPart 2: Thread detached fom the cylinder. This expression, however, still does not represent the maximum velocity attained by $P$. In the frame of reference of the free end of the thread, the mass continues to rotate about the end of the thread. The velocity of $P$, relative to Earth, reaches maximum in the moment when the thread reaches right angle with $\\mathrm{X}$-axis, i.e. the transverse component of velocity of $P$ aligns with $\\vec{u}$ :\n\n$$\nv_{\\max }=u(2 \\pi N+1)\n$$', ""\n\nRotating system of reference\n\n\n\nFigure 2: Mechanical accelerator\n\nPart 1: Thread in contact with the cylinder. Consider a point $Q$ on the end of the thread being pulled that coincided with $P$ at the moment when it touched the cylinder. Consider motion of the thread in a system of reference (SR), which rotates at angular speed $\\omega=u / R$ around the center of the cylinder. In that SR the part of the thread in contact with the cylinder is at rest and the point $Q$ rotates around the cylinder with the angular velocity $\\omega_{Q}=-\\omega$ (see the figure).\nSince the middle part of the thread is at rest, energy of the mass $P$ is conserved. For the same reason, the velocity $v_{P}$ of $P$ is perpendicular to the thread. Therefore the kinetic energy acquired by the mass in the rotating frame is equal to the decrease of its centrifugal potential energy:\n\n$$\n\\frac{1}{2} m v_{P}^{2}=-\\frac{1}{2} m \\omega^{2} R^{2}+\\frac{1}{2} m \\omega^{2} r^{2}=\\frac{1}{2} m \\omega^{2} l_{P}^{2}\n$$\n\nwhere $l_{P}$ is the length of the unwound part of the thread on the side of the mass $P$ (see the figure). Therefore, the mass $P$ rotates around the fixture point of the thread with a velocity:\n\n$$\nv_{P}=\\frac{u}{R} l_{P}\n$$\n\nand a constant angular velocity:\n\n$$\n\\omega_{P}=\\frac{u}{R}=\\omega\n$$\n\nSince $\\omega_{P}=-\\omega_{Q}$, in the rotating SR the two ends of the thread will unwind symmetrically and the lengths of the two straight parts of the string will be equal at any moment of time. Therefore, at the moment of detachment:\n\n$$\nl_{P}=\\frac{1}{2}(2 \\pi R N)=\\pi N R\n$$\n\nand the detachment velocity of $P$ is, respectively:\n\n$$\nv_{P}=\\pi N u\n$$\n\nWhen transforming the velocity of $P$ to the Earth's SR, the velocity $\\vec{v}_{P}$ should be added to the rotational velocity $\\vec{\\omega} \\times \\vec{r}$. It is easy to establish that the result for the transverse component of $P$ is:\n\n$$\nv_{\\perp}=2 v_{P}=2 \\pi N u\n$$\n\nPart 2: Thread detached fom the cylinder. This expression, however, still does not represent the maximum velocity attained by $P$. In the frame of reference of the free end of the thread, the mass continues to rotate about the end of the thread. The velocity of $P$, relative to Earth, reaches maximum in the moment when the thread reaches right angle with $\\mathrm{X}$-axis, i.e. the transverse component of velocity of $P$ aligns with $\\vec{u}$ :\n\n$$\nv_{\\max }=u(2 \\pi N+1)\n$$"", ""\n\nRotating system of reference\n\n\n\nFigure 2: Mechanical accelerator\n\nPart 1: The thread in contact with the cylinder\n\nLike in Solution I we decompose the velocity into longitudinal and transverse components, and come to the conclusion that the longitudinal component is $v_{l}=$ $-u$. Afterwards, the acceleration of $P$ is expressed. In this case, however, we consider the longitudinal (centripetal) component of the acceleration:\n\n$$\na_{l}=-v_{\\perp}^{2} / l \\equiv-v_{\\perp} \\frac{d \\phi}{d t}\n$$\n\nFrom the second Newton's law we obtain the tension $F$ of the thread:\n\n$$\nF=-m v_{\\perp} \\frac{d \\phi}{d t}\n$$\n\nThe rate of change of the kinetic energy of the mass is equal to the power of the tension force:\n\n$$\n\\frac{d E_{k}}{d t}=F v_{l}=+m v_{\\perp} \\frac{d \\phi}{d t} u\n$$\n\nTaking into account that:\n\n$$\nE_{k}=\\frac{1}{2} m\\left(u^{2}+v_{\\perp}^{2}\\right)\n$$\n\n\nand taking the first derivative from that expression, we obtain: $m v_{\\perp} d v_{\\perp} / d t=m v_{\\perp} d \\phi / d t u$, or:\n\n$$\n\\frac{d v_{\\perp}}{d t}=u \\frac{d \\phi}{d t}\n$$\n\nPart 2: Thread detached fom the cylinder. This expression, however, still does not represent the maximum velocity attained by $P$. In the frame of reference of the free end of the thread, the mass continues to rotate about the end of the thread. The velocity of $P$, relative to Earth, reaches maximum in the moment when the thread reaches right angle with $\\mathrm{X}$-axis, i.e. the transverse component of velocity of $P$ aligns with $\\vec{u}$ :\n\n$$\nv_{\\max }=u(2 \\pi N+1)\n$$""]",['$v_{\\max }=u(2 \\pi N+1)$'],False,,Expression, 1501,Optics,,"## T3: Cat eyes You may have noticed that in darkness, when a cat is within the light beam of a headlamp, its eyes appear very bright, see the photo below (left). This phenomenon can be modelled by a lens setup, see the photo on right, and the diagram beneath the photos. The photo on right was taken by a digital single-lens reflex camera. The light intensity at the camera sensor pixels marked by a red line (in the photo) is shown in the graph below: the log base 10 of the light intensity (measured as the number of photons caught by each pixel) is plotted against the $x$-coordinate, with the pixels' side length serving as the unit length. The lens modelling cat eyes can be treated as an ideal thin lens of focal length $f=55 \mathrm{~mm}$ and diameter $D=$ $39 \mathrm{~mm}$; however, you should keep in mind that the given graph shows real measurement data, and the lens has certain non-ideal features. Most importantly, partial reflections of brightly lit areas from the lens surfaces may decrease the contrast: dark areas seen through the lens appear less dark than they actually are; this effect can be neglected for the camera lens, but not so for the lens serving as a model of a cat's eye. Based on the given data, estimate (with the accuracy of ca 20\%) the distance $h$ between the axis of the camera and the axis of the lamp (which can be considered as a point source) if the distance of the camera from the paper sheet was $L=4.8 \mathrm{~m}$.","['When you look at the photo of the lens and/or the graph provided, four regions with different brightness levels can be distinguished. The brightest region represents the magnified image of the blur spot created by the lamp through the lens. The blur spot is created because the distance from the lens to the white sheet beneath it is slightly larger than the focal distance; as we can see from the graph, the blur spot is of almost constant brightness (a flat plateau at $\\log _{10} I=4.4$ ), so we can say that the entire luminous flux falling from the lamp onto the lens is distributed evenly over the blur spot. Note that the blur spot has no sharp edges, though, as you would expect from in such case. This is because the image of this bright disc is situated between the lens and the camera, and is at a fairly big distance away from the plane which is sharp at the image sensor (as seen from the photo, the camera is focused onto the lens). Because of that, the enlarged image of the blur spot has blurred edges in the photo (at the blurred edges, $\\log _{10} I$ varies from 3.4 to 4.4). The second-brightest region (with $\\log _{10} I=3.4$ ) represents the scattered light from the brightest region: in that region, we are still looking through the lens, and see the area next to the bright blur spot on the sheet. Ideally, its should be darker than the sheet seen in those places where it is not obstructed by the lens, because the lens is shading the light from the lamp. However, the glass elements of this big lens are non-ideal (and there are many glass elements inside the lens!), so the light from the lamp and the bright blur spot is scattered towards the camera giving rise to an increased apparent brightness. In the area where we see the blur spot, this light is insignificant (much weaker than the light from the blur spot), but not so in this dark area: here, the scatteredfrom-the-glass-surfaces dominates heavily over the light coming from the paper sheet. As a matter of fact, this fact could be used to improve the accuracy of the calculations: we could subtract the contribution of the scattered light $\\left(10^{3.4}\\right)$ from the total intensity of the light at the brightest spot $\\left(10^{4.4}\\right)$ to obtain the contribution coming from the blur spot on the sheet. The darkest regions (with $\\log _{10} I<1.75$ ) represent the interior black painting of the lens seen through the big front glass element of the lens, which absorbs most of the incident light, and the region with $x>420$ and $\\log _{10} I=1.95$ represents the white sheet illuminated by the lamp. The ratio between the measured light intensity of the brightest region and that of the region with $x>420$ can be utilized to find the distance of the sheet (the blur spot) from the lens to the paper sheet $d_{0}$, see below.\n\nFrom the data given in the problem text we know that $L \\gg f$; from the photo of the lens, it is also clear that $d_{0}$ is of the same order of magnitude as $f$. Because of that, the illuminance $E$ (luminous flux per unit area) near the lens can be assumed to be the same as at the paper sheet, The luminous flux per solid angle and unit area of a lightscattering (or radiating) surface is called the luminance $\\mathscr{L}$; since all these directions under which the scattered light enters the lens aperture are close to the surface nor$\\mathrm{mal}$, we may assume the luminance of the paper sheet to be constant over all these directions. With the small\n\n\n\nangle approximation, the light intensity $I$ (illuminance, luminous flux $\\Phi$ per unit area) at the camera sensor is proportional to $\\mathscr{L}$ (see Explanation 1).\n\nThe luminance of the blur spot on the sheet $\\mathscr{L}_{B S}$ is $1 / k$ larger than the luminance $\\mathscr{L}_{s}$ of the paper sheet, where $k$ equals the ratio between the area of the bright dot (the blur spot) on the paper sheet and the area of the lens, because all the light received by the lens is ""compressed"" into the tiny blur spot.\n\nSmall angle approximation is also used to show that luminance of the image of the blur spot $\\mathscr{L}_{I}$ equals to the luminance of the blur spot $\\mathscr{L}_{B S}$ (see Explanation 2). Therefore, the light intensity at the sensor cells corresponding to the brightest area (where we see the image of the blur spot) $I_{I}=I_{s} / k$, where $I_{s}$ stands for the intensity at the cells corresponding to unobscured paper sheet. So, from the graph, we can deduce the value of $k$, and knowing $k$ we can calculate $d_{0}$. Let the distance along the axis between the image of the bright region through the lens and the lens itself be denoted as $d_{S}$; according to the Newton\'s lens formula, $\\left(d_{S}-f\\right)\\left(d_{0}-f\\right)=$ $f^{2}$. Hence,\n\n$$\nd_{S}=f+\\frac{f^{2}}{d_{0}-f}=\\frac{d_{0} f}{\\left(d_{0}-f\\right)}\n$$\n\ncan be also determined.\n\nHypothesize that $d_{0}-f \\ll f$. Let us calculate the diameter of the image of the blur spot\n\n$$\nD_{I}=\\frac{D_{B S} d_{S}}{d_{0}}=\\frac{D_{B S} f}{d_{0}-f}\n$$\n\nwhere the diameter of the blur spot on the sheet\n\n$$\nD_{B S}=\\frac{D\\left(d_{0}-f-s\\right)}{f+s} \\approx \\frac{D\\left(d_{0}-f-s\\right)}{f}\n$$\n\nand $s$ denotes the distance of the image of the point source from the focal plane. Using Newton\'s lens formula, $s=f^{2} /\\left(L-f-d_{0}\\right) \\approx f^{2} / L$, This leads us to\n\n$$\nD_{B S} \\approx D\\left(\\frac{d_{0}-f}{f}-\\frac{f}{L}\\right)\n$$\n\nand therefore\n\n$$\nD_{I} \\approx D\\left[1-\\frac{f^{2}}{L\\left(d_{0}-f\\right)}\\right]\n$$\n\nKeeping in mind that $d_{0}-f=\\frac{d_{0} f}{d_{S}} \\approx \\frac{f^{2}}{d_{S}}$, we obtain\n\n$$\nD_{I} \\approx D\\left(1-\\frac{d_{S}}{L}\\right)=\\frac{D\\left(L-d_{S}\\right)}{L} .\n$$\n\nThis means that as seen from the position of the camera, the angular size of the image of the blur spot $\\theta_{B S}=$ $D_{I} /\\left(L-d_{S}\\right)$ equals to the angular size of the lens aperture $\\theta_{L}=D / L$. This fact is easily confirmed from the photo and is an important observation for two reasons. First, it means that based on the angular diameter of the image of the blur spot on the photo, it is impossible to figure out the distance $d_{0}$ (and hence, $d_{S}$ ). Second, it allows us to measure instead of the angular distance $\\theta$ between the centre of the lens and the centre of the image of the blur spot (as seen from the position of the cameraline), the respective distance between the edges of the respective circles. Equality of these two angular sizes is also easily seen from the geometric construction, see the figure. Ineed, consider blue lines $S A G$ and $S B F$ which arrive from the lamp $S$ to the edges of the blur spot. Image of point $F$, denoted by $J$, is now easily found as the intersection point of the ray $S B F$ with the ray $F O$ (passing through the centre of the lens); image $H$ of the other edge of the blur spot is found in the same way. From this construction, it becomes clear that the angular size of the image of the blur spot and the lens, as seen from the camera, are exactly equal, without any approximation. Due to the smallness of the distance $h$, these angular sizes remain almost constant when the observation point is moved from $S$ to $C$.\n\nGiven the images are approximately circular, the area ratio $k$ equals $\\left(D_{B S} / D\\right)^{2}$, or\n\n$$\n\\pm \\sqrt{k}=\\frac{1}{f}\\left(d_{0}-\\frac{L f}{L-f}\\right)=\\frac{d_{0}}{f}-\\frac{L}{L-f}\n$$\n\nIn the above equation, the \\pm sign represents the two cases where the paper sheet is behind or in front of the image of the lamp. From the graph, the ratio between the intensity of the brightest region and the dark region with $x>420$ is $10^{4.4-1.95} \\approx 282$, which equals $1 / k$. Then, $d_{0} / f$ can be found to be $\\pm \\sqrt{k}+1+f / L$, which gives two solutions $d_{0} / f \\approx 1.07$ and $d_{0} / f \\approx 0.95$. According to the experimental settings given in the problem text, $d_{0}$ is greater than $f$, and thus we obtain $d_{0} / f \\approx 1.07$ and $d_{S} \\approx 15.03 f \\approx 83 \\mathrm{~cm}$. This also verifies the hypothesis that $d_{0}-f \\ll f$.\n\nThe centre of the image of the blur spot is positioned at the height $h^{\\prime}=h \\frac{d_{S}}{L}$ above the direction to the centre of the lens (this expression from similarity of the triangles $O Q P$ and $O C S$ ) which means that $\\theta=h^{\\prime} /\\left(L-d_{S}\\right)$; meanwhile, the angular diameter of the lens $\\theta_{L}=D / L$. Therefore,\n\n$$\n\\frac{\\theta}{\\theta_{L}}=\\frac{h d_{S}}{D\\left(L-d_{S}\\right)}\n$$\n\nThe ratio of the angular distances is easily measured from the figure as the ratio of the width $d_{c r}$ of the crescent-shaped second-brightest region to the diameter of the lense\'s aperture $D^{\\prime}$ :\n\n$$\nh=\\frac{d_{c r}}{D^{\\prime}} \\frac{D\\left(L-d_{S}\\right)}{d_{S}}\n$$\n\nBased on the graph, $d_{c r} \\approx 90$ pixels (midpoint of the blurry edge is around $x \\approx 120 \\mathrm{px}$, and the left edge of the aperture (in the graph) is at $x \\approx 30 \\mathrm{px}$; the right edge of the lens aperture is at $x \\approx 240 \\mathrm{px}$ corresponding to $D^{\\prime}=210 \\mathrm{px}$ and yielding $h \\approx 80 \\mathrm{~mm}$.\n\nRemark 1. In order to obtain the final answer with a reasonably good accuracy, it is not strictly speaking necessary to show that the apparent angular diameters of the lens and of the image of the blur spot are equal. All the other calculations remain the same, just one needs to match a circle with the circular segment of the visible edge of the blur spot, and measure directly $h^{\\prime}$, the distance between the centre of the lens and the centre of the blur spot, together with the diameter of the lens aperture $D^{\\prime}$ (see the small figure).\n\nRemark 2. The width of the crescent-shaped secondbrightest area can be also measured from the photo of\n\n\n\nthe lens with the required accuracy; however, measuring in pixels from the graph is more accurate.\n\nExplanation 1: Consider a small light source of luminance $\\mathcal{L}$ and surface area $S$ at a large distance $\\uparrow$ from the camera. The illuminance (the luminous flux per unit area) at the position of the camera is proportional to $\\uparrow^{-2}$ and so is the total luminous flux received by the whole sensor. Meanwhile, all this light energy is focused onto a small area $S^{\\prime}$ on the sensor - onto the image of the light source, and this area is also proportional to $\\uparrow^{-2}$. Therefore, the illuminance $I$ at the position of those sensor pixels which are covered by the image is independent of the distance $\\uparrow$.\n\nExplanation 2: Consider a very narrow cone of light of solid angle $\\omega$, starting from a very small area $S$ at the blur spot in a direction close to the surface normal, and carrying a total luminous flux $\\Phi$. Since the cone is narrow, this light beam is entirely caught by the lens at distance $d_{0}$, and focused onto the image of surface area $S^{\\prime}=S\\left(d_{S} / d_{0}\\right)^{2}$ at distance $d_{S}$ from the lens. The light rays of this beam traverse the focus and form another light cone of solid angle $\\omega^{\\prime}$ departing from the image. It is easy to see from similar triangles that $\\omega / \\omega^{\\prime}=\\left(d_{S} / d_{0}\\right)^{2}$. Then, the luminance of the image $\\mathcal{L}_{I}=\\Phi /\\left(S^{\\prime} \\omega^{\\prime}\\right)=\\Phi /(S \\omega)$, i.e. equal to the luminance of the blur spot.\n\n\n\n\n\nRemark 3: After having derived Explanation 1 and Explanation 2 and calculated $d_{0}$, the geometrical optics aspect can also be tackled by considering the image of the camera through the lens. The region bounded by the darkest ring in the image represents the area on the paper sheet that is observable by the camera through the lens. This area can be approximated by a circular spot with diameter $d$ (represented by $D^{\\prime}$ in the graph) depending negligibly on the size of the lens of the camera. Given small angles and that the lamp and the camera have equal distances to the lens, the bright spot on the paper sheet also has a diameter of $d$, which means $d=D \\sqrt{k}$. The angular distance between the lamp and the camera as seen from the center of the lens is $h /\\left(L-d_{0}\\right)$, and therefore, the distances between the two spots\' centers, and also thus their boundaries (represented by $d_{c r}$ in the graph), on the paper sheet are $d_{0} h /\\left(L-d_{0}\\right)$. We then obtain an equivalent equation to that above:\n\n$$\n\\frac{d_{0} h /\\left(L-d_{0}\\right)}{D \\sqrt{k}}=\\frac{d_{c r}}{D^{\\prime}} \\Leftrightarrow h=\\frac{d_{c r}}{D^{\\prime}} D \\sqrt{k} \\frac{L-d_{0}}{d_{0}} \\approx 80 \\mathrm{~mm} .\n$$\n\nIt should be noted that, compared to the original analysis, the deviation in this calculation caused by $f / L$ (in finding $d_{0}$ and $d_{S}$ ) on the final result reduces drastically (from approximately $20 \\%$ down to approximately $1 \\%$ ). Even if $d_{0} \\approx f$ is assumed, the result is only deviated by approximately $7 \\%$.']",['$80$'],False,mm,Numerical,16e0 1501,Optics,,"## T3: Cat eyes You may have noticed that in darkness, when a cat is within the light beam of a headlamp, its eyes appear very bright, see the photo below (left). This phenomenon can be modelled by a lens setup, see the photo on right, and the diagram beneath the photos. ![](https://cdn.mathpix.com/cropped/2023_12_21_ac051a2ec661e03601f7g-1.jpg?height=690&width=928&top_left_y=484&top_left_x=1044) The photo on right was taken by a digital single-lens reflex camera. The light intensity at the camera sensor pixels marked by a red line (in the photo) is shown in the graph below: the log base 10 of the light intensity (measured as the number of photons caught by each pixel) is plotted against the $x$-coordinate, with the pixels' side length serving as the unit length. ![](https://cdn.mathpix.com/cropped/2023_12_21_ac051a2ec661e03601f7g-1.jpg?height=631&width=928&top_left_y=1489&top_left_x=1044) The lens modelling cat eyes can be treated as an ideal thin lens of focal length $f=55 \mathrm{~mm}$ and diameter $D=$ $39 \mathrm{~mm}$; however, you should keep in mind that the given graph shows real measurement data, and the lens has certain non-ideal features. Most importantly, partial reflections of brightly lit areas from the lens surfaces may decrease the contrast: dark areas seen through the lens appear less dark than they actually are; this effect can be neglected for the camera lens, but not so for the lens serving as a model of a cat's eye. Based on the given data, estimate (with the accuracy of ca 20\%) the distance $h$ between the axis of the camera and the axis of the lamp (which can be considered as a point source) if the distance of the camera from the paper sheet was $L=4.8 \mathrm{~m}$.","['When you look at the photo of the lens and/or the graph provided, four regions with different brightness levels can be distinguished. The brightest region represents the magnified image of the blur spot created by the lamp through the lens. The blur spot is created because the distance from the lens to the white sheet beneath it is slightly larger than the focal distance; as we can see from the graph, the blur spot is of almost constant brightness (a flat plateau at $\\log _{10} I=4.4$ ), so we can say that the entire luminous flux falling from the lamp onto the lens is distributed evenly over the blur spot. Note that the blur spot has no sharp edges, though, as you would expect from in such case. This is because the image of this bright disc is situated between the lens and the camera, and is at a fairly big distance away from the plane which is sharp at the image sensor (as seen from the photo, the camera is focused onto the lens). Because of that, the enlarged image of the blur spot has blurred edges in the photo (at the blurred edges, $\\log _{10} I$ varies from 3.4 to 4.4). The second-brightest region (with $\\log _{10} I=3.4$ ) represents the scattered light from the brightest region: in that region, we are still looking through the lens, and see the area next to the bright blur spot on the sheet. Ideally, its should be darker than the sheet seen in those places where it is not obstructed by the lens, because the lens is shading the light from the lamp. However, the glass elements of this big lens are non-ideal (and there are many glass elements inside the lens!), so the light from the lamp and the bright blur spot is scattered towards the camera giving rise to an increased apparent brightness. In the area where we see the blur spot, this light is insignificant (much weaker than the light from the blur spot), but not so in this dark area: here, the scatteredfrom-the-glass-surfaces dominates heavily over the light coming from the paper sheet. As a matter of fact, this fact could be used to improve the accuracy of the calculations: we could subtract the contribution of the scattered light $\\left(10^{3.4}\\right)$ from the total intensity of the light at the brightest spot $\\left(10^{4.4}\\right)$ to obtain the contribution coming from the blur spot on the sheet. The darkest regions (with $\\log _{10} I<1.75$ ) represent the interior black painting of the lens seen through the big front glass element of the lens, which absorbs most of the incident light, and the region with $x>420$ and $\\log _{10} I=1.95$ represents the white sheet illuminated by the lamp. The ratio between the measured light intensity of the brightest region and that of the region with $x>420$ can be utilized to find the distance of the sheet (the blur spot) from the lens to the paper sheet $d_{0}$, see below.\n\nFrom the data given in the problem text we know that $L \\gg f$; from the photo of the lens, it is also clear that $d_{0}$ is of the same order of magnitude as $f$. Because of that, the illuminance $E$ (luminous flux per unit area) near the lens can be assumed to be the same as at the paper sheet, The luminous flux per solid angle and unit area of a lightscattering (or radiating) surface is called the luminance $\\mathscr{L}$; since all these directions under which the scattered light enters the lens aperture are close to the surface nor$\\mathrm{mal}$, we may assume the luminance of the paper sheet to be constant over all these directions. With the small\n\n\n\nangle approximation, the light intensity $I$ (illuminance, luminous flux $\\Phi$ per unit area) at the camera sensor is proportional to $\\mathscr{L}$ (see Explanation 1).\n\nThe luminance of the blur spot on the sheet $\\mathscr{L}_{B S}$ is $1 / k$ larger than the luminance $\\mathscr{L}_{s}$ of the paper sheet, where $k$ equals the ratio between the area of the bright dot (the blur spot) on the paper sheet and the area of the lens, because all the light received by the lens is ""compressed"" into the tiny blur spot.\n\nSmall angle approximation is also used to show that luminance of the image of the blur spot $\\mathscr{L}_{I}$ equals to the luminance of the blur spot $\\mathscr{L}_{B S}$ (see Explanation 2). Therefore, the light intensity at the sensor cells corresponding to the brightest area (where we see the image of the blur spot) $I_{I}=I_{s} / k$, where $I_{s}$ stands for the intensity at the cells corresponding to unobscured paper sheet. So, from the graph, we can deduce the value of $k$, and knowing $k$ we can calculate $d_{0}$. Let the distance along the axis between the image of the bright region through the lens and the lens itself be denoted as $d_{S}$; according to the Newton\'s lens formula, $\\left(d_{S}-f\\right)\\left(d_{0}-f\\right)=$ $f^{2}$. Hence,\n\n$$\nd_{S}=f+\\frac{f^{2}}{d_{0}-f}=\\frac{d_{0} f}{\\left(d_{0}-f\\right)}\n$$\n\ncan be also determined.\n\nHypothesize that $d_{0}-f \\ll f$. Let us calculate the diameter of the image of the blur spot\n\n$$\nD_{I}=\\frac{D_{B S} d_{S}}{d_{0}}=\\frac{D_{B S} f}{d_{0}-f}\n$$\n\nwhere the diameter of the blur spot on the sheet\n\n$$\nD_{B S}=\\frac{D\\left(d_{0}-f-s\\right)}{f+s} \\approx \\frac{D\\left(d_{0}-f-s\\right)}{f}\n$$\n\nand $s$ denotes the distance of the image of the point source from the focal plane. Using Newton\'s lens formula, $s=f^{2} /\\left(L-f-d_{0}\\right) \\approx f^{2} / L$, This leads us to\n\n$$\nD_{B S} \\approx D\\left(\\frac{d_{0}-f}{f}-\\frac{f}{L}\\right)\n$$\n\nand therefore\n\n$$\nD_{I} \\approx D\\left[1-\\frac{f^{2}}{L\\left(d_{0}-f\\right)}\\right]\n$$\n\nKeeping in mind that $d_{0}-f=\\frac{d_{0} f}{d_{S}} \\approx \\frac{f^{2}}{d_{S}}$, we obtain\n\n$$\nD_{I} \\approx D\\left(1-\\frac{d_{S}}{L}\\right)=\\frac{D\\left(L-d_{S}\\right)}{L} .\n$$\n\nThis means that as seen from the position of the camera, the angular size of the image of the blur spot $\\theta_{B S}=$ $D_{I} /\\left(L-d_{S}\\right)$ equals to the angular size of the lens aperture $\\theta_{L}=D / L$. This fact is easily confirmed from the photo and is an important observation for two reasons. First, it means that based on the angular diameter of the image of the blur spot on the photo, it is impossible to figure out the distance $d_{0}$ (and hence, $d_{S}$ ). Second, it allows us to measure instead of the angular distance $\\theta$ between the centre of the lens and the centre of the image of the blur spot (as seen from the position of the cameraline), the respective distance between the edges of the respective circles. Equality of these two angular sizes is also easily seen from the geometric construction, see the figure. Ineed, consider blue lines $S A G$ and $S B F$ which arrive from the lamp $S$ to the edges of the blur spot. Image of point $F$, denoted by $J$, is now easily found as the intersection point of the ray $S B F$ with the ray $F O$ (passing through the centre of the lens); image $H$ of the other edge of the blur spot is found in the same way. From this construction, it becomes clear that the angular size of the image of the blur spot and the lens, as seen from the camera, are exactly equal, without any approximation. Due to the smallness of the distance $h$, these angular sizes remain almost constant when the observation point is moved from $S$ to $C$.\n\nGiven the images are approximately circular, the area ratio $k$ equals $\\left(D_{B S} / D\\right)^{2}$, or\n\n$$\n\\pm \\sqrt{k}=\\frac{1}{f}\\left(d_{0}-\\frac{L f}{L-f}\\right)=\\frac{d_{0}}{f}-\\frac{L}{L-f}\n$$\n\nIn the above equation, the \\pm sign represents the two cases where the paper sheet is behind or in front of the image of the lamp. From the graph, the ratio between the intensity of the brightest region and the dark region with $x>420$ is $10^{4.4-1.95} \\approx 282$, which equals $1 / k$. Then, $d_{0} / f$ can be found to be $\\pm \\sqrt{k}+1+f / L$, which gives two solutions $d_{0} / f \\approx 1.07$ and $d_{0} / f \\approx 0.95$. According to the experimental settings given in the problem text, $d_{0}$ is greater than $f$, and thus we obtain $d_{0} / f \\approx 1.07$ and $d_{S} \\approx 15.03 f \\approx 83 \\mathrm{~cm}$. This also verifies the hypothesis that $d_{0}-f \\ll f$.\n\nThe centre of the image of the blur spot is positioned at the height $h^{\\prime}=h \\frac{d_{S}}{L}$ above the direction to the centre of the lens (this expression from similarity of the triangles $O Q P$ and $O C S$ ) which means that $\\theta=h^{\\prime} /\\left(L-d_{S}\\right)$; meanwhile, the angular diameter of the lens $\\theta_{L}=D / L$. Therefore,\n\n$$\n\\frac{\\theta}{\\theta_{L}}=\\frac{h d_{S}}{D\\left(L-d_{S}\\right)}\n$$\n\nThe ratio of the angular distances is easily measured from the figure as the ratio of the width $d_{c r}$ of the crescent-shaped second-brightest region to the diameter of the lense\'s aperture $D^{\\prime}$ :\n\n$$\nh=\\frac{d_{c r}}{D^{\\prime}} \\frac{D\\left(L-d_{S}\\right)}{d_{S}}\n$$\n\nBased on the graph, $d_{c r} \\approx 90$ pixels (midpoint of the blurry edge is around $x \\approx 120 \\mathrm{px}$, and the left edge of the aperture (in the graph) is at $x \\approx 30 \\mathrm{px}$; the right edge of the lens aperture is at $x \\approx 240 \\mathrm{px}$ corresponding to $D^{\\prime}=210 \\mathrm{px}$ and yielding $h \\approx 80 \\mathrm{~mm}$.\n\nRemark 1. In order to obtain the final answer with a reasonably good accuracy, it is not strictly speaking necessary to show that the apparent angular diameters of the lens and of the image of the blur spot are equal. All the other calculations remain the same, just one needs to match a circle with the circular segment of the visible edge of the blur spot, and measure directly $h^{\\prime}$, the distance between the centre of the lens and the centre of the blur spot, together with the diameter of the lens aperture $D^{\\prime}$ (see the small figure).\n\nRemark 2. The width of the crescent-shaped secondbrightest area can be also measured from the photo of\n\n\n\nthe lens with the required accuracy; however, measuring in pixels from the graph is more accurate.\n\nExplanation 1: Consider a small light source of luminance $\\mathcal{L}$ and surface area $S$ at a large distance $\\uparrow$ from the camera. The illuminance (the luminous flux per unit area) at the position of the camera is proportional to $\\uparrow^{-2}$ and so is the total luminous flux received by the whole sensor. Meanwhile, all this light energy is focused onto a small area $S^{\\prime}$ on the sensor - onto the image of the light source, and this area is also proportional to $\\uparrow^{-2}$. Therefore, the illuminance $I$ at the position of those sensor pixels which are covered by the image is independent of the distance $\\uparrow$.\n\nExplanation 2: Consider a very narrow cone of light of solid angle $\\omega$, starting from a very small area $S$ at the blur spot in a direction close to the surface normal, and carrying a total luminous flux $\\Phi$. Since the cone is narrow, this light beam is entirely caught by the lens at distance $d_{0}$, and focused onto the image of surface area $S^{\\prime}=S\\left(d_{S} / d_{0}\\right)^{2}$ at distance $d_{S}$ from the lens. The light rays of this beam traverse the focus and form another light cone of solid angle $\\omega^{\\prime}$ departing from the image. It is easy to see from similar triangles that $\\omega / \\omega^{\\prime}=\\left(d_{S} / d_{0}\\right)^{2}$. Then, the luminance of the image $\\mathcal{L}_{I}=\\Phi /\\left(S^{\\prime} \\omega^{\\prime}\\right)=\\Phi /(S \\omega)$, i.e. equal to the luminance of the blur spot.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_5c3bc2296f9c6fb2e76fg-1.jpg?height=865&width=928&top_left_y=321&top_left_x=1044)\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_5c3bc2296f9c6fb2e76fg-1.jpg?height=757&width=1879&top_left_y=1226&top_left_x=91)\n\nRemark 3: After having derived Explanation 1 and Explanation 2 and calculated $d_{0}$, the geometrical optics aspect can also be tackled by considering the image of the camera through the lens. The region bounded by the darkest ring in the image represents the area on the paper sheet that is observable by the camera through the lens. This area can be approximated by a circular spot with diameter $d$ (represented by $D^{\\prime}$ in the graph) depending negligibly on the size of the lens of the camera. Given small angles and that the lamp and the camera have equal distances to the lens, the bright spot on the paper sheet also has a diameter of $d$, which means $d=D \\sqrt{k}$. The angular distance between the lamp and the camera as seen from the center of the lens is $h /\\left(L-d_{0}\\right)$, and therefore, the distances between the two spots\' centers, and also thus their boundaries (represented by $d_{c r}$ in the graph), on the paper sheet are $d_{0} h /\\left(L-d_{0}\\right)$. We then obtain an equivalent equation to that above:\n\n$$\n\\frac{d_{0} h /\\left(L-d_{0}\\right)}{D \\sqrt{k}}=\\frac{d_{c r}}{D^{\\prime}} \\Leftrightarrow h=\\frac{d_{c r}}{D^{\\prime}} D \\sqrt{k} \\frac{L-d_{0}}{d_{0}} \\approx 80 \\mathrm{~mm} .\n$$\n\nIt should be noted that, compared to the original analysis, the deviation in this calculation caused by $f / L$ (in finding $d_{0}$ and $d_{S}$ ) on the final result reduces drastically (from approximately $20 \\%$ down to approximately $1 \\%$ ). Even if $d_{0} \\approx f$ is assumed, the result is only deviated by approximately $7 \\%$.']",['$80$'],False,mm,Numerical,16e0 1502,Thermodynamics,"A solid, uniform cylinder of height $h=10 \mathrm{~cm}$ and base area $s=100 \mathrm{~cm}^{2}$ floats in a cylindrical beaker of height $H=20 \mathrm{~cm}$ and inner bottom area $S=102 \mathrm{~cm}^{2}$ filled with a liquid. The ratio between the density of the cylinder and that of the liquid is $\gamma=0.70$. The bottom of the cylinder is above the bottom of the beaker by a few centimeters. The cylinder is oscillating vertically, so that its axis always coincides with that of the beaker. The amplitude of the liquid level oscillations is $A=1 \mathrm{~mm}$.",Find the period of the motion $T$. Neglect the viscosity of the liquid.,"['Denote the density of the liquid by $\\varrho$, so the density of the cylinder is $\\gamma \\varrho$. In equilibrium (i.e. when the net force acting on the cylinder is zero) the immersed part of the cylinder has height $\\gamma h$.\n\nConsider the system in a moment when the cylinder is displaced by distance $x_{1}$ downward and moves down with velocity $v_{1}$. As a result of the motion of cylinder the liquid level rises by some height $x_{2}$, and the liquid flows in the gap between the cylinder and beaker with some velocity $v_{2}$ upwards (see Fig. 1).\n\n\n\nFig. 1\n\nThe relation between the aforementioned displacements and velocities are given by the continuity law:\n\n$$\nx_{1} s=x_{2}(S-s), \\quad v_{1} s=v_{2}(S-s)\n$$\n\nIn the following we express the potential and kinetic energy of the system. Compared to the equilibrium position the cylinder of mass $\\gamma \\varrho s h$ sunk by $x_{1}$, while the potential energy change caused by the redistribution of liquid can be imagined as the center of mass of liquid with mass $\\varrho s x_{1}$ rises by distance $\\gamma h+$ $x_{1} / 2+x_{2} / 2$. Taken the potential energy in the equilibrium state to be zero, the potential energy in the state indicated in the right figure can be written as\n\n$$\nE_{\\mathrm{pot}}=-\\gamma \\varrho \\operatorname{shg} x_{1}+\\varrho s x_{1} g\\left(\\gamma h+\\frac{x_{1}+x_{2}}{2}\\right) .\n$$\n\nAfter opening the bracket the first two terms cancel each other:\n\n$$\nE_{\\mathrm{pot}}=\\frac{1}{2} \\varrho \\operatorname{sg} x_{1}\\left(x_{1}+x_{2}\\right)\n$$\n\nAfter expressing $x_{2}$ from continuity law and some simplification we get a quadratic expression for the potential energy:\n\n$$\nE_{\\mathrm{pot}}=\\frac{1}{2} \\varrho s g x_{1}\\left(x_{1}+\\frac{s}{S-s} x_{1}\\right)=\\frac{1}{2} \\varrho \\frac{s S}{S-s} g x_{1}^{2} .\n$$\n\nNow let us calculate the kinetic energy of the system. The contribution from the cylinder is straightforward, $\\gamma \\varrho s h v_{1}^{2} / 2$, but the motion of the liquid is more complicated.\n\nNote. We may notice that since $s /(S-s)=50$, the speed $v_{2}$ of the liquid in the narrow gap is 50 times larger than the typical speed of the liquid below the cylinder (which can be estimated to be in the range of $v_{1}$ ). And while the mass of the liquid below the cylinder is much larger than the mass of liquid inside the gap (the ratio is ca. 25 if the „few centimeters"" in the problem text is taken to be $3.5 \\mathrm{~cm}$ ), the kinetic energy is proportional to the square of the velocity, so the kinetic energy of the liquid inside the gap is roughly 100 times larger than the kinetic energy of the liquid below the cylinder.\n\nSince the kinetic energy of the liquid below the cylinder is negligible, we can write the total kinetic energy of the system as:\n\n$$\nE_{\\text {kin }}=\\underbrace{\\frac{1}{2} \\gamma \\varrho s h v_{1}^{2}}_{\\text {cylinder }}+\\underbrace{\\frac{1}{2} \\varrho(S-s)\\left(\\gamma h+x_{1}+x_{2}\\right) v_{2}^{2}}_{\\text {liquid }} .\n$$\n\nHere $x_{1}, x_{2} \\ll \\gamma h$, so we shall keep only the term containing $\\gamma h$ in the second bracket:\n\n$$\nE_{\\text {kin }}=\\frac{1}{2} \\gamma \\varrho s h v_{1}^{2}+\\frac{1}{2} \\varrho(S-s) \\gamma h v_{2}^{2}\n$$\n\nExpressing $v_{2}$ from continuity law gives the following:\n\n$$\nE_{\\text {kin }}=\\frac{1}{2} \\gamma \\varrho s h v_{1}^{2}+\\frac{1}{2} \\varrho \\gamma h \\frac{s^{2}}{S-s} v_{1}^{2}=\\frac{1}{2} \\varrho \\gamma h \\frac{s S}{S-s} v_{1}^{2} .\n$$\n\nThe potential and kinetic energies can be written in the form\n\n$$\nE_{\\mathrm{pot}}=\\frac{1}{2} k_{\\mathrm{eff}} x_{1}^{2}, \\quad E_{\\mathrm{kin}}=\\frac{1}{2} m_{\\mathrm{eff}} v_{1}^{2},\n$$\n\nwhere the effective spring constant and effective mass are given by\n\n$$\nk_{\\mathrm{eff}}=\\varrho \\frac{s S}{S-s} g, \\quad m_{\\mathrm{eff}}=\\varrho \\gamma h \\frac{s S}{S-s}\n$$\n\nSo the oscillation is indeed harmonic, thus the angular frequency and the period are:\n\n$$\n\\omega=\\sqrt{\\frac{k_{\\mathrm{eff}}}{m_{\\mathrm{eff}}}}=\\sqrt{\\frac{g}{\\gamma h}}, \\quad T=2 \\pi \\sqrt{\\frac{\\gamma h}{g}}=0.53 \\mathrm{~s} .\n$$\n\nNote. The static restoring force, acting on the cylinder is due to the change (relative to the equilibrium position) of the hydrostatic pressure at its lower base:\n\n$$\nF=-\\operatorname{seg}\\left(x_{1}+x_{2}\\right)=-\\frac{s S}{S-s} \\rho g x_{1}\n$$\n\nThis immediately gives effective stiffness of the system $k_{\\text {eff }}=\\frac{s S}{S-s} \\rho g$.\n\nAlternatively, one may wish to integrate $\\int F \\mathrm{~d} x_{1}$ to get the potential energy\n\n$$\nE_{\\mathrm{pot}}=\\frac{s S}{S-s} \\frac{\\rho g}{2} x_{1}^{2}\n$$', ""When the cylinder is displaced from its equilibrium position downwards by distance $x_{1}$, the net restoring force (pointing up) can be calculated as the sum of the weight of the cylinder and the force from the difference of pressures at the top $\\left(p_{0}\\right)$ and bottom $(p)$ of the cylinder. As a result of the net force, the cylinder accelerates upwards with $a_{1}$, and at the same time, the liquid located in the gap between the cylinder and the wall of the beaker accelerates down with $a_{2}$. The relation between the magnitudes of $a_{1}$ and $a_{2}$ is given by the continuity law:\n\n$$\ns a_{1}=(S-s) a_{2}\n$$\n\n\n\nFig. 2\n\nIf the liquid in the gap was not accelerating, the pressure difference $p-p_{0}$ would be equal to the hydrostatic pressure of the liquid column in the gap. Due to the acceleration of the liquid, $p-p_{0}$ can be expressed from Newton's 2nd law applied for the liquid column of unit area located in the gap:\n\n$$\np_{0}-p+\\varrho g\\left(\\gamma h+x_{1}+x_{2}\\right)=\\varrho\\left(\\gamma h+x_{1}+x_{2}\\right) a_{2}\n$$\n\nwhere we used the notations of Solution I, and the downward direction was taken as positive.\n\nNewton's 2nd law for the cylinder reads as\n\n$$\n\\left(p-p_{0}\\right) s-\\gamma \\varrho s h g=\\gamma \\varrho s h a_{1}\n$$\n\nAfter expressing $p-p_{0}$ from the previous equation, and then substituting it here we get:\n\n$\\varrho g\\left(\\gamma h+x_{1}+x_{2}\\right) s-\\varrho\\left(\\gamma h+x_{1}+x_{2}\\right) a_{2} s-\\gamma \\varrho s h g=\\gamma \\varrho s h a_{1}$.\n\nSince the amplitude of the liquid level is small, the terms containing $a_{2} x_{1}$ and $a_{2} x_{2}$ can be neglected. After rearranging we get:\n\n$$\n\\varrho g s\\left(x_{1}+x_{2}\\right)=\\gamma \\varrho s h\\left(a_{1}+a_{2}\\right) .\n$$\n\nUsing the relations between the displacements and accelerations we finally get:\n\n$$\na_{1}=\\frac{g}{\\gamma h} x_{1}\n$$\n\nTaking into account the opposite directions of $x_{1}$ and $a_{1}$, this is the dynamical condition of a simple harmonic motion with angular frequency and period\n\n$$\n\\omega=\\sqrt{\\frac{g}{\\gamma h}}, \\quad T=2 \\pi \\sqrt{\\frac{\\gamma h}{g}}=0.53 \\mathrm{~s} .\n$$""]",['0.53'],False,s,Numerical,5e-2 1503,Electromagnetism,"A resistor is made of a material which undergoes a phase transition so that its resistance takes one of the two values, $R_{1}$ if its temperature is smaller than $T_{c}$, and $R_{2}>R_{1}$ if the temperature is larger than $T_{c}$. This resistor is connected to a voltage source through an inductor of inductance $L$. It appears that if the applied voltage $V$ is between two critical values, $V_{1}T_{c}$ if $R=R_{2}$. Solving for $V$, we have\n\n$$\nV=\\sqrt{R_{j} \\alpha\\left(T_{\\mathrm{eq}}-T_{0}\\right)}\n\\tag{1}\n$$\n\nThe critical values therefore are\n\n$$\nV_{1}=\\sqrt{R_{1} \\alpha\\left(T_{c}-T_{0}\\right)} \\quad \\text { and } \\quad V_{2}=\\sqrt{R_{2} \\alpha\\left(T_{c}-T_{0}\\right)}\n\\tag{2}\n$$']","['$V_{1}=\\sqrt{R_{1} \\alpha\\left(T_{c}-T_{0}\\right)}$ , $V_{2}=\\sqrt{R_{2} \\alpha\\left(T_{c}-T_{0}\\right)}$']",True,,Expression, 1504,Electromagnetism,"A resistor is made of a material which undergoes a phase transition so that its resistance takes one of the two values, $R_{1}$ if its temperature is smaller than $T_{c}$, and $R_{2}>R_{1}$ if the temperature is larger than $T_{c}$. This resistor is connected to a voltage source through an inductor of inductance $L$. It appears that if the applied voltage $V$ is between two critical values, $V_{1}\n\nFig. 1\n\nTogether with (3), we see that the temperature behaves like in Figure 2.\n\n\n\nFig. 2\n\nThe maximum and minimum temperatures will be attained just after the phase transitions occur. We obtain that\n\n$$\n\\frac{T_{\\max }-T_{0}}{T_{\\min }-T_{0}}=\\frac{R_{2} I_{1}^{2}}{R_{1} I_{2}^{2}}=\\frac{R_{2}^{2}}{R_{1}^{2}}\n\\tag{4}\n$$']",['$\\frac{R_{2}^{2}}{R_{1}^{2}}$'],False,,Expression, 1505,Electromagnetism,"A resistor is made of a material which undergoes a phase transition so that its resistance takes one of the two values, $R_{1}$ if its temperature is smaller than $T_{c}$, and $R_{2}>R_{1}$ if the temperature is larger than $T_{c}$. This resistor is connected to a voltage source through an inductor of inductance $L$. It appears that if the applied voltage $V$ is between two critical values, $V_{1}0$, i.e. counterclockwise rotation, and the ""-"" sign —to clockwise rotation. In either case, the initial velocity should point to the negative $Y$ direction:\n\n$$\n\\vec{v}_{0}=-\\frac{q d B}{2 m} \\hat{\\jmath}\n$$']","['$\\vec{v}_{0}=-\\frac{q d B}{2 m} \\hat{\\jmath}$ and $R_{c}=\\frac{q B d}{2 m|\\omega_{0}|}$ and $(x_{c}, y_{c})=( \\pm R_{c}, 0)$']",False,,Need_human_evaluate, 1508,Electromagnetism,"Two small balls of mass $m$ each with charges $+q$ and $-q$ respectively, connected by a rigid massless rod of length $d$, form a dipole. The dipole is parallel to plane $X Y$ and is placed in a uniform magnetic field $\vec{B}$ perpendicular to $X Y$. Initially, the dipole is aligned with the direction $X$ and has initial angular velocity $\omega_{0}$ in plane $X Y$, as shown. Its center of mass is initially located at origin and given initial velocity $\vec{v}_{0}$ parallel to $X Y$, as well. Consider three distinct scenarios (a, b, c-d): Context question: (a) Find $\omega_{0}$ and the direction of $\vec{v}_{0}$, so that the center of mass will move with the constant velocity $\vec{v}=\vec{v}_{0}$ ? Context answer: $\vec{v}_{0} \| Y$ , $\omega_{0}=0$ Context question: (b) Given $\omega_{0}$, find such $\vec{v}_{0}$ (direction and magnitude), so that the center of mass will travel in a circle. Find the circle radius $R_{c}$ and the coordinates $x_{c}, y_{c}$ of its center. You don't need to prove the uniqueness of the solution. Context answer: $\vec{v}_{0}=-\frac{q d B}{2 m} \hat{\jmath}$ and $R_{c}=\frac{q B d}{2 m|\omega_{0}|}$ and $(x_{c}, y_{c})=( \pm R_{c}, 0)$ ","(c) Given $\vec{v}_{0}=0$, find the minimal $\omega_{0}=\omega_{\text {min }}$ necessary for the dipole to reverse its orientation during the motion.","['In (10) we have shown that the net force:\n\n$$\n\\vec{F}=2 q(\\vec{\\omega} \\times \\vec{r}) \\times \\vec{B}=(\\vec{\\omega} \\times \\vec{p}) \\times \\vec{B}\n$$\n\nSince the dipole moment $\\vec{p}$ rotates with angular velocity $\\vec{\\omega}$, its time derivative:\n\n$$\n\\frac{d \\vec{p}}{d t}=\\vec{\\omega} \\times \\vec{p}\n$$\n\nFrom Newton\'s second law:\n\n$$\n2 m \\frac{d \\vec{v}}{d t}=\\vec{F}=\\frac{d \\vec{p}}{d t} \\times \\vec{B}\n$$\n\nBy integrating the equation, we arrive at an additional conservation law in the system (conservation of the so called ""generalized momentum""):\n\n$$\n2 m \\vec{v}-\\vec{p} \\times \\vec{B}=\\mathrm{const}\n$$\n\nThus, if $\\vec{p}$ has reversed its direction from $\\vec{p}_{0}$ to $\\vec{p}_{1}=-\\vec{p}_{0}$, then the velocity:\n\n$$\n\\vec{v}_{1}=\\vec{v}_{0}+\\frac{\\left(\\vec{p}_{1}-\\vec{p}_{0}\\right) \\times \\vec{B}}{2 m}=-\\frac{\\vec{p}_{0} \\times \\vec{B}}{m}\n$$\n\n\n\nSince the magnetic field does not perform work on moving electric charges, the kinetic energy of the dipole is conserved:\n\n$$\n\\frac{I}{2} \\omega_{0}^{2}=\\frac{I}{2} \\omega_{1}^{2}+\\frac{2 m}{2} v_{1}^{2}\n$$\n\nHere, $I=2 \\times m(d / 2)^{2}=m d^{2} / 2$ is the moment of inertia of the dipole with respect to its center-of-mass. Since $v_{1}$ doesn\'t depend on angular velocities, $\\omega_{0}$ is minimal when $\\omega_{1}=0$. Finally,\n\n$$\n\\omega_{\\min }=v_{1} \\sqrt{\\frac{2 m}{I}}=\\frac{p_{0} B}{m} \\sqrt{\\frac{4}{d^{2}}}=\\frac{2 q B}{m}\n$$\n\nAlternatively, we can introduce $\\theta$ to be the angle between the dipole moment and the axis $X\\left(\\theta_{0}=0\\right)$ and rewrite the equations of translational motion in coordinates using $\\omega=\\dot{\\theta}$ :\n\n$$\n\\dot{v}_{x}=\\dot{\\theta} \\frac{q B d}{2 m} \\cos \\theta, \\quad \\dot{v}_{y}=\\dot{\\theta} \\frac{q B d}{2 m} \\sin \\theta\n$$\n\nBy integrating these equations, given zero initial velocity, we find how velocity depends on $\\theta$ :\n\n$$\nv_{x}=\\frac{q B d}{2 m} \\sin \\theta, \\quad v_{y}=\\frac{q B d}{2 m}(1-\\cos \\theta) .\n$$\n\nUsing the expression (9) for the torque, we can write the equation of rotational motion as:\n\n$$\n\\begin{gathered}\n\\ddot{\\theta}=\\tau=-2 q B\\left(r_{x} v_{x}+r_{y} v_{y}\\right)=-\\frac{q^{2} B^{2} d^{2}}{2 m} \\sin \\theta \\\\\n\\ddot{\\theta}+\\frac{q^{2} B^{2}}{m^{2}} \\sin \\theta=0\n\\end{gathered}\n$$\n\nThis is the equation of a mathematical pendulum of length $L$ in gravitational field $g=L(q B / m)^{2}$. And the equivalent question becomes what is the minimal push $\\dot{\\theta}_{0}$ required in the bottom position for the pendulum to reach the top position. Kinetic energy of the pendulum $K=\\frac{1}{2} m L^{2} \\dot{\\theta}_{0}^{2}$ will be transfered to the potential energy $U=2 m g L$, from which we find:\n\n$$\n\\omega_{\\min }=\\dot{\\theta}_{0}=\\sqrt{4 \\frac{g}{L}}=2 \\frac{q B}{m}\n$$\n\nNote. Due to symmetry, both clockwise and counterclockwise initial rotation with absolute value of $\\left|\\omega_{0}\\right|$ will work.']",['$2 \\frac{q B}{m}$'],False,,Expression, 1509,Electromagnetism,"Two small balls of mass $m$ each with charges $+q$ and $-q$ respectively, connected by a rigid massless rod of length $d$, form a dipole. The dipole is parallel to plane $X Y$ and is placed in a uniform magnetic field $\vec{B}$ perpendicular to $X Y$. Initially, the dipole is aligned with the direction $X$ and has initial angular velocity $\omega_{0}$ in plane $X Y$, as shown. Its center of mass is initially located at origin and given initial velocity $\vec{v}_{0}$ parallel to $X Y$, as well. Consider three distinct scenarios (a, b, c-d): Context question: (a) Find $\omega_{0}$ and the direction of $\vec{v}_{0}$, so that the center of mass will move with the constant velocity $\vec{v}=\vec{v}_{0}$ ? Context answer: $\vec{v}_{0} \| Y$ , $\omega_{0}=0$ Context question: (b) Given $\omega_{0}$, find such $\vec{v}_{0}$ (direction and magnitude), so that the center of mass will travel in a circle. Find the circle radius $R_{c}$ and the coordinates $x_{c}, y_{c}$ of its center. You don't need to prove the uniqueness of the solution. Context answer: $\vec{v}_{0}=-\frac{q d B}{2 m} \hat{\jmath}$ and $R_{c}=\frac{q B d}{2 m|\omega_{0}|}$ and $(x_{c}, y_{c})=( \pm R_{c}, 0)$ Context question: (c) Given $\vec{v}_{0}=0$, find the minimal $\omega_{0}=\omega_{\text {min }}$ necessary for the dipole to reverse its orientation during the motion. Context answer: \boxed{$2 \frac{q B}{m}$} ","(d) (1 pt) If the dipole starts with $\vec{v}_{0}=0$ and $\omega_{0}=\omega_{\text {min }}$ found in part (c), the trajectory of its center of mass has an asymptote. Find the distance $D$ from the origin to the asymptote.","['If dipole\'s trajectory has an asymptote, then its movement along the asymptote is uniform. Indeed, if there is a linear motion with acceleration, the dipole $\\vec{p}$ should be always aligned with the direction of motion, thus, not rotating. and as we found in part (a), the absence of rotation can only be maintained if $\\vec{v}=$ const and $\\vec{v} \\perp \\vec{p}$.\n\nThe uniform linear motion requires $\\omega=0$, and this happens in the limit when the orientation is reversed $\\vec{p}_{1}=-\\vec{p}_{0}$. According to (11), in the limit, the dipole is travelling with the speed $\\vec{v}_{1}=p_{0} B \\hat{\\jmath} / m$. Thus the asymptote is parallel to $\\mathrm{Y}$ axis: $x=D$ (for counter-clockwise initial rotation).\n\n\n\nIf $\\vec{R}_{+}$and $\\vec{R}_{-}$are absolute positions of the charges, we can write equation for the angular momentum around the origin $L_{O}$ :\n\n$$\n\\begin{aligned}\n& \\frac{d \\vec{L}_{O}}{d t}=\\vec{R}_{+} \\times\\left(q \\dot{\\vec{R}}_{+} \\times \\vec{B}\\right)+\\vec{R}_{-} \\times\\left(-q \\dot{\\vec{R}}_{-} \\times \\vec{B}\\right)= \\\\\n& -q \\vec{B}\\left(\\vec{R}_{+} \\cdot \\dot{\\vec{R}}_{+}-\\vec{R}_{-} \\cdot \\dot{\\vec{R}}_{-}\\right)=-\\frac{q \\vec{B}}{2} \\frac{d}{d t}\\left(R_{+}^{2}-R_{-}^{2}\\right)\n\\end{aligned}\n$$\n\nAfter integration, we find one more conservation law (conservation of the ""generalized angular momentum""):\n\n$$\n\\begin{aligned}\n\\vec{L}_{O}+\\frac{q \\vec{B}}{2}\\left(R_{+}^{2}-R_{-}^{2}\\right)=\\vec{L}_{O}+ & \\frac{q \\vec{B}}{2}\\left(\\left(\\vec{R}_{+}+\\vec{R}_{-}\\right) \\cdot\\left(\\vec{R}_{+}-\\vec{R}_{-}\\right)\\right) \\\\\n& =\\vec{L}_{O}+\\vec{B}(\\vec{R} \\cdot \\vec{p})=\\mathrm{const},\n\\end{aligned}\n$$\n\nwhere $\\vec{R}=\\frac{1}{2}\\left(\\vec{R}_{+}+\\vec{R}_{-}\\right)$is the position of center of mass. We also used the fact that $q\\left(\\vec{R}_{+}-\\vec{R}_{-}\\right)=2 q \\vec{r}=\\vec{p}$.\n\nInitially, centre of mass coincides with origin $\\left(\\vec{R}_{0}=0\\right)$ :\n\n$$\nL_{O}(0)=I \\omega_{0}=2 m \\frac{d^{2}}{4} 2 \\frac{q B}{m}=q B d^{2}\n$$\n\nAt asymptote, the dipole has reversed direction $\\vec{p}_{1}=-\\vec{p}_{0}$ and charges are travelling along parallel lines $x=D \\pm r$ with the velocity $\\vec{v}_{1}$ :\n\n$$\n\\begin{aligned}\nL_{O}(\\infty)+ & B\\left(\\vec{R}_{1} \\cdot \\vec{p}_{1}\\right)=m(D-r) v_{1}+m(D+r) v_{1}-B D p_{0} \\\\\n& =2 m D \\frac{p_{0} B}{m}-B D p_{0}=B D p_{0}=B D q d\n\\end{aligned}\n$$\n\nSince (13) equals (14), we conclude that $D=d$.\n\n\n\nWe can arrive to the same conclusion differently. Notice that we are interested in the $x$ coordinate of $C$ at infinity:\n\n$$\nD=x_{\\infty}=\\int_{0}^{\\infty} v_{x} d t=\\frac{q B d}{2 m} \\int_{0}^{\\infty} \\sin \\theta d t\n$$\n\nFrom (12), we can express $\\sin \\theta$ :\n\n$$\n\\begin{aligned}\n\\int_{0}^{\\infty} \\sin \\theta d t= & -\\frac{m^{2}}{q^{2} B^{2}} \\int_{0}^{\\infty} \\ddot{\\theta} d t= \\\\\n& -\\frac{m^{2}}{q^{2} B^{2}}\\left(\\dot{\\theta}_{1}-\\dot{\\theta}_{0}\\right)=\\frac{m^{2}}{q^{2} B^{2}} \\omega_{\\min }=\\frac{2 m}{q B} .\n\\end{aligned}\n$$\n\nFinally,\n\n$$\nD=\\frac{q B d}{2 m} \\frac{2 m}{q B}=d\n$$\n\nNote. If initial rotation is clockwise $\\left(\\omega_{0}<0\\right)$, the asymptote has an equation $x=-D$, but the distance to the origin remains the same.']",['$D=d$'],False,,Expression, 1510,Thermodynamics,"## 1 Ice pellets An interesting weather phenomenon can occur when the temperature profile in the atmosphere shows an inversion. The solid blue line in figure 1 shows such a temperature profile. The inversion occurs at heights between $1 \mathrm{~km}$ and $2 \mathrm{~km}$. Under these conditions snow falling through the atmosphere (partially) melts in the warmer layer and (partially) freezes again before reaching the ground in the form of ""ice pellets"". Figure 1: Atmospheric temperature $T$ vs. height $h$ above the ground. Assume that a small, spherical ice droplet almost completely melts while falling through the atmospheric layer between $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$ where the temperature is above freezing point.",a. Determine the mass fraction of the droplet that freezes before reaching the ground.,"['With the given assumptions of constant droplet size and constant atmospheric density the drag force leads to a constant velocity $v$ for the fall of the droplet through the atmosphere.\n\nWe can assume that the droplet temperature $T_{\\mathrm{d}}$ above $h_{\\mathrm{A}}$ follows the atmosphere temperature profile (also see below) and remains constant and equal to $T_{0}=0^{\\circ} \\mathrm{C}$ during the melting below that height.\n\nFor small temperature differences $\\Delta T=T-T_{\\mathrm{d}}$ the heat exchange rate is proportional to this difference $\\mathrm{d} Q / \\mathrm{d} t=\\kappa \\Delta T$. The factor $\\kappa$ depends on the droplet geometry and its velocity, as well as on the air density. Since these are constant $\\kappa=$ const.\n\nIn the region between $h_{\\mathrm{A}}$ and $h_{\\mathrm{B}}$ the droplet is therefore heated at a rate\n\n$$\n\\mathrm{d} Q=\\kappa\\left(T-T_{0}\\right) \\mathrm{d} t=-\\kappa\\left(T-T_{0}\\right) \\frac{\\mathrm{d} h}{v}, \\tag{1}\n$$\n\nThe total heat transfer between $h_{\\mathrm{A}}$ and $h_{\\mathrm{B}}$ is\n\n$$\nQ=\\frac{\\kappa A}{v}=m L \\tag{2}\n$$\n\nwhere $A=5.0 \\mathrm{~km}{ }^{\\circ} \\mathrm{C}$ is the area between the temperature curve and the height-axis in the region between heights $h_{\\mathrm{A}}$ and $h_{\\mathrm{B}}$. The right hand side equates the heat with the latent heat necessary to completely melt the ice droplet of mass $m$.\n\nIn the region below $h_{\\mathrm{B}}$ the liquid droplet partially freezes again. During this process the temperature is again constant. The mass fraction $\\eta$ of liquid freezing before reaching the ground can again be derived from the area $A^{\\prime}=4.0 \\mathrm{~km}{ }^{\\circ} \\mathrm{C}$ between the curve and the height-axis.\n\n$$\nQ^{\\prime}=-\\frac{\\kappa A^{\\prime}}{v}=-\\eta m L \\tag{3}\n$$\n\nDividing (3) by (2) gives the mass fraction\n\n$$\n\\eta=\\frac{A^{\\prime}}{A}=\\frac{4}{5}=0.80 \\tag{4}\n$$']",['$\\frac{4}{5}$'],False,,Numerical,0 1511,Thermodynamics,"## 1 Ice pellets An interesting weather phenomenon can occur when the temperature profile in the atmosphere shows an inversion. The solid blue line in figure 1 shows such a temperature profile. The inversion occurs at heights between $1 \mathrm{~km}$ and $2 \mathrm{~km}$. Under these conditions snow falling through the atmosphere (partially) melts in the warmer layer and (partially) freezes again before reaching the ground in the form of ""ice pellets"". Figure 1: Atmospheric temperature $T$ vs. height $h$ above the ground. Assume that a small, spherical ice droplet almost completely melts while falling through the atmospheric layer between $h_{\mathrm{A}}$ and $h_{\mathrm{B}}$ where the temperature is above freezing point. Context question: a. Determine the mass fraction of the droplet that freezes before reaching the ground. Context answer: \boxed{$\frac{4}{5}$} ","b. Find, as precisely as possible, the temperature of the droplet at ground level if there were no inversion and the temperature profile followed the dashed line below a height of $2 \mathrm{~km}$. Neglect evaporation, condensation and size changes of the droplet. Assume that water and ice have very high thermal conductivity and that the density of the atmosphere is constant with height. Use $c_{\text {water }}=4.2 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ for the the specific heat of water and $c_{\text {ice }}=2.1 \mathrm{~kJ} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ for that of ice. The specific latent heat for the melting of ice is $L=334 \mathrm{~kJ} \mathrm{~kg}^{-1}$.","['If the temperature profile follows the dashed line the droplet will completely melt and the heat transferred from the atmosphere will heat it up. Since the latent heat of melting is much bigger than the specific heat of water times some degrees of temperature variation the temperature of the liquid droplet will closely follow the temperature of the atmosphere eventually. Its temperature at ground level should therefore be close to $8{ }^{\\circ} \\mathrm{C}$.\n\nFor a better estimate let us introduce a new coordinate $x$ whose origin is at the height, where the droplet is completely molten (somewhat higher than $\\left.h_{\\mathrm{B}}\\right)$, and which is oriented downwards. For the change in temperature of the droplet we then have\n\n$$\nm c_{\\text {water }} \\frac{\\mathrm{d} T_{\\mathrm{d}}}{\\mathrm{d} t}=m c_{\\text {water }} v \\frac{\\mathrm{d} T_{\\mathrm{d}}}{\\mathrm{d} x}=\\kappa\\left(T-T_{\\mathrm{d}}\\right), \\tag{5}\n$$\n\nwhere the temperature of the atmosphere is given by $T(x)=T(x=0)+2.0^{\\circ} \\mathrm{Ckm}^{-1} x=: T(x=0)+b x$.\nFor the difference $\\Delta T=T-T_{\\mathrm{d}}$ between the atmospheric temperature and the droplet temperature we have:\n\n$$\n\\frac{\\mathrm{d} \\Delta T}{\\mathrm{~d} x}=b-\\frac{\\mathrm{d} T_{\\mathrm{d}}}{\\mathrm{d} x}=b-\\frac{\\kappa}{m c_{\\text {water }} v} \\Delta T . \\tag{6}\n$$\n\nThis differential equation is solved by\n\n$$\n\\Delta T=b x_{0}+\\text { const } \\cdot \\exp \\left(-x / x_{0}\\right), \\tag{7}\n$$\n\nwhere $x_{0}=m c_{\\text {water }} v / \\kappa$. From equation (2) we find that $\\frac{m v}{\\kappa}=\\frac{A}{L}$ such that $x_{0}=\\frac{c_{\\text {water }}}{L} A \\approx 0.063 \\mathrm{~km}$. Therefore the exponential factor in the above equation is negligible at ground level (the constant is close to $4^{\\circ} \\mathrm{C}$ ) and we arrive at a temperature difference between the droplet temperature and the atmospheric temperature at ground level of\n\n$$\n\\Delta T \\approx 0.13^{\\circ} \\mathrm{C}, \\quad \\text { and therefore } \\quad T_{\\mathrm{d}} \\approx 7.9^{\\circ} \\mathrm{C} \\text {. } \\tag{8}\n$$\n\nAnother phenomenon, which effectively shifts atmosphere temperature as it is ""felt"" by the droplet is due to viscous dissipation. It can be estimated by equating the droplet potential energy loss rate $m g v$ to thermal power carried away by the circumfluent air $\\kappa \\Delta T^{*}$. This gives $\\Delta T^{*} \\sim m g v / \\kappa=g A / L=0.17^{\\circ} \\mathrm{C}$. If both corrections were taken into account, then the droplet temperature before hitting the ground would again be $T_{\\mathrm{d}} \\approx 8.0^{\\circ} \\mathrm{C}$.']",['$8$'],False,$^{\circ} \mathrm{C}$,Numerical,2e-1 1512,Electromagnetism,,(i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$.,['An electron with charge $-e(e>0)$ experiences the Lorentz force due to the perpendicular magnetic field and the electric force\n\n$$\nm^{*} \\frac{d \\vec{v}}{d t}=-e(\\vec{v} \\times \\vec{B}+\\vec{E})\n$$\n\nwhere $\\vec{v}$ is the velocity of the electron.'],['$$\nm^{*} \\frac{d \\vec{v}}{d t}=-e(\\vec{v} \\times \\vec{B}+\\vec{E})\n$$\nwhere $\\vec{v}$ is the velocity of the electron.'],False,,Need_human_evaluate, 1513,Electromagnetism,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. ",(ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case.,"['In the stationary regime, the acceleration vanishes. Hence\n\n$$\n\\vec{v}_{s} \\times \\vec{B}+\\vec{E}=0\n$$\n\nThe velocity can be expressed as\n\n$$\n\\vec{v}_{s}=\\frac{\\vec{E} \\times \\vec{B}}{B^{2}}\n$$\n\nwhose magnitude is simply $v_{s}=E / B$.']",['$E / B$.'],False,,Expression, 1515,Electromagnetism,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} ","(b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively.","['The Hall voltage $V_{H}=E_{y} W$. The current in the $-x$ direction is\n\n$$\nI=\\frac{\\Delta Q}{\\Delta t}=\\frac{N e}{L / v_{s}}=\\frac{N e}{L} \\frac{E_{y}}{B}=e \\frac{N}{\\phi} V_{H}\n$$\n\nTherefore,\n\n$$\nR_{H}=\\frac{V_{H}}{I}=\\frac{1}{e} \\frac{\\phi}{N}\n$$']",['$R_{H}=\\frac{1}{e} \\frac{\\phi}{N}$'],False,,Expression, 1516,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} ","(c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$,","['The Hall resistance can be rewritten as\n\n$$\nR_{H}=\\frac{1}{e} \\frac{\\phi}{N}=\\frac{h}{e^{2}} \\frac{\\phi /(h / e)}{N}=\\frac{h}{e^{2}} \\frac{N_{\\phi}}{N}\n$$\n\nAt the plateau, $\\nu=N / N_{\\phi}=1 / 3$.']",['$\\frac{1}{3}$'],False,,Numerical,0 1517,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} ","(d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$.","['The average distance between electrons can be written as $f l_{0}$, where\n\n$$\nl_{0}=\\sqrt{\\frac{L W}{N}}=\\sqrt{\\frac{\\phi}{N B}}=\\sqrt{\\frac{h}{\\nu e B}}\n$$\n\nand $f$ is a dimensionless constant that is determined by the electron distribution (or, quantum mechanically, wave function). Binding multiple vortices with an electron effectively reduces the probability of other electrons getting close. Therefore, the electrons optimize their distribution in such a way that their average distance increases from $f_{1} l_{0}$ to $f_{2} l_{0}\\left(f_{1} Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} Context question: (d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$. Context answer: \boxed{$\frac{1}{2}$} ","(e) As the magnetic field deviates from the exact filling $v=1 / n$ to a higher field, more vortices (whirlpools in the electron sea) are being created. They are not bound to electrons and behave like particles carrying effectively positive charges, hence known as quasiholes, compared to the negatively charged electrons. The amount of charge deficit in any of these quasiholes amounts to exactly $1 / n$ of an electronic charge. An analogous argument can be made for magnetic fields slightly below $v$ and the creation of quasielectrons of negative charge $e^{*}=-e / n$. At the quantized Hall plateau of $R_{\mathrm{H}}=3 h / e^{2}$, calculate the amount of change in $B$ that corresponds to the introduction of exactly one fractionally charged quasihole. (When their density is low, the quasiparticles are confined by the random potential generated by impurities and imperfections, hence the Hall resistance remains quantized for a finite range of $B$.)","['The flux change due to the change of the magnetic field is\n\n$$\n\\Delta \\phi=\\Delta B(W L)=\\frac{h}{e}\n$$\n\nTherefore, $\\Delta B=h /(e W L)$.']",['$\\Delta B=\\frac{h }{e W L}$'],False,,Expression, 1519,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} Context question: (d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$. Context answer: \boxed{$\frac{1}{2}$} Context question: (e) As the magnetic field deviates from the exact filling $v=1 / n$ to a higher field, more vortices (whirlpools in the electron sea) are being created. They are not bound to electrons and behave like particles carrying effectively positive charges, hence known as quasiholes, compared to the negatively charged electrons. The amount of charge deficit in any of these quasiholes amounts to exactly $1 / n$ of an electronic charge. An analogous argument can be made for magnetic fields slightly below $v$ and the creation of quasielectrons of negative charge $e^{*}=-e / n$. At the quantized Hall plateau of $R_{\mathrm{H}}=3 h / e^{2}$, calculate the amount of change in $B$ that corresponds to the introduction of exactly one fractionally charged quasihole. (When their density is low, the quasiparticles are confined by the random potential generated by impurities and imperfections, hence the Hall resistance remains quantized for a finite range of $B$.) Context answer: \boxed{$\Delta B=\frac{h }{e W L}$} Extra Supplementary Reading Materials: (f) In Tsui et al. experiment, the magnetic field corresponding to the center of the quantized Hall plateau $$ R_{\mathrm{H}}=3 h / e^{2}, B_{1 / 3}=15 \text { Tesla, } $$ the effective mass of an electron in GaAs, $m^{*}=0.067 m_{e}$, the electron mass, $m_{e}=9.1 \times 10^{-31} \mathrm{~kg}$, Coulomb's constant, $k=9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$, the vacuum permittivity, $\varepsilon_{0}=1 / 4 \pi k=8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}$, the relative permittivity (the ratio of the permittivity of a substance to the vacuum permittivity) of GaAs, $\varepsilon_{r}=13$, the elementary charge, $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant, $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$, and Boltzmann's constant, $k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$. In our analysis, we have neglected several factors, whose corresponding energy scales, compared to $\Delta U(B)$ discussed in (d), are either too large to excite or too small to be relevant.",(i) Calculate the thermal energy $E_{\text {th }}$ at temperature $T=1.0 \mathrm{~K}$.,['The thermal energy\n\n$$\nE_{t h}=k_{B} T=1.38 \\times 10^{-23} \\times 1.0=1.38 \\times 10^{-23} \\mathrm{~J}\n$$'],['$1.38 \\times 10^{-23}$'],False,J,Numerical,2e-25 1520,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} Context question: (d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$. Context answer: \boxed{$\frac{1}{2}$} Context question: (e) As the magnetic field deviates from the exact filling $v=1 / n$ to a higher field, more vortices (whirlpools in the electron sea) are being created. They are not bound to electrons and behave like particles carrying effectively positive charges, hence known as quasiholes, compared to the negatively charged electrons. The amount of charge deficit in any of these quasiholes amounts to exactly $1 / n$ of an electronic charge. An analogous argument can be made for magnetic fields slightly below $v$ and the creation of quasielectrons of negative charge $e^{*}=-e / n$. At the quantized Hall plateau of $R_{\mathrm{H}}=3 h / e^{2}$, calculate the amount of change in $B$ that corresponds to the introduction of exactly one fractionally charged quasihole. (When their density is low, the quasiparticles are confined by the random potential generated by impurities and imperfections, hence the Hall resistance remains quantized for a finite range of $B$.) Context answer: \boxed{$\Delta B=\frac{h }{e W L}$} Extra Supplementary Reading Materials: (f) In Tsui et al. experiment, the magnetic field corresponding to the center of the quantized Hall plateau $$ R_{\mathrm{H}}=3 h / e^{2}, B_{1 / 3}=15 \text { Tesla, } $$ the effective mass of an electron in GaAs, $m^{*}=0.067 m_{e}$, the electron mass, $m_{e}=9.1 \times 10^{-31} \mathrm{~kg}$, Coulomb's constant, $k=9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$, the vacuum permittivity, $\varepsilon_{0}=1 / 4 \pi k=8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}$, the relative permittivity (the ratio of the permittivity of a substance to the vacuum permittivity) of GaAs, $\varepsilon_{r}=13$, the elementary charge, $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant, $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$, and Boltzmann's constant, $k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$. In our analysis, we have neglected several factors, whose corresponding energy scales, compared to $\Delta U(B)$ discussed in (d), are either too large to excite or too small to be relevant. Context question: (i) Calculate the thermal energy $E_{\text {th }}$ at temperature $T=1.0 \mathrm{~K}$. Context answer: \boxed{$1.38 \times 10^{-23}$} ","(ii) The electrons spatially confined in the whirlpools (or vortices) have a large kinetic energy. Using the uncertainty relation, estimate the order of magnitude of the kinetic energy. (This amount would also be the additional energy penalty if we put two electrons in the same whirlpool, instead of in two separate whirlpools, due to Pauli exclusion principle.)","['The size of a vortex is of order\n\n$$\nl_{0}=\\sqrt{\\frac{h}{e B}}=\\sqrt{\\frac{6.626 \\times 10^{-34}}{1.6 \\times 10^{-19} \\times 15}}=1.66 \\times 10^{-8} \\mathrm{~m}\n$$\n\nAccording to the uncertainty relation, $p \\sim \\Delta p \\sim h / l_{0}$. Therefore, the kinetic energy is\n\n$$\n\\begin{aligned}\n\\frac{p^{2}}{2 m^{*}} & =\\frac{h^{2}}{2 m^{*}} \\frac{e B}{h}=\\frac{h}{2} \\frac{e B}{m^{*}} \\\\\n& =\\frac{6.626 \\times 10^{-34} \\times 1.6 \\times 10^{-19} \\times 15}{2 \\times 0.067 \\times 9.1 \\times 10^{-31}} \\\\\n& =1.3 \\times 10^{-20} \\mathrm{~J}\n\\end{aligned}\n$$']",['$1.3 \\times 10^{-20}$'],False,J,Numerical,5e-22 1521,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} Context question: (d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$. Context answer: \boxed{$\frac{1}{2}$} Context question: (e) As the magnetic field deviates from the exact filling $v=1 / n$ to a higher field, more vortices (whirlpools in the electron sea) are being created. They are not bound to electrons and behave like particles carrying effectively positive charges, hence known as quasiholes, compared to the negatively charged electrons. The amount of charge deficit in any of these quasiholes amounts to exactly $1 / n$ of an electronic charge. An analogous argument can be made for magnetic fields slightly below $v$ and the creation of quasielectrons of negative charge $e^{*}=-e / n$. At the quantized Hall plateau of $R_{\mathrm{H}}=3 h / e^{2}$, calculate the amount of change in $B$ that corresponds to the introduction of exactly one fractionally charged quasihole. (When their density is low, the quasiparticles are confined by the random potential generated by impurities and imperfections, hence the Hall resistance remains quantized for a finite range of $B$.) Context answer: \boxed{$\Delta B=\frac{h }{e W L}$} Extra Supplementary Reading Materials: (f) In Tsui et al. experiment, the magnetic field corresponding to the center of the quantized Hall plateau $$ R_{\mathrm{H}}=3 h / e^{2}, B_{1 / 3}=15 \text { Tesla, } $$ the effective mass of an electron in GaAs, $m^{*}=0.067 m_{e}$, the electron mass, $m_{e}=9.1 \times 10^{-31} \mathrm{~kg}$, Coulomb's constant, $k=9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$, the vacuum permittivity, $\varepsilon_{0}=1 / 4 \pi k=8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}$, the relative permittivity (the ratio of the permittivity of a substance to the vacuum permittivity) of GaAs, $\varepsilon_{r}=13$, the elementary charge, $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant, $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$, and Boltzmann's constant, $k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$. In our analysis, we have neglected several factors, whose corresponding energy scales, compared to $\Delta U(B)$ discussed in (d), are either too large to excite or too small to be relevant. Context question: (i) Calculate the thermal energy $E_{\text {th }}$ at temperature $T=1.0 \mathrm{~K}$. Context answer: \boxed{$1.38 \times 10^{-23}$} Context question: (ii) The electrons spatially confined in the whirlpools (or vortices) have a large kinetic energy. Using the uncertainty relation, estimate the order of magnitude of the kinetic energy. (This amount would also be the additional energy penalty if we put two electrons in the same whirlpool, instead of in two separate whirlpools, due to Pauli exclusion principle.) Context answer: \boxed{$1.3 \times 10^{-20}$} Extra Supplementary Reading Materials: (g) There are also a series of plateaus at $R_{\mathrm{H}}=h / i e^{2}$, where $i=1,2,3, \ldots$ in Tsui et al. experiment, as shown in Figure 1(b). These plateaus, known as the integer quantum Hall effect (IQHE), were reported previously by K. von Klitzing in 1980. Repeating (c)-(f) for the integer plateaus, one realizes that the novelty of the FQHE lies critically in the existence of fractionally charged quasiparticles. R. de-Picciotto et al. and L. Saminadayar et al. independently reported the observation of fractional charges at the $v=1 / 3$ filling in 1997. In the experiments, they measured the noise in the charge current across a narrow constriction, the so-called quantum point contact (QPC). In a simple statistical model, carriers with discrete charge $e^{*}$ tunnel across the QPC and generate charge current $I_{\mathrm{B}}$ (on top of a trivial background). The number of the carriers $n_{\tau}$ arriving at the electrode during a sufficiently small time interval $\tau$ obeys Poisson probability distribution with parameter $\lambda$ $$ \mathrm{P}\left(n_{\tau}=k\right)=\frac{\lambda^{k} \mathrm{e}^{-\lambda}}{k !} $$ where $k$ ! is the factorial of $k$. You may need the following summation $$ \mathrm{e}^{\lambda}=\sum_{\mathrm{k}=0}^{\infty} \frac{\lambda^{k}}{k !} $$","(i) Determine the charge current $I_{\mathrm{B}}$, which measures total charge per unit of time, in terms of $\lambda$ and $\tau$.","['The current can be calculated by the ratio of the total charge carried by the averaged $n_{\\tau}$ quasiparticles to the time interval $\\tau$.\n\n$$\n\\begin{aligned}\n\\left\\langle n_{\\tau}\\right\\rangle=\\sum_{k=1}^{\\infty} k P(k) & =\\sum_{k=1}^{\\infty} \\frac{\\lambda^{k} e^{-\\lambda}}{(k-1) !} \\\\\n& =\\lambda \\sum_{k=0}^{\\infty} P(k) \\\\\n& =\\lambda\n\\end{aligned}\n$$\n\nwhere we have used $\\sum_{k} P(k)=1$. Therefore,\n\n$$\nI_{B}=\\frac{\\left\\langle n_{\\tau}\\right\\rangle e^{*}}{\\tau}=\\frac{\\nu e \\lambda}{\\tau}\n$$']",['$I_{B}=\\frac{\\nu e \\lambda}{\\tau}$'],False,,Expression, 1522,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} Context question: (d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$. Context answer: \boxed{$\frac{1}{2}$} Context question: (e) As the magnetic field deviates from the exact filling $v=1 / n$ to a higher field, more vortices (whirlpools in the electron sea) are being created. They are not bound to electrons and behave like particles carrying effectively positive charges, hence known as quasiholes, compared to the negatively charged electrons. The amount of charge deficit in any of these quasiholes amounts to exactly $1 / n$ of an electronic charge. An analogous argument can be made for magnetic fields slightly below $v$ and the creation of quasielectrons of negative charge $e^{*}=-e / n$. At the quantized Hall plateau of $R_{\mathrm{H}}=3 h / e^{2}$, calculate the amount of change in $B$ that corresponds to the introduction of exactly one fractionally charged quasihole. (When their density is low, the quasiparticles are confined by the random potential generated by impurities and imperfections, hence the Hall resistance remains quantized for a finite range of $B$.) Context answer: \boxed{$\Delta B=\frac{h }{e W L}$} Extra Supplementary Reading Materials: (f) In Tsui et al. experiment, the magnetic field corresponding to the center of the quantized Hall plateau $$ R_{\mathrm{H}}=3 h / e^{2}, B_{1 / 3}=15 \text { Tesla, } $$ the effective mass of an electron in GaAs, $m^{*}=0.067 m_{e}$, the electron mass, $m_{e}=9.1 \times 10^{-31} \mathrm{~kg}$, Coulomb's constant, $k=9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$, the vacuum permittivity, $\varepsilon_{0}=1 / 4 \pi k=8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}$, the relative permittivity (the ratio of the permittivity of a substance to the vacuum permittivity) of GaAs, $\varepsilon_{r}=13$, the elementary charge, $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant, $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$, and Boltzmann's constant, $k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$. In our analysis, we have neglected several factors, whose corresponding energy scales, compared to $\Delta U(B)$ discussed in (d), are either too large to excite or too small to be relevant. Context question: (i) Calculate the thermal energy $E_{\text {th }}$ at temperature $T=1.0 \mathrm{~K}$. Context answer: \boxed{$1.38 \times 10^{-23}$} Context question: (ii) The electrons spatially confined in the whirlpools (or vortices) have a large kinetic energy. Using the uncertainty relation, estimate the order of magnitude of the kinetic energy. (This amount would also be the additional energy penalty if we put two electrons in the same whirlpool, instead of in two separate whirlpools, due to Pauli exclusion principle.) Context answer: \boxed{$1.3 \times 10^{-20}$} Extra Supplementary Reading Materials: (g) There are also a series of plateaus at $R_{\mathrm{H}}=h / i e^{2}$, where $i=1,2,3, \ldots$ in Tsui et al. experiment, as shown in Figure 1(b). These plateaus, known as the integer quantum Hall effect (IQHE), were reported previously by K. von Klitzing in 1980. Repeating (c)-(f) for the integer plateaus, one realizes that the novelty of the FQHE lies critically in the existence of fractionally charged quasiparticles. R. de-Picciotto et al. and L. Saminadayar et al. independently reported the observation of fractional charges at the $v=1 / 3$ filling in 1997. In the experiments, they measured the noise in the charge current across a narrow constriction, the so-called quantum point contact (QPC). In a simple statistical model, carriers with discrete charge $e^{*}$ tunnel across the QPC and generate charge current $I_{\mathrm{B}}$ (on top of a trivial background). The number of the carriers $n_{\tau}$ arriving at the electrode during a sufficiently small time interval $\tau$ obeys Poisson probability distribution with parameter $\lambda$ $$ \mathrm{P}\left(n_{\tau}=k\right)=\frac{\lambda^{k} \mathrm{e}^{-\lambda}}{k !} $$ where $k$ ! is the factorial of $k$. You may need the following summation $$ \mathrm{e}^{\lambda}=\sum_{\mathrm{k}=0}^{\infty} \frac{\lambda^{k}}{k !} $$ Context question: (i) Determine the charge current $I_{\mathrm{B}}$, which measures total charge per unit of time, in terms of $\lambda$ and $\tau$. Context answer: \boxed{$I_{B}=\frac{\nu e \lambda}{\tau}$} ",(ii) Current noise is defined as the charge fluctuations per unit of time. One can analyze the noise by measuring the mean square deviation of the number of current-carrying charges. Determine the current noise $S_{I}$ due to the discreteness of the current-carrying charges in terms of $\lambda$ and $\tau$.,"['Similarly, the noise can be related to the averaged charge fluctuations during the time interval $\\tau$.\n\n$$\n\\begin{aligned}\n\\left\\langle\\left(n_{\\tau}-\\left\\langle n_{\\tau}\\right)^{2}\\right\\rangle\\right. & =\\left\\langle n_{\\tau}^{2}\\right\\rangle-\\left\\langle n_{\\tau}\\right\\rangle^{2} \\\\\n& =\\sum_{k=1}^{\\infty} k^{2} P(k)-\\lambda^{2} \\\\\n& =\\left[\\lambda^{2} \\sum_{k=0}^{\\infty} P(k)+\\sum_{k=1}^{\\infty} k P(k)\\right]-\\lambda^{2} \\\\\n& =\\lambda\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\nS_{I}=\\frac{\\left\\langle\\left(n_{\\tau}-\\left\\langle n_{\\tau}\\right)^{2}\\right\\rangle\\left(e^{*}\\right)^{2}\\right.}{\\tau}=\\frac{(\\nu e)^{2} \\lambda}{\\tau}\n$$']",['$S_{I}=\\frac{(\\nu e)^{2} \\lambda}{\\tau}$'],False,,Expression, 1523,Modern Physics,"The fractional quantum Hall effect (FQHE) was discovered by D. C. Tsui and $\mathrm{H}$. Stormer at Bell Labs in 1981. In the experiment electrons were confined in two dimensions on the GaAs side by the interface potential of a GaAs/AlGaAs heterojunction fabricated by A. C. Gossard (here we neglect the thickness of the two-dimensional electron layer). A strong uniform magnetic field $B$ was applied perpendicular to the two-dimensional electron system. As illustrated in Figure 1, when a current $I$ was passing through the sample, the voltage $V_{\mathrm{H}}$ across the current path exhibited an unexpected quantized plateau (corresponding to a Hall resistance $R_{\mathrm{H}}=$ $3 h / e^{2}$ ) at sufficiently low temperatures. The appearance of the plateau would imply the presence of fractionally charged quasiparticles in the system, which we analyze below. For simplicity, we neglect the scattering of the electrons by random potential, as well as the electron spin. Figure 1 (a) Figure 1 (b) Figure 1: (a) Sketch of the experimental setup for the observation of the FQHE. As indicated, a current $I$ is passing through a two-dimensional electron system in the longitudinal direction with an effective length $L$. The Hall voltage $V_{\mathrm{H}}$ is measured in the transverse direction with an effective width $W$. In addition, a uniform magnetic field $B$ is applied perpendicular to the plane. The direction of the current is given for illustrative purpose only, which may not be correct. (b) Hall resistance $R_{\mathrm{H}}$ versus $B$ at four different temperatures (curves shifted for clarity) in the original publication on the FQHE. The features at $R_{\mathrm{H}}=3 h / e^{2}$ are due to the FQHE. (a) In a classical model, two-dimensional electrons behave like charged billiard balls on a table. In the GaAs/AlGaAs sample, however, the mass of the electrons is reduced to an effective mass $m^{*}$ due to their interaction with ions. Context question: (i) Write down the equation of motion of an electron in perpendicular electric field $\vec{E}=-E_{y} \hat{y}$ and magnetic field $\vec{B}=B \hat{z}$. Context answer: $$ m^{*} \frac{d \vec{v}}{d t}=-e(\vec{v} \times \vec{B}+\vec{E}) $$ where $\vec{v}$ is the velocity of the electron. Context question: (ii) Determine the magnitude of the velocity $v_{\mathrm{s}}$ of the electrons in the stationary case. Context answer: \boxed{$E / B$.} Context question: (iii) Which direction is the velocity pointing at? Context answer: \boxed{-x} Context question: (b) The Hall resistance is defined as $R_{\mathrm{H}}=V_{\mathrm{H}} / I$. In the classical model, find $R_{\mathrm{H}}$ as a function of the number of the electrons $N$ and the magnetic flux $\phi=B A=B W L$, where $A$ is the area of the sample, and $W$ and $L$ the effective width and length of the sample, respectively. Context answer: \boxed{$R_{H}=\frac{1}{e} \frac{\phi}{N}$} Context question: (c) We know that electrons move in circular orbits in the magnetic field. In the quantum mechanical picture, the impinging magnetic field $B$ could be viewed as creating tiny whirlpools, so-called vortices, in the sea of electrons-one whirlpool for each flux quantum $h / e$ of the magnetic field, where $h$ is the Planck's constant and $e$ the elementary charge of an electron. For the case of $R_{\mathrm{H}}=3 h / e^{2}$, which was discovered by Tsui and Stormer, derive the ratio of the number of the electrons $N$ to the number of the flux quanta $N_{\phi}$, Context answer: \boxed{$\frac{1}{3}$} Context question: (d) It turns out that binding an integer number of vortices $(n>1)$ with each electron generates a bigger surrounding whirlpool, hence pushes away all other electrons. Therefore, the system can considerably reduce its electrostatic Coulomb energy at the corresponding filling factor. Determine the scaling exponent $\alpha$ of the amount of energy gain for each electron $\Delta U(B) \propto B^{\alpha}$. Context answer: \boxed{$\frac{1}{2}$} Context question: (e) As the magnetic field deviates from the exact filling $v=1 / n$ to a higher field, more vortices (whirlpools in the electron sea) are being created. They are not bound to electrons and behave like particles carrying effectively positive charges, hence known as quasiholes, compared to the negatively charged electrons. The amount of charge deficit in any of these quasiholes amounts to exactly $1 / n$ of an electronic charge. An analogous argument can be made for magnetic fields slightly below $v$ and the creation of quasielectrons of negative charge $e^{*}=-e / n$. At the quantized Hall plateau of $R_{\mathrm{H}}=3 h / e^{2}$, calculate the amount of change in $B$ that corresponds to the introduction of exactly one fractionally charged quasihole. (When their density is low, the quasiparticles are confined by the random potential generated by impurities and imperfections, hence the Hall resistance remains quantized for a finite range of $B$.) Context answer: \boxed{$\Delta B=\frac{h }{e W L}$} Extra Supplementary Reading Materials: (f) In Tsui et al. experiment, the magnetic field corresponding to the center of the quantized Hall plateau $$ R_{\mathrm{H}}=3 h / e^{2}, B_{1 / 3}=15 \text { Tesla, } $$ the effective mass of an electron in GaAs, $m^{*}=0.067 m_{e}$, the electron mass, $m_{e}=9.1 \times 10^{-31} \mathrm{~kg}$, Coulomb's constant, $k=9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$, the vacuum permittivity, $\varepsilon_{0}=1 / 4 \pi k=8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}$, the relative permittivity (the ratio of the permittivity of a substance to the vacuum permittivity) of GaAs, $\varepsilon_{r}=13$, the elementary charge, $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant, $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$, and Boltzmann's constant, $k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$. In our analysis, we have neglected several factors, whose corresponding energy scales, compared to $\Delta U(B)$ discussed in (d), are either too large to excite or too small to be relevant. Context question: (i) Calculate the thermal energy $E_{\text {th }}$ at temperature $T=1.0 \mathrm{~K}$. Context answer: \boxed{$1.38 \times 10^{-23}$} Context question: (ii) The electrons spatially confined in the whirlpools (or vortices) have a large kinetic energy. Using the uncertainty relation, estimate the order of magnitude of the kinetic energy. (This amount would also be the additional energy penalty if we put two electrons in the same whirlpool, instead of in two separate whirlpools, due to Pauli exclusion principle.) Context answer: \boxed{$1.3 \times 10^{-20}$} Extra Supplementary Reading Materials: (g) There are also a series of plateaus at $R_{\mathrm{H}}=h / i e^{2}$, where $i=1,2,3, \ldots$ in Tsui et al. experiment, as shown in Figure 1(b). These plateaus, known as the integer quantum Hall effect (IQHE), were reported previously by K. von Klitzing in 1980. Repeating (c)-(f) for the integer plateaus, one realizes that the novelty of the FQHE lies critically in the existence of fractionally charged quasiparticles. R. de-Picciotto et al. and L. Saminadayar et al. independently reported the observation of fractional charges at the $v=1 / 3$ filling in 1997. In the experiments, they measured the noise in the charge current across a narrow constriction, the so-called quantum point contact (QPC). In a simple statistical model, carriers with discrete charge $e^{*}$ tunnel across the QPC and generate charge current $I_{\mathrm{B}}$ (on top of a trivial background). The number of the carriers $n_{\tau}$ arriving at the electrode during a sufficiently small time interval $\tau$ obeys Poisson probability distribution with parameter $\lambda$ $$ \mathrm{P}\left(n_{\tau}=k\right)=\frac{\lambda^{k} \mathrm{e}^{-\lambda}}{k !} $$ where $k$ ! is the factorial of $k$. You may need the following summation $$ \mathrm{e}^{\lambda}=\sum_{\mathrm{k}=0}^{\infty} \frac{\lambda^{k}}{k !} $$ Context question: (i) Determine the charge current $I_{\mathrm{B}}$, which measures total charge per unit of time, in terms of $\lambda$ and $\tau$. Context answer: \boxed{$I_{B}=\frac{\nu e \lambda}{\tau}$} Context question: (ii) Current noise is defined as the charge fluctuations per unit of time. One can analyze the noise by measuring the mean square deviation of the number of current-carrying charges. Determine the current noise $S_{I}$ due to the discreteness of the current-carrying charges in terms of $\lambda$ and $\tau$. Context answer: \boxed{$S_{I}=\frac{(\nu e)^{2} \lambda}{\tau}$} ","(iii) Calculate the noise-to-current ratio $S_{I} / I_{\mathrm{B}}$, which was verified by R. de-Picciotto et al. and L. Saminadayar et al. in 1997. (One year later, Tsui and Stormer shared the Nobel Prize in Physics with R. B. Laughlin, who proposed an elegant ansatz for the ground state wave function at $v=1 / 3$.)",['The noise-to-current ratio $S_{I} / I_{B}=e^{*}=\\nu e$.'],['$\\nu e$'],False,,Expression, 1524,Electromagnetism,,"(i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. ![](https://cdn.mathpix.com/cropped/2023_12_21_4f93d3ceab09e952853dg-1.jpg?height=283&width=574&top_left_y=818&top_left_x=724) Figure 6","['The electron motion equation\n\n$$\nm \\frac{\\mathrm{d} \\vec{v}}{\\mathrm{~d} t}=-e \\vec{v} \\times \\vec{B}\n\\tag{1}\n$$\n\n(1) \n\nLet $\\vec{\\omega}_{c}=-e \\vec{B} / m$, we have\n\n$$\n\\frac{\\mathrm{d} \\vec{v}}{\\mathrm{~d} t}=-\\vec{\\omega}_{c} \\times \\vec{v}\n\\tag{2}\n$$\n\nSince $\\vec{B}=B \\hat{z}$, thus Equation (2) can be written to be\n\n$$\n\\left\\{\\begin{array}{c}\n\\frac{\\mathrm{d} v_{x}}{\\mathrm{~d} t}=\\omega_{c} v_{y} \\\\\n\\frac{\\mathrm{d} v_{y}}{\\mathrm{~d} t}=-\\omega_{c} v_{x} \\\\\n\\frac{\\mathrm{d} v_{z}}{\\mathrm{~d} t}=0\n\\end{array}\\right.\n\\tag{3}\n$$\n\nThe general solutions of Equation (3) are\n\n$$\n\\left\\{\\begin{array}{c}\nv_{x}=v_{\\perp} \\cos \\left(\\omega_{c} t+\\varphi_{y}\\right) \\\\\nv_{y}=v_{\\perp} \\cos \\left(\\omega_{c} t+\\varphi_{y}\\right) \\\\\nv_{z}=v_{z}(t=0)\n\\end{array}\\right.\n\\tag{4}\n$$\n\nwhere $\\varphi_{x, y}$ are the initial phases. Due to $v_{z}(t=0)=0$, it is indicated that the motion of an electron remains to be perpendicular to the magnetic field. By further solving Equation (4), we obtain the motion trajectory of the electron,\n\n$$\n\\left\\{\\begin{array}{c}\nx=r_{c} \\sin \\omega_{c} t \\\\\ny=r_{c} \\cos \\omega_{c} t \\\\\n\\quad z=0\n\\end{array}\\right.\n\\tag{5}\n$$\n\n\nor\n\n$$\n\\left\\{\\begin{array}{l}\nx^{2}+y^{2}=r_{c}^{2} \\\\\nz=0\n\\end{array}\\right.\n$$\n\nwhere $\\left|r_{c}\\right|=\\frac{v_{\\perp}}{\\left|\\omega_{c}\\right|}=\\frac{m v_{\\perp}}{e B}$ is the gyro radius.']",['$$\n\\left\\{\\begin{array}{l}\nx^{2}+y^{2}=r_{c}^{2} \\\\\nz=0\n\\end{array}\\right.\n$$\n\nwhere $\\left|r_{c}\\right|=\\frac{v_{\\perp}}{\\left|\\omega_{c}\\right|}=\\frac{m v_{\\perp}}{e B}$ is the gyro radius.'],False,,Need_human_evaluate, 1525,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. ","(ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current.","['The electric current generated by the circular motion of an electron are\n\n\n\n$$\nI=\\frac{e}{T}=\\frac{e \\omega_{c}}{2 \\pi}=\\frac{e^{2} B}{2 \\pi m}\n$$\n\n\n\nBased on the definition of the magnetic moment, we have\n\n$$\n\\vec{\\mu}=I \\vec{A}=-\\frac{e^{2} \\vec{B}}{2 \\pi m} \\times \\pi r_{c}^{2}=-\\frac{e^{2} \\vec{B}}{2 \\pi m} \\times \\frac{\\pi m^{2} v_{\\perp}^{2}}{e^{2} B^{2}}=-\\frac{m v_{\\perp}^{2}}{2 B^{2}} \\vec{B} 。\n$$']","['$I=\\frac{e^{2} B}{2 \\pi m}$ , $\\vec{\\mu}=-\\frac{m v_{\\perp}^{2}}{2 B^{2}} \\vec{B}$']",True,,Expression, 1526,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. Context question: (ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current. Context answer: \boxed{$I=\frac{e^{2} B}{2 \pi m}$ , $\vec{\mu}=-\frac{m v_{\perp}^{2}}{2 B^{2}} \vec{B}$} ","(iii) If the initial electron velocity $\vec{v}$ is not perpendicular to the uniform magnetic field, i.e, the angle $\theta$ between $\vec{B}$ and $\vec{v}$ is $0^{0}<\theta<90^{\circ}$, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory.","['In the case, we have $v_{z}=v \\cos \\theta, v_{\\perp}=v \\sin \\theta$. Since the velocity of the electron in $\\mathrm{z}$ is not zero, the solution (5) in (i) becomes\n\n$$\n\\begin{equation}\n\\left\\{\\begin{array} { l } \n{ x = r _ { c } \\operatorname { s i n } \\omega _ { c } t } \\\\\n{ y = r _ { c } \\operatorname { c o s } \\omega _ { c } t } \\\\\n{ z = v t \\operatorname { c o s } \\theta }\n\\end{array} \\text { , or } \\left\\{\\begin{array}{l}\nx^2+y^2=r_c^2 \\\\\nz=v t \\cos \\theta\n\\end{array}\\right.\\right. \\text {. }\n\\end{equation}\n\\tag{6}\n$$\n\nThe orbit equation (6) indicates that the electron has a spiral trajectory. The screw pitch is\n\n$$\nh=v_{z} T=v \\cos \\theta \\frac{2 \\pi}{\\omega_{c}}=2 \\pi \\frac{m v}{e B} \\cos \\theta\n$$']",['$2 \\pi \\frac{m v}{e B} \\cos \\theta$'],False,,Expression, 1527,Electromagnetism,,(i) Please explain the generation mechanism of the electric current by a schematic drawing.,"['Since the magnetic field and the plasma are uniform $z$, the orbits of ions and electrons can project into in the x-y plane. From the results of (a), we know that an ion has a left-hand circular motion and an electron has a right-hand circular motion. Due to the linear increase of the plasma density in $x$, the number of ions with upward motion is less than that with downward motion at a given $x$ position, which leads a net upward ion flow. Similarly, electrons have a net downward flow. Combining the ion and election flows, we have a net upward electric current as illustrated below in schematic drawing.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_337a5d5ddb5cdc22a0ddg-1.jpg?height=639&width=577&top_left_y=2005&top_left_x=334)']",['开放性回答'],False,,Need_human_evaluate, 1528,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. Context question: (ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current. Context answer: \boxed{$I=\frac{e^{2} B}{2 \pi m}$ , $\vec{\mu}=-\frac{m v_{\perp}^{2}}{2 B^{2}} \vec{B}$} Context question: (iii) If the initial electron velocity $\vec{v}$ is not perpendicular to the uniform magnetic field, i.e, the angle $\theta$ between $\vec{B}$ and $\vec{v}$ is $0^{0}<\theta<90^{\circ}$, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory. Context answer: \boxed{$2 \pi \frac{m v}{e B} \cos \theta$} Extra Supplementary Reading Materials: (b) In the uniform background magnetic field as shown in Figure 6, theplasma density isnonuniform in $x$. For simplicity, we assume that the temperature and the distribution of the ions andelectrons are the same. Thus, the plasma pressure can be expressed as $$ p(x)=k T\left[n_{i}(x)+n_{e}(x)\right]=2 k T n(x)=2 k T\left(n_{0}+\alpha x\right), $$ Where $B, T, k, n_{0}$, and $\alpha$ are positive constants, $n_{i}(x)$ and $n_{e}(x)$ are the number densities of the ions and electrons. Context question: (i) Please explain the generation mechanism of the electric current by a schematic drawing. Context answer: 开放性回答 ","(ii) If both the ions and electrons have a Maxwellian distribution, the ion distribution is $f_{i}\left(x, v_{\perp}, v_{\|}\right)=n_{i}(x)\left(\frac{m_{i}}{2 \pi k T}\right)^{3 / 2} e^{-m_{i}\left(v_{\perp}^{2}+v_{\|}^{2}\right) / 2 k T}$, please calculate the constant $\beta$ in the magnetization $M=\beta n(x) \frac{k T}{B}$, where the magnetization M is the magnetic moment per unit volume. (Hint: We have $\int_{0}^{\infty} x \exp (-x) d x=1$ and $\left.\int_{-\infty}^{\infty} \exp \left(-x^{2}\right) d x=\sqrt{\pi} \cdot\right)$","['Based on $\\vec{\\mu}=-\\frac{m v_{\\perp}{ }^{2}}{2 B^{2}} \\vec{B}$ from Problem (a), the total magnetic moments per unit volume (i.e., the magnetization) for ions and electrons are\n\n$$\n\\begin{aligned}\n& M_{i, e}=\\iiint f_{i, e}(x, v) \\mu_{i, e} \\mathrm{~d}^{3} \\vec{v} \\\\\n& =-\\int_{0}^{\\infty} \\int_{-\\infty}^{\\infty} n(x)\\left(\\frac{m_{i, e}}{2 \\pi k T}\\right)^{3 / 2} e^{-m_{i, e}\\left(v_{\\perp}^{2}+v_{\\square}^{2}\\right) / 2 k T} \\frac{m_{i, e} v_{\\perp}^{2}}{2 B} 2 \\pi v_{\\perp} \\mathrm{d} v_{\\perp} \\mathrm{d} v_{\\square} \\\\\n& =-n(x) \\int_{0}^{\\infty} \\frac{m_{i, e}}{2 \\pi k T} e^{-m_{i, e} v_{\\perp}^{2} / 2 k T} \\frac{m_{i, e} v_{\\perp}^{2}}{2 B} 2 \\pi v_{\\perp} \\mathrm{d} v_{\\perp} \\mathrm{d} v_{\\square} \\int_{-\\infty}^{\\infty}\\left(\\frac{m_{i, e}}{2 \\pi k T}\\right)^{1 / 2} e^{-m_{i, e} v_{\\square}^{2} / 2 k T} \\mathrm{~d} v_{\\square} \\\\\n& =-\\frac{k T}{B} n(x)\n\\end{aligned}\n$$\n\nThe total magnetization for a plasma:\n\n$$\nM=M_{i}+M_{e}=-\\frac{2 k T}{B} n(x)=-\\frac{p(x)}{B}\n$$\n\nTherefore, we have\n\n$$\n\\beta=-2 .\n$$']",['$\\beta=-2$'],False,,Numerical,1e-8 1529,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. Context question: (ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current. Context answer: \boxed{$I=\frac{e^{2} B}{2 \pi m}$ , $\vec{\mu}=-\frac{m v_{\perp}^{2}}{2 B^{2}} \vec{B}$} Context question: (iii) If the initial electron velocity $\vec{v}$ is not perpendicular to the uniform magnetic field, i.e, the angle $\theta$ between $\vec{B}$ and $\vec{v}$ is $0^{0}<\theta<90^{\circ}$, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory. Context answer: \boxed{$2 \pi \frac{m v}{e B} \cos \theta$} Extra Supplementary Reading Materials: (b) In the uniform background magnetic field as shown in Figure 6, theplasma density isnonuniform in $x$. For simplicity, we assume that the temperature and the distribution of the ions andelectrons are the same. Thus, the plasma pressure can be expressed as $$ p(x)=k T\left[n_{i}(x)+n_{e}(x)\right]=2 k T n(x)=2 k T\left(n_{0}+\alpha x\right), $$ Where $B, T, k, n_{0}$, and $\alpha$ are positive constants, $n_{i}(x)$ and $n_{e}(x)$ are the number densities of the ions and electrons. Context question: (i) Please explain the generation mechanism of the electric current by a schematic drawing. Context answer: 开放性回答 Context question: (ii) If both the ions and electrons have a Maxwellian distribution, the ion distribution is $f_{i}\left(x, v_{\perp}, v_{\|}\right)=n_{i}(x)\left(\frac{m_{i}}{2 \pi k T}\right)^{3 / 2} e^{-m_{i}\left(v_{\perp}^{2}+v_{\|}^{2}\right) / 2 k T}$, please calculate the constant $\beta$ in the magnetization $M=\beta n(x) \frac{k T}{B}$, where the magnetization M is the magnetic moment per unit volume. (Hint: We have $\int_{0}^{\infty} x \exp (-x) d x=1$ and $\left.\int_{-\infty}^{\infty} \exp \left(-x^{2}\right) d x=\sqrt{\pi} \cdot\right)$ Context answer: \boxed{$\beta=-2$} ","(c) Now let's go back to the Earth's dipole magnetic field. Please apply the result from Question(b) to calculate the ratio of the diamagnetic field and the Earth's dipole magnetic field in Equation (1) at the position $\left(x=10 \mathrm{R}_{\mathrm{E}}, y=0, z=1 \mathrm{R}_{\mathrm{E}}\right)$. The plasma pressure is assumed to be $p(z)=p_{0} e^{-(z / a)^{2}}$, where $p_{0}=3 \times 10^{-10} \mathrm{pa}$ and $a=2 R_{E}$. The magnetic field around this position is also assumed to be uniform. Be aware of the difference in the coordinate systemsin Questions (b) and (c). (Hint: The diamagnetic field is given by $B_{m x}=\mu_{o} M$.)","[""Using the equation for the magnetization field $B_{m x}=\\mu_{o} M$ and the $\\mathrm{x}$ component of the Earth's dipole magnetic field at $\\left(x=10 R_{E}, y=0, z=1 R_{E}\\right)$, we have\n\n$$\n\\left|B_{m x} / B_{d}\\right|=\\left.\\mu_{0} \\frac{p(z)}{B_{d}^{2}\\left(10 R_{E}, 0, z\\right)}\\right|_{z=R_{E}}=\\frac{\\mu_{0} e^{-0.25} P_{0}}{28.5 \\times 10^{-8} B_{0}^{2}} \\approx 1.0\n$$""]",['1.0'],False,,Numerical,1e-1 1530,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. Context question: (ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current. Context answer: \boxed{$I=\frac{e^{2} B}{2 \pi m}$ , $\vec{\mu}=-\frac{m v_{\perp}^{2}}{2 B^{2}} \vec{B}$} Context question: (iii) If the initial electron velocity $\vec{v}$ is not perpendicular to the uniform magnetic field, i.e, the angle $\theta$ between $\vec{B}$ and $\vec{v}$ is $0^{0}<\theta<90^{\circ}$, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory. Context answer: \boxed{$2 \pi \frac{m v}{e B} \cos \theta$} Extra Supplementary Reading Materials: (b) In the uniform background magnetic field as shown in Figure 6, theplasma density isnonuniform in $x$. For simplicity, we assume that the temperature and the distribution of the ions andelectrons are the same. Thus, the plasma pressure can be expressed as $$ p(x)=k T\left[n_{i}(x)+n_{e}(x)\right]=2 k T n(x)=2 k T\left(n_{0}+\alpha x\right), $$ Where $B, T, k, n_{0}$, and $\alpha$ are positive constants, $n_{i}(x)$ and $n_{e}(x)$ are the number densities of the ions and electrons. Context question: (i) Please explain the generation mechanism of the electric current by a schematic drawing. Context answer: 开放性回答 Context question: (ii) If both the ions and electrons have a Maxwellian distribution, the ion distribution is $f_{i}\left(x, v_{\perp}, v_{\|}\right)=n_{i}(x)\left(\frac{m_{i}}{2 \pi k T}\right)^{3 / 2} e^{-m_{i}\left(v_{\perp}^{2}+v_{\|}^{2}\right) / 2 k T}$, please calculate the constant $\beta$ in the magnetization $M=\beta n(x) \frac{k T}{B}$, where the magnetization M is the magnetic moment per unit volume. (Hint: We have $\int_{0}^{\infty} x \exp (-x) d x=1$ and $\left.\int_{-\infty}^{\infty} \exp \left(-x^{2}\right) d x=\sqrt{\pi} \cdot\right)$ Context answer: \boxed{$\beta=-2$} Context question: (c) Now let's go back to the Earth's dipole magnetic field. Please apply the result from Question(b) to calculate the ratio of the diamagnetic field and the Earth's dipole magnetic field in Equation (1) at the position $\left(x=10 \mathrm{R}_{\mathrm{E}}, y=0, z=1 \mathrm{R}_{\mathrm{E}}\right)$. The plasma pressure is assumed to be $p(z)=p_{0} e^{-(z / a)^{2}}$, where $p_{0}=3 \times 10^{-10} \mathrm{pa}$ and $a=2 R_{E}$. The magnetic field around this position is also assumed to be uniform. Be aware of the difference in the coordinate systemsin Questions (b) and (c). (Hint: The diamagnetic field is given by $B_{m x}=\mu_{o} M$.) Context answer: \boxed{1.0} Extra Supplementary Reading Materials: (d) From Figures 2, 3, and 5, it can be clearly seen that the Earth's magnetic fieldstrength along a magnetic field line is the largest at the poles and the smallest in the equatorial plane.Since the Earth's dipole magnetic field is axially symmetric and slowly varying along a magnetic field line, it can for simplicitybe treated as a magnetic-mirror field as shown in Figure 7.The magnetic field strength along a magnetic field line is the smallest $\left(B_{0}\right)$ at the point "" $\mathrm{P}_{2}$ "" and the largest $\left(B_{m}\right)$ at the points "" $\mathrm{P}_{1}$ "" and "" $\mathrm{P}_{3}$ "". An electron with an initial velocity $\vec{v}$ is located at the point "" $\mathrm{P}_{2}$ "" and drifts towards the point "" $\mathrm{P}_{3}$ "". The angle between the initial velocity $\vec{v}$ and the magnetic field at the point $"" \mathrm{P}_{2} ""$ is $0^{0}<\theta<90^{\circ}$. For the magnetic-mirror field $\vec{B}=B_{r} \hat{r}+B_{z} \hat{z}$ (with $B_{r}< Figure 7","(i) Please give the gyro-averaged magnetic-field force along the magnetic field lines on an electron and show that the magnetic moment is a motion constant, i.e., $\frac{d \mu}{d t}=0$, based on the law of the total kinetic energy conservation.","[""Under the cylindrical coordinate system, the motion of an electron can project into the $r$ and $\\theta$ plane, the Lorentz force is $F_{z}=e v_{\\theta} B_{r}$.\n\nBy taking a gyro average, the averaged Lorentz force becomes\n\n$$\n\\left\\langle F_{z}\\right\\rangle=\\frac{1}{2} e v_{\\perp} r_{c} \\frac{d B}{d z}=-\\mu \\frac{d B}{d s}\n$$\n\nSince $\\frac{d B}{d z}=\\frac{d B}{d s}$, we have\n\n$$\n\\left\\langle F_{\\|}\\right\\rangle=\\left\\langle F_{z}\\right\\rangle=-\\mu \\frac{d B}{d s} .\n$$\n\n\nThe acceleration of the guiding center of the electron can be obtained by the Newton's law, i.e.,\n\n$$\nm \\frac{\\mathrm{d} v_{\\| /}}{\\mathrm{d} t}=-\\mu \\frac{d B}{d s}\n\\tag{7}\n$$\n\nwhere $v_{\\|}=\\mathrm{d} s / \\mathrm{d} t$. Equation (7) can be written to be\n\n$$\n\\frac{\\mathrm{d}}{\\mathrm{d} t}\\left(\\frac{1}{2} m v_{\\|}^{2}\\right)=-\\mu \\frac{\\mathrm{d} B}{\\mathrm{~d} t}\n\\tag{8}\n$$\n\nSince the total kinetic energy is conserved, i.e., $\\frac{\\mathrm{d}}{\\mathrm{d} t}\\left(\\frac{1}{2} m v_{/ /}^{2}+\\frac{1}{2} m v_{\\perp}^{2}\\right)=0$, we can obtain\n\n$$\n\\mu \\frac{\\mathrm{d} B}{\\mathrm{~d} t}=-\\frac{\\mathrm{d}}{\\mathrm{d} t}\\left(\\frac{1}{2} m v_{\\| / 2}^{2}\\right)=\\frac{\\mathrm{d}}{\\mathrm{d} t}\\left(\\frac{1}{2} m v_{\\perp}^{2}\\right)=\\frac{\\mathrm{d}}{\\mathrm{d} t}(\\mu B)=\\mu \\frac{\\mathrm{d} B}{\\mathrm{~d} t}+B \\frac{\\mathrm{d} \\mu}{\\mathrm{d} t}\n\\tag{9}\n$$\n\nCombining Equations (8) and (9), we get\n\n$$\n\\frac{\\mathrm{d} \\mu}{\\mathrm{d} t}=0\n\\tag{10}\n$$""]",['$-\\mu \\frac{d B}{d s}$'],False,,Expression, 1531,Electromagnetism,,"(ii) Based on the motion constant of magnetic moment, please determine what the condition should be satisfied for the angle $\theta$ between the initial electron velocity $\vec{v}$ and the magnetic field at the point "" $\mathrm{P}_{2}$ ""ifan electron will not escape from the magnetic mirror field.","['From the motion constant of the magnetic moment of an electron, the perpendicular velocity increases with increase of the magnetic field, which means the parallel velocity decreases due to the conservation of the total kinetic energy. When the electron arrives at the point "" $\\mathrm{P}_{3}$ "", its parallel velocity decreases to zero, then the electron will not escape from the magnetic mirror field. Thus, the initial velocityshould be\n\n$$\n\\frac{v_{\\perp 0}^{2}}{B_{0}} \\geq \\frac{v^{2}}{B_{m}}\n$$\n\nSince $v_{\\perp 0}=v \\sin \\theta$, we obtain\n\n$$\n\\theta_{c r} \\geq \\arcsin \\left(\\sqrt{\\frac{B_{0}}{B_{m}}}\\right)\n$$\n\ni.e., this is the condition for the electron confined in the magnetic mirror field.']",['$\\theta_{c r} \\geq \\arcsin \\left(\\sqrt{\\frac{B_{0}}{B_{m}}}\\right)$'],False,,Need_human_evaluate, 1532,Electromagnetism,,"(e) Earth's dipole magnetic field lines (blue lines) are shown in Figure 8. The spiral trajectory of a charged particle (red curve) is assumed to be confined in the $y=0$ plane since the gradient and the curvature of the magnetic field can be ignored. If a charged particle with the mass $m$, charge $q$, and velocity $\vec{v}$ is initially located at the equatorial position $\left[x=6 \mathrm{R}_{\mathrm{E}}, y=0, z=0\right]$ and the angle between the electron velocity $\vec{v}$ and the magnetic field is $\theta$ initially, please determine what the condition should be satisfied for $\theta$ if the charged particle arrives below $200 \mathrm{~km}$ of its altitude at the latitude $60^{\circ}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ae61c9b6b1a575bacd50g-1.jpg?height=545&width=908&top_left_y=2103&top_left_x=583) Figure 8","[""Since the guiding center of the electron motion is always confined in the $\\mathrm{y}=0$ plane, the Earth's dipole magnetic field $\\vec{B}=\\frac{B_{0} R_{E}^{3}}{r^{5}}\\left(-3 x z \\hat{x}+\\left(x^{2}-2 z^{2}\\right) \\hat{z}\\right)$. At the initial position $\\left(6 R_{E}, 0,0\\right)$ of an electron, the magnetic field strength\n\n\n\n$$\nB_{i}=\\frac{B_{0} R_{E}^{3}}{r^{5}}\\left(x^{2}\\right)=\\frac{B_{0}}{216}\n$$\n\n\n\nAt the altitude $\\mathrm{H}=200 \\mathrm{~km}$ and the latitude $\\theta_{L}=60^{\\circ}$, thus\n\n$$\nB_{L}=\\left(B_{x}^{2}+B_{z}^{2}\\right)^{1 / 2}=\\frac{\\sqrt{13}}{2} B_{0}\n$$\n\nWith the same argument in (d) (ii), the condition for the electron to arrive in below the altitude $\\mathrm{H}=200 \\mathrm{~km}$ with the latitude $\\theta_{L}=60^{\\circ}$ is\n\n$$\n\\theta<0.05=2.8^{\\circ} .\n$$""]",['$\\theta<2.8^{\\circ}$'],False,,Need_human_evaluate, 1533,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. Context question: (ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current. Context answer: \boxed{$I=\frac{e^{2} B}{2 \pi m}$ , $\vec{\mu}=-\frac{m v_{\perp}^{2}}{2 B^{2}} \vec{B}$} Context question: (iii) If the initial electron velocity $\vec{v}$ is not perpendicular to the uniform magnetic field, i.e, the angle $\theta$ between $\vec{B}$ and $\vec{v}$ is $0^{0}<\theta<90^{\circ}$, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory. Context answer: \boxed{$2 \pi \frac{m v}{e B} \cos \theta$} Extra Supplementary Reading Materials: (b) In the uniform background magnetic field as shown in Figure 6, theplasma density isnonuniform in $x$. For simplicity, we assume that the temperature and the distribution of the ions andelectrons are the same. Thus, the plasma pressure can be expressed as $$ p(x)=k T\left[n_{i}(x)+n_{e}(x)\right]=2 k T n(x)=2 k T\left(n_{0}+\alpha x\right), $$ Where $B, T, k, n_{0}$, and $\alpha$ are positive constants, $n_{i}(x)$ and $n_{e}(x)$ are the number densities of the ions and electrons. Context question: (i) Please explain the generation mechanism of the electric current by a schematic drawing. Context answer: 开放性回答 Context question: (ii) If both the ions and electrons have a Maxwellian distribution, the ion distribution is $f_{i}\left(x, v_{\perp}, v_{\|}\right)=n_{i}(x)\left(\frac{m_{i}}{2 \pi k T}\right)^{3 / 2} e^{-m_{i}\left(v_{\perp}^{2}+v_{\|}^{2}\right) / 2 k T}$, please calculate the constant $\beta$ in the magnetization $M=\beta n(x) \frac{k T}{B}$, where the magnetization M is the magnetic moment per unit volume. (Hint: We have $\int_{0}^{\infty} x \exp (-x) d x=1$ and $\left.\int_{-\infty}^{\infty} \exp \left(-x^{2}\right) d x=\sqrt{\pi} \cdot\right)$ Context answer: \boxed{$\beta=-2$} Context question: (c) Now let's go back to the Earth's dipole magnetic field. Please apply the result from Question(b) to calculate the ratio of the diamagnetic field and the Earth's dipole magnetic field in Equation (1) at the position $\left(x=10 \mathrm{R}_{\mathrm{E}}, y=0, z=1 \mathrm{R}_{\mathrm{E}}\right)$. The plasma pressure is assumed to be $p(z)=p_{0} e^{-(z / a)^{2}}$, where $p_{0}=3 \times 10^{-10} \mathrm{pa}$ and $a=2 R_{E}$. The magnetic field around this position is also assumed to be uniform. Be aware of the difference in the coordinate systemsin Questions (b) and (c). (Hint: The diamagnetic field is given by $B_{m x}=\mu_{o} M$.) Context answer: \boxed{1.0} Extra Supplementary Reading Materials: (d) From Figures 2, 3, and 5, it can be clearly seen that the Earth's magnetic fieldstrength along a magnetic field line is the largest at the poles and the smallest in the equatorial plane.Since the Earth's dipole magnetic field is axially symmetric and slowly varying along a magnetic field line, it can for simplicitybe treated as a magnetic-mirror field as shown in Figure 7.The magnetic field strength along a magnetic field line is the smallest $\left(B_{0}\right)$ at the point "" $\mathrm{P}_{2}$ "" and the largest $\left(B_{m}\right)$ at the points "" $\mathrm{P}_{1}$ "" and "" $\mathrm{P}_{3}$ "". An electron with an initial velocity $\vec{v}$ is located at the point "" $\mathrm{P}_{2}$ "" and drifts towards the point "" $\mathrm{P}_{3}$ "". The angle between the initial velocity $\vec{v}$ and the magnetic field at the point $"" \mathrm{P}_{2} ""$ is $0^{0}<\theta<90^{\circ}$. For the magnetic-mirror field $\vec{B}=B_{r} \hat{r}+B_{z} \hat{z}$ (with $B_{r}< Figure 7 Context question: (i) Please give the gyro-averaged magnetic-field force along the magnetic field lines on an electron and show that the magnetic moment is a motion constant, i.e., $\frac{d \mu}{d t}=0$, based on the law of the total kinetic energy conservation. Context answer: \boxed{$-\mu \frac{d B}{d s}$} Context question: (ii) Based on the motion constant of magnetic moment, please determine what the condition should be satisfied for the angle $\theta$ between the initial electron velocity $\vec{v}$ and the magnetic field at the point "" $\mathrm{P}_{2}$ ""ifan electron will not escape from the magnetic mirror field. Context answer: $\theta_{c r} \geq \arcsin \left(\sqrt{\frac{B_{0}}{B_{m}}}\right)$ Context question: (e) Earth's dipole magnetic field lines (blue lines) are shown in Figure 8. The spiral trajectory of a charged particle (red curve) is assumed to be confined in the $y=0$ plane since the gradient and the curvature of the magnetic field can be ignored. If a charged particle with the mass $m$, charge $q$, and velocity $\vec{v}$ is initially located at the equatorial position $\left[x=6 \mathrm{R}_{\mathrm{E}}, y=0, z=0\right]$ and the angle between the electron velocity $\vec{v}$ and the magnetic field is $\theta$ initially, please determine what the condition should be satisfied for $\theta$ if the charged particle arrives below $200 \mathrm{~km}$ of its altitude at the latitude $60^{\circ}$. Figure 8 Context answer: $\theta<2.8^{\circ}$ Extra Supplementary Reading Materials: (f) As shown in Figure 5, when magnetic reconnection takes place at the dayside magnetopause, reconnected magnetic field lines drift towards the nightside region because the solar wind flows tailward. Thus, some solar wind electrons in the region ""A"" also move towards the magnetotail in the region "" $\mathrm{P}$ "". After the electrons arrive in the region "" $\mathrm{P}$ "", some electrons can be accelerated to around $1 \mathrm{keV}$. If energetic electrons drift down to the thermosphere (The altitude of the thermosphere is about $85 \mathrm{~km}-800 \mathrm{~km}$.), energetic electrons can collide with the neutral atoms, which could cause the neutral atoms to jump into excited states. A photon is emitted when the higher excited state of aneutral atom returnsto its lower excited state or ground state. Splendid aurora (Figure 1) is generated in the aurora oval due to photons with different wavelengths. It is found that the aurorais mainly resulted fromphotonsemittedby oxygen atoms. The energy levels in the first and second excited states relative to the ground stateare $1.96 \mathrm{eV}$ and $4.17 \mathrm{eV}$, respectively. The lifetimes of the two excited states of an oxygen atomare 110s and 0.8s as shown in Figure 9. Figure 9","(i) Please give the atmospheric density as a function of the altitude and the ratio of the oxygen density at the altitudes $\mathrm{H}=160 \mathrm{~km}$ and $\mathrm{H}=220 \mathrm{~km}$. For simplicity, we assume that the atmospheric temperature is independent of the altitude and the air is an ideal gas. ( $\rho_{0} g / P_{0}=0.13 / \mathrm{km}$, where $\rho_{0}$ and $P_{0}$ are the atmospheric density and pressure at sea level.)","['\n\nBased on the balance between the gravitational force and the force from the difference of the atmospheric pressure, we obtain\n\n$$\nA[p(r+d r)-p(r)]=A d p=-\\frac{G M(\\rho A d r)}{r^{2}}\n$$\n\nor\n\n$$\nd p=-\\frac{G M(\\rho d r)}{r^{2}}\n\\tag{11}\n$$\n\nFor the ideal gas, we have\n\n$$\np=\\frac{N k T}{V}=\\frac{N m k T}{V m}=\\frac{\\rho k T}{m} .\n\\tag{12}\n$$\n\nSince the atmospheric temperature is constant, Equation (12) can written to be\n\n\n\n$$\np=\\frac{p_{0}}{\\rho_{0}} \\rho\n\\tag{13}\n$$\n\nInserting Equation (13) into (11), we have\n\n$$\n\\frac{d \\rho}{\\rho}=-\\frac{\\rho_{0} G M d r}{p_{0}}\n\\tag{14}\n$$\n\nBy integrating Equation (14), we obtain\n\n$$\n\\rho=\\rho_{0} e^{\\frac{\\rho_{0} G M}{P_{0}}\\left(\\frac{1}{r}-\\frac{1}{R_{E}}\\right)}=\\rho_{0} e^{-\\frac{\\rho_{0} g R_{E}}{P_{0}}\\left(1-\\frac{R_{E}}{r}\\right)} \\approx \\rho_{0} e^{-\\frac{\\rho_{0} g H}{P_{0}}} .\n\\tag{15}\n$$\n\nThus, the ration of the atmospheric density at the altitudes $\\mathrm{H}=160 \\mathrm{~km}$ and $\\mathrm{H}=220 \\mathrm{~km}$ is\n\n$$\n\\frac{\\rho\\left(H_{1}=160 \\mathrm{~km}\\right)}{\\rho\\left(H_{2}=220 \\mathrm{~km}\\right)} \\approx e^{\\frac{\\rho_{0} g\\left(H_{2}-H_{1}\\right)}{P_{0}}} \\approx 2.44 \\times 10^{3} .\n\\tag{16}\n$$']","['$\\rho_{0} e^{-\\frac{\\rho_{0} g R_{E}}{P_{0}}\\left(1-\\frac{R_{E}}{r}\\right)} $', '$ \\rho_{0} e^{-\\frac{\\rho_{0} g H}{P_{0}}}$ , $2.44 \\times 10^{3}$']",True,,"Expression,Numerical",",5e1" 1534,Electromagnetism,,"(ii) Please give the colors of auroras at the altitudes $\mathrm{H}=160 \mathrm{~km}$ and $\mathrm{H}=220 \mathrm{~km}$. (Hint: The dependence of the collision frequency of atmosphericmolecules on the atmospheric density is $v=v_{0} \rho / \rho_{0}$, where $v_{0} \approx 10^{9} / \mathrm{s}$ is the collision frequency of atmosphericmolecules at sea level.The excited oxygen atom will lose a part of its energy when it collides with other neutral molecules. )","['At the altitude $H=160 \\mathrm{~km}$, the collision frequency of atmosphericmolecules is\n\n$$\nv=\\frac{v_{0} \\rho}{\\rho_{0}}=v_{0} e^{-\\frac{\\rho_{0} g H}{P_{0}}} \\approx 0.93 / s\n\\tag{17}\n$$\n\n\n\nSince the lifetime of the oxygen atom in the first excited state lasts about 110s, the oxygen atom will collide with other molecules over one hundred times. Thus, the high-frequent collision between particles can de-excite oxygen atoms before they have a chance to radiate. But, the oxygen atoms in the second excited state will emit photons since their lifetime is so short. Thus, aurora at $\\mathrm{H}=160 \\mathrm{~km}$ is resulted from emission of the oxygen atoms in the second excited state. The wavelength is\n\n$$\n\\lambda=\\frac{h c}{e \\Delta V}=\\frac{\\left(6.626 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)\\left(2.998 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)}{\\left(1.6 \\times 10^{-19} \\mathrm{~J} / \\mathrm{eV}\\right)(2.21 \\mathrm{eV})}=562 \\mathrm{~nm} .\n$$\n\nThe color for this wavelength is green.\n\n\n\nAt the altitude $H=220 \\mathrm{~km}$, the collision frequency of atmosphericmolecules is\n\n$$\nv=\\frac{v_{0} \\rho}{\\rho_{0}}=v_{0} e^{-\\frac{\\rho_{0} g H}{P_{0}}} \\approx 3.8 \\times 10^{-4} / \\mathrm{s}\n\\tag{18}\n$$\n\nAll oxygen atoms in the first and second excited states have a chance to radiate since the collision frequency is so low. Because the number of the oxygen atoms in the first excited state is much larger than in the second excited state, we observe the aurora color is from the emission of the oxygen atoms in the first excited state. The wavelength is\n\n$$\n\\lambda=\\frac{h c}{e \\Delta V}=\\frac{\\left(6.626 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)\\left(2.998 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)}{\\left(1.6 \\times 10^{-19} \\mathrm{~J} / \\mathrm{eV}\\right)(1.96 \\mathrm{eV})}=633 \\mathrm{~nm}\n$$\n\n\n\nThe color for this wavelength is red.']","['when H=160km, the color is green\nwhen H=220km, the color is red']",True,,Need_human_evaluate, 1535,Electromagnetism," Figure 1 The following questions are designed to guide you to find the answer one step by one step. Background information of the interaction between the solar wind and the Earth's magnetic field It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, whichis a high-speed stream of hot plasma. (The plasma is the quasi-neutralionized gas.)The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3. Figure 2 Figure 3 The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock andfront of the Earth's magnetic field is called themagnetosheath. The region surroundedby the solar wind is called the magnetosphere.The Earth's magnetic field largely prevents the solar wind from enteringthe magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles ofthe solar wind in the region ""A"" to move into the magnetotail ""P"" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.Therefore, it isimportant to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field. Figure 4 Figure 5 Numerical values of physical constants and the Earth's dipole magnetic field: Speed of light in vacuum: $c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$; Permittivity in vacuum: $\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)$; Permeability in vacuum: $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~N} / \mathrm{A}^{2}$; Charge of a proton: $e=1.6 \times 10^{-19} \mathrm{C}$; Mass of an electron: $m=9.1 \times 10^{-31} \mathrm{~kg}$; Mass of a proton: $m_{p}=1.67 \times 10^{-27} \mathrm{~kg}$; Boltzmann's constant: $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$; Gravitational acceleration: $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$; Planck's constant: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$ Earth's radius $R_{E}=6.4 \times 10^{6} \mathrm{~m}$. The Earth's dipole magnetic field can be expressed as $$ \vec{B}_{d}=\frac{B_{0} R_{E}^{3}}{r^{5}}\left[3 x z \hat{x}-3 y z \hat{y}+\left(x^{2}+y^{2}-2 z^{2}\right) \hat{z}\right] \quad,\left(r \geq R_{E}\right) \tag{1} $$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}, B_{0}=3.1 \times 10^{-5} \mathrm{~T}$, and $\hat{x}, \hat{y}, \hat{z}$ are the unit vectors in the $x, y, z$ directions, respectively. Questions: (a) Context question: (i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field $\vec{B}$. When the initial electron velocity $\vec{v}$ is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at $(\mathrm{x}, \mathrm{y}, \mathrm{z})=(0,0,0)$. Figure 6 Context answer: $$ \left\{\begin{array}{l} x^{2}+y^{2}=r_{c}^{2} \\ z=0 \end{array}\right. $$ where $\left|r_{c}\right|=\frac{v_{\perp}}{\left|\omega_{c}\right|}=\frac{m v_{\perp}}{e B}$ is the gyro radius. Context question: (ii) Please determine the electric current of the electron motion and calculate the magnetic moment $\vec{\mu}=I \vec{A}$, where $\stackrel{\mathrm{I}}{A}$ is the area of the electron circular orbit and the direction of $\vec{A}$ is determined by the right-hand rule of theelectric current. Context answer: \boxed{$I=\frac{e^{2} B}{2 \pi m}$ , $\vec{\mu}=-\frac{m v_{\perp}^{2}}{2 B^{2}} \vec{B}$} Context question: (iii) If the initial electron velocity $\vec{v}$ is not perpendicular to the uniform magnetic field, i.e, the angle $\theta$ between $\vec{B}$ and $\vec{v}$ is $0^{0}<\theta<90^{\circ}$, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory. Context answer: \boxed{$2 \pi \frac{m v}{e B} \cos \theta$} Extra Supplementary Reading Materials: (b) In the uniform background magnetic field as shown in Figure 6, theplasma density isnonuniform in $x$. For simplicity, we assume that the temperature and the distribution of the ions andelectrons are the same. Thus, the plasma pressure can be expressed as $$ p(x)=k T\left[n_{i}(x)+n_{e}(x)\right]=2 k T n(x)=2 k T\left(n_{0}+\alpha x\right), $$ Where $B, T, k, n_{0}$, and $\alpha$ are positive constants, $n_{i}(x)$ and $n_{e}(x)$ are the number densities of the ions and electrons. Context question: (i) Please explain the generation mechanism of the electric current by a schematic drawing. Context answer: 开放性回答 Context question: (ii) If both the ions and electrons have a Maxwellian distribution, the ion distribution is $f_{i}\left(x, v_{\perp}, v_{\|}\right)=n_{i}(x)\left(\frac{m_{i}}{2 \pi k T}\right)^{3 / 2} e^{-m_{i}\left(v_{\perp}^{2}+v_{\|}^{2}\right) / 2 k T}$, please calculate the constant $\beta$ in the magnetization $M=\beta n(x) \frac{k T}{B}$, where the magnetization M is the magnetic moment per unit volume. (Hint: We have $\int_{0}^{\infty} x \exp (-x) d x=1$ and $\left.\int_{-\infty}^{\infty} \exp \left(-x^{2}\right) d x=\sqrt{\pi} \cdot\right)$ Context answer: \boxed{$\beta=-2$} Context question: (c) Now let's go back to the Earth's dipole magnetic field. Please apply the result from Question(b) to calculate the ratio of the diamagnetic field and the Earth's dipole magnetic field in Equation (1) at the position $\left(x=10 \mathrm{R}_{\mathrm{E}}, y=0, z=1 \mathrm{R}_{\mathrm{E}}\right)$. The plasma pressure is assumed to be $p(z)=p_{0} e^{-(z / a)^{2}}$, where $p_{0}=3 \times 10^{-10} \mathrm{pa}$ and $a=2 R_{E}$. The magnetic field around this position is also assumed to be uniform. Be aware of the difference in the coordinate systemsin Questions (b) and (c). (Hint: The diamagnetic field is given by $B_{m x}=\mu_{o} M$.) Context answer: \boxed{1.0} Extra Supplementary Reading Materials: (d) From Figures 2, 3, and 5, it can be clearly seen that the Earth's magnetic fieldstrength along a magnetic field line is the largest at the poles and the smallest in the equatorial plane.Since the Earth's dipole magnetic field is axially symmetric and slowly varying along a magnetic field line, it can for simplicitybe treated as a magnetic-mirror field as shown in Figure 7.The magnetic field strength along a magnetic field line is the smallest $\left(B_{0}\right)$ at the point "" $\mathrm{P}_{2}$ "" and the largest $\left(B_{m}\right)$ at the points "" $\mathrm{P}_{1}$ "" and "" $\mathrm{P}_{3}$ "". An electron with an initial velocity $\vec{v}$ is located at the point "" $\mathrm{P}_{2}$ "" and drifts towards the point "" $\mathrm{P}_{3}$ "". The angle between the initial velocity $\vec{v}$ and the magnetic field at the point $"" \mathrm{P}_{2} ""$ is $0^{0}<\theta<90^{\circ}$. For the magnetic-mirror field $\vec{B}=B_{r} \hat{r}+B_{z} \hat{z}$ (with $B_{r}< Figure 7 Context question: (i) Please give the gyro-averaged magnetic-field force along the magnetic field lines on an electron and show that the magnetic moment is a motion constant, i.e., $\frac{d \mu}{d t}=0$, based on the law of the total kinetic energy conservation. Context answer: \boxed{$-\mu \frac{d B}{d s}$} Context question: (ii) Based on the motion constant of magnetic moment, please determine what the condition should be satisfied for the angle $\theta$ between the initial electron velocity $\vec{v}$ and the magnetic field at the point "" $\mathrm{P}_{2}$ ""ifan electron will not escape from the magnetic mirror field. Context answer: $\theta_{c r} \geq \arcsin \left(\sqrt{\frac{B_{0}}{B_{m}}}\right)$ Context question: (e) Earth's dipole magnetic field lines (blue lines) are shown in Figure 8. The spiral trajectory of a charged particle (red curve) is assumed to be confined in the $y=0$ plane since the gradient and the curvature of the magnetic field can be ignored. If a charged particle with the mass $m$, charge $q$, and velocity $\vec{v}$ is initially located at the equatorial position $\left[x=6 \mathrm{R}_{\mathrm{E}}, y=0, z=0\right]$ and the angle between the electron velocity $\vec{v}$ and the magnetic field is $\theta$ initially, please determine what the condition should be satisfied for $\theta$ if the charged particle arrives below $200 \mathrm{~km}$ of its altitude at the latitude $60^{\circ}$. Figure 8 Context answer: $\theta<2.8^{\circ}$ Extra Supplementary Reading Materials: (f) As shown in Figure 5, when magnetic reconnection takes place at the dayside magnetopause, reconnected magnetic field lines drift towards the nightside region because the solar wind flows tailward. Thus, some solar wind electrons in the region ""A"" also move towards the magnetotail in the region "" $\mathrm{P}$ "". After the electrons arrive in the region "" $\mathrm{P}$ "", some electrons can be accelerated to around $1 \mathrm{keV}$. If energetic electrons drift down to the thermosphere (The altitude of the thermosphere is about $85 \mathrm{~km}-800 \mathrm{~km}$.), energetic electrons can collide with the neutral atoms, which could cause the neutral atoms to jump into excited states. A photon is emitted when the higher excited state of aneutral atom returnsto its lower excited state or ground state. Splendid aurora (Figure 1) is generated in the aurora oval due to photons with different wavelengths. It is found that the aurorais mainly resulted fromphotonsemittedby oxygen atoms. The energy levels in the first and second excited states relative to the ground stateare $1.96 \mathrm{eV}$ and $4.17 \mathrm{eV}$, respectively. The lifetimes of the two excited states of an oxygen atomare 110s and 0.8s as shown in Figure 9. Figure 9 Context question: (i) Please give the atmospheric density as a function of the altitude and the ratio of the oxygen density at the altitudes $\mathrm{H}=160 \mathrm{~km}$ and $\mathrm{H}=220 \mathrm{~km}$. For simplicity, we assume that the atmospheric temperature is independent of the altitude and the air is an ideal gas. ( $\rho_{0} g / P_{0}=0.13 / \mathrm{km}$, where $\rho_{0}$ and $P_{0}$ are the atmospheric density and pressure at sea level.) Context answer: \boxed{$\rho_{0} e^{-\frac{\rho_{0} g R_{E}}{P_{0}}\left(1-\frac{R_{E}}{r}\right)} $} Context question: (ii) Please give the colors of auroras at the altitudes $\mathrm{H}=160 \mathrm{~km}$ and $\mathrm{H}=220 \mathrm{~km}$. (Hint: The dependence of the collision frequency of atmosphericmolecules on the atmospheric density is $v=v_{0} \rho / \rho_{0}$, where $v_{0} \approx 10^{9} / \mathrm{s}$ is the collision frequency of atmosphericmolecules at sea level.The excited oxygen atom will lose a part of its energy when it collides with other neutral molecules. ) Context answer: when H=160km, the color is green when H=220km, the color is red ","(g) As mentioned above, a powerful solar wind can push the dayside magnetopause tovery close to the Earth, which could cause a high-orbit satelliteto be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite.For simplicity, the Earth's dipole magnetic field is assumed to remain unchanged when the solar wind compresses it and that the plasma density is ignorable in the magnetosphere. Please give the minimum solar wind speed to cause a damage of a geosynchronous satellite if the magnetic field strength and the plasma density of the solar wind are $B_{s}=5 \times 10^{-9} \mathrm{~T}$ and $\rho_{\mathrm{s}}=50$ proton $/ \mathrm{cm}^{3}$, respectively. (Hint: The force per unit area associated with the magnetic field is $f=B^{2} / 2 \mu_{0}$. We only consider the variation in $x$ for all physical quantities, i.e., the physical quantities are independent of $y$ and $z$.","[""At the geosynchronous orbit, we have\n\n$$\nG \\frac{m M}{\\left(h+R_{E}\\right)^{2}}=m \\frac{v^{2}}{h+R_{E}}=m\\left(h+R_{E}\\right) \\omega^{2}\n$$\n\nor\n\n$$\n\\left(h+R_{E}\\right)^{3}=\\frac{G M}{\\omega^{2}}=\\frac{g R_{E}^{2}}{\\omega^{2}}=\\frac{g}{\\omega^{2} R_{E}} R_{E}^{3}\n$$\n\nUsing $\\omega=2 \\pi /(24 \\times 3600 s)$, we have\n\n$$\nh=5.6 R_{E}\n$$\n\nUsing $\\Delta w=F \\Delta x=\\Delta E_{k}$, we have\n\n$$\n\\left(f_{m}-f_{s}\\right) \\Delta s \\Delta x=\\frac{1}{2} \\Delta m v_{s}^{2} .\n$$\n\nor\n\n$$\n\\frac{1}{2} \\rho v_{s}^{2}+\\frac{B_{s}^{2}}{2 \\mu_{0}}=\\frac{B_{m}^{2}}{2 \\mu_{0}}\n\\tag{19}\n$$\n\nwhere $\\rho$ is the mass density. At the position $x=R_{E}+h=6.6 R_{E}, y=z=0$, the Earth's dipole magnetic field strength is\n\n$$\nB_{d}=\\frac{B_{0}}{290} \\sim 100 n T\n$$\n\nFrom Equation (19), we obtain\n\n$$\nv=330 \\mathrm{~km} / \\mathrm{s}\n$$\n\nWith this speed of the solar wind, the position of the dayside magetopause is $6.6 \\mathrm{R}_{\\mathrm{E}}$ at the geosynchronous orbit. Thus, a geosynchronous satellite could be damaged by the solar wind.""]",['330'],False,km/s,Numerical,5e0 1536,Optics,,"(a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane.","['The transmittivity of the F-P etalon is given by:\n\n$$\nT=\\frac{1}{1+F \\sin ^{2} \\frac{\\delta}{2}}\n$$\n\nFor bright fringes, we have\n\n$$\n\\begin{aligned}\n& T=1 \\text { i.e. } \\sin ^{2} \\frac{\\delta}{2}=0 \\\\\n& \\frac{\\delta}{2}=m \\pi \\\\\n& 2 n t \\cos \\theta=m \\lambda\n\\end{aligned}\n$$\n\nFor $\\mathrm{n}=1.0, \\mathrm{t}=1 \\mathrm{~cm}, \\lambda=589.6 \\mathrm{~nm}$, thus:\n\n$$\n\\cos \\theta_{i}=\\frac{m_{i}}{2 n t / \\lambda}=\\frac{m_{i}}{33921.3} \\quad \\text { (a1)}\n$$\n\nBecause of $\\cos \\theta \\leq 1$, so the orders of the first three fringes are:\n\n$$\nm_{1}=33921, m_{2}=33920, m_{3}=33919 \\quad \\text { (a2) }\n$$\n\nThe incident angles of the first three fringes are:\n\n$$\n\\theta_{1}=0.241^{0}, \\theta_{2}=0.502^{0}, \\theta_{3}=0.667^{0} \\quad \\text { (a3) }\n$$\n\nThe fringe diameter is given by:\n\n$$\nD_{i}=2 f \\tan \\theta_{i} \\approx 2 f \\theta_{i} \\quad \\text { (a4) }\n$$\n\nFor the focal length $\\mathrm{f}=30 \\mathrm{~cm}$, thus:\n\n$$\nD_{1}=2.52 \\mathrm{~mm}, D_{2}=5.26 \\mathrm{~mm}, D_{3}=6.99 \\mathrm{~mm} \\text { (a5) }\n$$']","['$m_{1}=33921, m_{2}=33920, m_{3}=33919 \\quad \\text { (a2) }\n$\\theta_{1}=0.241^{0}, \\theta_{2}=0.502^{0}, \\theta_{3}=0.667^{0} \\quad \\text { (a3) }$\n$D_{1}=2.52 \\mathrm{~mm}, D_{2}=5.26 \\mathrm{~mm}, D_{3}=6.99 \\mathrm{~mm} \\text { (a5) }$']",True,,Need_human_evaluate, 1537,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2",(i) the width $\varepsilon$ of the spectral line.,"['The half maximum occurs at:\n\n$$\n\\delta=2 m \\pi \\pm \\frac{\\varepsilon}{2} \\text { (b1) }\n$$\n\nGiven that $T=0.5$, thus:\n\n$$\nF \\sin ^{2} \\frac{\\delta}{2}=1 \\quad \\text { (b2) }\n$$\n\n$\\varepsilon=\\frac{4}{\\sqrt{F}}=\\frac{2(1-R)}{\\sqrt{R}}=\\frac{2(1-0.9)}{\\sqrt{0.9}}=0.21 \\mathrm{rad}$( or 12.03degree) (b3)']",['0.21'],False,rad,Numerical,1e-2 1538,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} ",(ii) the resolution $\lambda / \Delta \lambda$ of the etalon.,"['The phase shift $\\delta$ is given by:\n\n$$\n\\delta=\\frac{4 \\pi n t \\cos \\theta}{\\lambda}\n$$\nFor a small $\\Delta \\lambda$, thus:\n\n$$\n\\Delta \\delta=-\\frac{4 \\pi n t \\cos \\theta}{\\lambda^{2}} \\Delta \\lambda \\quad \\text { (b4) }\n$$\n\nFor $\\Delta \\delta=\\varepsilon$ and $\\lambda=589.6 \\mathrm{~nm}$, we get:\n\n$\\frac{\\lambda}{\\Delta \\lambda}=\\frac{\\pi n t \\sqrt{F} \\cos \\theta}{\\lambda}=\\frac{3.14 \\times 1.0 \\times 1.0 \\times 10^{-2} \\times \\sqrt{360} \\times 1.0}{589.6 \\times 10^{-9}}=1.01 \\times 10^{6}$ (b5)']",['$1.01 \\times 10^{6}$'],False,,Numerical,1e4 1539,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} Context question: (ii) the resolution $\lambda / \Delta \lambda$ of the etalon. Context answer: \boxed{$1.01 \times 10^{6}$} ","(c) As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure.","['From Question (a), we know that the order of the 1st fringe near the center of ring patterns is $m=33921$ at the vacuum case $(\\mathrm{n}=1.0)$. When the air pressure reaches the standard atmospheric pressure, the order of the 1 st fringe becomes $\\mathrm{m}+10$, so we have:\n\n$n_{\\text {air }}=\\frac{m+10}{2 t / \\lambda}=\\frac{33931}{33921}=1.00029$. (c1)']",['$1.00029$'],False,,Numerical,3e-4 1540,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} Context question: (ii) the resolution $\lambda / \Delta \lambda$ of the etalon. Context answer: \boxed{$1.01 \times 10^{6}$} Context question: (c) As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure. Context answer: \boxed{$1.00029$} ","(d) Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$. Figure 3","['The frequency of D1 line $\\left({ }^{2} \\mathrm{P}_{1 / 2}\\right.$ to $\\left.{ }^{2} \\mathrm{~S}_{1 / 2}\\right)$ is given by: $v_{0}=c / \\lambda(\\lambda=589.6 \\mathrm{~nm})$\n\nWhen magnetic field $\\mathrm{B}$ is applied, the frequency of the line a,b,c,d are expressed as:\n\n1) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=-1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=1 / 2)$ : frequency of (a)): $v_{a}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$; \n2) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=1 / 2)$ : frequency of $(\\mathrm{b}): v_{b}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$; \n3) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=-1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=-1 / 2)$ : frequency of (c): $v_{c}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$; \n4) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=-1 / 2)$ : frequency of $(\\mathrm{d}): v_{d}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$;']","['$v_{a}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$,$v_{b}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$,$v_{c}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$,$v_{d}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$']",True,,Expression, 1540,Optics,,"(d) Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$. ![](https://cdn.mathpix.com/cropped/2023_12_21_aeb204c8d13bb4047ac3g-1.jpg?height=525&width=640&top_left_y=1085&top_left_x=731) Figure 3","['The frequency of D1 line $\\left({ }^{2} \\mathrm{P}_{1 / 2}\\right.$ to $\\left.{ }^{2} \\mathrm{~S}_{1 / 2}\\right)$ is given by: $v_{0}=c / \\lambda(\\lambda=589.6 \\mathrm{~nm})$\n\nWhen magnetic field $\\mathrm{B}$ is applied, the frequency of the line a,b,c,d are expressed as:\n\n1) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=-1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=1 / 2)$ : frequency of (a)): $v_{a}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$; \n2) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=1 / 2)$ : frequency of $(\\mathrm{b}): v_{b}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$; \n3) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=-1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=-1 / 2)$ : frequency of (c): $v_{c}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$; \n4) ${ }^{2} \\mathrm{P}_{1 / 2}(\\mathrm{mj}=1 / 2) \\rightarrow{ }^{2} \\mathrm{~S}_{1 / 2}(\\mathrm{mj}=-1 / 2)$ : frequency of $(\\mathrm{d}): v_{d}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$;']","['$v_{a}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$,$v_{b}=v_{0}-\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$,$v_{c}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{2}-\\Delta E_{1}\\right)$,$v_{d}=v_{0}+\\frac{1}{2 h}\\left(\\Delta E_{1}+\\Delta E_{2}\\right)$']",True,,Expression, 1541,Optics,,"(e) As shown in Fig. 4, when the magnetic field is turned on, each fringe of the D1 line will split into four sub-fringes $(1,2,3$, and 4$)$. The diameter of the four sub-fringes near the center is measured as $D_{1}, D_{2}, D_{3}$, and $D_{4}$. Please give the expression of the splitting energy gap $\Delta \mathrm{E}_{1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_aeb204c8d13bb4047ac3g-1.jpg?height=539&width=808&top_left_y=2055&top_left_x=630) Figure 4","['$\\quad \\theta_{m}<<1, \\cos \\theta_{m}=1-\\frac{\\theta_{m}^{2}}{2}$, (e1)\n\n\n\n$2 n t \\cos \\theta_{m}=m \\lambda, \\quad 1-\\frac{\\theta_{m}^{2}}{2}=\\frac{m \\lambda}{2 n t}$, (e2)\n\n\n\n$\\lambda \\rightarrow \\lambda+\\Delta \\lambda, \\theta_{m} \\rightarrow \\theta_{m}^{\\prime}$\n\n$1-\\frac{\\theta_{m}^{\\prime 2}}{2}=\\frac{m(\\lambda+\\Delta \\lambda)}{2 n t}$,\n\n$\\frac{\\theta_{m}^{2}-\\theta_{m}^{\\prime 2}}{2}=\\frac{m \\Delta \\lambda}{2 n t}$ (e3)\n\n\n\n$2 f \\theta_{m}=D_{m}, \\quad \\frac{D_{m}^{2}-D_{m}^{\\prime 2}}{8 f^{2}}=\\frac{m \\Delta \\lambda}{2 n t}=\\frac{\\Delta \\lambda}{\\lambda}$\n\n$\\Delta \\lambda=\\lambda \\frac{D_{m}^{2}-D_{m}^{\\prime 2}}{8 f^{2}}$ (e4) \n\n\n\nThe lines a, b, c, and d correspond to sub-fringe 1, 2, 3, and 4. From Question (d), we have.\n\nThe wavelength difference of the spectral line $a$ and $b$ is given by:\n\n$\\Delta \\lambda_{1}=\\lambda \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$\n\n\n\n$\\Delta E_{1}=h\\left(v_{b}-v_{a}\\right), \\Delta E_{2}=h\\left(v_{d}-v_{b}\\right)$\n\nor $\\Delta E_{1}=h\\left(v_{d}-v_{c}\\right), \\Delta E_{2}=h\\left(v_{c}-v_{a}\\right)$ (e5)\n\nThe wavelength difference of the spectral line $a$ and $b$ is given by:\n\n$\\Delta \\lambda_{1}=\\lambda \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$\n\nThen we obtain\n\n$\\Delta E_{1}=h \\Delta v_{1}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$\n\n( or $\\Delta E_{1}=h \\Delta v_{1}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}$ ) (e6)\n\nSimilarly, for $\\Delta \\mathrm{E}_{2}$, we get\n\n$\\Delta \\lambda_{2}=\\lambda \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$\n\n$\\Delta E_{2}=h \\Delta v_{2}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$\n\n( or $\\Delta E_{1}=h \\Delta v_{1}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}$ ) (e7)']","['$\\Delta E_{1}=\\frac{h c}{\\lambda} \\cdot \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$,$\\Delta E_{2}=\\frac{h c}{\\lambda} \\cdot \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$']",True,,Expression, 1541,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} Context question: (ii) the resolution $\lambda / \Delta \lambda$ of the etalon. Context answer: \boxed{$1.01 \times 10^{6}$} Context question: (c) As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure. Context answer: \boxed{$1.00029$} Context question: (d) Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$. Figure 3 Context answer: \boxed{$v_{a}=v_{0}-\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$,$v_{b}=v_{0}-\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{c}=v_{0}+\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{d}=v_{0}+\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$} ","(e) As shown in Fig. 4, when the magnetic field is turned on, each fringe of the D1 line will split into four sub-fringes $(1,2,3$, and 4$)$. The diameter of the four sub-fringes near the center is measured as $D_{1}, D_{2}, D_{3}$, and $D_{4}$. Please give the expression of the splitting energy gap $\Delta \mathrm{E}_{1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. Figure 4","['$\\quad \\theta_{m}<<1, \\cos \\theta_{m}=1-\\frac{\\theta_{m}^{2}}{2}$, (e1)\n\n\n\n$2 n t \\cos \\theta_{m}=m \\lambda, \\quad 1-\\frac{\\theta_{m}^{2}}{2}=\\frac{m \\lambda}{2 n t}$, (e2)\n\n\n\n$\\lambda \\rightarrow \\lambda+\\Delta \\lambda, \\theta_{m} \\rightarrow \\theta_{m}^{\\prime}$\n\n$1-\\frac{\\theta_{m}^{\\prime 2}}{2}=\\frac{m(\\lambda+\\Delta \\lambda)}{2 n t}$,\n\n$\\frac{\\theta_{m}^{2}-\\theta_{m}^{\\prime 2}}{2}=\\frac{m \\Delta \\lambda}{2 n t}$ (e3)\n\n\n\n$2 f \\theta_{m}=D_{m}, \\quad \\frac{D_{m}^{2}-D_{m}^{\\prime 2}}{8 f^{2}}=\\frac{m \\Delta \\lambda}{2 n t}=\\frac{\\Delta \\lambda}{\\lambda}$\n\n$\\Delta \\lambda=\\lambda \\frac{D_{m}^{2}-D_{m}^{\\prime 2}}{8 f^{2}}$ (e4) \n\n\n\nThe lines a, b, c, and d correspond to sub-fringe 1, 2, 3, and 4. From Question (d), we have.\n\nThe wavelength difference of the spectral line $a$ and $b$ is given by:\n\n$\\Delta \\lambda_{1}=\\lambda \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$\n\n\n\n$\\Delta E_{1}=h\\left(v_{b}-v_{a}\\right), \\Delta E_{2}=h\\left(v_{d}-v_{b}\\right)$\n\nor $\\Delta E_{1}=h\\left(v_{d}-v_{c}\\right), \\Delta E_{2}=h\\left(v_{c}-v_{a}\\right)$ (e5)\n\nThe wavelength difference of the spectral line $a$ and $b$ is given by:\n\n$\\Delta \\lambda_{1}=\\lambda \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$\n\nThen we obtain\n\n$\\Delta E_{1}=h \\Delta v_{1}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$\n\n( or $\\Delta E_{1}=h \\Delta v_{1}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}$ ) (e6)\n\nSimilarly, for $\\Delta \\mathrm{E}_{2}$, we get\n\n$\\Delta \\lambda_{2}=\\lambda \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$\n\n$\\Delta E_{2}=h \\Delta v_{2}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$\n\n( or $\\Delta E_{1}=h \\Delta v_{1}=\\left|-\\frac{h c}{\\lambda} \\bullet \\frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}\\right|=\\frac{h c}{\\lambda} \\bullet \\frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}$ ) (e7)']","['$\\Delta E_{1}=\\frac{h c}{\\lambda} \\cdot \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$,$\\Delta E_{2}=\\frac{h c}{\\lambda} \\cdot \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$']",True,,Expression, 1542,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} Context question: (ii) the resolution $\lambda / \Delta \lambda$ of the etalon. Context answer: \boxed{$1.01 \times 10^{6}$} Context question: (c) As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure. Context answer: \boxed{$1.00029$} Context question: (d) Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$. Figure 3 Context answer: \boxed{$v_{a}=v_{0}-\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$,$v_{b}=v_{0}-\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{c}=v_{0}+\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{d}=v_{0}+\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$} Context question: (e) As shown in Fig. 4, when the magnetic field is turned on, each fringe of the D1 line will split into four sub-fringes $(1,2,3$, and 4$)$. The diameter of the four sub-fringes near the center is measured as $D_{1}, D_{2}, D_{3}$, and $D_{4}$. Please give the expression of the splitting energy gap $\Delta \mathrm{E}_{1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. Figure 4 Context answer: \boxed{$\Delta E_{1}=\frac{h c}{\lambda} \cdot \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$,$\Delta E_{2}=\frac{h c}{\lambda} \cdot \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$} ","(f) For the magnetic field $B=0.1 \mathrm{~T}$, the diameter of four sub-fringes is measured as: $D_{1}=3.88 \mathrm{~mm}, D_{2}=4.05 \mathrm{~mm}, D_{3}=4.35 \mathrm{~mm}$, and $D_{4}=4.51 \mathrm{~mm}$. Please calculate the Landé factor $\mathrm{g}_{\mathrm{k} 1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\mathrm{g}_{\mathrm{k} 2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$.","['Given that $\\mathrm{B}=0.1 \\mathrm{~T}$, so we have:\n\n$\\mu_{B} B=\\frac{h e B}{4 \\pi m_{e}}=\\frac{6.626 \\times 10^{-34} \\times 0.1}{4 \\times 3.14 \\times 9.1 \\times 10^{-31}}=5.79 \\times 10^{-6} \\mathrm{eV} \\quad$ (f1)\n\n$\\Delta E_{1}=g_{k 1} \\mu_{b} B=\\frac{h c}{\\lambda} \\bullet \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}} ;$\n\n(or, $\\Delta E_{1}=g_{k 1} \\mu_{b} B=\\frac{h c}{\\lambda} \\bullet \\frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}$ )\n\nFor the D1 spectral line, $\\lambda=589.6 \\mathrm{~nm}$, so we can get:\n\n$$\n\\frac{h c}{\\lambda}=\\frac{6.626 \\times 10^{-34} \\times 3 \\times 10^{8}}{5.896 \\times 10^{-7} \\times 1.6 \\times 10^{-19}}=2.11 \\mathrm{eV}, \\quad \\text { (f3)}\n$$\n\nthus:\n\n$g_{k 1}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{\\left(4.05 \\times 10^{-3}\\right)^{2}-\\left(3.88 \\times 10^{-3}\\right)^{2}}{8 \\times 0.3 \\times 0.3}=0.68 ;$\n\n(or $g_{k 1}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{D_{4}^{2}-D_{3}^{2}}{8 f^{2}}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{\\left(4.51 \\times 10^{-3}\\right)^{2}-\\left(4.35 \\times 10^{-3}\\right)^{2}}{8 \\times 0.3 \\times 0.3}=0.72$ )\n\nSimilarly, we get:\n\n$g_{k 2}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}=\\frac{2.11}{5.79 \\times 10^{-6}} \\cdot \\frac{\\left(4.51 \\times 10^{-3}\\right)^{2}-\\left(4.05 \\times 10^{-3}\\right)^{2}}{8 \\times 0.3 \\times 0.3}=1.99 $\n\n( or $g_{k 2}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{D_{3}^{2}-D_{1}^{2}}{8 f^{2}}=\\frac{2.11}{5.79 \\times 10^{-6}} \\bullet \\frac{\\left(4.35 \\times 10^{-3}\\right)^{2}-\\left(3.88 \\times 10^{-3}\\right)^{2}}{8 \\times 0.3 \\times 0.3}=1.95$ )']","['0.68, 1.99']",True,,Numerical,5e-2 1543,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} Context question: (ii) the resolution $\lambda / \Delta \lambda$ of the etalon. Context answer: \boxed{$1.01 \times 10^{6}$} Context question: (c) As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure. Context answer: \boxed{$1.00029$} Context question: (d) Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$. Figure 3 Context answer: \boxed{$v_{a}=v_{0}-\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$,$v_{b}=v_{0}-\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{c}=v_{0}+\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{d}=v_{0}+\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$} Context question: (e) As shown in Fig. 4, when the magnetic field is turned on, each fringe of the D1 line will split into four sub-fringes $(1,2,3$, and 4$)$. The diameter of the four sub-fringes near the center is measured as $D_{1}, D_{2}, D_{3}$, and $D_{4}$. Please give the expression of the splitting energy gap $\Delta \mathrm{E}_{1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. Figure 4 Context answer: \boxed{$\Delta E_{1}=\frac{h c}{\lambda} \cdot \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$,$\Delta E_{2}=\frac{h c}{\lambda} \cdot \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$} Context question: (f) For the magnetic field $B=0.1 \mathrm{~T}$, the diameter of four sub-fringes is measured as: $D_{1}=3.88 \mathrm{~mm}, D_{2}=4.05 \mathrm{~mm}, D_{3}=4.35 \mathrm{~mm}$, and $D_{4}=4.51 \mathrm{~mm}$. Please calculate the Landé factor $\mathrm{g}_{\mathrm{k} 1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\mathrm{g}_{\mathrm{k} 2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. Context answer: \boxed{0.68, 1.99} ","(g) The magnetic field on the sun can be determined by measuring the Zeeman effect of the Sodium D1 line on some special regions of the sun. One observes that, in the four split lines, the wavelength difference between the shortest and longest wavelength is $0.012 \mathrm{~nm}$ by a solar spectrograph. What is the magnetic field B in this region of the sun?",['We have $\\Delta E_{1}=g_{k 1} \\mu_{B} B$ and $\\Delta E_{2}=g_{k 2} \\mu_{B} B$;\n\nThe line a has the longest wavelength and the line $\\mathrm{d}$ has the shortest wavelength line. The energy difference of the line $a$ and $d$ is\n\n$$\n\\begin{aligned}\n& \\Delta E=\\Delta E_{1}+\\Delta E_{2}=\\left(g_{k 1}+g_{k 2}\\right) \\mu_{B} B .(g1) \\\\\n& \\Delta v=\\left|-\\frac{c \\Delta \\lambda}{\\lambda^{2}}\\right|=\\frac{c \\Delta \\lambda}{\\lambda^{2}} \\quad \\text { (g2) } \\\\\n& \\Delta v=\\frac{\\left(g_{k 1}+g_{k 2}\\right) \\mu_{B} B}{h} \\quad \\text { (g3) } \\\\\n& \\mu_{B}=\\frac{h e}{4 \\pi m_{e}}\n\\end{aligned}\n$$\n\nSo the magnetic field $\\mathrm{B}$ is given by:\n\n$$\n\\begin{aligned}\n& B=\\frac{4 \\pi m_{e} \\Delta \\lambda c}{\\lambda^{2}\\left(g_{k 1}+g_{k 2}\\right) e} \\\\\n& =\\frac{4 \\times 3.14 \\times 9.1 \\times 10^{-31} \\times 0.012 \\times 10^{-9} \\times 3 \\times 10^{8}}{\\left(589.6 \\times 10^{-9}\\right)^{2} \\times 2.67 \\times 1.6 \\times 10^{-19}} T \\\\\n& = 0.2772 \\mathrm{~T}\\\\\n&=2772.1 Gauss\n\\end{aligned}\n$$'],['0.2772'],False,T,Numerical,3e-3 1544,Optics,"Figure 1 shows a Fabry-Perot (F-P) etalon, in which air pressure is tunable. The F-P etalon consists of two glass plates with high-reflectivity inner surfaces. The two plates form a cavity in which light can be reflected back and forth. The outer surfaces of the plates are generally not parallel to the inner ones and do not affect the back-and-forth reflection. The air density in the etalon can be controlled. Light from a Sodium lamp is collimated by the lens L1 and then passes through the F-P etalon. The transmitivity of the etalon is given by $T=\frac{1}{1+F \sin ^{2}(\delta / 2)}$, where $F=\frac{4 R}{(1-R)^{2}}, \mathrm{R}$ is the reflectivity of the inner surfaces, $\delta=\frac{4 \pi n t \cos \theta}{\lambda}$ is the phase shift of two neighboring rays, $\mathrm{n}$ is the refractive index of the gas, $\mathrm{t}$ is the spacing of inner surfaces, $\theta$ is the incident angle, and $\lambda$ is the light wavelength. Figure 1 The Sodium lamp emits D1 $(\lambda=589.6 \mathrm{~nm})$ and D2 $(589 \mathrm{~nm})$ spectral lines and is located in a tunable uniform magnetic field. For simplicity, an optical filter F1 is assumed to only allow the D1 line to pass through. The D1 line is then collimated to the F-P etalon by the lens L1. Circular interference fringes will be present on the focal plane of the lens L2 with a focal length $\mathrm{f}=30 \mathrm{~cm}$. Different fringes have the different incident angle $\theta$. A microscope is used to observe the fringes. We take the reflectivity $\mathrm{R}=90 \%$ and the inner-surface spacing $\mathrm{t}=1 \mathrm{~cm}$. Some physical constants: $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}, e=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}, c=3.0 \times 10^{8} \mathrm{~ms}^{-1}$. Context question: (a) The D1 line $(\lambda=589.6 \mathrm{~nm})$ is collimated to the F-P etalon. For the vacuum case ( $\mathrm{n}=1.0)$, please calculate (i) interference orders $m_{i}$, (ii) incidence angle $\theta_{i}$ and (iii) diameter $D_{i}$ for the first three $(\mathrm{i}=1,2,3)$ fringes from the center of the ring patterns on the focal plane. Context answer: $m_{1}=33921, m_{2}=33920, m_{3}=33919 \quad \text { (a2) } $\theta_{1}=0.241^{0}, \theta_{2}=0.502^{0}, \theta_{3}=0.667^{0} \quad \text { (a3) }$ $D_{1}=2.52 \mathrm{~mm}, D_{2}=5.26 \mathrm{~mm}, D_{3}=6.99 \mathrm{~mm} \text { (a5) }$ Extra Supplementary Reading Materials: (b) As shown in Fig. 2, the width $\varepsilon$ of the spectral line is defined as the full width of half maximum (FWHM) of light transmitivity T regarding the phase shift $\delta$. The resolution of the F-P etalon is defined as follows: for two wavelengths $\lambda$ and $\lambda+\Delta \lambda$, when the central phase difference $\Delta \delta$ of both spectral lines is larger than $\varepsilon$, they are thought to be resolvable; then the etalon resolution is $\lambda / \Delta \lambda$ when $\Delta \delta=\varepsilon$. For the vacuum case, the $\mathrm{D} 1$ line ( $\lambda=589.6 \mathrm{~nm}$ ), and because of the incident angle $\theta \approx 0$, take $\cos \theta \approx 1.0$, please calculate: Figure 2 Context question: (i) the width $\varepsilon$ of the spectral line. Context answer: \boxed{0.21} Context question: (ii) the resolution $\lambda / \Delta \lambda$ of the etalon. Context answer: \boxed{$1.01 \times 10^{6}$} Context question: (c) As shown in Fig. 1, the initial air pressure is zero. By slowly tuning the pin valve, air is gradually injected into the F-P etalon and finally the air pressure reaches the standard atmospheric pressure. On the same time, ten new fringes are observed to produce from the center of the ring patterns on the focal plane. Based on this phenomenon, calculate the refractive index of air $n_{\text {air }}$ at the standard atmospheric pressure. Context answer: \boxed{$1.00029$} Context question: (d) Energy levels splitting of Sodium atoms occurs when they are placed in a magnetic field. This is called as the Zeeman effect. The energy shift given by $\Delta E=m_{j} g_{k} \mu_{B} B$, where the quantum number $\mathrm{m}_{\mathrm{j}}$ can be $\mathrm{J}, \mathrm{J}-1, \ldots,-\mathrm{J}+1,-\mathrm{J}, \mathrm{J}$ is the total angular quantum number, $\mathrm{g}_{\mathrm{k}}$ is the Landé factor, $\mu_{B}=\frac{h e}{4 \pi m_{e}}$ is Bohr magneton, $\mathrm{h}$ is the Plank constant, e is the electron charge, $m_{e}$ is the electron mass, B is the magnetic field. As shown in Fig. 3, the D1 spectral line is emitted when Sodium atoms jump from the energy level ${ }^{2} \mathrm{P}_{1 / 2}$ down to ${ }^{2} \mathrm{~S}_{1 / 2}$. We have $J=\frac{1}{2}$ for both ${ }^{2} \mathrm{P}_{1 / 2}$ and ${ }^{2} \mathrm{~S}_{1 / 2}$. Therefore, in the magnetic field, each energy level will be split into two levels. We define the energy gap of two splitting levels as $\Delta \mathrm{E}_{1}$ for ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ for ${ }^{2} \mathrm{~S}_{1 / 2}$ respectively $\left(\Delta \mathrm{E}_{1}\right.$ $<\Delta \mathrm{E}_{2}$ ). As a result, the D1 line is split into 4 spectral lines (a, b, c, and d), as showed in Fig. 3. Please write down the expression of the frequency $(v)$ of four lines $a, b, c$, and $d$. Figure 3 Context answer: \boxed{$v_{a}=v_{0}-\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$,$v_{b}=v_{0}-\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{c}=v_{0}+\frac{1}{2 h}\left(\Delta E_{2}-\Delta E_{1}\right)$,$v_{d}=v_{0}+\frac{1}{2 h}\left(\Delta E_{1}+\Delta E_{2}\right)$} Context question: (e) As shown in Fig. 4, when the magnetic field is turned on, each fringe of the D1 line will split into four sub-fringes $(1,2,3$, and 4$)$. The diameter of the four sub-fringes near the center is measured as $D_{1}, D_{2}, D_{3}$, and $D_{4}$. Please give the expression of the splitting energy gap $\Delta \mathrm{E}_{1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\Delta \mathrm{E}_{2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. Figure 4 Context answer: \boxed{$\Delta E_{1}=\frac{h c}{\lambda} \cdot \frac{D_{2}^{2}-D_{1}^{2}}{8 f^{2}}$,$\Delta E_{2}=\frac{h c}{\lambda} \cdot \frac{D_{4}^{2}-D_{2}^{2}}{8 f^{2}}$} Context question: (f) For the magnetic field $B=0.1 \mathrm{~T}$, the diameter of four sub-fringes is measured as: $D_{1}=3.88 \mathrm{~mm}, D_{2}=4.05 \mathrm{~mm}, D_{3}=4.35 \mathrm{~mm}$, and $D_{4}=4.51 \mathrm{~mm}$. Please calculate the Landé factor $\mathrm{g}_{\mathrm{k} 1}$ of ${ }^{2} \mathrm{P}_{1 / 2}$ and $\mathrm{g}_{\mathrm{k} 2}$ of ${ }^{2} \mathrm{~S}_{1 / 2}$. Context answer: \boxed{0.68, 1.99} Context question: (g) The magnetic field on the sun can be determined by measuring the Zeeman effect of the Sodium D1 line on some special regions of the sun. One observes that, in the four split lines, the wavelength difference between the shortest and longest wavelength is $0.012 \mathrm{~nm}$ by a solar spectrograph. What is the magnetic field B in this region of the sun? Context answer: \boxed{0.2772} ","(h) A Light- Emitting Diode (LED) source with a central wavelength $\lambda=650 \mathrm{~nm}$ and spectral width $\Delta \lambda=20 \mathrm{~nm}$ is normally incident $(\theta=0)$ into the F-P etalon shown in Fig. 1. For the vacuum case, find (i) the number of lines in transmitted spectrum and (ii) the frequency width $\Delta v$ of each line?","['The wavelength of transmitted spectral lines is given by:\n\n$$\n\\begin{aligned}\n& 2 n t=m \\lambda_{m} \\quad \\text{(h1)} \\\\\n& v_{m}=\\frac{c}{\\lambda_{m}} \\\\\n& v_{m}=\\frac{m c}{2 n t} \\\\\n& \\Delta v_{m}=\\frac{c}{2 n t}=1.5 \\times 10^{10} \\mathrm{~Hz} \\quad \\text { (h2) }\n\\end{aligned}\n$$\n\nThe frequency width of the input LED is:\n\n$$\n\\begin{aligned}\n& \\Delta v_{s}=\\left|-\\frac{c \\Delta \\lambda}{\\lambda^{2}}\\right| \\text{(h3)}\\\\\n& =\\frac{3 \\times 10^{8} \\times 20 \\times 10^{-9}}{\\left(650 \\times 10^{-9}\\right)^{2}}=1.42 \\times 10^{13} \\mathrm{~Hz}\n\\end{aligned}\n$$\n\nSo we have the number of transmitted spectral line:\n\n$$\n\\begin{aligned}\n& N=\\frac{\\Delta v_{s}}{\\Delta v_{m}} \\text{(h4)}\\\\\n& =\\frac{1.42 \\times 10^{13}}{1.5 \\times 10^{10}}=946\n\\end{aligned}\n$$\n\nThe spectral width of transmitted spectral line is $\\Delta \\lambda=\\frac{\\lambda^{2}}{\\pi n t \\sqrt{F}}$, then we have\n\n$$\n\\begin{aligned}\n& \\Delta v=\\frac{c}{\\pi n t \\sqrt{F}} \\text{(h5)}\\\\\n& =\\frac{3 \\times 10^{8}}{3.14 \\times 1.0 \\times 10 \\times 10^{-3} \\times \\sqrt{360}}=5.0 \\times 10^{8} \\mathrm{~Hz}\n\\end{aligned}\n$$']","['946 , $5.0 \\times 10^{8}$']",True,",Hz",Numerical,"1e0,5e7" 1545,Mechanics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. ","(a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.)","[""As the problem is symmetric about the equator, we need only consider the northern hemisphere as shown below.\n\n\n\nConservation of angular momentum about the Earth's spin axis implies that:\n\n$$\n\\begin{aligned}\n& \\Omega a^{2}=\\left(\\Omega a \\cos \\varphi_{d}+u_{Y}\\right) a \\cos \\varphi_{d} \\\\\n& u_{Y}=\\Omega a\\left(\\frac{1}{\\cos \\varphi_{d}}-\\cos \\varphi_{d}\\right)\n\\end{aligned}\n$$""]",['$u_{Y}=\\Omega a\\left(\\frac{1}{\\cos \\varphi_{d}}-\\cos \\varphi_{d}\\right)$'],False,,Expression, 1546,Mechanics,,"(b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity.","['(I),(II)']","['(I),(II)']",True,,Need_human_evaluate, 1548,Thermodynamics,,"(f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.)","['Since there is less and less air above as one climbs upward in the atmosphere, atmospheric pressure must decrease upwards.\n\nSo,\n\n$$\np_{A}>p_{B}>p_{C} \\quad \\text { and } \\quad p_{E}>p_{D}>p_{C}\n$$\n\nThe process EA represents an isothermal expansion as heat is gained from the surface. So,\n\n$$\np_{E}>p_{A}\n$$\n\nSince the total heat gain must equal the total heat loss, more heat must be lost in the isothermal compression CD than in the isothermal expansion BC. So net heat loss occurs from $\\mathrm{B}$ to $\\mathrm{D}$ and hence\n\n$$\np_{D}>p_{B}\n$$\n\nSo with the values of the pressure at $A$ and $D$, we deduce that:\n\n$$\np_{A}>p_{D}\n$$\n\nCollecting all inequalities together,\n\n$$\np_{E}>p_{A}>p_{D}>p_{B}>p_{C}\n$$']",['$p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$'],False,,Need_human_evaluate, 1549,Thermodynamics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. Context question: (a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.) Context answer: \boxed{$u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$} Context question: (b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity. Context answer: (I),(II) Extra Supplementary Reading Materials: Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_{r}$ and the descending branches are located at $\varphi_{n}$ and $\varphi_{s}$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e). Context question: (c) Assume that there is no east-west wind velocity around the point $\mathrm{Z}$. Given that $\varphi_{r}=-8^{\circ}, \varphi_{n}=28^{\circ}$ and $\varphi_{s}=-20^{\circ}$, what are the east-west wind velocities $u_{P}, u_{Q}$ and $u_{R}$ respectively at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ ? (The radius of the Earth is a $=6370 \mathrm{~km}$.) Hence, which hemisphere below has a stronger atmospheric jet stream? (I) Winter Hemisphere (II) Summer Hemisphere (III) Both hemispheres have equally strong jet streams. Context answer: \boxed{(I)} Context question: (d) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the eastwest component of the Coriolis force acting on the tropical air mass (A) north of the equator; (B) south of the equator. Context answer: Context question: (e) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice. Context answer: Extra Supplementary Reading Materials: Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from $\mathrm{D}$ to $\mathrm{E}$ are adiabatic, while that from $\mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to $\mathrm{D}$ and from $\mathrm{E}$ to $\mathrm{A}$ are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space. Context question: (f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.) Context answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$ ","(g) Let the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\mathrm{A}$ and $\mathrm{E}$ is $20 \mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \mathrm{~K}$. Note that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\left(c_{p}\right)$ for air, $\kappa$, is $2 / 7$.","['Since $p_{E}>p_{A}$ and $p_{A}=1000 \\mathrm{hPa}$, we have $p_{E}=1020 \\mathrm{hPa}$.\n\nFrom the adiabatic compression from $\\mathrm{D}$ to $\\mathrm{E}$, we have:\n\n$$\n\\begin{aligned}\n& p_{E}^{-\\kappa} T_{H}=p_{D}^{-\\kappa} T_{C} \\\\\n& T_{C}=\\left(\\frac{p_{D}}{p_{E}}\\right)^{\\kappa} \\times T_{H}=\\left(\\frac{225}{1020}\\right)^{2 / 7} \\times 300 K=195 K\n\\end{aligned}\n$$']",['195'],False,K,Numerical,5e-1 1550,Thermodynamics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. Context question: (a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.) Context answer: \boxed{$u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$} Context question: (b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity. Context answer: (I),(II) Extra Supplementary Reading Materials: Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_{r}$ and the descending branches are located at $\varphi_{n}$ and $\varphi_{s}$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e). Context question: (c) Assume that there is no east-west wind velocity around the point $\mathrm{Z}$. Given that $\varphi_{r}=-8^{\circ}, \varphi_{n}=28^{\circ}$ and $\varphi_{s}=-20^{\circ}$, what are the east-west wind velocities $u_{P}, u_{Q}$ and $u_{R}$ respectively at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ ? (The radius of the Earth is a $=6370 \mathrm{~km}$.) Hence, which hemisphere below has a stronger atmospheric jet stream? (I) Winter Hemisphere (II) Summer Hemisphere (III) Both hemispheres have equally strong jet streams. Context answer: \boxed{(I)} Context question: (d) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the eastwest component of the Coriolis force acting on the tropical air mass (A) north of the equator; (B) south of the equator. Context answer: Context question: (e) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice. Context answer: Extra Supplementary Reading Materials: Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from $\mathrm{D}$ to $\mathrm{E}$ are adiabatic, while that from $\mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to $\mathrm{D}$ and from $\mathrm{E}$ to $\mathrm{A}$ are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space. Context question: (f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.) Context answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$ Context question: (g) Let the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\mathrm{A}$ and $\mathrm{E}$ is $20 \mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \mathrm{~K}$. Note that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\left(c_{p}\right)$ for air, $\kappa$, is $2 / 7$. Context answer: \boxed{195} ",(h) Calculate the pressure $p_{B}$.,"['From the adiabatic expansion $\\mathrm{AB}$ and adiabatic compression $\\mathrm{DE}$,\n\n$$\n\\begin{aligned}\n& \\left.\\begin{array}{l}\np_{A}^{-\\kappa} T_{H}=p_{B}^{-\\kappa} T_{C} \\\\\np_{E}^{-\\kappa} T_{H}=p_{D}^{-\\kappa} T_{C}\n\\end{array}\\right\\} \\quad \\frac{p_{A}}{p_{E}}=\\frac{p_{B}}{p_{D}} \\\\\n& \\therefore p_{B}=\\frac{p_{A}}{p_{E}} p_{D}=\\frac{1000}{1020} 225 h P a=220 h P a\n\\end{aligned}\n$$']",['220'],False,hPa,Numerical,5e0 1551,Thermodynamics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. Context question: (a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.) Context answer: \boxed{$u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$} Context question: (b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity. Context answer: (I),(II) Extra Supplementary Reading Materials: Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_{r}$ and the descending branches are located at $\varphi_{n}$ and $\varphi_{s}$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e). Context question: (c) Assume that there is no east-west wind velocity around the point $\mathrm{Z}$. Given that $\varphi_{r}=-8^{\circ}, \varphi_{n}=28^{\circ}$ and $\varphi_{s}=-20^{\circ}$, what are the east-west wind velocities $u_{P}, u_{Q}$ and $u_{R}$ respectively at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ ? (The radius of the Earth is a $=6370 \mathrm{~km}$.) Hence, which hemisphere below has a stronger atmospheric jet stream? (I) Winter Hemisphere (II) Summer Hemisphere (III) Both hemispheres have equally strong jet streams. Context answer: \boxed{(I)} Context question: (d) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the eastwest component of the Coriolis force acting on the tropical air mass (A) north of the equator; (B) south of the equator. Context answer: Context question: (e) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice. Context answer: Extra Supplementary Reading Materials: Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from $\mathrm{D}$ to $\mathrm{E}$ are adiabatic, while that from $\mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to $\mathrm{D}$ and from $\mathrm{E}$ to $\mathrm{A}$ are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space. Context question: (f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.) Context answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$ Context question: (g) Let the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\mathrm{A}$ and $\mathrm{E}$ is $20 \mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \mathrm{~K}$. Note that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\left(c_{p}\right)$ for air, $\kappa$, is $2 / 7$. Context answer: \boxed{195} Context question: (h) Calculate the pressure $p_{B}$. Context answer: \boxed{220} ","(i) For an air mass moving once around the winter Hadley circulation, using the molar gas constant, $R$, and the quantities defined above, obtain expressions for (A) the net work done per unit mole $W_{\text {net }}$ ignoring surface friction;","['Work done per mole in an isothermal process is generally given by $W=\\int p d V=\\int p d\\left(\\frac{R T}{p}\\right)=-R T \\int p^{-1} d p=-R T \\ln p+$ const. \n\nWork done per mole in processes EA and BCD are respectively,\n\n$$\n\\begin{aligned}\n& W_{E A}=-R T_{H} \\ln p_{A}+R T_{H} \\ln p_{E}=R T_{H} \\ln \\left(\\frac{p_{E}}{p_{A}}\\right) \\\\\n& W_{B C D}=R T_{C} \\ln \\left(\\frac{p_{B}}{p_{D}}\\right)\n\\end{aligned}\n$$\n\nWork done in an adiabatic process is used entirely to raise the internal energy of the air mass. Since the decrease in internal energy in process AB exactly cancels the increase in internal energy in process $\\mathrm{DE}$ because the respective decrease and increase in temperature cancel, no net work is done in the adiabatic processes.\n\n\n\nSo the net work done per mole on the air mass is:\n\n$$\n\\begin{aligned}\nW_{\\text {net }} & =W_{E A}+W_{B C D} \\\\\n& =R T_{H} \\ln \\left(\\frac{p_{E}}{p_{A}}\\right)+R T_{C} \\ln \\left(\\frac{p_{B}}{p_{D}}\\right) \\\\\n& =R\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{E}}{p_{A}}\\right)+R T_{C} \\ln \\left(\\frac{p_{B}}{p_{D}} \\frac{p_{E}}{p_{A}}\\right) \\\\\n& =R\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{E}}{p_{A}}\\right) \\quad \\text { or } \\quad R\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{D}}{p_{B}}\\right)\n\\end{aligned}\n$$\n\nusing equation $(*)$ in part (h)']","['$R\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{E}}{p_{A}}\\right)$', '$R\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{D}}{p_{B}}\\right)$']",False,,Expression, 1552,Thermodynamics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. Context question: (a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.) Context answer: \boxed{$u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$} Context question: (b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity. Context answer: (I),(II) Extra Supplementary Reading Materials: Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_{r}$ and the descending branches are located at $\varphi_{n}$ and $\varphi_{s}$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e). Context question: (c) Assume that there is no east-west wind velocity around the point $\mathrm{Z}$. Given that $\varphi_{r}=-8^{\circ}, \varphi_{n}=28^{\circ}$ and $\varphi_{s}=-20^{\circ}$, what are the east-west wind velocities $u_{P}, u_{Q}$ and $u_{R}$ respectively at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ ? (The radius of the Earth is a $=6370 \mathrm{~km}$.) Hence, which hemisphere below has a stronger atmospheric jet stream? (I) Winter Hemisphere (II) Summer Hemisphere (III) Both hemispheres have equally strong jet streams. Context answer: \boxed{(I)} Context question: (d) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the eastwest component of the Coriolis force acting on the tropical air mass (A) north of the equator; (B) south of the equator. Context answer: Context question: (e) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice. Context answer: Extra Supplementary Reading Materials: Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from $\mathrm{D}$ to $\mathrm{E}$ are adiabatic, while that from $\mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to $\mathrm{D}$ and from $\mathrm{E}$ to $\mathrm{A}$ are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space. Context question: (f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.) Context answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$ Context question: (g) Let the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\mathrm{A}$ and $\mathrm{E}$ is $20 \mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \mathrm{~K}$. Note that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\left(c_{p}\right)$ for air, $\kappa$, is $2 / 7$. Context answer: \boxed{195} Context question: (h) Calculate the pressure $p_{B}$. Context answer: \boxed{220} Context question: (i) For an air mass moving once around the winter Hadley circulation, using the molar gas constant, $R$, and the quantities defined above, obtain expressions for (A) the net work done per unit mole $W_{\text {net }}$ ignoring surface friction; Context answer: \boxed{$R\left(T_{H}-T_{C}\right) \ln \left(\frac{p_{E}}{p_{A}}\right)$} ",(B) the heat loss per unit mole $Q_{\text {loss }}$ at the top of the atmosphere.,['The heat loss per mole at the top of the atmosphere is the same as the work done per mole on the air mass because there is no change in internal energy for an isothermal process.\n\n$$\n\\begin{array}{rlr}\nQ_{\\text {loss }} & =W_{C D} & \\text { } \\\\\n& =R T_{C} \\ln \\left(\\frac{p_{D}}{p_{C}}\\right) &\n\\end{array}\n$$'],['$R T_{C} \\ln \\left(\\frac{p_{D}}{p_{C}}\\right)$'],False,,Expression, 1553,Thermodynamics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. Context question: (a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.) Context answer: \boxed{$u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$} Context question: (b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity. Context answer: (I),(II) Extra Supplementary Reading Materials: Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_{r}$ and the descending branches are located at $\varphi_{n}$ and $\varphi_{s}$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e). Context question: (c) Assume that there is no east-west wind velocity around the point $\mathrm{Z}$. Given that $\varphi_{r}=-8^{\circ}, \varphi_{n}=28^{\circ}$ and $\varphi_{s}=-20^{\circ}$, what are the east-west wind velocities $u_{P}, u_{Q}$ and $u_{R}$ respectively at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ ? (The radius of the Earth is a $=6370 \mathrm{~km}$.) Hence, which hemisphere below has a stronger atmospheric jet stream? (I) Winter Hemisphere (II) Summer Hemisphere (III) Both hemispheres have equally strong jet streams. Context answer: \boxed{(I)} Context question: (d) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the eastwest component of the Coriolis force acting on the tropical air mass (A) north of the equator; (B) south of the equator. Context answer: Context question: (e) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice. Context answer: Extra Supplementary Reading Materials: Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from $\mathrm{D}$ to $\mathrm{E}$ are adiabatic, while that from $\mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to $\mathrm{D}$ and from $\mathrm{E}$ to $\mathrm{A}$ are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space. Context question: (f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.) Context answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$ Context question: (g) Let the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\mathrm{A}$ and $\mathrm{E}$ is $20 \mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \mathrm{~K}$. Note that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\left(c_{p}\right)$ for air, $\kappa$, is $2 / 7$. Context answer: \boxed{195} Context question: (h) Calculate the pressure $p_{B}$. Context answer: \boxed{220} Context question: (i) For an air mass moving once around the winter Hadley circulation, using the molar gas constant, $R$, and the quantities defined above, obtain expressions for (A) the net work done per unit mole $W_{\text {net }}$ ignoring surface friction; Context answer: \boxed{$R\left(T_{H}-T_{C}\right) \ln \left(\frac{p_{E}}{p_{A}}\right)$} Context question: (B) the heat loss per unit mole $Q_{\text {loss }}$ at the top of the atmosphere. Context answer: \boxed{$R T_{C} \ln \left(\frac{p_{D}}{p_{C}}\right)$} ",(j) What is the value of the ideal thermodynamic efficiency $\varepsilon_{i}$ for the winter Hadley circulation?,['$$\n\\begin{array}{rlr}\n\\varepsilon_{i} & =1-\\frac{T_{C}}{T_{H}} \\\\\n& =1-\\frac{195}{300}=0.35\n\\end{array}\n$$'],['0.35'],False,,Numerical,1e-2 1554,Thermodynamics,"The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale. Context question: (a) Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east. (The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.) Context answer: \boxed{$u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$} Context question: (b) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation? Tick the correct answer(s). There can be more than one correct answer. (I) There is friction from the Earth's surface. (II) There is turbulence in the lower atmosphere, where different layers of air are mixed (III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth. (IV) The air is moist at the lower levels causing retardation to the wind velocity. Context answer: (I),(II) Extra Supplementary Reading Materials: Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_{r}$ and the descending branches are located at $\varphi_{n}$ and $\varphi_{s}$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e). Context question: (c) Assume that there is no east-west wind velocity around the point $\mathrm{Z}$. Given that $\varphi_{r}=-8^{\circ}, \varphi_{n}=28^{\circ}$ and $\varphi_{s}=-20^{\circ}$, what are the east-west wind velocities $u_{P}, u_{Q}$ and $u_{R}$ respectively at the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ ? (The radius of the Earth is a $=6370 \mathrm{~km}$.) Hence, which hemisphere below has a stronger atmospheric jet stream? (I) Winter Hemisphere (II) Summer Hemisphere (III) Both hemispheres have equally strong jet streams. Context answer: \boxed{(I)} Context question: (d) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the eastwest component of the Coriolis force acting on the tropical air mass (A) north of the equator; (B) south of the equator. Context answer: Context question: (e) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice. Context answer: Extra Supplementary Reading Materials: Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from $\mathrm{D}$ to $\mathrm{E}$ are adiabatic, while that from $\mathrm{B}$ to $\mathrm{C}, \mathrm{C}$ to $\mathrm{D}$ and from $\mathrm{E}$ to $\mathrm{A}$ are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space. Context question: (f) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities. (Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.) Context answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$ Context question: (g) Let the temperature next to the surface and at the top of the atmosphere be $T_{H}$ and $T_{C}$ respectively. Given that the pressure difference between points $\mathrm{A}$ and $\mathrm{E}$ is $20 \mathrm{hPa}$, calculate $T_{C}$ for $T_{H}=300 \mathrm{~K}$. Note that the ratio of molar gas constant $(R)$ to molar heat capacity at constant pressure $\left(c_{p}\right)$ for air, $\kappa$, is $2 / 7$. Context answer: \boxed{195} Context question: (h) Calculate the pressure $p_{B}$. Context answer: \boxed{220} Context question: (i) For an air mass moving once around the winter Hadley circulation, using the molar gas constant, $R$, and the quantities defined above, obtain expressions for (A) the net work done per unit mole $W_{\text {net }}$ ignoring surface friction; Context answer: \boxed{$R\left(T_{H}-T_{C}\right) \ln \left(\frac{p_{E}}{p_{A}}\right)$} Context question: (B) the heat loss per unit mole $Q_{\text {loss }}$ at the top of the atmosphere. Context answer: \boxed{$R T_{C} \ln \left(\frac{p_{D}}{p_{C}}\right)$} Context question: (j) What is the value of the ideal thermodynamic efficiency $\varepsilon_{i}$ for the winter Hadley circulation? Context answer: \boxed{0.35} ","(k) Prove that the actual thermodynamic efficiency $\varepsilon$ for the winter Hadley circulation is always smaller than $\varepsilon_{i}$, showing all mathematical steps.",['$$\n\\begin{aligned}\n\\varepsilon & =\\frac{W_{\\text {net }}}{Q_{\\text {loss }}+W_{\\text {net }}} \\\\\n\\frac{1}{\\varepsilon}-1 & =\\frac{Q_{\\text {loss }}}{W_{\\text {net }}}=\\frac{R T_{C} \\ln \\left(\\frac{p_{D}}{p_{C}}\\right)}{R\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{E}}{p_{A}}\\right)} \\\\\n& =\\frac{T_{C} \\ln \\left(\\frac{p_{D}}{p_{B}} \\times \\frac{p_{B}}{p_{C}}\\right)}{\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{E}}{p_{A}}\\right)} \\\\\n& >\\frac{T_{C} \\ln \\left(\\frac{p_{D}}{p_{B}}\\right)}{\\left(T_{H}-T_{C}\\right) \\ln \\left(\\frac{p_{E}}{p_{A}}\\right)} \\quad \\text { as } \\frac{p_{B}}{p_{C}}>1 \\\\\n& =\\frac{T_{C}}{T_{H}-T_{C}} \\quad \\text { using equation (*) in part (h) } \\\\\n\\frac{1}{\\varepsilon} & >1+\\frac{T_{C}}{T_{H}-T_{C}}=\\frac{T_{H}}{T_{H}-T_{C}} \\\\\n\\varepsilon & <\\frac{T_{H}-T_{C}}{T_{H}}=\\varepsilon_{i}\n\\end{aligned}\n$$'],,False,,, 1555,Thermodynamics,,"(l) Which of the following statements best explains why $\varepsilon$ is less than the ideal value? Tick the correct answer(s). There can be more than one correct answer. (I) We have ignored work done against surface friction. (II) Condensation occurs at a temperature lower than the temperature of the heat source. (III) There is irreversible evaporation of water at the surface. (IV) The ideal efficiency is applicable only when there is no phase change of water.",['(II) and (III)'],['(II) and (III)'],False,,Need_human_evaluate, 1556,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$.","(a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.)","['Writing out $|\\mathrm{E}|=\\frac{\\lambda}{2 \\pi \\epsilon_{0} r}=-\\frac{\\partial}{\\partial r} V(r)$\n\n$$\n=-\\frac{\\partial}{\\partial r} \\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln \\frac{b}{r}\n$$\n\nNote that\n\n$$\nV(r)=\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln \\frac{b}{r}(=0 \\text { at the mesh })\n$$\n\nAlso at the edge of the filament, $V_{a}=V(r=a)$, so\n\n$$\nV_{a}=\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln \\frac{b}{a}\n$$\n\nGiving together\n\n$$\nV(r)=V_{a} \\frac{\\ln (b / r)}{\\ln (b / a)} \\text { where } r=\\sqrt{x^{2}+z^{2}}\n$$\n\n']",['$V(r)=V_{a} \\frac{\\ln (b / \\sqrt{x^{2}+z^{2}})}{\\ln (b / a)}$'],False,,Expression, 1557,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ ","(b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant.","['There are several ways to work out the solution:\n\nA charge in an electric field will experience a force and hence a change in momentum. Note that potential energy of the electron (charge $=-e_{0}$ ) is $-e_{0} V(r)$. Using impulse acting on the electron due to the electric field, \n\n$$\n\\begin{aligned}\n\\text { Impulse } & =\\left.\\frac{1}{v_{z}} \\int_{-\\infty}^{\\infty}\\left(-e_{0}\\right)\\left(-\\frac{\\partial V\\left(x, z^{\\prime}\\right)}{\\partial x}\\right) d z^{\\prime}\\right|_{x=a} \\\\\n& =-\\left.\\frac{1}{v_{z}} \\int_{-\\infty}^{\\infty} \\frac{-e_{0} V_{a} x}{\\left(x^{2}+z^{\\prime 2}\\right) \\ln \\frac{b}{a}} d z^{\\prime}\\right|_{x=a} \\\\\n& =\\frac{e_{0} V_{a} \\pi}{v_{z} \\ln \\frac{b}{a}} \\\\\n\\Rightarrow k_{x} & =\\frac{e_{0} V_{a} \\pi}{\\hbar v_{z} \\ln \\frac{b}{a}}\n\\end{aligned}\n$$']",['$k_{x} =\\frac{e_{0} V_{a} \\pi}{\\hbar v_{z} \\ln \\frac{b}{a}}$'],False,,Expression, 1558,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ Context question: (b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant. Context answer: \boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$} Extra Supplementary Reading Materials: (c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$,",(i) assuming relativistic effects can be ignored,['Equating the kinetic energy to $e V_{0}$ \n\n$$\n\\begin{aligned}\n& \\frac{h}{\\lambda}=\\sqrt{2 m\\left|-e_{0}\\right| V_{0}} \\\\\n& \\lambda=\\frac{h}{\\sqrt{2 m e_{0} V_{0}}}\n\\end{aligned}\n$$'],['$\\frac{h}{\\sqrt{2 m e_{0} V_{0}}}$'],False,,Expression, 1559,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ Context question: (b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant. Context answer: \boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$} Extra Supplementary Reading Materials: (c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$, Context question: (i) assuming relativistic effects can be ignored Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$} ",(ii) taking relativistic effects into consideration.,['Consider\n\n$$\n\\begin{aligned}\nE^{2} & =(p c)^{2}+\\left(m c^{2}\\right)^{2} \\\\\n& =\\left(\\frac{h}{\\lambda} c\\right)^{2}+\\left(m c^{2}\\right)^{2} \\\\\n\\frac{h^{2} c^{2}}{\\lambda^{2}} & =\\left(m c^{2}+\\left|-e_{0}\\right| V_{0}\\right)^{2}-\\left(m c^{2}\\right)^{2} \\\\\n& =2 m c^{2} e_{0} V_{0}\\left(1+\\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\\right) \\\\\n\\lambda & =\\frac{h}{\\sqrt{2 m e_{0} V_{0}\\left(1+\\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\\right)}}\n\\end{aligned}\n$$'],['$\\frac{h}{\\sqrt{2 m e_{0} V_{0}\\left(1+\\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\\right)}}$'],False,,Expression, 1560,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ Context question: (b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant. Context answer: \boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$} Extra Supplementary Reading Materials: (c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$, Context question: (i) assuming relativistic effects can be ignored Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$} Context question: (ii) taking relativistic effects into consideration. Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}$} Extra Supplementary Reading Materials: (d) In Tonomura et al experiment, $$ \begin{array}{ll} v_{z} & =c / 2, \\ V_{a} & =10 \mathrm{~V}, \\ V_{0} & =50 \mathrm{kV}, \\ a & =0.5 \mu \mathrm{m}, \\ b & =5 \mathrm{~mm}, \\ \ell & =25 \mathrm{~cm}, \\ L & =1.5 \mathrm{~m}, \\ h & =6.6 \times 10^{-34} \mathrm{Js}, \\ \text { electron charge, } e=1.6 \times 10^{-19} \mathrm{C}, \\ \text { mass of electron, } \mathrm{m}_{0}=9.1 \times 10^{-31} \mathrm{~kg}, \\ \text { and the speed of light in vacuo, } c=3 \times 10^{8} \mathrm{~ms}^{-1} \end{array} $$","(i) calculate the value of $k_{x}$,",['Previous equation:\n\n$$\nk_{x}=\\frac{e_{0} V_{a} \\pi}{\\hbar v_{z} \\ln \\frac{b}{a}}\n$$\n\nPlugging the relevant numbers into the equation gives:\n\n$$\nk_{x}=\\frac{\\pi}{907} \\AA^{-1} \\text { or } 3.46 \\times 10^{7} m^{-1}\n$$'],['$3.46 \\times 10^{7} $'],False,$m^{-1}$,Numerical,5e5 1561,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ Context question: (b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant. Context answer: \boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$} Extra Supplementary Reading Materials: (c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$, Context question: (i) assuming relativistic effects can be ignored Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$} Context question: (ii) taking relativistic effects into consideration. Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}$} Extra Supplementary Reading Materials: (d) In Tonomura et al experiment, $$ \begin{array}{ll} v_{z} & =c / 2, \\ V_{a} & =10 \mathrm{~V}, \\ V_{0} & =50 \mathrm{kV}, \\ a & =0.5 \mu \mathrm{m}, \\ b & =5 \mathrm{~mm}, \\ \ell & =25 \mathrm{~cm}, \\ L & =1.5 \mathrm{~m}, \\ h & =6.6 \times 10^{-34} \mathrm{Js}, \\ \text { electron charge, } e=1.6 \times 10^{-19} \mathrm{C}, \\ \text { mass of electron, } \mathrm{m}_{0}=9.1 \times 10^{-31} \mathrm{~kg}, \\ \text { and the speed of light in vacuo, } c=3 \times 10^{8} \mathrm{~ms}^{-1} \end{array} $$ Context question: (i) calculate the value of $k_{x}$, Context answer: \boxed{$3.46 \times 10^{7} $} ","(ii) determine the fringe separation of the interference pattern on the screen,",['Fringe separation is given by $\\frac{1}{2} \\frac{2 \\pi}{k_{x}}=907 \\AA$'],['$907$'],False,$\AA$,Numerical,7e0 1563,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ Context question: (b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant. Context answer: \boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$} Extra Supplementary Reading Materials: (c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$, Context question: (i) assuming relativistic effects can be ignored Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$} Context question: (ii) taking relativistic effects into consideration. Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}$} Extra Supplementary Reading Materials: (d) In Tonomura et al experiment, $$ \begin{array}{ll} v_{z} & =c / 2, \\ V_{a} & =10 \mathrm{~V}, \\ V_{0} & =50 \mathrm{kV}, \\ a & =0.5 \mu \mathrm{m}, \\ b & =5 \mathrm{~mm}, \\ \ell & =25 \mathrm{~cm}, \\ L & =1.5 \mathrm{~m}, \\ h & =6.6 \times 10^{-34} \mathrm{Js}, \\ \text { electron charge, } e=1.6 \times 10^{-19} \mathrm{C}, \\ \text { mass of electron, } \mathrm{m}_{0}=9.1 \times 10^{-31} \mathrm{~kg}, \\ \text { and the speed of light in vacuo, } c=3 \times 10^{8} \mathrm{~ms}^{-1} \end{array} $$ Context question: (i) calculate the value of $k_{x}$, Context answer: \boxed{$3.46 \times 10^{7} $} Context question: (ii) determine the fringe separation of the interference pattern on the screen, Context answer: \boxed{$907$} Context question: (iii) If the electron wave is a spherical wave instead of a plane wave, is the fringe spacing larger, the same or smaller than the fringe spacing calculated in (ii)? Context answer: \boxed{larger} ","(iv) In part (c), determine the percentage error in the wavelength of the electron using non-relativistic approximation.",['Non-relativistic:\n\n$$\n\\begin{aligned}\n\\frac{h}{\\lambda} & =\\sqrt{2 m e_{0} V_{0}} \\\\\n\\lambda_{\\text {nonrel }} & =\\frac{h}{\\sqrt{2 m e_{0} V_{0}}} \\\\\n& =5.4697 \\times 10^{-12} \\mathrm{~m}\n\\end{aligned}\n$$\n\nRelativistic:\n\n$$\n\\begin{aligned}\n\\lambda_{\\text {rel }} & =\\frac{h}{\\sqrt{2 m e V_{0}\\left(1+\\frac{e V_{0}}{2 m_{0} c^{2}}\\right)}} \\\\\n& =5.3408 \\times 10^{-12} \\mathrm{~m}\n\\end{aligned}\n$$\n\nPercentage error:\n\n$$\n\\begin{aligned}\n\\text { Error }= & \\frac{\\lambda_{\\text {nonrel }}-\\lambda_{\\text {rel }}}{\\lambda_{\\text {rel }}} \\\\\n= & 0.024 \\\\\n& \\text { or } 2.4 \\text { percent. }\n\\end{aligned}\n$$'],['2.4'],False,percent,Numerical,1e-1 1564,Electromagnetism,"The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron ""biprism"" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram). The electron ""biprism"" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the ""biprism"" is $\ell$, and the distance between the distance between the ""biprism"" and the screen is $L$. Context question: (a) Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.) Context answer: $V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$ Context question: (b) An incoming electron plane wave with wave vector $k_{z}$ is deflected by the ""biprism"" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the ""biprism"" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant. Context answer: \boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$} Extra Supplementary Reading Materials: (c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$, Context question: (i) assuming relativistic effects can be ignored Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$} Context question: (ii) taking relativistic effects into consideration. Context answer: \boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}$} Extra Supplementary Reading Materials: (d) In Tonomura et al experiment, $$ \begin{array}{ll} v_{z} & =c / 2, \\ V_{a} & =10 \mathrm{~V}, \\ V_{0} & =50 \mathrm{kV}, \\ a & =0.5 \mu \mathrm{m}, \\ b & =5 \mathrm{~mm}, \\ \ell & =25 \mathrm{~cm}, \\ L & =1.5 \mathrm{~m}, \\ h & =6.6 \times 10^{-34} \mathrm{Js}, \\ \text { electron charge, } e=1.6 \times 10^{-19} \mathrm{C}, \\ \text { mass of electron, } \mathrm{m}_{0}=9.1 \times 10^{-31} \mathrm{~kg}, \\ \text { and the speed of light in vacuo, } c=3 \times 10^{8} \mathrm{~ms}^{-1} \end{array} $$ Context question: (i) calculate the value of $k_{x}$, Context answer: \boxed{$3.46 \times 10^{7} $} Context question: (ii) determine the fringe separation of the interference pattern on the screen, Context answer: \boxed{$907$} Context question: (iii) If the electron wave is a spherical wave instead of a plane wave, is the fringe spacing larger, the same or smaller than the fringe spacing calculated in (ii)? Context answer: \boxed{larger} Context question: (iv) In part (c), determine the percentage error in the wavelength of the electron using non-relativistic approximation. Context answer: \boxed{2.4} ",(v) the distance $d$ between the apparent double slits.,"['The double slit formula is given by\n\n$$\ny=\\frac{m \\lambda(\\ell+L)}{d}\n$$\n\nwhere $m$ is the order and $y$ is the distance for maximum intensity from the central fringe.\n\nIn this case, since the fringe spacing is $907 \\AA$,\n\n$$\nd=1.03 \\times 10^{-4} \\mathrm{~m}\n$$']",['$1.03 \\times 10^{-4}$'],False,m,Numerical,3e-6 1565,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.",1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.,"['Using Gauss law\n\n$$\n\\oint \\mathbf{E} \\cdot \\mathbf{d} \\mathbf{A}=\\frac{q}{\\epsilon_{0}} .\n\\tag{1}\n$$\n\nFrom symetry we know that the electric field only has radial component. Choose a cylinder (with a line charge as the axis) as the Gaussian surface, we obtain\n\n$$\nE .2 \\pi r l=\\frac{\\lambda l}{\\epsilon_{0}}\n$$\n\nSimplify to obtain\n\n$$\n\\mathbf{E}=\\hat{r} \\frac{\\lambda}{2 \\pi \\epsilon_{0} r}\n\\tag{2}\n$$\n\n']",['$\\mathbf{E}=\\hat{r} \\frac{\\lambda}{2 \\pi \\epsilon_{0} r}$'],False,,Expression, 1566,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$ ","2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$.","['The potential is given by\n\n$$\n\\begin{aligned}\nV & =-\\int_{\\mathrm{ref}}^{r} \\mathbf{E} \\cdot \\mathbf{d} \\mathbf{l} \\\\\n& =-\\int_{\\mathrm{ref}}^{r} E \\cdot d r \\\\\nV & =-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r+K,\n\\end{aligned}\n\\tag{3}\n$$\n\nso $f(r)=-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r$. where $K$ is a constant.']",['$f(r)=-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r$'],False,,Expression, 1567,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$ Context question: 2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$. Context answer: \boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$} Context question: 3. Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces. Context answer: Extra Supplementary Reading Materials: For the following questions, ignore any edge effects.","4. Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin.","['From eq.(5) and eq.(6), we see that for any arbitrary potential $V$, the equipotential surfaces of these two equal but opposite lines charge, are cylindrical surfaces. From this observation, we can choose the specific position for each line charge in both cylinders so that the surface of each cylinder is an equipotential surface.\n\nConsider the following figure\n\n\n\nFigure 3: Two line charges with its equipotential surfaces\n\nWe would like to find a cylindrical equipotential surface enclose one line charge, let say the $-\\lambda$ (if we could find the surface, by symmetry, we surely can find the identical one that enclose the line $\\lambda$ ). The potential is given by\n\n$$\n\\begin{aligned}\nV & =-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r_{1}+\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln r_{2} \\\\\n& =-\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\ln \\left(l_{1}^{2}+R^{2}-2 l_{1} R \\cos \\phi\\right)+\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\ln \\left(l_{2}^{2}+R^{2}-2 l_{2} R \\cos \\phi\\right)\n\\end{aligned}\n\\tag{7}\n$$\n\nSince the surface of the cylinder has to be the equipotential surface, so the potential should not depend on $\\phi$, i.e. $\\frac{\\partial V}{\\partial \\phi}=0$.\n\n$$\n-\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\frac{2 l_{1} R \\sin \\phi}{l_{1}^{2}+R^{2}-2 l_{1} R \\cos \\phi} +\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\frac{2 l_{2} R \\sin \\phi}{l_{2}^{2}+R^{2}-2 l_{2} R \\cos \\phi}=0\n\\tag{8}\n$$\n\n$$\n\\begin{aligned}\n\\frac{l_{1}}{l_{1}^{2}+R^{2}-2 l_{1} R \\cos \\phi} & =\\frac{l_{2}}{l_{2}^{2}+R^{2}-2 l_{2} R \\cos \\phi} \\\\\nl_{1}^{2} l_{2}+R^{2} l_{2}-2 l_{1} l_{2} R \\cos \\phi & =l_{1} l_{2}^{2}+R^{2} l_{1}-2 l_{1} l_{2} R \\cos \\phi \\\\\nl_{1} l_{2}\\left(l_{1}-l_{2}\\right) & =R^{2}\\left(l_{1}-l_{2}\\right).\n\\end{aligned}\n$$\n\n$$\nl_{1} l_{2} =R^{2}\n\\tag{9}\n$$\n\nFrom the data in the problem, we have\n\n$$\nl_{1}+l_{2} =10 a\n\\tag{10}\n$$\n$$\nl_{1} l_{2} =9 a^{2}\n\\tag{11}\n$$\n\nSolve this quadratic equation to get\n\n$$\nl_{1}=5 a \\pm 4 a\n\\tag{12}\n$$\n\nHowever, since $l_{1}>l_{2}$, we have\n\n$$\nl_{1}=9 a\n\\tag{13}\n$$\n$$\nl_{2}=a\n\\tag{14}\n$$\n\nUsing this results on eq.(5), we have\n\n$$\nV=\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\ln \\frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}\n\\tag{15}\n$$\n\nThis is the potential in all region except inside both cylinders. For cylinders at $x=-5 a$, the potential is constant and equal to\n\n$$\nV(x=-2 a, y=0)=\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\ln \\frac{(4 a+2 a)^{2}+0^{2}}{(4 a-2 a)^{2}+0^{2}}=\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln 3\n\\tag{16}\n$$\n\nFor cylinders at $x=5 a$, the potential is constant and equal to\n\n$$\nV(x=2 a, y=0)=\\frac{\\lambda}{4 \\pi \\epsilon_{0}} \\ln \\frac{(4 a-2 a)^{2}+0^{2}}{(4 a+2 a)^{2}+0^{2}}=-\\frac{\\lambda}{2 \\pi \\epsilon_{0}} \\ln 3\n\\tag{17}\n$$\n\nThe potential difference between both cylinders are\n\n$$\n\\Delta V=\\frac{\\lambda}{\\pi \\epsilon_{0}} \\ln 3 \\equiv V_{0}\n\\tag{18}\n$$\n\n\n\nSubstituting this results in the potential equation, the potential outside the two cylinders are:\n\n$$\nV=\\frac{V_{0}}{4 \\ln 3} \\ln \\frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}\n\\tag{19}\n$$\n\nAnd the potential inside the cylinders are:\n\nThe potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$.\n\nThe potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$.\n\n']","['$\\frac{V_{0}}{4 \\ln 3} \\ln \\frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$, $-V_{0} / 2$, $V_{0} / 2$']",True,,Expression, 1568,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$ Context question: 2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$. Context answer: \boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$} Context question: 3. Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces. Context answer: Extra Supplementary Reading Materials: For the following questions, ignore any edge effects. Context question: 4. Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin. Context answer: the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$ The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$. The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$. ",5. Calculate the capacitance $C$ of the system.,"['From eq.(18), we have\n\n$$\nV_{0}=\\frac{q}{l \\pi \\epsilon_{0}} \\ln 3\n\\tag{20}\n$$\n\nso we get\n\n$$\nC=\\frac{q}{V_{0}}=\\frac{l \\pi \\epsilon_{0}}{\\ln 3}\n\\tag{21}\n$$']",['$\\frac{l \\pi \\epsilon_{0}}{\\ln 3}$'],False,,Expression, 1569,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$ Context question: 2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$. Context answer: \boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$} Context question: 3. Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces. Context answer: Extra Supplementary Reading Materials: For the following questions, ignore any edge effects. Context question: 4. Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin. Context answer: the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$ The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$. The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$. Context question: 5. Calculate the capacitance $C$ of the system. Context answer: \boxed{$\frac{l \pi \epsilon_{0}}{\ln 3}$} ","6. Now both cylinders are totally immersed in a weakly conducting liquid with conductivity $\sigma$. Calculate the total current that flows between both cylinders. Assume the permittivity of the liquid is equal to that of vacuum, $\epsilon=\epsilon_{0}$.","['The electric field produces by both cylinders are\n\n$$\nE_{x}=\\frac{V_{0}}{2 \\ln 3}\\left(\\frac{4 a+x}{(4 a+x)^{2}+y^{2}}+\\frac{4 a-x}{(4 a-x)^{2}+y^{2}}\\right) .\n\\tag{22}\n$$\n$$\nE_{y}=\\frac{V_{0}}{2 \\ln 3}\\left(\\frac{y}{(4 a+x)^{2}+y^{2}}-\\frac{y}{(4 a-x)^{2}+y^{2}}\\right) .\n\\tag{23}\n$$\n\nThe volume current density is given by\n\n$$\n\\mathbf{J}=\\sigma \\mathbf{E}\n\\tag{24}\n$$\n\nTo calculate the total current, we may choose to calculate the current that flow through the $x=0$ plane. On this plane, there is no current in the $y$ direction. The total current is given by\n\n$$\n\\begin{aligned}\nI & =\\int \\mathbf{J} \\cdot \\mathbf{d} \\mathbf{A} \\\\\n& =\\int \\sigma E_{x} l d y \\\\\n& =\\sigma l \\frac{8 a V_{0}}{2 \\ln 3} \\int_{\\infty}^{\\infty} \\frac{d y}{(4 a)^{2}+y^{2}}\n\\end{aligned}\n\\tag{25}\n$$\n\n$$\nI =\\frac{V_{0} \\pi \\sigma l}{\\ln 3}\n\\tag{26}\n$$']",['$I=\\frac{V_{0} \\pi \\sigma l}{\\ln 3}$'],False,,Expression, 1570,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$ Context question: 2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$. Context answer: \boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$} Context question: 3. Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces. Context answer: Extra Supplementary Reading Materials: For the following questions, ignore any edge effects. Context question: 4. Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin. Context answer: the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$ The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$. The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$. Context question: 5. Calculate the capacitance $C$ of the system. Context answer: \boxed{$\frac{l \pi \epsilon_{0}}{\ln 3}$} Context question: 6. Now both cylinders are totally immersed in a weakly conducting liquid with conductivity $\sigma$. Calculate the total current that flows between both cylinders. Assume the permittivity of the liquid is equal to that of vacuum, $\epsilon=\epsilon_{0}$. Context answer: \boxed{$I=\frac{V_{0} \pi \sigma l}{\ln 3}$} ",7. Calculate the resistance $R$ of the system. Calculate $R C$ of the system.,['The resistance is given by\n\n$$\nR=\\frac{V_{0}}{I}=\\frac{\\ln 3}{\\pi \\sigma l}\n\\tag{27}\n$$\n\nand therefore\n\n$$\nR C=\\frac{\\epsilon_{0}}{\\sigma}\n\\tag{28}\n$$'],['$R C=\\frac{\\epsilon_{0}}{\\sigma}$'],False,,Expression, 1571,Electromagnetism,"A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$ Context question: 2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$. Context answer: \boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$} Context question: 3. Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces. Context answer: Extra Supplementary Reading Materials: For the following questions, ignore any edge effects. Context question: 4. Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin. Context answer: the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$ The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$. The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$. Context question: 5. Calculate the capacitance $C$ of the system. Context answer: \boxed{$\frac{l \pi \epsilon_{0}}{\ln 3}$} Context question: 6. Now both cylinders are totally immersed in a weakly conducting liquid with conductivity $\sigma$. Calculate the total current that flows between both cylinders. Assume the permittivity of the liquid is equal to that of vacuum, $\epsilon=\epsilon_{0}$. Context answer: \boxed{$I=\frac{V_{0} \pi \sigma l}{\ln 3}$} Context question: 7. Calculate the resistance $R$ of the system. Calculate $R C$ of the system. Context answer: \boxed{$R C=\frac{\epsilon_{0}}{\sigma}$} ","8. Calculate the magnetic field due to the current in question 6. Assume that the permeability of the liquid is equal to that of vacuum $\mu=\mu_{0}$. Notes $\int \frac{\alpha d x}{\alpha^{2}+x^{2}}=\arctan \frac{x}{a}+$ const","[""Since the system has a high symmetry, we may use Ampere's law. The magnetic field should not have any $z$ dependence, since the current has no $z$ dependence.\n\nFigure 4 shows the current density $\\mathbf{J}$ flow from one cylinder to the other cylinder. Choose an Ampere loop on a constant $x$ plane in a symmetrical way, so that the first path is pointing in the positive $z$ direction with constant $y$ coordinate, the second path is pointing to the negative $y$ direction with constant $z$ coordinate. The third path is pointing to the negative $z$ direction, but with constant $-y$ coordinate. The fourth path is pointing in the positive $y$ direction with constant $-z$ coordinate.\n\nHaving this path, we need to calculate the current that flow through the loop\n\n$$\n\\begin{aligned}\nI & =\\int \\mathbf{J} \\cdot \\mathbf{d} \\mathbf{A} \\\\\n& =\\int J_{x} l d y \\\\\n& =\\frac{V_{0} \\sigma l}{2 \\ln 3} \\int_{-y}^{y}\\left(\\frac{4 a+x}{(4 a+x)^{2}+y^{2}}+\\frac{4 a-x}{(4 a-x)^{2}+y^{2}}\\right) d y\n\\end{aligned}\n$$\n\n\n\nFigure 4: The Ampere loop\n\n$$\nI=\\frac{V_{0} \\sigma l}{\\ln 3}\\left(\\arctan \\frac{y}{4 a+x}+\\arctan \\frac{y}{4 a-x}\\right)\n\\tag{29}\n$$\n\nUsing the Ampere's law\n\n$$\n\\oint \\mathbf{B} \\cdot \\mathbf{d} \\mathbf{l}=\\mu_{0} I\n\\tag{30}\n$$\n\n$$\n\\begin{aligned}\n2 B_{z} l & =\\frac{\\mu_{0} V_{0} \\sigma l}{\\ln 3}\\left(\\arctan \\frac{y}{4 a+x}+\\arctan \\frac{y}{4 a-x}\\right) \\\\\nB_{z} & =\\mu_{0} \\frac{V_{0} \\sigma}{2 \\ln 3}\\left(\\arctan \\frac{y}{4 a+x}+\\arctan \\frac{y}{4 a-x}\\right)\n\\end{aligned}\n\\tag{31}\n$$\n\n\n\ntherefore\n\n$$\n\\mathbf{B}=\\hat{z} \\frac{\\mu_{0} V_{0} \\sigma}{2 \\ln 3}\\left(\\arctan \\frac{y}{4 a+x}+\\arctan \\frac{y}{4 a-x}\\right)\n\\tag{32}\n$$""]",['$\\mathbf{B}=\\hat{z} \\frac{\\mu_{0} V_{0} \\sigma}{2 \\ln 3}\\left(\\arctan \\frac{y}{4 a+x}+\\arctan \\frac{y}{4 a-x}\\right)$'],False,,Expression, 1572,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.","1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).",['The equation of motion is given by\n\n$$\n\\begin{aligned}\nF & =\\frac{d}{d t}(\\gamma m v) \\\\\n& =\\frac{m c \\dot{\\beta}}{\\left(1-\\beta^{2}\\right)^{\\frac{3}{2}}}\n\\end{aligned}\n\\tag{1}\n$$\n\n$$\nF=\\gamma^{3} m a\n\\tag{2}\n$$\n\nwhere $\\gamma=\\frac{1}{\\sqrt{1-\\beta^{2}}}$ and $\\beta=\\frac{v}{c}$. So the acceleration is given by\n\n$$\na=\\frac{F}{\\gamma^{3} m}\n\\tag{3}\n$$'],['$a=\\frac{F}{\\gamma^{3} m}$'],False,,Expression, 1573,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} ","2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.",['Eq.(3) can be rewritten as\n\n$$\n\\begin{aligned}\nc \\frac{d \\beta}{d t} & =\\frac{F}{\\gamma^{3} m} \\\\\n\\int_{0}^{\\beta} \\frac{d \\beta}{\\left(1-\\beta^{2}\\right)^{\\frac{3}{2}}} & =\\frac{F}{m c} \\int_{0}^{t} d t \\\\\n\\frac{\\beta}{\\sqrt{1-\\beta^{2}}} & =\\frac{F t}{m c}\n\\end{aligned}\n\\tag{4}\n$$\n\n$$\n\\beta=\\frac{\\frac{F t}{m c}}{\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}}\n\\tag{5}\n$$'],['$\\beta=\\frac{\\frac{F t}{m c}}{\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}}$'],False,,Expression, 1574,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} ","3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.","['Using Eq.(5), we get\n\n$$\n\\begin{aligned}\n\\int_{0}^{x} d x & =\\int_{0}^{t} \\frac{F t d t}{m \\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}} \\\\\nx & =\\frac{m c^{2}}{F}\\left(\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}-1\\right) .\n\\end{aligned}\n\\tag{6}\n$$']",['$x=\\frac{m c^{2}}{F}\\left(\\sqrt{1+\\left(\\frac{F t}{m c}\\right)^{2}}-1\\right)$'],False,,Expression, 1575,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} ","4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame.","[""Consider the following systems, a frame $S^{\\prime}$ is moving with respect to another frame $S$, with velocity $u$ in the $x$ direction. If a particle is moving in the $\\mathrm{S}^{\\prime}$ frame with velocity $v^{\\prime}$ also in $x$ direction, then the particle velocity in the $\\mathrm{S}$ frame is given by\n\n$$\nv=\\frac{u+v^{\\prime}}{1+\\frac{u v^{\\prime}}{c^{2}}}\n\\tag{7}\n$$\n\n\n\nIf the particles velocity changes with respect to the $S$ ' frame, then the velocity in the $S$ frame is also change according to\n\n$$\n\\begin{aligned}\nd v & =\\frac{d v^{\\prime}}{1+\\frac{u v^{\\prime}}{c^{2}}}-\\frac{u+v^{\\prime}}{\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)^{2}} \\frac{u d v^{\\prime}}{c^{2}} \\\\\nd v & =\\frac{1}{\\gamma^{2}} \\frac{d v^{\\prime}}{\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)^{2}}\n\\end{aligned}\n\\tag{8}\n$$\n\nThe time in the $\\mathrm{S}^{\\prime}$ frame is $t^{\\prime}$, so the time in the $\\mathrm{S}$ frame is given by\n\n$$\nt=\\gamma\\left(t^{\\prime}+\\frac{u x^{\\prime}}{c^{2}}\\right)\n\\tag{9}\n$$\n\nso the time change in the $S^{\\prime}$ frame will give a time change in the $\\mathrm{S}$ frame as follow\n\n$$\nd t=\\gamma d t^{\\prime}\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)\n\\tag{10}\n$$\n\nThe acceleration in the $\\mathrm{S}$ frame is given by\n\n$$\na=\\frac{d v}{d t}=\\frac{a^{\\prime}}{\\gamma^{3}} \\frac{1}{\\left(1+\\frac{u v^{\\prime}}{c^{2}}\\right)^{3}}\n\\tag{11}\n$$\n\nIf the $\\mathrm{S}^{\\prime}$ frame is the proper frame, then by definition the velocity $v^{\\prime}=0$. Substitute this to the last equation, we get\n\n$$\na=\\frac{a^{\\prime}}{\\gamma^{3}}\n\\tag{12}\n$$\n\nCombining Eq.(3) and Eq.(12), we get\n\n$$\na^{\\prime}=\\frac{F}{m} \\equiv g\n\\tag{13}\n$$""]",,False,,, 1576,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} ","5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.",['Eq.(3) can also be rewritten as\n\n$$\nc \\frac{d \\beta}{\\gamma d \\tau}=\\frac{g}{\\gamma^{3}}\n\\tag{14}\n$$\n\n$$\n\\int_{0}^{\\beta} \\frac{d \\beta}{1-\\beta^{2}} =\\frac{g}{c} \\int_{0}^{\\tau} d \\tau\n$$\n$$\n\\ln \\left(\\frac{1}{\\sqrt{1-\\beta^{2}}}+\\frac{\\beta}{\\sqrt{1-\\beta^{2}}}\\right) =\\frac{g \\tau}{c}\n\\tag{15}\n$$\n$$\n\\sqrt{\\frac{1+\\beta}{1-\\beta}} =e^{\\frac{g \\tau}{c}}\n$$\n\n$$\n\\beta\\left(e^{\\frac{g \\tau}{c}}+e^{-\\frac{g \\tau}{c}}\\right) =e^{\\frac{g \\tau}{c}}-e^{-\\frac{g \\tau}{c}}\n$$\n\n$$\n\\beta =\\tanh \\frac{g \\tau}{c} .\n\\tag{16}\n$$'],['$\\beta=\\tanh \\frac{g \\tau}{c}$'],False,,Expression, 1577,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} ","6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.","['The time dilation relation is\n\n$$\nd t=\\gamma d \\tau\n\\tag{17}\n$$\n\nFrom eq.(16), we have\n\n$$\n\\gamma=\\frac{1}{\\sqrt{1-\\beta^{2}}}=\\cosh \\frac{g \\tau}{c}\n\\tag{18}\n$$\n\nCombining this equations, we get\n\n$$\n\\begin{aligned}\n\\int_{0}^{t} d t & =\\int_{0}^{\\tau} d \\tau \\cosh \\frac{g \\tau}{c} \\\\\nt & =\\frac{c}{g} \\sinh \\frac{g \\tau}{c} .\n\\end{aligned}\n\\tag{19}\n$$']",['$t=\\frac{c}{g} \\sinh \\frac{g \\tau}{c}$'],False,,Expression, 1578,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A.","1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value?","[""When the clock in the origin time is equal to $t_{0}$, it emits a signal that contain the information of its time. This signal will arrive at the particle at time $t$, while the particle position is at $x(t)$. We have\n\n$$\nc\\left(t-t_{0}\\right) =x(t)\n\\tag{20}\n$$\n$$\nt-t_{0}=\\frac{c}{g}\\left(\\sqrt{1+\\left(\\frac{g t}{c}\\right)^{2}}-1\\right)\n$$\n$$\nt =\\frac{t_{0}}{2} \\frac{2-\\frac{g t_{0}}{c}}{1-\\frac{g t_{0}}{c}} .\n\\tag{21}\n$$\n\nWhen the information arrive at the particle, the particle's clock has a reading according to eq.(19). So we get\n\n$$\n\\begin{aligned}\n\\frac{c}{g} \\sinh \\frac{g \\tau}{c} & =\\frac{t_{0}}{2} \\frac{2-\\frac{g t_{0}}{c}}{1-\\frac{g t_{0}}{c}} \\\\\n0 & =\\frac{1}{2}\\left(\\frac{g t_{0}}{c}\\right)^{2}-\\frac{g t_{0}}{c}\\left(1+\\sinh \\frac{g \\tau}{c}\\right)+\\sinh \\frac{g \\tau}{c} .\n\\end{aligned}\n$$\n\n$$\n\\frac{g t_{0}}{c} =1+\\sinh \\frac{g \\tau}{c} \\pm \\cosh \\frac{g \\tau}{c}\n\\tag{22}\n$$\n\nUsing initial condition $t=0$ when $\\tau=0$, we choose the negative sign\n\n$$\n\\begin{aligned}\n\\frac{g t_{0}}{c} & =1+\\sinh \\frac{g \\tau}{c}-\\cosh \\frac{g \\tau}{c} \\\\\nt_{0} & =\\frac{c}{g}\\left(1-e^{-\\frac{g \\tau}{c}}\\right) .\n\\end{aligned}\n\\tag{23}\n$$\n\nAs $\\tau \\rightarrow \\infty, t_{0}=\\frac{c}{g}$. So the clock reading will freeze at this value.""]",,False,,, 1579,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} ","2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value?","[""When the particles clock has a reading $\\tau_{0}$, its position is given by eq.(6), and the time $t_{0}$ is given by eq.(19). Combining this two equation, we get\n\n$$\nx=\\frac{c^{2}}{g}\\left(\\sqrt{1+\\sinh ^{2} \\frac{g \\tau_{0}}{c}}-1\\right)\n\\tag{24}\n$$\n\nThe particle's clock reading is then sent to the observer at the origin. The total time needed for the information to arrive is given by\n\n$$\nt =\\frac{c}{g} \\sinh \\frac{g \\tau_{0}}{c}+\\frac{x}{c}\n\\tag{25}\n$$\n\n$$\n=\\frac{c}{g}\\left(\\sinh \\frac{g \\tau_{0}}{c}+\\cosh \\frac{g \\tau_{0}}{c}-1\\right)\n$$\n\n$$\nt =\\frac{c}{g}\\left(e^{\\frac{g \\tau_{0}}{c}}-1\\right)\n\\tag{26}\n$$\n\n$$\n\\tau_{0} =\\frac{c}{g} \\ln \\left(\\frac{g t}{c}+1\\right)\n\\tag{27}\n$$\n\nThe time will not freeze.""]",,False,,, 1580,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$","1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame.","[""The figure below show the setting of the problem.\n\nThe line $\\mathrm{AB}$ represents the stick with proper length equal $L$ in the $\\mathrm{S}$ frame.\n\nThe length $\\mathrm{AB}$ is equal to $\\sqrt{\\frac{1-\\beta^{2}}{1+\\beta^{2}}} L$ in the $\\mathrm{S}$ ' frame.\n\nThe stick length in the $\\mathrm{S}^{\\prime}$ frame is represented by the line $\\mathrm{AC}$\n\n\n\nFigure 1: Minkowski Diagram\n\n$$\n\\mathrm{AC}=\\frac{\\mathrm{AB}}{\\cos \\theta}=\\sqrt{1-\\beta^{2}} L\n\\tag{28}\n$$""]",['$\\sqrt{1-\\beta^{2}} L$'],False,,Expression, 1581,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} ",2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$.,['The position of the particle is given by eq.(6).\n\n\n\n\nFigure 2: Minkowski Diagram'],,False,,, 1582,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part.","1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame.",['$\\tau_{B}=\\tau_{A}$'],['$\\tau_{B}=\\tau_{A}$'],False,,Expression, 1583,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} ","2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$.","['From the diagram, we have\n\n$$\n\\tan \\theta=\\beta=\\frac{c t_{2}-c t_{1}}{x_{2}-x_{1}}\n\\tag{29}\n$$\n\nUsing eq.(6), and eq.(19) along with the initial condition, we get\n\n$$\nx_{1}=\\frac{c^{2}}{g}\\left(\\cosh \\frac{g \\tau_{1}}{c}-1\\right),\n\\tag{30}\n$$\n\n$$\nx_{2}=\\frac{c^{2}}{g}\\left(\\cosh \\frac{g \\tau_{2}}{c}-1\\right)+L .\n\\tag{31}\n$$\n\nUsing eq.(16), eq.(19), eq.(30) and eq.(31), we obtain\n\n$$\n\\begin{aligned}\n\\tanh \\frac{g \\tau_{1}}{c} & =\\frac{c\\left(\\frac{c}{g} \\sinh \\frac{g \\tau_{2}}{c}-\\frac{c}{g} \\sinh \\frac{g \\tau_{1}}{c}\\right)}{L+\\frac{c^{2}}{g}\\left(\\cosh \\frac{g \\tau_{2}}{c}-1\\right)-\\frac{c^{2}}{g}\\left(\\cosh \\frac{g \\tau_{1}}{c}-1\\right)} \\\\\n& =\\frac{\\sinh \\frac{g \\tau_{2}}{c}-\\sinh \\frac{g \\tau_{1}}{c}}{\\frac{g L}{c^{2}}+\\cosh \\frac{g \\tau_{2}}{c}-\\cosh \\frac{g \\tau_{1}}{c}} \\\\\n\\frac{g L}{c^{2}} \\sinh \\frac{g \\tau_{1}}{c} & =\\sinh \\frac{g \\tau_{2}}{c} \\cosh \\frac{g \\tau_{1}}{c}-\\cosh \\frac{g \\tau_{2}}{c} \\sinh \\frac{g \\tau_{1}}{c} \\\\\n\\frac{g L}{c^{2}} \\sinh \\frac{g \\tau_{1}}{c} & =\\sinh \\frac{g}{c}\\left(\\tau_{2}-\\tau_{1}\\right) .\n\\end{aligned}\n\\tag{32}\n$$\n\nSo $C_{1}=\\frac{g L}{c^{2}}$.\n\n\n\n\n\nFigure 3: Minkowski Diagram for two particles']",['$C_{1}=\\frac{g L}{c^{2}}$'],False,,Expression, 1584,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} Context question: 2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$. Context answer: \boxed{$C_{1}=\frac{g L}{c^{2}}$} ","3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is $$ \frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)} $$ where $C_{2}$ is a constant. Determine $C_{2}$.","['From the length contraction, we have\n\n$$\nL^{\\prime} =\\frac{x_{2}-x_{1}}{\\gamma_{1}}\n\\tag{33}\n$$\n\n$$\n\\frac{d L^{\\prime}}{d \\tau_{1}} =\\left(\\frac{d x_{2}}{d \\tau_{2}} \\frac{d \\tau_{2}}{d \\tau_{1}}-\\frac{d x_{1}}{d \\tau_{1}}\\right) \\frac{1}{\\gamma_{1}}-\\frac{x_{2}-x_{1}}{\\gamma_{1}^{2}} \\frac{d \\gamma_{1}}{d \\tau_{1}}\n\\tag{34}\n$$\n\nTake derivative of eq.(30), eq.(31) and eq.(32), we get\n\n$$\n\\frac{d x_{1}}{d \\tau_{1}} =c \\sinh \\frac{g \\tau_{1}}{c}\n\\tag{35}\n$$\n\n$$\n\\frac{d x_{2}}{d \\tau_{2}} =c \\sinh \\frac{g \\tau_{2}}{c}\n\\tag{36}\n$$\n\n$$\n\\frac{g L}{c^{2}} \\cosh \\frac{g \\tau_{1}}{c} =\\cosh \\frac{g}{c}\\left(\\tau_{2}-\\tau_{1}\\right)\\left(\\frac{d \\tau_{2}}{d \\tau_{1}}-1\\right) .\n\\tag{37}\n$$\n\nThe last equation can be rearrange to get\n\n$$\n\\frac{d \\tau_{2}}{d \\tau_{1}}=\\frac{\\frac{g L}{c^{2}} \\cosh \\frac{g \\tau_{1}}{c}}{\\cosh \\frac{g}{c}\\left(\\tau_{2}-\\tau_{1}\\right)}+1\n\\tag{38}\n$$\n\n\n\nFrom eq.(29), we have\n\n$$\nx_{2}-x_{1}=\\frac{c\\left(t_{2}-t_{1}\\right)}{\\beta_{1}}=\\frac{c}{\\tanh \\frac{g \\tau_{1}}{c}}\\left(\\frac{c}{g} \\sinh \\frac{g \\tau_{2}}{c}-\\frac{c}{g} \\sinh \\frac{g \\tau_{1}}{c}\\right)\n\\tag{39}\n$$\n\nCombining all these equations, we get\n\n$$\n\\begin{aligned}\n& \\frac{d L_{1}}{d \\tau_{1}}=\\left(c \\sinh \\frac{g \\tau_{2}}{c} \\frac{\\frac{g L}{c^{2}} \\cosh \\frac{g \\tau_{1}}{c}}{\\cosh \\frac{g}{c}\\left(\\tau_{2}-\\tau_{1}\\right)}+c \\sinh \\frac{g \\tau_{2}}{c}-c \\sinh \\frac{g \\tau_{1}}{c}\\right) \\frac{1}{\\cosh \\frac{g \\tau_{1}}{c}} \\\\\n& -\\frac{c^{2}}{g}\\left(\\sinh \\frac{g \\tau_{2}}{c}-\\sinh \\frac{g \\tau_{1}}{c}\\right) \\frac{1}{\\tanh \\frac{g \\tau_{1}}{c}} \\frac{1}{\\cosh ^{2} \\frac{g \\tau_{1}}{c}} \\frac{g}{c} \\sinh \\frac{g \\tau_{1}}{c} \\\\\n& \\frac{d L_{1}}{d \\tau_{1}}=\\frac{g L}{c} \\frac{\\sinh \\frac{g \\tau_{2}}{c}}{\\cosh \\frac{g}{c}\\left(\\tau_{2}-\\tau_{1}\\right)}\n\\end{aligned}\n\\tag{40}\n$$\n\nSo $C_{2}=\\frac{g L}{c}$.']",['$C_{2}=\\frac{g L}{c}$'],False,,Expression, 1585,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} Context question: 2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$. Context answer: \boxed{$C_{1}=\frac{g L}{c^{2}}$} Context question: 3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is $$ \frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)} $$ where $C_{2}$ is a constant. Determine $C_{2}$. Context answer: \boxed{$C_{2}=\frac{g L}{c}$} Extra Supplementary Reading Materials: Part E. Uniformly Accelerated Frame In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$.","1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$.","[""Distance from a certain point $x_{p}$ according to the particle's frame is\n\n$$\nL^{\\prime} =\\frac{x-x_{p}}{\\gamma}\n\\tag{41}\n$$\n\n$$\nL^{\\prime} =\\frac{\\frac{c^{2}}{g_{1}}\\left(\\cosh \\frac{g_{1} \\tau}{c}-1\\right)-x_{p}}{\\cosh \\frac{g_{1} \\tau}{c}}\n$$\n\n$$\nL^{\\prime} =\\frac{c^{2}}{g_{1}}-\\frac{\\frac{c^{2}}{g_{1}}+x_{p}}{\\cosh \\frac{g_{1} \\tau}{c}} .\n\\tag{42}\n$$\n\nFor $L^{\\prime}$ equal constant, we need $x_{p}=-\\frac{c^{2}}{g_{1}}$.""]",['$x_{p}=-\\frac{c^{2}}{g_{1}}$'],False,,Expression, 1586,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} Context question: 2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$. Context answer: \boxed{$C_{1}=\frac{g L}{c^{2}}$} Context question: 3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is $$ \frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)} $$ where $C_{2}$ is a constant. Determine $C_{2}$. Context answer: \boxed{$C_{2}=\frac{g L}{c}$} Extra Supplementary Reading Materials: Part E. Uniformly Accelerated Frame In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$. Context question: 1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$. Context answer: \boxed{$x_{p}=-\frac{c^{2}}{g_{1}}$} ","2. Given the proper acceleration of the first particle is $g_{1}$, determine the proper acceleration of the second particle $g_{2}$, so that the distance between the two particles are constant according to the first particle.","[""First method: If the distance in the $\\mathrm{S}$ ' frame is constant $=L$, then in the $\\mathrm{S}$ frame the length is\n\n$$\nL_{s}=L \\sqrt{\\frac{1+\\beta^{2}}{1-\\beta^{2}}}\n\\tag{43}\n$$\n\nSo the position of the second particle is\n\n$$\nx_{2}=x_{1}+L_{s} \\cos \\theta\n\\tag{44}\n$$\n\n$$\n\\begin{aligned}\n& =\\frac{c^{2}}{g_{1}}\\left(\\sqrt{1+\\left(\\frac{g_{1} t_{1}}{c}\\right)^{2}}-1\\right)+L \\sqrt{1+\\left(\\frac{g_{1} t_{1}}{c}\\right)} \\\\\nx_{2} & =\\left(\\frac{c^{2}}{g_{1}}+L\\right) \\sqrt{1+\\left(\\frac{g_{1} t_{1}}{c}\\right)^{2}}-\\frac{c^{2}}{g_{1}} .\n\\end{aligned}\n\\tag{45}\n$$\n\n\n\n\n\nFigure 4: Minkowski Diagram for two particles\n\nThe time of the second particle is\n\n$$\n\\begin{aligned}\nc t_{2} & =c t_{1}+L_{s} \\sin \\theta \\\\\n& =c t_{1}+L \\sqrt{\\frac{1+\\beta^{2}}{1-\\beta^{2}}} \\frac{\\beta}{\\sqrt{1+\\beta^{2}}}\n\\end{aligned}\n\\tag{46}\n$$\n\n$$\nc t_{2}=t_{1}\\left(c+\\frac{g_{1} L}{c}\\right) .\n\\tag{47}\n$$\n\nSubstitute eq.(47) to eq.(45) to get\n\n$$\nx_{2}=\\left(\\frac{c^{2}}{g_{1}}+L\\right) \\sqrt{1+\\left(\\frac{g_{1}}{c} \\frac{t_{2}}{1+\\frac{g_{1} L}{c^{2}}}\\right)^{2}}-\\frac{c^{2}}{g_{1}}\n$$\n\n$$\nx_{2}=\\left(\\frac{c^{2}}{g_{1}}+L\\right) \\sqrt{1+\\left(\\frac{g_{1}}{1+\\frac{g_{1} L}{c^{2}}} \\frac{t_{2}}{c}\\right)^{2}}-\\frac{c^{2}}{g_{1}} .\n\\tag{48}\n$$\n\nFrom the last equation, we can identify\n\n$$\ng_{2} \\equiv \\frac{g_{1}}{1+\\frac{g_{1} L}{c^{2}}}\n\\tag{49}\n$$\n\n\n\nAs for confirmation, we can subsitute this relation to the second particle position to get\n\n$$\nx_{2}=\\frac{c^{2}}{g_{2}} \\sqrt{1+\\left(\\frac{g_{2} t_{2}}{c}\\right)^{2}}-\\frac{c^{2}}{g_{1}}\n\\tag{50}\n$$\n\nSecond method: In this method, we will choose $g_{2}$ such that the special point like the one descirbe in the question 1 is exactly the same as the similar point for the proper acceleration $g_{1}$.\n\nFor first particle, we have $x_{p 1} g_{1}=c^{2}$\n\nFor second particle, we have $\\left(L+x_{p 1}\\right) g_{2}=c^{2}$\n\nCombining this two equations, we get\n\n$$\n\\begin{aligned}\ng_{2} & =\\frac{c^{2}}{L+\\frac{c^{2}}{g_{1}}} \\\\\ng_{2} & =\\frac{g_{1}}{1+\\frac{g_{1} L}{c^{2}}} .\n\\end{aligned}\n\\tag{51}\n$$""]",['$g_{2}=\\frac{g_{1}}{1+\\frac{g_{1} L}{c^{2}}}$'],False,,Expression, 1587,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} Context question: 2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$. Context answer: \boxed{$C_{1}=\frac{g L}{c^{2}}$} Context question: 3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is $$ \frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)} $$ where $C_{2}$ is a constant. Determine $C_{2}$. Context answer: \boxed{$C_{2}=\frac{g L}{c}$} Extra Supplementary Reading Materials: Part E. Uniformly Accelerated Frame In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$. Context question: 1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$. Context answer: \boxed{$x_{p}=-\frac{c^{2}}{g_{1}}$} Context question: 2. Given the proper acceleration of the first particle is $g_{1}$, determine the proper acceleration of the second particle $g_{2}$, so that the distance between the two particles are constant according to the first particle. Context answer: \boxed{$g_{2}=\frac{g_{1}}{1+\frac{g_{1} L}{c^{2}}}$} ","3. What is the ratio of time rate of the second particle to the first particle $\frac{d \tau_{2}}{d \tau_{1}}$, according to the first particle.",['The relation between the time in the two particles is given by eq.(47)\n\n$$\n\\begin{aligned}\nt_{2} & =t_{1}\\left(1+\\frac{g_{1} L}{c^{2}}\\right) \\\\\n\\frac{c^{2}}{g_{2}} \\sinh \\frac{g_{2} \\tau_{2}}{c} & =\\frac{c^{2}}{g_{1}} \\sinh \\frac{g_{1} \\tau_{1}}{c}\\left(1+\\frac{g_{1} L}{c^{2}}\\right) \\\\\n\\sinh \\frac{g_{2} \\tau_{2}}{c} & =\\sinh \\frac{g_{1} \\tau_{1}}{c}\n\\end{aligned}\n$$\n\n$$\ng_{2} \\tau_{2} =g_{1} \\tau_{1}\n\\tag{52}\n$$\n\n$$\n\\frac{d \\tau_{2}}{d \\tau_{1}} =\\frac{g_{1}}{g_{2}}=1+\\frac{g_{1} L}{c^{2}}\n\\tag{53}\n$$'],['$1+\\frac{g_{1} L}{c^{2}}$'],False,,Expression, 1588,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} Context question: 2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$. Context answer: \boxed{$C_{1}=\frac{g L}{c^{2}}$} Context question: 3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is $$ \frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)} $$ where $C_{2}$ is a constant. Determine $C_{2}$. Context answer: \boxed{$C_{2}=\frac{g L}{c}$} Extra Supplementary Reading Materials: Part E. Uniformly Accelerated Frame In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$. Context question: 1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$. Context answer: \boxed{$x_{p}=-\frac{c^{2}}{g_{1}}$} Context question: 2. Given the proper acceleration of the first particle is $g_{1}$, determine the proper acceleration of the second particle $g_{2}$, so that the distance between the two particles are constant according to the first particle. Context answer: \boxed{$g_{2}=\frac{g_{1}}{1+\frac{g_{1} L}{c^{2}}}$} Context question: 3. What is the ratio of time rate of the second particle to the first particle $\frac{d \tau_{2}}{d \tau_{1}}$, according to the first particle. Context answer: \boxed{$1+\frac{g_{1} L}{c^{2}}$} Extra Supplementary Reading Materials: Part F. Correction for GPS Part E indicates that the time rate of clocks at different altitude will not be the same, even though there is no relative movement between those clocks. According to the equivalence principle in general relativity, an observer in a small closed room could not tell the difference between a gravity pull $g$ and the fictitious force from accelerated frame with acceleration $g$. So we can conclude that two clocks at different gravitational potential will have different rate. Now let consider a GPS satellite that orbiting the Earth with a period of 12 hours.","1. If the gravitational acceleration on the Earth's surface is $9.78 \mathrm{~m} \cdot \mathrm{s}^{-2}$, and the Earth's radius is $6380 \mathrm{~km}$, what is the radius of the GPS satellite orbit? What is the velocity of the satellite? Calculate the numerical values of the radius and the velocity.",['From Newtons Law\n\n\n$$\n\\frac{G M m}{r^{2}} =m \\omega^{2} r\n\\tag{54}\n$$\n\n$$\n\\begin{aligned}\nr & =\\left(\\frac{g R^{2} T^{2}}{4 \\pi^{2}}\\right)^{\\frac{1}{3}} \\\\\nr & =2.66 \\times 10^{7} \\mathrm{~m} .\n\\end{aligned}\n\\tag{55}\n$$\n\nThe velocity is given by\n\n$$\n\\begin{aligned}\nv & =\\omega r=\\left(\\frac{2 \\pi g R^{2}}{T}\\right)^{\\frac{1}{3}} \\\\\n& =3.87 \\times 10^{3} \\mathrm{~m} / \\mathrm{s} .\n\\end{aligned}\n\\tag{56}\n$$'],['$r =2.66 \\times 10^{7}$ and $v=3.87 \\times 10^{3}$'],True,"m, m/s",Numerical,"5e6,2e2" 1589,Modern Physics,,"2. After one day, the clock reading on the Earth surface and the satellite will differ due to both special and general relativistic effects. Calculate the difference due to each effect for one day. Calculate the total difference for one day. Which clock is faster, a clock on the Earth's surface or the satellite's clock?","[""The general relativity effect is\n\n$$\n\\begin{aligned}\n\\frac{d \\tau_{g}}{d t} & =1+\\frac{\\Delta U}{m c^{2}} \\\\\n\\frac{d \\tau_{g}}{d t} & =1+\\frac{g R^{2}}{c^{2}} \\frac{R-r}{R r}\n\\end{aligned}\n$$\n\nAfter one day, the difference is\n\n$$\n\\begin{aligned}\n\\Delta \\tau_{g} & =\\frac{g R^{2}}{c^{2}} \\frac{R-r}{R r} \\Delta T \\\\\n& =4.55 \\times 10^{-5} \\mathrm{~s} .\n\\end{aligned}\n$$\n\nThe special relativity effect is\n\n$$\n\\begin{aligned}\n\\frac{d \\tau_{s}}{d t} & =\\sqrt{1-\\frac{v^{2}}{c^{2}}} \\\\\n& =\\sqrt{1-\\left(\\left(\\frac{2 \\pi g R^{2}}{T}\\right)^{\\frac{2}{3}}\\right) \\frac{1}{c^{2}}} \\\\\n& \\approx 1-\\frac{1}{2}\\left(\\left(\\frac{2 \\pi g R^{2}}{T}\\right)^{\\frac{2}{3}}\\right) \\frac{1}{c^{2}} .\n\\end{aligned}\n$$\n\nAfter one day, the difference is\n\n$$\n\\begin{aligned}\n\\Delta \\tau_{s} & =-\\frac{1}{2}\\left(\\left(\\frac{2 \\pi g R^{2}}{T}\\right)^{\\frac{2}{3}}\\right) \\frac{1}{c^{2}} \\Delta T \\\\\n& =-7.18 \\times 10^{-6} \\mathrm{~s} .\n\\end{aligned}\n$$\n\nThe satelite's clock is faster with total $\\Delta \\tau=\\Delta \\tau_{g}+\\Delta \\tau_{s}=3.83 \\times 10^{-5} \\mathrm{~s}$.""]","[""by general relativity effect, after one day, the difference is $4.55 \\times 10^{-5} \\mathrm{~s}$\nby special relativity effect, after one day, the difference is $-7.18 \\times 10^{-6} \\mathrm{~s}$\nThe satelite's clock is faster with total $\\Delta \\tau=\\Delta \\tau_{g}+\\Delta \\tau_{s}=3.83 \\times 10^{-5} \\mathrm{~s}$""]",True,,Need_human_evaluate, 1590,Modern Physics,"Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$} Context question: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$. Context answer: \boxed{$t=\frac{c}{g} \sinh \frac{g \tau}{c}$} Extra Supplementary Reading Materials: Part B. Flight Time The first part has not taken into account the flight time of the information to arrive to the observer. This part is the only part in the whole problem where the flight time is considered. The particle moves as in part A. Context question: 1. At a certain moment, the time experienced by the particle is $\tau$. What reading $t_{0}$ on a stationary clock located at $x=0$ will be observed by the particle? After a long period of time, does the observed reading $t_{0}$ approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Context question: 2. Now consider the opposite point of view. If an observer at the initial point $(x=0)$ is observing the particle's clock when the observer's time is $t$, what is the reading of the particle's clock $\tau_{0}$ ? After a long period of time, will this reading approach a certain value? If so, what is the value? Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part C. Minkowski Diagram In many occasion, it is very useful to illustrate relativistic events using a diagram, called as Minkowski Diagram. To make the diagram, we just need to use Lorentz transformation between the rest frame $S$ and the moving frame $S^{\prime}$ that move with velocity $v=\beta c$ with respect to the rest frame. $$ \begin{aligned} & x=\gamma\left(x^{\prime}+\beta c t^{\prime}\right), \\ & c t=\gamma\left(c t^{\prime}+\beta x^{\prime}\right), \\ & x^{\prime}=\gamma(x-\beta c t), \\ & c t^{\prime}=\gamma(c t-\beta x) . \\ & \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{aligned} $$ Let's choose $x$ and $c t$ as the orthogonal axes. A point $\left(x^{\prime}, c t^{\prime}\right)=(1,0)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma, \gamma \beta)$ in the rest frame $S$. The line connecting this point and the origin defines the $x^{\prime}$ axis. Another point $\left(x^{\prime}, c t^{\prime}\right)=(0,1)$ in the moving frame $S^{\prime}$ has a coordinate $(x, c t)=(\gamma \beta, \gamma)$ in the rest frame $S$. The line connecting this point and the origin defines the $c t^{\prime}$ axis. The angle between the $x$ and $\mathrm{x}^{\prime}$ axis is $\theta$, where $\tan \theta=\beta$. A unit length in the moving frame $S^{\prime}$ is equal to $\gamma \sqrt{1+\beta^{2}}=\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}}$ in the rest frame S. To get a better understanding of Minkowski diagram, let us take a look at this example. Consider a stick of proper length $L$ in a moving frame S'. We would like to find the length of the stick in the rest frame S. Consider the figure below. The stick is represented by the segment AC. The length $\mathrm{AC}$ is equal to $\sqrt{\frac{1+\beta^{2}}{1-\beta^{2}}} L$ in the $S$ frame. The stick length in the $S$ frame is represented by the line $A B$. $$ \begin{aligned} \mathrm{AB} & =\mathrm{AD}-\mathrm{BD} \\ & =A C \cos \theta-A C \sin \theta \tan \theta \\ & =L \sqrt{1-\beta^{2}} \end{aligned} $$ Context question: 1. Using a Minkowski diagram, calculate the length of a stick with proper length $L$ in the rest frame, as measured in the moving frame. Context answer: \boxed{$\sqrt{1-\beta^{2}} L$} Context question: 2. Now consider the case in part A. Plot the time ct versus the position $x$ of the particle. Draw the $x^{\prime}$ axis and $c t^{\prime}$ axis when $\frac{g t}{c}=1$ in the same graph using length scale $x\left(c^{2} / g\right)$ and $c t\left(c^{2} / g\right)$. Context answer: \boxed{证明题} Extra Supplementary Reading Materials: Part D. Two Accelerated Particles For this part, we will consider two accelerated particles, both of them have the same proper acceleration $g$ in the positive $x$ direction, but the first particle starts from $x=0$, while the second particle starts from $x=L$. Remember, DO NOT consider the flight time in this part. Context question: 1. After a while, an observer in the rest frame make an observation. The first particle's clock shows time at $\tau_{A}$. What is the reading of the second clock $\tau_{B}$, according to the observer in the rest frame. Context answer: \boxed{$\tau_{B}=\tau_{A}$} Context question: 2. Now consider the observation from the first particle's frame. At a certain moment, an observer that move together with the first particle observed that the reading of his own clock is $\tau_{1}$. At the same time, he observed the second particle's clock, and the reading is $\tau_{2}$. Show that $$ \sinh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)=C_{1} \sinh \frac{g \tau_{1}}{c} $$ where $C_{1}$ is a constant. Determine $C_{1}$. Context answer: \boxed{$C_{1}=\frac{g L}{c^{2}}$} Context question: 3. The first particle will see the second particle move away from him. Show that the rate of change of the distance between the two particles according to the first particle is $$ \frac{d L^{\prime}}{d \tau_{1}}=C_{2} \frac{\sinh \frac{g \tau_{2}}{c}}{\cosh \frac{g}{c}\left(\tau_{2}-\tau_{1}\right)} $$ where $C_{2}$ is a constant. Determine $C_{2}$. Context answer: \boxed{$C_{2}=\frac{g L}{c}$} Extra Supplementary Reading Materials: Part E. Uniformly Accelerated Frame In this part we will arrange the proper acceleration of the particles, so that the distance between both particles are constant according to each particle. Initially both particles are at rest, the first particle is at $x=0$, while the second particle is at $x=L$. Context question: 1. The first particle has a proper acceleration $g_{1}$ in the positive $x$ direction. When it is being accelerated, there exists a fixed point in the rest frame at $x=x_{\mathrm{p}}$ that has a constant distance from the first particle, according to the first particle thoughout the motion. Determine $x_{p}$. Context answer: \boxed{$x_{p}=-\frac{c^{2}}{g_{1}}$} Context question: 2. Given the proper acceleration of the first particle is $g_{1}$, determine the proper acceleration of the second particle $g_{2}$, so that the distance between the two particles are constant according to the first particle. Context answer: \boxed{$g_{2}=\frac{g_{1}}{1+\frac{g_{1} L}{c^{2}}}$} Context question: 3. What is the ratio of time rate of the second particle to the first particle $\frac{d \tau_{2}}{d \tau_{1}}$, according to the first particle. Context answer: \boxed{$1+\frac{g_{1} L}{c^{2}}$} Extra Supplementary Reading Materials: Part F. Correction for GPS Part E indicates that the time rate of clocks at different altitude will not be the same, even though there is no relative movement between those clocks. According to the equivalence principle in general relativity, an observer in a small closed room could not tell the difference between a gravity pull $g$ and the fictitious force from accelerated frame with acceleration $g$. So we can conclude that two clocks at different gravitational potential will have different rate. Now let consider a GPS satellite that orbiting the Earth with a period of 12 hours. Context question: 1. If the gravitational acceleration on the Earth's surface is $9.78 \mathrm{~m} \cdot \mathrm{s}^{-2}$, and the Earth's radius is $6380 \mathrm{~km}$, what is the radius of the GPS satellite orbit? What is the velocity of the satellite? Calculate the numerical values of the radius and the velocity. Context answer: \boxed{$r =2.66 \times 10^{7}$ and $v=3.87 \times 10^{3}$} Context question: 2. After one day, the clock reading on the Earth surface and the satellite will differ due to both special and general relativistic effects. Calculate the difference due to each effect for one day. Calculate the total difference for one day. Which clock is faster, a clock on the Earth's surface or the satellite's clock? Context answer: by general relativity effect, after one day, the difference is $4.55 \times 10^{-5} \mathrm{~s}$ by special relativity effect, after one day, the difference is $-7.18 \times 10^{-6} \mathrm{~s}$ The satelite's clock is faster with total $\Delta \tau=\Delta \tau_{g}+\Delta \tau_{s}=3.83 \times 10^{-5} \mathrm{~s}$ ","3. After one day, estimate the error in position due to this effect?",['$\\Delta L=c \\Delta \\tau=1.15 \\times 10^{4} \\mathrm{~m}=11.5 \\mathrm{~km}$.'],['11.5'],False,km,Numerical,1e-1 1591,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession","1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.)","['From the two equations given in the text, we obtain the relation\n\n$$\n\\frac{d \\boldsymbol{\\mu}}{d t}=-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B}\n\\tag{1}\n$$\n\nTaking the dot product of eq (1). with $\\boldsymbol{\\mu}$, we can prove that\n\n$$\n\\begin{aligned}\n\\boldsymbol{\\mu} \\cdot \\frac{d \\boldsymbol{\\mu}}{d t} & =-\\gamma \\boldsymbol{\\mu} \\cdot(\\boldsymbol{\\mu} \\times \\mathbf{B}) \\\\\n\\frac{d|\\boldsymbol{\\mu}|^{2}}{d t} & =0 \\\\\n\\mu=|\\boldsymbol{\\mu}| & =\\text { const. }\n\\end{aligned}\n\\tag{2}\n$$\n\nTaking the dot product of eq. (1) with $\\mathbf{B}$, we also prove that\n\n$$\n\\begin{aligned}\n\\mathbf{B} \\cdot \\frac{d \\boldsymbol{\\mu}}{d t} & =-\\gamma \\mathbf{B} \\cdot(\\boldsymbol{\\mu} \\times \\mathbf{B}), \\\\\n\\mathbf{B} \\cdot \\frac{d \\boldsymbol{\\mu}}{d t} & =0, \\\\\n\\mathbf{B} \\cdot \\boldsymbol{\\mu} & =\\text { const. }\n\\end{aligned}\n\\tag{3}\n$$\n\nAn acute reader will notice that our master equation in (1) is identical to the equation of motion for a charged particle in a magnetic field\n\n$$\n\\frac{d \\mathbf{v}}{d t}=\\frac{q}{m} \\mathbf{v} \\times \\mathbf{B}\n\\tag{4}\n$$\n\nHence, the same argument for a charged particle in magnetic field can be applied in this case.']",,False,,, 1592,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} ","2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.","['For a magnetic moment making an angle of $\\phi$ with $\\mathbf{B}$,\n\n$$\n\\begin{aligned}\n\\frac{d \\boldsymbol{\\mu}}{d t} & =-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B} \\\\\n|\\boldsymbol{\\mu}| \\sin \\phi \\frac{d \\theta}{d t} & =\\gamma|\\boldsymbol{\\mu}| B_{0} \\sin \\phi \\\\\n\\omega_{0}=\\frac{d \\theta}{d t} & =\\gamma B_{0}\n\\end{aligned}\n\\tag{5}\n$$']",['$\\omega_{0}=\\gamma B_{0}$'],False,,Expression, 1593,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$.","1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field.","['Using the relation given in the text, it is easily shown that\n\n$$\n\\begin{aligned}\n\\left(\\frac{d \\boldsymbol{\\mu}}{d t}\\right)_{\\mathrm{rot}} & =\\left(\\frac{d \\boldsymbol{\\mu}}{d t}\\right)_{\\mathrm{lab}}-\\boldsymbol{\\omega} \\times \\boldsymbol{\\mu} \\\\\n& =-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B}-\\omega \\mathbf{k}^{\\prime} \\times \\boldsymbol{\\mu} \\\\\n& =-\\gamma \\boldsymbol{\\mu} \\times\\left(\\mathbf{B}-\\frac{\\omega}{\\gamma} \\mathbf{k}^{\\prime}\\right) \\\\\n& =-\\gamma \\boldsymbol{\\mu} \\times \\mathbf{B}_{\\mathrm{eff}} .\n\\end{aligned}\n\\tag{6}\n$$\n\nNote that $\\mathbf{k}$ is equal to $\\mathbf{k}^{\\prime}$ as observed in the rotating frame.']",,False,,, 1594,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} ","2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?",['The new precession frequency as viewed on the rotating frame $S^{\\prime}$ is\n\n$$\n\\begin{aligned}\n\\vec{\\Delta} & =\\left(\\omega_{0}-\\omega\\right) \\mathbf{k}^{\\prime} \\\\\n\\Delta & =\\gamma B_{0}-\\omega .\n\\end{aligned}\n\\tag{7}\n$$'],['$\\Delta =\\gamma B_{0}-\\omega$'],False,,Expression, 1595,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} ","3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$","['Since the magnetic field as viewed in the rotating frame is $\\mathbf{B}=B_{0} \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}$,\n\n$$\n\\mathbf{B}_{\\mathrm{eff}}=\\mathbf{B}-\\omega / \\gamma \\mathbf{k}^{\\prime}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}\n$$\n\nand\n\n$$\n\\begin{aligned}\n\\Omega & =\\gamma\\left|\\mathbf{B}_{\\mathrm{eff}}\\right| \\\\\n& =\\gamma \\sqrt{\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right)^{2}+b^{2}}\n\\end{aligned}\n\\tag{8}\n$$']",,False,,, 1596,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} ","4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?","['In this case, the effective magnetic field becomes\n\n$$\n\\begin{aligned}\n\\mathbf{B}_{\\mathrm{eff}} & =\\mathbf{B}-\\omega / \\gamma \\mathbf{k}^{\\prime} \\\\\n& =\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b\\left(\\cos 2 \\omega t \\mathbf{i}^{\\prime}-\\sin 2 \\omega t \\mathbf{j}^{\\prime}\\right)\n\\end{aligned}\n\\tag{9}\n$$\n\nwhich has a time average of $\\overline{\\mathbf{B}_{\\mathrm{eff}}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}$.']","['$\\mathbf{B}_{\\mathrm{eff}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b\\left(\\cos 2 \\omega t \\mathbf{i}^{\\prime}-\\sin 2 \\omega t \\mathbf{j}^{\\prime}\\right)$ , $\\overline{\\mathbf{B}_{\\mathrm{eff}}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}$']",True,,Expression, 1597,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$","1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ?","['The oscillating field can be considered as a superposition of two oppositely rotating field:\n\n$$\n2 b \\cos \\omega_{0} t \\mathbf{i}=b\\left(\\cos \\omega_{0} t \\mathbf{i}+\\sin \\omega_{0} t \\mathbf{j}\\right)+b\\left(\\cos \\omega_{0} t \\mathbf{i}-\\sin \\omega_{0} t \\mathbf{j}\\right)\n$$\n\nwhich gives an effective field of (with $\\omega=\\omega_{0}=\\gamma B_{0}$ ):\n\n$$\n\\mathbf{B}_{\\mathrm{eff}}=\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}+b\\left(\\cos 2 \\omega_{0} t \\mathbf{i}^{\\prime}-\\sin 2 \\omega_{0} t \\mathbf{j}^{\\prime}\\right)\n$$\n\nSince $\\omega_{0} \\gg \\gamma b$, the rotation of the term $b\\left(\\cos 2 \\omega_{0} t \\mathbf{i}^{\\prime}-\\sin 2 \\omega_{0} t \\mathbf{j}^{\\prime}\\right)$ is so fast compared to the frequency $\\gamma b$. This means that we can take the approximation\n\n$$\n\\mathbf{B}_{\\mathrm{eff}} \\approx\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right) \\mathbf{k}^{\\prime}+b \\mathbf{i}^{\\prime}=b \\mathbf{i}^{\\prime}\n\\tag{10}\n$$\n\nwhere the magnetic moment precesses with frequency $\\Omega=\\gamma b$.\n\nAs $\\Omega=\\gamma b \\ll \\omega_{0}$, the magnetic moment does not ""feel"" the rotating term $b\\left(\\cos 2 \\omega_{0} t \\mathbf{i}^{\\prime}-\\sin 2 \\omega_{0} t \\mathbf{j}^{\\prime}\\right)$ which averaged to zero.']",['$\\Omega=\\gamma b$'],False,,Expression, 1598,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$ Context question: 1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ? Context answer: \boxed{$\Omega=\gamma b$} ","2. Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as $$ M(t)=N \mu(\cos \Omega t) . $$","['Since the angle $\\alpha$ that $\\boldsymbol{\\mu}$ makes with $\\mathbf{B}_{\\text {eff }}$ stays constant and $\\boldsymbol{\\mu}$ is initially oriented along the $z$ axis, $\\alpha$ is also the angle between $\\mathbf{B}_{\\text {eff }}$ and the $z$ axis which is\n\n$$\n\\tan \\alpha=\\frac{b}{B_{0}-\\frac{\\omega}{\\gamma}}\n\\tag{11}\n$$\n\n\n\nFrom the geometry of the system, we can show that $\\left(\\cos \\theta=\\mu_{z} / \\mu\\right)$ :\n\n$$\n\\begin{aligned}\n2 \\mu \\sin \\frac{\\theta}{2} & =2 \\mu \\sin \\alpha \\sin \\frac{\\Omega t}{2} \\\\\n\\sin ^{2} \\frac{\\theta}{2} & =\\sin ^{2} \\alpha \\sin ^{2} \\frac{\\Omega t}{2} \\\\\n\\frac{1-\\cos \\theta}{2} & =\\sin ^{2} \\alpha \\frac{1-\\cos \\Omega t}{2} \\\\\n\\cos \\theta & =1-\\sin ^{2} \\alpha+\\sin ^{2} \\alpha \\cos \\Omega t \\\\\n\\cos \\theta & =\\cos ^{2} \\alpha+\\sin ^{2} \\alpha \\cos \\Omega t .\n\\end{aligned}\n$$\n\nSo, the projected magnetic moment along the $z$ axis is $\\mu_{z}(t)=\\mu \\cos \\theta$ and the magnetization is\n\n$$\nM=N \\mu_{z}=N \\mu\\left(\\cos ^{2} \\alpha+\\sin ^{2} \\alpha \\cos \\Omega t\\right) .\n$$\n\nNote that the magnetization does not depend on the reference frame $S$ or $S^{\\prime}$ ( $\\mu_{z}$ has the same value viewed in both frames).\n\nTaking $\\omega=\\omega_{0}=\\gamma B_{0}$, the angle $\\alpha$ is $90^{\\circ}$ and $M=N \\mu \\cos \\Omega t$.']",,False,,, 1599,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$ Context question: 1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ? Context answer: \boxed{$\Omega=\gamma b$} Context question: 2. Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as $$ M(t)=N \mu(\cos \Omega t) . $$ Context answer: \boxed{证明题} ","3. Under the application of magnetic field described above, determine the fractional population of each spin up $P_{\uparrow}=N_{\uparrow} / N$ and spin down $P_{\downarrow}=N_{\downarrow} / N$ as a function of time. Plot $P_{\uparrow}(t)$ and $P_{\downarrow}(t)$ on the same graph vs. time $t$. The alternating spin up and spin down population as a function of time is called Rabi oscillation.","['From the relations\n\n$$\n\\begin{aligned}\nP_{\\uparrow}-P_{\\downarrow} & =\\frac{\\mu_{z}}{\\mu}=\\cos \\theta, \\\\\nP_{\\uparrow}+P_{\\downarrow} & =1,\n\\end{aligned}\n$$\n\n\n\nwe obtain the results $\\left(\\omega=\\omega_{0}\\right)$\n\n$$\n\\begin{aligned}\nP_{\\downarrow} & =\\frac{1-\\cos \\theta}{2} \\\\\n& =\\frac{1-\\cos ^{2} \\alpha-\\sin ^{2} \\alpha \\cos \\Omega t}{2} \\\\\n& =\\sin ^{2} \\alpha \\frac{1-\\cos \\Omega t}{2} \\\\\n& =\\frac{b^{2}}{\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right)^{2}+b^{2}} \\sin ^{2} \\frac{\\Omega t}{2} \\\\\n& =\\sin ^{2} \\frac{\\Omega t}{2},\n\\end{aligned}\n\\tag{13}\n$$\n\nand\n\n$$\nP_{\\uparrow}=\\frac{b^{2}}{\\left(B_{0}-\\frac{\\omega}{\\gamma}\\right)^{2}+b^{2}} \\cos ^{2} \\frac{\\Omega t}{2}=\\cos ^{2} \\frac{\\Omega t}{2}\n\\tag{14}\n$$\n\n']","['$P_{\\downarrow}=\\sin ^{2} \\frac{\\Omega t}{2}$ , $P_{\\uparrow}=\\cos ^{2} \\frac{\\Omega t}{2}$']",True,,Expression, 1600,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$ Context question: 1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ? Context answer: \boxed{$\Omega=\gamma b$} Context question: 2. Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as $$ M(t)=N \mu(\cos \Omega t) . $$ Context answer: \boxed{证明题} Context question: 3. Under the application of magnetic field described above, determine the fractional population of each spin up $P_{\uparrow}=N_{\uparrow} / N$ and spin down $P_{\downarrow}=N_{\downarrow} / N$ as a function of time. Plot $P_{\uparrow}(t)$ and $P_{\downarrow}(t)$ on the same graph vs. time $t$. The alternating spin up and spin down population as a function of time is called Rabi oscillation. Context answer: \boxed{$P_{\downarrow}=\sin ^{2} \frac{\Omega t}{2}$ , $P_{\uparrow}=\cos ^{2} \frac{\Omega t}{2}$} Extra Supplementary Reading Materials: Part D. Measurement incompatibility Spin is in fact a vector quantity; but due to its quantum properties, we cannot measure each of its components simultaneously (i.e. we can know both $|\boldsymbol{\mu}|$ and $\mu_{z}$ as in above problems; but not all $|\boldsymbol{\mu}|, \mu_{x}, \mu_{y}$, and $\mu_{z}$ simultaneously). In this problem, we will do a calculation based on the Heisenberg uncertainty principle (using the relation $\Delta p_{q} \Delta q \geq \hbar$ ) to show how these measurements are incompatible with each other.","1. Let us consider an oven source of silver atoms, which has a small opening. The atoms stream out of the opening along $-y$ direction (see Figure below) and experience a spatial varying field $\boldsymbol{B}_{1}$. The field $\boldsymbol{B}_{1}$ has strong bias field component in the $z$ direction, where the atoms with different magnetic moment $\mu_{z}= \pm \gamma \hbar$ are split in the $z$ direction. At a distance $D$ from the oven source, a screen $S C_{1}$ is put to allow only spin up atoms to pass (blocking spin down atoms). Thus, at the instant after passing the screen, the atoms are prepared in spin up states. After the screen, the atoms enter a region of nonhomogenous field $\boldsymbol{B}_{2}$ where the atoms feel a force $$ F_{x}=\mu_{x} C $$ The field $\boldsymbol{B}_{2}$ has strong bias field component in the $x$ direction, where the atoms have magnetic moment $\mu_{x}= \pm \gamma \hbar$. In order to determine $\mu_{x}$ by observing the splitting in $x$ direction, show that the following condition must be fulfilled: $$ \frac{1}{\hbar}\left|\mu_{x}\right| \Delta x C t \gg 1 $$ where $t$ is the duration after leaving the screen $S C_{1}$ and $\Delta x$ is the opening width on $S C_{1}$.","['In the $x$ direction, the uncertainty in position due to the screen opening is $\\Delta x$. According to the uncertainty principle, the atom momentum uncertainty $\\Delta p_{x}$ is given by\n\n$$\n\\Delta p_{x} \\approx \\frac{\\hbar}{\\Delta x},\n$$\n\nwhich translates into an uncertainty in the $x$ velocity of the atom,\n\n$$\nv_{x} \\approx \\frac{\\hbar}{m \\Delta x}\n$$\n\nConsequently, during the time of flight $t$ of the atoms through the device, the uncertainty in the width of the beam will grow by an amount $\\delta x$ given by\n\n$$\n\\delta x=\\Delta v_{x} t \\approx \\frac{\\hbar}{m \\Delta x} t\n$$\n\n\n\nSo, the width of the beams is growing linearly in time. Meanwhile, the two beams are separating at a rate determined by the force $F_{x}$ and the separation between the beams after a time $t$ becomes\n\n$$\nd_{x}=2 \\times \\frac{1}{2} \\frac{F_{x}}{m} t^{2}=\\frac{1}{m}\\left|\\mu_{x}\\right| C t^{2}\n$$\n\nIn order to be able to distinguish which beam a particle belongs to, the separation of the two beams must be greater than the widths of the beams; otherwise the two beams will overlap and it will be impossible to know what the $x$ component of the atom spin is. Thus, the condition must be satisfied is\n\n$$\n\\begin{aligned}\nd_{x} & \\gg \\delta x \\\\\n\\frac{1}{m}\\left|\\mu_{x}\\right| C t^{2} & \\gg \\frac{\\hbar}{m \\Delta x} t, \\\\\n\\frac{1}{\\hbar}\\left|\\mu_{x}\\right| \\Delta x C t & \\gg 1 .\n\\end{aligned}\n\\tag{15}\n$$']",,False,,, 1600,Modern Physics,,"1. Let us consider an oven source of silver atoms, which has a small opening. The atoms stream out of the opening along $-y$ direction (see Figure below) and experience a spatial varying field $\boldsymbol{B}_{1}$. The field $\boldsymbol{B}_{1}$ has strong bias field component in the $z$ direction, where the atoms with different magnetic moment $\mu_{z}= \pm \gamma \hbar$ are split in the $z$ direction. At a distance $D$ from the oven source, a screen $S C_{1}$ is put to allow only spin up atoms to pass (blocking spin down atoms). Thus, at the instant after passing the screen, the atoms are prepared in spin up states. After the screen, the atoms enter a region of nonhomogenous field $\boldsymbol{B}_{2}$ where the atoms feel a force $$ F_{x}=\mu_{x} C $$ The field $\boldsymbol{B}_{2}$ has strong bias field component in the $x$ direction, where the atoms have magnetic moment $\mu_{x}= \pm \gamma \hbar$. ![](https://cdn.mathpix.com/cropped/2023_12_21_a0eec930b34aa88d323eg-1.jpg?height=437&width=597&top_left_y=347&top_left_x=798) In order to determine $\mu_{x}$ by observing the splitting in $x$ direction, show that the following condition must be fulfilled: $$ \frac{1}{\hbar}\left|\mu_{x}\right| \Delta x C t \gg 1 $$ where $t$ is the duration after leaving the screen $S C_{1}$ and $\Delta x$ is the opening width on $S C_{1}$.","['In the $x$ direction, the uncertainty in position due to the screen opening is $\\Delta x$. According to the uncertainty principle, the atom momentum uncertainty $\\Delta p_{x}$ is given by\n\n$$\n\\Delta p_{x} \\approx \\frac{\\hbar}{\\Delta x},\n$$\n\nwhich translates into an uncertainty in the $x$ velocity of the atom,\n\n$$\nv_{x} \\approx \\frac{\\hbar}{m \\Delta x}\n$$\n\nConsequently, during the time of flight $t$ of the atoms through the device, the uncertainty in the width of the beam will grow by an amount $\\delta x$ given by\n\n$$\n\\delta x=\\Delta v_{x} t \\approx \\frac{\\hbar}{m \\Delta x} t\n$$\n\n\n\nSo, the width of the beams is growing linearly in time. Meanwhile, the two beams are separating at a rate determined by the force $F_{x}$ and the separation between the beams after a time $t$ becomes\n\n$$\nd_{x}=2 \\times \\frac{1}{2} \\frac{F_{x}}{m} t^{2}=\\frac{1}{m}\\left|\\mu_{x}\\right| C t^{2}\n$$\n\nIn order to be able to distinguish which beam a particle belongs to, the separation of the two beams must be greater than the widths of the beams; otherwise the two beams will overlap and it will be impossible to know what the $x$ component of the atom spin is. Thus, the condition must be satisfied is\n\n$$\n\\begin{aligned}\nd_{x} & \\gg \\delta x \\\\\n\\frac{1}{m}\\left|\\mu_{x}\\right| C t^{2} & \\gg \\frac{\\hbar}{m \\Delta x} t, \\\\\n\\frac{1}{\\hbar}\\left|\\mu_{x}\\right| \\Delta x C t & \\gg 1 .\n\\end{aligned}\n\\tag{15}\n$$']",['证明题'],False,,Need_human_evaluate, 1601,Modern Physics,"All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the ""intrinsic"" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term ""frequency"" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: ""up"" and ""down"". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$ Context question: 1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ? Context answer: \boxed{$\Omega=\gamma b$} Context question: 2. Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as $$ M(t)=N \mu(\cos \Omega t) . $$ Context answer: \boxed{证明题} Context question: 3. Under the application of magnetic field described above, determine the fractional population of each spin up $P_{\uparrow}=N_{\uparrow} / N$ and spin down $P_{\downarrow}=N_{\downarrow} / N$ as a function of time. Plot $P_{\uparrow}(t)$ and $P_{\downarrow}(t)$ on the same graph vs. time $t$. The alternating spin up and spin down population as a function of time is called Rabi oscillation. Context answer: \boxed{$P_{\downarrow}=\sin ^{2} \frac{\Omega t}{2}$ , $P_{\uparrow}=\cos ^{2} \frac{\Omega t}{2}$} Extra Supplementary Reading Materials: Part D. Measurement incompatibility Spin is in fact a vector quantity; but due to its quantum properties, we cannot measure each of its components simultaneously (i.e. we can know both $|\boldsymbol{\mu}|$ and $\mu_{z}$ as in above problems; but not all $|\boldsymbol{\mu}|, \mu_{x}, \mu_{y}$, and $\mu_{z}$ simultaneously). In this problem, we will do a calculation based on the Heisenberg uncertainty principle (using the relation $\Delta p_{q} \Delta q \geq \hbar$ ) to show how these measurements are incompatible with each other. Context question: 1. Let us consider an oven source of silver atoms, which has a small opening. The atoms stream out of the opening along $-y$ direction (see Figure below) and experience a spatial varying field $\boldsymbol{B}_{1}$. The field $\boldsymbol{B}_{1}$ has strong bias field component in the $z$ direction, where the atoms with different magnetic moment $\mu_{z}= \pm \gamma \hbar$ are split in the $z$ direction. At a distance $D$ from the oven source, a screen $S C_{1}$ is put to allow only spin up atoms to pass (blocking spin down atoms). Thus, at the instant after passing the screen, the atoms are prepared in spin up states. After the screen, the atoms enter a region of nonhomogenous field $\boldsymbol{B}_{2}$ where the atoms feel a force $$ F_{x}=\mu_{x} C $$ The field $\boldsymbol{B}_{2}$ has strong bias field component in the $x$ direction, where the atoms have magnetic moment $\mu_{x}= \pm \gamma \hbar$. In order to determine $\mu_{x}$ by observing the splitting in $x$ direction, show that the following condition must be fulfilled: $$ \frac{1}{\hbar}\left|\mu_{x}\right| \Delta x C t \gg 1 $$ where $t$ is the duration after leaving the screen $S C_{1}$ and $\Delta x$ is the opening width on $S C_{1}$. Context answer: \boxed{证明题} ","2. The atoms are initially prepared in the spin up states right after leaving the screen, where $\mu_{z}=\gamma \hbar=\left|\mu_{x}\right|$. This means the atoms will precess at rates covering a range of values $\Delta \omega$ with respect to the $x$ component of $\boldsymbol{B}_{2}$, specifically $B_{2 x}=B_{0}+C x$. Prove that the spread in the precession angle $\Delta \omega t$ is so large and hence we cannot measure both $\mu_{x}$ and $\mu_{z}$ simultaneously. In other words, the measurement of $\mu_{x}$ destroys the information on $\mu_{z}$.","['As the atoms pass through the screen, the variation of magnetic field strength across the beam width experienced by the atoms is\n\n$$\n\\Delta B=\\Delta x \\frac{d B}{d x}=C \\Delta x\n$$\n\nThis means the atoms will precess at rates covering a range of values $\\Delta \\omega$ given by\n\n$$\n\\Delta \\omega=\\gamma \\Delta B=\\frac{\\mu_{z}}{\\hbar} \\Delta B=\\frac{\\left|\\mu_{x}\\right|}{\\hbar} C \\Delta x\n$$\n\nand, if previous condition in measuring $\\mu_{x}$ is satisfied,\n\n$$\n\\Delta \\omega t \\gg 1 .\n\\tag{16}\n$$\n\nIn other words, the spread in the angle $\\Delta \\omega t$ through which the magnetic moments precess is so large that the $z$ component of the spin is completely randomized or the measurement uncertainty is very large.']",,False,,, 1602,Geometry,,"Let $P$ and $P^{\prime}$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\ell$ through $O$ the segment $\ell \cap P$ is strictly longer than the segment $\ell \cap P^{\prime}$. Is it possible that the ratio of the area of $P^{\prime}$ to the area of $P$ is greater than $1.9 ?$","['The answer is in the affirmative: Given a positive $\\epsilon<2$, the ratio in question may indeed be greater than $2-\\epsilon$.\n\n\n\nTo show this, consider a square $A B C D$ centred at $O$, and let $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ be the reflections of $O$ in $A, B$, and $C$, respectively. Notice that, if $\\ell$ is a line through $O$, then the segments $\\ell \\cap A B C D$ and $\\ell \\cap A^{\\prime} B^{\\prime} C^{\\prime}$ have equal lengths, unless $\\ell$ is the line $A C$.\n\n\n\nNext, consider the points $M$ and $N$ on the segments $B^{\\prime} A^{\\prime}$ and $B^{\\prime} C^{\\prime}$, respectively, such that $B^{\\prime} M / B^{\\prime} A^{\\prime}=B^{\\prime} N / B^{\\prime} C^{\\prime}=(1-\\epsilon / 4)^{1 / 2}$. Finally, let $P^{\\prime}$ be the image of the convex quadrangle $B^{\\prime} M O N$ under the homothety of ratio $(1-\\epsilon / 4)^{1 / 4}$ centred at $O$. Clearly, the quadrangles $P \\equiv A B C D$ and $P^{\\prime}$ satisfy the conditions in the statement, and the ratio of the area of $P^{\\prime}$ to the area of $P$ is exactly $2-\\epsilon / 2$.\n\n\n\n']",,True,,, 1603,Number Theory,,"Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that: $$ x=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor $$ Prove that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$","['We prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.\n\n\n\nLet $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,\n\n\n\n$$\n\n\\sum_{b \\in B, b \\leq c}\\left\\lfloor\\sqrt[k]{\\frac{c}{b}}\\right\\rfloor=c\n\n$$\n\n\n\nTo this end, notice that every positive integer has a unique representation as a product of an element in $B$ and a $k$ th power. Consequently, the set of all positive integers less than or equal to $c$ splits into\n\n\n\n$$\n\nC_{b}=\\left\\{x: x \\in \\mathbb{Z}_{>0}, x \\leq c, \\text { and } x / b \\text { is a } k \\text { th power }\\right\\}, \\quad b \\in B, b \\leq c\n\n$$\n\n\n\nClearly, $\\left|C_{b}\\right|=\\lfloor\\sqrt[k]{c / b}\\rfloor$, whence the desired equality.\n\n\n\nFinally, enumerate $B$ according to the natural order: $1=b_{1}b_{n-1}=a_{n-1}$ and\n\n\n\n$$\n\nb_{n}=\\sum_{i=1}^{n}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{b_{i}}}\\right\\rfloor=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{b_{i}}}\\right\\rfloor+1=\\sum_{i=1}^{n-1}\\left\\lfloor\\sqrt[k]{\\frac{b_{n}}{a_{i}}}\\right\\rfloor+1\n\n$$\n\n\n\nthe definition of $a_{n}$ forces $a_{n} \\leq b_{n}$. Were $a_{n}0$ then $f_{n}(x) \\leq f_{n}(x-1)$; this means that such an $x$ cannot equal $a_{n}$. Thus $a_{j} / a_{i}$ is never the $k$ th power of an integer if $j>i$.\n\n\n\nNow we are prepared to prove by induction on $n$ that $a_{1}, a_{2}, \\ldots, a_{n}$ are exactly all $k$ thpower-free integers in $\\left[1, a_{n}\\right]$. The base case $n=1$ is trivial.\n\n\n\n\n\n\n\nAssume that all the $k$ th-power-free integers on $\\left[1, a_{n}\\right]$ are exactly $a_{1}, \\ldots, a_{n}$. Let $b$ be the least integer larger than $a_{n}$ such that $g_{n}(b)=0$. We claim that: (1) $b=a_{n+1}$; and (2) $b$ is the least $k$ th-power-free number greater than $a_{n}$.\n\n\n\nTo prove (1), notice first that all the numbers of the form $a_{j} / a_{i}$ with $1 \\leq i1$ be the greatest integer such that $y^{k} \\mid b$; then $b / y^{k}$ is $k$ th-power-free and hence $b / y^{k}=a_{i}$ for some $i \\leq n$. So $b / a_{i}$ is the $k$ th power of an integer, which contradicts the definition of $b$.\n\n\n\nThus $a_{1}, a_{2}, \\ldots$ are exactly all $k$ th-power-free positive integers; consequently all primes are contained in this sequence.']",,True,,, 1604,Combinatorics,,"$2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exactly once. Prove that some edge has never been chosen.","['Step 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i\n\n\n\nFirst Approach. Since $\\mathrm{O}_{1} \\mathrm{O}_{2} \\mathrm{O}_{3} \\mathrm{O}_{4}$ is a parallelogram, $\\overrightarrow{F_{1} F_{2}}+\\overrightarrow{F_{3} F_{4}}=\\mathbf{0}$ and $\\overrightarrow{F_{2} F_{3}}+\\overrightarrow{F_{4} F_{1}}=\\mathbf{0}$; this still holds in the degenerate case, for if the $O_{i}$ are collinear, then they all lie on the line $T_{1} T_{4}$, and each $O_{i}$ is the midpoint of the segment $T_{i} T_{i+1}$. Consequently,\n\n\n\n$$\n\n\\begin{aligned}\n\n\\overrightarrow{P Q}-\\overrightarrow{R S}+\\overrightarrow{T U}-\\overrightarrow{V W}= & \\left(\\overrightarrow{P F_{1}}+\\overrightarrow{F_{1} F_{2}}+\\overrightarrow{F_{2} Q}\\right)-\\left(\\overrightarrow{R F_{2}}+\\overrightarrow{F_{2} F_{3}}+\\overrightarrow{F_{3} S}\\right) \\\\\n\n& +\\left(\\overrightarrow{T F_{3}}+\\overrightarrow{F_{3} F_{4}}+\\overrightarrow{F_{4} U}\\right)-\\left(\\overrightarrow{V F_{4}}+\\overrightarrow{F_{4} F_{1}}+\\overrightarrow{F_{1} W}\\right) \\\\\n\n= & \\left(\\overrightarrow{P F_{1}}-\\overrightarrow{F_{1} W}\\right)-\\left(\\overrightarrow{R F_{2}}-\\overrightarrow{F_{2} Q}\\right)+\\left(\\overrightarrow{T F_{3}}-\\overrightarrow{F_{3} S}\\right)-\\left(\\overrightarrow{V F_{4}}-\\overrightarrow{F_{4} U}\\right) \\\\\n\n& +\\left(\\overrightarrow{F_{1} F_{2}}+\\overrightarrow{F_{3} F_{4}}\\right)-\\left(\\overrightarrow{F_{2} F_{3}}+\\overrightarrow{F_{4} F_{1}}\\right)=\\mathbf{0} .\n\n\\end{aligned}\n\n$$\n\n\n\nAlternatively, but equivalently, $\\overrightarrow{P Q}+\\overrightarrow{T U}=\\overrightarrow{R S}+\\overrightarrow{V W}$, as required.\n\n\n\nSecond Approach. This is merely another way of reading the previous argument. Fix an orientation of the line $P W$, say, from $P$ towards $W$, and use a lower case letter to denote the coordinate of a point labelled by the corresponding upper case letter.\n\n\n\nSince the diagonals of a parallelogram bisect one another, $f_{1}+f_{3}=f_{2}+f_{4}$, the common value being twice the coordinate of the projection to $P W$ of the point where $O_{1} O_{3}$ and $O_{2} O_{4}$ cross; the relation clearly holds in the degenerate case as well.\n\n\n\n\n\n\n\nPlug $f_{1}=\\frac{1}{2}(p+w), f_{2}=\\frac{1}{2}(q+r), f_{3}=\\frac{1}{2}(s+t)$ and $f_{4}=\\frac{1}{2}(u+v)$ into the above equality to get $p+w+s+t=q+r+u+v$. Alternatively, but equivalently, $(q-p)+(u-t)=(s-r)+(w-v)$, that is, $P Q+T U=R Q+V W$, as required.']",,True,,, 1606,Combinatorics,,"Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding 5000. Then she fixes 20 distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of positive integers not exceeding 20 , and she tells him back the set $\left\{a_{k}: k \in S\right\}$ without spelling out which number corresponds to which index. How many moves does Sergey need to determine for sure the number Xenia thought of?","[""Sergey can determine Xenia's number in 2 but not fewer moves.\n\n\n\nWe first show that 2 moves are sufficient. Let Sergey provide the set $\\{17,18\\}$ on his first move, and the set $\\{18,19\\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able to identify $a_{17}, a_{18}$, and $a_{19}$, and thence the residue of $N$ modulo $17 \\cdot 18 \\cdot 19=5814>5000$, by the Chinese Remainder Theorem. This means that the given range contains a single number satisfying all congruences, and Sergey achieves his goal.\n\n\n\nTo show that 1 move is not sufficient, let $M=\\operatorname{lcm}(1,2, \\ldots, 10)=2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7=2520$. Notice that $M$ is divisible by the greatest common divisor of every pair of distinct positive integers not exceeding 20. Let Sergey provide the set $S=\\left\\{s_{1}, s_{2}, \\ldots, s_{k}\\right\\}$. We show that there exist pairwise distinct positive integers $b_{1}, b_{2}, \\ldots, b_{k}$ such that $1 \\equiv b_{i}\\left(\\bmod s_{i}\\right)$ and $M+1 \\equiv b_{i-1}\\left(\\bmod s_{i}\\right)$ (indices are reduced modulo $k$ ). Thus, if in response Xenia provides the set $\\left\\{b_{1}, b_{2}, \\ldots, b_{k}\\right\\}$, then Sergey will be unable to distinguish 1 from $M+1$, as desired.\n\n\n\nTo this end, notice that, for each $i$, the numbers of the form $1+m s_{i}, m \\in \\mathbb{Z}$, cover all residues modulo $s_{i+1}$ which are congruent to $1(\\equiv M+1)$ modulo $\\operatorname{gcd}\\left(s_{i}, s_{i+1}\\right) \\mid M$. Xenia can therefore choose a positive integer $b_{i}$ such that $b_{i} \\equiv 1\\left(\\bmod s_{i}\\right)$ and $b_{i} \\equiv M+1\\left(\\bmod s_{i+1}\\right)$. Clearly, such choices can be performed so as to make the $b_{i}$ pairwise distinct, as required.""]",['2'],False,,Numerical, 1607,Number Theory,,A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4 . Prove that the number of such ways to assign the leaders is divisible by 17 .,"[""Assume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.\n\n\n\nConsider the variables $x_{1}, x_{2}, \\ldots, x_{17}$ corresponding to the workers. Assign each brigade (from the $i$-th through the $j$-th worker) the polynomial $f_{i j}=x_{i}+x_{i+1}+\\cdots+x_{j}$, and form the product $f=\\prod_{1 \\leq i \\leq j \\leq 17} f_{i j}$. The number $N$ is the sum $\\Sigma(f)$ of the coefficients of all monomials $x_{1}^{\\alpha_{1}} x_{2}^{\\alpha_{2}} \\ldots x_{17}^{\\alpha_{17}}$ in the expansion of $f$, where the $\\alpha_{i}$ are all congruent to 1 modulo 4 . For any polynomial $P$, let $\\Sigma(P)$ denote the corresponding sum. From now on, all polynomials are considered with coefficients in the finite field $\\mathbb{F}_{17}$.\n\n\n\nRecall that for any positive integer $n$, and any integers $a_{1}, a_{2}, \\ldots, a_{n}$, there exist indices $i \\leq j$ such that $a_{i}+a_{i+1}+\\cdots+a_{j}$ is divisible by $n$. Consequently, $f\\left(a_{1}, a_{2}, \\ldots, a_{17}\\right)=0$ for all $a_{1}, a_{2}, \\ldots, a_{17}$ in $\\mathbb{F}_{17}$.\n\n\n\nNow, if some monomial in the expansion of $f$ is divisible by $x_{i}^{17}$, replace that $x_{i}^{17}$ by $x_{i}$; this does not alter the above overall vanishing property (by Fermat's Little Theorem), and preserves $\\Sigma(f)$. After several such changes, $f$ transforms into a polynomial $g$ whose degree in each variable does not exceed 16 , and $g\\left(a_{1}, a_{2}, \\ldots, a_{17}\\right)=0$ for all $a_{1}, a_{2}, \\ldots, a_{17}$ in $\\mathbb{F}_{17}$. For such a polynomial, an easy induction on the number of variables shows that it is identically zero. Consequently, $\\Sigma(g)=0$, so $\\Sigma(f)=0$ as well, as desired.""]",,True,,, 1608,Geometry,,"Let $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$.","['Let the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ coincide, so $J_{b} J_{c} \\perp A D$.\n\n\n\nNow let $O$ be the circumcentre of the triangle $A J_{b} J_{c}$. Then $\\angle J_{b} A O=\\pi / 2-\\angle A O J_{b} / 2=$ $\\pi / 2-\\angle A J_{c} J_{b}=\\angle X A J_{c}=\\frac{1}{2} \\angle D A C$. Therefore, $\\angle B A O=\\angle B A J_{b}+\\angle J_{b} A O=\\frac{1}{2} \\angle B A D+$ $\\frac{1}{2} \\angle D A C=\\frac{1}{2} \\angle B A C$, and the conclusion follows.\n\n\n\n\n\n\n\nFig. 1']",,True,,, 1609,Number Theory,,"Let $p \geq 5$ be a prime number. For a positive integer $k$ we denote by $R(k)$ the remainder of $k$ when divided by $p$. Determine all positive integers $aa $$ for every $m=1,2, \ldots, p-1$.","['The required integers are $p-1$ along with all the numbers of the form $\\lfloor p / q\\rfloor, q=$ $2, \\ldots, p-1$. In other words, these are $p-1$, along with the numbers $1,2, \\ldots,\\lfloor\\sqrt{p}\\rfloor$, and also the (distinct) numbers $\\lfloor p / q\\rfloor, q=2, \\ldots,\\left\\lfloor\\sqrt{p}-\\frac{1}{2}\\right\\rfloor$.\n\n\n\nWe begin by showing that these numbers satisfy the conditions in the statement. It is readily checked that $p-1$ satisfies the required inequalities, since $m+R(m(p-1))=m+(p-m)=$ $p>p-1$ for all $m=1, \\ldots, p-1$.\n\n\n\nNow, consider any number $a$ of the form $a=\\lfloor p / q\\rfloor$, where $q$ is an integer greater than 1 but less than $p$; then $p=a q+r$ with $0r$ and $y \\geq 1$. Thus $a$ satisfies the required condition.\n\n\n\nFinally, we show that if an integer $a \\in(0, p-1)$ satisfies the required condition then $a$ is indeed of the form $a=\\lfloor p / q\\rfloor$ for some integer $q \\in(0, p)$. This is clear for $a=1$, so we may (and will) assume that $a \\geq 2$.\n\n\n\nWrite $p=a q+r$ with $q, r \\in \\mathbb{Z}$ and $0\\frac{1}{2 n+2}$, let $U=(0,1) \\times(0,1)$, choose a small enough positive $\\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\\left(\\frac{i}{n+1} \\pm \\epsilon\\right) \\times\\left(\\frac{1}{2} \\pm \\epsilon\\right), i=1, \\ldots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at $\\operatorname{most}\\left(\\frac{1}{n+1}+\\epsilon\\right) \\cdot\\left(\\frac{1}{2}+\\epsilon\\right)<\\mu$ if $\\epsilon$ is small enough.\n\n\n\nWe now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\\mu_{0}=\\frac{2}{|C|+4}$.\n\n\n\nTo prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.\n\n\n\nLemma 1. Let $k$ be a positive integer, and let $\\lambda<\\frac{1}{\\lfloor k / 2\\rfloor+1}$ be a positive real number. If $t_{1}, \\ldots, t_{k}$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\\lambda$.\n\n\n\nLemma 2. Given an integer $k \\geq 2$ and positive integers $m_{1}, \\ldots, m_{k}$,\n\n\n\n$$\n\n\\left\\lfloor\\frac{m_{1}}{2}\\right\\rfloor+\\sum_{i=1}^{k}\\left\\lfloor\\frac{m_{i}}{2}\\right\\rfloor+\\left\\lfloor\\frac{m_{k}}{2}\\right\\rfloor \\leq \\sum_{i=1}^{k} m_{i}-k+2\n\n$$\n\n\n\nBack to the problem, let $U=(0,1) \\times(0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_{1}<\\cdots\\left(\\left\\lfloor m_{i} / 2\\right\\rfloor+1\\right) \\mu_{0}$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \\cap \\ell_{i}$ from the other ones by an open subinterval $x_{i} \\times J$ of $x_{i} \\times(0,1)$ whose length is greater than or equal to $\\mu_{0} /\\left(x_{i+1}-x_{i-1}\\right)$. Consequently, $\\left(x_{i-1}, x_{i+1}\\right) \\times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\\mu_{0}$.\n\n\n\nNext, we rule out the case $x_{i+1}-x_{i-1} \\leq\\left(\\left\\lfloor m_{i} / 2\\right\\rfloor+1\\right) \\mu_{0}$ for all indices $i$. If this were the case, notice that necessarily $k>1$; also, $x_{1}-x_{0}0$, and even if $n \\leq 0$. The integer $w(n)=\\lfloor\\ell / 2\\rfloor$ is called the weight of $n$.\n\n\n\nExistence once proved, uniqueness follows from the fact that there are as many such representations as integers in the range $-2^{M}+1$ through $2^{M}$, namely, $2^{M+1}$.\n\n\n\nTo prove existence, notice that the base case $M=0$ is clear, so let $M \\geq 1$ and let $n$ be an integer in the range $-2^{M}+1$ through $2^{M}$.\n\n\n\nIf $-2^{M}+1 \\leq n \\leq-2^{M-1}$, then $1 \\leq n+2^{M} \\leq 2^{M-1}$, so $n+2^{M}=\\sum_{j=1}^{2 k+1}(-1)^{j-1} 2^{m_{j}}$ for some integers $k \\geq 0$ and $0 \\leq m_{1}<\\cdots3$, then $f_{n}(x) \\equiv 0(\\bmod n+1)$ for all integers $x$, since $f_{n}(0)=-n !=-1 \\cdot 2 \\cdot \\cdots$. $\\frac{n+1}{2} \\cdot \\cdots \\cdot n \\equiv 0(\\bmod\\ n+1)$.\n\n\n\nFinally, let $P=f_{n}+n X+1$ if $n$ is odd, and let $P=f_{n-1}+n X+1$ if $n$ is even. In either case, $P$ is strictly increasing, hence injective, on the integers, and $P(x) \\equiv 1(\\bmod n)$ for all integers $x$.']",['2'],False,,Numerical, 1614,Combinatorics,,"Let $n$ be an integer greater than 1 and let $X$ be an $n$-element set. A non-empty collection of subsets $A_{1}, \ldots, A_{k}$ of $X$ is tight if the union $A_{1} \cup \cdots \cup A_{k}$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_{i}$ s. Find the largest cardinality of a collection of proper non-empty subsets of $X$, no non-empty subcollection of which is tight. Note. A subset $A$ of $X$ is proper if $A \neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.","[""The required maximum is $2 n-2$. To describe a $(2 n-2)$-element collection satisfying the required conditions, write $X=\\{1,2, \\ldots, n\\}$ and set $B_{k}=\\{1,2, \\ldots, k\\}$, $k=1,2, \\ldots, n-1$, and $B_{k}=\\{k-n+2, k-n+3, \\ldots, n\\}, k=n, n+1, \\ldots, 2 n-2$. To show that no subcollection of the $B_{k}$ is tight, consider a subcollection $\\mathcal{C}$ whose union $U$ is a proper subset of $X$, let $m$ be an element in $X \\backslash U$, and notice that $\\mathcal{C}$ is a subcollection of $\\left\\{B_{1}, \\ldots, B_{m-1}, B_{m+n-1}, \\ldots, B_{2 n-2}\\right\\}$, since the other $B$ 's are precisely those containing $m$. If $U$ contains elements less than $m$, let $k$ be the greatest such and notice that $B_{k}$ is the only member of $\\mathcal{C}$ containing $k$; and if $U$ contains elements greater than $m$, let $k$ be the least such and notice that $B_{k+n-2}$ is the only member of $\\mathcal{C}$ containing $k$. Consequently, $\\mathcal{C}$ is not tight.\n\n\n\nWe now proceed to show by induction on $n \\geq 2$ that the cardinality of a collection of proper non-empty subsets of $X$, no subcollection of which is tight, does not exceed $2 n-2$. The base case $n=2$ is clear, so let $n>2$ and suppose, if possible, that $\\mathcal{B}$ is a collection of $2 n-1$ proper non-empty subsets of $X$ containing no tight subcollection.\n\n\n\nTo begin, notice that $\\mathcal{B}$ has an empty intersection: if the members of $\\mathcal{B}$ shared an element $x$, then $\\mathcal{B}^{\\prime}=\\{B \\backslash\\{x\\}: B \\in \\mathcal{B}, B \\neq\\{x\\}\\}$ would be a collection of at least $2 n-2$ proper non-empty subsets of $X \\backslash\\{x\\}$ containing no tight subcollection, and the induction hypothesis would be contradicted.\n\n\n\nNow, for every $x$ in $X$, let $\\mathcal{B}_{x}$ be the (non-empty) collection of all members of $\\mathcal{B}$ not containing $x$. Since no subcollection of $\\mathcal{B}$ is tight, $\\mathcal{B}_{x}$ is not tight, and since the union of $\\mathcal{B}_{x}$ does not contain $x$, some $x^{\\prime}$ in $X$ is covered by a single member of $\\mathcal{B}_{x}$. In other words, there is a single set in $\\mathcal{B}$ covering $x^{\\prime}$ but not $x$. In this case, draw an arrow from $x$ to $x^{\\prime}$. Since there is at least one arrow from each $x$ in $X$, some of these arrows form a (minimal) cycle $x_{1} \\rightarrow x_{2} \\rightarrow \\cdots \\rightarrow x_{k} \\rightarrow x_{k+1}=x_{1}$ for some suitable integer $k \\geq 2$. Let $A_{i}$ be the unique member of $\\mathcal{B}$ containing $x_{i+1}$ but not $x_{i}$, and let $X^{\\prime}=\\left\\{x_{1}, x_{2}, \\ldots, x_{k}\\right\\}$.\n\n\n\nRemove $A_{1}, A_{2}, \\ldots, A_{k}$ from $\\mathcal{B}$ to obtain a collection $\\mathcal{B}^{\\prime}$ each member of which either contains or is disjoint from $X^{\\prime}$ : for if a member $B$ of $\\mathcal{B}^{\\prime}$ contained some but not all elements of $X^{\\prime}$, then $B$ should contain $x_{i+1}$ but not $x_{i}$ for some $i$, and $B=A_{i}$, a contradiction. This rules out the case $k=n$, for otherwise $\\mathcal{B}=\\left\\{A_{1}, A_{2}, \\ldots, A_{n}\\right\\}$, so $|\\mathcal{B}|<2 n-1$.\n\n\n\nTo rule out the case $k2$.\n\n\n\nFirstly, we perform a different modification of $\\mathcal{B}$. Choose any $x \\in X$, and consider the subcollection $\\mathcal{B}_{x}=\\{B: B \\in \\mathcal{B}, x \\notin B\\}$. By our assumption, $\\mathcal{B}_{x}$ is not tight. As the union of sets in $\\mathcal{B}_{x}$ is distinct from $X$, either this collection is empty, or there exists an element $y \\in X$ contained in a unique member $A_{x}$ of $\\mathcal{B}_{x}$. In the former case, we add the set $B_{x}=X \\backslash\\{x\\}$ to $\\mathcal{B}$, and in the latter we replace $A_{x}$ by $B_{x}$, to form a new collection $\\mathcal{B}^{\\prime}$. (Notice that if $B_{x} \\in \\mathcal{B}$, then $B_{x} \\in \\mathcal{B}_{x}$ and $y \\in B_{x}$, so $B_{x}=A_{x}$.)\n\n\n\nWe claim that the collection $\\mathcal{B}^{\\prime}$ is also good. Indeed, if $\\mathcal{B}^{\\prime}$ has a tight subcollection $\\mathcal{T}$, then $B_{x}$ should lie in $\\mathcal{T}$. Then, as the union of the sets in $\\mathcal{T}$ is distinct from $X$, we should have $\\mathcal{T} \\subseteq \\mathcal{B}_{x} \\cup\\left\\{B_{x}\\right\\}$. But in this case an element $y$ is contained in a unique member of $\\mathcal{T}$, namely $B_{x}$, so $\\mathcal{T}$ is not tight - a contradiction.\n\n\n\nPerform this procedure for every $x \\in X$, to get a good collection $\\mathcal{B}$ containing the sets $B_{x}=X \\backslash\\{x\\}$ for all $x \\in X$. Consider now an element $x \\in X$ such that $\\left|\\mathcal{B}_{x}\\right|$ is maximal. As we have mentioned before, there exists an element $y \\in X$ belonging to a unique member (namely, $B_{x}$ ) of $\\mathcal{B}_{x}$. Thus, $\\mathcal{B}_{x} \\backslash\\left\\{B_{x}\\right\\} \\subset \\mathcal{B}_{y}$; also, $B_{y} \\in \\mathcal{B}_{y} \\backslash \\mathcal{B}_{x}$. Thus we get $\\left|\\mathcal{B}_{y}\\right| \\geq\\left|\\mathcal{B}_{x}\\right|$, which by the maximality assumption yields the equality, which in turn means that $\\mathcal{B}_{y}=\\left(\\mathcal{B}_{x} \\backslash\\left\\{B_{x}\\right\\}\\right) \\cup\\left\\{B_{y}\\right\\}$.\n\n\n\nTherefore, each set in $\\mathcal{B} \\backslash\\left\\{B_{x}, B_{y}\\right\\}$ contains either both $x$ and $y$, or none of them. Collapsing $\\{x, y\\}$ to singleton $x^{*}$, we get a new collection of $|\\mathcal{B}|-2$ subsets of $(X \\backslash\\{x, y\\}) \\cup\\left\\{x^{*}\\right\\}$ containing no tight subcollection. This contradicts minimality of $n$.']",['$2n-2$'],False,,Expression, 1615,Geometry,,"Given a triangle $A B C$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $A H$. Let further $\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\ell$, respectively. Prove that $K P+K Q \geq B C$.","['Fix the origin at $O$ and the real axis along $\\ell$. A lower case letter denotes the complex coordinate of the corresponding point in the configuration. For convenience, let $|a|=|b|=|c|=1$.\n\n\n\nClearly, $k=a+\\frac{1}{2}(b+c), p=a+\\frac{1}{2}\\left(b+\\frac{1}{b}\\right)$ and $q=a+\\frac{1}{2}\\left(c+\\frac{1}{c}\\right)$.\n\n\n\nThen $|k-p|=\\left|a+\\frac{1}{2}\\left(c-\\frac{1}{b}\\right)\\right|=\\frac{1}{2}|2 a b+b c-1|$, since $|b|=1$.\n\n\n\nSimilarly, $|k-q|=\\frac{1}{2}|2 a c+b c-1|$, so, since $|a|=1$,\n\n\n\n$$\n\n\\begin{gathered}\n\n|k-p|+|k-q|=\\frac{1}{2}|2 a b+b c-1|+\\frac{1}{2}|2 a c+b c-1| \\\\\n\n\\geq \\frac{1}{2}|2 a(b-c)|=|b-c|,\n\n\\end{gathered}\n\n$$\n\n\n\nas required.', 'Let $M$ be the midpoint of $B C$, and let $R$ be the projection of $M$ onto $\\ell$. In other words, $R$ is the midpoint of $P Q$. Since $\\angle B P O=\\angle B M O=$ $90^{\\circ}$, the points $B, P, O$, and $M$ are concyclic, so $\\angle(O M, O B)=\\angle(P M, P B)=\\angle(P M, M R)$, so the right triangles $M R P$ and $O M B$ are similar and have different orientation. Similarly, the triangles $M R Q$ and $O M C$ are similar and have different orientation, hence so are the triangles $O B C$ and $M P Q$.\n\n\n\n\n\n\n\nRecall that $\\overrightarrow{A H}=2 \\overrightarrow{O M}$, so $\\overrightarrow{O M}=\\overrightarrow{A K}$. Hence $A O M K$ is a parallelogram, so $M K=O A=O B=$ $O C$.\n\n\n\nConsider the rotation through $\\angle(\\overrightarrow{O C}, \\overrightarrow{O B})$ about $M$. It maps $P$ to $Q$; let it map $K$ to some point $L$. Then $M K=M L=O B=O C$ and $\\angle L M K=\\angle B O C$, so the triangles $O B C$ and $M K L$ are congruent. Hence $B C=K L \\leq K Q+L Q=$ $K Q+K P$, as required.', ""Let $\\alpha=\\angle(P B, B C)=\\angle(Q C, B C)$. Since $P$ lies on the circle of diameter $O B$, $\\angle(O P, O M)=\\alpha$. Since also $Q$ lies on the circle of diameter $O C$, it immediately follows that $M P=M Q=R \\sin \\alpha$ by sine theorem in triangles $\\triangle O P M$ and $\\triangle O Q M$.\n\n\n\nBecause $P Q$ is the projection of $B C$ on line $\\ell$, it follows that $P Q=B C \\sin \\alpha$. Just like in the first solution, $K M=A O=R$ (the circumradius of triangle $\\triangle A B C$ ).\n\n\n\nNow apply Ptolemy's inequality for the quadrilateral $K P M Q: K P \\cdot M Q+K Q \\cdot M P \\geq P Q \\cdot K M$, and now substitute the relations from above, leading to\n\n\n\n$$\n\nR \\sin \\alpha(K P+K Q) \\geq R \\sin \\alpha \\cdot B C\n\n$$\n\n\n\nwhich is precisely the conclusion whenever $\\sin \\alpha \\neq 0$. The case when $\\sin \\alpha=0$ can be treated either directly, or via a limit argument."", 'Denote by $R$ and $O$ the circumradius and the circumcentre of triangle $A B C$, respectively. As in Solution 1, we see that $M K=R$.\n\n\n\nAssume now that $\\ell$ is fixed, while $A$ moves along the fixed circle $(A B C)$. Then $K$ will move along a cricle centred at $M$ with radius $R$. We must show that for each point $K$ on this circle we have $B C \\leq K P+K Q$. In doing so, we prove that the afore-mentioned circle contains an ellipse with foci at $Q$ and $P$ with distance $B C$.\n\n\n\nLet $S$ be the foot of the perpendicular from $M$ to $P Q$, it is easy to verify that $S$ is the center of the ellipse. We shall then consider it as the origin. Let $u=\\frac{B C}{2}$ and $t=\\frac{P Q}{2}$; notice that $u$ is the major semi-axis of the ellipse and $\\sqrt{u^{2}-t^{2}}$ is the minor one. Assume $X(x, y)$ is a point on this ellipse. We now need to prove $M X \\leq R$.\n\n\n\nSince $X$ is on the ellipse, we can write $(x, y)=$ $\\left(u \\cos \\theta, \\sqrt{u^{2}-t^{2}} \\sin \\theta\\right)$, for some $\\theta \\in(0,2 \\pi)$. Since $M X^{2}=x^{2}+(y+M S)^{2}$, we can expand and obtain\n\n\n\n$M X^{2}=u^{2}+M S^{2}-t^{2} \\cdot \\sin ^{2} \\theta+2 M S \\cdot \\sqrt{u^{2}-t^{2}} \\cdot \\sin \\theta$.\n\n\n\nAdd and subtract $M S^{2}\\left(u^{2}-t^{2}\\right) / t^{2}$ in order to obtain a square on the right hand side: $M X^{2}=u^{2}+$ $M S^{2}+\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}-\\left(t \\sin \\theta-\\frac{M S \\sqrt{u^{2}-t^{2}}}{t}\\right)^{2}$. It now suffices to show that $u^{2}+M S^{2}+\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}=$ $R^{2}$, since then it would immediately follow that $M X^{2} \\leq R^{2}$.\n\n\n\nApplying Pythagorean theorem in triangles $O B M$ and $O S M$, we obtain $R^{2}=u^{2}+O M^{2}$ and $O M^{2}=M S^{2}+O S^{2}$, so it remains to prove that $O S^{2}=\\frac{M S^{2}\\left(u^{2}-t^{2}\\right)}{t^{2}}$. Let $\\alpha=\\angle(O P, B M)$, then $O S / M S=\\tan \\alpha$ and $t / u=\\cos \\alpha$, so $O S^{2}=$ $M S^{2} \\tan ^{2} \\alpha=M S^{2}\\left(\\frac{1-\\cos ^{2} \\alpha}{\\cos ^{2} \\alpha}\\right)=M S^{2} \\cdot \\frac{u^{2}-t^{2}}{t^{2}}$, which is the desired result.']",,True,,, 1616,Algebra,,"Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$. Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$.","['Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\\operatorname{deg} Q \\mid \\operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically, as desired.\n\n\n\nChoose the polynomials $T(x)$ and $M(x)$ such that\n\n\n\n$$\n\nS(x)=T(Q(x))+M(x) \\tag{*}\n\n$$\n\n\n\nwhere $\\operatorname{deg} M$ is minimised; if $M=0$, then we get the desired result. For the sake of contradiction, suppose $M \\neq 0$. Then $q \\nmid m=\\operatorname{deg} M$; otherwise, $M(x)=\\beta Q(x)^{m / q}+M_{1}(x)$, where $\\beta$ is some number and $\\operatorname{deg} M_{1}<\\operatorname{deg} M$, contradicting the choice of $M$. In particular, $01$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that $\\operatorname{deg}\\left(P(x)-U(x)^{c}\\right)<(c-1) b$ and $\\operatorname{deg}(R(x)-$ $\\left.V(x)^{c}\\right)<(c-1) a$. Then $\\operatorname{deg}\\left(F(x)-U(Q(x))^{c}\\right)=$ $\\operatorname{deg}\\left(P(Q(x))-U(Q(x))^{c}\\right)<(c-1) a b d, \\operatorname{deg}(F(x)-$ $\\left.V(S(x))^{c}\\right)=\\operatorname{deg}\\left(R(S(x))-V(S(x))^{c}\\right)<(c-1) a b d$, so $\\operatorname{deg}\\left(U(Q(x))^{c}-V(S(x))^{c}\\right)=\\operatorname{deg}((F(x)-$ $\\left.\\left.V(S(x))^{c}\\right)-\\left(F(x)-U(Q(x))^{c}\\right)\\right)<(c-1) a b d$.\n\n\n\nOn the other hand, $U(Q(x))^{c}-V(S(x))^{c}=$ $(U(Q(x))-V(S(x)))\\left(U(Q(x))^{c-1}+\\cdots+\\right.$ $\\left.V(S(x))^{c-1}\\right)$.\n\n\n\nBy the preceding, the degree of the left-hand member is (strictly) less than $(c-1) a b d$ which is precisely the degree of the second factor in the righthand member. This forces $U(Q(x))=V(S(x))$, so $U(Q(x))=V(S(x))$ has degree $a b d1$. Let $V$ denote the vertex set of $\\Gamma$, and let $T_{r}, T_{g}$, and $T_{b}$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.\n\n\n\nCase 1: There exists a partition $V=A \\sqcup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.\n\n\n\nSince $\\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.\n\n\n\nAssume that $T$, one of the three trees, does not contain $e$. Then the graph $T \\cup\\{e\\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e^{\\prime}$ connecting $A$ and $B$; the edge $e^{\\prime}$ is also blue. Replace $e^{\\prime}$ by $e$ in $T$ to get another tree $T^{\\prime}$ with the same number of edges of each colour as in $T$, but containing $e$.\n\n\n\nPerforming such an operation to all three trees, we arrive at the situation where the three trees $T_{r}^{\\prime}$, $T_{g}^{\\prime}$, and $T_{b}^{\\prime}$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\\Gamma^{*}$, and set $r^{*}=r, g^{*}=g$, and $b^{*}=b-1$. The new graph satisfies the conditions in the statement for those new values - indeed, under the shrinking, each of the trees $T_{r}^{\\prime}, T_{g}^{\\prime}$, and $T_{b}^{\\prime}$ loses a blue edge. So $\\Gamma^{*}$ has a spanning tree with exactly $r$ red, exactly $g$ green, and exactly $b-1$ blue edges. Finally, pass back to $\\Gamma$ by restoring $e$, to obtain the a desired spanning tree in $\\Gamma$.\n\n\n\nCase 2: There is no such a partition.\n\n\n\nConsider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \\cup G \\cup B$ has the smallest number $k$ of components. If $k=1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).\n\n\n\nAssume now that $k \\geq 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \\backslash V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.\n\n\n\nLet $e^{\\prime}$ be a red edge in $C$ and set $R^{\\prime}=R \\backslash\\left\\{e^{\\prime}\\right\\} \\cup$ $\\{e\\}$. Then $\\left(R^{\\prime}, G, B\\right)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.', 'For a spanning tree $T$ in $\\Gamma$, denote by $r(T), g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.\n\n\n\nAssume that $\\mathcal{C}$ is some collection of spanning trees in $\\Gamma$. Write\n\n\n\n$$\n\\begin{aligned}\n\nr(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} r(T), & g(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} g(T), \\\\\n\nb(\\mathcal{C}) & =\\min _{T \\in \\mathcal{C}} b(T), & R(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}}(T), \\\\\n\nG(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}} g(T), & B(\\mathcal{C}) & =\\max _{T \\in \\mathcal{C}} b(T) .\n\n\\end{aligned}\n$$\n\n\n\nSay that a collection $\\mathcal{C}$ is good if $r \\in[r(\\mathcal{C}, R(\\mathcal{C})]$, $g \\in[g(\\mathcal{C}, G(\\mathcal{C})]$, and $b \\in[b(\\mathcal{C}, B(\\mathcal{C})]$. By the problem conditions, the collection of all spanning trees in $\\Gamma$ is good.\n\n\n\nFor a good collection $\\mathcal{C}$, say that an edge $e$ of $\\Gamma$ is suspicious if $e$ belongs to some tree in $\\mathcal{C}$ but not to all trees in $\\mathcal{C}$. Choose now a good collection $\\mathcal{C}$ minimizing the number of suspicious edges. If $\\mathcal{C}$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(\\mathcal{C})g$.\n\n\n\nWe now distinguish two cases.\n\n\n\nCase 1: $B(\\mathcal{C})=b$.\n\n\n\nLet $T^{0}$ be a tree in $\\mathcal{C}$ with $g\\left(T^{0}\\right)=g(\\mathcal{C}) \\leq g$. Since $G(\\mathcal{C})>g$, there exists a green edge $e$ contained in some tree in $\\mathcal{C}$ but not in $T^{0}$; clearly, $e$ is suspicious. Fix one such green edge $e$.\n\n\n\nNow, for every $T$ in $\\mathcal{C}$, define a spanning tree $T_{1}$ of $\\Gamma$ as follows. If $T$ does not contain $e$, then $T_{1}=T$; in particular, $\\left(T^{0}\\right)_{1}=T^{0}$. Otherwise, the graph $T \\backslash\\{e\\}$ falls into two components. The tree $T_{0}$ contains some edge $e^{\\prime}$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_{1}=T \\backslash\\{e\\} \\cup\\left\\{e^{\\prime}\\right\\}$.\n\n\n\nLet $\\mathcal{C}_{1}=\\left\\{T_{1}: T \\in \\mathcal{C}\\right\\}$. All edges suspicious for $\\mathcal{C}_{1}$ are also suspicious for $\\mathcal{C}$, but no tree in $\\mathcal{C}_{1}$ con- tains $e$. So the number of suspicious edges for $\\mathcal{C}_{1}$ is strictly smaller than that for $\\mathcal{C}$.\n\n\n\nWe now show that $\\mathcal{C}_{1}$ is good, reaching thereby a contradiction with the choice of $\\mathcal{C}$. For every $T$ in $\\mathcal{C}$, the tree $T_{1}$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g\\left(\\mathcal{C}_{1}\\right) \\leq g(\\mathcal{C}) \\leq g, G\\left(\\mathcal{C}_{1}\\right) \\geq G(\\mathcal{C})-1 \\geq g$, $R\\left(\\mathcal{C}_{1}\\right) \\geq R(\\mathcal{C}) \\geq r, r\\left(\\mathcal{C}_{1}\\right) \\leq r(\\mathcal{C})+1 \\leq r$, and $B\\left(\\mathcal{C}_{1}\\right) \\geq B(\\mathcal{C}) \\geq b$. Finally, we get $b\\left(T^{0}\\right) \\leq$ $B(\\mathcal{C})=b$; since $\\mathcal{C}_{1}$ contains $T^{0}$, it follows that $b\\left(\\mathcal{C}_{1}\\right) \\leq b\\left(T^{0}\\right) \\leq b$, which concludes the proof.\n\n\n\nCase 2: $B(\\mathcal{C})>b$.\n\n\n\nConsider a tree $T^{0}$ in $\\mathcal{C}$ satisfying $r\\left(T^{0}\\right)=$ $R(\\mathcal{C}) \\geq r$. Since $r(\\mathcal{C})\n\n\n\nWe will now construct a suitable $g$ as a sum of squares. This means that, if we write $g=g_{1}^{2}+g_{2}^{2}+$ $\\cdots+g_{m}^{2}$, then $g=0$ if and only if $g_{1}=\\cdots=g_{m}=0$, and that if their degrees are $d_{1}, \\ldots, d_{m}$, then $g$ has degree at most $2 \\max \\left(d_{1}, \\ldots, d_{m}\\right)$.\n\n\n\nThus, it is sufficient to exhibit several polynomials, all of degree at most $n$, such that $2 n$ points with zero sum are the vertices of a regular $2 n$-gon if and only if the polynomials are all zero at those points.\n\n\n\n\n\n\n\nFirst, we will impose the constraints that all $\\left|A_{i}\\right|^{2}=x_{i}^{2}+y_{i}^{2}$ are equal. This uses multiple degree 2 constraints.\n\n\n\nNow, we may assume that the points $A_{1}, \\ldots, A_{2 n}$ all lie on a circle with centre 0 , and $A_{1}+\\cdots+A_{2 n}=0$. If this circle has radius 0 , then all $A_{i}$ coincide, and we may ignore this case.\n\n\n\nOtherwise, the circle has positive radius. We will use the following lemma.\n\n\n\nLemma. Suppose that $a_{1}, \\ldots, a_{2 n}$ are complex numbers of the same non-zero magnitude, and suppose that $a_{1}^{k}+\\cdots+a_{2 n}^{k}=0, k=1, \\ldots, n$. Then $a_{1}, \\ldots, a_{2 n}$ form a regular $2 n$-gon centred at the origin. (Conversely, this is easily seen to be sufficient.)\n\n\n\nProof. Since all the hypotheses are homogenous, we may assume (mostly for convenience) that $a_{1}, \\ldots, a_{2 n}$ lie on the unit circle. By Newton's sums, the $k$-th symmetric sums of $a_{1}, \\ldots, a_{2 n}$ are all zero for $k$ in the range $1, \\ldots, n$.\n\n\n\nTaking conjugates yields $a_{1}^{-k}+\\cdots+a_{2 n}^{-k}=0$, $k=1, \\ldots, n$. Thus, we can repeat the above logic to obtain that the $k$-th symmetric sums of $a_{1}^{-1}, \\ldots, a_{2 n}^{-1}$ are also all zero for $k=1, \\ldots, n$. However, these are simply the $(2 n-k)$-th symmetric sums of $a_{1}, \\ldots, a_{2 n}$ (divided by $a_{1} \\cdots a_{2 n}$ ), so the first $2 n-1$ symmetric sums of $a_{1}, \\ldots, a_{2 n}$ are all zero. This implies that $a_{1}, \\ldots, a_{2 n}$ form a regular $2 n$-gon centred at the origin.\n\n\n\nWe will encode all of these constraints into our polynomial. More explicitly, write $a_{r}=x_{r}+y_{r} i$; then the constraint $a_{1}^{k}+\\cdots+a_{2 n}^{k}=0$ can be expressed as $p_{k}+q_{k} i=0$, where $p_{k}$ and $q_{k}$ are real polynomials in the coordinates. To incorporate this, simply impose the constraints $p_{k}=0$ and $q_{k}=0$; these are conditions of degree $k \\leq n$, so their squares are all of degree at most $2 n$.\n\n\n\nTo recap, taking the sum of squares of all of these constraints gives a polynomial $f$ of degree at most $2 n$ which works whenever $A_{1}+\\cdots+A_{2 n}=0$. Finally, the centroid-shifting trick gives a polynomial which works in general, as wanted.""]",['$2n$'],False,,Expression, 1621,Algebra,,Let $x$ and $y$ be positive real numbers such that $x+y^{2016} \geq 1$. Prove that $x^{2016}+y>$ $1-1 / 100$.,"[""If $x \\geq 1-1 /(100 \\cdot 2016)$, then\n\n\n\n$$\n\nx^{2016} \\geq\\left(1-\\frac{1}{100 \\cdot 2016}\\right)^{2016}>1-2016 \\cdot \\frac{1}{100 \\cdot 2016}=1-\\frac{1}{100}\n\n$$\n\n\n\nby Bernoulli's inequality, whence the conclusion.\n\n\n\nIf $x<1-1 /(100 \\cdot 2016)$, then $y \\geq(1-x)^{1 / 2016}>(100 \\cdot 2016)^{-1 / 2016}$, and it is sufficient to show that the latter is greater than $1-1 / 100=99 / 100$; alternatively, but equivalently, that\n\n\n\n$$\n\n\\left(1+\\frac{1}{99}\\right)^{2016}>100 \\cdot 2016\n\n$$\n\n\n\nTo establish the latter, refer again to Bernoulli's inequality to write\n\n\n\n$$\n\n\\left(1+\\frac{1}{99}\\right)^{2016}>\\left(1+\\frac{1}{99}\\right)^{99 \\cdot 20}>\\left(1+99 \\cdot \\frac{1}{99}\\right)^{20}=2^{20}>100 \\cdot 2016 .\n\n$$""]",,True,,, 1622,Geometry,,"A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the $\operatorname{arc} A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$. Prove that $R \geq r_{1}+r_{2}+r_{3}$.","['Let $\\ell_{1}$ be the tangent to $\\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\\omega_{1}$. The tangents $\\ell_{2}$ and $\\ell_{3}$ are defined similarly. The lines $\\ell_{1}$ and $\\ell_{2}, \\ell_{2}$ and $\\ell_{3}, \\ell_{3}$ and $\\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.\n\n\n\nEach of the triangles $\\Delta_{1}=\\triangle X S_{1} T_{1}, \\Delta_{2}=\\triangle T_{2} X S_{2}$, and $\\Delta_{3}=\\triangle S_{3} T_{3} X$ is similar to $\\Delta=\\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\\Delta_{i}$ and $\\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\\rho_{i}$ is the inradius of $\\Delta_{i}$, then $\\rho_{1}+\\rho_{2}+\\rho_{3}=R$.\n\n\n\nFinally, notice that $\\omega_{i}$ is interior to $\\Delta_{i}$, so $r_{i} \\leq \\rho_{i}$, and the conclusion follows by the preceding.\n\n\n\n\n\nFig. 1\n\n\n\n\n\n\n\nFig. 2']",,True,,, 1623,Geometry,,"A convex hexagon $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3}$ is inscribed in a circle $\Omega$ of radius $R$. The diagonals $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ concur at $X$. For $i=1,2,3$, let $\omega_{i}$ be the circle tangent to the segments $X A_{i}$ and $X B_{i}$, and to the $\operatorname{arc} A_{i} B_{i}$ of $\Omega$ not containing other vertices of the hexagon; let $r_{i}$ be the radius of $\omega_{i}$. If $R=r_{1}+r_{2}+r_{3}$, prove that the six points where the circles $\omega_{i}$ touch the diagonals $A_{1} B_{2}, A_{2} B_{3}, A_{3} B_{1}$ are concyclic.","['Let $\\ell_{1}$ be the tangent to $\\Omega$ parallel to $A_{2} B_{3}$, lying on the same side of $A_{2} B_{3}$ as $\\omega_{1}$. The tangents $\\ell_{2}$ and $\\ell_{3}$ are defined similarly. The lines $\\ell_{1}$ and $\\ell_{2}, \\ell_{2}$ and $\\ell_{3}, \\ell_{3}$ and $\\ell_{1}$ meet at $C_{3}, C_{1}, C_{2}$, respectively (see Fig. 1). Finally, the line $C_{2} C_{3}$ meets the rays $X A_{1}$ and $X B_{1}$ emanating from $X$ at $S_{1}$ and $T_{1}$, respectively; the points $S_{2}, T_{2}$, and $S_{3}, T_{3}$ are defined similarly.\n\n\n\nEach of the triangles $\\Delta_{1}=\\triangle X S_{1} T_{1}, \\Delta_{2}=\\triangle T_{2} X S_{2}$, and $\\Delta_{3}=\\triangle S_{3} T_{3} X$ is similar to $\\Delta=\\triangle C_{1} C_{2} C_{3}$, since their corresponding sides are parallel. Let $k_{i}$ be the ratio of similitude of $\\Delta_{i}$ and $\\Delta$ (e.g., $k_{1}=X S_{1} / C_{1} C_{2}$ and the like). Since $S_{1} X=C_{2} T_{3}$ and $X T_{2}=S_{3} C_{1}$, it follows that $k_{1}+k_{2}+k_{3}=1$, so, if $\\rho_{i}$ is the inradius of $\\Delta_{i}$, then $\\rho_{1}+\\rho_{2}+\\rho_{3}=R$.\n\n\n\nFinally, notice that $\\omega_{i}$ is interior to $\\Delta_{i}$, so $r_{i} \\leq \\rho_{i}$, and the conclusion follows by the preceding.\n\n\n\n\n\nFig. 1\n\n\n\n\n\n\n\nFig. 2\n\nBy above, the equality $R=r_{1}+r_{2}+r_{3}$ holds if and only if $r_{i}=\\rho_{i}$ for all $i$, which implies in turn that $\\omega_{i}$ is the incircle of $\\Delta_{i}$. Let $K_{i}, L_{i}, M_{i}$ be the points where $\\omega_{i}$ touches the sides $X S_{i}, X T_{i}, S_{i} T_{i}$, respectively. We claim that the six points $K_{i}$ and $L_{i}(i=1,2,3)$ are equidistant from $X$.\n\n\n\nClearly, $X K_{i}=X L_{i}$, and we are to prove that $X K_{2}=X L_{1}$ and $X K_{3}=X L_{2}$. By similarity, $\\angle T_{1} M_{1} L_{1}=\\angle C_{3} M_{1} M_{2}$ and $\\angle S_{2} M_{2} K_{2}=\\angle C_{3} M_{2} M_{1}$, so the points $M_{1}, M_{2}, L_{1}, K_{2}$ are collinear. Consequently, $\\angle X K_{2} L_{1}=\\angle C_{3} M_{1} M_{2}=\\angle C_{3} M_{2} M_{1}=\\angle X L_{1} K_{2}$, so $X K_{2}=X L_{1}$. Similarly, $X K_{3}=X L_{2}$.']",,True,,, 1624,Geometry,,"A set of $n$ points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets $\mathcal{A}$ and $\mathcal{B}$. An $\mathcal{A B}$-tree is a configuration of $n-1$ segments, each of which has an endpoint in $\mathcal{A}$ and the other in $\mathcal{B}$, and such that no segments form a closed polyline. An $\mathcal{A B}$-tree is transformed into another as follows: choose three distinct segments $A_{1} B_{1}, B_{1} A_{2}$ and $A_{2} B_{2}$ in the $\mathcal{A B}$-tree such that $A_{1}$ is in $\mathcal{A}$ and $A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and remove the segment $A_{1} B_{1}$ to replace it by the segment $A_{1} B_{2}$. Given any $\mathcal{A B}$-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps.","['The configurations of segments under consideration are all bipartite geometric trees on the points $n$ whose vertex-parts are $\\mathcal{A}$ and $\\mathcal{B}$, and transforming one into another preserves the degree of any vertex in $\\mathcal{A}$, but not necessarily that of a vertex in $\\mathcal{B}$.\n\n\n\nThe idea is to devise a strict semi-invariant of the process, i.e., assign each $\\mathcal{A B}$-tree a real number strictly decreasing under a transformation. Since the number of trees on the $n$ points is finite, the conclusion follows.\n\n\n\nTo describe the assignment, consider an $\\mathcal{A B}$-tree $\\mathcal{T}=(\\mathcal{A} \\sqcup \\mathcal{B}, \\mathcal{E})$. Removal of an edge $e$ of $\\mathcal{T}$ splits the graph into exactly two components. Let $p_{\\mathcal{T}}(e)$ be the number of vertices in $\\mathcal{A}$ lying in the component of $\\mathcal{T}-e$ containing the $\\mathcal{A}$-endpoint of $e$; since $\\mathcal{T}$ is a tree, $p_{\\mathcal{T}}(e)$ counts the number of paths in $\\mathcal{T}-e$ from the $\\mathcal{A}$-endpoint of $e$ to vertices in $\\mathcal{A}$ (including the one-vertex path). Define $f(\\mathcal{T})=\\sum_{e \\in \\mathcal{E}} p_{\\mathcal{T}}(e)|e|$, where $|e|$ is the Euclidean length of $e$.\n\n\n\nWe claim that $f$ strictly decreases under a transformation. To prove this, let $\\mathcal{T}^{\\prime}$ be obtained from $\\mathcal{T}$ by a transformation involving the polyline $A_{1} B_{1} A_{2} B_{2}$; that is, $A_{1}$ and $A_{2}$ are in $\\mathcal{A}, B_{1}$ and $B_{2}$ are in $\\mathcal{B}, A_{1} B_{1}+A_{2} B_{2}>A_{1} B_{2}+A_{2} B_{1}$, and $\\mathcal{T}^{\\prime}=\\mathcal{T}-A_{1} B_{1}+A_{1} B_{2}$. It is readily checked that $p_{\\mathcal{T}^{\\prime}}(e)=p_{\\mathcal{T}}(e)$ for every edge $e$ of $\\mathcal{T}$ different from $A_{1} B_{1}, A_{2} B_{1}$ and $A_{2} B_{2}, p_{\\mathcal{T}^{\\prime}}\\left(A_{1} B_{2}\\right)=$ $p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right), p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{1}\\right)=p_{\\mathcal{T}}\\left(A_{2} B_{1}\\right)+p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)$, and $p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B b_{2}\\right)=p_{\\mathcal{T}}\\left(A_{2} B_{2}\\right)-p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)$. Consequently,\n\n\n\n$$\n\n\\begin{aligned}\n\nf\\left(\\mathcal{T}^{\\prime}\\right)-f(\\mathcal{T})= & p_{\\mathcal{T}^{\\prime}}\\left(A_{1} B_{2}\\right) \\cdot A_{1} B_{2}+\\left(p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{1}\\right)-p_{\\mathcal{T}}\\left(A_{2} B_{1}\\right)\\right) \\cdot A_{2} B_{1}+ \\\\\n\n& \\left(p_{\\mathcal{T}^{\\prime}}\\left(A_{2} B_{2}\\right)-p_{\\mathcal{T}}\\left(A_{2} B_{2}\\right)\\right) \\cdot A_{2} B_{2}-p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right) \\cdot A_{1} B_{1} \\\\\n\n= & p_{\\mathcal{T}}\\left(A_{1} B_{1}\\right)\\left(A_{1} B_{2}+A_{2} B_{1}-A_{2} B_{2}-A_{1} B_{1}\\right)<0\n\n\\end{aligned}\n\n$$']",,True,,, 1625,Algebra,,"Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$. Prove that there are infinitely many positive integers $n$ such that $\lambda(n)=\lambda(n+1)=+1$","['Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nThe Pell equation $x^{2}-6 y^{2}=1$ has infinitely many solutions in positive integers; all solutions are given by $\\left(x_{n}, y_{n}\\right)$, where $x_{n}+y_{n} \\sqrt{6}=(5+2 \\sqrt{6})^{n}$. Since $\\lambda\\left(6 y^{2}\\right)=1$ and also $\\lambda\\left(6 y^{2}+1\\right)=\\lambda\\left(x^{2}\\right)=1$, the thesis is proven.', 'Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nTake any existing pair with $\\lambda(n)=$ $\\lambda(n+1)=1$. Then $\\lambda\\left((2 n+1)^{2}-1\\right)=\\lambda\\left(4 n^{2}+4 n\\right)=\\lambda(4) \\cdot \\lambda(n)$. $\\lambda(n+1)=1$, and also $\\lambda\\left((2 n+1)^{2}\\right)=\\lambda(2 n+1)^{2}=1$, so we have built a larger $(1,1)$ pair.']",,True,,, 1626,Algebra,,"Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$. Prove that there are infinitely many positive integers $n$ such that $\lambda(n)=\lambda(n+1)=-1$.","['Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nThe equation $3 x^{2}-2 y^{2}=1$ (again Pell theory) has also infinitely many solutions in positive integers, given by $\\left(x_{n}, y_{n}\\right)$, where $x_{n} \\sqrt{3}+y_{n} \\sqrt{2}=(\\sqrt{3}+\\sqrt{2})^{2 n+1}$. Since $\\lambda\\left(2 y^{2}\\right)=$ -1 and $\\lambda\\left(2 y^{2}+1\\right)=\\lambda\\left(3 x^{2}\\right)=-1$, the thesis is proven.', 'Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nAssume $(\\lambda(n-1), \\lambda(n))$ is the largest $(-1,-1)$ pair, therefore $\\lambda(n+1)=1$ and $\\lambda\\left(n^{2}+n\\right)=\\lambda(n)$. $\\lambda(n+1)=-1$, therefore again $\\lambda\\left(n^{2}+n+1\\right)=1$. But then $\\lambda\\left(n^{3}-1\\right)=\\lambda(n-1) \\cdot \\lambda\\left(n^{2}+n+1\\right)=-1$, and also $\\lambda\\left(n^{3}\\right)=$ $\\lambda(n)^{3}=-1$, so we found yet a larger such pair than the one we started with, contradiction.', 'Notice that we have $\\Omega(m n)=\\Omega(m)+\\Omega(n)$ for all positive integers $m, n$ ( $\\Omega$ is a completely additive arithmetic function), translating into $\\lambda(m n)=\\lambda(m) \\cdot \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p)=-1$ for any prime $p$, and $\\lambda\\left(k^{2}\\right)=\\lambda(k)^{2}=+1$ for all positive integers $k$.\n\n\n\nThe start (first 100 terms) of the sequence $\\mathfrak{S}=(\\lambda(n))_{n \\geq 1}$ is\n+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1\n\nAssume the pairs of consecutive terms $(-1,-1)$ in $\\mathfrak{S}$ are finitely many. Then from some rank on we only have subsequences $(1,-1,1,1, \\ldots, 1,-1,1)$. By ""doubling"" such a subsequence (like at point ii)), we will produce\n\n\n\n$$\n\n(-1, ?, 1, ?,-1, ?,-1, ?, \\ldots, ?,-1, ?, 1, ?,-1)\n\n$$\n\n\n\nAccording with our assumption, all ?-terms ought to be 1, hence the produced subsequence is\n\n\n\n$$\n\n(-1,1,1,1,-1,1,-1,1, \\ldots, 1,-1,1,1,1,-1)\n\n$$\n\n\n\nand so the ""separating packets"" of l\'s contain either one or three terms. Now assume some far enough $(1,1,1,1)$ or $(-1,1,1,-1)$ subsequence of $\\mathfrak{S}$ were to exist. Since it lies within some ""doubled"" subsequence, it contradicts the structure described above, which thus is the only prevalent from some rank on. But then all the positions of the $(-1)$ terms will have the same parity. However though, we have $\\lambda(p)=\\lambda\\left(2 p^{2}\\right)=-1$ for all odd primes $p$, and these terms have different parity of their positions. A contradiction has been reached.']",,True,,, 1627,Combinatorics,,"For every $n \geq 3$, determine all the configurations of $n$ distinct points $X_{1}, X_{2}, \ldots, X_{n}$ in the plane, with the property that for any pair of distinct points $X_{i}, X_{j}$ there exists a permutation $\sigma$ of the integers $\{1, \ldots, n\}$, such that $\mathrm{d}\left(X_{i}, X_{k}\right)=\mathrm{d}\left(X_{j}, X_{\sigma(k)}\right)$ for all $1 \leq k \leq n$. (We write $\mathrm{d}(X, Y)$ to denote the distance between points $X$ and $Y$.)","['Let us first prove that the points must be concyclic. Assign to each point $X_{k}$ the vector $x_{k}$ in a system of orthogonal coordinates whose origin is the point of mass of the configuration, thus $\\frac{1}{n} \\sum_{k=1}^{n} x_{k}=0$.\n\n\n\nThen $\\mathrm{d}^{2}\\left(X_{i}, X_{k}\\right)=\\left\\|x_{i}-x_{k}\\right\\|^{2}=\\left\\langle x_{i}-x_{k}, x_{i}-x_{k}\\right\\rangle=$ $\\left\\|x_{i}\\right\\|^{2}-2\\left\\langle x_{i}, x_{k}\\right\\rangle+\\left\\|x_{k}\\right\\|^{2}$, hence $\\sum_{k=1}^{n} \\mathrm{~d}^{2}\\left(X_{i}, X_{k}\\right)=n\\left\\|x_{i}\\right\\|^{2}-2\\left\\langle x_{i}, \\sum_{k=1}^{n} x_{k}\\right\\rangle+\\sum_{k=1}^{n}\\left\\|x_{k}\\right\\|^{2}=n\\left\\|x_{i}\\right\\|^{2}+\\sum_{k=1}^{n}\\left\\|x_{k}\\right\\|^{2}=n\\left\\|x_{j}\\right\\|^{2}+$ $\\sum_{k=1}^{n}\\left\\|x_{\\sigma(k)}\\right\\|^{2}=\\sum_{k=1}^{n} \\mathrm{~d}^{2}\\left(X_{j}, X_{\\sigma(k)}\\right)$, therefore $\\left\\|x_{i}\\right\\|=\\left\\|x_{j}\\right\\|$ for all pairs $(i, j)$. The points are thus concyclic (lying on a circle centred at $O(0,0))$.\n\n\n\nLet now $m$ be the least angular distance between any two points. Two points situated at angular distance $m$ must be adjacent on the circle. Let us connect each pair of such two points with an edge. The graph $G$ obtained must be regular, of degree $\\operatorname{deg}(G)=1$ or 2 . If $n$ is odd, since $\\sum_{k=1}^{n} \\operatorname{deg}\\left(X_{k}\\right)=$ $n \\operatorname{deg}(G)=2|E|$, we must have $\\operatorname{deg}(G)=2$, hence the configuration is a regular $n$-gon.\n\n\n\nIf $n$ is even, we may have the configuration of a regular $n$-gon, but we also may have $\\operatorname{deg}(G)=1$. In that case, let $M$ be the next least angular distance between any two points; such points must also be adjacent on the circle. Let us connect each pair of such two points with an edge, in order to get a graph $G^{\\prime}$. A similar reasoning yields $\\operatorname{deg}\\left(G^{\\prime}\\right)=1$, thus the configuration is that of an equiangular $n$-gon (with alternating equal side-lengths).']",['the configuration is that of an equiangular $n$-gon when $n$ is even and the configuration is a regular $n$-gon when $n$ is odd'],False,,Need_human_evaluate, 1628,Combinatorics,,"Consider an integer $n \geq 2$ and write the numbers $1,2, \ldots, n$ down on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. For all integers $n \geq 2$, determine whether it is possible to be left with exactly two numbers on the board after a finite number of moves.","['The answer is in the affirmative for all $n \\geq 2$. Induct on $n$. Leaving aside the trivial case $n=2$, deal first with particular cases $n=5$ and $n=6$.\n\n\n\nIf $n=5$, remove first the pair $(2,5)$, notice that $3=|2-5|$ is already on the board, so $7=2+5$ alone is written down. Removal of the pair $(3,4)$ then leaves exactly two numbers on the board, 1 and 7 , since $|3 \\pm 4|$ are both already there.\n\n\n\nIf $n=6$, remove first the pair $(1,6)$, notice that $5=|1-6|$ is already on the board, so $7=1+6$ alone is written down. Next, remove the pair $(2,5)$ and notice that $|2 \\pm 5|$ are both already on the board, so no new number is written down. Finally, removal of the pair $(3,4)$ provides a single number to be written down, $1=|3-4|$, since $7=3+4$ is already on the board. At this stage, the process comes to an end: 1 and 7 are the two numbers left.\n\n\n\nIn the remaining cases, the problem for $n$ is brought down to the corresponding problem for $[n / 2\\rceilb$ on the board, we can replace them by $a+b$ and $a-b$, and then, performing a move on the two new numbers, by $(a+b)+(a-b)=2 a$ and $(a+b)-(a-b)=2 b$. So we can double any two numbers on the board.\n\n\n\nWe now show that, if the board contains two even numbers $a$ and $b$, we can divide them both by 2 , while keeping the other numbers unchanged. If $k$ is even, split the other numbers into pairs to multiply each pair by 2 ; then clear out the common factor 2 . If $k$ is odd, split all numbers but $a$ into pairs to multiply each by 2 ; then do the same for all numbers but $b$; finally, clear out the common factor 4.\n\n\n\nBack to the problem, if two of the numbers $a_{1}, \\ldots, a_{k}$ are even, reduce them both by 2 to get a set with a smaller sum, which is impossible. Otherwise, two numbers, say, $a_{1}\n\n\n\nUsing the Claim above, we now show that the citizens may be placed at the $3 n^{2}+3 n+1$ tile vertices.\n\n\n\nConsider any tile $T_{1}$; its vertices are at least 1 apart from each other. Moreover, let $B A C$ be a part of the boundary of some tile $T$, and let $X$ be any point of the boundary of $T$, lying outside the half-open intervals $[A, B)$ and $[A, C)$ (in this case, we say that $X$ is not adjacent to $A$ ). Then $A X \\geq \\sqrt{3} / 2$.\n\n\n\nNow consider any two tile vertices $A$ and $B$. If they are vertices of the same tile we already know $A B \\geq 1$; otherwise, the segment $A B$ crosses the boundaries of some tiles containing $A$ and $B$ at some points $X$ and $Y$ not adjacent to $A$ and $B$, respectively. Hence $A B \\geq A X+Y B \\geq \\sqrt{3}>1$.']",,True,,, 1630,Algebra,,"Initially, a non-constant polynomial $S(x)$ with real coefficients is written down on a board. Whenever the board contains a polynomial $P(x)$, not necessarily alone, one can write down on the board any polynomial of the form $P(C+x)$ or $C+P(x)$, where $C$ is a real constant. Moreover, if the board contains two (not necessarily distinct) polynomials $P(x)$ and $Q(x)$, one can write $P(Q(x))$ and $P(x)+Q(x)$ down on the board. No polynomial is ever erased from the board. Given two sets of real numbers, $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ and $B=\left\{b_{1}, b_{2}, \ldots, b_{n}\right\}$, a polynomial $f(x)$ with real coefficients is $(A, B)$-nice if $f(A)=B$, where $f(A)=\left\{f\left(a_{i}\right): i=1,2, \ldots, n\right\}$. Determine all polynomials $S(x)$ that can initially be written down on the board such that, for any two finite sets $A$ and $B$ of real numbers, with $|A|=|B|$, one can produce an $(A, B)$-nice polynomial in a finite number of steps.","['The required polynomials are all polynomials of an even degree $d \\geq 2$, and all polynomials of odd degree $d \\geq 3$ with negative leading coefficient.\n\n\n\nPart I. We begin by showing that any (non-constant) polynomial $S(x)$ not listed above is not $(A, B)$-nice for some pair $(A, B)$ with either $|A|=|B|=2$, or $|A|=|B|=3$.\n\n\n\nIf $S(x)$ is linear, then so are all the polynomials appearing on the board. Therefore, none of them will be $(A, B)$-nice, say, for $A=\\{1,2,3\\}$ and $B=\\{1,2,4\\}$, as desired.\n\n\n\nOtherwise, $\\operatorname{deg} S=d \\geq 3$ is odd, and the leading coefficient is positive. In this case, we make use of the following technical fact, whose proof is presented at the end of the solution.\n\n\n\nClaim. There exists a positive constant $T$ such that $S(x)$ satisfies the following condition:\n\n\n\n$$\n\nS(b)-S(a) \\geq b-a \\quad \\text { whenever } \\quad b-a \\geq T \\text {. } \\tag{*}\n\n$$\n\n\n\nFix a constant $T$ provided by the Claim. Then, an immediate check shows that all newly appearing polynomials on the board also satisfy $(*)$ (with the same value of $T$ ). Therefore, none of them will be $(A, B)$-nice, say, for $A=\\{0, T\\}$ and $B=\\{0, T / 2\\}$, as desired.\n\n\n\nPart II. We show that the polynomials listed in the Answer satisfy the requirements. We will show that for any $a_{1}\\max _{x \\in \\Delta} S(x)$. Therefore, for any $x, y, z$ with $x \\leq \\alpha \\leq y \\leq \\beta \\leq z$ we get $S(x) \\leq S(\\alpha) \\leq$ $S(y) \\leq S(\\beta) \\leq S(z)$.\n\n\n\nWe may decrease $\\alpha$ and increase $\\beta$ (preserving the condition above) so that, in addition, $S^{\\prime}(x)>3$ for all $x \\notin[\\alpha, \\beta]$. Now we claim that the number $T=3(\\beta-\\alpha)$ fits the bill.\n\n\n\nIndeed, take any $a$ and $b$ with $b-a \\geq T$. Even if the segment $[a, b]$ crosses $[\\alpha, \\beta]$, there still is a segment $\\left[a^{\\prime}, b^{\\prime}\\right] \\subseteq[a, b] \\backslash(\\alpha, \\beta)$ of length $b^{\\prime}-a^{\\prime} \\geq(b-a) / 3$. Then\n\n\n\n$$\n\nS(b)-S(a) \\geq S\\left(b^{\\prime}\\right)-S\\left(a^{\\prime}\\right)=\\left(b^{\\prime}-a^{\\prime}\\right) \\cdot S^{\\prime}(\\xi) \\geq 3\\left(b^{\\prime}-a^{\\prime}\\right) \\geq b-a\n\n$$\n\n\n\nfor some $\\xi \\in\\left(a^{\\prime}, b^{\\prime}\\right)$.\n\n\n\nProof of Lemma 1. If $S(x)$ has an even degree, then the polynomial $T(x)=S\\left(x+a_{2}\\right)-S\\left(x+a_{1}\\right)$ has an odd degree, hence there exists $x_{0}$ with $T\\left(x_{0}\\right)=S\\left(x_{0}+a_{2}\\right)-S\\left(x_{0}+a_{1}\\right)=b_{2}-b_{1}$. Setting $G(x)=S\\left(x+x_{0}\\right)$, we see that $G\\left(a_{2}\\right)-G\\left(a_{1}\\right)=b_{2}-b_{1}$, so a suitable shift $F(x)=G(x)+\\left(b_{1}-G\\left(a_{1}\\right)\\right)$ fits the bill.\n\n\n\nAssume now that $S(x)$ has odd degree and a negative leading coefficient. Notice that the polynomial $S^{2}(x):=S(S(x))$ has an odd degree and a positive leading coefficient. So, the polynomial $S^{2}\\left(x+a_{2}\\right)-S^{2}\\left(x+a_{1}\\right)$ attains all sufficiently large positive values, while $S\\left(x+a_{2}\\right)-$ $S\\left(x+a_{1}\\right)$ attains all sufficiently large negative values. Therefore, the two-variable polynomial $S^{2}\\left(x+a_{2}\\right)-S^{2}\\left(x+a_{1}\\right)+S\\left(y+a_{2}\\right)-S\\left(y+a_{1}\\right)$ attains all real values; in particular, there exist $x_{0}$ and $y_{0}$ with $S^{2}\\left(x_{0}+a_{2}\\right)+S\\left(y_{0}+a_{2}\\right)-S^{2}\\left(x_{0}+a_{1}\\right)-S\\left(y_{0}+a_{1}\\right)=b_{2}-b_{1}$. Setting $G(x)=S^{2}\\left(x+x_{0}\\right)+S\\left(x+y_{0}\\right)$, we see that $G\\left(a_{2}\\right)-G\\left(a_{1}\\right)=b_{2}-b_{1}$, so a suitable shift of $G$ fits the bill.\n\n\n\nProof of Lemma 2. Let $\\Delta$ denote the segment $\\left[a_{1} ; a_{n}\\right]$. We modify the proof of Lemma 1 in order to obtain a polynomial $F$ convex (or concave) on $\\Delta$ such that $F\\left(a_{1}\\right)=F\\left(a_{2}\\right)$; then $F$ is a desired polynomial. Say that a polynomial $H(x)$ is good if $H$ is convex on $\\Delta$.\n\n\n\nIf $\\operatorname{deg} S$ is even, and its leading coefficient is positive, then $S(x+c)$ is good for all sufficiently large negative $c$, and $S\\left(a_{2}+c\\right)-S\\left(a_{1}+c\\right)$ attains all sufficiently large negative values for such $c$. Similarly, $S(x+c)$ is good for all sufficiently large positive $c$, and $S\\left(a_{2}+c\\right)-S\\left(a_{1}+c\\right)$ attains all sufficiently large positive values for such $c$. Therefore, there exist large $c_{1}<01$ then $2^{m}-1 \\mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \\ldots, y_{k}$ are primes for some $k \\geq 1$ then $a=x_{1}$ is also prime.\n\n\n\nNow we claim that for every odd prime $a$ at least one of the numbers $y_{1}, y_{2}, y_{3}$ is composite (and thus $k<3$ ). Assume, to the contrary, that $y_{1}, y_{2}$, and $y_{3}$ are primes; then $x_{1}, x_{2}, x_{3}$ are primes as well. Since $x_{1} \\geq 3$ is odd, we have $x_{2}>3$ and $x_{2} \\equiv 3(\\bmod 4)$; consequently, $x_{3} \\equiv 7$ $(\\bmod 8)$. This implies that 2 is a quadratic residue modulo $p=x_{3}$, so $2 \\equiv s^{2}(\\bmod p)$ for some integer $s$, and hence $2^{x_{2}}=2^{(p-1) / 2} \\equiv s^{p-1} \\equiv 1(\\bmod p)$. This means that $p \\mid y_{2}$, thus $2^{x_{2}}-1=x_{3}=2 x_{2}+1$. But it is easy to show that $2^{t}-1>2 t+1$ for all integer $t>3$. A contradiction.\n\n\n\nFinally, if $a=2$, then the numbers $y_{1}=3$ and $y_{2}=31$ are primes, while $y_{3}=2^{11}-1$ is divisible by 23 ; in this case we may choose $k=2$ but not $k=3$.']",['2'],False,,Numerical, 1632,Algebra,,"Does there exist a pair $(g, h)$ of functions $g, h: \mathbb{R} \rightarrow \mathbb{R}$ such that the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \in \mathbb{R}$ is the identity function $f(x) \equiv x$ ?","['Such a tester pair exists. We may biject $\\mathbb{R}$ with the closed unit interval, so it suffices to find a tester pair for that instead. We give an explicit example: take some positive real numbers $\\alpha, \\beta$ (which we will specify further later). Take\n\n\n\n$$\n\ng(x)=\\max (x-\\alpha, 0) \\quad \\text { and } \\quad h(x)=\\min (x+\\beta, 1)\n\n$$15\n\n\n\nSay a set $S \\subseteq[0,1]$ is invariant if $f(S) \\subseteq S$ for all functions $f$ commuting with both $g$ and $h$. Note that intersections and unions of invariant sets are invariant. Preimages of invariant sets under $g$ and $h$ are also invariant; indeed, if $S$ is invariant and, say, $T=g^{-1}(S)$, then $g(f(T))=$ $f(g(T)) \\subseteq f(S) \\subseteq S$, thus $f(T) \\subseteq T$.\n\n\n\nWe claim that (if we choose $\\alpha+\\beta<1$ ) the intervals $[0, n \\alpha-m \\beta]$ are invariant where $n$ and $m$ are nonnegative integers with $0 \\leq n \\alpha-m \\beta \\leq 1$. We prove this by induction on $m+n$.\n\n\n\nThe set $\\{0\\}$ is invariant, as for any $f$ commuting with $g$ we have $g(f(0))=f(g(0))=f(0)$, so $f(0)$ is a fixed point of $g$. This gives that $f(0)=0$, thus the induction base is established.\n\n\n\nSuppose now we have some $m, n$ such that $\\left[0, n^{\\prime} \\alpha-m^{\\prime} \\beta\\right]$ is invariant whenever $m^{\\prime}+n^{\\prime}<$ $m+n$. At least one of the numbers $(n-1) \\alpha-m \\beta$ and $n \\alpha-(m-1) \\beta$ lies in $(0,1)$. Note however that in the first case $[0, n \\alpha-m \\beta]=g^{-1}([0,(n-1) \\alpha-m \\beta])$, so $[0, n \\alpha-m \\beta]$ is invariant. In the second case $[0, n \\alpha-m \\beta]=h^{-1}([0, n \\alpha-(m-1) \\beta])$, so again $[0, n \\alpha-m \\beta]$ is invariant. This completes the induction.\n\n\n\nWe claim that if we choose $\\alpha+\\beta<1$, where $0<\\alpha \\notin \\mathbb{Q}$ and $\\beta=1 / k$ for some integer $k>1$, then all intervals $[0, \\delta]$ are invariant for $0 \\leq \\delta<1$. This occurs, as by the previous claim, for all nonnegative integers $n$ we have $[0,(n \\alpha \\bmod 1)]$ is invariant. The set of $n \\alpha \\bmod 1$ is dense in $[0,1]$, so in particular\n\n\n\n$$\n\n[0, \\delta]=\\bigcap_{(n \\alpha \\bmod 1)>\\delta}[0,(n \\alpha \\bmod 1)]\n\n$$\n\n\n\nis invariant.\n\n\n\nA similar argument establishes that $[\\delta, 1]$ is invariant, so by intersecting these $\\{\\delta\\}$ is invariant for $0<\\delta<1$. Yet we also have $\\{0\\},\\{1\\}$ both invariant, which proves $f$ to be the identity.', ""Let us agree that a sequence $\\mathbf{x}=\\left(x_{n}\\right)_{n=1,2, \\ldots}$ is cofinally non-constant if for every index $m$ there exists an index $n>m$ such that $x_{m} \\neq x_{n}$.\n\n\n\nBiject $\\mathbb{R}$ with the set of cofinally non-constant sequences of 0's and 1's, and define $g$ and $h$ by\n\n\n\n$$\n\ng(\\epsilon, \\mathbf{x})=\\left\\{\\begin{array}{ll}\n\n\\epsilon, \\mathbf{x} & \\text { if } \\epsilon=0 \\\\\n\n\\mathbf{x} & \\text { else }\n\n\\end{array} \\quad \\text { and } \\quad h(\\epsilon, \\mathbf{x})= \\begin{cases}\\epsilon, \\mathbf{x} & \\text { if } \\epsilon=1 \\\\\n\n\\mathbf{x} & \\text { else }\\end{cases}\\right.\n\n$$\n\n\n\nwhere $\\epsilon, \\mathbf{x}$ denotes the sequence formed by appending $\\mathbf{x}$ to the single-element sequence $\\epsilon$. Note that $g$ fixes precisely those sequences beginning with 0 , and $h$ fixes precisely those beginning with 1.\n\n\n\nNow assume that $f$ commutes with both $f$ and $g$. To prove that $f(\\mathbf{x})=\\mathbf{x}$ for all $\\mathbf{x}$ we show that $\\mathbf{x}$ and $f(\\mathbf{x})$ share the same first $n$ terms, by induction on $n$.\n\n\n\nThe base case $n=1$ is simple, as we have noticed above that the set of sequences beginning with a 0 is precisely the set of $g$-fixed points, so is preserved by $f$, and similarly for the set of sequences starting with 1 .\n\n\n\n\n\n\n\nSuppose that $f(\\mathbf{x})$ and $\\mathbf{x}$ agree for the first $n$ terms, whatever $\\mathbf{x}$. Consider any sequence, and write it as $\\mathbf{x}=\\epsilon, \\mathbf{y}$. Without loss of generality, we may (and will) assume that $\\epsilon=0$, so $f(\\mathbf{x})=0, \\mathbf{y}^{\\prime}$ by the base case. Yet then $f(\\mathbf{y})=f(h(\\mathbf{x}))=h(f(\\mathbf{x}))=h\\left(0, \\mathbf{y}^{\\prime}\\right)=\\mathbf{y}^{\\prime}$. Consequently, $f(\\mathbf{x})=0, f(\\mathbf{y})$, so $f(\\mathbf{x})$ and $\\mathbf{x}$ agree for the first $n+1$ terms by the inductive hypothesis.\n\n\n\nThus $f$ fixes all of cofinally non-constant sequences, and the conclusion follows."", 'We will show that there exists a tester pair of bijective functions $g$ and $h$.\n\n\n\nFirst of all, let us find out when a pair of functions is a tester pair. Let $g, h: \\mathbb{R} \\rightarrow \\mathbb{R}$ be arbitrary functions. We construct a directed graph $G_{g, h}$ with $\\mathbb{R}$ as the set of vertices, its edges being painted with two colors: for every vertex $x \\in \\mathbb{R}$, we introduce a red edge $x \\rightarrow g(x)$ and a blue edge $x \\rightarrow h(x)$.\n\n\n\nNow, assume that the function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies $f(g(x))=g(f(x))$ and $f(h(x))=h(f(x))$ for all $x \\in \\mathbb{R}$. This means exactly that if there exists an edge $x \\rightarrow y$, then there also exists an edge $f(x) \\rightarrow f(y)$ of the same color; that is $-f$ is an endomorphism of $G_{g, h}$.\n\n\n\nThus, a pair $(g, h)$ is a tester pair if and only if the graph $G_{g, h}$ admits no nontrivial endomorphisms. Notice that each endomorphism maps a component into a component. Thus, to construct a tester pair, it suffices to construct a continuum of components with no nontrivial endomorphisms and no homomorphisms from one to another. It can be done in many ways; below we present one of them.\n\n\n\nLet $g(x)=x+1$; the construction of $h$ is more involved. For every $x \\in[0,1)$ we define the set $S_{x}=x+\\mathbb{Z}$; the sets $S_{x}$ will be exactly the components of $G_{g, h}$. Now we will construct these components.\n\n\n\nLet us fix any $x \\in[0,1)$; let $x=0 . x_{1} x_{2} \\ldots$ be the binary representation of $x$. Define $h(x-n)=x-n+1$ for every $n>3$. Next, let $h(x-3)=x, h(x)=x-2, h(x-2)=x-1$, and $h(x-1)=x+1$ (that would be a ""marker"" which fixes a point in our component).\n\n\n\nNext, for every $i=1,2, \\ldots$, we define\n\n\n\n(1) $h(x+3 i-2)=x+3 i-1, h(x+3 i-1)=x+3 i$, and $h(x+3 i)=x+3 i+1$, if $x_{i}=0$;\n\n\n\n(2) $h(x+3 i-2)=x+3 i, h(x+3 i)=3 i-1$, and $h(x+3 i-1)=x+3 i+1$, if $x_{i}=1$.\n\n\n\nClearly, $h$ is a bijection mapping each $S_{x}$ to itself. Now we claim that the graph $G_{g, h}$ satisfies the desired conditions.\n\n\n\nConsider any homomorphism $f_{x}: S_{x} \\rightarrow S_{y}$ ( $x$ and $y$ may coincide). Since $g$ is a bijection, consideration of the red edges shows that $f_{x}(x+n)=x+n+k$ for a fixed real $k$. Next, there exists a blue edge $(x-3) \\rightarrow x$, and the only blue edge of the form $(y+m-3) \\rightarrow(y+m)$ is $(y-3) \\rightarrow y$; thus $f_{x}(x)=y$, and $k=0$.\n\n\n\nNext, if $x_{i}=0$ then there exists a blue edge $(x+3 i-2) \\rightarrow(x+3 i-1)$; then the edge $(y+3 i-2) \\rightarrow(y+3 i-1)$ also should exist, so $y_{i}=0$. Analogously, if $x_{i}=1$ then there exists a blue edge $(x+3 i-2) \\rightarrow(x+3 i)$; then the edge $(y+3 i-2) \\rightarrow(y+3 i)$ also should exist, so $y_{i}=1$. We conclude that $x=y$, and $f_{x}$ is the identity mapping, as required.']",,True,,, 1633,Geometry,,"Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$. The lines $A B$ and $C D$ meet at $P$, the lines $A D$ and $B C$ meet at $Q$, and the diagonals $A C$ and $B D$ meet at $R$. Let $M$ be the midpoint of the segment $P Q$, and let $K$ be the common point of the segment $M R$ and the circle $\omega$. Prove that the circumcircle of the triangle $K P Q$ and $\omega$ are tangent to one another.","['Let $O$ be the centre of $\\omega$. Notice that the points $P, Q$, and $R$ are the poles (with respect to $\\omega$ ) of the lines $Q R, R P$, and $P Q$, respectively. Hence we have $O P \\perp Q R, O Q \\perp R P$, and $O R \\perp P Q$, thus $R$ is the orthocentre of the triangle $O P Q$. Now, if $M R \\perp P Q$, then the points $P$ and $Q$ are the reflections of one another in the line $M R=M O$, and the triangle $P Q K$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\n\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $M R$, and let $U$ be the common point of the lines $O V$ and $P Q$. Since $U$ lies on the polar line of $R$ and $O U \\perp M R$, we obtain that $U$ is the pole of $M R$. Therefore, the line $U K$ is tangent to $\\omega$. Hence it is enough to prove that $U K^{2}=U P \\cdot U Q$, since this relation implies that $U K$ is also tangent to the circle $K P Q$.\n\n\n\nFrom the rectangular triangle $O K U$, we get $U K^{2}=U V \\cdot U O$. Let $\\Omega$ be the circumcircle of triangle $O P Q$, and let $R^{\\prime}$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $P Q$. It is well known that $R^{\\prime}$ is the point of $\\Omega$ opposite to $O$, hence $O R^{\\prime}$ is the diameter of $\\Omega$. Finally, since $\\angle O V R^{\\prime}=90^{\\circ}$, the point $V$ also lies on $\\Omega$, hence $U P \\cdot U Q=U V \\cdot U O=U K^{2}$, as required.\n\n\n\n']",,True,,, 1634,Algebra,,"Does there exist an infinite sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$ ?","['The answer is in the affirmative.\n\n\n\nThe idea is to consider a sequence of pairwise distinct primes $p_{1}, p_{2}, p_{3}, \\ldots$, cover the positive integers by a sequence of finite non-empty sets $I_{n}$ such that $I_{m}$ and $I_{n}$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_{n}=\\prod_{i \\in I_{n}} p_{i}, n=1,2,3, \\ldots$\n\n\n\nOne possible way of finding such sets is the following. For all positive integers $n$, let\n\n\n\n$$\n\n\\begin{aligned}\n\n& 2 n \\in I_{k} \\quad \\text { for all } k=n, n+3, n+5, n+7, \\ldots ; \\quad \\text { and } \\\\\n\n& 2 n-1 \\in I_{k} \\quad \\text { for all } k=n, n+2, n+4, n+6, \\ldots\n\n\\end{aligned}\n\n$$\n\n\n\nClearly, each $I_{k}$ is finite, since it contains none of the numbers greater than $2 k$. Next, the number $p_{2 n}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i}$, while the number $p_{2 n-1}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i+1}$ for $i=1,2, \\ldots$. Finally, none of the indices appears in two consecutive sets.']",,True,,, 1635,Combinatorics,,"For an integer $n \geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another counter. A move is legal if the area of the triangle formed by the counters is strictly greater after the move than before. The players take turns to make legal moves, and if a player cannot make a legal move, that player loses. For which values of $n$ does the player making the first move have a winning strategy?","['We shall prove that the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd.\n\n\n\nSince the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins od Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).\n\n\n\nLemma. Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.\n\n\n\nProof. Label the positions of the counters $X, Y, Z$ so that the $\\operatorname{arc} Y Z$ of the circumcircle is shortest and the $\\operatorname{arc} Z X$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $Y Z X$ until it forms an isosceles triangle $X Y Z^{\\prime}$ with apex at $Y$ (note that the $\\operatorname{arc} X Y$ is less than half the circle, so that $Z$ does not jump over the counter at $X$ ). If this configuration is a Loss, we are done.\n\n\n\nIf instead this configuration is a Win, then the counters can be moved legally from triangle $X Y Z^{\\prime}$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z^{\\prime}$ to a new location $Z^{\\prime \\prime}$. But then the counter at $Z$ could have been moved to $Z^{\\prime \\prime}$ in the first place, so the original configuration was a Win as well.\n\n\n\nFor every nonzero integer $x$, denote by $v_{2}(x)$ the exponent of 2 in the prime decomposition of $x$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices having lengths $a, a, b$ respectively (in appropriate units so that $2 a+b=n$ ), we show that the configuration is a Win if and only if $a \\neq b$ and $v_{2}(a-b)$ is odd.\n\n\n\nWrite $b=a \\pm|a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \\pm|a-b| / 2, a \\pm|a-b| / 2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.\n\n\n\nIf instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \\pm|a-b| / 2, a \\pm|a-b| / 2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \\pm|a-b| / 2, a \\pm|a-b| / 2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b| / 2$, the conclusion follows inductively once the exceptional and trivial case $a=b$ is dealt with.\n\n\n\nAs an immediate corollary, the configuration with arc lengths 1, 1, $n-2$ (the starting configuration of the question) is a Win if and only if $v_{2}(n-3)$ is odd.']",['the first player wins if and only the exponent of 2 in the prime decomposition of $n-3$ is odd'],False,,Need_human_evaluate, 1636,Combinatorics,,"A finite list of rational numbers is written on a blackboard. In an operation, we choose any two numbers $a, b$, erase them, and write down one of the numbers $$ a+b, a-b, b-a, a \times b, a / b(\text { if } b \neq 0), b / a(\text { if } a \neq 0) $$ Prove that, for every integer $n>100$, there are only finitely many integers $k \geq 0$, such that, starting from the list $$ k+1, k+2, \ldots, k+n $$ it is possible to obtain, after $n-1$ operations, the value $n$ !.","['We prove the problem statement even for all positive integer $n$.\n\n\n\nThere are only finitely many ways of constructing a number from $n$ pairwise distinct numbers $x_{1}, \\ldots, x_{n}$ only using the four elementary arithmetic operations, and each $x_{k}$ exactly once. Each such formula for $k>1$ is obtained by an elementary operation from two such formulas on two disjoint sets of the $x_{i}$.\n\n\n\nA straightforward induction on $n$ shows that the outcome of each such construction is a number of the form\n\n\n\n$$\n\n\\frac{\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in\\{0,1\\}} a_{\\alpha_{1}, \\ldots, \\alpha_{n}} x_{1}^{\\alpha_{1}} \\cdots x_{n}^{\\alpha_{n}}}{\\sum_{\\alpha_{1}, \\ldots, \\alpha_{n} \\in\\{0,1\\}} b_{\\alpha_{1}, \\ldots, \\alpha_{n}} x_{1}^{\\alpha_{1}} \\cdots x_{n}^{\\alpha_{n}}},\n\\tag{*}\n$$\n\n\n\nwhere the $a_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ and $b_{\\alpha_{1}, \\ldots, \\alpha_{n}}$ are all in the set $\\{0, \\pm 1\\}$, not all zero of course, $a_{0, \\ldots, 0}=b_{1, \\ldots, 1}=0$, and also $a_{\\alpha_{1}, \\ldots, \\alpha_{n}} \\cdot b_{\\alpha_{1}, \\ldots, \\alpha_{n}}=0$ for every set of indices.\n\n\n\nSince $\\left|a_{\\alpha_{1}, \\ldots, \\alpha_{n}}\\right| \\leq 1$, and $a_{0,0, \\ldots, 0}=0$, the absolute value of the numerator does not exceed $\\left(1+\\left|x_{1}\\right|\\right) \\cdots\\left(1+\\left|x_{n}\\right|\\right)-1$; in particular, if $c$ is an integer in the range $-n, \\ldots,-1$, and $x_{k}=c+k$, $k=1, \\ldots, n$, then the absolute value of the numerator is at most $(-c) !(n+c+1) !-1 \\leq n !-10>p_{2}$. The graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ cross at distinct points $A$ and $B$. The four tangents to $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ at $A$ and $B$ form a convex quadrilateral which has an inscribed circle. Prove that the graphs $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ have the same axis of symmetry.","['Let $\\mathcal{A}_{i}$ and $\\mathcal{B}_{i}$ be the tangents to $\\mathcal{G}_{i}$ at $A$ and $B$, respectively, and let $C_{i}=\\mathcal{A}_{i} \\cap \\mathcal{B}_{i}$. Since $f_{1}(x)$ is convex and $f_{2}(x)$ is concave, the convex quadrangle formed by the four tangents is exactly $A C_{1} B C_{2}$.\n\n\n\nLemma. If $C A$ and $C B$ are the tangents drawn from a point $C$ to the graph $\\mathcal{G}$ of a quadratic trinomial $f(x)=p x^{2}+q x+r, A, B \\in \\mathcal{G}, A \\neq B$, then the abscissa of $C$ is the arithmetic mean of the abscissae of $A$ and $B$.\n\n\n\nProof. Assume, without loss of generality, that $C$ is at the origin, so the equations of the two tangents have the form $y=k_{a} x$ and $y=k_{b} x$. Next, the abscissae $x_{A}$ and $x_{B}$ of the tangency points $A$ and $B$, respectively, are multiple roots of the polynomials $f(x)-k_{a} x$ and $f(x)-k_{b} x$, respectively. By the Vieta theorem, $x_{A}^{2}=r / p=x_{B}^{2}$, so $x_{A}=-x_{B}$, since the case $x_{A}=x_{B}$ is ruled out by $A \\neq B$.\n\n\n\n\n\nThe Lemma shows that the line $C_{1} C_{2}$ is parallel to the $y$-axis and the points $A$ and $B$ are equidistant from this line.\n\n\n\nSuppose, if possible, that the incentre $O$ of the quadrangle $A C_{1} B C_{2}$ does not lie on the line $C_{1} C_{2}$. Assume, without loss of generality, that $O$ lies inside the triangle $A C_{1} C_{2}$ and let $A^{\\prime}$ be the reflection of $A$ in the line $C_{1} C_{2}$. Then the ray $C_{i} B$ emanating from $C_{i}$ lies inside the angle $A C_{i} A^{\\prime}$, so $B$ lies inside the quadrangle $A C_{1} A^{\\prime} C_{2}$, whence $A$ and $B$ are not equidistant from $C_{1} C_{2}$ - a contradiction.\n\n\n\nThus $O$ lies on $C_{1} C_{2}$, so the lines $A C_{i}$ and $B C_{i}$ are reflections of one another in the line $C_{1} C_{2}$, and $B=A^{\\prime}$. Hence $y_{A}=y_{B}$, and since $f_{i}(x)=y_{A}+p_{i}\\left(x-x_{A}\\right)\\left(x-x_{B}\\right)$, the line $C_{1} C_{2}$ is the axis of symmetry of both parabolas, as required.', 'Use the standard equation of a tangent to a smooth curve in the plane, to deduce that the tangents at two distinct points $A$ and $B$ on the parabola of equation $y=p x^{2}+q x+r$,\n\n\n\n\n\n\n\n$p \\neq 0$, meet at some point $C$ whose coordinates are\n\n\n\n$$\n\nx_{C}=\\frac{1}{2}\\left(x_{A}+x_{B}\\right) \\quad \\text { and } \\quad y_{C}=p x_{A} x_{B}+q \\cdot \\frac{1}{2}\\left(x_{A}+x_{B}\\right)+r\n\n$$\n\n\n\nUsage of the standard formula for Euclidean distance yields\n\n\n\n$$\n\nC A=\\frac{1}{2}\\left|x_{B}-x_{A}\\right| \\sqrt{1+\\left(2 p x_{A}+q\\right)^{2}} \\quad \\text { and } \\quad C B=\\frac{1}{2}\\left|x_{B}-x_{A}\\right| \\sqrt{1+\\left(2 p x_{B}+q\\right)^{2}},\n\n$$\n\n\n\nso, after obvious manipulations,\n\n\n\n$$\n\nC B-C A=\\frac{2 p\\left(x_{B}-x_{A}\\right)\\left|x_{B}-x_{A}\\right|\\left(p\\left(x_{A}+x_{B}\\right)+q\\right)}{\\sqrt{1+\\left(2 p x_{A}+q\\right)^{2}}+\\sqrt{1+\\left(2 p x_{B}+q\\right)^{2}}}\n\n$$\n\n\n\nNow, write the condition in the statement in the form $C_{1} B-C_{1} A=C_{2} B-C_{2} A$, apply the above formula and clear common factors to get\n\n\n\n$$\n\n\\frac{p_{1}\\left(p_{1}\\left(x_{A}+x_{B}\\right)+q_{1}\\right)}{\\sqrt{1+\\left(2 p_{1} x_{A}+q_{1}\\right)^{2}}+\\sqrt{1+\\left(2 p_{1} x_{B}+q_{1}\\right)^{2}}}=\\frac{p_{2}\\left(p_{2}\\left(x_{A}+x_{B}\\right)+q_{2}\\right)}{\\sqrt{1+\\left(2 p_{2} x_{A}+q_{2}\\right)^{2}}+\\sqrt{1+\\left(2 p_{2} x_{B}+q_{2}\\right)^{2}}}\n\n$$\n\n\n\nNext, use the fact that $x_{A}$ and $x_{B}$ are the solutions of the quadratic equation $\\left(p_{1}-p_{2}\\right) x^{2}+$ $\\left(q_{1}-q_{2}\\right) x+r_{1}-r_{2}=0$, so $x_{A}+x_{B}=-\\left(q_{1}-q_{2}\\right) /\\left(p_{1}-p_{2}\\right)$, to obtain\n\n\n\n$$\n\n\\frac{p_{1}\\left(p_{1} q_{2}-p_{2} q_{1}\\right)}{\\sqrt{1+\\left(2 p_{1} x_{A}+q_{1}\\right)^{2}}+\\sqrt{1+\\left(2 p_{1} x_{B}+q_{1}\\right)^{2}}}=\\frac{p_{2}\\left(p_{1} q_{2}-p_{2} q_{1}\\right)}{\\sqrt{1+\\left(2 p_{2} x_{A}+q_{2}\\right)^{2}}+\\sqrt{1+\\left(2 p_{2} x_{B}+q_{2}\\right)^{2}}}\n\n$$\n\n\n\nFinally, since $p_{1} p_{2}<0$ and the denominators above are both positive, the last equality forces $p_{1} q_{2}-p_{2} q_{1}=0$; that is, $q_{1} / p_{1}=q_{2} / p_{2}$, so the two parabolas have the same axis.']",,True,,, 1638,Combinatorics,,"Fix an integer $n \geq 2$. An $n \times n$ sieve is an $n \times n$ array with $n$ cells removed so that exactly one cell is removed from every row and every column. A stick is a $1 \times k$ or $k \times 1$ array for any positive integer $k$. For any sieve $A$, let $m(A)$ be the minimal number of sticks required to partition $A$. Find all possible values of $m(A)$, as $A$ varies over all possible $n \times n$ sieves.","['Given $A, m(A)=2 n-2$, and it is achieved, for instance, by dissecting $A$ along all horizontal (or vertical) grid lines. It remains to prove that $m(A) \\geq 2 n-2$ for every $A$.\n\n\n\nBy holes we mean the cells which are cut out from the board. The cross of a hole in $A$ is the union of the row and the column through that hole.\n\n\n\nArguing indirectly, consider a dissection of $A$ into $2 n-3$ or fewer sticks. Horizontal sticks are all labelled $h$, and vertical sticks are labelled $v ; 1 \\times 1$ sticks are both horizontal and vertical, and labelled arbitrarily. Each cell of $A$ inherits the label of the unique containing stick.\n\n\n\nAssign each stick in the dissection to the cross of the unique hole on its row, if the stick is horizontal; on its column, if the stick is vertical.\n\n\n\nSince there are at most $2 n-3$ sticks and exactly $n$ crosses, there are two crosses each of which is assigned to at most one stick in the dissection. Let the crosses be $c$ and $d$, centred at $a=\\left(x_{a}, y_{a}\\right)$ and $b=\\left(x_{b}, y_{b}\\right)$, respectively, and assume, without loss of generality, $x_{a}n$, in which case the sticks in $S$ span all $n$ columns, and notice that we are again done if $|S| \\leq n$, to assume further $|S|>n$.\n\n\n\nLet $S^{\\prime}=G_{h} \\backslash S$, let $T$ be set of all neighbours of $S$, and let $T^{\\prime}=G_{v} \\backslash T$. Since the sticks in $S$ span all $n$ columns, $|T| \\geq n$, so $\\left|T^{\\prime}\\right| \\leq n-2$. Transposition of the above argument (replace $S$ by $T^{\\prime}$, shows that $\\left|T^{\\prime}\\right| \\leq\\left|S^{\\prime}\\right|$, so $|S| \\leq|T|$."", 'Yet another proof of the estimate $m(A) \\geq 2 n-2$. We use the induction on $n$. Now we need the base cases $n=2,3$ which can be completed by hands.\n\n\n\nAssume now that $n>3$ and consider any dissection of $A$ into sticks. Define the cross of a hole as in Solution 1, and notice that each stick is contained in some cross. Thus, if the dissection contains more than $n$ sticks, then there exists a cross containing at least two sticks. In this case, remove this cross from the sieve to obtain an $(n-1) \\times(n-1)$ sieve. The dissection of the original sieve induces a dissection of the new array: even if a stick is partitioned into two by the removed cross, then the remaining two parts form a stick in the new array. After this operation has been performed, the number of sticks decreases by at least 2 , and by the induction hypothesis the number of sticks in the new dissection is at least $2 n-4$. Hence, the initial dissection contains at least $(2 n-4)+2=2 n-2$ sticks, as required.\n\n\n\nIt remains to rule out the case when the dissection contains at most $n$ sticks. This can be done in many ways, one of which is removal a cross containing some stick. The resulting dissection of an $(n-1) \\times(n-1)$ array contains at most $n-1$ sticks, which is impossible by the induction hypothesis since $n-1<2(n-1)-2$.']",['$m(A) \\geq 2n-2$ for every $A$'],False,,Need_human_evaluate, 1639,Geometry,,"Let $A B C D$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $A B, B C, C D$, and $D A$, respectively. It is given that the segments $P R$ and $Q S$ dissect $A B C D$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.","[""We start with a lemma which holds even in a more general setup.\n\n\n\nLemma 1. Let $P Q R S$ be a convex quadrangle whose diagonals meet at $O$. Let $\\omega_{1}$ and $\\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\\ell$ be their radical axis. Finally, choose the points $A, B$, and $C$ outside this quadrangle so that: the point $P$ (respectively, $Q$ ) lies on the segment $A B$ (respectively, $B C$ ); and $A O \\perp P S, B O \\perp P Q$, and $C O \\perp Q R$. Then the three lines $A C, P Q$, and $\\ell$ are concurrent or parallel.\n\n\n\nProof. Assume first that the lines $P R$ and $Q S$ are not perpendicular. Let $H_{1}$ and $H_{2}$ be the orthocentres of the triangles $O S P$ and $O Q R$, respectively; notice that $H_{1}$ and $H_{2}$ do not coincide.\n\n\n\nSince $H_{1}$ is the radical centre of the circles on diameters $R S, S P$, and $P Q$, it lies on $\\ell$. Similarly, $H_{2}$ lies on $\\ell$, so the lines $H_{1} H_{2}$ and $\\ell$ coincide.\n\n\n\nThe corresponding sides of the triangles $A P H_{1}$ and $C Q H_{2}$ meet at $O, B$, and the orthocentre of the triangle $O P Q$ (which lies on $O B$ ). By Desargues' theorem, the lines $A C, P Q$ and $\\ell$ are concurrent or parallel.\n\n\n\n\n\n\n\nThe case when $P R \\perp Q S$ may be considered as a limit case, since the configuration in the statement of the lemma allows arbitrarily small perturbations. The lemma is proved.\n\n\n\nBack to the problem, let the segments $P R$ and $Q S$ cross at $O$, let $\\omega_{1}$ and $\\omega_{2}$ be the circles on diameters $P Q$ and $R S$, respectively, and let $\\ell$ be their radical axis. By the Lemma, the three lines $A C, \\ell$, and $P Q$ are concurrent or parallel, and similarly so are the three lines $A C, \\ell$, and $R S$. Thus, if the lines $A C$ and $\\ell$ are distinct, all four lines are concurrent or pairwise parallel.\n\n\n\nThis is clearly the case when the lines $P S$ and $Q R$ are not parallel (since $\\ell$ crosses $O A$ and $O C$ at the orthocentres of $O S P$ and $O Q R$, these orthocentres being distinct from $A$ and $C$ ). In this case, denote the concurrency point by $T$. If $T$ is not ideal, then we have $T P \\cdot T Q=T R \\cdot T S$ (as $T \\in \\ell$ ), so $P Q R S$ is cyclic. If $T$ is ideal (i.e., all four lines are parallel), then the segments $P Q$ and $R S$ have the same perpendicular bisector (namely, the line of centers of $\\omega_{1}$ and $\\omega_{2}$ ), and $P Q R S$ is cyclic again.\n\n\n\nAssume now $P S$ and $Q R$ parallel. By symmetry, $P Q$ and $R S$ may also be assumed parallel: otherwise, the preceding argument goes through after relabelling. In this case, we need to prove that the parallelogram $P Q R S$ is a rectangle.\n\n\n\n\n\n\n\n\n\n\n\nSuppose, by way of contradiction, that $O P>O Q$. Let the line through $O$ and parallel to $P Q$ meet $A B$ at $M$, and $C B$ at $N$. Since $O P>O Q$, the angle $S P Q$ is acute and the angle $P Q R$ is obtuse, so the angle $A O B$ is obtuse, the angle $B O C$ is acute, $M$ lies on the segment $A B$, and $N$ lies on the extension of the segment $B C$ beyond $C$. Therefore: $O A>O M$, since the angle $O M A$ is obtuse; $O M>O N$, since $O M: O N=K P: K Q$, where $K$ is the projection of $O$ onto $P Q$; and $O N>O C$, since the angle $O C N$ is obtuse. Consequently, $O A>O C$.\n\n\n\nSimilarly, $O R>O S$ yields $O C>O A$ : a contradiction. Consequently, $O P=O Q$ and $P Q R S$ is a rectangle. This ends the proof."", 'To begin, we establish a useful lemma.\n\n\n\nLemma 2. If $P$ is a point on the side $A B$ of a triangle $O A B$, then\n\n\n\n$$\n\n\\frac{\\sin A O P}{O B}+\\frac{\\sin P O B}{O A}=\\frac{\\sin A O B}{O P}\n\n$$\n\n\n\nProof. Let $[X Y Z]$ denote the area of a triangle $X Y Z$, to write\n\n\n\n$0=2([A O B]-[P O B]-[P O C])=O A \\cdot O B \\cdot \\sin A O B-O B \\cdot O P \\cdot \\sin P O B-O P \\cdot O A \\cdot \\sin A O P$, and divide by $O A \\cdot O B \\cdot O P$ to get the required identity.\n\n\n\nA similar statement remains valid if the point $C$ lies on the line $A B$; the proof is obtained by using signed areas and directed lengths.\n\n\n\nWe now turn to the solution. We first prove some sort of a converse statement, namely:\n\n\n\nClaim. Let $P Q R S$ be a cyclic quadrangle with $O=P R \\cap Q S$; assume that no its diagonal is perpendicular to a side. Let $\\ell_{A}, \\ell_{B}, \\ell_{C}$, and $\\ell_{D}$ be the lines through $O$ perpendicular to $S P$, $P Q, Q R$, and $R S$, respectively. Choose any point $A \\in \\ell_{A}$ and successively define $B=A P \\cap \\ell_{B}$, $C=B Q \\cap \\ell_{C}, D=C R \\cap \\ell_{D}$, and $A^{\\prime}=D S \\cap \\ell_{A}$. Then $A^{\\prime}=A$.\n\n\n\nProof. We restrict ourselves to the case when the points $A, B, C, D$, and $A^{\\prime}$ lie on $\\ell_{A}, \\ell_{B}$, $\\ell_{C}, \\ell_{D}$, and $\\ell_{A}$ on the same side of $O$ as their points of intersection with the respective sides of the quadrilateral $P Q R S$. Again, a general case is obtained by suitable consideration of directed lengths.\n\n\n\n\n\n\n\n\n\n\n\nDenote\n\n\n\n$$\n\n\\begin{gathered}\n\n\\alpha=\\angle Q P R=\\angle Q S R=\\pi / 2-\\angle P O B=\\pi / 2-\\angle D O S, \\\\\n\n\\beta=\\angle R P S=\\angle R Q S=\\pi / 2-\\angle A O P=\\pi / 2-\\angle Q O C, \\\\\n\n\\gamma=\\angle S Q P=\\angle S R P=\\pi / 2-\\angle B O Q=\\pi / 2-\\angle R O D, \\\\\n\n\\delta=\\angle P R Q=\\angle P S Q=\\pi / 2-\\angle C O R=\\pi / 2-\\angle S O A .\n\n\\end{gathered}\n\n$$\n\n\n\nBy Lemma 2 applied to the lines $A P B, P Q C, C R D$, and $D S A^{\\prime}$, we get\n\n\n\n$$\n\n\\begin{aligned}\n\n& \\frac{\\sin (\\alpha+\\beta)}{O P}=\\frac{\\cos \\alpha}{O A}+\\frac{\\cos \\beta}{O B}, \\quad \\frac{\\sin (\\beta+\\gamma)}{O Q}=\\frac{\\cos \\beta}{O B}+\\frac{\\cos \\gamma}{O C} \\\\\n\n& \\frac{\\sin (\\gamma+\\delta)}{O R}=\\frac{\\cos \\gamma}{O C}+\\frac{\\cos \\delta}{O D}, \\quad \\frac{\\sin (\\delta+\\alpha)}{O S}=\\frac{\\cos \\delta}{O D}+\\frac{\\cos \\alpha}{O A^{\\prime}}\n\n\\end{aligned}\n\n$$\n\n\n\nAdding the two equalities on the left and subtracting the two on the right, we see that the required equality $A=A^{\\prime}$ (i.e., $\\cos \\alpha / O A=\\cos \\alpha / O A^{\\prime}$, in view of $\\cos \\alpha \\neq 0$ ) is equivalent to the relation\n\n\n\n$$\n\n\\frac{\\sin Q P S}{O P}+\\frac{\\sin S R Q}{O R}=\\frac{\\sin P Q R}{O Q}+\\frac{\\sin R S P}{O S}\n\n$$\n\n\n\nLet $d$ denote the circumdiameter of $P Q R S$, so $\\sin Q P S=\\sin S R Q=Q S / d$ and $\\sin R S P=$ $\\sin P Q R=P R / d$. Thus the required relation reads\n\n\n\n$$\n\n\\frac{Q S}{O P}+\\frac{Q S}{O R}=\\frac{P R}{O S}+\\frac{P R}{O Q}, \\quad \\text { or } \\quad \\frac{Q S \\cdot P R}{O P \\cdot O R}=\\frac{P R \\cdot Q S}{O S \\cdot O Q}\n\n$$\n\n\n\nThe last relation is trivial, due again to cyclicity.\n\n\n\nFinally, it remains to derive the problem statement from our Claim. Assume that $P Q R S$ is not cyclic, e.g., that $O P \\cdot O R>O Q \\cdot O S$, where $O=P R \\cap Q S$. Mark the point $S^{\\prime}$ on the ray $O S$ so that $O P \\cdot O R=O Q \\cdot O S^{\\prime}$. Notice that no diagonal of $P Q R S$ is perpendicular to a side, so the quadrangle $P Q R S^{\\prime}$ satisfies the conditions of the claim.\n\n\n\nLet $\\ell_{A}^{\\prime}$ and $\\ell_{D}^{\\prime}$ be the lines through $O$ perpendicular to $P S^{\\prime}$ and $R S^{\\prime}$, respectively. Then $\\ell_{A}^{\\prime}$ and $\\ell_{D}^{\\prime}$ cross the segments $A P$ and $R D$, respectively, at some points $A^{\\prime}$ and $D^{\\prime}$. By the Claim, the line $A^{\\prime} D^{\\prime}$ passes through $S^{\\prime}$. This is impossible, because the segment $A^{\\prime} D^{\\prime}$ crosses the segment $O S$ at some interior point, while $S^{\\prime}$ lies on the extension of this segment. This contradiction completes the proof.']",,True,,, 1640,Algebra,,"Let $a, b, c, d$ be positive integers such that $a d \neq b c$ and $\operatorname{gcd}(a, b, c, d)=1$. Prove that, as $n$ runs through the positive integers, the values $\operatorname{gcd}(a n+b, c n+d)$ may achieve form the set of all positive divisors of some integer.","['We extend the problem statement by allowing $a$ and $c$ take non-negative integer values, and allowing $b$ and $d$ to take arbitrary integer values. (As usual, the greatest common divisor of two integers is non-negative.) Without loss of generality, we assume $0 \\leq a \\leq c$. Let $S(a, b, c, d)=\\left\\{\\operatorname{gcd}(a n+b, c n+d): n \\in \\mathbb{Z}_{>0}\\right\\}$.\n\nNow we induct on $a$. We first deal with the inductive step, leaving the base case $a=0$ to the end of the solution. So, assume that $a>0$; we intend to find a 4 -tuple $\\left(a^{\\prime}, b^{\\prime}, c^{\\prime}, d^{\\prime}\\right)$ satisfying the requirements of the extended problem, such that $S\\left(a^{\\prime}, b^{\\prime}, c^{\\prime}, d^{\\prime}\\right)=S(a, b, c, d)$ and $0 \\leq a^{\\prime}1)$. We thus $\\operatorname{get} \\operatorname{gcd}\\left(b^{\\prime}, c\\right)=1$ and $S\\left(0, b^{\\prime}, c, d\\right)=S(0, b, c, d)$.\n\nFinally, it is easily seen that $S\\left(0, b^{\\prime}, c, d\\right)$ is the set of all positive divisors of $b^{\\prime}$. Each member of $S\\left(0, b^{\\prime}, c, d\\right)$ is clearly a divisor of $b^{\\prime}$. Conversely, if $\\delta$ is a positive divisor of $b^{\\prime}$, then $c n+d \\equiv \\delta$ $\\left(\\bmod b^{\\prime}\\right)$ for some $n$, since $b^{\\prime}$ and $c$ are coprime, so $\\delta$ is indeed a member of $S\\left(0, b^{\\prime}, c, d\\right)$.', 'For positive integers $s$ and $t$ and prime $p$, we will denote by $\\operatorname{gcd}_{p}(s, t)$ the greatest common $p$-power divisor of $s$ and $t$.\n\nClaim 1. For any positive integer $n, \\operatorname{gcd}(a n+b, c n+d) \\mid a d-b c$.\n\nProof. This is clear from the identity\n\n$$\na(c n+d)-c(a n+b)=a d-b c . \\tag{\\dagger}\n$$\n\nClaim 2. The set of values taken by $\\operatorname{gcd}(a n+b, c n+d)$ is exactly the set of values taken by the product\n\n$$\n\\prod_{p \\mid a d-b c} \\operatorname{gcd}_{p}\\left(a n_{p}+b, c n_{p}+d\\right)\n$$\n\nas the $\\left(n_{p}\\right)_{p \\mid a d-b c}$ each range over positive integers.\n\n\n\nProof. From the identity\n\n$$\n\\operatorname{gcd}(a n+b, c n+d)=\\prod_{p \\mid a d-b c} \\operatorname{gcd}_{p}(a n+b, c n+d),\n$$\n\nit is clear that every value taken by $\\operatorname{gcd}(a n+b, c n+d)$ is also a value taken by the product (with all $\\left.n_{p}=n\\right)$. Conversely, it suffices to show that, given any positive integers $\\left(n_{p}\\right)_{p \\mid a d-b c}$, there is a positive integer $n$ such that $\\operatorname{gcd}_{p}(a n+b, c n+d)=\\operatorname{gcd}_{p}\\left(a n_{p}+b, c n_{p}+d\\right)$ for each $p \\mid a d-b c$. This can be achieved by requiring that $n$ be congruent to $n_{p}$ modulo a sufficiently large ${ }^{1}$ power of $p$ (using the Chinese Remainder Theorem).\n\nUsing Claim 2, it suffices to determine the sets of values taken by $\\operatorname{gcd}_{p}(a n+b, c n+d)$ as $n$ ranges over all positive integers. There are two cases.\n\nClaim 3. If $p \\mid a, c$, then $\\operatorname{gcd}_{p}(a n+b, c n+d)=1$ for all $n$.\n\nProof. If $p \\mid a n+b, c n+d$, then we would have $p \\mid a, b, c, d$, which is not the case.\n\nClaim 4. If $p \\nmid a$ or $p \\nmid c$, then the values taken by $\\operatorname{gcd}_{p}(a n+b, c n+d)$ are exactly the $p$-power divisors of $a d-b c$.\n\nProof. Assume without loss of generality that $p \\nmid a$. Then from identity $(\\dagger)$ we have $\\operatorname{gcd}_{p}(a n+$ $b, c n+d)=\\operatorname{gcd}_{p}(a n+b, a d-b c)$. But since $p \\nmid a$, the arithmetic progression $a n+b$ takes all possible values modulo the highest $p$-power divisor of $a d-b c$, and in particular the values taken by $\\operatorname{gcd}_{p}(a n+b, a d-b c)$ are exactly the $p$-power divisors of $a d-b c$.\n\nConclusion. Using claims 2,3 and 4 , we see that the set of values taken by $\\operatorname{gcd}(a n+b, c n+d)$ is equal to the set of products of $p$-power divisors of $a d-b c$, where we only consider those primes $p$ not dividing $\\operatorname{gcd}(a, c)$. Thus the set of values of $\\operatorname{gcd}(a n+b, c n+d)$ is equal to the set of divisors of the largest factor of $a d-b c$ coprime to $\\operatorname{gcd}(a, c)$.']",,True,,, 1641,Combinatorics,,"Let $n$ be a positive integer and fix $2 n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{A B}$ and $\overrightarrow{C D}$ such that $A B C D$ is a convex quadrangle oriented clockwise. Determine the number of good configurations.","['The required number is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. To prove this, trace the circumference counterclockwise to label the points $a_{1}, a_{2}, \\ldots, a_{2 n}$.\n\nLet $\\mathcal{C}$ be any good configuration and let $O(\\mathcal{C})$ be the set of all points from which arrows emerge. We claim that every $n$-element subset $S$ of $\\left\\{a_{1}, \\ldots, a_{2 n}\\right\\}$ is an $O$-image of a unique good configuration; clearly, this provides the answer.\n\nTo prove the claim induct on $n$. The base case $n=1$ is clear. For the induction step, consider any $n$-element subset $S$ of $\\left\\{a_{1}, \\ldots, a_{2 n}\\right\\}$, and assume that $S=O(\\mathcal{C})$ for some good configuration $\\mathcal{C}$. Take any index $k$ such that $a_{k} \\in S$ and $a_{k+1} \\notin S$ (assume throughout that indices are cyclic modulo $2 n$, i.e., $a_{2 n+1}=a_{1}$ etc.).\n\nIf the arrow from $a_{k}$ points to some $a_{\\ell}, k+1<\\ell(<2 n+k)$, then the arrow pointing to $a_{k+1}$ emerges from some $a_{m}, m$ in the range $k+2$ through $\\ell-1$, since these two arrows do not cross. Then the arrows $a_{k} \\rightarrow a_{\\ell}$ and $a_{m} \\rightarrow a_{k+1}$ form a prohibited quadrangle. Hence, $\\mathcal{C}$ contains an arrow $a_{k} \\rightarrow a_{k+1}$.\n\nOn the other hand, if any configuration $\\mathcal{C}$ contains the arrow $a_{k} \\rightarrow a_{k+1}$, then this arrow cannot cross other arrows, neither can it occur in prohibited quadrangles.\n\nThus, removing the points $a_{k}, a_{k+1}$ from $\\left\\{a_{1}, \\ldots, a_{2 n}\\right\\}$ and the point $a_{k}$ from $S$, we may apply the induction hypothesis to find a unique good configuration $\\mathcal{C}^{\\prime}$ on $2 n-2$ points compatible with the new set of sources (i.e., points from which arrows emerge). Adjunction of the arrow $a_{k} \\rightarrow a_{k+1}$ to $\\mathcal{C}^{\\prime}$ yields a unique good configuration on $2 n$ points, as required.', 'Use the counterclockwise labelling $a_{1}, a_{2}, \\ldots, a_{2 n}$ in the solution above.\n\nLetting $D_{n}$ be the number of good configurations on $2 n$ points, we establish a recurrence relation for the $D_{n}$. To this end, let $C_{n}=\\frac{(2 n) !}{n !(n+1) !}$ the $n$th Catalan number; it is well-known that $C_{n}$ is the number of ways to connect $2 n$ given points on the circumference by $n$ pairwise disjoint chords.\n\nSince no two arrows cross, in any good configuration the vertex $a_{1}$ is connected to some $a_{2 k}$. Fix $k$ in the range 1 through $n$ and count the number of good configurations containing the arrow $a_{1} \\rightarrow a_{2 k}$. Let $\\mathcal{C}$ be any such configuration.\n\nIn $\\mathcal{C}$, the vertices $a_{2}, \\ldots, a_{2 k-1}$ are paired off with one other, each arrow pointing from the smaller to the larger index, for otherwise it would form a prohibited quadrangle with $a_{1} \\rightarrow a_{2 k}$. Consequently, there are $C_{k-1}$ ways of drawing such arrows between $a_{2}, \\ldots, a_{2 k-1}$.\n\nOn the other hand, the arrows between $a_{2 k+1}, \\ldots, a_{2 n}$ also form a good configuration, which can be chosen in $D_{n-k}$ ways. Finally, it is easily seen that any configuration of the first kind and any configuration of the second kind combine together to yield an overall good configuration.\n\nThus the number of good configurations containing the arrow $a_{1} \\rightarrow a_{2 k}$ is $C_{k-1} D_{n-k}$. Clearly, this is also the number of good configurations containing the arrow $a_{2(n-k+1)} \\rightarrow a_{1}$, so\n\n$$\nD_{n}=2 \\sum_{k=1}^{n} C_{k-1} D_{n-k} \\tag{*}\n$$\n\nTo find an explicit formula for $D_{n}$, let $d(x)=\\sum_{n=0}^{\\infty} D_{n} x^{n}$ and let $c(x)=\\sum_{n=0}^{\\infty} C_{n} x^{n}=$ $\\frac{1-\\sqrt{1-4 x}}{2 x}$ be the generating functions of the $D_{n}$ and the $C_{n}$, respectively. Since $D_{0}=1$, relation $(*)$\n\n\n\nyields $d(x)=2 x c(x) d(x)+1$, so\n\n$$\n\\begin{aligned}\nd(x)=\\frac{1}{1-2 x c(x)}=(1-4 x)^{-1 / 2} & =\\sum_{n \\geq 0}\\left(-\\frac{1}{2}\\right)\\left(-\\frac{3}{2}\\right) \\ldots\\left(-\\frac{2 n-1}{2}\\right) \\frac{(-4 x)^{n}}{n !} \\\\\n& =\\sum_{n \\geq 0} \\frac{2^{n}(2 n-1) ! !}{n !} x^{n}=\\sum_{n \\geq 0}\\left(\\begin{array}{c}\n2 n \\\\\nn\n\\end{array}\\right) x^{n} .\n\\end{aligned}\n$$\n\nConsequently, $D_{n}=\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$.\n\n### solution_2\nLet $C_{n}=\\frac{1}{n+1}\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ denote the $n$th Catalan number and recall that there are exactly $C_{n}$ ways to join $2 n$ distinct points on a circumference by $n$ pairwise disjoint chords. Such a configuration of chords will be referred to as a Catalan n-configuration. An orientation of the chords in a Catalan configuration $\\mathcal{C}$ making it into a good configuration (in the sense defined in the statement of the problem) will be referred to as a good orientation for $\\mathcal{C}$.\n\nWe show by induction on $n$ that there are exactly $n+1$ good orientations for any Catalan $n$-configuration, so there are exactly $(n+1) C_{n}=\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ good configurations on $2 n$ points. The base case $n=1$ is clear.\n\nFor the induction step, let $n>1$, let $\\mathcal{C}$ be a Catalan $n$-configuration, and let $a b$ be a chord of minimal length in $\\mathcal{C}$. By minimality, the endpoints of the other chords in $\\mathcal{C}$ all lie on the major arc $a b$ of the circumference.\n\nLabel the $2 n$ endpoints $1,2, \\ldots, 2 n$ counterclockwise so that $\\{a, b\\}=\\{1,2\\}$, and notice that the good orientations for $\\mathcal{C}$ fall into two disjoint classes: Those containing the arrow $1 \\rightarrow 2$, and those containing the opposite arrow.\n\nSince the arrow $1 \\rightarrow 2$ cannot be involved in a prohibited quadrangle, the induction hypothesis applies to the Catalan $(n-1)$-configuration formed by the other chords to show that the first class contains exactly $n$ good orientations.\n\nFinally, the second class consists of a single orientation, namely, $2 \\rightarrow 1$, every other arrow emerging from the smaller endpoint of the respective chord; a routine verification shows that this is indeed a good orientation. This completes the induction step and ends the proof.\n\n### solution_3\nWe intend to count the number of good orientations of a Catalan $n$-configuration.\n\nFor each such configuration, we consider its dual graph $T$ whose vertices are finite regions bounded by chords and the circle, and an edge connects two regions sharing a boundary segment. This graph $T$ is a plane tree with $n$ edges and $n+1$ vertices.\n\nThere is a canonical bijection between orientations of chords and orientations of edges of $T$ in such a way that each chord crosses an edge of $T$ from the right to the left of the arrow on that edge. A good orientation of chords corresponds to an orientation of the tree containing no two edges oriented towards each other. Such an orientation is defined uniquely by its source vertex, i.e., the unique vertex having no in-arrows.\n\nTherefore, for each tree $T$ on $n+1$ vertices, there are exactly $n+1$ ways to orient it so that the source vertex is unique - one for each choice of the source. Thus, the answer is obtained in the same way as above.']",['$\\binom{2n}{n}$'],False,,Expression, 1642,Geometry,,"Fix a circle $\Gamma$, a line $\ell$ tangent to $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ cross $\ell$ at $Y$ and $Z$. Prove that, as $X$ traces $\Omega$, the circle $X Y Z$ is tangent to two fixed circles.","['Assume $\\Gamma$ of unit radius and invert with respect to $\\Gamma$. No reference will be made to the original configuration, so images will be denoted by the same letters. Letting $\\Gamma$ be centred at $G$, notice that inversion in $\\Gamma$ maps tangents to $\\Gamma$ to circles of unit diameter through $G$ (hence internally tangent to $\\Gamma$ ). Under inversion, the statement reads as follows:\n\nFix a circle $\\Gamma$ of unit radius centred at $G$, a circle $\\ell$ of unit diameter through $G$, and a circle $\\Omega$ inside $\\ell$ disjoint from $\\ell$. The circles $\\eta$ and $\\zeta$ of unit diameter, through $G$ and a variable point $X$ on $\\Omega$, cross $\\ell$ again at $Y$ and $Z$, respectively. Prove that, as $X$ traces $\\Omega$, the circle $X Y Z$ is tangent to two fixed circles.\n\n\n\nSince $\\eta$ and $\\zeta$ are the reflections of the circumcircle $\\ell$ of the triangle $G Y Z$ in its sidelines $G Y$ and $G Z$, respectively, they pass through the orthocentre of this triangle. And since $\\eta$ and $\\zeta$ cross again at $X$, the latter is the orthocentre of the triangle $G Y Z$. Hence the circle $\\xi$ through $X, Y, Z$ is the reflection of $\\ell$ in the line $Y Z$; in particular, $\\xi$ is also of unit diameter.\n\nLet $O$ and $L$ be the centres of $\\Omega$ and $\\ell$, respectively, and let $R$ be the (variable) centre of $\\xi$. Let $G X \\operatorname{cross} \\xi$ again at $X^{\\prime}$; then $G$ and $X^{\\prime}$ are reflections of one another in the line $Y Z$, so $G L R X^{\\prime}$ is an isosceles trapezoid. Then $L R \\| G X$ and $\\angle(L G, G X)=\\angle\\left(G X^{\\prime}, X^{\\prime} R\\right)=\\angle(R X, X G)$, i.e., $L G \\| R X$; this means that $G L R X$ is a parallelogram, so $\\overrightarrow{X R}=\\overrightarrow{G L}$ is constant.\n\nFinally, consider the fixed point $N$ defined by $\\overrightarrow{O N}=\\overrightarrow{G L}$. Then $X R N O$ is a parallelogram, so the distance $R N=O X$ is constant. Consequently, $\\xi$ is tangent to the fixed circles centred at $N$ of radii $|1 / 2-O X|$ and $1 / 2+O X$.\n\nOne last check is needed to show that the inverse images of the two obtained circles are indeed circles and not lines. This might happen if one of them contained $G$; we show that this is impossible. Indeed, since $\\Omega$ lies inside $\\ell$, we have $O L<1 / 2-O X$, so\n\n$$\nN G=|\\overrightarrow{G L}+\\overrightarrow{L O}+\\overrightarrow{O N}|=|2 \\overrightarrow{G L}+\\overrightarrow{L O}| \\geq 2|\\overrightarrow{G L}|-|\\overrightarrow{L O}|>1-(1 / 2-O X)=1 / 2+O X\n$$\n\nthis shows that $G$ is necessarily outside the obtained circles.']",,True,,, 1643,Geometry,,"Let $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$. The circle $A B D$ meets the segment $A C$ again at an interior point $E$. The circle $A C D$ meets the segment $A B$ again at an interior point $F$. Let $A^{\prime}$ be the reflection of $A$ in the line $B C$. The lines $A^{\prime} C$ and $D E$ meet at $P$, and the lines $A^{\prime} B$ and $D F$ meet at $Q$. Prove that the lines $A D, B P$ and $C Q$ are concurrent (or all parallel).","['Let $\\sigma$ denote reflection in the line $B C$. Since $\\angle B D F=\\angle B A C=$ $\\angle C D E$, by concyclicity, the lines $D E$ and $D F$ are images of one another under $\\sigma$, so the lines $A C$ and $D F$ meet at $P^{\\prime}=\\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\\prime}=\\sigma(Q)$. Consequently, the lines $P Q$ and $P^{\\prime} Q^{\\prime}=\\sigma(P Q)$ meet at some (possibly ideal) point $R$ on the line $B C$.\n\n\n\nSince the pairs of lines $(C A, Q D),(A B, D P),(B C, P Q)$ meet at three collinear points, namely $P^{\\prime}, Q^{\\prime}, R$ respectively, the triangles $A B C$ and $D P Q$ are perspective, i.e., the lines $A D, B P, C Q$ are concurrent, by the Desargues theorem.\n\n\n\n\n\n\n\nFig. 1\n\n\n\n\n\n\n\nFig. 2', 'As in the first solution, $\\sigma$ denotes reflection in the line $B C$, the lines $D E$ and $D F$ are images of one another under $\\sigma$, the lines $A C$ and $D F$ meet at $P^{\\prime}=\\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\\prime}=\\sigma(Q)$.\n\n\n\nLet the line $A D$ meet the circle $A B C$ again at $M$. Letting $M^{\\prime}=\\sigma(M)$, it is sufficient to prove that the lines $D M^{\\prime}=\\sigma(A D), B P^{\\prime}=\\sigma(B P)$ and $C Q^{\\prime}=\\sigma(C Q)$ are concurrent.\n\n\n\n\n\n\n\nBegin by noticing that $\\angle\\left(B M^{\\prime}, M^{\\prime} D\\right)=-\\angle(B M, M A)=-\\angle(B C, C A)=\\angle(B F, F D)$, to infer that $M^{\\prime}$ lies on the circle $B D F$. Similarly, $M^{\\prime}$ lies on the circle $C D E$, so the line $D M^{\\prime}$ is the radical axis of the circles $B D F$ and $C D E$.\n\n\n\nSince $P^{\\prime}$ lies on the lines $A C$ and $D F$, it is the radical centre of the circles $A B C, A D C$, and $B D F$; hence the line $B P^{\\prime}$ is the radical axis of the circles $B D F$ and $A B C$. Similarly, the line $C Q^{\\prime}$ is the radical axis of the circles $C D E$ and $A B C$. So the conclusion follows: the lines $D M^{\\prime}$, $B P^{\\prime}$ and $C Q^{\\prime}$ are concurrent at the radical centre of the circles $A B C, B D F$ and $C D E$; thus the lines $D M, B P^{\\prime}$ and $C Q^{\\prime}$ are also concurrent.', 'As in the previous solutions, $\\sigma$ denotes reflection in the line $B C$. Let the lines $B E$ and $C F$ meet at $X$. Due to the circles $B D E A$ and $C D F A$, we have $\\angle X B D=$ $\\angle E A D=\\angle X F D$, so the quadrilateral $B F X D$ is cyclic; similarly, the quadrilateral $C E X D$ is cyclic. Hence $\\angle X D B=\\angle C F A=\\angle C D A$, the lines $D X$ and $D A$ are therefore images of one another under $\\sigma$, and $X^{\\prime}=\\sigma(X)$ lies on the line $A D$. Let $E^{\\prime}=\\sigma(E)$ and $F^{\\prime}=\\sigma(F)$, and apply the Pappus theorem to the hexagon $B P F^{\\prime} C Q E^{\\prime}$ to infer that $X^{\\prime}, D$, and $B P \\cap C Q$ are collinear. The conclusion follows.\n\n\n\n\n\n\n\nFig. 3']",,True,,, 1644,Combinatorics,,"Given positive integers $m$ and $n \geq m$, determine the largest number of dominoes $(1 \times 2$ or $2 \times 1$ rectangles) that can be placed on a rectangular board with $m$ rows and $2 n$ columns consisting of cells $(1 \times 1$ squares $)$ so that: (i) each domino covers exactly two adjacent cells of the board; (ii) no two dominoes overlap; (iii) no two form a $2 \times 2$ square; and (iv) the bottom row of the board is completely covered by $n$ dominoes.","['The required maximum is $m n-\\lfloor m / 2\\rfloor$ and is achieved by the brick-like vertically symmetric arrangement of blocks of $n$ and $n-1$ horizontal dominoes placed on alternate rows, so that the bottom row of the board is completely covered by $n$ dominoes.\n\n\n\nTo show that the number of dominoes in an arrangement satisfying the conditions in the statement does not exceed $m n-\\lfloor m / 2\\rfloor$, label the rows upwards $0,1, \\ldots, m-1$, and, for each $i$ in this range, draw a vertically symmetric block of $n-i$ fictitious horizontal dominoes in the $i$-th row (so the block on the $i$-th row leaves out $i$ cells on either side) - Figure 4 illustrates the case $m=n=6$. A fictitious domino is good if it is completely covered by a domino in the arrangement; otherwise, it is bad.\n\n\n\nIf the fictitious dominoes are all good, then the dominoes in the arrangement that cover no fictitious domino, if any, all lie in two triangular regions of side-length $m-1$ at the upper-left and upper-right corners of the board. Colour the cells of the board chess-like and notice that in each of the two triangular regions the number of black cells and the number of white cells differ by $\\lfloor m / 2\\rfloor$. Since each domino covers two cells of different colours, at least $\\lfloor m / 2\\rfloor$ cells are not covered in each of these regions, and the conclusion follows.\n\n\n\n\n\n\n\nFig. 4\n\n\n\n\n\nFig. 5\n\n\n\nTo deal with the remaining case where bad fictitious dominoes are present, we show that an arrangement satisfying the conditions in the statement can be transformed into another such with at least as many dominoes, but fewer bad fictitious dominoes. A finite number of such transformations eventually leads to an arrangement of at least as many dominoes all of whose fictitious dominoes are good, and the conclusion follows by the preceding.\n\n\n\nConsider the row of minimal rank containing bad fictitious dominoes - this is certainly not the bottom row - and let $D$ be one such. Let $\\ell$, respectively $r$, be the left, respectively right, cell of $D$ and notice that the cell below $\\ell$, respectively $r$, is the right, respectively left, cell of a domino $D_{1}$, respectively $D_{2}$, in the arrangement.\n\n\n\nIf $\\ell$ is covered by a domino $D_{\\ell}$ in the arrangement, since $D$ is bad and no two dominoes in the arrangement form a square, it follows that $D_{\\ell}$ is vertical. If $r$ were also covered by a domino $D_{r}$ in the arrangement, then $D_{r}$ would also be vertical, and would therefore form a square with $D_{\\ell}-$ a contradiction. Hence $r$ is not covered, and there is room for $D_{\\ell}$ to be placed so as to cover $D$, to obtain a new arrangement satisfying the conditions in the statement; the latter has as many dominoes as the former, but fewer bad fictitious dominoes. The case where $r$ is covered is dealt with similarly.\n\n\n\nFinally, if neither cell of $D$ is covered, addition of an extra domino to cover $D$ and, if necessary, removal of the domino above $D$ to avoid formation of a square, yields a new arrangement satisfying the conditions in the statement; the latter has at least as many dominoes as the former, but fewer bad fictitious dominoes. (Figure 5 illustrates the two cases.)', ""We present an alternative proof of the bound.\n\n\n\nLabel the rows upwards $0,1, \\ldots, m-1$, and the columns from the left to the right by $0,1, \\ldots, 2 n-1$; label each cell by the pair of its column's and row's numbers, so that $(1,0)$ is the second left cell in the bottom row. Colour the cells chess-like so that $(0,0)$ is white. For $0 \\leq i \\leq n-1$, we say that the $i$ th white diagonal is the set of cells of the form $(2 i+k, k)$, where $k$ ranges over all appropriate indices. Similarly, the ith black diagonal is the set of cells of the form $(2 i+1-k, k)$. (Notice that the white cells in the upper-left corner and the black cells in the upper-right corner are not covered by these diagonals.)\n\n\n\nClaim. Assume that $K$ lowest cells of some white diagonal are all covered by dominoes. Then all these $K$ dominoes face right or up from the diagonal. (In other words, the black cell of any such\n\n\n\n\n\n\n\ndomino is to the right or to the top of its white cell.) Similarly, if $K$ lowest cells of some black diagonal are covered by dominoes, then all these dominoes face left or up from the diagonal.\n\n\n\nProof. By symmetry, it suffices to prove the first statement. Assume that $K$ lowest cells of the $i$ th white diagonal is completely covered. We prove by induction on $k1$, then $(n-1)^{2}', 'Let $Q$ be the isogonal conjugate of $P$ with respect to $\\triangle A E D$, so $\\angle(Q A, A D)=$ $\\angle(E A, A P)=\\angle(E B, B A)$ and $\\angle(Q D, D A)=\\angle(E D, D P)=\\angle(E C, C D)$. Now our aim is to prove that $Q E \\| C D$; this will yield that $\\angle(E C, C D)=\\angle(A E, E Q)=\\angle(P E, E D)$, whence $P E$ is tangent to $\\Omega$.\n\n\n\nLet $D Q$ meet $A B$ at $X$. Then we have $\\angle(X D, D A)=\\angle(E C, C D)=\\angle(E A, A B)$ and $\\angle(D A, A X)=\\angle(A B, B C)$, hence the triangles $D A X$ and $A B C$ are similar. Since $\\angle(A B, B E)=$ $\\angle(D A, A Q)$, the points $Q$ and $E$ correspond to each other in these triangles, hence $Q$ is the midpoint of $D X$. This yields that the points $Q$ and $E$ lie on the midline of the trapezoid parallel to $C D$, as desired.\n\n\n\n\n\n\n\n', 'Let $O$ be the intersection of the diagonals $A C$ and $B D$. Let $F$ be the midpoint of $B D$. Let $S$ be the second intersection point of the circumcircles of triangles $A O F$ and $D O E$. We will prove that $S D$ and $S E$ are tangent to $\\Omega$; the symmetric argument would then imply also that $S A$ and $S F$ are tangent to $\\Gamma$. Thus $S=P$ and the claimed tangency holds.\n\n\n\nWe first prove that $O S$ is parallel to $A B$ and $D C$. Compute the powers of the points $A, B$ with respect to the circumcircles of $A O F$ and $D O E$ :\n\n\n\n$$\n\n\\begin{gathered}\n\nd(A, A O F)=0, \\quad d(A, D O E)=A O \\cdot A E \\\\\n\nd(B, A O F)=B O \\cdot B F, \\quad d(B, D O E)=B O \\cdot B D=2 B O \\cdot B F\n\n\\end{gathered}\n\n$$\n\n\n\nAnd therefore\n\n\n\n$$\n\nd(B, D O E)-d(B, A O F)=B O \\cdot B F=A O \\cdot A E=d(A, D O E)-d(A, A O F)\n\n$$\n\n\n\nThus both $A$ and $B$ belong to a locus of the form\n\n\n\n$$\n\nd(X, D O E)-d(X, A O F)=\\mathrm{const}\n\n$$\n\n\n\nwhich is always a lines parallel to the radical axis of the respective circles. Since this radical axis is $O S$ by definition, it follows that $A B$ is parallel to $O S$, as claimed.\n\n\n\nNow by angle chasing in the cyclic quadrilateral $D S O E$, we find\n\n\n\n$$\n\n\\begin{aligned}\n\n\\angle(S D, D E) & =\\angle(S O, O E)=\\angle(D C, C E), \\\\\n\n\\angle(S E, E D)=\\angle(S O, O D) & =\\angle(D C, D B)=\\angle(A C, C D)=\\angle(E C, C D),\n\n\\end{aligned}\n\n$$\n\n\n\nand these angle equalities are exactly the conditions of $S D, S E$ being tanget to $\\Omega$, as claimed.']",,True,,, 1648,Algebra,,"Given any positive real number $\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\varepsilon) v$ edges has two distinct simple cycles of equal lengths. (Recall that the notion of a simple cycle does not allow repetition of vertices in a cycle.)","['Fix a positive real number $\\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\\varepsilon) v$ edges, all of whose simple cycles have pairwise distinct lengths.\n\n\n\nAssuming $\\varepsilon^{2} v \\geq 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of simple cycles in $G$, showing thereby that $v$ cannot be arbitrarily large, whence the conclusion.\n\n\n\nSince a simple cycle in $G$ has at most $v$ vertices, and each length class contains at most one such, $G$ has at most $v$ pairwise distinct simple cycles. This establishes the desired upper bound.\n\n\n\nFor the lower bound, consider a spanning tree for each component of $G$, and collect them all together to form a spanning forest $F$. Let $A$ be the set of edges of $F$, and let $B$ be the set of all other edges of $G$. Clearly, $|A| \\leq v-1$, so $|B| \\geq(1+\\varepsilon) v-|A| \\geq(1+\\varepsilon) v-(v-1)=\\varepsilon v+1>\\varepsilon v$.\n\n\n\nFor each edge $b$ in $B$, adjoining $b$ to $F$ produces a unique simple cycle $C_{b}$ through $b$. Let $S_{b}$ be the set of edges in $A$ along $C_{b}$. Since the $C_{b}$ have pairwise distinct lengths, $\\sum_{b \\in B}\\left|S_{b}\\right| \\geq$ $2+\\cdots+(|B|+1)=|B|(|B|+3) / 2>|B|^{2} / 2>\\varepsilon^{2} v^{2} / 2$.\n\n\n\nConsequently, some edge in $A$ lies in more than $\\varepsilon^{2} v^{2} /(2 v)=\\varepsilon^{2} v / 2$ of the $S_{b}$. Fix such an edge $a$ in $A$, and let $B^{\\prime}$ be the set of all edges $b$ in $B$ whose corresponding $S_{b}$ contain $a$, so $\\left|B^{\\prime}\\right|>\\varepsilon^{2} v / 2$.\n\n\n\nFor each 2-edge subset $\\left\\{b_{1}, b_{2}\\right\\}$ of $B^{\\prime}$, the union $C_{b_{1}} \\cup C_{b_{2}}$ of the cycles $C_{b_{1}}$ and $C_{b_{2}}$ forms a $\\theta$-graph, since their common part is a path in $F$ through $a$; and since neither of the $b_{i}$ lies along this path, $C_{b_{1}} \\cup C_{b_{2}}$ contains a third simple cycle $C_{b_{1}, b_{2}}$ through both $b_{1}$ and $b_{2}$. Finally, since $B^{\\prime} \\cap C_{b_{1}, b_{2}}=\\left\\{b_{1}, b_{2}\\right\\}$, the assignment $\\left\\{b_{1}, b_{2}\\right\\} \\mapsto C_{b_{1}, b_{2}}$ is injective, so the total number of simple cycles in $G$ is at least $\\left(\\begin{array}{c}\\left|B^{\\prime}\\right| \\\\ 2\\end{array}\\right)>\\left(\\begin{array}{c}\\varepsilon^{2} v / 2 \\\\ 2\\end{array}\\right)$. This establishes the desired lower bound and concludes the proof.', 'Recall that the girth of a graph $G$ is the minimal length of a (simple) cycle in this graph.\n\n\n\nLemma. For any fixed positive $\\delta$, a graph on $v$ vertices whose girth is at least $\\delta v$ has at most $v+o(v)$ edges.\n\n\n\nProof. Define $f(v)$ to be the maximal number $f$ such that a graph on $v$ vertices whose girth is at least $\\delta v$ may have $v+f$ edges. We are interested in the recursive estimates for $f$.\n\n\n\nLet $G$ be a graph on $v$ vertices whose gifth is at least $\\delta v$ containing $v+f(v)$ edges. If $G$ contains a leaf (i.e., a vertex of degree 1), then one may remove this vertex along with its edge, obtaining a graph with at most $v-1+f(v-1)$ edges. Thus, in this case $f(v) \\leq f(v-1)$.\n\n\n\nDefine an isolated path of length $k$ to be a sequence of vertices $v_{0}, v_{1}, \\ldots, v_{k}$, such that $v_{i}$ is connected to $v_{i+1}$, and each of the vertices $v_{1}, \\ldots, v_{k-1}$ has degree 2 (so, these vertices are connected only to their neighbors in the path). If $G$ contains an isolated path $v_{0}, \\ldots, v_{k}$ of length, say, $k>\\sqrt{v}$, then one may remove all its middle vertices $v_{1}, \\ldots, v_{k-1}$, along with all their $k$ edges. We obtain a graph on $v-k+1$ vertices with at most $(v-k+1)+f(v-k+1)$ edges. Thus, in this case $f(v) \\leq f(v-k+1)+1$.\n\n\n\nAssume now that the lengths of all isolated paths do not exceed $\\sqrt{v}$; we show that in this case $v$ is bounded from above. For that purpose, we replace each maximal isolated path by an edge between its endpoints, removing all middle vertices. We obtain a graph $H$ whose girth is at least $\\delta v / \\sqrt{v}=\\delta \\sqrt{v}$. Each vertex of $H$ has degree at least 3. By the girth condition, the neighborhood of any vertex $x$ of radius $r=\\lfloor(\\delta \\sqrt{v}-1) / 2\\rfloor$ is a tree rooted at $x$. Any vertex at level $i\\sqrt{v}$. This easily yields $f(v)=o(v)$, as desired.\n\n\n\nNow we proceed to the problem. Consider a graph on $v$ vertices containing no two simple cycles of the same length. Take its $\\lfloor\\varepsilon v / 2\\rfloor$ shortest cycles (or all its cycles, if their total number is smaller) and remove an edge from each. We get a graph of girth at least $\\varepsilon v / 2$. By the lemma, the number of edges in the obtained graph is at most $v+o(v)$, so the number of edges in the initial graph is at most $v+\\varepsilon v / 2+o(v)$, which is smaller than $(1+\\varepsilon) v$ if $v$ is large enough.']",,True,,, 1649,Algebra,,"Prove that there exist two functions $$ f, g: \mathbb{R} \rightarrow \mathbb{R} $$ such that $f \circ g$ is strictly decreasing, while $g \circ f$ is strictly increasing.","['Let\n\n\n\n$$\n\n\\begin{aligned}\n\n& \\text { - } A=\\bigcup_{k \\in \\mathbb{Z}}\\left(\\left[-2^{2 k+1},-2^{2 k}\\right) \\bigcup\\left(2^{2 k}, 2^{2 k+1}\\right]\\right) ; \\\\\n\n& \\text { - } B=\\bigcup_{k \\in \\mathbb{Z}}\\left(\\left[-2^{2 k},-2^{2 k-1}\\right) \\bigcup\\left(2^{2 k-1}, 2^{2 k}\\right]\\right) .\n\n\\end{aligned}\n\n$$\n\n\n\nThus $A=2 B, B=2 A, A=-A, B=-B, A \\cap B=\\varnothing$, and finally $A \\cup B \\cup\\{0\\}=\\mathbb{R}$. Let us take\n\n\n\n$$\n\nf(x)=\\left\\{\\begin{array}{lll}\n\nx & \\text { for } & x \\in A \\\\\n\n-x & \\text { for } & x \\in B \\\\\n\n0 & \\text { for } & x=0\n\n\\end{array}\\right.\n\n$$\n\n\n\nTake $g(x)=2 f(x)$. Thus $f(g(x))=f(2 f(x))=-2 x$ and $g(f(x))=2 f(f(x))=2 x$.']",,True,,, 1650,Number Theory,,"Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties: (1) for each integer $k$, the number $f(k)$ is an integer if and only if $k$ is not divisible by $n$; (2) the degree of $f$ is less than $n$","['We will show that such polynomial exists if and only if $n=1$ or $n$ is a power of a prime.\n\n\n\nWe will use two known facts stated in Lemmata 1 and 2.\n\n\n\nLEMMA 1. If $p^{a}$ is a power of a prime and $k$ is an integer, then $\\frac{(k-1)(k-2) \\ldots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right) !}$ is divisible by $p$ if and only if $k$ is not divisible by $p^{a}$.\n\n\n\nProof. First suppose that $p^{a} \\mid k$ and consider\n\n\n\n$$\n\n\\frac{(k-1)(k-2) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right) !}=\\frac{k-1}{p^{a}-1} \\cdot \\frac{k-2}{p^{a}-2} \\cdots \\frac{k-p^{a}+1}{1} .\n\n$$\n\n\n\nIn every fraction on the right-hand side, $p$ has the same maximal exponent in the numerator as in the denominator.\n\nTherefore, the product (which is an integer) is not divisible by $p$.\n\n\n\nNow suppose that $p^{a} \\nmid k$. We have\n\n\n\n$$\n\n\\frac{(k-1)(k-2) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}-1\\right) !}=\\frac{p^{a}}{k} \\cdot \\frac{k(k-1) \\cdots\\left(k-p^{a}+1\\right)}{\\left(p^{a}\\right) !} .\n\n$$\n\n\n\nThe last fraction is an integer. In the fraction $\\frac{p^{a}}{k}$, the denominator $k$ is not divisible by $p^{a}$.\n\n\n\nLEMMA 2. If $g(x)$ is a polynomial with degree less than $n$ then\n\n\n\n$$\n\n\\sum_{\\ell=0}^{n}(-1)^{\\ell}\\left(\\begin{array}{l}\n\nn \\\\\n\n\\ell\n\n\\end{array}\\right) g(x+n-\\ell)=0\n\n$$\n\n\n\nProof. Apply induction on $n$. For $n=1$ then $g(x)$ is a constant and\n\n\n\n$$\n\n\\left(\\begin{array}{l}\n\n1 \\\\\n\n0\n\n\\end{array}\\right) g(x+1)-\\left(\\begin{array}{l}\n\n1 \\\\\n\n1\n\n\\end{array}\\right) g(x)=g(x+1)-g(x)=0 .\n\n$$\n\n\n\nNow assume that $n>1$ and the Lemma holds for $n-1$. Let $h(x)=g(x+1)-g(x)$; the degree of $h$ is less than the degree of $g$, so the induction hypothesis applies for $g$ and $n-1$ :\n\n\n\n$$\n\n\\begin{gathered}\n\n\\sum_{\\ell=0}^{n-1}(-1)^{\\ell}\\left(\\begin{array}{c}\n\nn-1 \\\\\n\n\\ell\n\n\\end{array}\\right) h(x+n-1-\\ell)=0 \\\\\n\n\\sum_{\\ell=0}^{n-1}(-1)^{\\ell}\\left(\\begin{array}{c}\n\nn-1 \\\\\n\n\\ell\n\n\\end{array}\\right)(g(x+n-\\ell)-g(x+n-1-\\ell))=0 \\\\\n\n\\left(\\begin{array}{c}\n\nn-1 \\\\\n\n0\n\n\\end{array}\\right) g(x+n)+\\sum_{\\ell=1}^{n-1}(-1)^{\\ell}\\left(\\left(\\begin{array}{c}\n\nn-1 \\\\\n\n\\ell-1\n\n\\end{array}\\right)+\\right. \\\\\n\n\\left.\\left(\\begin{array}{c}\n\nn-1 \\\\\n\n\\ell\n\n\\end{array}\\right)\\right) g(x+n-\\ell)-(-1)^{n-1}\\left(\\begin{array}{l}\n\nn-1 \\\\\n\nn-1\n\n\\end{array}\\right) g(x)=0 \\\\\n\n\\sum_{\\ell=0}^{n}(-1)^{\\ell}\\left(\\begin{array}{l}\n\nn \\\\\n\n\\ell\n\n\\end{array}\\right) g(x+n-\\ell)=0 .\n\n\\end{gathered}\n\n$$\n\n\n\nLEMmA 3. If $n$ has at least two distinct prime divisors then the greatest common divisor of $\\left(\\begin{array}{l}n \\\\ 1\\end{array}\\right),\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right), \\ldots,\\left(\\begin{array}{c}n \\\\ n-1\\end{array}\\right)$ is 1 .\n\n\n\nProof. Suppose to the contrary that $p$ is a common prime divisor of $\\left(\\begin{array}{l}n \\\\ 1\\end{array}\\right), \\ldots,\\left(\\begin{array}{c}n \\\\ n-1\\end{array}\\right)$. In particular, $p \\mid\\left(\\begin{array}{l}n \\\\ 1\\end{array}\\right)=n$. Let $a$ be the exponent of $p$ in the prime factorization of $n$. Since $n$ has at least two prime divisors, we have $1\n\n### 分析']",,True,,, 1653,Algebra,,"Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying $$ P(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20} $$","['The answer is in the negative. Comparing the degrees of both sides in $(1)$ we get $\\operatorname{deg} P=21 n$ and $\\operatorname{deg} Q=10 n$ for some positive integer $n$. Take the derivative of $(1)$ to obtain\n\n$$\nP^{\\prime} P^{8}(10 P+9)=Q^{\\prime} Q^{19}(21 Q+20)\n$$\n\nSince $\\operatorname{gcd}(10 P+9, P)=\\operatorname{gcd}(10 P+9, P+1)=1$, it follows that $\\operatorname{gcd}\\left(10 P+9, P^{9}(P+1)\\right)=1$, so $\\operatorname{gcd}(10 P+9, Q)=1$, by $(1)$. Thus $(2)$ yields $10 P+9 \\mid Q^{\\prime}(21 Q+20)$, which is impossible since $0<\\operatorname{deg}\\left(Q^{\\prime}(21 Q+20)\\right)=20 n-1<21 n=\\operatorname{deg}(10 P+9)$. A contradiction.', 'Letting $r$ and $s$ be integers such that $r \\geq 2$ and $s \\geq 2 r$, we show that if $P^{r}+P^{r-1}=$ $Q^{s}+Q^{s-1}$, then $Q$ is constant.\n\nLet $m=\\operatorname{deg} P$ and $n=\\operatorname{deg} Q$. A degree inspection in the given relation shows that $m \\geq 2 n$.\n\nWe will prove that $P(P+1)$ has at least $m+1$ distinct complex roots. Assuming this for the moment, notice that $Q$ takes on one of the values 0 or -1 at each of those roots. Since $m+1 \\geq 2 n+1$, it follows that $Q$ takes on one of the values 0 and -1 at more than $n$ distinct points, so $Q$ must be constant.\n\nFinally, we prove that $P(P+1)$ has at least $m+1$ distinct complex roots. This can be done either by referring to the Mason-Stothers theorem or directly, in terms of multiplicities of the roots in question.\n\nSince $P$ and $P+1$ are relatively prime, the Mason-Stothers theorem implies that the number of distinct roots of $P(P+1)$ is greater than $m$, hence at least $m+1$.\n\nFor a direct proof, let $z_{1}, \\ldots, z_{t}$ be the distinct complex roots of $P(P+1)$, and let $z_{k}$ have multiplicity $\\alpha_{k}, k=1, \\ldots, t$. Since $P$ and $P+1$ have no roots in common, and $P^{\\prime}=(P+1)^{\\prime}$, it follows that $P^{\\prime}$ has a root of multiplicity $\\alpha_{k}-1$ at $z_{k}$. Consequently, $m-1=\\operatorname{deg} P^{\\prime} \\geq$ $\\sum_{k=1}^{t}\\left(\\alpha_{k}-1\\right)=\\sum_{k=1}^{t} \\alpha_{k}-t=2 m-t$; that is, $t \\geq m+1$. This completes the prof.\n\n### 分析']",,True,,, 1654,Combinatorics,,"Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win?","[""The answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.\n\nWe say that two unit grid-segments form a low-left corner (or LL-corner) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An up-right corner (or UR-corner) is defined similarly. The common endpoint of two unit grid-segments at a corner is the joint of that corner.\n\nFix a vertical line on the grid and call it the midline; the unit grid-segments along the midline are called middle segments. The unit grid-segments lying to the left/right of the midline are called left/right segments. Partition all left segments into LL-corners, and all right segments into UR-corners.\n\nWe now describe Ann's strategy. Her first move consists in orienting some middle segment arbitrarily. Assume that at some stage, Bob orients some segment $s$. If $s$ is a middle segment, Ann orients any free middle segment arbitrarily. Otherwise, $s$ forms a corner in the partition with some other segment $t$. Then Ann orients $t$ so that the joint of the corner is either the source of both arrows, or the target of both. Notice that after any move of Ann's, each corner in the partition is either completely oriented or completely not oriented. This means that Ann can always make a required move.\n\nAssume that Bob wins at some stage, i.e., an oriented cycle $C$ occurs. Let $X$ be the lowest of the leftmost points of $C$, and let $Y$ be the topmost of the rightmost points of $C$. If $X$ lies (strictly) to the left of the midline, then $X$ is the joint of some corner whose segments are both oriented. But, according to Ann's strategy, they are oriented so that they cannot occur in a cycle - a contradiction. Otherwise, $Y$ lies to the right of the midline, and a similar argument applies. Thus, Bob will never win, as desired.\n\n\n### 分析""]",,True,,, 1655,Algebra,,"Given a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of real numbers. For each $i(1 \leq i \leq n)$ define $$ d_{i}=\max \left\{a_{j}: 1 \leq j \leq i\right\}-\min \left\{a_{j}: i \leq j \leq n\right\} $$ and let $$ d=\max \left\{d_{i}: 1 \leq i \leq n\right\} $$ Prove that for arbitrary real numbers $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$, $$ \max \left\{\left|x_{i}-a_{i}\right|: 1 \leq i \leq n\right\} \geq \frac{d}{2} \tag{1} $$","['Let $1 \\leq p \\leq q \\leq r \\leq n$ be indices for which\n\n$$\nd=d_{q}, \\quad a_{p}=\\max \\left\\{a_{j}: 1 \\leq j \\leq q\\right\\}, \\quad a_{r}=\\min \\left\\{a_{j}: q \\leq j \\leq n\\right\\}\n$$\n\nand thus $d=a_{p}-a_{r}$. (These indices are not necessarily unique.)\n\n\n\nFor arbitrary real numbers $x_{1} \\leq x_{2} \\leq \\ldots \\leq x_{n}$, consider just the two quantities $\\left|x_{p}-a_{p}\\right|$ and $\\left|x_{r}-a_{r}\\right|$. Since\n\n$$\n\\left(a_{p}-x_{p}\\right)+\\left(x_{r}-a_{r}\\right)=\\left(a_{p}-a_{r}\\right)+\\left(x_{r}-x_{p}\\right) \\geq a_{p}-a_{r}=d\n$$\n\nwe have either $a_{p}-x_{p} \\geq \\frac{d}{2}$ or $x_{r}-a_{r} \\geq \\frac{d}{2}$. Hence,\n\n$$\n\\max \\left\\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\right\\} \\geq \\max \\left\\{\\left|x_{p}-a_{p}\\right|,\\left|x_{r}-a_{r}\\right|\\right\\} \\geq \\max \\left\\{a_{p}-x_{p}, x_{r}-a_{r}\\right\\} \\geq \\frac{d}{2}\n$$']",,True,,, 1656,Algebra,,"Given a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of real numbers. For each $i(1 \leq i \leq n)$ define $$ d_{i}=\max \left\{a_{j}: 1 \leq j \leq i\right\}-\min \left\{a_{j}: i \leq j \leq n\right\} $$ and let $$ d=\max \left\{d_{i}: 1 \leq i \leq n\right\} $$ Show that there exists a sequence $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$ of real numbers such that we have equality in (1).","['Define the sequence $\\left(x_{k}\\right)$ as\n\n$$\nx_{1}=a_{1}-\\frac{d}{2}, \\quad x_{k}=\\max \\left\\{x_{k-1}, a_{k}-\\frac{d}{2}\\right\\} \\quad \\text { for } 2 \\leq k \\leq n\n$$\n\nWe show that we have equality in (1) for this sequence.\n\nBy the definition, sequence $\\left(x_{k}\\right)$ is non-decreasing and $x_{k}-a_{k} \\geq-\\frac{d}{2}$ for all $1 \\leq k \\leq n$. Next we prove that\n\n$$\nx_{k}-a_{k} \\leq \\frac{d}{2} \\quad \\text { for all } 1 \\leq k \\leq n\\tag{2}\n$$\n\nConsider an arbitrary index $1 \\leq k \\leq n$. Let $\\ell \\leq k$ be the smallest index such that $x_{k}=x_{\\ell}$. We have either $\\ell=1$, or $\\ell \\geq 2$ and $x_{\\ell}>x_{\\ell-1}$. In both cases,\n\n$$\nx_{k}=x_{\\ell}=a_{\\ell}-\\frac{d}{2}\\tag{3}\n$$\n\nSince\n\n$$\na_{\\ell}-a_{k} \\leq \\max \\left\\{a_{j}: 1 \\leq j \\leq k\\right\\}-\\min \\left\\{a_{j}: k \\leq j \\leq n\\right\\}=d_{k} \\leq d\n$$\n\nequality (3) implies\n\n$$\nx_{k}-a_{k}=a_{\\ell}-a_{k}-\\frac{d}{2} \\leq d-\\frac{d}{2}=\\frac{d}{2}\n$$\n\nWe obtained that $-\\frac{d}{2} \\leq x_{k}-a_{k} \\leq \\frac{d}{2}$ for all $1 \\leq k \\leq n$, so\n\n$$\n\\max \\left\\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\right\\} \\leq \\frac{d}{2}\n$$\n\nWe have equality because $\\left|x_{1}-a_{1}\\right|=\\frac{d}{2}$.']",,True,,, 1657,Algebra,,"Consider those functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the condition $$ f(m+n) \geq f(m)+f(f(n))-1\tag{1} $$ for all $m, n \in \mathbb{N}$. Find all possible values of $f(2007)$. ( $\mathbb{N}$ denotes the set of all positive integers.)","['Suppose that a function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies (1). For arbitrary positive integers $m>n$, by (1) we have\n\n$$\nf(m)=f(n+(m-n)) \\geq f(n)+f(f(m-n))-1 \\geq f(n),\n$$\n\nso $f$ is nondecreasing.\n\nFunction $f \\equiv 1$ is an obvious solution. To find other solutions, assume that $f \\not \\equiv 1$ and take the smallest $a \\in \\mathbb{N}$ such that $f(a)>1$. Then $f(b) \\geq f(a)>1$ for all integer $b \\geq a$.\n\nSuppose that $f(n)>n$ for some $n \\in \\mathbb{N}$. Then we have\n\n$$\nf(f(n))=f((f(n)-n)+n) \\geq f(f(n)-n)+f(f(n))-1\n$$\n\nso $f(f(n)-n) \\leq 1$ and hence $f(n)-n0 \\Longleftrightarrow h(t)y$ for all $y \\in \\mathbb{R}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y) \\neq y$ immediately. If $f(y)f(y),\n$$\n\ncontradiction. Therefore $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$.\n\nFor $x \\in \\mathbb{R}^{+}$define $g(x)=f(x)-x$; then $f(x)=g(x)+x$ and, as we have seen, $g(x)>0$. Transforming (1) for function $g(x)$ and setting $t=x+y$,\n\n$$\n\\begin{aligned}\nf(t+g(y)) & =f(t)+f(y) \\\\\ng(t+g(y))+t+g(y) & =(g(t)+t)+(g(y)+y)\n\\end{aligned}\n$$\n\nand therefore\n\n$$\ng(t+g(y))=g(t)+y \\quad \\text { for all } t>y>0 \\tag{2}\n$$\n\nNext we prove that function $g(x)$ is injective. Suppose that $g\\left(y_{1}\\right)=g\\left(y_{2}\\right)$ for some numbers $y_{1}, y_{2} \\in \\mathbb{R}^{+}$. Then by $(2)$,\n\n$$\ng(t)+y_{1}=g\\left(t+g\\left(y_{1}\\right)\\right)=g\\left(t+g\\left(y_{2}\\right)\\right)=g(t)+y_{2}\n$$\n\nfor all $t>\\max \\left\\{y_{1}, y_{2}\\right\\}$. Hence, $g\\left(y_{1}\\right)=g\\left(y_{2}\\right)$ is possible only if $y_{1}=y_{2}$.\n\nNow let $u, v$ be arbitrary positive numbers and $t>u+v$. Applying (2) three times,\n\n$$\ng(t+g(u)+g(v))=g(t+g(u))+v=g(t)+u+v=g(t+g(u+v)) \\text {. }\n$$\n\nBy the injective property we conclude that $t+g(u)+g(v)=t+g(u+v)$, hence\n\n$$\ng(u)+g(v)=g(u+v)\\tag{3}\n$$\n\nSince function $g(v)$ is positive, equation (3) also shows that $g$ is an increasing function.\n\nFinally we prove that $g(x)=x$. Combining (2) and (3), we obtain\n\n$$\ng(t)+y=g(t+g(y))=g(t)+g(g(y))\n$$\n\nand hence\n\n$$\ng(g(y))=y\n$$\n\nSuppose that there exists an $x \\in \\mathbb{R}^{+}$such that $g(x) \\neq x$. By the monotonicity of $g$, if $x>g(x)$ then $g(x)>g(g(x))=x$. Similarly, if $xy$ and introduce function $g(x)=f(x)-x>0$ in the same way as in Solution 1.\n\nFor arbitrary $t>y>0$, substitute $x=t-y$ into (1) to obtain\n\n$$\nf(t+g(y))=f(t)+f(y)\n$$\n\nwhich, by induction, implies\n\n$$\nf(t+n g(y))=f(t)+n f(y) \\quad \\text { for all } t>y>0, n \\in \\mathbb{N} \\tag{4}\n$$\n\nTake two arbitrary positive reals $y$ and $z$ and a third fixed number $t>\\max \\{y, z\\}$. For each positive integer $k$, let $\\ell_{k}=\\left\\lfloor k \\frac{g(y)}{g(z)}\\right\\rfloor$. Then $t+k g(y)-\\ell_{k} g(z) \\geq t>z$ and, applying (4) twice,\n\n$$\n\\begin{gathered}\nf\\left(t+k g(y)-\\ell_{k} g(z)\\right)+\\ell_{k} f(z)=f(t+k g(y))=f(t)+k f(y), \\\\\n0<\\frac{1}{k} f\\left(t+k g(y)-\\ell_{k} g(z)\\right)=\\frac{f(t)}{k}+f(y)-\\frac{\\ell_{k}}{k} f(z) .\n\\end{gathered}\n$$\n\nAs $k \\rightarrow \\infty$ we get\n\n$$\n0 \\leq \\lim _{k \\rightarrow \\infty}\\left(\\frac{f(t)}{k}+f(y)-\\frac{\\ell_{k}}{k} f(z)\\right)=f(y)-\\frac{g(y)}{g(z)} f(z)=f(y)-\\frac{f(y)-y}{f(z)-z} f(z)\n$$\n\nand therefore\n\n$$\n\\frac{f(y)}{y} \\leq \\frac{f(z)}{z}\n$$\n\nExchanging variables $y$ and $z$, we obtain the reverse inequality. Hence, $\\frac{f(y)}{y}=\\frac{f(z)}{z}$ for arbitrary $y$ and $z$; so function $\\frac{f(x)}{x}$ is constant, $f(x)=c x$.\n\nSubstituting back into (1), we find that $f(x)=c x$ is a solution if and only if $c=2$. So the only solution for the problem is $f(x)=2 x$.']",['$f(x)=2 x$'],False,,Expression, 1660,Algebra,,"Let $c>2$, and let $a(1), a(2), \ldots$ be a sequence of nonnegative real numbers such that $$ a(m+n) \leq 2 a(m)+2 a(n) \text { for all } m, n \geq 1\tag{1} $$ and $$ a\left(2^{k}\right) \leq \frac{1}{(k+1)^{c}} \quad \text { for all } k \geq 0\tag{2} $$ Prove that the sequence $a(n)$ is bounded.","['For convenience, define $a(0)=0$; then condition (1) persists for all pairs of nonnegative indices.\n\nLemma 1. For arbitrary nonnegative indices $n_{1}, \\ldots, n_{k}$, we have\n\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq \\sum_{i=1}^{k} 2^{i} a\\left(n_{i}\\right)\\tag{3}\n$$\n\nand\n\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq 2 k \\sum_{i=1}^{k} a\\left(n_{i}\\right)\\tag{4}\n$$\n\nProof. Inequality (3) is proved by induction on $k$. The base case $k=1$ is trivial, while the induction step is provided by\n\n$$\na\\left(\\sum_{i=1}^{k+1} n_{i}\\right)=a\\left(n_{1}+\\sum_{i=2}^{k+1} n_{i}\\right) \\leq 2 a\\left(n_{1}\\right)+2 a\\left(\\sum_{i=1}^{k} n_{i+1}\\right) \\leq 2 a\\left(n_{1}\\right)+2 \\sum_{i=1}^{k} 2^{i} a\\left(n_{i+1}\\right)=\\sum_{i=1}^{k+1} 2^{i} a\\left(n_{i}\\right)\n$$\n\nTo establish (4), first the inequality\n$$\na\\left(\\sum_{i=1}^{2^{d}} n_{i}\\right) \\leq 2^{d} \\sum_{i=1}^{2^{d}} a\\left(n_{i}\\right)\n$$\ncan be proved by an obvious induction on $d$. Then, turning to (4), we find an integer $d$ such that $2^{d-1}d$, and take some positive integer $f$ such that $M_{f}>d$. Applying (3), we get\n$$\na(n)=a\\left(\\sum_{k=1}^{f} \\sum_{M_{k-1} \\leq i2$. We can assume that $s_{1} \\leq s_{2} \\leq \\cdots \\leq s_{k}$. Note that\n$$\n\\sum_{i=1}^{k-1} 2^{-s_{i}} \\leq 1-2^{-s_{k-1}}\n$$\nsince the left-hand side is a fraction with the denominator $2^{s_{k-1}}$, and this fraction is less than 1. Define $s_{k-1}^{\\prime}=s_{k-1}-1$ and $n_{k-1}^{\\prime}=n_{k-1}+n_{k}$; then we have\n$$\n\\sum_{i=1}^{k-2} 2^{-s_{i}}+2^{-s_{k-1}^{\\prime}} \\leq\\left(1-2 \\cdot 2^{-s_{k-1}}\\right)+2^{1-s_{k-1}}=1 .\n$$\nNow, the induction hypothesis can be applied to achieve\n$$\n\\begin{aligned}\na\\left(\\sum_{i=1}^{k} n_{i}\\right)=a\\left(\\sum_{i=1}^{k-2} n_{i}+n_{k-1}^{\\prime}\\right) & \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}^{\\prime}} a\\left(n_{k-1}^{\\prime}\\right) \\\\\n& \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}-1} \\cdot 2\\left(a\\left(n_{k-1}\\right)+a\\left(n_{k}\\right)\\right) \\\\\n& \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}} a\\left(n_{k-1}\\right)+2^{s_{k}} a\\left(n_{k}\\right) .\n\\end{aligned}\n$$\n\n\nLet $q=c / 2>1$. Take an arbitrary positive integer $n$ and write\n$$\nn=\\sum_{i=1}^{k} 2^{u_{i}}, \\quad 0 \\leq u_{1}1$.']",,True,,, 1661,Algebra,,"Let $a_{1}, a_{2}, \ldots, a_{100}$ be nonnegative real numbers such that $a_{1}^{2}+a_{2}^{2}+\ldots+a_{100}^{2}=1$. Prove that $$ a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\ldots+a_{100}^{2} a_{1}<\frac{12}{25} $$","['Let $S=\\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$. (As usual, we consider the indices modulo 100, e.g. we set $a_{101}=a_{1}$ and $a_{102}=a_{2}$.)\n\nApplying the Cauchy-Schwarz inequality to sequences $\\left(a_{k+1}\\right)$ and $\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)$, and then the AM-GM inequality to numbers $a_{k+1}^{2}$ and $a_{k+2}^{2}$,\n$$\n\\begin{aligned}\n(3 S)^{2} & =\\left(\\sum_{k=1}^{100} a_{k+1}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)\\right)^{2} \\leq\\left(\\sum_{k=1}^{100} a_{k+1}^{2}\\right)\\left(\\sum_{k=1}^{100}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)^{2}\\right) \\\\\n& =1 \\cdot \\sum_{k=1}^{100}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)^{2}=\\sum_{k=1}^{100}\\left(a_{k}^{4}+4 a_{k}^{2} a_{k+1} a_{k+2}+4 a_{k+1}^{2} a_{k+2}^{2}\\right) \\\\\n& \\leq \\sum_{k=1}^{100}\\left(a_{k}^{4}+2 a_{k}^{2}\\left(a_{k+1}^{2}+a_{k+2}^{2}\\right)+4 a_{k+1}^{2} a_{k+2}^{2}\\right)=\\sum_{k=1}^{100}\\left(a_{k}^{4}+6 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\\right) .\n\\end{aligned}\n$$\nApplying the trivial estimates\n$$\n\\sum_{k=1}^{100}\\left(a_{k}^{4}+2 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\\right) \\leq\\left(\\sum_{k=1}^{100} a_{k}^{2}\\right)^{2} \\quad \\text { and } \\quad \\sum_{k=1}^{100} a_{k}^{2} a_{k+1}^{2} \\leq\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}\\right)\\left(\\sum_{j=1}^{50} a_{2 j}^{2}\\right)\n$$\nwe obtain that\n$$\n(3 S)^{2} \\leq\\left(\\sum_{k=1}^{100} a_{k}^{2}\\right)^{2}+4\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}\\right)\\left(\\sum_{j=1}^{50} a_{2 j}^{2}\\right) \\leq 1+\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}+\\sum_{j=1}^{50} a_{2 j}^{2}\\right)^{2}=2,\n$$\nhence\n$$\nS \\leq \\frac{\\sqrt{2}}{3} \\approx 0.4714<\\frac{12}{25}=0.48\n$$']",,True,,, 1662,Combinatorics,,"Let $n>1$ be an integer. In the space, consider the set $$ S=\{(x, y, z) \mid x, y, z \in\{0,1, \ldots, n\}, x+y+z>0\} $$ Find the smallest number of planes that jointly contain all $(n+1)^{3}-1$ points of $S$ but none of them passes through the origin.","['It is easy to find $3 n$ such planes. For example, planes $x=i, y=i$ or $z=i$ $(i=1,2, \\ldots, n)$ cover the set $S$ but none of them contains the origin. Another such collection consists of all planes $x+y+z=k$ for $k=1,2, \\ldots, 3 n$.\n\nWe show that $3 n$ is the smallest possible number.\n\nLemma 1. Consider a nonzero polynomial $P\\left(x_{1}, \\ldots, x_{k}\\right)$ in $k$ variables. Suppose that $P$ vanishes at all points $\\left(x_{1}, \\ldots, x_{k}\\right)$ such that $x_{1}, \\ldots, x_{k} \\in\\{0,1, \\ldots, n\\}$ and $x_{1}+\\cdots+x_{k}>0$, while $P(0,0, \\ldots, 0) \\neq 0$. Then $\\operatorname{deg} P \\geq k n$.\n\nProof. We use induction on $k$. The base case $k=0$ is clear since $P \\neq 0$. Denote for clarity $y=x_{k}$.\n\nLet $R\\left(x_{1}, \\ldots, x_{k-1}, y\\right)$ be the residue of $P$ modulo $Q(y)=y(y-1) \\ldots(y-n)$. Polynomial $Q(y)$ vanishes at each $y=0,1, \\ldots, n$, hence $P\\left(x_{1}, \\ldots, x_{k-1}, y\\right)=R\\left(x_{1}, \\ldots, x_{k-1}, y\\right)$ for all $x_{1}, \\ldots, x_{k-1}, y \\in\\{0,1, \\ldots, n\\}$. Therefore, $R$ also satisfies the condition of the Lemma; moreover, $\\operatorname{deg}_{y} R \\leq n$. Clearly, $\\operatorname{deg} R \\leq \\operatorname{deg} P$, so it suffices to prove that $\\operatorname{deg} R \\geq n k$.\n\nNow, expand polynomial $R$ in the powers of $y$ :\n$$\nR\\left(x_{1}, \\ldots, x_{k-1}, y\\right)=R_{n}\\left(x_{1}, \\ldots, x_{k-1}\\right) y^{n}+R_{n-1}\\left(x_{1}, \\ldots, x_{k-1}\\right) y^{n-1}+\\cdots+R_{0}\\left(x_{1}, \\ldots, x_{k-1}\\right)\n$$\nWe show that polynomial $R_{n}\\left(x_{1}, \\ldots, x_{k-1}\\right)$ satisfies the condition of the induction hypothesis.\n\nConsider the polynomial $T(y)=R(0, \\ldots, 0, y)$ of degree $\\leq n$. This polynomial has $n$ roots $y=1, \\ldots, n$; on the other hand, $T(y) \\not \\equiv 0$ since $T(0) \\neq 0$. Hence $\\operatorname{deg} T=n$, and its leading coefficient is $R_{n}(0,0, \\ldots, 0) \\neq 0$. In particular, in the case $k=1$ we obtain that coefficient $R_{n}$ is nonzero.\n\nSimilarly, take any numbers $a_{1}, \\ldots, a_{k-1} \\in\\{0,1, \\ldots, n\\}$ with $a_{1}+\\cdots+a_{k-1}>0$. Substituting $x_{i}=a_{i}$ into $R\\left(x_{1}, \\ldots, x_{k-1}, y\\right)$, we get a polynomial in $y$ which vanishes at all points $y=0, \\ldots, n$ and has degree $\\leq n$. Therefore, this polynomial is null, hence $R_{i}\\left(a_{1}, \\ldots, a_{k-1}\\right)=0$ for all $i=0,1, \\ldots, n$. In particular, $R_{n}\\left(a_{1}, \\ldots, a_{k-1}\\right)=0$.\n\nThus, the polynomial $R_{n}\\left(x_{1}, \\ldots, x_{k-1}\\right)$ satisfies the condition of the induction hypothesis. So, we have $\\operatorname{deg} R_{n} \\geq(k-1) n$ and $\\operatorname{deg} P \\geq \\operatorname{deg} R \\geq \\operatorname{deg} R_{n}+n \\geq k n$.\n\nNow we can finish the solution. Suppose that there are $N$ planes covering all the points of $S$ but not containing the origin. Let their equations be $a_{i} x+b_{i} y+c_{i} z+d_{i}=0$. Consider the polynomial\n$$\nP(x, y, z)=\\prod_{i=1}^{N}\\left(a_{i} x+b_{i} y+c_{i} z+d_{i}\\right)\n$$\nIt has total degree $N$. This polynomial has the property that $P\\left(x_{0}, y_{0}, z_{0}\\right)=0$ for any $\\left(x_{0}, y_{0}, z_{0}\\right) \\in S$, while $P(0,0,0) \\neq 0$. Hence by Lemma 1 we get $N=\\operatorname{deg} P \\geq 3 n$, as desired.', 'We present a different proof of the main Lemma 1. Here we confine ourselves to the case $k=3$, which is applied in the solution, and denote the variables by $x, y$ and $z$. (The same proof works for the general statement as well.)\n\nThe following fact is known with various proofs; we provide one possible proof for the completeness.\n\nLemma 2. For arbitrary integers $0 \\leq m1$ rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of some rectangle. Prove that in this dissection, there exists a rectangle having no point on the boundary of the square.","['Call the directions of the sides of the square horizontal and vertical. A horizontal or vertical line, which intersects the interior of the square but does not intersect the interior of any rectangle, will be called a splitting line. A rectangle having no point on the boundary of the square will be called an interior rectangle.\n\nSuppose, to the contrary, that there exists a dissection of the square into more than one rectangle, such that no interior rectangle and no splitting line appear. Consider such a dissection with the least possible number of rectangles. Notice that this number of rectangles is greater than 2, otherwise their common side provides a splitting line.\n\nIf there exist two rectangles having a common side, then we can replace them by their union (see Figure 1). The number of rectangles was greater than 2, so in a new dissection it is greater than 1. Clearly, in the new dissection, there is also no splitting line as well as no interior rectangle. This contradicts the choice of the original dissection.\n\nDenote the initial square by $A B C D$, with $A$ and $B$ being respectively the lower left and lower right vertices. Consider those two rectangles $a$ and $b$ containing vertices $A$ and $B$, respectively. (Note that $a \\neq b$, otherwise its top side provides a splitting line.) We can assume that the height of $a$ is not greater than that of $b$. Then consider the rectangle $c$ neighboring to the lower right corner of $a$ (it may happen that $c=b$ ). By aforementioned, the heights of $a$ and $c$ are distinct. Then two cases are possible.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nCase 1. The height of $c$ is less than that of $a$. Consider the rectangle $d$ which is adjacent to both $a$ and $c$, i. e. the one containing the angle marked in Figure 2. This rectangle has no common point with $B C$ (since $a$ is not higher than $b$ ), as well as no common point with $A B$ or with $A D$ (obviously). Then $d$ has a common point with $C D$, and its left side provides a splitting line. Contradiction.\n\nCase 2. The height of $c$ is greater than that of $a$. Analogously, consider the rectangle $d$ containing the angle marked on Figure 3. It has no common point with $A D$ (otherwise it has a common side with $a$ ), as well as no common point with $A B$ or with $B C$ (obviously). Then $d$ has a common point with $C D$. Hence its right side provides a splitting line, and we get the contradiction again.', 'Again, we suppose the contrary. Consider an arbitrary counterexample. Then we know that each rectangle is attached to at least one side of the square. Observe that a rectangle cannot be attached to two opposite sides, otherwise one of its sides lies on a splitting line.\n\nWe say that two rectangles are opposite if they are attached to opposite sides of $A B C D$. We claim that there exist two opposite rectangles having a common point.\n\nConsider the union $L$ of all rectangles attached to the left. Assume, to the contrary, that $L$ has no common point with the rectangles attached to the right. Take a polygonal line $p$ connecting the top and the bottom sides of the square and passing close from the right to the boundary of $L$ (see Figure 4). Then all its points belong to the rectangles attached either to the top or to the bottom. Moreover, the upper end-point of $p$ belongs to a rectangle attached to the top, and the lower one belongs to an other rectangle attached to the bottom. Hence, there is a point on $p$ where some rectangles attached to the top and to the bottom meet each other. So, there always exists a pair of neighboring opposite rectangles.\n\n\n\nFigure 4\n\n\n\nFigure 5\n\n\n\nFigure 6\n\nNow, take two opposite neighboring rectangles $a$ and $b$. We can assume that $a$ is attached to the left and $b$ is attached to the right. Let $X$ be their common point. If $X$ belongs to their horizontal sides (in particular, $X$ may appear to be a common vertex of $a$ and $b$ ), then these sides provide a splitting line (see Figure 5). Otherwise, $X$ lies on the vertical sides. Let $\\ell$ be the line containing these sides.\n\nSince $\\ell$ is not a splitting line, it intersects the interior of some rectangle. Let $c$ be such a rectangle, closest to $X$; we can assume that $c$ lies above $X$. Let $Y$ be the common point of $\\ell$ and the bottom side of $c$ (see Figure 6). Then $Y$ is also a vertex of two rectangles lying below $c$.\n\nSo, let $Y$ be the upper-right and upper-left corners of the rectangles $a^{\\prime}$ and $b^{\\prime}$, respectively. Then $a^{\\prime}$ and $b^{\\prime}$ are situated not lower than $a$ and $b$, respectively (it may happen that $a=a^{\\prime}$ or $b=b^{\\prime}$ ). We claim that $a^{\\prime}$ is attached to the left. If $a=a^{\\prime}$ then of course it is. If $a \\neq a^{\\prime}$ then $a^{\\prime}$ is above $a$, below $c$ and to the left from $b^{\\prime}$. Hence, it can be attached to the left only.\n\nAnalogously, $b^{\\prime}$ is attached to the right. Now, the top sides of these two rectangles pass through $Y$, hence they provide a splitting line again. This last contradiction completes the proof.']",,True,,, 1664,Combinatorics,,"Find all positive integers $n$, for which the numbers in the set $S=\{1,2, \ldots, n\}$ can be colored red and blue, with the following condition being satisfied: the set $S \times S \times S$ contains exactly 2007 ordered triples $(x, y, z)$ such that (i) $x, y, z$ are of the same color and (ii) $x+y+z$ is divisible by $n$.","['Suppose that the numbers $1,2, \\ldots, n$ are colored red and blue. Denote by $R$ and $B$ the sets of red and blue numbers, respectively; let $|R|=r$ and $|B|=b=n-r$. Call a triple $(x, y, z) \\in S \\times S \\times S$ monochromatic if $x, y, z$ have the same color, and bichromatic otherwise. Call a triple $(x, y, z)$ divisible if $x+y+z$ is divisible by $n$. We claim that there are exactly $r^{2}-r b+b^{2}$ divisible monochromatic triples.\n\nFor any pair $(x, y) \\in S \\times S$ there exists a unique $z_{x, y} \\in S$ such that the triple $\\left(x, y, z_{x, y}\\right)$ is divisible; so there are exactly $n^{2}$ divisible triples. Furthermore, if a divisible triple $(x, y, z)$ is bichromatic, then among $x, y, z$ there are either one blue and two red numbers, or vice versa. In both cases, exactly one of the pairs $(x, y),(y, z)$ and $(z, x)$ belongs to the set $R \\times B$. Assign such pair to the triple $(x, y, z)$.\n\nConversely, consider any pair $(x, y) \\in R \\times B$, and denote $z=z_{x, y}$. Since $x \\neq y$, the triples $(x, y, z),(y, z, x)$ and $(z, x, y)$ are distinct, and $(x, y)$ is assigned to each of them. On the other hand, if $(x, y)$ is assigned to some triple, then this triple is clearly one of those mentioned above. So each pair in $R \\times B$ is assigned exactly three times.\n\nThus, the number of bichromatic divisible triples is three times the number of elements in $R \\times B$, and the number of monochromatic ones is $n^{2}-3 r b=(r+b)^{2}-3 r b=r^{2}-r b+b^{2}$, as claimed.\n\nSo, to find all values of $n$ for which the desired coloring is possible, we have to find all $n$, for which there exists a decomposition $n=r+b$ with $r^{2}-r b+b^{2}=2007$. Therefore, $9 \\mid r^{2}-r b+b^{2}=(r+b)^{2}-3 r b$. From this it consequently follows that $3|r+b, 3| r b$, and then $3|r, 3| b$. Set $r=3 s, b=3 c$. We can assume that $s \\geq c$. We have $s^{2}-s c+c^{2}=223$.\n\nFurthermore,\n$$\n892=4\\left(s^{2}-s c+c^{2}\\right)=(2 c-s)^{2}+3 s^{2} \\geq 3 s^{2} \\geq 3 s^{2}-3 c(s-c)=3\\left(s^{2}-s c+c^{2}\\right)=669\n$$\nso $297 \\geq s^{2} \\geq 223$ and $17 \\geq s \\geq 15$. If $s=15$ then\n$$\nc(15-c)=c(s-c)=s^{2}-\\left(s^{2}-s c+c^{2}\\right)=15^{2}-223=2\n$$\nwhich is impossible for an integer $c$. In a similar way, if $s=16$ then $c(16-c)=33$, which is also impossible. Finally, if $s=17$ then $c(17-c)=66$, and the solutions are $c=6$ and $c=11$. Hence, $(r, b)=(51,18)$ or $(r, b)=(51,33)$, and the possible values of $n$ are $n=51+18=69$ and $n=51+33=84$.']","['$69$,$84$']",True,,Numerical, 1665,Combinatorics,,"Let $A_{0}=\left(a_{1}, \ldots, a_{n}\right)$ be a finite sequence of real numbers. For each $k \geq 0$, from the sequence $A_{k}=\left(x_{1}, \ldots, x_{n}\right)$ we construct a new sequence $A_{k+1}$ in the following way. 1. We choose a partition $\{1, \ldots, n\}=I \cup J$, where $I$ and $J$ are two disjoint sets, such that the expression $$ \left|\sum_{i \in I} x_{i}-\sum_{j \in J} x_{j}\right| $$ attains the smallest possible value. (We allow the sets $I$ or $J$ to be empty; in this case the corresponding sum is 0 .) If there are several such partitions, one is chosen arbitrarily. 2. We set $A_{k+1}=\left(y_{1}, \ldots, y_{n}\right)$, where $y_{i}=x_{i}+1$ if $i \in I$, and $y_{i}=x_{i}-1$ if $i \in J$. Prove that for some $k$, the sequence $A_{k}$ contains an element $x$ such that $|x| \geq n / 2$.","['Lemma. Suppose that all terms of the sequence $\\left(x_{1}, \\ldots, x_{n}\\right)$ satisfy the inequality $\\left|x_{i}\\right|1)$. By the induction hypothesis there exists a splitting $\\{1, \\ldots, n-1\\}=I^{\\prime} \\cup J^{\\prime}$ such that\n$$\n\\left|\\sum_{i \\in I^{\\prime}} x_{i}-\\sum_{j \\in J^{\\prime}} x_{j}\\right|n-2 \\cdot \\frac{n}{2}=0\n$$\n\nThus we obtain $S_{q}>S_{q-1}>\\cdots>S_{p}$. This is impossible since $A_{p}=A_{q}$ and hence $S_{p}=S_{q}$.']",,True,,, 1666,Combinatorics,,"In the Cartesian coordinate plane define the strip $S_{n}=\{(x, y) \mid n \leq xb$. Then $a_{1}>b_{1} \\geq 1$, so $a_{1} \\geq 3$.\n\nChoose integers $k$ and $\\ell$ such that $k a_{1}-\\ell b_{1}=1$ and therefore $k a-\\ell b=d$. Since $a_{1}$ and $b_{1}$ are odd, $k+\\ell$ is odd as well. Hence, for every integer $n$, strips $S_{n}$ and $S_{n+k a-\\ell b}=S_{n+d}$ have opposite colors. This also implies that the coloring is periodic with period $2 d$, i.e. strips $S_{n}$ and $S_{n+2 d}$ have the same color for every $n$.\n\n\n\nFigure 1\n\nWe will construct the desired rectangle $A B C D$ with $A B=C D=a$ and $B C=A D=b$ in a position such that vertex $A$ lies on the $x$-axis, and the projection of side $A B$ onto the $x$-axis is of length $2 d$ (see Figure 1). This is possible since $a=a_{1} d>2 d$. The coordinates of the vertices will have the forms\n$$\nA=(t, 0), \\quad B=\\left(t+2 d, y_{1}\\right), \\quad C=\\left(u+2 d, y_{2}\\right), \\quad D=\\left(u, y_{3}\\right)\n$$\nLet $\\varphi=\\sqrt{a_{1}^{2}-4}$. By Pythagoras' theorem,\n$$\ny_{1}=B B_{0}=\\sqrt{a^{2}-4 d^{2}}=d \\sqrt{a_{1}^{2}-4}=d \\varphi .\n$$\nSo, by the similar triangles $A D D_{0}$ and $B A B_{0}$, we have the constraint\n$$\nu-t=A D_{0}=\\frac{A D}{A B} \\cdot B B_{0}=\\frac{b d}{a} \\varphi \\tag{1}\n$$\n\nfor numbers $t$ and $u$. Computing the numbers $y_{2}$ and $y_{3}$ is not required since they have no effect to the colors.\n\nObserve that the number $\\varphi$ is irrational, because $\\varphi^{2}$ is an integer, but $\\varphi$ is not: $a_{1}>\\varphi \\geq$ $\\sqrt{a_{1}^{2}-2 a_{1}+2}>a_{1}-1$.\n\nBy the periodicity, points $A$ and $B$ have the same color; similarly, points $C$ and $D$ have the same color. Furthermore, these colors depend only on the values of $t$ and $u$. So it is sufficient to choose numbers $t$ and $u$ such that vertices $A$ and $D$ have the same color.\n\nLet $w$ be the largest positive integer such that there exist $w$ consecutive strips $S_{n_{0}}, S_{n_{0}+1}, \\ldots$, $S_{n_{0}+w-1}$ with the same color, say red. (Since $S_{n_{0}+d}$ must be blue, we have $w \\leq d$.) We will choose $t$ from the interval $\\left(n_{0}, n_{0}+w\\right)$.\n\n\n\nFigure 2\n\nConsider the interval $I=\\left(n_{0}+\\frac{b d}{a} \\varphi, n_{0}+\\frac{b d}{a} \\varphi+w\\right)$ on the $x$-axis (see Figure 2). Its length is $w$, and the end-points are irrational. Therefore, this interval intersects $w+1$ consecutive strips. Since at most $w$ consecutive strips may have the same color, interval $I$ must contain both red and blue points. Choose $u \\in I$ such that the line $x=u$ is red and set $t=u-\\frac{b d}{a} \\varphi$, according to the constraint (1). Then $t \\in\\left(n_{0}, n_{0}+w\\right)$ and $A=(t, 0)$ is red as well as $D=\\left(u, y_{3}\\right)$.\n\nHence, variables $u$ and $t$ can be set such that they provide a rectangle with four red vertices.""]",,True,,, 1667,Combinatorics,,"In a mathematical competition some competitors are friends; friendship is always mutual. Call a group of competitors a clique if each two of them are friends. The number of members in a clique is called its size. It is known that the largest size of cliques is even. Prove that the competitors can be arranged in two rooms such that the largest size of cliques in one room is the same as the largest size of cliques in the other room.","['We present an algorithm to arrange the competitors. Let the two rooms be Room A and Room B. We start with an initial arrangement, and then we modify it several times by sending one person to the other room. At any state of the algorithm, $A$ and $B$ denote the sets of the competitors in the rooms, and $c(A)$ and $c(B)$ denote the largest sizes of cliques in the rooms, respectively.\n\nStep 1. Let $M$ be one of the cliques of largest size, $|M|=2 m$. Send all members of $M$ to Room $A$ and all other competitors to Room $B$.\n\nSince $M$ is a clique of the largest size, we have $c(A)=|M| \\geq c(B)$.\n\nStep 2. While $c(A)>c(B)$, send one person from Room $A$ to Room $B$.\n\n\n\nNote that $c(A)>c(B)$ implies that Room $A$ is not empty.\n\nIn each step, $c(A)$ decreases by one and $c(B)$ increases by at most one. So at the end we have $c(A) \\leq c(B) \\leq c(A)+1$.\n\nWe also have $c(A)=|A| \\geq m$ at the end. Otherwise we would have at least $m+1$ members of $M$ in Room $B$ and at most $m-1$ in Room $A$, implying $c(B)-c(A) \\geq(m+1)-(m-1)=2$.\n\nStep 3. Let $k=c(A)$. If $c(B)=k$ then $S T O P$.\n\nIf we reached $c(A)=c(B)=k$ then we have found the desired arrangement.\n\nIn all other cases we have $c(B)=k+1$.\n\nFrom the estimate above we also know that $k=|A|=|A \\cap M| \\geq m$ and $|B \\cap M| \\leq m$.\n\nStep 4. If there exists a competitor $x \\in B \\cap M$ and a clique $C \\subset B$ such that $|C|=k+1$ and $x \\notin C$, then move $x$ to Room $A$ and STOP.\n\n\n\nAfter moving $x$ back to Room $A$, we will have $k+1$ members of $M$ in Room $A$, thus $c(A)=k+1$. Due to $x \\notin C, c(B)=|C|$ is not decreased, and after this step we have $c(A)=c(B)=k+1$.\n\n\n\nIf there is no such competitor $x$, then in Room $B$, all cliques of size $k+1$ contain $B \\cap M$ as a subset.\n\nStep 5. While $c(B)=k+1$, choose a clique $C \\subset B$ such that $|C|=k+1$ and move one member of $C \\backslash M$ to Room $A$.\n\n\n\nNote that $|C|=k+1>m \\geq|B \\cap M|$, so $C \\backslash M$ cannot be empty.\n\nEvery time we move a single person from Room $B$ to Room $A$, so $c(B)$ decreases by at most 1 . Hence, at the end of this loop we have $c(B)=k$.\n\nIn Room $A$ we have the clique $A \\cap M$ with size $|A \\cap M|=k$ thus $c(A) \\geq k$. We prove that there is no clique of larger size there. Let $Q \\subset A$ be an arbitrary clique. We show that $|Q| \\leq k$.\n\n\n\nIn Room $A$, and specially in set $Q$, there can be two types of competitors:\n\n- Some members of $M$. Since $M$ is a clique, they are friends with all members of $B \\cap M$.\n- Competitors which were moved to Room $A$ in Step 5 . Each of them has been in a clique with $B \\cap M$ so they are also friends with all members of $B \\cap M$.\n\nHence, all members of $Q$ are friends with all members of $B \\cap M$. Sets $Q$ and $B \\cap M$ are cliques themselves, so $Q \\cup(B \\cap M)$ is also a clique. Since $M$ is a clique of the largest size,\n$$\n|M| \\geq|Q \\cup(B \\cap M)|=|Q|+|B \\cap M|=|Q|+|M|-|A \\cap M|\n$$\ntherefore\n$$\n|Q| \\leq|A \\cap M|=k\n$$\nFinally, after Step 5 we have $c(A)=c(B)=k$.']",,True,,, 1668,Combinatorics,,"Let $\alpha<\frac{3-\sqrt{5}}{2}$ be a positive real number. Prove that there exist positive integers $n$ and $p>\alpha \cdot 2^{n}$ for which one can select $2 p$ pairwise distinct subsets $S_{1}, \ldots, S_{p}, T_{1}, \ldots, T_{p}$ of the set $\{1,2, \ldots, n\}$ such that $S_{i} \cap T_{j} \neq \varnothing$ for all $1 \leq i, j \leq p$.","['Let $k$ and $m$ be positive integers (to be determined later) and set $n=k m$. Decompose the set $\\{1,2, \\ldots, n\\}$ into $k$ disjoint subsets, each of size $m$; denote these subsets by $A_{1}, \\ldots, A_{k}$. Define the following families of sets:\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\left\\{S \\subset\\{1,2, \\ldots, n\\}: \\forall i S \\cap A_{i} \\neq \\varnothing\\right\\} \\\\\n\\mathcal{T}_{1} & =\\left\\{T \\subset\\{1,2, \\ldots, n\\}: \\exists i A_{i} \\subset T\\right\\}, \\quad \\mathcal{T}=\\mathcal{T}_{1} \\backslash \\mathcal{S} .\n\\end{aligned}\n$$\nFor each set $T \\in \\mathcal{T} \\subset \\mathcal{T}_{1}$, there exists an index $1 \\leq i \\leq k$ such that $A_{i} \\subset T$. Then for all $S \\in \\mathcal{S}$, $S \\cap T \\supset S \\cap A_{i} \\neq \\varnothing$. Hence, each $S \\in \\mathcal{S}$ and each $T \\in \\mathcal{T}$ have at least one common element.\n\nBelow we show that the numbers $m$ and $k$ can be chosen such that $|\\mathcal{S}|,|\\mathcal{T}|>\\alpha \\cdot 2^{n}$. Then, choosing $p=\\min \\{|\\mathcal{S}|,|\\mathcal{T}|\\}$, one can select the desired $2 p$ sets $S_{1}, \\ldots, S_{p}$ and $T_{1}, \\ldots, T_{p}$ from families $\\mathcal{S}$ and $\\mathcal{T}$, respectively. Since families $\\mathcal{S}$ and $\\mathcal{T}$ are disjoint, sets $S_{i}$ and $T_{j}$ will be pairwise distinct.\n\nTo count the sets $S \\in \\mathcal{S}$, observe that each $A_{i}$ has $2^{m}-1$ nonempty subsets so we have $2^{m}-1$ choices for $S \\cap A_{i}$. These intersections uniquely determine set $S$, so\n$$\n|\\mathcal{S}|=\\left(2^{m}-1\\right)^{k}\n$$\nSimilarly, if a set $H \\subset\\{1,2, \\ldots, n\\}$ does not contain a certain set $A_{i}$ then we have $2^{m}-1$ choices for $H \\cap A_{i}$ : all subsets of $A_{i}$, except $A_{i}$ itself. Therefore, the complement of $\\mathcal{T}_{1}$ contains $\\left(2^{m}-1\\right)^{k}$ sets and\n$$\n\\left|\\mathcal{T}_{1}\\right|=2^{k m}-\\left(2^{m}-1\\right)^{k} .\n$$\nNext consider the family $\\mathcal{S} \\backslash \\mathcal{T}_{1}$. If a set $S$ intersects all $A_{i}$ but does not contain any of them, then there exists $2^{m}-2$ possible values for each $S \\cap A_{i}$ : all subsets of $A_{i}$ except $\\varnothing$ and $A_{i}$. Therefore the number of such sets $S$ is $\\left(2^{m}-2\\right)^{k}$, so\n$$\n\\left|\\mathcal{S} \\backslash \\mathcal{T}_{1}\\right|=\\left(2^{m}-2\\right)^{k}\n$$\nFrom (1), (2), and (3) we obtain\n$$\n|\\mathcal{T}|=\\left|\\mathcal{T}_{1}\\right|-\\left|\\mathcal{S} \\cap \\mathcal{T}_{1}\\right|=\\left|\\mathcal{T}_{1}\\right|-\\left(|\\mathcal{S}|-\\left|\\mathcal{S} \\backslash \\mathcal{T}_{1}\\right|\\right)=2^{k m}-2\\left(2^{m}-1\\right)^{k}+\\left(2^{m}-2\\right)^{k}\n$$\nLet $\\delta=\\frac{3-\\sqrt{5}}{2}$ and $k=k(m)=\\left[2^{m} \\log \\frac{1}{\\delta}\\right]$. Then\n$$\n\\lim _{m \\rightarrow \\infty} \\frac{|\\mathcal{S}|}{2^{k m}}=\\lim _{m \\rightarrow \\infty}\\left(1-\\frac{1}{2^{m}}\\right)^{k}=\\exp \\left(-\\lim _{m \\rightarrow \\infty} \\frac{k}{2^{m}}\\right)=\\delta\n$$\nand similarly\n$$\n\\lim _{m \\rightarrow \\infty} \\frac{|\\mathcal{T}|}{2^{k m}}=1-2 \\lim _{m \\rightarrow \\infty}\\left(1-\\frac{1}{2^{m}}\\right)^{k}+\\lim _{m \\rightarrow \\infty}\\left(1-\\frac{2}{2^{m}}\\right)^{k}=1-2 \\delta+\\delta^{2}=\\delta\n$$\nHence, if $m$ is sufficiently large then $\\frac{|\\mathcal{S}|}{2^{m k}}$ and $\\frac{|\\mathcal{T}|}{2^{m k}}$ are greater than $\\alpha$ (since $\\alpha<\\delta$ ). So $|\\mathcal{S}|,|\\mathcal{T}|>\\alpha \\cdot 2^{m k}=\\alpha \\cdot 2^{n}$.']",,True,,, 1669,Combinatorics,,"Given a convex $n$-gon $P$ in the plane. For every three vertices of $P$, consider the triangle determined by them. Call such a triangle good if all its sides are of unit length. Prove that there are not more than $\frac{2}{3} n$ good triangles.","['Consider all good triangles containing a certain vertex $A$. The other two vertices of any such triangle lie on the circle $\\omega_{A}$ with unit radius and center $A$. Since $P$ is convex, all these vertices lie on an arc of angle less than $180^{\\circ}$. Let $L_{A} R_{A}$ be the shortest such arc, oriented clockwise (see Figure 1). Each of segments $A L_{A}$ and $A R_{A}$ belongs to a unique good triangle. We say that the good triangle with side $A L_{A}$ is assigned counterclockwise to $A$, and the second one, with side $A R_{A}$, is assigned clockwise to $A$. In those cases when there is a single good triangle containing vertex $A$, this triangle is assigned to $A$ twice.\n\nThere are at most two assignments to each vertex of the polygon. (Vertices which do not belong to any good triangle have no assignment.) So the number of assignments is at most $2 n$.\n\nConsider an arbitrary good triangle $A B C$, with vertices arranged clockwise. We prove that $A B C$ is assigned to its vertices at least three times. Then, denoting the number of good triangles by $t$, we obtain that the number $K$ of all assignments is at most $2 n$, while it is not less than $3 t$. Then $3 t \\leq K \\leq 2 n$, as required.\n\nActually, we prove that triangle $A B C$ is assigned either counterclockwise to $C$ or clockwise to $B$. Then, by the cyclic symmetry of the vertices, we obtain that triangle $A B C$ is assigned either counterclockwise to $A$ or clockwise to $C$, and either counterclockwise to $B$ or clockwise to $A$, providing the claim.\n\n\nFigure 1\n\n\n\nFigure 2\n\nAssume, to the contrary, that $L_{C} \\neq A$ and $R_{B} \\neq A$. Denote by $A^{\\prime}, B^{\\prime}, C^{\\prime}$ the intersection points of circles $\\omega_{A}, \\omega_{B}$ and $\\omega_{C}$, distinct from $A, B, C$ (see Figure 2). Let $C L_{C} L_{C}^{\\prime}$ be the good triangle containing $C L_{C}$. Observe that the angle of arc $L_{C} A$ is less than $120^{\\circ}$. Then one of the points $L_{C}$ and $L_{C}^{\\prime}$ belongs to $\\operatorname{arc} B^{\\prime} A$ of $\\omega_{C}$; let this point be $X$. In the case when $L_{C}=B^{\\prime}$ and $L_{C}^{\\prime}=A$, choose $X=B^{\\prime}$.\n\nAnalogously, considering the good triangle $B R_{B}^{\\prime} R_{B}$ which contains $B R_{B}$ as an edge, we see that one of the points $R_{B}$ and $R_{B}^{\\prime}$ lies on $\\operatorname{arc} A C^{\\prime}$ of $\\omega_{B}$. Denote this point by $Y, Y \\neq A$. Then angles $X A Y, Y A B, B A C$ and $C A X$ (oriented clockwise) are not greater than $180^{\\circ}$. Hence, point $A$ lies in quadrilateral $X Y B C$ (either in its interior or on segment $X Y$ ). This is impossible, since all these five points are vertices of $P$.\n\nHence, each good triangle has at least three assignments, and the statement is proved.']",,True,,, 1670,Geometry,,"In triangle $A B C$, the angle bisector at vertex $C$ intersects the circumcircle and the perpendicular bisectors of sides $B C$ and $C A$ at points $R, P$, and $Q$, respectively. The midpoints of $B C$ and $C A$ are $S$ and $T$, respectively. Prove that triangles $R Q T$ and $R P S$ have the same area.","['. If $A C=B C$ then triangle $A B C$ is isosceles, triangles $R Q T$ and $R P S$ are symmetric about the bisector $C R$ and the statement is trivial. If $A C \\neq B C$ then it can be assumed without loss of generality that $A C\n\nDenote the circumcenter by $O$. The right triangles $C T Q$ and $C S P$ have equal angles at vertex $C$, so they are similar, $\\angle C P S=\\angle C Q T=\\angle O Q P$ and\n$$\n\\frac{Q T}{P S}=\\frac{C Q}{C P}\n$$\nLet $\\ell$ be the perpendicular bisector of chord $C R$; of course, $\\ell$ passes through the circumcenter $O$. Due to the equal angles at $P$ and $Q$, triangle $O P Q$ is isosceles with $O P=O Q$. Then line $\\ell$ is the axis of symmetry in this triangle as well. Therefore, points $P$ and $Q$ lie symmetrically on line segment $C R$,\n$$\nR P=C Q \\quad \\text { and } \\quad R Q=C P .\n$$\nTriangles $R Q T$ and $R P S$ have equal angles at vertices $Q$ and $P$, respectively. Then\n$$\n\\frac{\\operatorname{area}(R Q T)}{\\operatorname{area}(R P S)}=\\frac{\\frac{1}{2} \\cdot R Q \\cdot Q T \\cdot \\sin \\angle R Q T}{\\frac{1}{2} \\cdot R P \\cdot P S \\cdot \\sin \\angle R P S}=\\frac{R Q}{R P} \\cdot \\frac{Q T}{P S}\n$$\nSubstituting (1) and (2),\n$$\n\\frac{\\operatorname{area}(R Q T)}{\\operatorname{area}(R P S)}=\\frac{R Q}{R P} \\cdot \\frac{Q T}{P S}=\\frac{C P}{C Q} \\cdot \\frac{C Q}{C P}=1\n$$\nHence, $\\operatorname{area}(R Q T)=\\operatorname{area}(R S P)$.', '. Assume again $A C\n\nConsider the rotation around point $O$ by angle $\\gamma$. This transform moves $A$ to $R, R$ to $B$ and $Q$ to $P$; hence triangles $R Q A$ and $B P R$ are congruent and they have the same area.\n\nTriangles $R Q T$ and $R Q A$ have $R Q$ as a common side, so the ratio between their areas is\n$$\n\\frac{\\operatorname{area}(R Q T)}{\\operatorname{area}(R Q A)}=\\frac{d(T, C R)}{d(A, C R)}=\\frac{C T}{C A}=\\frac{1}{2}\n$$\n$(d(X, Y Z)$ denotes the distance between point $X$ and line $Y Z)$.\n\nIt can be obtained similarly that\n$$\n\\frac{\\operatorname{area}(R P S)}{\\operatorname{area}(B P R)}=\\frac{C S}{C B}=\\frac{1}{2}\n$$\nNow the proof can be completed as\n$$\n\\operatorname{area}(R Q T)=\\frac{1}{2} \\operatorname{area}(R Q A)=\\frac{1}{2} \\operatorname{area}(B P R)=\\operatorname{area}(R P S)\n$$']",,True,,, 1671,Geometry,,"Given an isosceles triangle $A B C$ with $A B=A C$. The midpoint of side $B C$ is denoted by $M$. Let $X$ be a variable point on the shorter arc $M A$ of the circumcircle of triangle $A B M$. Let $T$ be the point in the angle domain $B M A$, for which $\angle T M X=90^{\circ}$ and $T X=B X$. Prove that $\angle M T B-\angle C T M$ does not depend on $X$.","['. Let $N$ be the midpoint of segment $B T$ (see Figure 1). Line $X N$ is the axis of symmetry in the isosceles triangle $B X T$, thus $\\angle T N X=90^{\\circ}$ and $\\angle B X N=\\angle N X T$. Moreover, in triangle $B C T$, line $M N$ is the midline parallel to $C T$; hence $\\angle C T M=\\angle N M T$.\n\nDue to the right angles at points $M$ and $N$, these points lie on the circle with diameter $X T$. Therefore,\n$$\n\\angle M T B=\\angle M T N=\\angle M X N \\quad \\text { and } \\quad \\angle C T M=\\angle N M T=\\angle N X T=\\angle B X N .\n$$\nHence\n$$\n\\angle M T B-\\angle C T M=\\angle M X N-\\angle B X N=\\angle M X B=\\angle M A B\n$$\nwhich does not depend on $X$.\n\n\n\nFigure 1\n\n\n\nFigure 2', '. Let $S$ be the reflection of point $T$ over $M$ (see Figure 2). Then $X M$ is the perpendicular bisector of $T S$, hence $X B=X T=X S$, and $X$ is the circumcenter of triangle $B S T$. Moreover, $\\angle B S M=\\angle C T M$ since they are symmetrical about $M$. Then\n$$\n\\angle M T B-\\angle C T M=\\angle S T B-\\angle B S T=\\frac{\\angle S X B-\\angle B X T}{2} .\n$$\nObserve that $\\angle S X B=\\angle S X T-\\angle B X T=2 \\angle M X T-\\angle B X T$, so\n$$\n\\angle M T B-\\angle C T M=\\frac{2 \\angle M X T-2 \\angle B X T}{2}=\\angle M X B=\\angle M A B\n$$\nwhich is constant.']",,True,,, 1672,Geometry,,"The diagonals of a trapezoid $A B C D$ intersect at point $P$. Point $Q$ lies between the parallel lines $B C$ and $A D$ such that $\angle A Q D=\angle C Q B$, and line $C D$ separates points $P$ and $Q$. Prove that $\angle B Q P=\angle D A Q$.","['Let $t=\\frac{A D}{B C}$. Consider the homothety $h$ with center $P$ and scale $-t$. Triangles $P D A$ and $P B C$ are similar with ratio $t$, hence $h(B)=D$ and $h(C)=A$.\n\n\n\nLet $Q^{\\prime}=h(Q)$ (see Figure 1). Then points $Q, P$ and $Q^{\\prime}$ are obviously collinear. Points $Q$ and $P$ lie on the same side of $A D$, as well as on the same side of $B C$; hence $Q^{\\prime}$ and $P$ are also on the same side of $h(B C)=A D$, and therefore $Q$ and $Q^{\\prime}$ are on the same side of $A D$. Moreover, points $Q$ and $C$ are on the same side of $B D$, while $Q^{\\prime}$ and $A$ are on the opposite side (see Figure above).\n\nBy the homothety, $\\angle A Q^{\\prime} D=\\angle C Q B=\\angle A Q D$, hence quadrilateral $A Q^{\\prime} Q D$ is cyclic. Then\n$$\n\\angle D A Q=\\angle D Q^{\\prime} Q=\\angle D Q^{\\prime} P=\\angle B Q P\n$$\n(the latter equality is valid by the homothety again).']",,True,,, 1673,Geometry,,"Consider five points $A, B, C, D, E$ such that $A B C D$ is a parallelogram and $B C E D$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$, and let $\ell$ intersect segment $D C$ and line $B C$ at points $F$ and $G$, respectively. Suppose that $E F=E G=E C$. Prove that $\ell$ is the bisector of angle $D A B$.","['If $C F=C G$, then $\\angle F G C=\\angle G F C$, hence $\\angle G A B=\\angle G F C=\\angle F G C=\\angle F A D$, and $\\ell$ is a bisector.\n\nAssume that $C F\\sqrt{E C^{2}-L C^{2}}=L E .\n$$\nSince quadrilateral $B C E D$ is cyclic, we have $\\angle E D C=\\angle E B C$, so the right triangles $B E L$ and $D E K$ are similar. Then $K E>L E$ implies $D K>B L$, and hence\n$$\nD F=D K-K F>B L-L C=B C=A D .\n$$\nBut triangles $A D F$ and $G C F$ are similar, so we have $1>\\frac{A D}{D F}=\\frac{G C}{C F}$; this contradicts our assumption.\n\nThe case $C F>G C$ is completely similar. We consequently obtain the converse inequalities $K F>L C, K E']",,True,,, 1674,Geometry,,"Let $A B C$ be a fixed triangle, and let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $B C, C A, A B$, respectively. Let $P$ be a variable point on the circumcircle. Let lines $P A_{1}, P B_{1}, P C_{1}$ meet the circumcircle again at $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Assume that the points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$ are distinct, and lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ form a triangle. Prove that the area of this triangle does not depend on $P$.","[""Let $A_{0}, B_{0}, C_{0}$ be the points of intersection of the lines $A A^{\\prime}, B B^{\\prime}$ and $C C^{\\prime}$ (see Figure). We claim that area $\\left(A_{0} B_{0} C_{0}\\right)=\\frac{1}{2} \\operatorname{area}(A B C)$, hence it is constant.\n\nConsider the inscribed hexagon $A B C C^{\\prime} P A^{\\prime}$. By Pascal's theorem, the points of intersection of its opposite sides (or of their extensions) are collinear. These points are $A B \\cap C^{\\prime} P=C_{1}$, $B C \\cap P A^{\\prime}=A_{1}, C C^{\\prime} \\cap A^{\\prime} A=B_{0}$. So point $B_{0}$ lies on the midline $A_{1} C_{1}$ of triangle $A B C$. Analogously, points $A_{0}$ and $C_{0}$ lie on lines $B_{1} C_{1}$ and $A_{1} B_{1}$, respectively.\n\nLines $A C$ and $A_{1} C_{1}$ are parallel, so triangles $B_{0} C_{0} A_{1}$ and $A C_{0} B_{1}$ are similar; hence we have\n$$\n\\frac{B_{0} C_{0}}{A C_{0}}=\\frac{A_{1} C_{0}}{B_{1} C_{0}}\n$$\nAnalogously, from $B C \\| B_{1} C_{1}$ we obtain\n$$\n\\frac{A_{1} C_{0}}{B_{1} C_{0}}=\\frac{B C_{0}}{A_{0} C_{0}}\n$$\nCombining these equalities, we get\n$$\n\\frac{B_{0} C_{0}}{A C_{0}}=\\frac{B C_{0}}{A_{0} C_{0}}\n$$\nor\n$$\nA_{0} C_{0} \\cdot B_{0} C_{0}=A C_{0} \\cdot B C_{0}\n$$\nHence we have\n\n\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\frac{1}{2} A_{0} C_{0} \\cdot B_{0} C_{0} \\sin \\angle A_{0} C_{0} B_{0}=\\frac{1}{2} A C_{0} \\cdot B C_{0} \\sin \\angle A C_{0} B=\\operatorname{area}\\left(A B C_{0}\\right) \\text {. }\n$$\nSince $C_{0}$ lies on the midline, we have $d\\left(C_{0}, A B\\right)=\\frac{1}{2} d(C, A B)$ (we denote by $d(X, Y Z)$ the distance between point $X$ and line $Y Z)$. Then we obtain\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\operatorname{area}\\left(A B C_{0}\\right)=\\frac{1}{2} A B \\cdot d\\left(C_{0}, A B\\right)=\\frac{1}{4} A B \\cdot d(C, A B)=\\frac{1}{2} \\operatorname{area}(A B C)\n$$"", '. Again, we prove that $\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\frac{1}{2} \\operatorname{area}(A B C)$.\n\nWe can assume that $P$ lies on arc $A C$. Mark a point $L$ on side $A C$ such that $\\angle C B L=$ $\\angle P B A$; then $\\angle L B A=\\angle C B A-\\angle C B L=\\angle C B A-\\angle P B A=\\angle C B P$. Note also that $\\angle B A L=\\angle B A C=\\angle B P C$ and $\\angle L C B=\\angle A P B$. Hence, triangles $B A L$ and $B P C$ are similar, and so are triangles $L C B$ and $A P B$.\n\nAnalogously, mark points $K$ and $M$ respectively on the extensions of sides $C B$ and $A B$ beyond point $B$, such that $\\angle K A B=\\angle C A P$ and $\\angle B C M=\\angle P C A$. For analogous reasons, $\\angle K A C=\\angle B A P$ and $\\angle A C M=\\angle P C B$. Hence $\\triangle A B K \\sim \\triangle A P C \\sim \\triangle M B C, \\triangle A C K \\sim$ $\\triangle A P B$, and $\\triangle M A C \\sim \\triangle B P C$. From these similarities, we have $\\angle C M B=\\angle K A B=\\angle C A P$, while we have seen that $\\angle C A P=\\angle C B P=\\angle L B A$. Hence, $A K\\|B L\\| C M$.\n\n\n\n\n\nLet line $C C^{\\prime}$ intersect $B L$ at point $X$. Note that $\\angle L C X=\\angle A C C^{\\prime}=\\angle A P C^{\\prime}=\\angle A P C_{1}$, and $P C_{1}$ is a median in triangle $A P B$. Since triangles $A P B$ and $L C B$ are similar, $C X$ is a median in triangle $L C B$, and $X$ is a midpoint of $B L$. For the same reason, $A A^{\\prime}$ passes through this midpoint, so $X=B_{0}$. Analogously, $A_{0}$ and $C_{0}$ are the midpoints of $A K$ and $C M$.\n\nNow, from $A A_{0} \\| C C_{0}$, we have\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\operatorname{area}\\left(A C_{0} A_{0}\\right)-\\operatorname{area}\\left(A B_{0} A_{0}\\right)=\\operatorname{area}\\left(A C A_{0}\\right)-\\operatorname{area}\\left(A B_{0} A_{0}\\right)=\\operatorname{area}\\left(A C B_{0}\\right)\n$$\nFinally,\n$$\n\\operatorname{area}\\left(A_{0} B_{0} C_{0}\\right)=\\operatorname{area}\\left(A C B_{0}\\right)=\\frac{1}{2} B_{0} L \\cdot A C \\sin A L B_{0}=\\frac{1}{4} B L \\cdot A C \\sin A L B=\\frac{1}{2} \\operatorname{area}(A B C)\n$$']",,True,,, 1675,Geometry,,"Determine the smallest positive real number $k$ with the following property. Let $A B C D$ be a convex quadrilateral, and let points $A_{1}, B_{1}, C_{1}$ and $D_{1}$ lie on sides $A B, B C$, $C D$ and $D A$, respectively. Consider the areas of triangles $A A_{1} D_{1}, B B_{1} A_{1}, C C_{1} B_{1}$, and $D D_{1} C_{1}$; let $S$ be the sum of the two smallest ones, and let $S_{1}$ be the area of quadrilateral $A_{1} B_{1} C_{1} D_{1}$. Then we always have $k S_{1} \geq S$.","['Throughout the solution, triangles $A A_{1} D_{1}, B B_{1} A_{1}, C C_{1} B_{1}$, and $D D_{1} C_{1}$ will be referred to as border triangles. We will denote by $[\\mathcal{R}]$ the area of a region $\\mathcal{R}$.\n\nFirst, we show that $k \\geq 1$. Consider a triangle $A B C$ with unit area; let $A_{1}, B_{1}, K$ be the midpoints of its sides $A B, B C, A C$, respectively. Choose a point $D$ on the extension of $B K$, close to $K$. Take points $C_{1}$ and $D_{1}$ on sides $C D$ and $D A$ close to $D$ (see Figure 1). We have $\\left[B B_{1} A_{1}\\right]=\\frac{1}{4}$. Moreover, as $C_{1}, D_{1}, D \\rightarrow K$, we get $\\left[A_{1} B_{1} C_{1} D_{1}\\right] \\rightarrow\\left[A_{1} B_{1} K\\right]=\\frac{1}{4}$, $\\left[A A_{1} D_{1}\\right] \\rightarrow\\left[A A_{1} K\\right]=\\frac{1}{4},\\left[C C_{1} B_{1}\\right] \\rightarrow\\left[C K B_{1}\\right]=\\frac{1}{4}$ and $\\left[D D_{1} C_{1}\\right] \\rightarrow 0$. Hence, the sum of the two smallest areas of border triangles tends to $\\frac{1}{4}$, as well as $\\left[A_{1} B_{1} C_{1} D_{1}\\right]$; therefore, their ratio tends to 1 , and $k \\geq 1$.\n\nWe are left to prove that $k=1$ satisfies the desired property.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nLemma. Let points $A_{1}, B_{1}, C_{1}$ lie respectively on sides $B C, C A, A B$ of a triangle $A B C$. Then $\\left[A_{1} B_{1} C_{1}\\right] \\geq \\min \\left\\{\\left[A C_{1} B_{1}\\right],\\left[B A_{1} C_{1}\\right],\\left[C B_{1} A_{1}\\right]\\right\\}$.\n\nProof. Let $A^{\\prime}, B^{\\prime}, C^{\\prime}$ be the midpoints of sides $B C, C A$ and $A B$, respectively.\n\nSuppose that two of points $A_{1}, B_{1}, C_{1}$ lie in one of triangles $A C^{\\prime} B^{\\prime}, B A^{\\prime} C^{\\prime}$ and $C B^{\\prime} A^{\\prime}$ (for convenience, let points $B_{1}$ and $C_{1}$ lie in triangle $A C^{\\prime} B^{\\prime}$; see Figure 2). Let segments $B_{1} C_{1}$ and $A A_{1}$ intersect at point $X$. Then $X$ also lies in triangle $A C^{\\prime} B^{\\prime}$. Hence $A_{1} X \\geq A X$, and we have\n$$\n\\frac{\\left[A_{1} B_{1} C_{1}\\right]}{\\left[A C_{1} B_{1}\\right]}=\\frac{\\frac{1}{2} A_{1} X \\cdot B_{1} C_{1} \\cdot \\sin \\angle A_{1} X C_{1}}{\\frac{1}{2} A X \\cdot B_{1} C_{1} \\cdot \\sin \\angle A X B_{1}}=\\frac{A_{1} X}{A X} \\geq 1\n$$\nas required.\n\nOtherwise, each one of triangles $A C^{\\prime} B^{\\prime}, B A^{\\prime} C^{\\prime}, C B^{\\prime} A^{\\prime}$ contains exactly one of points $A_{1}$, $B_{1}, C_{1}$, and we can assume that $B A_{1}1$; also, lines $A_{1} C^{\\prime}$ and $C A$ intersect at a point $Z$ on the extension of $C A$ beyond point $A$, hence $\\frac{\\left[A_{1} B_{1} C^{\\prime}\\right]}{\\left[A_{1} B^{\\prime} C^{\\prime}\\right]}=\\frac{B_{1} Z}{B^{\\prime} Z}>1$. Finally, since $A_{1} A^{\\prime} \\| B^{\\prime} C^{\\prime}$, we have $\\left[A_{1} B_{1} C_{1}\\right]>\\left[A_{1} B_{1} C^{\\prime}\\right]>\\left[A_{1} B^{\\prime} C^{\\prime}\\right]=\\left[A^{\\prime} B^{\\prime} C^{\\prime}\\right]=\\frac{1}{4}[A B C]$.\n\n\n\nNow, from $\\left[A_{1} B_{1} C_{1}\\right]+\\left[A C_{1} B_{1}\\right]+\\left[B A_{1} C_{1}\\right]+\\left[C B_{1} A_{1}\\right]=[A B C]$ we obtain that one of the remaining triangles $A C_{1} B_{1}, B A_{1} C_{1}, C B_{1} A_{1}$ has an area less than $\\frac{1}{4}[A B C]$, so it is less than $\\left[A_{1} B_{1} C_{1}\\right]$.\n\nNow we return to the problem. We say that triangle $A_{1} B_{1} C_{1}$ is small if $\\left[A_{1} B_{1} C_{1}\\right]$ is less than each of $\\left[B B_{1} A_{1}\\right]$ and $\\left[C C_{1} B_{1}\\right]$; otherwise this triangle is big (the similar notion is introduced for triangles $B_{1} C_{1} D_{1}, C_{1} D_{1} A_{1}, D_{1} A_{1} B_{1}$ ). If both triangles $A_{1} B_{1} C_{1}$ and $C_{1} D_{1} A_{1}$ are big, then $\\left[A_{1} B_{1} C_{1}\\right]$ is not less than the area of some border triangle, and $\\left[C_{1} D_{1} A_{1}\\right]$ is not less than the area of another one; hence, $S_{1}=\\left[A_{1} B_{1} C_{1}\\right]+\\left[C_{1} D_{1} A_{1}\\right] \\geq S$. The same is valid for the pair of $B_{1} C_{1} D_{1}$ and $D_{1} A_{1} B_{1}$. So it is sufficient to prove that in one of these pairs both triangles are big.\n\nSuppose the contrary. Then there is a small triangle in each pair. Without loss of generality, assume that triangles $A_{1} B_{1} C_{1}$ and $D_{1} A_{1} B_{1}$ are small. We can assume also that $\\left[A_{1} B_{1} C_{1}\\right] \\leq$ $\\left[D_{1} A_{1} B_{1}\\right]$. Note that in this case ray $D_{1} C_{1}$ intersects line $B C$.\n\nConsider two cases.\n\n\n\nFigure 4\n\n\n\nFigure 5\n\nCase 1. Ray $C_{1} D_{1}$ intersects line $A B$ at some point $K$. Let ray $D_{1} C_{1}$ intersect line $B C$ at point $L$ (see Figure 4). Then we have $\\left[A_{1} B_{1} C_{1}\\right]<\\left[C C_{1} B_{1}\\right]<\\left[L C_{1} B_{1}\\right],\\left[A_{1} B_{1} C_{1}\\right]<\\left[B B_{1} A_{1}\\right]$ (both - since $\\left[A_{1} B_{1} C_{1}\\right]$ is small), and $\\left[A_{1} B_{1} C_{1}\\right] \\leq\\left[D_{1} A_{1} B_{1}\\right]<\\left[A A_{1} D_{1}\\right]<\\left[K A_{1} D_{1}\\right]<\\left[K A_{1} C_{1}\\right]$ (since triangle $D_{1} A_{1} B_{1}$ is small). This contradicts the Lemma, applied for triangle $A_{1} B_{1} C_{1}$ inside $L K B$.\n\nCase 2. Ray $C_{1} D_{1}$ does not intersect $A B$. Then choose a ""sufficiently far"" point $K$ on ray $B A$ such that $\\left[K A_{1} C_{1}\\right]>\\left[A_{1} B_{1} C_{1}\\right]$, and that ray $K C_{1}$ intersects line $B C$ at some point $L$ (see Figure 5). Since ray $C_{1} D_{1}$ does not intersect line $A B$, the points $A$ and $D_{1}$ are on different sides of $K L$; then $A$ and $D$ are also on different sides, and $C$ is on the same side as $A$ and $B$. Then analogously we have $\\left[A_{1} B_{1} C_{1}\\right]<\\left[C C_{1} B_{1}\\right]<\\left[L C_{1} B_{1}\\right]$ and $\\left[A_{1} B_{1} C_{1}\\right]<\\left[B B_{1} A_{1}\\right]$ since triangle $A_{1} B_{1} C_{1}$ is small. This (together with $\\left[A_{1} B_{1} C_{1}\\right]<\\left[K A_{1} C_{1}\\right]$ ) contradicts the Lemma again.']",['$k=1$'],False,,Numerical, 1676,Geometry,,"Given an acute triangle $A B C$ with angles $\alpha, \beta$ and $\gamma$ at vertices $A, B$ and $C$, respectively, such that $\beta>\gamma$. Point $I$ is the incenter, and $R$ is the circumradius. Point $D$ is the foot of the altitude from vertex $A$. Point $K$ lies on line $A D$ such that $A K=2 R$, and $D$ separates $A$ and $K$. Finally, lines $D I$ and $K I$ meet sides $A C$ and $B C$ at $E$ and $F$, respectively. Prove that if $I E=I F$ then $\beta \leq 3 \gamma$.","['We first prove that\n$$\n\\angle K I D=\\frac{\\beta-\\gamma}{2} \\tag{1}\n$$\neven without the assumption that $I E=I F$. Then we will show that the statement of the problem is a consequence of this fact.\n\nDenote the circumcenter by $O$. On the circumcircle, let $P$ be the point opposite to $A$, and let the angle bisector $A I$ intersect the circle again at $M$. Since $A K=A P=2 R$, triangle $A K P$ is isosceles. It is known that $\\angle B A D=\\angle C A O$, hence $\\angle D A I=\\angle B A I-\\angle B A D=\\angle C A I-$ $\\angle C A O=\\angle O A I$, and $A M$ is the bisector line in triangle $A K P$. Therefore, points $K$ and $P$ are symmetrical about $A M$, and $\\angle A M K=\\angle A M P=90^{\\circ}$. Thus, $M$ is the midpoint of $K P$, and $A M$ is the perpendicular bisector of $K P$.\n\n\n\nDenote the perpendicular feet of incenter $I$ on lines $B C, A C$, and $A D$ by $A_{1}, B_{1}$, and $T$, respectively. Quadrilateral $D A_{1} I T$ is a rectangle, hence $T D=I A_{1}=I B_{1}$.\n\nDue to the right angles at $T$ and $B_{1}$, quadrilateral $A B_{1} I T$ is cyclic. Hence $\\angle B_{1} T I=$ $\\angle B_{1} A I=\\angle C A M=\\angle B A M=\\angle B P M$ and $\\angle I B_{1} T=\\angle I A T=\\angle M A K=\\angle M A P=$ $\\angle M B P$. Therefore, triangles $B_{1} T I$ and $B P M$ are similar and $\\frac{I T}{I B_{1}}=\\frac{M P}{M B}$.\n\nIt is well-known that $M B=M C=M I$. Then right triangles $I T D$ and $K M I$ are also\n\n\n\nsimilar, because $\\frac{I T}{T D}=\\frac{I T}{I B_{1}}=\\frac{M P}{M B}=\\frac{K M}{M I}$. Hence, $\\angle K I M=\\angle I D T=\\angle I D A$, and\n$$\n\\angle K I D=\\angle M I D-\\angle K I M=(\\angle I A D+\\angle I D A)-\\angle I D A=\\angle I A D .\n$$\nFinally, from the right triangle $A D B$ we can compute\n$$\n\\angle K I D=\\angle I A D=\\angle I A B-\\angle D A B=\\frac{\\alpha}{2}-\\left(90^{\\circ}-\\beta\\right)=\\frac{\\alpha}{2}-\\frac{\\alpha+\\beta+\\gamma}{2}+\\beta=\\frac{\\beta-\\gamma}{2} .\n$$\nNow let us turn to the statement and suppose that $I E=I F$. Since $I A_{1}=I B_{1}$, the right triangles $I E B_{1}$ and $I F A_{1}$ are congruent and $\\angle I E B_{1}=\\angle I F A_{1}$. Since $\\beta>\\gamma, A_{1}$ lies in the interior of segment $C D$ and $F$ lies in the interior of $A_{1} D$. Hence, $\\angle I F C$ is acute. Then two cases are possible depending on the order of points $A, C, B_{1}$ and $E$.\n\n\nIf point $E$ lies between $C$ and $B_{1}$ then $\\angle I F C=\\angle I E A$, hence quadrilateral $C E I F$ is cyclic and $\\angle F C E=180^{\\circ}-\\angle E I F=\\angle K I D$. By (1), in this case we obtain $\\angle F C E=\\gamma=\\angle K I D=$ $\\frac{\\beta-\\gamma}{2}$ and $\\beta=3 \\gamma$.\n\nOtherwise, if point $E$ lies between $A$ and $B_{1}$, quadrilateral $C E I F$ is a deltoid such that $\\angle I E C=\\angle I F C<90^{\\circ}$. Then we have $\\angle F C E>180^{\\circ}-\\angle E I F=\\angle K I D$. Therefore, $\\angle F C E=\\gamma>\\angle K I D=\\frac{\\beta-\\gamma}{2}$ and $\\beta<3 \\gamma$.', '$$\n\\angle K I D=\\frac{\\beta-\\gamma}{2} \\tag{1}\n$$\nWe give a different proof for (1). Then the solution can be finished in the same way as above.\n\nDefine points $M$ and $P$ again; it can be proved in the same way that $A M$ is the perpendicular bisector of $K P$. Let $J$ be the center of the excircle touching side $B C$. It is well-known that points $B, C, I, J$ lie on a circle with center $M$; denote this circle by $\\omega_{1}$.\n\nLet $B^{\\prime}$ be the reflection of point $B$ about the angle bisector $A M$. By the symmetry, $B^{\\prime}$ is the second intersection point of circle $\\omega_{1}$ and line $A C$. Triangles $P B A$ and $K B^{\\prime} A$ are symmetrical\n\n\n\nwith respect to line $A M$, therefore $\\angle K B^{\\prime} A=\\angle P B A=90^{\\circ}$. By the right angles at $D$ and $B^{\\prime}$, points $K, D, B^{\\prime}, C$ are concyclic and\n$$\nA D \\cdot A K=A B^{\\prime} \\cdot A C\n$$\nFrom the cyclic quadrilateral $I J C B^{\\prime}$ we obtain $A B^{\\prime} \\cdot A C=A I \\cdot A J$ as well, therefore\n$$\nA D \\cdot A K=A B^{\\prime} \\cdot A C=A I \\cdot A J\n$$\nand points $I, J, K, D$ are also concyclic. Denote circle $I D K J$ by $\\omega_{2}$.\n\n\n\nLet $N$ be the point on circle $\\omega_{2}$ which is opposite to $K$. Since $\\angle N D K=90^{\\circ}=\\angle C D K$, point $N$ lies on line $B C$. Point $M$, being the center of circle $\\omega_{1}$, is the midpoint of segment $I J$, and $K M$ is perpendicular to $I J$. Therefore, line $K M$ is the perpendicular bisector of $I J$ and hence it passes through $N$.\n\nFrom the cyclic quadrilateral $I D K N$ we obtain\n$$\n\\angle K I D=\\angle K N D=90^{\\circ}-\\angle D K N=90^{\\circ}-\\angle A K M=\\angle M A K=\\frac{\\beta-\\gamma}{2}\n$$']",,True,,, 1677,Geometry,,"Point $P$ lies on side $A B$ of a convex quadrilateral $A B C D$. Let $\omega$ be the incircle of triangle $C P D$, and let $I$ be its incenter. Suppose that $\omega$ is tangent to the incircles of triangles $A P D$ and $B P C$ at points $K$ and $L$, respectively. Let lines $A C$ and $B D$ meet at $E$, and let lines $A K$ and $B L$ meet at $F$. Prove that points $E, I$, and $F$ are collinear.","['Let $\\Omega$ be the circle tangent to segment $A B$ and to rays $A D$ and $B C$; let $J$ be its center. We prove that points $E$ and $F$ lie on line $I J$.\n\n\n\nDenote the incircles of triangles $A D P$ and $B C P$ by $\\omega_{A}$ and $\\omega_{B}$. Let $h_{1}$ be the homothety with a negative scale taking $\\omega$ to $\\Omega$. Consider this homothety as the composition of two homotheties: one taking $\\omega$ to $\\omega_{A}$ (with a negative scale and center $K$ ), and another one taking $\\omega_{A}$ to $\\Omega$ (with a positive scale and center $A$ ). It is known that in such a case the three centers of homothety are collinear (this theorem is also referred to as the theorem on the three similitude centers). Hence, the center of $h_{1}$ lies on line $A K$. Analogously, it also lies on $B L$, so this center is $F$. Hence, $F$ lies on the line of centers of $\\omega$ and $\\Omega$, i. e. on $I J$ (if $I=J$, then $F=I$ as well, and the claim is obvious).\n\nConsider quadrilateral $A P C D$ and mark the equal segments of tangents to $\\omega$ and $\\omega_{A}$ (see the figure below to the left). Since circles $\\omega$ and $\\omega_{A}$ have a common point of tangency with $P D$, one can easily see that $A D+P C=A P+C D$. So, quadrilateral $A P C D$ is circumscribed; analogously, circumscribed is also quadrilateral $B C D P$. Let $\\Omega_{A}$ and $\\Omega_{B}$ respectively be their incircles.\n\n\n\n\nConsider the homothety $h_{2}$ with a positive scale taking $\\omega$ to $\\Omega$. Consider $h_{2}$ as the composition of two homotheties: taking $\\omega$ to $\\Omega_{A}$ (with a positive scale and center $C$ ), and taking $\\Omega_{A}$ to $\\Omega$ (with a positive scale and center $A$ ), respectively. So the center of $h_{2}$ lies on line $A C$. By analogous reasons, it lies also on $B D$, hence this center is $E$. Thus, $E$ also lies on the line of centers $I J$, and the claim is proved.']",,True,,, 1678,Number Theory,,"Find all pairs $(k, n)$ of positive integers for which $7^{k}-3^{n}$ divides $k^{4}+n^{2}$.","['Suppose that a pair $(k, n)$ satisfies the condition of the problem. Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even, hence $k$ and $n$ have the same parity. If $k$ and $n$ are odd, then $k^{4}+n^{2} \\equiv 1+1=2(\\bmod 4)$, while $7^{k}-3^{n} \\equiv 7-3 \\equiv 0(\\bmod 4)$, so $k^{4}+n^{2}$ cannot be divisible by $7^{k}-3^{n}$. Hence, both $k$ and $n$ must be even.\n\nWrite $k=2 a, n=2 b$. Then $7^{k}-3^{n}=7^{2 a}-3^{2 b}=\\frac{7^{a}-3^{b}}{2} \\cdot 2\\left(7^{a}+3^{b}\\right)$, and both factors are integers. So $2\\left(7^{a}+3^{b}\\right) \\mid 7^{k}-3^{n}$ and $7^{k}-3^{n} \\mid k^{4}+n^{2}=2\\left(8 a^{4}+2 b^{2}\\right)$, hence\n$$\n7^{a}+3^{b} \\leq 8 a^{4}+2 b^{2}\n$$\nWe prove by induction that $8 a^{4}<7^{a}$ for $a \\geq 4,2 b^{2}<3^{b}$ for $b \\geq 1$ and $2 b^{2}+9 \\leq 3^{b}$ for $b \\geq 3$. In the initial cases $a=4, b=1, b=2$ and $b=3$ we have $8 \\cdot 4^{4}=2048<7^{4}=2401,2<3$, $2 \\cdot 2^{2}=8<3^{2}=9$ and $2 \\cdot 3^{2}+9=3^{3}=27$, respectively.\n\nIf $8 a^{4}<7^{a}(a \\geq 4)$ and $2 b^{2}+9 \\leq 3^{b}(b \\geq 3)$, then\n$$\n\\begin{aligned}\n8(a+1)^{4} & =8 a^{4}\\left(\\frac{a+1}{a}\\right)^{4}<7^{a}\\left(\\frac{5}{4}\\right)^{4}=7^{a} \\frac{625}{256}<7^{a+1} \\quad \\text { and } \\\\\n2(b+1)^{2}+9 & <\\left(2 b^{2}+9\\right)\\left(\\frac{b+1}{b}\\right)^{2} \\leq 3^{b}\\left(\\frac{4}{3}\\right)^{2}=3^{b} \\frac{16}{9}<3^{b+1},\n\\end{aligned}\n$$\nas desired.\n\nFor $a \\geq 4$ we obtain $7^{a}+3^{b}>8 a^{4}+2 b^{2}$ and inequality (1) cannot hold. Hence $a \\leq 3$, and three cases are possible.\n\nCase 1: $a=1$. Then $k=2$ and $8+2 b^{2} \\geq 7+3^{b}$, thus $2 b^{2}+1 \\geq 3^{b}$. This is possible only if $b \\leq 2$. If $b=1$ then $n=2$ and $\\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\\frac{2^{4}+2^{2}}{7^{2}-3^{2}}=\\frac{1}{2}$, which is not an integer. If $b=2$ then $n=4$ and $\\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\\frac{2^{4}+4^{2}}{7^{2}-3^{4}}=-1$, so $(k, n)=(2,4)$ is a solution.\n\nCase 2: $a=2$. Then $k=4$ and $k^{4}+n^{2}=256+4 b^{2} \\geq\\left|7^{4}-3^{n}\\right|=\\left|49-3^{b}\\right| \\cdot\\left(49+3^{b}\\right)$. The smallest value of the first factor is 22 , attained at $b=3$, so $128+2 b^{2} \\geq 11\\left(49+3^{b}\\right)$, which is impossible since $3^{b}>2 b^{2}$.\n\nCase 3: $a=3$. Then $k=6$ and $k^{4}+n^{2}=1296+4 b^{2} \\geq\\left|7^{6}-3^{n}\\right|=\\left|343-3^{b}\\right| \\cdot\\left(343+3^{b}\\right)$. Analogously, $\\left|343-3^{b}\\right| \\geq 100$ and we have $324+b^{2} \\geq 25\\left(343+3^{b}\\right)$, which is impossible again.\n\nWe find that there exists a unique solution $(k, n)=(2,4)$.']","['$(2,4)$']",False,,Tuple, 1679,Number Theory,,"Let $b, n>1$ be integers. Suppose that for each $k>1$ there exists an integer $a_{k}$ such that $b-a_{k}^{n}$ is divisible by $k$. Prove that $b=A^{n}$ for some integer $A$.","['Let the prime factorization of $b$ be $b=p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$, where $p_{1}, \\ldots, p_{s}$ are distinct primes. Our goal is to show that all exponents $\\alpha_{i}$ are divisible by $n$, then we can set $A=p_{1}^{\\alpha_{1} / n} \\ldots p_{s}^{\\alpha_{s} / n}$.\n\nApply the condition for $k=b^{2}$. The number $b-a_{k}^{n}$ is divisible by $b^{2}$ and hence, for each $1 \\leq i \\leq s$, it is divisible by $p_{i}^{2 \\alpha_{i}}>p_{i}^{\\alpha_{i}}$ as well. Therefore\n$$\na_{k}^{n} \\equiv b \\equiv 0 \\quad\\left(\\bmod p_{i}^{\\alpha_{i}}\\right)\n$$\nand\n$$\na_{k}^{n} \\equiv b \\not \\equiv 0 \\quad\\left(\\bmod p_{i}^{\\alpha_{i}+1}\\right)\n$$\nwhich implies that the largest power of $p_{i}$ dividing $a_{k}^{n}$ is $p_{i}^{\\alpha_{i}}$. Since $a_{k}^{n}$ is a complete $n$th power, this implies that $\\alpha_{i}$ is divisible by $n$.']",,True,,, 1680,Number Theory,,"For every integer $k \geq 2$, prove that $2^{3 k}$ divides the number $$ \left(\begin{array}{c} 2^{k+1} \\ 2^{k} \end{array}\right)-\left(\begin{array}{c} 2^{k} \\ 2^{k-1} \end{array}\right) \tag{1} $$ but $2^{3 k+1}$ does not.","['We use the notation $(2 n-1) ! !=1 \\cdot 3 \\cdots(2 n-1)$ and $(2 n) ! !=2 \\cdot 4 \\cdots(2 n)=2^{n} n !$ for any positive integer $n$. Observe that $(2 n) !=(2 n) ! !(2 n-1) ! !=2^{n} n !(2 n-1) ! !$.\n\nFor any positive integer $n$ we have\n$$\n\\begin{aligned}\n& \\left(\\begin{array}{c}\n4 n \\\\\n2 n\n\\end{array}\\right)=\\frac{(4 n) !}{(2 n) !^{2}}=\\frac{2^{2 n}(2 n) !(4 n-1) ! !}{(2 n) !^{2}}=\\frac{2^{2 n}}{(2 n) !}(4 n-1) ! ! \\\\\n& \\left(\\begin{array}{c}\n2 n \\\\\nn\n\\end{array}\\right)=\\frac{1}{(2 n) !}\\left(\\frac{(2 n) !}{n !}\\right)^{2}=\\frac{1}{(2 n) !}\\left(2^{n}(2 n-1) ! !\\right)^{2}=\\frac{2^{2 n}}{(2 n) !}(2 n-1) ! !^{2} .\n\\end{aligned}\n$$\nThen expression (1) can be rewritten as follows:\n$$\n\\begin{aligned}\n\\left(\\begin{array}{c}\n2^{k+1} \\\\\n2^{k}\n\\end{array}\\right) & -\\left(\\begin{array}{c}\n2^{k} \\\\\n2^{k-1}\n\\end{array}\\right)=\\frac{2^{2^{k}}}{\\left(2^{k}\\right) !}\\left(2^{k+1}-1\\right) ! !-\\frac{2^{2^{k}}}{\\left(2^{k}\\right) !}\\left(2^{k}-1\\right) ! !^{2} \\\\\n& =\\frac{2^{2^{k}}\\left(2^{k}-1\\right) ! !}{\\left(2^{k}\\right) !} \\cdot\\left(\\left(2^{k}+1\\right)\\left(2^{k}+3\\right) \\ldots\\left(2^{k}+2^{k}-1\\right)-\\left(2^{k}-1\\right)\\left(2^{k}-3\\right) \\ldots\\left(2^{k}-2^{k}+1\\right)\\right) .\n\\end{aligned} \\tag{2}\n$$\nWe compute the exponent of 2 in the prime decomposition of each factor (the first one is a rational number but not necessarily an integer; it is not important).\n\nFirst, we show by induction on $n$ that the exponent of 2 in $\\left(2^{n}\\right)$ ! is $2^{n}-1$. The base case $n=1$ is trivial. Suppose that $\\left(2^{n}\\right) !=2^{2^{n}-1}(2 d+1)$ for some integer $d$. Then we have\n$$\n\\left(2^{n+1}\\right) !=2^{2^{n}}\\left(2^{n}\\right) !\\left(2^{n+1}-1\\right) ! !=2^{2^{n}} 2^{2^{n}-1} \\cdot(2 d+1)\\left(2^{n+1}-1\\right) ! !=2^{2^{n+1}-1} \\cdot(2 q+1)\n$$\nfor some integer $q$. This finishes the induction step.\n\nHence, the exponent of 2 in the first factor in $(2)$ is $2^{k}-\\left(2^{k}-1\\right)=1$.\n\nThe second factor in (2) can be considered as the value of the polynomial\n$$\nP(x)=(x+1)(x+3) \\ldots\\left(x+2^{k}-1\\right)-(x-1)(x-3) \\ldots\\left(x-2^{k}+1\\right) .\\tag{3}\n$$\nat $x=2^{k}$. Now we collect some information about $P(x)$.\n\nObserve that $P(-x)=-P(x)$, since $k \\geq 2$. So $P(x)$ is an odd function, and it has nonzero coefficients only at odd powers of $x$. Hence $P(x)=x^{3} Q(x)+c x$, where $Q(x)$ is a polynomial with integer coefficients.\n\nCompute the exponent of 2 in $c$. We have\n$$\n\\begin{aligned}\nc & =2\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}} \\frac{1}{2 i-1}=\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}}\\left(\\frac{1}{2 i-1}+\\frac{1}{2^{k}-2 i+1}\\right) \\\\\n& =\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}} \\frac{2^{k}}{(2 i-1)\\left(2^{k}-2 i+1\\right)}=2^{k} \\sum_{i=1}^{2^{k-1}} \\frac{\\left(2^{k}-1\\right) ! !}{(2 i-1)\\left(2^{k}-2 i+1\\right)}=2^{k} S .\n\\end{aligned}\n$$\n\nFor any integer $i=1, \\ldots, 2^{k-1}$, denote by $a_{2 i-1}$ the residue inverse to $2 i-1$ modulo $2^{k}$. Clearly, when $2 i-1$ runs through all odd residues, so does $a_{2 i-1}$, hence\n$$\n\\begin{gathered}\nS=\\sum_{i=1}^{2^{k-1}} \\frac{\\left(2^{k}-1\\right) ! !}{(2 i-1)\\left(2^{k}-2 i+1\\right)} \\equiv-\\sum_{i=1}^{2^{k-1}} \\frac{\\left(2^{k}-1\\right) ! !}{(2 i-1)^{2}} \\equiv-\\sum_{i=1}^{2^{k-1}}\\left(2^{k}-1\\right) ! ! a_{2 i-1}^{2} \\\\\n=-\\left(2^{k}-1\\right) ! ! \\sum_{i=1}^{2^{k-1}}(2 i-1)^{2}=-\\left(2^{k}-1\\right) ! ! \\frac{2^{k-1}\\left(2^{2 k}-1\\right)}{3} \\quad\\left(\\bmod 2^{k}\\right) .\n\\end{gathered}\n$$\nTherefore, the exponent of 2 in $S$ is $k-1$, so $c=2^{k} S=2^{2 k-1}(2 t+1)$ for some integer $t$.\n\nFinally we obtain that\n$$\nP\\left(2^{k}\\right)=2^{3 k} Q\\left(2^{k}\\right)+2^{k} c=2^{3 k} Q\\left(2^{k}\\right)+2^{3 k-1}(2 t+1),\n$$\nwhich is divisible exactly by $2^{3 k-1}$. Thus, the exponent of 2 in $(2)$ is $1+(3 k-1)=3 k$.']",,True,,, 1681,Number Theory,,"Find all surjective functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for every $m, n \in \mathbb{N}$ and every prime $p$, the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p$. ( $\mathbb{N}$ is the set of all positive integers.)","[""Suppose that function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies the problem conditions.\n\nLemma. For any prime $p$ and any $x, y \\in \\mathbb{N}$, we have $x \\equiv y(\\bmod p)$ if and only if $f(x) \\equiv f(y)$ $(\\bmod p)$. Moreover, $p \\mid f(x)$ if and only if $p \\mid x$.\n\nProof. Consider an arbitrary prime $p$. Since $f$ is surjective, there exists some $x \\in \\mathbb{N}$ such that $p \\mid f(x)$. Let\n$$\nd=\\min \\{x \\in \\mathbb{N}: p \\mid f(x)\\}\n$$\nBy induction on $k$, we obtain that $p \\mid f(k d)$ for all $k \\in \\mathbb{N}$. The base is true since $p \\mid f(d)$. Moreover, if $p \\mid f(k d)$ and $p \\mid f(d)$ then, by the problem condition, $p \\mid f(k d+d)=f((k+1) d)$ as required.\n\nSuppose that there exists an $x \\in \\mathbb{N}$ such that $d \\not x$ but $p \\mid f(x)$. Let\n$$\ny=\\min \\{x \\in \\mathbb{N}: d \\nmid x, p \\mid f(x)\\} .\n$$\nBy the choice of $d$, we have $y>d$, and $y-d$ is a positive integer not divisible by $d$. Then $p \\nmid f(y-d)$, while $p \\mid f(d)$ and $p \\mid f(d+(y-d))=f(y)$. This contradicts the problem condition. Hence, there is no such $x$, and\n$$\np|f(x) \\Longleftrightarrow d| x .\\tag{1}\n$$\nTake arbitrary $x, y \\in \\mathbb{N}$ such that $x \\equiv y(\\bmod d)$. We have $p \\mid f(x+(2 x d-x))=f(2 x d)$; moreover, since $d \\mid 2 x d+(y-x)=y+(2 x d-x)$, we get $p \\mid f(y+(2 x d-x))$. Then by the problem condition $p|f(x)+f(2 x d-x), p| f(y)+f(2 x d-x)$, and hence $f(x) \\equiv-f(2 x d-x) \\equiv f(y)$ $(\\bmod p)$.\n\nOn the other hand, assume that $f(x) \\equiv f(y)(\\bmod p)$. Again we have $p \\mid f(x)+f(2 x d-x)$ which by our assumption implies that $p \\mid f(x)+f(2 x d-x)+(f(y)-f(x))=f(y)+f(2 x d-x)$. Hence by the problem condition $p \\mid f(y+(2 x d-x))$. Using (1) we get $0 \\equiv y+(2 x d-x) \\equiv y-x$ $(\\bmod d)$.\n\nThus, we have proved that\n$$\nx \\equiv y \\quad(\\bmod d) \\Longleftrightarrow f(x) \\equiv f(y) \\quad(\\bmod p)\\tag{2}\n$$\nWe are left to show that $p=d$ : in this case (1) and (2) provide the desired statements.\n\nThe numbers $1,2, \\ldots, d$ have distinct residues modulo $d$. By (2), numbers $f(1), f(2), \\ldots$, $f(d)$ have distinct residues modulo $p$; hence there are at least $d$ distinct residues, and $p \\geq d$. On the other hand, by the surjectivity of $f$, there exist $x_{1}, \\ldots, x_{p} \\in \\mathbb{N}$ such that $f\\left(x_{i}\\right)=i$ for any $i=1,2, \\ldots, p$. By (2), all these $x_{i}$ 's have distinct residues modulo $d$. For the same reasons, $d \\geq p$. Hence, $d=p$.\n\nNow we prove that $f(n)=n$ by induction on $n$. If $n=1$ then, by the Lemma, $p \\nmid f(1)$ for any prime $p$, so $f(1)=1$, and the base is established. Suppose that $n>1$ and denote $k=f(n)$. Note that there exists a prime $q \\mid n$, so by the Lemma $q \\mid k$ and $k>1$.\n\nIf $k>n$ then $k-n+1>1$, and there exists a prime $p \\mid k-n+1$; we have $k \\equiv n-1$ $(\\bmod p)$. By the induction hypothesis we have $f(n-1)=n-1 \\equiv k=f(n)(\\bmod p)$. Now, by the Lemma we obtain $n-1 \\equiv n(\\bmod p)$ which cannot be true.\n\n\n\nAnalogously, if $k1$, so there exists a prime $p \\mid n-k+1$ and $n \\equiv k-1(\\bmod p)$. By the Lemma again, $k=f(n) \\equiv$ $f(k-1)=k-1(\\bmod p)$, which is also false. The only remaining case is $k=n$, so $f(n)=n$.\n\nFinally, the function $f(n)=n$ obviously satisfies the condition.""]",['$f(n)=n$'],False,,Expression, 1682,Number Theory,,Let $k$ be a positive integer. Prove that the number $\left(4 k^{2}-1\right)^{2}$ has a positive divisor of the form $8 k n-1$ if and only if $k$ is even.,"['The statement follows from the following fact.\n\nLemma. For arbitrary positive integers $x$ and $y$, the number $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ if and only if $x=y$.\n\nProof. If $x=y$ then $4 x y-1=4 x^{2}-1$ obviously divides $\\left(4 x^{2}-1\\right)^{2}$ so it is sufficient to consider the opposite direction.\n\nCall a pair $(x, y)$ of positive integers bad if $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ but $x \\neq y$. In order to prove that bad pairs do not exist, we present two properties of them which provide an infinite descent.\n\nProperty (i). If $(x, y)$ is a bad pair and $xr$. Then $\\nu_{p}\\left(q p^{k}+r\\right)=\\nu_{p}\\left(q p^{k}\\right)+\\nu_{p}(r)$.\n\nProof. We claim that $\\operatorname{ord}_{p}\\left(q p^{k}+i\\right)=\\operatorname{ord}_{p}(i)$ for all $01$. By the construction of the sequence, $p_{i}^{n_{\\ell_{1}}}$ divides $n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}$; clearly, $p_{i}^{n_{\\ell_{1}}}>n_{\\ell_{1}}$ for all $1 \\leq i \\leq k$. Therefore the Lemma can be applied for $p=p_{i}, k=r=n_{\\ell_{1}}$ and $q p^{k}=n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}$ to obtain\n$$\nf_{i}\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f_{i}\\left(n_{\\ell_{1}}\\right)+f_{i}\\left(n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right) \\quad \\text { for all } 1 \\leq i \\leq k\n$$\nand hence\n$$\nf\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}\\right)+\\ldots+f\\left(n_{\\ell_{m}}\\right)\n$$\nby the induction hypothesis.\n\nNow consider the values $f\\left(n_{1}\\right), f\\left(n_{2}\\right), \\ldots$ There exist finitely many possible values of $f$. Hence, there exists an infinite sequence of indices $\\ell_{1}<\\ell_{2}<\\ldots$ such that $f\\left(n_{\\ell_{1}}\\right)=f\\left(n_{\\ell_{2}}\\right)=\\ldots$ and thus\n$$\nf\\left(n_{\\ell_{m+1}}+n_{\\ell_{m+2}}+\\ldots+n_{\\ell_{m+d}}\\right)=f\\left(n_{\\ell_{m+1}}\\right)+\\ldots+f\\left(n_{\\ell_{m+d}}\\right)=d \\cdot f\\left(n_{\\ell_{1}}\\right)=(\\overline{0}, \\ldots, \\overline{0})\n$$\nfor all $m$. We have found infinitely many suitable numbers.', 'We use the same Lemma and definition of the function $f$.\n\nLet $S=\\{f(n): n \\in \\mathbb{N}\\}$. Obviously, set $S$ is finite. For every $s \\in S$ choose the minimal $n_{s}$ such that $f\\left(n_{s}\\right)=s$. Denote $N=\\max _{s \\in S} n_{s}$. Moreover, let $g$ be an integer such that $p_{i}^{g}>N$ for each $i=1,2, \\ldots, k$. Let $P=\\left(p_{1} p_{2} \\ldots p_{k}\\right)^{g}$.\n\nWe claim that\n$$\n\\{f(n) \\mid n \\in[m P, m P+N]\\}=S \\tag{1}\n$$\nfor every positive integer $m$. In particular, since $(\\overline{0}, \\ldots, \\overline{0})=f(1) \\in S$, it follows that for an arbitrary $m$ there exists $n \\in[m P, m P+N]$ such that $f(n)=(\\overline{0}, \\ldots, \\overline{0})$. So there are infinitely many suitable numbers.\n\nTo prove (1), let $a_{i}=f_{i}(m P)$. Consider all numbers of the form $n_{m, s}=m P+n_{s}$ with $s=\\left(s_{1}, \\ldots, s_{k}\\right) \\in S$ (clearly, all $n_{m, s}$ belong to $[m P, m P+N]$ ). Since $n_{s} \\leq Na_{1}+a_{3}$ and $a_{3}+a_{4}>a_{1}+a_{2}$. Hence $a_{2}+a_{4}$ and $a_{3}+a_{4}$ do not divide $s_{A}$. This proves $p_{A} \\leq 4$.\n\nNow suppose $p_{A}=4$. By the previous argument we have\n\n$$\n\\begin{array}{lll}\na_{1}+a_{4} \\mid a_{2}+a_{3} & \\text { and } & a_{2}+a_{3} \\mid a_{1}+a_{4}, \\\\\na_{1}+a_{2} \\mid a_{3}+a_{4} & \\text { and } & a_{3}+a_{4} \\nmid a_{1}+a_{2}, \\\\\na_{1}+a_{3} \\mid a_{2}+a_{4} & \\text { and } & a_{2}+a_{4} \\nmid a_{1}+a_{3} .\n\\end{array}\n$$\n\nHence, there exist positive integers $m$ and $n$ with $m>n \\geq 2$ such that\n\n$$\n\\left\\{\\begin{array}{l}\na_{1}+a_{4}=a_{2}+a_{3} \\\\\nm\\left(a_{1}+a_{2}\\right)=a_{3}+a_{4} \\\\\nn\\left(a_{1}+a_{3}\\right)=a_{2}+a_{4}\n\\end{array}\\right.\n$$\n\nAdding up the first equation and the third one, we get $n\\left(a_{1}+a_{3}\\right)=2 a_{2}+a_{3}-a_{1}$. If $n \\geq 3$, then $n\\left(a_{1}+a_{3}\\right)>3 a_{3}>2 a_{2}+a_{3}>2 a_{2}+a_{3}-a_{1}$. This is a contradiction. Therefore $n=2$. If we multiply by 2 the sum of the first equation and the third one, we obtain\n\n$$\n6 a_{1}+2 a_{3}=4 a_{2},\n$$\n\nwhile the sum of the first one and the second one is\n\n$$\n(m+1) a_{1}+(m-1) a_{2}=2 a_{3}\n$$\n\nAdding up the last two equations we get\n\n$$\n(m+7) a_{1}=(5-m) a_{2}\n$$\n\n\n\nIt follows that $5-m \\geq 1$, because the left-hand side of the last equation and $a_{2}$ are positive. Since we have $m>n=2$, the integer $m$ can be equal only to either 3 or 4 . Substituting $(3,2)$ and $(4,2)$ for $(m, n)$ and solving the previous system of equations, we find the families of solutions $\\{d, 5 d, 7 d, 11 d\\}$ and $\\{d, 11 d, 19 d, 29 d\\}$, where $d$ is any positive integer.']","['$\\{d, 5 d, 7 d, 11 d\\}$ and $\\{d, 11 d, 19 d, 29 d\\}$, where $d$ is any positive integer. For all these sets $p_{A}$ is 4']",True,,Need_human_evaluate, 1685,Algebra,,"Determine all sequences $\left(x_{1}, x_{2}, \ldots, x_{2011}\right)$ of positive integers such that for every positive integer $n$ there is an integer $a$ with $$ x_{1}^{n}+2 x_{2}^{n}+\cdots+2011 x_{2011}^{n}=a^{n+1}+1 \text {. } $$","['Throughout this solution, the set of positive integers will be denoted by $\\mathbb{Z}_{+}$.\n\nPut $k=2+3+\\cdots+2011=2023065$. We have\n\n$$\n1^{n}+2 k^{n}+\\cdots 2011 k^{n}=1+k \\cdot k^{n}=k^{n+1}+1\n$$\n\nfor all $n$, so $(1, k, \\ldots, k)$ is a valid sequence. We shall prove that it is the only one.\n\nLet a valid sequence $\\left(x_{1}, \\ldots, x_{2011}\\right)$ be given. For each $n \\in \\mathbb{Z}_{+}$we have some $y_{n} \\in \\mathbb{Z}_{+}$with\n\n$$\nx_{1}^{n}+2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}=y_{n}^{n+1}+1 \\text {. }\n$$\n\nNote that $x_{1}^{n}+2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}<\\left(x_{1}+2 x_{2}+\\cdots+2011 x_{2011}\\right)^{n+1}$, which implies that the sequence $\\left(y_{n}\\right)$ is bounded. In particular, there is some $y \\in \\mathbb{Z}_{+}$with $y_{n}=y$ for infinitely many $n$.\n\nLet $m$ be the maximum of all the $x_{i}$. Grouping terms with equal $x_{i}$ together, the $\\operatorname{sum} x_{1}^{n}+$ $2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}$ can be written as\n\n$$\nx_{1}^{n}+2 x_{2}^{n}+\\cdots+x_{2011}^{n}=a_{m} m^{n}+a_{m-1}(m-1)^{n}+\\cdots+a_{1}\n$$\n\nwith $a_{i} \\geq 0$ for all $i$ and $a_{1}+\\cdots+a_{m}=1+2+\\cdots+2011$. So there exist arbitrarily large values of $n$, for which\n\n$$\na_{m} m^{n}+\\cdots+a_{1}-1-y \\cdot y^{n}=0 .\n\\tag{1}\n$$\n\nThe following lemma will help us to determine the $a_{i}$ and $y$ :\n\nLemma. Let integers $b_{1}, \\ldots, b_{N}$ be given and assume that there are arbitrarily large positive integers $n$ with $b_{1}+b_{2} 2^{n}+\\cdots+b_{N} N^{n}=0$. Then $b_{i}=0$ for all $i$.\n\nProof. Suppose that not all $b_{i}$ are zero. We may assume without loss of generality that $b_{N} \\neq 0$.\n\n\n\nDividing through by $N^{n}$ gives\n\n$$\n\\left|b_{N}\\right|=\\left|b_{N-1}\\left(\\frac{N-1}{N}\\right)^{n}+\\cdots+b_{1}\\left(\\frac{1}{N}\\right)^{n}\\right| \\leq\\left(\\left|b_{N-1}\\right|+\\cdots+\\left|b_{1}\\right|\\right)\\left(\\frac{N-1}{N}\\right)^{n} .\n$$\n\nThe expression $\\left(\\frac{N-1}{N}\\right)^{n}$ can be made arbitrarily small for $n$ large enough, contradicting the assumption that $b_{N}$ be non-zero.\n\nWe obviously have $y>1$. Applying the lemma to (1) we see that $a_{m}=y=m, a_{1}=1$, and all the other $a_{i}$ are zero. This implies $\\left(x_{1}, \\ldots, x_{2011}\\right)=(1, m, \\ldots, m)$. But we also have $1+m=a_{1}+\\cdots+a_{m}=1+\\cdots+2011=1+k$ so $m=k$, which is what we wanted to show.']","['$$\n\\left(x_{1}, \\ldots, x_{2011}\\right)=(1, k, \\ldots, k) \\quad \\text { with } k=2+3+\\cdots+2011=2023065\n$$']",False,,Need_human_evaluate, 1686,Algebra,,"Determine all pairs $(f, g)$ of functions from the set of real numbers to itself that satisfy $$ g(f(x+y))=f(x)+(2 x+y) g(y) $$ for all real numbers $x$ and $y$.","['Clearly all these pairs of functions satisfy the functional equation in question, so it suffices to verify that there cannot be any further ones. Substituting $-2 x$ for $y$ in the given functional equation we obtain\n\n$$\ng(f(-x))=f(x)\n\\tag{1}\n$$\n\nUsing this equation for $-x-y$ in place of $x$ we obtain\n\n$$\nf(-x-y)=g(f(x+y))=f(x)+(2 x+y) g(y) .\n\\tag{2}\n$$\n\nNow for any two real numbers $a$ and $b$, setting $x=-b$ and $y=a+b$ we get\n\n$$\nf(-a)=f(-b)+(a-b) g(a+b) .\n$$\n\nIf $c$ denotes another arbitrary real number we have similarly\n\n$$\nf(-b)=f(-c)+(b-c) g(b+c)\n$$\n\nas well as\n\n$$\nf(-c)=f(-a)+(c-a) g(c+a) .\n$$\n\nAdding all these equations up, we obtain\n\n$$\n((a+c)-(b+c)) g(a+b)+((a+b)-(a+c)) g(b+c)+((b+c)-(a+b)) g(a+c)=0 \\text {. }\n$$\n\nNow given any three real numbers $x, y$, and $z$ one may determine three reals $a, b$, and $c$ such that $x=b+c, y=c+a$, and $z=a+b$, so that we get\n\n$$\n(y-x) g(z)+(z-y) g(x)+(x-z) g(y)=0 .\n$$\n\nThis implies that the three points $(x, g(x)),(y, g(y))$, and $(z, g(z))$ from the graph of $g$ are collinear. Hence that graph is a line, i.e., $g$ is either a constant or a linear function.\n\n\n\nLet us write $g(x)=A x+B$, where $A$ and $B$ are two real numbers. Substituting $(0,-y)$ for $(x, y)$ in (2) and denoting $C=f(0)$, we have $f(y)=A y^{2}-B y+C$. Now, comparing the coefficients of $x^{2}$ in (1) we see that $A^{2}=A$, so $A=0$ or $A=1$.\n\nIf $A=0$, then (11) becomes $B=-B x+C$ and thus $B=C=0$, which provides the first of the two solutions mentioned above.\n\nNow suppose $A=1$. Then (11) becomes $x^{2}-B x+C+B=x^{2}-B x+C$, so $B=0$. Thus, $g(x)=x$ and $f(x)=x^{2}+C$, which is the second solution from above.']","['Either both $f$ and $g$ vanish identically, or there exists a real number $C$ such that $f(x)=x^{2}+C$ and $g(x)=x$ for all real numbers $x$']",True,,Need_human_evaluate, 1687,Algebra,,"Determine all pairs $(f, g)$ of functions from the set of positive integers to itself that satisfy $$ f^{g(n)+1}(n)+g^{f(n)}(n)=f(n+1)-g(n+1)+1 $$ for every positive integer $n$. Here, $f^{k}(n)$ means $\underbrace{f(f(\ldots f}_{k}(n) \ldots))$.","['The given relation implies\n\n$$\nf\\left(f^{g(n)}(n)\\right)1$, then (1) reads $f\\left(f^{g(x-1)}(x-1)\\right)n$. Substituting $x-1$ into (1) we have $f\\left(f^{g(x-1)}(x-1)\\right)n$. So $b=n$, and hence $y_{n}=n$, which proves (ii) ${ }_{n}$. Next, from (i) ${ }_{n}$ we now get $f(k)=n \\Longleftrightarrow k=n$, so removing all the iterations of $f$ in (3) we obtain $x-1=b=n$, which proves $(\\mathrm{i})_{n+1}$.\n\nSo, all the statements in (2) are valid and hence $f(n)=n$ for all $n$. The given relation between $f$ and $g$ now reads $n+g^{n}(n)=n+1-g(n+1)+1$ or $g^{n}(n)+g(n+1)=2$, from which it immediately follows that we have $g(n)=1$ for all $n$.\n\n']","['$f(n)=n$, $g(n)=1$']",False,,Expression, 1688,Algebra,,"Prove that for every positive integer $n$, the set $\{2,3,4, \ldots, 3 n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle.","['Throughout the solution, we denote by $[a, b]$ the set $\\{a, a+1, \\ldots, b\\}$. We say that $\\{a, b, c\\}$ is an obtuse triple if $a, b, c$ are the sides of some obtuse triangle.\n\nWe prove by induction on $n$ that there exists a partition of $[2,3 n+1]$ into $n$ obtuse triples $A_{i}$ $(2 \\leq i \\leq n+1)$ having the form $A_{i}=\\left\\{i, a_{i}, b_{i}\\right\\}$. For the base case $n=1$, one can simply set $A_{2}=\\{2,3,4\\}$. For the induction step, we need the following simple lemma.\n\nLemma. Suppose that the numbers $a(c-b)(c+b)>a^{2}$.\n\nNow we turn to the induction step. Let $n>1$ and put $t=\\lfloor n / 2\\rfloor\\frac{n}{2} \\cdot \\frac{9 n}{2} \\geq(n+1)^{2} \\geq i^{2}$,\n\nso this triangle is obtuse. The proof is completed.']",,True,,, 1689,Algebra,,"Let $f$ be a function from the set of real numbers to itself that satisfies $$ f(x+y) \leq y f(x)+f(f(x)) \tag{1} $$ for all real numbers $x$ and $y$. Prove that $f(x)=0$ for all $x \leq 0$.","['Substituting $y=t-x$, we rewrite (11) as\n\n$$\nf(t) \\leq t f(x)-x f(x)+f(f(x)) .\n\\tag{2}\n$$\n\nConsider now some real numbers $a, b$ and use (2) with $t=f(a), x=b$ as well as with $t=f(b)$, $x=a$. We get\n\n$$\n\\begin{aligned}\n& f(f(a))-f(f(b)) \\leq f(a) f(b)-b f(b) \\\\\n& f(f(b))-f(f(a)) \\leq f(a) f(b)-a f(a)\n\\end{aligned}\n$$\n\nAdding these two inequalities yields\n\n$$\n2 f(a) f(b) \\geq a f(a)+b f(b)\n$$\n\nNow, substitute $b=2 f(a)$ to obtain $2 f(a) f(b) \\geq a f(a)+2 f(a) f(b)$, or $a f(a) \\leq 0$. So, we get\n\n$$\nf(a) \\geq 0 \\text { for all } a<0\n\\tag{3}\n$$\n\nNow suppose $f(x)>0$ for some real number $x$. From (2) we immediately get that for every $t<\\frac{x f(x)-f(f(x))}{f(x)}$ we have $f(t)<0$. This contradicts (3); therefore\n\n$$\nf(x) \\leq 0 \\text { for all real } x\n\\tag{4}\n$$\n\nand by (3) again we get $f(x)=0$ for all $x<0$.\n\nWe are left to find $f(0)$. Setting $t=x<0$ in (2) we get\n\n$$\n0 \\leq 0-0+f(0)\n$$\n\nso $f(0) \\geq 0$. Combining this with (4) we obtain $f(0)=0$.', 'We will also use the condition of the problem in form (2). For clarity we divide the argument into four steps.\n\n\n\nStep 1. We begin by proving that $f$ attains nonpositive values only. Assume that there exist some real number $z$ with $f(z)>0$. Substituting $x=z$ into (2) and setting $A=f(z)$, $B=-z f(z)-f(f(z))$ we get $f(t) \\leq A t+B$ for all real $t$. Hence, if for any positive real number $t$ we substitute $x=-t, y=t$ into (1), we get\n\n$$\n\\begin{aligned}\nf(0) & \\leq t f(-t)+f(f(-t)) \\leq t(-A t+B)+A f(-t)+B \\\\\n& \\leq-t(A t-B)+A(-A t+B)+B=-A t^{2}-\\left(A^{2}-B\\right) t+(A+1) B .\n\\end{aligned}\n$$\n\nBut surely this cannot be true if we take $t$ to be large enough. This contradiction proves that we have indeed $f(x) \\leq 0$ for all real numbers $x$. Note that for this reason (1) entails\n\n$$\nf(x+y) \\leq y f(x)\n\\tag{5}\n$$\n\nfor all real numbers $x$ and $y$.\n\nStep 2. We proceed by proving that $f$ has at least one zero. If $f(0)=0$, we are done. Otherwise, in view of Step 1 we get $f(0)<0$. Observe that (5) tells us now $f(y) \\leq y f(0)$ for all real numbers $y$. Thus we can specify a positive real number $a$ that is so large that $f(a)^{2}>-f(0)$. Put $b=f(a)$ and substitute $x=b$ and $y=-b$ into (5); we learn $-b^{2}\sqrt{2}$ and $a^{2}+b^{2}+c^{2}=3$. Prove that $$ \frac{a}{(b+c-a)^{2}}+\frac{b}{(c+a-b)^{2}}+\frac{c}{(a+b-c)^{2}} \geq \frac{3}{(a b c)^{2}} \tag{1} $$ Throughout both solutions, we denote the sums of the form $f(a, b, c)+f(b, c, a)+f(c, a, b)$ by $\sum f(a, b, c)$.","[""The condition $b+c>\\sqrt{2}$ implies $b^{2}+c^{2}>1$, so $a^{2}=3-\\left(b^{2}+c^{2}\\right)<2$, i.e. $a<\\sqrt{2}0$, and also $c+a-b>0$ and $a+b-c>0$ for similar reasons.\n\nWe will use the variant of HÖLDER's inequality\n\n$$\n\\frac{x_{1}^{p+1}}{y_{1}^{p}}+\\frac{x_{1}^{p+1}}{y_{1}^{p}}+\\ldots+\\frac{x_{n}^{p+1}}{y_{n}^{p}} \\geq \\frac{\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)^{p+1}}{\\left(y_{1}+y_{2}+\\ldots+y_{n}\\right)^{p}}\n$$\n\nwhich holds for all positive real numbers $p, x_{1}, x_{2}, \\ldots, x_{n}, y_{1}, y_{2}, \\ldots, y_{n}$. Applying it to the left-hand side of (1) with $p=2$ and $n=3$, we get\n\n$$\n\\sum \\frac{a}{(b+c-a)^{2}}=\\sum \\frac{\\left(a^{2}\\right)^{3}}{a^{5}(b+c-a)^{2}} \\geq \\frac{\\left(a^{2}+b^{2}+c^{2}\\right)^{3}}{\\left(\\sum a^{5 / 2}(b+c-a)\\right)^{2}}=\\frac{27}{\\left(\\sum a^{5 / 2}(b+c-a)\\right)^{2}}\n\\tag{2}\n$$\n\nTo estimate the denominator of the right-hand part, we use an instance of ScHUR's inequality, namely\n\n$$\n\\sum a^{3 / 2}(a-b)(a-c) \\geq 0\n$$\n\nwhich can be rewritten as\n\n$$\n\\sum a^{5 / 2}(b+c-a) \\leq a b c(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})\n$$\n\nMoreover, by the inequality between the arithmetic mean and the fourth power mean we also have\n\n$$\n\\left(\\frac{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{3}\\right)^{4} \\leq \\frac{a^{2}+b^{2}+c^{2}}{3}=1\n$$\n\ni.e., $\\sqrt{a}+\\sqrt{b}+\\sqrt{c} \\leq 3$. Hence, (2) yields\n\n$$\n\\sum \\frac{a}{(b+c-a)^{2}} \\geq \\frac{27}{(a b c(\\sqrt{a}+\\sqrt{b}+\\sqrt{c}))^{2}} \\geq \\frac{3}{a^{2} b^{2} c^{2}}\n$$\n\nthus solving the problem."", 'we mention that all the numbers $b+c-a, a+c-b, a+b-c$ are positive. We will use only this restriction and the condition\n\n$$\na^{5}+b^{5}+c^{5} \\geq 3\n\\tag{3}\n$$\n\nwhich is weaker than the given one. Due to the symmetry we may assume that $a \\geq b \\geq c$.\n\nIn view of (3), it suffices to prove the inequality\n\n$$\n\\sum \\frac{a^{3} b^{2} c^{2}}{(b+c-a)^{2}} \\geq \\sum a^{5}\n$$\n\nor, moving all the terms into the left-hand part,\n\n$$\n\\sum \\frac{a^{3}}{(b+c-a)^{2}}\\left((b c)^{2}-(a(b+c-a))^{2}\\right) \\geq 0\n\\tag{4}\n$$\n\nNote that the signs of the expressions $(y z)^{2}-(x(y+z-x))^{2}$ and $y z-x(y+z-x)=(x-y)(x-z)$ are the same for every positive $x, y, z$ satisfying the triangle inequality. So the terms in (4) corresponding to $a$ and $c$ are nonnegative, and hence it is sufficient to prove that the sum of the terms corresponding to $a$ and $b$ is nonnegative. Equivalently, we need the relation\n\n$$\n\\frac{a^{3}}{(b+c-a)^{2}}(a-b)(a-c)(b c+a(b+c-a)) \\geq \\frac{b^{3}}{(a+c-b)^{2}}(a-b)(b-c)(a c+b(a+c-b)) .\n$$\n\nObviously, we have\n\n$$\na^{3} \\geq b^{3} \\geq 0, \\quad 0(a-b)^{2}\n$$\n\nwhich holds since $c>a-b \\geq 0$ and $a+b>a-b \\geq 0$.']",,True,,, 1691,Combinatorics,,"Let $n>0$ be an integer. We are given a balance and $n$ weights of weight $2^{0}, 2^{1}, \ldots, 2^{n-1}$. In a sequence of $n$ moves we place all weights on the balance. In the first move we choose a weight and put it on the left pan. In each of the following moves we choose one of the remaining weights and we add it either to the left or to the right pan. Compute the number of ways in which we can perform these $n$ moves in such a way that the right pan is never heavier than the left pan.","['Assume $n \\geq 2$. We claim\n\n$$\nf(n)=(2 n-1) f(n-1) .\n\\tag{1}\n$$\n\nFirstly, note that after the first move the left pan is always at least 1 heavier than the right one. Hence, any valid way of placing the $n$ weights on the scale gives rise, by not considering weight 1 , to a valid way of placing the weights $2,2^{2}, \\ldots, 2^{n-1}$.\n\nIf we divide the weight of each weight by 2 , the answer does not change. So these $n-1$ weights can be placed on the scale in $f(n-1)$ valid ways. Now we look at weight 1 . If it is put on the scale in the first move, then it has to be placed on the left side, otherwise it can be placed either on the left or on the right side, because after the first move the difference between the weights on the left pan and the weights on the right pan is at least 2 . Hence, there are exactly $2 n-1$ different ways of inserting weight 1 in each of the $f(n-1)$ valid sequences for the $n-1$ weights in order to get a valid sequence for the $n$ weights. This proves the claim.\n\nSince $f(1)=1$, by induction we obtain for all positive integers $n$\n\n$$\nf(n)=(2 n-1) ! !=1 \\cdot 3 \\cdot 5 \\cdot \\ldots \\cdot(2 n-1)\n$$', 'We present a different way of obtaining (1). Set $f(0)=1$. Firstly, we find a recurrent formula for $f(n)$.\n\nComment. It is useful to remark that the answer is the same for any set of weights where each weight is heavier than the sum of the lighter ones. Indeed, in such cases the given condition is equivalent to asking that during the process the heaviest weight on the balance is always on the left pan.\n\nAssume $n \\geq 1$. Suppose that weight $2^{n-1}$ is placed on the balance in the $i$-th move with $1 \\leq i \\leq n$. This weight has to be put on the left pan. For the previous moves we have $\\left(\\begin{array}{c}n-1 \\\\ i-1\\end{array}\\right)$ choices of the weights and from Comment. there are $f(i-1)$ valid ways of placing them on the balance. For later moves there is no restriction on the way in which the weights are to be put on the pans. Therefore, all $(n-i) ! 2^{n-i}$ ways are possible. This gives\n\n$$\nf(n)=\\sum_{i=1}^{n}\\left(\\begin{array}{c}\nn-1 \\\\\ni-1\n\\end{array}\\right) f(i-1)(n-i) ! 2^{n-i}=\\sum_{i=1}^{n} \\frac{(n-1) ! f(i-1) 2^{n-i}}{(i-1) !}\n\\tag{2}\n$$\n\nNow we are ready to prove (1). Using $n-1$ instead of $n$ in (2) we get\n\n$$\nf(n-1)=\\sum_{i=1}^{n-1} \\frac{(n-2) ! f(i-1) 2^{n-1-i}}{(i-1) !}\n$$\n\nHence, again from (2) we get\n\n$$\n\\begin{aligned}\nf(n)=2(n-1) \\sum_{i=1}^{n-1} & \\frac{(n-2) ! f(i-1) 2^{n-1-i}}{(i-1) !}+f(n-1) \\\\\n& =(2 n-2) f(n-1)+f(n-1)=(2 n-1) f(n-1),\n\\end{aligned}\n$$\n\nQED.']","['The number $f(n)$ of ways of placing the $n$ weights is equal to the product of all odd positive integers less than or equal to $2 n-1$, i.e. $f(n)=(2 n-1) ! !=1 \\cdot 3 \\cdot 5 \\cdot \\ldots \\cdot(2 n-1)$']",True,,Need_human_evaluate, 1692,Combinatorics,,"Suppose that 1000 students are standing in a circle. Prove that there exists an integer $k$ with $100 \leq k \leq 300$ such that in this circle there exists a contiguous group of $2 k$ students, for which the first half contains the same number of girls as the second half.","['Number the students consecutively from 1 to 1000. Let $a_{i}=1$ if the $i$ th student is a girl, and $a_{i}=0$ otherwise. We expand this notion for all integers $i$ by setting $a_{i+1000}=$ $a_{i-1000}=a_{i}$. Next, let\n\n$$\nS_{k}(i)=a_{i}+a_{i+1}+\\cdots+a_{i+k-1}\n$$\n\nNow the statement of the problem can be reformulated as follows:\n\nThere exist an integer $k$ with $100 \\leq k \\leq 300$ and an index $i$ such that $S_{k}(i)=S_{k}(i+k)$.\n\nAssume now that this statement is false. Choose an index $i$ such that $S_{100}(i)$ attains the maximal possible value. In particular, we have $S_{100}(i-100)-S_{100}(i)<0$ and $S_{100}(i)-S_{100}(i+100)>0$, for if we had an equality, then the statement would hold. This means that the function $S(j)-$ $S(j+100)$ changes sign somewhere on the segment $[i-100, i]$, so there exists some index $j \\in$ $[i-100, i-1]$ such that\n\n$$\nS_{100}(j) \\leq S_{100}(j+100)-1, \\quad \\text { but } \\quad S_{100}(j+1) \\geq S_{100}(j+101)+1\n\\tag{1}\n$$\n\nSubtracting the first inequality from the second one, we get $a_{j+100}-a_{j} \\geq a_{j+200}-a_{j+100}+2$, so\n\n$$\na_{j}=0, \\quad a_{j+100}=1, \\quad a_{j+200}=0\n$$\n\nSubstituting this into the inequalities of (1), we also obtain $S_{99}(j+1) \\leq S_{99}(j+101) \\leq S_{99}(j+1)$, which implies\n\n$$\nS_{99}(j+1)=S_{99}(j+101) .\n\\tag{2}\n$$\n\nNow let $k$ and $\\ell$ be the least positive integers such that $a_{j-k}=1$ and $a_{j+200+\\ell}=1$. By symmetry, we may assume that $k \\geq \\ell$. If $k \\geq 200$ then we have $a_{j}=a_{j-1}=\\cdots=a_{j-199}=0$, so $S_{100}(j-199)=S_{100}(j-99)=0$, which contradicts the initial assumption. Hence $\\ell \\leq k \\leq 199$. Finally, we have\n\n$$\n\\begin{gathered}\nS_{100+\\ell}(j-\\ell+1)=\\left(a_{j-\\ell+1}+\\cdots+a_{j}\\right)+S_{99}(j+1)+a_{j+100}=S_{99}(j+1)+1 \\\\\nS_{100+\\ell}(j+101)=S_{99}(j+101)+\\left(a_{j+200}+\\cdots+a_{j+200+\\ell-1}\\right)+a_{j+200+\\ell}=S_{99}(j+101)+1\n\\end{gathered}\n$$\n\nComparing with (2) we get $S_{100+\\ell}(j-\\ell+1)=S_{100+\\ell}(j+101)$ and $100+\\ell \\leq 299$, which again contradicts our assumption.']",,True,,, 1693,Combinatorics,,"Let $\mathcal{S}$ be a finite set of at least two points in the plane. Assume that no three points of $\mathcal{S}$ are collinear. By a windmill we mean a process as follows. Start with a line $\ell$ going through a point $P \in \mathcal{S}$. Rotate $\ell$ clockwise around the pivot $P$ until the line contains another point $Q$ of $\mathcal{S}$. The point $Q$ now takes over as the new pivot. This process continues indefinitely, with the pivot always being a point from $\mathcal{S}$. Show that for a suitable $P \in \mathcal{S}$ and a suitable starting line $\ell$ containing $P$, the resulting windmill will visit each point of $\mathcal{S}$ as a pivot infinitely often.","['Give the rotating line an orientation and distinguish its sides as the oranje side and the blue side. Notice that whenever the pivot changes from some point $T$ to another point $U$, after the change, $T$ is on the same side as $U$ was before. Therefore, the number of elements of $\\mathcal{S}$ on the oranje side and the number of those on the blue side remain the same throughout the whole process (except for those moments when the line contains two points).\n\n\nFirst consider the case that $|\\mathcal{S}|=2 n+1$ is odd. We claim that through any point $T \\in \\mathcal{S}$, there is a line that has $n$ points on each side. To see this, choose an oriented line through $T$ containing no other point of $\\mathcal{S}$ and suppose that it has $n+r$ points on its oranje side. If $r=0$ then we have established the claim, so we may assume that $r \\neq 0$. As the line rotates through $180^{\\circ}$ around $T$, the number of points of $\\mathcal{S}$ on its oranje side changes by 1 whenever the line passes through a point; after $180^{\\circ}$, the number of points on the oranje side is $n-r$. Therefore there is an intermediate stage at which the oranje side, and thus also the blue side, contains $n$ points.\n\nNow select the point $P$ arbitrarily, and choose a line through $P$ that has $n$ points of $\\mathcal{S}$ on each side to be the initial state of the windmill. We will show that during a rotation over $180^{\\circ}$, the line of the windmill visits each point of $\\mathcal{S}$ as a pivot. To see this, select any point $T$ of $\\mathcal{S}$ and select a line $\\boldsymbol{\\ell}$ through $T$ that separates $\\mathcal{S}$ into equal halves. The point $T$ is the unique point of $\\mathcal{S}$ through which a line in this direction can separate the points of $\\mathcal{S}$ into equal halves (parallel translation would disturb the balance). Therefore, when the windmill line is parallel to $\\ell$, it must be $\\ell$ itself, and so pass through $T$.\n\nNext suppose that $|\\mathcal{S}|=2 n$. Similarly to the odd case, for every $T \\in \\mathcal{S}$ there is an oriented\n\n\n\nline through $T$ with $n-1$ points on its oranje side and $n$ points on its blue side. Select such an oriented line through an arbitrary $P$ to be the initial state of the windmill.\n\nWe will now show that during a rotation over $360^{\\circ}$, the line of the windmill visits each point of $\\mathcal{S}$ as a pivot. To see this, select any point $T$ of $\\mathcal{S}$ and an oriented line $\\ell$ through $T$ that separates $\\mathcal{S}$ into two subsets with $n-1$ points on its oranje and $n$ points on its blue side. Again, parallel translation would change the numbers of points on the two sides, so when the windmill line is parallel to $\\ell$ with the same orientation, the windmill line must pass through $T$.']",,True,,, 1694,Combinatorics,,"Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \ldots, A_{k}$ such that for all integers $n \geq 15$ and all $i \in\{1,2, \ldots, k\}$ there exist two distinct elements of $A_{i}$ whose sum is $n$.","['There are various examples showing that $k=3$ does indeed have the property under consideration. E.g. one can take\n\n$$\n\\begin{gathered}\nA_{1}=\\{1,2,3\\} \\cup\\{3 m \\mid m \\geq 4\\} \\\\\nA_{2}=\\{4,5,6\\} \\cup\\{3 m-1 \\mid m \\geq 4\\} \\\\\nA_{3}=\\{7,8,9\\} \\cup\\{3 m-2 \\mid m \\geq 4\\}\n\\end{gathered}\n$$\n\nTo check that this partition fits, we notice first that the sums of two distinct elements of $A_{i}$ obviously represent all numbers $n \\geq 1+12=13$ for $i=1$, all numbers $n \\geq 4+11=15$ for $i=2$, and all numbers $n \\geq 7+10=17$ for $i=3$. So, we are left to find representations of the numbers 15 and 16 as sums of two distinct elements of $A_{3}$. These are $15=7+8$ and $16=7+9$.\n\nLet us now suppose that for some $k \\geq 4$ there exist sets $A_{1}, A_{2}, \\ldots, A_{k}$ satisfying the given property. Obviously, the sets $A_{1}, A_{2}, A_{3}, A_{4} \\cup \\cdots \\cup A_{k}$ also satisfy the same property, so one may assume $k=4$.\n\nPut $B_{i}=A_{i} \\cap\\{1,2, \\ldots, 23\\}$ for $i=1,2,3,4$. Now for any index $i$ each of the ten numbers $15,16, \\ldots, 24$ can be written as sum of two distinct elements of $B_{i}$. Therefore this set needs to contain at least five elements. As we also have $\\left|B_{1}\\right|+\\left|B_{2}\\right|+\\left|B_{3}\\right|+\\left|B_{4}\\right|=23$, there has to be some index $j$ for which $\\left|B_{j}\\right|=5$. Let $B_{j}=\\left\\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\\right\\}$. Finally, now the sums of two distinct elements of $A_{j}$ representing the numbers $15,16, \\ldots, 24$ should be exactly all the pairwise sums of the elements of $B_{j}$. Calculating the sum of these numbers in two different ways, we reach\n\n$$\n4\\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\\right)=15+16+\\ldots+24=195\n$$\n\nThus the number 195 should be divisible by 4, which is false. This contradiction completes our solution.', 'Again we only prove that $k \\leq 3$. Assume that $A_{1}, A_{2}, \\ldots, A_{k}$ is a partition satisfying the given property. We construct a graph $\\mathcal{G}$ on the set $V=\\{1,2, \\ldots, 18\\}$ of vertices as follows. For each $i \\in\\{1,2, \\ldots, k\\}$ and each $d \\in\\{15,16,17,19\\}$ we choose one pair of distinct elements $a, b \\in A_{i}$ with $a+b=d$, and we draw an $e d g e$ in the $i^{\\text {th }}$ color connecting $a$ with $b$. By hypothesis, $\\mathcal{G}$ has exactly 4 edges of each color.\n\nClaim. The graph $\\mathcal{G}$ contains at most one circuit.\n\nProof. Note that all the connected components of $\\mathcal{G}$ are monochromatic and hence contain at most four edges. Thus also all circuits of $\\mathcal{G}$ are monochromatic and have length at most four. Moreover, each component contains at most one circuit since otherwise it should contain at least five edges.\n\nSuppose that there is a 4-cycle in $\\mathcal{G}$, say with vertices $a, b, c$, and $d$ in order. Then $\\{a+b, b+$ $c, c+d, d+a\\}=\\{15,16,17,19\\}$. Taking sums we get $2(a+b+c+d)=15+16+17+19$ which is impossible for parity reasons. Thus all circuits of $\\mathcal{G}$ are triangles.\n\nNow if the vertices $a, b$, and $c$ form such a triangle, then by a similar reasoning the set $\\{a+b, b+$ $c, c+a\\}$ coincides with either $\\{15,16,17\\}$, or $\\{15,16,19\\}$, or $\\{16,17,19\\}$, or $\\{15,17,19\\}$. The last of these alternatives can be excluded for parity reasons again, whilst in the first three cases the set $\\{a, b, c\\}$ appears to be either $\\{7,8,9\\}$, or $\\{6,9,10\\}$, or $\\{7,9,10\\}$, respectively. Thus, a component containing a circuit should contain 9 as a vertex. Therefore there is at most one such component and hence at most one circuit.\n\nBy now we know that $\\mathcal{G}$ is a graph with $4 k$ edges, at least $k$ components and at most one circuit. Consequently, $\\mathcal{G}$ must have at least $4 k+k-1$ vertices. Thus $5 k-1 \\leq 18$, and $k \\leq 3$.']",['3'],False,,Numerical, 1695,Combinatorics,,"Let $m$ be a positive integer and consider a checkerboard consisting of $m$ by $m$ unit squares. At the midpoints of some of these unit squares there is an ant. At time 0, each ant starts moving with speed 1 parallel to some edge of the checkerboard. When two ants moving in opposite directions meet, they both turn $90^{\circ}$ clockwise and continue moving with speed 1 . When more than two ants meet, or when two ants moving in perpendicular directions meet, the ants continue moving in the same direction as before they met. When an ant reaches one of the edges of the checkerboard, it falls off and will not re-appear. Considering all possible starting positions, determine the latest possible moment at which the last ant falls off the checkerboard or prove that such a moment does not necessarily exist.","['For $m=1$ the answer is clearly correct, so assume $m>1$. In the sequel, the word collision will be used to denote meeting of exactly two ants, moving in opposite directions.\n\nIf at the beginning we place an ant on the southwest corner square facing east and an ant on the southeast corner square facing west, then they will meet in the middle of the bottom row at time $\\frac{m-1}{2}$. After the collision, the ant that moves to the north will stay on the board for another $m-\\frac{1}{2}$ time units and thus we have established an example in which the last ant falls off at time $\\frac{m-1}{2}+m-\\frac{1}{2}=\\frac{3 m}{2}-1$. So, we are left to prove that this is the latest possible moment.\n\nConsider any collision of two ants $a$ and $a^{\\prime}$. Let us change the rule for this collision, and enforce these two ants to turn anticlockwise. Then the succeeding behavior of all the ants does not change; the only difference is that $a$ and $a^{\\prime}$ swap their positions. These arguments may be applied to any collision separately, so we may assume that at any collision, either both ants rotate clockwise or both of them rotate anticlockwise by our own choice.\n\nFor instance, we may assume that there are only two types of ants, depending on their initial direction: NE-ants, which move only north or east, and $S W$-ants, moving only south and west. Then we immediately obtain that all ants will have fallen off the board after $2 m-1$ time units. However, we can get a better bound by considering the last moment at which a given ant collides with another ant.\n\nChoose a coordinate system such that the corners of the checkerboard are $(0,0),(m, 0),(m, m)$ and $(0, m)$. At time $t$, there will be no NE-ants in the region $\\{(x, y): x+y2 m-t-1\\}$. So if two ants collide at $(x, y)$ at time $t$, we have\n\n$$\nt+1 \\leq x+y \\leq 2 m-t-1\n\\tag{1}\n$$\n\n\n\nAnalogously, we may change the rules so that each ant would move either alternatingly north and west, or alternatingly south and east. By doing so, we find that apart from (11) we also have $|x-y| \\leq m-t-1$ for each collision at point $(x, y)$ and time $t$.\n\nTo visualize this, put\n\n$$\nB(t)=\\left\\{(x, y) \\in[0, m]^{2}: t+1 \\leq x+y \\leq 2 m-t-1 \\text { and }|x-y| \\leq m-t-1\\right\\}\n$$\n\nAn ant can thus only collide with another ant at time $t$ if it happens to be in the region $B(t)$. The following figure displays $B(t)$ for $t=\\frac{1}{2}$ and $t=\\frac{7}{2}$ in the case $m=6$ :\n\n\n\nNow suppose that an NE-ant has its last collision at time $t$ and that it does so at the point $(x, y)$ (if the ant does not collide at all, it will fall off the board within $m-\\frac{1}{2}<\\frac{3 m}{2}-1$ time units, so this case can be ignored). Then we have $(x, y) \\in B(t)$ and thus $x+y \\geq t+1$ and $x-y \\geq-(m-t-1)$. So we get\n\n$$\nx \\geq \\frac{(t+1)-(m-t-1)}{2}=t+1-\\frac{m}{2}\n$$\n\nBy symmetry we also have $y \\geq t+1-\\frac{m}{2}$, and hence $\\min \\{x, y\\} \\geq t+1-\\frac{m}{2}$. After this collision, the ant will move directly to an edge, which will take at $\\operatorname{most} m-\\min \\{x, y\\}$ units of time. In sum, the total amount of time the ant stays on the board is at most\n\n$$\nt+(m-\\min \\{x, y\\}) \\leq t+m-\\left(t+1-\\frac{m}{2}\\right)=\\frac{3 m}{2}-1\n$$\n\nBy symmetry, the same bound holds for SW-ants as well.']",['$\\frac{3 m}{2}-1$'],False,,Expression, 1696,Combinatorics,,"Let $n$ be a positive integer and let $W=\ldots x_{-1} x_{0} x_{1} x_{2} \ldots$ be an infinite periodic word consisting of the letters $a$ and $b$. Suppose that the minimal period $N$ of $W$ is greater than $2^{n}$. A finite nonempty word $U$ is said to appear in $W$ if there exist indices $k \leq \ell$ such that $U=x_{k} x_{k+1} \ldots x_{\ell}$. A finite word $U$ is called ubiquitous if the four words $U a, U b, a U$, and $b U$ all appear in $W$. Prove that there are at least $n$ ubiquitous finite nonempty words.","['Throughout the solution, all the words are nonempty. For any word $R$ of length $m$, we call the number of indices $i \\in\\{1,2, \\ldots, N\\}$ for which $R$ coincides with the subword $x_{i+1} x_{i+2} \\ldots x_{i+m}$ of $W$ the multiplicity of $R$ and denote it by $\\mu(R)$. Thus a word $R$ appears in $W$ if and only if $\\mu(R)>0$. Since each occurrence of a word in $W$ is both succeeded by either the letter $a$ or the letter $b$ and similarly preceded by one of those two letters, we have\n\n$$\n\\mu(R)=\\mu(R a)+\\mu(R b)=\\mu(a R)+\\mu(b R)\n\\tag{1}\n$$\n\nfor all words $R$.\n\nWe claim that the condition that $N$ is in fact the minimal period of $W$ guarantees that each word of length $N$ has multiplicity 1 or 0 depending on whether it appears or not. Indeed, if the words $x_{i+1} x_{i+2} \\ldots x_{i+N}$ and $x_{j+1} \\ldots x_{j+N}$ are equal for some $1 \\leq i2^{n}$, at least one of the two words $a$ and $b$ has a multiplicity that is strictly larger than $2^{n-1}$.\n\nFor each $k=0,1, \\ldots, n-1$, let $U_{k}$ be a subword of $W$ whose multiplicity is strictly larger than $2^{k}$ and whose length is maximal subject to this property. Note that such a word exists in view of the two observations made in the two previous paragraphs.\n\nFix some index $k \\in\\{0,1, \\ldots, n-1\\}$. Since the word $U_{k} b$ is longer than $U_{k}$, its multiplicity can be at most $2^{k}$, so in particular $\\mu\\left(U_{k} b\\right)<\\mu\\left(U_{k}\\right)$. Therefore, the word $U_{k} a$ has to appear by (1). For a similar reason, the words $U_{k} b, a U_{k}$, and $b U_{k}$ have to appear as well. Hence, the word $U_{k}$ is ubiquitous. Moreover, if the multiplicity of $U_{k}$ were strictly greater than $2^{k+1}$, then by (11) at least one of the two words $U_{k} a$ and $U_{k} b$ would have multiplicity greater than $2^{k}$ and would thus violate the maximality condition imposed on $U_{k}$.\n\nSo we have $\\mu\\left(U_{0}\\right) \\leq 2<\\mu\\left(U_{1}\\right) \\leq 4<\\ldots \\leq 2^{n-1}<\\mu\\left(U_{n-1}\\right)$, which implies in particular that the words $U_{0}, U_{1}, \\ldots, U_{n-1}$ have to be distinct. As they have been proved to be ubiquitous as well, the problem is solved.']",,True,,, 1697,Combinatorics,,"On a square table of 2011 by 2011 cells we place a finite number of napkins that each cover a square of 52 by 52 cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is the largest value of $k$ ?","['Let $m=39$, then $2011=52 m-17$. We begin with an example showing that there can exist 3986729 cells carrying the same positive number.\n\n\n\nTo describe it, we number the columns from the left to the right and the rows from the bottom to the top by $1,2, \\ldots, 2011$. We will denote each napkin by the coordinates of its lowerleft cell. There are four kinds of napkins: first, we take all napkins $(52 i+36,52 j+1)$ with $0 \\leq j \\leq i \\leq m-2$; second, we use all napkins $(52 i+1,52 j+36)$ with $0 \\leq i \\leq j \\leq m-2$; third, we use all napkins $(52 i+36,52 i+36)$ with $0 \\leq i \\leq m-2$; and finally the napkin $(1,1)$. Different groups of napkins are shown by different types of hatchings in the picture.\n\nNow except for those squares that carry two or more different hatchings, all squares have the number 1 written into them. The number of these exceptional cells is easily computed to be $\\left(52^{2}-35^{2}\\right) m-17^{2}=57392$.\n\nWe are left to prove that 3986729 is an upper bound for the number of cells containing the same number. Consider any configuration of napkins and any positive integer $M$. Suppose there are $g$ cells with a number different from $M$. Then it suffices to show $g \\geq 57392$. Throughout the solution, a line will mean either a row or a column.\n\nConsider any line $\\ell$. Let $a_{1}, \\ldots, a_{52 m-17}$ be the numbers written into its consecutive cells. For $i=1,2, \\ldots, 52$, let $s_{i}=\\sum_{t \\equiv i(\\bmod 52)} a_{t}$. Note that $s_{1}, \\ldots, s_{35}$ have $m$ terms each, while $s_{36}, \\ldots, s_{52}$ have $m-1$ terms each. Every napkin intersecting $\\ell$ contributes exactly 1 to each $s_{i}$;\n\n\n\nhence the number $s$ of all those napkins satisfies $s_{1}=\\cdots=s_{52}=s$. Call the line $\\ell$ rich if $s>(m-1) M$ and poor otherwise.\n\nSuppose now that $\\ell$ is rich. Then in each of the sums $s_{36}, \\ldots, s_{52}$ there exists a term greater than $M$; consider all these terms and call the corresponding cells the rich bad cells for this line. So, each rich line contains at least 17 cells that are bad for this line.\n\nIf, on the other hand, $\\ell$ is poor, then certainly $s\n\nLet $\\alpha=\\angle C A B$. The angles $\\angle C A B$ and $\\angle C^{\\prime} O B^{\\prime}$ are inscribed into the two circles with centers $O$ and $L$, respectively, so $\\angle C O B=2 \\angle C A B=2 \\alpha$ and $2 \\angle C^{\\prime} O B^{\\prime}=360^{\\circ}-\\angle C^{\\prime} L B^{\\prime}$. From the kite $A B^{\\prime} L C^{\\prime}$ we have $\\angle C^{\\prime} L B^{\\prime}=180^{\\circ}-\\angle C^{\\prime} A B^{\\prime}=180^{\\circ}-\\alpha$. Combining these, we get\n\n$$\n2 \\alpha=\\angle C O B<\\angle C^{\\prime} O B^{\\prime}=\\frac{360^{\\circ}-\\angle C^{\\prime} L B^{\\prime}}{2}=\\frac{360^{\\circ}-\\left(180^{\\circ}-\\alpha\\right)}{2}=90^{\\circ}+\\frac{\\alpha}{2}\n$$\n\nso\n\n$$\n\\alpha<60^{\\circ}\n$$\n\nLet $O^{\\prime}$ be the reflection of $O$ in the line $B C$. In the quadrilateral $A B O^{\\prime} C$ we have\n\n$$\n\\angle C O^{\\prime} B+\\angle C A B=\\angle C O B+\\angle C A B=2 \\alpha+\\alpha<180^{\\circ},\n$$\n\nso the point $O^{\\prime}$ is outside the circle $A B C$. Hence, $O$ and $O^{\\prime}$ are two points of $\\omega$ such that one of them lies inside the circumcircle, while the other one is located outside. Therefore, the two circles intersect.']",,True,,, 1699,Geometry,,"Let $A_{1} A_{2} A_{3} A_{4}$ be a non-cyclic quadrilateral. Let $O_{1}$ and $r_{1}$ be the circumcenter and the circumradius of the triangle $A_{2} A_{3} A_{4}$. Define $O_{2}, O_{3}, O_{4}$ and $r_{2}, r_{3}, r_{4}$ in a similar way. Prove that $$ \frac{1}{O_{1} A_{1}^{2}-r_{1}^{2}}+\frac{1}{O_{2} A_{2}^{2}-r_{2}^{2}}+\frac{1}{O_{3} A_{3}^{2}-r_{3}^{2}}+\frac{1}{O_{4} A_{4}^{2}-r_{4}^{2}}=0 $$","['Let $M$ be the point of intersection of the diagonals $A_{1} A_{3}$ and $A_{2} A_{4}$. On each diagonal choose a direction and let $x, y, z$, and $w$ be the signed distances from $M$ to the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$, respectively.\n\nLet $\\omega_{1}$ be the circumcircle of the triangle $A_{2} A_{3} A_{4}$ and let $B_{1}$ be the second intersection point of $\\omega_{1}$ and $A_{1} A_{3}$ (thus, $B_{1}=A_{3}$ if and only if $A_{1} A_{3}$ is tangent to $\\omega_{1}$ ). Since the expression $O_{1} A_{1}^{2}-r_{1}^{2}$ is the power of the point $A_{1}$ with respect to $\\omega_{1}$, we get\n\n$$\nO_{1} A_{1}^{2}-r_{1}^{2}=A_{1} B_{1} \\cdot A_{1} A_{3} .\n$$\n\nOn the other hand, from the equality $M B_{1} \\cdot M A_{3}=M A_{2} \\cdot M A_{4}$ we obtain $M B_{1}=y w / z$. Hence, we have\n\n$$\nO_{1} A_{1}^{2}-r_{1}^{2}=\\left(\\frac{y w}{z}-x\\right)(z-x)=\\frac{z-x}{z}(y w-x z) .\n$$\n\nSubstituting the analogous expressions into the sought sum we get\n\n$$\n\\sum_{i=1}^{4} \\frac{1}{O_{i} A_{i}^{2}-r_{i}^{2}}=\\frac{1}{y w-x z}\\left(\\frac{z}{z-x}-\\frac{w}{w-y}+\\frac{x}{x-z}-\\frac{y}{y-w}\\right)=0\n$$\n\nas desired.', 'Introduce a Cartesian coordinate system in the plane. Every circle has an equation of the form $p(x, y)=x^{2}+y^{2}+l(x, y)=0$, where $l(x, y)$ is a polynomial of degree at most 1 . For any point $A=\\left(x_{A}, y_{A}\\right)$ we have $p\\left(x_{A}, y_{A}\\right)=d^{2}-r^{2}$, where $d$ is the distance from $A$ to the center of the circle and $r$ is the radius of the circle.\n\nFor each $i$ in $\\{1,2,3,4\\}$ let $p_{i}(x, y)=x^{2}+y^{2}+l_{i}(x, y)=0$ be the equation of the circle with center $O_{i}$ and radius $r_{i}$ and let $d_{i}$ be the distance from $A_{i}$ to $O_{i}$. Consider the equation\n\n$$\n\\sum_{i=1}^{4} \\frac{p_{i}(x, y)}{d_{i}^{2}-r_{i}^{2}}=1\n\\tag{1}\n$$\n\n\n\nSince the coordinates of the points $A_{1}, A_{2}, A_{3}$, and $A_{4}$ satisfy (1) but these four points do not lie on a circle or on an line, equation (1) defines neither a circle, nor a line. Hence, the equation is an identity and the coefficient of the quadratic term $x^{2}+y^{2}$ also has to be zero, i.e.\n\n$$\n\\sum_{i=1}^{4} \\frac{1}{d_{i}^{2}-r_{i}^{2}}=0\n$$']",,True,,, 1700,Geometry,,"Let $A B C D$ be a convex quadrilateral whose sides $A D$ and $B C$ are not parallel. Suppose that the circles with diameters $A B$ and $C D$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_{E}$ be the circle through the feet of the perpendiculars from $E$ to the lines $A B, B C$, and $C D$. Let $\omega_{F}$ be the circle through the feet of the perpendiculars from $F$ to the lines $C D, D A$, and $A B$. Prove that the midpoint of the segment $E F$ lies on the line through the two intersection points of $\omega_{E}$ and $\omega_{F}$.","['Denote by $P, Q, R$, and $S$ the projections of $E$ on the lines $D A, A B, B C$, and $C D$ respectively. The points $P$ and $Q$ lie on the circle with diameter $A E$, so $\\angle Q P E=\\angle Q A E$; analogously, $\\angle Q R E=\\angle Q B E$. So $\\angle Q P E+\\angle Q R E=\\angle Q A E+\\angle Q B E=90^{\\circ}$. By similar reasons, we have $\\angle S P E+\\angle S R E=90^{\\circ}$, hence we get $\\angle Q P S+\\angle Q R S=90^{\\circ}+90^{\\circ}=180^{\\circ}$, and the quadrilateral $P Q R S$ is inscribed in $\\omega_{E}$. Analogously, all four projections of $F$ onto the sides of $A B C D$ lie on $\\omega_{F}$.\n\nDenote by $K$ the meeting point of the lines $A D$ and $B C$. Due to the arguments above, there is no loss of generality in assuming that $A$ lies on segment $D K$. Suppose that $\\angle C K D \\geq 90^{\\circ}$; then the circle with diameter $C D$ covers the whole quadrilateral $A B C D$, so the points $E, F$ cannot lie inside this quadrilateral. Hence our assumption is wrong. Therefore, the lines $E P$ and $B C$ intersect at some point $P^{\\prime}$, while the lines $E R$ and $A D$ intersect at some point $R^{\\prime}$.\n\n\n\nFigure 1\n\nWe claim that the points $P^{\\prime}$ and $R^{\\prime}$ also belong to $\\omega_{E}$. Since the points $R, E, Q, B$ are concyclic, $\\angle Q R K=\\angle Q E B=90^{\\circ}-\\angle Q B E=\\angle Q A E=\\angle Q P E$. So $\\angle Q R K=\\angle Q P P^{\\prime}$, which means that the point $P^{\\prime}$ lies on $\\omega_{E}$. Analogously, $R^{\\prime}$ also lies on $\\omega_{E}$.\n\nIn the same manner, denote by $M$ and $N$ the projections of $F$ on the lines $A D$ and $B C$\n\n\n\nrespectively, and let $M^{\\prime}=F M \\cap B C, N^{\\prime}=F N \\cap A D$. By the same arguments, we obtain that the points $M^{\\prime}$ and $N^{\\prime}$ belong to $\\omega_{F}$.\n\n\n\nFigure 2\n\nNow we concentrate on Figure 2, where all unnecessary details are removed. Let $U=N N^{\\prime} \\cap$ $P P^{\\prime}, V=M M^{\\prime} \\cap R R^{\\prime}$. Due to the right angles at $N$ and $P$, the points $N, N^{\\prime}, P, P^{\\prime}$ are concyclic, so $U N \\cdot U N^{\\prime}=U P \\cdot U P^{\\prime}$ which means that $U$ belongs to the radical axis $g$ of the circles $\\omega_{E}$ and $\\omega_{F}$. Analogously, $V$ also belongs to $g$.\n\nFinally, since $E U F V$ is a parallelogram, the radical axis $U V$ of $\\omega_{E}$ and $\\omega_{F}$ bisects $E F$.']",,True,,, 1701,Geometry,,"Let $A B C$ be an acute triangle with circumcircle $\Omega$. Let $B_{0}$ be the midpoint of $A C$ and let $C_{0}$ be the midpoint of $A B$. Let $D$ be the foot of the altitude from $A$, and let $G$ be the centroid of the triangle $A B C$. Let $\omega$ be a circle through $B_{0}$ and $C_{0}$ that is tangent to the circle $\Omega$ at a point $X \neq A$. Prove that the points $D, G$, and $X$ are collinear.","['If $A B=A C$, then the statement is trivial. So without loss of generality we may assume $A B\n\nDenote by $T$ the second intersection point of $\\Omega$ and the line $D X$. Note that $O$ belongs to $\\Omega_{1}$. Using the circles $\\gamma$ and $\\Omega$, we find $\\angle D A T=\\angle A D X-\\angle A T D=\\frac{1}{2}\\left(360^{\\circ}-\\angle A W X\\right)-\\frac{1}{2} \\angle A O X=$ $180^{\\circ}-\\frac{1}{2}(\\angle A W X+\\angle A O X)=90^{\\circ}$. So, $A D \\perp A T$, and hence $A T \\| B C$. Thus, $A T C B$ is an isosceles trapezoid inscribed in $\\Omega$.\n\nDenote by $A_{0}$ the midpoint of $B C$, and consider the image of $A T C B$ under the homothety $h$ with center $G$ and factor $-\\frac{1}{2}$. We have $h(A)=A_{0}, h(B)=B_{0}$, and $h(C)=C_{0}$. From the\n\n\n\nsymmetry about $B_{0} C_{0}$, we have $\\angle T C B=\\angle C B A=\\angle B_{0} C_{0} A=\\angle D C_{0} B_{0}$. Using $A T \\| D A_{0}$, we conclude $h(T)=D$. Hence the points $D, G$, and $T$ are collinear, and $X$ lies on the same line.', 'We define the points $A_{0}, O$, and $W$ as in the previous solution and we concentrate on the case $A B\n\nTo complete the argument, we note that the homothety centered at $G$ sending the triangle $A B C$ to the triangle $A_{0} B_{0} C_{0}$ maps the altitude $A D$ to the altitude $A_{0} Q$. Therefore it maps $D$ to $Q$, so the points $D, G$, and $Q$ are collinear. Hence $G$ lies on $\\ell$ as well.']",,True,,, 1702,Geometry,,"Let $A B C$ be a triangle with incenter $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with the lines $A I$ and $B I$, respectively. The chord $D E$ meets $A C$ at a point $F$, and $B C$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $A D$ and the line through $G$ parallel to $B E$. Suppose that the tangents to $\omega$ at $A$ and at $B$ meet at a point $K$. Prove that the three lines $A E, B D$, and $K P$ are either parallel or concurrent.","['Since\n\n$$\n\\angle I A F=\\angle D A C=\\angle B A D=\\angle B E D=\\angle I E F\n$$\n\nthe quadrilateral $A I F E$ is cyclic. Denote its circumcircle by $\\omega_{1}$. Similarly, the quadrilateral $B D G I$ is cyclic; denote its circumcircle by $\\omega_{2}$.\n\nThe line $A E$ is the radical axis of $\\omega$ and $\\omega_{1}$, and the line $B D$ is the radical axis of $\\omega$ and $\\omega_{2}$. Let $t$ be the radical axis of $\\omega_{1}$ and $\\omega_{2}$. These three lines meet at the radical center of the three circles, or they are parallel to each other. We will show that $t$ is in fact the line $P K$.\n\nLet $L$ be the second intersection point of $\\omega_{1}$ and $\\omega_{2}$, so $t=I L$. (If the two circles are tangent to each other then $L=I$ and $t$ is the common tangent.)\n\n\n\nLet the line $t$ meet the circumcircles of the triangles $A B L$ and $F G L$ at $K^{\\prime} \\neq L$ and $P^{\\prime} \\neq L$, respectively. Using oriented angles we have\n\n$$\n\\angle\\left(A B, B K^{\\prime}\\right)=\\angle\\left(A L, L K^{\\prime}\\right)=\\angle(A L, L I)=\\angle(A E, E I)=\\angle(A E, E B)=\\angle(A B, B K),\n$$\n\n\n\nso $B K^{\\prime} \\| B K$. Similarly we have $A K^{\\prime} \\| A K$, and therefore $K^{\\prime}=K$. Next, we have\n\n$$\n\\angle\\left(P^{\\prime} F, F G\\right)=\\angle\\left(P^{\\prime} L, L G\\right)=\\angle(I L, L G)=\\angle(I D, D G)=\\angle(A D, D E)=\\angle(P F, F G),\n$$\n\nhence $P^{\\prime} F \\| P F$ and similarly $P^{\\prime} G \\| P G$. Therefore $P^{\\prime}=P$. This means that $t$ passes through $K$ and $P$, which finishes the proof.', ""Let $M$ be the intersection point of the tangents to $\\omega$ at $D$ and $E$, and let the lines $A E$ and $B D$ meet at $T$; if $A E$ and $B D$ are parallel, then let $T$ be their common ideal point. It is well-known that the points $K$ and $M$ lie on the line $T I$ (as a consequence of PASCAL's theorem, applied to the inscribed degenerate hexagons $A A D B B E$ and $A D D B E E$ ).\n\nThe lines $A D$ and $B E$ are the angle bisectors of the angles $\\angle C A B$ and $\\angle A B C$, respectively, so $D$ and $E$ are the midpoints of the arcs $B C$ and $C A$ of the circle $\\omega$, respectively. Hence, $D M$ is parallel to $B C$ and $E M$ is parallel to $A C$.\n\nApply PASCAL's theorem to the degenerate hexagon $C A D D E B$. By the theorem, the points $C A \\cap D E=F, A D \\cap E B=I$ and the common ideal point of lines $D M$ and $B C$ are collinear, therefore $F I$ is parallel to $B C$ and $D M$. Analogously, the line $G I$ is parallel to $A C$ and $E M$.\n\n\n\nNow consider the homothety with scale factor $-\\frac{F G}{E D}$ which takes $E$ to $G$ and $D$ to $F$. Since the triangles $E D M$ and $G F I$ have parallel sides, the homothety takes $M$ to $I$. Similarly, since the triangles $D E I$ and $F G P$ have parallel sides, the homothety takes $I$ to $P$. Hence, the points $M, I, P$ and the homothety center $H$ must lie on the same line. Therefore, the point $P$ also lies on the line $T K I M$.""]",,True,,, 1703,Geometry,,"Let $A B C$ be a triangle with $A B=A C$, and let $D$ be the midpoint of $A C$. The angle bisector of $\angle B A C$ intersects the circle through $D, B$, and $C$ in a point $E$ inside the triangle $A B C$. The line $B D$ intersects the circle through $A, E$, and $B$ in two points $B$ and $F$. The lines $A F$ and $B E$ meet at a point $I$, and the lines $C I$ and $B D$ meet at a point $K$. Show that $I$ is the incenter of triangle $K A B$.","[""Let $D^{\\prime}$ be the midpoint of the segment $A B$, and let $M$ be the midpoint of $B C$. By symmetry at line $A M$, the point $D^{\\prime}$ has to lie on the circle $B C D$. Since the $\\operatorname{arcs} D^{\\prime} E$ and $E D$ of that circle are equal, we have $\\angle A B I=\\angle D^{\\prime} B E=\\angle E B D=I B K$, so $I$ lies on the angle bisector of $\\angle A B K$. For this reason it suffices to prove in the sequel that the ray $A I$ bisects the angle $\\angle B A K$.\n\nFrom\n\n$$\n\\angle D F A=180^{\\circ}-\\angle B F A=180^{\\circ}-\\angle B E A=\\angle M E B=\\frac{1}{2} \\angle C E B=\\frac{1}{2} \\angle C D B\n$$\n\nwe derive $\\angle D F A=\\angle D A F$ so the triangle $A F D$ is isosceles with $A D=D F$.\n\n\n\nApplying Menelaus's theorem to the triangle $A D F$ with respect to the line $C I K$, and applying the angle bisector theorem to the triangle $A B F$, we infer\n\n$$\n1=\\frac{A C}{C D} \\cdot \\frac{D K}{K F} \\cdot \\frac{F I}{I A}=2 \\cdot \\frac{D K}{K F} \\cdot \\frac{B F}{A B}=2 \\cdot \\frac{D K}{K F} \\cdot \\frac{B F}{2 \\cdot A D}=\\frac{D K}{K F} \\cdot \\frac{B F}{A D}\n$$\n\nand therefore\n\n$$\n\\frac{B D}{A D}=\\frac{B F+F D}{A D}=\\frac{B F}{A D}+1=\\frac{K F}{D K}+1=\\frac{D F}{D K}=\\frac{A D}{D K}\n$$\n\n\n\nIt follows that the triangles $A D K$ and $B D A$ are similar, hence $\\angle D A K=\\angle A B D$. Then\n\n$$\n\\angle I A B=\\angle A F D-\\angle A B D=\\angle D A F-\\angle D A K=\\angle K A I\n$$\n\nshows that the point $K$ is indeed lying on the angle bisector of $\\angle B A K$."", 'It can be shown in the same way as in the first solution that $I$ lies on the angle bisector of $\\angle A B K$. Here we restrict ourselves to proving that $K I$ bisects $\\angle A K B$.\n\n\n\nDenote the circumcircle of triangle $B C D$ and its center by $\\omega_{1}$ and by $O_{1}$, respectively. Since the quadrilateral $A B F E$ is cyclic, we have $\\angle D F E=\\angle B A E=\\angle D A E$. By the same reason, we have $\\angle E A F=\\angle E B F=\\angle A B E=\\angle A F E$. Therefore $\\angle D A F=\\angle D F A$, and hence $D F=D A=D C$. So triangle $A F C$ is inscribed in a circle $\\omega_{2}$ with center $D$.\n\nDenote the circumcircle of triangle $A B D$ by $\\omega_{3}$, and let its center be $O_{3}$. Since the arcs $B E$ and $E C$ of circle $\\omega_{1}$ are equal, and the triangles $A D E$ and $F D E$ are congruent, we have $\\angle A O_{1} B=2 \\angle B D E=\\angle B D A$, so $O_{1}$ lies on $\\omega_{3}$. Hence $\\angle O_{3} O_{1} D=\\angle O_{3} D O_{1}$.\n\nThe line $B D$ is the radical axis of $\\omega_{1}$ and $\\omega_{3}$. Point $C$ belongs to the radical axis of $\\omega_{1}$ and $\\omega_{2}$, and $I$ also belongs to it since $A I \\cdot I F=B I \\cdot I E$. Hence $K=B D \\cap C I$ is the radical center of $\\omega_{1}$, $\\omega_{2}$, and $\\omega_{3}$, and $A K$ is the radical axis of $\\omega_{2}$ and $\\omega_{3}$. Now, the radical axes $A K, B K$ and $I K$ are perpendicular to the central lines $O_{3} D, O_{3} O_{1}$ and $O_{1} D$, respectively. By $\\angle O_{3} O_{1} D=\\angle O_{3} D O_{1}$, we get that $K I$ is the angle bisector of $\\angle A K B$.', 'Again, let $M$ be the midpoint of $B C$. As in the previous solutions, we can deduce $\\angle A B I=\\angle I B K$. We show that the point $I$ lies on the angle bisector of $\\angle K A B$.\n\nLet $G$ be the intersection point of the circles $A F C$ and $B C D$, different from $C$. The lines\n\n\n\n$C G, A F$, and $B E$ are the radical axes of the three circles $A G F C, C D B$, and $A B F E$, so $I=A F \\cap B E$ is the radical center of the three circles and $C G$ also passes through $I$.\n\n\n\nThe angle between line $D E$ and the tangent to the circle $B C D$ at $E$ is equal to $\\angle E B D=$ $\\angle E A F=\\angle A B E=\\angle A F E$. As the tangent at $E$ is perpendicular to $A M$, the line $D E$ is perpendicular to $A F$. The triangle $A F E$ is isosceles, so $D E$ is the perpendicular bisector of $A F$ and thus $A D=D F$. Hence, the point $D$ is the center of the circle $A F C$, and this circle passes through $M$ as well since $\\angle A M C=90^{\\circ}$.\n\nLet $B^{\\prime}$ be the reflection of $B$ in the point $D$, so $A B C B^{\\prime}$ is a parallelogram. Since $D C=D G$ we have $\\angle G C D=\\angle D B C=\\angle K B^{\\prime} A$. Hence, the quadrilateral $A K C B^{\\prime}$ is cyclic and thus $\\angle C A K=\\angle C B^{\\prime} K=\\angle A B D=2 \\angle M A I$. Then\n\n$$\n\\angle I A B=\\angle M A B-\\angle M A I=\\frac{1}{2} \\angle C A B-\\frac{1}{2} \\angle C A K=\\frac{1}{2} \\angle K A B\n$$\n\nand therefore $A I$ is the angle bisector of $\\angle K A B$.']",,True,,, 1704,Geometry,,Let $A B C D E F$ be a convex hexagon all of whose sides are tangent to a circle $\omega$ with center $O$. Suppose that the circumcircle of triangle $A C E$ is concentric with $\omega$. Let $J$ be the foot of the perpendicular from $B$ to $C D$. Suppose that the perpendicular from $B$ to $D F$ intersects the line $E O$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $D E$. Prove that $D J=D L$.,"['Since $\\omega$ and the circumcircle of triangle $A C E$ are concentric, the tangents from $A$, $C$, and $E$ to $\\omega$ have equal lengths; that means that $A B=B C, C D=D E$, and $E F=F A$. Moreover, we have $\\angle B C D=\\angle D E F=\\angle F A B$.\n\n\n\nConsider the rotation around point $D$ mapping $C$ to $E$; let $B^{\\prime}$ and $L^{\\prime}$ be the images of the points $B$ and $J$, respectively, under this rotation. Then one has $D J=D L^{\\prime}$ and $B^{\\prime} L^{\\prime} \\perp D E$; moreover, the triangles $B^{\\prime} E D$ and $B C D$ are congruent. Since $\\angle D E O<90^{\\circ}$, the lines $E O$ and $B^{\\prime} L^{\\prime}$ intersect at some point $K^{\\prime}$. We intend to prove that $K^{\\prime} B \\perp D F$; this would imply $K=K^{\\prime}$, therefore $L=L^{\\prime}$, which proves the problem statement.\n\nAnalogously, consider the rotation around $F$ mapping $A$ to $E$; let $B^{\\prime \\prime}$ be the image of $B$ under this rotation. Then the triangles $F A B$ and $F E B^{\\prime \\prime}$ are congruent. We have $E B^{\\prime \\prime}=A B=B C=$ $E B^{\\prime}$ and $\\angle F E B^{\\prime \\prime}=\\angle F A B=\\angle B C D=\\angle D E B^{\\prime}$, so the points $B^{\\prime}$ and $B^{\\prime \\prime}$ are symmetrical with respect to the angle bisector $E O$ of $\\angle D E F$. So, from $K^{\\prime} B^{\\prime} \\perp D E$ we get $K^{\\prime} B^{\\prime \\prime} \\perp E F$. From these two relations we obtain\n\n$$\nK^{\\prime} D^{2}-K^{\\prime} E^{2}=B^{\\prime} D^{2}-B^{\\prime} E^{2} \\text { and } K^{\\prime} E^{2}-K^{\\prime} F^{2}=B^{\\prime \\prime} E^{2}-B^{\\prime \\prime} F^{2} \\text {. }\n$$\n\nAdding these equalities and taking into account that $B^{\\prime} E=B^{\\prime \\prime} E$ we obtain\n\n$$\nK^{\\prime} D^{2}-K^{\\prime} F^{2}=B^{\\prime} D^{2}-B^{\\prime \\prime} F^{2}=B D^{2}-B F^{2},\n$$\n\n\n\nwhich means exactly that $K^{\\prime} B \\perp D F$.', 'Let us denote the points of tangency of $A B, B C, C D, D E, E F$, and $F A$ to $\\omega$ by $R, S, T, U, V$, and $W$, respectively. As in the previous solution, we mention that $A R=$ $A W=C S=C T=E U=E V$.\n\nThe reflection in the line $B O$ maps $R$ to $S$, therefore $A$ to $C$ and thus also $W$ to $T$. Hence, both lines $R S$ and $W T$ are perpendicular to $O B$, therefore they are parallel. On the other hand, the lines $U V$ and $W T$ are not parallel, since otherwise the hexagon $A B C D E F$ is symmetric with respect to the line $B O$ and the lines defining the point $K$ coincide, which contradicts the conditions of the problem. Therefore we can consider the intersection point $Z$ of $U V$ and $W T$.\n\n\n\nNext, we recall a well-known fact that the points $D, F, Z$ are collinear. Actually, $D$ is the pole of the line $U T, F$ is the pole of $V W$, and $Z=T W \\cap U V$; so all these points belong to the polar line of $T U \\cap V W$.\n\n\n\nNow, we put $O$ into the origin, and identify each point (say $X$ ) with the vector $\\overrightarrow{O X}$. So, from now on all the products of points refer to the scalar products of the corresponding vectors.\n\nSince $O K \\perp U Z$ and $O B \\perp T Z$, we have $K \\cdot(Z-U)=0=B \\cdot(Z-T)$. Next, the condition $B K \\perp D Z$ can be written as $K \\cdot(D-Z)=B \\cdot(D-Z)$. Adding these two equalities we get\n\n$$\nK \\cdot(D-U)=B \\cdot(D-T)\n$$\n\nBy symmetry, we have $D \\cdot(D-U)=D \\cdot(D-T)$. Subtracting this from the previous equation, we obtain $(K-D) \\cdot(D-U)=(B-D) \\cdot(D-T)$ and rewrite it in vector form as\n\n$$\n\\overrightarrow{D K} \\cdot \\overrightarrow{U D}=\\overrightarrow{D B} \\cdot \\overrightarrow{T D}\n$$\n\nFinally, projecting the vectors $\\overrightarrow{D K}$ and $\\overrightarrow{D B}$ onto the lines $U D$ and $T D$ respectively, we can rewrite this equality in terms of segment lengths as $D L \\cdot U D=D J \\cdot T D$, thus $D L=D J$.']",,True,,, 1705,Geometry,,"Let $A B C$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $B C, C A$, and $A B$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is tangent to the circle $\omega$. To avoid a large case distinction, we will use the notion of oriented angles. Namely, for two lines $\ell$ and $m$, we denote by $\angle(\ell, m)$ the angle by which one may rotate $\ell$ anticlockwise to obtain a line parallel to $m$. Thus, all oriented angles are considered modulo $180^{\circ}$. ![]","[""Denote by $T$ the point of tangency of $t$ and $\\omega$. Let $A^{\\prime}=t_{b} \\cap t_{c}, B^{\\prime}=t_{a} \\cap t_{c}$, $C^{\\prime}=t_{a} \\cap t_{b}$. Introduce the point $A^{\\prime \\prime}$ on $\\omega$ such that $T A=A A^{\\prime \\prime}\\left(A^{\\prime \\prime} \\neq T\\right.$ unless $T A$ is a diameter). Define the points $B^{\\prime \\prime}$ and $C^{\\prime \\prime}$ in a similar way.\n\nSince the points $C$ and $B$ are the midpoints of arcs $T C^{\\prime \\prime}$ and $T B^{\\prime \\prime}$, respectively, we have\n\n$$\n\\begin{aligned}\n\\angle\\left(t, B^{\\prime \\prime} C^{\\prime \\prime}\\right) & =\\angle\\left(t, T C^{\\prime \\prime}\\right)+\\angle\\left(T C^{\\prime \\prime}, B^{\\prime \\prime} C^{\\prime \\prime}\\right)=2 \\angle(t, T C)+2 \\angle\\left(T C^{\\prime \\prime}, B C^{\\prime \\prime}\\right) \\\\\n& =2(\\angle(t, T C)+\\angle(T C, B C))=2 \\angle(t, B C)=\\angle\\left(t, t_{a}\\right) .\n\\end{aligned}\n$$\n\nIt follows that $t_{a}$ and $B^{\\prime \\prime} C^{\\prime \\prime}$ are parallel. Similarly, $t_{b} \\| A^{\\prime \\prime} C^{\\prime \\prime}$ and $t_{c} \\| A^{\\prime \\prime} B^{\\prime \\prime}$. Thus, either the triangles $A^{\\prime} B^{\\prime} C^{\\prime}$ and $A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime}$ are homothetic, or they are translates of each other. Now we will prove that they are in fact homothetic, and that the center $K$ of the homothety belongs\n\n\n\nto $\\omega$. It would then follow that their circumcircles are also homothetic with respect to $K$ and are therefore tangent at this point, as desired.\n\nWe need the two following claims.\n\nClaim 1. The point of intersection $X$ of the lines $B^{\\prime \\prime} C$ and $B C^{\\prime \\prime}$ lies on $t_{a}$.\n\nProof. Actually, the points $X$ and $T$ are symmetric about the line $B C$, since the lines $C T$ and $C B^{\\prime \\prime}$ are symmetric about this line, as are the lines $B T$ and $B C^{\\prime \\prime}$.\n\nClaim 2. The point of intersection $I$ of the lines $B B^{\\prime}$ and $C C^{\\prime}$ lies on the circle $\\omega$.\n\nProof. We consider the case that $t$ is not parallel to the sides of $A B C$; the other cases may be regarded as limit cases. Let $D=t \\cap B C, E=t \\cap A C$, and $F=t \\cap A B$.\n\nDue to symmetry, the line $D B$ is one of the angle bisectors of the lines $B^{\\prime} D$ and $F D$; analogously, the line $F B$ is one of the angle bisectors of the lines $B^{\\prime} F$ and $D F$. So $B$ is either the incenter or one of the excenters of the triangle $B^{\\prime} D F$. In any case we have $\\angle(B D, D F)+\\angle(D F, F B)+$ $\\angle\\left(B^{\\prime} B, B^{\\prime} D\\right)=90^{\\circ}$, so\n\n$$\n\\angle\\left(B^{\\prime} B, B^{\\prime} C^{\\prime}\\right)=\\angle\\left(B^{\\prime} B, B^{\\prime} D\\right)=90^{\\circ}-\\angle(B C, D F)-\\angle(D F, B A)=90^{\\circ}-\\angle(B C, A B) .\n$$\n\nAnalogously, we get $\\angle\\left(C^{\\prime} C, B^{\\prime} C^{\\prime}\\right)=90^{\\circ}-\\angle(B C, A C)$. Hence,\n\n$$\n\\angle(B I, C I)=\\angle\\left(B^{\\prime} B, B^{\\prime} C^{\\prime}\\right)+\\angle\\left(B^{\\prime} C^{\\prime}, C^{\\prime} C\\right)=\\angle(B C, A C)-\\angle(B C, A B)=\\angle(A B, A C),\n$$\n\nwhich means exactly that the points $A, B, I, C$ are concyclic.\n\nNow we can complete the proof. Let $K$ be the second intersection point of $B^{\\prime} B^{\\prime \\prime}$ and $\\omega$. Applying PASCAL's theorem to hexagon $K B^{\\prime \\prime} C I B C^{\\prime \\prime}$ we get that the points $B^{\\prime}=K B^{\\prime \\prime} \\cap I B$ and $X=B^{\\prime \\prime} C \\cap B C^{\\prime \\prime}$ are collinear with the intersection point $S$ of $C I$ and $C^{\\prime \\prime} K$. So $S=$ $C I \\cap B^{\\prime} X=C^{\\prime}$, and the points $C^{\\prime}, C^{\\prime \\prime}, K$ are collinear. Thus $K$ is the intersection point of $B^{\\prime} B^{\\prime \\prime}$ and $C^{\\prime} C^{\\prime \\prime}$ which implies that $K$ is the center of the homothety mapping $A^{\\prime} B^{\\prime} C^{\\prime}$ to $A^{\\prime \\prime} B^{\\prime \\prime} C^{\\prime \\prime}$, and it belongs to $\\omega$."", ""Define the points $T, A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $B C, C A$, and $A B$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $A B C$, therefore the points $X, Y, Z$ are also collinear. Moreover, we have $X \\in B^{\\prime} C^{\\prime}, Y \\in C^{\\prime} A^{\\prime}$, $Z \\in A^{\\prime} B^{\\prime}$.\n\nDenote $\\alpha=\\angle(t, T C)=\\angle(B T, B C)$. Using the symmetry in the lines $A C$ and $B C$, we get\n\n$$\n\\angle(B C, B X)=\\angle(B T, B C)=\\alpha \\quad \\text { and } \\quad \\angle\\left(X C, X C^{\\prime}\\right)=\\angle(t, T C)=\\angle\\left(Y C, Y C^{\\prime}\\right)=\\alpha \\text {. }\n$$\n\nSince $\\angle\\left(X C, X C^{\\prime}\\right)=\\angle\\left(Y C, Y C^{\\prime}\\right)$, the points $X, Y, C, C^{\\prime}$ lie on some circle $\\omega_{c}$. Define the circles $\\omega_{a}$ and $\\omega_{b}$ analogously. Let $\\omega^{\\prime}$ be the circumcircle of triangle $A^{\\prime} B^{\\prime} C^{\\prime}$.\n\n\n\nNow, applying MiqueL's theorem to the four lines $A^{\\prime} B^{\\prime}, A^{\\prime} C^{\\prime}, B^{\\prime} C^{\\prime}$, and $X Y$, we obtain that the circles $\\omega^{\\prime}, \\omega_{a}, \\omega_{b}, \\omega_{c}$ intersect at some point $K$. We will show that $K$ lies on $\\omega$, and that the tangent lines to $\\omega$ and $\\omega^{\\prime}$ at this point coincide; this implies the problem statement.\n\nDue to symmetry, we have $X B=T B=Z B$, so the point $B$ is the midpoint of one of the $\\operatorname{arcs} X Z$ of circle $\\omega_{b}$. Therefore $\\angle(K B, K X)=\\angle(X Z, X B)$. Analogously, $\\angle(K X, K C)=$ $\\angle(X C, X Y)$. Adding these equalities and using the symmetry in the line $B C$ we get\n\n$$\n\\angle(K B, K C)=\\angle(X Z, X B)+\\angle(X C, X Z)=\\angle(X C, X B)=\\angle(T B, T C) .\n$$\n\nTherefore, $K$ lies on $\\omega$.\n\nNext, let $k$ be the tangent line to $\\omega$ at $K$. We have\n\n$$\n\\begin{aligned}\n\\angle\\left(k, K C^{\\prime}\\right) & =\\angle(k, K C)+\\angle\\left(K C, K C^{\\prime}\\right)=\\angle(K B, B C)+\\angle\\left(X C, X C^{\\prime}\\right) \\\\\n& =(\\angle(K B, B X)-\\angle(B C, B X))+\\alpha=\\angle\\left(K B^{\\prime}, B^{\\prime} X\\right)-\\alpha+\\alpha=\\angle\\left(K B^{\\prime}, B^{\\prime} C^{\\prime}\\right),\n\\end{aligned}\n$$\n\nwhich means exactly that $k$ is tangent to $\\omega^{\\prime}$.\n\n""]",,True,,, 1706,Number Theory,,"For any integer $d>0$, let $f(d)$ be the smallest positive integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16$, and $f(6)=12$ ). Prove that for every integer $k \geq 0$ the number $f\left(2^{k}\right)$ divides $f\left(2^{k+1}\right)$.","['For any positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. Let $n=\\prod_{p} p^{a(p)}$ be the prime factorization of $n$ where $p$ ranges over the prime numbers, the integers $a(p)$ are nonnegative and all but finitely many $a(p)$ are zero. Then we have $d(n)=\\prod_{p}(a(p)+1)$. Thus, $d(n)$ is a power of 2 if and only if for every prime $p$ there is a nonnegative integer $b(p)$ with $a(p)=2^{b(p)}-1=1+2+2^{2}+\\cdots+2^{b(p)-1}$. We then have\n\n$$\nn=\\prod_{p} \\prod_{i=0}^{b(p)-1} p^{2^{i}}, \\quad \\text { and } \\quad d(n)=2^{k} \\quad \\text { with } \\quad k=\\sum_{p} b(p)\n$$\n\nLet $\\mathcal{S}$ be the set of all numbers of the form $p^{2^{r}}$ with $p$ prime and $r$ a nonnegative integer. Then we deduce that $d(n)$ is a power of 2 if and only if $n$ is the product of the elements of some finite subset $\\mathcal{T}$ of $\\mathcal{S}$ that satisfies the following condition: for all $t \\in \\mathcal{T}$ and $s \\in \\mathcal{S}$ with $s \\mid t$ we have $s \\in \\mathcal{T}$. Moreover, if $d(n)=2^{k}$ then the corresponding set $\\mathcal{T}$ has $k$ elements.\n\nNote that the $\\operatorname{set} \\mathcal{T}_{k}$ consisting of the smallest $k$ elements from $\\mathcal{S}$ obviously satisfies the condition above. Thus, given $k$, the smallest $n$ with $d(n)=2^{k}$ is the product of the elements of $\\mathcal{T}_{k}$. This $n$ is $f\\left(2^{k}\\right)$. Since obviously $\\mathcal{T}_{k} \\subset \\mathcal{T}_{k+1}$, it follows that $f\\left(2^{k}\\right) \\mid f\\left(2^{k+1}\\right)$.']",,True,,, 1707,Number Theory,,"Consider a polynomial $P(x)=\left(x+d_{1}\right)\left(x+d_{2}\right) \cdot \ldots \cdot\left(x+d_{9}\right)$, where $d_{1}, d_{2}, \ldots, d_{9}$ are nine distinct integers. Prove that there exists an integer $N$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20 .","['Note that the statement of the problem is invariant under translations of $x$; hence without loss of generality we may suppose that the numbers $d_{1}, d_{2}, \\ldots, d_{9}$ are positive.\n\nThe key observation is that there are only eight primes below 20, while $P(x)$ involves more than eight factors.\n\nWe shall prove that $N=d^{8}$ satisfies the desired property, where $d=\\max \\left\\{d_{1}, d_{2}, \\ldots, d_{9}\\right\\}$. Suppose for the sake of contradiction that there is some integer $x \\geq N$ such that $P(x)$ is composed of primes below 20 only. Then for every index $i \\in\\{1,2, \\ldots, 9\\}$ the number $x+d_{i}$ can be expressed as product of powers of the first 8 primes.\n\nSince $x+d_{i}>x \\geq d^{8}$ there is some prime power $f_{i}>d$ that divides $x+d_{i}$. Invoking the pigeonhole principle we see that there are two distinct indices $i$ and $j$ such that $f_{i}$ and $f_{j}$ are powers of the same prime number. For reasons of symmetry, we may suppose that $f_{i} \\leq f_{j}$. Now both of the numbers $x+d_{i}$ and $x+d_{j}$ are divisible by $f_{i}$ and hence so is their difference $d_{i}-d_{j}$. But as\n\n$$\n0<\\left|d_{i}-d_{j}\\right| \\leq \\max \\left(d_{i}, d_{j}\\right) \\leq dD_{i}$ the numerator of the fraction we thereby get cannot be 1 , and hence it has to be divisible by some prime number $p_{i}<20$.\n\nBy the pigeonhole principle, there are a prime number $p$ and two distinct indices $i$ and $j$ such that $p_{i}=p_{j}=p$. Let $p^{\\alpha_{i}}$ and $p^{\\alpha_{j}}$ be the greatest powers of $p$ dividing $x+d_{i}$ and $x+d_{j}$, respectively. Due to symmetry we may suppose $\\alpha_{i} \\leq \\alpha_{j}$. But now $p^{\\alpha_{i}}$ divides $d_{i}-d_{j}$ and hence also $D_{i}$, which means that all occurrences of $p$ in the numerator of the fraction $\\left(x+d_{i}\\right) / D_{i}$ cancel out, contrary to the choice of $p=p_{i}$. This contradiction proves our claim.', 'Given a nonzero integer $N$ as well as a prime number $p$ we write $v_{p}(N)$ for the exponent with which $p$ occurs in the prime factorization of $|N|$.\n\nEvidently, if the statement of the problem were not true, then there would exist an infinite sequence $\\left(x_{n}\\right)$ of positive integers tending to infinity such that for each $n \\in \\mathbb{Z}_{+}$the integer $P\\left(x_{n}\\right)$ is not divisible by any prime number $>20$. Observe that the numbers $-d_{1},-d_{2}, \\ldots,-d_{9}$ do not appear in this sequence.\n\nNow clearly there exists a prime $p_{1}<20$ for which the sequence $v_{p_{1}}\\left(x_{n}+d_{1}\\right)$ is not bounded; thinning out the sequence $\\left(x_{n}\\right)$ if necessary we may even suppose that\n\n$$\nv_{p_{1}}\\left(x_{n}+d_{1}\\right) \\longrightarrow \\infty\n$$\n\nRepeating this argument eight more times we may similarly choose primes $p_{2}, \\ldots, p_{9}<20$ and suppose that our sequence $\\left(x_{n}\\right)$ has been thinned out to such an extent that $v_{p_{i}}\\left(x_{n}+d_{i}\\right) \\longrightarrow \\infty$ holds for $i=2, \\ldots, 9$ as well. In view of the pigeonhole principle, there are distinct indices $i$ and $j$ as well as a prime $p<20$ such that $p_{i}=p_{j}=p$. Setting $k=v_{p}\\left(d_{i}-d_{j}\\right)$ there now has to be some $n$ for which both $v_{p}\\left(x_{n}+d_{i}\\right)$ and $v_{p}\\left(x_{n}+d_{j}\\right)$ are greater than $k$. But now the numbers $x_{n}+d_{i}$ and $x_{n}+d_{j}$ are divisible by $p^{k+1}$ whilst their difference $d_{i}-d_{j}$ is not - a contradiction.']",,True,,, 1708,Number Theory,,Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^{n}-y^{n}$.,"['Obviously, all functions in the answer satisfy the condition of the problem. We will show that there are no other functions satisfying that condition.\n\nLet $f$ be a function satisfying the given condition. For each integer $n$, the function $g$ defined by $g(x)=f(x)+n$ also satisfies the same condition. Therefore, by subtracting $f(0)$ from $f(x)$ we may assume that $f(0)=0$.\n\nFor any prime $p$, the condition on $f$ with $(x, y)=(p, 0)$ states that $f(p)$ divides $p^{n}$. Since the set of primes is infinite, there exist integers $d$ and $\\varepsilon$ with $0 \\leq d \\leq n$ and $\\varepsilon \\in\\{1,-1\\}$ such that for infinitely many primes $p$ we have $f(p)=\\varepsilon p^{d}$. Denote the set of these primes by $P$. Since a function $g$ satisfies the given condition if and only if $-g$ satisfies the same condition, we may suppose $\\varepsilon=1$.\n\nThe case $d=0$ is easily ruled out, because 0 does not divide any nonzero integer. Suppose $d \\geq 1$ and write $n$ as $m d+r$, where $m$ and $r$ are integers such that $m \\geq 1$ and $0 \\leq r \\leq d-1$. Let $x$ be an arbitrary integer. For each prime $p$ in $P$, the difference $f(p)-f(x)$ divides $p^{n}-x^{n}$. Using the equality $f(p)=p^{d}$, we get\n\n$$\np^{n}-x^{n}=p^{r}\\left(p^{d}\\right)^{m}-x^{n} \\equiv p^{r} f(x)^{m}-x^{n} \\equiv 0 \\quad\\left(\\bmod p^{d}-f(x)\\right)\n$$\n\nSince we have $r8$. For each positive integer $n$, there exists an $i \\in\\{0,1, \\ldots, a-5\\}$ such that $n+i=2 d$ for some odd $d$. We get\n\n$$\nt(n+i)=d \\not \\equiv d+2=t(n+i+4) \\quad(\\bmod 4)\n$$\n\nand\n\n$$\nt(n+a+i)=n+a+i \\equiv n+a+i+4=t(n+a+i+4) \\quad(\\bmod 4)\n$$\n\nTherefore, the integers $t(n+a+i)-t(n+i)$ and $t(n+a+i+4)-t(n+i+4)$ cannot be both divisible by 4 , and therefore there are no winning pairs in this case.\n\nCase 3: $a=7$. For each positive integer $n$, there exists an $i \\in\\{0,1, \\ldots, 6\\}$ such that $n+i$ is either of the form $8 k+3$ or of the form $8 k+6$, where $k$ is a nonnegative integer. But we have\n\n$$\nt(8 k+3) \\equiv 3 \\not \\equiv 1 \\equiv 4 k+5=t(8 k+3+7) \\quad(\\bmod 4)\n$$\n\nand\n\n$$\nt(8 k+6)=4 k+3 \\equiv 3 \\not \\equiv 1 \\equiv t(8 k+6+7) \\quad(\\bmod 4)\n$$\n\nHence, there are no winning pairs of the form $(7, n)$.']","['1,3,5']",True,,Numerical, 1710,Number Theory,,"Let $f$ be a function from the set of integers to the set of positive integers. Suppose that for any two integers $m$ and $n$, the difference $f(m)-f(n)$ is divisible by $f(m-n)$. Prove that for all integers $m, n$ with $f(m) \leq f(n)$ the number $f(n)$ is divisible by $f(m)$.","['Suppose that $x$ and $y$ are two integers with $f(x)0\n$$\n\nso $f(x-y) \\leq f(y)-f(x)f(1)$. Note that such a number exists due to the symmetry of $f$ obtained in Claim 2 .\n\n\n\nClaim 3. $f(n) \\neq f(1)$ if and only if $a \\mid n$.\n\nProof. Since $f(1)=\\cdots=f(a-1)0$, so $f(n+a) \\leq$ $f(a)-f(n)0$ this implies $M \\leq 3^{P(m+a x-b y)}-1$. But $P(m+a x-b y)$ is listed among $P(1), P(2), \\ldots, P(d)$, so\n\n$$\nM<3^{P(m+a x-b y)} \\leq 3^{\\max \\{P(1), P(2), \\ldots, P(d)\\}}\n$$\n\n\n\nwhich contradicts (1).""]",,True,,, 1712,Number Theory,,"Let $p$ be an odd prime number. For every integer $a$, define the number $$ S_{a}=\frac{a}{1}+\frac{a^{2}}{2}+\cdots+\frac{a^{p-1}}{p-1} $$ Let $m$ and $n$ be integers such that $$ S_{3}+S_{4}-3 S_{2}=\frac{m}{n} $$ Prove that $p$ divides $m$.","[""For rational numbers $p_{1} / q_{1}$ and $p_{2} / q_{2}$ with the denominators $q_{1}, q_{2}$ not divisible by $p$, we write $p_{1} / q_{1} \\equiv p_{2} / q_{2}(\\bmod p)$ if the numerator $p_{1} q_{2}-p_{2} q_{1}$ of their difference is divisible by $p$.\n\nWe start with finding an explicit formula for the residue of $S_{a}$ modulo $p$. Note first that for every $k=1, \\ldots, p-1$ the number $\\left(\\begin{array}{l}p \\\\ k\\end{array}\\right)$ is divisible by $p$, and\n\n$$\n\\frac{1}{p}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right)=\\frac{(p-1)(p-2) \\cdots(p-k+1)}{k !} \\equiv \\frac{(-1) \\cdot(-2) \\cdots(-k+1)}{k !}=\\frac{(-1)^{k-1}}{k} \\quad(\\bmod p)\n$$\n\nTherefore, we have\n\n$$\nS_{a}=-\\sum_{k=1}^{p-1} \\frac{(-a)^{k}(-1)^{k-1}}{k} \\equiv-\\sum_{k=1}^{p-1}(-a)^{k} \\cdot \\frac{1}{p}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right) \\quad(\\bmod p)\n$$\n\nThe number on the right-hand side is integer. Using the binomial formula we express it as\n\n$$\n-\\sum_{k=1}^{p-1}(-a)^{k} \\cdot \\frac{1}{p}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right)=-\\frac{1}{p}\\left(-1-(-a)^{p}+\\sum_{k=0}^{p}(-a)^{k}\\left(\\begin{array}{l}\np \\\\\nk\n\\end{array}\\right)\\right)=\\frac{(a-1)^{p}-a^{p}+1}{p}\n$$\n\nsince $p$ is odd. So, we have\n\n$$\nS_{a} \\equiv \\frac{(a-1)^{p}-a^{p}+1}{p} \\quad(\\bmod p)\n$$\n\nFinally, using the obtained formula we get\n\n$$\n\\begin{aligned}\nS_{3}+S_{4}-3 S_{2} & \\equiv \\frac{\\left(2^{p}-3^{p}+1\\right)+\\left(3^{p}-4^{p}+1\\right)-3\\left(1^{p}-2^{p}+1\\right)}{p} \\\\\n& =\\frac{4 \\cdot 2^{p}-4^{p}-4}{p}=-\\frac{\\left(2^{p}-2\\right)^{2}}{p} \\quad(\\bmod p) .\n\\end{aligned}\n$$\n\nBy Fermat's theorem, $p \\mid 2^{p}-2$, so $p^{2} \\mid\\left(2^{p}-2\\right)^{2}$ and hence $S_{3}+S_{4}-3 S_{2} \\equiv 0(\\bmod p)$."", ""One may solve the problem without finding an explicit formula for $S_{a}$. It is enough to find the following property.\n\nLemma. For every integer $a$, we have $S_{a+1} \\equiv S_{-a}(\\bmod p)$.\n\nProof. We expand $S_{a+1}$ using the binomial formula as\n\n$$\nS_{a+1}=\\sum_{k=1}^{p-1} \\frac{1}{k} \\sum_{j=0}^{k}\\left(\\begin{array}{c}\nk \\\\\nj\n\\end{array}\\right) a^{j}=\\sum_{k=1}^{p-1}\\left(\\frac{1}{k}+\\sum_{j=1}^{k} a^{j} \\cdot \\frac{1}{k}\\left(\\begin{array}{c}\nk \\\\\nj\n\\end{array}\\right)\\right)=\\sum_{k=1}^{p-1} \\frac{1}{k}+\\sum_{j=1}^{p-1} a^{j} \\sum_{k=j}^{p-1} \\frac{1}{k}\\left(\\begin{array}{l}\nk \\\\\nj\n\\end{array}\\right) a^{k} .\n$$\n\nNote that $\\frac{1}{k}+\\frac{1}{p-k}=\\frac{p}{k(p-k)} \\equiv 0(\\bmod p)$ for all $1 \\leq k \\leq p-1$; hence the first sum vanishes modulo $p$. For the second sum, we use the relation $\\frac{1}{k}\\left(\\begin{array}{c}k \\\\ j\\end{array}\\right)=\\frac{1}{j}\\left(\\begin{array}{c}k-1 \\\\ j-1\\end{array}\\right)$ to obtain\n\n$$\nS_{a+1} \\equiv \\sum_{j=1}^{p-1} \\frac{a^{j}}{j} \\sum_{k=1}^{p-1}\\left(\\begin{array}{c}\nk-1 \\\\\nj-1\n\\end{array}\\right) \\quad(\\bmod p)\n$$\n\nFinally, from the relation\n\n$$\n\\sum_{k=1}^{p-1}\\left(\\begin{array}{c}\nk-1 \\\\\nj-1\n\\end{array}\\right)=\\left(\\begin{array}{c}\np-1 \\\\\nj\n\\end{array}\\right)=\\frac{(p-1)(p-2) \\ldots(p-j)}{j !} \\equiv(-1)^{j} \\quad(\\bmod p)\n$$\n\nwe obtain\n\n$$\nS_{a+1} \\equiv \\sum_{j=1}^{p-1} \\frac{a^{j}(-1)^{j}}{j !}=S_{-a}\n$$\n\nNow we turn to the problem. Using the lemma we get\n\n$$\nS_{3}-3 S_{2} \\equiv S_{-2}-3 S_{2}=\\sum_{\\substack{1 \\leq k \\leq p-1 \\\\ k \\text { is even }}} \\frac{-2 \\cdot 2^{k}}{k}+\\sum_{\\substack{1 \\leq k \\leq p-1 \\\\ k \\text { is odd }}} \\frac{-4 \\cdot 2^{k}}{k}(\\bmod p)\n\\tag{1}\n$$\n\nThe first sum in (11) expands as\n\n$$\n\\sum_{\\ell=1}^{(p-1) / 2} \\frac{-2 \\cdot 2^{2 \\ell}}{2 \\ell}=-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{4^{\\ell}}{\\ell}\n$$\n\nNext, using Fermat's theorem, we expand the second sum in (11) as\n\n$$\n-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{2^{2 \\ell+1}}{2 \\ell-1} \\equiv-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{2^{p+2 \\ell}}{p+2 \\ell-1}=-\\sum_{m=(p+1) / 2}^{p-1} \\frac{2 \\cdot 4^{m}}{2 m}=-\\sum_{m=(p+1) / 2}^{p-1} \\frac{4^{m}}{m}(\\bmod p)\n$$\n\n(here we set $m=\\ell+\\frac{p-1}{2}$ ). Hence,\n\n$$\nS_{3}-3 S_{2} \\equiv-\\sum_{\\ell=1}^{(p-1) / 2} \\frac{4^{\\ell}}{\\ell}-\\sum_{m=(p+1) / 2}^{p-1} \\frac{4^{m}}{m}=-S_{4} \\quad(\\bmod p)\n$$""]",,True,,, 1713,Number Theory,,"Let $k$ be a positive integer and set $n=2^{k}+1$. Prove that $n$ is a prime number if and only if the following holds: there is a permutation $a_{1}, \ldots, a_{n-1}$ of the numbers $1,2, \ldots, n-1$ and a sequence of integers $g_{1}, g_{2}, \ldots, g_{n-1}$ such that $n$ divides $g_{i}^{a_{i}}-a_{i+1}$ for every $i \in\{1,2, \ldots, n-1\}$, where we set $a_{n}=a_{1}$.","['Let $N=\\{1,2, \\ldots, n-1\\}$. For $a, b \\in N$, we say that $b$ follows $a$ if there exists an integer $g$ such that $b \\equiv g^{a}(\\bmod n)$ and denote this property as $a \\rightarrow b$. This way we have a directed graph with $N$ as set of vertices. If $a_{1}, \\ldots, a_{n-1}$ is a permutation of $1,2, \\ldots, n-1$ such that $a_{1} \\rightarrow a_{2} \\rightarrow \\ldots \\rightarrow a_{n-1} \\rightarrow a_{1}$ then this is a Hamiltonian cycle in the graph.\n\nStep I. First consider the case when $n$ is composite. Let $n=p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$ be its prime factorization. All primes $p_{i}$ are odd.\n\nSuppose that $\\alpha_{i}>1$ for some $i$. For all integers $a, g$ with $a \\geq 2$, we have $g^{a} \\not \\equiv p_{i}\\left(\\bmod p_{i}^{2}\\right)$, because $g^{a}$ is either divisible by $p_{i}^{2}$ or it is not divisible by $p_{i}$. It follows that in any Hamiltonian cycle $p_{i}$ comes immediately after 1 . The same argument shows that $2 p_{i}$ also should come immediately after 1, which is impossible. Hence, there is no Hamiltonian cycle in the graph.\n\nNow suppose that $n$ is square-free. We have $n=p_{1} p_{2} \\ldots p_{s}>9$ and $s \\geq 2$. Assume that there exists a Hamiltonian cycle. There are $\\frac{n-1}{2}$ even numbers in this cycle, and each number which follows one of them should be a quadratic residue modulo $n$. So, there should be at least $\\frac{n-1}{2}$ nonzero quadratic residues modulo $n$. On the other hand, for each $p_{i}$ there exist exactly $\\frac{p_{i}+1}{2}$ quadratic residues modulo $p_{i}$; by the Chinese Remainder Theorem, the number of quadratic residues modulo $n$ is exactly $\\frac{p_{1}+1}{2} \\cdot \\frac{p_{2}+1}{2} \\cdot \\ldots \\cdot \\frac{p_{s}+1}{2}$, including 0 . Then we have a contradiction by\n\n$$\n\\frac{p_{1}+1}{2} \\cdot \\frac{p_{2}+1}{2} \\cdot \\ldots \\cdot \\frac{p_{s}+1}{2} \\leq \\frac{2 p_{1}}{3} \\cdot \\frac{2 p_{2}}{3} \\cdot \\ldots \\cdot \\frac{2 p_{s}}{3}=\\left(\\frac{2}{3}\\right)^{s} n \\leq \\frac{4 n}{9}<\\frac{n-1}{2}\n$$\n\nThis proves the ""if""-part of the problem.\n\nStep II. Now suppose that $n$ is prime. For any $a \\in N$, denote by $\\nu_{2}(a)$ the exponent of 2 in the prime factorization of $a$, and let $\\mu(a)=\\max \\left\\{t \\in[0, k] \\mid 2^{t} \\rightarrow a\\right\\}$.\n\nLemma. For any $a, b \\in N$, we have $a \\rightarrow b$ if and only if $\\nu_{2}(a) \\leq \\mu(b)$.\n\nProof. Let $\\ell=\\nu_{2}(a)$ and $m=\\mu(b)$.\n\nSuppose $\\ell \\leq m$. Since $b$ follows $2^{m}$, there exists some $g_{0}$ such that $b \\equiv g_{0}^{2^{m}}(\\bmod n)$. By $\\operatorname{gcd}(a, n-1)=2^{\\ell}$ there exist some integers $p$ and $q$ such that $p a-q(n-1)=2^{\\ell}$. Choosing $g=g_{0}^{2^{m-\\ell} p}$ we have $g^{a}=g_{0}^{2^{m-\\ell} p a}=g_{0}^{2^{m}+2^{m-\\ell} q(n-1)} \\equiv g_{0}^{2^{m}} \\equiv b(\\bmod n)$ by FERMAT\'s theorem. Hence, $a \\rightarrow b$.\n\nTo prove the reverse statement, suppose that $a \\rightarrow b$, so $b \\equiv g^{a}(\\bmod n)$ with some $g$. Then $b \\equiv\\left(g^{a / 2^{\\ell}}\\right)^{2^{\\ell}}$, and therefore $2^{\\ell} \\rightarrow b$. By the definition of $\\mu(b)$, we have $\\mu(b) \\geq \\ell$. The lemma is\n\n\n\nproved.\n\nNow for every $i$ with $0 \\leq i \\leq k$, let\n\n$$\n\\begin{aligned}\nA_{i} & =\\left\\{a \\in N \\mid \\nu_{2}(a)=i\\right\\}, \\\\\nB_{i} & =\\{a \\in N \\mid \\mu(a)=i\\}, \\\\\n\\text { and } C_{i} & =\\{a \\in N \\mid \\mu(a) \\geq i\\}=B_{i} \\cup B_{i+1} \\cup \\ldots \\cup B_{k} .\n\\end{aligned}\n$$\n\nWe claim that $\\left|A_{i}\\right|=\\left|B_{i}\\right|$ for all $0 \\leq i \\leq k$. Obviously we have $\\left|A_{i}\\right|=2^{k-i-1}$ for all $i=$ $0, \\ldots, k-1$, and $\\left|A_{k}\\right|=1$. Now we determine $\\left|C_{i}\\right|$. We have $\\left|C_{0}\\right|=n-1$ and by Fermat\'s theorem we also have $C_{k}=\\{1\\}$, so $\\left|C_{k}\\right|=1$. Next, notice that $C_{i+1}=\\left\\{x^{2} \\bmod n \\mid x \\in C_{i}\\right\\}$. For every $a \\in N$, the relation $x^{2} \\equiv a(\\bmod n)$ has at most two solutions in $N$. Therefore we have $2\\left|C_{i+1}\\right| \\leq\\left|C_{i}\\right|$, with the equality achieved only if for every $y \\in C_{i+1}$, there exist distinct elements $x, x^{\\prime} \\in C_{i}$ such that $x^{2} \\equiv x^{\\prime 2} \\equiv y(\\bmod n)$ (this implies $\\left.x+x^{\\prime}=n\\right)$. Now, since $2^{k}\\left|C_{k}\\right|=\\left|C_{0}\\right|$, we obtain that this equality should be achieved in each step. Hence $\\left|C_{i}\\right|=2^{k-i}$ for $0 \\leq i \\leq k$, and therefore $\\left|B_{i}\\right|=2^{k-i-1}$ for $0 \\leq i \\leq k-1$ and $\\left|B_{k}\\right|=1$.\n\nFrom the previous arguments we can see that for each $z \\in C_{i}(0 \\leq i0$. Finally, if $\\lambda=k-1$, then $C$ contains $2^{k-1}$ which is the only element of $A_{k-1}$. Since $B_{k-1}=\\left\\{2^{k}\\right\\}=A_{k}$ and $B_{k}=\\{1\\}$, the cycle $C$ contains the path $2^{k-1} \\rightarrow 2^{k} \\rightarrow 1$ and it contains an odd number again. This completes the proof of the ""only if""-part of the problem.']",,True,,, 1714,Algebra,,"Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$ f([x] y)=f(x)[f(y)] \tag{1} $$ holds for all $x, y \in \mathbb{R}$. Here, by $[x]$ we denote the greatest integer not exceeding $x$.","['First, setting $x=0$ in (1) we get\n\n$$\nf(0)=f(0)[f(y)] \\tag{2}\n$$\n\nfor all $y \\in \\mathbb{R}$. Now, two cases are possible.\n\nCase 1. Assume that $f(0) \\neq 0$. Then from $(2)$ we conclude that $[f(y)]=1$ for all $y \\in \\mathbb{R}$. Therefore, equation (1) becomes $f([x] y)=f(x)$, and substituting $y=0$ we have $f(x)=f(0)=C \\neq 0$. Finally, from $[f(y)]=1=[C]$ we obtain that $1 \\leq C<2$.\n\nCase 2. Now we have $f(0)=0$. Here we consider two subcases.\n\nSubcase 2a. Suppose that there exists $0<\\alpha<1$ such that $f(\\alpha) \\neq 0$. Then setting $x=\\alpha$ in (1) we obtain $0=f(0)=f(\\alpha)[f(y)]$ for all $y \\in \\mathbb{R}$. Hence, $[f(y)]=0$ for all $y \\in \\mathbb{R}$. Finally, substituting $x=1$ in (1) provides $f(y)=0$ for all $y \\in \\mathbb{R}$, thus contradicting the condition $f(\\alpha) \\neq 0$.\n\nSubcase 2b. Conversely, we have $f(\\alpha)=0$ for all $0 \\leq \\alpha<1$. Consider any real $z$; there exists an integer $N$ such that $\\alpha=\\frac{z}{N} \\in[0,1)$ (one may set $N=[z]+1$ if $z \\geq 0$ and $N=[z]-1$ otherwise). Now, from (1) we get $f(z)=f([N] \\alpha)=f(N)[f(\\alpha)]=0$ for all $z \\in \\mathbb{R}$.\n\nFinally, a straightforward check shows that all the obtained functions satisfy (1).', 'Assume that $[f(y)]=0$ for some $y$; then the substitution $x=1$ provides $f(y)=f(1)[f(y)]=0$. Hence, if $[f(y)]=0$ for all $y$, then $f(y)=0$ for all $y$. This function obviously satisfies the problem conditions.\n\nSo we are left to consider the case when $[f(a)] \\neq 0$ for some $a$. Then we have\n\n$$\nf([x] a)=f(x)[f(a)], \\quad \\text { or } \\quad f(x)=\\frac{f([x] a)}{[f(a)]}\\tag{3}\n$$\n\nThis means that $f\\left(x_{1}\\right)=f\\left(x_{2}\\right)$ whenever $\\left[x_{1}\\right]=\\left[x_{2}\\right]$, hence $f(x)=f([x])$, and we may assume that $a$ is an integer.\n\nNow we have\n\n$$\nf(a)=f\\left(2 a \\cdot \\frac{1}{2}\\right)=f(2 a)\\left[f\\left(\\frac{1}{2}\\right)\\right]=f(2 a)[f(0)] ;\n$$\n\nthis implies $[f(0)] \\neq 0$, so we may even assume that $a=0$. Therefore equation (3) provides\n\n$$\nf(x)=\\frac{f(0)}{[f(0)]}=C \\neq 0\n$$\n\nfor each $x$. Now, condition (1) becomes equivalent to the equation $C=C[C]$ which holds exactly when $[C]=1$.']","['$f(x)=$ const $=C$, where $C=0$ or $1 \\leq C<2$']",True,,Need_human_evaluate, 1715,Algebra,,"Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that $$ 36 \leq 4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)-\left(a^{4}+b^{4}+c^{4}+d^{4}\right) \leq 48 . $$","['Observe that\n\n$$\n\\begin{gathered}\n4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)-\\left(a^{4}+b^{4}+c^{4}+d^{4}\\right)=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right) \\\\\n+6\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-4(a+b+c+d)+4 \\\\\n=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right)+52\n\\end{gathered}\n$$\n\nNow, introducing $x=a-1, y=b-1, z=c-1, t=d-1$, we need to prove the inequalities\n\n$$\n16 \\geq x^{4}+y^{4}+z^{4}+t^{4} \\geq 4\n$$\n\nunder the constraint\n\n$$\nx^{2}+y^{2}+z^{2}+t^{2}=\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-2(a+b+c+d)+4=4\n\\tag{1}\n$$\n\n(we will not use the value of $x+y+z+t$ though it can be found).\n\nNow the rightmost inequality in (1) follows from the power mean inequality:\n\n$$\nx^{4}+y^{4}+z^{4}+t^{4} \\geq \\frac{\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}}{4}=4\n$$\n\nFor the other one, expanding the brackets we note that\n\n$$\n\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}=\\left(x^{4}+y^{4}+z^{4}+t^{4}\\right)+q,\n$$\n\nwhere $q$ is a nonnegative number, so\n\n$$\nx^{4}+y^{4}+z^{4}+t^{4} \\leq\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}=16\n$$\n\nand we are done.', 'First, we claim that $0 \\leq a, b, c, d \\leq 3$. Actually, we have\n\n$$\na+b+c=6-d, \\quad a^{2}+b^{2}+c^{2}=12-d^{2}\n$$\n\nhence the power mean inequality\n\n$$\na^{2}+b^{2}+c^{2} \\geq \\frac{(a+b+c)^{2}}{3}\n$$\n\nrewrites as\n\n$$\n12-d^{2} \\geq \\frac{(6-d)^{2}}{3} \\quad \\Longleftrightarrow \\quad 2 d(d-3) \\leq 0\n$$\n\n\n\nwhich implies the desired inequalities for $d$; since the conditions are symmetric, we also have the same estimate for the other variables.\n\nNow, to prove the rightmost inequality, we use the obvious inequality $x^{2}(x-2)^{2} \\geq 0$ for each real $x$; this inequality rewrites as $4 x^{3}-x^{4} \\leq 4 x^{2}$. It follows that\n\n$$\n\\left(4 a^{3}-a^{4}\\right)+\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\leq 4\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=48\n$$\n\nas desired.\n\nNow we prove the leftmost inequality in an analogous way. For each $x \\in[0,3]$, we have $(x+1)(x-1)^{2}(x-3) \\leq 0$ which is equivalent to $4 x^{3}-x^{4} \\geq 2 x^{2}+4 x-3$. This implies that $\\left(4 a^{3}-a^{4}\\right)+\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\geq 2\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)+4(a+b+c+d)-12=36$, as desired.', 'First, expanding $48=4\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)$ and applying the AM-GM inequality, we have\n\n$$\n\\begin{aligned}\na^{4}+b^{4}+c^{4}+d^{4}+48 & =\\left(a^{4}+4 a^{2}\\right)+\\left(b^{4}+4 b^{2}\\right)+\\left(c^{4}+4 c^{2}\\right)+\\left(d^{4}+4 d^{2}\\right) \\\\\n& \\geq 2\\left(\\sqrt{a^{4} \\cdot 4 a^{2}}+\\sqrt{b^{4} \\cdot 4 b^{2}}+\\sqrt{c^{4} \\cdot 4 c^{2}}+\\sqrt{d^{4} \\cdot 4 d^{2}}\\right) \\\\\n& =4\\left(\\left|a^{3}\\right|+\\left|b^{3}\\right|+\\left|c^{3}\\right|+\\left|d^{3}\\right|\\right) \\geq 4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right),\n\\end{aligned}\n$$\n\nwhich establishes the rightmost inequality.\n\nTo prove the leftmost inequality, we first show that $a, b, c, d \\in[0,3]$ as in the previous solution. Moreover, we can assume that $0 \\leq a \\leq b \\leq c \\leq d$. Then we have $a+b \\leq b+c \\leq$ $\\frac{2}{3}(b+c+d) \\leq \\frac{2}{3} \\cdot 6=4$.\n\nNext, we show that $4 b-b^{2} \\leq 4 c-c^{2}$. Actually, this inequality rewrites as $(c-b)(b+c-4) \\leq 0$, which follows from the previous estimate. The inequality $4 a-a^{2} \\leq 4 b-b^{2}$ can be proved analogously.\n\nFurther, the inequalities $a \\leq b \\leq c$ together with $4 a-a^{2} \\leq 4 b-b^{2} \\leq 4 c-c^{2}$ allow us to apply the Chebyshev inequality obtaining\n\n$$\n\\begin{aligned}\na^{2}\\left(4 a-a^{2}\\right)+b^{2}\\left(4 b-b^{2}\\right)+c^{2}\\left(4 c-c^{2}\\right) & \\geq \\frac{1}{3}\\left(a^{2}+b^{2}+c^{2}\\right)\\left(4(a+b+c)-\\left(a^{2}+b^{2}+c^{2}\\right)\\right) \\\\\n& =\\frac{\\left(12-d^{2}\\right)\\left(4(6-d)-\\left(12-d^{2}\\right)\\right)}{3} .\n\\end{aligned}\n$$\n\nThis implies that\n\n$$\n\\begin{aligned}\n\\left(4 a^{3}-a^{4}\\right) & +\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\geq \\frac{\\left(12-d^{2}\\right)\\left(d^{2}-4 d+12\\right)}{3}+4 d^{3}-d^{4} \\\\\n& =\\frac{144-48 d+16 d^{3}-4 d^{4}}{3}=36+\\frac{4}{3}(3-d)(d-1)\\left(d^{2}-3\\right) .\n\\end{aligned}\n\\tag{2}\n$$\n\nFinally, we have $d^{2} \\geq \\frac{1}{4}\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=3$ (which implies $d>1$ ); so, the expression $\\frac{4}{3}(3-d)(d-1)\\left(d^{2}-3\\right)$ in the right-hand part of $(2)$ is nonnegative, and the desired inequality is proved.']",,True,,, 1716,Algebra,,"Let $x_{1}, \ldots, x_{100}$ be nonnegative real numbers such that $x_{i}+x_{i+1}+x_{i+2} \leq 1$ for all $i=1, \ldots, 100$ (we put $x_{101}=x_{1}, x_{102}=x_{2}$ ). Find the maximal possible value of the sum $$ S=\sum_{i=1}^{100} x_{i} x_{i+2} $$","[""Let $x_{2 i}=0, x_{2 i-1}=\\frac{1}{2}$ for all $i=1, \\ldots, 50$. Then we have $S=50 \\cdot\\left(\\frac{1}{2}\\right)^{2}=\\frac{25}{2}$. So, we are left to show that $S \\leq \\frac{25}{2}$ for all values of $x_{i}$ 's satisfying the problem conditions.\n\nConsider any $1 \\leq i \\leq 50$. By the problem condition, we get $x_{2 i-1} \\leq 1-x_{2 i}-x_{2 i+1}$ and $x_{2 i+2} \\leq 1-x_{2 i}-x_{2 i+1}$. Hence by the AM-GM inequality we get\n\n$$\n\\begin{aligned}\nx_{2 i-1} x_{2 i+1} & +x_{2 i} x_{2 i+2} \\leq\\left(1-x_{2 i}-x_{2 i+1}\\right) x_{2 i+1}+x_{2 i}\\left(1-x_{2 i}-x_{2 i+1}\\right) \\\\\n& =\\left(x_{2 i}+x_{2 i+1}\\right)\\left(1-x_{2 i}-x_{2 i+1}\\right) \\leq\\left(\\frac{\\left(x_{2 i}+x_{2 i+1}\\right)+\\left(1-x_{2 i}-x_{2 i+1}\\right)}{2}\\right)^{2}=\\frac{1}{4} .\n\\end{aligned}\n$$\n\nSumming up these inequalities for $i=1,2, \\ldots, 50$, we get the desired inequality\n\n$$\n\\sum_{i=1}^{50}\\left(x_{2 i-1} x_{2 i+1}+x_{2 i} x_{2 i+2}\\right) \\leq 50 \\cdot \\frac{1}{4}=\\frac{25}{2}\n$$"", 'We present another proof of the estimate. From the problem condition, we get\n\n$$\n\\begin{aligned}\nS=\\sum_{i=1}^{100} x_{i} x_{i+2} \\leq \\sum_{i=1}^{100} x_{i}\\left(1-x_{i}-x_{i+1}\\right) & =\\sum_{i=1}^{100} x_{i}-\\sum_{i=1}^{100} x_{i}^{2}-\\sum_{i=1}^{100} x_{i} x_{i+1} \\\\\n& =\\sum_{i=1}^{100} x_{i}-\\frac{1}{2} \\sum_{i=1}^{100}\\left(x_{i}+x_{i+1}\\right)^{2}\n\\end{aligned}\n$$\n\nBy the AM-QM inequality, we have $\\sum\\left(x_{i}+x_{i+1}\\right)^{2} \\geq \\frac{1}{100}\\left(\\sum\\left(x_{i}+x_{i+1}\\right)\\right)^{2}$, so\n\n$$\n\\begin{aligned}\nS \\leq \\sum_{i=1}^{100} x_{i}-\\frac{1}{200}\\left(\\sum_{i=1}^{100}\\left(x_{i}+x_{i+1}\\right)\\right)^{2} & =\\sum_{i=1}^{100} x_{i}-\\frac{2}{100}\\left(\\sum_{i=1}^{100} x_{i}\\right)^{2} \\\\\n& =\\frac{2}{100}\\left(\\sum_{i=1}^{100} x_{i}\\right)\\left(\\frac{100}{2}-\\sum_{i=1}^{100} x_{i}\\right) .\n\\end{aligned}\n$$\n\nAnd finally, by the AM-GM inequality\n\n$$\nS \\leq \\frac{2}{100} \\cdot\\left(\\frac{1}{2}\\left(\\sum_{i=1}^{100} x_{i}+\\frac{100}{2}-\\sum_{i=1}^{100} x_{i}\\right)\\right)^{2}=\\frac{2}{100} \\cdot\\left(\\frac{100}{4}\\right)^{2}=\\frac{25}{2}\n$$']",['$\\frac{25}{2}$'],False,,Numerical, 1717,Algebra,,"A sequence $x_{1}, x_{2}, \ldots$ is defined by $x_{1}=1$ and $x_{2 k}=-x_{k}, x_{2 k-1}=(-1)^{k+1} x_{k}$ for all $k \geq 1$. Prove that $x_{1}+x_{2}+\cdots+x_{n} \geq 0$ for all $n \geq 1$.","['We start with some observations. First, from the definition of $x_{i}$ it follows that for each positive integer $k$ we have\n\n$$\nx_{4 k-3}=x_{2 k-1}=-x_{4 k-2} \\quad \\text { and } \\quad x_{4 k-1}=x_{4 k}=-x_{2 k}=x_{k} \\text {. }\n\\tag{1}\n$$\n\nHence, denoting $S_{n}=\\sum_{i=1}^{n} x_{i}$, we have\n\n$$\nS_{4 k}=\\sum_{i=1}^{k}\\left(\\left(x_{4 k-3}+x_{4 k-2}\\right)+\\left(x_{4 k-1}+x_{4 k}\\right)\\right)=\\sum_{i=1}^{k}\\left(0+2 x_{k}\\right)=2 S_{k}\\tag{2}\n$$\n$$\nS_{4 k+2}=S_{4 k}+\\left(x_{4 k+1}+x_{4 k+2}\\right)=S_{4 k}\\tag{3}\n$$\n\nObserve also that $S_{n}=\\sum_{i=1}^{n} x_{i} \\equiv \\sum_{i=1}^{n} 1=n(\\bmod 2)$.\n\nNow we prove by induction on $k$ that $S_{i} \\geq 0$ for all $i \\leq 4 k$. The base case is valid since $x_{1}=x_{3}=x_{4}=1, x_{2}=-1$. For the induction step, assume that $S_{i} \\geq 0$ for all $i \\leq 4 k$. Using the relations (1)-(3), we obtain\n\n$$\nS_{4 k+4}=2 S_{k+1} \\geq 0, \\quad S_{4 k+2}=S_{4 k} \\geq 0, \\quad S_{4 k+3}=S_{4 k+2}+x_{4 k+3}=\\frac{S_{4 k+2}+S_{4 k+4}}{2} \\geq 0\n$$\n\nSo, we are left to prove that $S_{4 k+1} \\geq 0$. If $k$ is odd, then $S_{4 k}=2 S_{k} \\geq 0$; since $k$ is odd, $S_{k}$ is odd as well, so we have $S_{4 k} \\geq 2$ and hence $S_{4 k+1}=S_{4 k}+x_{4 k+1} \\geq 1$.\n\nConversely, if $k$ is even, then we have $x_{4 k+1}=x_{2 k+1}=x_{k+1}$, hence $S_{4 k+1}=S_{4 k}+x_{4 k+1}=$ $2 S_{k}+x_{k+1}=S_{k}+S_{k+1} \\geq 0$. The step is proved.', 'We will use the notation of $S_{n}$ and the relations (1)-(3) from the previous solution.\n\nAssume the contrary and consider the minimal $n$ such that $S_{n+1}<0$; surely $n \\geq 1$, and from $S_{n} \\geq 0$ we get $S_{n}=0, x_{n+1}=-1$. Hence, we are especially interested in the set $M=\\left\\{n: S_{n}=0\\right\\}$; our aim is to prove that $x_{n+1}=1$ whenever $n \\in M$ thus coming to a contradiction.\n\nFor this purpose, we first describe the set $M$ inductively. We claim that (i) $M$ consists only of even numbers, (ii) $2 \\in M$, and (iii) for every even $n \\geq 4$ we have $n \\in M \\Longleftrightarrow[n / 4] \\in M$. Actually, (i) holds since $S_{n} \\equiv n(\\bmod 2)$, (ii) is straightforward, while (iii) follows from the relations $S_{4 k+2}=S_{4 k}=2 S_{k}$.\n\nNow, we are left to prove that $x_{n+1}=1$ if $n \\in M$. We use the induction on $n$. The base case is $n=2$, that is, the minimal element of $M$; here we have $x_{3}=1$, as desired.\n\nFor the induction step, consider some $4 \\leq n \\in M$ and let $m=[n / 4] \\in M$; then $m$ is even, and $x_{m+1}=1$ by the induction hypothesis. We prove that $x_{n+1}=x_{m+1}=1$. If $n=4 m$ then we have $x_{n+1}=x_{2 m+1}=x_{m+1}$ since $m$ is even; otherwise, $n=4 m+2$, and $x_{n+1}=-x_{2 m+2}=x_{m+1}$, as desired. The proof is complete.']",,True,,, 1718,Algebra,,"Denote by $\mathbb{Q}^{+}$the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ which satisfy the following equation for all $x, y \in \mathbb{Q}^{+}$: $$ f\left(f(x)^{2} y\right)=x^{3} f(x y) \tag{1} $$","['By substituting $y=1$, we get\n\n$$\nf\\left(f(x)^{2}\\right)=x^{3} f(x)\\tag{2}\n$$\n\nThen, whenever $f(x)=f(y)$, we have\n\n$$\nx^{3}=\\frac{f\\left(f(x)^{2}\\right)}{f(x)}=\\frac{f\\left(f(y)^{2}\\right)}{f(y)}=y^{3}\n$$\n\nwhich implies $x=y$, so the function $f$ is injective.\n\nNow replace $x$ by $x y$ in (2), and apply (1) twice, second time to $\\left(y, f(x)^{2}\\right)$ instead of $(x, y)$ :\n\n$$\nf\\left(f(x y)^{2}\\right)=(x y)^{3} f(x y)=y^{3} f\\left(f(x)^{2} y\\right)=f\\left(f(x)^{2} f(y)^{2}\\right)\n$$\n\nSince $f$ is injective, we get\n\n$$\n\\begin{aligned}\nf(x y)^{2} & =f(x)^{2} f(y)^{2} \\\\\nf(x y) & =f(x) f(y) .\n\\end{aligned}\n$$\n\nTherefore, $f$ is multiplicative. This also implies $f(1)=1$ and $f\\left(x^{n}\\right)=f(x)^{n}$ for all integers $n$.\n\nThen the function equation (1) can be re-written as\n\n$$\n\\begin{aligned}\nf(f(x))^{2} f(y) & =x^{3} f(x) f(y), \\\\\nf(f(x)) & =\\sqrt{x^{3} f(x)} .\n\\end{aligned}\n\\tag{3}\n$$\n\nLet $g(x)=x f(x)$. Then, by (3), we have\n\n$$\n\\begin{aligned}\ng(g(x)) & =g(x f(x))=x f(x) \\cdot f(x f(x))=x f(x)^{2} f(f(x))= \\\\\n& =x f(x)^{2} \\sqrt{x^{3} f(x)}=(x f(x))^{5 / 2}=(g(x))^{5 / 2},\n\\end{aligned}\n$$\n\nand, by induction,\n\n$$\n\\underbrace{g(g(\\ldots g}_{n+1}(x) \\ldots))=(g(x))^{(5 / 2)^{n}}\n\\tag{4}\n$$\n\nfor every positive integer $n$.\n\nConsider (4) for a fixed $x$. The left-hand side is always rational, so $(g(x))^{(5 / 2)^{n}}$ must be rational for every $n$. We show that this is possible only if $g(x)=1$. Suppose that $g(x) \\neq 1$, and let the prime factorization of $g(x)$ be $g(x)=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}$ where $p_{1}, \\ldots, p_{k}$ are distinct primes and $\\alpha_{1}, \\ldots, \\alpha_{k}$ are nonzero integers. Then the unique prime factorization of (4) is\n\n$$\n\\underbrace{g(g(\\ldots g}_{n+1}(x) \\ldots))=(g(x))^{(5 / 2)^{n}}=p_{1}^{(5 / 2)^{n} \\alpha_{1}} \\ldots p_{k}^{(5 / 2)^{n} \\alpha_{k}}\n$$\n\n\n\nwhere the exponents should be integers. But this is not true for large values of $n$, for example $\\left(\\frac{5}{2}\\right)^{n} \\alpha_{1}$ cannot be a integer number when $2^{n} \\nmid \\alpha_{1}$. Therefore, $g(x) \\neq 1$ is impossible.\n\nHence, $g(x)=1$ and thus $f(x)=\\frac{1}{x}$ for all $x$.\n\nThe function $f(x)=\\frac{1}{x}$ satisfies the equation (1):\n\n$$\nf\\left(f(x)^{2} y\\right)=\\frac{1}{f(x)^{2} y}=\\frac{1}{\\left(\\frac{1}{x}\\right)^{2} y}=\\frac{x^{3}}{x y}=x^{3} f(x y)\n$$']",['$f(x)=\\frac{1}{x}$'],False,,Expression, 1719,Algebra,,Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n))=f(n)+1$ and $g(f(n))=$ $g(n)+1$ hold for all positive integers. Prove that $f(n)=g(n)$ for all positive integer $n$.,"['Throughout the solution, by $\\mathbb{N}$ we denote the set of all positive integers. For any function $h: \\mathbb{N} \\rightarrow \\mathbb{N}$ and for any positive integer $k$, define $h^{k}(x)=\\underbrace{h(h(\\ldots h}_{k}(x) \\ldots)$ ) (in particular, $\\left.h^{0}(x)=x\\right)$.\n\nObserve that $f\\left(g^{k}(x)\\right)=f\\left(g^{k-1}(x)\\right)+1=\\cdots=f(x)+k$ for any positive integer $k$, and similarly $g\\left(f^{k}(x)\\right)=g(x)+k$. Now let $a$ and $b$ are the minimal values attained by $f$ and $g$, respectively; say $f\\left(n_{f}\\right)=a, g\\left(n_{g}\\right)=b$. Then we have $f\\left(g^{k}\\left(n_{f}\\right)\\right)=a+k, g\\left(f^{k}\\left(n_{g}\\right)\\right)=b+k$, so the function $f$ attains all values from the set $N_{f}=\\{a, a+1, \\ldots\\}$, while $g$ attains all the values from the set $N_{g}=\\{b, b+1, \\ldots\\}$.\n\nNext, note that $f(x)=f(y)$ implies $g(x)=g(f(x))-1=g(f(y))-1=g(y)$; surely, the converse implication also holds. Now, we say that $x$ and $y$ are similar (and write $x \\sim y$ ) if $f(x)=f(y)$ (equivalently, $g(x)=g(y)$ ). For every $x \\in \\mathbb{N}$, we define $[x]=\\{y \\in \\mathbb{N}: x \\sim y\\}$; surely, $y_{1} \\sim y_{2}$ for all $y_{1}, y_{2} \\in[x]$, so $[x]=[y]$ whenever $y \\in[x]$.\n\nNow we investigate the structure of the sets $[x]$.\n\nClaim 1. Suppose that $f(x) \\sim f(y)$; then $x \\sim y$, that is, $f(x)=f(y)$. Consequently, each class $[x]$ contains at most one element from $N_{f}$, as well as at most one element from $N_{g}$.\n\nProof. If $f(x) \\sim f(y)$, then we have $g(x)=g(f(x))-1=g(f(y))-1=g(y)$, so $x \\sim y$. The second statement follows now from the sets of values of $f$ and $g$.\n\nNext, we clarify which classes do not contain large elements.\n\nClaim 2. For any $x \\in \\mathbb{N}$, we have $[x] \\subseteq\\{1,2, \\ldots, b-1\\}$ if and only if $f(x)=a$. Analogously, $[x] \\subseteq\\{1,2, \\ldots, a-1\\}$ if and only if $g(x)=b$.\n\nProof. We will prove that $[x] \\nsubseteq\\{1,2, \\ldots, b-1\\} \\Longleftrightarrow f(x)>a$; the proof of the second statement is similar.\n\nNote that $f(x)>a$ implies that there exists some $y$ satisfying $f(y)=f(x)-1$, so $f(g(y))=$ $f(y)+1=f(x)$, and hence $x \\sim g(y) \\geq b$. Conversely, if $b \\leq c \\sim x$ then $c=g(y)$ for some $y \\in \\mathbb{N}$, which in turn follows $f(x)=f(g(y))=f(y)+1 \\geq a+1$, and hence $f(x)>a$.\n\nClaim 2 implies that there exists exactly one class contained in $\\{1, \\ldots, a-1\\}$ (that is, the class $\\left[n_{g}\\right]$ ), as well as exactly one class contained in $\\{1, \\ldots, b-1\\}$ (the class $\\left[n_{f}\\right]$ ). Assume for a moment that $a \\leq b$; then $\\left[n_{g}\\right]$ is contained in $\\{1, \\ldots, b-1\\}$ as well, hence it coincides with $\\left[n_{g}\\right]$. So, we get that\n\n$$\nf(x)=a \\Longleftrightarrow g(x)=b \\Longleftrightarrow x \\sim n_{f} \\sim n_{g} .\n\\tag{1}\n$$\n\nClaim 3. $a=b$.\n\nProof. By Claim 2, we have $[a] \\neq\\left[n_{f}\\right]$, so $[a]$ should contain some element $a^{\\prime} \\geq b$ by Claim 2 again. If $a \\neq a^{\\prime}$, then $[a]$ contains two elements $\\geq a$ which is impossible by Claim 1 . Therefore, $a=a^{\\prime} \\geq b$. Similarly, $b \\geq a$.\n\nNow we are ready to prove the problem statement. First, we establish the following\n\nClaim 4. For every integer $d \\geq 0, f^{d+1}\\left(n_{f}\\right)=g^{d+1}\\left(n_{f}\\right)=a+d$.\n\nProof. Induction on $d$. For $d=0$, the statement follows from (1) and Claim 3. Next, for $d>1$ from the induction hypothesis we have $f^{d+1}\\left(n_{f}\\right)=f\\left(f^{d}\\left(n_{f}\\right)\\right)=f\\left(g^{d}\\left(n_{f}\\right)\\right)=f\\left(n_{f}\\right)+d=a+d$. The equality $g^{d+1}\\left(n_{f}\\right)=a+d$ is analogous.\n\n\n\nFinally, for each $x \\in \\mathbb{N}$, we have $f(x)=a+d$ for some $d \\geq 0$, so $f(x)=f\\left(g^{d}\\left(n_{f}\\right)\\right)$ and hence $x \\sim g^{d}\\left(n_{f}\\right)$. It follows that $g(x)=g\\left(g^{d}\\left(n_{f}\\right)\\right)=g^{d+1}\\left(n_{f}\\right)=a+d=f(x)$ by Claim 4 .', ""We start with the same observations, introducing the relation $\\sim$ and proving Claim 1 from the previous solution.\n\nNote that $f(a)>a$ since otherwise we have $f(a)=a$ and hence $g(a)=g(f(a))=g(a)+1$, which is false.\n\nClaim $2^{\\prime} . \\quad a=b$.\n\nProof. We can assume that $a \\leq b$. Since $f(a) \\geq a+1$, there exists some $x \\in \\mathbb{N}$ such that $f(a)=f(x)+1$, which is equivalent to $f(a)=f(g(x))$ and $a \\sim g(x)$. Since $g(x) \\geq b \\geq a$, by Claim 1 we have $a=g(x) \\geq b$, which together with $a \\leq b$ proves the Claim.\n\nNow, almost the same method allows to find the values $f(a)$ and $g(a)$.\n\nClaim $3^{\\prime} . \\quad f(a)=g(a)=a+1$.\n\nProof. Assume the contrary; then $f(a) \\geq a+2$, hence there exist some $x, y \\in \\mathbb{N}$ such that $f(x)=f(a)-2$ and $f(y)=g(x)($ as $g(x) \\geq a=b)$. Now we get $f(a)=f(x)+2=f\\left(g^{2}(x)\\right)$, so $a \\sim g^{2}(x) \\geq a$, and by Claim 1 we get $a=g^{2}(x)=g(f(y))=1+g(y) \\geq 1+a$; this is impossible. The equality $g(a)=a+1$ is similar.\n\nNow, we are prepared for the proof of the problem statement. First, we prove it for $n \\geq a$. Claim 4'. For each integer $x \\geq a$, we have $f(x)=g(x)=x+1$.\n\nProof. Induction on $x$. The base case $x=a$ is provided by Claim $3^{\\prime}$, while the induction step follows from $f(x+1)=f(g(x))=f(x)+1=(x+1)+1$ and the similar computation for $g(x+1)$.\n\nFinally, for an arbitrary $n \\in \\mathbb{N}$ we have $g(n) \\geq a$, so by Claim $4^{\\prime}$ we have $f(n)+1=$ $f(g(n))=g(n)+1$, hence $f(n)=g(n)$.""]",,True,,, 1720,Algebra,,"Let $a_{1}, \ldots, a_{r}$ be positive real numbers. For $n>r$, we inductively define $$ a_{n}=\max _{1 \leq k \leq n-1}\left(a_{k}+a_{n-k}\right) \tag{1} $$ Prove that there exist positive integers $\ell \leq r$ and $N$ such that $a_{n}=a_{n-\ell}+a_{\ell}$ for all $n \geq N$.","['First, from the problem conditions we have that each $a_{n}(n>r)$ can be expressed as $a_{n}=a_{j_{1}}+a_{j_{2}}$ with $j_{1}, j_{2}r$ then we can proceed in the same way with $a_{j_{1}}$, and so on. Finally, we represent $a_{n}$ in a form\n\n$$\na_{n}=a_{i_{1}}+\\cdots+a_{i_{k}}\\tag{2}\n$$\n$$\n1 \\leq i_{j} \\leq r, \\quad i_{1}+\\cdots+i_{k}=n .\n\n\\tag{3}\n$$\n\nMoreover, if $a_{i_{1}}$ and $a_{i_{2}}$ are the numbers in (2) obtained on the last step, then $i_{1}+i_{2}>r$. Hence we can adjust (3) as\n\n$$\n1 \\leq i_{j} \\leq r, \\quad i_{1}+\\cdots+i_{k}=n, \\quad i_{1}+i_{2}>r\n\\tag{4}\n$$\n\nOn the other hand, suppose that the indices $i_{1}, \\ldots, i_{k}$ satisfy the conditions (4). Then, denoting $s_{j}=i_{1}+\\cdots+i_{j}$, from (1) we have\n\n$$\na_{n}=a_{s_{k}} \\geq a_{s_{k-1}}+a_{i_{k}} \\geq a_{s_{k-2}}+a_{i_{k-1}}+a_{i_{k}} \\geq \\cdots \\geq a_{i_{1}}+\\cdots+a_{i_{k}}\n$$\n\nSummarizing these observations we get the following\n\nClaim. For every $n>r$, we have\n\n$$\na_{n}=\\max \\left\\{a_{i_{1}}+\\cdots+a_{i_{k}}: \\text { the collection }\\left(i_{1}, \\ldots, i_{k}\\right) \\text { satisfies }(4)\\right\\}\n$$\n\nNow we denote\n\n$$\ns=\\max _{1 \\leq i \\leq r} \\frac{a_{i}}{i}\n$$\n\nand fix some index $\\ell \\leq r$ such that $s=\\frac{a_{\\ell}}{\\ell}$.\n\nConsider some $n \\geq r^{2} \\ell+2 r$ and choose an expansion of $a_{n}$ in the form (2), (4). Then we have $n=i_{1}+\\cdots+i_{k} \\leq r k$, so $k \\geq n / r \\geq r \\ell+2$. Suppose that none of the numbers $i_{3}, \\ldots, i_{k}$ equals $\\ell$. Then by the pigeonhole principle there is an index $1 \\leq j \\leq r$ which appears among $i_{3}, \\ldots, i_{k}$ at least $\\ell$ times, and surely $j \\neq \\ell$. Let us delete these $\\ell$ occurrences of $j$ from $\\left(i_{1}, \\ldots, i_{k}\\right)$, and add $j$ occurrences of $\\ell$ instead, obtaining a sequence $\\left(i_{1}, i_{2}, i_{3}^{\\prime}, \\ldots, i_{k^{\\prime}}^{\\prime}\\right)$ also satisfying (4). By Claim, we have\n\n$$\na_{i_{1}}+\\cdots+a_{i_{k}}=a_{n} \\geq a_{i_{1}}+a_{i_{2}}+a_{i_{3}^{\\prime}}+\\cdots+a_{i_{k^{\\prime}}^{\\prime}}\n$$\n\nor, after removing the coinciding terms, $\\ell a_{j} \\geq j a_{\\ell}$, so $\\frac{a_{\\ell}}{\\ell} \\leq \\frac{a_{j}}{j}$. By the definition of $\\ell$, this means that $\\ell a_{j}=j a_{\\ell}$, hence\n\n$$\na_{n}=a_{i_{1}}+a_{i_{2}}+a_{i_{3}^{\\prime}}+\\cdots+a_{i_{k^{\\prime}}^{\\prime}}\n$$\n\nThus, for every $n \\geq r^{2} \\ell+2 r$ we have found a representation of the form (2), (4) with $i_{j}=\\ell$ for some $j \\geq 3$. Rearranging the indices we may assume that $i_{k}=\\ell$.\n\nFinally, observe that in this representation, the indices $\\left(i_{1}, \\ldots, i_{k-1}\\right)$ satisfy the conditions (4) with $n$ replaced by $n-\\ell$. Thus, from the Claim we get\n\n$$\na_{n-\\ell}+a_{\\ell} \\geq\\left(a_{i_{1}}+\\cdots+a_{i_{k-1}}\\right)+a_{\\ell}=a_{n}\n$$\n\nwhich by (1) implies\n\n$$\na_{n}=a_{n-\\ell}+a_{\\ell} \\quad \\text { for each } n \\geq r^{2} \\ell+2 r,\n$$\n\nas desired.', ""As in the previous solution, we involve the expansion (2), (3), and we fix some index $1 \\leq \\ell \\leq r$ such that\n\n$$\n\\frac{a_{\\ell}}{\\ell}=s=\\max _{1 \\leq i \\leq r} \\frac{a_{i}}{i}\n$$\n\nNow, we introduce the sequence $\\left(b_{n}\\right)$ as $b_{n}=a_{n}-s n$; then $b_{\\ell}=0$.\n\nWe prove by induction on $n$ that $b_{n} \\leq 0$, and $\\left(b_{n}\\right)$ satisfies the same recurrence relation as $\\left(a_{n}\\right)$. The base cases $n \\leq r$ follow from the definition of $s$. Now, for $n>r$ from the induction hypothesis we have\n\n$$\nb_{n}=\\max _{1 \\leq k \\leq n-1}\\left(a_{k}+a_{n-k}\\right)-n s=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}+n s\\right)-n s=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}\\right) \\leq 0\n$$\n\nas required.\n\nNow, if $b_{k}=0$ for all $1 \\leq k \\leq r$, then $b_{n}=0$ for all $n$, hence $a_{n}=s n$, and the statement is trivial. Otherwise, define\n\n$$\nM=\\max _{1 \\leq i \\leq r}\\left|b_{i}\\right|, \\quad \\varepsilon=\\min \\left\\{\\left|b_{i}\\right|: 1 \\leq i \\leq r, b_{i}<0\\right\\}\n$$\n\nThen for $n>r$ we obtain\n\n$$\nb_{n}=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}\\right) \\geq b_{\\ell}+b_{n-\\ell}=b_{n-\\ell}\n$$\n\nSo\n\n$$\n0 \\geq b_{n} \\geq b_{n-\\ell} \\geq b_{n-2 \\ell} \\geq \\cdots \\geq-M\n$$\n\nThus, in view of the expansion (2), (3) applied to the sequence $\\left(b_{n}\\right)$, we get that each $b_{n}$ is contained in a set\n\n$$\nT=\\left\\{b_{i_{1}}+b_{i_{2}}+\\cdots+b_{i_{k}}: i_{1}, \\ldots, i_{k} \\leq r\\right\\} \\cap[-M, 0]\n$$\n\nWe claim that this set is finite. Actually, for any $x \\in T$, let $x=b_{i_{1}}+\\cdots+b_{i_{k}}\\left(i_{1}, \\ldots, i_{k} \\leq r\\right)$. Then among $b_{i_{j}}$ 's there are at most $\\frac{M}{\\varepsilon}$ nonzero terms (otherwise $x<\\frac{M}{\\varepsilon} \\cdot(-\\varepsilon)<-M$ ). Thus $x$ can be expressed in the same way with $k \\leq \\frac{M}{\\varepsilon}$, and there is only a finite number of such sums.\n\nFinally, for every $t=1,2, \\ldots, \\ell$ we get that the sequence\n\n$$\nb_{r+t}, b_{r+t+\\ell}, b_{r+t+2 \\ell}, \\ldots\n$$\n\nis non-decreasing and attains the finite number of values; therefore it is constant from some index. Thus, the sequence $\\left(b_{n}\\right)$ is periodic with period $\\ell$ from some index $N$, which means that\n\n$$\nb_{n}=b_{n-\\ell}=b_{n-\\ell}+b_{\\ell} \\quad \\text { for all } n>N+\\ell\n$$\n\nand hence\n\n$$\na_{n}=b_{n}+n s=\\left(b_{n-\\ell}+(n-\\ell) s\\right)+\\left(b_{\\ell}+\\ell s\\right)=a_{n-\\ell}+a_{\\ell} \\quad \\text { for all } n>N+\\ell,\n$$\n\nas desired.""]",,True,,, 1721,Combinatorics,,"Given six positive numbers $a, b, c, d, e, f$ such that $a\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} . \tag{1} $$","['We define also $\\sigma=a c+c e+a e, \\tau=b d+b f+d f$. The idea of the solution is to interpret (1) as a natural inequality on the roots of an appropriate polynomial.\n\nActually, consider the polynomial\n\n$$\n\\begin{array}{r}\nP(x)=(b+d+f)(x-a)(x-c)(x-e)+(a+c+e)(x-b)(x-d)(x-f) \\\\\n=T\\left(x^{3}-S x^{2}+\\sigma x-a c e\\right)+S\\left(x^{3}-T x^{2}+\\tau x-b d f\\right) .\n\\end{array}\n\\tag{2}\n$$\n\nSurely, $P$ is cubic with leading coefficient $S+T>0$. Moreover, we have\n\n$$\n\\begin{array}{ll}\nP(a)=S(a-b)(a-d)(a-f)<0, & P(c)=S(c-b)(c-d)(c-f)>0 \\\\\nP(e)=S(e-b)(e-d)(e-f)<0, & P(f)=T(f-a)(f-c)(f-e)>0\n\\end{array}\n$$\n\nHence, each of the intervals $(a, c),(c, e),(e, f)$ contains at least one root of $P(x)$. Since there are at most three roots at all, we obtain that there is exactly one root in each interval (denote them by $\\alpha \\in(a, c), \\beta \\in(c, e), \\gamma \\in(e, f))$. Moreover, the polynomial $P$ can be factorized as\n\n$$\nP(x)=(T+S)(x-\\alpha)(x-\\beta)(x-\\gamma) \\text {. }\n\\tag{3}\n$$\n\nEquating the coefficients in the two representations (2) and (3) of $P(x)$ provides\n\n$$\n\\alpha+\\beta+\\gamma=\\frac{2 T S}{T+S}, \\quad \\alpha \\beta+\\alpha \\gamma+\\beta \\gamma=\\frac{S \\tau+T \\sigma}{T+S}\n$$\n\nNow, since the numbers $\\alpha, \\beta, \\gamma$ are distinct, we have\n\n$$\n0<(\\alpha-\\beta)^{2}+(\\alpha-\\gamma)^{2}+(\\beta-\\gamma)^{2}=2(\\alpha+\\beta+\\gamma)^{2}-6(\\alpha \\beta+\\alpha \\gamma+\\beta \\gamma)\n$$\n\nwhich implies\n\n$$\n\\frac{4 S^{2} T^{2}}{(T+S)^{2}}=(\\alpha+\\beta+\\gamma)^{2}>3(\\alpha \\beta+\\alpha \\gamma+\\beta \\gamma)=\\frac{3(S \\tau+T \\sigma)}{T+S}\n$$\n\nor\n\n$$\n4 S^{2} T^{2}>3(T+S)(T \\sigma+S \\tau)\n$$\n\nwhich is exactly what we need.', 'Let\n\n$$\nU=\\frac{1}{2}\\left((e-a)^{2}+(c-a)^{2}+(e-c)^{2}\\right)=S^{2}-3(a c+a e+c e)\n$$\n\nand\n\n$$\nV=\\frac{1}{2}\\left((f-b)^{2}+(f-d)^{2}+(d-b)^{2}\\right)=T^{2}-3(b d+b f+d f)\n$$\n\nThen\n\n$$\n\\begin{aligned}\n& \\text { (L.H.S. })^{2}-(\\text { R.H.S. })^{2}=(2 S T)^{2}-(S+T)(S \\cdot 3(b d+b f+d f)+T \\cdot 3(a c+a e+c e))= \\\\\n& \\quad=4 S^{2} T^{2}-(S+T)\\left(S\\left(T^{2}-V\\right)+T\\left(S^{2}-U\\right)\\right)=(S+T)(S V+T U)-S T(T-S)^{2},\n\\end{aligned}\n$$\n\nand the statement is equivalent with\n\n$$\n(S+T)(S V+T U)>S T(T-S)^{2} .\\tag{4}\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n(S+T)(T U+S V) \\geq(\\sqrt{S \\cdot T U}+\\sqrt{T \\cdot S V})^{2}=S T(\\sqrt{U}+\\sqrt{V})^{2} .\n\\tag{5}\n$$\n\nEstimate the quantities $\\sqrt{U}$ and $\\sqrt{V}$ by the QM-AM inequality with the positive terms $(e-c)^{2}$ and $(d-b)^{2}$ being omitted:\n\n$$\n\\begin{aligned}\n\\sqrt{U}+\\sqrt{V} & >\\sqrt{\\frac{(e-a)^{2}+(c-a)^{2}}{2}}+\\sqrt{\\frac{(f-b)^{2}+(f-d)^{2}}{2}} \\\\\n& >\\frac{(e-a)+(c-a)}{2}+\\frac{(f-b)+(f-d)}{2}=\\left(f-\\frac{d}{2}-\\frac{b}{2}\\right)+\\left(\\frac{e}{2}+\\frac{c}{2}-a\\right) \\\\\n& =(T-S)+\\frac{3}{2}(e-d)+\\frac{3}{2}(c-b)>T-S .\n\\end{aligned}\\tag{6}\n$$\n\nThe estimates (5) and (6) prove (4) and hence the statement.', 'We define $\\sigma=a c+c e+a e, \\tau=b d+b f+d f$.\nMoreover, let $s=c+e$. Note that\n\n$$\n(c-b)(c-d)+(e-f)(e-d)+(e-f)(c-b)<0\n$$\n\nsince each summand is negative. This rewrites as\n\n$$\n\\begin{gathered}\n(b d+b f+d f)-(a c+c e+a e)<(c+e)(b+d+f-a-c-e), \\text { or } \\\\\n\\tau-\\sigma\\sqrt{3(S+T)(S(b d+b f+d f)+T(a c+a e+c e))} .\n$$', 'We introduce the expressions $\\sigma$ and $\\tau$ as in the previous solutions. The idea of the solution is to change the values of variables $a, \\ldots, f$ keeping the left-hand side unchanged and increasing the right-hand side; it will lead to a simpler inequality which can be proved in a direct way.\n\nNamely, we change the variables (i) keeping the (non-strict) inequalities $a \\leq b \\leq c \\leq d \\leq$ $e \\leq f$; (ii) keeping the values of sums $S$ and $T$ unchanged; and finally (iii) increasing the values of $\\sigma$ and $\\tau$. Then the left-hand side of (1) remains unchanged, while the right-hand side increases. Hence, the inequality (1) (and even a non-strict version of (1)) for the changed values would imply the same (strict) inequality for the original values.\n\nFirst, we find the sufficient conditions for (ii) and (iii) to be satisfied.\n\nLemma. Let $x, y, z>0$; denote $U(x, y, z)=x+y+z, v(x, y, z)=x y+x z+y z$. Suppose that $x^{\\prime}+y^{\\prime}=x+y$ but $|x-y| \\geq\\left|x^{\\prime}-y^{\\prime}\\right|$; then we have $U\\left(x^{\\prime}, y^{\\prime}, z\\right)=U(x, y, z)$ and $v\\left(x^{\\prime}, y^{\\prime}, z\\right) \\geq$ $v(x, y, z)$ with equality achieved only when $|x-y|=\\left|x^{\\prime}-y^{\\prime}\\right|$.\n\nProof. The first equality is obvious. For the second, we have\n\n$$\n\\begin{aligned}\nv\\left(x^{\\prime}, y^{\\prime}, z\\right)=z\\left(x^{\\prime}+y^{\\prime}\\right)+x^{\\prime} y^{\\prime} & =z\\left(x^{\\prime}+y^{\\prime}\\right)+\\frac{\\left(x^{\\prime}+y^{\\prime}\\right)^{2}-\\left(x^{\\prime}-y^{\\prime}\\right)^{2}}{4} \\\\\n& \\geq z(x+y)+\\frac{(x+y)^{2}-(x-y)^{2}}{4}=v(x, y, z),\n\\end{aligned}\n$$\n\nwith the equality achieved only for $\\left(x^{\\prime}-y^{\\prime}\\right)^{2}=(x-y)^{2} \\Longleftrightarrow\\left|x^{\\prime}-y^{\\prime}\\right|=|x-y|$, as desired.\n\nNow, we apply Lemma several times making the following changes. For each change, we denote the new values by the same letters to avoid cumbersome notations.\n\n1. Let $k=\\frac{d-c}{2}$. Replace $(b, c, d, e)$ by $(b+k, c+k, d-k, e-k)$. After the change we have $a2^{N-2}$. Consider the collection of all $2^{N-2}$ flags having yellow first squares and blue second ones. Obviously, both colors appear on the diagonal of each $N \\times N$ square formed by these flags.\n\nWe are left to show that $M_{N} \\leq 2^{N-2}+1$, thus obtaining the desired answer. We start with establishing this statement for $N=4$.\n\nSuppose that we have 5 flags of length 4 . We decompose each flag into two parts of 2 squares each; thereby, we denote each flag as $L R$, where the $2 \\times 1$ flags $L, R \\in \\mathcal{S}=\\{\\mathrm{BB}, \\mathrm{BY}, \\mathrm{YB}, \\mathrm{YY}\\}$ are its left and right parts, respectively. First, we make two easy observations on the flags $2 \\times 1$ which can be checked manually.\n\n(i) For each $A \\in \\mathcal{S}$, there exists only one $2 \\times 1$ flag $C \\in \\mathcal{S}$ (possibly $C=A$ ) such that $A$ and $C$ cannot form a $2 \\times 2$ square with monochrome diagonal (for part $\\mathrm{BB}$, that is $\\mathrm{YY}$, and for $\\mathrm{BY}$ that is $\\mathrm{YB)}$.\n\n(ii) Let $A_{1}, A_{2}, A_{3} \\in \\mathcal{S}$ be three distinct elements; then two of them can form a $2 \\times 2$ square with yellow diagonal, and two of them can form a $2 \\times 2$ square with blue diagonal (for all parts but $\\mathrm{BB}$, a pair (BY, YB) fits for both statements, while for all parts but BY, these pairs are $(\\mathrm{YB}, \\mathrm{YY})$ and $(\\mathrm{BB}, \\mathrm{YB}))$.\n\nNow, let $\\ell$ and $r$ be the numbers of distinct left and right parts of our 5 flags, respectively. The total number of flags is $5 \\leq r \\ell$, hence one of the factors (say, $r$ ) should be at least 3 . On the other hand, $\\ell, r \\leq 4$, so there are two flags with coinciding right part; let them be $L_{1} R_{1}$ and $L_{2} R_{1}\\left(L_{1} \\neq L_{2}\\right)$.\n\nNext, since $r \\geq 3$, there exist some flags $L_{3} R_{3}$ and $L_{4} R_{4}$ such that $R_{1}, R_{3}, R_{4}$ are distinct. Let $L^{\\prime} R^{\\prime}$ be the remaining flag. By (i), one of the pairs $\\left(L^{\\prime}, L_{1}\\right)$ and $\\left(L^{\\prime}, L_{2}\\right)$ can form a $2 \\times 2$ square with monochrome diagonal; we can assume that $L^{\\prime}, L_{2}$ form a square with a blue diagonal. Finally, the right parts of two of the flags $L_{1} R_{1}, L_{3} R_{3}, L_{4} R_{4}$ can also form a $2 \\times 2$ square with a blue diagonal by (ii). Putting these $2 \\times 2$ squares on the diagonal of a $4 \\times 4$ square, we find a desired arrangement of four flags.\n\nWe are ready to prove the problem statement by induction on $N$; actually, above we have proved the base case $N=4$. For the induction step, assume that $N>4$, consider any $2^{N-2}+1$ flags of length $N$, and arrange them into a large flag of size $\\left(2^{N-2}+1\\right) \\times N$. This flag contains a non-monochrome column since the flags are distinct; we may assume that this column is the first one. By the pigeonhole principle, this column contains at least $\\left\\lceil\\frac{2^{N-2}+1}{2}\\right\\rceil=2^{N-3}+1$ squares of one color (say, blue). We call the flags with a blue first square good.\n\nConsider all the good flags and remove the first square from each of them. We obtain at least $2^{N-3}+1 \\geq M_{N-1}$ flags of length $N-1$; by the induction hypothesis, $N-1$ of them\n\n\n\ncan form a square $Q$ with the monochrome diagonal. Now, returning the removed squares, we obtain a rectangle $(N-1) \\times N$, and our aim is to supplement it on the top by one more flag.\n\nIf $Q$ has a yellow diagonal, then we can take each flag with a yellow first square (it exists by a choice of the first column; moreover, it is not used in $Q$ ). Conversely, if the diagonal of $Q$ is blue then we can take any of the $\\geq 2^{N-3}+1-(N-1)>0$ remaining good flags. So, in both cases we get a desired $N \\times N$ square.', ""We present a different proof of the estimate $M_{N} \\leq 2^{N-2}+1$. We do not use the induction, involving Hall's lemma on matchings instead.\n\nConsider arbitrary $2^{N-2}+1$ distinct flags and arrange them into a large $\\left(2^{N-2}+1\\right) \\times N$ flag. Construct two bipartite graphs $G_{\\mathrm{y}}=\\left(V \\cup V^{\\prime}, E_{\\mathrm{y}}\\right)$ and $G_{\\mathrm{b}}=\\left(V \\cup V^{\\prime}, E_{\\mathrm{b}}\\right)$ with the common set of vertices as follows. Let $V$ and $V^{\\prime}$ be the set of columns and the set of flags under consideration, respectively. Next, let the edge $(c, f)$ appear in $E_{\\mathrm{y}}$ if the intersection of column $c$ and flag $f$ is yellow, and $(c, f) \\in E_{\\mathrm{b}}$ otherwise. Then we have to prove exactly that one of the graphs $G_{\\mathrm{y}}$ and $G_{\\mathrm{b}}$ contains a matching with all the vertices of $V$ involved.\n\nAssume that these matchings do not exist. By Hall's lemma, it means that there exist two sets of columns $S_{\\mathrm{y}}, S_{\\mathrm{b}} \\subset V$ such that $\\left|E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)\\right| \\leq\\left|S_{\\mathrm{y}}\\right|-1$ and $\\left|E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right| \\leq\\left|S_{\\mathrm{b}}\\right|-1$ (in the left-hand sides, $E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)$ and $E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)$ denote respectively the sets of all vertices connected to $S_{\\mathrm{y}}$ and $S_{\\mathrm{b}}$ in the corresponding graphs). Our aim is to prove that this is impossible. Note that $S_{\\mathrm{y}}, S_{\\mathrm{b}} \\neq V$ since $N \\leq 2^{N-2}+1$.\n\nFirst, suppose that $S_{\\mathrm{y}} \\cap S_{\\mathrm{b}} \\neq \\varnothing$, so there exists some $c \\in S_{\\mathrm{y}} \\cap S_{\\mathrm{b}}$. Note that each flag is connected to $c$ either in $G_{\\mathrm{y}}$ or in $G_{\\mathrm{b}}$, hence $E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right) \\cup E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)=V^{\\prime}$. Hence we have $2^{N-2}+1=\\left|V^{\\prime}\\right| \\leq\\left|E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)\\right|+\\left|E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right| \\leq\\left|S_{\\mathrm{y}}\\right|+\\left|S_{\\mathrm{b}}\\right|-2 \\leq 2 N-4$; this is impossible for $N \\geq 4$.\n\nSo, we have $S_{\\mathrm{y}} \\cap S_{\\mathrm{b}}=\\varnothing$. Let $y=\\left|S_{\\mathrm{y}}\\right|, b=\\left|S_{\\mathrm{b}}\\right|$. From the construction of our graph, we have that all the flags in the set $V^{\\prime \\prime}=V^{\\prime} \\backslash\\left(E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right) \\cup E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right)$ have blue squares in the columns of $S_{\\mathrm{y}}$ and yellow squares in the columns of $S_{\\mathrm{b}}$. Hence the only undetermined positions in these flags are the remaining $N-y-b$ ones, so $2^{N-y-b} \\geq\\left|V^{\\prime \\prime}\\right| \\geq\\left|V^{\\prime}\\right|-\\left|E_{\\mathrm{y}}\\left(S_{\\mathrm{y}}\\right)\\right|-\\left|E_{\\mathrm{b}}\\left(S_{\\mathrm{b}}\\right)\\right| \\geq$ $2^{N-2}+1-(y-1)-(b-1)$, or, denoting $c=y+b, 2^{N-c}+c>2^{N-2}+2$. This is impossible since $N \\geq c \\geq 2$.""]",['$M=2^{N-2}+1$'],False,,Expression, 1724,Combinatorics,,"2500 chess kings have to be placed on a $100 \times 100$ chessboard so that (i) no king can capture any other one (i.e. no two kings are placed in two squares sharing a common vertex); (ii) each row and each column contains exactly 25 kings. Find the number of such arrangements. (Two arrangements differing by rotation or symmetry are supposed to be different.)","[""Suppose that we have an arrangement satisfying the problem conditions. Divide the board into $2 \\times 2$ pieces; we call these pieces blocks. Each block can contain not more than one king (otherwise these two kings would attack each other); hence, by the pigeonhole principle each block must contain exactly one king.\n\nNow assign to each block a letter $\\mathrm{T}$ or $\\mathrm{B}$ if a king is placed in its top or bottom half, respectively. Similarly, assign to each block a letter $\\mathrm{L}$ or $\\mathrm{R}$ if a king stands in its left or right half. So we define T-blocks, B-blocks, L-blocks, and $R$-blocks. We also combine the letters; we call a block $a T L$-block if it is simultaneously T-block and L-block. Similarly we define TR-blocks, $B L$-blocks, and BR-blocks. The arrangement of blocks determines uniquely the arrangement of kings; so in the rest of the solution we consider the $50 \\times 50$ system of blocks (see Fig. 1). We identify the blocks by their coordinate pairs; the pair $(i, j)$, where $1 \\leq i, j \\leq 50$, refers to the $j$ th block in the $i$ th row (or the $i$ th block in the $j$ th column). The upper-left block is $(1,1)$.\n\nThe system of blocks has the following properties..\n\n$\\left(\\mathrm{i}^{\\prime}\\right)$ If $(i, j)$ is a B-block then $(i+1, j)$ is a B-block: otherwise the kings in these two blocks can take each other. Similarly: if $(i, j)$ is a T-block then $(i-1, j)$ is a T-block; if $(i, j)$ is an L-block then $(i, j-1)$ is an L-block; if $(i, j)$ is an R-block then $(i, j+1)$ is an R-block.\n\n(ii') Each column contains exactly 25 L-blocks and 25 R-blocks, and each row contains exactly 25 T-blocks and 25 B-blocks. In particular, the total number of L-blocks (or R-blocks, or T-blocks, or B-blocks) is equal to $25 \\cdot 50=1250$.\n\nConsider any B-block of the form $(1, j)$. By $\\left(\\mathrm{i}^{\\prime}\\right)$, all blocks in the $j$ th column are B-blocks; so we call such a column $B$-column. By (ii'), we have 25 B-blocks in the first row, so we obtain 25 B-columns. These $25 \\mathrm{~B}$-columns contain $1250 \\mathrm{~B}$-blocks, hence all blocks in the remaining columns are T-blocks, and we obtain 25 T-columns. Similarly, there are exactly 25 L-rows and exactly 25 -rows.\n\nNow consider an arbitrary pair of a T-column and a neighboring B-column (columns with numbers $j$ and $j+1$ ).\n\n\n\nFig. 1\n\n\n\nFig. 2\n\nCase 1. Suppose that the $j$ th column is a T-column, and the $(j+1)$ th column is a Bcolumn. Consider some index $i$ such that the $i$ th row is an L-row; then $(i, j+1)$ is a BL-block. Therefore, $(i+1, j)$ cannot be a TR-block (see Fig. 2), hence $(i+1, j)$ is a TL-block, thus the $(i+1)$ th row is an L-row. Now, choosing the $i$ th row to be the topmost L-row, we successively obtain that all rows from the $i$ th to the 50th are L-rows. Since we have exactly 25 L-rows, it follows that the rows from the 1st to the 25th are R-rows, and the rows from the 26th to the 50th are L-rows.\n\nNow consider the neighboring R-row and L-row (that are the rows with numbers 25 and 26). Replacing in the previous reasoning rows by columns and vice versa, the columns from the 1 st to the 25th are T-columns, and the columns from the 26th to the 50th are B-columns. So we have a unique arrangement of blocks that leads to the arrangement of kings satisfying the condition of the problem (see Fig. 3).\n\n\n\nFig. 3\n\n\n\nFig. 4\n\nCase 2. Suppose that the $j$ th column is a B-column, and the $(j+1)$ th column is a T-column. Repeating the arguments from Case 1, we obtain that the rows from the 1st to the 25th are L-rows (and all other rows are R-rows), the columns from the 1st to the 25th are B-columns (and all other columns are T-columns), so we find exactly one more arrangement of kings (see Fig. 4).""]",['2'],False,,Numerical, 1726,Combinatorics,,"$n \geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players bad if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that there is no bad company in this tournament. Let $w_{i}$ and $\ell_{i}$ be respectively the number of wins and losses of the $i$ th player. Prove that $$ \sum_{i=1}^{n}\left(w_{i}-\ell_{i}\right)^{3} \geq 0 \tag{1} $$","['For any tournament $T$ satisfying the problem condition, denote by $S(T)$ sum under consideration, namely\n\n$$\nS(T)=\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3}\n$$\n\nFirst, we show that the statement holds if a tournament $T$ has only 4 players. Actually, let $A=\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)$ be the number of wins of the players; we may assume that $a_{1} \\geq a_{2} \\geq a_{3} \\geq a_{4}$. We have $a_{1}+a_{2}+a_{3}+a_{4}=\\left(\\begin{array}{l}4 \\\\ 2\\end{array}\\right)=6$, hence $a_{4} \\leq 1$. If $a_{4}=0$, then we cannot have $a_{1}=a_{2}=a_{3}=2$, otherwise the company of all players is bad. Hence we should have $A=(3,2,1,0)$, and $S(T)=3^{3}+1^{3}+(-1)^{3}+(-3)^{3}=0$. On the other hand, if $a_{4}=1$, then only two possibilities, $A=(3,1,1,1)$ and $A=(2,2,1,1)$ can take place. In the former case we have $S(T)=3^{3}+3 \\cdot(-2)^{3}>0$, while in the latter one $S(T)=1^{3}+1^{3}+(-1)^{3}+(-1)^{3}=0$, as desired.\n\nNow we turn to the general problem. Consider a tournament $T$ with no bad companies and enumerate the players by the numbers from 1 to $n$. For every 4 players $i_{1}, i_{2}, i_{3}, i_{4}$ consider a ""sub-tournament"" $T_{i_{1} i_{2} i_{3} i_{4}}$ consisting of only these players and the games which they performed with each other. By the abovementioned, we have $S\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right) \\geq 0$. Our aim is to prove that\n\n$$\nS(T)=\\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right)\n\\tag{2}\n$$\n\nwhere the sum is taken over all 4 -tuples of distinct numbers from the set $\\{1, \\ldots, n\\}$. This way the problem statement will be established.\n\nWe interpret the number $\\left(w_{i}-\\ell_{i}\\right)^{3}$ as following. For $i \\neq j$, let $\\varepsilon_{i j}=1$ if the $i$ th player wins against the $j$ th one, and $\\varepsilon_{i j}=-1$ otherwise. Then\n\n$$\n\\left(w_{i}-\\ell_{i}\\right)^{3}=\\left(\\sum_{j \\neq i} \\varepsilon_{i j}\\right)^{3}=\\sum_{j_{1}, j_{2}, j_{3} \\neq i} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}\n$$\n\nHence,\n\n$$\nS(T)=\\sum_{i \\notin\\left\\{j_{1}, j_{2}, j_{3}\\right\\}} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}\n$$\n\nTo simplify this expression, consider all the terms in this sum where two indices are equal. If, for instance, $j_{1}=j_{2}$, then the term contains $\\varepsilon_{i j_{1}}^{2}=1$, so we can replace this term by $\\varepsilon_{i j_{3}}$. Make such replacements for each such term; obviously, after this change each term of the form $\\varepsilon_{i j_{3}}$ will appear $P(T)$ times, hence\n\n$$\nS(T)=\\sum_{\\left|\\left\\{i, j_{1}, j_{2}, j_{3}\\right\\}\\right|=4} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}+P(T) \\sum_{i \\neq j} \\varepsilon_{i j}=S_{1}(T)+P(T) S_{2}(T)\n$$\n\n\n\nWe show that $S_{2}(T)=0$ and hence $S(T)=S_{1}(T)$ for each tournament. Actually, note that $\\varepsilon_{i j}=-\\varepsilon_{j i}$, and the whole sum can be split into such pairs. Since the sum in each pair is 0 , so is $S_{2}(T)$.\n\nThus the desired equality (2) rewrites as\n\n$$\nS_{1}(T)=\\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S_{1}\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right)\n\\tag{3}\n$$\n\nNow, if all the numbers $j_{1}, j_{2}, j_{3}$ are distinct, then the set $\\left\\{i, j_{1}, j_{2}, j_{3}\\right\\}$ is contained in exactly one 4-tuple, hence the term $\\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}$ appears in the right-hand part of (3) exactly once, as well as in the left-hand part. Clearly, there are no other terms in both parts, so the equality is established.', ""Similarly to the first solution, we call the subsets of players as companies, and the $k$-element subsets will be called as $k$-companies.\n\nIn any company of the players, call a player the local champion of the company if he defeated all other members of the company. Similarly, if a player lost all his games against the others in the company then call him the local loser of the company. Obviously every company has at most one local champion and at most one local loser. By the condition of the problem, whenever a 4-company has a local loser, then this company has a local champion as well.\n\nSuppose that $k$ is some positive integer, and let us count all cases when a player is the local champion of some $k$-company. The $i$ th player won against $w_{i}$ other player. To be the local champion of a $k$-company, he must be a member of the company, and the other $k-1$ members must be chosen from those whom he defeated. Therefore, the $i$ th player is the local champion of $\\left(\\begin{array}{c}w_{i} \\\\ k-1\\end{array}\\right) k$-companies. Hence, the total number of local champions of all $k$-companies is $\\sum_{i=1}^{n}\\left(\\begin{array}{c}w_{i} \\\\ k-1\\end{array}\\right)$.\n\nSimilarly, the total number of local losers of the $k$-companies is $\\sum_{i=1}^{n}\\left(\\begin{array}{c}\\ell_{i} \\\\ k-1\\end{array}\\right)$.\n\nNow apply this for $k=2,3$ and 4 .\n\nSince every game has a winner and a loser, we have $\\sum_{i=1}^{n} w_{i}=\\sum_{i=1}^{n} \\ell_{i}=\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$, and hence\n\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)=0\n\\tag{4}\n$$\n\nIn every 3-company, either the players defeated one another in a cycle or the company has both a local champion and a local loser. Therefore, the total number of local champions and local losers in the 3 -companies is the same, $\\sum_{i=1}^{n}\\left(\\begin{array}{c}w_{i} \\\\ 2\\end{array}\\right)=\\sum_{i=1}^{n}\\left(\\begin{array}{c}\\ell_{i} \\\\ 2\\end{array}\\right)$. So we have\n\n$$\n\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n2\n\\end{array}\\right)\\right)=0\\tag{5}\n$$\n\nIn every 4-company, by the problem's condition, the number of local losers is less than or equal to the number of local champions. Then the same holds for the total numbers of local\n\n\n\nchampions and local losers in all 4-companies, so $\\sum_{i=1}^{n}\\left(\\begin{array}{c}w_{i} \\\\ 3\\end{array}\\right) \\geq \\sum_{i=1}^{n}\\left(\\begin{array}{c}\\ell_{i} \\\\ 3\\end{array}\\right)$. Hence,\n\n$$\n\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n3\n\\end{array}\\right)\\right) \\geq 0\\tag{6}\n$$\n\nNow we establish the problem statement (1) as a linear combination of (4), (5) and (6). It is easy check that\n\n$$\n(x-y)^{3}=24\\left(\\left(\\begin{array}{l}\nx \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{l}\ny \\\\\n3\n\\end{array}\\right)\\right)+24\\left(\\left(\\begin{array}{l}\nx \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{l}\ny \\\\\n2\n\\end{array}\\right)\\right)-\\left(3(x+y)^{2}-4\\right)(x-y)\n$$\n\nApply this identity to $x=w_{1}$ and $y=\\ell_{i}$. Since every player played $n-1$ games, we have $w_{i}+\\ell_{i}=n-1$, and thus\n\n$$\n\\left(w_{i}-\\ell_{i}\\right)^{3}=24\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n3\n\\end{array}\\right)\\right)+24\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n2\n\\end{array}\\right)\\right)-\\left(3(n-1)^{2}-4\\right)\\left(w_{i}-\\ell_{i}\\right)\n$$\n\nThen\n\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3}=24 \\underbrace{\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n3\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n3\n\\end{array}\\right)\\right)}_{\\geq 0}+24 \\underbrace{\\sum_{i=1}^{n}\\left(\\left(\\begin{array}{c}\nw_{i} \\\\\n2\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\ell_{i} \\\\\n2\n\\end{array}\\right)\\right)}_{0}-\\left(3(n-1)^{2}-4\\right) \\underbrace{\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)}_{0} \\geq 0 .\n$$""]",,True,,, 1727,Combinatorics,,"Given a positive integer $k$ and other two integers $b>w>1$. There are two strings of pearls, a string of $b$ black pearls and a string of $w$ white pearls. The length of a string is the number of pearls on it. One cuts these strings in some steps by the following rules. In each step: (i) The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then $k$ first ones (if they consist of more than one pearl) are chosen; if there are less than $k$ strings longer than 1 , then one chooses all of them. (ii) Next, one cuts each chosen string into two parts differing in length by at most one. (For instance, if there are strings of 5, 4, 4, 2 black pearls, strings of $8,4,3$ white pearls and $k=4$, then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts $(4,4),(3,2),(2,2)$ and $(2,2)$, respectively.) The process stops immediately after the step when a first isolated white pearl appears. Prove that at this stage, there will still exist a string of at least two black pearls.","['Denote the situation after the $i$ th step by $A_{i}$; hence $A_{0}$ is the initial situation, and $A_{i-1} \\rightarrow A_{i}$ is the $i$ th step. We call a string containing $m$ pearls an $m$-string; it is an $m$-w-string or a $m$-b-string if it is white or black, respectively.\n\nWe continue the process until every string consists of a single pearl. We will focus on three moments of the process: (a) the first stage $A_{s}$ when the first 1-string (no matter black or white) appears; (b) the first stage $A_{t}$ where the total number of strings is greater than $k$ (if such moment does not appear then we put $t=\\infty$ ); and (c) the first stage $A_{f}$ when all black pearls are isolated. It is sufficient to prove that in $A_{f-1}$ (or earlier), a 1-w-string appears.\n\nWe start with some easy properties of the situations under consideration. Obviously, we have $s \\leq f$. Moreover, all b-strings from $A_{f-1}$ become single pearls in the $f$ th step, hence all of them are 1 - or 2 -b-strings.\n\nNext, observe that in each step $A_{i} \\rightarrow A_{i+1}$ with $i \\leq t-1$, all $(>1)$-strings were cut since there are not more than $k$ strings at all; if, in addition, $i1$ as $s-1 \\leq \\min \\{s, t\\}$. Now, if $s=f$, then in $A_{s-1}$, there is no $1-\\mathrm{w}$-string as well as no $(>2)$-b-string. That is, $2=B_{s-1} \\geq W_{s-1} \\geq b_{s-1} \\geq w_{s-1}>1$, hence all these numbers equal 2 . This means that in $A_{s-1}$, all strings contain 2 pearls, and there are $2^{s-1}$ black and $2^{s-1}$ white strings, which means $b=2 \\cdot 2^{s-1}=w$. This contradicts the problem conditions.\n\nHence we have $s \\leq f-1$ and thus $s \\leq t$. Therefore, in the $s$ th step each string is cut into two parts. Now, if a 1-b-string appears in this step, then from $w_{s-1} \\leq b_{s-1}$ we see that a 1-w-string appears as well; so, in each case in the sth step a 1-w-string appears, while not all black pearls become single, as desired.\n\nCase 2. Now assume that $t+1 \\leq s$ and $t+2 \\leq f$. Then in $A_{t}$ we have exactly $2^{t}$ white and $2^{t}$ black strings, all being larger than 1 , and $2^{t+1}>k \\geq 2^{t}$ (the latter holds since $2^{t}$ is the total number of strings in $\\left.A_{t-1}\\right)$. Now, in the $(t+1)$ st step, exactly $k$ strings are cut, not more than $2^{t}$ of them being black; so the number of w-strings in $A_{t+1}$ is at least $2^{t}+\\left(k-2^{t}\\right)=k$. Since the number of w-strings does not decrease in our process, in $A_{f-1}$ we have at least $k$ white strings as well.\n\nFinally, in $A_{f-1}$, all b-strings are not larger than 2, and at least one 2-b-string is cut in the $f$ th step. Therefore, at most $k-1$ white strings are cut in this step, hence there exists a w-string $\\mathcal{W}$ which is not cut in the $f$ th step. On the other hand, since a 2 -b-string is cut, all $(\\geq 2)$-w-strings should also be cut in the $f$ th step; hence $\\mathcal{W}$ should be a single pearl. This is exactly what we needed.', 'We use the same notations as introduced in the first paragraph of the previous solution. We claim that at every stage, there exist a $u$-b-string and a $v$-w-string such that either\n\n(i) $u>v \\geq 1$, or\n\n(ii) $2 \\leq u \\leq v<2 u$, and there also exist $k-1$ of $(>v / 2)$-strings other than considered above.\n\nFirst, we notice that this statement implies the problem statement. Actually, in both cases (i) and (ii) we have $u>1$, so at each stage there exists a $(\\geq 2)$-b-string, and for the last stage it is exactly what we need.\n\nNow, we prove the claim by induction on the number of the stage. Obviously, for $A_{0}$ the condition (i) holds since $b>w$. Further, we suppose that the statement holds for $A_{i}$, and prove it for $A_{i+1}$. Two cases are possible.\n\nCase 1. Assume that in $A_{i}$, there are a $u$-b-string and a $v$-w-string with $u>v$. We can assume that $v$ is the length of the shortest w-string in $A_{i}$; since we are not at the final stage, we have $v \\geq 2$. Now, in the $(i+1)$ st step, two subcases may occur.\n\nSubcase 1a. Suppose that either no $u$-b-string is cut, or both some $u$-b-string and some $v$-w-string are cut. Then in $A_{i+1}$, we have either a $u$-b-string and a $(\\leq v)$-w-string (and (i) is valid), or we have a $\\lceil u / 2\\rceil$-b-string and a $\\lfloor v / 2\\rfloor$-w-string. In the latter case, from $u>v$ we get $\\lceil u / 2\\rceil>\\lfloor v / 2\\rfloor$, and (i) is valid again.\n\nSubcase 1b. Now, some $u$-b-string is cut, and no $v$-w-string is cut (and hence all the strings which are cut are longer than $v$ ). If $u^{\\prime}=\\lceil u / 2\\rceil>v$, then the condition (i) is satisfied since we have a $u^{\\prime}$-b-string and a $v$-w-string in $A_{i+1}$. Otherwise, notice that the inequality $u>v \\geq 2$ implies $u^{\\prime} \\geq 2$. Furthermore, besides a fixed $u$-b-string, other $k-1$ of $(\\geq v+1)$-strings should be cut in the $(i+1)$ st step, hence providing at least $k-1$ of $(\\geq\\lceil(v+1) / 2\\rceil)$-strings, and $\\lceil(v+1) / 2\\rceil>v / 2$. So, we can put $v^{\\prime}=v$, and we have $u^{\\prime} \\leq vv$, so each one results in a $(>v / 2)$ string. Hence in $A_{i+1}$, there exist $k \\geq k-1$ of $(>v / 2)$-strings other than the considered $u$ - and $v$-strings, and the condition (ii) is satisfied.\n\nSubcase 2c. In the remaining case, all $u$-b-strings are cut. This means that all $(\\geq u)$-strings are cut as well, hence our $v$-w-string is cut. Therefore in $A_{i+1}$ there exists a $[u / 2]$-b-string together with a $\\lfloor v / 2\\rfloor$-w-string. Now, if $u^{\\prime}=\\lceil u / 2\\rceil>\\lfloor v / 2\\rfloor=v^{\\prime}$ then the condition (i) is fulfilled. Otherwise, we have $u^{\\prime} \\leq v^{\\prime}$ $|f(W)|$; if there is no such set then we set $W=\\varnothing$. Denote $W^{\\prime}=f(W), U=V \\backslash W, U^{\\prime}=V^{\\prime} \\backslash W^{\\prime}$.\n\n\n\nBy our assumption and the Lemma condition, $|f(V)|=\\left|V^{\\prime}\\right| \\geq|V|$, hence $W \\neq V$ and $U \\neq \\varnothing$. Permuting the coordinates, we can assume that $U^{\\prime}=\\left\\{v_{i j}: 1 \\leq i \\leq \\ell\\right\\}, W^{\\prime}=\\left\\{v_{i j}: \\ell+1 \\leq i \\leq k\\right\\}$.\n\nConsider the induced subgraph $G^{\\prime}$ of $G$ on the vertices $U \\cup U^{\\prime}$. We claim that for every $X \\subset U$, we get $\\left|f(X) \\cap U^{\\prime}\\right| \\geq|X|$ (so $G^{\\prime}$ satisfies the conditions of Hall's lemma). Actually, we have $|W| \\geq|f(W)|$, so if $|X|>\\left|f(X) \\cap U^{\\prime}\\right|$ for some $X \\subset U$, then we have\n\n$$\n|W \\cup X|=|W|+|X|>|f(W)|+\\left|f(X) \\cap U^{\\prime}\\right|=\\left|f(W) \\cup\\left(f(X) \\cap U^{\\prime}\\right)\\right|=|f(W \\cup X)|\n$$\n\nThis contradicts the maximality of $|W|$.\n\nThus, applying Hall's lemma, we can assign to each $L \\in U$ some vertex $v_{i j} \\in U^{\\prime}$ so that to distinct elements of $U$, distinct vertices of $U^{\\prime}$ are assigned. In this situation, we say that $L \\in U$ corresponds to the $i$ th axis, and write $g(L)=i$. Since there are $n_{i}-1$ vertices of the form $v_{i j}$, we get that for each $1 \\leq i \\leq \\ell$, not more than $n_{i}-1$ subgrids correspond to the $i$ th axis.\n\nFinally, we are ready to present the desired point. Since $W \\neq V$, there exists a point $b=\\left(b_{1}, b_{2}, \\ldots, b_{k}\\right) \\in N \\backslash\\left(\\cup_{L \\in W} L\\right)$. On the other hand, for every $1 \\leq i \\leq \\ell$, consider any subgrid $L \\in U$ with $g(L)=i$. This means exactly that $L$ is orthogonal to the $i$ th axis, and hence all its elements have the same $i$ th coordinate $c_{L}$. Since there are at most $n_{i}-1$ such subgrids, there exists a number $0 \\leq a_{i} \\leq n_{i}-1$ which is not contained in a set $\\left\\{c_{L}: g(L)=i\\right\\}$. Choose such number for every $1 \\leq i \\leq \\ell$. Now we claim that point $a=\\left(a_{1}, \\ldots, a_{\\ell}, b_{\\ell+1}, \\ldots, b_{k}\\right)$ is not covered, hence contradicting the Lemma condition.\n\nSurely, point $a$ cannot lie in some $L \\in U$, since all the points in $L$ have $g(L)$ th coordinate $c_{L} \\neq a_{g(L)}$. On the other hand, suppose that $a \\in L$ for some $L \\in W$; recall that $b \\notin L$. But the points $a$ and $b$ differ only at first $\\ell$ coordinates, so $L$ should be orthogonal to at least one of the first $\\ell$ axes, and hence our graph contains some edge $\\left(L, v_{i j}\\right)$ for $i \\leq \\ell$. It contradicts the definition of $W^{\\prime}$. The Lemma is proved.\n\nNow we turn to the problem. Let $d_{j}$ be the step of the progression $P_{j}$. Note that since $n=$ l.c.m. $\\left(d_{1}, \\ldots, d_{s}\\right)$, for each $1 \\leq i \\leq k$ there exists an index $j(i)$ such that $p_{i}^{\\alpha_{i}} \\mid d_{j(i)}$. We assume that $n>1$; otherwise the problem statement is trivial.\n\nFor each $0 \\leq m \\leq n-1$ and $1 \\leq i \\leq k$, let $m_{i}$ be the residue of $m$ modulo $p_{i}^{\\alpha_{i}}$, and let $m_{i}=\\overline{r_{i \\alpha_{i}} \\ldots r_{i 1}}$ be the base $p_{i}$ representation of $m_{i}$ (possibly, with some leading zeroes). Now, we put into correspondence to $m$ the sequence $r(m)=\\left(r_{11}, \\ldots, r_{1 \\alpha_{1}}, r_{21}, \\ldots, r_{k \\alpha_{k}}\\right)$. Hence $r(m)$ lies in a $\\underbrace{p_{1} \\times \\cdots \\times p_{1}}_{\\alpha_{1} \\text { times }} \\times \\cdots \\times \\underbrace{p_{k} \\times \\cdots \\times p_{k}}_{\\alpha_{k} \\text { times }} \\operatorname{grid} N$.\n\nSurely, if $r(m)=r\\left(m^{\\prime}\\right)$ then $p_{i}^{\\alpha_{i}} \\mid m_{i}-m_{i}^{\\prime}$, which follows $p_{i}^{\\alpha_{i}} \\mid m-m^{\\prime}$ for all $1 \\leq i \\leq k$; consequently, $n \\mid m-m^{\\prime}$. So, when $m$ runs over the set $\\{0, \\ldots, n-1\\}$, the sequences $r(m)$ do not repeat; since $|N|=n$, this means that $r$ is a bijection between $\\{0, \\ldots, n-1\\}$ and $N$. Now we will show that for each $1 \\leq i \\leq s$, the set $L_{i}=\\left\\{r(m): m \\in P_{i}\\right\\}$ is a subgrid, and that for each axis there exists a subgrid orthogonal to this axis. Obviously, these subgrids cover $N$, and the condition (ii') follows directly from (ii). Hence the Lemma provides exactly the estimate we need.\n\nConsider some $1 \\leq j \\leq s$ and let $d_{j}=p_{1}^{\\gamma_{1}} \\ldots p_{k}^{\\gamma_{k}}$. Consider some $q \\in P_{j}$ and let $r(q)=$ $\\left(r_{11}, \\ldots, r_{k \\alpha_{k}}\\right)$. Then for an arbitrary $q^{\\prime}$, setting $r\\left(q^{\\prime}\\right)=\\left(r_{11}^{\\prime}, \\ldots, r_{k \\alpha_{k}}^{\\prime}\\right)$ we have\n\n$$\nq^{\\prime} \\in P_{j} \\quad \\Longleftrightarrow \\quad p_{i}^{\\gamma_{i}} \\mid q-q^{\\prime} \\text { for each } 1 \\leq i \\leq k \\quad \\Longleftrightarrow \\quad r_{i, t}=r_{i, t}^{\\prime} \\text { for all } t \\leq \\gamma_{i} \\text {. }\n$$\n\nHence $L_{j}=\\left\\{\\left(r_{11}^{\\prime}, \\ldots, r_{k \\alpha_{k}}^{\\prime}\\right) \\in N: r_{i, t}=r_{i, t}^{\\prime}\\right.$ for all $\\left.t \\leq \\gamma_{i}\\right\\}$ which means that $L_{j}$ is a subgrid containing $r(q)$. Moreover, in $L_{j(i)}$, all the coordinates corresponding to $p_{i}$ are fixed, so it is orthogonal to all of their axes, as desired."", ""We start with introducing some notation. For positive integer $r$, we denote $[r]=\\{1,2, \\ldots, r\\}$. Next, we say that a set of progressions $\\mathcal{P}=\\left\\{P_{1}, \\ldots, P_{s}\\right\\}$ cover $\\mathbb{Z}$ if each integer belongs to some of them; we say that this covering is minimal if no proper subset of $\\mathcal{P}$ covers $\\mathbb{Z}$. Obviously, each covering contains a minimal subcovering.\n\nNext, for a minimal covering $\\left\\{P_{1}, \\ldots, P_{s}\\right\\}$ and for every $1 \\leq i \\leq s$, let $d_{i}$ be the step of progression $P_{i}$, and $h_{i}$ be some number which is contained in $P_{i}$ but in none of the other progressions. We assume that $n>1$, otherwise the problem is trivial. This implies $d_{i}>1$, otherwise the progression $P_{i}$ covers all the numbers, and $n=1$.\n\nWe will prove a more general statement, namely the following\n\nClaim. Assume that the progressions $P_{1}, \\ldots, P_{s}$ and number $n=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}>1$ are chosen as in the problem statement. Moreover, choose some nonempty set of indices $I=\\left\\{i_{1}, \\ldots, i_{t}\\right\\} \\subseteq[k]$ and some positive integer $\\beta_{i} \\leq \\alpha_{i}$ for every $i \\in I$. Consider the set of indices\n\n$$\nT=\\left\\{j: 1 \\leq j \\leq s, \\text { and } p_{i}^{\\alpha_{i}-\\beta_{i}+1} \\mid d_{j} \\text { for some } i \\in I\\right\\} .\n$$\n\nThen\n\n$$\n|T| \\geq 1+\\sum_{i \\in I} \\beta_{i}\\left(p_{i}-1\\right)\n\\tag{2}\n$$\n\nObserve that the Claim for $I=[k]$ and $\\beta_{i}=\\alpha_{i}$ implies the problem statement, since the left-hand side in (2) is not greater than $s$. Hence, it suffices to prove the Claim.\n\n1. First, we prove the Claim assuming that all $d_{j}$ 's are prime numbers. If for some $1 \\leq i \\leq k$ we have at least $p_{i}$ progressions with the step $p_{i}$, then they do not intersect and hence cover all the integers; it means that there are no other progressions, and $n=p_{i}$; the Claim is trivial in this case.\n\nNow assume that for every $1 \\leq i \\leq k$, there are not more than $p_{i}-1$ progressions with step $p_{i}$; each such progression covers the numbers with a fixed residue modulo $p_{i}$, therefore there exists a residue $q_{i} \\bmod p_{i}$ which is not touched by these progressions. By the Chinese Remainder Theorem, there exists a number $q$ such that $q \\equiv q_{i}\\left(\\bmod p_{i}\\right)$ for all $1 \\leq i \\leq k$; this number cannot be covered by any progression with step $p_{i}$, hence it is not covered at all. A contradiction.\n\n\n\n2. Now, we assume that the general Claim is not valid, and hence we consider a counterexample $\\left\\{P_{1}, \\ldots, P_{s}\\right\\}$ for the Claim; we can choose it to be minimal in the following sense:\n\n- the number $n$ is minimal possible among all the counterexamples; of $n$.\n- the sum $\\sum_{i} d_{i}$ is minimal possible among all the counterexamples having the chosen value\n\nAs was mentioned above, not all numbers $d_{i}$ are primes; hence we can assume that $d_{1}$ is composite, say $p_{1} \\mid d_{1}$ and $d_{1}^{\\prime}=\\frac{d_{1}}{p_{1}}>1$. Consider a progression $P_{1}^{\\prime}$ having the step $d_{1}^{\\prime}$, and containing $P_{1}$. We will focus on two coverings constructed as follows.\n\n(i) Surely, the progressions $P_{1}^{\\prime}, P_{2}, \\ldots, P_{s}$ cover $\\mathbb{Z}$, though this covering in not necessarily minimal. So, choose some minimal subcovering $\\mathcal{P}^{\\prime}$ in it; surely $P_{1}^{\\prime} \\in \\mathcal{P}^{\\prime}$ since $h_{1}$ is not covered by $P_{2}, \\ldots, P_{s}$, so we may assume that $\\mathcal{P}^{\\prime}=\\left\\{P_{1}^{\\prime}, P_{2}, \\ldots, P_{s^{\\prime}}\\right\\}$ for some $s^{\\prime} \\leq s$. Furthermore, the period of the covering $\\mathcal{P}^{\\prime}$ can appear to be less than $n$; so we denote this period by\n\n$$\nn^{\\prime}=p_{1}^{\\alpha_{1}-\\sigma_{1}} \\ldots p_{k}^{\\alpha_{k}-\\sigma_{k}}=\\text { l.c.m. }\\left(d_{1}^{\\prime}, d_{2}, \\ldots, d_{s^{\\prime}}\\right)\n$$\n\nObserve that for each $P_{j} \\notin \\mathcal{P}^{\\prime}$, we have $h_{j} \\in P_{1}^{\\prime}$, otherwise $h_{j}$ would not be covered by $\\mathcal{P}$.\n\n(ii) On the other hand, each nonempty set of the form $R_{i}=P_{i} \\cap P_{1}^{\\prime}(1 \\leq i \\leq s)$ is also a progression with a step $r_{i}=$ l.c.m. $\\left(d_{i}, d_{1}^{\\prime}\\right)$, and such sets cover $P_{1}^{\\prime}$. Scaling these progressions with the ratio $1 / d_{1}^{\\prime}$, we obtain the progressions $Q_{i}$ with steps $q_{i}=r_{i} / d_{1}^{\\prime}$ which cover $\\mathbb{Z}$. Now we choose a minimal subcovering $\\mathcal{Q}$ of this covering; again we should have $Q_{1} \\in \\mathcal{Q}$ by the reasons of $h_{1}$. Now, denote the period of $\\mathcal{Q}$ by\n\n$$\nn^{\\prime \\prime}=\\text { l.c.m. }\\left\\{q_{i}: Q_{i} \\in \\mathcal{Q}\\right\\}=\\frac{\\text { l.c.m. }\\left\\{r_{i}: Q_{i} \\in \\mathcal{Q}\\right\\}}{d_{1}^{\\prime}}=\\frac{p_{1}^{\\gamma_{1}} \\ldots p_{k}^{\\gamma_{k}}}{d_{1}^{\\prime}} \\text {. }\n$$\n\nNote that if $h_{j} \\in P_{1}^{\\prime}$, then the image of $h_{j}$ under the scaling can be covered by $Q_{j}$ only; so, in this case we have $Q_{j} \\in \\mathcal{Q}$.\n\nOur aim is to find the desired number of progressions in coverings $\\mathcal{P}$ and $\\mathcal{Q}$. First, we have $n \\geq n^{\\prime}$, and the sum of the steps in $\\mathcal{P}^{\\prime}$ is less than that in $\\mathcal{P}$; hence the Claim is valid for $\\mathcal{P}^{\\prime}$. We apply it to the set of indices $I^{\\prime}=\\left\\{i \\in I: \\beta_{i}>\\sigma_{i}\\right\\}$ and the exponents $\\beta_{i}^{\\prime}=\\beta_{i}-\\sigma_{i}$; hence the set under consideration is\n\n$$\nT^{\\prime}=\\left\\{j: 1 \\leq j \\leq s^{\\prime}, \\text { and } p_{i}^{\\left(\\alpha_{i}-\\sigma_{i}\\right)-\\beta_{i}^{\\prime}+1}=p_{i}^{\\alpha_{i}-\\beta_{i}+1} \\mid d_{j} \\text { for some } i \\in I^{\\prime}\\right\\} \\subseteq T \\cap\\left[s^{\\prime}\\right] \\text {, }\n$$\n\nand we obtain that\n\n$$\n\\left|T \\cap\\left[s^{\\prime}\\right]\\right| \\geq\\left|T^{\\prime}\\right| \\geq 1+\\sum_{i \\in I^{\\prime}}\\left(\\beta_{i}-\\sigma_{i}\\right)\\left(p_{i}-1\\right)=1+\\sum_{i \\in I}\\left(\\beta_{i}-\\sigma_{i}\\right)_{+}\\left(p_{i}-1\\right)\n$$\n\nwhere $(x)_{+}=\\max \\{x, 0\\}$; the latter equality holds as for $i \\notin I^{\\prime}$ we have $\\beta_{i} \\leq \\sigma_{i}$.\n\nObserve that $x=(x-y)_{+}+\\min \\{x, y\\}$ for all $x, y$. So, if we find at least\n\n$$\nG=\\sum_{i \\in I} \\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}\\left(p_{i}-1\\right)\n$$\n\nindices in $T \\cap\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}$, then we would have\n\n$|T|=\\left|T \\cap\\left[s^{\\prime}\\right]\\right|+\\left|T \\cap\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}\\right| \\geq 1+\\sum_{i \\in I}\\left(\\left(\\beta_{i}-\\sigma_{i}\\right)_{+}+\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}\\right)\\left(p_{i}-1\\right)=1+\\sum_{i \\in I} \\beta_{i}\\left(p_{i}-1\\right)$,\n\nthus leading to a contradiction with the choice of $\\mathcal{P}$. We will find those indices among the indices of progressions in $\\mathcal{Q}$.\n\n\n\n3. Now denote $I^{\\prime \\prime}=\\left\\{i \\in I: \\sigma_{i}>0\\right\\}$ and consider some $i \\in I^{\\prime \\prime}$; then $p_{i}^{\\alpha_{i}} \\nmid n^{\\prime}.$. On the other hand, there exists an index $j(i)$ such that $p_{i}^{\\alpha_{i}} \\mid d_{j(i)}$; this means that $d_{j(i)} \\nmid n^{\\prime}$ and hence $P_{j(i)}$ cannot appear in $\\mathcal{P}^{\\prime}$, so $j(i)>s^{\\prime}$. Moreover, we have observed before that in this case $h_{j(i)} \\in P_{1}^{\\prime}$, hence $Q_{j(i)} \\in \\mathcal{Q}$. This means that $q_{j(i)} \\mid n^{\\prime \\prime}$, therefore $\\gamma_{i}=\\alpha_{i}$ for each $i \\in I^{\\prime \\prime}$ (recall here that $q_{i}=r_{i} / d_{1}^{\\prime}$ and hence $\\left.d_{j(i)}\\left|r_{j(i)}\\right| d_{1}^{\\prime} n^{\\prime \\prime}\\right)$.\n\nLet $d_{1}^{\\prime}=p_{1}^{\\tau_{1}} \\ldots p_{k}^{\\tau_{k}}$. Then $n^{\\prime \\prime}=p_{1}^{\\gamma_{1}-\\tau_{1}} \\ldots p_{k}^{\\gamma_{i}-\\tau_{i}}$. Now, if $i \\in I^{\\prime \\prime}$, then for every $\\beta$ the condition $p_{i}^{\\left(\\gamma_{i}-\\tau_{i}\\right)-\\beta+1} \\mid q_{j}$ is equivalent to $p_{i}^{\\alpha_{i}-\\beta+1} \\mid r_{j}$.\n\nNote that $n^{\\prime \\prime} \\leq n / d_{1}^{\\prime}0$. So, the set under consideration is\n\n$$\n\\begin{aligned}\nT^{\\prime \\prime} & =\\left\\{j: Q_{j} \\in \\mathcal{Q}, \\text { and } p_{i}^{\\left(\\gamma_{i}-\\tau_{i}\\right)-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid q_{j} \\text { for some } i \\in I^{\\prime \\prime}\\right\\} \\\\\n& =\\left\\{j: Q_{j} \\in \\mathcal{Q}, \\text { and } p_{i}^{\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid r_{j} \\text { for some } i \\in I^{\\prime \\prime}\\right\\},\n\\end{aligned}\n$$\n\nand we obtain $\\left|T^{\\prime \\prime}\\right| \\geq 1+G$. Finally, we claim that $T^{\\prime \\prime} \\subseteq T \\cap\\left(\\{1\\} \\cup\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}\\right)$; then we will obtain $\\left|T \\cap\\left\\{s^{\\prime}+1, \\ldots, s\\right\\}\\right| \\geq G$, which is exactly what we need.\n\nTo prove this, consider any $j \\in T^{\\prime \\prime}$. Observe first that $\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1>\\alpha_{i}-\\sigma_{i} \\geq \\tau_{i}$, hence from $p_{i}^{\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid r_{j}=$ l.c.m. $\\left(d_{1}^{\\prime}, d_{j}\\right)$ we have $p_{i}^{\\alpha_{i}-\\min \\left\\{\\beta_{i}, \\sigma_{i}\\right\\}+1} \\mid d_{j}$, which means that $j \\in T$. Next, the exponent of $p_{i}$ in $d_{j}$ is greater than that in $n^{\\prime}$, which means that $P_{j} \\notin \\mathcal{P}^{\\prime}$. This may appear only if $j=1$ or $j>s^{\\prime}$, as desired. This completes the proof.""]",,True,,, 1729,Geometry,,"Let $A B C$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $B C, C A, A B$ respectively. One of the intersection points of the line $E F$ and the circumcircle is $P$. The lines $B P$ and $D F$ meet at point $Q$. Prove that $A P=A Q$.","['The line $E F$ intersects the circumcircle at two points. Depending on the choice of $P$, there are two different cases to consider.\n\nCase 1: The point $P$ lies on the ray $E F$ (see Fig. 1).\n\nLet $\\angle C A B=\\alpha, \\angle A B C=\\beta$ and $\\angle B C A=\\gamma$. The quadrilaterals $B C E F$ and $C A F D$ are cyclic due to the right angles at $D, E$ and $F$. So,\n\n$$\n\\begin{aligned}\n& \\angle B D F=180^{\\circ}-\\angle F D C=\\angle C A F=\\alpha, \\\\\n& \\angle A F E=180^{\\circ}-\\angle E F B=\\angle B C E=\\gamma, \\\\\n& \\angle D F B=180^{\\circ}-\\angle A F D=\\angle D C A=\\gamma .\n\\end{aligned}\n$$\n\nSince $P$ lies on the arc $A B$ of the circumcircle, $\\angle P B A<\\angle B C A=\\gamma$. Hence, we have\n\n$$\n\\angle P B D+\\angle B D F=\\angle P B A+\\angle A B D+\\angle B D F<\\gamma+\\beta+\\alpha=180^{\\circ}\n$$\n\nand the point $Q$ must lie on the extensions of $B P$ and $D F$ beyond the points $P$ and $F$, respectively.\n\nFrom the cyclic quadrilateral $A P B C$ we get\n\n$$\n\\angle Q P A=180^{\\circ}-\\angle A P B=\\angle B C A=\\gamma=\\angle D F B=\\angle Q F A .\n$$\n\nHence, the quadrilateral $A Q P F$ is cyclic. Then $\\angle A Q P=180^{\\circ}-\\angle P F A=\\angle A F E=\\gamma$.\n\nWe obtained that $\\angle A Q P=\\angle Q P A=\\gamma$, so the triangle $A Q P$ is isosceles, $A P=A Q$.\n\n\n\nFig. 1\n\n\n\nFig. 2\n\n\n\nCase 2: The point $P$ lies on the ray $F E$ (see Fig. 2). In this case the point $Q$ lies inside the segment $F D$.\n\nSimilarly to the first case, we have\n\n$$\n\\angle Q P A=\\angle B C A=\\gamma=\\angle D F B=180^{\\circ}-\\angle A F Q .\n$$\n\nHence, the quadrilateral $A F Q P$ is cyclic.\n\nThen $\\angle A Q P=\\angle A F P=\\angle A F E=\\gamma=\\angle Q P A$. The triangle $A Q P$ is isosceles again, $\\angle A Q P=\\angle Q P A$ and thus $A P=A Q$.', 'For arbitrary points $X, Y$ on the circumcircle, denote by $\\overparen{X Y}$ the central angle of the $\\operatorname{arc} X Y$.\n\nLet $P$ and $P^{\\prime}$ be the two points where the line $E F$ meets the circumcircle; let $P$ lie on the $\\operatorname{arc} A B$ and let $P^{\\prime}$ lie on the $\\operatorname{arc} C A$. Let $B P$ and $B P^{\\prime}$ meet the line $D F$ and $Q$ and $Q^{\\prime}$, respectively (see Fig. 3). We will prove that $A P=A P^{\\prime}=A Q=A Q^{\\prime}$.\n\n\n\nFig. 3\n\nLike in the first solution, we have $\\angle A F E=\\angle B F P=\\angle D F B=\\angle B C A=\\gamma$ from the cyclic quadrilaterals $B C E F$ and $C A F D$.\n\nBy $\\overparen{P B}+\\overparen{P^{\\prime} A}=2 \\angle A F P^{\\prime}=2 \\gamma=2 \\angle B C A=\\overparen{A P}+\\overparen{P B}$, we have\n\n$$\n\\overparen{A P}=\\overparen{P^{\\prime} A}, \\quad \\angle P B A=\\angle A B P^{\\prime} \\quad \\text { and } \\quad A P=A P^{\\prime} \\text {. }\\tag{1}\n$$\n\nDue to $\\overparen{A P}=\\overparen{P^{\\prime} A}$, the lines $B P$ and $B Q^{\\prime}$ are symmetrical about line $A B$.\n\nSimilarly, by $\\angle B F P=\\angle Q^{\\prime} F B$, the lines $F P$ and $F Q^{\\prime}$ are symmetrical about $A B$. It follows that also the points $P$ and $P^{\\prime}$ are symmetrical to $Q^{\\prime}$ and $Q$, respectively. Therefore,\n\n$$\nA P=A Q^{\\prime} \\quad \\text { and } \\quad A P^{\\prime}=A Q .\\tag{2}\n$$\n\nThe relations (1) and (2) together prove $A P=A P^{\\prime}=A Q=A Q^{\\prime}$.']",,True,,, 1730,Geometry,,"Point $P$ lies inside triangle $A B C$. Lines $A P, B P, C P$ meet the circumcircle of $A B C$ again at points $K, L, M$, respectively. The tangent to the circumcircle at $C$ meets line $A B$ at $S$. Prove that $S C=S P$ if and only if $M K=M L$.","['We assume that $C A>C B$, so point $S$ lies on the ray $A B$.\n\nFrom the similar triangles $\\triangle P K M \\sim \\triangle P C A$ and $\\triangle P L M \\sim \\triangle P C B$ we get $\\frac{P M}{K M}=\\frac{P A}{C A}$ and $\\frac{L M}{P M}=\\frac{C B}{P B}$. Multiplying these two equalities, we get\n\n$$\n\\frac{L M}{K M}=\\frac{C B}{C A} \\cdot \\frac{P A}{P B}\n$$\n\nHence, the relation $M K=M L$ is equivalent to $\\frac{C B}{C A}=\\frac{P B}{P A}$.\n\nDenote by $E$ the foot of the bisector of angle $B$ in triangle $A B C$. Recall that the locus of points $X$ for which $\\frac{X A}{X B}=\\frac{C A}{C B}$ is the Apollonius circle $\\Omega$ with the center $Q$ on the line $A B$, and this circle passes through $C$ and $E$. Hence, we have $M K=M L$ if and only if $P$ lies on $\\Omega$, that is $Q P=Q C$.\n\n\n\nFig. 1\n\nNow we prove that $S=Q$, thus establishing the problem statement. We have $\\angle C E S=$ $\\angle C A E+\\angle A C E=\\angle B C S+\\angle E C B=\\angle E C S$, so $S C=S E$. Hence, the point $S$ lies on $A B$ as well as on the perpendicular bisector of $C E$ and therefore coincides with $Q$.', 'As in the previous solution, we assume that $S$ lies on the ray $A B$.\n\n1. Let $P$ be an arbitrary point inside both the circumcircle $\\omega$ of the triangle $A B C$ and the angle $A S C$, the points $K, L, M$ defined as in the problem. We claim that $S P=S C$ implies $M K=M L$.\n\nLet $E$ and $F$ be the points of intersection of the line $S P$ with $\\omega$, point $E$ lying on the segment $S P$ (see Fig. 2).\n\n\n\n\n\nFig. 2\n\nWe have $S P^{2}=S C^{2}=S A \\cdot S B$, so $\\frac{S P}{S B}=\\frac{S A}{S P}$, and hence $\\triangle P S A \\sim \\triangle B S P$. Then $\\angle B P S=\\angle S A P$. Since $2 \\angle B P S=\\overparen{B E}+\\overparen{L F}$ and $2 \\angle S A P=\\overparen{B E}+\\overparen{E K}$ we have\n\n$$\n\\overparen{L F}=\\overparen{E K} .\\tag{1}\n$$\n\nOn the other hand, from $\\angle S P C=\\angle S C P$ we have $\\overparen{E C}+\\overparen{M F}=\\overparen{E C}+\\overparen{E M}$, or\n\n$$\n\\overparen{M F}=\\overparen{E M}\\tag{2}\n$$\n\nFrom (1) and (2) we get $\\overparen{M F L}=\\overparen{M F}+\\overparen{F L}=\\overparen{M E}+\\overparen{E K}=\\overparen{M E K}$ and hence $M K=M L$. The claim is proved.\n\n2. We are left to prove the converse. So, assume that $M K=M L$, and introduce the points $E$ and $F$ as above. We have $S C^{2}=S E \\cdot S F$; hence, there exists a point $P^{\\prime}$ lying on the segment $E F$ such that $S P^{\\prime}=S C$ (see Fig. 3).\n\n\n\nFig. 3\n\n\n\nAssume that $P \\neq P^{\\prime}$. Let the lines $A P^{\\prime}, B P^{\\prime}, C P^{\\prime}$ meet $\\omega$ again at points $K^{\\prime}, L^{\\prime}, M^{\\prime}$ respectively. Now, if $P^{\\prime}$ lies on the segment $P F$ then by the first part of the solution we have $\\overparen{M^{\\prime} F L^{\\prime}}=\\overparen{M^{\\prime} E K^{\\prime}}$. On the other hand, we have $\\overparen{M F L}>\\overparen{M^{\\prime} F L^{\\prime}}=\\overparen{M^{\\prime} E K^{\\prime}}>\\overparen{M E K}$, therefore $\\overparen{M F L}>\\overparen{M E K}$ which contradicts $M K=M L$.\n\nSimilarly, if point $P^{\\prime}$ lies on the segment $E P$ then we get $\\overparen{M F L}<\\overparen{M E K}$ which is impossible. Therefore, the points $P$ and $P^{\\prime}$ coincide and hence $S P=S P^{\\prime}=S C$.', 'We present a different proof of the converse direction, that is, $M K=M L \\Rightarrow$ $S P=S C$. As in the previous solutions we assume that $C A>C B$, and the line $S P$ meets $\\omega$ at $E$ and $F$.\n\nFrom $M L=M K$ we get $\\overparen{M E K}=\\overparen{M F L}$. Now we claim that $\\overparen{M E}=\\overparen{M F}$ and $\\overparen{E K}=\\overparen{F L}$.\n\nTo the contrary, suppose first that $\\overparen{M E}>\\overparen{M F}$; then $\\overparen{E K}=\\overparen{M E K}-\\overparen{M E}<\\overparen{M F L}-\\overparen{M F}=$ $\\overparen{F L}$. Now, the inequality $\\overparen{M E}>\\overparen{M F}$ implies $2 \\angle S C M=\\overparen{E C}+\\overparen{M E}>\\overparen{E C}+\\overparen{M F}=2 \\angle S P C$ and hence $S P>S C$. On the other hand, the inequality $\\overparen{E K}<\\overparen{F L}$ implies $2 \\angle S P K=$ $\\overparen{E K}+\\overparen{A F}<\\overparen{F L}+\\overparen{A F}=2 \\angle A B L$, hence\n\n$$\n\\angle S P A=180^{\\circ}-\\angle S P K>180^{\\circ}-\\angle A B L=\\angle S B P .\n$$\n\n\n\nFig. 4\n\nConsider the point $A^{\\prime}$ on the ray $S A$ for which $\\angle S P A^{\\prime}=\\angle S B P$; in our case, this point lies on the segment $S A$ (see Fig. 4). Then $\\triangle S B P \\sim \\triangle S P A^{\\prime}$ and $S P^{2}=S B \\cdot S A^{\\prime}S C$.\n\nSimilarly, one can prove that the inequality $\\overparen{M E}<\\overparen{M F}$ is also impossible. So, we get $\\overparen{M E}=\\overparen{M F}$ and therefore $2 \\angle S C M=\\overparen{E C}+\\overparen{M E}=\\overparen{E C}+\\overparen{M F}=2 \\angle S P C$, which implies $S C=S P$.']",,True,,, 1731,Geometry,,"Let $A_{1} A_{2} \ldots A_{n}$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_{1}, \ldots, P_{n}$ onto lines $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_{1}, \ldots, X_{n}$ on sides $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively, $$ \max \left\{\frac{X_{1} X_{2}}{P_{1} P_{2}}, \ldots, \frac{X_{n} X_{1}}{P_{n} P_{1}}\right\} \geq 1 $$","['Denote $P_{n+1}=P_{1}, X_{n+1}=X_{1}, A_{n+1}=A_{1}$.\n\nLemma. Let point $Q$ lies inside $A_{1} A_{2} \\ldots A_{n}$. Then it is contained in at least one of the circumcircles of triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$.\n\nProof. If $Q$ lies in one of the triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$, the claim is obvious. Otherwise $Q$ lies inside the polygon $X_{1} X_{2} \\ldots X_{n}$ (see Fig. 1). Then we have\n\n$$\n\\begin{aligned}\n& \\left(\\angle X_{1} A_{2} X_{2}+\\angle X_{1} Q X_{2}\\right)+\\cdots+\\left(\\angle X_{n} A_{1} X_{1}+\\angle X_{n} Q X_{1}\\right) \\\\\n& \\quad \\quad=\\left(\\angle X_{1} A_{1} X_{2}+\\cdots+\\angle X_{n} A_{1} X_{1}\\right)+\\left(\\angle X_{1} Q X_{2}+\\cdots+\\angle X_{n} Q X_{1}\\right)=(n-2) \\pi+2 \\pi=n \\pi\n\\end{aligned}\n$$\n\nhence there exists an index $i$ such that $\\angle X_{i} A_{i+1} X_{i+1}+\\angle X_{i} Q X_{i+1} \\geq \\frac{\\pi n}{n}=\\pi$. Since the quadrilateral $Q X_{i} A_{i+1} X_{i+1}$ is convex, this means exactly that $Q$ is contained the circumcircle of $\\triangle X_{i} A_{i+1} X_{i+1}$, as desired.\n\nNow we turn to the solution. Applying lemma, we get that $P$ lies inside the circumcircle of triangle $X_{i} A_{i+1} X_{i+1}$ for some $i$. Consider the circumcircles $\\omega$ and $\\Omega$ of triangles $P_{i} A_{i+1} P_{i+1}$ and $X_{i} A_{i+1} X_{i+1}$ respectively (see Fig. 2); let $r$ and $R$ be their radii. Then we get $2 r=A_{i+1} P \\leq 2 R$ (since $P$ lies inside $\\Omega$ ), hence\n\n$$\nP_{i} P_{i+1}=2 r \\sin \\angle P_{i} A_{i+1} P_{i+1} \\leq 2 R \\sin \\angle X_{i} A_{i+1} X_{i+1}=X_{i} X_{i+1},\n$$\n\nQED.\n\n\n\nFig. 1\n\n\n\nFig. 2', 'we assume that all indices of points are considered modulo $n$. We will prove a bit stronger inequality, namely\n\n$$\n\\max \\left\\{\\frac{X_{1} X_{2}}{P_{1} P_{2}} \\cos \\alpha_{1}, \\ldots, \\frac{X_{n} X_{1}}{P_{n} P_{1}} \\cos \\alpha_{n}\\right\\} \\geq 1\n$$\n\nwhere $\\alpha_{i}(1 \\leq i \\leq n)$ is the angle between lines $X_{i} X_{i+1}$ and $P_{i} P_{i+1}$. We denote $\\beta_{i}=\\angle A_{i} P_{i} P_{i-1}$ and $\\gamma_{i}=\\angle A_{i+1} P_{i} P_{i+1}$ for all $1 \\leq i \\leq n$.\n\nSuppose that for some $1 \\leq i \\leq n$, point $X_{i}$ lies on the segment $A_{i} P_{i}$, while point $X_{i+1}$ lies on the segment $P_{i+1} A_{i+2}$. Then the projection of the segment $X_{i} X_{i+1}$ onto the line $P_{i} P_{i+1}$ contains segment $P_{i} P_{i+1}$, since $\\gamma_{i}$ and $\\beta_{i+1}$ are acute angles (see Fig. 3). Therefore, $X_{i} X_{i+1} \\cos \\alpha_{i} \\geq$ $P_{i} P_{i+1}$, and in this case the statement is proved.\n\nSo, the only case left is when point $X_{i}$ lies on segment $P_{i} A_{i+1}$ for all $1 \\leq i \\leq n$ (the case when each $X_{i}$ lies on segment $A_{i} P_{i}$ is completely analogous).\n\nNow, assume to the contrary that the inequality\n\n$$\nX_{i} X_{i+1} \\cos \\alpha_{i}P_{i+1} Y_{i+1}^{\\prime}$ (again since $\\gamma_{i}$ and $\\beta_{i+1}$ are acute; see Fig. 4). Hence, we have\n\n$$\nX_{i} P_{i} \\cos \\gamma_{i}>X_{i+1} P_{i+1} \\cos \\beta_{i+1}, \\quad 1 \\leq i \\leq n\n$$\n\nMultiplying these inequalities, we get\n\n$$\n\\cos \\gamma_{1} \\cos \\gamma_{2} \\cdots \\cos \\gamma_{n}>\\cos \\beta_{1} \\cos \\beta_{2} \\cdots \\cos \\beta_{n}\n\\tag{2}\n$$\n\nOn the other hand, the sines theorem applied to triangle $P P_{i} P_{i+1}$ provides\n\n$$\n\\frac{P P_{i}}{P P_{i+1}}=\\frac{\\sin \\left(\\frac{\\pi}{2}-\\beta_{i+1}\\right)}{\\sin \\left(\\frac{\\pi}{2}-\\gamma_{i}\\right)}=\\frac{\\cos \\beta_{i+1}}{\\cos \\gamma_{i}}\n$$\n\nMultiplying these equalities we get\n\n$$\n1=\\frac{\\cos \\beta_{2}}{\\cos \\gamma_{1}} \\cdot \\frac{\\cos \\beta_{3}}{\\cos \\gamma_{2}} \\cdots \\frac{\\cos \\beta_{1}}{\\cos \\gamma_{n}}\n$$\n\nwhich contradicts (2).\n\n\n\nFig. 3\n\n\n\nFig. 4']",,True,,, 1732,Geometry,,"Let $I$ be the incenter of a triangle $A B C$ and $\Gamma$ be its circumcircle. Let the line $A I$ intersect $\Gamma$ at a point $D \neq A$. Let $F$ and $E$ be points on side $B C$ and $\operatorname{arc} B D C$ respectively such that $\angle B A F=\angle C A E<\frac{1}{2} \angle B A C$. Finally, let $G$ be the midpoint of the segment $I F$. Prove that the lines $D G$ and $E I$ intersect on $\Gamma$.","['Let $X$ be the second point of intersection of line $E I$ with $\\Gamma$, and $L$ be the foot of the bisector of angle $B A C$. Let $G^{\\prime}$ and $T$ be the points of intersection of segment $D X$ with lines $I F$ and $A F$, respectively. We are to prove that $G=G^{\\prime}$, or $I G^{\\prime}=G^{\\prime} F$. By the Menelaus theorem applied to triangle $A I F$ and line $D X$, it means that we need the relation\n\n$$\n1=\\frac{G^{\\prime} F}{I G^{\\prime}}=\\frac{T F}{A T} \\cdot \\frac{A D}{I D}, \\quad \\text { or } \\quad \\frac{T F}{A T}=\\frac{I D}{A D}\n$$\n\nLet the line $A F$ intersect $\\Gamma$ at point $K \\neq A$ (see Fig. 1); since $\\angle B A K=\\angle C A E$ we have $\\overparen{B K}=\\overparen{C E}$, hence $K E \\| B C$. Notice that $\\angle I A T=\\angle D A K=\\angle E A D=\\angle E X D=\\angle I X T$, so the points $I, A, X, T$ are concyclic. Hence we have $\\angle I T A=\\angle I X A=\\angle E X A=\\angle E K A$, so $I T\\|K E\\| B C$. Therefore we obtain $\\frac{T F}{A T}=\\frac{I L}{A I}$.\n\nSince $C I$ is the bisector of $\\angle A C L$, we get $\\frac{I L}{A I}=\\frac{C L}{A C}$. Furthermore, $\\angle D C L=\\angle D C B=$ $\\angle D A B=\\angle C A D=\\frac{1}{2} \\angle B A C$, hence the triangles $D C L$ and $D A C$ are similar; therefore we get $\\frac{C L}{A C}=\\frac{D C}{A D}$. Finally, it is known that the midpoint $D$ of $\\operatorname{arc} B C$ is equidistant from points $I$, $B, C$, hence $\\frac{D C}{A D}=\\frac{I D}{A D}$.\n\nSummarizing all these equalities, we get\n\n$$\n\\frac{T F}{A T}=\\frac{I L}{A I}=\\frac{C L}{A C}=\\frac{D C}{A D}=\\frac{I D}{A D}\n$$\n\nas desired.\n\n\n\nFig. 1\n\n\n\nFig. 2', 'As in the previous solution, we introduce the points $X, T$ and $K$ and note that it suffice to prove the equality\n\n$$\n\\frac{T F}{A T}=\\frac{D I}{A D} \\quad \\Longleftrightarrow \\quad \\frac{T F+A T}{A T}=\\frac{D I+A D}{A D} \\quad \\Longleftrightarrow \\quad \\frac{A T}{A D}=\\frac{A F}{D I+A D} .\n$$\n\nSince $\\angle F A D=\\angle E A I$ and $\\angle T D A=\\angle X D A=\\angle X E A=\\angle I E A$, we get that the triangles $A T D$ and $A I E$ are similar, therefore $\\frac{A T}{A D}=\\frac{A I}{A E}$.\n\nNext, we also use the relation $D B=D C=D I$. Let $J$ be the point on the extension of segment $A D$ over point $D$ such that $D J=D I=D C$ (see Fig. 2). Then $\\angle D J C=$ $\\angle J C D=\\frac{1}{2}(\\pi-\\angle J D C)=\\frac{1}{2} \\angle A D C=\\frac{1}{2} \\angle A B C=\\angle A B I$. Moreover, $\\angle B A I=\\angle J A C$, hence triangles $A B I$ and $A J C$ are similar, so $\\frac{A B}{A J}=\\frac{A I}{A C}$, or $A B \\cdot A C=A J \\cdot A I=(D I+A D) \\cdot A I$.\n\nOn the other hand, we get $\\angle A B F=\\angle A B C=\\angle A E C$ and $\\angle B A F=\\angle C A E$, so triangles $A B F$ and $A E C$ are also similar, which implies $\\frac{A F}{A C}=\\frac{A B}{A E}$, or $A B \\cdot A C=A F \\cdot A E$.\n\nSummarizing we get\n\n$$\n(D I+A D) \\cdot A I=A B \\cdot A C=A F \\cdot A E \\quad \\Rightarrow \\quad \\frac{A I}{A E}=\\frac{A F}{A D+D I} \\quad \\Rightarrow \\quad \\frac{A T}{A D}=\\frac{A F}{A D+D I}\n$$\n\nas desired.']",,True,,, 1733,Geometry,,"Let $A B C D E$ be a convex pentagon such that $B C \| A E, A B=B C+A E$, and $\angle A B C=$ $\angle C D E$. Let $M$ be the midpoint of $C E$, and let $O$ be the circumcenter of triangle $B C D$. Given that $\angle D M O=90^{\circ}$, prove that $2 \angle B D A=\angle C D E$.","['Choose point $T$ on ray $A E$ such that $A T=A B$; then from $A E \\| B C$ we have $\\angle C B T=\\angle A T B=\\angle A B T$, so $B T$ is the bisector of $\\angle A B C$. On the other hand, we have $E T=A T-A E=A B-A E=B C$, hence quadrilateral $B C T E$ is a parallelogram, and the midpoint $M$ of its diagonal $C E$ is also the midpoint of the other diagonal $B T$.\n\nNext, let point $K$ be symmetrical to $D$ with respect to $M$. Then $O M$ is the perpendicular bisector of segment $D K$, and hence $O D=O K$, which means that point $K$ lies on the circumcircle of triangle $B C D$. Hence we have $\\angle B D C=\\angle B K C$. On the other hand, the angles $B K C$ and $T D E$ are symmetrical with respect to $M$, so $\\angle T D E=\\angle B K C=\\angle B D C$.\n\nTherefore, $\\angle B D T=\\angle B D E+\\angle E D T=\\angle B D E+\\angle B D C=\\angle C D E=\\angle A B C=180^{\\circ}-$ $\\angle B A T$. This means that the points $A, B, D, T$ are concyclic, and hence $\\angle A D B=\\angle A T B=$ $\\frac{1}{2} \\angle A B C=\\frac{1}{2} \\angle C D E$, as desired.\n', 'Let $\\angle C B D=\\alpha, \\angle B D C=\\beta, \\angle A D E=\\gamma$, and $\\angle A B C=\\angle C D E=2 \\varphi$. Then we have $\\angle A D B=2 \\varphi-\\beta-\\gamma, \\angle B C D=180^{\\circ}-\\alpha-\\beta, \\angle A E D=360^{\\circ}-\\angle B C D-\\angle C D E=$ $180^{\\circ}-2 \\varphi+\\alpha+\\beta$, and finally $\\angle D A E=180^{\\circ}-\\angle A D E-\\angle A E D=2 \\varphi-\\alpha-\\beta-\\gamma$.\n\n\nLet $N$ be the midpoint of $C D$; then $\\angle D N O=90^{\\circ}=\\angle D M O$, hence points $M, N$ lie on the circle with diameter $O D$. Now, if points $O$ and $M$ lie on the same side of $C D$, we have $\\angle D M N=\\angle D O N=\\frac{1}{2} \\angle D O C=\\alpha ;$ in the other case, we have $\\angle D M N=180^{\\circ}-\\angle D O N=\\alpha ;$\n\n\n\nso, in both cases $\\angle D M N=\\alpha$ (see Figures). Next, since $M N$ is a midline in triangle $C D E$, we have $\\angle M D E=\\angle D M N=\\alpha$ and $\\angle N D M=2 \\varphi-\\alpha$.\n\nNow we apply the sine rule to the triangles $A B D, A D E$ (twice), $B C D$ and $M N D$ obtaining\n\n$$\n\\begin{gathered}\n\\frac{A B}{A D}=\\frac{\\sin (2 \\varphi-\\beta-\\gamma)}{\\sin (2 \\varphi-\\alpha)}, \\quad \\frac{A E}{A D}=\\frac{\\sin \\gamma}{\\sin (2 \\varphi-\\alpha-\\beta)}, \\quad \\frac{D E}{A D}=\\frac{\\sin (2 \\varphi-\\alpha-\\beta-\\gamma)}{\\sin (2 \\varphi-\\alpha-\\beta)} \\\\\n\\frac{B C}{C D}=\\frac{\\sin \\beta}{\\sin \\alpha}, \\quad \\frac{C D}{D E}=\\frac{C D / 2}{D E / 2}=\\frac{N D}{N M}=\\frac{\\sin \\alpha}{\\sin (2 \\varphi-\\alpha)}\n\\end{gathered}\n$$\n\nwhich implies\n\n$$\n\\frac{B C}{A D}=\\frac{B C}{C D} \\cdot \\frac{C D}{D E} \\cdot \\frac{D E}{A D}=\\frac{\\sin \\beta \\cdot \\sin (2 \\varphi-\\alpha-\\beta-\\gamma)}{\\sin (2 \\varphi-\\alpha) \\cdot \\sin (2 \\varphi-\\alpha-\\beta)}\n$$\n\nHence, the condition $A B=A E+B C$, or equivalently $\\frac{A B}{A D}=\\frac{A E+B C}{A D}$, after multiplying by the common denominator rewrites as\n\n$$\n\\begin{aligned}\n& \\sin (2 \\varphi-\\alpha-\\beta) \\cdot \\sin (2 \\varphi-\\beta-\\gamma)=\\sin \\gamma \\cdot \\sin (2 \\varphi-\\alpha)+\\sin \\beta \\cdot \\sin (2 \\varphi-\\alpha-\\beta-\\gamma) \\\\\n& \\Longleftrightarrow \\cos (\\gamma-\\alpha)-\\cos (4 \\varphi-2 \\beta-\\alpha-\\gamma)=\\cos (2 \\varphi-\\alpha-2 \\beta-\\gamma)-\\cos (2 \\varphi+\\gamma-\\alpha) \\\\\n& \\Longleftrightarrow \\cos (\\gamma-\\alpha)+\\cos (2 \\varphi+\\gamma-\\alpha)=\\cos (2 \\varphi-\\alpha-2 \\beta-\\gamma)+\\cos (4 \\varphi-2 \\beta-\\alpha-\\gamma) \\\\\n& \\Longleftrightarrow \\cos \\varphi \\cdot \\cos (\\varphi+\\gamma-\\alpha)=\\cos \\varphi \\cdot \\cos (3 \\varphi-2 \\beta-\\alpha-\\gamma) \\\\\n& \\Longleftrightarrow \\cos \\varphi \\cdot(\\cos (\\varphi+\\gamma-\\alpha)-\\cos (3 \\varphi-2 \\beta-\\alpha-\\gamma))=0 \\\\\n& \\Longleftrightarrow \\cos \\varphi \\cdot \\sin (2 \\varphi-\\beta-\\alpha) \\cdot \\sin (\\varphi-\\beta-\\gamma)=0 .\n\\end{aligned}\n$$\n\nSince $2 \\varphi-\\beta-\\alpha=180^{\\circ}-\\angle A E D<180^{\\circ}$ and $\\varphi=\\frac{1}{2} \\angle A B C<90^{\\circ}$, it follows that $\\varphi=\\beta+\\gamma$, hence $\\angle B D A=2 \\varphi-\\beta-\\gamma=\\varphi=\\frac{1}{2} \\angle C D E$, as desired.']",,True,,, 1734,Geometry,,"The vertices $X, Y, Z$ of an equilateral triangle $X Y Z$ lie respectively on the sides $B C$, $C A, A B$ of an acute-angled triangle $A B C$. Prove that the incenter of triangle $A B C$ lies inside triangle $X Y Z$.","['We will prove a stronger fact; namely, we will show that the incenter $I$ of triangle $A B C$ lies inside the incircle of triangle $X Y Z$ (and hence surely inside triangle $X Y Z$ itself). We denote by $d(U, V W)$ the distance between point $U$ and line $V W$.\n\nDenote by $O$ the incenter of $\\triangle X Y Z$ and by $r, r^{\\prime}$ and $R^{\\prime}$ the inradii of triangles $A B C, X Y Z$ and the circumradius of $X Y Z$, respectively. Then we have $R^{\\prime}=2 r^{\\prime}$, and the desired inequality is $O I \\leq r^{\\prime}$. We assume that $O \\neq I$; otherwise the claim is trivial.\n\nLet the incircle of $\\triangle A B C$ touch its sides $B C, A C, A B$ at points $A_{1}, B_{1}, C_{1}$ respectively. The lines $I A_{1}, I B_{1}, I C_{1}$ cut the plane into 6 acute angles, each one containing one of the points $A_{1}, B_{1}, C_{1}$ on its border. We may assume that $O$ lies in an angle defined by lines $I A_{1}$, $I C_{1}$ and containing point $C_{1}$ (see Fig. 1). Let $A^{\\prime}$ and $C^{\\prime}$ be the projections of $O$ onto lines $I A_{1}$ and $I C_{1}$, respectively.\n\nSince $O X=R^{\\prime}$, we have $d(O, B C) \\leq R^{\\prime}$. Since $O A^{\\prime} \\| B C$, it follows that $d\\left(A^{\\prime}, B C\\right)=$ $A^{\\prime} I+r \\leq R^{\\prime}$, or $A^{\\prime} I \\leq R^{\\prime}-r$. On the other hand, the incircle of $\\triangle X Y Z$ lies inside $\\triangle A B C$, hence $d(O, A B) \\geq r^{\\prime}$, and analogously we get $d(O, A B)=C^{\\prime} C_{1}=r-I C^{\\prime} \\geq r^{\\prime}$, or $I C^{\\prime} \\leq r-r^{\\prime}$.\n\n\n\nFig. 1\n\n\n\nFig. 2\n\nFinally, the quadrilateral $I A^{\\prime} O C^{\\prime}$ is circumscribed due to the right angles at $A^{\\prime}$ and $C^{\\prime}$ (see Fig. 2). On its circumcircle, we have $\\overparen{A^{\\prime} O C^{\\prime}}=2 \\angle A^{\\prime} I C^{\\prime}<180^{\\circ}=\\overparen{O C^{\\prime} I \\text {, hence } 180^{\\circ} \\geq}$ $\\overparen{I C^{\\prime}}>\\overparen{A^{\\prime} O}$. This means that $I C^{\\prime}>A^{\\prime} O$. Finally, we have $O I \\leq I A^{\\prime}+A^{\\prime} O90^{\\circ}$ thus leading to a contradiction.\n\nNote that $\\omega$ intersects each of the segments $X Y$ and $Y Z$ at two points; let $U, U^{\\prime}$ and $V$, $V^{\\prime}$ be the points of intersection of $\\omega$ with $X Y$ and $Y Z$, respectively $\\left(U Y>U^{\\prime} Y, V Y>V^{\\prime} Y\\right.$; see Figs. 3 and 4). Note that $60^{\\circ}=\\angle X Y Z=\\frac{1}{2}\\left(\\overparen{U V}-\\overparen{U^{\\prime} V^{\\prime}}\\right) \\leq \\frac{1}{2} \\overparen{U V}$, hence $\\overparen{U V} \\geq 120^{\\circ}$.\n\n\n\nOn the other hand, since $I$ lies in $\\triangle A Y Z$, we get $\\sqrt{U V^{\\prime}}<180^{\\circ}$, hence $\\sqrt{U A_{1} U^{\\prime}} \\leq \\sqrt{U A_{1} V^{\\prime}}<$ $180^{\\circ}-\\overparen{U V} \\leq 60^{\\circ}$.\n\nNow, two cases are possible due to the order of points $Y, B_{1}$ on segment $A C$.\n\n\n\nFig. 3\n\n\n\nFig. 4\n\nCase 1. Let point $Y$ lie on the segment $A B_{1}$ (see Fig. 3). Then we have $\\angle Y X C=$ $\\frac{1}{2}\\left(\\overparen{A_{1} U^{\\prime}}-\\overparen{A_{1} U}\\right) \\leq \\frac{1}{2} \\overparen{U A_{1} U^{\\prime}}<30^{\\circ}$; analogously, we get $\\angle X Y C \\leq \\frac{1}{2} \\overparen{U A_{1} U^{\\prime}}<30^{\\circ}$. Therefore, $\\angle Y C X=180^{\\circ}-\\angle Y X C-\\angle X Y C>120^{\\circ}$, as desired.\n\nCase 2. Now let point $Y$ lie on the segment $C B_{1}$ (see Fig. 4). Analogously, we obtain $\\angle Y X C<30^{\\circ}$. Next, $\\angle I Y X>\\angle Z Y X=60^{\\circ}$, but $\\angle I Y X<\\angle I Y B_{1}$, since $Y B_{1}$ is a tangent and $Y X$ is a secant line to circle $\\omega$ from point $Y$. Hence, we get $120^{\\circ}<\\angle I Y B_{1}+\\angle I Y X=$ $\\angle B_{1} Y X=\\angle Y X C+\\angle Y C X<30^{\\circ}+\\angle Y C X$, hence $\\angle Y C X>120^{\\circ}-30^{\\circ}=90^{\\circ}$, as desired.']",,True,,, 1735,Geometry,,"Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that $\operatorname{arc} \gamma_{2}$ lies between the $\operatorname{arcs} \gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be three rays starting at $B$, lying in the same half-plane, $h_{2}$ being between $h_{1}$ and $h_{3}$. For $i, j=1,2,3$, denote by $V_{i j}$ the point of intersection of $h_{i}$ and $\gamma_{j}$ (see the Figure below). Denote by $\overparen{V_{i j} V_{k j}} \overparen{k_{k \ell} V_{i \ell}}$ the curved quadrilateral, whose sides are the segments $V_{i j} V_{i \ell}, V_{k j} V_{k \ell}$ and $\operatorname{arcs} V_{i j} V_{k j}$ and $V_{i \ell} V_{k \ell}$. We say that this quadrilateral is circumscribed if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\sqrt{V_{11} V_{21}} \sqrt{V_{22} V_{12}}, \sqrt{V_{12} V_{22}} \sqrt{V_{23} V_{13}}, \sqrt{V_{21} V_{31}} \sqrt{V_{32} V_{22}}$ are circumscribed, then the curved quadrilateral $\overparen{V_{22} V_{32}} \overparen{V_{33} V_{23}}$ is circumscribed, too. Fig. 1","['Denote by $O_{i}$ and $R_{i}$ the center and the radius of $\\gamma_{i}$, respectively. Denote also by $H$ the half-plane defined by $A C$ which contains the whole configuration. For every point $P$ in the half-plane $H$, denote by $d(P)$ the distance between $P$ and line $A C$. Furthermore, for any $r>0$, denote by $\\Omega(P, r)$ the circle with center $P$ and radius $r$.\n\nLemma 1. For every $1 \\leq iR_{i}$ and $O_{j} P0$; then the\n\n\n\n\n\nFig. 2\n\n\n\nFig. 3\n\ncircle $\\Omega(P, r)$ touches $\\gamma_{i}$ externally and touches $\\gamma_{j}$ internally, so $P$ belongs to the locus under investigation.\n\n(b) Let $\\vec{\\rho}=\\overrightarrow{A P}, \\vec{\\rho}_{i}=\\overrightarrow{A O_{i}}$, and $\\vec{\\rho}_{j}=\\overrightarrow{A O_{j}}$; let $d_{i j}=O_{i} O_{j}$, and let $\\vec{v}$ be a unit vector orthogonal to $A C$ and directed toward $H$. Then we have $\\left|\\vec{\\rho}_{i}\\right|=R_{i},\\left|\\vec{\\rho}_{j}\\right|=R_{j},\\left|\\overrightarrow{O_{i} P}\\right|=$ $\\left|\\vec{\\rho}-\\vec{\\rho}_{i}\\right|=R_{i}+r,\\left|\\overrightarrow{O_{j} P}\\right|=\\left|\\vec{\\rho}-\\vec{\\rho}_{j}\\right|=R_{j}-r$, hence\n\n$$\n\\begin{gathered}\n\\left(\\vec{\\rho}-\\vec{\\rho}_{i}\\right)^{2}-\\left(\\vec{\\rho}-\\vec{\\rho}_{j}\\right)^{2}=\\left(R_{i}+r\\right)^{2}-\\left(R_{j}-r\\right)^{2} \\\\\n\\left(\\vec{\\rho}_{i}^{2}-\\vec{\\rho}_{j}^{2}\\right)+2 \\vec{\\rho} \\cdot\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right)=\\left(R_{i}^{2}-R_{j}^{2}\\right)+2 r\\left(R_{i}+R_{j}\\right) \\\\\nd_{i j} \\cdot d(P)=d_{i j} \\vec{v} \\cdot \\vec{\\rho}=\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right) \\cdot \\vec{\\rho}=r\\left(R_{i}+R_{j}\\right)\n\\end{gathered}\n$$\n\nTherefore,\n\n$$\nr=\\frac{d_{i j}}{R_{i}+R_{j}} \\cdot d(P)\n$$\n\nand the value $v_{i j}=\\frac{d_{i j}}{R_{i}+R_{j}}$ does not depend on $P$.\n\nLemma 3. The curved quadrilateral $\\mathcal{Q}_{i j}=\\overparen{V_{i, j} V_{i+1}, j} \\overparen{V_{i+1, j+1} V_{i, j+1}}$ is circumscribed if and only if $u_{i, i+1}=v_{j, j+1}$.\n\nProof. First suppose that the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed and $\\Omega(P, r)$ is its inscribed circle. By Lemma 1 and Lemma 2 we have $r=u_{i, i+1} \\cdot d(P)$ and $r=v_{j, j+1} \\cdot d(P)$ as well. Hence, $u_{i, i+1}=v_{j, j+1}$.\n\nTo prove the opposite direction, suppose $u_{i, i+1}=v_{j, j+1}$. Let $P$ be the intersection of the angle bisector $\\beta_{i, i+1}$ and the ellipse arc $\\varepsilon_{j, j+1}$. Choose $r=u_{i, i+1} \\cdot d(P)=v_{j, j+1} \\cdot d(P)$. Then the circle $\\Omega(P, r)$ is tangent to the half lines $h_{i}$ and $h_{i+1}$ by Lemma 1 , and it is tangent to the $\\operatorname{arcs} \\gamma_{j}$ and $\\gamma_{j+1}$ by Lemma 2. Hence, the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed.\n\nBy Lemma 3, the statement of the problem can be reformulated to an obvious fact: If the equalities $u_{12}=v_{12}, u_{12}=v_{23}$, and $u_{23}=v_{12}$ hold, then $u_{23}=v_{23}$ holds as well.']",,True,,, 1735,Geometry,,"Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that $\operatorname{arc} \gamma_{2}$ lies between the $\operatorname{arcs} \gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be three rays starting at $B$, lying in the same half-plane, $h_{2}$ being between $h_{1}$ and $h_{3}$. For $i, j=1,2,3$, denote by $V_{i j}$ the point of intersection of $h_{i}$ and $\gamma_{j}$ (see the Figure below). Denote by $\overparen{V_{i j} V_{k j}} \overparen{k_{k \ell} V_{i \ell}}$ the curved quadrilateral, whose sides are the segments $V_{i j} V_{i \ell}, V_{k j} V_{k \ell}$ and $\operatorname{arcs} V_{i j} V_{k j}$ and $V_{i \ell} V_{k \ell}$. We say that this quadrilateral is circumscribed if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\sqrt{V_{11} V_{21}} \sqrt{V_{22} V_{12}}, \sqrt{V_{12} V_{22}} \sqrt{V_{23} V_{13}}, \sqrt{V_{21} V_{31}} \sqrt{V_{32} V_{22}}$ are circumscribed, then the curved quadrilateral $\overparen{V_{22} V_{32}} \overparen{V_{33} V_{23}}$ is circumscribed, too. ![](https://cdn.mathpix.com/cropped/2023_12_21_b6518374d76811a403d3g-1.jpg?height=494&width=814&top_left_y=795&top_left_x=655) Fig. 1","['Denote by $O_{i}$ and $R_{i}$ the center and the radius of $\\gamma_{i}$, respectively. Denote also by $H$ the half-plane defined by $A C$ which contains the whole configuration. For every point $P$ in the half-plane $H$, denote by $d(P)$ the distance between $P$ and line $A C$. Furthermore, for any $r>0$, denote by $\\Omega(P, r)$ the circle with center $P$ and radius $r$.\n\nLemma 1. For every $1 \\leq iR_{i}$ and $O_{j} P0$; then the\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3a433bd7e6b1edc7f4d5g-1.jpg?height=682&width=539&top_left_y=367&top_left_x=353)\n\nFig. 2\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3a433bd7e6b1edc7f4d5g-1.jpg?height=771&width=714&top_left_y=277&top_left_x=954)\n\nFig. 3\n\ncircle $\\Omega(P, r)$ touches $\\gamma_{i}$ externally and touches $\\gamma_{j}$ internally, so $P$ belongs to the locus under investigation.\n\n(b) Let $\\vec{\\rho}=\\overrightarrow{A P}, \\vec{\\rho}_{i}=\\overrightarrow{A O_{i}}$, and $\\vec{\\rho}_{j}=\\overrightarrow{A O_{j}}$; let $d_{i j}=O_{i} O_{j}$, and let $\\vec{v}$ be a unit vector orthogonal to $A C$ and directed toward $H$. Then we have $\\left|\\vec{\\rho}_{i}\\right|=R_{i},\\left|\\vec{\\rho}_{j}\\right|=R_{j},\\left|\\overrightarrow{O_{i} P}\\right|=$ $\\left|\\vec{\\rho}-\\vec{\\rho}_{i}\\right|=R_{i}+r,\\left|\\overrightarrow{O_{j} P}\\right|=\\left|\\vec{\\rho}-\\vec{\\rho}_{j}\\right|=R_{j}-r$, hence\n\n$$\n\\begin{gathered}\n\\left(\\vec{\\rho}-\\vec{\\rho}_{i}\\right)^{2}-\\left(\\vec{\\rho}-\\vec{\\rho}_{j}\\right)^{2}=\\left(R_{i}+r\\right)^{2}-\\left(R_{j}-r\\right)^{2} \\\\\n\\left(\\vec{\\rho}_{i}^{2}-\\vec{\\rho}_{j}^{2}\\right)+2 \\vec{\\rho} \\cdot\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right)=\\left(R_{i}^{2}-R_{j}^{2}\\right)+2 r\\left(R_{i}+R_{j}\\right) \\\\\nd_{i j} \\cdot d(P)=d_{i j} \\vec{v} \\cdot \\vec{\\rho}=\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right) \\cdot \\vec{\\rho}=r\\left(R_{i}+R_{j}\\right)\n\\end{gathered}\n$$\n\nTherefore,\n\n$$\nr=\\frac{d_{i j}}{R_{i}+R_{j}} \\cdot d(P)\n$$\n\nand the value $v_{i j}=\\frac{d_{i j}}{R_{i}+R_{j}}$ does not depend on $P$.\n\nLemma 3. The curved quadrilateral $\\mathcal{Q}_{i j}=\\overparen{V_{i, j} V_{i+1}, j} \\overparen{V_{i+1, j+1} V_{i, j+1}}$ is circumscribed if and only if $u_{i, i+1}=v_{j, j+1}$.\n\nProof. First suppose that the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed and $\\Omega(P, r)$ is its inscribed circle. By Lemma 1 and Lemma 2 we have $r=u_{i, i+1} \\cdot d(P)$ and $r=v_{j, j+1} \\cdot d(P)$ as well. Hence, $u_{i, i+1}=v_{j, j+1}$.\n\nTo prove the opposite direction, suppose $u_{i, i+1}=v_{j, j+1}$. Let $P$ be the intersection of the angle bisector $\\beta_{i, i+1}$ and the ellipse arc $\\varepsilon_{j, j+1}$. Choose $r=u_{i, i+1} \\cdot d(P)=v_{j, j+1} \\cdot d(P)$. Then the circle $\\Omega(P, r)$ is tangent to the half lines $h_{i}$ and $h_{i+1}$ by Lemma 1 , and it is tangent to the $\\operatorname{arcs} \\gamma_{j}$ and $\\gamma_{j+1}$ by Lemma 2. Hence, the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed.\n\nBy Lemma 3, the statement of the problem can be reformulated to an obvious fact: If the equalities $u_{12}=v_{12}, u_{12}=v_{23}$, and $u_{23}=v_{12}$ hold, then $u_{23}=v_{23}$ holds as well.']",['证明题，略'],True,,Need_human_evaluate, 1736,Number Theory,,"Find the least positive integer $n$ for which there exists a set $\left\{s_{1}, s_{2}, \ldots, s_{n}\right\}$ consisting of $n$ distinct positive integers such that $$ \left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010} $$","['Suppose that for some $n$ there exist the desired numbers; we may assume that $s_{1}1$ since otherwise $1-\\frac{1}{s_{1}}=0$. So we have $2 \\leq s_{1} \\leq s_{2}-1 \\leq \\cdots \\leq s_{n}-(n-1)$, hence $s_{i} \\geq i+1$ for each $i=1, \\ldots, n$. Therefore\n\n$$\n\\begin{aligned}\n\\frac{51}{2010} & =\\left(1-\\frac{1}{s_{1}}\\right)\\left(1-\\frac{1}{s_{2}}\\right) \\ldots\\left(1-\\frac{1}{s_{n}}\\right) \\\\\n& \\geq\\left(1-\\frac{1}{2}\\right)\\left(1-\\frac{1}{3}\\right) \\ldots\\left(1-\\frac{1}{n+1}\\right)=\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots \\frac{n}{n+1}=\\frac{1}{n+1}\n\\end{aligned}\n$$\n\nwhich implies\n\n$$\nn+1 \\geq \\frac{2010}{51}=\\frac{670}{17}>39\n$$\n\nso $n \\geq 39$.\n\nNow we are left to show that $n=39$ fits. Consider the set $\\{2,3, \\ldots, 33,35,36, \\ldots, 40,67\\}$ which contains exactly 39 numbers. We have\n\n$$\n\\frac{1}{2} \\cdot \\frac{2}{3} \\cdots \\frac{32}{33} \\cdot \\frac{34}{35} \\cdots \\frac{39}{40} \\cdot \\frac{66}{67}=\\frac{1}{33} \\cdot \\frac{34}{40} \\cdot \\frac{66}{67}=\\frac{17}{670}=\\frac{51}{2010}\n\\tag{1}\n$$\n\nhence for $n=39$ there exists a desired example.']",['39'],False,,Numerical, 1737,Number Theory,,"Find all pairs $(m, n)$ of nonnegative integers for which $$ m^{2}+2 \cdot 3^{n}=m\left(2^{n+1}-1\right) \tag{1} $$","['For fixed values of $n$, the equation (1) is a simple quadratic equation in $m$. For $n \\leq 5$ the solutions are listed in the following table.\n\n| case | equation | discriminant | integer roots |\n| :--- | :--- | :--- | :--- |\n| $n=0$ | $m^{2}-m+2=0$ | -7 | none |\n| $n=1$ | $m^{2}-3 m+6=0$ | -15 | none |\n| $n=2$ | $m^{2}-7 m+18=0$ | -23 | none |\n| $n=3$ | $m^{2}-15 m+54=0$ | 9 | $m=6$ and $m=9$ |\n| $n=4$ | $m^{2}-31 m+162=0$ | 313 | none |\n| $n=5$ | $m^{2}-63 m+486=0$ | $2025=45^{2}$ | $m=9$ and $m=54$ |\n\nWe prove that there is no solution for $n \\geq 6$.\n\nSuppose that $(m, n)$ satisfies (1) and $n \\geq 6$. Since $m \\mid 2 \\cdot 3^{n}=m\\left(2^{n+1}-1\\right)-m^{2}$, we have $m=3^{p}$ with some $0 \\leq p \\leq n$ or $m=2 \\cdot 3^{q}$ with some $0 \\leq q \\leq n$.\n\nIn the first case, let $q=n-p$; then\n\n$$\n2^{n+1}-1=m+\\frac{2 \\cdot 3^{n}}{m}=3^{p}+2 \\cdot 3^{q}\n$$\n\nIn the second case let $p=n-q$. Then\n\n$$\n2^{n+1}-1=m+\\frac{2 \\cdot 3^{n}}{m}=2 \\cdot 3^{q}+3^{p}\n$$\n\nHence, in both cases we need to find the nonnegative integer solutions of\n\n$$\n3^{p}+2 \\cdot 3^{q}=2^{n+1}-1, \\quad p+q=n .\n\\tag{2}\n$$\n\nNext, we prove bounds for $p, q$. From (2) we get\n\n$$\n3^{p}<2^{n+1}=8^{\\frac{n+1}{3}}<9^{\\frac{n+1}{3}}=3^{\\frac{2(n+1)}{3}}\n$$\n\nand\n\n$$\n2 \\cdot 3^{q}<2^{n+1}=2 \\cdot 8^{\\frac{n}{3}}<2 \\cdot 9^{\\frac{n}{3}}=2 \\cdot 3^{\\frac{2 n}{3}}<2 \\cdot 3^{\\frac{2(n+1)}{3}}\n$$\n\nso $p, q<\\frac{2(n+1)}{3}$. Combining these inequalities with $p+q=n$, we obtain\n\n$$\n\\frac{n-2}{3}\\frac{n-2}{3}$; in particular, we have $h>1$. On the left-hand side of (2), both terms are divisible by $3^{h}$, therefore $9\\left|3^{h}\\right| 2^{n+1}-1$. It is easy check that $\\operatorname{ord}_{9}(2)=6$, so $9 \\mid 2^{n+1}-1$ if and only if $6 \\mid n+1$. Therefore, $n+1=6 r$ for some positive integer $r$, and we can write\n\n$$\n2^{n+1}-1=4^{3 r}-1=\\left(4^{2 r}+4^{r}+1\\right)\\left(2^{r}-1\\right)\\left(2^{r}+1\\right)\n\\tag{4}\n$$\n\n\n\nNotice that the factor $4^{2 r}+4^{r}+1=\\left(4^{r}-1\\right)^{2}+3 \\cdot 4^{r}$ is divisible by 3 , but it is never divisible by 9 . The other two factors in (4), $2^{r}-1$ and $2^{r}+1$ are coprime: both are odd and their difference is 2 . Since the whole product is divisible by $3^{h}$, we have either $3^{h-1} \\mid 2^{r}-1$ or $3^{h-1} \\mid 2^{r}+1$. In any case, we have $3^{h-1} \\leq 2^{r}+1$. Then\n\n$$\n\\begin{gathered}\n3^{h-1} \\leq 2^{r}+1 \\leq 3^{r}=3^{\\frac{n+1}{6}} \\\\\n\\frac{n-2}{3}-10$ of the system of equations\n(i) $\\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}$\n(ii) $\\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}$\n(iii) $\\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2}$.\n\nWe will show that such a solution does not exist.\n\nAssume the contrary and consider a solution with minimal $m$. Note that if an integer $x$ is odd then $x^{2} \\equiv 1(\\bmod 8)$. Otherwise (i.e., if $x$ is even) we have $x^{2} \\equiv 0(\\bmod 8)$ or $x^{2} \\equiv 4$ $(\\bmod 8)$. Hence, by $(\\mathrm{i})$, we get that $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are even. Similarly, by (ii), we get that $y_{1}, y_{2}, y_{3}$ and $y_{4}$ are even. Thus the LHS of (iii) is divisible by 4 and $m$ is also even. It follows that $\\left(\\frac{x_{1}}{2}, \\frac{y_{1}}{2}, \\frac{x_{2}}{2}, \\frac{y_{2}}{2}, \\frac{x_{3}}{2}, \\frac{y_{3}}{2}, \\frac{x_{4}}{2}, \\frac{y_{4}}{2}, \\frac{m}{2}\\right)$ is a solution of the system of equations (i), (ii) and (iii), which contradicts the minimality of $m$.']",['5'],False,,Numerical, 1739,Number Theory,,"Let $a, b$ be integers, and let $P(x)=a x^{3}+b x$. For any positive integer $n$ we say that the pair $(a, b)$ is $n$-good if $n \mid P(m)-P(k)$ implies $n \mid m-k$ for all integers $m, k$. We say that $(a, b)$ is very good if $(a, b)$ is $n$-good for infinitely many positive integers $n$. Show that all 2010-good pairs are very good.","['We will show that if a pair $(a, b)$ is $2010-\\operatorname{good}$ then $(a, b)$ is $67^{i}$-good for all positive integer $i$.\n\nClaim 1. If $(a, b)$ is $2010-\\operatorname{good}$ then $(a, b)$ is 67 -good.\n\nProof. Assume that $P(m)=P(k)(\\bmod 67)$. Since 67 and 30 are coprime, there exist integers $m^{\\prime}$ and $k^{\\prime}$ such that $k^{\\prime} \\equiv k(\\bmod 67), k^{\\prime} \\equiv 0(\\bmod 30)$, and $m^{\\prime} \\equiv m(\\bmod 67), m^{\\prime} \\equiv 0$ $(\\bmod 30)$. Then we have $P\\left(m^{\\prime}\\right) \\equiv P(0) \\equiv P\\left(k^{\\prime}\\right)(\\bmod 30)$ and $P\\left(m^{\\prime}\\right) \\equiv P(m) \\equiv P(k) \\equiv P\\left(k^{\\prime}\\right)$ $(\\bmod 67)$, hence $P\\left(m^{\\prime}\\right) \\equiv P\\left(k^{\\prime}\\right)(\\bmod 2010)$. This implies $m^{\\prime} \\equiv k^{\\prime}(\\bmod 2010)$ as $(a, b)$ is 2010 -good. It follows that $m \\equiv m^{\\prime} \\equiv k^{\\prime} \\equiv k(\\bmod 67)$. Therefore, $(a, b)$ is 67 -good.\n\nClaim 2. If $(a, b)$ is 67 -good then $67 \\mid a$.\n\nProof. Suppose that $67 \\nmid a$. Consider the sets $\\left\\{a t^{2}(\\bmod 67): 0 \\leq t \\leq 33\\right\\}$ and $\\left\\{-3 a s^{2}-b\\right.$ $\\bmod 67: 0 \\leq s \\leq 33\\}$. Since $a \\not \\equiv 0(\\bmod 67)$, each of these sets has 34 elements. Hence they have at least one element in common. If $a t^{2} \\equiv-3 a s^{2}-b(\\bmod 67)$ then for $m=t \\pm s, k=\\mp 2 s$ we have\n\n$$\n\\begin{aligned}\nP(m)-P(k)=a\\left(m^{3}-k^{3}\\right)+b(m-k) & =(m-k)\\left(a\\left(m^{2}+m k+k^{2}\\right)+b\\right) \\\\\n& =(t \\pm 3 s)\\left(a t^{2}+3 a s^{2}+b\\right) \\equiv 0 \\quad(\\bmod 67)\n\\end{aligned}\n$$\n\nSince $(a, b)$ is 67 -good, we must have $m \\equiv k(\\bmod 67)$ in both cases, that is, $t \\equiv 3 s(\\bmod 67)$ and $t \\equiv-3 s(\\bmod 67)$. This means $t \\equiv s \\equiv 0(\\bmod 67)$ and $b \\equiv-3 a s^{2}-a t^{2} \\equiv 0(\\bmod 67)$. But then $67 \\mid P(7)-P(2)=67 \\cdot 5 a+5 b$ and $67 / 17-2$, contradicting that $(a, b)$ is 67 -good.\n\nClaim 3. If $(a, b)$ is $2010-\\operatorname{good}$ then $(a, b)$ is $67^{i}-\\operatorname{good}$ all $i \\geq 1$.\n\nProof. By Claim 2 , we have $67 \\mid a$. If $67 \\mid b$, then $P(x) \\equiv P(0)(\\bmod 67)$ for all $x$, contradicting that $(a, b)$ is 67 -good. Hence, $67 /\\langle b$.\n\nSuppose that $67^{i} \\mid P(m)-P(k)=(m-k)\\left(a\\left(m^{2}+m k+k^{2}\\right)+b\\right)$. Since $67 \\mid a$ and $67 \\not \\nmid b$, the second factor $a\\left(m^{2}+m k+k^{2}\\right)+b$ is coprime to 67 and hence $67^{i} \\mid m-k$. Therefore, $(a, b)$ is $67^{i}$-good.']",,True,,, 1740,Number Theory,,"Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the number $(f(m)+n)(m+f(n))$ is a square for all $m, n \in \mathbb{N}$.","['First, it is clear that all functions of the form $f(n)=n+c$ with a constant nonnegative integer $c$ satisfy the problem conditions since $(f(m)+n)(f(n)+m)=(n+m+c)^{2}$ is a square.\n\nWe are left to prove that there are no other functions. We start with the following Lemma. Suppose that $p \\mid f(k)-f(\\ell)$ for some prime $p$ and positive integers $k, \\ell$. Then $p \\mid k-\\ell$. Proof. Suppose first that $p^{2} \\mid f(k)-f(\\ell)$, so $f(\\ell)=f(k)+p^{2} a$ for some integer $a$. Take some positive integer $D>\\max \\{f(k), f(\\ell)\\}$ which is not divisible by $p$ and set $n=p D-f(k)$. Then the positive numbers $n+f(k)=p D$ and $n+f(\\ell)=p D+(f(\\ell)-f(k))=p(D+p a)$ are both divisible by $p$ but not by $p^{2}$. Now, applying the problem conditions, we get that both the numbers $(f(k)+n)(f(n)+k)$ and $(f(\\ell)+n)(f(n)+\\ell)$ are squares divisible by $p$ (and thus by $p^{2}$ ); this means that the multipliers $f(n)+k$ and $f(n)+\\ell$ are also divisible by $p$, therefore $p \\mid(f(n)+k)-(f(n)+\\ell)=k-\\ell$ as well.\n\nOn the other hand, if $f(k)-f(\\ell)$ is divisible by $p$ but not by $p^{2}$, then choose the same number $D$ and set $n=p^{3} D-f(k)$. Then the positive numbers $f(k)+n=p^{3} D$ and $f(\\ell)+n=$ $p^{3} D+(f(\\ell)-f(k))$ are respectively divisible by $p^{3}$ (but not by $p^{4}$ ) and by $p$ (but not by $p^{2}$ ). Hence in analogous way we obtain that the numbers $f(n)+k$ and $f(n)+\\ell$ are divisible by $p$, therefore $p \\mid(f(n)+k)-(f(n)+\\ell)=k-\\ell$.\n\nWe turn to the problem. First, suppose that $f(k)=f(\\ell)$ for some $k, \\ell \\in \\mathbb{N}$. Then by Lemma we have that $k-\\ell$ is divisible by every prime number, so $k-\\ell=0$, or $k=\\ell$. Therefore, the function $f$ is injective.\n\nNext, consider the numbers $f(k)$ and $f(k+1)$. Since the number $(k+1)-k=1$ has no prime divisors, by Lemma the same holds for $f(k+1)-f(k)$; thus $|f(k+1)-f(k)|=1$.\n\nNow, let $f(2)-f(1)=q,|q|=1$. Then we prove by induction that $f(n)=f(1)+q(n-1)$. The base for $n=1,2$ holds by the definition of $q$. For the step, if $n>1$ we have $f(n+1)=$ $f(n) \\pm q=f(1)+q(n-1) \\pm q$. Since $f(n) \\neq f(n-2)=f(1)+q(n-2)$, we get $f(n)=f(1)+q n$, as desired.\n\nFinally, we have $f(n)=f(1)+q(n-1)$. Then $q$ cannot be -1 since otherwise for $n \\geq f(1)+1$ we have $f(n) \\leq 0$ which is impossible. Hence $q=1$ and $f(n)=(f(1)-1)+n$ for each $n \\in \\mathbb{N}$, and $f(1)-1 \\geq 0$, as desired.']","['All functions of the form $f(n)=n+c$, where $c \\in \\mathbb{N} \\cup\\{0\\}$']",True,,Need_human_evaluate, 1741,Number Theory,,"The rows and columns of a $2^{n} \times 2^{n}$ table are numbered from 0 to $2^{n}-1$. The cells of the table have been colored with the following property being satisfied: for each $0 \leq i, j \leq 2^{n}-1$, the $j$ th cell in the $i$ th row and the $(i+j)$ th cell in the $j$ th row have the same color. (The indices of the cells in a row are considered modulo $2^{n}$.) Prove that the maximal possible number of colors is $2^{n}$.","['Throughout the solution we denote the cells of the table by coordinate pairs; $(i, j)$ refers to the $j$ th cell in the $i$ th row.\n\nConsider the directed graph, whose vertices are the cells of the board, and the edges are the arrows $(i, j) \\rightarrow(j, i+j)$ for all $0 \\leq i, j \\leq 2^{n}-1$. From each vertex $(i, j)$, exactly one edge passes (to $\\left(j, i+j \\bmod 2^{n}\\right)$ ); conversely, to each cell $(j, k)$ exactly one edge is directed (from the cell $\\left.\\left(k-j \\bmod 2^{n}, j\\right)\\right)$. Hence, the graph splits into cycles.\n\nNow, in any coloring considered, the vertices of each cycle should have the same color by the problem condition. On the other hand, if each cycle has its own color, the obtained coloring obviously satisfies the problem conditions. Thus, the maximal possible number of colors is the same as the number of cycles, and we have to prove that this number is $2^{n}$.\n\nNext, consider any cycle $\\left(i_{1}, j_{1}\\right),\\left(i_{2}, j_{2}\\right), \\ldots$; we will describe it in other terms. Define a sequence $\\left(a_{0}, a_{1}, \\ldots\\right)$ by the relations $a_{0}=i_{1}, a_{1}=j_{1}, a_{n+1}=a_{n}+a_{n-1}$ for all $n \\geq 1$ (we say that such a sequence is a Fibonacci-type sequence). Then an obvious induction shows that $i_{k} \\equiv a_{k-1}\\left(\\bmod 2^{n}\\right), j_{k} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$. Hence we need to investigate the behavior of Fibonacci-type sequences modulo $2^{n}$.\n\nDenote by $F_{0}, F_{1}, \\ldots$ the Fibonacci numbers defined by $F_{0}=0, F_{1}=1$, and $F_{n+2}=$ $F_{n+1}+F_{n}$ for $n \\geq 0$. We also set $F_{-1}=1$ according to the recurrence relation.\n\nFor every positive integer $m$, denote by $\\nu(m)$ the exponent of 2 in the prime factorization of $m$, i.e. for which $2^{\\nu(m)} \\mid m$ but $2^{\\nu(m)+1} \\mid\\langle m$.\n\nLemma 1. For every Fibonacci-type sequence $a_{0}, a_{1}, a_{2}, \\ldots$, and every $k \\geq 0$, we have $a_{k}=$ $F_{k-1} a_{0}+F_{k} a_{1}$.\n\nProof. Apply induction on $k$. The base cases $k=0,1$ are trivial. For the step, from the induction hypothesis we get\n\n$$\na_{k+1}=a_{k}+a_{k-1}=\\left(F_{k-1} a_{0}+F_{k} a_{1}\\right)+\\left(F_{k-2} a_{0}+F_{k-1} a_{1}\\right)=F_{k} a_{0}+F_{k+1} a_{1}\n$$\n\nLemma 2. For every $m \\geq 3$,\n\n(a) we have $\\nu\\left(F_{3 \\cdot 2^{m-2}}\\right)=m$;\n\n(b) $d=3 \\cdot 2^{m-2}$ is the least positive index for which $2^{m} \\mid F_{d}$;\n\n(c) $F_{3 \\cdot 2^{m-2}+1} \\equiv 1+2^{m-1}\\left(\\bmod 2^{m}\\right)$.\n\nProof. Apply induction on $m$. In the base case $m=3$ we have $\\nu\\left(F_{3 \\cdot 2^{m-2}}\\right)=F_{6}=8$, so $\\nu\\left(F_{3 \\cdot 2^{m-2}}\\right)=\\nu(8)=3$, the preceding Fibonacci-numbers are not divisible by 8 , and indeed $F_{3 \\cdot 2^{m-2}+1}=F_{7}=13 \\equiv 1+4(\\bmod 8)$.\n\nNow suppose that $m>3$ and let $k=3 \\cdot 2^{m-3}$. By applying Lemma 1 to the Fibonacci-type sequence $F_{k}, F_{k+1}, \\ldots$ we get\n\n$$\n\\begin{gathered}\nF_{2 k}=F_{k-1} F_{k}+F_{k} F_{k+1}=\\left(F_{k+1}-F_{k}\\right) F_{k}+F_{k+1} F_{k}=2 F_{k+1} F_{k}-F_{k}^{2}, \\\\\nF_{2 k+1}=F_{k} \\cdot F_{k}+F_{k+1} \\cdot F_{k+1}=F_{k}^{2}+F_{k+1}^{2} .\n\\end{gathered}\n$$\n\nBy the induction hypothesis, $\\nu\\left(F_{k}\\right)=m-1$, and $F_{k+1}$ is odd. Therefore we get $\\nu\\left(F_{k}^{2}\\right)=$ $2(m-1)>(m-1)+1=\\nu\\left(2 F_{k} F_{k+1}\\right)$, which implies $\\nu\\left(F_{2 k}\\right)=m$, establishing statement (a).\n\n\n\nMoreover, since $F_{k+1}=1+2^{m-2}+a 2^{m-1}$ for some integer $a$, we get\n\n$$\nF_{2 k+1}=F_{k}^{2}+F_{k+1}^{2} \\equiv 0+\\left(1+2^{m-2}+a 2^{m-1}\\right)^{2} \\equiv 1+2^{m-1} \\quad\\left(\\bmod 2^{m}\\right)\n$$\n\nas desired in statement (c).\n\nWe are left to prove that $2^{m} \\nmid F_{\\ell}$ for $\\ell<2 k$. Assume the contrary. Since $2^{m-1} \\mid F_{\\ell}$, from the induction hypothesis it follows that $\\ell>k$. But then we have $F_{\\ell}=F_{k-1} F_{\\ell-k}+F_{k} F_{\\ell-k+1}$, where the second summand is divisible by $2^{m-1}$ but the first one is not (since $F_{k-1}$ is odd and $\\ell-k0$ such that $a_{k+p} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$ for all $k \\geq 0$.\n\nLemma 3. Let $A=\\left(a_{0}, a_{1}, \\ldots\\right)$ be a Fibonacci-type sequence such that $\\mu(A)=k0$ )\n\n$$\n\\left\\lfloor a_{i+1}\\right\\rfloor \\leq a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle<\\left\\lfloor a_{i}\\right\\rfloor\n$$\n\nthe sequence $\\left\\lfloor a_{i}\\right\\rfloor$ is strictly decreasing as long as its terms are in $[1, \\infty)$. Eventually there appears a number from the interval $[0,1)$ and all subsequent terms are 0 .\n\nNow pass to the more interesting situation where $a_{0}<0$; then all $a_{i} \\leq 0$. Suppose the sequence never hits 0 . Then we have $\\left\\lfloor a_{i}\\right\\rfloor \\leq-1$ for all $i$, and so\n\n$$\n1+\\left\\lfloor a_{i+1}\\right\\rfloor>a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle>\\left\\lfloor a_{i}\\right\\rfloor ;\n$$\n\nthis means that the sequence $\\left\\lfloor a_{i}\\right\\rfloor$ is nondecreasing. And since all its terms are integers from $(-\\infty,-1]$, this sequence must be constant from some term on:\n\n$$\n\\left\\lfloor a_{i}\\right\\rfloor=c \\quad \\text { for } \\quad i \\geq i_{0} ; \\quad c \\text { a negative integer. }\n$$\n\nThe defining formula becomes\n\n$$\na_{i+1}=c \\cdot\\left\\langle a_{i}\\right\\rangle=c\\left(a_{i}-c\\right)=c a_{i}-c^{2}\n$$\n\nConsider the sequence\n\n$$\nb_{i}=a_{i}-\\frac{c^{2}}{c-1} \\tag{1}\n$$\n\nIt satisfies the recursion rule\n\n$$\nb_{i+1}=a_{i+1}-\\frac{c^{2}}{c-1}=c a_{i}-c^{2}-\\frac{c^{2}}{c-1}=c b_{i} \n$$\n\nimplying\n\n$$\nb_{i}=c^{i-i_{0}} b_{i_{0}} \\quad \\text { for } \\quad i \\geq i_{0} \\tag{2}\n$$\n\nSince all the numbers $a_{i}$ (for $i \\geq i_{0}$ ) lie in $\\left[c, c+1\\right.$ ), the sequence $\\left(b_{i}\\right)$ is bounded. The equation (2) can be satisfied only if either $b_{i_{0}}=0$ or $|c|=1$, i.e., $c=-1$.\n\n\n\nIn the first case, $b_{i}=0$ for all $i \\geq i_{0}$, so that\n\n$$\na_{i}=\\frac{c^{2}}{c-1} \\quad \\text { for } \\quad i \\geq i_{0}\n$$\n\nIn the second case, $c=-1$, equations (1) and (2) say that\n\n$$\na_{i}=-\\frac{1}{2}+(-1)^{i-i_{0}} b_{i_{0}}= \\begin{cases}a_{i_{0}} & \\text { for } i=i_{0}, i_{0}+2, i_{0}+4, \\ldots \\\\ 1-a_{i_{0}} & \\text { for } i=i_{0}+1, i_{0}+3, i_{0}+5, \\ldots\\end{cases}\n$$\n\nSummarising, we see that (from some point on) the sequence $\\left(a_{i}\\right)$ either is constant or takes alternately two values from the interval $(-1,0)$. The result follows.']",,True,,, 1743,Algebra,,"The sequence of real numbers $a_{0}, a_{1}, a_{2}, \ldots$ is defined recursively by $$ a_{0}=-1, \quad \sum_{k=0}^{n} \frac{a_{n-k}}{k+1}=0 \quad \text { for } \quad n \geq 1 $$ Show that $a_{n}>0$ for $n \geq 1$.","['The proof goes by induction. For $n=1$ the formula yields $a_{1}=1 / 2$. Take $n \\geq 1$, assume $a_{1}, \\ldots, a_{n}>0$ and write the recurrence formula for $n$ and $n+1$, respectively as\n\n$$\n\\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1}=0 \\quad \\text { and } \\quad \\sum_{k=0}^{n+1} \\frac{a_{k}}{n-k+2}=0\n$$\n\nSubtraction yields\n\n$$\n\\begin{aligned}\n0=(n+2) \\sum_{k=0}^{n+1} \\frac{a_{k}}{n-k+2}-(n+1) \\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1} \\\\\n=(n+2) a_{n+1}+\\sum_{k=0}^{n}\\left(\\frac{n+2}{n-k+2}-\\frac{n+1}{n-k+1}\\right) a_{k} .\n\\end{aligned}\n$$\n\nThe coefficient of $a_{0}$ vanishes, so\n\n$$\na_{n+1}=\\frac{1}{n+2} \\sum_{k=1}^{n}\\left(\\frac{n+1}{n-k+1}-\\frac{n+2}{n-k+2}\\right) a_{k}=\\frac{1}{n+2} \\sum_{k=1}^{n} \\frac{k}{(n-k+1)(n-k+2)} a_{k} .\n$$\n\nThe coefficients of $a_{1}, \\ldots, a_{n}$ are all positive. Therefore, $a_{1}, \\ldots, a_{n}>0$ implies $a_{n+1}>0$.']",,True,,, 1744,Algebra,,"The sequence $c_{0}, c_{1}, \ldots, c_{n}, \ldots$ is defined by $c_{0}=1, c_{1}=0$ and $c_{n+2}=c_{n+1}+c_{n}$ for $n \geq 0$. Consider the set $S$ of ordered pairs $(x, y)$ for which there is a finite set $J$ of positive integers such that $x=\sum_{j \in J} c_{j}, y=\sum_{j \in J} c_{j-1}$. Prove that there exist real numbers $\alpha, \beta$ and $m, M$ with the following property: An ordered pair of nonnegative integers $(x, y)$ satisfies the inequality $$ m<\alpha x+\beta y1$ and $-1<\\psi<0$, this implies $\\alpha \\varphi+\\beta=0$.\n\nTo satisfy $\\alpha \\varphi+\\beta=0$, one can set for instance $\\alpha=\\psi, \\beta=1$. We now find the required $m$ and $M$ for this choice of $\\alpha$ and $\\beta$.\n\nNote first that the above displayed equation gives $c_{n} \\psi+c_{n-1}=\\psi^{n-1}, n \\geq 1$. In the sequel, we denote the pairs in $S$ by $\\left(a_{J}, b_{J}\\right)$, where $J$ is a finite subset of the set $\\mathbb{N}$ of positive integers and $a_{J}=\\sum_{j \\in J} c_{j}, b_{J}=\\sum_{j \\in J} c_{j-1}$. Since $\\psi a_{J}+b_{J}=\\sum_{j \\in J}\\left(c_{j} \\psi+c_{j-1}\\right)$, we obtain\n\n$$\n\\psi a_{J}+b_{J}=\\sum_{j \\in J} \\psi^{j-1} \\quad \\text { for each }\\left(a_{J}, b_{J}\\right) \\in S \\tag{1}\n$$\n\nOn the other hand, in view of $-1<\\psi<0$,\n\n$$\n-1=\\frac{\\psi}{1-\\psi^{2}}=\\sum_{j=0}^{\\infty} \\psi^{2 j+1}<\\sum_{j \\in J} \\psi^{j-1}<\\sum_{j=0}^{\\infty} \\psi^{2 j}=\\frac{1}{1-\\psi^{2}}=1-\\psi=\\varphi\n$$\n\nTherefore, according to (1),\n\n$$\n-1<\\psi a_{J}+b_{J}<\\varphi \\quad \\text { for each }\\left(a_{J}, b_{J}\\right) \\in S\n$$\n\nThus $m=-1$ and $M=\\varphi$ is an appropriate choice.\n\nConversely, we prove that if an ordered pair of nonnegative integers $(x, y)$ satisfies the inequality $-1<\\psi x+y<\\varphi$ then $(x, y) \\in S$.\n\n\n\nLemma. Let $x, y$ be nonnegative integers such that $-1<\\psi x+y<\\varphi$. Then there exists a subset $J$ of $\\mathbb{N}$ such that\n\n$$\n\\psi x+y=\\sum_{j \\in J} \\psi^{j-1} \\tag{2}\n$$\n\nProof. For $x=y=0$ it suffices to choose the empty subset of $\\mathbb{N}$ as $J$, so let at least one of $x, y$ be nonzero. There exist representations of $\\psi x+y$ of the form\n\n$$\n\\psi x+y=\\psi^{i_{1}}+\\cdots+\\psi^{i_{k}}\n$$\n\nwhere $i_{1} \\leq \\cdots \\leq i_{k}$ is a sequence of nonnegative integers, not necessarily distinct. For instance, we can take $x$ summands $\\psi^{1}=\\psi$ and $y$ summands $\\psi^{0}=1$. Consider all such representations of minimum length $k$ and focus on the ones for which $i_{1}$ has the minimum possible value $j_{1}$. Among them, consider the representations where $i_{2}$ has the minimum possible value $j_{2}$. Upon choosing $j_{3}, \\ldots, j_{k}$ analogously, we obtain a sequence $j_{1} \\leq \\cdots \\leq j_{k}$ which clearly satisfies $\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}$. To prove the lemma, it suffices to show that $j_{1}, \\ldots, j_{k}$ are pairwise distinct.\n\nSuppose on the contrary that $j_{r}=j_{r+1}$ for some $r=1, \\ldots, k-1$. Let us consider the case $j_{r} \\geq 2$ first. Observing that $2 \\psi^{2}=1+\\psi^{3}$, we replace $j_{r}$ and $j_{r+1}$ by $j_{r}-2$ and $j_{r}+1$, respectively. Since\n\n$$\n\\psi^{j_{r}}+\\psi^{j_{r+1}}=2 \\psi^{j_{r}}=\\psi^{j_{r}-2}\\left(1+\\psi^{3}\\right)=\\psi^{j_{r}-2}+\\psi^{j_{r}+1}\n$$\n\nthe new sequence also represents $\\psi x+y$ as needed, and the value of $i_{r}$ in it contradicts the minimum choice of $j_{r}$.\n\nLet $j_{r}=j_{r+1}=0$. Then the sum $\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}$ contains at least two summands equal to $\\psi^{0}=1$. On the other hand $j_{s} \\neq 1$ for all $s$, because the equality $1+\\psi=\\psi^{2}$ implies that a representation of minimum length cannot contain consecutive $i_{r}$ 's. It follows that\n\n$$\n\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}>2+\\psi^{3}+\\psi^{5}+\\psi^{7}+\\cdots=2-\\psi^{2}=\\varphi\n$$\n\ncontradicting the condition of the lemma.\n\nLet $j_{r}=j_{r+1}=1$; then $\\sum_{r=1}^{k} \\psi^{j_{r}}$ contains at least two summands equal to $\\psi^{1}=\\psi$. Like in the case $j_{r}=j_{r+1}=0$, we also infer that $j_{s} \\neq 0$ and $j_{s} \\neq 2$ for all $s$. Therefore\n\n$$\n\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}<2 \\psi+\\psi^{4}+\\psi^{6}+\\psi^{8}+\\cdots=2 \\psi-\\psi^{3}=-1\n$$\n\nwhich is a contradiction again. The conclusion follows.\n\nNow let the ordered pair $(x, y)$ satisfy $-1<\\psi x+y<\\varphi$; hence the lemma applies to $(x, y)$. Let $J \\subset \\mathbb{N}$ be such that (2) holds. Comparing (1) and (2), we conclude that $\\psi x+y=\\psi a_{J}+b_{J}$. Now, $x, y, a_{J}$ and $b_{J}$ are integers, and $\\psi$ is irrational. So the last equality implies $x=a_{J}$ and $y=b_{J}$. This shows that the numbers $\\alpha=\\psi, \\beta=1, m=-1, M=\\varphi$ meet the requirements.""]",,True,,, 1745,Algebra,,"Prove the inequality $$ \sum_{i\\sqrt{a+b}>\\sqrt{c}$.\n\nLet $x=\\sqrt{b}+\\sqrt{c}-\\sqrt{a}, y=\\sqrt{c}+\\sqrt{a}-\\sqrt{b}$ and $z=\\sqrt{a}+\\sqrt{b}-\\sqrt{c}$. Then\n\n$b+c-a=\\left(\\frac{z+x}{2}\\right)^{2}+\\left(\\frac{x+y}{2}\\right)^{2}-\\left(\\frac{y+z}{2}\\right)^{2}=\\frac{x^{2}+x y+x z-y z}{2}=x^{2}-\\frac{1}{2}(x-y)(x-z)$\n\nand\n\n$$\n\\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}=\\sqrt{1-\\frac{(x-y)(x-z)}{2 x^{2}}} \\leq 1-\\frac{(x-y)(x-z)}{4 x^{2}}\n$$\n\napplying $\\sqrt{1+2 u} \\leq 1+u$ in the last step. Similarly we obtain\n\n$$\n\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}} \\leq 1-\\frac{(z-x)(z-y)}{4 z^{2}} \\quad \\text { and } \\quad \\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 1-\\frac{(y-z)(y-x)}{4 y^{2}}\n$$\n\nSubstituting these quantities into the statement, it is sufficient to prove that\n\n$$\n\\frac{(x-y)(x-z)}{x^{2}}+\\frac{(y-z)(y-x)}{y^{2}}+\\frac{(z-x)(z-y)}{z^{2}} \\geq 0 \\tag{1}\n$$\n\nBy symmetry we can assume $x \\leq y \\leq z$. Then\n\n$$\n\\begin{gathered}\n\\frac{(x-y)(x-z)}{x^{2}}=\\frac{(y-x)(z-x)}{x^{2}} \\geq \\frac{(y-x)(z-y)}{y^{2}}=-\\frac{(y-z)(y-x)}{y^{2}} \\\\\n\\frac{(z-x)(z-y)}{z^{2}} \\geq 0\n\\end{gathered}\n$$\n\nand (1) follows.', 'Due to the symmetry of variables, it can be assumed that $a \\geq b \\geq c$. We claim that\n\n$$\n\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 1 \\quad \\text { and } \\quad \\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}+\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}} \\leq 2\n$$\n\nThe first inequality follows from\n\n$$\n\\sqrt{a+b-c}-\\sqrt{a}=\\frac{(a+b-c)-a}{\\sqrt{a+b-c}+\\sqrt{a}} \\leq \\frac{b-c}{\\sqrt{b}+\\sqrt{c}}=\\sqrt{b}-\\sqrt{c}\n$$\n\nFor proving the second inequality, let $p=\\sqrt{a}+\\sqrt{b}$ and $q=\\sqrt{a}-\\sqrt{b}$. Then $a-b=p q$ and the inequality becomes\n\n$$\n\\frac{\\sqrt{c-p q}}{\\sqrt{c}-q}+\\frac{\\sqrt{c+p q}}{\\sqrt{c}+q} \\leq 2\n$$\n\nFrom $a \\geq b \\geq c$ we have $p \\geq 2 \\sqrt{c}$. Applying the Cauchy-Schwarz inequality,\n\n$$\n\\begin{gathered}\n\\left(\\frac{\\sqrt{c-p q}}{\\sqrt{c}-q}+\\frac{\\sqrt{c+p q}}{\\sqrt{c}+q}\\right)^{2} \\leq\\left(\\frac{c-p q}{\\sqrt{c}-q}+\\frac{c+p q}{\\sqrt{c}+q}\\right)\\left(\\frac{1}{\\sqrt{c}-q}+\\frac{1}{\\sqrt{c}+q}\\right) \\\\\n=\\frac{2\\left(c \\sqrt{c}-p q^{2}\\right)}{c-q^{2}} \\cdot \\frac{2 \\sqrt{c}}{c-q^{2}}=4 \\cdot \\frac{c^{2}-\\sqrt{c} p q^{2}}{\\left(c-q^{2}\\right)^{2}} \\leq 4 \\cdot \\frac{c^{2}-2 c q^{2}}{\\left(c-q^{2}\\right)^{2}} \\leq 4 .\n\\end{gathered}\n$$']",,True,,, 1747,Algebra,,"Determine the smallest number $M$ such that the inequality $$ \left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2} $$ holds for all real numbers $a, b, c$.","['We first consider the cubic polynomial\n\n$$\nP(t)=t b\\left(t^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c t\\left(c^{2}-t^{2}\\right) .\n$$\n\nIt is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore\n\n$$\nP(t)=(b-c)(t-b)(t-c)(t+b+c)\n$$\n\nsince the cubic coefficient is $b-c$. The left-hand side of the proposed inequality can therefore be written in the form\n\n$$\n\\left|a b\\left(a^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c a\\left(c^{2}-a^{2}\\right)\\right|=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)| .\n$$\n\nThe problem comes down to finding the smallest number $M$ that satisfies the inequality\n\n$$\n|(b-c)(a-b)(a-c)(a+b+c)| \\leq M \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{2} . \\tag{1}\n$$\n\nNote that this expression is symmetric, and we can therefore assume $a \\leq b \\leq c$ without loss of generality. With this assumption,\n\n$$\n|(a-b)(b-c)|=(b-a)(c-b) \\leq\\left(\\frac{(b-a)+(c-b)}{2}\\right)^{2}=\\frac{(c-a)^{2}}{4} \\tag{2}\n$$\n\nwith equality if and only if $b-a=c-b$, i.e. $2 b=a+c$. Also\n\n$$\n\\left(\\frac{(c-b)+(b-a)}{2}\\right)^{2} \\leq \\frac{(c-b)^{2}+(b-a)^{2}}{2} \\tag{3}\n$$\n\nor equivalently,\n\n$$\n3(c-a)^{2} \\leq 2 \\cdot\\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\\right]\n$$\n\nagain with equality only for $2 b=a+c$. From (2) and (3) we get\n\n$$\n\\begin{aligned}\n& |(b-c)(a-b)(a-c)(a+b+c)| \\\\\n\\leq & \\frac{1}{4} \\cdot\\left|(c-a)^{3}(a+b+c)\\right| \\\\\n= & \\frac{1}{4} \\cdot \\sqrt{(c-a)^{6}(a+b+c)^{2}} \\\\\n\\leq & \\frac{1}{4} \\cdot \\sqrt{\\left(\\frac{2 \\cdot\\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\\right]}{3}\\right)^{3} \\cdot(a+b+c)^{2}} \\\\\n= & \\frac{\\sqrt{2}}{2} \\cdot\\left(\\sqrt[4]{\\left(\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\\right)^{3} \\cdot(a+b+c)^{2}}\\right)^{2} .\n\\end{aligned}\n$$\n\n\n\nBy the weighted AM-GM inequality this estimate continues as follows:\n\n$$\n\\begin{aligned}\n& |(b-c)(a-b)(a-c)(a+b+c)| \\\\\n\\leq & \\frac{\\sqrt{2}}{2} \\cdot\\left(\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\\right)^{2} \\\\\n= & \\frac{9 \\sqrt{2}}{32} \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{2} .\n\\end{aligned}\n$$\n\nWe see that the inequality (1) is satisfied for $M=\\frac{9}{32} \\sqrt{2}$, with equality if and only if $2 b=a+c$ and\n\n$$\n\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2}\n$$\n\nPlugging $b=(a+c) / 2$ into the last equation, we bring it to the equivalent form\n\n$$\n2(c-a)^{2}=9(a+c)^{2} .\n$$\n\nThe conditions for equality can now be restated as\n\n$$\n2 b=a+c \\quad \\text { and } \\quad(c-a)^{2}=18 b^{2} .\n$$\n\nSetting $b=1$ yields $a=1-\\frac{3}{2} \\sqrt{2}$ and $c=1+\\frac{3}{2} \\sqrt{2}$. We see that $M=\\frac{9}{32} \\sqrt{2}$ is indeed the smallest constant satisfying the inequality, with equality for any triple $(a, b, c)$ proportional to $\\left(1-\\frac{3}{2} \\sqrt{2}, 1,1+\\frac{3}{2} \\sqrt{2}\\right)$, up to permutation.\n\n']",['$M=\\frac{9}{32} \\sqrt{2}$'],False,,Numerical, 1748,Combinatorics,,"We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: - if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$ ) are in the same state, then $L_{i}$ is switched off; - otherwise, $L_{i}$ is switched on. Initially all the lamps are off except the leftmost one which is on. Prove that there are infinitely many integers $n$ for which all the lamps will eventually be off.","[""Experiments with small $n$ lead to the guess that every $n$ of the form $2^{k}$ should be good. This is indeed the case, and more precisely: let $A_{k}$ be the $2^{k} \\times 2^{k}$ matrix whose rows represent the evolution of the system, with entries 0,1 (for off and on respectively). The top row shows the initial state $[1,0,0, \\ldots, 0]$; the bottom row shows the state after $2^{k}-1$ steps. The claim is that:\n\n$$\n\\text { The bottom row of } A_{k} \\text { is }[1,1,1, \\ldots, 1] \\text {. }\n$$\n\nThis will of course suffice because one more move then produces $[0,0,0, \\ldots, 0]$, as required.\n\nThe proof is by induction on $k$. The base $k=1$ is obvious. Assume the claim to be true for a\n\n$k \\geq 1$ and write the matrix $A_{k+1}$ in the block form $\\left(\\begin{array}{cc}A_{k} & O_{k} \\\\ B_{k} & C_{k}\\end{array}\\right)$ with four $2^{k} \\times 2^{k}$ matrices. After $m$ steps, the last 1 in a row is at position $m+1$. Therefore $O_{k}$ is the zero matrix. According to the induction hypothesis, the bottom row of $\\left[A_{k} O_{k}\\right]$ is $[1, \\ldots, 1,0, \\ldots, 0]$, with $2^{k}$ ones and $2^{k}$ zeros. The next row is thus\n\n$$\n[\\underbrace{0, \\ldots, 0}_{2^{k}-1}, 1,1, \\underbrace{0, \\ldots, 0}_{2^{k}-1}]\n$$\n\nIt is symmetric about its midpoint, and this symmetry is preserved in all subsequent rows because the procedure described in the problem statement is left/right symmetric. Thus $B_{k}$ is the mirror image of $C_{k}$. In particular, the rightmost column of $B_{k}$ is identical with the leftmost column of $C_{k}$.\n\nImagine the matrix $C_{k}$ in isolation from the rest of $A_{k+1}$. Suppose it is subject to evolution as defined in the problem: the first (leftmost) term in a row depends only on the two first terms in the preceding row, according as they are equal or not. Now embed $C_{k}$ again in $A_{k}$. The 'leftmost' terms in the rows of $C_{k}$ now have neighbours on their left side - but these neighbours are their exact copies. Consequently the actual evolution within $C_{k}$ is the same, whether or not $C_{k}$ is considered as a piece of $A_{k+1}$ or in isolation. And since the top row of $C_{k}$ is $[1,0, \\ldots, 0]$, it follows that $C_{k}$ is identical with $A_{k}$.\n\n\n\nThe bottom row of $A_{k}$ is $[1,1, \\ldots, 1]$; the same is the bottom row of $C_{k}$, hence also of $B_{k}$, which mirrors $C_{k}$. So the bottom row of $A_{k+1}$ consists of ones only and the induction is complete.""]",,True,,, 1749,Combinatorics,,"We have $n \geq 2$ lamps $L_{1}, \ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: - if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$ ) are in the same state, then $L_{i}$ is switched off; - otherwise, $L_{i}$ is switched on. Initially all the lamps are off except the leftmost one which is on. Prove that there are infinitely many integers $n$ for which the lamps will never be all off.","['There are many ways to produce an infinite sequence of those $n$ for which the state $[0,0, \\ldots, 0]$ will never be achieved. As an example, consider $n=2^{k}+1$ (for $k \\geq 1$ ). The evolution of the system can be represented by a matrix $\\mathcal{A}$ of width $2^{k}+1$ with infinitely many rows. The top $2^{k}$ rows form the matrix $A_{k}$ discussed above, with one column of zeros attached at its right.\n\nIn the next row we then have the vector $[0,0, \\ldots, 0,1,1]$. But this is just the second row of $\\mathcal{A}$ reversed. Subsequent rows will be mirror copies of the foregoing ones, starting from the second one. So the configuration $[1,1,0, \\ldots, 0,0]$, i.e. the second row of $\\mathcal{A}$, will reappear. Further rows will periodically repeat this pattern and there will be no row of zeros.']",,True,,, 1750,Combinatorics,,"A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals. Suppose the 2006-gon has been dissected into triangles by 2003 nonintersecting diagonals. Find the maximum possible number of isosceles triangles with two odd sides.","['Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity.\n\nLemma. Let $A B$ be one of dissecting diagonals and let $\\mathcal{L}$ be the shorter part of the boundary of the 2006-gon with endpoints $A, B$. Suppose that $\\mathcal{L}$ consists of $n$ segments. Then the number of iso-odd triangles with vertices on $\\mathcal{L}$ does not exceed $n / 2$.\n\nProof. This is obvious for $n=2$. Take $n$ with $20$ and assume the statement for less than $n$ points. Take a set $S$ of $n$ points.\n\nLet $C$ be the set of vertices of the convex hull of $S$, let $m=|C|$.\n\nLet $X \\subset C$ be an arbitrary nonempty set. For any convex polygon $P$ with vertices in the set $S \\backslash X$, we have $b(P)$ points of $S$ outside $P$. Excluding the points of $X-$ all outside $P$ - the set $S \\backslash X$ contains exactly $b(P)-|X|$ of them. Writing $1-x=y$, by the induction hypothesis\n\n$$\n\\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)-|X|}=1\n$$\n\n(where $P \\subset S \\backslash X$ means that the vertices of $P$ belong to the set $S \\backslash X$ ). Therefore\n\n$$\n\\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)}=y^{|X|}\n$$\n\nAll convex polygons appear at least once, except the convex hull $C$ itself. The convex hull adds $x^{m}$. We can use the inclusion-exclusion principle to compute the sum of the other terms:\n\n$$\n\\begin{gathered}\n\\sum_{P \\neq C} x^{a(P)} y^{b(P)}=\\sum_{k=1}^{m}(-1)^{k-1} \\sum_{|X|=k} \\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)}=\\sum_{k=1}^{m}(-1)^{k-1} \\sum_{|X|=k} y^{k} \\\\\n=\\sum_{k=1}^{m}(-1)^{k-1}\\left(\\begin{array}{c}\nm \\\\\nk\n\\end{array}\\right) y^{k}=-\\left((1-y)^{m}-1\\right)=1-x^{m}\n\\end{gathered}\n$$\n\nand then\n\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{P=C}+\\sum_{P \\neq C}=x^{m}+\\left(1-x^{m}\\right)=1\n$$']",,True,,, 1752,Combinatorics,,"A cake has the form of an $n \times n$ square composed of $n^{2}$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\mathcal{A}$. Let $\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\mathcal{B}$ than of arrangement $\mathcal{A}$. Prove that arrangement $\mathcal{B}$ can be obtained from $\mathcal{A}$ by performing a number of switches, defined as follows: A switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.","['We use capital letters to denote unit squares; $O$ is the top left corner square. For any two squares $X$ and $Y$ let $[X Y]$ be the smallest grid rectangle containing these two squares. Strawberries lie on some squares in arrangement $\\mathcal{A}$. Put a plum on each square of the target configuration $\\mathcal{B}$. For a square $X$ denote by $a(X)$ and $b(X)$ respectively the number of strawberries and the number of plums in $[O X]$. By hypothesis $a(X) \\leq b(X)$ for each $X$, with strict inequality for some $X$ (otherwise the two arrangements coincide and there is nothing to prove).\n\nThe idea is to show that by a legitimate switch one can obtain an arrangement $\\mathcal{A}^{\\prime}$ such that\n\n$$\na(X) \\leq a^{\\prime}(X) \\leq b(X) \\quad \\text { for each } X ; \\quad \\sum_{X} a(X)<\\sum_{X} a^{\\prime}(X) \\tag{1}\n$$\n\n(with $a^{\\prime}(X)$ defined analogously to $a(X)$; the sums range over all unit squares $X$ ). This will be enough because the same reasoning then applies to $\\mathcal{A}^{\\prime}$, giving rise to a new arrangement $\\mathcal{A}^{\\prime \\prime}$, and so on (induction). Since $\\sum a(X)<\\sum a^{\\prime}(X)<\\sum a^{\\prime \\prime}(X)<\\ldots$ and all these sums do not exceed $\\sum b(X)$, we eventually obtain a sum with all summands equal to the respective $b(X) \\mathrm{s}$; all strawberries will meet with plums.\n\nConsider the uppermost row in which the plum and the strawberry lie on different squares $P$ and $S$ (respectively); clearly $P$ must be situated left to $S$. In the column passing through $P$, let $T$ be the top square and $B$ the bottom square. The strawberry in that column lies below the plum (because there is no plum in that column above $P$, and the positions of strawberries and plums coincide everywhere above the row of $P$ ). Hence there is at least one strawberry in the region $[B S]$ below $[P S]$. Let $V$ be the position of the uppermost strawberry in that region.\n\n\n\n\n\nDenote by $W$ the square at the intersection of the row through $V$ with the column through $S$ and let $R$ be the square vertex-adjacent to $W$ up-left. We claim that\n\n$$\na(X)\n\nIf $\\Gamma$ is above line $\\ell$ then so is $\\Gamma+\\Delta$, which contradicts the maximum choice of $\\Delta$. If $\\Gamma$ contains cells from row $B$, observe that $\\Gamma+\\Delta$ contains $u$. Let $s$ be the size of $\\Gamma+\\Delta$. Being full of $H \\backslash\\{u\\}, \\Gamma+\\Delta$ contains $s$ holes other than $u$. But it also contains $u$, contradicting the assumption that $H$ is spread out.\n\nThe claim follows, showing that $b_{1}=w$ is an appropriate choice for $a_{1}=u$ and $a_{2}=v$. As explained above, this is enough to complete the induction.']",,True,,, 1755,Combinatorics,,"Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it. Call a pair of points of the polyhedron antipodal if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes. Let $A$ be the number of antipodal pairs of vertices, and let $B$ be the number of antipodal pairs of midpoints of edges. Determine the difference $A-B$ in terms of the numbers of vertices, edges and faces.","['Denote the polyhedron by $\\Gamma$; let its vertices, edges and faces be $V_{1}, V_{2}, \\ldots, V_{n}$, $E_{1}, E_{2}, \\ldots, E_{m}$ and $F_{1}, F_{2}, \\ldots, F_{\\ell}$, respectively. Denote by $Q_{i}$ the midpoint of edge $E_{i}$.\n\nLet $S$ be the unit sphere, the set of all unit vectors in three-dimensional space. Map the boundary elements of $\\Gamma$ to some objects on $S$ as follows.\n\nFor a face $F_{i}$, let $S^{+}\\left(F_{i}\\right)$ and $S^{-}\\left(F_{i}\\right)$ be the unit normal vectors of face $F_{i}$, pointing outwards from $\\Gamma$ and inwards to $\\Gamma$, respectively. These points are diametrically opposite.\n\nFor an edge $E_{j}$, with neighbouring faces $F_{i_{1}}$ and $F_{i_{2}}$, take all support planes of $\\Gamma$ (planes which have a common point with $\\Gamma$ but do not intersect it) containing edge $E_{j}$, and let $S^{+}\\left(E_{j}\\right)$ be the set of their outward normal vectors. The set $S^{+}\\left(E_{j}\\right)$ is an arc of a great circle on $S$. Arc $S^{+}\\left(E_{j}\\right)$ is perpendicular to edge $E_{j}$ and it connects points $S^{+}\\left(F_{i_{1}}\\right)$ and $S^{+}\\left(F_{i_{2}}\\right)$.\n\nDefine also the set of inward normal vectors $S^{-}\\left(E_{i}\\right)$ which is the reflection of $S^{+}\\left(E_{i}\\right)$ across the origin.\n\nFor a vertex $V_{k}$, which is the common endpoint of edges $E_{j_{1}}, \\ldots, E_{j_{h}}$ and shared by faces $F_{i_{1}}, \\ldots, F_{i_{h}}$, take all support planes of $\\Gamma$ through point $V_{k}$ and let $S^{+}\\left(V_{k}\\right)$ be the set of their outward normal vectors. This is a region on $S$, a spherical polygon with vertices $S^{+}\\left(F_{i_{1}}\\right), \\ldots, S^{+}\\left(F_{i_{h}}\\right)$ bounded by $\\operatorname{arcs} S^{+}\\left(E_{j_{1}}\\right), \\ldots, S^{+}\\left(E_{j_{h}}\\right)$. Let $S^{-}\\left(V_{k}\\right)$ be the reflection of $S^{+}\\left(V_{k}\\right)$, the set of inward normal vectors.\n\nNote that region $S^{+}\\left(V_{k}\\right)$ is convex in the sense that it is the intersection of several half spheres.\n\n\nNow translate the conditions on $\\Gamma$ to the language of these objects.\n\n(a) Polyhedron $\\Gamma$ has no parallel edges - the great circles of $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{j}\\right)$ are different for all $i \\neq j$.\n\n(b) If an edge $E_{i}$ does not belong to a face $F_{j}$ then they are not parallel - the great circle which contains $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{i}\\right)$ does not pass through points $S^{+}\\left(F_{j}\\right)$ and $S^{-}\\left(F_{j}\\right)$.\n\n(c) Polyhedron $\\Gamma$ has no parallel faces - points $S^{+}\\left(F_{i}\\right)$ and $S^{-}\\left(F_{j}\\right)$ are pairwise distinct.\n\nThe regions $S^{+}\\left(V_{k}\\right)$, arcs $S^{+}\\left(E_{j}\\right)$ and points $S^{+}\\left(F_{i}\\right)$ provide a decomposition of the surface of the sphere. Regions $S^{-}\\left(V_{k}\\right)$, arcs $S^{-}\\left(E_{j}\\right)$ and points $S^{-}\\left(F_{i}\\right)$ provide the reflection of this decomposition. These decompositions are closely related to the problem.\n\n\n\nLemma 1. For any $1 \\leq i, j \\leq n$, regions $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right)$ overlap if and only if vertices $V_{i}$ and $V_{j}$ are antipodal.\n\nLemma 2. For any $1 \\leq i, j \\leq m$, arcs $S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect if and only if the midpoints $Q_{i}$ and $Q_{j}$ of edges $E_{i}$ and $E_{j}$ are antipodal.\n\nProof of lemma 1. First note that by properties (a,b,c) above, the two regions cannot share only a single point or an arc. They are either disjoint or they overlap.\n\nAssume that the two regions have a common interior point $u$. Let $P_{1}$ and $P_{2}$ be two parallel support planes of $\\Gamma$ through points $V_{i}$ and $V_{j}$, respectively, with normal vector $u$. By the definition of regions $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right), u$ is the inward normal vector of $P_{1}$ and the outward normal vector of $P_{2}$. Therefore polyhedron $\\Gamma$ lies between the two planes; vertices $V_{i}$ and $V_{j}$ are antipodal.\n\nTo prove the opposite direction, assume that $V_{i}$ and $V_{j}$ are antipodal. Then there exist two parallel support planes $P_{1}$ and $P_{2}$ through $V_{i}$ and $V_{j}$, respectively, such that $\\Gamma$ is between them. Let $u$ be the inward normal vector of $P_{1}$; then $u$ is the outward normal vector of $P_{2}$, therefore $u \\in S^{-}\\left(V_{i}\\right) \\cap S^{+}\\left(V_{j}\\right)$. The two regions have a common point, so they overlap.\n\nProof of lemma 2. Again, by properties (a,b) above, the endpoints of arc $S^{-}\\left(E_{i}\\right)$ cannot belong to $S^{+}\\left(E_{j}\\right)$ and vice versa. The two arcs are either disjoint or intersecting.\n\nAssume that arcs $S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect at point $u$. Let $P_{1}$ and $P_{2}$ be the two support planes through edges $E_{i}$ and $E_{j}$, respectively, with normal vector $u$. By the definition of $\\operatorname{arcs} S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$, vector $u$ points inwards from $P_{1}$ and outwards from $P_{2}$. Therefore $\\Gamma$ is between the planes. Since planes $P_{1}$ and $P_{2}$ pass through $Q_{i}$ and $Q_{j}$, these points are antipodal.\n\nFor the opposite direction, assume that points $Q_{i}$ and $Q_{j}$ are antipodal. Let $P_{1}$ and $P_{2}$ be two support planes through these points, respectively. An edge cannot intersect a support plane, therefore $E_{i}$ and $E_{j}$ lie in the planes $P_{1}$ and $P_{2}$, respectively. Let $u$ be the inward normal vector of $P_{1}$, which is also the outward normal vector of $P_{2}$. Then $u \\in S^{-}\\left(E_{i}\\right) \\cap S^{+}\\left(E_{j}\\right)$. So the two arcs are not disjoint; they therefore intersect.\n\nNow create a new decomposition of sphere $S$. Draw all $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{j}\\right)$ on sphere $S$ and put a knot at each point where two arcs meet. We have $\\ell$ knots at points $S^{+}\\left(F_{i}\\right)$ and another $\\ell$ knots at points $S^{-}\\left(F_{i}\\right)$, corresponding to the faces of $\\Gamma$; by property (c) they are different. We also have some pairs $1 \\leq i, j \\leq m$ where $\\operatorname{arcs} S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect. By Lemma 2, each antipodal pair $\\left(Q_{i}, Q_{j}\\right)$ gives rise to two such intersections; hence, the number of all intersections is $2 B$ and we have $2 \\ell+2 B$ knots in all.\n\nEach intersection knot splits two arcs, increasing the number of arcs by 2 . Since we started with $2 m$ arcs, corresponding the edges of $\\Gamma$, the number of the resulting curve segments is $2 m+4 B$.\n\nThe network of these curve segments divides the sphere into some ""new"" regions. Each new region is the intersection of some overlapping sets $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right)$. Due to the convexity, the intersection of two overlapping regions is convex and thus contiguous. By Lemma 1, each pair of overlapping regions corresponds to an antipodal vertex pair and each antipodal vertex pair gives rise to two different overlaps, which are symmetric with respect to the origin. So the number of new regions is $2 A$.\n\nThe result now follows from Euler\'s polyhedron theorem. We have $n+l=m+2$ and\n\n$$\n(2 \\ell+2 B)+2 A=(2 m+4 B)+2,\n$$\n\ntherefore\n\n$$\nA-B=m-\\ell+1=n-1 \\text {. }\n$$\n\nTherefore $A-B$ is by one less than the number of vertices of $\\Gamma$.', ""Use the same notations for the polyhedron and its vertices, edges and faces as in Solution 1. We regard points as vectors starting from the origin. Polyhedron $\\Gamma$ is regarded as a closed convex set, including its interior. In some cases the edges and faces of $\\Gamma$ are also regarded as sets of points. The symbol $\\partial$ denotes the boundary of the certain set; e.g. $\\partial \\Gamma$ is the surface of $\\Gamma$.\n\nLet $\\Delta=\\Gamma-\\Gamma=\\{U-V: U, V \\in \\Gamma\\}$ be the set of vectors between arbitrary points of $\\Gamma$. Then $\\Delta$, being the sum of two bounded convex sets, is also a bounded convex set and, by construction, it is also centrally symmetric with respect to the origin. We will prove that $\\Delta$ is also a polyhedron and express the numbers of its faces, edges and vertices in terms $n, m, \\ell, A$ and $B$.\n\nLemma 1. For points $U, V \\in \\Gamma$, point $W=U-V$ is a boundary point of $\\Delta$ if and only if $U$ and $V$ are antipodal. Moreover, for each boundary point $W \\in \\partial \\Delta$ there exists exactly one pair of points $U, V \\in \\Gamma$ such that $W=U-V$.\n\nProof. Assume first that $U$ and $V$ are antipodal points of $\\Gamma$. Let parallel support planes $P_{1}$ and $P_{2}$ pass through them such that $\\Gamma$ is in between. Consider plane $P=P_{1}-U=$ $P_{2}-V$. This plane separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. After reflecting one of the sets, e.g. $\\Gamma-V$, the sets $\\Gamma-U$ and $-\\Gamma+V$ lie in the same half space bounded by $P$. Then $(\\Gamma-U)+(-\\Gamma+V)=\\Delta-W$ lies in that half space, so $0 \\in P$ is a boundary point of the set $\\Delta-W$. Translating by $W$ we obtain that $W$ is a boundary point of $\\Delta$.\n\nTo prove the opposite direction, let $W=U-V$ be a boundary point of $\\Delta$, and let $\\Psi=$ $(\\Gamma-U) \\cap(\\Gamma-V)$. We claim that $\\Psi=\\{0\\}$. Clearly $\\Psi$ is a bounded convex set and $0 \\in \\Psi$. For any two points $X, Y \\in \\Psi$, we have $U+X, V+Y \\in \\Gamma$ and $W+(X-Y)=(U+X)-(V+Y) \\in \\Delta$. Since $W$ is a boundary point of $\\Delta$, the vector $X-Y$ cannot have the same direction as $W$. This implies that the interior of $\\Psi$ is empty. Now suppose that $\\Psi$ contains a line segment $S$. Then $S+U$ and $S+V$ are subsets of some faces or edges of $\\Gamma$ and these faces/edges are parallel to $S$. In all cases, we find two faces, two edges, or a face and an edge which are parallel, contradicting the conditions of the problem. Therefore, $\\Psi=\\{0\\}$ indeed.\n\nSince $\\Psi=(\\Gamma-U) \\cap(\\Gamma-V)$ consists of a single point, the interiors of bodies $\\Gamma-U$ and $\\Gamma-V$ are disjoint and there exists a plane $P$ which separates them. Let $u$ be the normal vector of $P$ pointing into that half space bounded by $P$ which contains $\\Gamma-U$. Consider the planes $P+U$ and $P+V$; they are support planes of $\\Gamma$, passing through $U$ and $V$, respectively. From plane $P+U$, the vector $u$ points into that half space which contains $\\Gamma$. From plane $P+V$, vector $u$ points into the opposite half space containing $\\Gamma$. Therefore, we found two proper support through points $U$ and $V$ such that $\\Gamma$ is in between.\n\nFor the uniqueness part, assume that there exist points $U_{1}, V_{1} \\in \\Gamma$ such that $U_{1}-V_{1}=U-V$. The points $U_{1}-U$ and $V_{1}-V$ lie in the sets $\\Gamma-U$ and $\\Gamma-V$ separated by $P$. Since $U_{1}-U=V_{1}-V$, this can happen only if both are in $P$; but the only such point is 0 . Therefore, $U_{1}-V_{1}=U-V$ implies $U_{1}=U$ and $V_{1}=V$. The lemma is complete.\n\nLemma 2. Let $U$ and $V$ be two antipodal points and assume that plane $P$, passing through 0 , separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. Let $\\Psi_{1}=(\\Gamma-U) \\cap P$ and $\\Psi_{2}=(\\Gamma-V) \\cap P$. Then $\\Delta \\cap(P+U-V)=\\Psi_{1}-\\Psi_{2}+U-V$.\n\nProof. The sets $\\Gamma-U$ and $-\\Gamma+V$ lie in the same closed half space bounded by $P$. Therefore, for any points $X \\in(\\Gamma-U)$ and $Y \\in(-\\Gamma+V)$, we have $X+Y \\in P$ if and only if $X, Y \\in P$. Then\n\n$(\\Delta-(U-V)) \\cap P=((\\Gamma-U)+(-\\Gamma+V)) \\cap P=((\\Gamma-U) \\cap P)+((-\\Gamma+V) \\cap P)=\\Psi_{1}-\\Psi_{2}$.\n\nNow a translation by $(U-V)$ completes the lemma.\n\n\n\nNow classify the boundary points $W=U-V$ of $\\Delta$, according to the types of points $U$ and $V$. In all cases we choose a plane $P$ through 0 which separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. We will use the notation $\\Psi_{1}=(\\Gamma-U) \\cap P$ and $\\Psi_{2}=(\\Gamma-V) \\cap P$ as well.\n\nCase 1: Both $U$ and $V$ are vertices of $\\Gamma$. Bodies $\\Gamma-U$ and $\\Gamma-V$ have a common vertex which is 0 . Choose plane $P$ in such a way that $\\Psi_{1}=\\Psi_{2}=\\{0\\}$. Then Lemma 2 yields $\\Delta \\cap(P+W)=\\{W\\}$. Therefore $P+W$ is a support plane of $\\Delta$ such that they have only one common point so no line segment exists on $\\partial \\Delta$ which would contain $W$ in its interior.\n\nSince this case occurs for antipodal vertex pairs and each pair is counted twice, the number of such boundary points on $\\Delta$ is $2 A$.\n\nCase 2: Point $U$ is an interior point of an edge $E_{i}$ and $V$ is a vertex of $\\Gamma$. Choose plane $P$ such that $\\Psi_{1}=E_{i}-U$ and $\\Psi_{2}=\\{0\\}$. By Lemma $2, \\Delta \\cap(P+W)=E_{i}-V$. Hence there exists a line segment in $\\partial \\Delta$ which contains $W$ in its interior, but there is no planar region in $\\partial \\Delta$ with the same property.\n\nWe obtain a similar result if $V$ belongs to an edge of $\\Gamma$ and $U$ is a vertex.\n\nCase 3: Points $U$ and $V$ are interior points of edges $E_{i}$ and $E_{j}$, respectively. Let $P$ be the plane of $E_{i}-U$ and $E_{j}-V$. Then $\\Psi_{1}=E_{i}-U, \\Psi_{2}=E_{j}-V$ and $\\Delta \\cap(P+W)=E_{i}-E_{j}$. Therefore point $W$ belongs to a parallelogram face on $\\partial \\Delta$.\n\nThe centre of the parallelogram is $Q_{i}-Q_{j}$, the vector between the midpoints. Therefore an edge pair $\\left(E_{i}, E_{j}\\right)$ occurs if and only if $Q_{i}$ and $Q_{j}$ are antipodal which happens $2 B$ times.\n\nCase 4: Point $U$ lies in the interior of a face $F_{i}$ and $V$ is a vertex of $\\Gamma$. The only choice for $P$ is the plane of $F_{i}-U$. Then we have $\\Psi_{1}=F_{i}-U, \\Psi_{2}=\\{0\\}$ and $\\Delta \\cap(P+W)=F_{i}-V$. This is a planar face of $\\partial \\Delta$ which is congruent to $F_{i}$.\n\nFor each face $F_{i}$, the only possible vertex $V$ is the farthest one from the plane of $F_{i}$.\n\nIf $U$ is a vertex and $V$ belongs to face $F_{i}$ then we obtain the same way that $W$ belongs to a face $-F_{i}+U$ which is also congruent to $F_{i}$. Therefore, each face of $\\Gamma$ has two copies on $\\partial \\Delta$, a translated and a reflected copy.\n\nCase 5: Point $U$ belongs to a face $F_{i}$ of $\\Gamma$ and point $V$ belongs to an edge or a face $G$. In this case objects $F_{i}$ and $G$ must be parallel which is not allowed.\n\n\nNow all points in $\\partial \\Delta$ belong to some planar polygons (cases 3 and 4 ), finitely many line segments (case 2) and points (case 1). Therefore $\\Delta$ is indeed a polyhedron. Now compute the numbers of its vertices, edges and faces.\n\nThe vertices are obtained in case 1 , their number is $2 A$.\n\nFaces are obtained in cases 3 and 4 . Case 3 generates $2 B$ parallelogram faces. Case 4 generates $2 \\ell$ faces.\n\nWe compute the number of edges of $\\Delta$ from the degrees (number of sides) of faces of $\\Gamma$. Let $d_{i}$ be the the degree of face $F_{i}$. The sum of degrees is twice as much as the number of edges, so $d_{1}+d_{2}+\\ldots+d_{l}=2 m$. The sum of degrees of faces of $\\Delta$ is $2 B \\cdot 4+2\\left(d_{1}+d_{2}+\\cdots+d_{l}\\right)=8 B+4 m$, so the number of edges on $\\Delta$ is $4 B+2 m$.\n\nApplying Euler's polyhedron theorem on $\\Gamma$ and $\\Delta$, we have $n+l=m+2$ and $2 A+(2 B+2 \\ell)=$ $(4 B+2 m)+2$. Then the conclusion follows:\n\n$$\nA-B=m-\\ell+1=n-1 \\text {. }\n$$""]",,True,,, 1756,Geometry,,"Let $A B C$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies $$ \angle P B A+\angle P C A=\angle P B C+\angle P C B . $$ Show that $A P \geq A I$ and that equality holds if and only if $P$ coincides with $I$.","['Let $\\angle A=\\alpha, \\angle B=\\beta, \\angle C=\\gamma$. Since $\\angle P B A+\\angle P C A+\\angle P B C+\\angle P C B=\\beta+\\gamma$, the condition from the problem statement is equivalent to $\\angle P B C+\\angle P C B=(\\beta+\\gamma) / 2$, i. e. $\\angle B P C=90^{\\circ}+\\alpha / 2$.\n\nOn the other hand $\\angle B I C=180^{\\circ}-(\\beta+\\gamma) / 2=90^{\\circ}+\\alpha / 2$. Hence $\\angle B P C=\\angle B I C$, and since $P$ and $I$ are on the same side of $B C$, the points $B, C, I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\\omega$ of triangle $B C I$.\n\n\n\nLet $\\Omega$ be the circumcircle of triangle $A B C$. It is a well-known fact that the centre of $\\omega$ is the midpoint $M$ of the $\\operatorname{arc} B C$ of $\\Omega$. This is also the point where the angle bisector $A I$ intersects $\\Omega$.\n\nFrom triangle $A P M$ we have\n\n$$\nA P+P M \\geq A M=A I+I M=A I+P M\n$$\n\nTherefore $A P \\geq A I$. Equality holds if and only if $P$ lies on the line segment $A I$, which occurs if and only if $P=I$.']",,True,,, 1757,Geometry,,"Let $A B C D$ be a trapezoid with parallel sides $A B>C D$. Points $K$ and $L$ lie on the line segments $A B$ and $C D$, respectively, so that $A K / K B=D L / L C$. Suppose that there are points $P$ and $Q$ on the line segment $K L$ satisfying $$ \angle A P B=\angle B C D \quad \text { and } \quad \angle C Q D=\angle A B C \text {. } $$ Prove that the points $P, Q, B$ and $C$ are concyclic.","['Because $A B \\| C D$, the relation $A K / K B=D L / L C$ readily implies that the lines $A D, B C$ and $K L$ have a common point $S$.\n\n\n\nConsider the second intersection points $X$ and $Y$ of the line $S K$ with the circles $(A B P)$ and $(C D Q)$, respectively. Since $A P B X$ is a cyclic quadrilateral and $A B \\| C D$, one has\n\n$$\n\\angle A X B=180^{\\circ}-\\angle A P B=180^{\\circ}-\\angle B C D=\\angle A B C \\text {. }\n$$\n\nThis shows that $B C$ is tangent to the circle $(A B P)$ at $B$. Likewise, $B C$ is tangent to the circle $(C D Q)$ at $C$. Therefore $S P \\cdot S X=S B^{2}$ and $S Q \\cdot S Y=S C^{2}$.\n\nLet $h$ be the homothety with centre $S$ and ratio $S C / S B$. Since $h(B)=C$, the above conclusion about tangency implies that $h$ takes circle $(A B P)$ to circle $(C D Q)$. Also, $h$ takes $A B$ to $C D$, and it easily follows that $h(P)=Y, h(X)=Q$, yielding $S P / S Y=S B / S C=S X / S Q$.\n\nEqualities $S P \\cdot S X=S B^{2}$ and $S Q / S X=S C / S B$ imply $S P \\cdot S Q=S B \\cdot S C$, which is equivalent to $P, Q, B$ and $C$ being concyclic.', ""The case where $P=Q$ is trivial. Thus assume that $P$ and $Q$ are two distinct points. As in the first solution, notice that the lines $A D, B C$ and $K L$ concur at a point $S$.\n\n\n\nLet the lines $A P$ and $D Q$ meet at $E$, and let $B P$ and $C Q$ meet at $F$. Then $\\angle E P F=\\angle B C D$ and $\\angle F Q E=\\angle A B C$ by the condition of the problem. Since the angles $B C D$ and $A B C$ add up to $180^{\\circ}$, it follows that $P E Q F$ is a cyclic quadrilateral.\n\nApplying Menelaus' theorem, first to triangle $A S P$ and line $D Q$ and then to triangle $B S P$ and line $C Q$, we have\n\n$$\n\\frac{A D}{D S} \\cdot \\frac{S Q}{Q P} \\cdot \\frac{P E}{E A}=1 \\quad \\text { and } \\quad \\frac{B C}{C S} \\cdot \\frac{S Q}{Q P} \\cdot \\frac{P F}{F B}=1\n$$\n\nThe first factors in these equations are equal, as $A B \\| C D$. Thus the last factors are also equal, which implies that $E F$ is parallel to $A B$ and $C D$. Using this and the cyclicity of $P E Q F$, we obtain\n\n$$\n\\angle B C D=\\angle B C F+\\angle F C D=\\angle B C Q+\\angle E F Q=\\angle B C Q+\\angle E P Q .\n$$\n\nOn the other hand,\n\n$$\n\\angle B C D=\\angle A P B=\\angle E P F=\\angle E P Q+\\angle Q P F,\n$$\n\nand consequently $\\angle B C Q=\\angle Q P F$. The latter angle either coincides with $\\angle Q P B$ or is supplementary to $\\angle Q P B$, depending on whether $Q$ lies between $K$ and $P$ or not. In either case it follows that $P, Q, B$ and $C$ are concyclic.""]",,True,,, 1758,Geometry,,"Let $A B C D E$ be a convex pentagon such that $$ \angle B A C=\angle C A D=\angle D A E \quad \text { and } \quad \angle A B C=\angle A C D=\angle A D E \text {. } $$ The diagonals $B D$ and $C E$ meet at $P$. Prove that the line $A P$ bisects the side $C D$.","[""Let the diagonals $A C$ and $B D$ meet at $Q$, the diagonals $A D$ and $C E$ meet at $R$, and let the ray $A P$ meet the side $C D$ at $M$. We want to prove that $C M=M D$ holds.\n\n\n\nThe idea is to show that $Q$ and $R$ divide $A C$ and $A D$ in the same ratio, or more precisely\n\n$$\n\\frac{A Q}{Q C}=\\frac{A R}{R D} \\tag{1}\n$$\n\n(which is equivalent to $Q R \\| C D$ ). The given angle equalities imply that the triangles $A B C$, $A C D$ and $A D E$ are similar. We therefore have\n\n$$\n\\frac{A B}{A C}=\\frac{A C}{A D}=\\frac{A D}{A E}\n$$\n\nSince $\\angle B A D=\\angle B A C+\\angle C A D=\\angle C A D+\\angle D A E=\\angle C A E$, it follows from $A B / A C=$ $A D / A E$ that the triangles $A B D$ and $A C E$ are also similar. Their angle bisectors in $A$ are $A Q$ and $A R$, respectively, so that\n\n$$\n\\frac{A B}{A C}=\\frac{A Q}{A R}\n$$\n\nBecause $A B / A C=A C / A D$, we obtain $A Q / A R=A C / A D$, which is equivalent to (1). Now Ceva's theorem for the triangle $A C D$ yields\n\n$$\n\\frac{A Q}{Q C} \\cdot \\frac{C M}{M D} \\cdot \\frac{D R}{R A}=1\n$$\n\nIn view of (1), this reduces to $C M=M D$, which completes the proof.""]",,True,,, 1759,Geometry,,"A point $D$ is chosen on the side $A C$ of a triangle $A B C$ with $\angle C<\angle A<90^{\circ}$ in such a way that $B D=B A$. The incircle of $A B C$ is tangent to $A B$ and $A C$ at points $K$ and $L$, respectively. Let $J$ be the incentre of triangle $B C D$. Prove that the line $K L$ intersects the line segment $A J$ at its midpoint.","['Denote by $P$ be the common point of $A J$ and $K L$. Let the parallel to $K L$ through $J$ meet $A C$ at $M$. Then $P$ is the midpoint of $A J$ if and only if $A M=2 \\cdot A L$, which we are about to show.\n\n\n\nDenoting $\\angle B A C=2 \\alpha$, the equalities $B A=B D$ and $A K=A L$ imply $\\angle A D B=2 \\alpha$ and $\\angle A L K=90^{\\circ}-\\alpha$. Since $D J$ bisects $\\angle B D C$, we obtain $\\angle C D J=\\frac{1}{2} \\cdot\\left(180^{\\circ}-\\angle A D B\\right)=90^{\\circ}-\\alpha$. Also $\\angle D M J=\\angle A L K=90^{\\circ}-\\alpha$ since $J M \\| K L$. It follows that $J D=J M$.\n\nLet the incircle of triangle $B C D$ touch its side $C D$ at $T$. Then $J T \\perp C D$, meaning that $J T$ is the altitude to the base $D M$ of the isosceles triangle $D M J$. It now follows that $D T=M T$, and we have\n\n$$\nD M=2 \\cdot D T=B D+C D-B C \\text {. }\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\nA M & =A D+(B D+C D-B C) \\\\\n& =A D+A B+D C-B C \\\\\n& =A C+A B-B C \\\\\n& =2 \\cdot A L,\n\\end{aligned}\n$$\n\nwhich completes the proof.']",,True,,, 1760,Geometry,,"In triangle $A B C$, let $J$ be the centre of the excircle tangent to side $B C$ at $A_{1}$ and to the extensions of sides $A C$ and $A B$ at $B_{1}$ and $C_{1}$, respectively. Suppose that the lines $A_{1} B_{1}$ and $A B$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $D J$. Determine the angles $\angle B E A_{1}$ and $\angle A E B_{1}$.","['Let $K$ be the intersection point of lines $J C$ and $A_{1} B_{1}$. Obviously $J C \\perp A_{1} B_{1}$ and since $A_{1} B_{1} \\perp A B$, the lines $J K$ and $C_{1} D$ are parallel and equal. From the right triangle $B_{1} C J$ we obtain $J C_{1}^{2}=J B_{1}^{2}=J C \\cdot J K=J C \\cdot C_{1} D$ from which we infer that $D C_{1} / C_{1} J=C_{1} J / J C$ and the right triangles $D C_{1} J$ and $C_{1} J C$ are similar. Hence $\\angle C_{1} D J=\\angle J C_{1} C$, which implies that the lines $D J$ and $C_{1} C$ are perpendicular, i.e. the points $C_{1}, E, C$ are collinear.\n\n\n\nSince $\\angle C A_{1} J=\\angle C B_{1} J=\\angle C E J=90^{\\circ}$, points $A_{1}, B_{1}$ and $E$ lie on the circle of diameter $C J$. Then $\\angle D B A_{1}=\\angle A_{1} C J=\\angle D E A_{1}$, which implies that quadrilateral $B E A_{1} D$ is cyclic; therefore $\\angle A_{1} E B=90^{\\circ}$.\n\nQuadrilateral $A D E B_{1}$ is also cyclic because $\\angle E B_{1} A=\\angle E J C=\\angle E D C_{1}$, therefore we obtain $\\angle A E B_{1}=\\angle A D B=90^{\\circ}$.\n\n', 'Consider the circles $\\omega_{1}, \\omega_{2}$ and $\\omega_{3}$ of diameters $C_{1} D, A_{1} B$ and $A B_{1}$, respectively. Line segments $J C_{1}, J B_{1}$ and $J A_{1}$ are tangents to those circles and, due to the right angle at $D$, $\\omega_{2}$ and $\\omega_{3}$ pass through point $D$. Since $\\angle C_{1} E D$ is a right angle, point $E$ lies on circle $\\omega_{1}$, therefore\n\n$$\nJ C_{1}^{2}=J D \\cdot J E\n$$\n\nSince $J A_{1}=J B_{1}=J C_{1}$ are all radii of the excircle, we also have\n\n$$\nJ A_{1}^{2}=J D \\cdot J E \\quad \\text { and } \\quad J B_{1}^{2}=J D \\cdot J E .\n$$\n\nThese equalities show that $E$ lies on circles $\\omega_{2}$ and $\\omega_{3}$ as well, so $\\angle B E A_{1}=\\angle A E B_{1}=90^{\\circ}$.\n\n### solution_2\nFirst note that $A_{1} B_{1}$ is perpendicular to the external angle bisector $C J$ of $\\angle B C A$ and parallel to the internal angle bisector of that angle. Therefore, $A_{1} B_{1}$ is perpendicular to $A B$ if and only if triangle $A B C$ is isosceles, $A C=B C$. In that case the external bisector $C J$ is parallel to $A B$.\n\nTriangles $A B C$ and $B_{1} A_{1} J$ are similar, as their corresponding sides are perpendicular. In particular, we have $\\angle D A_{1} J=\\angle C_{1} B A_{1}$; moreover, from cyclic deltoid $J A_{1} B C_{1}$,\n\n$$\n\\angle C_{1} A_{1} J=\\angle C_{1} B J=\\frac{1}{2} \\angle C_{1} B A_{1}=\\frac{1}{2} \\angle D A_{1} J\n$$\n\nTherefore, $A_{1} C_{1}$ bisects angle $\\angle D A_{1} J$.\n\n\n\nIn triangle $B_{1} A_{1} J$, line $J C_{1}$ is the external bisector at vertex $J$. The point $C_{1}$ is the intersection of two external angle bisectors (at $A_{1}$ and $J$ ) so $C_{1}$ is the centre of the excircle $\\omega$, tangent to side $A_{1} J$, and to the extension of $B_{1} A_{1}$ at point $D$.\n\nNow consider the similarity transform $\\varphi$ which moves $B_{1}$ to $A, A_{1}$ to $B$ and $J$ to $C$. This similarity can be decomposed into a rotation by $90^{\\circ}$ around a certain point $O$ and a homothety from the same centre. This similarity moves point $C_{1}$ (the centre of excircle $\\omega$ ) to $J$ and moves $D$ (the point of tangency) to $C_{1}$.\n\nSince the rotation angle is $90^{\\circ}$, we have $\\angle X O \\varphi(X)=90^{\\circ}$ for an arbitrary point $X \\neq O$. For $X=D$ and $X=C_{1}$ we obtain $\\angle D O C_{1}=\\angle C_{1} O J=90^{\\circ}$. Therefore $O$ lies on line segment $D J$ and $C_{1} O$ is perpendicular to $D J$. This means that $O=E$.\n\nFor $X=A_{1}$ and $X=B_{1}$ we obtain $\\angle A_{1} O B=\\angle B_{1} O A=90^{\\circ}$, i.e.\n\n$$\n\\angle B E A_{1}=\\angle A E B_{1}=90^{\\circ} .\n$$']","['$\\angle B E A_{1}=90$,$\\angle A E B_{1}=90$']",True,^{\circ},Numerical, 1761,Geometry,,"Circles $\omega_{1}$ and $\omega_{2}$ with centres $O_{1}$ and $O_{2}$ are externally tangent at point $D$ and internally tangent to a circle $\omega$ at points $E$ and $F$, respectively. Line $t$ is the common tangent of $\omega_{1}$ and $\omega_{2}$ at $D$. Let $A B$ be the diameter of $\omega$ perpendicular to $t$, so that $A, E$ and $O_{1}$ are on the same side of $t$. Prove that lines $A O_{1}, B O_{2}, E F$ and $t$ are concurrent.","[""Point $E$ is the centre of a homothety $h$ which takes circle $\\omega_{1}$ to circle $\\omega$. The radii $O_{1} D$ and $O B$ of these circles are parallel as both are perpendicular to line $t$. Also, $O_{1} D$ and $O B$ are on the same side of line $E O$, hence $h$ takes $O_{1} D$ to $O B$. Consequently, points $E$, $D$ and $B$ are collinear. Likewise, points $F, D$ and $A$ are collinear as well.\n\nLet lines $A E$ and $B F$ intersect at $C$. Since $A F$ and $B E$ are altitudes in triangle $A B C$, their common point $D$ is the orthocentre of this triangle. So $C D$ is perpendicular to $A B$, implying that $C$ lies on line $t$. Note that triangle $A B C$ is acute-angled. We mention the well-known fact that triangles $F E C$ and $A B C$ are similar in ratio $\\cos \\gamma$, where $\\gamma=\\angle A C B$. In addition, points $C, E, D$ and $F$ lie on the circle with diameter $C D$.\n\n\n\nLet $P$ be the common point of lines $E F$ and $t$. We are going to prove that $P$ lies on line $A O_{1}$. Denote by $N$ the second common point of circle $\\omega_{1}$ and $A C$; this is the point of $\\omega_{1}$ diametrically opposite to $D$. By Menelaus' theorem for triangle $D C N$, points $A, O_{1}$ and $P$ are collinear if and only if\n\n$$\n\\frac{C A}{A N} \\cdot \\frac{N O_{1}}{O_{1} D} \\cdot \\frac{D P}{P C}=1\n$$\n\nBecause $N O_{1}=O_{1} D$, this reduces to $C A / A N=C P / P D$. Let line $t$ meet $A B$ at $K$. Then $C A / A N=C K / K D$, so it suffices to show that\n\n$$\n\\frac{C P}{P D}=\\frac{C K}{K D}\n$$\n\nTo verify (1), consider the circumcircle $\\Omega$ of triangle $A B C$. Draw its diameter $C U$ through $C$, and let $C U$ meet $A B$ at $V$. Extend $C K$ to meet $\\Omega$ at $L$. Since $A B$ is parallel to $U L$, we have $\\angle A C U=\\angle B C L$. On the other hand $\\angle E F C=\\angle B A C, \\angle F E C=\\angle A B C$ and $E F / A B=\\cos \\gamma$, as stated above. So reflection in the bisector of $\\angle A C B$ followed by a homothety with centre $C$ and ratio $1 / \\cos \\gamma$ takes triangle $F E C$ to triangle $A B C$. Consequently, this transformation\n\n\n\ntakes $C D$ to $C U$, which implies $C P / P D=C V / V U$. Next, we have $K L=K D$, because $D$ is the orthocentre of triangle $A B C$. Hence $C K / K D=C K / K L$. Finally, $C V / V U=C K / K L$ because $A B$ is parallel to $U L$. Relation (1) follows, proving that $P$ lies on line $A O_{1}$. By symmetry, $P$ also lies on line $A O_{2}$ which completes the solution.\n\nSolution 2. We proceed as in the first solution to define a triangle $A B C$ with orthocentre $D$, in which $A F$ and $B E$ are altitudes.\n\nDenote by $M$ the midpoint of $C D$. The quadrilateral $C E D F$ is inscribed in a circle with centre $M$, hence $M C=M E=M D=M F$.\n\n\n\nConsider triangles $A B C$ and $O_{1} O_{2} M$. Lines $O_{1} O_{2}$ and $A B$ are parallel, both of them being perpendicular to line $t$. Next, $M O_{1}$ is the line of centres of circles $(C E F)$ and $\\omega_{1}$ whose common chord is $D E$. Hence $M O_{1}$ bisects $\\angle D M E$ which is the external angle at $M$ in the isosceles triangle $C E M$. It follows that $\\angle D M O_{1}=\\angle D C A$, so that $M O_{1}$ is parallel to $A C$. Likewise, $M O_{2}$ is parallel to $B C$.\n\nThus the respective sides of triangles $A B C$ and $O_{1} O_{2} M$ are parallel; in addition, these triangles are not congruent. Hence there is a homothety taking $A B C$ to $O_{1} O_{2} M$. The lines $A O_{1}$, $B O_{2}$ and $C M=t$ are concurrent at the centre $Q$ of this homothety.\n\nFinally, apply Pappus' theorem to the triples of collinear points $A, O, B$ and $O_{2}, D, O_{1}$. The theorem implies that the points $A D \\cap O O_{2}=F, A O_{1} \\cap B O_{2}=Q$ and $O O_{1} \\cap B D=E$ are collinear. In other words, line $E F$ passes through the common point $Q$ of $A O_{1}, B O_{2}$ and $t$.\n\nComment. Relation (1) from Solution 1 expresses the well-known fact that points $P$ and $K$ are harmonic conjugates with respect to points $C$ and $D$. It is also easy to justify it by direct computation. Denoting $\\angle C A B=\\alpha, \\angle A B C=\\beta$, it is straightforward to obtain $C P / P D=C K / K D=\\tan \\alpha \\tan \\beta$.""]",,True,,, 1762,Geometry,,"In a triangle $A B C$, let $M_{a}, M_{b}, M_{c}$ be respectively the midpoints of the sides $B C, C A$, $A B$ and $T_{a}, T_{b}, T_{c}$ be the midpoints of the $\operatorname{arcs} B C, C A, A B$ of the circumcircle of $A B C$, not containing the opposite vertices. For $i \in\{a, b, c\}$, let $\omega_{i}$ be the circle with $M_{i} T_{i}$ as diameter. Let $p_{i}$ be the common external tangent to $\omega_{j}, \omega_{k}(\{i, j, k\}=\{a, b, c\})$ such that $\omega_{i}$ lies on the opposite side of $p_{i}$ than $\omega_{j}, \omega_{k}$ do. Prove that the lines $p_{a}, p_{b}, p_{c}$ form a triangle similar to $A B C$ and find the ratio of similitude.","['Let $T_{a} T_{b}$ intersect circle $\\omega_{b}$ at $T_{b}$ and $U$, and let $T_{a} T_{c}$ intersect circle $\\omega_{c}$ at $T_{c}$ and $V$. Further, let $U X$ be the tangent to $\\omega_{b}$ at $U$, with $X$ on $A C$, and let $V Y$ be the tangent to $\\omega_{c}$ at $V$, with $Y$ on $A B$. The homothety with centre $T_{b}$ and ratio $T_{b} T_{a} / T_{b} U$ maps the circle $\\omega_{b}$ onto the circumcircle of $A B C$ and the line $U X$ onto the line tangent to the circumcircle at $T_{a}$, which is parallel to $B C$; thus $U X \\| B C$. The same is true of $V Y$, so that $U X\\|B C\\| V Y$.\n\nLet $T_{a} T_{b}$ cut $A C$ at $P$ and let $T_{a} T_{c}$ cut $A B$ at $Q$. The point $X$ lies on the hypotenuse $P M_{b}$ of the right triangle $P U M_{b}$ and is equidistant from $U$ and $M_{b}$. So $X$ is the midpoint of $M_{b} P$. Similarly $Y$ is the midpoint of $M_{c} Q$.\n\nDenote the incentre of triangle $A B C$ as usual by $I$. It is a known fact that $T_{a} I=T_{a} B$ and $T_{c} I=T_{c} B$. Therefore the points $B$ and $I$ are symmetric across $T_{a} T_{c}$, and consequently $\\angle Q I B=\\angle Q B I=\\angle I B C$. This implies that $B C$ is parallel to the line $I Q$, and likewise, to $I P$. In other words, $P Q$ is the line parallel to $B C$ passing through $I$.\n\n\n\nClearly $M_{b} M_{c} \\| B C$. So $P M_{b} M_{c} Q$ is a trapezoid and the segment $X Y$ connects the midpoints of its nonparallel sides; hence $X Y \\| B C$. This combined with the previously established relations $U X\\|B C\\| V Y$ shows that all the four points $U, X, Y, V$ lie on a line which is the common tangent to circles $\\omega_{b}, \\omega_{c}$. Since it leaves these two circles on one side and the circle $\\omega_{a}$ on the other, this line is just the line $p_{a}$ from the problem statement.\n\nLine $p_{a}$ runs midway between $I$ and $M_{b} M_{c}$. Analogous conclusions hold for the lines $p_{b}$ and $p_{c}$. So these three lines form a triangle homothetic from centre $I$ to triangle $M_{a} M_{b} M_{c}$ in ratio $1 / 2$, hence similar to $A B C$ in ratio $1 / 4$.']",,True,,, 1763,Geometry,,"Let $A B C D$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that $$ \angle P A B+\angle P D C \leq 90^{\circ} \quad \text { and } \quad \angle P B A+\angle P C D \leq 90^{\circ} \text {. } $$ Prove that $A B+C D \geq B C+A D$.","['We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $A B C D$. Then:\n\n\n\n$$\n\\text{Circles $(B C T)$ and $(D A T)$ are tangent at $T$}\n\\newline\n\\text { if and only if } \\angle A D T+\\angle B C T=\\angle A T B \\text {. } \\tag{1}\n$$\n\nIndeed, if the two circles touch each other then their common tangent at $T$ intersects the segment $A B$ at a point $Z$, and so $\\angle A D T=\\angle A T Z, \\angle B C T=\\angle B T Z$, by the tangent-chord theorem. Thus $\\angle A D T+\\angle B C T=\\angle A T Z+\\angle B T Z=\\angle A T B$.\n\nAnd conversely, if $\\angle A D T+\\angle B C T=\\angle A T B$ then one can draw from $T$ a ray $T Z$ with $Z$ on $A B$ so that $\\angle A D T=\\angle A T Z, \\angle B C T=\\angle B T Z$. The first of these equalities implies that $T Z$ is tangent to the circle $(D A T)$; by the second equality, $T Z$ is tangent to the circle $(B C T)$, so the two circles are tangent at $T$.\n\n\n\nSo the equivalence (1) is settled. It will be used later on. Now pass to the actual solution. Its key idea is to introduce the circumcircles of triangles $A B P$ and $C D P$ and to consider their second intersection $Q$ (assume for the moment that they indeed meet at two distinct points $P$ and $Q$ ).\n\nSince the point $A$ lies outside the circle $(B C P)$, we have $\\angle B C P+\\angle B A P<180^{\\circ}$. Therefore the point $C$ lies outside the circle $(A B P)$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same $\\operatorname{arc} C D$ of the circle $(B C P)$.\n\n\n\n\n\nBy symmetry, $P$ and $Q$ lie on the same arc $A B$ of the circle $(A B P)$. Thus the point $Q$ lies either inside the angle $B P C$ or inside the angle $A P D$. Without loss of generality assume that $Q$ lies inside the angle $B P C$. Then\n\n$$\n\\angle A Q D=\\angle P Q A+\\angle P Q D=\\angle P B A+\\angle P C D \\leq 90^{\\circ}\\tag{2}\n$$\n\nby the condition of the problem.\n\nIn the cyclic quadrilaterals $A P Q B$ and $D P Q C$, the angles at vertices $A$ and $D$ are acute. So their angles at $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $B P C$ but in fact inside the triangle $B P C$, hence also inside the quadrilateral $A B C D$.\n\nNow an argument similar to that used in deriving (2) shows that\n\n$$\n\\angle B Q C=\\angle P A B+\\angle P D C \\leq 90^{\\circ} .\\tag{3}\n$$\n\nMoreover, since $\\angle P C Q=\\angle P D Q$, we get\n\n$$\n\\angle A D Q+\\angle B C Q=\\angle A D P+\\angle P D Q+\\angle B C P-\\angle P C Q=\\angle A D P+\\angle B C P .\n$$\n\nThe last sum is equal to $\\angle A P B$, according to the observation (1) applied to $T=P$. And because $\\angle A P B=\\angle A Q B$, we obtain\n\n$$\n\\angle A D Q+\\angle B C Q=\\angle A Q B\n$$\n\nApplying now (1) to $T=Q$ we conclude that the circles $(B C Q)$ and $(D A Q)$ are externally tangent at $Q$. (We have assumed $P \\neq Q$; but if $P=Q$ then the last conclusion holds trivially.)\n\nFinally consider the halfdiscs with diameters $B C$ and $D A$ constructed inwardly to the quadrilateral $A B C D$. They have centres at $M$ and $N$, the midpoints of $B C$ and $D A$ respectively. In view of (2) and (3), these two halfdiscs lie entirely inside the circles $(B Q C)$ and $(A Q D)$; and since these circles are tangent, the two halfdiscs cannot overlap. Hence $M N \\geq \\frac{1}{2} B C+\\frac{1}{2} D A$.\n\nOn the other hand, since $\\overrightarrow{M N}=\\frac{1}{2}(\\overrightarrow{B A}+\\overrightarrow{C D})$, we have $M N \\leq \\frac{1}{2}(A B+C D)$. Thus indeed $A B+C D \\geq B C+D A$, as claimed.']",,True,,, 1764,Geometry,,"Points $A_{1}, B_{1}, C_{1}$ are chosen on the sides $B C, C A, A B$ of a triangle $A B C$, respectively. The circumcircles of triangles $A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$ intersect the circumcircle of triangle $A B C$ again at points $A_{2}, B_{2}, C_{2}$, respectively $\left(A_{2} \neq A, B_{2} \neq B, C_{2} \neq C\right)$. Points $A_{3}, B_{3}, C_{3}$ are symmetric to $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of the sides $B C, C A, A B$ respectively. Prove that the triangles $A_{2} B_{2} C_{2}$ and $A_{3} B_{3} C_{3}$ are similar.","['We will work with oriented angles between lines. For two straight lines $\\ell, m$ in the plane, $\\angle(\\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\\pi$ are identified, so that\n\n$$\n\\angle(\\ell, m)=-\\angle(m, \\ell), \\quad \\angle(\\ell, m)+\\angle(m, n)=\\angle(\\ell, n)\n$$\n\nIf $\\ell$ is the line through points $K, L$ and $m$ is the line through $M, N$, one writes $\\angle(K L, M N)$ for $\\angle(\\ell, m)$; the characters $K, L$ are freely interchangeable; and so are $M, N$.\n\nThe counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K, L, M, N$ are four noncollinear points in the plane then\n\n$$\nK, L, M, N \\text { are concyclic if and only if } \\angle(K M, L M)=\\angle(K N, L N) \\text {. }\\tag{1}\n$$\n\nPassing to the solution proper, we first show that the three circles $\\left(A B_{1} C_{1}\\right),\\left(B C_{1} A_{1}\\right)$, $\\left(C A_{1} B_{1}\\right)$ have a common point. So, let $\\left(A B_{1} C_{1}\\right)$ and $\\left(B C_{1} A_{1}\\right)$ intersect at the points $C_{1}$ and $P$. Then by $(1)$\n\n$$\n\\begin{gathered}\n\\angle\\left(P A_{1}, C A_{1}\\right)=\\angle\\left(P A_{1}, B A_{1}\\right)=\\angle\\left(P C_{1}, B C_{1}\\right) \\\\\n=\\angle\\left(P C_{1}, A C_{1}\\right)=\\angle\\left(P B_{1}, A B_{1}\\right)=\\angle\\left(P B_{1}, C B_{1}\\right) .\n\\end{gathered}\n$$\n\nDenote this angle by $\\varphi$.\n\nThe equality between the outer terms shows, again by (1), that the points $A_{1}, B_{1}, P, C$ are concyclic. Thus $P$ is the common point of the three mentioned circles.\n\nFrom now on the basic property (1) will be used without explicit reference. We have\n\n$$\n\\varphi=\\angle\\left(P A_{1}, B C\\right)=\\angle\\left(P B_{1}, C A\\right)=\\angle\\left(P C_{1}, A B\\right)\\tag{2}\n$$\n\n\n\n\n\nLet lines $A_{2} P, B_{2} P, C_{2} P$ meet the circle $(A B C)$ again at $A_{4}, B_{4}, C_{4}$, respectively. As\n\n$$\n\\angle\\left(A_{4} A_{2}, A A_{2}\\right)=\\angle\\left(P A_{2}, A A_{2}\\right)=\\angle\\left(P C_{1}, A C_{1}\\right)=\\angle\\left(P C_{1}, A B\\right)=\\varphi\n$$\n\nwe see that line $A_{2} A$ is the image of line $A_{2} A_{4}$ under rotation about $A_{2}$ by the angle $\\varphi$. Hence the point $A$ is the image of $A_{4}$ under rotation by $2 \\varphi$ about $O$, the centre of $(A B C)$. The same rotation sends $B_{4}$ to $B$ and $C_{4}$ to $C$. Triangle $A B C$ is the image of $A_{4} B_{4} C_{4}$ in this map. Thus\n\n$$\n\\angle\\left(A_{4} B_{4}, A B\\right)=\\angle\\left(B_{4} C_{4}, B C\\right)=\\angle\\left(C_{4} A_{4}, C A\\right)=2 \\varphi\n$$\n\nSince the rotation by $2 \\varphi$ about $O$ takes $B_{4}$ to $B$, we have $\\angle\\left(A B_{4}, A B\\right)=\\varphi$. Hence by $(2)$\n\n$$\n\\angle\\left(A B_{4}, P C_{1}\\right)=\\angle\\left(A B_{4}, A B\\right)+\\angle\\left(A B, P C_{1}\\right)=\\varphi+(-\\varphi)=0\n$$\n\nwhich means that $A B_{4} \\| P C_{1}$.\n\n\nLet $C_{5}$ be the intersection of lines $P C_{1}$ and $A_{4} B_{4}$; define $A_{5}, B_{5}$ analogously. So $A B_{4} \\| C_{1} C_{5}$ and, by (3) and (2),\n\n$$\n\\angle\\left(A_{4} B_{4}, P C_{1}\\right)=\\angle\\left(A_{4} B_{4}, A B\\right)+\\angle\\left(A B, P C_{1}\\right)=2 \\varphi+(-\\varphi)=\\varphi ;\n$$\n\ni.e., $\\angle\\left(B_{4} C_{5}, C_{5} C_{1}\\right)=\\varphi$. This combined with $\\angle\\left(C_{5} C_{1}, C_{1} A\\right)=\\angle\\left(P C_{1}, A B\\right)=\\varphi$ (see (2)) proves that the quadrilateral $A B_{4} C_{5} C_{1}$ is an isosceles trapezoid with $A C_{1}=B_{4} C_{5}$.\n\nInterchanging the roles of $A$ and $B$ we infer that also $B C_{1}=A_{4} C_{5}$. And since $A C_{1}+B C_{1}=$ $A B=A_{4} B_{4}$, it follows that the point $C_{5}$ lies on the line segment $A_{4} B_{4}$ and partitions it into segments $A_{4} C_{5}, B_{4} C_{5}$ of lengths $B C_{1}\\left(=A C_{3}\\right)$ and $A C_{1}$ (=BC $C_{3}$ ). In other words, the rotation which maps triangle $A_{4} B_{4} C_{4}$ onto $A B C$ carries $C_{5}$ onto $C_{3}$. Likewise, it sends $A_{5}$ to $A_{3}$ and $B_{5}$ to $B_{3}$. So the triangles $A_{3} B_{3} C_{3}$ and $A_{5} B_{5} C_{5}$ are congruent. It now suffices to show that the latter is similar to $A_{2} B_{2} C_{2}$.\n\nLines $B_{4} C_{5}$ and $P C_{5}$ coincide respectively with $A_{4} B_{4}$ and $P C_{1}$. Thus by (4)\n\n$$\n\\angle\\left(B_{4} C_{5}, P C_{5}\\right)=\\varphi\n$$\n\nAnalogously (by cyclic shift) $\\varphi=\\angle\\left(C_{4} A_{5}, P A_{5}\\right)$, which rewrites as\n\n$$\n\\varphi=\\angle\\left(B_{4} A_{5}, P A_{5}\\right)\n$$\n\n\n\nThese relations imply that the points $P, B_{4}, C_{5}, A_{5}$ are concyclic. Analogously, $P, C_{4}, A_{5}, B_{5}$ and $P, A_{4}, B_{5}, C_{5}$ are concyclic quadruples. Therefore\n\n$$\n\\angle\\left(A_{5} B_{5}, C_{5} B_{5}\\right)=\\angle\\left(A_{5} B_{5}, P B_{5}\\right)+\\angle\\left(P B_{5}, C_{5} B_{5}\\right)=\\angle\\left(A_{5} C_{4}, P C_{4}\\right)+\\angle\\left(P A_{4}, C_{5} A_{4}\\right) .\n$$\n\nOn the other hand, since the points $A_{2}, B_{2}, C_{2}, A_{4}, B_{4}, C_{4}$ all lie on the circle $(A B C)$, we have\n\n$$\n\\angle\\left(A_{2} B_{2}, C_{2} B_{2}\\right)=\\angle\\left(A_{2} B_{2}, B_{4} B_{2}\\right)+\\angle\\left(B_{4} B_{2}, C_{2} B_{2}\\right)=\\angle\\left(A_{2} A_{4}, B_{4} A_{4}\\right)+\\angle\\left(B_{4} C_{4}, C_{2} C_{4}\\right)\n$$\n\nBut the lines $A_{2} A_{4}, B_{4} A_{4}, B_{4} C_{4}, C_{2} C_{4}$ coincide respectively with $P A_{4}, C_{5} A_{4}, A_{5} C_{4}, P C_{4}$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\\angle\\left(A_{5} B_{5}, C_{5} B_{5}\\right)=\\angle\\left(A_{2} B_{2}, C_{2} B_{2}\\right)$. Hence (by cyclic shift, once more) also $\\angle\\left(B_{5} C_{5}, A_{5} C_{5}\\right)=\\angle\\left(B_{2} C_{2}, A_{2} C_{2}\\right)$ and $\\angle\\left(C_{5} A_{5}, B_{5} A_{5}\\right)=\\angle\\left(C_{2} A_{2}, B_{2} A_{2}\\right)$. This means that the triangles $A_{5} B_{5} C_{5}$ and $A_{2} B_{2} C_{2}$ have their corresponding angles equal, and consequently they are similar.']",,True,,, 1765,Geometry,,To each side $a$ of a convex polygon we assign the maximum area of a triangle contained in the polygon and having $a$ as one of its sides. Show that the sum of the areas assigned to all sides of the polygon is not less than twice the area of the polygon.,"['Lemma. Every convex $(2 n)$-gon, of area $S$, has a side and a vertex that jointly span a triangle of area not less than $S / n$.\n\nProof. By main diagonals of the (2n)-gon we shall mean those which partition the (2n)-gon into two polygons with equally many sides. For any side $b$ of the $(2 n)$-gon denote by $\\Delta_{b}$ the triangle $A B P$ where $A, B$ are the endpoints of $b$ and $P$ is the intersection point of the main diagonals $A A^{\\prime}, B B^{\\prime}$. We claim that the union of triangles $\\Delta_{b}$, taken over all sides, covers the whole polygon.\n\nTo show this, choose any side $A B$ and consider the main diagonal $A A^{\\prime}$ as a directed segment. Let $X$ be any point in the polygon, not on any main diagonal. For definiteness, let $X$ lie on the left side of the ray $A A^{\\prime}$. Consider the sequence of main diagonals $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}, \\ldots$, where $A, B, C, \\ldots$ are consecutive vertices, situated right to $A A^{\\prime}$.\n\nThe $n$-th item in this sequence is the diagonal $A^{\\prime} A$ (i.e. $A A^{\\prime}$ reversed), having $X$ on its right side. So there are two successive vertices $K, L$ in the sequence $A, B, C, \\ldots$ before $A^{\\prime}$ such that $X$ still lies to the left of $K K^{\\prime}$ but to the right of $L L^{\\prime}$. And this means that $X$ is in the triangle $\\Delta_{\\ell^{\\prime}}, \\ell^{\\prime}=K^{\\prime} L^{\\prime}$. Analogous reasoning applies to points $X$ on the right of $A A^{\\prime}$ (points lying on main diagonals can be safely ignored). Thus indeed the triangles $\\Delta_{b}$ jointly cover the whole polygon.\n\nThe sum of their areas is no less than $S$. So we can find two opposite sides, say $b=A B$ and $b^{\\prime}=A^{\\prime} B^{\\prime}$ (with $A A^{\\prime}, B B^{\\prime}$ main diagonals) such that $\\left[\\Delta_{b}\\right]+\\left[\\Delta_{b^{\\prime}}\\right] \\geq S / n$, where $[\\cdots]$ stands for the area of a region. Let $A A^{\\prime}, B B^{\\prime}$ intersect at $P$; assume without loss of generality that $P B \\geq P B^{\\prime}$. Then\n\n$$\n\\left[A B A^{\\prime}\\right]=[A B P]+\\left[P B A^{\\prime}\\right] \\geq[A B P]+\\left[P A^{\\prime} B^{\\prime}\\right]=\\left[\\Delta_{b}\\right]+\\left[\\Delta_{b^{\\prime}}\\right] \\geq S / n\n$$\n\nproving the lemma.\n\nNow, let $\\mathcal{P}$ be any convex polygon, of area $S$, with $m$ sides $a_{1}, \\ldots, a_{m}$. Let $S_{i}$ be the area of the greatest triangle in $\\mathcal{P}$ with side $a_{i}$. Suppose, contrary to the assertion, that\n\n$$\n\\sum_{i=1}^{m} \\frac{S_{i}}{S}<2\n$$\n\nThen there exist rational numbers $q_{1}, \\ldots, q_{m}$ such that $\\sum q_{i}=2$ and $q_{i}>S_{i} / S$ for each $i$.\n\nLet $n$ be a common denominator of the $m$ fractions $q_{1}, \\ldots, q_{m}$. Write $q_{i}=k_{i} / n$; so $\\sum k_{i}=2 n$. Partition each side $a_{i}$ of $\\mathcal{P}$ into $k_{i}$ equal segments, creating a convex (2n)-gon of area $S$ (with some angles of size $180^{\\circ}$ ), to which we apply the lemma. Accordingly, this refined polygon has a side $b$ and a vertex $H$ spanning a triangle $T$ of area $[T] \\geq S / n$. If $b$ is a piece of a side $a_{i}$ of $\\mathcal{P}$, then the triangle $W$ with base $a_{i}$ and summit $H$ has area\n\n$$\n[W]=k_{i} \\cdot[T] \\geq k_{i} \\cdot S / n=q_{i} \\cdot S>S_{i}\n$$\n\nin contradiction with the definition of $S_{i}$. This ends the proof.', 'As in the first solution, we allow again angles of size $180^{\\circ}$ at some vertices of the convex polygons considered.\n\nTo each convex $n$-gon $\\mathcal{P}=A_{1} A_{2} \\ldots A_{n}$ we assign a centrally symmetric convex $(2 n)$-gon $\\mathcal{Q}$ with side vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}, 1 \\leq i \\leq n$. The construction is as follows. Attach the $2 n$ vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}$ at a common origin and label them $\\overrightarrow{\\mathbf{b}_{1}}, \\overrightarrow{\\mathbf{b}_{2}}, \\ldots, \\overrightarrow{\\mathbf{b}_{2 n}}$ in counterclockwise direction; the choice of the first vector $\\overrightarrow{\\mathbf{b}_{1}}$ is irrelevant. The order of labelling is well-defined if $\\mathcal{P}$ has neither parallel sides nor angles equal to $180^{\\circ}$. Otherwise several collinear vectors with the same direction are labelled consecutively $\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, \\overrightarrow{\\mathbf{b}_{j+r}}$. One can assume that in such cases the respective opposite vectors occur in the order $-\\overrightarrow{\\mathbf{b}_{j}},-\\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots,-\\overrightarrow{\\mathbf{b}_{j+r}}$, ensuring that $\\overrightarrow{\\mathbf{b}_{j+n}}=-\\overrightarrow{\\mathbf{b}_{j}}$ for $j=1, \\ldots, 2 n$. Indices are taken cyclically here and in similar situations below.\n\nChoose points $B_{1}, B_{2}, \\ldots, B_{2 n}$ satisfying $\\overrightarrow{B_{j} B_{j+1}}=\\overrightarrow{\\mathbf{b}_{j}}$ for $j=1, \\ldots, 2 n$. The polygonal line $\\mathcal{Q}=B_{1} B_{2} \\ldots B_{2 n}$ is closed, since $\\sum_{j=1}^{2 n} \\overrightarrow{\\mathbf{b}_{j}}=\\overrightarrow{0}$. Moreover, $\\mathcal{Q}$ is a convex $(2 n)$-gon due to the arrangement of the vectors $\\overrightarrow{\\mathbf{b}_{j}}$, possibly with $180^{\\circ}$-angles. The side vectors of $\\mathcal{Q}$ are $\\pm \\overrightarrow{A_{i} A_{i+1}}$, $1 \\leq i \\leq n$. So in particular $\\mathcal{Q}$ is centrally symmetric, because it contains as side vectors $\\overrightarrow{A_{i} A_{i+1}}$ and $-\\overrightarrow{A_{i} A_{i+1}}$ for each $i=1, \\ldots, n$. Note that $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are opposite sides of $\\mathcal{Q}$, $1 \\leq j \\leq n$. We call $\\mathcal{Q}$ the associate of $\\mathcal{P}$.\n\nLet $S_{i}$ be the maximum area of a triangle with side $A_{i} A_{i+1}$ in $\\mathcal{P}, 1 \\leq i \\leq n$. We prove that\n\n$$\n\\left[B_{1} B_{2} \\ldots B_{2 n}\\right]=2 \\sum_{i=1}^{n} S_{i} \\tag{1}\n$$\n\nand\n\n$$\n\\left[B_{1} B_{2} \\ldots B_{2 n}\\right] \\geq 4\\left[A_{1} A_{2} \\ldots A_{n}\\right] \\tag{2}\n$$\n\nIt is clear that (1) and (2) imply the conclusion of the original problem.\n\nLemma. For a side $A_{i} A_{i+1}$ of $\\mathcal{P}$, let $h_{i}$ be the maximum distance from a point of $\\mathcal{P}$ to line $A_{i} A_{i+1}$, $i=1, \\ldots, n$. Denote by $B_{j} B_{j+1}$ the side of $\\mathcal{Q}$ such that $\\overrightarrow{A_{i} A_{i+1}}=\\overrightarrow{B_{j} B_{j+1}}$. Then the distance between $B_{j} B_{j+1}$ and its opposite side in $\\mathcal{Q}$ is equal to $2 h_{i}$.\n\nProof. Choose a vertex $A_{k}$ of $\\mathcal{P}$ at distance $h_{i}$ from line $A_{i} A_{i+1}$. Let $\\mathbf{u}$ be the unit vector perpendicular to $A_{i} A_{i+1}$ and pointing inside $\\mathcal{P}$. Denoting by $\\mathbf{x} \\cdot \\mathbf{y}$ the dot product of vectors $\\mathbf{x}$ and $\\mathbf{y}$, we have\n\n$$\nh=\\mathbf{u} \\cdot \\overrightarrow{A_{i} A_{k}}=\\mathbf{u} \\cdot\\left(\\overrightarrow{A_{i} A_{i+1}}+\\cdots+\\overrightarrow{A_{k-1} A_{k}}\\right)=\\mathbf{u} \\cdot\\left(\\overrightarrow{A_{i} A_{i-1}}+\\cdots+\\overrightarrow{A_{k+1} A_{k}}\\right)\n$$\n\nIn $\\mathcal{Q}$, the distance $H_{i}$ between the opposite sides $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ is given by\n\n$$\nH_{i}=\\mathbf{u} \\cdot\\left(\\overrightarrow{B_{j} B_{j+1}}+\\cdots+\\overrightarrow{B_{j+n-1} B_{j+n}}\\right)=\\mathbf{u} \\cdot\\left(\\overrightarrow{\\mathbf{b}_{j}}+\\overrightarrow{\\mathbf{b}_{j+1}}+\\cdots+\\overrightarrow{\\mathbf{b}_{j+n-1}}\\right) .\n$$\n\nThe choice of vertex $A_{k}$ implies that the $n$ consecutive vectors $\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, \\overrightarrow{\\mathbf{b}_{j+n-1}}$ are precisely $\\overrightarrow{A_{i} A_{i+1}}, \\ldots, \\overrightarrow{A_{k-1} A_{k}}$ and $\\overrightarrow{A_{i} A_{i-1}}, \\ldots, \\overrightarrow{A_{k+1} A_{k}}$, taken in some order. This implies $H_{i}=2 h_{i}$.\n\nFor a proof of (1), apply the lemma to each side of $\\mathcal{P}$. If $O$ the centre of $\\mathcal{Q}$ then, using the notation of the lemma,\n\n$$\n\\left[B_{j} B_{j+1} O\\right]=\\left[B_{j+n} B_{j+n+1} O\\right]=\\left[A_{i} A_{i+1} A_{k}\\right]=S_{i}\n$$\n\nSummation over all sides of $\\mathcal{P}$ yields (1).\n\nSet $d(\\mathcal{P})=[\\mathcal{Q}]-4[\\mathcal{P}]$ for a convex polygon $\\mathcal{P}$ with associate $\\mathcal{Q}$. Inequality $(2)$ means that $d(\\mathcal{P}) \\geq 0$ for each convex polygon $\\mathcal{P}$. The last inequality will be proved by induction on the\n\n\n\nnumber $\\ell$ of side directions of $\\mathcal{P}$, i. e. the number of pairwise nonparallel lines each containing a side of $\\mathcal{P}$.\n\nWe choose to start the induction with $\\ell=1$ as a base case, meaning that certain degenerate polygons are allowed. More exactly, we regard as degenerate convex polygons all closed polygonal lines of the form $X_{1} X_{2} \\ldots X_{k} Y_{1} Y_{2} \\ldots Y_{m} X_{1}$, where $X_{1}, X_{2}, \\ldots, X_{k}$ are points in this order on a line segment $X_{1} Y_{1}$, and so are $Y_{m}, Y_{m-1}, \\ldots, Y_{1}$. The initial construction applies to degenerate polygons; their associates are also degenerate, and the value of $d$ is zero. For the inductive step, consider a convex polygon $\\mathcal{P}$ which determines $\\ell$ side directions, assuming that $d(\\mathcal{P}) \\geq 0$ for polygons with smaller values of $\\ell$.\n\nSuppose first that $\\mathcal{P}$ has a pair of parallel sides, i. e. sides on distinct parallel lines. Let $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$ be such a pair, and let $A_{i} A_{i+1} \\leq A_{j} A_{j+1}$. Remove from $\\mathcal{P}$ the parallelogram $R$ determined by vectors $\\overrightarrow{A_{i} A_{i+1}}$ and $\\overrightarrow{A_{i} A_{j+1}}$. Two polygons are obtained in this way. Translating one of them by vector $\\overrightarrow{A_{i} A_{i+1}}$ yields a new convex polygon $\\mathcal{P}^{\\prime}$, of area $[\\mathcal{P}]-[R]$ and with value of $\\ell$ not exceeding the one of $\\mathcal{P}$. The construction just described will be called operation A.\n\n\nThe associate of $\\mathcal{P}^{\\prime}$ is obtained from $\\mathcal{Q}$ upon decreasing the lengths of two opposite sides by an amount of $2 A_{i} A_{i+1}$. By the lemma, the distance between these opposite sides is twice the distance between $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$. Thus operation $\\mathbf{A}$ decreases $[\\mathcal{Q}]$ by the area of a parallelogram with base and respective altitude twice the ones of $R$, i. e. by $4[R]$. Hence $\\mathbf{A}$ leaves the difference $d(\\mathcal{P})=[\\mathcal{Q}]-4[\\mathcal{P}]$ unchanged.\n\nNow, if $\\mathcal{P}^{\\prime}$ also has a pair of parallel sides, apply operation $\\mathbf{A}$ to it. Keep doing so with the subsequent polygons obtained for as long as possible. Now, A decreases the number $p$ of pairs of parallel sides in $\\mathcal{P}$. Hence its repeated applications gradually reduce $p$ to 0 , and further applications of $\\mathbf{A}$ will be impossible after several steps. For clarity, let us denote by $\\mathcal{P}$ again the polygon obtained at that stage.\n\nThe inductive step is complete if $\\mathcal{P}$ is degenerate. Otherwise $\\ell>1$ and $p=0$, i. e. there are no parallel sides in $\\mathcal{P}$. Observe that then $\\ell \\geq 3$. Indeed, $\\ell=2$ means that the vertices of $\\mathcal{P}$ all lie on the boundary of a parallelogram, implying $p>0$.\n\nFurthermore, since $\\mathcal{P}$ has no parallel sides, consecutive collinear vectors in the sequence $\\left(\\overrightarrow{\\mathbf{b}_{k}}\\right)$ (if any) correspond to consecutive $180^{\\circ}$-angles in $\\mathcal{P}$. Removing the vertices of such angles, we obtain a convex polygon with the same value of $d(\\mathcal{P})$.\n\nIn summary, if operation $\\mathbf{A}$ is impossible for a nondegenerate polygon $\\mathcal{P}$, then $\\ell \\geq 3$. In addition, one may assume that $\\mathcal{P}$ has no angles of size $180^{\\circ}$.\n\nThe last two conditions then also hold for the associate $\\mathcal{Q}$ of $\\mathcal{P}$, and we perform the following construction. Since $\\ell \\geq 3$, there is a side $B_{j} B_{j+1}$ of $\\mathcal{Q}$ such that the sum of the angles at $B_{j}$ and $B_{j+1}$ is greater than $180^{\\circ}$. (Such a side exists in each convex $k$-gon for $k>4$.) Naturally, $B_{j+n} B_{j+n+1}$ is a side with the same property. Extend the pairs of sides $B_{j-1} B_{j}, B_{j+1} B_{j+2}$\n\n\n\nand $B_{j+n-1} B_{j+n}, B_{j+n+1} B_{j+n+2}$ to meet at $U$ and $V$, respectively. Let $\\mathcal{Q}^{\\prime}$ be the centrally symmetric convex 2(n+1)-gon obtained from $\\mathcal{Q}$ by inserting $U$ and $V$ into the sequence $B_{1}, \\ldots, B_{2 n}$ as new vertices between $B_{j}, B_{j+1}$ and $B_{j+n}, B_{j+n+1}$, respectively. Informally, we adjoin to $\\mathcal{Q}$ the congruent triangles $B_{j} B_{j+1} U$ and $B_{j+n} B_{j+n+1} V$. Note that $B_{j}, B_{j+1}, B_{j+n}$ and $B_{j+n+1}$ are kept as vertices of $\\mathcal{Q}^{\\prime}$, although $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are no longer its sides.\n\nLet $A_{i} A_{i+1}$ be the side of $\\mathcal{P}$ such that $\\overrightarrow{A_{i} A_{i+1}}=\\overrightarrow{B_{j} B_{j+1}}=\\overrightarrow{\\mathbf{b}_{j}}$. Consider the point $W$ such that triangle $A_{i} A_{i+1} W$ is congruent to triangle $B_{j} B_{j+1} U$ and exterior to $\\mathcal{P}$. Insert $W$ into the sequence $A_{1}, A_{2}, \\ldots, A_{n}$ as a new vertex between $A_{i}$ and $A_{i+1}$ to obtain an $(n+1)$-gon $\\mathcal{P}^{\\prime}$. We claim that $\\mathcal{P}^{\\prime}$ is convex and its associate is $\\mathcal{Q}^{\\prime}$.\n\n\nVectors $\\overrightarrow{A_{i} W}$ and $\\overrightarrow{\\mathbf{b}_{j-1}}$ are collinear and have the same direction, as well as vectors $\\overrightarrow{W A_{i+1}}$ and $\\overrightarrow{\\mathbf{b}_{j+1}}$. Since $\\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}$ are consecutive terms in the sequence $\\left(\\overrightarrow{\\mathbf{b}_{k}}\\right)$, the angle inequalities $\\angle\\left(\\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{\\mathbf{b}_{j}}\\right) \\leq \\angle\\left(\\overrightarrow{A_{i-1} A_{i}}, \\overrightarrow{\\mathbf{b}_{j}}\\right)$ and $\\angle\\left(\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}\\right) \\leq \\angle\\left(\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{A_{i+1} A_{i+2}}\\right)$ hold true. They show that $\\mathcal{P}^{\\prime}$ is a convex polygon. To construct its associate, vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}= \\pm \\overrightarrow{\\mathbf{b}_{j}}$ must be deleted from the defining sequence $\\left(\\overrightarrow{\\mathbf{b}_{k}}\\right)$ of $\\mathcal{Q}$, and the vectors $\\pm \\overrightarrow{A_{i} W}, \\pm \\overrightarrow{W A_{i+1}}$ must be inserted appropriately into it. The latter can be done as follows:\n\n$$\n\\ldots, \\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{A_{i} W}, \\overrightarrow{W A_{i+1}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots,-\\overrightarrow{\\mathbf{b}_{j-1}},-\\overrightarrow{A_{i} W},-\\overrightarrow{W A_{i+1}},-\\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots\n$$\n\nThis updated sequence produces $\\mathcal{Q}^{\\prime}$ as the associate of $\\mathcal{P}^{\\prime}$.\n\nIt follows from the construction that $\\left[\\mathcal{P}^{\\prime}\\right]=[\\mathcal{P}]+\\left[A_{i} A_{i+1} W\\right]$ and $\\left[\\mathcal{Q}^{\\prime}\\right]=[\\mathcal{Q}]+2\\left[A_{i} A_{i+1} W\\right]$. Therefore $d\\left(\\mathcal{P}^{\\prime}\\right)=d(\\mathcal{P})-2\\left[A_{i} A_{i+1} W\\right]0$; without loss of generality confine attention to $y>0$. The equation rewritten as\n\n$$\n2^{x}\\left(1+2^{x+1}\\right)=(y-1)(y+1)\n$$\n\nshows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by 4 . Hence $x \\geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^{x}$. So\n\n$$\ny=2^{x-1} m+\\epsilon, \\quad m \\text { odd }, \\quad \\epsilon= \\pm 1\\tag{1}\n$$\n\nPlugging this into the original equation we obtain\n\n$$\n2^{x}\\left(1+2^{x+1}\\right)=\\left(2^{x-1} m+\\epsilon\\right)^{2}-1=2^{2 x-2} m^{2}+2^{x} m \\epsilon,\n$$\n\nor, equivalently\n\n$$\n1+2^{x+1}=2^{x-2} m^{2}+m \\epsilon .\n$$\n\nTherefore\n\n$$\n1-\\epsilon m=2^{x-2}\\left(m^{2}-8\\right) .\\tag{2}\n$$\n\nFor $\\epsilon=1$ this yields $m^{2}-8 \\leq 0$, i.e., $m=1$, which fails to satisfy (2).\n\nFor $\\epsilon=-1$ equation (2) gives us\n\n$$\n1+m=2^{x-2}\\left(m^{2}-8\\right) \\geq 2\\left(m^{2}-8\\right),\n$$\n\nimplying $2 m^{2}-m-17 \\leq 0$. Hence $m \\leq 3$; on the other hand $m$ cannot be 1 by $(2)$. Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From (1) we get $y=23$. These values indeed satisfy the given equation. Recall that then $y=-23$ is also good. Thus we have the complete list of solutions $(x, y):(0,2),(0,-2),(4,23),(4,-23)$.']","['$(0,2),(0,-2),(4,23),(4,-23)$']",True,,Tuple, 1767,Number Theory,,"For $x \in(0,1)$ let $y \in(0,1)$ be the number whose $n$th digit after the decimal point is the $\left(2^{n}\right)$ th digit after the decimal point of $x$. Show that if $x$ is rational then so is $y$.","[""Since $x$ is rational, its digits repeat periodically starting at some point. We wish to show that this is also true for the digits of $y$, implying that $y$ is rational.\n\nLet $d$ be the length of the period of $x$ and let $d=2^{u} \\cdot v$, where $v$ is odd. There is a positive integer $w$ such that\n\n$$\n2^{w} \\equiv 1 \\quad(\\bmod v)\n$$\n\n(For instance, one can choose $w$ to be $\\varphi(v)$, the value of Euler's function at $v$.) Therefore\n\n$$\n2^{n+w}=2^{n} \\cdot 2^{w} \\equiv 2^{n} \\quad(\\bmod v)\n$$\n\nfor each $n$. Also, for $n \\geq u$ we have\n\n$$\n2^{n+w} \\equiv 2^{n} \\equiv 0 \\quad\\left(\\bmod 2^{u}\\right) .\n$$\n\nIt follows that, for all $n \\geq u$, the relation\n\n$$\n2^{n+w} \\equiv 2^{n} \\quad(\\bmod d)\n$$\n\nholds. Thus, for $n$ sufficiently large, the $2^{n+w}$ th digit of $x$ is in the same spot in the cycle of $x$ as its $2^{n}$ th digit, and so these digits are equal. Hence the $(n+w)$ th digit of $y$ is equal to its $n$th digit. This means that the digits of $y$ repeat periodically with period $w$ from some point on, as required.""]",,True,,, 1768,Number Theory,,"The sequence $f(1), f(2), f(3), \ldots$ is defined by $$ f(n)=\frac{1}{n}\left(\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\cdots+\left\lfloor\frac{n}{n}\right\rfloor\right) $$ where $\lfloor x\rfloor$ denotes the integer part of $x$. (a) Prove that $f(n+1)>f(n)$ infinitely often. (b) Prove that $f(n+1)f(n)$ and $d(n+1)1, d(n) \\geq 2$ holds, with equality if and only if $n$ is prime. Since $f(6)=7 / 3>2$, it follows that $f(n)>2$ holds for all $n \\geq 6$.\n\nSince there are infinitely many primes, $d(n+1)=2$ holds for infinitely many values of $n$, and for each such $n \\geq 6$ we have $d(n+1)=2\\max \\{d(1), d(2), \\ldots, d(n)\\}$ for infinitely many $n$. For all such $n$, we have $d(n+1)>f(n)$. This completes the solution.']",,True,,, 1769,Number Theory,,"Let $P$ be a polynomial of degree $n>1$ with integer coefficients and let $k$ be any positive integer. Consider the polynomial $Q(x)=P(P(\ldots P(P(x)) \ldots))$, with $k$ pairs of parentheses. Prove that $Q$ has no more than $n$ integer fixed points, i.e. integers satisfying the equation $Q(x)=x$.","['The claim is obvious if every integer fixed point of $Q$ is a fixed point of $P$ itself. For the sequel assume that this is not the case. Take any integer $x_{0}$ such that $Q\\left(x_{0}\\right)=x_{0}$, $P\\left(x_{0}\\right) \\neq x_{0}$ and define inductively $x_{i+1}=P\\left(x_{i}\\right)$ for $i=0,1,2, \\ldots ;$ then $x_{k}=x_{0}$.\n\nIt is evident that\n\n$$\nP(u)-P(v) \\text { is divisible by } u-v \\text { for distinct integers } u, v \\text {. }\\tag{1}\n$$\n\n(Indeed, if $P(x)=\\sum a_{i} x^{i}$ then each $a_{i}\\left(u^{i}-v^{i}\\right)$ is divisible by $u-v$.) Therefore each term in the chain of (nonzero) differences\n\n$$\nx_{0}-x_{1}, \\quad x_{1}-x_{2}, \\quad \\ldots, \\quad x_{k-1}-x_{k}, \\quad x_{k}-x_{k+1}\\tag{2}\n$$\n\nis a divisor of the next one; and since $x_{k}-x_{k+1}=x_{0}-x_{1}$, all these differences have equal absolute values. For $x_{m}=\\min \\left(x_{1}, \\ldots, x_{k}\\right)$ this means that $x_{m-1}-x_{m}=-\\left(x_{m}-x_{m+1}\\right)$. Thus $x_{m-1}=x_{m+1}\\left(\\neq x_{m}\\right)$. It follows that consecutive differences in the sequence (2) have opposite signs. Consequently, $x_{0}, x_{1}, x_{2}, \\ldots$ is an alternating sequence of two distinct values. In other words, every integer fixed point of $Q$ is a fixed point of the polynomial $P(P(x))$. Our task is to prove that there are at most $n$ such points.\n\nLet $a$ be one of them so that $b=P(a) \\neq a$ (we have assumed that such an $a$ exists); then $a=P(b)$. Take any other integer fixed point $\\alpha$ of $P(P(x))$ and let $P(\\alpha)=\\beta$, so that $P(\\beta)=\\alpha$; the numbers $\\alpha$ and $\\beta$ need not be distinct ( $\\alpha$ can be a fixed point of $P$ ), but each of $\\alpha, \\beta$ is different from each of $a, b$. Applying property (1) to the four pairs of integers $(\\alpha, a),(\\beta, b)$, $(\\alpha, b),(\\beta, a)$ we get that the numbers $\\alpha-a$ and $\\beta-b$ divide each other, and also $\\alpha-b$ and $\\beta-a$ divide each other. Consequently\n\n$$\n\\alpha-b= \\pm(\\beta-a), \\quad \\alpha-a= \\pm(\\beta-b)\\tag{3}\n$$\n\nSuppose we have a plus in both instances: $\\alpha-b=\\beta-a$ and $\\alpha-a=\\beta-b$. Subtraction yields $a-b=b-a$, a contradiction, as $a \\neq b$. Therefore at least one equality in (3) holds with a minus sign. For each of them this means that $\\alpha+\\beta=a+b$; equivalently $a+b-\\alpha-P(\\alpha)=0$.\n\nDenote $a+b$ by $C$. We have shown that every integer fixed point of $Q$ other that $a$ and $b$ is a root of the polynomial $F(x)=C-x-P(x)$. This is of course true for $a$ and $b$ as well. And since $P$ has degree $n>1$, the polynomial $F$ has the same degree, so it cannot have more than $n$ roots. Hence the result.']",,True,,, 1770,Number Theory,,"Find all integer solutions of the equation $$ \frac{x^{7}-1}{x-1}=y^{5}-1 $$","[""The equation has no integer solutions. To show this, we first prove a lemma.\n\nLemma. If $x$ is an integer and $p$ is a prime divisor of $\\frac{x^{7}-1}{x-1}$ then either $p \\equiv 1(\\bmod 7)$ or $p=7$.\n\nProof. Both $x^{7}-1$ and $x^{p-1}-1$ are divisible by $p$, by hypothesis and by Fermat's little theorem, respectively. Suppose that 7 does not divide $p-1$. Then $\\operatorname{gcd}(p-1,7)=1$, so there exist integers $k$ and $m$ such that $7 k+(p-1) m=1$. We therefore have\n\n$$\nx \\equiv x^{7 k+(p-1) m} \\equiv\\left(x^{7}\\right)^{k} \\cdot\\left(x^{p-1}\\right)^{m} \\equiv 1 \\quad(\\bmod p)\n$$\n\nand so\n\n$$\n\\frac{x^{7}-1}{x-1}=1+x+\\cdots+x^{6} \\equiv 7 \\quad(\\bmod p)\n$$\n\nIt follows that $p$ divides 7 , hence $p=7$ must hold if $p \\equiv 1(\\bmod 7)$ does not, as stated.\n\nThe lemma shows that each positive divisor $d$ of $\\frac{x^{7}-1}{x-1}$ satisfies either $d \\equiv 0(\\bmod 7)$ or $d \\equiv 1(\\bmod 7)$.\n\nNow assume that $(x, y)$ is an integer solution of the original equation. Notice that $y-1>0$, because $\\frac{x^{7}-1}{x-1}>0$ for all $x \\neq 1$. Since $y-1$ divides $\\frac{x^{7}-1}{x-1}=y^{5}-1$, we have $y \\equiv 1(\\bmod 7)$ or $y \\equiv 2(\\bmod 7)$ by the previous paragraph. In the first case, $1+y+y^{2}+y^{3}+y^{4} \\equiv 5(\\bmod 7)$, and in the second $1+y+y^{2}+y^{3}+y^{4} \\equiv 3(\\bmod 7)$. Both possibilities contradict the fact that the positive divisor $1+y+y^{2}+y^{3}+y^{4}$ of $\\frac{x^{7}-1}{x-1}$ is congruent to 0 or 1 modulo 7 . So the given equation has no integer solutions.""]",['the given equation has no integer solutions'],False,,Need_human_evaluate, 1771,Number Theory,,"Let $a>b>1$ be relatively prime positive integers. Define the weight of an integer $c$, denoted by $w(c)$, to be the minimal possible value of $|x|+|y|$ taken over all pairs of integers $x$ and $y$ such that $$ a x+b y=c . $$ An integer $c$ is called a local champion if $w(c) \geq w(c \pm a)$ and $w(c) \geq w(c \pm b)$. Find all local champions and determine their number.","[""Call the pair of integers $(x, y)$ a representation of $c$ if $a x+b y=c$ and $|x|+|y|$ has the smallest possible value, i.e. $|x|+|y|=w(c)$.\n\nWe characterise the local champions by the following three observations.\n\nLemma 1. If $(x, y)$ a representation of a local champion $c$ then $x y<0$.\n\nProof. Suppose indirectly that $x \\geq 0$ and $y \\geq 0$ and consider the values $w(c)$ and $w(c+a)$. All representations of the numbers $c$ and $c+a$ in the form $a u+b v$ can be written as\n\n$$\nc=a(x-k b)+b(y+k a), \\quad c+a=a(x+1-k b)+b(y+k a)\n$$\n\nwhere $k$ is an arbitrary integer.\n\nSince $|x|+|y|$ is minimal, we have\n\n$$\nx+y=|x|+|y| \\leq|x-k b|+|y+k a|\n$$\n\nfor all $k$. On the other hand, $w(c+a) \\leq w(c)$, so there exists a $k$ for which\n\n$$\n|x+1-k b|+|y+k a| \\leq|x|+|y|=x+y .\n$$\n\nThen\n\n$$\n(x+1-k b)+(y+k a) \\leq|x+1-k b|+|y+k a| \\leq x+y \\leq|x-k b|+|y+k a| \\text {. }\n$$\n\nComparing the first and the third expressions, we find $k(a-b)+1 \\leq 0$ implying $k<0$. Comparing the second and fourth expressions, we get $|x+1-k b| \\leq|x-k b|$, therefore $k b>x$; this is a contradiction.\n\nIf $x, y \\leq 0$ then we can switch to $-c,-x$ and $-y$.\n\nFrom this point, write $c=a x-b y$ instead of $c=a x+b y$ and consider only those cases where $x$ and $y$ are nonzero and have the same sign. By Lemma 1, there is no loss of generality in doing so.\n\nLemma 2. Let $c=a x-b y$ where $|x|+|y|$ is minimal and $x, y$ have the same sign. The number $c$ is a local champion if and only if $|x|0$.\n\nThe numbers $c-a$ and $c+b$ can be written as\n\n$$\nc-a=a(x-1)-b y \\quad \\text { and } \\quad c+b=a x-b(y-1)\n$$\n\nand trivially $w(c-a) \\leq(x-1)+y0$. We prove that we can choose $k=1$.\n\nConsider the function $f(t)=|x+1-b t|+|y-a t|-(x+y)$. This is a convex function and we have $f(0)=1$ and $f(k) \\leq 0$. By Jensen's inequality, $f(1) \\leq\\left(1-\\frac{1}{k}\\right) f(0)+\\frac{1}{k} f(k)<1$. But $f(1)$ is an integer. Therefore $f(1) \\leq 0$ and\n\n$$\n|x+1-b|+|y-a| \\leq x+y\n$$\n\nKnowing $c=a(x-b)-b(y-a)$, we also have\n\n$$\nx+y \\leq|x-b|+|y-a| \\text {. }\n$$\n\nCombining the two inequalities yields $|x+1-b| \\leq|x-b|$ which is equivalent to $xb$, we also have $0N$ and\n\n$$\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d)\n$$\n\nThis yields the claim for $m=b_{0}$.\n\nThe base case $d=1$ is trivial. Take an $a>1$ and assume that the statement holds for all $d\\max \\left(2^{M}, N\\right)$ and\n\n$$\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d) \\quad \\text { for } \\quad i=0,1,2, \\ldots, d-1 \\tag{1}\n$$\n\nFor each $i=0,1, \\ldots, d-1$ consider the sequence\n\n$$\n2^{b_{i}}+b_{i}, \\quad 2^{b_{i}+k}+\\left(b_{i}+k\\right), \\ldots, 2^{b_{i}+\\left(a^{\\prime}-1\\right) k}+\\left(b_{i}+\\left(a^{\\prime}-1\\right) k\\right)\\tag{2}\n$$\n\nModulo $a$, these numbers are congruent to\n\n$$\n2^{b_{i}}+b_{i}, 2^{b_{i}}+\\left(b_{i}+k\\right), \\ldots, 2^{b_{i}}+\\left(b_{i}+\\left(a^{\\prime}-1\\right) k\\right),\n$$\n\nrespectively. The $d$ sequences contain $a^{\\prime} d=a$ numbers altogether. We shall now prove that no two of these numbers are congruent modulo $a$.\n\nSuppose that\n\n$$\n2^{b_{i}}+\\left(b_{i}+m k\\right) \\equiv 2^{b_{j}}+\\left(b_{j}+n k\\right) \\quad(\\bmod a)\\tag{3}\n$$\n\nfor some values of $i, j \\in\\{0,1, \\ldots, d-1\\}$ and $m, n \\in\\left\\{0,1, \\ldots, a^{\\prime}-1\\right\\}$. Since $d$ is a divisor of $a$, we also have\n\n$$\n2^{b_{i}}+\\left(b_{i}+m k\\right) \\equiv 2^{b_{j}}+\\left(b_{j}+n k\\right) \\quad(\\bmod d) .\n$$\n\nBecause $d$ is a divisor of $k$ and in view of $(1)$, we obtain $i \\equiv j(\\bmod d)$. As $i, j \\in\\{0,1, \\ldots, d-1\\}$, this just means that $i=j$. Substituting this into (3) yields $m k \\equiv n k(\\bmod a)$. Therefore $m k^{\\prime} \\equiv n k^{\\prime}\\left(\\bmod a^{\\prime}\\right)$; and since $a^{\\prime}$ and $k^{\\prime}$ are coprime, we get $m \\equiv n\\left(\\bmod a^{\\prime}\\right)$. Hence also $m=n$\n\nIt follows that the $a$ numbers that make up the $d$ sequences (2) satisfy all the requirements; they are certainly all greater than $N$ because we chose each $b_{i}>\\max \\left(2^{M}, N\\right)$. So the statement holds for $a$, completing the induction.']",,True,,, 1773,Algebra,,"Let $n$ be an integer, and let $A$ be a subset of $\left\{0,1,2,3, \ldots, 5^{n}\right\}$ consisting of $4 n+2$ numbers. Prove that there exist $a, b, c \in A$ such that $a3 b$.","['(By contradiction) Suppose that there exist $4 n+2$ non-negative integers $x_{0}<$ $x_{1}<\\cdots5^{n} \\cdot 1\n$$', 'Denote the maximum element of $A$ by $c$. For $k=0, \\ldots, 4 n-1$ let\n\n$$\nA_{k}=\\left\\{x \\in A:\\left(1-(2 / 3)^{k}\\right) c \\leqslant x<\\left(1-(2 / 3)^{k+1}\\right) c\\right\\} \\text {. }\n$$\n\nNote that\n\n$$\n\\left(1-(2 / 3)^{4 n}\\right) c=c-(16 / 81)^{n} c>c-(1 / 5)^{n} c \\geqslant c-1 \\text {, }\n$$\n\nwhich shows that the sets $A_{0}, A_{1}, \\ldots, A_{4 n-1}$ form a partition of $A \\backslash\\{c\\}$. Since $A \\backslash\\{c\\}$ has $4 n+1$ elements, by the pigeonhole principle some set $A_{k}$ does contain at least two elements of $A \\backslash\\{c\\}$. Denote these two elements $a$ and $b$ and assume $a3 b\n$$\n\nas desired.']",,True,,, 1774,Algebra,,"For every integer $n \geqslant 1$ consider the $n \times n$ table with entry $\left\lfloor\frac{i j}{n+1}\right\rfloor$ at the intersection of row $i$ and column $j$, for every $i=1, \ldots, n$ and $j=1, \ldots, n$. Determine all integers $n \geqslant 1$ for which the sum of the $n^{2}$ entries in the table is equal to $\frac{1}{4} n^{2}(n-1)$.","['First, observe that every pair $x, y$ of real numbers for which the sum $x+y$ is integer satisfies\n\n$$\n\\lfloor x\\rfloor+\\lfloor y\\rfloor \\geqslant x+y-1\n\\tag{1}\n$$\n\nThe inequality is strict if $x$ and $y$ are integers, and it holds with equality otherwise.\n\nWe estimate the sum $S$ as follows.\n\n$$\n\\begin{array}{r}\n2 S=\\sum_{1 \\leqslant i, j \\leqslant n}\\left(\\left\\lfloor\\frac{i j}{n+1}\\right\\rfloor+\\left\\lfloor\\frac{i j}{n+1}\\right\\rfloor\\right)=\\sum_{1 \\leqslant i, j \\leqslant n}\\left(\\left\\lfloor\\frac{i j}{n+1}\\right\\rfloor+\\left\\lfloor\\frac{(n+1-i) j}{n+1}\\right\\rfloor\\right) \\\\\n\\geqslant \\sum_{1 \\leqslant i, j \\leqslant n}(j-1)=\\frac{(n-1) n^{2}}{2} .\n\\end{array}\n$$\n\nThe inequality in the last line follows from (1) by setting $x=i j /(n+1)$ and $y=(n+1-$ i) $j /(n+1)$, so that $x+y=j$ is integral.\n\nNow $S=\\frac{1}{4} n^{2}(n-1)$ if and only if the inequality in the last line holds with equality, which means that none of the values $i j /(n+1)$ with $1 \\leqslant i, j \\leqslant n$ may be integral.\n\nHence, if $n+1$ is composite with factorisation $n+1=a b$ for $2 \\leqslant a, b \\leqslant n$, one gets a strict inequality for $i=a$ and $j=b$. If $n+1$ is a prime, then $i j /(n+1)$ is never integral and $S=\\frac{1}{4} n^{2}(n-1)$.', 'To simplify the calculation with indices, extend the table by adding a phantom column of index 0 with zero entries (which will not change the sum of the table). Fix a row $i$ with $1 \\leqslant i \\leqslant n$, and let $d:=\\operatorname{gcd}(i, n+1)$ and $k:=(n+1) / d$. For columns $j=0, \\ldots, n$, define the remainder $r_{j}:=i j \\bmod (n+1)$. We first prove the following\n\nClaim. For every integer $g$ with $1 \\leqslant g \\leqslant d$, the remainders $r_{j}$ with indices $j$ in the range\n\n$$\n(g-1) k \\leqslant j \\leqslant g k-1\n\\tag{2}\n$$\n\nform a permutation of the $k$ numbers $0 \\cdot d, 1 \\cdot d, 2 \\cdot d, \\ldots,(k-1) \\cdot d$.\n\nProof. If $r_{j^{\\prime}}=r_{j}$ holds for two indices $j^{\\prime}$ and $j$ in (2), then $i\\left(j^{\\prime}-j\\right) \\equiv 0 \\bmod (n+1)$, so that $j^{\\prime}-j$ is a multiple of $k$; since $\\left|j^{\\prime}-j\\right| \\leqslant k-1$, this implies $j^{\\prime}=j$. Hence, the $k$ remainders are pairwise distinct. Moreover, each remainder $r_{j}=i j \\bmod (n+1)$ is a multiple of $d=\\operatorname{gcd}(i, n+1)$. This proves the claim.\n\nWe then have\n\n$$\n\\sum_{j=0}^{n} r_{j}=\\sum_{g=1}^{d} \\sum_{\\ell=0}^{(n+1) / d-1} \\ell d=d^{2} \\cdot \\frac{1}{2}\\left(\\frac{n+1}{d}-1\\right) \\frac{n+1}{d}=\\frac{(n+1-d)(n+1)}{2}\n\\tag{3}\n$$\n\nBy using (3), compute the sum $S_{i}$ of row $i$ as follows:\n\n$$\n\\begin{aligned}\nS_{i}=\\sum_{j=0}^{n}\\left\\lfloor\\frac{i j}{n+1}\\right\\rfloor & =\\sum_{j=0}^{n} \\frac{i j-r_{j}}{n+1}=\\frac{i}{n+1} \\sum_{j=0}^{n} j-\\frac{1}{n+1} \\sum_{j=0}^{n} r_{j} \\\\\n& =\\frac{i}{n+1} \\cdot \\frac{n(n+1)}{2}-\\frac{1}{n+1} \\cdot \\frac{(n+1-d)(n+1)}{2}=\\frac{(i n-n-1+d)}{2} .\n\\end{aligned}\n\\tag{4}\n$$\n\n\n\nEquation (4) yields the following lower bound on the row sum $S_{i}$, which holds with equality if and only if $d=\\operatorname{gcd}(i, n+1)=1$ :\n\n$$\nS_{i} \\geqslant \\frac{(i n-n-1+1)}{2}=\\frac{n(i-1)}{2}\n\\tag{5}\n$$\n\nBy summing up the bounds (5) for the rows $i=1, \\ldots, n$, we get the following lower bound for the sum of all entries in the table\n\n$$\n\\sum_{i=1}^{n} S_{i} \\geqslant \\sum_{i=1}^{n} \\frac{n}{2}(i-1)=\\frac{n^{2}(n-1)}{4}\n\\tag{6}\n$$\n\nIn (6) we have equality if and only if equality holds in (5) for each $i=1, \\ldots, n$, which happens if and only if $\\operatorname{gcd}(i, n+1)=1$ for each $i=1, \\ldots, n$, which is equivalent to the fact that $n+1$ is a prime. Thus the sum of the table entries is $\\frac{1}{4} n^{2}(n-1)$ if and only if $n+1$ is a prime.']",['All integers $n$ for which $n+1$ is a prime'],False,,Need_human_evaluate, 1775,Algebra,,"Given a positive integer $n$, find the smallest value of $\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{n}}{n}\right\rfloor$ over all permutations $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $(1,2, \ldots, n)$.","['Suppose that $2^{k} \\leqslant n<2^{k+1}$ with some nonnegative integer $k$. First we show a permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ such that $\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor=k+1$; then we will prove that $\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor \\geqslant k+1$ for every permutation. Hence, the minimal possible value will be $k+1$.\n\nI. Consider the permutation\n\n$$\n\\begin{gathered}\n\\left(a_{1}\\right)=(1), \\quad\\left(a_{2}, a_{3}\\right)=(3,2), \\quad\\left(a_{4}, a_{5}, a_{6}, a_{7}\\right)=(7,4,5,6), \\\\\n\\left(a_{2^{k-1}}, \\ldots, a_{2^{k}-1}\\right)=\\left(2^{k}-1,2^{k-1}, 2^{k-1}+1, \\ldots, 2^{k}-2\\right), \\\\\n\\left(a_{2^{k}}, \\ldots, a_{n}\\right)=\\left(n, 2^{k}, 2^{k}+1, \\ldots, n-1\\right) .\n\\end{gathered}\n$$\n\nThis permutation consists of $k+1$ cycles. In every cycle $\\left(a_{p}, \\ldots, a_{q}\\right)=(q, p, p+1, \\ldots, q-1)$ we have $q<2 p$, so\n\n$$\n\\sum_{i=p}^{q}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor=\\left\\lfloor\\frac{q}{p}\\right\\rfloor+\\sum_{i=p+1}^{q}\\left\\lfloor\\frac{i-1}{i}\\right\\rfloor=1\n$$\n\nThe total sum over all cycles is precisely $k+1$.\n\nII. In order to establish the lower bound, we prove a more general statement.\n\nClaim. If $b_{1}, \\ldots, b_{2^{k}}$ are distinct positive integers then\n\n$$\n\\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant k+1\n$$\n\nFrom the Claim it follows immediately that $\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant k+1$.\n\nProof of the Claim. Apply induction on $k$. For $k=1$ the claim is trivial, $\\left\\lfloor\\frac{b_{1}}{1}\\right\\rfloor \\geqslant 1$. Suppose the Claim holds true for some positive integer $k$, and consider $k+1$.\n\nIf there exists an index $j$ such that $2^{k}k\n$$\n\nAs the left-hand side is an integer, it must be at least $k+1$.']",['$\\left\\lfloor\\log _{2} n\\right\\rfloor+1$'],False,,Expression, 1776,Algebra,,"Show that for all real numbers $x_{1}, \ldots, x_{n}$ the following inequality holds: $$ \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\left|x_{i}-x_{j}\right|} \leqslant \sum_{i=1}^{n} \sum_{j=1}^{n} \sqrt{\left|x_{i}+x_{j}\right|} $$","['If we add $t$ to all the variables then the left-hand side remains constant and the right-hand side becomes\n\n$$\nH(t):=\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{\\left|x_{i}+x_{j}+2 t\\right|}\n$$\n\nLet $T$ be large enough such that both $H(-T)$ and $H(T)$ are larger than the value $L$ of the lefthand side of the inequality we want to prove. Not necessarily distinct points $p_{i, j}:=-\\left(x_{i}+x_{j}\\right) / 2$ together with $T$ and $-T$ split the real line into segments and two rays such that on each of these segments and rays the function $H(t)$ is concave since $f(t):=\\sqrt{|\\ell+2 t|}$ is concave on both intervals $(-\\infty,-\\ell / 2]$ and $[-\\ell / 2,+\\infty)$. Let $[a, b]$ be the segment containing zero. Then concavity implies $H(0) \\geqslant \\min \\{H(a), H(b)\\}$ and, since $H( \\pm T)>L$, it suffices to prove the inequalities $H\\left(-\\left(x_{i}+x_{j}\\right) / 2\\right) \\geqslant L$, that is to prove the original inequality in the case when all numbers are shifted in such a way that two variables $x_{i}$ and $x_{j}$ add up to zero. In the following we denote the shifted variables still by $x_{i}$.\n\nIf $i=j$, i.e. $x_{i}=0$ for some index $i$, then we can remove $x_{i}$ which will decrease both sides by $2 \\sum_{k} \\sqrt{\\left|x_{k}\\right|}$. Similarly, if $x_{i}+x_{j}=0$ for distinct $i$ and $j$ we can remove both $x_{i}$ and $x_{j}$ which decreases both sides by\n\n$$\n2 \\sqrt{2\\left|x_{i}\\right|}+2 \\cdot \\sum_{k \\neq i, j}\\left(\\sqrt{\\left|x_{k}+x_{i}\\right|}+\\sqrt{\\left|x_{k}+x_{j}\\right|}\\right)\n$$\n\nIn either case we reduced our inequality to the case of smaller $n$. It remains to note that for $n=0$ and $n=1$ the inequality is trivial.', 'For real $p$ consider the integral\n\n$$\nI(p)=\\int_{0}^{\\infty} \\frac{1-\\cos (p x)}{x \\sqrt{x}} d x\n$$\n\nwhich clearly converges to a strictly positive number. By changing the variable $y=|p| x$ one notices that $I(p)=\\sqrt{|p|} I(1)$. Hence, by using the trigonometric formula $\\cos (\\alpha-\\beta)-\\cos (\\alpha+$ $\\beta)=2 \\sin \\alpha \\sin \\beta$ we obtain\n\n$\\sqrt{|a+b|}-\\sqrt{|a-b|}=\\frac{1}{I(1)} \\int_{0}^{\\infty} \\frac{\\cos ((a-b) x)-\\cos ((a+b) x)}{x \\sqrt{x}} d x=\\frac{1}{I(1)} \\int_{0}^{\\infty} \\frac{2 \\sin (a x) \\sin (b x)}{x \\sqrt{x}} d x$,\n\nfrom which our inequality immediately follows:\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{\\left|x_{i}+x_{j}\\right|}-\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{\\left|x_{i}-x_{j}\\right|}=\\frac{2}{I(1)} \\int_{0}^{\\infty} \\frac{\\left(\\sum_{i=1}^{n} \\sin \\left(x_{i} x\\right)\\right)^{2}}{x \\sqrt{x}} d x \\geqslant 0\n$$']",,True,,, 1777,Algebra,,"Let $n \geqslant 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove that $$ \sum_{k=1}^{n} \frac{a_{k}}{1-a_{k}}\left(a_{1}+a_{2}+\cdots+a_{k-1}\right)^{2}<\frac{1}{3} $$","['For all $k \\leqslant n$, let\n\n$$\ns_{k}=a_{1}+a_{2}+\\cdots+a_{k} \\quad \\text { and } \\quad b_{k}=\\frac{a_{k} s_{k-1}^{2}}{1-a_{k}}\n$$\n\nwith the convention that $s_{0}=0$. Note that $b_{k}$ is exactly a summand in the sum we need to estimate. We shall prove the inequality\n\n$$\nb_{k}<\\frac{s_{k}^{3}-s_{k-1}^{3}}{3}\n\\tag{1}\n$$\n\nIndeed, it suffices to check that\n\n$$\n\\begin{aligned}\n(1) & \\Longleftrightarrow 0<\\left(1-a_{k}\\right)\\left(\\left(s_{k-1}+a_{k}\\right)^{3}-s_{k-1}^{3}\\right)-3 a_{k} s_{k-1}^{2} \\\\\n& \\Longleftrightarrow 0<\\left(1-a_{k}\\right)\\left(3 s_{k-1}^{2}+3 s_{k-1} a_{k}+a_{k}^{2}\\right)-3 s_{k-1}^{2} \\\\\n& \\Longleftrightarrow 0<-3 a_{k} s_{k-1}^{2}+3\\left(1-a_{k}\\right) s_{k-1} a_{k}+\\left(1-a_{k}\\right) a_{k}^{2} \\\\\n& \\Longleftrightarrow 0<3\\left(1-a_{k}-s_{k-1}\\right) s_{k-1} a_{k}+\\left(1-a_{k}\\right) a_{k}^{2}\n\\end{aligned}\n$$\n\nwhich holds since $a_{k}+s_{k-1}=s_{k} \\leqslant 1$ and $a_{k} \\in(0,1)$.\n\nThus, adding inequalities (1) for $k=1, \\ldots, n$, we conclude that\n\n$$\nb_{1}+b_{2}+\\cdots+b_{n}<\\frac{s_{n}^{3}-s_{0}^{3}}{3}=\\frac{1}{3}\n$$\n\nas desired.', 'First, let us define\n\n$$\nS\\left(a_{1}, \\ldots, a_{n}\\right):=\\sum_{k=1}^{n} \\frac{a_{k}}{1-a_{k}}\\left(a_{1}+a_{2}+\\cdots+a_{k-1}\\right)^{2}\n$$\n\n\n\nFor some index $i$, denote $a_{1}+\\cdots+a_{i-1}$ by $s$. If we replace $a_{i}$ with two numbers $a_{i} / 2$ and $a_{i} / 2$, i.e. replace the tuple $\\left(a_{1}, \\ldots, a_{n}\\right)$ with $\\left(a_{1}, \\ldots, a_{i-1}, a_{i} / 2, a_{i} / 2, a_{i+1}, \\ldots, a_{n}\\right)$, the sum will increase by\n\n$$\n\\begin{aligned}\nS\\left(a_{1}, \\ldots, a_{i} / 2, a_{i} / 2, \\ldots, a_{n}\\right)-S\\left(a_{1}, \\ldots, a_{n}\\right) & =\\frac{a_{i} / 2}{1-a_{i} / 2}\\left(s^{2}+\\left(s+a_{i} / 2\\right)^{2}\\right)-\\frac{a_{i}}{1-a_{i}} s^{2} \\\\\n& =a_{i} \\frac{\\left(1-a_{i}\\right)\\left(2 s^{2}+s a_{i}+a_{i}^{2} / 4\\right)-\\left(2-a_{i}\\right) s^{2}}{\\left(2-a_{i}\\right)\\left(1-a_{i}\\right)} \\\\\n& =a_{i} \\frac{\\left(1-a_{i}-s\\right) s a_{i}+\\left(1-a_{i}\\right) a_{i}^{2} / 4}{\\left(2-a_{i}\\right)\\left(1-a_{i}\\right)},\n\\end{aligned}\n$$\n\nwhich is strictly positive. So every such replacement strictly increases the sum. By repeating this process and making maximal number in the tuple tend to zero, we keep increasing the sum which will converge to\n\n$$\n\\int_{0}^{1} x^{2} d x=\\frac{1}{3}\n$$\n\nThis completes the proof.', 'We sketch a probabilistic version of the first solution. Let $x_{1}, x_{2}, x_{3}$ be drawn uniformly and independently at random from the segment [0,1]. Let $I_{1} \\cup I_{2} \\cup \\cdots \\cup I_{n}$ be a partition of $[0,1]$ into segments of length $a_{1}, a_{2}, \\ldots, a_{n}$ in this order. Let $J_{k}:=I_{1} \\cup \\cdots \\cup I_{k-1}$ for $k \\geqslant 2$ and $J_{1}:=\\varnothing$. Then\n\n$$\n\\begin{aligned}\n\\frac{1}{3}= & \\sum_{k=1}^{n} \\mathbb{P}\\left\\{x_{1} \\geqslant x_{2}, x_{3} ; x_{1} \\in I_{k}\\right\\} \\\\\n= & \\sum_{k=1}^{n}\\left(\\mathbb{P}\\left\\{x_{1} \\in I_{k} ; x_{2}, x_{3} \\in J_{k}\\right\\}+2 \\cdot \\mathbb{P}\\left\\{x_{1} \\geqslant x_{2} ; x_{1}, x_{2} \\in I_{k} ; x_{3} \\in J_{k}\\right\\}\\right. \\\\\n& \\left.\\quad+\\mathbb{P}\\left\\{x_{1} \\geqslant x_{2}, x_{3} ; x_{1}, x_{2}, x_{3} \\in I_{k}\\right\\}\\right) \\\\\n= & \\sum_{k=1}^{n}\\left(a_{k}\\left(a_{1}+\\cdots+a_{k-1}\\right)^{2}+2 \\cdot \\frac{a_{k}^{2}}{2} \\cdot\\left(a_{1}+\\cdots+a_{k-1}\\right)+\\frac{a_{k}^{3}}{3}\\right) \\\\\n> & \\sum_{k=1}^{n}\\left(a_{k}\\left(a_{1}+\\cdots+a_{k-1}\\right)^{2}+a_{k}^{2}\\left(a_{1}+\\cdots+a_{k-1}\\right) \\cdot \\frac{a_{1}+\\cdots+a_{k-1}}{1-a_{k}}\\right)\n\\end{aligned}\n$$\n\nwhere for the last inequality we used that $1-a_{k} \\geqslant a_{1}+\\cdots+a_{k-1}$. This completes the proof since\n\n$$\na_{k}+\\frac{a_{k}^{2}}{1-a_{k}}=\\frac{a_{k}}{1-a_{k}}\n$$']",,True,,, 1778,Algebra,,"Let $A$ be a finite set of (not necessarily positive) integers, and let $m \geqslant 2$ be an integer. Assume that there exist non-empty subsets $B_{1}, B_{2}, B_{3}, \ldots, B_{m}$ of $A$ whose elements add up to the sums $m^{1}, m^{2}, m^{3}, \ldots, m^{m}$, respectively. Prove that $A$ contains at least $m / 2$ elements.","['Let $A=\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$. Assume that, on the contrary, $k=|A|\left(\frac{2 n}{3}\right)^{3 / 2} . $$","[""Lemma 1.1. If $a, b, c$ are non-negative numbers such that $a b-c^{2} \\geqslant 1$, then\n\n$$\n(a+2 b)^{2} \\geqslant(b+2 c)^{2}+6\n$$\n\nProof. $(a+2 b)^{2}-(b+2 c)^{2}=(a-b)^{2}+2(b-c)^{2}+6\\left(a b-c^{2}\\right) \\geqslant 6$.\n\nLemma 1.2. $\\sqrt{1}+\\cdots+\\sqrt{n}>\\frac{2}{3} n^{3 / 2}$.\n\nProof. Bernoulli's inequality $(1+t)^{3 / 2}>1+\\frac{3}{2} t$ for $0>t \\geqslant-1$ (or, alternatively, a straightforward check) gives\n\n$$\n(k-1)^{3 / 2}=k^{3 / 2}\\left(1-\\frac{1}{k}\\right)^{3 / 2}>k^{3 / 2}\\left(1-\\frac{3}{2 k}\\right)=k^{3 / 2}-\\frac{3}{2} \\sqrt{k}\n\\tag{*}\n$$\n\nSumming up (*) over $k=1,2, \\ldots, n$ yields\n\n$$\n0>n^{3 / 2}-\\frac{3}{2}(\\sqrt{1}+\\cdots+\\sqrt{n}) .\n$$\n\nNow put $y_{i}:=2 x_{i}+x_{i+1}$ for $i=0,1, \\ldots, n$. We get $y_{0} \\geqslant 0$ and $y_{i}^{2} \\geqslant y_{i-1}^{2}+6$ for $i=1,2, \\ldots, n$ by Lemma 1.1. Thus, an easy induction on $i$ gives $y_{i} \\geqslant \\sqrt{6 i}$. Using this estimate and Lemma 1.2 we get\n\n$$\n3\\left(x_{0}+\\ldots+x_{n+1}\\right) \\geqslant y_{1}+\\ldots+y_{n} \\geqslant \\sqrt{6}(\\sqrt{1}+\\sqrt{2}+\\ldots+\\sqrt{n})>\\sqrt{6} \\cdot \\frac{2}{3} n^{3 / 2}=3\\left(\\frac{2 n}{3}\\right)^{3 / 2}\n$$"", 'Say that an index $i \\in\\{0,1, \\ldots, n+1\\}$ is $\\operatorname{good}$, if $x_{i} \\geqslant \\sqrt{\\frac{2}{3}} i$, otherwise call the index $i$ bad.\n\nLemma 2.1. There are no two consecutive bad indices.\n\nProof. Assume the contrary and consider two bad indices $j, j+1$ with minimal possible $j$. Since 0 is good, we get $j>0$, thus by minimality $j-1$ is a good index and we have\n\n$$\n\\frac{2}{3} \\sqrt{j(j+1)}>x_{j} x_{j+1} \\geqslant x_{j-1}^{2}+1 \\geqslant \\frac{2}{3}(j-1)+1=\\frac{2}{3} \\cdot \\frac{j+(j+1)}{2}\n$$\n\nthat contradicts the AM-GM inequality for numbers $j$ and $j+1$.\n\nLemma 2.2. If an index $j \\leqslant n-1$ is good, then\n\n$$\nx_{j+1}+x_{j+2} \\geqslant \\sqrt{\\frac{2}{3}}(\\sqrt{j+1}+\\sqrt{j+2})\n$$\n\nProof. We have\n\n$$\nx_{j+1}+x_{j+2} \\geqslant 2 \\sqrt{x_{j+1} x_{j+2}} \\geqslant 2 \\sqrt{x_{j}^{2}+1} \\geqslant 2 \\sqrt{\\frac{2}{3} j+1} \\geqslant \\sqrt{\\frac{2}{3} j+\\frac{2}{3}}+\\sqrt{\\frac{2}{3} j+\\frac{4}{3}},\n$$\n\nthe last inequality follows from concavity of the square root function, or, alternatively, from the AM-QM inequality for the numbers $\\sqrt{\\frac{2}{3} j+\\frac{2}{3}}$ and $\\sqrt{\\frac{2}{3} j+\\frac{4}{3}}$.\n\n\n\nLet $S_{i}=x_{1}+\\ldots+x_{i}$ and $T_{i}=\\sqrt{\\frac{2}{3}}(\\sqrt{1}+\\ldots+\\sqrt{i})$.\n\nLemma 2.3. If an index $i$ is good, then $S_{i} \\geqslant T_{i}$.\n\nProof. Induction on $i$. The base case $i=0$ is clear. Assume that the claim holds for good indices less than $i$ and prove it for a good index $i>0$.\n\nIf $i-1$ is good, then by the inductive hypothesis we get $S_{i}=S_{i-1}+x_{i} \\geqslant T_{i-1}+\\sqrt{\\frac{2}{3} i}=T_{i}$.\n\nIf $i-1$ is bad, then $i>1$, and $i-2$ is good by Lemma 2.1. Then using Lemma 2.2 and the inductive hypothesis we get\n\n$$\nS_{i}=S_{i-2}+x_{i-1}+x_{i} \\geqslant T_{i-2}+\\sqrt{\\frac{2}{3}}(\\sqrt{i-1}+\\sqrt{i})=T_{i}\n$$\n\nSince either $n$ or $n+1$ is good by Lemma 2.1, Lemma 2.3 yields in both cases $S_{n+1} \\geqslant T_{n}$']",,True,,, 1780,Algebra,,"Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy $$ (f(a)-f(b))(f(b)-f(c))(f(c)-f(a))=f\left(a b^{2}+b c^{2}+c a^{2}\right)-f\left(a^{2} b+b^{2} c+c^{2} a\right) $$ for all real numbers $a, b, c$.","['It is straightforward to check that above functions satisfy the equation. Now let $f(x)$ satisfy the equation, which we denote $E(a, b, c)$. Then clearly $f(x)+C$ also does; therefore, we may suppose without loss of generality that $f(0)=0$.We start with proving\n\nLemma. Either $f(x) \\equiv 0$ or $f$ is injective.\n\nProof. Denote by $\\Theta \\subseteq \\mathbb{R}^{2}$ the set of points $(a, b)$ for which $f(a)=f(b)$. Let $\\Theta^{*}=\\{(x, y) \\in \\Theta$ : $x \\neq y\\}$. The idea is that if $(a, b) \\in \\Theta$, then by $E(a, b, x)$ we get\n\n$$\nH_{a, b}(x):=\\left(a b^{2}+b x^{2}+x a^{2}, a^{2} b+b^{2} x+x^{2} a\\right) \\in \\Theta\n$$\n\nfor all real $x$. Reproducing this argument starting with $(a, b) \\in \\Theta^{*}$, we get more and more points in $\\Theta$. There are many ways to fill in the details, we give below only one of them.\n\nAssume that $(a, b) \\in \\Theta^{*}$. Note that\n\n$$\ng_{-}(x):=\\left(a b^{2}+b x^{2}+x a^{2}\\right)-\\left(a^{2} b+b^{2} x+x^{2} a\\right)=(a-b)(b-x)(x-a)\n$$\n\nand\n\n$$\ng_{+}(x):=\\left(a b^{2}+b x^{2}+x a^{2}\\right)+\\left(a^{2} b+b^{2} x+x^{2} a\\right)=\\left(x^{2}+a b\\right)(a+b)+x\\left(a^{2}+b^{2}\\right) .\n$$\n\nHence, there exists $x$ for which both $g_{-}(x) \\neq 0$ and $g_{+}(x) \\neq 0$. This gives a point $(\\alpha, \\beta)=$ $H_{a, b}(x) \\in \\Theta^{*}$ for which $\\alpha \\neq-\\beta$. Now compare $E(\\alpha, 1,0)$ and $E(\\beta, 1,0)$. The left-hand side expressions coincide, on right-hand side we get $f(\\alpha)-f\\left(\\alpha^{2}\\right)=f(\\beta)-f\\left(\\beta^{2}\\right)$, respectively. Hence, $f\\left(\\alpha^{2}\\right)=f\\left(\\beta^{2}\\right)$ and we get a point $\\left(\\alpha_{1}, \\beta_{1}\\right):=\\left(\\alpha^{2}, \\beta^{2}\\right) \\in \\Theta^{*}$ with both coordinates $\\alpha_{1}, \\beta_{1}$ non-negative. Continuing squaring the coordinates, we get a point $(\\gamma, \\delta) \\in \\Theta^{*}$ for which $\\delta>5 \\gamma \\geqslant 0$. Our nearest goal is to get a point $(0, r) \\in \\Theta^{*}$. If $\\gamma=0$, this is already done. If $\\gamma>0$, denote by $x$ a real root of the quadratic equation $\\delta \\gamma^{2}+\\gamma x^{2}+x \\delta^{2}=0$, which exists since the discriminant $\\delta^{4}-4 \\delta \\gamma^{3}$ is positive. Also $x<0$ since this equation cannot have non-negative root. For the point $H_{\\delta, \\gamma}(x)=:(0, r) \\in \\Theta$ the first coordinate is 0 . The difference of coordinates equals $-r=(\\delta-\\gamma)(\\gamma-x)(x-\\delta)<0$, so $r \\neq 0$ as desired.\n\nNow, let $(0, r) \\in \\Theta^{*}$. We get $H_{0, r}(x)=\\left(r x^{2}, r^{2} x\\right) \\in \\Theta$. Thus $f\\left(r x^{2}\\right)=f\\left(r^{2} x\\right)$ for all $x \\in \\mathbb{R}$. Replacing $x$ to $-x$ we get $f\\left(r x^{2}\\right)=f\\left(r^{2} x\\right)=f\\left(-r^{2} x\\right)$, so $f$ is even: $(a,-a) \\in \\Theta$ for all $a$. Then $H_{a,-a}(x)=\\left(a^{3}-a x^{2}+x a^{2},-a^{3}+a^{2} x+x^{2} a\\right) \\in \\Theta$ for all real $a, x$. Putting $x=\\frac{1+\\sqrt{5}}{2} a$ we obtain $\\left(0,(1+\\sqrt{5}) a^{3}\\right) \\in \\Theta$ which means that $f(y)=f(0)=0$ for every real $y$.\n\nHereafter we assume that $f$ is injective and $f(0)=0$. By $E(a, b, 0)$ we get\n\n$$\nf(a) f(b)(f(a)-f(b))=f\\left(a^{2} b\\right)-f\\left(a b^{2}\\right) .\n\\tag{1}\n$$\n\nLet $\\kappa:=f(1)$ and note that $\\kappa=f(1) \\neq f(0)=0$ by injectivity. Putting $b=1$ in ( 8 ) we get\n\n$$\n\\kappa f(a)(f(a)-\\kappa)=f\\left(a^{2}\\right)-f(a)\n\\tag{2}\n$$\n\nSubtracting the same equality for $-a$ we get\n\n$$\n\\kappa(f(a)-f(-a))(f(a)+f(-a)-\\kappa)=f(-a)-f(a) .\n$$\n\n\n\nNow, if $a \\neq 0$, by injectivity we get $f(a)-f(-a) \\neq 0$ and thus\n\n$$\nf(a)+f(-a)=\\kappa-\\kappa^{-1}=: \\lambda\n\\tag{3}\n$$\n\nIt follows that\n\n$$\nf(a)-f(b)=f(-b)-f(-a)\n$$\n\nfor all non-zero $a, b$. Replace non-zero numbers $a, b$ in ( $\\mathcal{( )}$ with $-a,-b$, respectively, and add the two equalities. Due to $(3)$ we get\n\n$$\n(f(a)-f(b))(f(a) f(b)-f(-a) f(-b))=0,\n$$\n\nthus $f(a) f(b)=f(-a) f(-b)=(\\lambda-f(a))(\\lambda-f(b))$ for all non-zero $a \\neq b$. If $\\lambda \\neq 0$, this implies $f(a)+f(b)=\\lambda$ that contradicts injectivity when we vary $b$ with fixed $a$. Therefore, $\\lambda=0$ and $\\kappa= \\pm 1$. Thus $f$ is odd. Replacing $f$ with $-f$ if necessary (this preserves the original equation) we may suppose that $f(1)=1$.\n\nNow, (2) yields $f\\left(a^{2}\\right)=f^{2}(a)$. Summing relations $(1)$ for pairs $(a, b)$ and $(a,-b)$, we get $-2 f(a) f^{2}(b)=-2 f\\left(a b^{2}\\right)$, i.e. $f(a) f\\left(b^{2}\\right)=f\\left(a b^{2}\\right)$. Putting $b=\\sqrt{x}$ for each non-negative $x$ we get $f(a x)=f(a) f(x)$ for all real $a$ and non-negative $x$. Since $f$ is odd, this multiplicativity relation is true for all $a, x$. Also, from $f\\left(a^{2}\\right)=f^{2}(a)$ we see that $f(x) \\geqslant 0$ for $x \\geqslant 0$. Next, $f(x)>0$ for $x>0$ by injectivity.\n\nAssume that $f(x)$ for $x>0$ does not have the form $f(x)=x^{\\tau}$ for a constant $\\tau$. The known property of multiplicative functions yields that the graph of $f$ is dense on $(0, \\infty)^{2}$. In particular, we may find positive $b<1 / 10$ for which $f(b)>1$. Also, such $b$ can be found if $f(x)=x^{\\tau}$ for some $\\tau<0$. Then for all $x$ we have $x^{2}+x b^{2}+b \\geqslant 0$ and so $E(1, b, x)$ implies that\n\n$$\nf\\left(b^{2}+b x^{2}+x\\right)=f\\left(x^{2}+x b^{2}+b\\right)+(f(b)-1)(f(x)-f(b))(f(x)-1) \\geqslant 0-\\left((f(b)-1)^{3} / 4\\right.\n$$\n\nis bounded from below (the quadratic trinomial bound $(t-f(1))(t-f(b)) \\geqslant-(f(b)-1)^{2} / 4$ for $t=f(x)$ is used). Hence, $f$ is bounded from below on $\\left(b^{2}-\\frac{1}{4 b},+\\infty\\right)$, and since $f$ is odd it is bounded from above on $\\left(0, \\frac{1}{4 b}-b^{2}\\right)$. This is absurd if $f(x)=x^{\\tau}$ for $\\tau<0$, and contradicts to the above dense graph condition otherwise.\n\nTherefore, $f(x)=x^{\\tau}$ for $x>0$ and some constant $\\tau>0$. Dividing $E(a, b, c)$ by $(a-b)(b-$ $c)(c-a)=\\left(a b^{2}+b c^{2}+c a^{2}\\right)-\\left(a^{2} b+b^{2} c+c^{2} a\\right)$ and taking a limit when $a, b, c$ all go to 1 (the divided ratios tend to the corresponding derivatives, say, $\\frac{a^{\\tau}-b^{\\tau}}{a-b} \\rightarrow\\left(x^{\\tau}\\right)_{x=1}^{\\prime}=\\tau$ ), we get $\\tau^{3}=\\tau \\cdot 3^{\\tau-1}, \\tau^{2}=3^{\\tau-1}, F(\\tau):=3^{\\tau / 2-1 / 2}-\\tau=0$. Since function $F$ is strictly convex, it has at most two roots, and we get $\\tau \\in\\{1,3\\}$.']","['$f(x)=\\alpha x+\\beta$ or $f(x)=\\alpha x^{3}+\\beta$ where $\\alpha \\in\\{-1,0,1\\}$ and $\\beta \\in \\mathbb{R}$']",True,,Need_human_evaluate, 1781,Algebra,,"Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a, b, c, d \in S$ with $\operatorname{gcd}(a, b) \neq \operatorname{gcd}(c, d)$. Prove that there exist three pairwise distinct $x, y, z \in S$ such that $\operatorname{gcd}(x, y)=\operatorname{gcd}(y, z) \neq \operatorname{gcd}(z, x)$.","['There exists $\\alpha \\in S$ so that $\\{\\operatorname{gcd}(\\alpha, s) \\mid s \\in S, s \\neq \\alpha\\}$ contains at least two elements. Since $\\alpha$ has only finitely many divisors, there is a $d \\mid \\alpha$ such that the set $B=\\{\\beta \\in$ $S \\mid \\operatorname{gcd}(\\alpha, \\beta)=d\\}$ is infinite. Pick $\\gamma \\in S$ so that $\\operatorname{gcd}(\\alpha, \\gamma) \\neq d$. Pick $\\beta_{1}, \\beta_{2} \\in B$ so that $\\operatorname{gcd}\\left(\\beta_{1}, \\gamma\\right)=\\operatorname{gcd}\\left(\\beta_{2}, \\gamma\\right)=: d^{\\prime}$. If $d=d^{\\prime}$, then $\\operatorname{gcd}\\left(\\alpha, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\gamma, \\beta_{1}\\right) \\neq \\operatorname{gcd}(\\alpha, \\gamma)$. If $d \\neq d^{\\prime}$, then either $\\operatorname{gcd}\\left(\\alpha, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\alpha, \\beta_{2}\\right)=d$ and $\\operatorname{gcd}\\left(\\beta_{1}, \\beta_{2}\\right) \\neq d \\operatorname{or} \\operatorname{gcd}\\left(\\gamma, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\gamma, \\beta_{2}\\right)=d^{\\prime}$ and $\\operatorname{gcd}\\left(\\beta_{1}, \\beta_{2}\\right) \\neq d^{\\prime}$.']",,True,,, 1782,Combinatorics,,"Let $n \geqslant 3$ be an integer. An integer $m \geqslant n+1$ is called $n$-colourful if, given infinitely many marbles in each of $n$ colours $C_{1}, C_{2}, \ldots, C_{n}$, it is possible to place $m$ of them around a circle so that in any group of $n+1$ consecutive marbles there is at least one marble of colour $C_{i}$ for each $i=1, \ldots, n$. Prove that there are only finitely many positive integers which are not $n$-colourful. Find the largest among them.","['First suppose that there are $n(n-1)-1$ marbles. Then for one of the colours, say blue, there are at most $n-2$ marbles, which partition the non-blue marbles into at most $n-2$ groups with at least $(n-1)^{2}>n(n-2)$ marbles in total. Thus one of these groups contains at least $n+1$ marbles and this group does not contain any blue marble.\n\nNow suppose that the total number of marbles is at least $n(n-1)$. Then we may write this total number as $n k+j$ with some $k \\geqslant n-1$ and with $0 \\leqslant j \\leqslant n-1$. We place around a circle $k-j$ copies of the colour sequence $[1,2,3, \\ldots, n]$ followed by $j$ copies of the colour sequence $[1,1,2,3, \\ldots, n]$.']",['$m_{\\max }=n^{2}-n-1$'],False,,Expression, 1783,Combinatorics,,"A thimblerigger has 2021 thimbles numbered from 1 through 2021. The thimbles are arranged in a circle in arbitrary order. The thimblerigger performs a sequence of 2021 moves; in the $k^{\text {th }}$ move, he swaps the positions of the two thimbles adjacent to thimble $k$. Prove that there exists a value of $k$ such that, in the $k^{\text {th }}$ move, the thimblerigger swaps some thimbles $a$ and $b$ such that $ak \geqslant 1$. There are $2 n+1$ students standing in a circle. Each student $S$ has $2 k$ neighbours - namely, the $k$ students closest to $S$ on the right, and the $k$ students closest to $S$ on the left. Suppose that $n+1$ of the students are girls, and the other $n$ are boys. Prove that there is a girl with at least $k$ girls among her neighbours.","[""We replace the girls by 1's, and the boys by 0's, getting the numbers $a_{1}, a_{2}, \\ldots, a_{2 n+1}$ arranged in a circle. We extend this sequence periodically by letting $a_{2 n+1+k}=a_{k}$ for all $k \\in \\mathbb{Z}$. We get an infinite periodic sequence\n\n$$\n\\ldots, a_{1}, a_{2}, \\ldots, a_{2 n+1}, a_{1}, a_{2}, \\ldots, a_{2 n+1}, \\ldots\n$$\n\nConsider the numbers $b_{i}=a_{i}+a_{i-k-1}-1 \\in\\{-1,0,1\\}$ for all $i \\in \\mathbb{Z}$. We know that\n\n$$\nb_{m+1}+b_{m+2}+\\cdots+b_{m+2 n+1}=1 \\quad(m \\in \\mathbb{Z})\n\\tag{1}\n$$\n\nin particular, this yields that there exists some $i_{0}$ with $b_{i_{0}}=1$. Now we want to find an index $i$ such that\n\n$$\nb_{i}=1 \\quad \\text { and } \\quad b_{i+1}+b_{i+2}+\\cdots+b_{i+k} \\geqslant 0 .\n\\tag{2}\n$$\n\nThis will imply that $a_{i}=1$ and\n\n$$\n\\left(a_{i-k}+a_{i-k+1}+\\cdots+a_{i-1}\\right)+\\left(a_{i+1}+a_{i+2}+\\cdots+a_{i+k}\\right) \\geqslant k,\n$$\n\nas desired.\n\nSuppose, to the contrary, that for every index $i$ with $b_{i}=1$ the sum $b_{i+1}+b_{i+2}+\\cdots+b_{i+k}$ is negative. We start from some index $i_{0}$ with $b_{i_{0}}=1$ and construct a sequence $i_{0}, i_{1}, i_{2}, \\ldots$, where $i_{j}(j>0)$ is the smallest possible index such that $i_{j}>i_{j-1}+k$ and $b_{i_{j}}=1$. We can choose two numbers among $i_{0}, i_{1}, \\ldots, i_{2 n+1}$ which are congruent modulo $2 n+1$ (without loss of generality, we may assume that these numbers are $i_{0}$ and $i_{T}$ ).\n\nOn the one hand, for every $j$ with $0 \\leqslant j \\leqslant T-1$ we have\n\n$$\nS_{j}:=b_{i_{j}}+b_{i_{j}+1}+b_{i_{j}+2}+\\cdots+b_{i_{j+1}-1} \\leqslant b_{i_{j}}+b_{i_{j}+1}+b_{i_{j}+2}+\\cdots+b_{i_{j}+k} \\leqslant 0\n$$\n\nsince $b_{i_{j}+k+1}, \\ldots, b_{i_{j+1}-1} \\leqslant 0$. On the other hand, since $\\left(i_{T}-i_{0}\\right) \\mid(2 n+1)$, from (1) we deduce\n\n$$\nS_{0}+\\cdots+S_{T-1}=\\sum_{i=i_{0}}^{i_{T}-1} b_{i}=\\frac{i_{T}-i_{0}}{2 n+1}>0\n$$\n\nThis contradiction finishes the solution.""]",,True,,, 1786,Combinatorics,,"A hunter and an invisible rabbit play a game on an infinite square grid. First the hunter fixes a colouring of the cells with finitely many colours. The rabbit then secretly chooses a cell to start in. Every minute, the rabbit reports the colour of its current cell to the hunter, and then secretly moves to an adjacent cell that it has not visited before (two cells are adjacent if they share a side). The hunter wins if after some finite time either - the rabbit cannot move; or - the hunter can determine the cell in which the rabbit started. Decide whether there exists a winning strategy for the hunter.","[""A central idea is that several colourings $C_{1}, C_{2}, \\ldots, C_{k}$ can be merged together into a single product colouring $C_{1} \\times C_{2} \\times \\cdots \\times C_{k}$ as follows: the colours in the product colouring are ordered tuples $\\left(c_{1}, \\ldots, c_{n}\\right)$ of colours, where $c_{i}$ is a colour used in $C_{i}$, so that each cell gets a tuple consisting of its colours in the individual colourings $C_{i}$. This way, any information which can be determined from one of the individual colourings can also be determined from the product colouring.\n\nNow let the hunter merge the following colourings:\n\n- The first two colourings $C_{1}$ and $C_{2}$ allow the tracking of the horizontal and vertical movements of the rabbit.\n\nThe colouring $C_{1}$ colours the cells according to the residue of their $x$-coordinates modulo 3 , which allows to determine whether the rabbit moves left, moves right, or moves vertically. Similarly, the colouring $C_{2}$ uses the residues of the $y$-coordinates modulo 3 , which allows to determine whether the rabbit moves up, moves down, or moves horizontally.\n\n- Under the condition that the rabbit's $x$-coordinate is unbounded, colouring $C_{3}$ allows to determine the exact value of the $x$-coordinate:\n\nIn $C_{3}$, the columns are coloured white and black so that the gaps between neighboring black columns are pairwise distinct. As the rabbit's $x$-coordinate is unbounded, it will eventually visit two black cells in distinct columns. With the help of colouring $C_{1}$ the hunter can catch that moment, and determine the difference of $x$-coordinates of those two black cells, hence deducing the precise column.\n\nSymmetrically, under the condition that the rabbit's $y$-coordinate is unbounded, there is a colouring $C_{4}$ that allows the hunter to determine the exact value of the $y$-coordinate.\n\n- Finally, under the condition that the sum $x+y$ of the rabbit's coordinates is unbounded, colouring $C_{5}$ allows to determine the exact value of this sum: The diagonal lines $x+y=$ const are coloured black and white, so that the gaps between neighboring black diagonals are pairwise distinct.\n\nUnless the rabbit gets stuck, at least two of the three values $x, y$ and $x+y$ must be unbounded as the rabbit keeps moving. Hence the hunter can eventually determine two of these three values; thus he does know all three. Finally the hunter works backwards with help of the colourings $C_{1}$ and $C_{2}$ and computes the starting cell of the rabbit.""]",,True,,, 1787,Combinatorics,,"Consider a checkered $3 m \times 3 m$ square, where $m$ is an integer greater than 1 . A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F$. The frog can hop from any cell to either the next cell to the right or the next cell upwards. Some cells can be sticky, and the frog gets trapped once it hops on such a cell. A set $X$ of cells is called blocking if the frog cannot reach $F$ from $S$ when all the cells of $X$ are sticky. A blocking set is minimal if it does not contain a smaller blocking set. (a) Prove that there exists a minimal blocking set containing at least $3 m^{2}-3 m$ cells.","['In the following example the square is divided into $m$ stripes of size $3 \\times 3 \\mathrm{~m}$. It is easy to see that $X$ is a minimal blocking set. The first and the last stripe each contains $3 m-1$ cells from the set $X$; every other stripe contains $3 m-2$ cells, see Figure 1 . The total number of cells in the set $X$ is $3 m^{2}-2 m+2$.\n\n\n\nFigure 1']",,True,,, 1788,Combinatorics,,"Consider a checkered $3 m \times 3 m$ square, where $m$ is an integer greater than 1 . A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F$. The frog can hop from any cell to either the next cell to the right or the next cell upwards. Some cells can be sticky, and the frog gets trapped once it hops on such a cell. A set $X$ of cells is called blocking if the frog cannot reach $F$ from $S$ when all the cells of $X$ are sticky. A blocking set is minimal if it does not contain a smaller blocking set. Prove that every minimal blocking set contains at most $3 m^{2}$ cells. Note. An example of a minimal blocking set for $m=2$ is shown below. Cells of the set $X$ are marked by letters $x$. | | | | | | $F$ | | :--- | :--- | :--- | :--- | :--- | :--- | | $x$ | $x$ | | | | | | | | $x$ | | | | | | | | $x$ | | | | | | | | $x$ | | | $S$ | | $x$ | | | |","['For a given blocking set $X$, say that a non-sticky cell is red if the frog can reach it from $S$ via some hops without entering set $X$. We call a non-sticky cell blue if the frog can reach $F$ from that cell via hops without entering set $X$. One can regard the blue cells as those reachable from $F$ by anti-hops, i.e. moves downwards and to the left. We also colour all cells in $X$ green. It follows from the definition of the blocking set that no cell will be coloured twice. In Figure 2 we show a sample of a blocking set and the corresponding colouring.\n\nNow assume that $X$ is a minimal blocking set. We denote by $R$ (resp., $B$ and $G$ ) be the total number of red (resp., blue and green) cells.\n\nWe claim that $G \\leqslant R+1$ and $G \\leqslant B+1$. Indeed, there are at most $2 R$ possible frog hops from red cells. Every green or red cell (except for $S$ ) is accessible by such hops. Hence $2 R \\geqslant G+(R-1)$, or equivalently $G \\leqslant R+1$. In order to prove the inequality $G \\leqslant B+1$, we turn over the board and apply the similar arguments.\n\nTherefore we get $9 m^{2} \\geqslant B+R+G \\geqslant 3 G-2$, so $G \\leqslant 3 m^{2}$.\n\n\n\n\n\nFigure 2 (a)\n\n\n\nFigure 2 (b)', 'We shall use the same colouring as in the above solution. Again, assume that $X$ is a minimal blocking set.\n\nNote that any $2 \\times 2$ square cannot contain more than 2 green cells. Indeed, on Figure 3(a) the cell marked with ""?"" does not block any path, while on Figure 3(b) the cell marked with ""?"" should be coloured red and blue simultaneously. So we can split all green cells into chains consisting of three types of links shown on Figure 4 (diagonal link in the other direction is not allowed, corresponding green cells must belong to different chains). For example, there are 3 chains in Figure 2(b).\n\n\nFigure 3\n\n\n\n\nand\n\n\nFigure 4\n\n\n\nand\n\n\n\nFigure 5\n\nWe will inscribe green chains in disjoint axis-aligned rectangles so that the number of green cells in each rectangle will not exceed $1 / 3$ of the area of the rectangle. This will give us the bound $G \\leqslant 3 m^{2}$. Sometimes the rectangle will be the minimal bounding rectangle of the chain, sometimes minimal bounding rectangles will be expanded in one or two directions in order to have sufficiently large area.\n\nNote that for any two consecutive cells in the chain the colouring of some neighbouring cells is uniquely defined (see Figure 5). In particular, this observation gives a corresponding rectangle for the chains of height (or width) 1 (see Figure 6(a)). A separate green cell can be inscribed in $1 \\times 3$ or $3 \\times 1$ rectangle with one red and one blue cell, see Figure $6(\\mathrm{~b})-(\\mathrm{c})$, otherwise we get one of impossible configurations shown in Figure 3.\n\n\n\n$(a)$\n\n\n\n(b)\n\n\n\n$(c)$\n\nFigure 6\n\n\n\n(a)\n\n\n\n(b)\n\nFigure 7\n\nAny diagonal chain of length 2 is always inscribed in a $2 \\times 3$ or $3 \\times 2$ rectangle without another green cells. Indeed, one of the squares marked with ""?"" in Figure 7(a) must be red. If it is the bottom question mark, then the remaining cell in the corresponding $2 \\times 3$ rectangle must have the same colour, see Figure 7(b).\n\nA longer chain of height (or width) 2 always has a horizontal (resp., vertical) link and can be inscribed into a $3 \\times a$ rectangle. In this case we expand the minimal bounding rectangle across the long side which touches the mentioned link. On Figure 8(a) the corresponding expansion of the minimal bounding rectangle is coloured in light blue. The upper right corner cell must be also blue. Indeed it cannot be red or green. If it is not coloured in blue, see Figure 8(b), then all anti-hop paths from $F$ to ""?"" are blocked with green cells. And these green cells are surrounded by blue ones, what is impossible. In this case the green chain contains $a$ cells, which is exactly $1 / 3$ of the area of the rectangle.\n\n\n\n\n\nFigure 8 (a)\n\n\n\nFigure 8 (b)\n\nIn the remaining case the minimal bounding rectangle of the chain is of size $a \\times b$ where $a, b \\geqslant 3$. Denote by $\\ell$ the length of the chain (i.e. the number of cells in the chain).\n\nIf the chain has at least two diagonal links (see Figure 9), then $\\ell \\leqslant a+b-3 \\leqslant a b / 3$.\n\nIf the chain has only one diagonal link then $\\ell=a+b-2$. In this case the chain has horizontal and vertical end-links, and we expand the minimal bounding rectangle in two directions to get an $(a+1) \\times(b+1)$ rectangle. On Figure 10 a corresponding expansion of the minimal bounding rectangle is coloured in light red. Again the length of the chain does not exceed $1 / 3$ of the rectangle\'s area: $\\ell \\leqslant a+b-2 \\leqslant(a+1)(b+1) / 3$.\n\nOn the next step we will use the following statement: all cells in constructed rectangles are coloured red, green or blue (the cells upwards and to the right of green cells are blue; the cells downwards and to the left of green cells are red). The proof repeats the same arguments as before (see Figure 8(b).)\n\n\n\nFigure 9\n\n\n\nFigure 10\n\n\n\n(a)\n\n\n\n(b)\n\nFigure 11\n\nNote that all constructed rectangles are disjoint. Indeed, assume that two rectangles have a common cell. Using the above statement, one can see that the only such cell can be a common corner cell, as shown in Figure 11. Moreover, in this case both rectangles should be expanded, otherwise they would share a green corner cell.\n\nIf they were expanded along the same axis (see Figure 11(a)), then again the common corner cannot be coloured correctly. If they were expanded along different axes (see Figure 11(b)) then the two chains have a common point and must be connected in one chain. (These arguments work for $2 \\times 3$ and $1 \\times 3$ rectangles in a similar manner.)']",,True,,, 1789,Combinatorics,,"Determine the largest $N$ for which there exists a table $T$ of integers with $N$ rows and 100 columns that has the following properties: (i) Every row contains the numbers 1,2, ., 100 in some order. (ii) For any two distinct rows $r$ and $s$, there is a column $c$ such that $|T(r, c)-T(s, c)| \geqslant 2$. Here $T(r, c)$ means the number at the intersection of the row $r$ and the column $c$.","['Non-existence of a larger table. Let us consider some fixed row in the table, and let us replace (for $k=1,2, \\ldots, 50$ ) each of two numbers $2 k-1$ and $2 k$ respectively by the symbol $x_{k}$. The resulting pattern is an arrangement of 50 symbols $x_{1}, x_{2}, \\ldots, x_{50}$, where every symbol occurs exactly twice. Note that there are $N=100 ! / 2^{50}$ distinct patterns $P_{1}, \\ldots, P_{N}$.\n\nIf two rows $r \\neq s$ in the table have the same pattern $P_{i}$, then $|T(r, c)-T(s, c)| \\leqslant 1$ holds for all columns $c$. As this violates property (ii) in the problem statement, different rows have different patterns. Hence there are at most $N=100 ! / 2^{50}$ rows.\n\nExistence of a table with $N$ rows. We construct the table by translating every pattern $P_{i}$ into a corresponding row with the numbers $1,2, \\ldots, 100$. We present a procedure that inductively replaces the symbols by numbers. The translation goes through steps $k=1,2, \\ldots, 50$ in increasing order and at step $k$ replaces the two occurrences of symbol $x_{k}$ by $2 k-1$ and $2 k$.\n\n- The left occurrence of $x_{1}$ is replaced by 1 , and its right occurrence is replaced by 2 .\n- For $k \\geqslant 2$, we already have the number $2 k-2$ somewhere in the row, and now we are looking for the places for $2 k-1$ and $2 k$. We make the three numbers $2 k-2,2 k-1,2 k$ show up (ordered from left to right) either in the order $2 k-2,2 k-1,2 k$, or as $2 k, 2 k-2,2 k-1$, or as $2 k-1,2 k, 2 k-2$. This is possible, since the number $2 k-2$ has been placed in the preceding step, and shows up before / between / after the two occurrences of the symbol $x_{k}$.\n\nWe claim that the $N$ rows that result from the $N$ patterns yield a table with the desired property (ii). Indeed, consider the $r$-th and the $s$-th row $(r \\neq s)$, which by construction result from patterns $P_{r}$ and $P_{s}$. Call a symbol $x_{i}$ aligned, if it occurs in the same two columns in $P_{r}$ and in $P_{s}$. Let $k$ be the largest index, for which symbol $x_{k}$ is not aligned. Note that $k \\geqslant 2$. Consider the column $c^{\\prime}$ with $T\\left(r, c^{\\prime}\\right)=2 k$ and the column $c^{\\prime \\prime}$ with $T\\left(s, c^{\\prime \\prime}\\right)=2 k$. Then $T\\left(r, c^{\\prime \\prime}\\right) \\leqslant 2 k$ and $T\\left(s, c^{\\prime}\\right) \\leqslant 2 k$, as all symbols $x_{i}$ with $i \\geqslant k+1$ are aligned.\n\n- If $T\\left(r, c^{\\prime \\prime}\\right) \\leqslant 2 k-2$, then $\\left|T\\left(r, c^{\\prime \\prime}\\right)-T\\left(s, c^{\\prime \\prime}\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(s, c^{\\prime}\\right) \\leqslant 2 k-2$, then $\\left|T\\left(r, c^{\\prime}\\right)-T\\left(s, c^{\\prime}\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(r, c^{\\prime \\prime}\\right)=2 k-1$ and $T\\left(s, c^{\\prime}\\right)=2 k-1$, then the symbol $x_{k}$ is aligned; contradiction.\n\nIn the only remaining case we have $c^{\\prime}=c^{\\prime \\prime}$, so that $T\\left(r, c^{\\prime}\\right)=T\\left(s, c^{\\prime}\\right)=2 k$ holds. Now let us consider the columns $d^{\\prime}$ and $d^{\\prime \\prime}$ with $T\\left(r, d^{\\prime}\\right)=2 k-1$ and $T\\left(s, d^{\\prime \\prime}\\right)=2 k-1$. Then $d \\neq d^{\\prime \\prime}$ (as the symbol $x_{k}$ is not aligned), and $T\\left(r, d^{\\prime \\prime}\\right) \\leqslant 2 k-2$ and $T\\left(s, d^{\\prime}\\right) \\leqslant 2 k-2$ (as all symbols $x_{i}$ with $i \\geqslant k+1$ are aligned).\n\n- If $T\\left(r, d^{\\prime \\prime}\\right) \\leqslant 2 k-3$, then $\\left|T\\left(r, d^{\\prime \\prime}\\right)-T\\left(s, d^{\\prime \\prime}\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(s, c^{\\prime}\\right) \\leqslant 2 k-3$, then $\\left|T\\left(r, d^{\\prime}\\right)-T\\left(s, d^{\\prime}\\right)\\right| \\geqslant 2$ as desired.\n\nIn the only remaining case we have $T\\left(r, d^{\\prime \\prime}\\right)=2 k-2$ and $T\\left(s, d^{\\prime}\\right)=2 k-2$. Now the row $r$ has the numbers $2 k-2,2 k-1,2 k$ in the three columns $d^{\\prime}, d^{\\prime \\prime}, c^{\\prime}$. As one of these triples violates the ordering property of $2 k-2,2 k-1,2 k$, we have the final contradiction.']",['$\\frac{100!}{2^{50}}$'],False,,Expression, 1790,Geometry,,"Let $A B C D$ be a parallelogram such that $A C=B C$. A point $P$ is chosen on the extension of the segment $A B$ beyond $B$. The circumcircle of the triangle $A C D$ meets the segment $P D$ again at $Q$, and the circumcircle of the triangle $A P Q$ meets the segment $P C$ again at $R$. Prove that the lines $C D, A Q$, and $B R$ are concurrent.","['Since $A C=B C=A D$, we have $\\angle A B C=\\angle B A C=\\angle A C D=\\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n\n$$\n\\angle C R A=180^{\\circ}-\\angle A R P=180^{\\circ}-\\angle A Q P=\\angle D Q A=\\angle D C A=\\angle C B A,\n$$\n\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\nIntroduce the point $X=A Q \\cap C D$; we need to prove that $B, R$ and $X$ are collinear.\n\nBy means of the circle $(A P R Q)$ we have\n\n$$\n\\angle R Q X=180^{\\circ}-\\angle A Q R=\\angle R P A=\\angle R C X\n$$\n\n(the last equality holds in view of $A B \\| C D$ ), which means that the points $C, Q, R$, and $X$ also lie on some circle $\\delta$.\n\nUsing the circles $\\delta$ and $\\gamma$ we finally obtain\n\n$$\n\\angle X R C=\\angle X Q C=180^{\\circ}-\\angle C Q A=\\angle A D C=\\angle B A C=180^{\\circ}-\\angle C R B,\n$$\n\nthat proves the desired collinearity.\n\n', ""Since $A C=B C=A D$, we have $\\angle A B C=\\angle B A C=\\angle A C D=\\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n\n$$\n\\angle C R A=180^{\\circ}-\\angle A R P=180^{\\circ}-\\angle A Q P=\\angle D Q A=\\angle D C A=\\angle C B A,\n$$\n\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\n\nLet $\\alpha$ denote the circle $(A P R Q)$. Since\n\n$$\n\\angle C A P=\\angle A C D=\\angle A Q D=180^{\\circ}-\\angle A Q P\n$$\n\nthe line $A C$ is tangent to $\\alpha$.\n\nNow, let $A D$ meet $\\alpha$ again at a point $Y$ (which necessarily lies on the extension of $D A$ beyond $A$ ). Using the circle $\\gamma$, along with the fact that $A C$ is tangent to $\\alpha$, we have\n\n$$\n\\angle A R Y=\\angle C A D=\\angle A C B=\\angle A R B\n$$\n\nso the points $Y, B$, and $R$ are collinear.\n\nApplying Pascal's theorem to the hexagon $A A Y R P Q$ (where $A A$ is regarded as the tangent to $\\alpha$ at $A$ ), we see that the points $A A \\cap R P=C, A Y \\cap P Q=D$, and $Y R \\cap Q A$ are collinear. Hence the lines $C D, A Q$, and $B R$ are concurrent."", 'Since $A C=B C=A D$, we have $\\angle A B C=\\angle B A C=\\angle A C D=\\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n\n$$\n\\angle C R A=180^{\\circ}-\\angle A R P=180^{\\circ}-\\angle A Q P=\\angle D Q A=\\angle D C A=\\angle C B A,\n$$\n\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\n\nwe introduce the point $X=A Q \\cap C D$ and aim at proving that the points $B, R$, and $X$ are collinear. we denote $\\alpha=(A P Q R)$; but now we define $Y$ to be the second meeting point of $R B$ with $\\alpha$.\n\nUsing the circle $\\alpha$ and noticing that $C D$ is tangent to $\\gamma$, we obtain\n\n$$\n\\angle R Y A=\\angle R P A=\\angle R C X=\\angle R B C .\n$$\n\nSo $A Y \\| B C$, and hence $Y$ lies on $D A$.\n\nNow the chain of equalities (1) shows also that $\\angle R Y D=\\angle R C X$, which implies that the points $C, D, Y$, and $R$ lie on some circle $\\beta$. Hence, the lines $C D, A Q$, and $Y B R$ are the pairwise radical axes of the circles $(A Q C D), \\alpha$, and $\\beta$, so those lines are concurrent.']",,True,,, 1791,Geometry,,"Let $A B C D$ be a convex quadrilateral circumscribed around a circle with centre $I$. Let $\omega$ be the circumcircle of the triangle $A C I$. The extensions of $B A$ and $B C$ beyond $A$ and $C$ meet $\omega$ at $X$ and $Z$, respectively. The extensions of $A D$ and $C D$ beyond $D$ meet $\omega$ at $Y$ and $T$, respectively. Prove that the perimeters of the (possibly self-intersecting) quadrilaterals $A D T X$ and $C D Y Z$ are equal.","['The point $I$ is the intersection of the external bisector of the angle $T C Z$ with the circumcircle $\\omega$ of the triangle $T C Z$, so $I$ is the midpoint of the $\\operatorname{arc} T C Z$ and $I T=I Z$. Similarly, $I$ is the midpoint of the $\\operatorname{arc} Y A X$ and $I X=I Y$. Let $O$ be the centre of $\\omega$. Then $X$ and $T$ are the reflections of $Y$ and $Z$ in $I O$, respectively. So $X T=Y Z$.\n\n\n\nLet the incircle of $A B C D$ touch $A B, B C, C D$, and $D A$ at points $P, Q, R$, and $S$, respectively.\n\nThe right triangles $I X P$ and $I Y S$ are congruent, since $I P=I S$ and $I X=I Y$. Similarly, the right triangles $I R T$ and $I Q Z$ are congruent. Therefore, $X P=Y S$ and $R T=Q Z$.\n\nDenote the perimeters of $A D T X$ and $C D Y Z$ by $P_{A D T X}$ and $P_{C D Y Z}$ respectively. Since $A S=A P, C Q=R C$, and $S D=D R$, we obtain\n\n$$\n\\begin{aligned}\nP_{A D T X}=X T+X A+A S+ & S D+D T=X T+X P+R T \\\\\n& =Y Z+Y S+Q Z=Y Z+Y D+D R+R C+C Z=P_{C D Y Z}\n\\end{aligned}\n$$\n\nas required.']",,True,,, 1792,Geometry,,"Version 1. Let $n$ be a fixed positive integer, and let $\mathrm{S}$ be the set of points $(x, y)$ on the Cartesian plane such that both coordinates $x$ and $y$ are nonnegative integers smaller than $2 n$ (thus $|\mathrm{S}|=4 n^{2}$ ). Assume that $\mathcal{F}$ is a set consisting of $n^{2}$ quadrilaterals such that all their vertices lie in $S$, and each point in $S$ is a vertex of exactly one of the quadrilaterals in $\mathcal{F}$. Determine the largest possible sum of areas of all $n^{2}$ quadrilaterals in $\mathcal{F}$.","['the area of a polygon $P$ will be denoted by $[P]$.\n\nWe say that a polygon is legal if all its vertices belong to S. Let $O=\\left(n-\\frac{1}{2}, n-\\frac{1}{2}\\right)$ be the centre of $\\mathrm{S}$. We say that a legal square is central if its centre is situated at $O$. Finally, say that a set $\\mathcal{F}$ of polygons is acceptable if it satisfies the problem requirements, i.e. if all polygons in $\\mathcal{F}$ are legal, and each point in $\\mathrm{S}$ is a vertex of exactly one polygon in $\\mathcal{F}$. For an acceptable set $\\mathcal{F}$, we denote by $\\Sigma(\\mathcal{F})$ the sum of areas of polygons in $\\mathcal{F}$.\n\n\nEach point in $S$ is a vertex of a unique central square. Thus the set $\\mathcal{G}$ of central squares is acceptable. We will show that\n\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\Sigma(\\mathcal{G})=\\Sigma(n)\n\\tag{1}\n$$\n\nthus establishing the answer.\n\nWe will use the following key lemma.\n\nLemma 1. Let $P=A_{1} A_{2} \\ldots A_{m}$ be a polygon, and let $O$ be an arbitrary point in the plane. Then\n\n$$\n[P] \\leqslant \\frac{1}{2} \\sum_{i=1}^{m} O A_{i}^{2}\n\\tag{2}\n$$\n\nmoreover, if $P$ is a square centred at $O$, then the inequality (2) turns into an equality.\n\nProof. Put $A_{n+1}=A_{1}$. For each $i=1,2, \\ldots, m$, we have\n\n$$\n\\left[O A_{i} A_{i+1}\\right] \\leqslant \\frac{O A_{i} \\cdot O A_{i+1}}{2} \\leqslant \\frac{O A_{i}^{2}+O A_{i+1}^{2}}{4}\n$$\n\nTherefore,\n\n$$\n[P] \\leqslant \\sum_{i=1}^{m}\\left[O A_{i} A_{i+1}\\right] \\leqslant \\frac{1}{4} \\sum_{i=1}^{m}\\left(O A_{i}^{2}+O A_{i+1}^{2}\\right)=\\frac{1}{2} \\sum_{i=1}^{m} O A_{i}^{2}\n$$\n\nwhich proves (2). Finally, all the above inequalities turn into equalities when $P$ is a square centred at $O$.\n\nBack to the problem, consider an arbitrary acceptable set $\\mathcal{F}$. Applying Lemma 1 to each element in $\\mathcal{F}$ and to each element in $\\mathcal{G}$ (achieving equality in the latter case), we obtain\n\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\frac{1}{2} \\sum_{A \\in S} O A^{2}=\\Sigma(\\mathcal{G})\n$$\n\nwhich establishes the left inequality in (1).\n\n\n\nIt remains to compute $\\Sigma(\\mathcal{G})$. We have\n\n$$\n\\begin{aligned}\n\\Sigma(\\mathcal{G})=\\frac{1}{2} & \\sum_{A \\in S} O A^{2}=\\frac{1}{2} \\sum_{i=0}^{2 n-1} \\sum_{j=0}^{2 n-1}\\left(\\left(n-\\frac{1}{2}-i\\right)^{2}+\\left(n-\\frac{1}{2}-j\\right)^{2}\\right) \\\\\n& =\\frac{1}{8} \\cdot 4 \\cdot 2 n \\sum_{i=0}^{n-1}(2 n-2 i-1)^{2}=n \\sum_{j=0}^{n-1}(2 j+1)^{2}=n\\left(\\sum_{j=1}^{2 n} j^{2}-\\sum_{j=1}^{n}(2 j)^{2}\\right) \\\\\n& =n\\left(\\frac{2 n(2 n+1)(4 n+1)}{6}-4 \\cdot \\frac{n(n+1)(2 n+1)}{6}\\right)=\\frac{n^{2}(2 n+1)(2 n-1)}{3}=\\Sigma(n) .\n\\end{aligned}\n$$', 'the area of a polygon $P$ will be denoted by $[P]$.\n\nWe say that a polygon is legal if all its vertices belong to S. Let $O=\\left(n-\\frac{1}{2}, n-\\frac{1}{2}\\right)$ be the centre of $\\mathrm{S}$. We say that a legal square is central if its centre is situated at $O$. Finally, say that a set $\\mathcal{F}$ of polygons is acceptable if it satisfies the problem requirements, i.e. if all polygons in $\\mathcal{F}$ are legal, and each point in $\\mathrm{S}$ is a vertex of exactly one polygon in $\\mathcal{F}$. For an acceptable set $\\mathcal{F}$, we denote by $\\Sigma(\\mathcal{F})$ the sum of areas of polygons in $\\mathcal{F}$.\n\nLet $\\mathcal{F}$ be an accessible set of quadrilaterals. For every quadrilateral $A B C D$ in $\\mathcal{F}$ write\n\n$$\n[A B C D]=\\frac{A C \\cdot B D}{2} \\sin \\phi \\leqslant \\frac{A C^{2}+B D^{2}}{4}\n\\tag{3}\n$$\n\nwhere $\\phi$ is the angle between $A C$ and $B D$. Applying this estimate to all members in $\\mathcal{F}$ we obtain\n\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\frac{1}{4} \\sum_{i=1}^{2 n^{2}} A_{i} B_{i}^{2}\n$$\n\nwhere $A_{1}, A_{2}, \\ldots, A_{2 n^{2}}, B_{1}, B_{2}, \\ldots, B_{2 n^{2}}$ is some permutation of S. For brevity, denote\n\n$$\nf\\left(\\left(A_{i}\\right),\\left(B_{i}\\right)\\right):=\\sum_{i=1}^{2 n^{2}} A_{i} B_{i}^{2}\n$$\n\nThe rest of the solution is based on the following lemma.\n\nLemma 2. The maximal value of $f\\left(\\left(A_{i}\\right),\\left(B_{i}\\right)\\right)$ over all permutations of $S$ equals $\\frac{4}{3} n^{2}\\left(4 n^{2}-1\\right)$ and is achieved when $A_{i}$ is symmetric to $B_{i}$ with respect to $O$, for every $i=1,2, \\ldots, 2 n^{2}$.\n\nProof. Let $A_{i}=\\left(p_{i}, q_{i}\\right)$ and $B_{i}=\\left(r_{i}, s_{i}\\right)$, for $i=1,2, \\ldots, 2 n^{2}$. We have\n\n$$\nf\\left(\\left(A_{i}\\right),\\left(B_{i}\\right)\\right)=\\sum_{i=1}^{2 n^{2}}\\left(p_{i}-r_{i}\\right)^{2}+\\sum_{i=1}^{2 n^{2}}\\left(q_{i}-s_{i}\\right)^{2}\n$$\n\n\n\nit suffices to bound the first sum, the second is bounded similarly. This can be done, e.g., by means of the QM-AM inequality as follows:\n\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{2 n^{2}}\\left(p_{i}-r_{i}\\right)^{2}=\\sum_{i=1}^{2 n^{2}}\\left(2 p_{i}^{2}+2 r_{i}^{2}-\\left(p_{i}+r_{i}\\right)^{2}\\right)=4 n \\sum_{j=0}^{2 n-1} j^{2}-\\sum_{i=1}^{2 n^{2}}\\left(p_{i}+r_{i}\\right)^{2} \\\\\n& \\leqslant 4 n \\sum_{j=0}^{2 n-1} j^{2}-\\frac{1}{2 n^{2}}\\left(\\sum_{i=1}^{2 n^{2}}\\left(p_{i}+r_{i}\\right)\\right)^{2}=4 n \\sum_{j=0}^{2 n-1} j^{2}-\\frac{1}{2 n^{2}}\\left(2 n \\cdot \\sum_{j=0}^{2 n-1} j\\right)^{2} \\\\\n& =4 n \\cdot \\frac{2 n(2 n-1)(4 n-1)}{6}-2 n^{2}(2 n-1)^{2}=\\frac{2 n^{2}(2 n-1)(2 n+1)}{3} .\n\\end{aligned}\n$$\n\nAll the estimates are sharp if $p_{i}+r_{i}=2 n-1$ for all $i$. Thus,\n\n$$\nf\\left(\\left(A_{i}\\right),\\left(B_{i}\\right)\\right) \\leqslant \\frac{4 n^{2}\\left(4 n^{2}-1\\right)}{3}\n$$\n\nand the estimate is sharp when $p_{i}+r_{i}=q_{i}+s_{i}=2 n-1$ for all $i$, i.e. when $A_{i}$ and $B_{i}$ are symmetric with respect to $O$.\n\nLemma 2 yields\n\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\frac{1}{4} \\cdot \\frac{4 n^{2}\\left(4 n^{2}-1\\right)}{3}=\\frac{n^{2}(2 n-1)(2 n+1)}{3}\n$$\n\nFinally, all estimates are achieved simultaneously on the set $\\mathcal{G}$ of central squares.']",['$\\Sigma(n):=\\frac{1}{3} n^{2}(2 n+1)(2 n-1)$'],False,,Need_human_evaluate, 1793,Geometry,,"Let $A B C D$ be a quadrilateral inscribed in a circle $\Omega$. Let the tangent to $\Omega$ at $D$ intersect the rays $B A$ and $B C$ at points $E$ and $F$, respectively. A point $T$ is chosen inside the triangle $A B C$ so that $T E \| C D$ and $T F \| A D$. Let $K \neq D$ be a point on the segment $D F$ such that $T D=T K$. Prove that the lines $A C, D T$ and $B K$ intersect at one point.","['Let the segments $T E$ and $T F$ cross $A C$ at $P$ and $Q$, respectively. Since $P E \\| C D$ and $E D$ is tangent to the circumcircle of $A B C D$, we have\n\n$$\n\\angle E P A=\\angle D C A=\\angle E D A\n$$\n\nand so the points $A, P, D$, and $E$ lie on some circle $\\alpha$. Similarly, the points $C, Q, D$, and $F$ lie on some circle $\\gamma$.\n\nWe now want to prove that the line $D T$ is tangent to both $\\alpha$ and $\\gamma$ at $D$. Indeed, since $\\angle F C D+\\angle E A D=180^{\\circ}$, the circles $\\alpha$ and $\\gamma$ are tangent to each other at $D$. To prove that $T$ lies on their common tangent line at $D$ (i.e., on their radical axis), it suffices to check that $T P \\cdot T E=T Q \\cdot T F$, or that the quadrilateral $P E F Q$ is cyclic. This fact follows from\n\n$$\n\\angle Q F E=\\angle A D E=\\angle A P E .\n$$\n\nSince $T D=T K$, we have $\\angle T K D=\\angle T D K$. Next, as $T D$ and $D E$ are tangent to $\\alpha$ and $\\Omega$, respectively, we obtain\n\n$$\n\\angle T K D=\\angle T D K=\\angle E A D=\\angle B D E\n$$\n\nwhich implies $T K \\| B D$.\n\nNext we prove that the five points $T, P, Q, D$, and $K$ lie on some circle $\\tau$. Indeed, since $T D$ is tangent to the circle $\\alpha$ we have\n\n$$\n\\angle E P D=\\angle T D F=\\angle T K D\n$$\n\nwhich means that the point $P$ lies on the circle (TDK). Similarly, we have $Q \\in(T D K)$.\n\nFinally, we prove that $P K \\| B C$. Indeed, using the circles $\\tau$ and $\\gamma$ we conclude that\n\n$$\n\\angle P K D=\\angle P Q D=\\angle D F C\n$$\n\nwhich means that $P K \\| B C$.\n\nTriangles $T P K$ and $D C B$ have pairwise parallel sides, which implies the fact that $T D, P C$ and $K B$ are concurrent, as desired.\n\n', 'Consider the spiral similarity $\\phi$ centred at $D$ which maps $B$ to $F$. Recall that for any two points $X$ and $Y$, the triangles $D X \\phi(X)$ and $D Y \\phi(Y)$ are similar.\n\nDefine $T^{\\prime}=\\phi(E)$. Then\n\n$$\n\\angle C D F=\\angle F B D=\\angle \\phi(B) B D=\\angle \\phi(E) E D=\\angle T^{\\prime} E D,\n$$\n\nso $C D \\| T^{\\prime} E$. Using the fact that $D E$ is tangent to $(A B D)$ and then applying $\\phi$ we infer\n\n$$\n\\angle A D E=\\angle A B D=\\angle T^{\\prime} F D\n$$\n\nso $A D \\| T^{\\prime} F$; hence $T^{\\prime}$ coincides with $T$. Therefore,\n\n$$\n\\angle B D E=\\angle F D T=\\angle D K T\n$$\n\nwhence $T K \\| B D$.\n\nLet $B K \\cap T D=X, A C \\cap T D=Y$, and $A C \\cap T F=Q$. Notice that $T K \\| B D$ implies\n\n$$\n\\frac{T X}{X D}=\\frac{T K}{B D}=\\frac{T D}{B D}\n$$\n\nSo we wish to prove that $\\frac{T Y}{Y D}$ is equal to the same ratio.\n\nWe first show that $\\phi(A)=Q$. Indeed,\n\n$$\n\\angle D A \\phi(A)=\\angle D B F=\\angle D A C\n$$\n\nand so $\\phi(A) \\in A C$. Together with $\\phi(A) \\in \\phi(E B)=T F$ this yields $\\phi(A)=Q$. It follows that\n\n$$\n\\frac{TQ}{AE}=\\frac{TD}{ED}\n$$\n\n\n\n\nNow, since $T F \\| A D$ and $\\triangle E A D \\sim \\triangle E D B$, we have\n\n$$\n\\frac{T Y}{Y D}=\\frac{T Q}{A D}=\\frac{T Q}{A E} \\cdot \\frac{A E}{A D}=\\frac{T D}{E D} \\cdot \\frac{E D}{B D}=\\frac{T D}{B D}\n$$\n\nwhich completes the proof.']",,True,,, 1794,Geometry,,"Let $A B C D$ be a cyclic quadrilateral whose sides have pairwise different lengths. Let $O$ be the circumcentre of $A B C D$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $B_{1}$ and $D_{1}$, respectively. Let $O_{B}$ be the centre of the circle which passes through $B$ and is tangent to $A C$ at $D_{1}$. Similarly, let $O_{D}$ be the centre of the circle which passes through $D$ and is tangent to $A C$ at $B_{1}$. Assume that $B D_{1} \| D B_{1}$. Prove that $O$ lies on the line $O_{B} O_{D}$.","['We introduce some objects and establish some preliminary facts common for all solutions below.\n\nLet $\\Omega$ denote the circle $(A B C D)$, and let $\\gamma_{B}$ and $\\gamma_{D}$ denote the two circles from the problem statement (their centres are $O_{B}$ and $O_{D}$, respectively). Clearly, all three centres $O, O_{B}$, and $O_{D}$ are distinct.\n\nAssume, without loss of generality, that $A B>B C$. Suppose that $A D>D C$, and let $H=A C \\cap B D$. Then the rays $B B_{1}$ and $D D_{1}$ lie on one side of $B D$, as they contain the midpoints of the arcs $A D C$ and $A B C$, respectively. However, if $B D_{1} \\| D B_{1}$, then $B_{1}$ and $D_{1}$ should be separated by $H$. This contradiction shows that $A D\ncentres $O_{B} O$; likewise, $D T_{D} \\perp O_{D} O$. Therefore, $O \\in O_{B} O_{D} \\Longleftrightarrow O_{B} O\\left\\|O_{D} O \\Longleftrightarrow B T_{B}\\right\\|$ $D T_{D}$, and the problem reduces to showing that\n\n$$\nB T_{B} \\| D T_{D}\n\\tag{1}\n$$\n\n\nLet the diagonals $A C$ and $B D$ cross at $H$. Consider the homothety $h$ centred at $H$ and mapping $B$ to $D$. Since $B D_{1} \\| D B_{1}$, we have $h\\left(D_{1}\\right)=B_{1}$.\n\nLet the tangents to $\\Omega$ at $B$ and $D$ meet $A C$ at $L_{B}$ and $L_{D}$, respectively. We have\n\n$$\n\\angle L_{B} B B_{1}=\\angle L_{B} B C+\\angle C B B_{1}=\\angle B A L_{B}+\\angle B_{1} B A=\\angle B B_{1} L_{B},\n$$\n\nwhich means that the triangle $L_{B} B B_{1}$ is isosceles, $L_{B} B=L_{B} B_{1}$. The powers of $L_{B}$ with respect to $\\Omega$ and $\\gamma_{D}$ are $L_{B} B^{2}$ and $L_{B} B_{1}^{2}$, respectively; so they are equal, whence $L_{B}$ lies on the radical axis $T_{D} D$ of those two circles. Similarly, $L_{D}$ lies on the radical axis $T_{B} B$ of $\\Omega$ and $\\gamma_{B}$.\n\nBy the sine rule in the triangle $B H L_{B}$, we obtain\n\n$$\n\\frac{H L_{B}}{\\sin \\angle H B L_{B}}=\\frac{B L_{B}}{\\sin \\angle B H L_{B}}=\\frac{B_{1} L_{B}}{\\sin \\angle B H L_{B}}\n\\tag{2}\n$$\n\nsimilarly,\n\n$$\n\\frac{H L_{D}}{\\sin \\angle H D L_{D}}=\\frac{D L_{D}}{\\sin \\angle D H L_{D}}=\\frac{D_{1} L_{D}}{\\sin \\angle D H L_{D}}\n\\tag{3}\n$$\n\nClearly, $\\angle B H L_{B}=\\angle D H L_{D}$. In the circle $\\Omega$, tangent lines $B L_{B}$ and $D L_{D}$ form equal angles with the chord $B D$, so $\\sin \\angle H B L_{B}=\\sin \\angle H D L_{D}$ (this equality does not depend on the picture). Thus, dividing (2) by (3) we get\n\n$$\n\\frac{H L_{B}}{H L_{D}}=\\frac{B_{1} L_{B}}{D_{1} L_{D}}, \\quad \\text { and hence } \\quad \\frac{H L_{B}}{H L_{D}}=\\frac{H L_{B}-B_{1} L_{B}}{H L_{D}-D_{1} L_{D}}=\\frac{H B_{1}}{H D_{1}}\n$$\n\nSince $h\\left(D_{1}\\right)=B_{1}$, the obtained relation yields $h\\left(L_{D}\\right)=L_{B}$, so $h$ maps the line $L_{D} B$ to $L_{B} D$, and these lines are parallel, as desired.\n\n\n\n', ""Let $B D_{1}$ and $T_{B} D_{1}$ meet $\\Omega$ again at $X_{B}$ and $Y_{B}$, respectively. Then\n\n$$\n\\angle B D_{1} C=\\angle B T_{B} D_{1}=\\angle B T_{B} Y_{B}=\\angle B X_{B} Y_{B}\n$$\n\nwhich shows that $X_{B} Y_{B} \\| A C$. Similarly, let $D B_{1}$ and $T_{D} B_{1}$ meet $\\Omega$ again at $X_{D}$ and $Y_{D}$, respectively; then $X_{D} Y_{D} \\| A C$.\n\nLet $M_{D}$ and $M_{B}$ be the midpoints of the arcs $A B C$ and $A D C$, respectively; then the points $D_{1}$ and $B_{1}$ lie on $D M_{D}$ and $B M_{B}$, respectively. Let $K$ be the midpoint of $A C$ (which lies on $M_{B} M_{D}$ ). Applying Pascal's theorem to $M_{D} D X_{D} X_{B} B M_{B}$, we obtain that the points $D_{1}=M_{D} D \\cap X_{B} B, B_{1}=D X_{D} \\cap B M_{B}$, and $X_{D} X_{B} \\cap M_{B} M_{D}$ are collinear, which means that $X_{B} X_{D}$ passes through $K$. Due to symmetry, the diagonals of an isosceles trapezoid $X_{B} Y_{B} X_{D} Y_{D}$ cross at $K$.\n\n\n\nLet $b$ and $d$ denote the distances from the lines $X_{B} Y_{B}$ and $X_{D} Y_{D}$, respectively, to $A C$. Then we get\n\n$$\n\\frac{X_{B} Y_{B}}{X_{D} Y_{D}}=\\frac{b}{d}=\\frac{D_{1} X_{B}}{B_{1} X_{D}}\n$$\n\nwhere the second equation holds in view of $D_{1} X_{B} \\| B_{1} X_{D}$. Therefore, the triangles $D_{1} X_{B} Y_{B}$ and $B_{1} X_{D} Y_{D}$ are similar. The triangles $D_{1} T_{B} B$ and $B_{1} T_{D} D$ are similar to them and hence to each other. Since $B D_{1} \\| D B_{1}$, these triangles are also homothetical. This yields $B T_{B} \\| D T_{D}$, as desired.""]",,True,,, 1795,Geometry,,Determine all integers $n \geqslant 3$ satisfying the following property: every convex $n$-gon whose sides all have length 1 contains an equilateral triangle of side length 1.,"['First we show that for every even $n \\geqslant 4$ there exists a polygon violating the required statement. Consider a regular $k$-gon $A_{0} A_{1}, \\ldots A_{k-1}$ with side length 1 . Let $B_{1}, B_{2}, \\ldots, B_{n / 2-1}$ be the points symmetric to $A_{1}, A_{2}, \\ldots, A_{n / 2-1}$ with respect to the line $A_{0} A_{n / 2}$. Then $P=$ $A_{0} A_{1} A_{2} \\ldots A_{n / 2-1} A_{n / 2} B_{n / 2-1} B_{n / 2-2} \\ldots B_{2} B_{1}$ is a convex $n$-gon whose sides all have length 1 . If $k$ is big enough, $P$ is contained in a strip of width $1 / 2$, which clearly does not contain any equilateral triangle of side length 1.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3cb109cf6fb51c3d6552g-1.jpg?height=169&width=1353&top_left_y=835&top_left_x=357)\n\nAssume now that $n=2 k+1$. As the case $k=1$ is trivially true, we assume $k \\geqslant 2$ henceforth. Consider a convex $(2 k+1)$-gon $P$ whose sides all have length 1 . Let $d$ be its longest diagonal. The endpoints of $d$ split the perimeter of $P$ into two polylines, one of which has length at least $k+1$. Hence we can label the vertices of $P$ so that $P=A_{0} A_{1} \\ldots A_{2 k}$ and $d=A_{0} A_{\\ell}$ with $\\ell \\geqslant k+1$. We will show that, in fact, the polygon $A_{0} A_{1} \\ldots A_{\\ell}$ contains an equilateral triangle of side length 1.\n\nSuppose that $\\angle A_{\\ell} A_{0} A_{1} \\geqslant 60^{\\circ}$. Since $d$ is the longest diagonal, we have $A_{1} A_{\\ell} \\leqslant A_{0} A_{\\ell}$, so $\\angle A_{0} A_{1} A_{\\ell} \\geqslant \\angle A_{\\ell} A_{0} A_{1} \\geqslant 60^{\\circ}$. It follows that there exists a point $X$ inside the triangle $A_{0} A_{1} A_{\\ell}$ such that the triangle $A_{0} A_{1} X$ is equilateral, and this triangle is contained in $P$. Similar arguments apply if $\\angle A_{\\ell-1} A_{\\ell} A_{0} \\geqslant 60^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3cb109cf6fb51c3d6552g-1.jpg?height=220&width=734&top_left_y=1529&top_left_x=661)\n\nFrom now on, assume $\\angle A_{\\ell} A_{0} A_{1}<60^{\\circ}$ and $A_{\\ell-1} A_{\\ell} A_{0}<60^{\\circ}$.\n\nConsider an isosceles trapezoid $A_{0} Y Z A_{\\ell}$ such that $A_{0} A_{\\ell} \\| Y Z, A_{0} Y=Z A_{\\ell}=1$, and $\\angle A_{\\ell} A_{0} Y=\\angle Z A_{\\ell} A_{0}=60^{\\circ}$. Suppose that $A_{0} A_{1} \\ldots A_{\\ell}$ is contained in $A_{0} Y Z A_{\\ell}$. Note that the perimeter of $A_{0} A_{1} \\ldots A_{\\ell}$ equals $\\ell+A_{0} A_{\\ell}$ and the perimeter of $A_{0} Y Z A_{\\ell}$ equals $2 A_{0} A_{\\ell}+1$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3cb109cf6fb51c3d6552g-1.jpg?height=209&width=1328&top_left_y=1957&top_left_x=364)\n\nRecall a well-known fact stating that if a convex polygon $P_{1}$ is contained in a convex polygon $P_{2}$, then the perimeter of $P_{1}$ is at most the perimeter of $P_{2}$. Hence we obtain\n\n$$\n\\ell+A_{0} A_{\\ell} \\leqslant 2 A_{0} A_{\\ell}+1, \\quad \\text { i.e. } \\quad \\ell-1 \\leqslant A_{0} A_{\\ell}\n$$\n\nOn the other hand, the triangle inequality yields\n\n$$\nA_{0} A_{\\ell}1 / 2$ and $P A_{\\ell}>1 / 2$. Choose points $Q \\in A_{0} P, R \\in P A_{\\ell}$, and $S \\in P A_{m}$ such that $P Q=P R=1 / 2$ and $P S=\\sqrt{3} / 2$. Then $Q R S$ is an equilateral triangle of side length 1 contained in $A_{0} A_{1} \\ldots A_{\\ell}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_464cf4ff32688fa149fbg-1.jpg?height=332&width=1330&top_left_y=408&top_left_x=363)']",['All odd $n \\geqslant 3$'],False,,Need_human_evaluate, 1796,Geometry,,"A point $D$ is chosen inside an acute-angled triangle $A B C$ with $A B>A C$ so that $\angle B A D=\angle D A C$. A point $E$ is constructed on the segment $A C$ so that $\angle A D E=\angle D C B$. Similarly, a point $F$ is constructed on the segment $A B$ so that $\angle A D F=\angle D B C$. A point $X$ is chosen on the line $A C$ so that $C X=B X$. Let $O_{1}$ and $O_{2}$ be the circumcentres of the triangles $A D C$ and $D X E$. Prove that the lines $B C, E F$, and $O_{1} O_{2}$ are concurrent.","['Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\\angle B A D=\\angle D A C$, the point $Q$ lies on $A D$. Then $\\angle Q B A=\\angle D B C=\\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \\cdot A B=A D \\cdot A Q=A E \\cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.\n\n\n\nLet $T$ be the intersection of $B C$ and $F E$.\n\nClaim. $T D^{2}=T B \\cdot T C=T F \\cdot T E$.\n\nProof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get\n\n$$\n\\begin{aligned}\n& \\angle B D F=\\angle A F D-\\angle A B D=\\left(180^{\\circ}-\\angle F A D-\\angle F D A\\right)-(\\angle A B C-\\angle D B C) \\\\\n= & 180^{\\circ}-\\angle F A D-\\angle A B C=180^{\\circ}-\\angle D A E-\\angle F E A=\\angle F E D+\\angle A D E=\\angle F E D+\\angle D C B,\n\\end{aligned}\n$$\n\nwhich implies the desired tangency.\n\nSince the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \\cdot T F=T B \\cdot T C$.\n\n\n\nLet $T A$ intersect the circle $(A B C)$ again at $M$. Due to the circles $(B C E F)$ and $(A M C B)$, and using the above Claim, we get $T M \\cdot T A=T F \\cdot T E=T B \\cdot T C=T D^{2}$; in particular, the points $A, M, E$, and $F$ are concyclic.\n\nUnder the inversion with centre $T$ and radius $T D$, the point $M$ maps to $A$, and $B$ maps to $C$, which implies that the circle $(M B D)$ maps to the circle $(A D C)$. Their common point $D$ lies on the circle of the inversion, so the second intersection point $K$ also lies on that circle, which means $T K=T D$. It follows that the point $T$ and the centres of the circles $(K D E)$ and $(A D C)$ lie on the perpendicular bisector of $K D$.\n\nSince the center of $(A D C)$ is $O_{1}$, it suffices to show now that the points $D, K, E$, and $X$ are concyclic (the center of the corresponding circle will be $O_{2}$ ).\n\nThe lines $B M, D K$, and $A C$ are the pairwise radical axes of the circles $(A B C M),(A C D K)$ and $(B M D K)$, so they are concurrent at some point $P$. Also, $M$ lies on the circle $(A E F)$, thus\n\n$$\n\\begin{aligned}\n\\sphericalangle(E X, X B) & =\\sphericalangle(C X, X B)=\\sphericalangle(X C, B C)+\\sphericalangle(B C, B X)=2 \\sphericalangle(A C, C B) \\\\\n& =\\sphericalangle(A C, C B)+\\sphericalangle(E F, F A)=\\sphericalangle(A M, B M)+\\sphericalangle(E M, M A)=\\sphericalangle(E M, B M),\n\\end{aligned}\n$$\n\nso the points $M, E, X$, and $B$ are concyclic. Therefore, $P E \\cdot P X=P M \\cdot P B=P K \\cdot P D$, so the points $E, K, D$, and $X$ are concyclic, as desired.\n\n\n\n', 'Common remarks. Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\\angle B A D=\\angle D A C$, the point $Q$ lies on $A D$. Then $\\angle Q B A=\\angle D B C=\\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \\cdot A B=A D \\cdot A Q=A E \\cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.\n\n\n\nLet $T$ be the intersection of $B C$ and $F E$.\n\nClaim. $T D^{2}=T B \\cdot T C=T F \\cdot T E$.\n\nProof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get\n\n$$\n\\begin{aligned}\n& \\angle B D F=\\angle A F D-\\angle A B D=\\left(180^{\\circ}-\\angle F A D-\\angle F D A\\right)-(\\angle A B C-\\angle D B C) \\\\\n= & 180^{\\circ}-\\angle F A D-\\angle A B C=180^{\\circ}-\\angle D A E-\\angle F E A=\\angle F E D+\\angle A D E=\\angle F E D+\\angle D C B,\n\\end{aligned}\n$$\n\nwhich implies the desired tangency.\n\nSince the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \\cdot T F=T B \\cdot T C$.\n\n\n\n\nWe use only the first part of the Common remarks, namely, the facts that the tuples $(C, D, Q, E)$ and $(B, C, E, F)$ are both concyclic. We also introduce the point $T=$ $B C \\cap E F$. Let the circle $(C D E)$ meet $B C$ again at $E_{1}$. Since $\\angle E_{1} C Q=\\angle D C E$, the arcs $D E$ and $Q E_{1}$ of the circle $(C D Q)$ are equal, so $D Q \\| E E_{1}$.\n\nSince $B F E C$ is cyclic, the line $A D$ forms equal angles with $B C$ and $E F$, hence so does $E E_{1}$. Therefore, the triangle $E E_{1} T$ is isosceles, $T E=T E_{1}$, and $T$ lies on the common perpendicular bisector of $E E_{1}$ and $D Q$.\n\nLet $U$ and $V$ be the centres of circles $(A D E)$ and $(C D Q E)$, respectively. Then $U O_{1}$ is the perpendicular bisector of $A D$. Moreover, the points $U, V$, and $O_{2}$ belong to the perpendicular bisector of $D E$. Since $U O_{1} \\| V T$, in order to show that $O_{1} O_{2}$ passes through $T$, it suffices to show that\n\n$$\n\\frac{O_{2} U}{O_{2} V}=\\frac{O_{1} U}{T V}\n\\tag{1}\n$$\n\nDenote angles $A, B$, and $C$ of the triangle $A B C$ by $\\alpha, \\beta$, and $\\gamma$, respectively. Projecting onto $A C$ we obtain\n\n$$\n\\frac{O_{2} U}{O_{2} V}=\\frac{(X E-A E) / 2}{(X E+E C) / 2}=\\frac{A X}{C X}=\\frac{A X}{B X}=\\frac{\\sin (\\gamma-\\beta)}{\\sin \\alpha}\n\\tag{2}\n$$\n\nThe projection of $O_{1} U$ onto $A C$ is $(A C-A E) / 2=C E / 2$; the angle between $O_{1} U$ and $A C$ is $90^{\\circ}-\\alpha / 2$, so\n\n$$\n\\frac{O_{1} U}{E C}=\\frac{1}{2 \\sin (\\alpha / 2)}\n\\tag{3}\n$$\n\nNext, we claim that $E, V, C$, and $T$ are concyclic. Indeed, the point $V$ lies on the perpendicular bisector of $C E$, as well as on the internal angle bisector of $\\angle C T F$. Therefore, $V$ coincides with the midpoint of the arc $C E$ of the circle $(T C E)$.\n\nNow we have $\\angle E V C=2 \\angle E E_{1} C=180^{\\circ}-(\\gamma-\\beta)$ and $\\angle V E T=\\angle V E_{1} T=90^{\\circ}-\\angle E_{1} E C=$ $90^{\\circ}-\\alpha / 2$. Therefore,\n\n$$\n\\frac{E C}{T V}=\\frac{\\sin \\angle E T C}{\\sin \\angle V E T}=\\frac{\\sin (\\gamma-\\beta)}{\\cos (\\alpha / 2)}\n\\tag{4}\n$$\n\n\n\n\n\nRecalling (2) and multiplying (3) and (4) we establish (1):\n\n$$\n\\frac{O_{2} U}{O_{2} V}=\\frac{\\sin (\\gamma-\\beta)}{\\sin \\alpha}=\\frac{1}{2 \\sin (\\alpha / 2)} \\cdot \\frac{\\sin (\\gamma-\\beta)}{\\cos (\\alpha / 2)}=\\frac{O_{1} U}{E C} \\cdot \\frac{E C}{T V}=\\frac{O_{1} U}{T V}\n$$', 'Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\\angle B A D=\\angle D A C$, the point $Q$ lies on $A D$. Then $\\angle Q B A=\\angle D B C=\\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \\cdot A B=A D \\cdot A Q=A E \\cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.\n\n\n\nLet $T$ be the intersection of $B C$ and $F E$.\n\nClaim. $T D^{2}=T B \\cdot T C=T F \\cdot T E$.\n\nProof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get\n\n$$\n\\begin{aligned}\n& \\angle B D F=\\angle A F D-\\angle A B D=\\left(180^{\\circ}-\\angle F A D-\\angle F D A\\right)-(\\angle A B C-\\angle D B C) \\\\\n= & 180^{\\circ}-\\angle F A D-\\angle A B C=180^{\\circ}-\\angle D A E-\\angle F E A=\\angle F E D+\\angle A D E=\\angle F E D+\\angle D C B,\n\\end{aligned}\n$$\n\nwhich implies the desired tangency.\n\nSince the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \\cdot T F=T B \\cdot T C$.\n\n\nNotice that $\\angle A Q E=\\angle Q C B$ and $\\angle A Q F=\\angle Q B C$; so, if we replace the point $D$ with $Q$ in the problem set up, the points $E, F$, and $T$ remain the same. So, by the Claim, we have $T Q^{2}=T B \\cdot T C=T D^{2}$.\n\nThus, there exists a circle $\\Gamma$ centred at $T$ and passing through $D$ and $Q$. We denote the second meeting point of the circles $\\Gamma$ and $(A D C)$ by $K$. Let the line $A C$ meet the circle $(D E K)$ again at $Y$; we intend to prove that $Y=X$. this will yield that the point $T$, as well as the centres $O_{1}$ and $O_{2}$, all lie on the perpendicular bisector of $D K$.\n\nLet $L=A D \\cap B C$. We perform an inversion centred at $C$; the images of the points will be denoted by primes, e.g., $A^{\\prime}$ is the image of $A$. We obtain the following configuration, constructed in a triangle $A^{\\prime} C L^{\\prime}$.\n\nThe points $D^{\\prime}$ and $Q^{\\prime}$ are chosen on the circumcircle $\\Omega$ of $A^{\\prime} L^{\\prime} C$ such that $\\sphericalangle\\left(L^{\\prime} C, D^{\\prime} C\\right)=$ $\\sphericalangle\\left(Q^{\\prime} C, A^{\\prime} C\\right)$, which means that $A^{\\prime} L^{\\prime} \\| D^{\\prime} Q^{\\prime}$. The lines $D^{\\prime} Q^{\\prime}$ and $A^{\\prime} C$ meet at $E^{\\prime}$.\n\nA circle $\\Gamma^{\\prime}$ centred on $C L^{\\prime}$ passes through $D^{\\prime}$ and $Q^{\\prime}$. Notice here that $B^{\\prime}$ lies on the segment $C L^{\\prime}$, and that $\\angle A^{\\prime} B^{\\prime} C=\\angle B A C=2 \\angle L A C=2 \\angle A^{\\prime} L^{\\prime} C$, so that $B^{\\prime} L^{\\prime}=B^{\\prime} A^{\\prime}$, and $B^{\\prime}$ lies on the perpendicular bisector of $A^{\\prime} L^{\\prime}$ (which coincides with that of $D^{\\prime} Q^{\\prime}$ ). All this means that $B^{\\prime}$ is the centre of $\\Gamma^{\\prime}$.\n\nFinally, $K^{\\prime}$ is the second meeting point of $A^{\\prime} D^{\\prime}$ and $\\Gamma^{\\prime}$, and $Y^{\\prime}$ is the second meeting point of the circle $\\left(D^{\\prime} K^{\\prime} E^{\\prime}\\right)$ and the line $A^{\\prime} E^{\\prime}$, We have $\\sphericalangle\\left(Y^{\\prime} K^{\\prime}, K^{\\prime} A^{\\prime}\\right)=\\sphericalangle\\left(Y^{\\prime} E^{\\prime}, E^{\\prime} D^{\\prime}\\right)=$ $\\sphericalangle\\left(Y^{\\prime} A^{\\prime}, A^{\\prime} L^{\\prime}\\right)$, so $A^{\\prime} L^{\\prime}$ is tangent to the circumcircle $\\omega$ of the triangle $Y^{\\prime} A^{\\prime} K^{\\prime}$.\n\nLet $O$ and $O^{*}$ be the centres of $\\Omega$ and $\\omega$, respectively. Then $O^{*} A^{\\prime} \\perp A^{\\prime} L^{\\prime} \\perp B^{\\prime} O$. The projections of vectors $\\overrightarrow{O^{*} A^{\\prime}}$ and $\\overrightarrow{B^{\\prime} O}$ onto $K^{\\prime} D^{\\prime}$ are equal to $\\overrightarrow{K^{\\prime} A^{\\prime}} / 2=\\overrightarrow{K^{\\prime} D^{\\prime}} / 2-\\overrightarrow{A^{\\prime} D^{\\prime}} / 2$. So $\\overrightarrow{O^{*} A^{\\prime}}=\\overrightarrow{B^{\\prime} O}$, or equivalently $\\overrightarrow{A^{\\prime} O}=\\overrightarrow{O^{*} B^{\\prime}}$. Projecting this equality onto $A^{\\prime} C$, we see that the projection of $\\overrightarrow{O^{*} B^{\\prime}}$ equals $\\overrightarrow{A^{\\prime} C} / 2$. Since $O^{*}$ is projected to the midpoint of $A^{\\prime} Y^{\\prime}$, this yields that $B^{\\prime}$ is projected to the midpoint of $C Y^{\\prime}$, i.e., $B^{\\prime} Y^{\\prime}=B^{\\prime} C$ and $\\angle B^{\\prime} Y^{\\prime} C=\\angle B^{\\prime} C Y^{\\prime}$. In the original figure, this rewrites as $\\angle C B Y=\\angle B C Y$, so $Y$ lies on the perpendicular bisector of $B C$, as desired.\n\n']",,True,,, 1797,Geometry,,"Let $\omega$ be the circumcircle of a triangle $A B C$, and let $\Omega_{A}$ be its excircle which is tangent to the segment $B C$. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_{A}$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_{A}$ at $X$ and $Y$, respectively. The tangent line at $P$ to the circumcircle of the triangle $A P X$ intersects the tangent line at $Q$ to the circumcircle of the triangle $A Q Y$ at a point $R$. Prove that $A R \perp B C$.","['Let $D$ be the point of tangency of $B C$ and $\\Omega_{A}$. Let $D^{\\prime}$ be the point such that $D D^{\\prime}$ is a diameter of $\\Omega_{A}$. Let $R^{\\prime}$ be (the unique) point such that $A R^{\\prime} \\perp B C$ and $R^{\\prime} D^{\\prime} \\| B C$. We shall prove that $R^{\\prime}$ coincides with $R$.\n\nLet $P X$ intersect $A B$ and $D^{\\prime} R^{\\prime}$ at $S$ and $T$, respectively. Let $U$ be the ideal common point of the parallel lines $B C$ and $D^{\\prime} R^{\\prime}$. Note that the (degenerate) hexagon $A S X T U C$ is circumscribed around $\\Omega_{A}$, hence by the Brianchon theorem $A T, S U$, and $X C$ concur at a point which we denote by $V$. Then $V S \\| B C$. It follows that $\\sphericalangle(S V, V X)=\\sphericalangle(B C, C X)=$ $\\sphericalangle(B A, A X)$, hence $A X S V$ is cyclic. Therefore, $\\sphericalangle(P X, X A)=\\sphericalangle(S V, V A)=\\sphericalangle\\left(R^{\\prime} T, T A\\right)$. Since $\\angle A P T=\\angle A R^{\\prime} T=90^{\\circ}$, the quadrilateral $A P R^{\\prime} T$ is cyclic. Hence,\n\n$$\n\\sphericalangle(X A, A P)=90^{\\circ}-\\sphericalangle(P X, X A)=90^{\\circ}-\\sphericalangle\\left(R^{\\prime} T, T A\\right)=\\sphericalangle\\left(T A, A R^{\\prime}\\right)=\\sphericalangle\\left(T P, P R^{\\prime}\\right)\n$$\n\nIt follows that $P R^{\\prime}$ is tangent to the circle $(A P X)$.\n\nAnalogous argument shows that $Q R^{\\prime}$ is tangent to the circle $(A Q Y)$. Therefore, $R=R^{\\prime}$ and $A R \\perp B C$.\n\n', ""Let $J$ and $r$ be the center and the radius of $\\Omega_{A}$. Denote the diameter of $\\omega$ by $d$ and its center by $O$. By Euler's formula, $O J^{2}=(d / 2)^{2}+d r$, so the power of $J$ with respect to $\\omega$ equals $d r$.\n\n\n\nLet $J X$ intersect $\\omega$ again at $L$. Then $J L=d$. Let $L K$ be a diameter of $\\omega$ and let $M$ be the midpoint of $J K$. Since $J L=L K$, we have $\\angle L M K=90^{\\circ}$, so $M$ lies on $\\omega$. Let $R^{\\prime}$ be the point such that $R^{\\prime} P$ is tangent to the circle $(A P X)$ and $A R^{\\prime} \\perp B C$. Note that the line $A R^{\\prime}$ is symmetric to the line $A O$ with respect to $A J$.\n\n\nLemma. Let $M$ be the midpoint of the side $J K$ in a triangle $A J K$. Let $X$ be a point on the circle $(A M K)$ such that $\\angle J X K=90^{\\circ}$. Then there exists a point $T$ on the line $K X$ such that the triangles $A K J$ and $A J T$ are similar and equioriented.\n\nProof. Note that $M X=M K$. We construct a parallelogram $A J N K$. Let $T$ be a point on $K X$ such that $\\sphericalangle(N J, J A)=\\sphericalangle(K J, J T)$. Then\n\n$$\n\\sphericalangle(J N, N A)=\\sphericalangle(K A, A M)=\\sphericalangle(K X, X M)=\\sphericalangle(M K, K X)=\\sphericalangle(J K, K T) .\n$$\n\nSo there exists a spiral similarity with center $J$ mapping the triangle $A J N$ to the triangle $T J K$. Therefore, the triangles $N J K$ and $A J T$ are similar and equioriented. It follows that the triangles $A K J$ and $A J T$ are similar and equioriented.\n\n\n\n\n\nBack to the problem, we construct a point $T$ as in the lemma. We perform the composition $\\phi$ of inversion with centre $A$ and radius $A J$ and reflection in $A J$. It is known that every triangle $A E F$ is similar and equioriented to $A \\phi(F) \\phi(E)$.\n\nSo $\\phi(K)=T$ and $\\phi(T)=K$. Let $P^{*}=\\phi(P)$ and $R^{*}=\\phi\\left(R^{\\prime}\\right)$. Observe that $\\phi(T K)$ is a circle with diameter $A P^{*}$. Let $A A^{\\prime}$ be a diameter of $\\omega$. Then $P^{*} K \\perp A K \\perp A^{\\prime} K$, so $A^{\\prime}$ lies on $P^{*} K$. The triangles $A R^{\\prime} P$ and $A P^{*} R^{*}$ are similar and equioriented, hence\n\n$\\sphericalangle\\left(A A^{\\prime}, A^{\\prime} P^{*}\\right)=\\sphericalangle\\left(A A^{\\prime}, A^{\\prime} K\\right)=\\sphericalangle(A X, X P)=\\sphericalangle(A X, X P)=\\sphericalangle\\left(A P, P R^{\\prime}\\right)=\\sphericalangle\\left(A R^{*}, R^{*} P^{*}\\right)$,\n\nso $A, A^{\\prime}, R^{*}$, and $P^{*}$ are concyclic. Since $A^{\\prime}$ and $R^{*}$ lie on $A O$, we obtain $R^{*}=A^{\\prime}$. So $R^{\\prime}=\\phi\\left(A^{\\prime}\\right)$, and $\\phi\\left(A^{\\prime}\\right) P$ is tangent to the circle $(A P X)$.\n\nAn identical argument shows that $\\phi\\left(A^{\\prime}\\right) Q$ is tangent to the circle $(A Q Y)$. Therefore, $R=$ $\\phi\\left(A^{\\prime}\\right)$ and $A R \\perp B C$.""]",,True,,, 1798,Number Theory,,"Determine all integers $n \geqslant 1$ for which there exists a pair of positive integers $(a, b)$ such that no cube of a prime divides $a^{2}+b+3$ and $$ \frac{a b+3 b+8}{a^{2}+b+3}=n $$","['As $b \\equiv-a^{2}-3\\left(\\bmod a^{2}+b+3\\right)$, the numerator of the given fraction satisfies\n\n$$\na b+3 b+8 \\equiv a\\left(-a^{2}-3\\right)+3\\left(-a^{2}-3\\right)+8 \\equiv-(a+1)^{3} \\quad\\left(\\bmod a^{2}+b+3\\right)\n$$\n\nAs $a^{2}+b+3$ is not divisible by $p^{3}$ for any prime $p$, if $a^{2}+b+3$ divides $(a+1)^{3}$ then it does also divide $(a+1)^{2}$. Since\n\n$$\n0<(a+1)^{2}<2\\left(a^{2}+b+3\\right)\n$$\n\nwe conclude $(a+1)^{2}=a^{2}+b+3$. This yields $b=2(a-1)$ and $n=2$. The choice $(a, b)=(2,2)$ with $a^{2}+b+3=9$ shows that $n=2$ indeed is a solution.']",['2'],False,,Numerical, 1799,Number Theory,,"Let $n \geqslant 100$ be an integer. The numbers $n, n+1, \ldots, 2 n$ are written on $n+1$ cards, one number per card. The cards are shuffled and divided into two piles. Prove that one of the piles contains two cards such that the sum of their numbers is a perfect square.","['To solve the problem it suffices to find three squares and three cards with numbers $a, b, c$ on them such that pairwise sums $a+b, b+c, a+c$ are equal to the chosen squares. By choosing the three consecutive squares $(2 k-1)^{2},(2 k)^{2},(2 k+1)^{2}$ we arrive at the triple\n\n$$\n(a, b, c)=\\left(2 k^{2}-4 k, \\quad 2 k^{2}+1, \\quad 2 k^{2}+4 k\\right)\n$$\n\nWe need a value for $k$ such that\n\n$$\nn \\leqslant 2 k^{2}-4 k, \\quad \\text { and } \\quad 2 k^{2}+4 k \\leqslant 2 n \\text {. }\n$$\n\nA concrete $k$ is suitable for all $n$ with\n\n$$\nn \\in\\left[k^{2}+2 k, 2 k^{2}-4 k+1\\right]=: I_{k} .\n$$\n\nFor $k \\geqslant 9$ the intervals $I_{k}$ and $I_{k+1}$ overlap because\n\n$$\n(k+1)^{2}+2(k+1) \\leqslant 2 k^{2}-4 k+1 .\n$$\n\nHence $I_{9} \\cup I_{10} \\cup \\ldots=[99, \\infty)$, which proves the statement for $n \\geqslant 99$.']",,True,,, 1800,Number Theory,,"Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $\left(d_{1}, d_{2}, \ldots, d_{k}\right)$ such that for every $i=1,2, \ldots, k$, the number $d_{1}+\cdots+d_{i}$ is a perfect square.","['For $i=1,2, \\ldots, k$ let $d_{1}+\\ldots+d_{i}=s_{i}^{2}$, and define $s_{0}=0$ as well. Obviously $0=s_{0}2 i>d_{i}>d_{i-1}>\\ldots>d_{1}$, so $j \\leqslant i$ is not possible. The only possibility is $j=i+1$.\n\nHence,\n\n$$\n\\begin{gathered}\ns_{i+1}+i=d_{i+1}=s_{i+1}^{2}-s_{i}^{2}=s_{i+1}^{2}-i^{2} \\\\\ns_{i+1}^{2}-s_{i+1}=i(i+1) .\n\\end{gathered}\n$$\n\nBy solving this equation we get $s_{i+1}=i+1$ and $d_{i+1}=2 i+1$, that finishes the proof.\n\nNow we know that the positive divisors of the number $n$ are $1,3,5, \\ldots, n-2, n$. The greatest divisor is $d_{k}=2 k-1=n$ itself, so $n$ must be odd. The second greatest divisor is $d_{k-1}=n-2$; then $n-2$ divides $n=(n-2)+2$, so $n-2$ divides 2 . Therefore, $n$ must be 1 or 3 .\n\nThe numbers $n=1$ and $n=3$ obviously satisfy the requirements: for $n=1$ we have $k=1$ and $d_{1}=1^{2}$; for $n=3$ we have $k=2, d_{1}=1^{2}$ and $d_{1}+d_{2}=1+3=2^{2}$.']","['1,3']",True,,Numerical, 1801,Number Theory,,"Alice is given a rational number $r>1$ and a line with two points $B \neq R$, where point $R$ contains a red bead and point $B$ contains a blue bead. Alice plays a solitaire game by performing a sequence of moves. In every move, she chooses a (not necessarily positive) integer $k$, and a bead to move. If that bead is placed at point $X$, and the other bead is placed at $Y$, then Alice moves the chosen bead to point $X^{\prime}$ with $\overrightarrow{Y X^{\prime}}=r^{k} \overrightarrow{Y X}$. Alice's goal is to move the red bead to the point $B$. Find all rational numbers $r>1$ such that Alice can reach her goal in at most 2021 moves.","[""Denote the red and blue beads by $\\mathcal{R}$ and $\\mathcal{B}$, respectively. Introduce coordinates on the line and identify the points with their coordinates so that $R=0$ and $B=1$. Then, during the game, the coordinate of $\\mathcal{R}$ is always smaller than the coordinate of $\\mathcal{B}$. Moreover, the distance between the beads always has the form $r^{\\ell}$ with $\\ell \\in \\mathbb{Z}$, since it only multiplies by numbers of this form. Denote the value of the distance after the $m^{\\text {th }}$ move by $d_{m}=r^{\\alpha_{m}}$, $m=0,1,2, \\ldots$ (after the $0^{\\text {th }}$ move we have just the initial position, so $\\left.\\alpha_{0}=0\\right)$.\n\nIf some bead is moved in two consecutive moves, then Alice could instead perform a single move (and change the distance from $d_{i}$ directly to $d_{i+2}$ ) which has the same effect as these two moves. So, if Alice can achieve her goal, then she may as well achieve it in fewer (or the same) number of moves by alternating the moves of $\\mathcal{B}$ and $\\mathcal{R}$. In the sequel, we assume that Alice alternates the moves, and that $\\mathcal{R}$ is shifted altogether $t$ times.\n\nIf $\\mathcal{R}$ is shifted in the $m^{\\text {th }}$ move, then its coordinate increases by $d_{m}-d_{m+1}$. Therefore, the total increment of $\\mathcal{R}$ 's coordinate, which should be 1 , equals\n\n$$\n\\begin{aligned}\n& \\text { either } \\quad \\begin{aligned}\n\\left(d_{0}-d_{1}\\right)+\\left(d_{2}-d_{3}\\right)+\\cdots+\\left(d_{2 t-2}-d_{2 t-1}\\right) & =1+\\sum_{i=1}^{t-1} r^{\\alpha_{2 i}}-\\sum_{i=1}^{t} r^{\\alpha_{2 i-1}}, \\\\\n\\quad \\text { or } \\quad\\left(d_{1}-d_{2}\\right)+\\left(d_{3}-d_{4}\\right)+\\cdots+\\left(d_{2 t-1}-d_{2 t}\\right) & =\\sum_{i=1}^{t} r^{\\alpha_{2 i-1}}-\\sum_{i=1}^{t} r^{\\alpha_{2 i}}\n\\end{aligned}\n\\end{aligned}\n$$\n\ndepending on whether $\\mathcal{R}$ or $\\mathcal{B}$ is shifted in the first move. Moreover, in the former case we should have $t \\leqslant 1011$, while in the latter one we need $t \\leqslant 1010$. So both cases reduce to an equation\n\n$$\n\\sum_{i=1}^{n} r^{\\beta_{i}}=\\sum_{i=1}^{n-1} r^{\\gamma_{i}}, \\quad \\beta_{i}, \\gamma_{i} \\in \\mathbb{Z}\n\\tag{1}\n$$\n\nfor some $n \\leqslant 1011$. Thus, if Alice can reach her goal, then this equation has a solution for $n=1011$ (we can add equal terms to both sums in order to increase $n$ ).\n\nConversely, if (1) has a solution for $n=1011$, then Alice can compose a corresponding sequence of distances $d_{0}, d_{1}, d_{2}, \\ldots, d_{2021}$ and then realise it by a sequence of moves. So the problem reduces to the solvability of (1) for $n=1011$.\n\nAssume that, for some rational $r$, there is a solution of (1). Write $r$ in lowest terms as $r=a / b$. Substitute this into (1), multiply by the common denominator, and collect all terms on the left hand side to get\n\n$$\n\\sum_{i=1}^{2 n-1}(-1)^{i} a^{\\mu_{i}} b^{N-\\mu_{i}}=0, \\quad \\mu_{i} \\in\\{0,1, \\ldots, N\\}\n\\tag{2}\n$$\n\nfor some $N \\geqslant 0$. We assume that there exist indices $j_{-}$and $j_{+}$such that $\\mu_{j_{-}}=0$ and $\\mu_{j_{+}}=N$.\n\n\n\nReducing (2) modulo $a-b$ (so that $a \\equiv b$ ), we get\n\n$$\n0=\\sum_{i=1}^{2 n-1}(-1)^{i} a^{\\mu_{i}} b^{N-\\mu_{i}} \\equiv \\sum_{i=1}^{2 n-1}(-1)^{i} b^{\\mu_{i}} b^{N-\\mu_{i}}=-b^{N} \\quad \\bmod (a-b)\n$$\n\nSince $\\operatorname{gcd}(a-b, b)=1$, this is possible only if $a-b=1$.\n\nReducing (2) modulo $a+b$ (so that $a \\equiv-b$ ), we get\n\n$$\n0=\\sum_{i=1}^{2 n-1}(-1)^{i} a^{\\mu_{i}} b^{N-\\mu_{i}} \\equiv \\sum_{i=1}^{2 n-1}(-1)^{i}(-1)^{\\mu_{i}} b^{\\mu_{i}} b^{N-\\mu_{i}}=S b^{N} \\quad \\bmod (a+b)\n$$\n\nfor some odd (thus nonzero) $S$ with $|S| \\leqslant 2 n-1$. Since $\\operatorname{gcd}(a+b, b)=1$, this is possible only if $a+b \\mid S$. So $a+b \\leqslant 2 n-1$, and hence $b=a-1 \\leqslant n-1=1010$.\n\nThus we have shown that any sought $r$ has the form indicated in the answer. It remains to show that for any $b=1,2, \\ldots, 1010$ and $a=b+1$, Alice can reach the goal. For this purpose, in (1) we put $n=a, \\beta_{1}=\\beta_{2}=\\cdots=\\beta_{a}=0$, and $\\gamma_{1}=\\gamma_{2}=\\cdots=\\gamma_{b}=1$.""]","['All $r=(b+1) / b$ with $b=1, \\ldots, 1010$']",True,,Need_human_evaluate, 1802,Number Theory,,"Prove that there are only finitely many quadruples $(a, b, c, n)$ of positive integers such that $$ n !=a^{n-1}+b^{n-1}+c^{n-1} . $$","[""For fixed $n$ there are clearly finitely many solutions; we will show that there is no solution with $n>100$. So, assume $n>100$. By the AM-GM inequality,\n\n$$\n\\begin{aligned}\nn ! & =2 n(n-1)(n-2)(n-3) \\cdot(3 \\cdot 4 \\cdots(n-4)) \\\\\n& \\leqslant 2(n-1)^{4}\\left(\\frac{3+\\cdots+(n-4)}{n-6}\\right)^{n-6}=2(n-1)^{4}\\left(\\frac{n-1}{2}\\right)^{n-6}<\\left(\\frac{n-1}{2}\\right)^{n-1}\n\\end{aligned}\n$$\n\nthus $a, b, c<(n-1) / 2$.\n\nFor every prime $p$ and integer $m \\neq 0$, let $\\nu_{p}(m)$ denote the $p$-adic valuation of $m$; that is, the greatest non-negative integer $k$ for which $p^{k}$ divides $m$. Legendre's formula states that\n\n$$\n\\nu_{p}(n !)=\\sum_{s=1}^{\\infty}\\left\\lfloor\\frac{n}{p^{s}}\\right\\rfloor\n$$\n\nand a well-know corollary of this formula is that\n\n$$\n\\nu_{p}(n !)<\\sum_{s=1}^{\\infty} \\frac{n}{p^{s}}=\\frac{n}{p-1}\n\\tag{1}\n$$\n\nIf $n$ is odd then $a^{n-1}, b^{n-1}, c^{n-1}$ are squares, and by considering them modulo 4 we conclude that $a, b$ and $c$ must be even. Hence, $2^{n-1} \\mid n$ ! but that is impossible for odd $n$ because $\\nu_{2}(n !)=\\nu_{2}((n-1) !)a+b$. On the other hand, $p \\mid c$ implies that $p100$.""]",,True,,, 1803,Number Theory,,"Determine all integers $n \geqslant 2$ with the following property: every $n$ pairwise distinct integers whose sum is not divisible by $n$ can be arranged in some order $a_{1}, a_{2}, \ldots, a_{n}$ so that $n$ divides $1 \cdot a_{1}+2 \cdot a_{2}+\cdots+n \cdot a_{n}$.","['If $n=2^{k} a$, where $a \\geqslant 3$ is odd and $k$ is a positive integer, we can consider a set containing the number $2^{k}+1$ and $n-1$ numbers congruent to 1 modulo $n$. The sum of these numbers is congruent to $2^{k}$ modulo $n$ and therefore is not divisible by $n$; for any permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of these numbers\n\n$$\n1 \\cdot a_{1}+2 \\cdot a_{2}+\\cdots+n \\cdot a_{n} \\equiv 1+\\cdots+n \\equiv 2^{k-1} a\\left(2^{k} a+1\\right) \\not \\equiv 0 \\quad\\left(\\bmod 2^{k}\\right)\n$$\n\nand a fortiori $1 \\cdot a_{1}+2 \\cdot a_{2}+\\cdots+n \\cdot a_{n}$ is not divisible by $n$.\n\nFrom now on, we suppose that $n$ is either odd or a power of 2 . Let $S$ be the given set of integers, and $s$ be the sum of elements of $S$.\n\nLemma 1. If there is a permutation $\\left(a_{i}\\right)$ of $S$ such that $(n, s)$ divides $\\sum_{i=1}^{n} i a_{i}$, then there is a permutation $\\left(b_{i}\\right)$ of $S$ such that $n$ divides $\\sum_{i=1}^{n} i b_{i}$.\n\nProof. Let $r=\\sum_{i=1}^{n} i a_{i}$. Consider the permutation $\\left(b_{i}\\right)$ defined by $b_{i}=a_{i+x}$, where $a_{j+n}=a_{j}$. For this permutation, we have\n\n$$\n\\sum_{i=1}^{n} i b_{i}=\\sum_{i=1}^{n} i a_{i+x} \\equiv \\sum_{i=1}^{n}(i-x) a_{i} \\equiv r-s x \\quad(\\bmod n)\n$$\n\nSince $(n, s)$ divides $r$, the congruence $r-s x \\equiv 0(\\bmod n)$ admits a solution.\n\nLemma 2. Every set $T$ of $k m$ integers, $m>1$, can be partitioned into $m$ sets of $k$ integers so that in every set either the sum of elements is not divisible by $k$ or all the elements leave the same remainder upon division by $k$.\n\nProof. The base case, $m=2$. If $T$ contains $k$ elements leaving the same remainder upon division by $k$, we form one subset $A$ of these elements; the remaining elements form a subset $B$. If $k$ does not divide the sum of all elements of $B$, we are done. Otherwise it is enough to exchange any element of $A$ with any element of $B$ not congruent to it modulo $k$, thus making sums of both $A$ and $B$ not divisible by $k$. This cannot be done only when all the elements of $T$ are congruent modulo $k$; in this case any partition will do.\n\nIf no $k$ elements of $T$ have the same residue modulo $k$, there are three elements $a, b, c \\in T$ leaving pairwise distinct remainders upon division by $k$. Let $t$ be the sum of elements of $T$. It suffices to find $A \\subset T$ such that $|A|=k$ and $\\sum_{x \\in A} x \\not \\equiv 0, t(\\bmod k)$ : then neither the sum of elements of $A$ nor the sum of elements of $B=T \\backslash A$ is divisible by $k$. Consider $U^{\\prime} \\subset T \\backslash\\{a, b, c\\}$ with $\\left|U^{\\prime}\\right|=k-1$. The sums of elements of three sets $U^{\\prime} \\cup\\{a\\}, U^{\\prime} \\cup\\{b\\}, U^{\\prime} \\cup\\{c\\}$ leave three different remainders upon division by $k$, and at least one of them is not congruent either to 0 or to $t$.\n\nNow let $m>2$. If $T$ contains $k$ elements leaving the same remainder upon division by $k$, we form one subset $A$ of these elements and apply the inductive hypothesis to the remaining $k(m-1)$ elements. Otherwise, we choose any $U \\subset T,|U|=k-1$. Since all the remaining elements cannot be congruent modulo $k$, there is $a \\in T \\backslash U$ such that $a \\not \\equiv-\\sum_{x \\in U} x(\\bmod k)$. Now we can take $A=U \\cup\\{a\\}$ and apply the inductive hypothesis to $T \\backslash A$.\n\n\n\nNow we are ready to prove the statement of the problem for all odd $n$ and $n=2^{k}$. The proof is by induction.\n\nIf $n$ is prime, the statement follows immediately from Lemma 1 , since in this case $(n, s)=1$. Turning to the general case, we can find prime $p$ and an integer $t$ such that $p^{t} \\mid n$ and $p^{t} \\nmid s$. By Lemma 2, we can partition $S$ into $p$ sets of $\\frac{n}{p}=k$ elements so that in every set either the sum of numbers is not divisible by $k$ or all numbers have the same residue modulo $k$.\n\nFor sets in the first category, by the inductive hypothesis there is a permutation $\\left(a_{i}\\right)$ such that $k \\mid \\sum_{i=1}^{k} i a_{i}$.\n\nIf $n$ (and therefore $k$ ) is odd, then for each permutation $\\left(b_{i}\\right)$ of a set in the second category we have\n\n$$\n\\sum_{i=1}^{k} i b_{i} \\equiv b_{1} \\frac{k(k+1)}{2} \\equiv 0 \\quad(\\bmod k)\n$$\n\nBy combining such permutation for all sets of the partition, we get a permutation $\\left(c_{i}\\right)$ of $S$ such that $k \\mid \\sum_{i=1}^{n} i c_{i}$. Since this sum is divisible by $k$, and $k$ is divisible by $(n, s)$, we are done by Lemma 1 .\n\nIf $n=2^{s}$, we have $p=2$ and $k=2^{s-1}$. Then for each of the subsets there is a permutation $\\left(a_{1}, \\ldots, a_{k}\\right)$ such that $\\sum_{i=1}^{k} i a_{i}$ is divisible by $2^{s-2}=\\frac{k}{2}$ : if the subset belongs to the first category, the expression is divisible even by $k$, and if it belongs to the second one,\n\n$$\n\\sum_{i=1}^{k} i a_{i} \\equiv a_{1} \\frac{k(k+1)}{2} \\equiv 0\\left(\\bmod \\frac{k}{2}\\right)\n$$\n\nNow the numbers of each permutation should be multiplied by all the odd or all the even numbers not exceeding $n$ in increasing order so that the resulting sums are divisible by $k$ :\n\n$$\n\\sum_{i=1}^{k}(2 i-1) a_{i} \\equiv \\sum_{i=1}^{k} 2 i a_{i} \\equiv 2 \\sum_{i=1}^{k} i a_{i} \\equiv 0 \\quad(\\bmod k)\n$$\n\nCombining these two sums, we again get a permutation $\\left(c_{i}\\right)$ of $S$ such that $k \\mid \\sum_{i=1}^{n} i c_{i}$, and finish the case by applying Lemma 1.']",['All odd integers and all powers of 2'],False,,Need_human_evaluate, 1804,Number Theory,,"Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of positive integers such that $a_{n+2 m}$ divides $a_{n}+a_{n+m}$ for all positive integers $n$ and $m$. Prove that this sequence is eventually periodic, i.e. there exist positive integers $N$ and $d$ such that $a_{n}=a_{n+d}$ for all $n>N$.","['We will make repeated use of the following simple observation:\n\nLemma 1. If a positive integer $d$ divides $a_{n}$ and $a_{n-m}$ for some $m$ and $n>2 m$, it also divides $a_{n-2 m}$. If $d$ divides $a_{n}$ and $a_{n-2 m}$, it also divides $a_{n-m}$.\n\nProof. Both parts are obvious since $a_{n}$ divides $a_{n-2 m}+a_{n-m}$.\n\nClaim. The sequence $\\left(a_{n}\\right)$ is bounded.\n\nProof. Suppose the contrary. Then there exist infinitely many indices $n$ such that $a_{n}$ is greater than each of the previous terms $a_{1}, a_{2}, \\ldots, a_{n-1}$. Let $a_{n}=k$ be such a term, $n>10$. For each $s<\\frac{n}{2}$ the number $a_{n}=k$ divides $a_{n-s}+a_{n-2 s}<2 k$, therefore\n\n$$\na_{n-s}+a_{n-2 s}=k \\text {. }\n$$\n\nIn particular,\n\n$$\na_{n}=a_{n-1}+a_{n-2}=a_{n-2}+a_{n-4}=a_{n-4}+a_{n-8}\n$$\n\nthat is, $a_{n-1}=a_{n-4}$ and $a_{n-2}=a_{n-8}$. It follows from Lemma 1 that $a_{n-1}$ divides $a_{n-1-3 s}$ for $3 sN$, then $a_{j}=t$ for infinitely many $j$.\n\nClearly the sequence $\\left(a_{n+N}\\right)_{n>0}$ satisfies the divisibility condition, and it is enough to prove that this sequence is eventually periodic. Thus truncating the sequence if necessary, we can assume that each number appears infinitely many times in the sequence. Let $k$ be the maximum number appearing in the sequence.\n\nLemma 2. If a positive integer $d$ divides $a_{n}$ for some $n$, then the numbers $i$ such that $d$ divides $a_{i}$ form an arithmetical progression with an odd difference.\n\nProof. Let $i_{1}1$ and $\\operatorname{gcd}(a, b)=1$. Note that, since $\\operatorname{gcd}(a, b)=1$,\n\n$$\nf_{m, a b}=f_{m, a} f_{m, b}\n$$\n\nby the Chinese remainder theorem. Also, note that, if $f_{m, \\ell}=f_{m+1, \\ell}$, then $P$ permutes the image of $P^{m}$ on $\\mathbb{Z}_{\\ell}$, and therefore $f_{s, \\ell}=f_{m, \\ell}$ for all $s>m$. So, as $f_{m, a b}=1$ for sufficiently large $m$, we have for each $m$\n\n$$\nf_{m, a}>f_{m+1, a} \\quad \\text { or } \\quad f_{m, a}=1, \\quad f_{m, b}>f_{m+1, b} \\quad \\text { or } \\quad f_{m, b}=1\n$$\n\nChoose the smallest $m$ such that $f_{m+1, a}=1$ or $f_{m+1, b}=1$. Without loss of generality assume that $f_{m+1, a}=1$. Then $f_{m+1, a b}=f_{m+1, b}1$. Let us choose the smallest $k$ for which this is so. To each residue in $P\\left(S_{r}\\right)$ we assign its residue modulo $p^{k-1}$; denote the resulting set by $\\bar{P}(S, r)$. We have $|\\bar{P}(S, r)|=p^{k-2}$ by virtue of minimality of $k$. Then $\\left|P\\left(S_{r}\\right)\\right|1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $$ has no positive roots.","['We first prove that, for $x>0$,\n\n$$\na_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i}\n\\tag{1}\n$$\n\nwith equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$.\n\nIf $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields\n\n$$\n\\frac{(x+1)+\\overbrace{1+1+\\cdots+1}^{a_{i}-1 \\text { ones }}}{a_{i}} \\geqslant \\sqrt[a_{i}]{(x+1) \\cdot 1^{a_{i}-1}} \\Longrightarrow a_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i} .\n$$\n\nSince $x+1>1$, the inequality is strict for $a_{i}>1$.\n\nMultiplying the inequalities (1) for $i=1,2, \\ldots, n$ yields\n\n$$\n\\prod_{i=1}^{n} a_{i}(x+1)^{1 / a_{i}} \\leqslant \\prod_{i=1}^{n}\\left(x+a_{i}\\right) \\Longleftrightarrow M(x+1)^{\\sum_{i=1}^{n} 1 / a_{i}}-\\prod_{i=1}^{n}\\left(x+a_{i}\\right) \\leqslant 0 \\Longleftrightarrow P(x) \\leqslant 0\n$$\n\nwith equality iff $a_{i}=1$ for all $i \\in\\{1,2, \\ldots, n\\}$. But this implies $M=1$, which is not possible. Hence $P(x)<0$ for all $x \\in \\mathbb{R}^{+}$, and $P$ has no positive roots.', 'We will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$.\n\nIndeed, since $a_{j} \\geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \\ldots a_{n}=M>1$ ), we have $k=\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}1$, if (2) is true for a given $r2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$.","['Set $D=2017$. Denote\n\n$$\nM_{n}=\\max _{kD$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$.\n\n(i) There exist indices $p$ and $q$ such that $a_{n}=-\\left(a_{p}+a_{q}\\right)$ and $p+q=n$. Since $a_{p}, a_{q} \\leqslant M_{n}$, we have $a_{n} \\geqslant-2 M_{n}$.\n\n(ii) On the other hand, choose an index $kD$ is lucky if $m_{n} \\leqslant 2 M_{n}$. Two cases are possible.\n\nCase 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \\leqslant 2 M_{n}$ and $M_{n} \\leqslant M_{n+1} \\leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \\leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \\leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$.\n\nCase 2. Assume now that there is no lucky index, i.e., $2 M_{n}D$. Then (1) shows that for all $n>D$ we have $m_{n} \\leqslant m_{n+1} \\leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}a_{i} \\text { for each } i-2 \\ell, \\quad \\text { and } \\quad n>D\n\\tag{2}\n$$\n\nWe first show that there must be some good index $n$. By assumption, we may take an index $N$ such that $a_{N}>\\max \\{L,-2 \\ell\\}$. Choose $n$ minimally such that $a_{n}=\\max \\left\\{a_{1}, a_{2}, \\ldots, a_{N}\\right\\}$. Now, the first condition in (2) is satisfied because of the minimality of $n$, and the second and third conditions are satisfied because $a_{n} \\geqslant a_{N}>L,-2 \\ell$, and $L \\geqslant a_{i}$ for every $i$ such that $1 \\leqslant i \\leqslant D$.\n\n\n\nLet $n$ be a good index. We derive a contradiction. We have that\n\n$$\na_{n}+a_{u}+a_{v} \\leqslant 0\\tag{3}\n$$\n\nwhenever $u+v=n$.\n\nWe define the index $u$ to maximize $a_{u}$ over $1 \\leqslant u \\leqslant n-1$, and let $v=n-u$. Then, we note that $a_{u} \\geqslant a_{v}$ by the maximality of $a_{u}$.\n\nAssume first that $v \\leqslant D$. Then, we have that\n\n$$\na_{N}+2 \\ell \\leqslant 0\n$$\n\nbecause $a_{u} \\geqslant a_{v} \\geqslant \\ell$. But this contradicts our assumption that $a_{n}>-2 \\ell$ in the second criteria of (2).\n\nNow assume that $v>D$. Then, there exist some indices $w_{1}, w_{2}$ summing up to $v$ such that\n\n$$\na_{v}+a_{w_{1}}+a_{w_{2}}=0\n$$\n\nBut combining this with (3), we have\n\n$$\na_{n}+a_{u} \\leqslant a_{w_{1}}+a_{w_{2}}\n$$\n\nBecause $a_{n}>a_{u}$, we have that $\\max \\left\\{a_{w_{1}}, a_{w_{2}}\\right\\}>a_{u}$. But since each of the $w_{i}$ is less than $v$, this contradicts the maximality of $a_{u}$.']",,True,,, 1810,Algebra,,"An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have $$ \sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1 $$ Find the largest constant $K=K(n)$ such that $$ \sum_{1 \leqslant i\n\nNow, because the $z_{i}$ are Shiny, we have that (1) yields the following bound:\n\n$$\n\\sum_{i\\ell$.\n\nCase 1: $k>\\ell$.\n\nConsider all permutations $\\phi:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ such that $\\phi^{-1}(T)=\\{2,4, \\ldots, 2 \\ell\\}$. Note that there are $k ! \\ell$ ! such permutations $\\phi$. Define\n\n$$\nf(\\phi)=\\sum_{i=1}^{n-1} x_{\\phi(i)} x_{\\phi(i+1)}\n$$\n\nWe know that $f(\\phi) \\geqslant-1$ for every permutation $\\phi$ with the above property. Averaging $f(\\phi)$ over all $\\phi$ gives\n\n$$\n-1 \\leqslant \\frac{1}{k ! \\ell !} \\sum_{\\phi} f(\\phi)=\\frac{2 \\ell}{k \\ell} M+\\frac{2(k-\\ell-1)}{k(k-1)} K\n$$\n\nwhere the equality holds because there are $k \\ell$ products in $M$, of which $2 \\ell$ are selected for each $\\phi$, and there are $k(k-1) / 2$ products in $K$, of which $k-\\ell-1$ are selected for each $\\phi$. We now have\n\n$$\nK+L+M \\geqslant K+L+\\left(-\\frac{k}{2}-\\frac{k-\\ell-1}{k-1} K\\right)=-\\frac{k}{2}+\\frac{\\ell}{k-1} K+L .\n$$\n\nSince $k \\leqslant n-1$ and $K, L \\geqslant 0$, we get the desired inequality.\n\nCase 2: $k=\\ell=n / 2$.\n\nWe do a similar approach, considering all $\\phi:\\{1,2, \\ldots, n\\} \\rightarrow\\{1,2, \\ldots, n\\}$ such that $\\phi^{-1}(T)=$ $\\{2,4, \\ldots, 2 \\ell\\}$, and defining $f$ the same way. Analogously to Case 1 , we have\n\n$$\n-1 \\leqslant \\frac{1}{k ! \\ell !} \\sum_{\\phi} f(\\phi)=\\frac{2 \\ell-1}{k \\ell} M\n$$\n\nbecause there are $k \\ell$ products in $M$, of which $2 \\ell-1$ are selected for each $\\phi$. Now, we have that\n\n$$\nK+L+M \\geqslant M \\geqslant-\\frac{n^{2}}{4(n-1)} \\geqslant-\\frac{n-1}{2}\n$$\n\nwhere the last inequality holds because $n \\geqslant 4$.']",['$-(n-1) / 2$'],False,,Expression, 1811,Algebra,,"Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(f(x) f(y))+f(x+y)=f(x y) \tag{*} $$ for all $x, y \in \mathbb{R}$.","['An easy check shows that all the 3 above mentioned functions indeed satisfy the original equation $(*)$.\n\nIn order to show that these are the only solutions, first observe that if $f(x)$ is a solution then $-f(x)$ is also a solution. Hence, without loss of generality we may (and will) assume that $f(0) \\leqslant 0$ from now on. We have to show that either $f$ is identically zero or $f(x)=x-1$ $(\\forall x \\in \\mathbb{R})$.\n\nObserve that, for a fixed $x \\neq 1$, we may choose $y \\in \\mathbb{R}$ so that $x+y=x y \\Longleftrightarrow y=\\frac{x}{x-1}$, and therefore from the original equation $(*)$ we have\n\n$$\nf\\left(f(x) \\cdot f\\left(\\frac{x}{x-1}\\right)\\right)=0 \\quad(x \\neq 1) .\n\\tag{1}\n$$\n\nIn particular, plugging in $x=0$ in (1), we conclude that $f$ has at least one zero, namely $(f(0))^{2}$ :\n\n$$\nf\\left((f(0))^{2}\\right)=0 \\text {. }\n\\tag{2}\n$$\n\nWe analyze two cases (recall that $f(0) \\leqslant 0$ ):\n\nCase 1: $f(0)=0$.\n\nSetting $y=0$ in the original equation we get the identically zero solution:\n\n$$\nf(f(x) f(0))+f(x)=f(0) \\Longrightarrow f(x)=0 \\text { for all } x \\in \\mathbb{R}\n$$\n\nFrom now on, we work on the main\n\nCase 2: $f(0)<0$.\n\nWe begin with the following\n\nClaim 1.\n\n$$\nf(1)=0, \\quad f(a)=0 \\Longrightarrow a=1, \\quad \\text { and } \\quad f(0)=-1\n\\tag{3}\n$$\n\nProof. We need to show that 1 is the unique zero of $f$. First, observe that $f$ has at least one zero $a$ by (2); if $a \\neq 1$ then setting $x=a$ in (1) we get $f(0)=0$, a contradiction. Hence from (2) we get $(f(0))^{2}=1$. Since we are assuming $f(0)<0$, we conclude that $f(0)=-1$.\n\nSetting $y=1$ in the original equation $(*)$ we get\n\n$f(f(x) f(1))+f(x+1)=f(x) \\Longleftrightarrow f(0)+f(x+1)=f(x) \\Longleftrightarrow f(x+1)=f(x)+1 \\quad(x \\in \\mathbb{R})$.\n\nAn easy induction shows that\n\n$$\nf(x+n)=f(x)+n \\quad(x \\in \\mathbb{R}, n \\in \\mathbb{Z})\n\\tag{4}\n$$\n\n\n\nNow we make the following\n\nClaim 2. $f$ is injective.\n\nProof. Suppose that $f(a)=f(b)$ with $a \\neq b$. Then by (4), for all $N \\in \\mathbb{Z}$,\n\n$$\nf(a+N+1)=f(b+N)+1 .\n$$\n\nChoose any integer $N<-b$; then there exist $x_{0}, y_{0} \\in \\mathbb{R}$ with $x_{0}+y_{0}=a+N+1, x_{0} y_{0}=b+N$. Since $a \\neq b$, we have $x_{0} \\neq 1$ and $y_{0} \\neq 1$. Plugging in $x_{0}$ and $y_{0}$ in the original equation (*) we get\n\n$$\n\\begin{array}{rlrl}\nf\\left(f\\left(x_{0}\\right) f\\left(y_{0}\\right)\\right)+f(a+N+1)=f(b+N) & \\Longleftrightarrow f\\left(f\\left(x_{0}\\right) f\\left(y_{0}\\right)\\right)+1=0 \\\\\n& \\Longleftrightarrow f\\left(f\\left(x_{0}\\right) f\\left(y_{0}\\right)+1\\right)=0 \\quad & \\text { by (4) } \\\\\n& \\Longleftrightarrow f\\left(x_{0}\\right) f\\left(y_{0}\\right)=0 & \\text { by (3). }\n\\end{array}\n$$\n\nHowever, by Claim 1 we have $f\\left(x_{0}\\right) \\neq 0$ and $f\\left(y_{0}\\right) \\neq 0$ since $x_{0} \\neq 1$ and $y_{0} \\neq 1$, a contradiction.\n\nNow the end is near. For any $t \\in \\mathbb{R}$, plug in $(x, y)=(t,-t)$ in the original equation (*) to get\n\n$$\n\\begin{aligned}\nf(f(t) f(-t))+f(0)=f\\left(-t^{2}\\right) & \\Longleftrightarrow f(f(t) f(-t))=f\\left(-t^{2}\\right)+1 & & \\text { by }(3) \\\\\n& \\Longleftrightarrow f(f(t) f(-t))=f\\left(-t^{2}+1\\right) & & \\text { by }(4) \\\\\n& \\Longleftrightarrow f(t) f(-t)=-t^{2}+1 & & \\text { by injectivity of } f .\n\\end{aligned}\n$$\n\nSimilarly, plugging in $(x, y)=(t, 1-t)$ in $(*)$ we get\n\n$$\n\\begin{aligned}\nf(f(t) f(1-t))+f(1)=f(t(1-t)) & \\Longleftrightarrow f(f(t) f(1-t))=f(t(1-t)) \\text { by }(3) \\\\\n& \\Longleftrightarrow f(t) f(1-t)=t(1-t) \\quad \\text { by injectivity of } f .\n\\end{aligned}\n$$\n\nBut since $f(1-t)=1+f(-t)$ by $(4)$, we get\n\n$$\n\\begin{aligned}\nf(t) f(1-t)=t(1-t) & \\Longleftrightarrow f(t)(1+f(-t))=t(1-t) \\Longleftrightarrow f(t)+\\left(-t^{2}+1\\right)=t(1-t) \\\\\n& \\Longleftrightarrow f(t)=t-1,\n\\end{aligned}\n$$\n\nas desired.']",['There are 3 solutions:\n\n$$\nx \\mapsto 0 \\quad \\text { or } \\quad x \\mapsto x-1 \\quad \\text { or } \\quad x \\mapsto 1-x \\quad(x \\in \\mathbb{R}) .\n$$'],False,,Need_human_evaluate, 1812,Algebra,,"Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text { for } n=1,2, \ldots\right. $$ Prove that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to 2017 .","['The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$.\n\nLemma. We have $a_{n} \\geqslant 1$ for all $n \\geqslant 1$.\n\nProof. Let us suppose otherwise in order to obtain a contradiction. Let\n\n$$\nn \\geqslant 1 \\text { be the smallest integer with } a_{n} \\leqslant 0 \\text {. }\n\\tag{1}\n$$\n\nNote that $n \\geqslant 2$. It follows that $a_{n-1} \\geqslant 1$ and $a_{n-2} \\geqslant 0$. Thus we cannot have $a_{n}=$ $a_{n-1} b_{n-1}+a_{n-2}$, so we must have $a_{n}=a_{n-1} b_{n-1}-a_{n-2}$. Since $a_{n} \\leqslant 0$, we have $a_{n-1} \\leqslant a_{n-2}$. Thus we have $a_{n-2} \\geqslant a_{n-1} \\geqslant a_{n}$.\n\nLet\n\n$$\nr \\text { be the smallest index with } a_{r} \\geqslant a_{r+1} \\geqslant a_{r+2} \\text {. }\n\\tag{2}\n$$\n\nThen $r \\leqslant n-2$ by the above, but also $r \\geqslant 2$ : if $b_{1}=1$, then $a_{2}=a_{1}=1$ and $a_{3}=a_{2} b_{2}+a_{1}>a_{2}$; if $b_{1}>1$, then $a_{2}=b_{1}>1=a_{1}$.\n\nBy the minimal choice (2) of $r$, it follows that $a_{r-1}0$. In order to have $a_{r+1} \\geqslant a_{r+2}$, we must have $a_{r+2}=a_{r+1} b_{r+1}-a_{r}$ so that $b_{r} \\geqslant 2$. Putting everything together, we conclude that\n\n$$\na_{r+1}=a_{r} b_{r} \\pm a_{r-1} \\geqslant 2 a_{r}-a_{r-1}=a_{r}+\\left(a_{r}-a_{r-1}\\right)>a_{r},\n$$\n\nwhich contradicts (2).\n\nTo complete the problem, we prove that $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$ by induction. The cases $n=0,1$ are given. Assume it is true for all non-negative integers strictly less than $n$, where $n \\geqslant 2$. There are two cases:\n\nCase 1: $b_{n-1}=1$.\n\nThen $a_{n+1}=a_{n} b_{n}+a_{n-1}$. By the inductive assumption one of $a_{n-1}, a_{n}$ is at least $n-1$ and the other, by the lemma, is at least 1 . Hence\n\n$$\na_{n+1}=a_{n} b_{n}+a_{n-1} \\geqslant a_{n}+a_{n-1} \\geqslant(n-1)+1=n .\n$$\n\nThus $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$, as desired.\n\nCase 2: $b_{n-1}>1$.\n\nSince we defined $b_{0}=1$ there is an index $r$ with $1 \\leqslant r \\leqslant n-1$ such that\n\n$$\nb_{n-1}, b_{n-2}, \\ldots, b_{r} \\geqslant 2 \\quad \\text { and } \\quad b_{r-1}=1\n$$\n\nWe have $a_{r+1}=a_{r} b_{r}+a_{r-1} \\geqslant 2 a_{r}+a_{r-1}$. Thus $a_{r+1}-a_{r} \\geqslant a_{r}+a_{r-1}$.\n\nNow we claim that $a_{r}+a_{r-1} \\geqslant r$. Indeed, this holds by inspection for $r=1$; for $r \\geqslant 2$, one of $a_{r}, a_{r-1}$ is at least $r-1$ by the inductive assumption, while the other, by the lemma, is at least 1 . Hence $a_{r}+a_{r-1} \\geqslant r$, as claimed, and therefore $a_{r+1}-a_{r} \\geqslant r$ by the last inequality in the previous paragraph.\n\nSince $r \\geqslant 1$ and, by the lemma, $a_{r} \\geqslant 1$, from $a_{r+1}-a_{r} \\geqslant r$ we get the following two inequalities:\n\n$$\na_{r+1} \\geqslant r+1 \\quad \\text { and } \\quad a_{r+1}>a_{r} \\text {. }\n$$\n\n\n\nNow observe that\n\n$$\na_{m}>a_{m-1} \\Longrightarrow a_{m+1}>a_{m} \\text { for } m=r+1, r+2, \\ldots, n-1\n$$\n\nsince $a_{m+1}=a_{m} b_{m}-a_{m-1} \\geqslant 2 a_{m}-a_{m-1}=a_{m}+\\left(a_{m}-a_{m-1}\\right)>a_{m}$. Thus\n\n$$\na_{n}>a_{n-1}>\\cdots>a_{r+1} \\geqslant r+1 \\Longrightarrow a_{n} \\geqslant n .\n$$\n\nSo $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$, as desired.', 'We say that an index $n>1$ is bad if $b_{n-1}=1$ and $b_{n-2}>1$; otherwise $n$ is good. The value of $b_{0}$ is irrelevant to the definition of $\\left(a_{n}\\right)$ since $a_{0}=0$; so we assume that $b_{0}>1$.\n\nLemma 1. (a) $a_{n} \\geqslant 1$ for all $n>0$.\n\n(b) If $n>1$ is good, then $a_{n}>a_{n-1}$.\n\nProof. Induction on $n$. In the base cases $n=1,2$ we have $a_{1}=1 \\geqslant 1, a_{2}=b_{1} a_{1} \\geqslant 1$, and finally $a_{2}>a_{1}$ if 2 is good, since in this case $b_{1}>1$.\n\nNow we assume that the lemma statement is proved for $n=1,2, \\ldots, k$ with $k \\geqslant 2$, and prove it for $n=k+1$. Recall that $a_{k}$ and $a_{k-1}$ are positive by the induction hypothesis.\n\nCase 1: $k$ is bad.\n\nWe have $b_{k-1}=1$, so $a_{k+1}=b_{k} a_{k}+a_{k-1} \\geqslant a_{k}+a_{k-1}>a_{k} \\geqslant 1$, as required.\n\nCase 2: $k$ is good.\n\nWe already have $a_{k}>a_{k-1} \\geqslant 1$ by the induction hypothesis. We consider three easy subcases.\n\nSubcase 2.1: $b_{k}>1$.\n\nThen $a_{k+1} \\geqslant b_{k} a_{k}-a_{k-1} \\geqslant a_{k}+\\left(a_{k}-a_{k-1}\\right)>a_{k} \\geqslant 1$.\n\nSubcase 2.2: $b_{k}=b_{k-1}=1$.\n\nThen $a_{k+1}=a_{k}+a_{k-1}>a_{k} \\geqslant 1$.\n\nSubcase 2.3: $b_{k}=1$ but $b_{k-1}>1$.\n\nThen $k+1$ is bad, and we need to prove only (a), which is trivial: $a_{k+1}=a_{k}-a_{k-1} \\geqslant 1$.\n\nSo, in all three subcases we have verified the required relations.\n\nLemma 2. Assume that $n>1$ is bad. Then there exists a $j \\in\\{1,2,3\\}$ such that $a_{n+j} \\geqslant$ $a_{n-1}+j+1$, and $a_{n+i} \\geqslant a_{n-1}+i$ for all $1 \\leqslant i0: b_{n+i-1}>1\\right\\}\n$$\n\n(possibly $m=+\\infty$ ). We claim that $j=\\min \\{m, 3\\}$ works. Again, we distinguish several cases, according to the value of $m$; in each of them we use Lemma 1 without reference.\n\nCase 1: $m=1$, so $b_{n}>1$.\n\nThen $a_{n+1} \\geqslant 2 a_{n}+a_{n-1} \\geqslant a_{n-1}+2$, as required.\n\nCase 2: $m=2$, so $b_{n}=1$ and $b_{n+1}>1$.\n\nThen we successively get\n\n$$\n\\begin{gathered}\na_{n+1}=a_{n}+a_{n-1} \\geqslant a_{n-1}+1 \\\\\na_{n+2} \\geqslant 2 a_{n+1}+a_{n} \\geqslant 2\\left(a_{n-1}+1\\right)+a_{n}=a_{n-1}+\\left(a_{n-1}+a_{n}+2\\right) \\geqslant a_{n-1}+4\n\\end{gathered}\n$$\n\nwhich is even better than we need.\n\n\n\nCase 3: $m>2$, so $b_{n}=b_{n+1}=1$.\n\nThen we successively get\n\n$$\n\\begin{gathered}\na_{n+1}=a_{n}+a_{n-1} \\geqslant a_{n-1}+1, \\quad a_{n+2}=a_{n+1}+a_{n} \\geqslant a_{n-1}+1+a_{n} \\geqslant a_{n-1}+2, \\\\\na_{n+3} \\geqslant a_{n+2}+a_{n+1} \\geqslant\\left(a_{n-1}+1\\right)+\\left(a_{n-1}+2\\right) \\geqslant a_{n-1}+4\n\\end{gathered}\n$$\n\nas required.\n\nLemmas 1(b) and 2 provide enough information to prove that $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$ for all $n$ and, moreover, that $a_{n} \\geqslant n$ often enough. Indeed, assume that we have found some $n$ with $a_{n-1} \\geqslant n-1$. If $n$ is good, then by Lemma 1(b) we have $a_{n} \\geqslant n$ as well. If $n$ is bad, then Lemma 2 yields $\\max \\left\\{a_{n+i}, a_{n+i+1}\\right\\} \\geqslant a_{n-1}+i+1 \\geqslant n+i$ for all $0 \\leqslant i0$, we have $f(x)+y=f(y)+x$. Prove that $f(x)+y \leqslant f(y)+x$ whenever $x>y$.","['Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows:\n\nFor every $x, y \\in \\mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$.\n\nThis condition may in turn be rewritten in the following form:\n$$\n\\text{If }g(x) \\neq g(y)\\text{, then the number x+y lies (non-strictly)}\\text{ between }g(x)\\text{ and }g(y). \n\\tag{*}\n$$\nNotice here that the function $g_{1}(x)=-g(-x)$ also satisfies $(*)$, since\n\n$$\n\\begin{gathered}\ng_{1}(x) \\neq g_{1}(y) \\Longrightarrow \\quad g(-x) \\neq g(-y) \\Longrightarrow \\quad-(x+y) \\text { lies between } g(-x) \\text { and } g(-y) \\\\\n\\Longrightarrow \\quad x+y \\text { lies between } g_{1}(x) \\text { and } g_{1}(y) .\n\\end{gathered}\n$$\n\nOn the other hand, the relation we need to prove reads now as\n\n$$\ng(x) \\leqslant g(y) \\quad \\text { whenever } xX$. Similarly, if $X>2 x$, then $g$ attains at most two values on $[x ; X-x)$ - namely, $X$ and, possibly, some $YX$ and hence by $(*)$ we get $X \\leqslant a+x \\leqslant g(a)$.\n\nNow, for any $b \\in(X-x ; x)$ with $g(b) \\neq X$ we similarly get $b+x \\leqslant g(b)$. Therefore, the number $a+b$ (which is smaller than each of $a+x$ and $b+x$ ) cannot lie between $g(a)$ and $g(b)$, which by $(*)$ implies that $g(a)=g(b)$. Hence $g$ may attain only two values on $(X-x ; x]$, namely $X$ and $g(a)>X$.\n\nTo prove the second claim, notice that $g_{1}(-x)=-X<2 \\cdot(-x)$, so $g_{1}$ attains at most two values on $(-X+x,-x]$, i.e., $-X$ and, possibly, some $-Y>-X$. Passing back to $g$, we get what we need.\n\nLemma 2. If $X<2 x$, then $g$ is constant on $(X-x ; x)$. Similarly, if $X>2 x$, then $g$ is constant on $(x ; X-x)$.\n\nProof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist $a, b \\in(X-x ; x)$ with $g(a) \\neq g(b)$; by Lemma 1 , we may assume that $g(a)=X$ and $Y=g(b)>X$.\n\nNotice that $\\min \\{X-a, X-b\\}>X-x$, so there exists a $u \\in(X-x ; x)$ such that $u<\\min \\{X-a, X-b\\}$. By Lemma 1, we have either $g(u)=X$ or $g(u)=Y$. In the former case, by (*) we have $X \\leqslant u+b \\leqslant Y$ which contradicts $u2 x$, then $g(a)=X$ for all $a \\in(x ; X-x)$.\n\nProof. Again, we only prove the first claim.\n\nBy Lemmas 1 and 2, this claim may be violated only if $g$ takes on a constant value $Y>X$ on $(X-x, x)$. Choose any $a, b \\in(X-x ; x)$ with $a2 a$. Applying Lemma 2 to $a$ in place of $x$, we obtain that $g$ is constant on $(a, Y-a)$. By (2) again, we have $x \\leqslant Y-bg(y)$ for some $xc$. Now, for any $x$ with $g(x)=c$, by $(*)$ we have $c \\leqslant x+y \\leqslant g(y)$, or $c-y \\leqslant x \\leqslant g(y)-y$. Since $c$ and $y$ are constant, we get what we need.\n\nIf $g(y)-c$. By the above arguments, we obtain that all the solutions of $g_{1}(-x)=-c$ are bounded, which is equivalent to what we need.\n\nAs an immediate consequence, the function $g$ takes on infinitely many values, which shows that the next claim is indeed widely applicable.\n\nClaim 2. If $g(x)g(y)$ for some $x']",,True,,, 1815,Combinatorics,,"Let $n$ be a positive integer. Define a chameleon to be any sequence of $3 n$ letters, with exactly $n$ occurrences of each of the letters $a, b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3 n^{2} / 2$ swaps.","['To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps.\n\nFor any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \\geqslant d(X, Z)$ for any three chameleons $X, Y$, and $Z$.\n\nLemma. Consider two chameleons\n\n$$\nP=\\underbrace{a a \\ldots a}_{n} \\underbrace{b b \\ldots b}_{n} \\underbrace{c c \\ldots c}_{n} \\text { and } Q=\\underbrace{c c \\ldots c}_{n} \\underbrace{b b \\ldots b}_{n} \\underbrace{a a \\ldots a}_{n} .\n$$\n\nThen $d(P, Q) \\geqslant 3 n^{2}$.\n\nProof. For any chameleon $X$ and any pair of distinct letters $u, v \\in\\{a, b, c\\}$, we define $f_{u, v}(X)$ to be the number of pairs of positions in $X$ such that the left one is occupied by $u$, and the right one is occupied by $v$. Define $f(X)=f_{a, b}(X)+f_{a, c}(X)+f_{b, c}(X)$. Notice that $f_{a, b}(P)=f_{a, c}(P)=f_{b, c}(P)=n^{2}$ and $f_{a, b}(Q)=f_{a, c}(Q)=f_{b, c}(Q)=0$, so $f(P)=3 n^{2}$ and $f(Q)=0$.\n\nNow consider some swap changing a chameleon $X$ to $X^{\\prime}$; say, the letters $a$ and $b$ are swapped. Then $f_{a, b}(X)$ and $f_{a, b}\\left(X^{\\prime}\\right)$ differ by exactly 1 , while $f_{a, c}(X)=f_{a, c}\\left(X^{\\prime}\\right)$ and $f_{b, c}(X)=f_{b, c}\\left(X^{\\prime}\\right)$. This yields $\\left|f(X)-f\\left(X^{\\prime}\\right)\\right|=1$, i.e., on any swap the value of $f$ changes by 1 . Hence $d(X, Y) \\geqslant$ $|f(X)-f(Y)|$ for any two chameleons $X$ and $Y$. In particular, $d(P, Q) \\geqslant|f(P)-f(Q)|=3 n^{2}$, as desired.\n\nBack to the problem, take any chameleon $X$ and notice that $d(X, P)+d(X, Q) \\geqslant d(P, Q) \\geqslant$ $3 n^{2}$ by the lemma. Consequently, $\\max \\{d(X, P), d(X, Q)\\} \\geqslant \\frac{3 n^{2}}{2}$, which establishes the problem statement.', ""In any chameleon $X$, we enumerate the positions in it from left to right by $1,2, \\ldots, 3 n$. Define $s_{c}(X)$ as the sum of positions occupied by $c$. The value of $s_{c}$ changes by at most 1 on each swap, but this fact alone does not suffice to solve the problem; so we need an improvement.\n\nFor every chameleon $X$, denote by $X_{\\bar{c}}$ the sequence obtained from $X$ by removing all $n$ letters $c$. Enumerate the positions in $X_{\\bar{c}}$ from left to right by $1,2, \\ldots, 2 n$, and define $s_{\\bar{c}, b}(X)$ as the sum of positions in $X_{\\bar{c}}$ occupied by $b$. (In other words, here we consider the positions of the $b$ 's relatively to the $a$ 's only.) Finally, denote\n\n$$\nd^{\\prime}(X, Y):=\\left|s_{c}(X)-s_{c}(Y)\\right|+\\left|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(Y)\\right| .\n$$\n\n\n\nNow consider any swap changing a chameleon $X$ to $X^{\\prime}$. If no letter $c$ is involved into this swap, then $s_{c}(X)=s_{c}\\left(X^{\\prime}\\right)$; on the other hand, exactly one letter $b$ changes its position in $X_{\\bar{c}}$, so $\\left|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}\\left(X^{\\prime}\\right)\\right|=1$. If a letter $c$ is involved into a swap, then $X_{\\bar{c}}=X_{\\bar{c}}^{\\prime}$, so $s_{\\bar{c}, b}(X)=s_{\\bar{c}, b}\\left(X^{\\prime}\\right)$ and $\\left|s_{c}(X)-s_{c}\\left(X^{\\prime}\\right)\\right|=1$. Thus, in all cases we have $d^{\\prime}\\left(X, X^{\\prime}\\right)=1$.\n\nAs in the previous solution, this means that $d(X, Y) \\geqslant d^{\\prime}(X, Y)$ for any two chameleons $X$ and $Y$. Now, for any chameleon $X$ we will indicate a chameleon $Y$ with $d^{\\prime}(X, Y) \\geqslant 3 n^{2} / 2$, thus finishing the solution.\n\nThe function $s_{c}$ attains all integer values from $1+\\cdots+n=\\frac{n(n+1)}{2}$ to $(2 n+1)+\\cdots+3 n=$ $2 n^{2}+\\frac{n(n+1)}{2}$. If $s_{c}(X) \\leqslant n^{2}+\\frac{n(n+1)}{2}$, then we put the letter $c$ into the last $n$ positions in $Y$; otherwise we put the letter $c$ into the first $n$ positions in $Y$. In either case we already have $\\left|s_{c}(X)-s_{c}(Y)\\right| \\geqslant n^{2}$.\n\nSimilarly, $s_{\\bar{c}, b}$ ranges from $\\frac{n(n+1)}{2}$ to $n^{2}+\\frac{n(n+1)}{2}$. So, if $s_{\\bar{c}, b}(X) \\leqslant \\frac{n^{2}}{2}+\\frac{n(n+1)}{2}$, then we put the letter $b$ into the last $n$ positions in $Y$ which are still free; otherwise, we put the letter $b$ into the first $n$ such positions. The remaining positions are occupied by $a$. In any case, we have $\\left|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(Y)\\right| \\geqslant \\frac{n^{2}}{2}$, thus $d^{\\prime}(X, Y) \\geqslant n^{2}+\\frac{n^{2}}{2}=\\frac{3 n^{2}}{2}$, as desired.""]",,True,,, 1816,Combinatorics,,"Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: (1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell. (2) Choose two (not necessarily adjacent) cells with the same number in them; denote that number by $2^{j}$. Replace the number in one of the cells with $2^{j+1}$ and erase the number in the other cell. At the end of the game, one cell contains the number $2^{n}$, where $n$ is a given positive integer, while the other cells are empty. Determine the maximum number of moves that Sir Alex could have made, in terms of $n$.","[""We will solve a more general problem, replacing the row of 9 cells with a row of $k$ cells, where $k$ is a positive integer. Denote by $m(n, k)$ the maximum possible number of moves Sir Alex can make starting with a row of $k$ empty cells, and ending with one cell containing the number $2^{n}$ and all the other $k-1$ cells empty. Call an operation of type (1) an insertion, and an operation of type (2) a merge.\n\nOnly one move is possible when $k=1$, so we have $m(n, 1)=1$. From now on we consider $k \\geqslant 2$, and we may assume Sir Alex's last move was a merge. Then, just before the last move, there were exactly two cells with the number $2^{n-1}$, and the other $k-2$ cells were empty.\n\nPaint one of those numbers $2^{n-1}$ blue, and the other one red. Now trace back Sir Alex's moves, always painting the numbers blue or red following this rule: if $a$ and $b$ merge into $c$, paint $a$ and $b$ with the same color as $c$. Notice that in this backward process new numbers are produced only by reversing merges, since reversing an insertion simply means deleting one of the numbers. Therefore, all numbers appearing in the whole process will receive one of the two colors.\n\nSir Alex's first move is an insertion. Without loss of generality, assume this first number inserted is blue. Then, from this point on, until the last move, there is always at least one cell with a blue number.\n\nBesides the last move, there is no move involving a blue and a red number, since all merges involves numbers with the same color, and insertions involve only one number. Call an insertion of a blue number or merge of two blue numbers a blue move, and define a red move analogously.\n\nThe whole sequence of blue moves could be repeated on another row of $k$ cells to produce one cell with the number $2^{n-1}$ and all the others empty, so there are at most $m(n-1, k)$ blue moves.\n\nNow we look at the red moves. Since every time we perform a red move there is at least one cell occupied with a blue number, the whole sequence of red moves could be repeated on a row of $k-1$ cells to produce one cell with the number $2^{n-1}$ and all the others empty, so there are at most $m(n-1, k-1)$ red moves. This proves that\n\n$$\nm(n, k) \\leqslant m(n-1, k)+m(n-1, k-1)+1 .\n$$\n\nOn the other hand, we can start with an empty row of $k$ cells and perform $m(n-1, k)$ moves to produce one cell with the number $2^{n-1}$ and all the others empty, and after that perform $m(n-1, k-1)$ moves on those $k-1$ empty cells to produce the number $2^{n-1}$ in one of them, leaving $k-2$ empty. With one more merge we get one cell with $2^{n}$ and the others empty, proving that\n\n$$\nm(n, k) \\geqslant m(n-1, k)+m(n-1, k-1)+1 .\n$$\n\n\n\nIt follows that\n\n$$\nm(n, k)=m(n-1, k)+m(n-1, k-1)+1\n\\tag{1}\n$$\n\nfor $n \\geqslant 1$ and $k \\geqslant 2$.\n\nIf $k=1$ or $n=0$, we must insert $2^{n}$ on our first move and immediately get the final configuration, so $m(0, k)=1$ and $m(n, 1)=1$, for $n \\geqslant 0$ and $k \\geqslant 1$. These initial values, together with the recurrence relation (1), determine $m(n, k)$ uniquely.\n\nFinally, we show that\n\n$$\nm(n, k)=2 \\sum_{j=0}^{k-1}\\left(\\begin{array}{l}\nn \\\\\nj\n\\end{array}\\right)-1\n\\tag{2}\n$$\n\nfor all integers $n \\geqslant 0$ and $k \\geqslant 1$.\n\nWe use induction on $n$. Since $m(0, k)=1$ for $k \\geqslant 1,(2)$ is true for the base case. We make the induction hypothesis that (2) is true for some fixed positive integer $n$ and all $k \\geqslant 1$. We have $m(n+1,1)=1=2\\left(\\begin{array}{c}n+1 \\\\ 0\\end{array}\\right)-1$, and for $k \\geqslant 2$ the recurrence relation (1) and the induction hypothesis give us\n\n$$\n\\begin{aligned}\n& m(n+1, k)=m(n, k)+m(n, k-1)+1=2 \\sum_{j=0}^{k-1}\\left(\\begin{array}{c}\nn \\\\\nj\n\\end{array}\\right)-1+2 \\sum_{j=0}^{k-2}\\left(\\begin{array}{l}\nn \\\\\nj\n\\end{array}\\right)-1+1 \\\\\n& \\quad=2 \\sum_{j=0}^{k-1}\\left(\\begin{array}{c}\nn \\\\\nj\n\\end{array}\\right)+2 \\sum_{j=0}^{k-1}\\left(\\begin{array}{c}\nn \\\\\nj-1\n\\end{array}\\right)-1=2 \\sum_{j=0}^{k-1}\\left(\\left(\\begin{array}{c}\nn \\\\\nj\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn \\\\\nj-1\n\\end{array}\\right)\\right)-1=2 \\sum_{j=0}^{k-1}\\left(\\begin{array}{c}\nn+1 \\\\\nj\n\\end{array}\\right)-1,\n\\end{aligned}\n$$\n\nwhich completes the proof.""]",['$2 \\sum_{j=0}^{8}\\left(\\begin{array}{c}n \\\\ j\\end{array}\\right)-1$'],False,,Need_human_evaluate, 1817,Combinatorics,,"A hunter and an invisible rabbit play a game in the Euclidean plane. The hunter's starting point $H_{0}$ coincides with the rabbit's starting point $R_{0}$. In the $n^{\text {th }}$ round of the game $(n \geqslant 1)$, the following happens. (1) First the invisible rabbit moves secretly and unobserved from its current point $R_{n-1}$ to some new point $R_{n}$ with $R_{n-1} R_{n}=1$. (2) The hunter has a tracking device (e.g. dog) that returns an approximate position $R_{n}^{\prime}$ of the rabbit, so that $R_{n} R_{n}^{\prime} \leqslant 1$. (3) The hunter then visibly moves from point $H_{n-1}$ to a new point $H_{n}$ with $H_{n-1} H_{n}=1$. Is there a strategy for the hunter that guarantees that after $10^{9}$ such rounds the distance between the hunter and the rabbit is below 100 ?","['If the answer were ""yes"", the hunter would have a strategy that would ""work"", no matter how the rabbit moved or where the radar pings $R_{n}^{\\prime}$ appeared. We will show the opposite: with bad luck from the radar pings, there is no strategy for the hunter that guarantees that the distance stays below 100 in $10^{9}$ rounds.\n\nSo, let $d_{n}$ be the distance between the hunter and the rabbit after $n$ rounds. Of course, if $d_{n} \\geqslant 100$ for any $n<10^{9}$, the rabbit has won — it just needs to move straight away from the hunter, and the distance will be kept at or above 100 thereon.\n\nWe will now show that, while $d_{n}<100$, whatever given strategy the hunter follows, the rabbit has a way of increasing $d_{n}^{2}$ by at least $\\frac{1}{2}$ every 200 rounds (as long as the radar pings are lucky enough for the rabbit). This way, $d_{n}^{2}$ will reach $10^{4}$ in less than $2 \\cdot 10^{4} \\cdot 200=4 \\cdot 10^{6}<10^{9}$ rounds, and the rabbit wins.\n\nSuppose the hunter is at $H_{n}$ and the rabbit is at $R_{n}$. Suppose even that the rabbit reveals its position at this moment to the hunter (this allows us to ignore all information from previous radar pings). Let $r$ be the line $H_{n} R_{n}$, and $Y_{1}$ and $Y_{2}$ be points which are 1 unit away from $r$ and 200 units away from $R_{n}$, as in the figure below.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_52ff42def30603016bd3g-1.jpg?height=440&width=1140&top_left_y=1642&top_left_x=458)\n\nThe rabbit\'s plan is simply to choose one of the points $Y_{1}$ or $Y_{2}$ and hop 200 rounds straight towards it. Since all hops stay within 1 distance unit from $r$, it is possible that all radar pings stay on $r$. In particular, in this case, the hunter has no way of knowing whether the rabbit chose $Y_{1}$ or $Y_{2}$.\n\nLooking at such pings, what is the hunter going to do? If the hunter\'s strategy tells him to go 200 rounds straight to the right, he ends up at point $H^{\\prime}$ in the figure. Note that the hunter does not have a better alternative! Indeed, after these 200 rounds he will always end up at a point to the left of $H^{\\prime}$. If his strategy took him to a point above $r$, he would end up even further from $Y_{2}$; and if his strategy took him below $r$, he would end up even further from $Y_{1}$. In other words, no matter what strategy the hunter follows, he can never be sure his distance to the rabbit will be less than $y \\stackrel{\\text { def }}{=} H^{\\prime} Y_{1}=H^{\\prime} Y_{2}$ after these 200 rounds.\n\nTo estimate $y^{2}$, we take $Z$ as the midpoint of segment $Y_{1} Y_{2}$, we take $R^{\\prime}$ as a point 200 units to the right of $R_{n}$ and we define $\\varepsilon=Z R^{\\prime}$ (note that $H^{\\prime} R^{\\prime}=d_{n}$ ). Then\n\n\n\n$$\ny^{2}=1+\\left(H^{\\prime} Z\\right)^{2}=1+\\left(d_{n}-\\varepsilon\\right)^{2}\n$$\n\nwhere\n\n$$\n\\varepsilon=200-R_{n} Z=200-\\sqrt{200^{2}-1}=\\frac{1}{200+\\sqrt{200^{2}-1}}>\\frac{1}{400}\n$$\n\nIn particular, $\\varepsilon^{2}+1=400 \\varepsilon$, so\n\n$$\ny^{2}=d_{n}^{2}-2 \\varepsilon d_{n}+\\varepsilon^{2}+1=d_{n}^{2}+\\varepsilon\\left(400-2 d_{n}\\right)\n$$\n\nSince $\\varepsilon>\\frac{1}{400}$ and we assumed $d_{n}<100$, this shows that $y^{2}>d_{n}^{2}+\\frac{1}{2}$. So, as we claimed, with this list of radar pings, no matter what the hunter does, the rabbit might achieve $d_{n+200}^{2}>d_{n}^{2}+\\frac{1}{2}$. The wabbit wins.']","['There is no such strategy for the hunter. The rabbit ""wins"".']",False,,Need_human_evaluate, 1818,Combinatorics,,"Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed only once). This way, we get $3 n$ sets of colors, split into three groups according to the orientation. It happens that for every set in any group, the same set appears in both of the other groups. Determine, in terms of $n$, the maximal possible number of colors that are present.","['Call a $n \\times n \\times 1$ box an $x$-box, a $y$-box, or a $z$-box, according to the direction of its short side. Let $C$ be the number of colors in a valid configuration. We start with the upper bound for $C$.\n\nLet $\\mathcal{C}_{1}, \\mathcal{C}_{2}$, and $\\mathcal{C}_{3}$ be the sets of colors which appear in the big cube exactly once, exactly twice, and at least thrice, respectively. Let $M_{i}$ be the set of unit cubes whose colors are in $\\mathcal{C}_{i}$, and denote $n_{i}=\\left|M_{i}\\right|$.\n\nConsider any $x$-box $X$, and let $Y$ and $Z$ be a $y$ - and a $z$-box containing the same set of colors as $X$ does.\n\nClaim. $4\\left|X \\cap M_{1}\\right|+\\left|X \\cap M_{2}\\right| \\leqslant 3 n+1$.\n\nProof. We distinguish two cases.\n\nCase 1: $X \\cap M_{1} \\neq \\varnothing$.\n\nA cube from $X \\cap M_{1}$ should appear in all three boxes $X, Y$, and $Z$, so it should lie in $X \\cap Y \\cap Z$. Thus $X \\cap M_{1}=X \\cap Y \\cap Z$ and $\\left|X \\cap M_{1}\\right|=1$.\n\nConsider now the cubes in $X \\cap M_{2}$. There are at most $2(n-1)$ of them lying in $X \\cap Y$ or $X \\cap Z$ (because the cube from $X \\cap Y \\cap Z$ is in $M_{1}$ ). Let $a$ be some other cube from $X \\cap M_{2}$. Recall that there is just one other cube $a^{\\prime}$ sharing a color with $a$. But both $Y$ and $Z$ should contain such cube, so $a^{\\prime} \\in Y \\cap Z$ (but $a^{\\prime} \\notin X \\cap Y \\cap Z$ ). The map $a \\mapsto a^{\\prime}$ is clearly injective, so the number of cubes $a$ we are interested in does not exceed $|(Y \\cap Z) \\backslash X|=n-1$. Thus $\\left|X \\cap M_{2}\\right| \\leqslant 2(n-1)+(n-1)=3(n-1)$, and hence $4\\left|X \\cap M_{1}\\right|+\\left|X \\cap M_{2}\\right| \\leqslant 4+3(n-1)=3 n+1$.\n\nCase 2: $X \\cap M_{1}=\\varnothing$.\n\nIn this case, the same argument applies with several changes. Indeed, $X \\cap M_{2}$ contains at most $2 n-1$ cubes from $X \\cap Y$ or $X \\cap Z$. Any other cube $a$ in $X \\cap M_{2}$ corresponds to some $a^{\\prime} \\in Y \\cap Z$ (possibly with $a^{\\prime} \\in X$ ), so there are at most $n$ of them. All this results in $\\left|X \\cap M_{2}\\right| \\leqslant(2 n-1)+n=3 n-1$, which is even better than we need (by the assumptions of our case).\n\nSumming up the inequalities from the Claim over all $x$-boxes $X$, we obtain\n\n$$\n4 n_{1}+n_{2} \\leqslant n(3 n+1) .\n$$\n\nObviously, we also have $n_{1}+n_{2}+n_{3}=n^{3}$.\n\nNow we are prepared to estimate $C$. Due to the definition of the $M_{i}$, we have $n_{i} \\geqslant i\\left|\\mathcal{C}_{i}\\right|$, so\n\n$$\nC \\leqslant n_{1}+\\frac{n_{2}}{2}+\\frac{n_{3}}{3}=\\frac{n_{1}+n_{2}+n_{3}}{3}+\\frac{4 n_{1}+n_{2}}{6} \\leqslant \\frac{n^{3}}{3}+\\frac{3 n^{2}+n}{6}=\\frac{n(n+1)(2 n+1)}{6} .\n$$\n\nIt remains to present an example of an appropriate coloring in the above-mentioned number of colors. For each color, we present the set of all cubes of this color. These sets are:\n\n1. $n$ singletons of the form $S_{i}=\\{(i, i, i)\\}$ (with $1 \\leqslant i \\leqslant n$ );\n2. $3\\left(\\begin{array}{c}n \\\\ 2\\end{array}\\right)$ doubletons of the forms $D_{i, j}^{1}=\\{(i, j, j),(j, i, i)\\}, D_{i, j}^{2}=\\{(j, i, j),(i, j, i)\\}$, and $D_{i, j}^{3}=$ $\\{(j, j, i),(i, i, j)\\}$ (with $1 \\leqslant i0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \cdots)}_{B \text { appears } a \text { times }} . $$","['For any function $g: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ and any subset $X \\subset \\mathbb{Z}_{>0}$, we define $g(X)=$ $\\{g(x): x \\in X\\}$. We have that the image of $f_{X}$ is $f_{X}\\left(\\mathbb{Z}_{>0}\\right)=\\mathbb{Z}_{>0} \\backslash X$. We now show a general lemma about the operation $*$, with the goal of showing that $*$ is associative.\n\nLemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \\circ f_{Y}$ are equal.\n\nProof. We have\n\n$f_{X * Y}\\left(\\mathbb{Z}_{>0}\\right)=\\mathbb{Z}_{>0} \\backslash(X * Y)=\\left(\\mathbb{Z}_{>0} \\backslash X\\right) \\backslash f_{X}(Y)=f_{X}\\left(\\mathbb{Z}_{>0}\\right) \\backslash f_{X}(Y)=f_{X}\\left(\\mathbb{Z}_{>0} \\backslash Y\\right)=f_{X}\\left(f_{Y}\\left(\\mathbb{Z}_{>0}\\right)\\right)$.\n\nThus, the functions $f_{X * Y}$ and $f_{X} \\circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \\circ f_{Y}$.\n\nLemma 1 implies that $*$ is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting\n\n$$\n\\begin{gathered}\n\\mathbb{Z}_{>0} \\backslash((A * B) * C)=f_{(A * B) * C}\\left(\\mathbb{Z}_{>0}\\right)=f_{A * B}\\left(f_{C}\\left(\\mathbb{Z}_{>0}\\right)\\right)=f_{A}\\left(f_{B}\\left(f_{C}\\left(\\mathbb{Z}_{>0}\\right)\\right)\\right) \\\\\n=f_{A}\\left(f_{B * C}\\left(\\mathbb{Z}_{>0}\\right)=f_{A *(B * C)}\\left(\\mathbb{Z}_{>0}\\right)=\\mathbb{Z}_{>0} \\backslash(A *(B * C))\\right.\n\\end{gathered}\n$$\n\nIn light of the associativity of $*$, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation\n\n$$\nX^{* k}=\\underbrace{X *(X * \\cdots *(X *(X * X)) \\ldots)}_{X \\text { appears } k \\text { times }} .\n$$\n\nOur goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma.\n\nLemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$.\n\nProof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \\in X \\backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that\n\n$$\nf_{X}(s)=s+\\left|X \\cap\\left\\{1,2, \\ldots, f_{X}(s)\\right\\}\\right|\n\\tag{1}\n$$\n\nSince $f_{X}(s) \\geqslant s$, we have that\n\n$$\n\\left\\{f_{X}(s)+1, f_{X}(s)+2, \\ldots\\right\\} \\cap X=\\left\\{f_{X}(s)+1, f_{X}(s)+2, \\ldots\\right\\} \\cap Y\n$$\n\nwhich, together with the assumption that $|X|=|Y|$, gives\n\n$$\n\\left|X \\cap\\left\\{1,2, \\ldots, f_{X}(s)\\right\\}\\right|=\\left|Y \\cap\\left\\{1,2, \\ldots, f_{X}(s)\\right\\}\\right|\n\\tag{2}\n$$\n\n\n\nNow consider the equation\n\n$$\nt-|Y \\cap\\{1,2, \\ldots, t\\}|=s\n$$\n\nThis equation is satisfied only when $t \\in\\left[f_{Y}(s), f_{Y}(s+1)\\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \\notin X$ and $f_{X}(s) \\geqslant s$, we have that $f_{X}(s) \\notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$.\n\nFinally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \\notin X * Y$. However, since $s \\in X$, we have $f_{Y}(s) \\in Y * X$, a contradiction.\n\nWe are now ready to finish the proof. Note first of all that $\\left|A^{* b}\\right|=a b=\\left|B^{* a}\\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2 , we have $A^{* b}=B^{* a}$, as desired.']",,True,,, 1820,Combinatorics,,"Let $n$ be a given positive integer. In the Cartesian plane, each lattice point with nonnegative coordinates initially contains a butterfly, and there are no other butterflies. The neighborhood of a lattice point $c$ consists of all lattice points within the axis-aligned $(2 n+1) \times$ $(2 n+1)$ square centered at $c$, apart from $c$ itself. We call a butterfly lonely, crowded, or comfortable, depending on whether the number of butterflies in its neighborhood $N$ is respectively less than, greater than, or equal to half of the number of lattice points in $N$. Every minute, all lonely butterflies fly away simultaneously. This process goes on for as long as there are any lonely butterflies. Assuming that the process eventually stops, determine the number of comfortable butterflies at the final state.","['We always identify a butterfly with the lattice point it is situated at. For two points $p$ and $q$, we write $p \\geqslant q$ if each coordinate of $p$ is at least the corresponding coordinate of $q$. Let $O$ be the origin, and let $\\mathcal{Q}$ be the set of initially occupied points, i.e., of all lattice points with nonnegative coordinates. Let $\\mathcal{R}_{\\mathrm{H}}=\\{(x, 0): x \\geqslant 0\\}$ and $\\mathcal{R}_{\\mathrm{V}}=\\{(0, y): y \\geqslant 0\\}$ be the sets of the lattice points lying on the horizontal and vertical boundary rays of $\\mathcal{Q}$. Denote by $N(a)$ the neighborhood of a lattice point $a$.\n\n1. Initial observations. We call a set of lattice points up-right closed if its points stay in the set after being shifted by any lattice vector $(i, j)$ with $i, j \\geqslant 0$. Whenever the butterflies form a up-right closed set $\\mathcal{S}$, we have $|N(p) \\cap \\mathcal{S}| \\geqslant|N(q) \\cap \\mathcal{S}|$ for any two points $p, q \\in \\mathcal{S}$ with $p \\geqslant q$. So, since $\\mathcal{Q}$ is up-right closed, the set of butterflies at any moment also preserves this property. We assume all forthcoming sets of lattice points to be up-right closed.\n\nWhen speaking of some set $\\mathcal{S}$ of lattice points, we call its points lonely, comfortable, or crowded with respect to this set (i.e., as if the butterflies were exactly at all points of $\\mathcal{S}$ ). We call a set $\\mathcal{S} \\subset \\mathcal{Q}$ stable if it contains no lonely points. In what follows, we are interested only in those stable sets whose complements in $\\mathcal{Q}$ are finite, because one can easily see that only a finite number of butterflies can fly away on each minute.\n\nIf the initial set $\\mathcal{Q}$ of butterflies contains some stable set $\\mathcal{S}$, then, clearly no butterfly of this set will fly away. On the other hand, the set $\\mathcal{F}$ of all butterflies in the end of the process is stable. This means that $\\mathcal{F}$ is the largest (with respect to inclusion) stable set within $\\mathcal{Q}$, and we are about to describe this set.\n\n2. A description of a final set. The following notion will be useful. Let $\\mathcal{U}=\\left\\{\\vec{u}_{1}, \\vec{u}_{2}, \\ldots, \\vec{u}_{d}\\right\\}$ be a set of $d$ pairwise non-parallel lattice vectors, each having a positive $x$ - and a negative $y$-coordinate. Assume that they are numbered in increasing order according to slope. We now define a $\\mathcal{U}$-curve to be the broken line $p_{0} p_{1} \\ldots p_{d}$ such that $p_{0} \\in \\mathcal{R}_{\\mathrm{V}}, p_{d} \\in \\mathcal{R}_{\\mathrm{H}}$, and $\\vec{p}_{i-1} \\vec{p}_{i}=\\vec{u}_{i}$ for all $i=1,2, \\ldots, m$ (see the Figure below to the left).\n\n\n\nConstruction of $\\mathcal{U}$-curve\n\n\n\n\n\nConstruction of $\\mathcal{D}$\n\n\n\nNow, let $\\mathcal{K}_{n}=\\{(i, j): 1 \\leqslant i \\leqslant n,-n \\leqslant j \\leqslant-1\\}$. Consider all the rays emerging at $O$ and passing through a point from $\\mathcal{K}_{n}$; number them as $r_{1}, \\ldots, r_{m}$ in increasing order according to slope. Let $A_{i}$ be the farthest from $O$ lattice point in $r_{i} \\cap \\mathcal{K}_{n}$, set $k_{i}=\\left|r_{i} \\cap \\mathcal{K}_{n}\\right|$, let $\\vec{v}_{i}=\\overrightarrow{O A_{i}}$, and finally denote $\\mathcal{V}=\\left\\{\\vec{v}_{i}: 1 \\leqslant i \\leqslant m\\right\\}$; see the Figure above to the right. We will concentrate on the $\\mathcal{V}$-curve $d_{0} d_{1} \\ldots d_{m}$; let $\\mathcal{D}$ be the set of all lattice points $p$ such that $p \\geqslant p^{\\prime}$ for some (not necessarily lattice) point $p^{\\prime}$ on the $\\mathcal{V}$-curve. In fact, we will show that $\\mathcal{D}=\\mathcal{F}$.\n\nClearly, the $\\mathcal{V}$-curve is symmetric in the line $y=x$. Denote by $D$ the convex hull of $\\mathcal{D}$.\n\n3. We prove that the set $\\mathcal{D}$ contains all stable sets. Let $\\mathcal{S} \\subset \\mathcal{Q}$ be a stable set (recall that it is assumed to be up-right closed and to have a finite complement in $\\mathcal{Q}$ ). Denote by $S$ its convex hull; clearly, the vertices of $S$ are lattice points. The boundary of $S$ consists of two rays (horizontal and vertical ones) along with some $\\mathcal{V}_{*}$-curve for some set of lattice vectors $\\mathcal{V}_{*}$.\n\nClaim 1. For every $\\vec{v}_{i} \\in \\mathcal{V}$, there is a $\\vec{v}_{i}^{*} \\in \\mathcal{V}_{*}$ co-directed with $\\vec{v}$ with $\\left|\\vec{v}_{i}^{*}\\right| \\geqslant|\\vec{v}|$.\n\nProof. Let $\\ell$ be the supporting line of $S$ parallel to $\\vec{v}_{i}$ (i.e., $\\ell$ contains some point of $S$, and the set $S$ lies on one side of $\\ell$ ). Take any point $b \\in \\ell \\cap \\mathcal{S}$ and consider $N(b)$. The line $\\ell$ splits the set $N(b) \\backslash \\ell$ into two congruent parts, one having an empty intersection with $\\mathcal{S}$. Hence, in order for $b$ not to be lonely, at least half of the set $\\ell \\cap N(b)$ (which contains $2 k_{i}$ points) should lie in $S$. Thus, the boundary of $S$ contains a segment $\\ell \\cap S$ with at least $k_{i}+1$ lattice points (including $b$ ) on it; this segment corresponds to the required vector $\\vec{v}_{i}^{*} \\in \\mathcal{V}_{*}$.\n\n\n\nProof of Claim 1\n\n\n\nProof of Claim 2\n\nClaim 2. Each stable set $\\mathcal{S} \\subseteq \\mathcal{Q}$ lies in $\\mathcal{D}$.\n\nProof. To show this, it suffices to prove that the $\\mathcal{V}_{*}$-curve lies in $D$, i.e., that all its vertices do so. Let $p^{\\prime}$ be an arbitrary vertex of the $\\mathcal{V}_{*}$-curve; $p^{\\prime}$ partitions this curve into two parts, $\\mathcal{X}$ (being down-right of $p$ ) and $\\mathcal{Y}$ (being up-left of $p$ ). The set $\\mathcal{V}$ is split now into two parts: $\\mathcal{V}_{\\mathcal{X}}$ consisting of those $\\vec{v}_{i} \\in \\mathcal{V}$ for which $\\vec{v}_{i}^{*}$ corresponds to a segment in $\\mathcal{X}$, and a similar part $\\mathcal{V}_{\\mathcal{Y}}$. Notice that the $\\mathcal{V}$-curve consists of several segments corresponding to $\\mathcal{V}_{\\mathcal{X}}$, followed by those corresponding to $\\mathcal{V}_{\\mathcal{Y}}$. Hence there is a vertex $p$ of the $\\mathcal{V}$-curve separating $\\mathcal{V}_{\\mathcal{X}}$ from $\\mathcal{V}_{\\mathcal{Y}}$. Claim 1 now yields that $p^{\\prime} \\geqslant p$, so $p^{\\prime} \\in \\mathcal{D}$, as required.\n\nClaim 2 implies that the final set $\\mathcal{F}$ is contained in $\\mathcal{D}$.\n\n4. $\\mathcal{D}$ is stable, and its comfortable points are known. Recall the definitions of $r_{i}$; let $r_{i}^{\\prime}$ be the ray complementary to $r_{i}$. By our definitions, the set $N(O)$ contains no points between the rays $r_{i}$ and $r_{i+1}$, as well as between $r_{i}^{\\prime}$ and $r_{i+1}^{\\prime}$.\n\nClaim 3. In the set $\\mathcal{D}$, all lattice points of the $\\mathcal{V}$-curve are comfortable.\n\nProof. Let $p$ be any lattice point of the $\\mathcal{V}$-curve, belonging to some segment $d_{i} d_{i+1}$. Draw the line $\\ell$ containing this segment. Then $\\ell \\cap \\mathcal{D}$ contains exactly $k_{i}+1$ lattice points, all of which lie in $N(p)$ except for $p$. Thus, exactly half of the points in $N(p) \\cap \\ell$ lie in $\\mathcal{D}$. It remains to show that all points of $N(p)$ above $\\ell$ lie in $\\mathcal{D}$ (recall that all the points below $\\ell$ lack this property).\n\n\n\nNotice that each vector in $\\mathcal{V}$ has one coordinate greater than $n / 2$; thus the neighborhood of $p$ contains parts of at most two segments of the $\\mathcal{V}$-curve succeeding $d_{i} d_{i+1}$, as well as at most two of those preceding it.\n\nThe angles formed by these consecutive segments are obtained from those formed by $r_{j}$ and $r_{j-1}^{\\prime}$ (with $i-1 \\leqslant j \\leqslant i+2$ ) by shifts; see the Figure below. All the points in $N(p)$ above $\\ell$ which could lie outside $\\mathcal{D}$ lie in shifted angles between $r_{j}, r_{j+1}$ or $r_{j}^{\\prime}, r_{j-1}^{\\prime}$. But those angles, restricted to $N(p)$, have no lattice points due to the above remark. The claim is proved.\n\n\nProof of Claim 3\n\nClaim 4. All the points of $\\mathcal{D}$ which are not on the boundary of $D$ are crowded.\n\nProof. Let $p \\in \\mathcal{D}$ be such a point. If it is to the up-right of some point $p^{\\prime}$ on the curve, then the claim is easy: the shift of $N\\left(p^{\\prime}\\right) \\cap \\mathcal{D}$ by $\\overrightarrow{p^{\\prime} p}$ is still in $\\mathcal{D}$, and $N(p)$ contains at least one more point of $\\mathcal{D}$ - either below or to the left of $p$. So, we may assume that $p$ lies in a right triangle constructed on some hypothenuse $d_{i} d_{i+1}$. Notice here that $d_{i}, d_{i+1} \\in N(p)$.\n\nDraw a line $\\ell \\| d_{i} d_{i+1}$ through $p$, and draw a vertical line $h$ through $d_{i}$; see Figure below. Let $\\mathcal{D}_{\\mathrm{L}}$ and $\\mathcal{D}_{\\mathrm{R}}$ be the parts of $\\mathcal{D}$ lying to the left and to the right of $h$, respectively (points of $\\mathcal{D} \\cap h$ lie in both parts).\n\n\n\nNotice that the vectors $\\overrightarrow{d_{i} p}, \\overrightarrow{d_{i+1} d_{i+2}}, \\overrightarrow{d_{i} d_{i+1}}, \\overrightarrow{d_{i-1} d_{i}}$, and $\\overrightarrow{p d_{i+1}}$ are arranged in non-increasing order by slope. This means that $\\mathcal{D}_{\\mathrm{L}}$ shifted by $\\overrightarrow{d_{i} p}$ still lies in $\\mathcal{D}$, as well as $\\mathcal{D}_{\\mathrm{R}}$ shifted by $\\overrightarrow{d_{i+1} p}$. As we have seen in the proof of Claim 3, these two shifts cover all points of $N(p)$ above $\\ell$, along with those on $\\ell$ to the left of $p$. Since $N(p)$ contains also $d_{i}$ and $d_{i+1}$, the point $p$ is crowded.\n\nThus, we have proved that $\\mathcal{D}=\\mathcal{F}$, and have shown that the lattice points on the $\\mathcal{V}$-curve are exactly the comfortable points of $\\mathcal{D}$. It remains to find their number.\n\nRecall the definition of $\\mathcal{K}_{n}$ (see Figure on the first page of the solution). Each segment $d_{i} d_{i+1}$ contains $k_{i}$ lattice points different from $d_{i}$. Taken over all $i$, these points exhaust all the lattice points in the $\\mathcal{V}$-curve, except for $d_{1}$, and thus the number of lattice points on the $\\mathcal{V}$-curve is $1+\\sum_{i=1}^{m} k_{i}$. On the other hand, $\\sum_{i=1}^{m} k_{i}$ is just the number of points in $\\mathcal{K}_{n}$, so it equals $n^{2}$. Hence the answer to the problem is $n^{2}+1$.']",['$n^{2}+1$'],False,,Expression, 1821,Geometry,,"Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent.","['Throughout the solution, we refer to $\\angle A, \\angle B, \\angle C, \\angle D$, and $\\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \\perp C I$ and, similarly, $A C \\perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \\perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \\perp B C$.\n\nLines $I B$ and $I C$ bisect $\\angle B$ and $\\angle C$, respectively. Since $I A=I C, I B=I D$, and $A B=$ $B C=C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\\angle I A B=\\angle I C B=$ $\\angle C / 2=\\angle A / 2$, so the line $I A$ bisects $\\angle A$. Similarly, the line $I D$ bisects $\\angle D$. Finally, the line $I E$ bisects $\\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon.\n\nThe sum of the internal angles in a pentagon is $540^{\\circ}$, so\n\n$$\n\\angle E=540^{\\circ}-2 \\angle A+2 \\angle B \\text {. }\n$$\n\nIn quadrilateral $A B I E$,\n\n$$\n\\begin{aligned}\n\\angle B I E & =360^{\\circ}-\\angle E A B-\\angle A B I-\\angle A E I=360^{\\circ}-\\angle A-\\frac{1}{2} \\angle B-\\frac{1}{2} \\angle E \\\\\n& =360^{\\circ}-\\angle A-\\frac{1}{2} \\angle B-\\left(270^{\\circ}-\\angle A-\\angle B\\right) \\\\\n& =90^{\\circ}+\\frac{1}{2} \\angle B=90^{\\circ}+\\angle I B C,\n\\end{aligned}\n$$\n\nwhich means that $E I \\perp B C$, completing the proof.\n\n', ""We present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$.\n\nApplying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.""]",,True,,, 1822,Geometry,,"Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$.","['In the circles $\\Omega$ and $\\Gamma$ we have $\\angle J R S=\\angle J I S=\\angle A R^{\\prime} S$. On the other hand, since $R A$ is tangent to $\\Omega$, we get $\\angle S J R=\\angle S R A$. So the triangles $A R R^{\\prime}$ and $S J R$ are similar, and\n\n$$\n\\frac{R^{\\prime} R}{R J}=\\frac{A R^{\\prime}}{S R}=\\frac{A R^{\\prime}}{S R^{\\prime}}\n$$\n\nThe last relation, together with $\\angle A R^{\\prime} S=\\angle J R R^{\\prime}$, yields $\\triangle A S R^{\\prime} \\sim \\triangle R^{\\prime} J R$, hence $\\angle S A R^{\\prime}=\\angle R R^{\\prime} J$. It follows that $J R^{\\prime}$ is tangent to $\\Gamma$ at $R^{\\prime}$.\n\n', '\n\nWe notice that $\\angle J R S=\\angle J I S=\\angle A R^{\\prime} S$, so we have $R J \\| A R^{\\prime}$.\n\nLet $A^{\\prime}$ be the reflection of $A$ about $S$; then $A R A^{\\prime} R^{\\prime}$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $R A^{\\prime}$.\n\nFrom $\\angle S R^{\\prime} A^{\\prime}=\\angle S R A=\\angle S J R$ we get that the points $S, J, A^{\\prime}, R^{\\prime}$ are concyclic. This proves that $\\angle S R^{\\prime} J=\\angle S A^{\\prime} J=\\angle S A^{\\prime} R=\\angle S A R^{\\prime}$, so $J R^{\\prime}$ is tangent to $\\Gamma$ at $R^{\\prime}$.']",,True,,, 1823,Geometry,,"Let $O$ be the circumcenter of an acute scalene triangle $A B C$. Line $O A$ intersects the altitudes of $A B C$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $P Q H$ lies on a median of triangle $A B C$.","['Suppose, without loss of generality, that $A B\n\nLet $T$ be the center of $\\omega$ and let lines $A T$ and $B C$ meet at $M$. We will take advantage of the similarity between $A B C$ and $H P Q$ and the fact that $A H$ is tangent to $\\omega$ at $H$, with $A$ on line $P Q$. Consider the corresponding tangent $A S$ to $\\Omega$, with $S \\in B C$. Then $S$ and $A$ correspond to each other in $\\triangle A B C \\sim \\triangle H P Q$, and therefore $\\angle O S M=\\angle O A T=\\angle O A M$. Hence quadrilateral $S A O M$ is cyclic, and since the tangent line $A S$ is perpendicular to $A O$, $\\angle O M S=180^{\\circ}-\\angle O A S=90^{\\circ}$. This means that $M$ is the orthogonal projection of $O$ onto $B C$, which is its midpoint. So $T$ lies on median $A M$ of triangle $A B C$.']",,True,,, 1824,Geometry,,"In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$.","['Denote by $\\Omega$ the circle $A E F P Q$, and denote by $\\gamma$ the circle $P Q M$. Let the line $A D$ meet $\\omega$ again at $T \\neq D$. We will show that $\\gamma$ is tangent to $\\omega$ at $T$.\n\nWe first prove that points $P, Q, M, T$ are concyclic. Let $A^{\\prime}$ be the center of $\\omega$. Since $A^{\\prime} E \\perp A E$ and $A^{\\prime} F \\perp A F, A A^{\\prime}$ is a diameter in $\\Omega$. Let $N$ be the midpoint of $D T$; from $A^{\\prime} D=A^{\\prime} T$ we can see that $\\angle A^{\\prime} N A=90^{\\circ}$ and therefore $N$ also lies on the circle $\\Omega$. Now, from the power of $D$ with respect to the circles $\\gamma$ and $\\Omega$ we get\n\n$$\nD P \\cdot D Q=D A \\cdot D N=2 D M \\cdot \\frac{D T}{2}=D M \\cdot D T\n$$\n\nso $P, Q, M, T$ are concyclic.\n\nIf $E F \\| B C$, then $A B C$ is isosceles and the problem is now immediate by symmetry. Otherwise, let the tangent line to $\\omega$ at $T$ meet line $B C$ at point $R$. The tangent line segments $R D$ and $R T$ have the same length, so $A^{\\prime} R$ is the perpendicular bisector of $D T$; since $N D=N T$, $N$ lies on this perpendicular bisector.\n\nIn right triangle $A^{\\prime} R D, R D^{2}=R N \\cdot R A^{\\prime}=R P \\cdot R Q$, in which the last equality was obtained from the power of $R$ with respect to $\\Omega$. Hence $R T^{2}=R P \\cdot R Q$, which implies that $R T$ is also tangent to $\\gamma$. Because $R T$ is a common tangent to $\\omega$ and $\\gamma$, these two circles are tangent at $T$.\n\n', 'After proving that $P, Q, M, T$ are concyclic, we finish the problem in a different fashion. We only consider the case in which $E F$ and $B C$ are not parallel. Let lines $P Q$ and $E F$ meet at point $R$. Since $P Q$ and $E F$ are radical axes of $\\Omega, \\gamma$ and $\\omega, \\gamma$, respectively, $R$ is the radical center of these three circles.\n\nWith respect to the circle $\\omega$, the line $D R$ is the polar of $D$, and the line $E F$ is the polar of $A$. So the pole of line $A D T$ is $D R \\cap E F=R$, and therefore $R T$ is tangent to $\\omega$.\n\nFinally, since $T$ belongs to $\\gamma$ and $\\omega$ and $R$ is the radical center of $\\gamma, \\omega$ and $\\Omega$, line $R T$ is the radical axis of $\\gamma$ and $\\omega$, and since it is tangent to $\\omega$, it is also tangent to $\\gamma$. Because $R T$ is a common tangent to $\\omega$ and $\\gamma$, these two circles are tangent at $T$.', 'We give an alternative proof that the circles are tangent at the common point $T$. Again, we start from the fact that $P, Q, M, T$ are concyclic. Let point $O$ be the midpoint of diameter $A A^{\\prime}$. Then $M O$ is the midline of triangle $A D A^{\\prime}$, so $M O \\| A^{\\prime} D$. Since $A^{\\prime} D \\perp P Q$, $M O$ is perpendicular to $P Q$ as well.\n\nLooking at circle $\\Omega$, which has center $O, M O \\perp P Q$ implies that $M O$ is the perpendicular bisector of the chord $P Q$. Thus $M$ is the midpoint of arc $\\overparen{P Q}$ from $\\gamma$, and the tangent line $m$ to $\\gamma$ at $M$ is parallel to $P Q$.\n\n\n\nConsider the homothety with center $T$ and ratio $\\frac{T D}{T M}$. It takes $D$ to $M$, and the line $P Q$ to the line $m$. Since the circle that is tangent to a line at a given point and that goes through another given point is unique, this homothety also takes $\\omega$ (tangent to $P Q$ and going through $T$ ) to $\\gamma$ (tangent to $m$ and going through $T$ ). We conclude that $\\omega$ and $\\gamma$ are tangent at $T$.']",,True,,, 1825,Geometry,,"Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$.","['If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1}\n\nFinally, let $X$ be the second intersection point of $\\omega$ and the line $D E$. Since $B M$ is a diameter in $\\omega$, we have $\\angle B X M=90^{\\circ}$. Moreover,\n\n$$\n\\angle E X M=180^{\\circ}-\\angle M B E=180^{\\circ}-\\angle F B E=\\angle E D F,\n$$\n\nso $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.', 'Define point $M$ as the point opposite to $B$ on circle $\\omega$, and point $R$ as the intersection of lines $A C, A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\\angle A D C$, like in the first solution.\n\nSince $B$ is the midpoint of the arc $\\overparen{A E C}$, the line $B E R$ is the external bisector of $\\angle C E A$. Now we show that the internal angle bisectors of $\\angle A D C$ and $\\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\\angle C E A$, which is line $E M$, meet $A C$ at $S^{\\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\\angle A D C$ and $\\angle C E A$,\n\n$$\nA S: C S=A D: C D=A R: C R=A E: C E=A S^{\\prime}: C S^{\\prime}\n$$\n\nso indeed $S=S^{\\prime}$.\n\nBy $\\angle R D S=\\angle S E R=90^{\\circ}$ the points $R, S, D$ and $E$ are concyclic.\n\n\n\nNow let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\\angle E X B=\\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\\angle E D S=\\angle E R S$ in circle $S R D E$, and $\\angle E R S=\\angle E M B$ because $S R \\perp B M$ and $E R \\perp M E$. Therefore, $\\angle E X B=\\angle E M B$, so indeed, the point $X$ lies on $\\omega$.']",,True,,, 1826,Geometry,,"Let $n \geqslant 3$ be an integer. Two regular $n$-gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive. (That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary from the other vertices of $\mathcal{A}$.)","['By a polygon we always mean its interior together with its boundary.\n\nWe start with finding a regular $n$-gon $\\mathcal{C}$ which $(i)$ is inscribed into $\\mathcal{B}$ (that is, all vertices of $\\mathcal{C}$ lie on the perimeter of $\\mathcal{B}$ ); and $($ ii) is either a translation of $\\mathcal{A}$, or a homothetic image of $\\mathcal{A}$ with a positive factor.\n\nSuch a polygon may be constructed as follows. Let $O_{A}$ and $O_{B}$ be the centers of $\\mathcal{A}$ and $\\mathcal{B}$, respectively, and let $A$ be an arbitrary vertex of $\\mathcal{A}$. Let $\\overrightarrow{O_{B} C}$ be the vector co-directional to $\\overrightarrow{O_{A} A}$, with $C$ lying on the perimeter of $\\mathcal{B}$. The rotations of $C$ around $O_{B}$ by multiples of $2 \\pi / n$ form the required polygon. Indeed, it is regular, inscribed into $\\mathcal{B}$ (due to the rotational symmetry of $\\mathcal{B}$ ), and finally the translation/homothety mapping $\\overrightarrow{O_{A} A}$ to $\\overrightarrow{O_{B} C}$ maps $\\mathcal{A}$ to $\\mathcal{C}$.\n\nNow we separate two cases.\n\n\n\nConstruction of $\\mathcal{C}$\n\n\n\nCase 1: Translation\n\nCase 1: $\\mathcal{C}$ is a translation of $\\mathcal{A}$ by a vector $\\vec{v}$.\n\nDenote by $t$ the translation transform by vector $\\vec{v}$. We need to prove that the vertices of $\\mathcal{C}$ which stay in $\\mathcal{B}$ under $t$ are consecutive. To visualize the argument, we refer the plane to Cartesian coordinates so that the $x$-axis is co-directional with $\\vec{v}$. This way, the notions of right/left and top/bottom are also introduced, according to the $x$ - and $y$-coordinates, respectively.\n\nLet $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ be the top and the bottom vertices of $\\mathcal{B}$ (if several vertices are extremal, we take the rightmost of them). They split the perimeter of $\\mathcal{B}$ into the right part $\\mathcal{B}_{\\mathrm{R}}$ and the left part $\\mathcal{B}_{\\mathrm{L}}$ (the vertices $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ are assumed to lie in both parts); each part forms a connected subset of the perimeter of $\\mathcal{B}$. So the vertices of $\\mathcal{C}$ are also split into two parts $\\mathcal{C}_{\\mathrm{L}} \\subset \\mathcal{B}_{\\mathrm{L}}$ and $\\mathcal{C}_{\\mathrm{R}} \\subset \\mathcal{B}_{\\mathrm{R}}$, each of which consists of consecutive vertices.\n\nNow, all the points in $\\mathcal{B}_{\\mathrm{R}}$ (and hence in $\\mathcal{C}_{\\mathrm{R}}$ ) move out from $\\mathcal{B}$ under $t$, since they are the rightmost points of $\\mathcal{B}$ on the corresponding horizontal lines. It remains to prove that the vertices of $\\mathcal{C}_{\\mathrm{L}}$ which stay in $\\mathcal{B}$ under $t$ are consecutive.\n\nFor this purpose, let $C_{1}, C_{2}$, and $C_{3}$ be three vertices in $\\mathcal{C}_{\\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $t\\left(C_{1}\\right)$ and $t\\left(C_{3}\\right)$ lie in $\\mathcal{B}$; we need to prove that $t\\left(C_{2}\\right) \\in \\mathcal{B}$ as well. Let $A_{i}=t\\left(C_{i}\\right)$. The line through $C_{2}$ parallel to $\\vec{v}$ crosses the segment $C_{1} C_{3}$ to the right of $C_{2}$; this means that this line crosses $A_{1} A_{3}$ to the right of $A_{2}$, so $A_{2}$ lies inside the triangle $A_{1} C_{2} A_{3}$ which is contained in $\\mathcal{B}$. This yields the desired result.\n\nCase 2: $\\mathcal{C}$ is a homothetic image of $\\mathcal{A}$ centered at $X$ with factor $k>0$.\n\n\n\nDenote by $h$ the homothety mapping $\\mathcal{C}$ to $\\mathcal{A}$. We need now to prove that the vertices of $\\mathcal{C}$ which stay in $\\mathcal{B}$ after applying $h$ are consecutive. If $X \\in \\mathcal{B}$, the claim is easy. Indeed, if $k<1$, then the vertices of $\\mathcal{A}$ lie on the segments of the form $X C$ ( $C$ being a vertex of $\\mathcal{C})$ which lie in $\\mathcal{B}$. If $k>1$, then the vertices of $\\mathcal{A}$ lie on the extensions of such segments $X C$ beyond $C$, and almost all these extensions lie outside $\\mathcal{B}$. The exceptions may occur only in case when $X$ lies on the boundary of $\\mathcal{B}$, and they may cause one or two vertices of $\\mathcal{A}$ stay on the boundary of $\\mathcal{B}$. But even in this case those vertices are still consecutive.\n\nSo, from now on we assume that $X \\notin \\mathcal{B}$.\n\nNow, there are two vertices $B_{\\mathrm{T}}$ and $\\mathcal{B}_{\\mathrm{B}}$ of $\\mathcal{B}$ such that $\\mathcal{B}$ is contained in the angle $\\angle B_{\\mathrm{T}} X B_{\\mathrm{B}}$; if there are several options, say, for $B_{\\mathrm{T}}$, then we choose the farthest one from $X$ if $k>1$, and the nearest one if $k<1$. For the visualization purposes, we refer the plane to Cartesian coordinates so that the $y$-axis is co-directional with $\\overrightarrow{B_{\\mathrm{B}} B_{\\mathrm{T}}}$, and $X$ lies to the left of the line $B_{\\mathrm{T}} B_{\\mathrm{B}}$. Again, the perimeter of $\\mathcal{B}$ is split by $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ into the right part $\\mathcal{B}_{\\mathrm{R}}$ and the left part $\\mathcal{B}_{\\mathrm{L}}$, and the set of vertices of $\\mathcal{C}$ is split into two subsets $\\mathcal{C}_{\\mathrm{R}} \\subset \\mathcal{B}_{\\mathrm{R}}$ and $\\mathcal{C}_{\\mathrm{L}} \\subset \\mathcal{B}_{\\mathrm{L}}$.\n\n\n\nCase $2, X$ inside $\\mathcal{B}$\n\n\n\nSubcase 2.1: $k>1$\n\nSubcase 2.1: $k>1$.\n\nIn this subcase, all points from $\\mathcal{B}_{\\mathrm{R}}$ (and hence from $\\mathcal{C}_{\\mathrm{R}}$ ) move out from $\\mathcal{B}$ under $h$, because they are the farthest points of $\\mathcal{B}$ on the corresponding rays emanated from $X$. It remains to prove that the vertices of $\\mathcal{C}_{\\mathrm{L}}$ which stay in $\\mathcal{B}$ under $h$ are consecutive.\n\nAgain, let $C_{1}, C_{2}, C_{3}$ be three vertices in $\\mathcal{C}_{\\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $h\\left(C_{1}\\right)$ and $h\\left(C_{3}\\right)$ lie in $\\mathcal{B}$. Let $A_{i}=h\\left(C_{i}\\right)$. Then the ray $X C_{2}$ crosses the segment $C_{1} C_{3}$ beyond $C_{2}$, so this ray crosses $A_{1} A_{3}$ beyond $A_{2}$; this implies that $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\\mathcal{B}$.\n\n\n\nSubcase $2.2: k<1$\n\nSubcase 2.2: $k<1$.\n\nThis case is completely similar to the previous one. All points from $\\mathcal{B}_{\\mathrm{L}}$ (and hence from $\\mathcal{C}_{\\mathrm{L}}$ move out from $\\mathcal{B}$ under $h$, because they are the nearest points of $\\mathcal{B}$ on the corresponding\n\n\n\nrays emanated from $X$. Assume that $C_{1}, C_{2}$, and $C_{3}$ are three vertices in $\\mathcal{C}_{\\mathrm{R}}$ such that $C_{2}$ lies between $C_{1}$ and $C_{3}$, and $h\\left(C_{1}\\right)$ and $h\\left(C_{3}\\right)$ lie in $\\mathcal{B}$; let $A_{i}=h\\left(C_{i}\\right)$. Then $A_{2}$ lies on the segment $X C_{2}$, and the segments $X A_{2}$ and $A_{1} A_{3}$ cross each other. Thus $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\\mathcal{B}$.', 'By a polygon we always mean its interior together with its boundary.\n\nLet $O_{A}$ and $O_{B}$ be the centers of $\\mathcal{A}$ and $\\mathcal{B}$, respectively. Denote $[n]=\\{1,2, \\ldots, n\\}$.\n\nWe start with introducing appropriate enumerations and notations. Enumerate the sidelines of $\\mathcal{B}$ clockwise as $\\ell_{1}, \\ell_{2}, \\ldots, \\ell_{n}$. Denote by $\\mathcal{H}_{i}$ the half-plane of $\\ell_{i}$ that contains $\\mathcal{B}\\left(\\mathcal{H}_{i}\\right.$ is assumed to contain $\\ell_{i}$ ); by $B_{i}$ the midpoint of the side belonging to $\\ell_{i}$; and finally denote $\\overrightarrow{b_{i}}=\\overrightarrow{B_{i} O_{B}}$. (As usual, the numbering is cyclic modulo $n$, so $\\ell_{n+i}=\\ell_{i}$ etc.)\n\nNow, choose a vertex $A_{1}$ of $\\mathcal{A}$ such that the vector $\\overrightarrow{O_{A} A_{1}}$ points ""mostly outside $\\mathcal{H}_{1}$ ""; strictly speaking, this means that the scalar product $\\left\\langle\\overrightarrow{O_{A} A_{1}}, \\overrightarrow{b_{1}}\\right\\rangle$ is minimal. Starting from $A_{1}$, enumerate the vertices of $\\mathcal{A}$ clockwise as $A_{1}, A_{2}, \\ldots, A_{n}$; by the rotational symmetry, the choice of $A_{1}$ yields that the vector $\\overrightarrow{O_{A} A_{i}}$ points ""mostly outside $\\mathcal{H}_{i}$ "", i.e.,\n\n$$\n\\left\\langle\\overrightarrow{O_{A} A_{i}}, \\overrightarrow{b_{i}}\\right\\rangle=\\min _{j \\in[n]}\\left\\langle\\overrightarrow{O_{A} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle\n\\tag{1}\n$$\n\n\n\nEnumerations and notations\n\nWe intend to reformulate the problem in more combinatorial terms, for which purpose we introduce the following notion. Say that a subset $I \\subseteq[n]$ is connected if the elements of this set are consecutive in the cyclic order (in other words, if we join each $i$ with $i+1 \\bmod n$ by an edge, this subset is connected in the usual graph sense). Clearly, the union of two connected subsets sharing at least one element is connected too. Next, for any half-plane $\\mathcal{H}$ the indices of vertices of, say, $\\mathcal{A}$ that lie in $\\mathcal{H}$ form a connected set.\n\nTo access the problem, we denote\n\n$$\nM=\\left\\{j \\in[n]: A_{j} \\notin \\mathcal{B}\\right\\}, \\quad M_{i}=\\left\\{j \\in[n]: A_{j} \\notin \\mathcal{H}_{i}\\right\\} \\quad \\text { for } i \\in[n]\n$$\n\nWe need to prove that $[n] \\backslash M$ is connected, which is equivalent to $M$ being connected. On the other hand, since $\\mathcal{B}=\\bigcap_{i \\in[n]} \\mathcal{H}_{i}$, we have $M=\\bigcup_{i \\in[n]} M_{i}$, where the sets $M_{i}$ are easier to investigate. We will utilize the following properties of these sets; the first one holds by the definition of $M_{i}$, along with the above remark.\n\n\n\n\n\nThe sets $M_{i}$\n\nProperty 1: Each set $M_{i}$ is connected.\n\nProperty 2: If $M_{i}$ is nonempty, then $i \\in M_{i}$.\n\nProof. Indeed, we have\n\n$$\nj \\in M_{i} \\Longleftrightarrow A_{j} \\notin \\mathcal{H}_{i} \\Longleftrightarrow\\left\\langle\\overrightarrow{B_{i} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle<0 \\Longleftrightarrow\\left\\langle\\overrightarrow{O_{A} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle<\\left\\langle\\overrightarrow{O_{A} B_{i}}, \\overrightarrow{b_{i}}\\right\\rangle .\n\\tag{2}\n$$\n\nThe right-hand part of the last inequality does not depend on $j$. Therefore, if some $j$ lies in $M_{i}$, then by (1) so does $i$.\n\nIn view of Property 2, it is useful to define the set\n\n$$\nM^{\\prime}=\\left\\{i \\in[n]: i \\in M_{i}\\right\\}=\\left\\{i \\in[n]: M_{i} \\neq \\varnothing\\right\\}\n$$\n\nProperty 3: The set $M^{\\prime}$ is connected.\n\nProof. To prove this property, we proceed on with the investigation started in (2) to write\n\n$$\ni \\in M^{\\prime} \\Longleftrightarrow A_{i} \\in M_{i} \\Longleftrightarrow\\left\\langle\\overrightarrow{B_{i} A_{i}}, \\overrightarrow{b_{i}}\\right\\rangle<0 \\Longleftrightarrow\\left\\langle\\overrightarrow{O_{B} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle<\\left\\langle\\overrightarrow{O_{B} B_{i}}, \\overrightarrow{b_{i}}\\right\\rangle+\\left\\langle\\overrightarrow{A_{i} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle\n$$\n\nThe right-hand part of the obtained inequality does not depend on $i$, due to the rotational symmetry; denote its constant value by $\\mu$. Thus, $i \\in M^{\\prime}$ if and only if $\\left\\langle\\overrightarrow{O_{B} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle<\\mu$. This condition is in turn equivalent to the fact that $B_{i}$ lies in a certain (open) half-plane whose boundary line is orthogonal to $O_{B} O_{A}$; thus, it defines a connected set.\n\nNow we can finish the solution. Since $M^{\\prime} \\subseteq M$, we have\n\n$$\nM=\\bigcup_{i \\in[n]} M_{i}=M^{\\prime} \\cup \\bigcup_{i \\in[n]} M_{i}\n$$\n\nso $M$ can be obtained from $M^{\\prime}$ by adding all the sets $M_{i}$ one by one. All these sets are connected, and each nonempty $M_{i}$ contains an element of $M^{\\prime}$ (namely, $i$ ). Thus their union is also connected.']",,True,,, 1827,Geometry,,"A convex quadrilateral $A B C D$ has an inscribed circle with center $I$. Let $I_{a}, I_{b}, I_{c}$, and $I_{d}$ be the incenters of the triangles $D A B, A B C, B C D$, and $C D A$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common external tangents of the circles $B I_{a} I_{c}$ and $D I_{a} I_{c}$ meet at $Y$. Prove that $\angle X I Y=90^{\circ}$.","['Denote by $\\omega_{a}, \\omega_{b}, \\omega_{c}$ and $\\omega_{d}$ the circles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}, O_{b}, O_{c}$ and $O_{d}$, and let their radii be $r_{a}, r_{b}, r_{c}$ and $r_{d}$, respectively.\n\nClaim 1. $I_{b} I_{d} \\perp A C$ and $I_{a} I_{c} \\perp B D$.\n\nProof. Let the incircles of triangles $A B C$ and $A C D$ be tangent to the line $A C$ at $T$ and $T^{\\prime}$, respectively. (See the figure to the left.) We have $A T=\\frac{A B+A C-B C}{2}$ in triangle $A B C, A T^{\\prime}=$ $\\frac{A D+A C-C D}{2}$ in triangle $A C D$, and $A B-B C=A D-C D$ in quadrilateral $A B C D$, so\n\n$$\nA T=\\frac{A C+A B-B C}{2}=\\frac{A C+A D-C D}{2}=A T^{\\prime}\n$$\n\nThis shows $T=T^{\\prime}$. As an immediate consequence, $I_{b} I_{d} \\perp A C$.\n\nThe second statement can be shown analogously.\n\n\nClaim 2. The points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the lines $A I, B I, C I$ and $D I$, respectively.\n\nProof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.)\n\nNotice first that the incircles of triangles $A B C$ and $A C D$ can be obtained from the incircle of the quadrilateral $A B C D$ with homothety centers $B$ and $D$, respectively, and homothety factors less than 1 , therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $B I$ and $D I$, respectively.\n\nAs is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $A T$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\\angle I_{b} A T=\\angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $A B C$ and $A C D$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\\angle B A C$ and $\\angle C A D$, respectively. Then\n\n$$\n\\angle O_{a} A D=\\angle O_{a} A I_{d}+\\angle I_{d} A D=\\angle I_{b} A T+\\angle I_{d} A D=\\frac{1}{2} \\angle B A C+\\frac{1}{2} \\angle C A D=\\frac{1}{2} \\angle B A D\n$$\n\nso $O_{a}$ lies on the angle bisector of $\\angle B A D$, that is, on line $A I$.\n\nThe point $X$ is the external similitude center of $\\omega_{a}$ and $\\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\\omega_{a}$ and $\\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $A C$.\n\n\n\n\n\nFrom the similarity of the circles $\\omega_{a}$ and $\\omega_{c}$, from $O_{a} I_{b}=O_{a} I_{d}=O_{a} A=r_{a}$ and $O_{c} I_{b}=$ $O_{c} I_{d}=O_{c} C=r_{c}$, and from $A C \\| O_{a} O_{c}$ we can see that\n\n$$\n\\frac{O_{a} X}{O_{c} X}=\\frac{O_{a} U}{O_{c} U}=\\frac{r_{a}}{r_{c}}=\\frac{O_{a} I_{b}}{O_{c} I_{b}}=\\frac{O_{a} I_{d}}{O_{c} I_{d}}=\\frac{O_{a} A}{O_{c} C}=\\frac{O_{a} I}{O_{c} I}\n$$\n\nSo the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a}: r_{c}$. In this Apollonius circle $X U$ is a diameter, and the lines $I U$ and $I X$ are respectively the internal and external bisectors of $\\angle O_{a} I O_{c}=\\angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $U X$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $I X$ and $I U$ are the internal and external bisectors of $\\angle I_{b} I I_{d}=\\angle B I D$, respectively.\n\nRepeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $I Y$ is the internal bisector of $\\angle A I C$ and the external bisector of $\\angle B I D$. Therefore, the lines $I X$ and $I Y$ respectively are the internal and external bisectors of $\\angle B I D$, so they are perpendicular.']",,True,,, 1828,Geometry,,"There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn. Find all possible numbers of tangent segments when he stops drawing.","['First, consider a particular arrangement of circles $C_{1}, C_{2}, \\ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $\\left(i^{2}, 0\\right)$ and radius $i / 2$ works. Then the only tangent segments that can be drawn are between adjacent circles $C_{i}$ and $C_{i+1}$, and exactly three segments can be drawn for each pair. So Luciano will draw exactly $3(n-1)$ segments in this case.\n\n\n\nFor the general case, start from a final configuration (that is, an arrangement of circles and segments in which no further segments can be drawn). The idea of the solution is to continuously resize and move the circles around the plane, one by one (in particular, making sure we never have 4 circles with a common tangent line), and show that the number of segments drawn remains constant as the picture changes. This way, we can reduce any circle/segment configuration to the particular one mentioned above, and the final number of segments must remain at $3 n-3$.\n\nSome preliminary considerations: look at all possible tangent segments joining any two circles. A segment that is tangent to a circle $A$ can do so in two possible orientations - it may come out of $A$ in clockwise or counterclockwise orientation. Two segments touching the same circle with the same orientation will never intersect each other. Each pair $(A, B)$ of circles has 4 choices of tangent segments, which can be identified by their orientations - for example, $(A+, B-)$ would be the segment which comes out of $A$ in clockwise orientation and comes out of $B$ in counterclockwise orientation. In total, we have $2 n(n-1)$ possible segments, disregarding intersections.\n\nNow we pick a circle $C$ and start to continuously move and resize it, maintaining all existing tangent segments according to their identifications, including those involving $C$. We can keep our choice of tangent segments until the configuration reaches a transition. We lose nothing if we assume that $C$ is kept at least $\\varepsilon$ units away from any other circle, where $\\varepsilon$ is a positive, fixed constant; therefore at a transition either: (1) a currently drawn tangent segment $t$ suddenly becomes obstructed; or (2) a currently absent tangent segment $t$ suddenly becomes unobstructed and available.\n\nClaim. A transition can only occur when three circles $C_{1}, C_{2}, C_{3}$ are tangent to a common line $\\ell$ containing $t$, in a way such that the three tangent segments lying on $\\ell$ (joining the three circles pairwise) are not obstructed by any other circles or tangent segments (other than $C_{1}, C_{2}, C_{3}$ ). Proof. Since (2) is effectively the reverse of (1), it suffices to prove the claim for (1). Suppose $t$ has suddenly become obstructed, and let us consider two cases.\n\n\n\nCase 1: $t$ becomes obstructed by a circle\n\n\n\nThen the new circle becomes the third circle tangent to $\\ell$, and no other circles or tangent segments are obstructing $t$.\n\nCase 2: $t$ becomes obstructed by another tangent segment $t^{\\prime}$ \n\nWhen two segments $t$ and $t^{\\prime}$ first intersect each other, they must do so at a vertex of one of them. But if a vertex of $t^{\\prime}$ first crossed an interior point of $t$, the circle associated to this vertex was already blocking $t$ (absurd), or is about to (we already took care of this in case 1). So we only have to analyze the possibility of $t$ and $t^{\\prime}$ suddenly having a common vertex. However, if that happens, this vertex must belong to a single circle (remember we are keeping different circles at least $\\varepsilon$ units apart from each other throughout the moving/resizing process), and therefore they must have different orientations with respect to that circle.\n\n\nThus, at the transition moment, both $t$ and $t^{\\prime}$ are tangent to the same circle at a common point, that is, they must be on the same line $\\ell$ and hence we again have three circles simultaneously tangent to $\\ell$. Also no other circles or tangent segments are obstructing $t$ or $t^{\\prime}$ (otherwise, they would have disappeared before this transition).\n\nNext, we focus on the maximality of a configuration immediately before and after a transition, where three circles share a common tangent line $\\ell$. Let the three circles be $C_{1}, C_{2}, C_{3}$, ordered by their tangent points. The only possibly affected segments are the ones lying on $\\ell$, namely $t_{12}, t_{23}$ and $t_{13}$. Since $C_{2}$ is in the middle, $t_{12}$ and $t_{23}$ must have different orientations with respect to $C_{2}$. For $C_{1}, t_{12}$ and $t_{13}$ must have the same orientation, while for $C_{3}, t_{13}$ and $t_{23}$ must have the same orientation. The figure below summarizes the situation, showing alternative positions for $C_{1}$ (namely, $C_{1}$ and $C_{1}^{\\prime}$ ) and for $C_{3}\\left(C_{3}\\right.$ and $C_{3}^{\\prime}$ ).\n\n\n\n\n\nNow perturb the diagram slightly so the three circles no longer have a common tangent, while preserving the definition of $t_{12}, t_{23}$ and $t_{13}$ according to their identifications. First note that no other circles or tangent segments can obstruct any of these segments. Also recall that tangent segments joining the same circle at the same orientation will never obstruct each other.\n\nThe availability of the tangent segments can now be checked using simple diagrams.\n\nCase 1: $t_{13}$ passes through $C_{2}$\n\n\n\nIn this case, $t_{13}$ is not available, but both $t_{12}$ and $t_{23}$ are.\n\nCase 2: $t_{13}$ does not pass through $C_{2}$\n\n\n\nNow $t_{13}$ is available, but $t_{12}$ and $t_{23}$ obstruct each other, so only one can be drawn.\n\nIn any case, exactly 2 out of these 3 segments can be drawn. Thus the maximal number of segments remains constant as we move or resize the circles, and we are done.', ""First note that all tangent segments lying on the boundary of the convex hull of the circles are always drawn since they do not intersect anything else. Now in the final picture, aside from the $n$ circles, the blackboard is divided into regions. We can consider the picture as a plane (multi-)graph $G$ in which the circles are the vertices and the tangent segments are the edges. The idea of this solution is to find a relation between the number of edges and the number of regions in $G$; then, once we prove that $G$ is connected, we can use Euler's formula to finish the problem.\n\nThe boundary of each region consists of 1 or more (for now) simple closed curves, each made of arcs and tangent segments. The segment and the arc might meet smoothly (as in $S_{i}$, $i=1,2, \\ldots, 6$ in the figure below) or not (as in $P_{1}, P_{2}, P_{3}, P_{4}$; call such points sharp corners of the boundary). In other words, if a person walks along the border, her direction would suddenly turn an angle of $\\pi$ at a sharp corner.\n\n\n\n\n\nClaim 1. The outer boundary $B_{1}$ of any internal region has at least 3 sharp corners.\n\nProof. Let a person walk one lap along $B_{1}$ in the counterclockwise orientation. As she does so, she will turn clockwise as she moves along the circle arcs, and not turn at all when moving along the lines. On the other hand, her total rotation after one lap is $2 \\pi$ in the counterclockwise direction! Where could she be turning counterclockwise? She can only do so at sharp corners, and, even then, she turns only an angle of $\\pi$ there. But two sharp corners are not enough, since at least one arc must be present - so she must have gone through at least 3 sharp corners.\n\nClaim 2. Each internal region is simply connected, that is, has only one boundary curve.\n\nProof. Suppose, by contradiction, that some region has an outer boundary $B_{1}$ and inner boundaries $B_{2}, B_{3}, \\ldots, B_{m}(m \\geqslant 2)$. Let $P_{1}$ be one of the sharp corners of $B_{1}$.\n\nNow consider a car starting at $P_{1}$ and traveling counterclockwise along $B_{1}$. It starts in reverse, i.e., it is initially facing the corner $P_{1}$. Due to the tangent conditions, the car may travel in a way so that its orientation only changes when it is moving along an arc. In particular, this means the car will sometimes travel forward. For example, if the car approaches a sharp corner when driving in reverse, it would continue travel forward after the corner, instead of making an immediate half-turn. This way, the orientation of the car only changes in a clockwise direction since the car always travels clockwise around each arc.\n\nNow imagine there is a laser pointer at the front of the car, pointing directly ahead. Initially, the laser endpoint hits $P_{1}$, but, as soon as the car hits an arc, the endpoint moves clockwise around $B_{1}$. In fact, the laser endpoint must move continuously along $B_{1}$ ! Indeed, if the endpoint ever jumped (within $B_{1}$, or from $B_{1}$ to one of the inner boundaries), at the moment of the jump the interrupted laser would be a drawable tangent segment that Luciano missed (see figure below for an example).\n\n\n\n\n\nNow, let $P_{2}$ and $P_{3}$ be the next two sharp corners the car goes through, after $P_{1}$ (the previous lemma assures their existence). At $P_{2}$ the car starts moving forward, and at $P_{3}$ it will start to move in reverse again. So, at $P_{3}$, the laser endpoint is at $P_{3}$ itself. So while the car moved counterclockwise between $P_{1}$ and $P_{3}$, the laser endpoint moved clockwise between $P_{1}$ and $P_{3}$. That means the laser beam itself scanned the whole region within $B_{1}$, and it should have crossed some of the inner boundaries.\n\nClaim 3. Each region has exactly 3 sharp corners.\n\nProof. Consider again the car of the previous claim, with its laser still firmly attached to its front, traveling the same way as before and going through the same consecutive sharp corners $P_{1}, P_{2}$ and $P_{3}$. As we have seen, as the car goes counterclockwise from $P_{1}$ to $P_{3}$, the laser endpoint goes clockwise from $P_{1}$ to $P_{3}$, so together they cover the whole boundary. If there were a fourth sharp corner $P_{4}$, at some moment the laser endpoint would pass through it. But, since $P_{4}$ is a sharp corner, this means the car must be on the extension of a tangent segment going through $P_{4}$. Since the car is not on that segment itself (the car never goes through $P_{4}$ ), we would have 3 circles with a common tangent line, which is not allowed.\n\n\n\nWe are now ready to finish the solution. Let $r$ be the number of internal regions, and $s$ be the number of tangent segments. Since each tangent segment contributes exactly 2 sharp corners to the diagram, and each region has exactly 3 sharp corners, we must have $2 s=3 r$. Since the graph corresponding to the diagram is connected, we can use Euler's formula $n-s+r=1$ and find $s=3 n-3$ and $r=2 n-2$.""]",['6048'],False,,Numerical, 1829,Number Theory,,"The sequence $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers satisfies $$ a_{n+1}=\left\{\begin{array}{ll} \sqrt{a_{n}}, & \text { if } \sqrt{a_{n}} \text { is an integer } \\ a_{n}+3, & \text { otherwise } \end{array} \text { for every } n \geqslant 0 .\right. $$ Determine all values of $a_{0}>1$ for which there is at least one number $a$ such that $a_{n}=a$ for infinitely many values of $n$.","['Since the value of $a_{n+1}$ only depends on the value of $a_{n}$, if $a_{n}=a_{m}$ for two different indices $n$ and $m$, then the sequence is eventually periodic. So we look for the values of $a_{0}$ for which the sequence is eventually periodic.\n\nClaim 1. If $a_{n} \\equiv-1(\\bmod 3)$, then, for all $m>n, a_{m}$ is not a perfect square. It follows that the sequence is eventually strictly increasing, so it is not eventually periodic.\n\nProof. A square cannot be congruent to -1 modulo 3 , so $a_{n} \\equiv-1(\\bmod 3)$ implies that $a_{n}$ is not a square, therefore $a_{n+1}=a_{n}+3>a_{n}$. As a consequence, $a_{n+1} \\equiv a_{n} \\equiv-1(\\bmod 3)$, so $a_{n+1}$ is not a square either. By repeating the argument, we prove that, from $a_{n}$ on, all terms of the sequence are not perfect squares and are greater than their predecessors, which completes the proof.\n\nClaim 2. If $a_{n} \\not \\equiv-1(\\bmod 3)$ and $a_{n}>9$ then there is an index $m>n$ such that $a_{m}9, t$ is at least 3. The first square in the sequence $a_{n}, a_{n}+3, a_{n}+6, \\ldots$ will be $(t+1)^{2},(t+2)^{2}$ or $(t+3)^{2}$, therefore there is an index $m>n$ such that $a_{m} \\leqslant t+3n$ such that $a_{m}=3$.\n\nProof. First we notice that, by the definition of the sequence, a multiple of 3 is always followed by another multiple of 3 . If $a_{n} \\in\\{3,6,9\\}$ the sequence will eventually follow the periodic pattern $3,6,9,3,6,9, \\ldots$. If $a_{n}>9$, let $j$ be an index such that $a_{j}$ is equal to the minimum value of the set $\\left\\{a_{n+1}, a_{n+2}, \\ldots\\right\\}$. We must have $a_{j} \\leqslant 9$, otherwise we could apply Claim 2 to $a_{j}$ and get a contradiction on the minimality hypothesis. It follows that $a_{j} \\in\\{3,6,9\\}$, and the proof is complete.\n\nClaim 4. If $a_{n} \\equiv 1(\\bmod 3)$, then there is an index $m>n$ such that $a_{m} \\equiv-1(\\bmod 3)$.\n\nProof. In the sequence, 4 is always followed by $2 \\equiv-1(\\bmod 3)$, so the claim is true for $a_{n}=4$. If $a_{n}=7$, the next terms will be $10,13,16,4,2, \\ldots$ and the claim is also true. For $a_{n} \\geqslant 10$, we again take an index $j>n$ such that $a_{j}$ is equal to the minimum value of the set $\\left\\{a_{n+1}, a_{n+2}, \\ldots\\right\\}$, which by the definition of the sequence consists of non-multiples of 3 . Suppose $a_{j} \\equiv 1(\\bmod 3)$. Then we must have $a_{j} \\leqslant 9$ by Claim 2 and the minimality of $a_{j}$. It follows that $a_{j} \\in\\{4,7\\}$, so $a_{m}=2j$, contradicting the minimality of $a_{j}$. Therefore, we must have $a_{j} \\equiv-1(\\bmod 3)$.\n\nIt follows from the previous claims that if $a_{0}$ is a multiple of 3 the sequence will eventually reach the periodic pattern $3,6,9,3,6,9, \\ldots$; if $a_{0} \\equiv-1(\\bmod 3)$ the sequence will be strictly increasing; and if $a_{0} \\equiv 1(\\bmod 3)$ the sequence will be eventually strictly increasing.\n\nSo the sequence will be eventually periodic if, and only if, $a_{0}$ is a multiple of 3 .']",['All positive multiples of 3'],False,,Need_human_evaluate, 1830,Number Theory,,"Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices $i \in\{0,1, \ldots, p-1\}$ have been chosen. Then the following number is computed: $$ M=a_{0}+10 \cdot a_{1}+\cdots+10^{p-1} \cdot a_{p-1}=\sum_{j=0}^{p-1} a_{j} \cdot 10^{j} $$ The goal of Eduardo is to make the number $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy.","[""We say that a player makes the move $\\left(i, a_{i}\\right)$ if he chooses the index $i$ and then the element $a_{i}$ of the set $\\{0,1,2,3,4,5,6,7,8,9\\}$ in this move.\n\nIf $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of 10 .\n\nNow assume that the prime number $p$ does not belong to $\\{2,5\\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, $\\left(10^{(p-1) / 2}\\right)^{2}=10^{p-1} \\equiv 1(\\bmod p)$, so $p \\mid\\left(10^{(p-1) / 2}\\right)^{2}-1=\\left(10^{(p-1) / 2}+1\\right)\\left(10^{(p-1) / 2}-1\\right)$. Since $p$ is prime, either $p \\mid 10^{(p-1) / 2}+1$ or $p \\mid 10^{(p-1) / 2}-1$. Thus we have two cases:\n\nCase a: $10^{(p-1) / 2} \\equiv-1(\\bmod p)$\n\nIn this case, for each move $\\left(i, a_{i}\\right)$ of Fernando, Eduardo immediately makes the move $\\left(j, a_{j}\\right)=$ $\\left(i+\\frac{p-1}{2}, a_{i}\\right)$, if $0 \\leqslant i \\leqslant \\frac{p-3}{2}$, or $\\left(j, a_{j}\\right)=\\left(i-\\frac{p-1}{2}, a_{i}\\right)$, if $\\frac{p-1}{2} \\leqslant i \\leqslant p-2$. We will have $10^{j} \\equiv-10^{i}$ $(\\bmod p)$, and so $a_{j} \\cdot 10^{j}=a_{i} \\cdot 10^{j} \\equiv-a_{i} \\cdot 10^{i}(\\bmod p)$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\\{r, r+(p-1) / 2\\}$ with $0 \\leqslant r \\leqslant(p-3) / 2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_{k} \\cdot 10^{k}$ corresponding to the already chosen pairs $\\left(k, a_{k}\\right)$ divisible by $p$, and thus wins the game.\n\nCase b: $10^{(p-1) / 2} \\equiv 1(\\bmod p)$\n\nIn this case, for each move $\\left(i, a_{i}\\right)$ of Fernando, Eduardo immediately makes the move $\\left(j, a_{j}\\right)=$ $\\left(i+\\frac{p-1}{2}, 9-a_{i}\\right)$, if $0 \\leqslant i \\leqslant \\frac{p-3}{2}$, or $\\left(j, a_{j}\\right)=\\left(i-\\frac{p-1}{2}, 9-a_{i}\\right)$, if $\\frac{p-1}{2} \\leqslant i \\leqslant p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^{j} \\equiv 10^{i}$ $(\\bmod p)$, and so $a_{j} \\cdot 10^{j}+a_{i} \\cdot 10^{i} \\equiv\\left(a_{i}+a_{j}\\right) \\cdot 10^{i}=9 \\cdot 10^{i}(\\bmod p)$. Therefore, at the end of the game, the sum of all terms $a_{k} \\cdot 10^{k}$ will be congruent to\n\n$$\n\\sum_{i=0}^{\\frac{p-3}{2}} 9 \\cdot 10^{i}=10^{(p-1) / 2}-1 \\equiv 0 \\quad(\\bmod p)\n$$\n\nand Eduardo wins the game.""]",,True,,, 1831,Number Theory,,"Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. (We let $a_{i}=a_{i-n}$ when $i>n$.)","['Let us first show that, if $n=a b$, with $a, b \\geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \\leqslant k \\leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\\cdots+a_{n}=a \\cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \\leqslant i \\leqslant n$ be an arbitrary index. Taking $j=b$ if $1 \\leqslant i \\leqslant n-b$, and $j=b+1$ if $n-b0$ such that $10^{t} \\equiv 1$ $(\\bmod \\mathrm{cm})$ for some integer $c \\in C$, that is, the set of orders of 10 modulo $\\mathrm{cm}$. In other words,\n\n$$\nS(m)=\\left\\{\\operatorname{ord}_{c m}(10): c \\in C\\right\\}\n$$\n\nSince there are $4 \\cdot 201+3=807$ numbers $c$ with $1 \\leqslant c \\leqslant 2017$ and $\\operatorname{gcd}(c, 10)=1$, namely those such that $c \\equiv 1,3,7,9(\\bmod 10)$,\n\n$$\n|S(m)| \\leqslant|C|=807\n$$\n\nNow we find $m$ such that $|S(m)|=807$. Let\n\n$$\nP=\\{1

0}$. We will show that $k=c$.\n\nDenote by $\\nu_{p}(n)$ the number of prime factors $p$ in $n$, that is, the maximum exponent $\\beta$ for which $p^{\\beta} \\mid n$. For every $\\ell \\geqslant 1$ and $p \\in P$, the Lifting the Exponent Lemma provides\n\n$$\n\\nu_{p}\\left(10^{\\ell \\alpha}-1\\right)=\\nu_{p}\\left(\\left(10^{\\alpha}\\right)^{\\ell}-1\\right)=\\nu_{p}\\left(10^{\\alpha}-1\\right)+\\nu_{p}(\\ell)=\\nu_{p}(m)+\\nu_{p}(\\ell)\n$$\n\nso\n\n$$\n\\begin{aligned}\nc m \\mid 10^{k \\alpha}-1 & \\Longleftrightarrow \\forall p \\in P ; \\nu_{p}(\\mathrm{~cm}) \\leqslant \\nu_{p}\\left(10^{k \\alpha}-1\\right) \\\\\n& \\Longleftrightarrow \\forall p \\in P ; \\nu_{p}(m)+\\nu_{p}(c) \\leqslant \\nu_{p}(m)+\\nu_{p}(k) \\\\\n& \\Longleftrightarrow \\forall p \\in P ; \\nu_{p}(c) \\leqslant \\nu_{p}(k) \\\\\n& \\Longleftrightarrow c \\mid k .\n\\end{aligned}\n$$\n\nThe first such $k$ is $k=c$, so $\\operatorname{ord}_{c m}(10)=c \\alpha$.']",['807'],False,,Numerical, 1833,Number Theory,,"Find all pairs $(p, q)$ of prime numbers with $p>q$ for which the number $$ \frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1} $$ is an integer.","[""Let $M=(p+q)^{p-q}(p-q)^{p+q}-1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$.\n\nBy eliminating the term -1 in the numerator,\n\n$$\n(p+q)^{p+q}(p-q)^{p-q}-1 \\equiv(p+q)^{p-q}(p-q)^{p+q}-1 \\quad(\\bmod M)\\\\\n(p+q)^{2 q} \\equiv(p-q)^{2 q} \\quad(\\bmod M)\\tag{1}\n$$\n$$\n\\left((p+q) \\cdot(p-q)^{-1}\\right)^{2 q} \\equiv 1 \\quad(\\bmod M) .\n\\tag{2}\n$$\n\nCase 1: $q \\geqslant 5$.\n\nConsider an arbitrary prime divisor $r$ of $M$. Notice that $M$ is odd, so $r \\geqslant 3$. By (2), the multiplicative order of $\\left((p+q) \\cdot(p-q)^{-1}\\right)$ modulo $r$ is a divisor of the exponent $2 q$ in $(2)$, so it can be 1,2 , $q$ or $2 q$.\n\nBy Fermat's theorem, the order divides $r-1$. So, if the order is $q$ or $2 q$ then $r \\equiv 1(\\bmod q)$. If the order is 1 or 2 then $r \\mid(p+q)^{2}-(p-q)^{2}=4 p q$, so $r=p$ or $r=q$. The case $r=p$ is not possible, because, by applying Fermat's theorem,\n\n$M=(p+q)^{p-q}(p-q)^{p+q}-1 \\equiv q^{p-q}(-q)^{p+q}-1=\\left(q^{2}\\right)^{p}-1 \\equiv q^{2}-1=(q+1)(q-1) \\quad(\\bmod p)$\n\nand the last factors $q-1$ and $q+1$ are less than $p$ and thus $p \\nmid M$. Hence, all prime divisors of $M$ are either $q$ or of the form $k q+1$; it follows that all positive divisors of $M$ are congruent to 0 or 1 modulo $q$.\n\nNow notice that\n\n$$\nM=\\left((p+q)^{\\frac{p-q}{2}}(p-q)^{\\frac{p+q}{2}}-1\\right)\\left((p+q)^{\\frac{p-q}{2}}(p-q)^{\\frac{p+q}{2}}+1\\right)\n$$\n\nis the product of two consecutive positive odd numbers; both should be congruent to 0 or 1 modulo $q$. But this is impossible by the assumption $q \\geqslant 5$. So, there is no solution in Case 1 .\n\nCase 2: $q=2$.\n\nBy (1), we have $M \\mid(p+q)^{2 q}-(p-q)^{2 q}=(p+2)^{4}-(p-2)^{4}$, so\n\n$$\n\\begin{gathered}\n(p+2)^{p-2}(p-2)^{p+2}-1=M \\leqslant(p+2)^{4}-(p-2)^{4} \\leqslant(p+2)^{4}-1, \\\\\n(p+2)^{p-6}(p-2)^{p+2} \\leqslant 1 .\n\\end{gathered}\n$$\n\nIf $p \\geqslant 7$ then the left-hand side is obviously greater than 1 . For $p=5$ we have $(p+2)^{p-6}(p-2)^{p+2}=7^{-1} \\cdot 3^{7}$ which is also too large.\n\nThere remains only one candidate, $p=3$, which provides a solution:\n\n$$\n\\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}=\\frac{5^{5} \\cdot 1^{1}-1}{5^{1} \\cdot 1^{5}-1}=\\frac{3124}{4}=781\n$$\n\nSo in Case 2 the only solution is $(p, q)=(3,2)$.\n\n\n\nCase 3: $q=3$.\n\nSimilarly to Case 2, we have\n\n$$\nM \\mid(p+q)^{2 q}-(p-q)^{2 q}=64 \\cdot\\left(\\left(\\frac{p+3}{2}\\right)^{6}-\\left(\\frac{p-3}{2}\\right)^{6}\\right)\n$$\n\nSince $M$ is odd, we conclude that\n\n$$\nM \\mid\\left(\\frac{p+3}{2}\\right)^{6}-\\left(\\frac{p-3}{2}\\right)^{6}\n$$\n\nand\n\n$$\n\\begin{gathered}\n(p+3)^{p-3}(p-3)^{p+3}-1=M \\leqslant\\left(\\frac{p+3}{2}\\right)^{6}-\\left(\\frac{p-3}{2}\\right)^{6} \\leqslant\\left(\\frac{p+3}{2}\\right)^{6}-1 \\\\\n64(p+3)^{p-9}(p-3)^{p+3} \\leqslant 1\n\\end{gathered}\n$$\n\nIf $p \\geqslant 11$ then the left-hand side is obviously greater than 1 . If $p=7$ then the left-hand side is $64 \\cdot 10^{-2} \\cdot 4^{10}>1$. If $p=5$ then the left-hand side is $64 \\cdot 8^{-4} \\cdot 2^{8}=2^{2}>1$. Therefore, there is no solution in Case 3.""]","['$(3,2)$']",False,,Tuple, 1834,Number Theory,,"Find the smallest positive integer $n$, or show that no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ such that both $$ a_{1}+a_{2}+\cdots+a_{n} \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} $$ are integers.","['For $n=1, a_{1} \\in \\mathbb{Z}_{>0}$ and $\\frac{1}{a_{1}} \\in \\mathbb{Z}_{>0}$ if and only if $a_{1}=1$. Next we show that\n\n(i) There are finitely many $(x, y) \\in \\mathbb{Q}_{>0}^{2}$ satisfying $x+y \\in \\mathbb{Z}$ and $\\frac{1}{x}+\\frac{1}{y} \\in \\mathbb{Z}$\n\nWrite $x=\\frac{a}{b}$ and $y=\\frac{c}{d}$ with $a, b, c, d \\in \\mathbb{Z}_{>0}$ and $\\operatorname{gcd}(a, b)=\\operatorname{gcd}(c, d)=1$. Then $x+y \\in \\mathbb{Z}$ and $\\frac{1}{x}+\\frac{1}{y} \\in \\mathbb{Z}$ is equivalent to the two divisibility conditions\n\n$$\nb d \\mid a d+b c \\text { (1) and } \\quad a c \\mid a d+b c\n$$\n\nCondition (1) implies that $d|a d+b c \\Longleftrightarrow d| b c \\Longleftrightarrow d \\mid b$ since $\\operatorname{gcd}(c, d)=1$. Still from (1) we get $b|a d+b c \\Longleftrightarrow b| a d \\Longleftrightarrow b \\mid d$ since $\\operatorname{gcd}(a, b)=1$. From $b \\mid d$ and $d \\mid b$ we have $b=d$.\n\nAn analogous reasoning with condition (2) shows that $a=c$. Hence $x=\\frac{a}{b}=\\frac{c}{d}=y$, i.e., the problem amounts to finding all $x \\in \\mathbb{Q}_{>0}$ such that $2 x \\in \\mathbb{Z}_{>0}$ and $\\frac{2}{x} \\in \\mathbb{Z}_{>0}$. Letting $n=2 x \\in \\mathbb{Z}_{>0}$, we have that $\\frac{2}{x} \\in \\mathbb{Z}_{>0} \\Longleftrightarrow \\frac{4}{n} \\in \\mathbb{Z}_{>0} \\Longleftrightarrow n=1,2$ or 4 , and there are finitely many solutions, namely $(x, y)=\\left(\\frac{1}{2}, \\frac{1}{2}\\right)^{n},(1,1)$ or $(2,2)$.\n\n(ii) There are infinitely many triples $(x, y, z) \\in \\mathbb{Q}_{>0}^{2}$ such that $x+y+z \\in \\mathbb{Z}$ and $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\in \\mathbb{Z}$. We will look for triples such that $x+y+z=1$, so we may write them in the form\n\n$$\n(x, y, z)=\\left(\\frac{a}{a+b+c}, \\frac{b}{a+b+c}, \\frac{c}{a+b+c}\\right) \\quad \\text { with } a, b, c \\in \\mathbb{Z}_{>0}\n$$\n\nWe want these to satisfy\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{a+b+c}{a}+\\frac{a+b+c}{b}+\\frac{a+b+c}{c} \\in \\mathbb{Z} \\Longleftrightarrow \\frac{b+c}{a}+\\frac{a+c}{b}+\\frac{a+b}{c} \\in \\mathbb{Z}\n$$\n\nFixing $a=1$, it suffices to find infinitely many pairs $(b, c) \\in \\mathbb{Z}_{>0}^{2}$ such that\n\n$$\n\\frac{1}{b}+\\frac{1}{c}+\\frac{c}{b}+\\frac{b}{c}=3 \\Longleftrightarrow b^{2}+c^{2}-3 b c+b+c=0\n\\tag{*}\n$$\n\nTo show that equation (*) has infinitely many solutions, we use Vieta jumping (also known as root flipping): starting with $b=2, c=3$, the following algorithm generates infinitely many solutions. Let $c \\geqslant b$, and view $(*)$ as a quadratic equation in $b$ for $c$ fixed:\n\n$$\nb^{2}-(3 c-1) \\cdot b+\\left(c^{2}+c\\right)=0\n\\tag{**}\n$$\n\nThen there exists another root $b_{0} \\in \\mathbb{Z}$ of $(* *)$ which satisfies $b+b_{0}=3 c-1$ and $b \\cdot b_{0}=c^{2}+c$. Since $c \\geqslant b$ by assumption,\n\n$$\nb_{0}=\\frac{c^{2}+c}{b} \\geqslant \\frac{c^{2}+c}{c}>c\n$$\n\nHence from the solution $(b, c)$ we obtain another one $\\left(c, b_{0}\\right)$ with $b_{0}>c$, and we can then ""jump"" again, this time with $c$ as the ""variable"" in the quadratic (*). This algorithm will generate an infinite sequence of distinct solutions, whose first terms are\n\n$(2,3),(3,6),(6,14),(14,35),(35,90),(90,234),(234,611),(611,1598),(1598,4182), \\ldots$', 'Call the $n$-tuples $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right) \\in \\mathbb{Q}_{>0}^{n}$ satisfying the conditions of the problem statement good, and those for which\n\n$$\nf\\left(a_{1}, \\ldots, a_{n}\\right) \\stackrel{\\text { def }}{=}\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)\\left(\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}\\right)\n$$\n\nis an integer pretty. Then good $n$-tuples are pretty, and if $\\left(b_{1}, \\ldots, b_{n}\\right)$ is pretty then\n\n$$\n\\left(\\frac{b_{1}}{b_{1}+b_{2}+\\cdots+b_{n}}, \\frac{b_{2}}{b_{1}+b_{2}+\\cdots+b_{n}}, \\ldots, \\frac{b_{n}}{b_{1}+b_{2}+\\cdots+b_{n}}\\right)\n$$\n\nis good since the sum of its components is 1 , and the sum of the reciprocals of its components equals $f\\left(b_{1}, \\ldots, b_{n}\\right)$. We declare pretty $n$-tuples proportional to each other equivalent since they are precisely those which give rise to the same good $n$-tuple. Clearly, each such equivalence class contains exactly one $n$-tuple of positive integers having no common prime divisors. Call such $n$-tuple a primitive pretty tuple. Our task is to find infinitely many primitive pretty $n$-tuples.\n\nFor $n=1$, there is clearly a single primitive 1-tuple. For $n=2$, we have $f(a, b)=\\frac{(a+b)^{2}}{a b}$, which can be integral (for coprime $a, b \\in \\mathbb{Z}_{>0}$ ) only if $a=b=1$ (see for instance (i) in the first solution).\n\nNow we construct infinitely many primitive pretty triples for $n=3$. Fix $b, c, k \\in \\mathbb{Z}_{>0}$; we will try to find sufficient conditions for the existence of an $a \\in \\mathbb{Q}_{>0}$ such that $f(a, b, c)=k$. Write $\\sigma=b+c, \\tau=b c$. From $f(a, b, c)=k$, we have that $a$ should satisfy the quadratic equation\n\n$$\na^{2} \\cdot \\sigma+a \\cdot\\left(\\sigma^{2}-(k-1) \\tau\\right)+\\sigma \\tau=0\n$$\n\nwhose discriminant is\n\n$$\n\\Delta=\\left(\\sigma^{2}-(k-1) \\tau\\right)^{2}-4 \\sigma^{2} \\tau=\\left((k+1) \\tau-\\sigma^{2}\\right)^{2}-4 k \\tau^{2}\n$$\n\nWe need it to be a square of an integer, say, $\\Delta=M^{2}$ for some $M \\in \\mathbb{Z}$, i.e., we want\n\n$$\n\\left((k+1) \\tau-\\sigma^{2}\\right)^{2}-M^{2}=2 k \\cdot 2 \\tau^{2}\n$$\n\nso that it suffices to set\n\n$$\n(k+1) \\tau-\\sigma^{2}=\\tau^{2}+k, \\quad M=\\tau^{2}-k .\n$$\n\nThe first relation reads $\\sigma^{2}=(\\tau-1)(k-\\tau)$, so if $b$ and $c$ satisfy\n\n$$\n\\tau-1 \\mid \\sigma^{2} \\quad \\text { i.e. } \\quad b c-1 \\mid(b+c)^{2}\n$$\n\nthen $k=\\frac{\\sigma^{2}}{\\tau-1}+\\tau$ will be integral, and we find rational solutions to (1), namely\n\n$$\na=\\frac{\\sigma}{\\tau-1}=\\frac{b+c}{b c-1} \\quad \\text { or } \\quad a=\\frac{\\tau^{2}-\\tau}{\\sigma}=\\frac{b c \\cdot(b c-1)}{b+c}\n$$\n\n\n\nWe can now find infinitely many pairs $(b, c)$ satisfying (2) by Vieta jumping. For example, if we impose\n\n$$\n(b+c)^{2}=5 \\cdot(b c-1)\n$$\n\nthen all pairs $(b, c)=\\left(v_{i}, v_{i+1}\\right)$ satisfy the above condition, where\n\n$$\nv_{1}=2, v_{2}=3, \\quad v_{i+2}=3 v_{i+1}-v_{i} \\quad \\text { for } i \\geqslant 0\n$$\n\nFor $(b, c)=\\left(v_{i}, v_{i+1}\\right)$, one of the solutions to (1) will be $a=(b+c) /(b c-1)=5 /(b+c)=$ $5 /\\left(v_{i}+v_{i+1}\\right)$. Then the pretty triple $(a, b, c)$ will be equivalent to the integral pretty triple\n\n$$\n\\left(5, v_{i}\\left(v_{i}+v_{i+1}\\right), v_{i+1}\\left(v_{i}+v_{i+1}\\right)\\right)\n$$\n\nAfter possibly dividing by 5 , we obtain infinitely many primitive pretty triples, as required.']",['3'],False,,Numerical, 1835,Number Theory,,"Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for each $(x, y)$ in the set $S$. Note: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form $$ f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} . $$","[""First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \\pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\\left(x_{1}, y_{1}\\right)$ through $\\left(x_{n}, y_{n}\\right)$. If any two of these lattice points $\\left(x_{i}, y_{i}\\right)$ and $\\left(x_{j}, y_{j}\\right)$ lie on the same line through the origin, then $\\left(x_{j}, y_{j}\\right)=\\left(-x_{i},-y_{i}\\right)$ because both of the points are irreducible. We then have $f\\left(x_{j}, y_{j}\\right)= \\pm f\\left(x_{i}, y_{i}\\right)$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points.\n\nConsider the homogenous polynomials $\\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define\n\n$$\ng_{i}(x, y)=\\prod_{j \\neq i} \\ell_{j}(x, y)\n$$\n\nThen $\\ell_{i}\\left(x_{j}, y_{j}\\right)=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}\\left(x_{j}, y_{j}\\right)=0$ for all $j \\neq i$. Define $a_{i}=g_{i}\\left(x_{i}, y_{i}\\right)$, and note that $a_{i} \\neq 0$.\n\nNote that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties:\n\n1. $g_{i}\\left(x_{j}, y_{j}\\right)=0$ if $j \\neq i$.\n2. $g_{i}\\left(x_{i}, y_{i}\\right)=a_{i}$.\n\nFor any $N \\geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}\\left(x_{i}, y_{i}\\right)=1$, which exists since $\\left(x_{i}, y_{i}\\right)$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$.\n\nWe may now reduce the problem to the following claim:\n\nClaim: For each positive integer a, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1 , such that $f_{a}(x, y) \\equiv 1(\\bmod a)$ for all relatively prime $(x, y)$.\n\nTo see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}(1 \\leqslant i \\leqslant n)$. Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial.\n\nWe prove the claim by factoring $a$. First, if $a$ is a power of a prime $\\left(a=p^{k}\\right)$, then we may choose either:\n\n- $f_{a}(x, y)=\\left(x^{p-1}+y^{p-1}\\right)^{\\phi(a)}$ if $p$ is odd;\n- $f_{a}(x, y)=\\left(x^{2}+x y+y^{2}\\right)^{\\phi(a)}$ if $p=2$.\n\nNow suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \\cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that\n\n$$\n\\frac{a}{q_{i}} F_{q_{i}}(x, y) \\equiv \\frac{a}{q_{i}}(\\bmod a)\n$$\n\nfor any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\\frac{a}{q_{i}}$ that equals 1 . Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \\equiv 1$ $(\\bmod a)$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree."", ""As in the previous solution, label the irreducible lattice points $\\left(x_{1}, y_{1}\\right), \\ldots,\\left(x_{n}, y_{n}\\right)$ and assume without loss of generality that no two of the points are collinear with the origin. We induct on $n$ to construct a homogenous polynomial $f(x, y)$ such that $f\\left(x_{i}, y_{i}\\right)=1$ for all $1 \\leqslant i \\leqslant n$.\n\nIf $n=1$ : Since $x_{1}$ and $y_{1}$ are relatively prime, there exist some integers $c, d$ such that $c x_{1}+d y_{1}=1$. Then $f(x, y)=c x+d y$ is suitable.\n\nIf $n \\geqslant 2$ : By the induction hypothesis we already have a homogeneous polynomial $g(x, y)$ with $g\\left(x_{1}, y_{1}\\right)=\\ldots=g\\left(x_{n-1}, y_{n-1}\\right)=1$. Let $j=\\operatorname{deg} g$,\n\n$$\ng_{n}(x, y)=\\prod_{k=1}^{n-1}\\left(y_{k} x-x_{k} y\\right)\n$$\n\nand $a_{n}=g_{n}\\left(x_{n}, y_{n}\\right)$. By assumption, $a_{n} \\neq 0$. Take some integers $c, d$ such that $c x_{n}+d y_{n}=1$. We will construct $f(x, y)$ in the form\n\n$$\nf(x, y)=g(x, y)^{K}-C \\cdot g_{n}(x, y) \\cdot(c x+d y)^{L}\n$$\n\nwhere $K$ and $L$ are some positive integers and $C$ is some integer. We assume that $L=K j-n+1$ so that $f$ is homogenous.\n\nDue to $g\\left(x_{1}, y_{1}\\right)=\\ldots=g\\left(x_{n-1}, y_{n-1}\\right)=1$ and $g_{n}\\left(x_{1}, y_{1}\\right)=\\ldots=g_{n}\\left(x_{n-1}, y_{n-1}\\right)=0$, the property $f\\left(x_{1}, y_{1}\\right)=\\ldots=f\\left(x_{n-1}, y_{n-1}\\right)=1$ is automatically satisfied with any choice of $K, L$, and $C$.\n\nFurthermore,\n\n$$\nf\\left(x_{n}, y_{n}\\right)=g\\left(x_{n}, y_{n}\\right)^{K}-C \\cdot g_{n}\\left(x_{n}, y_{n}\\right) \\cdot\\left(c x_{n}+d y_{n}\\right)^{L}=g\\left(x_{n}, y_{n}\\right)^{K}-C a_{n}\n$$\n\nIf we have an exponent $K$ such that $g\\left(x_{n}, y_{n}\\right)^{K} \\equiv 1\\left(\\bmod a_{n}\\right)$, then we may choose $C$ such that $f\\left(x_{n}, y_{n}\\right)=1$. We now choose such a $K$.\n\nConsider an arbitrary prime divisor $p$ of $a_{n}$. By\n\n$$\np \\mid a_{n}=g_{n}\\left(x_{n}, y_{n}\\right)=\\prod_{k=1}^{n-1}\\left(y_{k} x_{n}-x_{k} y_{n}\\right)\n$$\n\nthere is some $1 \\leqslant k0$.)""]",,True,,, 1836,Number Theory,,"Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties: - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal to 1 ; - $f(a+b, b)=f(a, b)$ for any pair of relatively prime positive integers $(a, b)$. Prove that $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right) \geqslant \sqrt{2 p}-2 $$","['Denote by $\\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \\in \\mathbb{A}$ there exists a pair $(u, v) \\in \\mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $\\left(u_{0}, v_{0}\\right)$ is one such pair, then all such pairs are of the form $(u, v)=\\left(u_{0}+k b, v_{0}-k a\\right)$, where $k \\in \\mathbb{Z}$. So there exists a unique such pair $(u, v)$ with $-b / 20$.\n\nProof. We induct on $a+b$. The base case is $a+b=2$. In this case, we have that $a=b=1$, $g(a, b)=g(1,1)=(0,1)$ and $f(1,1)=0$, so the claim holds.\n\nAssume now that $a+b>2$, and so $a \\neq b$, since $a$ and $b$ are coprime. Two cases are possible. Case 1: $a>b$.\n\nNotice that $g(a-b, b)=(u, v+u)$, since $u(a-b)+(v+u) b=1$ and $u \\in(-b / 2, b / 2]$. Thus $f(a, b)=1 \\Longleftrightarrow f(a-b, b)=1 \\Longleftrightarrow u>0$ by the induction hypothesis.\n\nCase 2: $av b \\geqslant 1-\\frac{a b}{2}, \\quad \\text { so } \\quad \\frac{1+a}{2} \\geqslant \\frac{1}{b}+\\frac{a}{2}>v \\geqslant \\frac{1}{b}-\\frac{a}{2}>-\\frac{a}{2}\n$$\n\nThus $1+a>2 v>-a$, so $a \\geqslant 2 v>-a$, hence $a / 2 \\geqslant v>-a / 2$, and thus $g(b, a)=(v, u)$.\n\nObserve that $f(a, b)=1 \\Longleftrightarrow f(b, a)=0 \\Longleftrightarrow f(b-a, a)=0$. We know from Case 1 that $g(b-a, a)=(v, u+v)$. We have $f(b-a, a)=0 \\Longleftrightarrow v \\leqslant 0$ by the inductive hypothesis. Then, since $b>a \\geqslant 1$ and $u a+v b=1$, we have $v \\leqslant 0 \\Longleftrightarrow u>0$, and we are done.\n\nThe Lemma proves that, for all $(a, b) \\in \\mathbb{A}, f(a, b)=1$ if and only if the inverse of $a$ modulo $b$, taken in $\\{1,2, \\ldots, b-1\\}$, is at most $b / 2$. Then, for any odd prime $p$ and integer $n$ such that $n \\not \\equiv 0(\\bmod p), f\\left(n^{2}, p\\right)=1$ iff the inverse of $n^{2} \\bmod p$ is less than $p / 2$. Since $\\left\\{n^{2} \\bmod p: 1 \\leqslant n \\leqslant p-1\\right\\}=\\left\\{n^{-2} \\bmod p: 1 \\leqslant n \\leqslant p-1\\right\\}$, including multiplicities (two for each quadratic residue in each set), we conclude that the desired sum is twice the number of quadratic residues that are less than $p / 2$, i.e.,\n\n$$\n\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right)=2 \\mid\\left\\{k: 1 \\leqslant k \\leqslant \\frac{p-1}{2} \\text { and } k^{2} \\bmod p<\\frac{p}{2}\\right\\} \\mid .\n\\tag{1}\n$$\n\nSince the number of perfect squares in the interval $[1, p / 2)$ is $\\lfloor\\sqrt{p / 2}\\rfloor>\\sqrt{p / 2}-1$, we conclude that\n\n$$\n\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right)>2\\left(\\sqrt{\\frac{p}{2}}-1\\right)=\\sqrt{2 p}-2\n$$', 'We provide a different proof for the Lemma. For this purpose, we use continued fractions to find $g(a, b)=(u, v)$ explicitly.\n\nThe function $f$ is completely determined on $\\mathbb{A}$ by the following\n\nClaim. Represent $a / b$ as a continued fraction; that is, let $a_{0}$ be an integer and $a_{1}, \\ldots, a_{k}$ be positive integers such that $a_{k} \\geqslant 2$ and\n\n$$\n\\frac{a}{b}=a_{0}+\\frac{1}{a_{1}+\\frac{1}{a_{2}+\\frac{1}{\\cdots+\\frac{1}{a_{k}}}}}=\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{k}\\right] .\n$$\n\nThen $f(a, b)=0 \\Longleftrightarrow k$ is even.\n\nProof. We induct on $b$. If $b=1$, then $a / b=[a]$ and $k=0$. Then, for $a \\geqslant 1$, an easy induction shows that $f(a, 1)=f(1,1)=0$.\n\nNow consider the case $b>1$. Perform the Euclidean division $a=q b+r$, with $0 \\leqslant r0$ and define $q_{-1}=0$ if necessary. Then\n\n- $q_{k}=a_{k} q_{k-1}+q_{k-2}$, and\n- $a q_{k-1}-b p_{k-1}=p_{k} q_{k-1}-q_{k} p_{k-1}=(-1)^{k-1}$.\n\nAssume that $k>0$. Then $a_{k} \\geqslant 2$, and\n\n$$\nb=q_{k}=a_{k} q_{k-1}+q_{k-2} \\geqslant a_{k} q_{k-1} \\geqslant 2 q_{k-1} \\Longrightarrow q_{k-1} \\leqslant \\frac{b}{2}\n$$\n\nwith strict inequality for $k>1$, and\n\n$$\n(-1)^{k-1} q_{k-1} a+(-1)^{k} p_{k-1} b=1\n$$\n\nNow we finish the proof of the Lemma. It is immediate for $k=0$. If $k=1$, then $(-1)^{k-1}=1$, so\n\n$$\n-b / 2<0 \\leqslant(-1)^{k-1} q_{k-1} \\leqslant b / 2 .\n$$\n\nIf $k>1$, we have $q_{k-1}0$, we find that $g(a, b)=\\left((-1)^{k-1} q_{k-1},(-1)^{k} p_{k-1}\\right)$, and so\n\n$$\nf(a, b)=1 \\Longleftrightarrow k \\text { is odd } \\Longleftrightarrow u=(-1)^{k-1} q_{k-1}>0 .\n$$']",,True,,, 1837,Algebra,,"Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1\tag{1} $$","['We first show the following.\n\n- Claim. For any positive real numbers $x, y$ with $x y \\geqslant 1$, we have\n\n$$\n\\left(x^{2}+1\\right)\\left(y^{2}+1\\right) \\leqslant\\left(\\left(\\frac{x+y}{2}\\right)^{2}+1\\right)^{2}\\tag{2}\n$$\n\nProof. Note that $x y \\geqslant 1$ implies $\\left(\\frac{x+y}{2}\\right)^{2}-1 \\geqslant x y-1 \\geqslant 0$. We find that\n\n$$\n\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)=(x y-1)^{2}+(x+y)^{2} \\leqslant\\left(\\left(\\frac{x+y}{2}\\right)^{2}-1\\right)^{2}+(x+y)^{2}=\\left(\\left(\\frac{x+y}{2}\\right)^{2}+1\\right)^{2} .\n$$\n\nWithout loss of generality, assume $a \\geqslant b \\geqslant c$. This implies $a \\geqslant 1$. Let $d=\\frac{a+b+c}{3}$. Note that\n\n$$\na d=\\frac{a(a+b+c)}{3} \\geqslant \\frac{1+1+1}{3}=1 .\n$$\n\nThen we can apply (2) to the pair $(a, d)$ and the pair $(b, c)$. We get\n\n$$\n\\left(a^{2}+1\\right)\\left(d^{2}+1\\right)\\left(b^{2}+1\\right)\\left(c^{2}+1\\right) \\leqslant\\left(\\left(\\frac{a+d}{2}\\right)^{2}+1\\right)^{2}\\left(\\left(\\frac{b+c}{2}\\right)^{2}+1\\right)^{2} .\\tag{3}\n$$\n\nNext, from\n\n$$\n\\frac{a+d}{2} \\cdot \\frac{b+c}{2} \\geqslant \\sqrt{a d} \\cdot \\sqrt{b c} \\geqslant 1\n$$\n\nwe can apply (2) again to the pair $\\left(\\frac{a+d}{2}, \\frac{b+c}{2}\\right)$. Together with (3), we have\n\n$$\n\\left(a^{2}+1\\right)\\left(d^{2}+1\\right)\\left(b^{2}+1\\right)\\left(c^{2}+1\\right) \\leqslant\\left(\\left(\\frac{a+b+c+d}{4}\\right)^{2}+1\\right)^{4}=\\left(d^{2}+1\\right)^{4} .\n$$\n\nTherefore, $\\left(a^{2}+1\\right)\\left(b^{2}+1\\right)\\left(c^{2}+1\\right) \\leqslant\\left(d^{2}+1\\right)^{3}$, and (1) follows by taking cube root of both sides.']",,True,,, 1838,Algebra,,"Find the smallest real constant $C$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that $$ \left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C .\tag{1} $$","['We first show that $C \\leqslant \\frac{1}{2}$. For any positive real numbers $a_{1} \\leqslant a_{2} \\leqslant a_{3} \\leqslant a_{4} \\leqslant a_{5}$, consider the five fractions\n\n$$\n\\frac{a_{1}}{a_{2}}, \\frac{a_{3}}{a_{4}}, \\frac{a_{1}}{a_{5}}, \\frac{a_{2}}{a_{3}}, \\frac{a_{4}}{a_{5}}\\tag{2}\n$$\n\nEach of them lies in the interval $(0,1]$. Therefore, by the Pigeonhole Principle, at least three of them must lie in $\\left(0, \\frac{1}{2}\\right]$ or lie in $\\left(\\frac{1}{2}, 1\\right]$ simultaneously. In particular, there must be two consecutive terms in (2) which belong to an interval of length $\\frac{1}{2}$ (here, we regard $\\frac{a_{1}}{a_{2}}$ and $\\frac{a_{4}}{a_{5}}$ as consecutive). In other words, the difference of these two fractions is less than $\\frac{1}{2}$. As the indices involved in these two fractions are distinct, we can choose them to be $i, j, k, l$ and conclude that $C \\leqslant \\frac{1}{2}$.\n\nNext, we show that $C=\\frac{1}{2}$ is best possible. Consider the numbers $1,2,2,2, n$ where $n$ is a large real number. The fractions formed by two of these numbers in ascending order are $\\frac{1}{n}, \\frac{2}{n}, \\frac{1}{2}, \\frac{2}{2}, \\frac{2}{1}, \\frac{n}{2}, \\frac{n}{1}$. Since the indices $i, j, k, l$ are distinct, $\\frac{1}{n}$ and $\\frac{2}{n}$ cannot be chosen simultaneously. Therefore the minimum value of the left-hand side of (1) is $\\frac{1}{2}-\\frac{2}{n}$. When $n$ tends to infinity, this value approaches $\\frac{1}{2}$, and so $C$ cannot be less than $\\frac{1}{2}$.\n\nThese conclude that $C=\\frac{1}{2}$ is the smallest possible choice.']",['$\\frac{1}{2}$'],False,,Numerical, 1839,Algebra,,"Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either -1 or 1 , such that $$ \left|\sum_{k=1}^{n} x_{k} a_{k}\right|+\left|\sum_{k=1}^{n} x_{k} b_{k}\right| \leqslant 1 \tag{1} $$","['For any even integer $n \\geqslant 4$, we consider the case\n\n$$\na_{1}=a_{2}=\\cdots=a_{n-1}=b_{n}=0 \\quad \\text { and } \\quad b_{1}=b_{2}=\\cdots=b_{n-1}=a_{n}=1\n$$\n\nThe condition $\\left|a_{k}\\right|+\\left|b_{k}\\right|=1$ is satisfied for each $1 \\leqslant k \\leqslant n$. No matter how we choose each $x_{k}$, both sums $\\sum_{k=1}^{n} x_{k} a_{k}$ and $\\sum_{k=1}^{n} x_{k} b_{k}$ are odd integers. This implies $\\left|\\sum_{k=1}^{n} x_{k} a_{k}\\right| \\geqslant 1$ and $\\left|\\sum_{k=1}^{n} x_{k} b_{k}\\right| \\geqslant 1$, which shows (1) cannot hold.\n\nFor any odd integer $n \\geqslant 3$, we may assume without loss of generality $b_{k} \\geqslant 0$ for $1 \\leqslant k \\leqslant n$ (this can be done by flipping the pair $\\left(a_{k}, b_{k}\\right)$ to $\\left(-a_{k},-b_{k}\\right)$ and $x_{k}$ to $-x_{k}$ if necessary) and $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{m} \\geqslant 0>a_{m+1} \\geqslant \\cdots \\geqslant a_{n}$. We claim that the choice $x_{k}=(-1)^{k+1}$ for $1 \\leqslant k \\leqslant n$ will work. Define\n\n$$\ns=\\sum_{k=1}^{m} x_{k} a_{k} \\quad \\text { and } \\quad t=-\\sum_{k=m+1}^{n} x_{k} a_{k}\n$$\n\nNote that\n\n$$\ns=\\left(a_{1}-a_{2}\\right)+\\left(a_{3}-a_{4}\\right)+\\cdots \\geqslant 0\n$$\n\nby the assumption $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{m}$ (when $m$ is odd, there is a single term $a_{m}$ at the end, which is also positive). Next, we have\n\n$$\ns=a_{1}-\\left(a_{2}-a_{3}\\right)-\\left(a_{4}-a_{5}\\right)-\\cdots \\leqslant a_{1} \\leqslant 1 .\n$$\n\nSimilarly,\n\n$$\nt=\\left(-a_{n}+a_{n-1}\\right)+\\left(-a_{n-2}+a_{n-3}\\right)+\\cdots \\geqslant 0\n$$\n\nand\n\n$$\nt=-a_{n}+\\left(a_{n-1}-a_{n-2}\\right)+\\left(a_{n-3}-a_{n-4}\\right)+\\cdots \\leqslant-a_{n} \\leqslant 1 .\n$$\n\nFrom the condition, we have $a_{k}+b_{k}=1$ for $1 \\leqslant k \\leqslant m$ and $-a_{k}+b_{k}=1$ for $m+1 \\leqslant k \\leqslant n$. It follows that $\\sum_{k=1}^{n} x_{k} a_{k}=s-t$ and $\\sum_{k=1}^{n} x_{k} b_{k}=1-s-t$. Hence it remains to prove\n\n$$\n|s-t|+|1-s-t| \\leqslant 1\n$$\n\nunder the constraint $0 \\leqslant s, t \\leqslant 1$. By symmetry, we may assume $s \\geqslant t$. If $1-s-t \\geqslant 0$, then we have\n\n$$\n|s-t|+|1-s-t|=s-t+1-s-t=1-2 t \\leqslant 1\n$$\n\nIf $1-s-t \\leqslant 0$, then we have\n\n$$\n|s-t|+|1-s-t|=s-t-1+s+t=2 s-1 \\leqslant 1\n$$\n\nHence, the inequality is true in both cases.\n\nThese show $n$ can be any odd integer greater than or equal to 3 .', 'For the odd case, we prove by induction on $n$.\n\nFirstly, for $n=3$, we may assume without loss of generality $a_{1} \\geqslant a_{2} \\geqslant a_{3} \\geqslant 0$ and $b_{1}=a_{1}-1$ (if $b_{1}=1-a_{1}$, we may replace each $b_{k}$ by $-b_{k}$ ).\n\n- Case 1. $b_{2}=a_{2}-1$ and $b_{3}=a_{3}-1$, in which case we take $\\left(x_{1}, x_{2}, x_{3}\\right)=(1,-1,1)$.\n\nLet $c=a_{1}-a_{2}+a_{3}$ so that $0 \\leqslant c \\leqslant 1$. Then $\\left|b_{1}-b_{2}+b_{3}\\right|=\\left|a_{1}-a_{2}+a_{3}-1\\right|=1-c$ and hence $|c|+\\left|b_{1}-b_{2}+b_{3}\\right|=1$.\n\n- Case 2. $b_{2}=1-a_{2}$ and $b_{3}=1-a_{3}$, in which case we take $\\left(x_{1}, x_{2}, x_{3}\\right)=(1,-1,1)$.\n\nLet $c=a_{1}-a_{2}+a_{3}$ so that $0 \\leqslant c \\leqslant 1$. Since $a_{3} \\leqslant a_{2}$ and $a_{1} \\leqslant 1$, we have\n\n$$\nc-1 \\leqslant b_{1}-b_{2}+b_{3}=a_{1}+a_{2}-a_{3}-1 \\leqslant 1-c .\n$$\n\nThis gives $\\left|b_{1}-b_{2}+b_{3}\\right| \\leqslant 1-c$ and hence $|c|+\\left|b_{1}-b_{2}+b_{3}\\right| \\leqslant 1$.\n\n- Case 3. $b_{2}=a_{2}-1$ and $b_{3}=1-a_{3}$, in which case we take $\\left(x_{1}, x_{2}, x_{3}\\right)=(-1,1,1)$.\n\nLet $c=-a_{1}+a_{2}+a_{3}$. If $c \\geqslant 0$, then $a_{3} \\leqslant 1$ and $a_{2} \\leqslant a_{1}$ imply\n\n$$\nc-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}+a_{2}-a_{3}+1 \\leqslant 1-c .\n$$\n\nIf $c<0$, then $a_{1} \\leqslant a_{2}+1$ and $a_{3} \\geqslant 0$ imply\n\n$$\n-c-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}+a_{2}-a_{3}+1 \\leqslant 1+c .\n$$\n\nIn both cases, we get $\\left|-b_{1}+b_{2}+b_{3}\\right| \\leqslant 1-|c|$ and hence $|c|+\\left|-b_{1}+b_{2}+b_{3}\\right| \\leqslant 1$.\n\n- Case 4. $b_{2}=1-a_{2}$ and $b_{3}=a_{3}-1$, in which case we take $\\left(x_{1}, x_{2}, x_{3}\\right)=(-1,1,1)$.\n\nLet $c=-a_{1}+a_{2}+a_{3}$. If $c \\geqslant 0$, then $a_{2} \\leqslant 1$ and $a_{3} \\leqslant a_{1}$ imply\n\n$$\nc-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}-a_{2}+a_{3}+1 \\leqslant 1-c .\n$$\n\nIf $c<0$, then $a_{1} \\leqslant a_{3}+1$ and $a_{2} \\geqslant 0$ imply\n\n$$\n-c-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}-a_{2}+a_{3}+1 \\leqslant 1+c .\n$$\n\nIn both cases, we get $\\left|-b_{1}+b_{2}+b_{3}\\right| \\leqslant 1-|c|$ and hence $|c|+\\left|-b_{1}+b_{2}+b_{3}\\right| \\leqslant 1$.\n\nWe have found $x_{1}, x_{2}, x_{3}$ satisfying (1) in each case for $n=3$.\n\nNow, let $n \\geqslant 5$ be odd and suppose the result holds for any smaller odd cases. Again we may assume $a_{k} \\geqslant 0$ for each $1 \\leqslant k \\leqslant n$. By the Pigeonhole Principle, there are at least three indices $k$ for which $b_{k}=a_{k}-1$ or $b_{k}=1-a_{k}$. Without loss of generality, suppose $b_{k}=a_{k}-1$ for $k=1,2,3$. Again by the Pigeonhole Principle, as $a_{1}, a_{2}, a_{3}$ lies between 0 and 1 , the difference of two of them is at most $\\frac{1}{2}$. By changing indices if necessary, we may assume $0 \\leqslant d=a_{1}-a_{2} \\leqslant \\frac{1}{2}$.\n\nBy the inductive hypothesis, we can choose $x_{3}, x_{4}, \\ldots, x_{n}$ such that $a^{\\prime}=\\sum_{k=3}^{n} x_{k} a_{k}$ and $b^{\\prime}=\\sum_{k=3}^{n} x_{k} b_{k}$ satisfy $\\left|a^{\\prime}\\right|+\\left|b^{\\prime}\\right| \\leqslant 1$. We may further assume $a^{\\prime} \\geqslant 0$.\n\n\n\n- Case 1. $b^{\\prime} \\geqslant 0$, in which case we take $\\left(x_{1}, x_{2}\\right)=(-1,1)$.\n\nWe have $\\left|-a_{1}+a_{2}+a^{\\prime}\\right|+\\left|-\\left(a_{1}-1\\right)+\\left(a_{2}-1\\right)+b^{\\prime}\\right|=\\left|-d+a^{\\prime}\\right|+\\left|-d+b^{\\prime}\\right| \\leqslant$ $\\max \\left\\{a^{\\prime}+b^{\\prime}-2 d, a^{\\prime}-b^{\\prime}, b^{\\prime}-a^{\\prime}, 2 d-a^{\\prime}-b^{\\prime}\\right\\} \\leqslant 1$ since $0 \\leqslant a^{\\prime}, b^{\\prime}, a^{\\prime}+b^{\\prime} \\leqslant 1$ and $0 \\leqslant d \\leqslant \\frac{1}{2}$.\n\n- Case 2. $0>b^{\\prime} \\geqslant-a^{\\prime}$, in which case we take $\\left(x_{1}, x_{2}\\right)=(-1,1)$.\n\nWe have $\\left|-a_{1}+a_{2}+a^{\\prime}\\right|+\\left|-\\left(a_{1}-1\\right)+\\left(a_{2}-1\\right)+b^{\\prime}\\right|=\\left|-d+a^{\\prime}\\right|+\\left|-d+b^{\\prime}\\right|$. If $-d+a^{\\prime} \\geqslant 0$, this equals $a^{\\prime}-b^{\\prime}=\\left|a^{\\prime}\\right|+\\left|b^{\\prime}\\right| \\leqslant 1$. If $-d+a^{\\prime}<0$, this equals $2 d-a^{\\prime}-b^{\\prime} \\leqslant 2 d \\leqslant 1$.\n\n- Case 3. $b^{\\prime}<-a^{\\prime}$, in which case we take $\\left(x_{1}, x_{2}\\right)=(1,-1)$.\n\nWe have $\\left|a_{1}-a_{2}+a^{\\prime}\\right|+\\left|\\left(a_{1}-1\\right)-\\left(a_{2}-1\\right)+b^{\\prime}\\right|=\\left|d+a^{\\prime}\\right|+\\left|d+b^{\\prime}\\right|$. If $d+b^{\\prime} \\geqslant 0$, this equals $2 d+a^{\\prime}+b^{\\prime}<2 d \\leqslant 1$. If $d+b^{\\prime}<0$, this equals $a^{\\prime}-b^{\\prime}=\\left|a^{\\prime}\\right|+\\left|b^{\\prime}\\right| \\leqslant 1$.\n\nTherefore, we have found $x_{1}, x_{2}, \\ldots, x_{n}$ satisfying (1) in each case. By induction, the property holds for all odd integers $n \\geqslant 3$.']",['$n$ can be any odd integer greater than or equal to 3'],False,,Need_human_evaluate, 1840,Algebra,,"Denote by $\mathbb{R}^{+}$ the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $$ x f\left(x^{2}\right) f(f(y))+f(y f(x))=f(x y)\left(f\left(f\left(x^{2}\right)\right)+f\left(f\left(y^{2}\right)\right)\right) \tag{1} $$ for all positive real numbers $x$ and $y$.","['Taking $x=y=1$ in (1), we get $f(1) f(f(1))+f(f(1))=2 f(1) f(f(1))$ and hence $f(1)=1$. Swapping $x$ and $y$ in (1) and comparing with (1) again, we find\n\n$$\nx f\\left(x^{2}\\right) f(f(y))+f(y f(x))=y f\\left(y^{2}\\right) f(f(x))+f(x f(y)) .\n\\tag{2}\n$$\n\nTaking $y=1$ in (2), we have $x f\\left(x^{2}\\right)+f(f(x))=f(f(x))+f(x)$, that is,\n\n$$\nf\\left(x^{2}\\right)=\\frac{f(x)}{x}\n\\tag{3}\n$$\n\nTake $y=1$ in (1) and apply (3) to $x f\\left(x^{2}\\right)$. We get $f(x)+f(f(x))=f(x)\\left(f\\left(f\\left(x^{2}\\right)\\right)+1\\right)$, which implies\n\n$$\nf\\left(f\\left(x^{2}\\right)\\right)=\\frac{f(f(x))}{f(x)}\n\\tag{4}\n$$\n\nFor any $x \\in \\mathbb{R}^{+}$, we find that\n\n$$\nf\\left(f(x)^{2}\\right) \\stackrel{(3)}{=} \\frac{f(f(x))}{f(x)} \\stackrel{(4)}{=} f\\left(f\\left(x^{2}\\right)\\right) \\stackrel{(3)}{=} f\\left(\\frac{f(x)}{x}\\right)\n\\tag{5}\n$$\n\nIt remains to show the following key step.\n\n- Claim. The function $f$ is injective.\n\nProof. Using (3) and (4), we rewrite (1) as\n\n$$\nf(x) f(f(y))+f(y f(x))=f(x y)\\left(\\frac{f(f(x))}{f(x)}+\\frac{f(f(y))}{f(y)}\\right) .\n\\tag{6}\n$$\n\nTake $x=y$ in (6) and apply (3). This gives $f(x) f(f(x))+f(x f(x))=2 \\frac{f(f(x))}{x}$, which means\n\n$$\nf(x f(x))=f(f(x))\\left(\\frac{2}{x}-f(x)\\right)\n\\tag{7}\n$$\n\nUsing (3), equation (2) can be rewritten as\n\n$$\nf(x) f(f(y))+f(y f(x))=f(y) f(f(x))+f(x f(y)) .\n\\tag{8}\n$$\n\nSuppose $f(x)=f(y)$ for some $x, y \\in \\mathbb{R}^{+}$. Then (8) implies\n\n$$\nf(y f(y))=f(y f(x))=f(x f(y))=f(x f(x)) .\n$$\n\nUsing (7), this gives\n\n$$\nf(f(y))\\left(\\frac{2}{y}-f(y)\\right)=f(f(x))\\left(\\frac{2}{x}-f(x)\\right)\n$$\n\nNoting $f(x)=f(y)$, we find $x=y$. This establishes the injectivity.\n\n\n\nBy the Claim and (5), we get the only possible solution $f(x)=\\frac{1}{x}$. It suffices to check that this is a solution. Indeed, the left-hand side of (1) becomes\n\n$$\nx \\cdot \\frac{1}{x^{2}} \\cdot y+\\frac{x}{y}=\\frac{y}{x}+\\frac{x}{y}\n$$\n\nwhile the right-hand side becomes\n\n$$\n\\frac{1}{x y}\\left(x^{2}+y^{2}\\right)=\\frac{x}{y}+\\frac{y}{x}\n$$\n\nThe two sides agree with each other.', 'Taking $x=y=1$ in (1), we get $f(1) f(f(1))+f(f(1))=2 f(1) f(f(1))$ and hence $f(1)=1$. Putting $x=1$ in (1), we have $f(f(y))+f(y)=f(y)\\left(1+f\\left(f\\left(y^{2}\\right)\\right)\\right)$ so that\n\n$$\nf(f(y))=f(y) f\\left(f\\left(y^{2}\\right)\\right) .\n\\tag{9}\n$$\n\nPutting $y=1$ in (1), we get $x f\\left(x^{2}\\right)+f(f(x))=f(x)\\left(f\\left(f\\left(x^{2}\\right)\\right)+1\\right)$. Using (9), this gives\n\n$$\nx f\\left(x^{2}\\right)=f(x)\n\\tag{10}\n$$\n\nReplace $y$ by $\\frac{1}{x}$ in (1). Then we have\n\n$$\nx f\\left(x^{2}\\right) f\\left(f\\left(\\frac{1}{x}\\right)\\right)+f\\left(\\frac{f(x)}{x}\\right)=f\\left(f\\left(x^{2}\\right)\\right)+f\\left(f\\left(\\frac{1}{x^{2}}\\right)\\right) .\n$$\n\nThe relation (10) shows $f\\left(\\frac{f(x)}{x}\\right)=f\\left(f\\left(x^{2}\\right)\\right)$. Also, using (9) with $y=\\frac{1}{x}$ and using (10) again, the last equation reduces to\n\n$$\nf(x) f\\left(\\frac{1}{x}\\right)=1\n\\tag{11}\n$$\n\nReplace $x$ by $\\frac{1}{x}$ and $y$ by $\\frac{1}{y}$ in (1) and apply (11). We get\n\n$$\n\\frac{1}{x f\\left(x^{2}\\right) f(f(y))}+\\frac{1}{f(y f(x))}=\\frac{1}{f(x y)}\\left(\\frac{1}{f\\left(f\\left(x^{2}\\right)\\right)}+\\frac{1}{f\\left(f\\left(y^{2}\\right)\\right)}\\right) .\n$$\n\nClearing denominators, we can use (1) to simplify the numerators and obtain\n\n$$\nf(x y)^{2} f\\left(f\\left(x^{2}\\right)\\right) f\\left(f\\left(y^{2}\\right)\\right)=x f\\left(x^{2}\\right) f(f(y)) f(y f(x)) .\n$$\n\nUsing (9) and (10), this is the same as\n\n$$\nf(x y)^{2} f(f(x))=f(x)^{2} f(y) f(y f(x)) .\n\\tag{12}\n$$\n\nSubstitute $y=f(x)$ in (12) and apply (10) (with $x$ replaced by $f(x)$ ). We have\n\n$$\nf(x f(x))^{2}=f(x) f(f(x)) .\\tag{13}\n$$\n\nTaking $y=x$ in (12), squaring both sides, and using (10) and (13), we find that\n\n$$\nf(f(x))=x^{4} f(x)^{3}\\tag{14}\n$$\n\nFinally, we combine (9), (10) and (14) to get\n\n$$\ny^{4} f(y)^{3} \\stackrel{(14)}{=} f(f(y)) \\stackrel{(9)}{=} f(y) f\\left(f\\left(y^{2}\\right)\\right) \\stackrel{(14)}{=} f(y) y^{8} f\\left(y^{2}\\right)^{3} \\stackrel{(10)}{=} y^{5} f(y)^{4},\n$$\n\nwhich implies $f(y)=\\frac{1}{y}$.']",['$f(x)=\\frac{1}{x}$ for any $x \\in \\mathbb{R}^{+}$'],False,,Need_human_evaluate, 1841,Algebra,," Prove that for every positive integer $n$, there exists a fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $02016$ or $4 k2$ for $0 \\leqslant j \\leqslant 503$ in this case. So each term in the product lies strictly between 0 and 1 , and the whole product must be less than 1 , which is impossible.\n\n- Case 4. $4 k+20$, then $f(0)^{2}+2 f(0)=0$ gives no positive solution. If $f(0)<0$, then $f(0)^{2}+f(0)=0$ gives $f(0)=-1$. Putting $y=0$ in (1), we have $f(x)^{2}=-2 f(x)+f\\left(x^{2}\\right)$, which is the same as $(f(x)+1)^{2}=f\\left(x^{2}\\right)+1$. Let $g(x)=f(x)+1$. Then for any $x \\in \\mathbb{R}$, we have\n\n$$\ng\\left(x^{2}\\right)=g(x)^{2} \\geqslant 0\n\\tag{2}\n$$\n\nFrom (1), we find that $f(x+y)^{2} \\geqslant 2 f(x) f(y)+f\\left(x^{2}\\right)+f\\left(y^{2}\\right)$. In terms of $g$, this becomes $(g(x+y)-1)^{2} \\geqslant 2(g(x)-1)(g(y)-1)+g\\left(x^{2}\\right)+g\\left(y^{2}\\right)-2$. Using (2), this means\n\n$$\n(g(x+y)-1)^{2} \\geqslant(g(x)+g(y)-1)^{2}-1\n\\tag{3}\n$$\n\nPutting $x=1$ in (2), we get $g(1)=0$ or 1 . The two cases are handled separately.\n\n- Case 1. $g(1)=0$, which is the same as $f(1)=-1$.\n\nWe put $x=-1$ and $y=0$ in (1). This gives $f(-1)^{2}=-2 f(-1)-1$, which forces $f(-1)=-1$. Next, we take $x=-1$ and $y=1$ in (1) to get $1=2+\\max \\{-2, f(2)\\}$. This clearly implies $1=2+f(2)$ and hence $f(2)=-1$, that is, $g(2)=0$. From (2), we can prove inductively that $g\\left(2^{2^{n}}\\right)=g(2)^{2^{n}}=0$ for any $n \\in \\mathbb{N}$. Substitute $y=2^{2^{n}}-x$ in (3). We obtain\n\n$$\n\\left(g(x)+g\\left(2^{2^{n}}-x\\right)-1\\right)^{2} \\leqslant\\left(g\\left(2^{2^{n}}\\right)-1\\right)^{2}+1=2 .\n$$\n\nFor any fixed $x \\geqslant 0$, we consider $n$ to be sufficiently large so that $2^{2^{n}}-x>0$. From (2), this implies $g\\left(2^{2^{n}}-x\\right) \\geqslant 0$ so that $g(x) \\leqslant 1+\\sqrt{2}$. Using (2) again, we get\n\n$$\ng(x)^{2^{n}}=g\\left(x^{2^{n}}\\right) \\leqslant 1+\\sqrt{2}\n$$\n\nfor any $n \\in \\mathbb{N}$. Therefore, $|g(x)| \\leqslant 1$ for any $x \\geqslant 0$.\n\nIf there exists $a \\in \\mathbb{R}$ for which $g(a) \\neq 0$, then for sufficiently large $n$ we must have $g\\left(\\left(a^{2}\\right)^{\\frac{1}{2^{n}}}\\right)=g\\left(a^{2}\\right)^{\\frac{1}{2^{n}}}>\\frac{1}{2}$. By taking $x=-y=-\\left(a^{2}\\right)^{\\frac{1}{2^{n}}}$ in (1), we obtain\n\n$$\n\\begin{aligned}\n1 & =2 f(x) f(-x)+\\max \\left\\{2 f\\left(x^{2}\\right), f\\left(2 x^{2}\\right)\\right\\} \\\\\n& =2(g(x)-1)(g(-x)-1)+\\max \\left\\{2\\left(g\\left(x^{2}\\right)-1\\right), g\\left(2 x^{2}\\right)-1\\right\\} \\\\\n& \\leqslant 2\\left(-\\frac{1}{2}\\right)\\left(-\\frac{1}{2}\\right)+0=\\frac{1}{2}\n\\end{aligned}\n$$\n\nsince $|g(-x)|=|g(x)| \\in\\left(\\frac{1}{2}, 1\\right]$ by $(2)$ and the choice of $x$, and since $g(z) \\leqslant 1$ for $z \\geqslant 0$. This yields a contradiction and hence $g(x)=0$ must hold for any $x$. This means $f(x)=-1$ for any $x \\in \\mathbb{R}$, which clearly satisfies (1).\n\n\n\n- Case 2. $g(1)=1$, which is the same as $f(1)=0$.\n\nWe put $x=-1$ and $y=1$ in (1) to get $1=\\max \\{0, f(2)\\}$. This clearly implies $f(2)=1$ and hence $g(2)=2$. Setting $x=2 n$ and $y=2$ in (3), we have\n\n$$\n(g(2 n+2)-1)^{2} \\geqslant(g(2 n)+1)^{2}-1\n$$\n\nBy induction on $n$, it is easy to prove that $g(2 n) \\geqslant n+1$ for all $n \\in \\mathbb{N}$. For any real number $a>1$, we choose a large $n \\in \\mathbb{N}$ and take $k$ to be the positive integer such that $2 k \\leqslant a^{2^{n}}<2 k+2$. From (2) and (3), we have\n\n$$\n\\left(g(a)^{2^{n}}-1\\right)^{2}+1=\\left(g\\left(a^{2^{n}}\\right)-1\\right)^{2}+1 \\geqslant\\left(g(2 k)+g\\left(a^{2^{n}}-2 k\\right)-1\\right)^{2} \\geqslant k^{2}>\\frac{1}{4}\\left(a^{2^{n}}-2\\right)^{2}\n$$\n\nsince $g\\left(a^{2^{n}}-2 k\\right) \\geqslant 0$. For large $n$, this clearly implies $g(a)^{2^{n}}>1$. Thus,\n\n$$\n\\left(g(a)^{2^{n}}\\right)^{2}>\\left(g(a)^{2^{n}}-1\\right)^{2}+1>\\frac{1}{4}\\left(a^{2^{n}}-2\\right)^{2}\n$$\n\nThis yields\n\n$$\ng(a)^{2^{n}}>\\frac{1}{2}\\left(a^{2^{n}}-2\\right) .\n\\tag{4}\n$$\n\nNote that\n\n$$\n\\frac{a^{2^{n}}}{a^{2^{n}}-2}=1+\\frac{2}{a^{2^{n}}-2} \\leqslant\\left(1+\\frac{2}{2^{n}\\left(a^{2^{n}}-2\\right)}\\right)^{2^{n}}\n$$\n\nby binomial expansion. This can be rewritten as\n\n$$\n\\left(a^{2^{n}}-2\\right)^{\\frac{1}{2^{n}}} \\geqslant \\frac{a}{1+\\frac{2}{2^{n}\\left(a^{2^{n}}-2\\right)}}\n$$\n\nTogether with (4), we conclude $g(a) \\geqslant a$ by taking $n$ sufficiently large.\n\nConsider $x=n a$ and $y=a>1$ in (3). This gives $(g((n+1) a)-1)^{2} \\geqslant(g(n a)+g(a)-1)^{2}-1$. By induction on $n$, it is easy to show $g(n a) \\geqslant(n-1)(g(a)-1)+a$ for any $n \\in \\mathbb{N}$. We choose a large $n \\in \\mathbb{N}$ and take $k$ to be the positive integer such that $k a \\leqslant 2^{2^{n}}<(k+1) a$. Using (2) and (3), we have\n\n$2^{2^{n+1}}>\\left(2^{2^{n}}-1\\right)^{2}+1=\\left(g\\left(2^{2^{n}}\\right)-1\\right)^{2}+1 \\geqslant\\left(g\\left(2^{2^{n}}-k a\\right)+g(k a)-1\\right)^{2} \\geqslant((k-1)(g(a)-1)+a-1)^{2}$,\n\nfrom which it follows that\n\n$$\n2^{2^{n}} \\geqslant(k-1)(g(a)-1)+a-1>\\frac{2^{2^{n}}}{a}(g(a)-1)-2(g(a)-1)+a-1\n$$\n\nholds for sufficiently large $n$. Hence, we must have $\\frac{g(a)-1}{a} \\leqslant 1$, which implies $g(a) \\leqslant a+1$ for any $a>1$. Then for large $n \\in \\mathbb{N}$, from (3) and (2) we have\n\n$$\n4 a^{2^{n+1}}=\\left(2 a^{2^{n}}\\right)^{2} \\geqslant\\left(g\\left(2 a^{2^{n}}\\right)-1\\right)^{2} \\geqslant\\left(2 g\\left(a^{2^{n}}\\right)-1\\right)^{2}-1=\\left(2 g(a)^{2^{n}}-1\\right)^{2}-1 .\n$$\n\n\n\nThis implies\n\n$$\n2 a^{2^{n}}>\\frac{1}{2}\\left(1+\\sqrt{4 a^{2^{n+1}}+1}\\right) \\geqslant g(a)^{2^{n}}\n$$\n\nWhen $n$ tends to infinity, this forces $g(a) \\leqslant a$. Together with $g(a) \\geqslant a$, we get $g(a)=a$ for all real numbers $a>1$, that is, $f(a)=a-1$ for all $a>1$.\n\nFinally, for any $x \\in \\mathbb{R}$, we choose $y$ sufficiently large in (1) so that $y, x+y>1$. This gives $(x+y-1)^{2}=2 f(x)(y-1)+\\max \\left\\{f\\left(x^{2}\\right)+y^{2}-1, x^{2}+y^{2}-1\\right\\}$, which can be rewritten as\n\n$$\n2(x-1-f(x)) y=-x^{2}+2 x-2-2 f(x)+\\max \\left\\{f\\left(x^{2}\\right), x^{2}\\right\\}\n$$\n\nAs the right-hand side is fixed, this can only hold for all large $y$ when $f(x)=x-1$. We now check that this function satisfies (1). Indeed, we have\n\n$$\n\\begin{aligned}\nf(x+y)^{2} & =(x+y-1)^{2}=2(x-1)(y-1)+\\left(x^{2}+y^{2}-1\\right) \\\\\n& =2 f(x) f(y)+\\max \\left\\{f\\left(x^{2}\\right)+f\\left(y^{2}\\right), f\\left(x^{2}+y^{2}\\right)\\right\\} .\n\\end{aligned}\n$$', 'Taking $x=y=0$ in (1), we get $f(0)^{2}=2 f(0)^{2}+\\max \\{2 f(0), f(0)\\}$. If $f(0)>0$, then $f(0)^{2}+2 f(0)=0$ gives no positive solution. If $f(0)<0$, then $f(0)^{2}+f(0)=0$ gives $f(0)=-1$. Putting $y=0$ in (1), we have\n\n$$\nf(x)^{2}=-2 f(x)+f\\left(x^{2}\\right) .\n\\tag{5}\n$$\n\nReplace $x$ by $-x$ in (5) and compare with (5) again. We get $f(x)^{2}+2 f(x)=f(-x)^{2}+2 f(-x)$, which implies\n\n$$\nf(x)=f(-x) \\text { or } f(x)+f(-x)=-2 \\text {. }\n\\tag{6}\n$$\n\nTaking $x=y$ and $x=-y$ respectively in (1) and comparing the two equations obtained, we have\n\n$$\nf(2 x)^{2}-2 f(x)^{2}=1-2 f(x) f(-x) \\text {. }\n\\tag{7}\n$$\n\nCombining (6) and (7) to eliminate $f(-x)$, we find that $f(2 x)$ can be \\pm 1 (when $f(x)=f(-x)$ ) or $\\pm(2 f(x)+1)$ (when $f(x)+f(-x)=-2)$.\n\nWe prove the following.\n\n- Claim. $f(x)+f(-x)=-2$ for any $x \\in \\mathbb{R}$.\n\nProof. Suppose there exists $a \\in \\mathbb{R}$ such that $f(a)+f(-a) \\neq-2$. Then $f(a)=f(-a) \\neq-1$ and we may assume $a>0$. We first show that $f(a) \\neq 1$. Suppose $f(a)=1$. Consider $y=a$ in (7). We get $f(2 a)^{2}=1$. Taking $x=y=a$ in (1), we have $1=2+\\max \\left\\{2 f\\left(a^{2}\\right), f\\left(2 a^{2}\\right)\\right\\}$. From (5), $f\\left(a^{2}\\right)=3$ so that $1 \\geqslant 2+6$. This is impossible, and thus $f(a) \\neq 1$.\n\nAs $f(a) \\neq \\pm 1$, we have $f(a)= \\pm\\left(2 f\\left(\\frac{a}{2}\\right)+1\\right)$. Similarly, $f(-a)= \\pm\\left(2 f\\left(-\\frac{a}{2}\\right)+1\\right)$. These two expressions are equal since $f(a)=f(-a)$. If $f\\left(\\frac{a}{2}\\right)=f\\left(-\\frac{a}{2}\\right)$, then the above argument works when we replace $a$ by $\\frac{a}{2}$. In particular, we have $f(a)^{2}=f\\left(2 \\cdot \\frac{a}{2}\\right)^{2}=1$, which is a contradiction. Therefore, (6) forces $f\\left(\\frac{a}{2}\\right)+f\\left(-\\frac{a}{2}\\right)=-2$. Then we get\n\n$$\n\\pm\\left(2 f\\left(\\frac{a}{2}\\right)+1\\right)= \\pm\\left(-2 f\\left(\\frac{a}{2}\\right)-3\\right)\n$$\n\n\n\nFor any choices of the two signs, we either get a contradiction or $f\\left(\\frac{a}{2}\\right)=-1$, in which case $f\\left(\\frac{a}{2}\\right)=f\\left(-\\frac{a}{2}\\right)$ and hence $f(a)= \\pm 1$ again. Therefore, there is no such real number $a$ and the Claim follows.\n\nReplace $x$ and $y$ by $-x$ and $-y$ in (1) respectively and compare with (1). We get\n\n$$\nf(x+y)^{2}-2 f(x) f(y)=f(-x-y)^{2}-2 f(-x) f(-y) .\n$$\n\nUsing the Claim, this simplifies to $f(x+y)=f(x)+f(y)+1$. In addition, (5) can be rewritten as $(f(x)+1)^{2}=f\\left(x^{2}\\right)+1$. Therefore, the function $g$ defined by $g(x)=f(x)+1$ satisfies $g(x+y)=g(x)+g(y)$ and $g(x)^{2}=g\\left(x^{2}\\right)$. The latter relation shows $g(y)$ is nonnegative for $y \\geqslant 0$. For such a function satisfying the Cauchy Equation $g(x+y)=g(x)+g(y)$, it must be monotonic increasing and hence $g(x)=c x$ for some constant $c$.\n\nFrom $(c x)^{2}=g(x)^{2}=g\\left(x^{2}\\right)=c x^{2}$, we get $c=0$ or 1 , which corresponds to the two functions $f(x)=-1$ and $f(x)=x-1$ respectively']",['- $f(x)=-1$ for any $x \\in \\mathbb{R}$; or\n- $f(x)=x-1$ for any $x \\in \\mathbb{R}$'],False,,Need_human_evaluate, 1845,Algebra,,"Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}4 \\sum_{k=1}^{n} \\frac{k+1}{x_{k}}\n$$\n\nThis shows (1) holds for $a=\\frac{4}{9}$.\n\nNext, we show that $a=\\frac{4}{9}$ is the optimal choice. Consider the sequence defined by $x_{0}=0$ and $x_{k}=x_{k-1}+k(k+1)$ for $k \\geqslant 1$, that is, $x_{k}=\\frac{1}{3} k(k+1)(k+2)$. Then the left-hand side of (1) equals\n\n$$\n\\sum_{k=1}^{n} \\frac{1}{k(k+1)}=\\sum_{k=1}^{n}\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)=1-\\frac{1}{n+1}\n$$\n\nwhile the right-hand side equals\n\n$$\na \\sum_{k=1}^{n} \\frac{k+1}{x_{k}}=3 a \\sum_{k=1}^{n} \\frac{1}{k(k+2)}=\\frac{3}{2} a \\sum_{k=1}^{n}\\left(\\frac{1}{k}-\\frac{1}{k+2}\\right)=\\frac{3}{2}\\left(1+\\frac{1}{2}-\\frac{1}{n+1}-\\frac{1}{n+2}\\right) a .\n$$\n\nWhen $n$ tends to infinity, the left-hand side tends to 1 while the right-hand side tends to $\\frac{9}{4} a$. Therefore $a$ has to be at most $\\frac{4}{9}$.\n\nHence the largest value of $a$ is $\\frac{4}{9}$.', 'We shall give an alternative method to establish (1) with $a=\\frac{4}{9}$. We define $y_{k}=x_{k}-x_{k-1}>0$ for $1 \\leqslant k \\leqslant n$. By the Cauchy-Schwarz Inequality, for $1 \\leqslant k \\leqslant n$, we have\n\n$$\n\\left(y_{1}+y_{2}+\\cdots+y_{k}\\right)\\left(\\sum_{j=1}^{k} \\frac{1}{y_{j}}\\left(\\begin{array}{c}\nj+1 \\\\\n2\n\\end{array}\\right)^{2}\\right) \\geqslant\\left(\\left(\\begin{array}{l}\n2 \\\\\n2\n\\end{array}\\right)+\\left(\\begin{array}{l}\n3 \\\\\n2\n\\end{array}\\right)+\\cdots+\\left(\\begin{array}{c}\nk+1 \\\\\n2\n\\end{array}\\right)\\right)^{2}=\\left(\\begin{array}{c}\nk+2 \\\\\n3\n\\end{array}\\right)^{2} .\n$$\n\n\n\nThis can be rewritten as\n\n$$\n\\frac{k+1}{y_{1}+y_{2}+\\cdots+y_{k}} \\leqslant \\frac{36}{k^{2}(k+1)(k+2)^{2}}\\left(\\sum_{j=1}^{k} \\frac{1}{y_{j}}\\left(\\begin{array}{c}\nj+1 \\\\\n2\n\\end{array}\\right)^{2}\\right) .\n\\tag{3}\n$$\n\nSumming (3) over $k=1,2, \\ldots, n$, we get\n\n$$\n\\frac{2}{y_{1}}+\\frac{3}{y_{1}+y_{2}}+\\cdots+\\frac{n+1}{y_{1}+y_{2}+\\cdots+y_{n}} \\leqslant \\frac{c_{1}}{y_{1}}+\\frac{c_{2}}{y_{2}}+\\cdots+\\frac{c_{n}}{y_{n}}\n\\tag{4}\n$$\n\nwhere for $1 \\leqslant m \\leqslant n$,\n\n$$\n\\begin{aligned}\nc_{m} & =36\\left(\\begin{array}{c}\nm+1 \\\\\n2\n\\end{array}\\right)^{2} \\sum_{k=m}^{n} \\frac{1}{k^{2}(k+1)(k+2)^{2}} \\\\\n& =\\frac{9 m^{2}(m+1)^{2}}{4} \\sum_{k=m}^{n}\\left(\\frac{1}{k^{2}(k+1)^{2}}-\\frac{1}{(k+1)^{2}(k+2)^{2}}\\right) \\\\\n& =\\frac{9 m^{2}(m+1)^{2}}{4}\\left(\\frac{1}{m^{2}(m+1)^{2}}-\\frac{1}{(n+1)^{2}(n+2)^{2}}\\right)<\\frac{9}{4} .\n\\end{aligned}\n$$\n\nFrom (4), the inequality (1) holds for $a=\\frac{4}{9}$. This is also the upper bound as can be verified in the same way as']",['$\\frac{4}{9}$'],False,,Numerical, 1846,Combinatorics,,"The leader of an IMO team chooses positive integers $n$ and $k$ with $n>k$, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an $n$-digit binary string, and the deputy leader writes down all $n$-digit binary strings which differ from the leader's in exactly $k$ positions. (For example, if $n=3$ and $k=1$, and if the leader chooses 101, the deputy leader would write down 001, 111 and 100.) The contestant is allowed to look at the strings written by the deputy leader and guess the leader's string. What is the minimum number of guesses (in terms of $n$ and $k$ ) needed to guarantee the correct answer?","[""Let $X$ be the binary string chosen by the leader and let $X^{\\prime}$ be the binary string of length $n$ every digit of which is different from that of $X$. The strings written by the deputy leader are the same as those in the case when the leader's string is $X^{\\prime}$ and $k$ is changed to $n-k$. In view of this, we may assume $k \\geqslant \\frac{n}{2}$. Also, for the particular case $k=\\frac{n}{2}$, this argument shows that the strings $X$ and $X^{\\prime}$ cannot be distinguished, and hence in that case the contestant has to guess at least twice.\n\nIt remains to show that the number of guesses claimed suffices. Consider any string $Y$ which differs from $X$ in $m$ digits where $02$, the deputy leader would write down the same strings if the leader's string $X$ is replaced by $X^{\\prime}$ obtained by changing each digit of $X$. This shows at least 2 guesses are needed. We shall show that 2 guesses suffice in this case. Suppose the first two digits of the leader's string are the same. Then among the strings written by the deputy leader, the prefices 01 and 10 will occur $\\left(\\begin{array}{c}2 k-2 \\\\ k-1\\end{array}\\right)$ times each, while the prefices 00 and 11 will occur $\\left(\\begin{array}{c}2 k-2 \\\\ k\\end{array}\\right)$ times each. The two numbers are interchanged if the first two digits of the leader's string are different. Since $\\left(\\begin{array}{c}2 k-2 \\\\ k-1\\end{array}\\right) \\neq\\left(\\begin{array}{c}2 k-2 \\\\ k\\end{array}\\right)$, the contestant can tell whether the first two digits of the leader's string are the same or not. He can work out the relation of the first digit and the\n\n\n\nother digits in the same way and reduce the leader's string to only 2 possibilities. The proof is complete.""]",['2 if $n=2 k$ and 1 if $n \\neq 2 k$'],False,,Need_human_evaluate, 1847,Combinatorics,,"Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints: - each cell contains a distinct divisor; - the sums of all rows are equal; and - the sums of all columns are equal.","['Suppose all positive divisors of $n$ can be arranged into a rectangular table of size $k \\times l$ where the number of rows $k$ does not exceed the number of columns $l$. Let the sum of numbers in each column be $s$. Since $n$ belongs to one of the columns, we have $s \\geqslant n$, where equality holds only when $n=1$.\n\nFor $j=1,2, \\ldots, l$, let $d_{j}$ be the largest number in the $j$-th column. Without loss of generality, assume $d_{1}>d_{2}>\\cdots>d_{l}$. Since these are divisors of $n$, we have\n\n$$\nd_{l} \\leqslant \\frac{n}{l}\\tag{1}\n$$\n\nAs $d_{l}$ is the maximum entry of the $l$-th column, we must have\n\n$$\nd_{l} \\geqslant \\frac{s}{k} \\geqslant \\frac{n}{k}\\tag{2}\n$$\n\nThe relations (1) and (2) combine to give $\\frac{n}{l} \\geqslant \\frac{n}{k}$, that is, $k \\geqslant l$. Together with $k \\leqslant l$, we conclude that $k=l$. Then all inequalities in (1) and (2) are equalities. In particular, $s=n$ and so $n=1$, in which case the conditions are clearly satisfied.', 'Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \\cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1n$. Therefore, we have\n\n$$\n\\begin{aligned}\n\\left(r_{1}+1\\right)\\left(r_{2}+1\\right) \\cdots\\left(r_{t}+1\\right) & =k l \\leqslant l^{2}<\\left(\\frac{\\sigma(n)}{n}\\right)^{2} \\\\\n& =\\left(1+\\frac{1}{p_{1}}+\\cdots+\\frac{1}{p_{1}^{r_{1}}}\\right)^{2} \\cdots\\left(1+\\frac{1}{p_{t}}+\\cdots+\\frac{1}{p_{t}^{r_{t}}}\\right)^{2} .\n\\end{aligned}\n$$\n\nThis can be rewritten as\n\n$$\nf\\left(p_{1}, r_{1}\\right) f\\left(p_{2}, r_{2}\\right) \\cdots f\\left(p_{t}, r_{t}\\right)<1\n\\tag{3}\n$$\n\nwhere\n\n$$\nf(p, r)=\\frac{r+1}{\\left(1+\\frac{1}{p}+\\cdots+\\frac{1}{p^{r}}\\right)^{2}}=\\frac{(r+1)\\left(1-\\frac{1}{p}\\right)^{2}}{\\left(1-\\frac{1}{p^{r+1}}\\right)^{2}}\n$$\n\n\n\nDirect computation yields\n\n$$\nf(2,1)=\\frac{8}{9}, \\quad f(2,2)=\\frac{48}{49}, \\quad f(3,1)=\\frac{9}{8}\n$$\n\nAlso, we find that\n\n$$\n\\begin{aligned}\n& f(2, r) \\geqslant\\left(1-\\frac{1}{2^{r+1}}\\right)^{-2}>1 \\quad \\text { for } r \\geqslant 3, \\\\\n& f(3, r) \\geqslant \\frac{4}{3}\\left(1-\\frac{1}{3^{r+1}}\\right)^{-2}>\\frac{4}{3}>\\frac{9}{8} \\quad \\text { for } r \\geqslant 2, \\text { and } \\\\\n& f(p, r) \\geqslant \\frac{32}{25}\\left(1-\\frac{1}{p^{r+1}}\\right)^{-2}>\\frac{32}{25}>\\frac{9}{8} \\quad \\text { for } p \\geqslant 5 .\n\\end{aligned}\n$$\n\nFrom these values and bounds, it is clear that (3) holds only when $n=2$ or 4 . In both cases, it is easy to see that the conditions are not satisfied. Hence, the only possible $n$ is 1 .']",['1'],False,,Numerical, 1848,Combinatorics,,Let $n$ be a positive integer relatively prime to 6 . We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours.,"['For $k=1,2,3$, let $a_{k}$ be the number of isosceles triangles whose vertices contain exactly $k$ colours. Suppose on the contrary that $a_{3}=0$. Let $b, c, d$ be the number of vertices of the three different colours respectively. We now count the number of pairs $(\\triangle, E)$ where $\\triangle$ is an isosceles triangle and $E$ is a side of $\\triangle$ whose endpoints are of different colours.\n\nOn the one hand, since we have assumed $a_{3}=0$, each triangle in the pair must contain exactly two colours, and hence each triangle contributes twice. Thus the number of pairs is $2 a_{2}$.\n\nOn the other hand, if we pick any two vertices $A, B$ of distinct colours, then there are three isosceles triangles having these as vertices, two when $A B$ is not the base and one when $A B$ is the base since $n$ is odd. Note that the three triangles are all distinct as $(n, 3)=1$. In this way, we count the number of pairs to be $3(b c+c d+d b)$. However, note that $2 a_{2}$ is even while $3(b c+c d+d b)$ is odd, as each of $b, c, d$ is. This yields a contradiction and hence $a_{3} \\geqslant 1$.']",,True,,, 1849,Combinatorics,,"Find all positive integers $n$ for which we can fill in the entries of an $n \times n$ table with the following properties: - each entry can be one of $I, M$ and $O$; - in each row and each column, the letters $I, M$ and $O$ occur the same number of times; and - in any diagonal whose number of entries is a multiple of three, the letters $I, M$ and $O$ occur the same number of times.","[""We first show that such a table exists when $n$ is a multiple of 9 . Consider the following $9 \\times 9$ table.\n\n$$\n\\left(\\begin{array}{ccccccccc}\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M\n\\end{array}\\right)\\tag{1}\n$$\n\nIt is a direct checking that the table (1) satisfies the requirements. For $n=9 k$ where $k$ is a positive integer, we form an $n \\times n$ table using $k \\times k$ copies of (1). For each row and each column of the table of size $n$, since there are three $I$ 's, three $M$ 's and three $O$ 's for any nine consecutive entries, the numbers of $I, M$ and $O$ are equal. In addition, every diagonal of the large table whose number of entries is divisible by 3 intersects each copy of (1) at a diagonal with number of entries divisible by 3 (possibly zero). Therefore, every such diagonal also contains the same number of $I, M$ and $O$.\n\nNext, consider any $n \\times n$ table for which the requirements can be met. As the number of entries of each row should be a multiple of 3 , we let $n=3 k$ where $k$ is a positive integer. We divide the whole table into $k \\times k$ copies of $3 \\times 3$ blocks. We call the entry at the centre of such a $3 \\times 3$ square a vital entry. We also call any row, column or diagonal that contains at least one vital entry a vital line. We compute the number of pairs $(l, c)$ where $l$ is a vital line and $c$ is an entry belonging to $l$ that contains the letter $M$. We let this number be $N$.\n\nOn the one hand, since each vital line contains the same number of $I, M$ and $O$, it is obvious that each vital row and each vital column contain $k$ occurrences of $M$. For vital diagonals in either direction, we count there are exactly\n\n$$\n1+2+\\cdots+(k-1)+k+(k-1)+\\cdots+2+1=k^{2}\n$$\n\noccurrences of $M$. Therefore, we have $N=4 k^{2}$.\n\n\n\nOn the other hand, there are $3 k^{2}$ occurrences of $M$ in the whole table. Note that each entry belongs to exactly 1 or 4 vital lines. Therefore, $N$ must be congruent to $3 k^{2} \\bmod 3$.\n\nFrom the double counting, we get $4 k^{2} \\equiv 3 k^{2}(\\bmod 3)$, which forces $k$ to be a multiple of 3. Therefore, $n$ has to be a multiple of 9 and the proof is complete.""]",['$n$ can be any multiple of 9'],False,,Need_human_evaluate, 1850,Combinatorics,,"Let $n \geqslant 3$ be a positive integer. Find the maximum number of diagonals of a regular $n$-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other.","['We consider two cases according to the parity of $n$.\n\n- Case 1. $n$ is odd.\n\nWe first claim that no pair of diagonals is perpendicular. Suppose $A, B, C, D$ are vertices where $A B$ and $C D$ are perpendicular, and let $E$ be the vertex lying on the perpendicular bisector of $A B$. Let $E^{\\prime}$ be the opposite point of $E$ on the circumcircle of the regular polygon. Since $E C=E^{\\prime} D$ and $C, D, E$ are vertices of the regular polygon, $E^{\\prime}$ should also belong to the polygon. This contradicts the fact that a regular polygon with an odd number of vertices does not contain opposite points on the circumcircle.\n![](https://cdn.mathpix.com/cropped/2023_12_21_9ee4f50247737d9408edg-1.jpg?height=504&width=1248&top_left_y=1062&top_left_x=491)\n\nTherefore in the odd case we can only select diagonals which do not intersect. In the maximal case these diagonals should divide the regular $n$-gon into $n-2$ triangles, so we can select at most $n-3$ diagonals. This can be done, for example, by selecting all diagonals emanated from a particular vertex.\n\n- Case 2. $n$ is even.\n\nIf there is no intersection, then the proof in the odd case works. Suppose there are two perpendicular diagonals selected. We consider the set $S$ of all selected diagonals parallel to one of them which intersect with some selected diagonals. Suppose $S$ contains $k$ diagonals and the number of distinct endpoints of the $k$ diagonals is $l$.\n\nFirstly, consider the longest diagonal in one of the two directions in $S$. No other diagonal in $S$ can start from either endpoint of that diagonal, since otherwise it has to meet another longer diagonal in $S$. The same holds true for the other direction. Ignoring these two longest diagonals and their four endpoints, the remaining $k-2$ diagonals share $l-4$ endpoints where each endpoint can belong to at most two diagonals. This gives $2(l-4) \\geqslant 2(k-2)$, so that $k \\leqslant l-2$.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_76b69bb4e7ba5ecafb62g-1.jpg?height=420&width=1432&top_left_y=294&top_left_x=298)\n\nConsider a group of consecutive vertices of the regular $n$-gon so that each of the two outermost vertices is an endpoint of a diagonal in $S$, while the interior points are not. There are $l$ such groups. We label these groups $P_{1}, P_{2}, \\ldots, P_{l}$ in this order. We claim that each selected diagonal outside $S$ must connect vertices of the same group $P_{i}$. Consider any diagonal $d$ joining vertices from distinct groups $P_{i}$ and $P_{j}$. Let $d_{1}$ and $d_{2}$ be two diagonals in $S$ each having one of the outermost points of $P_{i}$ as endpoint. Then $d$ must meet either $d_{1}, d_{2}$ or a diagonal in $S$ which is perpendicular to both $d_{1}$ and $d_{2}$. In any case $d$ should belong to $S$ by definition, which is a contradiction.\n\nWithin the same group $P_{i}$, there are no perpendicular diagonals since the vertices belong to the same side of a diameter of the circumcircle. Hence there can be at most $\\left|P_{i}\\right|-2$ selected diagonals within $P_{i}$, including the one joining the two outermost points of $P_{i}$ when $\\left|P_{i}\\right|>2$. Therefore, the maximum number of diagonals selected is\n\n$$\n\\sum_{i=1}^{l}\\left(\\left|P_{i}\\right|-2\\right)+k=\\sum_{i=1}^{l}\\left|P_{i}\\right|-2 l+k=(n+l)-2 l+k=n-l+k \\leqslant n-2\n$$\n\nThis upper bound can be attained as follows. We take any vertex $A$ and let $A^{\\prime}$ be the vertex for which $A A^{\\prime}$ is a diameter of the circumcircle. If we select all diagonals emanated from $A$ together with the diagonal $d^{\\prime}$ joining the two neighbouring vertices of $A^{\\prime}$, then the only pair of diagonals that meet each other is $A A^{\\prime}$ and $d^{\\prime}$, which are perpendicular to each other. In total we can take $n-2$ diagonals.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_76b69bb4e7ba5ecafb62g-1.jpg?height=447&width=420&top_left_y=1804&top_left_x=798)', 'Instead of dealing separately with the case where $n$ is even, we shall prove by induction more generally that we can select at most $n-2$ diagonals for any cyclic $n$-gon with circumcircle $\\Gamma$.\n\n\n\nThe base case $n=3$ is trivial since there is no diagonal at all. Suppose the upper bound holds for any cyclic polygon with fewer than $n$ sides. For a cyclic $n$-gon, if there is a selected diagonal which does not intersect any other selected diagonal, then this diagonal divides the $n$-gon into an $m$-gon and an $l$-gon (with $m+l=n+2$ ) so that each selected diagonal belongs to one of them. Without loss of generality, we may assume the $m$-gon lies on the same side of a diameter of $\\Gamma$. Then no two selected diagonals of the $m$-gon can intersect, and hence we can select at most $m-3$ diagonals. Also, we can apply the inductive hypothesis to the $l$-gon. This shows the maximum number of selected diagonals is $(m-3)+(l-2)+1=n-2$.\n\nIt remains to consider the case when all selected diagonals meet at least one other selected diagonal. Consider a pair of selected perpendicular diagonals $d_{1}, d_{2}$. They divide the circumference of $\\Gamma$ into four arcs, each of which lies on the same side of a diameter of $\\Gamma$. If there are two selected diagonals intersecting each other and neither is parallel to $d_{1}$ or $d_{2}$, then their endpoints must belong to the same arc determined by $d_{1}, d_{2}$, and hence they cannot be perpendicular. This violates the condition, and hence all selected diagonals must have the same direction as one of $d_{1}, d_{2}$.\n![](https://cdn.mathpix.com/cropped/2023_12_21_02deb1202c246c7b11b9g-1.jpg?height=422&width=1220&top_left_y=1046&top_left_x=496)\n\nTake the longest selected diagonal in one of the two directions. We argue that its endpoints do not belong to any other selected diagonal. The same holds true for the longest diagonal in the other direction. Apart from these four endpoints, each of the remaining $n-4$ vertices can belong to at most two selected diagonals. Thus we can select at most $\\frac{1}{2}(2(n-4)+4)=n-2$ diagonals. Then the proof follows by induction.']",['$n-2$ if $n$ is even and $n-3$ if $n$ is odd'],False,,Need_human_evaluate, 1851,Combinatorics,,"There are $n \geqslant 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route. After each year, the ferry company will close a ferry route between some two islands $X$ and $Y$. At the same time, in order to maintain its service, the company will open new routes according to the following rule: for any island which is connected by a ferry route to exactly one of $X$ and $Y$, a new route between this island and the other of $X$ and $Y$ is added. Suppose at any moment, if we partition all islands into two nonempty groups in any way, then it is known that the ferry company will close a certain route connecting two islands from the two groups after some years. Prove that after some years there will be an island which is connected to all other islands by ferry routes.","['Initially, we pick any pair of islands $A$ and $B$ which are connected by a ferry route and put $A$ in set $\\mathcal{A}$ and $B$ in set $\\mathcal{B}$. From the condition, without loss of generality there must be another island which is connected to $A$. We put such an island $C$ in set $\\mathcal{B}$. We say that two sets of islands form a network if each island in one set is connected to each island in the other set.\n\nNext, we shall included all islands to $\\mathcal{A} \\cup \\mathcal{B}$ one by one. Suppose we have two sets $\\mathcal{A}$ and $\\mathcal{B}$ which form a network where $3 \\leqslant|\\mathcal{A} \\cup \\mathcal{B}|\n\nWithout loss of generality, we may assume the snails start at $A_{i}$ and $A_{j}$ respectively. Let $l_{i}$ intersect $l_{j}$ at $P$. Note that there is an odd number of points between $\\operatorname{arc} A_{i} A_{j}$. Each of these points belongs to a line $l_{k}$. Such a line $l_{k}$ must intersect exactly one of\n\n\n\nthe segments $A_{i} P$ and $A_{j} P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i} P$ and $A_{j} P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i} P$ and $A_{j} P$ (not counting $P$ ) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i} P$ and $A_{j} P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other.""]",,True,,, 1853,Combinatorics,,"Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps. Prove that if $n$ is even then there must be a moment when some two snails occupy the same intersection point no matter how Jeff places the snails.","[""We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$.\n\nFor even $n$, we consider any way that Jeff places the snails and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_{i}$ and $A_{j}$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_{i} B_{i}$ and $A_{j} B_{j}$. Then any other segment meeting one of the segments $A_{i} P$ and $A_{j} P$ must also meet the other one, and so the number of intersections on $A_{i} P$ and $A_{j} P$ are the same. This shows the snails starting from $A_{i}$ and $A_{j}$ will meet at $P$.""]",,True,,, 1854,Combinatorics,,"Let $n$ be a positive integer. Determine the smallest positive integer $k$ with the following property: it is possible to mark $k$ cells on a $2 n \times 2 n$ board so that there exists a unique partition of the board into $1 \times 2$ and $2 \times 1$ dominoes, none of which contains two marked cells.","['We first construct an example of marking $2 n$ cells satisfying the requirement. Label the rows and columns $1,2, \\ldots, 2 n$ and label the cell in the $i$-th row and the $j$-th column $(i, j)$.\n\nFor $i=1,2, \\ldots, n$, we mark the cells $(i, i)$ and $(i, i+1)$. We claim that the required partition exists and is unique. The two diagonals of the board divide the board into four regions. Note that the domino covering cell $(1,1)$ must be vertical. This in turn shows that each domino covering $(2,2),(3,3), \\ldots,(n, n)$ is vertical. By induction, the dominoes in the left region are all vertical. By rotational symmetry, the dominoes in the bottom region are horizontal, and so on. This shows that the partition exists and is unique.\n\n\nIt remains to show that this value of $k$ is the smallest possible. Assume that only $k<2 n$ cells are marked, and there exists a partition $P$ satisfying the requirement. It suffices to show there exists another desirable partition distinct from $P$. Let $d$ be the main diagonal of the board.\n\nConstruct the following graph with edges of two colours. Its vertices are the cells of the board. Connect two vertices with a red edge if they belong to the same domino of $P$. Connect two vertices with a blue edge if their reflections in $d$ are connected by a red edge. It is possible that two vertices are connected by edges of both colours. Clearly, each vertex has both red and blue degrees equal to 1 . Thus the graph splits into cycles where the colours of edges in each cycle alternate (a cycle may have length 2).\n\nConsider any cell $c$ lying on the diagonal $d$. Its two edges are symmetrical with respect to $d$. Thus they connect $c$ to different cells. This shows $c$ belongs to a cycle $C(c)$ of length at least 4. Consider a part of this cycle $c_{0}, c_{1}, \\ldots, c_{m}$ where $c_{0}=c$ and $m$ is the least positive integer such that $c_{m}$ lies on $d$. Clearly, $c_{m}$ is distinct from $c$. From the construction, the path symmetrical to this with respect to $d$ also lies in the graph, and so these paths together form $C(c)$. Hence, $C(c)$ contains exactly two cells from $d$. Then all $2 n$ cells in $d$ belong to $n$ cycles $C_{1}, C_{2}, \\ldots, C_{n}$, each has length at least 4.\n\nBy the Pigeonhole Principle, there exists a cycle $C_{i}$ containing at most one of the $k$ marked cells. We modify $P$ as follows. We remove all dominoes containing the vertices of $C_{i}$, which\n\n\n\ncorrespond to the red edges of $C_{i}$. Then we put the dominoes corresponding to the blue edges of $C_{i}$. Since $C_{i}$ has at least 4 vertices, the resultant partition $P^{\\prime}$ is different from $P$. Clearly, no domino in $P^{\\prime}$ contains two marked cells as $C_{i}$ contains at most one marked cell. This shows the partition is not unique and hence $k$ cannot be less than $2 n$.']",['$2 n$'],False,,Expression, 1855,Geometry,,"In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A .\tag{1 } $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.","['Denote the common angle in (1) by $\\theta$. As $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C}=\\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. From $E A=E D$, we get\n\n$$\n\\angle A F D=\\angle A B C=90^{\\circ}+\\theta=180^{\\circ}-\\frac{1}{2} \\angle A E D .\n$$\n\nHence, $F$ lies on the circle with centre $E$ and radius $E A$. In particular, $E F=E A=E D$. As $\\angle E F A=\\angle E A F=2 \\theta=\\angle B F C$, points $B, F, E$ are collinear.\n\nAs $\\angle E D A=\\angle M A D$, we have $E D / / A M$ and hence $E, D, X$ are collinear. As $M$ is the midpoint of $C F$ and $\\angle C B F=90^{\\circ}$, we get $M F=M B$. In the isosceles triangles $E F A$ and $M F B$, we have $\\angle E F A=\\angle M F B$ and $A F=B F$. Therefore, they are congruent to each other. Then we have $B M=A E=X M$ and $B E=B F+F E=A F+F M=A M=E X$. This shows $\\triangle E M B \\cong \\triangle E M X$. As $F$ and $D$ lie on $E B$ and $E X$ respectively and $E F=E D$, we know that lines $B D$ and $X F$ are symmetric with respect to $E M$. It follows that the three lines are concurrent.\n\n', ""From $\\angle C A D=\\angle E D A$, we have $A C / / E D$. Together with $A C / / E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\\theta$. From $\\triangle A B F \\sim \\triangle A C D$, we get $\\frac{A B}{A C}=\\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. This yields $\\angle A F D=\\angle A B C=90^{\\circ}+\\theta$ and hence $\\angle F D C=90^{\\circ}$, implying that $B C D F$ is cyclic. Let $\\Gamma_{1}$ be its circumcircle.\n\nNext, from $\\triangle A B F \\sim \\triangle A D E$, we have $\\frac{A B}{A D}=\\frac{A F}{A E}$ so that $\\triangle A B D \\sim \\triangle A F E$. Therefore,\n\n$$\n\\angle A F E=\\angle A B D=\\theta+\\angle F B D=\\theta+\\angle F C D=2 \\theta=180^{\\circ}-\\angle B F A .\n$$\n\nThis implies $B, F, E$ are collinear. Note that $F$ is the incentre of triangle $D A B$. Point $E$ lies on the internal angle bisector of $\\angle D B A$ and lies on the perpendicular bisector of $A D$. It follows that $E$ lies on the circumcircle $\\Gamma_{2}$ of triangle $A B D$, and $E A=E F=E D$.\n\nAlso, since $C F$ is a diameter of $\\Gamma_{1}$ and $M$ is the midpoint of $C F, M$ is the centre of $\\Gamma_{1}$ and hence $\\angle A M D=2 \\theta=\\angle A B D$. This shows $M$ lies on $\\Gamma_{2}$. Next, $\\angle M D X=\\angle M A E=\\angle D X M$ since $A M X E$ is a parallelogram. Hence $M D=M X$ and $X$ lies on $\\Gamma_{1}$.\n\n\n\nWe now have two ways to complete the solution.\n\n- Method 1. From $E F=E A=X M$ and $E X / / F M, E F M X$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by $\\Gamma_{3}$. Since $B D, E M, F X$ are the three radical axes of $\\Gamma_{1}, \\Gamma_{2}, \\Gamma_{3}$, they must be concurrent.\n- Method 2. As $\\angle D M F=2 \\theta=\\angle B F M$, we have $D M / / E B$. Also,\n\n$$\n\\angle B F D+\\angle X B F=\\angle B F C+\\angle C F D+90^{\\circ}-\\angle C B X=2 \\theta+\\left(90^{\\circ}-\\theta\\right)+90^{\\circ}-\\theta=180^{\\circ}\n$$\n\nimplies $D F / / X B$. These show the corresponding sides of triangles $D M F$ and $B E X$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $D B, M E, F X$ meet at a point."", 'Let the common angle in (1) be $\\theta$. From $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C}=\\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. Then $\\angle A D F=\\angle A C B=90^{\\circ}-2 \\theta=90^{\\circ}-\\angle B A D$ and hence $D F \\perp A B$. As $F A=F B$, this implies $\\triangle D A B$ is isosceles with $D A=D B$. Then $F$ is the incentre of $\\triangle D A B$.\n\nNext, from $\\angle A E D=180^{\\circ}-2 \\theta=180^{\\circ}-\\angle D B A$, points $A, B, D, E$ are concyclic. Since we also have $E A=E D$, this shows $E, F, B$ are collinear and $E A=E F=E D$.\n\n\n\nNote that $C$ lies on the internal angle bisector of $\\angle B A D$ and lies on the external angle bisector of $\\angle D B A$. It follows that it is the $A$-excentre of triangle $D A B$. As $M$ is the midpoint of $C F, M$ lies on the circumcircle of triangle $D A B$ and it is the centre of the circle passing through $D, F, B, C$. By symmetry, $D E F M$ is a rhombus. Then the midpoints of $A X, E M$ and $D F$ coincide, and it follows that $D A F X$ is a parallelogram.\n\nLet $P$ be the intersection of $B D$ and $E M$, and $Q$ be the intersection of $A D$ and $B E$. From $\\angle B A C=\\angle D C A$, we know that $D C, A B, E M$ are parallel. Thus we have $\\frac{D P}{P B}=\\frac{C M}{M A}$. This is further equal to $\\frac{A E}{B E}$ since $C M=D M=D E=A E$ and $M A=B E$. From $\\triangle A E Q \\sim \\triangle B E A$, we find that\n\n$$\n\\frac{D P}{P B}=\\frac{A E}{B E}=\\frac{A Q}{B A}=\\frac{Q F}{F B}\n$$\n\nby the Angle Bisector Theorem. This implies $Q D / / F P$ and hence $F, P, X$ are collinear, as desired.']",,True,,, 1856,Geometry,,Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$.,"['Let $A M$ meet $\\Gamma$ again at $Y$ and $X Y$ meet $B C$ at $D^{\\prime}$. It suffices to show $D^{\\prime}=D$. We shall apply the following fact.\n\n- Claim. For any cyclic quadrilateral $P Q R S$ whose diagonals meet at $T$, we have\n\n$$\n\\frac{Q T}{T S}=\\frac{P Q \\cdot Q R}{P S \\cdot S R}\n$$\n\nProof. We use $\\left[W_{1} W_{2} W_{3}\\right]$ to denote the area of $W_{1} W_{2} W_{3}$. Then\n\n$$\n\\frac{Q T}{T S}=\\frac{[P Q R]}{[P S R]}=\\frac{\\frac{1}{2} P Q \\cdot Q R \\sin \\angle P Q R}{\\frac{1}{2} P S \\cdot S R \\sin \\angle P S R}=\\frac{P Q \\cdot Q R}{P S \\cdot S R}\n$$\n\nApplying the Claim to $A B Y C$ and $X B Y C$ respectively, we have $1=\\frac{B M}{M C}=\\frac{A B \\cdot B Y}{A C \\cdot C Y}$ and $\\frac{B D^{\\prime}}{D^{\\prime} C}=\\frac{X B \\cdot B Y}{X C \\cdot C Y}$. These combine to give\n\n$$\n\\frac{B D^{\\prime}}{C D^{\\prime}}=\\frac{X B}{X C} \\cdot \\frac{B Y}{C Y}=\\frac{X B}{X C} \\cdot \\frac{A C}{A B}\n\\tag{1}\n$$\n\nNext, we use directed angles to find that $\\measuredangle X B F=\\measuredangle X B A=\\measuredangle X C A=\\measuredangle X C E$ and $\\measuredangle X F B=\\measuredangle X F A=\\measuredangle X E A=\\measuredangle X E C$. This shows triangles $X B F$ and $X C E$ are directly similar. In particular, we have\n\n$$\n\\frac{X B}{X C}=\\frac{B F}{C E}\n\\tag{2}\n$$\n\nIn the following, we give two ways to continue the proof.\n\n- Method 1. Here is a geometrical method. As $\\angle F I B=\\angle A I B-90^{\\circ}=\\frac{1}{2} \\angle A C B=\\angle I C B$ and $\\angle F B I=\\angle I B C$, the triangles $F B I$ and $I B C$ are similar. Analogously, triangles $E I C$ and $I B C$ are also similar. Hence, we get\n\n$$\n\\frac{F B}{I B}=\\frac{B I}{B C} \\quad \\text { and } \\quad \\frac{E C}{I C}=\\frac{I C}{B C}\n\\tag{3}\n$$\n\n\n\n\n\nNext, construct a line parallel to $B C$ and tangent to the incircle. Suppose it meets sides $A B$ and $A C$ at $B_{1}$ and $C_{1}$ respectively. Let the incircle touch $A B$ and $A C$ at $B_{2}$ and $C_{2}$ respectively. By homothety, the line $B_{1} I$ is parallel to the external angle bisector of $\\angle A B C$, and hence $\\angle B_{1} I B=90^{\\circ}$. Since $\\angle B B_{2} I=90^{\\circ}$, we get $B B_{2} \\cdot B B_{1}=B I^{2}$, and similarly $C C_{2} \\cdot C C_{1}=C I^{2}$. Hence,\n\n$$\n\\frac{B I^{2}}{C I^{2}}=\\frac{B B_{2} \\cdot B B_{1}}{C C_{2} \\cdot C C_{1}}=\\frac{B B_{1}}{C C_{1}} \\cdot \\frac{B D}{C D}=\\frac{A B}{A C} \\cdot \\frac{B D}{C D}\n\\tag{4}\n$$\n\nCombining (1), (2), (3) and (4), we conclude\n\n$$\n\\frac{B D^{\\prime}}{C D^{\\prime}}=\\frac{X B}{X C} \\cdot \\frac{A C}{A B}=\\frac{B F}{C E} \\cdot \\frac{A C}{A B}=\\frac{B I^{2}}{C I^{2}} \\cdot \\frac{A C}{A B}=\\frac{B D}{C D}\n$$\n\nso that $D^{\\prime}=D$. The result then follows.\n\n- Method 2. Let $\\beta=\\frac{1}{2} \\angle A B C$ and $\\gamma=\\frac{1}{2} \\angle A C B$. Observe that $\\angle F I B=\\angle A I B-90^{\\circ}=\\gamma$. Hence, $\\frac{B F}{F I}=\\frac{\\sin \\angle F I B}{\\sin \\angle I B F}=\\frac{\\sin \\gamma}{\\sin \\beta}$. Similarly, $\\frac{C E}{E I}=\\frac{\\sin \\beta}{\\sin \\gamma}$. As $F I=E I$, we get\n\n$$\n\\frac{B F}{C E}=\\frac{B F}{F I} \\cdot \\frac{E I}{C E}=\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}\n\\tag{5}\n$$\n\n\n\nTogether with (1) and (2), we find that\n\n$$\n\\frac{B D^{\\prime}}{C D^{\\prime}}=\\frac{A C}{A B} \\cdot\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}=\\frac{\\sin 2 \\beta}{\\sin 2 \\gamma} \\cdot\\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^{2}=\\frac{\\tan \\gamma}{\\tan \\beta}=\\frac{I D / C D}{I D / B D}=\\frac{B D}{C D}\n$$\n\nThis shows $D^{\\prime}=D$ and the result follows.', 'Let $\\omega_{A}$ be the $A$-mixtilinear incircle of triangle $A B C$. From the properties of mixtilinear incircles, $\\omega_{A}$ touches sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose $\\omega_{A}$ is tangent to $\\Gamma$ at $T$. Let $A M$ meet $\\Gamma$ again at $Y$, and let $D_{1}, T_{1}$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $B C$ respectively. It is well-known that $\\angle B A T=\\angle D_{1} A C$ so that $A, D_{1}, T_{1}$ are collinear.\n\n\n\nWe then show that $X, M, T_{1}$ are collinear. Let $R$ be the radical centre of $\\omega_{A}, \\Gamma$ and the circumcircle of triangle $A E F$. Then $R$ lies on $A X, E F$ and the tangent at $T$ to $\\Gamma$. Let $A T$ meet $\\omega_{A}$ again at $S$ and meet $E F$ at $P$. Obviously, $S F T E$ is a harmonic quadrilateral. Projecting from $T$, the pencil $(R, P ; F, E)$ is harmonic. We further project the pencil onto $\\Gamma$ from $A$, so that $X B T C$ is a harmonic quadrilateral. As $T T_{1} / / B C$, the projection from $T_{1}$ onto $B C$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $B C$, which is $M$. This shows $X, M, T_{1}$ are collinear.\n\nWe have two ways to finish the proof.\n\n- Method 1. Note that both $A Y$ and $X T_{1}$ are chords of $\\Gamma$ passing through the midpoint $M$ of the chord $B C$. By the Butterfly Theorem, $X Y$ and $A T_{1}$ cut $B C$ at a pair of symmetric points with respect to $M$, and hence $X, D, Y$ are collinear. The proof is thus complete.\n\n\n\n- Method 2. Here, we finish the proof without using the Butterfly Theorem. As DTT $D_{1}$ is an isosceles trapezoid, we have\n\n$$\n\\measuredangle Y T D=\\measuredangle Y T T_{1}+\\measuredangle T_{1} T D=\\measuredangle Y A T_{1}+\\measuredangle A D_{1} D=\\measuredangle Y M D\n$$\n\nso that $D, T, Y, M$ are concyclic. As $X, M, T_{1}$ are collinear, we have\n\n$$\n\\measuredangle A Y D=\\measuredangle M T D=\\measuredangle D_{1} T_{1} M=\\measuredangle A T_{1} X=\\measuredangle A Y X\n$$\n\nThis shows $X, D, Y$ are collinear.']",,True,,, 1857,Geometry,,"Let $B=(-1,0)$ and $C=(1,0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be nice if (i) there is a point $T$ in $S$ such that for every point $Q$ in $S$, the segment $T Q$ lies entirely in $S$; and (ii) for any triangle $P_{1} P_{2} P_{3}$, there exists a unique point $A$ in $S$ and a permutation $\sigma$ of the indices $\{1,2,3\}$ for which triangles $A B C$ and $P_{\sigma(1)} P_{\sigma(2)} P_{\sigma(3)}$ are similar. Prove that there exist two distinct nice subsets $S$ and $S^{\prime}$ of the set $\{(x, y): x \geqslant 0, y \geqslant 0\}$ such that if $A \in S$ and $A^{\prime} \in S^{\prime}$ are the unique choices of points in (ii), then the product $B A \cdot B A^{\prime}$ is a constant independent of the triangle $P_{1} P_{2} P_{3}$.","['If in the similarity of $\\triangle A B C$ and $\\triangle P_{\\sigma(1)} P_{\\sigma(2)} P_{\\sigma(3)}, B C$ corresponds to the longest side of $\\triangle P_{1} P_{2} P_{3}$, then we have $B C \\geqslant A B \\geqslant A C$. The condition $B C \\geqslant A B$ is equivalent to $(x+1)^{2}+y^{2} \\leqslant 4$, while $A B \\geqslant A C$ is trivially satisfied for any point in the first quadrant. Then we first define\n\n$$\nS=\\left\\{(x, y):(x+1)^{2}+y^{2} \\leqslant 4, x \\geqslant 0, y \\geqslant 0\\right\\}\n$$\n\nNote that $S$ is the intersection of a disk and the first quadrant, so it is bounded and convex, and we can choose any $T \\in S$ to meet condition (i). For any point $A$ in $S$, the relation $B C \\geqslant A B \\geqslant A C$ always holds. Therefore, the point $A$ in (ii) is uniquely determined, while its existence is guaranteed by the above construction.\n\n\n\nNext, if in the similarity of $\\triangle A^{\\prime} B C$ and $\\triangle P_{\\sigma(1)} P_{\\sigma(2)} P_{\\sigma(3)}, B C$ corresponds to the second longest side of $\\triangle P_{1} P_{2} P_{3}$, then we have $A^{\\prime} B \\geqslant B C \\geqslant A^{\\prime} C$. The two inequalities are equivalent to $(x+1)^{2}+y^{2} \\geqslant 4$ and $(x-1)^{2}+y^{2} \\leqslant 4$ respectively. Then we define\n\n$$\nS^{\\prime}=\\left\\{(x, y):(x+1)^{2}+y^{2} \\geqslant 4,(x-1)^{2}+y^{2} \\leqslant 4, x \\geqslant 0, y \\geqslant 0\\right\\}\n$$\n\n\n\nThe boundedness condition is satisfied while (ii) can be argued as in the previous case. For (i), note that $S^{\\prime}$ contains points inside the disk $(x-1)^{2}+y^{2} \\leqslant 4$ and outside the disk $(x+1)^{2}+y^{2} \\geqslant 4$. This shows we can take $T^{\\prime}=(1,2)$ in (i), which is the topmost point of the circle $(x-1)^{2}+y^{2}=4$.\n\nIt remains to check that the product $B A \\cdot B A^{\\prime}$ is a constant. Suppose we are given a triangle $P_{1} P_{2} P_{3}$ with $P_{1} P_{2} \\geqslant P_{2} P_{3} \\geqslant P_{3} P_{1}$. By the similarity, we have\n\n$$\nB A=B C \\cdot \\frac{P_{2} P_{3}}{P_{1} P_{2}} \\quad \\text { and } \\quad B A^{\\prime}=B C \\cdot \\frac{P_{1} P_{2}}{P_{2} P_{3}}\n$$\n\nThus $B A \\cdot B A^{\\prime}=B C^{2}=4$, which is certainly independent of the triangle $P_{1} P_{2} P_{3}$.']",,True,,, 1858,Geometry,,"Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.","['\n\nLet $\\Gamma$ be the circle with centre $E$ passing through $B$ and $C$. Since $E D \\perp A C$, the point $F$ symmetric to $C$ with respect to $D$ lies on $\\Gamma$. From $\\angle D C I=\\angle I C B=\\angle C B I$, the line $D C$ is a tangent to the circumcircle of triangle $I B C$. Let $J$ be the symmetric point of $I$ with respect to $D$. Using directed lengths, from\n\n$$\nD C \\cdot D F=-D C^{2}=-D I \\cdot D B=D J \\cdot D B\n$$\n\nthe point $J$ also lies on $\\Gamma$. Let $I^{\\prime}$ be the reflection of $I$ in $A C$. Since $I J$ and $C F$ bisect each other, $C J F I$ is a parallelogram. From $\\angle F I^{\\prime} C=\\angle C I F=\\angle F J C$, we find that $I^{\\prime}$ lies on $\\Gamma$. This gives $E I^{\\prime}=E B$.\n\nNote that $A C$ is the internal angle bisector of $\\angle B D I^{\\prime}$. This shows $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$ as $D E \\perp A C$. Together with $E I^{\\prime}=E B$, it is well-known that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.', 'Let $I^{\\prime}$ be the reflection of $I$ in $A C$ and let $S$ be the intersection of $I^{\\prime} C$ and $A I$. Using directed angles, we let $\\theta=\\measuredangle A C I=\\measuredangle I C B=\\measuredangle C B I$. We have\n\n$$\n\\measuredangle I^{\\prime} S E=\\measuredangle I^{\\prime} C A+\\measuredangle C A I=\\theta+\\left(\\frac{\\pi}{2}+2 \\theta\\right)=3 \\theta+\\frac{\\pi}{2}\n$$\n\nand\n\n$$\n\\measuredangle I^{\\prime} D E=\\measuredangle I^{\\prime} D C+\\frac{\\pi}{2}=\\measuredangle C D I+\\frac{\\pi}{2}=\\measuredangle D C B+\\measuredangle C B D+\\frac{\\pi}{2}=3 \\theta+\\frac{\\pi}{2} .\n$$\n\nThis shows $I^{\\prime}, D, E, S$ are concyclic.\n\nNext, we find $\\measuredangle I^{\\prime} S B=2 \\measuredangle I^{\\prime} S E=6 \\theta$ and $\\measuredangle I^{\\prime} D B=2 \\measuredangle C D I=6 \\theta$. Therefore, $I^{\\prime}, D, B, S$ are concyclic so that $I^{\\prime}, D, E, B, S$ lie on the same circle. The result then follows.\n\n\n\n', 'Let $I^{\\prime}$ be the reflection of $I$ in $A C$, and let $D^{\\prime}$ be the second intersection of $A I$ and the circumcircle of triangle $A B D$. Since $A D^{\\prime}$ bisects $\\angle B A D$, point $D^{\\prime}$ is the midpoint of the arc $B D$ and $D D^{\\prime}=B D^{\\prime}=C D^{\\prime}$. Obviously, $A, E, D^{\\prime}$ lie on $A I$ in this order.\n\n\n\nWe find that $\\angle E D^{\\prime} D=\\angle A D^{\\prime} D=\\angle A B D=\\angle I B C=\\angle I C B$. Next, since $D^{\\prime}$ is the circumcentre of triangle $B C D$, we have $\\angle E D D^{\\prime}=90^{\\circ}-\\angle D^{\\prime} D C=\\angle C B D=\\angle I B C$. The two relations show that triangles $E D^{\\prime} D$ and $I C B$ are similar. Therefore, we have\n\n$$\n\\frac{B C}{C I^{\\prime}}=\\frac{B C}{C I}=\\frac{D D^{\\prime}}{D^{\\prime} E}=\\frac{B D^{\\prime}}{D^{\\prime} E}\n$$\n\nAlso, we get\n\n$$\n\\angle B C I^{\\prime}=\\angle B C A+\\angle A C I^{\\prime}=\\angle B C A+\\angle I C A=\\angle B C A+\\angle D B C=\\angle B D A=\\angle B D^{\\prime} E .\n$$\n\nThese show triangles $B C I^{\\prime}$ and $B D^{\\prime} E$ are similar, and hence triangles $B C D^{\\prime}$ and $B I^{\\prime} E$ are similar. As $B C D^{\\prime}$ is isosceles, we obtain $B E=I^{\\prime} E$.\n\nAs $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$ and $E I^{\\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.', 'Let $A I$ and $B I$ meet the circumcircle of triangle $A B C$ again at $A^{\\prime}$ and $B^{\\prime}$ respectively, and let $E^{\\prime}$ be the reflection of $E$ in $A C$. From\n\n$$\n\\begin{aligned}\n\\angle B^{\\prime} A E^{\\prime} & =\\angle B^{\\prime} A D-\\angle E^{\\prime} A D=\\frac{\\angle A B C}{2}-\\frac{\\angle B A C}{2}=90^{\\circ}-\\angle B A C-\\frac{\\angle A B C}{2} \\\\\n& =90^{\\circ}-\\angle B^{\\prime} D A=\\angle B^{\\prime} D E^{\\prime},\n\\end{aligned}\n$$\n\npoints $B^{\\prime}, A, D, E^{\\prime}$ are concyclic. Then\n\n$$\n\\angle D B^{\\prime} E^{\\prime}=\\angle D A E^{\\prime}=\\frac{\\angle B A C}{2}=\\angle B A A^{\\prime}=\\angle D B^{\\prime} A^{\\prime}\n$$\n\nand hence $B^{\\prime}, E^{\\prime}, A^{\\prime}$ are collinear. It is well-known that $A^{\\prime} B^{\\prime}$ is the perpendicular bisector of $C I$, so that $C E^{\\prime}=I E^{\\prime}$. Let $I^{\\prime}$ be the reflection of $I$ in $A C$. This implies $B E=C E=I^{\\prime} E$. As $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$ and $E I^{\\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.\n\n', 'Let $F$ be the intersection of $C I$ and $A B$. Clearly, $F$ and $D$ are symmetric with respect to $A I$. Let $O$ be the circumcentre of triangle $B I F$, and let $I^{\\prime}$ be the reflection of $I$ in $A C$.\n\n\n\n\n\nFrom $\\angle B F O=90^{\\circ}-\\angle F I B=\\frac{1}{2} \\angle B A C=\\angle B A I$, we get $E I / / F O$. Also, from the relation $\\angle O I B=90^{\\circ}-\\angle B F I=90^{\\circ}-\\angle C D I=\\angle I^{\\prime} I D$, we know that $O, I, I^{\\prime}$ are collinear.\n\nNote that $E D / / O I$ since both are perpendicular to $A C$. Then $\\angle F E I=\\angle D E I=\\angle O I E$. Together with $E I / / F O$, the quadrilateral $E F O I$ is an isosceles trapezoid. Therefore, we find that $\\angle D I E=\\angle F I E=\\angle O E I$ so $O E / / I D$. Then $D E O I$ is a parallelogram. Hence, we have $D I^{\\prime}=D I=E O$, which shows $D E O I^{\\prime}$ is an isosceles trapezoid. In addition, $E D=O I=O B$ and $O E / / B D$ imply $E O B D$ is another isosceles trapezoid. In particular, both $D E O I^{\\prime}$ and $E O B D$ are cyclic. This shows $B, D, E, I^{\\prime}$ are concyclic.', 'Let $I^{\\prime}$ be the reflection of $I$ in $A C$. Denote by $T$ and $M$ the projections from $I$ to sides $A B$ and $B C$ respectively. Since $B I$ is the perpendicular bisector of $T M$, we have\n\n$$\nD T=D M\\tag{1}\n$$\n\nSince $\\angle A D E=\\angle A T I=90^{\\circ}$ and $\\angle D A E=\\angle T A I$, we have $\\triangle A D E \\sim \\triangle A T I$. This shows $\\frac{A D}{A E}=\\frac{A T}{A I}=\\frac{A T}{A I^{\\prime}}$. Together with $\\angle D A T=2 \\angle D A E=\\angle E A I^{\\prime}$, this yields $\\triangle D A T \\sim \\triangle E A I^{\\prime}$. In particular, we have\n\n$$\n\\frac{D T}{E I^{\\prime}}=\\frac{A T}{A I^{\\prime}}=\\frac{A T}{A I}\n\\tag{2}\n$$\n\nObviously, the right-angled triangles $A M B$ and $A T I$ are similar. Then we get\n\n$$\n\\frac{A M}{A B}=\\frac{A T}{A I}\\tag{3}\n$$\n\nNext, from $\\triangle A M B \\sim \\triangle A T I \\sim \\triangle A D E$, we have $\\frac{A M}{A B}=\\frac{A D}{A E}$ so that $\\triangle A D M \\sim \\triangle A E B$. It follows that\n\n$$\n\\frac{D M}{E B}=\\frac{A M}{A B}\\tag{4}\n$$\n\nCombining (1), (2), (3) and (4), we get $E B=E I^{\\prime}$. As $D E$ is the external angle bisector of $\\angle B D I^{\\prime}$, we know that $E$ lies on the circumcircle of triangle $B D I^{\\prime}$.\n\n']",,True,,, 1859,Geometry,,"Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.","['Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $Q$ be the midpoint of $A H$ and $N$ be the nine-point centre of triangle $A B C$. It is known that $Q$ lies on the nine-point circle of triangle $A B C, N$ is the midpoint of $Q M$ and that $Q M$ is parallel to $A O$.\n\nLet the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S^{\\prime}$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have\n\n$$\n\\begin{aligned}\n\\measuredangle S D Q & =\\measuredangle Q A S=\\measuredangle X A S-\\measuredangle X A Q=\\left(\\frac{\\pi}{2}-\\measuredangle A Y X\\right)-\\measuredangle B A P=\\measuredangle C B A-\\measuredangle A Y X \\\\\n& =(\\measuredangle C B A-\\measuredangle A C B)-\\measuredangle B C A-\\measuredangle A Y X=\\measuredangle P E M-(\\measuredangle B C A+\\measuredangle A Y X) \\\\\n& =\\measuredangle P Q M-\\angle(B C, X Y)=\\frac{\\pi}{2}-\\angle\\left(S^{\\prime} Q, B C\\right)-\\angle(B C, X Y)=\\measuredangle S S^{\\prime} Q .\n\\end{aligned}\n$$\n\nThis shows $D, S^{\\prime}, S, Q$ are concyclic.\n\n\n\nLet the perpendicular from $N$ to $B C$ intersect line $S S^{\\prime}$ at $O_{1}$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_{1}$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From\n\n$$\n\\measuredangle D S^{\\prime} O_{1}=\\measuredangle D Q S=\\measuredangle S Q A=\\angle(S Q, Q A)=\\angle\\left(O D, O_{1} N\\right)=\\measuredangle D N O_{1}\n$$\n\n\n\nsince $S Q / / O D$ and $Q A / / O_{1} N$, we know that $D, O_{1}, S^{\\prime}, N$ are concyclic. Therefore, we get\n\n$$\n\\measuredangle S D S^{\\prime}=\\measuredangle S Q S^{\\prime}=\\angle\\left(S Q, Q S^{\\prime}\\right)=\\angle\\left(N D, N S^{\\prime}\\right)=\\measuredangle D N S^{\\prime}\n$$\n\nso that $S D$ is a tangent to the circle through $D, O_{1}, S^{\\prime}, N$. Then we have\n\n$$\nS S^{\\prime} \\cdot S O_{1}=S D^{2}=S X^{2}\n\\tag{1}\n$$\n\nNext, we show that $S$ and $S^{\\prime}$ are symmetric with respect to $X Y$. By the Sine Law, we have\n\n$$\n\\frac{S S^{\\prime}}{\\sin \\angle S Q S^{\\prime}}=\\frac{S Q}{\\sin \\angle S S^{\\prime} Q}=\\frac{S Q}{\\sin \\angle S D Q}=\\frac{S Q}{\\sin \\angle S A Q}=\\frac{S A}{\\sin \\angle S Q A} .\n$$\n\nIt follows that\n\n$$\nS S^{\\prime}=S A \\cdot \\frac{\\sin \\angle S Q S^{\\prime}}{\\sin \\angle S Q A}=S A \\cdot \\frac{\\sin \\angle H O A}{\\sin \\angle O H A}=S A \\cdot \\frac{A H}{A O}=S A \\cdot 2 \\cos A,\n$$\n\nwhich is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\\angle S A C<\\angle O A C$ if and only if $\\angle Y X A>\\angle C B A$. This shows $S$ and $O_{1}$ lie on different sides of $X Y$. As $S^{\\prime}$ lies on ray $S O_{1}$, it follows that $S$ and $S^{\\prime}$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S^{\\prime}$ are symmetric with respect to $X Y$.\n\nLet $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S^{\\prime}$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S^{\\prime}=2 \\frac{S X^{2}}{d}$. It follows from (1) that $d=2 S O_{1}$. As $S O_{1}$ is the perpendicular bisector of $X Y$, point $O_{1}$ is the circumcentre of triangle $X S Y$.', 'Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_{1}$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S^{\\prime}$ and $S^{\\prime \\prime}$ and define the analogous points $X^{\\prime}, Y^{\\prime}, O_{1}^{\\prime}, X^{\\prime \\prime}, Y^{\\prime \\prime} O_{1}^{\\prime \\prime}$ accordingly. Note that $S, S^{\\prime}, S^{\\prime \\prime}$ lie on the perpendicular bisector of $A D$.\n\nAs $X X^{\\prime}$ and $Y Y^{\\prime}$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X^{\\prime} Y^{\\prime}$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X^{\\prime} Y^{\\prime}$. We find that\n\n$$\n\\measuredangle S X Y=\\frac{\\pi}{2}-\\measuredangle Y A X=\\frac{\\pi}{2}-\\measuredangle Y^{\\prime} A X^{\\prime}=\\measuredangle S^{\\prime} X^{\\prime} Y^{\\prime}\n$$\n\nand similarly $\\measuredangle S Y X=\\measuredangle S^{\\prime} Y^{\\prime} X^{\\prime}$. This shows triangles $S X Y$ and $S^{\\prime} X^{\\prime} Y^{\\prime}$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_{1}$ to $S^{\\prime}, X^{\\prime}, Y^{\\prime}, O_{1}^{\\prime}$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_{1}$ to $S^{\\prime \\prime}, X^{\\prime \\prime}, Y^{\\prime \\prime}, O_{1}^{\\prime \\prime}$. From these, we see that there is a spiral similarity taking the corresponding points $S, S^{\\prime}, S^{\\prime \\prime}$ to points $O_{1}, O_{1}^{\\prime}, O_{1}^{\\prime \\prime}$. In particular, $O_{1}, O_{1}^{\\prime}, O_{1}^{\\prime \\prime}$ are collinear.\n\n\n\n\n\nIt now suffices to show that $O_{1}$ lies on the perpendicular bisector of $P M$ for two special cases.\n\nFirstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_{1}$ is the nine-point centre and $O_{1} P=O_{1} M$. Indeed, $P$ and $M$ also lies on the nine-point circle.\n\nSecondly, we take $S^{\\prime}$ to be the midpoint of $A O$. Then $X^{\\prime}$ and $Y^{\\prime}$ are the midpoints of $A B$ and $A C$ respectively. Then $X^{\\prime} Y^{\\prime} / / B C$. Clearly, $S^{\\prime}$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X^{\\prime} Y^{\\prime}$ and $P M$ coincide. Hence, we must have $O_{1}^{\\prime} P=O_{1}^{\\prime} M$.\n']",,True,,, 1860,Geometry,,"Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segments $B M$ and $D M$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $X E$ and $Y F$. Prove that $P Q \perp A C$.","['\n\nLet $\\omega_{1}$ be the circumcircle of triangle $A B C$. We first prove that $Y$ lies on $\\omega_{1}$. Let $Y^{\\prime}$ be the point on ray $M D$ such that $M Y^{\\prime} \\cdot M D=M A^{2}$. Then triangles $M A Y^{\\prime}$ and $M D A$ are oppositely similar. Since $M C^{2}=M A^{2}=M Y^{\\prime} \\cdot M D$, triangles $M C Y^{\\prime}$ and $M D C$ are also oppositely similar. Therefore, using directed angles, we have\n\n$$\n\\measuredangle A Y^{\\prime} C=\\measuredangle A Y^{\\prime} M+\\measuredangle M Y^{\\prime} C=\\measuredangle M A D+\\measuredangle D C M=\\measuredangle C D A=\\measuredangle A B C\n$$\n\nso that $Y^{\\prime}$ lies on $\\omega_{1}$.\n\nLet $Z$ be the intersection point of lines $B C$ and $A D$. Since $\\measuredangle P D Z=\\measuredangle P B C=\\measuredangle P B Z$, point $Z$ lies on $\\omega$. In addition, from $\\measuredangle Y^{\\prime} B Z=\\measuredangle Y^{\\prime} B C=\\measuredangle Y^{\\prime} A C=\\measuredangle Y^{\\prime} A M=\\measuredangle Y^{\\prime} D Z$, we also know that $Y^{\\prime}$ lies on $\\omega$. Note that $\\angle A D C$ is acute implies $M A \\neq M D$ so $M Y^{\\prime} \\neq M D$. Therefore, $Y^{\\prime}$ is the second intersection of $D M$ and $\\omega$. Then $Y^{\\prime}=Y$ and hence $Y$ lies on $\\omega_{1}$.\n\nNext, by the Angle Bisector Theorem and the similar triangles, we have\n\n$$\n\\frac{F A}{F C}=\\frac{A D}{C D}=\\frac{A D}{A M} \\cdot \\frac{C M}{C D}=\\frac{Y A}{Y M} \\cdot \\frac{Y M}{Y C}=\\frac{Y A}{Y C}\n$$\n\nHence, $F Y$ is the internal angle bisector of $\\angle A Y C$.\n\nLet $B^{\\prime}$ be the second intersection of the internal angle bisector of $\\angle C B A$ and $\\omega_{1}$. Then $B^{\\prime}$ is the midpoint of arc $A C$ not containing $B$. Therefore, $Y B^{\\prime}$ is the external angle bisector of $\\angle A Y C$, so that $B^{\\prime} Y \\perp F Y$.\n\n\n\nDenote by $l$ the line through $P$ parallel to $A C$. Suppose $l$ meets line $B^{\\prime} Y$ at $S$. From\n\n$$\n\\begin{aligned}\n\\measuredangle P S Y & =\\measuredangle\\left(A C, B^{\\prime} Y\\right)=\\measuredangle A C Y+\\measuredangle C Y B^{\\prime}=\\measuredangle A C Y+\\measuredangle C A B^{\\prime}=\\measuredangle A C Y+\\measuredangle B^{\\prime} C A \\\\\n& =\\measuredangle B^{\\prime} C Y=\\measuredangle B^{\\prime} B Y=\\measuredangle P B Y,\n\\end{aligned}\n$$\n\nthe point $S$ lies on $\\omega$. Similarly, the line through $X$ perpendicular to $X E$ also passes through the second intersection of $l$ and $\\omega$, which is the point $S$. From $Q Y \\perp Y S$ and $Q X \\perp X S$, point $Q$ lies on $\\omega$ and $Q S$ is a diameter of $\\omega$. Therefore, $P Q \\perp P S$ so that $P Q \\perp A C$.', 'Denote by $\\omega_{1}$ and $\\omega_{2}$ the circumcircles of triangles $A B C$ and $A D C$ respectively. Since $\\angle A B C=\\angle A D C$, we know that $\\omega_{1}$ and $\\omega_{2}$ are symmetric with respect to the midpoint $M$ of $A C$.\n\nFirstly, we show that $X$ lies on $\\omega_{2}$. Let $X_{1}$ be the second intersection of ray $M B$ and $\\omega_{2}$ and $X^{\\prime}$ be its symmetric point with respect to $M$. Then $X^{\\prime}$ lies on $\\omega_{1}$ and $X^{\\prime} A X_{1} C$ is a parallelogram. Hence, we have\n\n$$\n\\begin{aligned}\n\\measuredangle D X_{1} B & =\\measuredangle D X_{1} A+\\measuredangle A X_{1} B=\\measuredangle D C A+\\measuredangle A X_{1} X^{\\prime}=\\measuredangle D C A+\\measuredangle C X^{\\prime} X_{1} \\\\\n& =\\measuredangle D C A+\\measuredangle C A B=\\angle(C D, A B) .\n\\end{aligned}\n$$\n\n\n\nAlso, we have\n\n$$\n\\measuredangle D P B=\\measuredangle P D C+\\angle(C D, A B)+\\measuredangle A B P=\\angle(C D, A B)\n$$\n\nThese yield $\\measuredangle D X_{1} B=\\measuredangle D P B$ and hence $X_{1}$ lies on $\\omega$. It follows that $X_{1}=X$ and $X$ lies on $\\omega_{2}$. Similarly, $Y$ lies on $\\omega_{1}$.\n\n\n\nNext, we prove that $Q$ lies on $\\omega$. Suppose the perpendicular bisector of $A C$ meet $\\omega_{1}$ at $B^{\\prime}$ and $M_{1}$ and meet $\\omega_{2}$ at $D^{\\prime}$ and $M_{2}$, so that $B, M_{1}$ and $D^{\\prime}$ lie on the same side of $A C$. Note that $B^{\\prime}$ lies on the angle bisector of $\\angle A B C$ and similarly $D^{\\prime}$ lies on $D P$.\n\nIf we denote the area of $W_{1} W_{2} W_{3}$ by $\\left[W_{1} W_{2} W_{3}\\right]$, then\n\n$$\n\\frac{B A \\cdot X^{\\prime} A}{B C \\cdot X^{\\prime} C}=\\frac{\\frac{1}{2} B A \\cdot X^{\\prime} A \\sin \\angle B A X^{\\prime}}{\\frac{1}{2} B C \\cdot X^{\\prime} C \\sin \\angle B C X^{\\prime}}=\\frac{\\left[B A X^{\\prime}\\right]}{\\left[B C X^{\\prime}\\right]}=\\frac{M A}{M C}=1\n$$\n\nAs $B E$ is the angle bisector of $\\angle A B C$, we have\n\n$$\n\\frac{E A}{E C}=\\frac{B A}{B C}=\\frac{X^{\\prime} C}{X^{\\prime} A}=\\frac{X A}{X C}\n$$\n\nTherefore, $X E$ is the angle bisector of $\\angle A X C$, so that $M_{2}$ lies on the line joining $X, E, Q$. Analogously, $M_{1}, F, Q, Y$ are collinear. Thus,\n\n$$\n\\begin{aligned}\n\\measuredangle X Q Y & =\\measuredangle M_{2} Q M_{1}=\\measuredangle Q M_{2} M_{1}+\\measuredangle M_{2} M_{1} Q=\\measuredangle X M_{2} D^{\\prime}+\\measuredangle B^{\\prime} M_{1} Y \\\\\n& =\\measuredangle X D D^{\\prime}+\\measuredangle B^{\\prime} B Y=\\measuredangle X D P+\\measuredangle P B Y=\\measuredangle X B P+\\measuredangle P B Y=\\measuredangle X B Y,\n\\end{aligned}\n$$\n\nwhich implies $Q$ lies on $\\omega$.\n\nFinally, as $M_{1}$ and $M_{2}$ are symmetric with respect to $M$, the quadrilateral $X^{\\prime} M_{2} X M_{1}$ is a parallelogram. Consequently,\n\n$$\n\\measuredangle X Q P=\\measuredangle X B P=\\measuredangle X^{\\prime} B B^{\\prime}=\\measuredangle X^{\\prime} M_{1} B^{\\prime}=\\measuredangle X M_{2} M_{1}\n$$\n\nThis shows $Q P / / M_{2} M_{1}$. As $M_{2} M_{1} \\perp A C$, we get $Q P \\perp A C$.', 'We first state two results which will be needed in our proof.\n\n- Claim 1. In $\\triangle X^{\\prime} Y^{\\prime} Z^{\\prime}$ with $X^{\\prime} Y^{\\prime} \\neq X^{\\prime} Z^{\\prime}$, let $N^{\\prime}$ be the midpoint of $Y^{\\prime} Z^{\\prime}$ and $W^{\\prime}$ be the foot of internal angle bisector from $X^{\\prime}$. Then $\\tan ^{2} \\measuredangle W^{\\prime} X^{\\prime} Z^{\\prime}=\\tan \\measuredangle N^{\\prime} X^{\\prime} W^{\\prime} \\tan \\measuredangle Z^{\\prime} W^{\\prime} X^{\\prime}$.\n\nProof.\n\n\n\nWithout loss of generality, assume $X^{\\prime} Y^{\\prime}>X^{\\prime} Z^{\\prime}$. Then $W^{\\prime}$ lies between $N^{\\prime}$ and $Z^{\\prime}$. The signs of both sides agree so it suffices to establish the relation for ordinary angles. Let $\\angle W^{\\prime} X^{\\prime} Z^{\\prime}=\\alpha, \\angle N^{\\prime} X^{\\prime} W^{\\prime}=\\beta$ and $\\angle Z^{\\prime} W^{\\prime} X^{\\prime}=\\gamma$. We have\n\n$$\n\\frac{\\sin (\\gamma-\\alpha)}{\\sin (\\alpha-\\beta)}=\\frac{N^{\\prime} X^{\\prime}}{N^{\\prime} Y^{\\prime}}=\\frac{N^{\\prime} X^{\\prime}}{N^{\\prime} Z^{\\prime}}=\\frac{\\sin (\\gamma+\\alpha)}{\\sin (\\alpha+\\beta)}\n$$\n\n\n\nThis implies\n\n$$\n\\frac{\\tan \\gamma-\\tan \\alpha}{\\tan \\gamma+\\tan \\alpha}=\\frac{\\sin \\gamma \\cos \\alpha-\\cos \\gamma \\sin \\alpha}{\\sin \\gamma \\cos \\alpha+\\cos \\gamma \\sin \\alpha}=\\frac{\\sin \\alpha \\cos \\beta-\\cos \\alpha \\sin \\beta}{\\sin \\alpha \\cos \\beta+\\cos \\alpha \\sin \\beta}=\\frac{\\tan \\alpha-\\tan \\beta}{\\tan \\alpha+\\tan \\beta}\n$$\n\nExpanding and simplifying, we get the desired result $\\tan ^{2} \\alpha=\\tan \\beta \\tan \\gamma$.\n\n- Claim 2. Let $A^{\\prime} B^{\\prime} C^{\\prime} D^{\\prime}$ be a quadrilateral inscribed in circle $\\Gamma$. Let diagonals $A^{\\prime} C^{\\prime}$ and $B^{\\prime} D^{\\prime}$ meet at $E^{\\prime}$, and $F^{\\prime}$ be the intersection of lines $A^{\\prime} B^{\\prime}$ and $C^{\\prime} D^{\\prime}$. Let $M^{\\prime}$ be the midpoint of $E^{\\prime} F^{\\prime}$. Then the power of $M^{\\prime}$ with respect to $\\Gamma$ is equal to $\\left(M^{\\prime} E^{\\prime}\\right)^{2}$.\n\nProof.\n\n\n\nLet $O^{\\prime}$ be the centre of $\\Gamma$ and let $\\Gamma^{\\prime}$ be the circle with centre $M^{\\prime}$ passing through $E^{\\prime}$. Let $F_{1}$ be the inversion image of $F^{\\prime}$ with respect to $\\Gamma$. It is well-known that $E^{\\prime}$ lies on the polar of $F^{\\prime}$ with respect to $\\Gamma$. This shows $E^{\\prime} F_{1} \\perp O^{\\prime} F^{\\prime}$ and hence $F_{1}$ lies on $\\Gamma^{\\prime}$. It follows that the inversion image of $\\Gamma^{\\prime}$ with respect to $\\Gamma$ is $\\Gamma^{\\prime}$ itself. This shows $\\Gamma^{\\prime}$ is orthogonal to $\\Gamma$, and thus the power of $M^{\\prime}$ with respect to $\\Gamma$ is the square of radius of $\\Gamma^{\\prime}$, which is $\\left(M^{\\prime} E^{\\prime}\\right)^{2}$.\n\nWe return to the main problem. Let $Z$ be the intersection of lines $A D$ and $B C$, and $W$ be the intersection of lines $A B$ and $C D$. Since $\\measuredangle P D Z=\\measuredangle P B C=\\measuredangle P B Z$, point $Z$ lies on $\\omega$. Similarly, $W$ lies on $\\omega$. Applying Claim 2 to the cyclic quadrilateral $Z B D W$, we know that the power of $M$ with respect to $\\omega$ is $M A^{2}$. Hence, $M X \\cdot M B=M A^{2}$.\n\nSuppose the line through $B$ perpendicular to $B E$ meets line $A C$ at $T$. Then $B E$ and $B T$ are the angle bisectors of $\\angle C B A$. This shows $(T, E ; A, C)$ is harmonic. Thus, we have $M E \\cdot M T=M A^{2}=M X \\cdot M B$. It follows that $E, T, B, X$ are concyclic.\n\n\n\n\n\nThe result is trivial for the special case $A D=C D$ since $P, Q$ lie on the perpendicular bisector of $A C$ in that case. Similarly, the case $A B=C B$ is trivial. It remains to consider the general cases where we can apply Claim 1 in the latter part of the proof.\n\nLet the projections from $P$ and $Q$ to $A C$ be $P^{\\prime}$ and $Q^{\\prime}$ respectively. Then $P Q \\perp A C$ if and only if $P^{\\prime}=Q^{\\prime}$ if and only if $\\frac{E P^{\\prime}}{F P^{\\prime}}=\\frac{E Q^{\\prime}}{F Q^{\\prime}}$ in terms of directed lengths. Note that\n\n$$\n\\frac{E P^{\\prime}}{F P^{\\prime}}=\\frac{\\tan \\measuredangle E F P}{\\tan \\measuredangle F E P}=\\frac{\\tan \\measuredangle A F D}{\\tan \\measuredangle A E B}\n$$\n\nNext, we have $\\frac{E Q^{\\prime}}{F Q^{\\prime}}=\\frac{\\tan \\measuredangle E F Q}{\\tan \\measuredangle F E Q}$ where $\\measuredangle F E Q=\\measuredangle T E X=\\measuredangle T B X=\\frac{\\pi}{2}+\\measuredangle E B M$ and by symmetry $\\measuredangle E F Q=\\frac{\\pi}{2}+\\measuredangle F D M$. Combining all these, it suffices to show\n\n$$\n\\frac{\\tan \\measuredangle A F D}{\\tan \\measuredangle A E B}=\\frac{\\tan \\measuredangle M B E}{\\tan \\measuredangle M D F}\n$$\n\nWe now apply Claim 1 twice to get\n\n$$\n\\tan \\measuredangle A F D \\tan \\measuredangle M D F=\\tan ^{2} \\measuredangle F D C=\\tan ^{2} \\measuredangle E B A=\\tan \\measuredangle M B E \\tan \\measuredangle A E B .\n$$\n\nThe result then follows.']",,True,,, 1861,Geometry,,"Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$. Prove that $P$ lies on line $O I$ where $O$ is the circumcentre of triangle $A B C$.","['Let $A^{\\prime}$ be the reflection of $A$ in $B C$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\\Gamma$ of triangle $A B C$. As triangles $A B A^{\\prime}$ and $A O C$ are isosceles with $\\angle A B A^{\\prime}=2 \\angle A B C=\\angle A O C$, they are similar to each other. Also, triangles $A B I_{A}$ and $A I C$ are similar. Therefore we have\n\n$$\n\\frac{A A^{\\prime}}{A I_{A}}=\\frac{A A^{\\prime}}{A B} \\cdot \\frac{A B}{A I_{A}}=\\frac{A C}{A O} \\cdot \\frac{A I}{A C}=\\frac{A I}{A O}\n$$\n\nTogether with $\\angle A^{\\prime} A I_{A}=\\angle I A O$, we find that triangles $A A^{\\prime} I_{A}$ and $A I O$ are similar.\n\n\n\nDenote by $P^{\\prime}$ the intersection of line $A P$ and line $O I$. Using directed angles, we have\n\n$$\n\\begin{aligned}\n\\measuredangle M A P^{\\prime} & =\\measuredangle I_{A}^{\\prime} A I_{A}=\\measuredangle I_{A}^{\\prime} A A^{\\prime}-\\measuredangle I_{A} A A^{\\prime}=\\measuredangle A A^{\\prime} I_{A}-\\measuredangle(A M, O M) \\\\\n& =\\measuredangle A I O-\\measuredangle A M O=\\measuredangle M O P^{\\prime} .\n\\end{aligned}\n$$\n\nThis shows $M, O, A, P^{\\prime}$ are concyclic.\n\n\n\nDenote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Then\n\n$$\nI P^{\\prime}=\\frac{I A \\cdot I M}{I O}=\\frac{I O^{2}-R^{2}}{I O}\n$$\n\nis independent of $A$. Hence, $B P$ also meets line $O I$ at the same point $P^{\\prime}$ so that $P^{\\prime}=P$, and $P$ lies on $O I$.', 'Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}^{\\prime} C, A I_{B} C, I_{A} B C$ and $I_{A}^{\\prime} B C$ are all similar. From $\\triangle A I_{B}^{\\prime} C \\sim \\triangle I_{A}^{\\prime} B C$, we get $\\triangle A I_{A}^{\\prime} C \\sim \\triangle I_{B}^{\\prime} B C$. From $\\measuredangle A B P=\\measuredangle I_{B}^{\\prime} B C=\\measuredangle A I_{A}^{\\prime} C$ and $\\measuredangle B A P=\\measuredangle I_{A}^{\\prime} A C$, the triangles $A B P$ and $A I_{A}^{\\prime} C$ are directly similar.\n\n\n\nConsider the inversion with centre $A$ and radius $\\sqrt{A B \\cdot A C}$ followed by the reflection in $A I$. Then $B$ and $C$ are mapped to each other, and $I$ and $I_{A}$ are mapped to each other.\n\n\n\nFrom the similar triangles obtained, we have $A P \\cdot A I_{A}^{\\prime}=A B \\cdot A C$ so that $P$ is mapped to $I_{A}^{\\prime}$ under the transformation. In addition, line $A O$ is mapped to the altitude from $A$, and hence $O$ is mapped to the reflection of $A$ in $B C$, which we call point $A^{\\prime}$. Note that $A A^{\\prime} I_{A} I_{A}^{\\prime}$ is an isosceles trapezoid, which shows it is inscribed in a circle. The preimage of this circle is a straight line, meaning that $O, I, P$ are collinear.']",,True,,, 1862,Geometry,,"Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$. Let one of the tangents from $P$ to the incircle of triangle $A B C$ meet the circumcircle at points $X$ and $Y$. Show that $\angle X I Y=120^{\circ}$.","[""By Poncelet's Porism, the other tangents to the incircle of triangle $A B C$ from $X$ and $Y$ meet at a point $Z$ on $\\Gamma$. Let $T$ be the touching point of the incircle to $X Y$, and let $D$ be the midpoint of $X Y$. We have\n\n$$\n\\begin{aligned}\nO D & =I T \\cdot \\frac{O P}{I P}=r\\left(1+\\frac{O I}{I P}\\right)=r\\left(1+\\frac{O I^{2}}{O I \\cdot I P}\\right)=r\\left(1+\\frac{R^{2}-2 R r}{R^{2}-I O^{2}}\\right) \\\\\n& =r\\left(1+\\frac{R^{2}-2 R r}{2 R r}\\right)=\\frac{R}{2}=\\frac{O X}{2} .\n\\end{aligned}\n$$"", 'Denote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Note that by the above transformation, we have $\\triangle A P O \\sim \\triangle A A^{\\prime} I_{A}^{\\prime}$ and $\\triangle A A^{\\prime} I_{A} \\sim \\triangle A I O$. Therefore, we find that\n\n$$\nP O=A^{\\prime} I_{A}^{\\prime} \\cdot \\frac{A O}{A I_{A}^{\\prime}}=A I_{A} \\cdot \\frac{A O}{A^{\\prime} I_{A}}=\\frac{A I_{A}}{A^{\\prime} I_{A}} \\cdot A O=\\frac{A O}{I O} \\cdot A O\n$$\n\nThis shows $P O \\cdot I O=R^{2}$, and it follows that $P$ and $I$ are mapped to each other under the inversion with respect to the circumcircle $\\Gamma$ of triangle $A B C$. Then $P X \\cdot P Y$, which is the power of $P$ with respect to $\\Gamma$, equals $P I \\cdot P O$. This yields $X, I, O, Y$ are concyclic.\n\nLet $T$ be the touching point of the incircle to $X Y$, and let $D$ be the midpoint of $X Y$. Then\n\n$$\nO D=I T \\cdot \\frac{P O}{P I}=r \\cdot \\frac{P O}{P O-I O}=r \\cdot \\frac{R^{2}}{R^{2}-I O^{2}}=r \\cdot \\frac{R^{2}}{2 R r}=\\frac{R}{2}\n$$\n\nThis shows $\\angle D O X=60^{\\circ}$ and hence $\\angle X I Y=\\angle X O Y=120^{\\circ}$.\nThis shows $\\angle X Z Y=60^{\\circ}$ and hence $\\angle X I Y=120^{\\circ}$.']",,True,,, 1863,Geometry,,"Let $A_{1}, B_{1}$ and $C_{1}$ be points on sides $B C, C A$ and $A B$ of an acute triangle $A B C$ respectively, such that $A A_{1}, B B_{1}$ and $C C_{1}$ are the internal angle bisectors of triangle $A B C$. Let $I$ be the incentre of triangle $A B C$, and $H$ be the orthocentre of triangle $A_{1} B_{1} C_{1}$. Show that $$ A H+B H+C H \geqslant A I+B I+C I \text {. } $$","['Without loss of generality, assume $\\alpha=\\angle B A C \\leqslant \\beta=\\angle C B A \\leqslant \\gamma=\\angle A C B$. Denote by $a, b, c$ the lengths of $B C, C A, A B$ respectively. We first show that triangle $A_{1} B_{1} C_{1}$ is acute.\n\nChoose points $D$ and $E$ on side $B C$ such that $B_{1} D / / A B$ and $B_{1} E$ is the internal angle bisector of $\\angle B B_{1} C$. As $\\angle B_{1} D B=180^{\\circ}-\\beta$ is obtuse, we have $B B_{1}>B_{1} D$. Thus,\n\n$$\n\\frac{B E}{E C}=\\frac{B B_{1}}{B_{1} C}>\\frac{D B_{1}}{B_{1} C}=\\frac{B A}{A C}=\\frac{B A_{1}}{A_{1} C}\n$$\n\nTherefore, $B E>B A_{1}$ and $\\frac{1}{2} \\angle B B_{1} C=\\angle B B_{1} E>\\angle B B_{1} A_{1}$. Similarly, $\\frac{1}{2} \\angle B B_{1} A>\\angle B B_{1} C_{1}$. It follows that\n\n$$\n\\angle A_{1} B_{1} C_{1}=\\angle B B_{1} A_{1}+\\angle B B_{1} C_{1}<\\frac{1}{2}\\left(\\angle B B_{1} C+\\angle B B_{1} A\\right)=90^{\\circ}\n$$\n\nis acute. By symmetry, triangle $A_{1} B_{1} C_{1}$ is acute.\n\nLet $B B_{1}$ meet $A_{1} C_{1}$ at $F$. From $\\alpha \\leqslant \\gamma$, we get $a \\leqslant c$, which implies\n\n$$\nB A_{1}=\\frac{c a}{b+c} \\leqslant \\frac{a c}{a+b}=B C_{1}\n$$\n\nand hence $\\angle B C_{1} A_{1} \\leqslant \\angle B A_{1} C_{1}$. As $B F$ is the internal angle bisector of $\\angle A_{1} B C_{1}$, this shows $\\angle B_{1} F C_{1}=\\angle B F A_{1} \\leqslant 90^{\\circ}$. Hence, $H$ lies on the same side of $B B_{1}$ as $C_{1}$. This shows $H$ lies inside triangle $B B_{1} C_{1}$. Similarly, from $\\alpha \\leqslant \\beta$ and $\\beta \\leqslant \\gamma$, we know that $H$ lies inside triangles $C C_{1} B_{1}$ and $A A_{1} C_{1}$.\n\n\n\n\n\nAs $\\alpha \\leqslant \\beta \\leqslant \\gamma$, we have $\\alpha \\leqslant 60^{\\circ} \\leqslant \\gamma$. Then $\\angle B I C \\leqslant 120^{\\circ} \\leqslant \\angle A I B$. Firstly, suppose $\\angle A I C \\geqslant 120^{\\circ}$.\n\nRotate points $B, I, H$ through $60^{\\circ}$ about $A$ to $B^{\\prime}, I^{\\prime}, H^{\\prime}$ so that $B^{\\prime}$ and $C$ lie on different sides of $A B$. Since triangle $A I^{\\prime} I$ is equilateral, we have\n\n$$\nA I+B I+C I=I^{\\prime} I+B^{\\prime} I^{\\prime}+I C=B^{\\prime} I^{\\prime}+I^{\\prime} I+I C .\n\\tag{1}\n$$\n\nSimilarly,\n\n$$\nA H+B H+C H=H^{\\prime} H+B^{\\prime} H^{\\prime}+H C=B^{\\prime} H^{\\prime}+H^{\\prime} H+H C \\text {. }\\tag{2}\n$$\n\nAs $\\angle A I I^{\\prime}=\\angle A I^{\\prime} I=60^{\\circ}, \\angle A I^{\\prime} B^{\\prime}=\\angle A I B \\geqslant 120^{\\circ}$ and $\\angle A I C \\geqslant 120^{\\circ}$, the quadrilateral $B^{\\prime} I^{\\prime} I C$ is convex and lies on the same side of $B^{\\prime} C$ as $A$.\n\nNext, since $H$ lies inside triangle $A C C_{1}, H$ lies outside $B^{\\prime} I^{\\prime} I C$. Also, $H$ lying inside triangle $A B I$ implies $H^{\\prime}$ lies inside triangle $A B^{\\prime} I^{\\prime}$. This shows $H^{\\prime}$ lies outside $B^{\\prime} I^{\\prime} I C$ and hence the convex quadrilateral $B^{\\prime} I^{\\prime} I C$ is contained inside the quadrilateral $B^{\\prime} H^{\\prime} H C$. It follows that the perimeter of $B^{\\prime} I^{\\prime} I C$ cannot exceed the perimeter of $B^{\\prime} H^{\\prime} H C$. From (1) and (2), we conclude that\n\n$$\nA H+B H+C H \\geqslant A I+B I+C I \\text {. }\n$$\n\nFor the case $\\angle A I C<120^{\\circ}$, we can rotate $B, I, H$ through $60^{\\circ}$ about $C$ to $B^{\\prime}, I^{\\prime}, H^{\\prime}$ so that $B^{\\prime}$ and $A$ lie on different sides of $B C$. The proof is analogous to the previous case and we still get the desired inequality.']",,True,,, 1864,Number Theory,,"For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geqslant 2016$, the integer $P(n)$ is positive and $$ S(P(n))=P(S(n))\tag{1} $$","[""We consider three cases according to the degree of $P$.\n\n- Case 1. $P(x)$ is a constant polynomial.\n\nLet $P(x)=c$ where $c$ is an integer constant. Then (1) becomes $S(c)=c$. This holds if and only if $1 \\leqslant c \\leqslant 9$.\n\n- Case 2. $\\operatorname{deg} P=1$.\n\nWe have the following observation. For any positive integers $m, n$, we have\n\n$$\nS(m+n) \\leqslant S(m)+S(n)\\tag{2}\n$$\n\nand equality holds if and only if there is no carry in the addition $m+n$.\n\nLet $P(x)=a x+b$ for some integers $a, b$ where $a \\neq 0$. As $P(n)$ is positive for large $n$, we must have $a \\geqslant 1$. The condition (1) becomes $S(a n+b)=a S(n)+b$ for all $n \\geqslant 2016$. Setting $n=2025$ and $n=2020$ respectively, we get\n\n$$\nS(2025 a+b)-S(2020 a+b)=(a S(2025)+b)-(a S(2020)+b)=5 a .\n$$\n\nOn the other hand, (2) implies\n\n$$\nS(2025 a+b)=S((2020 a+b)+5 a) \\leqslant S(2020 a+b)+S(5 a) .\n$$\n\nThese give $5 a \\leqslant S(5 a)$. As $a \\geqslant 1$, this holds only when $a=1$, in which case (1) reduces to $S(n+b)=S(n)+b$ for all $n \\geqslant 2016$. Then we find that\n\n$$\nS(n+1+b)-S(n+b)=(S(n+1)+b)-(S(n)+b)=S(n+1)-S(n) .\\tag{3}\n$$\n\nIf $b>0$, we choose $n$ such that $n+1+b=10^{k}$ for some sufficiently large $k$. Note that all the digits of $n+b$ are 9's, so that the left-hand side of (3) equals $1-9 k$. As $n$ is a positive integer less than $10^{k}-1$, we have $S(n)<9 k$. Therefore, the right-hand side of (3) is at least $1-(9 k-1)=2-9 k$, which is a contradiction.\n\nThe case $b<0$ can be handled similarly by considering $n+1$ to be a large power of 10 . Therefore, we conclude that $P(x)=x$, in which case (1) is trivially satisfied.\n\n\n\n- Case 3. $\\operatorname{deg} P \\geqslant 2$.\n\nSuppose the leading term of $P$ is $a_{d} n^{d}$ where $a_{d} \\neq 0$. Clearly, we have $a_{d}>0$. Consider $n=10^{k}-1$ in (1). We get $S(P(n))=P(9 k)$. Note that $P(n)$ grows asymptotically as fast as $n^{d}$, so $S(P(n))$ grows asymptotically as no faster than a constant multiple of $k$. On the other hand, $P(9 k)$ grows asymptotically as fast as $k^{d}$. This shows the two sides of the last equation cannot be equal for sufficiently large $k$ since $d \\geqslant 2$.\n\nTherefore, we conclude that $P(x)=c$ where $1 \\leqslant c \\leqslant 9$ is an integer, or $P(x)=x$."", ""Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\\cdots+a_{0}$. Clearly $a_{d}>0$. There exists an integer $m \\geqslant 1$ such that $\\left|a_{i}\\right|<10^{m}$ for all $0 \\leqslant i \\leqslant d$. Consider $n=9 \\times 10^{k}$ for a sufficiently large integer $k$ in (1). If there exists an index $0 \\leqslant i \\leqslant d-1$ such that $a_{i}<0$, then all digits of $P(n)$ in positions from $10^{i k+m+1}$ to $10^{(i+1) k-1}$ are all 9's. Hence, we have $S(P(n)) \\geqslant 9(k-m-1)$. On the other hand, $P(S(n))=P(9)$ is a fixed constant. Therefore, (1) cannot hold for large $k$. This shows $a_{i} \\geqslant 0$ for all $0 \\leqslant i \\leqslant d-1$.\n\nHence, $P(n)$ is an integer formed by the nonnegative integers $a_{d} \\times 9^{d}, a_{d-1} \\times 9^{d-1}, \\ldots, a_{0}$ by inserting some zeros in between. This yields\n\n$$\nS(P(n))=S\\left(a_{d} \\times 9^{d}\\right)+S\\left(a_{d-1} \\times 9^{d-1}\\right)+\\cdots+S\\left(a_{0}\\right) .\n$$\n\nCombining with (1), we have\n\n$$\nS\\left(a_{d} \\times 9^{d}\\right)+S\\left(a_{d-1} \\times 9^{d-1}\\right)+\\cdots+S\\left(a_{0}\\right)=P(9)=a_{d} \\times 9^{d}+a_{d-1} \\times 9^{d-1}+\\cdots+a_{0} .\n$$\n\nAs $S(m) \\leqslant m$ for any positive integer $m$, with equality when $1 \\leqslant m \\leqslant 9$, this forces each $a_{i} \\times 9^{i}$ to be a positive integer between 1 and 9 . In particular, this shows $a_{i}=0$ for $i \\geqslant 2$ and hence $d \\leqslant 1$. Also, we have $a_{1} \\leqslant 1$ and $a_{0} \\leqslant 9$. If $a_{1}=1$ and $1 \\leqslant a_{0} \\leqslant 9$, we take $n=10^{k}+\\left(10-a_{0}\\right)$ for sufficiently large $k$ in (1). This yields a contradiction since\n\n$$\nS(P(n))=S\\left(10^{k}+10\\right)=2 \\neq 11=P\\left(11-a_{0}\\right)=P(S(n)) .\n$$\n\nThe zero polynomial is also rejected since $P(n)$ is positive for large $n$. The remaining candidates are $P(x)=x$ or $P(x)=a_{0}$ where $1 \\leqslant a_{0} \\leqslant 9$, all of which satisfy (1), and hence are the only solutions.""]",['- $P(x)=c$ where $1 \\leqslant c \\leqslant 9$ is an integer; or\n- $P(x)=x$'],False,,Need_human_evaluate, 1865,Number Theory,,Let $\tau(n)$ be the number of positive divisors of $n$. Let $\tau_{1}(n)$ be the number of positive divisors of $n$ which have remainders 1 when divided by 3 . Find all possible integral values of the fraction $\frac{\tau(10 n)}{\tau_{1}(10 n)}$.,"['In this solution, we always use $p_{i}$ to denote primes congruent to $1 \\bmod 3$, and use $q_{j}$ to denote primes congruent to $2 \\bmod 3$. When we express a positive integer $m$ using its prime factorization, we also include the special case $m=1$ by allowing the exponents to be zeros. We first compute $\\tau_{1}(m)$ for a positive integer $m$.\n\n- Claim. Let $m=3^{x} p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{s}^{a_{s}} q_{1}^{b_{1}} q_{2}^{b_{2}} \\cdots q_{t}^{b_{t}}$ be the prime factorization of $m$. Then\n\n$$\n\\tau_{1}(m)=\\prod_{i=1}^{s}\\left(a_{i}+1\\right)\\left\\lceil\\frac{1}{2} \\prod_{j=1}^{t}\\left(b_{j}+1\\right)\\right\\rceil\\tag{1}\n$$\n\nProof. To choose a divisor of $m$ congruent to $1 \\bmod 3$, it cannot have the prime divisor 3 , while there is no restriction on choosing prime factors congruent to 1 mod 3. Also, we have to choose an even number of prime factors (counted with multiplicity) congruent to $2 \\bmod 3$.\n\nIf $\\prod_{j=1}^{t}\\left(b_{j}+1\\right)$ is even, then we may assume without loss of generality $b_{1}+1$ is even. We can choose the prime factors $q_{2}, q_{3}, \\ldots, q_{t}$ freely in $\\prod_{j=2}^{t}\\left(b_{j}+1\\right)$ ways. Then the parity of the number of $q_{1}$ is uniquely determined, and hence there are $\\frac{1}{2}\\left(b_{1}+1\\right)$ ways to choose the exponent of $q_{1}$. Hence (1) is verified in this case.\n\nIf $\\prod_{j=1}^{t}\\left(b_{j}+1\\right)$ is odd, we use induction on $t$ to count the number of choices. When $t=1$, there are $\\left\\lceil\\frac{b_{1}+1}{2}\\right\\rceil$ choices for which the exponent is even and $\\left\\lfloor\\frac{b_{1}+1}{2}\\right\\rfloor$ choices for which the exponent is odd. For the inductive step, we find that there are\n\n$$\n\\left\\lceil\\frac{1}{2} \\prod_{j=1}^{t-1}\\left(b_{j}+1\\right)\\right\\rceil \\cdot\\left\\lceil\\frac{b_{t}+1}{2}\\right\\rceil+\\left\\lfloor\\frac{1}{2} \\prod_{j=1}^{t-1}\\left(b_{j}+1\\right)\\right\\rfloor \\cdot\\left\\lfloor\\frac{b_{t}+1}{2}\\right\\rfloor=\\left\\lceil\\frac{1}{2} \\prod_{j=1}^{t}\\left(b_{j}+1\\right)\\right\\rceil\n$$\n\nchoices with an even number of prime factors and hence $\\left\\lfloor\\frac{1}{2} \\prod_{j=1}^{t}\\left(b_{j}+1\\right)\\right\\rfloor$ choices with an odd number of prime factors. Hence (1) is also true in this case.\n\nLet $n=3^{x} 2^{y} 5^{z} p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{s}^{a_{s}} q_{1}^{b_{1}} q_{2}^{b_{2}} \\cdots q_{t}^{b_{t}}$. Using the well-known formula for computing the divisor function, we get\n\n$$\n\\tau(10 n)=(x+1)(y+2)(z+2) \\prod_{i=1}^{s}\\left(a_{i}+1\\right) \\prod_{j=1}^{t}\\left(b_{j}+1\\right)\\tag{2}\n$$\n\nBy the Claim, we have\n\n$$\n\\tau_{1}(10 n)=\\prod_{i=1}^{s}\\left(a_{i}+1\\right)\\left\\lceil\\frac{1}{2}(y+2)(z+2) \\prod_{j=1}^{t}\\left(b_{j}+1\\right)\\right\\rceil\\tag{3}\n$$\n\n\n\nIf $c=(y+2)(z+2) \\prod_{j=1}^{t}\\left(b_{j}+1\\right)$ is even, then (2) and (3) imply\n\n$$\n\\frac{\\tau(10 n)}{\\tau_{1}(10 n)}=2(x+1)\n$$\n\nIn this case $\\frac{\\tau(10 n)}{\\tau_{1}(10 n)}$ can be any even positive integer as $x$ runs through all nonnegative integers. If $c$ is odd, which means $y, z$ are odd and each $b_{j}$ is even, then (2) and (3) imply\n\n$$\n\\frac{\\tau(10 n)}{\\tau_{1}(10 n)}=\\frac{2(x+1) c}{c+1}\\tag{4}\n$$\n\nFor this to be an integer, we need $c+1$ divides $2(x+1)$ since $c$ and $c+1$ are relatively prime. Let $2(x+1)=k(c+1)$. Then $(4)$ reduces to\n\n$$\n\\frac{\\tau(10 n)}{\\tau_{1}(10 n)}=k c=k(y+2)(z+2) \\prod_{j=1}^{t}\\left(b_{j}+1\\right)\n\\tag{5}\n$$\n\nNoting that $y, z$ are odd, the integers $y+2$ and $z+2$ are at least 3 . This shows the integer in this case must be composite. On the other hand, for any odd composite number $a b$ with $a, b \\geqslant 3$, we may simply take $n=3^{\\frac{a b-1}{2}} \\cdot 2^{a-2} \\cdot 5^{b-2}$ so that $\\frac{\\tau(10 n)}{\\tau_{1}(10 n)}=a b$ from (5).\n\nWe conclude that the fraction can be any even integer or any odd composite number. Equivalently, it can be 2 or any composite number.']",['All composite numbers together with 2'],False,,Need_human_evaluate, 1866,Number Theory,,"Define $P(n)=n^{2}+n+1$. For any positive integers $a$ and $b$, the set $$ \{P(a), P(a+1), P(a+2), \ldots, P(a+b)\} $$ is said to be fragrant if none of its elements is relatively prime to the product of the other elements. Determine the smallest size of a fragrant set.","['We have the following observations.\n\n(i) $(P(n), P(n+1))=1$ for any $n$.\n\nWe have $(P(n), P(n+1))=\\left(n^{2}+n+1, n^{2}+3 n+3\\right)=\\left(n^{2}+n+1,2 n+2\\right)$. Noting that $n^{2}+n+1$ is odd and $\\left(n^{2}+n+1, n+1\\right)=(1, n+1)=1$, the claim follows.\n\n(ii) $(P(n), P(n+2))=1$ for $n \\not \\equiv 2(\\bmod 7)$ and $(P(n), P(n+2))=7$ for $n \\equiv 2(\\bmod 7)$.\n\nFrom $(2 n+7) P(n)-(2 n-1) P(n+2)=14$ and the fact that $P(n)$ is odd, $(P(n), P(n+2))$ must be a divisor of 7 . The claim follows by checking $n \\equiv 0,1, \\ldots, 6(\\bmod 7)$ directly.\n\n(iii) $(P(n), P(n+3))=1$ for $n \\not \\equiv 1(\\bmod 3)$ and $3 \\mid(P(n), P(n+3))$ for $n \\equiv 1(\\bmod 3)$.\n\nFrom $(n+5) P(n)-(n-1) P(n+3)=18$ and the fact that $P(n)$ is odd, $(P(n), P(n+3))$ must be a divisor of 9 . The claim follows by checking $n \\equiv 0,1,2(\\bmod 3)$ directly.\n\nSuppose there exists a fragrant set with at most 5 elements. We may assume it contains exactly 5 elements $P(a), P(a+1), \\ldots, P(a+4)$ since the following argument also works with fewer elements. Consider $P(a+2)$. From (i), it is relatively prime to $P(a+1)$ and $P(a+3)$. Without loss of generality, assume $(P(a), P(a+2))>1$. From (ii), we have $a \\equiv 2(\\bmod 7)$. The same observation implies $(P(a+1), P(a+3))=1$. In order that the set is fragrant, $(P(a), P(a+3))$ and $(P(a+1), P(a+4))$ must both be greater than 1 . From (iii), this holds only when both $a$ and $a+1$ are congruent to $1 \\bmod 3$, which is a contradiction.\n\nIt now suffices to construct a fragrant set of size 6 . By the Chinese Remainder Theorem, we can take a positive integer $a$ such that\n\n$$\na \\equiv 7 \\quad(\\bmod 19), \\quad a+1 \\equiv 2 \\quad(\\bmod 7), \\quad a+2 \\equiv 1 \\quad(\\bmod 3)\n$$\n\nFor example, we may take $a=197$. From (ii), both $P(a+1)$ and $P(a+3)$ are divisible by 7. From (iii), both $P(a+2)$ and $P(a+5)$ are divisible by 3 . One also checks from $19 \\mid P(7)=57$ and $19 \\mid P(11)=133$ that $P(a)$ and $P(a+4)$ are divisible by 19 . Therefore, the set $\\{P(a), P(a+1), \\ldots, P(a+5)\\}$ is fragrant.\n\nTherefore, the smallest size of a fragrant set is 6 .']",['6'],False,,Numerical, 1867,Number Theory,,"Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^{k}+m n^{l}+1$ divides $n^{k+l}-1$. Prove that - $m=1$ and $l=2 k$; or - $l \mid k$ and $m=\frac{n^{k-l}-1}{n^{l}-1}$.","['It is given that\n\n$$\nn^{k}+m n^{l}+1 \\mid n^{k+l}-1\n\\tag{1}\n$$\n\nThis implies\n\n$$\nn^{k}+m n^{l}+1 \\mid\\left(n^{k+l}-1\\right)+\\left(n^{k}+m n^{l}+1\\right)=n^{k+l}+n^{k}+m n^{l} .\\tag{2}\n$$\n\nWe have two cases to discuss.\n\n- Case 1. $l \\geqslant k$.\n\nSince $\\left(n^{k}+m n^{l}+1, n\\right)=1,(2)$ yields\n\n$$\nn^{k}+m n^{l}+1 \\mid n^{l}+m n^{l-k}+1\n$$\n\nIn particular, we get $n^{k}+m n^{l}+1 \\leqslant n^{l}+m n^{l-k}+1$. As $n \\geqslant 2$ and $k \\geqslant 1,(m-1) n^{l}$ is at least $2(m-1) n^{l-k}$. It follows that the inequality cannot hold when $m \\geqslant 2$. For $m=1$, the above divisibility becomes\n\n$$\nn^{k}+n^{l}+1 \\mid n^{l}+n^{l-k}+1\n$$\n\nNote that $n^{l}+n^{l-k}+12 m n^{l}+1>n^{l}+m n^{l-k}+1$, it follows that $n^{k}+m n^{l}+1=n^{l}+m n^{l-k}+1$, that is,\n\n$$\nm\\left(n^{l}-n^{l-k}\\right)=n^{l}-n^{k} .\n$$\n\nIf $m \\geqslant 2$, then $m\\left(n^{l}-n^{l-k}\\right) \\geqslant 2 n^{l}-2 n^{l-k} \\geqslant 2 n^{l}-n^{l}>n^{l}-n^{k}$ gives a contradiction. Hence $m=1$ and $l-k=k$, which means $m=1$ and $l=2 k$.\n\n- Case 2. $l2 n^{k}+m>n^{k}+n^{k-l}+m$, it follows that $n^{k}+m n^{l}+1=n^{k}+n^{k-l}+m$. This gives $m=\\frac{n^{k-l}-1}{n^{l}-1}$. Note that $n^{l}-1 \\mid n^{k-l}-1$ implies $l \\mid k-l$ and hence $l \\mid k$. The proof is thus complete.']",,True,,, 1868,Number Theory,,"Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that $$ k=\frac{x^{2}-a}{x^{2}-y^{2}} \tag{1} $$ for some integers $x$ and $y$ with $x>\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with $0 \leqslant x<\sqrt{a}$. Prove that $A=B$.","[""We first prove the following preliminary result.\n\n- Claim. For fixed $k$, let $x, y$ be integers satisfying (1). Then the numbers $x_{1}, y_{1}$ defined by\n\n$$\nx_{1}=\\frac{1}{2}\\left(x-y+\\frac{(x-y)^{2}-4 a}{x+y}\\right), \\quad y_{1}=\\frac{1}{2}\\left(x-y-\\frac{(x-y)^{2}-4 a}{x+y}\\right)\n$$\n\nare integers and satisfy (1) (with $x, y$ replaced by $x_{1}, y_{1}$ respectively).\n\nProof. Since $x_{1}+y_{1}=x-y$ and\n\n$$\nx_{1}=\\frac{x^{2}-x y-2 a}{x+y}=-x+\\frac{2\\left(x^{2}-a\\right)}{x+y}=-x+2 k(x-y)\n$$\n\nboth $x_{1}$ and $y_{1}$ are integers. Let $u=x+y$ and $v=x-y$. The relation (1) can be rewritten as\n\n$$\nu^{2}-(4 k-2) u v+\\left(v^{2}-4 a\\right)=0 \\text {. }\n$$\n\nBy Vieta's Theorem, the number $z=\\frac{v^{2}-4 a}{u}$ satisfies\n\n$$\nv^{2}-(4 k-2) v z+\\left(z^{2}-4 a\\right)=0\n$$\n\nSince $x_{1}$ and $y_{1}$ are defined so that $v=x_{1}+y_{1}$ and $z=x_{1}-y_{1}$, we can reverse the process and verify (1) for $x_{1}, y_{1}$.\n\nWe first show that $B \\subset A$. Take any $k \\in B$ so that (1) is satisfied for some integers $x, y$ with $0 \\leqslant x<\\sqrt{a}$. Clearly, $y \\neq 0$ and we may assume $y$ is positive. Since $a$ is not a square, we have $k>1$. Hence, we get $0 \\leqslant x\\sqrt{a}\n$$\n\nThis implies $k \\in A$ and hence $B \\subset A$.\n\n\n\nNext, we shall show that $A \\subset B$. Take any $k \\in A$ so that (1) is satisfied for some integers $x, y$ with $x>\\sqrt{a}$. Again, we may assume $y$ is positive. Among all such representations of $k$, we choose the one with smallest $x+y$. Define\n\n$$\nx_{1}=\\frac{1}{2}\\left|x-y+\\frac{(x-y)^{2}-4 a}{x+y}\\right|, \\quad y_{1}=\\frac{1}{2}\\left(x-y-\\frac{(x-y)^{2}-4 a}{x+y}\\right) .\n$$\n\nBy the Claim, $x_{1}, y_{1}$ are integers satisfying (1). Since $k>1$, we get $x>y>\\sqrt{a}$. Therefore, we have $y_{1}>\\frac{4 a}{x+y}>0$ and $\\frac{4 a}{x+y}\\sqrt{a}$, we get a contradiction due to the minimality of $x+y$. Therefore, we must have $0 \\leqslant x_{1}<\\sqrt{a}$, which means $k \\in B$ so that $A \\subset B$.\n\nThe two subset relations combine to give $A=B$."", ""The relation (1) is equivalent to\n\n$$\nk y^{2}-(k-1) x^{2}=a .\n\\tag{2}\n$$\n\nMotivated by Pell's Equation, we prove the following, which is essentially the same as the Claim in"", '- Claim. If $\\left(x_{0}, y_{0}\\right)$ is a solution to (2), then $\\left((2 k-1) x_{0} \\pm 2 k y_{0},(2 k-1) y_{0} \\pm 2(k-1) x_{0}\\right)$ is also a solution to (2).\n\nProof. We check directly that\n\n$$\n\\begin{aligned}\n& k\\left((2 k-1) y_{0} \\pm 2(k-1) x_{0}\\right)^{2}-(k-1)\\left((2 k-1) x_{0} \\pm 2 k y_{0}\\right)^{2} \\\\\n= & \\left(k(2 k-1)^{2}-(k-1)(2 k)^{2}\\right) y_{0}^{2}+\\left(k(2(k-1))^{2}-(k-1)(2 k-1)^{2}\\right) x_{0}^{2} \\\\\n= & k y_{0}^{2}-(k-1) x_{0}^{2}=a\n\\end{aligned}\n$$\n\nIf (2) is satisfied for some $0 \\leqslant x<\\sqrt{a}$ and nonnegative integer $y$, then clearly (1) implies $y>x$. Also, we have $k>1$ since $a$ is not a square number. By the Claim, consider another solution to (2) defined by\n\n$$\nx_{1}=(2 k-1) x+2 k y, \\quad y_{1}=(2 k-1) y+2(k-1) x\n$$\n\nIt satisfies $x_{1} \\geqslant(2 k-1) x+2 k(x+1)=(4 k-1) x+2 k>x$. Then we can replace the old solution by a new one which has a larger value in $x$. After a finite number of replacements, we must get a solution with $x>\\sqrt{a}$. This shows $B \\subset A$.\n\nIf (2) is satisfied for some $x>\\sqrt{a}$ and nonnegative integer $y$, by the Claim we consider another solution to (2) defined by\n\n$$\nx_{1}=|(2 k-1) x-2 k y|, \\quad y_{1}=(2 k-1) y-2(k-1) x\n$$\n\n\n\nFrom (2), we get $\\sqrt{k} y>\\sqrt{k-1} x$. This implies $k y>\\sqrt{k(k-1)} x>(k-1) x$ and hence $(2 k-1) x-2 k yy$. Then it is clear that $(2 k-1) x-2 k y>-x$. These combine to give $x_{1}|x|$, and, in case of $x>\\sqrt{a}$, another solution $\\left(x_{2}, y_{2}\\right)$ with $\\left|x_{2}\\right|<|x|$.\n\nWithout loss of generality, assume $x, y \\geqslant 0$. Let $u=x+y$ and $v=x-y$. Then $u \\geqslant v$ and (1) becomes\n\n$$\nk=\\frac{(u+v)^{2}-4 a}{4 u v}\n\\tag{3}\n$$\n\nThis is the same as\n\n$$\nv^{2}+(2 u-4 k u) v+u^{2}-4 a=0\n$$\n\nLet $v_{1}=4 k u-2 u-v$. Then $u+v_{1}=4 k u-u-v \\geqslant 8 u-u-v>u+v$. By Vieta's Theorem, $v_{1}$ satisfies\n\n$$\nv_{1}^{2}+(2 u-4 k u) v_{1}+u^{2}-4 a=0\n$$\n\nThis gives $k=\\frac{\\left(u+v_{1}\\right)^{2}-4 a}{4 u v_{1}}$. As $k$ is an integer, $u+v_{1}$ must be even. Therefore, $x_{1}=\\frac{u+v_{1}}{2}$ and $y_{1}=\\frac{v_{1}-u}{2}$ are integers. By reversing the process, we can see that $\\left(x_{1}, y_{1}\\right)$ is a solution to (1), with $x_{1}=\\frac{u+v_{1}}{2}>\\frac{u+v}{2}=x \\geqslant 0$. This completes the first half of the proof.\n\nSuppose $x>\\sqrt{a}$. Then $u+v>2 \\sqrt{a}$ and (3) can be rewritten as\n\n$$\nu^{2}+(2 v-4 k v) u+v^{2}-4 a=0\n$$\n\nLet $u_{2}=4 k v-2 v-u$. By Vieta's Theorem, we have $u u_{2}=v^{2}-4 a$ and\n\n$$\nu_{2}^{2}+(2 v-4 k v) u_{2}+v^{2}-4 a=0\n\\tag{4}\n$$\n\nBy $u>0, u+v>2 \\sqrt{a}$ and (3), we have $v>0$. If $u_{2} \\geqslant 0$, then $v u_{2} \\leqslant u u_{2}=v^{2}-4 a0$ and $u_{2}+v\\nu_{p}\\left(A_{1} A_{m+1}^{2}\\right)$ for $2 \\leqslant m \\leqslant k-1$.\n\nProof. The case $m=2$ is obvious since $\\nu_{p}\\left(A_{1} A_{2}^{2}\\right) \\geqslant p^{t}>\\nu_{p}\\left(A_{1} A_{3}^{2}\\right)$ by the condition and the above assumption.\n\nSuppose $\\nu_{p}\\left(A_{1} A_{2}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{3}^{2}\\right)>\\cdots>\\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$ where $3 \\leqslant m \\leqslant k-1$. For the induction step, we apply Ptolemy's Theorem to the cyclic quadrilateral $A_{1} A_{m-1} A_{m} A_{m+1}$ to get\n\n$$\nA_{1} A_{m+1} \\times A_{m-1} A_{m}+A_{1} A_{m-1} \\times A_{m} A_{m+1}=A_{1} A_{m} \\times A_{m-1} A_{m+1}\n$$\n\nwhich can be rewritten as\n\n$$\n\\begin{aligned}\nA_{1} A_{m+1}^{2} \\times A_{m-1} A_{m}^{2}= & A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}+A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2} \\\\\n& -2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}\n\\end{aligned}\n\\tag{1}\n$$\n\nFrom this, $2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}$ is an integer. We consider the component of $p$ of each term in (1). By the inductive hypothesis, we have $\\nu_{p}\\left(A_{1} A_{m-1}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$. Also, we have $\\nu_{p}\\left(A_{m} A_{m+1}^{2}\\right) \\geqslant p^{t}>\\nu_{p}\\left(A_{m-1} A_{m+1}^{2}\\right)$. These give\n\n$$\n\\nu_{p}\\left(A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)\n\\tag{2}\n$$\n\nNext, we have $\\nu_{p}\\left(4 A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2} \\times A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)=\\nu_{p}\\left(A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}\\right)+$ $\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)>2 \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)$ from (2). This implies\n\n$$\n\\nu_{p}\\left(2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}\\right)>\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) .\n\\tag{3}\n$$\n\nCombining (1), (2) and (3), we conclude that\n\n$$\n\\nu_{p}\\left(A_{1} A_{m+1}^{2} \\times A_{m-1} A_{m}^{2}\\right)=\\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) .\n$$\n\nBy $\\nu_{p}\\left(A_{m-1} A_{m}^{2}\\right) \\geqslant p^{t}>\\nu_{p}\\left(A_{m-1} A_{m+1}^{2}\\right)$, we get $\\nu_{p}\\left(A_{1} A_{m+1}^{2}\\right)<\\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$. The Claim follows by induction.\n\n\n\nFrom the Claim, we get a chain of inequalities\n\n$$\np^{t}>\\nu_{p}\\left(A_{1} A_{3}^{2}\\right)>\\nu_{p}\\left(A_{1} A_{4}^{2}\\right)>\\cdots>\\nu_{p}\\left(A_{1} A_{k}^{2}\\right) \\geqslant p^{t}\n$$\n\nwhich yields a contradiction. Therefore, we can show by induction that $2 S$ is divisible by $n$.""]",,True,,, 1871,Number Theory,,"Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\frac{1}{2}<\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}<2$ and $\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}$ is the $d$-th power of a rational number for every pair of indices $i$ and $j$ with $1 \leqslant i, j \leqslant n$.","[""Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\\cdots+b_{0}$ and $B=\\max _{0 \\leqslant i \\leqslant d}\\left\\{\\left|b_{i}\\right|\\right\\}$ (where $b_{d-1}=0$ ).\n\nThe condition shows that for each $n \\geqslant 1$, there exist integers $y_{1}, y_{2}, \\ldots, y_{n}$ such that $\\frac{1}{2}<\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{j}\\right)}<2$ and $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{j}\\right)}$ is the $d$-th power of a rational number for $1 \\leqslant i, j \\leqslant n$. Since $n$ can be arbitrarily large, we may assume all $x_{i}$ 's and hence $y_{i}$ 's are integers larger than some absolute constant in the following.\n\nBy Dirichlet's Theorem, since $d$ is odd, we can find a sufficiently large prime $p$ such that $p \\equiv 2(\\bmod d)$. In particular, we have $(p-1, d)=1$. For this fixed $p$, we choose $n$ to be sufficiently large. Then by the Pigeonhole Principle, there must be $d+1$ of $y_{1}, y_{2}, \\ldots, y_{n}$ which are congruent $\\bmod p$. Without loss of generality, assume $y_{i} \\equiv y_{j}(\\bmod p)$ for $1 \\leqslant i, j \\leqslant d+1$. We shall establish the following.\n\n- Claim. $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}=\\frac{y_{i}^{d}}{y_{1}^{d}}$ for $2 \\leqslant i \\leqslant d+1$.\n\nProof. Let $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}=\\frac{l^{d}}{m^{d}}$ where $(l, m)=1$ and $l, m>0$. This can be rewritten in the expanded form\n\n$$\nb_{d}\\left(m^{d} y_{i}^{d}-l^{d} y_{1}^{d}\\right)=-\\sum_{j=0}^{d-2} b_{j}\\left(m^{d} y_{i}^{j}-l^{d} y_{1}^{j}\\right)\n\\tag{1}\n$$\n\nLet $c$ be the common denominator of $Q$, so that $c Q(k)$ is an integer for any integer $k$. Note that $c$ depends only on $P$ and so we may assume $(p, c)=1$. Then $y_{1} \\equiv y_{i}(\\bmod p)$ implies $c Q\\left(y_{1}\\right) \\equiv c Q\\left(y_{i}\\right)(\\bmod p)$.\n\n- Case 1. $p \\mid c Q\\left(y_{1}\\right)$.\n\nIn this case, there is a cancellation of $p$ in the numerator and denominator of $\\frac{c Q\\left(y_{i}\\right)}{c Q\\left(y_{1}\\right)}$, so that $m^{d} \\leqslant p^{-1}\\left|c Q\\left(y_{1}\\right)\\right|$. Noting $\\left|Q\\left(y_{1}\\right)\\right|<2 B y_{1}^{d}$ as $y_{1}$ is large, we get\n\n$$\nm \\leqslant p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} y_{1} .\n\\tag{2}\n$$\n\nFor large $y_{1}$ and $y_{i}$, the relation $\\frac{1}{2}<\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}<2$ implies\n\n$$\n\\frac{1}{3}<\\frac{y_{i}^{d}}{y_{1}^{d}}<3\n\\tag{3}\n$$\n\nWe also have\n\n$$\n\\frac{1}{2}<\\frac{l^{d}}{m^{d}}<2\n\\tag{4}\n$$\n\n\n\nNow, the left-hand side of (1) is\n\n$$\nb_{d}\\left(m y_{i}-l y_{1}\\right)\\left(m^{d-1} y_{i}^{d-1}+m^{d-2} y_{i}^{d-2} l y_{1}+\\cdots+l^{d-1} y_{1}^{d-1}\\right) .\n$$\n\nSuppose on the contrary that $m y_{i}-l y_{1} \\neq 0$. Then the absolute value of the above expression is at least $\\left|b_{d}\\right| m^{d-1} y_{i}^{d-1}$. On the other hand, the absolute value of the right-hand side of (1) is at most\n\n$$\n\\begin{aligned}\n\\sum_{j=0}^{d-2} B\\left(m^{d} y_{i}^{j}+l^{d} y_{1}^{j}\\right) & \\leqslant(d-1) B\\left(m^{d} y_{i}^{d-2}+l^{d} y_{1}^{d-2}\\right) \\\\\n& \\leqslant(d-1) B\\left(7 m^{d} y_{i}^{d-2}\\right) \\\\\n& \\leqslant 7(d-1) B\\left(p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} y_{1}\\right) m^{d-1} y_{i}^{d-2} \\\\\n& \\leqslant 21(d-1) B p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} m^{d-1} y_{i}^{d-1}\n\\end{aligned}\n$$\n\nby using successively (3), (4), (2) and again (3). This shows\n\n$$\n\\left|b_{d}\\right| m^{d-1} y_{i}^{d-1} \\leqslant 21(d-1) B p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} m^{d-1} y_{i}^{d-1},\n$$\n\nwhich is a contradiction for large $p$ as $b_{d}, B, c, d$ depend only on the polynomial $P$. Therefore, we have $m y_{i}-l y_{1}=0$ in this case.\n\n- Case 2. $\\left(p, c Q\\left(y_{1}\\right)\\right)=1$.\n\nFrom $c Q\\left(y_{1}\\right) \\equiv c Q\\left(y_{i}\\right)(\\bmod p)$, we have $l^{d} \\equiv m^{d}(\\bmod p)$. Since $(p-1, d)=1$, we use Fermat Little Theorem to conclude $l \\equiv m(\\bmod p)$. Then $p \\mid m y_{i}-l y_{1}$. Suppose on the contrary that $m y_{i}-l y_{1} \\neq 0$. Then the left-hand side of (1) has absolute value at least $\\left|b_{d}\\right| p m^{d-1} y_{i}^{d-1}$. Similar to Case 1, the right-hand side of (1) has absolute value at most\n\n$$\n21(d-1) B(2 c B)^{\\frac{1}{d}} m^{d-1} y_{i}^{d-1},\n$$\n\nwhich must be smaller than $\\left|b_{d}\\right| p m^{d-1} y_{i}^{d-1}$ for large $p$. Again this yields a contradiction and hence $m y_{i}-l y_{1}=0$.\n\nIn both cases, we find that $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}=\\frac{l^{d}}{m^{d}}=\\frac{y_{i}^{d}}{y_{1}^{d}}$.\n\nFrom the Claim, the polynomial $Q\\left(y_{1}\\right) y^{d}-y_{1}^{d} Q(y)$ has roots $y=y_{1}, y_{2}, \\ldots, y_{d+1}$. Since its degree is at most $d$, this must be the zero polynomial. Hence, $Q(y)=b_{d} y^{d}$. This implies $P(x)=a_{d}\\left(x+\\frac{a_{d-1}}{d a_{d}}\\right)^{d}$. Let $\\frac{a_{d-1}}{d a_{d}}=\\frac{s}{r}$ with integers $r, s$ where $r \\geqslant 1$ and $(r, s)=1$. Since $P$ has integer coefficients, we need $r^{d} \\mid a_{d}$. Let $a_{d}=r^{d} a$. Then $P(x)=a(r x+s)^{d}$. It is obvious that such a polynomial satisfies the conditions.""]","['$P(x)=a(r x+s)^{d}$ where $a, r, s$ are integers with $a \\neq 0, r \\geqslant 1$ and $(r, s)=1$']",True,,Need_human_evaluate, 1872,Algebra,,"Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$, $$ \sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant a_{n}(x-1)^{2}+x $$","['First of all, assume that $a_{n}0$, we should have\n\n$$\n\\frac{(1+t)^{2 N}+1}{2} \\leqslant\\left(1+t+a_{n} t^{2}\\right)^{N}\n$$\n\nExpanding the brackets we get\n\n$$\n\\left(1+t+a_{n} t^{2}\\right)^{N}-\\frac{(1+t)^{2 N}+1}{2}=\\left(N a_{n}-\\frac{N^{2}}{2}\\right) t^{2}+c_{3} t^{3}+\\ldots+c_{2 N} t^{2 N}\n\\tag{1}\n$$\n\nwith some coefficients $c_{3}, \\ldots, c_{2 N}$. Since $a_{n}0$ and $a+b+c+d=1$. Prove that $$ (a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d}<1 $$","['The weighted AM-GM inequality with weights $a, b, c, d$ gives\n\n$$\na^{a} b^{b} c^{c} d^{d} \\leqslant a \\cdot a+b \\cdot b+c \\cdot c+d \\cdot d=a^{2}+b^{2}+c^{2}+d^{2}\n$$\n\nso it suffices to prove that $(a+2 b+3 c+4 d)\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)<1=(a+b+c+d)^{3}$. This can be done in various ways, for example:\n\n$$\n\\begin{gathered}\n(a+b+c+d)^{3}>a^{2}(a+3 b+3 c+3 d)+b^{2}(3 a+b+3 c+3 d) \\\\\n\\quad+c^{2}(3 a+3 b+c+3 d)+d^{2}(3 a+3 b+3 c+d) \\\\\n\\geqslant\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right) \\cdot(a+2 b+3 c+4 d)\n\\end{gathered}\n$$', 'From $b \\geqslant d$ we get\n\n$$\na+2 b+3 c+4 d \\leqslant a+3 b+3 c+3 d=3-2 a .\n$$\n\nIf $a<\\frac{1}{2}$, then the statement can be proved by\n\n$$\n(a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d} \\leqslant(3-2 a) a^{a} a^{b} a^{c} a^{d}=(3-2 a) a=1-(1-a)(1-2 a)<1 .\n$$\n\nFrom now on we assume $\\frac{1}{2} \\leqslant a<1$.\n\nBy $b, c, d<1-a$ we have\n\n$$\nb^{b} c^{c} d^{d}<(1-a)^{b} \\cdot(1-a)^{c} \\cdot(1-a)^{d}=(1-a)^{1-a} .\n$$\n\nTherefore,\n\n$$\n(a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d}<(3-2 a) a^{a}(1-a)^{1-a} .\n$$\n\nFor $00,\n$$\n\nso $g$ is strictly convex on $(0,1)$.\n\nBy $g\\left(\\frac{1}{2}\\right)=\\log 2+2 \\cdot \\frac{1}{2} \\log \\frac{1}{2}=0$ and $\\lim _{x \\rightarrow 1-} g(x)=0$, we have $g(x) \\leqslant 0$ (and hence $f(x) \\leqslant 1$ ) for all $x \\in\\left[\\frac{1}{2}, 1\\right)$, and therefore\n\n$$\n(a+2 b+3 c+4 d) a^{a} b^{b} c^{c} d^{d}\\max _{z \\in \\mathcal{O}(0)}|z|$, this yields $f(a)=f(-a)$ and $f^{2 a^{2}}(0)=0$. Therefore, the sequence $\\left(f^{k}(0): k=0,1, \\ldots\\right)$ is purely periodic with a minimal period $T$ which divides $2 a^{2}$. Analogously, $T$ divides $2(a+1)^{2}$, therefore, $T \\mid \\operatorname{gcd}\\left(2 a^{2}, 2(a+1)^{2}\\right)=2$, i.e., $f(f(0))=0$ and $a(f(a)-f(-a))=f^{2 a^{2}}(0)=0$ for all $a$. Thus,\n\n$$\nf(a)=f(-a) \\quad \\text { for all } a \\neq 0\n\\tag{2}\n$$\n$$\n\\text { in particular, } \\quad f(1)=f(-1)=0\n\\tag{3}\n$$\n\nNext, for each $n \\in \\mathbb{Z}$, by $E(n, 1-n)$ we get\n\n$$\nn f(n)+(1-n) f(1-n)=f^{n^{2}+(1-n)^{2}}(1)=f^{2 n^{2}-2 n}(0)=0\n\\tag{4}\n$$\n\nAssume that there exists some $m \\neq 0$ such that $f(m) \\neq 0$. Choose such an $m$ for which $|m|$ is minimal possible. Then $|m|>1$ due to $(3) ; f(|m|) \\neq 0$ due to $(2)$; and $f(1-|m|) \\neq 0$ due to $(4)$ for $n=|m|$. This contradicts to the minimality assumption.\n\nSo, $f(n)=0$ for $n \\neq 0$. Finally, $f(0)=f^{3}(0)=f^{4}(2)=2 f(2)=0$. Clearly, the function $f(x) \\equiv 0$ satisfies the problem condition, which provides the first of the two answers.\n\nCase 2: All orbits are infinite.\n\nSince the orbits $\\mathcal{O}(a)$ and $\\mathcal{O}(a-1)$ differ by finitely many terms for all $a \\in \\mathbb{Z}$, each two orbits $\\mathcal{O}(a)$ and $\\mathcal{O}(b)$ have infinitely many common terms for arbitrary $a, b \\in \\mathbb{Z}$.\n\nFor a minute, fix any $a, b \\in \\mathbb{Z}$. We claim that all pairs $(n, m)$ of nonnegative integers such that $f^{n}(a)=f^{m}(b)$ have the same difference $n-m$. Arguing indirectly, we have $f^{n}(a)=f^{m}(b)$ and $f^{p}(a)=f^{q}(b)$ with, say, $n-m>p-q$, then $f^{p+m+k}(b)=f^{p+n+k}(a)=f^{q+n+k}(b)$, for all nonnegative integers $k$. This means that $f^{\\ell+(n-m)-(p-q)}(b)=f^{\\ell}(b)$ for all sufficiently large $\\ell$, i.e., that the sequence $\\left(f^{n}(b)\\right)$ is eventually periodic, so $\\mathcal{O}(b)$ is finite, which is impossible.\n\nNow, for every $a, b \\in \\mathbb{Z}$, denote the common difference $n-m$ defined above by $X(a, b)$. We have $X(a-1, a)=1$ by (1). Trivially, $X(a, b)+X(b, c)=X(a, c)$, as if $f^{n}(a)=f^{m}(b)$ and $f^{p}(b)=f^{q}(c)$, then $f^{p+n}(a)=f^{p+m}(b)=f^{q+m}(c)$. These two properties imply that $X(a, b)=b-a$ for all $a, b \\in \\mathbb{Z}$.\n\n\n\nBut (1) yields $f^{a^{2}+1}(f(a-1))=f^{a^{2}}(f(a))$, so\n\n$$\n1=X(f(a-1), f(a))=f(a)-f(a-1) \\quad \\text { for all } a \\in \\mathbb{Z}\n$$\n\nRecalling that $f(-1)=0$, we conclude by (two-sided) induction on $x$ that $f(x)=x+1$ for all $x \\in \\mathbb{Z}$.\n\nFinally, the obtained function also satisfies the assumption. Indeed, $f^{n}(x)=x+n$ for all $n \\geqslant 0$, so\n\n$$\nf^{a^{2}+b^{2}}(a+b)=a+b+a^{2}+b^{2}=a f(a)+b f(b) .\n$$']","['Either $f(x)=0$ for all $x \\in \\mathbb{Z}$, or $f(x)=x+1$ for all $x \\in \\mathbb{Z}$']",True,,Need_human_evaluate, 1878,Algebra,,"Let $n$ and $k$ be positive integers. Prove that for $a_{1}, \ldots, a_{n} \in\left[1,2^{k}\right]$ one has $$ \sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{1}^{2}+\ldots+a_{i}^{2}}} \leqslant 4 \sqrt{k n} $$","['Partition the set of indices $\\{1,2, \\ldots, n\\}$ into disjoint subsets $M_{1}, M_{2}, \\ldots, M_{k}$ so that $a_{\\ell} \\in\\left[2^{j-1}, 2^{j}\\right]$ for $\\ell \\in M_{j}$. Then, if $\\left|M_{j}\\right|=: p_{j}$, we have\n\n$$\n\\sum_{\\ell \\in M_{j}} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant \\sum_{i=1}^{p_{j}} \\frac{2^{j}}{2^{j-1} \\sqrt{i}}=2 \\sum_{i=1}^{p_{j}} \\frac{1}{\\sqrt{i}}\n$$\n\nwhere we used that $a_{\\ell} \\leqslant 2^{j}$ and in the denominator every index from $M_{j}$ contributes at least $\\left(2^{j-1}\\right)^{2}$. Now, using $\\sqrt{i}-\\sqrt{i-1}=\\frac{1}{\\sqrt{i}+\\sqrt{i-1}} \\geqslant \\frac{1}{2 \\sqrt{i}}$, we deduce that\n\n$$\n\\sum_{\\ell \\in M_{j}} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant 2 \\sum_{i=1}^{p_{j}} \\frac{1}{\\sqrt{i}} \\leqslant 2 \\sum_{i=1}^{p_{j}} 2(\\sqrt{i}-\\sqrt{i-1})=4 \\sqrt{p_{j}}\n$$\n\nTherefore, summing over $j=1, \\ldots, k$ and using the QM-AM inequality, we obtain\n\n$$\n\\sum_{\\ell=1}^{n} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant 4 \\sum_{j=1}^{k} \\sqrt{\\left|M_{j}\\right|} \\leqslant 4 \\sqrt{k \\sum_{j=1}^{k}\\left|M_{j}\\right|}=4 \\sqrt{k n}\n$$', 'Apply induction on $n$. The base $n \\leqslant 16$ is clear: our sum does not exceed $n \\leqslant 4 \\sqrt{n k}$. For the inductive step from $1, \\ldots, n-1$ to $n \\geqslant 17$ consider two similar cases.\n\nCase 1: $n=2 t$.\n\nLet $x_{\\ell}=\\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}}$. We have\n\n$$\n\\exp \\left(-x_{t+1}^{2}-\\ldots-x_{2 t}^{2}\\right) \\geqslant\\left(1-x_{t+1}^{2}\\right) \\ldots\\left(1-x_{2 t}^{2}\\right)=\\frac{a_{1}^{2}+\\ldots+a_{t}^{2}}{a_{1}^{2}+\\ldots+a_{2 t}^{2}} \\geqslant \\frac{1}{1+4^{k}}\n$$\n\nwhere we used that the product is telescopic and then an estimate $a_{t+i} \\leqslant 2^{k} a_{i}$ for $i=1, \\ldots, t$. Therefore, $x_{t+1}^{2}+\\ldots+x_{2 t}^{2} \\leqslant \\log \\left(4^{k}+1\\right) \\leqslant 2 k$, where log denotes the natural logarithm. This implies $x_{t+1}+\\ldots+x_{2 t} \\leqslant \\sqrt{2 k t}$. Hence, using the inductive hypothesis for $n=t$ we get\n\n$$\n\\sum_{\\ell=1}^{2 t} x_{\\ell} \\leqslant 4 \\sqrt{k t}+\\sqrt{2 k t} \\leqslant 4 \\sqrt{2 k t}\n$$\n\n\n\nCase 2: $n=2 t+1$.\n\nAnalogously, we get $x_{t+2}^{2}+\\ldots+x_{2 t+1}^{2} \\leqslant \\log \\left(4^{k}+1\\right) \\leqslant 2 k$ and\n\n$$\n\\sum_{\\ell=1}^{2 t+1} x_{\\ell} \\leqslant 4 \\sqrt{k(t+1)}+\\sqrt{2 k t} \\leqslant 4 \\sqrt{k(2 t+1)}\n$$\n\nThe last inequality is true for all $t \\geqslant 8$ since\n\n$$\n4 \\sqrt{2 t+1}-\\sqrt{2 t} \\geqslant 3 \\sqrt{2 t}=\\sqrt{18 t} \\geqslant \\sqrt{16 t+16}=4 \\sqrt{t+1} .\n$$']",,True,,, 1879,Algebra,,"Let $\mathbb{R}^{+}$be the set of positive real numbers. Determine all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that, for all positive real numbers $x$ and $y$, $$ f(x+f(x y))+y=f(x) f(y)+1 \tag{*} $$","['A straightforward check shows that $f(x)=x+1$ satisfies (*). We divide the proof of the converse statement into a sequence of steps.\n\nStep 1: $f$ is injective.\n\nPut $x=1$ in (*) and rearrange the terms to get\n\n$$\ny=f(1) f(y)+1-f(1+f(y))\n$$\n\nTherefore, if $f\\left(y_{1}\\right)=f\\left(y_{2}\\right)$, then $y_{1}=y_{2}$.\n\nStep 2: $f$ is (strictly) monotone increasing.\n\nFor any fixed $y \\in \\mathbb{R}^{+}$, the function\n\n$$\ng(x):=f(x+f(x y))=f(x) f(y)+1-y\n$$\n\nis injective by Step 1. Therefore, $x_{1}+f\\left(x_{1} y\\right) \\neq x_{2}+f\\left(x_{2} y\\right)$ for all $y, x_{1}, x_{2} \\in \\mathbb{R}^{+}$with $x_{1} \\neq x_{2}$. Plugging in $z_{i}=x_{i} y$, we arrive at\n\n$$\n\\frac{z_{1}-z_{2}}{y} \\neq f\\left(z_{2}\\right)-f\\left(z_{1}\\right), \\quad \\text { or } \\quad \\frac{1}{y} \\neq \\frac{f\\left(z_{2}\\right)-f\\left(z_{1}\\right)}{z_{1}-z_{2}}\n$$\n\nfor all $y, z_{1}, z_{2} \\in \\mathbb{R}^{+}$with $z_{1} \\neq z_{2}$. This means that the right-hand side of the rightmost relation is always non-positive, i.e., $f$ is monotone non-decreasing. Since $f$ is injective, it is strictly monotone.\n\nStep 3: There exist constants $a$ and $b$ such that $f(y)=a y+b$ for all $y \\in \\mathbb{R}^{+}$.\n\nSince $f$ is monotone and bounded from below by 0 , for each $x_{0} \\geqslant 0$, there exists a right limit $\\lim _{x \\searrow x_{0}} f(x) \\geqslant 0$. Put $p=\\lim _{x \\searrow 0} f(x)$ and $q=\\lim _{x \\searrow p} f(x)$.\n\nFix an arbitrary $y$ and take the limit of $(*)$ as $x \\searrow 0$. We have $f(x y) \\searrow p$ and hence $f(x+f(x y)) \\searrow q$; therefore, we obtain\n\n$$\nq+y=p f(y)+1, \\quad \\text { or } \\quad f(y)=\\frac{q+y-1}{p}\n$$\n\n(Notice that $p \\neq 0$, otherwise $q+y=1$ for all $y$, which is absurd.) The claim is proved.\n\nStep 4: $f(x)=x+1$ for all $x \\in \\mathbb{R}^{+}$.\n\nBased on the previous step, write $f(x)=a x+b$. Putting this relation into (*) we get\n\n$$\na(x+a x y+b)+b+y=(a x+b)(a y+b)+1,\n$$\n\nwhich can be rewritten as\n\n$$\n(a-a b) x+(1-a b) y+a b+b-b^{2}-1=0 \\quad \\text { for all } x, y \\in \\mathbb{R}^{+}\n$$\n\nThis identity may hold only if all the coefficients are 0 , i.e.,\n\n$$\na-a b=1-a b=a b+b-b^{2}-1=0 .\n$$\n\nHence, $a=b=1$.', 'We provide another proof that $f(x)=x+1$ is the only function satisfying $(*)$.\n\nPut $a=f(1)$. Define the function $\\phi: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}$ by\n\n$$\n\\phi(x)=f(x)-x-1\n$$\n\nThen equation $(*)$ reads as\n\n$$\n\\phi(x+f(x y))=f(x) f(y)-f(x y)-x-y .\n\\tag{1}\n$$\n\nSince the right-hand side of (1) is symmetric under swapping $x$ and $y$, we obtain\n\n$$\n\\phi(x+f(x y))=\\phi(y+f(x y))\n$$\n\nIn particular, substituting $(x, y)=(t, 1 / t)$ we get\n\n$$\n\\phi(a+t)=\\phi\\left(a+\\frac{1}{t}\\right), \\quad t \\in \\mathbb{R}^{+}\n\\tag{2}\n$$\n\nNotice that the function $f$ is bounded from below by a positive constant. Indeed, for each $y \\in \\mathbb{R}^{+}$, the relation $(*)$ yields $f(x) f(y)>y-1$, hence\n\n$$\nf(x)>\\frac{y-1}{f(y)} \\quad \\text { for all } x \\in \\mathbb{R}^{+}\n$$\n\nIf $y>1$, this provides a desired positive lower bound for $f(x)$.\n\nNow, let $M=\\inf _{x \\in \\mathbb{R}^{+}} f(x)>0$. Then, for all $y \\in \\mathbb{R}^{+}$,\n\n$$\nM \\geqslant \\frac{y-1}{f(y)}, \\quad \\text { or } \\quad f(y) \\geqslant \\frac{y-1}{M}\n\\tag{3}\n$$\n\nLemma 1. The function $f(x)$ (and hence $\\phi(x)$ ) is bounded on any segment $[p, q]$, where $0\\max \\left\\{C, \\frac{C}{y}\\right\\}$. Then all three numbers $x, x y$, and $x+f(x y)$ are greater than $C$, so $(*)$ reads as\n\n$$\n(x+x y+1)+1+y=(x+1) f(y)+1, \\text { hence } f(y)=y+1\n$$']",['$f(x)=x+1$'],False,,Expression, 1880,Combinatorics,,"Let $n$ be a positive integer. Find the number of permutations $a_{1}, a_{2}, \ldots, a_{n}$ of the sequence $1,2, \ldots, n$ satisfying $$ a_{1} \leqslant 2 a_{2} \leqslant 3 a_{3} \leqslant \ldots \leqslant n a_{n} \tag{*} $$","['Denote by $P_{n}$ the number of permutations that satisfy $(*)$. It is easy to see that $P_{1}=1$ and $P_{2}=2$.\n\nLemma 1. Let $n \\geqslant 3$. If a permutation $a_{1}, \\ldots, a_{n}$ satisfies $(*)$ then either $a_{n}=n$, or $a_{n-1}=n$ and $a_{n}=n-1$.\n\nProof. Let $k$ be the index for which $a_{k}=n$. If $k=n$ then we are done.\n\nIf $k=n-1$ then, by $(*)$, we have $n(n-1)=(n-1) a_{n-1} \\leqslant n a_{n}$, so $a_{n} \\geqslant n-1$. Since $a_{n} \\neq a_{n-1}=n$, the only choice for $a_{n}$ is $a_{n}=n-1$.\n\nNow suppose that $k \\leqslant n-2$. For every $kn k$, so $a_{n} \\geqslant k+1$. Now the $n-k+1$ numbers $a_{k}, a_{k+1}, \\ldots, a_{n}$ are all greater than $k$; but there are only $n-k$ such values; this is not possible.\n\nIf $a_{n}=n$ then $a_{1}, a_{2}, \\ldots, a_{n-1}$ must be a permutation of the numbers $1, \\ldots, n-1$ satisfying $a_{1} \\leqslant 2 a_{2} \\leqslant \\ldots \\leqslant(n-1) a_{n-1}$; there are $P_{n-1}$ such permutations. The last inequality in $(*)$, $(n-1) a_{n-1} \\leqslant n a_{n}=n^{2}$, holds true automatically.\n\nIf $\\left(a_{n-1}, a_{n}\\right)=(n, n-1)$, then $a_{1}, \\ldots, a_{n-2}$ must be a permutation of $1, \\ldots, n-2$ satisfying $a_{1} \\leqslant \\ldots \\leqslant(n-2) a_{n-2}$; there are $P_{n-2}$ such permutations. The last two inequalities in (*) hold true automatically by $(n-2) a_{n-2} \\leqslant(n-2)^{2}k$. If $t=k$ then we are done, so assume $t>k$.\n\nNotice that one of the numbers among the $t-k$ numbers $a_{k}, a_{k+1}, \\ldots, a_{t-1}$ is at least $t$, because there are only $t-k-1$ values between $k$ and $t$. Let $i$ be an index with $k \\leqslant ik+1$. Then the chain of inequalities $k t=k a_{k} \\leqslant \\ldots \\leqslant t a_{t}=k t$ should also turn into a chain of equalities. From this point we can find contradictions in several ways; for example by pointing to $a_{t-1}=\\frac{k t}{t-1}=k+\\frac{k}{t-1}$ which cannot be an integer, or considering\n\n\n\nthe product of the numbers $(k+1) a_{k+1}, \\ldots,(t-1) a_{t-1}$; the numbers $a_{k+1}, \\ldots, a_{t-1}$ are distinct and greater than $k$, so\n\n$$\n(k t)^{t-k-1}=(k+1) a_{k+1} \\cdot(k+2) a_{k+2} \\cdot \\ldots \\cdot(t-1) a_{t-1} \\geqslant((k+1)(k+2) \\cdot \\ldots \\cdot(t-1))^{2}\n$$\n\nNotice that $(k+i)(t-i)=k t+i(t-k-i)>k t$ for $1 \\leqslant i(k t)^{t-k-1}\n$$\n\nTherefore, the case $t>k+1$ is not possible.']","['The number of such permutations is $F_{n+1}$, where $F_{k}$ is the $k^{\\text {th }}$ Fibonacci number: $F_{1}=F_{2}=1, F_{n+1}=F_{n}+F_{n-1}$']",True,,Need_human_evaluate, 1881,Combinatorics,,"In a regular 100-gon, 41 vertices are colored black and the remaining 59 vertices are colored white. Prove that there exist 24 convex quadrilaterals $Q_{1}, \ldots, Q_{24}$ whose corners are vertices of the 100 -gon, so that - the quadrilaterals $Q_{1}, \ldots, Q_{24}$ are pairwise disjoint, and - every quadrilateral $Q_{i}$ has three corners of one color and one corner of the other color.","['Call a quadrilateral skew-colored, if it has three corners of one color and one corner of the other color. We will prove the following\n\nClaim. If the vertices of a convex $(4 k+1)$-gon $P$ are colored black and white such that each color is used at least $k$ times, then there exist $k$ pairwise disjoint skew-colored quadrilaterals whose vertices are vertices of $P$. (One vertex of $P$ remains unused.)\n\nThe problem statement follows by removing 3 arbitrary vertices of the 100-gon and applying the Claim to the remaining 97 vertices with $k=24$.\n\nProof of the Claim. We prove by induction. For $k=1$ we have a pentagon with at least one black and at least one white vertex. If the number of black vertices is even then remove a black vertex; otherwise remove a white vertex. In the remaining quadrilateral, there are an odd number of black and an odd number of white vertices, so the quadrilateral is skew-colored.\n\nFor the induction step, assume $k \\geqslant 2$. Let $b$ and $w$ be the numbers of black and white vertices, respectively; then $b, w \\geqslant k$ and $b+w=4 k+1$. Without loss of generality we may assume $w \\geqslant b$, so $k \\leqslant b \\leqslant 2 k$ and $2 k+1 \\leqslant w \\leqslant 3 k+1$.\n\nWe want to find four consecutive vertices such that three of them are white, the fourth one is black. Denote the vertices by $V_{1}, V_{2}, \\ldots, V_{4 k+1}$ in counterclockwise order, such that $V_{4 k+1}$ is black, and consider the following $k$ groups of vertices:\n\n$$\n\\left(V_{1}, V_{2}, V_{3}, V_{4}\\right),\\left(V_{5}, V_{6}, V_{7}, V_{8}\\right), \\ldots,\\left(V_{4 k-3}, V_{4 k-2}, V_{4 k-1}, V_{4 k}\\right)\n$$\n\nIn these groups there are $w$ white and $b-1$ black vertices. Since $w>b-1$, there is a group, $\\left(V_{i}, V_{i+1}, V_{i+2}, V_{i+3}\\right)$ that contains more white than black vertices. If three are white and one is black in that group, we are done. Otherwise, if $V_{i}, V_{i+1}, V_{i+2}, V_{i+3}$ are all white then let $V_{j}$ be the first black vertex among $V_{i+4}, \\ldots, V_{4 k+1}$ (recall that $V_{4 k+1}$ is black); then $V_{j-3}, V_{j-2}$ and $V_{j-1}$ are white and $V_{j}$ is black.\n\nNow we have four consecutive vertices $V_{i}, V_{i+1}, V_{i+2}, V_{i+3}$ that form a skew-colored quadrilateral. The remaining vertices form a convex $(4 k-3)$-gon; $w-3$ of them are white and $b-1$ are black. Since $b-1 \\geqslant k-1$ and $w-3 \\geqslant(2 k+1)-3>k-1$, we can apply the Claim with $k-1$.']",,True,,, 1882,Combinatorics,,"Let $n$ be an integer with $n \geqslant 2$. On a slope of a mountain, $n^{2}$ checkpoints are marked, numbered from 1 to $n^{2}$ from the bottom to the top. Each of two cable car companies, $A$ and $B$, operates $k$ cable cars numbered from 1 to $k$; each cable car provides a transfer from some checkpoint to a higher one. For each company, and for any $i$ and $j$ with $1 \leqslant im_{p-1}$.\n\nChoose $k$ to be the smallest index satisfying $m_{k}>m_{p-1}$; by our assumptions, we have $1\\left|R_{2}\\right|$. We will find a saddle subpair $\\left(R^{\\prime}, C^{\\prime}\\right)$ of $\\left(R_{1}, C_{1}\\right)$ with $\\left|R^{\\prime}\\right| \\leqslant\\left|R_{2}\\right|$; clearly, this implies the desired statement.\n\nStep 1: We construct maps $\\rho: R_{1} \\rightarrow R_{1}$ and $\\sigma: C_{1} \\rightarrow C_{1}$ such that $\\left|\\rho\\left(R_{1}\\right)\\right| \\leqslant\\left|R_{2}\\right|$, and $a\\left(\\rho\\left(r_{1}\\right), c_{1}\\right) \\geqslant a\\left(r_{1}, \\sigma\\left(c_{1}\\right)\\right)$ for all $r_{1} \\in R_{1}$ and $c_{1} \\in C_{1}$.\n\nSince $\\left(R_{1}, C_{1}\\right)$ is a saddle pair, for each $r_{2} \\in R_{2}$ there is $r_{1} \\in R_{1}$ such that $a\\left(r_{1}, c_{1}\\right) \\geqslant a\\left(r_{2}, c_{1}\\right)$ for all $c_{1} \\in C_{1}$; denote one such an $r_{1}$ by $\\rho_{1}\\left(r_{2}\\right)$. Similarly, we define four functions\n\n$$\n\\begin{array}{llllll}\n\\rho_{1}: R_{2} \\rightarrow R_{1} & \\text { such that } & a\\left(\\rho_{1}\\left(r_{2}\\right), c_{1}\\right) \\geqslant a\\left(r_{2}, c_{1}\\right) & \\text { for all } & r_{2} \\in R_{2}, & c_{1} \\in C_{1} ; \\\\\n\\rho_{2}: R_{1} \\rightarrow R_{2} & \\text { such that } & a\\left(\\rho_{2}\\left(r_{1}\\right), c_{2}\\right) \\geqslant a\\left(r_{1}, c_{2}\\right) & \\text { for all } & r_{1} \\in R_{1}, & c_{2} \\in C_{2} ; \\\\\n\\sigma_{1}: C_{2} \\rightarrow C_{1} & \\text { such that } & a\\left(r_{1}, \\sigma_{1}\\left(c_{2}\\right)\\right) \\leqslant a\\left(r_{1}, c_{2}\\right) & \\text { for all } & r_{1} \\in R_{1}, & c_{2} \\in C_{2} ; \\\\\n\\sigma_{2}: C_{1} \\rightarrow C_{2} & \\text { such that } & a\\left(r_{2}, \\sigma_{2}\\left(c_{1}\\right)\\right) \\leqslant a\\left(r_{2}, c_{1}\\right) & \\text { for all } & r_{2} \\in R_{2}, & c_{1} \\in C_{1} .\n\\end{array}\n\\tag{1}\n$$\n\nSet now $\\rho=\\rho_{1} \\circ \\rho_{2}: R_{1} \\rightarrow R_{1}$ and $\\sigma=\\sigma_{1} \\circ \\sigma_{2}: C_{1} \\rightarrow C_{1}$. We have\n\n$$\n\\left|\\rho\\left(R_{1}\\right)\\right|=\\left|\\rho_{1}\\left(\\rho_{2}\\left(R_{1}\\right)\\right)\\right| \\leqslant\\left|\\rho_{1}\\left(R_{2}\\right)\\right| \\leqslant\\left|R_{2}\\right| .\n$$\n\nMoreover, for all $r_{1} \\in R_{1}$ and $c_{1} \\in C_{1}$, we get\n\n$$\n\\begin{aligned}\n& a\\left(\\rho\\left(r_{1}\\right), c_{1}\\right)=a\\left(\\rho_{1}\\left(\\rho_{2}\\left(r_{1}\\right)\\right), c_{1}\\right) \\geqslant a\\left(\\rho_{2}\\left(r_{1}\\right), c_{1}\\right) \\geqslant a\\left(\\rho_{2}\\left(r_{1}\\right), \\sigma_{2}\\left(c_{1}\\right)\\right) \\\\\n& \\geqslant a\\left(r_{1}, \\sigma_{2}\\left(c_{1}\\right)\\right) \\geqslant a\\left(r_{1}, \\sigma_{1}\\left(\\sigma_{2}\\left(c_{1}\\right)\\right)\\right)=a\\left(r_{1}, \\sigma\\left(c_{1}\\right)\\right)\n\\end{aligned}\n\\tag{2}\n$$\n\nas desired.\n\nStep 2: Given maps $\\rho$ and $\\sigma$, we construct a proper saddle subpair $\\left(R^{\\prime}, C^{\\prime}\\right)$ of $\\left(R_{1}, C_{1}\\right)$.\n\nThe properties of $\\rho$ and $\\sigma$ yield that\n\n$$\na\\left(\\rho^{i}\\left(r_{1}\\right), c_{1}\\right) \\geqslant a\\left(\\rho^{i-1}\\left(r_{1}\\right), \\sigma\\left(c_{1}\\right)\\right) \\geqslant \\ldots \\geqslant a\\left(r_{1}, \\sigma^{i}\\left(c_{1}\\right)\\right)\n$$\n\nfor each positive integer $i$ and all $r_{1} \\in R_{1}, c_{1} \\in C_{1}$.\n\nConsider the images $R^{i}=\\rho^{i}\\left(R_{1}\\right)$ and $C^{i}=\\sigma^{i}\\left(C_{1}\\right)$. Clearly, $R_{1}=R^{0} \\supseteq R^{1} \\supseteq R^{2} \\supseteq \\ldots$ and $C_{1}=C^{0} \\supseteq C^{1} \\supseteq C^{2} \\supseteq \\ldots$. Since both chains consist of finite sets, there is an index $n$ such that $R^{n}=R^{n+1}=\\ldots$ and $C^{n}=C^{n+1}=\\ldots$ Then $\\rho^{n}\\left(R^{n}\\right)=R^{2 n}=R^{n}$, so $\\rho^{n}$ restricted to $R^{n}$ is a bijection. Similarly, $\\sigma^{n}$ restricted to $C^{n}$ is a bijection from $C^{n}$ to itself. Therefore, there exists a positive integer $k$ such that $\\rho^{n k}$ acts identically on $R^{n}$, and $\\sigma^{n k}$ acts identically on $C^{n}$.\n\nWe claim now that $\\left(R^{n}, C^{n}\\right)$ is a saddle subpair of $\\left(R_{1}, C_{1}\\right)$, with $\\left|R^{n}\\right| \\leqslant\\left|R^{1}\\right|=\\left|\\rho\\left(R_{1}\\right)\\right| \\leqslant$ $\\left|R_{2}\\right|$, which is what we needed. To check that this is a saddle pair, take any row $r^{\\prime}$; since $\\left(R_{1}, C_{1}\\right)$ is a saddle pair, there exists $r_{1} \\in R_{1}$ such that $a\\left(r_{1}, c_{1}\\right) \\geqslant a\\left(r^{\\prime}, c_{1}\\right)$ for all $c_{1} \\in C_{1}$. Set now $r_{*}=\\rho^{n k}\\left(r_{1}\\right) \\in R^{n}$. Then, for each $c \\in C^{n}$ we have $c=\\sigma^{n k}(c)$ and hence\n\n$$\na\\left(r_{*}, c\\right)=a\\left(\\rho^{n k}\\left(r_{1}\\right), c\\right) \\geqslant a\\left(r_{1}, \\sigma^{n k}(c)\\right)=a\\left(r_{1}, c\\right) \\geqslant a\\left(r^{\\prime}, c\\right)\n$$\n\nwhich establishes condition $(i)$. Condition (ii) is checked similarly.']",,True,,, 1887,Combinatorics,,"Players $A$ and $B$ play a game on a blackboard that initially contains 2020 copies of the number 1. In every round, player $A$ erases two numbers $x$ and $y$ from the blackboard, and then player $B$ writes one of the numbers $x+y$ and $|x-y|$ on the blackboard. The game terminates as soon as, at the end of some round, one of the following holds: (1) one of the numbers on the blackboard is larger than the sum of all other numbers; (2) there are only zeros on the blackboard. Player $B$ must then give as many cookies to player $A$ as there are numbers on the blackboard. Player $A$ wants to get as many cookies as possible, whereas player $B$ wants to give as few as possible. Determine the number of cookies that $A$ receives if both players play optimally.","[""For a positive integer $n$, we denote by $S_{2}(n)$ the sum of digits in its binary representation. We prove that, in fact, if a board initially contains an even number $n>1$ of ones, then $A$ can guarantee to obtain $S_{2}(n)$, but not more, cookies. The binary representation of 2020 is $2020=\\overline{11111100100}_{2}$, so $S_{2}(2020)=7$, and the answer follows.\n\nA strategy for A. At any round, while possible, A chooses two equal nonzero numbers on the board. Clearly, while $A$ can make such choice, the game does not terminate. On the other hand, $A$ can follow this strategy unless the game has already terminated. Indeed, if $A$ always chooses two equal numbers, then each number appearing on the board is either 0 or a power of 2 with non-negative integer exponent, this can be easily proved using induction on the number of rounds. At the moment when $A$ is unable to follow the strategy all nonzero numbers on the board are distinct powers of 2 . If the board contains at least one such power, then the largest of those powers is greater than the sum of the others. Otherwise there are only zeros on the blackboard, in both cases the game terminates.\n\nFor every number on the board, define its range to be the number of ones it is obtained from. We can prove by induction on the number of rounds that for any nonzero number $k$ written by $B$ its range is $k$, and for any zero written by $B$ its range is a power of 2 . Thus at the end of each round all the ranges are powers of two, and their sum is $n$. Since $S_{2}(a+b) \\leqslant S_{2}(a)+S_{2}(b)$ for any positive integers $a$ and $b$, the number $n$ cannot be represented as a sum of less than $S_{2}(n)$ powers of 2 . Thus at the end of each round the board contains at least $S_{2}(n)$ numbers, while $A$ follows the above strategy. So $A$ can guarantee at least $S_{2}(n)$ cookies for himself.\n\nA strategy for $B$. Denote $s=S_{2}(n)$.\n\nLet $x_{1}, \\ldots, x_{k}$ be the numbers on the board at some moment of the game after $B$ 's turn or at the beginning of the game. Say that a collection of $k \\operatorname{signs} \\varepsilon_{1}, \\ldots, \\varepsilon_{k} \\in\\{+1,-1\\}$ is balanced if\n\n$$\n\\sum_{i=1}^{k} \\varepsilon_{i} x_{i}=0\n$$\n\nWe say that a situation on the board is good if $2^{s+1}$ does not divide the number of balanced collections. An appropriate strategy for $B$ can be explained as follows: Perform a move so that the situation remains good, while it is possible. We intend to show that in this case $B$ will not lose more than $S_{2}(n)$ cookies. For this purpose, we prove several lemmas.\n\nFor a positive integer $k$, denote by $\\nu_{2}(k)$ the exponent of the largest power of 2 that divides $k$. Recall that, by Legendre's formula, $\\nu_{2}(n !)=n-S_{2}(n)$ for every positive integer $n$.\n\n\n\nLemma 1. The initial situation is good.\n\nProof. In the initial configuration, the number of balanced collections is equal to $\\left(\\begin{array}{c}n \\\\ n / 2\\end{array}\\right)$. We have\n\n$$\n\\nu_{2}\\left(\\left(\\begin{array}{c}\nn \\\\\nn / 2\n\\end{array}\\right)\\right)=\\nu_{2}(n !)-2 \\nu_{2}((n / 2) !)=\\left(n-S_{2}(n)\\right)-2\\left(\\frac{n}{2}-S_{2}(n / 2)\\right)=S_{2}(n)=s\n$$\n\nHence $2^{s+1}$ does not divide the number of balanced collections, as desired.\n\nLemma 2. B may play so that after each round the situation remains good.\n\nProof. Assume that the situation $\\left(x_{1}, \\ldots, x_{k}\\right)$ before a round is good, and that $A$ erases two numbers, $x_{p}$ and $x_{q}$.\n\nLet $N$ be the number of all balanced collections, $N_{+}$be the number of those having $\\varepsilon_{p}=\\varepsilon_{q}$, and $N_{-}$be the number of other balanced collections. Then $N=N_{+}+N_{-}$. Now, if $B$ replaces $x_{p}$ and $x_{q}$ by $x_{p}+x_{q}$, then the number of balanced collections will become $N_{+}$. If $B$ replaces $x_{p}$ and $x_{q}$ by $\\left|x_{p}-x_{q}\\right|$, then this number will become $N_{-}$. Since $2^{s+1}$ does not divide $N$, it does not divide one of the summands $N_{+}$and $N_{-}$, hence $B$ can reach a good situation after the round.\n\nLemma 3. Assume that the game terminates at a good situation. Then the board contains at most $s$ numbers.\n\nProof. Suppose, one of the numbers is greater than the sum of the other numbers. Then the number of balanced collections is 0 and hence divisible by $2^{s+1}$. Therefore, the situation is not good.\n\nThen we have only zeros on the blackboard at the moment when the game terminates. If there are $k$ of them, then the number of balanced collections is $2^{k}$. Since the situation is good, we have $k \\leqslant s$.\n\nBy Lemmas 1 and 2, $B$ may act in such way that they keep the situation good. By Lemma 3, when the game terminates, the board contains at most $s$ numbers. This is what we aimed to prove.""]",['7'],False,,Numerical, 1888,Geometry,,"Let $A B C$ be an isosceles triangle with $B C=C A$, and let $D$ be a point inside side $A B$ such that $A D\n\nFinally, both $O$ and $M$ lie on lines $\\ell$ and $C M$, therefore $O=M$, and $\\angle A C B=90^{\\circ}$ follows.', 'Like in the first solution, we conclude that points $C, P, Q, D, F$ and the midpoint $M$ of $A B$ lie on one circle $\\omega$ with diameter $C D$, and $M$ lies on $\\ell$, the perpendicular bisector of $P Q$.\n\nLet $B F$ and $C M$ meet at $G$ and let $\\alpha=\\angle A B F$. Then, since $E$ lies on $\\ell$, and the quadrilaterals $F C B A$ and $F C P Q$ are cyclic, we have\n\n$$\n\\angle C Q P=\\angle F P Q=\\angle F C Q=\\angle F C A=\\angle F B A=\\alpha\n$$\n\nSince points $P, E, F$ are collinear, we have\n\n$$\n\\angle F E M=\\angle F E Q+\\angle Q E M=2 \\alpha+\\left(90^{\\circ}-\\alpha\\right)=90^{\\circ}+\\alpha\n$$\n\nBut $\\angle F G M=90^{\\circ}+\\alpha$, so $F E G M$ is cyclic. Hence\n\n$$\n\\angle E G C=\\angle E F M=\\angle P F M=\\angle P C M .\n$$\n\nThus $G E \\| B C$. It follows that $\\angle F A C=\\angle C B F=\\angle E G F$, so $F E G A$ is cyclic, too. Hence $\\angle A C B=\\angle A F B=\\angle A F G=180^{\\circ}-\\angle A M G=90^{\\circ}$, that completes the proof.\n\n\n\n']",,True,,, 1889,Geometry,,"Let $A B C D$ be a convex quadrilateral. Suppose that $P$ is a point in the interior of $A B C D$ such that $$ \angle P A D: \angle P B A: \angle D P A=1: 2: 3=\angle C B P: \angle B A P: \angle B P C . $$ The internal bisectors of angles $A D P$ and $P C B$ meet at a point $Q$ inside the triangle $A B P$. Prove that $A Q=B Q$.","['Let $\\varphi=\\angle P A D$ and $\\psi=\\angle C B P$; then we have $\\angle P B A=2 \\varphi, \\angle D P A=3 \\varphi$, $\\angle B A P=2 \\psi$ and $\\angle B P C=3 \\psi$. Let $X$ be the point on segment $A D$ with $\\angle X P A=\\varphi$. Then\n\n$$\n\\angle P X D=\\angle P A X+\\angle X P A=2 \\varphi=\\angle D P A-\\angle X P A=\\angle D P X\n$$\n\nIt follows that triangle $D P X$ is isosceles with $D X=D P$ and therefore the internal angle bisector of $\\angle A D P$ coincides with the perpendicular bisector of $X P$. Similarly, if $Y$ is a point on $B C$ such that $\\angle B P Y=\\psi$, then the internal angle bisector of $\\angle P C B$ coincides with the perpendicular bisector of $P Y$. Hence, we have to prove that the perpendicular bisectors of $X P$, $P Y$, and $A B$ are concurrent.\n\n\n\nNotice that\n\n$$\n\\angle A X P=180^{\\circ}-\\angle P X D=180^{\\circ}-2 \\varphi=180^{\\circ}-\\angle P B A .\n$$\n\nHence the quadrilateral $A X P B$ is cyclic; in other words, $X$ lies on the circumcircle of triangle $A P B$. Similarly, $Y$ lies on the circumcircle of triangle $A P B$. It follows that the perpendicular bisectors of $X P, P Y$, and $A B$ all pass through the center of circle $(A B Y P X)$. This finishes the proof.', 'We define the angles $\\varphi=\\angle P A D, \\psi=\\angle C B P$ and use $\\angle P B A=2 \\varphi, \\angle D P A=$ $3 \\varphi, \\angle B A P=2 \\psi$ and $\\angle B P C=3 \\psi$ again. Let $O$ be the circumcenter of $\\triangle A P B$.\n\nNotice that $\\angle A D P=180^{\\circ}-\\angle P A D-\\angle D P A=180^{\\circ}-4 \\varphi$, which, in particular, means that $4 \\varphi<180^{\\circ}$. Further, $\\angle P O A=2 \\angle P B A=4 \\varphi=180^{\\circ}-\\angle A D P$, therefore the quadrilateral $A D P O$ is cyclic. By $A O=O P$, it follows that $\\angle A D O=\\angle O D P$. Thus $D O$ is the internal bisector of $\\angle A D P$. Similarly, $C O$ is the internal bisector of $\\angle P C B$.\n\n\n\n\n\nFinally, $O$ lies on the perpendicular bisector of $A B$ as it is the circumcenter of $\\triangle A P B$. Therefore the three given lines in the problem statement concur at point $O$.']",,True,,, 1890,Geometry,,"Let $A B C D$ be a convex quadrilateral with $\angle A B C>90^{\circ}, \angle C D A>90^{\circ}$, and $\angle D A B=\angle B C D$. Denote by $E$ and $F$ the reflections of $A$ in lines $B C$ and $C D$, respectively. Suppose that the segments $A E$ and $A F$ meet the line $B D$ at $K$ and $L$, respectively. Prove that the circumcircles of triangles $B E K$ and $D F L$ are tangent to each other.","['Denote by $A^{\\prime}$ the reflection of $A$ in $B D$. We will show that that the quadrilaterals $A^{\\prime} B K E$ and $A^{\\prime} D L F$ are cyclic, and their circumcircles are tangent to each other at point $A^{\\prime}$.\n\nFrom the symmetry about line $B C$ we have $\\angle B E K=\\angle B A K$, while from the symmetry in $B D$ we have $\\angle B A K=\\angle B A^{\\prime} K$. Hence $\\angle B E K=\\angle B A^{\\prime} K$, which implies that the quadrilateral $A^{\\prime} B K E$ is cyclic. Similarly, the quadrilateral $A^{\\prime} D L F$ is also cyclic.\n\n\n\nFor showing that circles $A^{\\prime} B K E$ and $A^{\\prime} D L F$ are tangent it suffices to prove that\n\n$$\n\\angle A^{\\prime} K B+\\angle A^{\\prime} L D=\\angle B A^{\\prime} D .\n$$\n\nIndeed, by $A K \\perp B C, A L \\perp C D$, and again the symmetry in $B D$ we have\n\n$$\n\\angle A^{\\prime} K B+\\angle A^{\\prime} L D=180^{\\circ}-\\angle K A^{\\prime} L=180^{\\circ}-\\angle K A L=\\angle B C D=\\angle B A D=\\angle B A^{\\prime} D,\n$$\n\nas required.', 'Note that $\\angle K A L=180^{\\circ}-\\angle B C D$, since $A K$ and $A L$ are perpendicular to $B C$ and $C D$, respectively. Reflect both circles $(B E K)$ and $(D F L)$ in $B D$. Since $\\angle K E B=\\angle K A B$ and $\\angle D F L=\\angle D A L$, the images are the circles $(K A B)$ and $(L A D)$, respectively; so they meet at $A$. We need to prove that those two reflections are tangent at $A$.\n\nFor this purpose, we observe that\n\n$$\n\\angle A K B+\\angle A L D=180^{\\circ}-\\angle K A L=\\angle B C D=\\angle B A D .\n$$\n\nThus, there exists a ray $A P$ inside angle $\\angle B A D$ such that $\\angle B A P=\\angle A K B$ and $\\angle D A P=$ $\\angle D L A$. Hence the line $A P$ is a common tangent to the circles $(K A B)$ and $(L A D)$, as desired.']",,True,,, 1891,Geometry,,"In the plane, there are $n \geqslant 6$ pairwise disjoint disks $D_{1}, D_{2}, \ldots, D_{n}$ with radii $R_{1} \geqslant R_{2} \geqslant \ldots \geqslant R_{n}$. For every $i=1,2, \ldots, n$, a point $P_{i}$ is chosen in disk $D_{i}$. Let $O$ be an arbitrary point in the plane. Prove that $$ O P_{1}+O P_{2}+\ldots+O P_{n} \geqslant R_{6}+R_{7}+\ldots+R_{n} $$","['We will make use of the following lemma.\n\nLemma. Let $D_{1}, \\ldots, D_{6}$ be disjoint disks in the plane with radii $R_{1}, \\ldots, R_{6}$. Let $P_{i}$ be a point in $D_{i}$, and let $O$ be an arbitrary point. Then there exist indices $i$ and $j$ such that $O P_{i} \\geqslant R_{j}$.\n\nProof. Let $O_{i}$ be the center of $D_{i}$. Consider six rays $O O_{1}, \\ldots, O O_{6}$ (if $O=O_{i}$, then the ray $O O_{i}$ may be assumed to have an arbitrary direction). These rays partition the plane into six angles (one of which may be non-convex) whose measures sum up to $360^{\\circ}$; hence one of the angles, say $\\angle O_{i} O O_{j}$, has measure at most $60^{\\circ}$. Then $O_{i} O_{j}$ cannot be the unique largest side in (possibly degenerate) triangle $O O_{i} O_{j}$, so, without loss of generality, $O O_{i} \\geqslant O_{i} O_{j} \\geqslant R_{i}+R_{j}$. Therefore, $O P_{i} \\geqslant O O_{i}-R_{i} \\geqslant\\left(R_{i}+R_{j}\\right)-R_{i}=R_{j}$, as desired.\n\nNow we prove the required inequality by induction on $n \\geqslant 5$. The base case $n=5$ is trivial. For the inductive step, apply the Lemma to the six largest disks, in order to find indices $i$ and $j$ such that $1 \\leqslant i, j \\leqslant 6$ and $O P_{i} \\geqslant R_{j} \\geqslant R_{6}$. Removing $D_{i}$ from the configuration and applying the inductive hypothesis, we get\n\n$$\n\\sum_{k \\neq i} O P_{k} \\geqslant \\sum_{\\ell \\geqslant 7} R_{\\ell}\n$$\n\nAdding up this inequality with $O P_{i} \\geqslant R_{6}$ we establish the inductive step.']",,True,,, 1892,Geometry,,"Let $A B C D$ be a cyclic quadrilateral with no two sides parallel. Let $K, L, M$, and $N$ be points lying on sides $A B, B C, C D$, and $D A$, respectively, such that $K L M N$ is a rhombus with $K L \| A C$ and $L M \| B D$. Let $\omega_{1}, \omega_{2}, \omega_{3}$, and $\omega_{4}$ be the incircles of triangles $A N K$, $B K L, C L M$, and $D M N$, respectively. Prove that the internal common tangents to $\omega_{1}$ and $\omega_{3}$ and the internal common tangents to $\omega_{2}$ and $\omega_{4}$ are concurrent.","['Let $I_{i}$ be the center of $\\omega_{i}$, and let $r_{i}$ be its radius for $i=1,2,3,4$. Denote by $T_{1}$ and $T_{3}$ the points of tangency of $\\omega_{1}$ and $\\omega_{3}$ with $N K$ and $L M$, respectively. Suppose that the internal common tangents to $\\omega_{1}$ and $\\omega_{3}$ meet at point $S$, which is the center of homothety $h$ with negative ratio (namely, with ratio $-\\frac{r_{3}}{r_{1}}$ ) mapping $\\omega_{1}$ to $\\omega_{3}$. This homothety takes $T_{1}$ to $T_{3}$ (since the tangents to $\\omega_{1}$ and $\\omega_{3}$ at $T_{1}$ to $T_{3}$ are parallel), hence $S$ is a point on the segment $T_{1} T_{3}$ with $T_{1} S: S T_{3}=r_{1}: r_{3}$.\n\nConstruct segments $S_{1} S_{3} \\| K L$ and $S_{2} S_{4} \\| L M$ through $S$ with $S_{1} \\in N K, S_{2} \\in K L$, $S_{3} \\in L M$, and $S_{4} \\in M N$. Note that $h$ takes $S_{1}$ to $S_{3}$, hence $I_{1} S_{1} \\| I_{3} S_{3}$, and $S_{1} S: S S_{3}=r_{1}: r_{3}$. We will prove that $S_{2} S: S S_{4}=r_{2}: r_{4}$ or, equivalently, $K S_{1}: S_{1} N=r_{2}: r_{4}$. This will yield the problem statement; indeed, applying similar arguments to the intersection point $S^{\\prime}$ of the internal common tangents to $\\omega_{2}$ and $\\omega_{4}$, we see that $S^{\\prime}$ satisfies similar relations, and there is a unique point inside $K L M N$ satisfying them. Therefore, $S^{\\prime}=S$.\n\n\nFurther, denote by $I_{A}, I_{B}, I_{C}, I_{D}$ and $r_{A}, r_{B}, r_{C}, r_{D}$ the incenters and inradii of triangles $D A B, A B C, B C D$, and $C D A$, respectively. One can shift triangle $C L M$ by $\\overrightarrow{L K}$ to glue it with triangle $A K N$ into a quadrilateral $A K C^{\\prime} N$ similar to $A B C D$. In particular, this shows that $r_{1}: r_{3}=r_{A}: r_{C}$; similarly, $r_{2}: r_{4}=r_{B}: r_{D}$. Moreover, the same shift takes $S_{3}$ to $S_{1}$, and it also takes $I_{3}$ to the incenter $I_{3}^{\\prime}$ of triangle $K C^{\\prime} N$. Since $I_{1} S_{1} \\| I_{3} S_{3}$, the points $I_{1}, S_{1}, I_{3}^{\\prime}$ are collinear. Thus, in order to complete the solution, it suffices to apply the following Lemma to quadrilateral $A K C^{\\prime} N$.\n\nLemma 1. Let $A B C D$ be a cyclic quadrilateral, and define $I_{A}, I_{C}, r_{B}$, and $r_{D}$ as above. Let $I_{A} I_{C}$ meet $B D$ at $X$; then $B X: X D=r_{B}: r_{D}$.\n\nProof. Consider an inversion centered at $X$; the images under that inversion will be denoted by primes, e.g., $A^{\\prime}$ is the image of $A$.\n\nBy properties of inversion, we have\n\n$$\n\\angle I_{C}^{\\prime} I_{A}^{\\prime} D^{\\prime}=\\angle X I_{A}^{\\prime} D^{\\prime}=\\angle X D I_{A}=\\angle B D A / 2=\\angle B C A / 2=\\angle A C I_{B} .\n$$\n\nWe obtain $\\angle I_{A}^{\\prime} I_{C}^{\\prime} D^{\\prime}=\\angle C A I_{B}$ likewise; therefore, $\\triangle I_{C}^{\\prime} I_{A}^{\\prime} D^{\\prime} \\sim \\triangle A C I_{B}$. In the same manner, we get $\\triangle I_{C}^{\\prime} I_{A}^{\\prime} B^{\\prime} \\sim \\triangle A C I_{D}$, hence the quadrilaterals $I_{C}^{\\prime} B^{\\prime} I_{A}^{\\prime} D^{\\prime}$ and $A I_{D} C I_{B}$ are also similar. But the diagonals $A C$ and $I_{B} I_{D}$ of quadrilateral $A I_{D} C I_{B}$ meet at a point $Y$ such that $I_{B} Y$ :\n\n\n\n$Y I_{D}=r_{B}: r_{D}$. By similarity, we get $D^{\\prime} X: B^{\\prime} X=r_{B}: r_{D}$ and hence $B X: X D=D^{\\prime} X:$ $B^{\\prime} X=r_{B}: r_{D}$.', ""This solution is based on the following general Lemma.\n\nLemma 2. Let $E$ and $F$ be distinct points, and let $\\omega_{i}, i=1,2,3,4$, be circles lying in the same halfplane with respect to $E F$. For distinct indices $i, j \\in\\{1,2,3,4\\}$, denote by $O_{i j}^{+}$ (respectively, $O_{i j}^{-}$) the center of homothety with positive (respectively, negative) ratio taking $\\omega_{i}$ to $\\omega_{j}$. Suppose that $E=O_{12}^{+}=O_{34}^{+}$and $F=O_{23}^{+}=O_{41}^{+}$. Then $O_{13}^{-}=O_{24}^{-}$.\n\nProof. Applying Monge's theorem to triples of circles $\\omega_{1}, \\omega_{2}, \\omega_{4}$ and $\\omega_{1}, \\omega_{3}, \\omega_{4}$, we get that both points $O_{24}^{-}$and $O_{13}^{-}$lie on line $E O_{14}^{-}$. Notice that this line is distinct from $E F$. Similarly we obtain that both points $O_{24}^{-}$and $O_{13}^{-}$lie on $\\mathrm{FO}_{34}^{-}$. Since the lines $E O_{14}^{-}$and $\\mathrm{FO}_{34}^{-}$are distinct, both points coincide with the meeting point of those lines.\n\n\n\n\nTurning back to the problem, let $A B$ intersect $C D$ at $E$ and let $B C$ intersect $D A$ at $F$. Assume, without loss of generality, that $B$ lies on segments $A E$ and $C F$. We will show that the points $E$ and $F$, and the circles $\\omega_{i}$ satisfy the conditions of Lemma 2, so the problem statement follows. In the sequel, we use the notation of $O_{i j}^{ \\pm}$from the statement of Lemma 2, applied to circles $\\omega_{1}, \\omega_{2}, \\omega_{3}$, and $\\omega_{4}$.\n\nUsing the relations $\\triangle E C A \\sim \\triangle E B D, K N \\| B D$, and $M N \\| A C$. we get\n\n$$\n\\frac{A N}{N D}=\\frac{A N}{A D} \\cdot \\frac{A D}{N D}=\\frac{K N}{B D} \\cdot \\frac{A C}{N M}=\\frac{A C}{B D}=\\frac{A E}{E D}\n$$\n\nTherefore, by the angle bisector theorem, point $N$ lies on the internal angle bisector of $\\angle A E D$. We prove similarly that $L$ also lies on that bisector, and that the points $K$ and $M$ lie on the internal angle bisector of $\\angle A F B$.\n\nSince $K L M N$ is a rhombus, points $K$ and $M$ are symmetric in line $E L N$. Hence, the convex quadrilateral determined by the lines $E K, E M, K L$, and $M L$ is a kite, and therefore it has an incircle $\\omega_{0}$. Applying Monge's theorem to $\\omega_{0}, \\omega_{2}$, and $\\omega_{3}$, we get that $O_{23}^{+}$lies on $K M$. On the other hand, $O_{23}^{+}$lies on $B C$, as $B C$ is an external common tangent to $\\omega_{2}$ and $\\omega_{3}$. It follows that $F=O_{23}^{+}$. Similarly, $E=O_{12}^{+}=O_{34}^{+}$, and $F=O_{41}^{+}$.""]",,True,,, 1893,Geometry,,"Let $I$ and $I_{A}$ be the incenter and the $A$-excenter of an acute-angled triangle $A B C$ with $A B\n\nTo show that $\\omega$ and $\\Omega$ are tangent at $T$, let $\\ell$ be the tangent to $\\omega$ at $T$, so that $\\sphericalangle\\left(T I_{A}, \\ell\\right)=$ $\\sphericalangle\\left(E I_{A}, E T\\right)$. Using circles $(B D E T)$ and $\\left(B I C I_{A}\\right)$, we get\n\n$$\n\\sphericalangle\\left(E I_{A}, E T\\right)=\\sphericalangle(E B, E T)=\\sphericalangle(D B, D T) .\n$$\n\nTherefore,\n\n$$\n\\sphericalangle(T I, \\ell)=90^{\\circ}+\\sphericalangle\\left(T I_{A}, \\ell\\right)=90^{\\circ}+\\sphericalangle(D B, D T)=\\sphericalangle(D I, D T)\n$$\n\nwhich shows that $\\ell$ is tangent to $\\Omega$ at $T$.', 'We use the notation of circles $\\Gamma, \\omega$, and $\\Omega$ as in the previous solution.\n\nLet $L$ be the point opposite to $I$ in circle $\\Omega$. Then $\\angle I A L=\\angle I D L=90^{\\circ}$, which means that $L$ is the foot of the external bisector of $\\angle A$ in triangle $A B C$. Let $L I$ cross $\\Gamma$ again at $M$.\n\n\n\nLet $T$ be the foot of the perpendicular from $I$ onto $I_{A} L$. Then $T$ is the second intersection point of $\\Gamma$ and $\\Gamma$. We will show that $T$ is the desired tangency point.\n\nFirst, we show that $T$ lies on circle $\\omega$. Notice that\n\n$$\n\\sphericalangle(L T, L M)=\\sphericalangle(A T, A I) \\text { and } \\quad \\sphericalangle(M T, M L)=\\sphericalangle(M T, M I)=\\sphericalangle\\left(I_{A} T, I_{A} I\\right) \\text {, }\n$$\n\nwhich shows that triangles $T M L$ and $T I_{A} A$ are similar and equioriented. So there exists a rotational homothety $\\tau$ mapping $T M L$ to $T I_{A} A$.\n\nSince $\\sphericalangle(M L, L D)=\\sphericalangle(A I, A D)$, we get $\\tau(B C)=A D$. Next, since\n\n$$\n\\sphericalangle(M B, M L)=\\sphericalangle(M B, M I)=\\sphericalangle\\left(I_{A} B, I_{A} I\\right)=\\sphericalangle\\left(I_{A} E, I_{A} A\\right)\n$$\n\nwe get $\\tau(B)=E$. Similarly, $\\tau(C)=F$. Since the points $M, B, C$, and $T$ are concyclic, so are their $\\tau$-images, which means that $T$ lies on $\\omega=\\tau(\\Gamma)$.\n\n\n\nFinally, since $\\tau(L)=A$ and $\\tau(B)=E$, triangles $A T L$ and $E T B$ are similar so that\n\n$$\n\\sphericalangle(A T, A L)=\\sphericalangle(E T, E B)=\\sphericalangle\\left(E I_{A}, E T\\right) .\n$$\n\nThis means that the tangents to $\\Omega$ and $\\omega$ at $T$ make the same angle with the line $I_{A} T L$, so the circles are indeed tangent at $T$.', 'We also use the notation of circles $\\omega$, and $\\Omega$ from the previous solutions.\n\nPerform an inversion centered at $D$. The images of the points will be denoted by primes, e.g., $A^{\\prime}$ is the image of $A$.\n\nFor convenience, we use the notation $\\angle B I D=\\beta, \\angle C I D=\\gamma$, and $\\alpha=180^{\\circ}-\\beta-\\gamma=$ $90^{\\circ}-\\angle B A I$. We start with computing angles appearing after inversion. We get\n\n$$\n\\begin{gathered}\n\\angle D B^{\\prime} I^{\\prime}=\\beta, \\quad \\angle D C^{\\prime} I^{\\prime}=\\gamma, \\quad \\text { and hence } \\angle B^{\\prime} I^{\\prime} C^{\\prime}=\\alpha ; \\\\\n\\angle E^{\\prime} I_{A}^{\\prime} F^{\\prime}=\\angle E^{\\prime} I_{A}^{\\prime} D-\\angle F^{\\prime} I_{A}^{\\prime} D=\\angle I_{A} E D-\\angle I_{A} F D=\\angle E I_{A} F=180^{\\circ}-\\alpha .\n\\end{gathered}\n$$\n\nNext, we have\n\n$$\n\\angle A^{\\prime} E^{\\prime} B^{\\prime}=\\angle D E^{\\prime} B^{\\prime}=\\angle D B E=\\beta=90^{\\circ}-\\frac{\\angle D B A}{2}=90^{\\circ}-\\frac{\\angle E^{\\prime} A^{\\prime} B^{\\prime}}{2},\n$$\n\nwhich yields that triangle $A^{\\prime} B^{\\prime} E^{\\prime}$ is isosceles with $A^{\\prime} B^{\\prime}=A^{\\prime} E^{\\prime}$. Similarly, $A^{\\prime} F^{\\prime}=A^{\\prime} C^{\\prime}$.\n\nFinally, we get\n\n$$\n\\begin{aligned}\n\\angle A^{\\prime} B^{\\prime} I^{\\prime}=\\angle I^{\\prime} B^{\\prime} D-\\angle A^{\\prime} B^{\\prime} D=\\beta- & \\angle B A D=\\beta-\\left(90^{\\circ}-\\alpha\\right)+\\angle I A D \\\\\n& =\\angle I C D+\\angle I A D=\\angle C^{\\prime} I^{\\prime} D+\\angle A^{\\prime} I^{\\prime} D=\\angle C^{\\prime} I^{\\prime} A^{\\prime}\n\\end{aligned}\n$$\n\nsimilarly, $\\angle A^{\\prime} C^{\\prime} I^{\\prime}=\\angle A^{\\prime} I^{\\prime} B^{\\prime}$, so that triangles $A^{\\prime} B^{\\prime} I^{\\prime}$ and $A^{\\prime} I^{\\prime} C^{\\prime}$ are similar. Therefore, $A^{\\prime} I^{2}=A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{\\prime}$.\n\nRecall that we need to prove the tangency of line $A^{\\prime} I^{\\prime}=\\Omega^{\\prime}$ with circle $\\left(E^{\\prime} F^{\\prime} I_{A}^{\\prime}\\right)=\\omega^{\\prime}$. A desired tangency point $T^{\\prime}$ must satisfy $A^{\\prime} T^{\\prime 2}=A^{\\prime} E^{\\prime} \\cdot A^{\\prime} F^{\\prime}$; the relations obtained above yield\n\n$$\nA^{\\prime} E^{\\prime} \\cdot A^{\\prime} F^{\\prime}=A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{\\prime}=A^{\\prime} I^{\\prime 2},\n$$\n\nso that $T^{\\prime}$ should be symmetric to $I^{\\prime}$ with respect to $A^{\\prime}$.\n\nThus, let us define a point $T^{\\prime}$ as the reflection of $I^{\\prime}$ in $A^{\\prime}$, and show that $T^{\\prime}$ lies on circle $\\Omega^{\\prime}$; the equalities above will then imply that $A^{\\prime} T^{\\prime}$ is tangent to $\\Omega^{\\prime}$, as desired.\n\n\nThe property that triangles $B^{\\prime} A^{\\prime} I^{\\prime}$ and $I^{\\prime} A^{\\prime} C^{\\prime}$ are similar means that quadrilateral $B^{\\prime} I^{\\prime} C^{\\prime} T^{\\prime}$ is harmonic. Indeed, let $C^{*}$ be the reflection of $C^{\\prime}$ in the perpendicular bisector of $I^{\\prime} T^{\\prime}$; then $C^{*}$ lies on $B^{\\prime} A^{\\prime}$ by $\\angle B^{\\prime} A^{\\prime} I^{\\prime}=\\angle A^{\\prime} I^{\\prime} C^{\\prime}=\\angle T^{\\prime} I^{\\prime} C^{*}$, and then $C^{*}$ lies on circle $\\left(I^{\\prime} B^{\\prime} T^{\\prime}\\right)$ since $A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{*}=A^{\\prime} B^{\\prime} \\cdot A^{\\prime} C^{\\prime}=A^{\\prime} I^{\\prime 2}=A^{\\prime} I^{\\prime} \\cdot A^{\\prime} T^{\\prime}$. Therefore, $C^{\\prime}$ also lies on that circle (and the circle is $\\left.\\left(B^{\\prime} I^{\\prime} C^{\\prime}\\right)=\\Gamma^{\\prime}\\right)$. Moreover, $B^{\\prime} C^{*}$ is a median in triangle $B^{\\prime} I^{\\prime} T^{\\prime}$, so $B^{\\prime} C^{\\prime}$ is its symmedian, which establishes harmonicity.\n\nNow we have $\\angle A^{\\prime} B^{\\prime} T^{\\prime}=\\angle I^{\\prime} B^{\\prime} C^{\\prime}=\\beta=\\angle A^{\\prime} B^{\\prime} E^{\\prime}$; which shows that $E^{\\prime}$ lies on $B^{\\prime} T^{\\prime}$. Similarly, $F^{\\prime}$ lies on $C^{\\prime} T^{\\prime}$. Hence, $\\angle E^{\\prime} T^{\\prime} F^{\\prime}=\\angle B^{\\prime} I^{\\prime} C^{\\prime}=180^{\\circ}-\\angle E^{\\prime} I_{A}^{\\prime} F^{\\prime}$, which establishes $T^{\\prime} \\in \\omega^{\\prime}$.']",,True,,, 1894,Geometry,,"Let $P$ be a point on the circumcircle of an acute-angled triangle $A B C$. Let $D$, $E$, and $F$ be the reflections of $P$ in the midlines of triangle $A B C$ parallel to $B C, C A$, and $A B$, respectively. Denote by $\omega_{A}, \omega_{B}$, and $\omega_{C}$ the circumcircles of triangles $A D P, B E P$, and $C F P$, respectively. Denote by $\omega$ the circumcircle of the triangle formed by the perpendicular bisectors of segments $A D, B E$ and $C F$. Show that $\omega_{A}, \omega_{B}, \omega_{C}$, and $\omega$ have a common point.","[""Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes in triangle $A B C$, and let $m_{A}, m_{B}$, and $m_{C}$ be the midlines parallel to $B C, C A$, and $A B$, respectively. We always denote by $\\sphericalangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\\circ}$.\n\nStep 1: Circles $\\omega_{A}, \\omega_{B}$, and $\\omega_{C}$ share a common point $Q$ different from $P$.\n\nNotice that $m_{A}$ is the perpendicular bisector of $P D$, so $\\omega_{A}$ is symmetric with respect to $m_{A}$. Since $A$ and $A_{1}$ are also symmetric to each other in $m_{A}$, the point $A_{1}$ lies on $\\omega_{A}$. Similarly, $B_{1}$ and $C_{1}$ lie on $\\omega_{B}$ and $\\omega_{C}$, respectively.\n\nLet $H$ be the orthocenter of $\\triangle A B C$. Quadrilaterals $A B A_{1} B_{1}$ and $B C B_{1} C_{1}$ are cyclic, so $A H \\cdot H A_{1}=B H \\cdot H B_{1}=C H \\cdot H C_{1}$. This means that $H$ lies on pairwise radical axes of $\\omega_{A}$, $\\omega_{B}$, and $\\omega_{C}$. Point $P$ also lies on those radical axes; hence the three circles have a common radical axis $\\ell=P H$, and the second meeting point $Q$ of $\\ell$ with $\\omega_{A}$ is the second common point of the three circles. Notice here that $H$ lies inside all three circles, hence $Q \\neq P$.\n\n\nStep 2: Point $Q$ lies on $\\omega$.\n\nLet $p_{A}, p_{B}$, and $p_{C}$ denote the perpendicular bisectors of $A D, B E$, and $C F$, respectively; denote by $\\Delta$ the triangle formed by those perpendicular bisectors. By Simson's theorem, in order to show that $Q$ lies on the circumcircle $\\omega$ of $\\Delta$, it suffices to prove that the projections of $Q$ onto the sidelines $p_{A}, p_{B}$, and $p_{C}$ are collinear. Alternatively, but equivalently, it suffices to prove that the reflections $Q_{A}, Q_{B}$, and $Q_{C}$ of $Q$ in those lines, respectively, are collinear. In fact, we will show that four points $P, Q_{A}, Q_{B}$, and $Q_{C}$ are collinear.\n\nSince $p_{A}$ is the common perpendicular bisector of $A D$ and $Q Q_{A}$, the point $Q_{A}$ lies on $\\omega_{A}$, and, moreover, $\\sphericalangle\\left(D A, D Q_{A}\\right)=\\sphericalangle(A Q, A D)$. Therefore,\n\n$$\n\\sphericalangle\\left(P A, P Q_{A}\\right)=\\sphericalangle\\left(D A, D Q_{A}\\right)=\\sphericalangle(A Q, A D)=\\sphericalangle(P Q, P D)=\\sphericalangle(P Q, B C)+90^{\\circ}\n$$\n\n\n\nSimilarly, we get $\\sphericalangle\\left(P B, P Q_{B}\\right)=\\sphericalangle(P Q, C A)+90^{\\circ}$. Therefore,\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(P Q_{A}, P Q_{B}\\right)=\\sphericalangle\\left(P Q_{A}, P A\\right) & +\\sphericalangle(P A, P B)+\\sphericalangle\\left(P B, P Q_{B}\\right) \\\\\n& =\\sphericalangle(B C, P Q)+90^{\\circ}+\\sphericalangle(C A, C B)+\\sphericalangle(P Q, C A)+90^{\\circ}=0,\n\\end{aligned}\n$$\n\nwhich shows that $P, Q_{A}$, and $Q_{B}$ are collinear. Similarly, $Q_{C}$ also lies on $P Q_{A}$.""]",,True,,, 1895,Geometry,,"Let $\Gamma$ and $I$ be the circumcircle and the incenter of an acute-angled triangle $A B C$. Two circles $\omega_{B}$ and $\omega_{C}$ passing through $B$ and $C$, respectively, are tangent at $I$. Let $\omega_{B}$ meet the shorter arc $A B$ of $\Gamma$ and segment $A B$ again at $P$ and $M$, respectively. Similarly, let $\omega_{C}$ meet the shorter arc $A C$ of $\Gamma$ and segment $A C$ again at $Q$ and $N$, respectively. The rays $P M$ and $Q N$ meet at $X$, and the tangents to $\omega_{B}$ and $\omega_{C}$ at $B$ and $C$, respectively, meet at $Y$. Prove that the points $A, X$, and $Y$ are collinear.","['Let $A I, B I$, and $C I$ meet $\\Gamma$ again at $D, E$, and $F$, respectively. Let $\\ell$ be the common tangent to $\\omega_{B}$ and $\\omega_{C}$ at $I$. We always denote by $\\sphericalangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\\circ}$.\n\nStep 1: We show that $Y$ lies on $\\Gamma$.\n\nRecall that any chord of a circle makes complementary directed angles with the tangents to the circle at its endpoints. Hence,\n\n$$\n\\sphericalangle(B Y, B I)+\\sphericalangle(C I, C Y)=\\sphericalangle(I B, \\ell)+\\sphericalangle(\\ell, I C)=\\sphericalangle(I B, I C) .\n$$\n\nTherefore,\n\n$$\n\\begin{array}{r}\n\\sphericalangle(B Y, B A)+\\sphericalangle(C A, C Y)=\\sphericalangle(B I, B A)+\\sphericalangle(B Y, B I)+\\sphericalangle(C I, C Y)+\\sphericalangle(C A, C I) \\\\\n=\\sphericalangle(B C, B I)+\\sphericalangle(I B, I C)+\\sphericalangle(C I, C B)=0,\n\\end{array}\n$$\n\nwhich yields $Y \\in \\Gamma$.\n\n\n\nStep 2: We show that $X=\\ell \\cap E F$.\n\nLet $X_{*}=\\ell \\cap E F$. To prove our claim, it suffices to show that $X_{*}$ lies on both $P M$ and $Q N$; this will yield $X_{*}=X$. Due to symmetry, it suffices to show $X_{*} \\in Q N$.\n\nNotice that\n\n$$\n\\sphericalangle\\left(I X_{*}, I Q\\right)=\\sphericalangle(C I, C Q)=\\sphericalangle(C F, C Q)=\\sphericalangle(E F, E Q)=\\sphericalangle\\left(E X_{*}, E Q\\right) ;\n$$\n\n\n\ntherefore, the points $X_{*}, I, Q$, and $E$ are concyclic (if $Q=E$, then the direction of $E Q$ is supposed to be the direction of a tangent to $\\Gamma$ at $Q$; in this case, the equality means that the circle $\\left(X_{*} I Q\\right)$ is tangent to $\\Gamma$ at $\\left.Q\\right)$. Then we have\n\n$$\n\\sphericalangle\\left(Q X_{*}, Q I\\right)=\\sphericalangle\\left(E X_{*}, E I\\right)=\\sphericalangle(E F, E B)=\\sphericalangle(C A, C F)=\\sphericalangle(C N, C I)=\\sphericalangle(Q N, Q I) \\text {, }\n$$\n\nwhich shows that $X_{*} \\in Q N$.\n\nStep 3: We finally show that $A, X$, and $Y$ are collinear.\n\nRecall that $I$ is the orthocenter of triangle $D E F$, and $A$ is symmetric to $I$ with respect to $E F$. Therefore,\n\n$$\n\\sphericalangle(A X, A E)=\\sphericalangle(I E, I X)=\\sphericalangle(B I, \\ell)=\\sphericalangle(B Y, B I)=\\sphericalangle(B Y, B E)=\\sphericalangle(A Y, A E) \\text {, }\n$$\n\nwhich yields the desired collinearity.', 'Perform an inversion centered at $I$; the images of the points are denoted by primes, e.g., $A^{\\prime}$ is the image of $A$.\n\nOn the inverted figure, $I$ and $\\Gamma^{\\prime}$ are the orthocenter and the circumcircle of triangle $A^{\\prime} B^{\\prime} C^{\\prime}$, respectively. The points $P^{\\prime}$ and $Q^{\\prime}$ lie on $\\Gamma^{\\prime}$ such that $B^{\\prime} P^{\\prime} \\| C^{\\prime} Q^{\\prime}$ (since $B^{\\prime} P^{\\prime}=\\omega_{B}^{\\prime}$ and $C^{\\prime} Q^{\\prime}=\\omega_{C}^{\\prime}$ ). The points $M^{\\prime}$ and $N^{\\prime}$ are the second intersections of lines $B^{\\prime} P^{\\prime}$ and $C^{\\prime} Q^{\\prime}$ with the circumcircles $\\gamma_{B}$ and $\\gamma_{C}$ of triangles $A^{\\prime} I B^{\\prime}$ and $A^{\\prime} I C^{\\prime}$, respectively. Notice here that $\\gamma_{C}$ is obtained from $\\gamma_{B}$ by the translation at $\\overrightarrow{B^{\\prime} C^{\\prime}}$; the same translation maps line $B^{\\prime} P^{\\prime}$ to $C^{\\prime} Q^{\\prime}$, and hence $M^{\\prime}$ to $N^{\\prime}$. In other words, $B^{\\prime} M^{\\prime} N^{\\prime} C^{\\prime}$ is a parallelogram, and $P^{\\prime} Q^{\\prime}$ partitions it into two isosceles trapezoids.\n\nPoint $X^{\\prime}$ is the second intersection point of circles $\\left(I P^{\\prime} M^{\\prime}\\right)$ and $\\left(I Q^{\\prime} N^{\\prime}\\right)$ that is - the reflection of $I$ in their line of centers. But the centers lie on the common perpendicular bisector $p$ of $P^{\\prime} M^{\\prime}$ and $Q^{\\prime} N^{\\prime}$, so $p$ is that line of centers. Hence, $I X^{\\prime} \\| B^{\\prime} P^{\\prime}$, as both lines are perpendicular to $p$.\n\nFinally, the point $Y$ satisfies $\\sphericalangle(B Y, B I)=\\sphericalangle(P B, P I)$ and $\\sphericalangle(C Y, C I)=\\sphericalangle(Q C, Q I)$, which yields $\\sphericalangle\\left(Y^{\\prime} B^{\\prime}, Y^{\\prime} I\\right)=\\sphericalangle\\left(B^{\\prime} P^{\\prime}, B^{\\prime} I\\right)$ and $\\sphericalangle\\left(Y^{\\prime} C^{\\prime}, Y^{\\prime} I\\right)=\\sphericalangle\\left(C^{\\prime} Q^{\\prime}, C^{\\prime} I\\right)$. Therefore,\n\n$$\n\\sphericalangle\\left(Y^{\\prime} B^{\\prime}, Y^{\\prime} C^{\\prime}\\right)=\\sphericalangle\\left(B^{\\prime} P^{\\prime}, B^{\\prime} I\\right)+\\sphericalangle\\left(C^{\\prime} I, C^{\\prime} Q^{\\prime}\\right)=\\sphericalangle\\left(C^{\\prime} I, B^{\\prime} I\\right)=\\sphericalangle\\left(A^{\\prime} B^{\\prime}, A^{\\prime} C^{\\prime}\\right)\n$$\n\nwhich shows that $Y^{\\prime} \\in \\Gamma^{\\prime}$.\n\nIn congruent circles $\\Gamma^{\\prime}$ and $\\gamma_{B}$, the chords $A^{\\prime} P^{\\prime}$ and $A^{\\prime} M^{\\prime}$ subtend the same angle $\\angle A^{\\prime} B^{\\prime} P^{\\prime}$; therefore, $A^{\\prime} P^{\\prime}=A^{\\prime} M^{\\prime}$, and hence $A^{\\prime} \\in p$. This yields $A^{\\prime} X^{\\prime}=A^{\\prime} I$, and hence $\\sphericalangle\\left(I A^{\\prime}, I X^{\\prime}\\right)=$ $\\sphericalangle\\left(X^{\\prime} I, X^{\\prime} A^{\\prime}\\right)$.\n\nFinally, we have\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(Y^{\\prime} I, Y^{\\prime} A^{\\prime}\\right) & =\\sphericalangle\\left(Y^{\\prime} I, Y^{\\prime} B^{\\prime}\\right)+\\sphericalangle\\left(Y^{\\prime} B^{\\prime}, Y^{\\prime} A^{\\prime}\\right) \\\\\n& =\\sphericalangle\\left(B^{\\prime} I, B^{\\prime} P^{\\prime}\\right)+\\sphericalangle\\left(I A^{\\prime}, I B^{\\prime}\\right)=\\sphericalangle\\left(I A^{\\prime}, B^{\\prime} P^{\\prime}\\right)=\\sphericalangle\\left(I A^{\\prime}, I X^{\\prime}\\right)=\\sphericalangle\\left(X^{\\prime} I, X^{\\prime} A^{\\prime}\\right),\n\\end{aligned}\n$$\n\nwhich yields that the points $A^{\\prime}, X^{\\prime}, Y^{\\prime}$, and $I$ are concyclic. This means exactly that $A, X$, and $Y$ are collinear.\n\n\n\n']",,True,,, 1896,Number Theory,,"Prove that there exists a positive constant $c$ such that the following statement is true: Assume that $n$ is an integer with $n \geqslant 2$, and let $\mathcal{S}$ be a set of $n$ points in the plane such that the distance between any two distinct points in $\mathcal{S}$ is at least 1 . Then there is a line $\ell$ separating $\mathcal{S}$ such that the distance from any point of $\mathcal{S}$ to $\ell$ is at least $c n^{-1 / 3}$. (A line $\ell$ separates a point set $\mathcal{S}$ if some segment joining two points in $\mathcal{S}$ crosses $\ell$.)","['We prove that the desired statement is true with $c=\\frac{1}{8}$. Set $\\delta=\\frac{1}{8} n^{-1 / 3}$. For any line $\\ell$ and any point $X$, let $X_{\\ell}$ denote the projection of $X$ to $\\ell$; a similar notation applies to sets of points.\n\nSuppose that, for some line $\\ell$, the set $\\mathcal{S}_{\\ell}$ contains two adjacent points $X$ and $Y$ with $X Y=2 d$. Then the line perpendicular to $\\ell$ and passing through the midpoint of segment $X Y$ separates $\\mathcal{S}$, and all points in $\\mathcal{S}$ are at least $d$ apart from $\\ell$. Thus, if $d \\geqslant \\delta$, then a desired line has been found. For the sake of contradiction, we assume that no such points exist, in any projection.\n\nChoose two points $A$ and $B$ in $\\mathcal{S}$ with the maximal distance $M=A B$ (i.e., $A B$ is a diameter of $\\mathcal{S})$; by the problem condition, $M \\geqslant 1$. Denote by $\\ell$ the line $A B$. The set $\\mathcal{S}$ is contained in the intersection of two disks $D_{A}$ and $D_{B}$ of radius $M$ centered at $A$ and $B$, respectively. Hence, the projection $\\mathcal{S}_{\\ell}$ is contained in the segment $A B$. Moreover, the points in $\\mathcal{S}_{\\ell}$ divide that segment into at most $n-1$ parts, each of length less than $2 \\delta$. Therefore,\n\n$$\nM\n\nChoose a point $H$ on segment $A B$ with $A H=\\frac{1}{2}$. Let $P$ be a strip between the lines $a$ and $h$ perpendicular to $A B$ and passing through $A$ and $H$, respectively; we assume that $P$ contains its boundary, which consists of lines $a$ and $h$. Set $\\mathcal{T}=P \\cap \\mathcal{S}$ and let $t=|\\mathcal{T}|$. By our assumption, segment $A H$ contains at least $\\left\\lceil\\frac{1}{2}:(2 \\delta)\\right\\rceil$ points of $S_{\\ell}$, which yields\n\n$$\nt \\geqslant \\frac{1}{4 \\delta}\\tag{2}\n$$\n\nNotice that $\\mathcal{T}$ is contained in $Q=P \\cap D_{B}$. The set $Q$ is a circular segment, and its projection $Q_{a}$ is a line segment of length\n\n$$\n2 \\sqrt{M^{2}-\\left(M-\\frac{1}{2}\\right)^{2}}<2 \\sqrt{M}\n$$\n\nOn the other hand, for any two points $X, Y \\in \\mathcal{T}$, we have $X Y \\geqslant 1$ and $X_{\\ell} Y_{\\ell} \\leqslant \\frac{1}{2}$, so $X_{a} Y_{a}=$ $\\sqrt{X Y^{2}-X_{\\ell} Y_{\\ell}^{2}} \\geqslant \\frac{\\sqrt{3}}{2}$. To summarize, $t$ points constituting $\\mathcal{T}_{a}$ lie on the segment of length less than $2 \\sqrt{M}$, and are at least $\\frac{\\sqrt{3}}{2}$ apart from each other. This yields $2 \\sqrt{M}>(t-1) \\frac{\\sqrt{3}}{2}$, or\n\n$$\nt<1+\\frac{4 \\sqrt{M}}{\\sqrt{3}}<4 \\sqrt{M}\\tag{3}\n$$\n\n\n\nas $M \\geqslant 1$.\n\nCombining the estimates (1), (2), and (3), we finally obtain\n\n$$\n\\frac{1}{4 \\delta} \\leqslant t<4 \\sqrt{M}<4 \\sqrt{2 n \\delta}, \\quad \\text { or } \\quad 512 n \\delta^{3}>1\n$$\n\nwhich does not hold for the chosen value of $\\delta$.']",,True,,, 1897,Number Theory,,"Given a positive integer $k$, show that there exists a prime $p$ such that one can choose distinct integers $a_{1}, a_{2}, \ldots, a_{k+3} \in\{1,2, \ldots, p-1\}$ such that $p$ divides $a_{i} a_{i+1} a_{i+2} a_{i+3}-i$ for all $i=1,2, \ldots, k$.","['First we choose distinct positive rational numbers $r_{1}, \\ldots, r_{k+3}$ such that\n\n$$\nr_{i} r_{i+1} r_{i+2} r_{i+3}=i \\text { for } 1 \\leqslant i \\leqslant k .\n$$\n\nLet $r_{1}=x, r_{2}=y, r_{3}=z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_{4}=\\frac{1}{r_{1} r_{2} r_{3}}$ and $r_{i+4}=\\frac{i+1}{i} r_{i}$. It follows that if $r_{i}$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \\equiv 1(\\bmod 4)$, by $y$ for $i \\equiv 2(\\bmod 4)$, by $z$ for $i \\equiv 3(\\bmod 4)$ and by none for $i \\equiv 0(\\bmod 4)$. Notice that $r_{i}3$ dividing a number of the form $x^{2}-x+1$ with integer $x$ there are two unconnected islands in $p$-Landia.\n\nFor brevity's sake, when a bridge connects the islands numbered $m$ and $n$, we shall speak simply that it connects $m$ and $n$.\n\nA bridge connects $m$ and $n$ if $n \\equiv m^{2}+1(\\bmod p)$ or $m \\equiv n^{2}+1(\\bmod p)$. If $m^{2}+1 \\equiv n$ $(\\bmod p)$, we draw an arrow starting at $m$ on the bridge connecting $m$ and $n$. Clearly only one arrow starts at $m$ if $m^{2}+1 \\not \\equiv m(\\bmod p)$, and no arrows otherwise. The total number of bridges does not exceed the total number of arrows.\n\nSuppose $x^{2}-x+1 \\equiv 0(\\bmod p)$. We may assume that $1 \\leqslant x \\leqslant p$; then there is no arrow starting at $x$. Since $(1-x)^{2}-(1-x)+1=x^{2}-x+1,(p+1-x)^{2}+1 \\equiv(p+1-x)(\\bmod p)$, and there is also no arrow starting at $p+1-x$. If $x=p+1-x$, that is, $x=\\frac{p+1}{2}$, then $4\\left(x^{2}-x+1\\right)=p^{2}+3$ and therefore $x^{2}-x+1$ is not divisible by $p$. Thus the islands $x$ and $p+1-x$ are different, and no arrows start at either of them. It follows that the total number of bridges in $p$-Landia does not exceed $p-2$.\n\nLet $1,2, \\ldots, p$ be the vertices of a graph $G_{p}$, where an edge connects $m$ and $n$ if and only if there is a bridge between $m$ and $n$. The number of vertices of $G_{p}$ is $p$ and the number of edges is less than $p-1$. This means that the graph is not connected, which means that there are two islands not connected by a chain of bridges.\n\nIt remains to prove that there are infinitely many primes $p$ dividing $x^{2}-x+1$ for some integer $x$. Let $p_{1}, p_{2}, \\ldots, p_{k}$ be any finite set of such primes. The number $\\left(p_{1} p_{2} \\cdot \\ldots \\cdot p_{k}\\right)^{2}-p_{1} p_{2} \\cdot \\ldots \\cdot p_{k}+1$ is greater than 1 and not divisible by any $p_{i}$; therefore it has another prime divisor with the required property."", 'One can show, by using only arithmetical methods, that for infinitely many $p$, the kingdom of $p$-Ladia contains two islands connected to no other island, except for each other.\n\nLet arrows between islands have the same meaning as in the previous solution. Suppose that positive $a3$. It follows that $a b \\equiv a(1-a) \\equiv 1$ $(\\bmod p)$. If an arrow goes from $t$ to $a$, then $t$ must satisfy the congruence $t^{2}+1 \\equiv a \\equiv a^{2}+1$ $(\\bmod p)$; the only such $t \\neq a$ is $p-a$. Similarly, the only arrow going to $b$ goes from $p-b$. If one of the numbers $p-a$ and $p-b$, say, $p-a$, is not at the end of any arrow, the pair $a, p-a$ is not connected with the rest of the islands. This is true if at least one of the congruences $x^{2}+1 \\equiv-a, x^{2}+1 \\equiv-b$ has no solutions, that is, either $-a-1$ or $-b-1$ is a quadratic non-residue modulo $p$.\n\nNote that $x^{2}-x+1 \\equiv x^{2}-(a+b) x+a b \\equiv(x-a)(x-b)(\\bmod p)$. Substituting $x=-1$ we get $(-1-a)(-1-b) \\equiv 3(\\bmod p)$. If 3 is a quadratic non-residue modulo $p$, so is one of the numbers $-1-a$ and $-1-b$.\n\nThus it is enough to find infinitely many primes $p>3$ dividing $x^{2}-x+1$ for some integer $x$ and such that 3 is a quadratic non-residue modulo $p$.\n\nIf $x^{2}-x+1 \\equiv 0(\\bmod p)$ then $(2 x-1)^{2} \\equiv-3(\\bmod p)$, that is, -3 is a quadratic residue modulo $p$, so 3 is a quadratic non-residue if and only if -1 is also a non-residue, in other words, $p \\equiv-1(\\bmod 4)$.\n\nSimilarly to the first solution, let $p_{1}, \\ldots, p_{k}$ be primes congruent to -1 modulo 4 and dividing numbers of the form $x^{2}-x+1$. The number $\\left(2 p_{1} \\cdot \\ldots \\cdot p_{k}\\right)^{2}-2 p_{1} \\cdot \\ldots \\cdot p_{k}+1$ is\n\n\n\nnot divisible by any $p_{i}$ and is congruent to -1 modulo 4 , therefore, it has some prime divisor $p \\equiv-1(\\bmod 4)$ which has the required properties.']",,True,,, 1899,Number Theory,,"Let $n$ be an integer with $n \geqslant 2$. Does there exist a sequence $\left(a_{1}, \ldots, a_{n}\right)$ of positive integers with not all terms being equal such that the arithmetic mean of every two terms is equal to the geometric mean of some (one or more) terms in this sequence?","['Suppose that $a_{1}, \\ldots, a_{n}$ satisfy the required properties. Let $d=\\operatorname{gcd}\\left(a_{1} \\ldots, a_{n}\\right)$. If $d>1$ then replace the numbers $a_{1}, \\ldots, a_{n}$ by $\\frac{a_{1}}{d}, \\ldots, \\frac{a_{n}}{d}$; all arithmetic and all geometric means will be divided by $d$, so we obtain another sequence satisfying the condition. Hence, without loss of generality, we can assume that $\\operatorname{gcd}\\left(a_{1} \\ldots, a_{n}\\right)=1$.\n\nWe show two numbers, $a_{m}$ and $a_{k}$ such that their arithmetic mean, $\\frac{a_{m}+a_{k}}{2}$ is different from the geometric mean of any (nonempty) subsequence of $a_{1} \\ldots, a_{n}$. That proves that there cannot exist such a sequence.\n\nChoose the index $m \\in\\{1, \\ldots, n\\}$ such that $a_{m}=\\max \\left(a_{1}, \\ldots, a_{n}\\right)$. Note that $a_{m} \\geqslant 2$, because $a_{1}, \\ldots, a_{n}$ are not all equal. Let $p$ be a prime divisor of $a_{m}$.\n\nLet $k \\in\\{1, \\ldots, n\\}$ be an index such that $a_{k}=\\max \\left\\{a_{i}: p \\nmid a_{i}\\right\\}$. Due to $\\operatorname{gcd}\\left(a_{1} \\ldots, a_{n}\\right)=1$, not all $a_{i}$ are divisible by $p$, so such a $k$ exists. Note that $a_{m}>a_{k}$ because $a_{m} \\geqslant a_{k}, p \\mid a_{m}$ and $p \\nmid a_{k}$.\n\nLet $b=\\frac{a_{m}+a_{k}}{2}$; we will show that $b$ cannot be the geometric mean of any subsequence of $a_{1}, \\ldots, a_{n}$.\n\nConsider the geometric mean, $g=\\sqrt[t]{a_{i_{1}} \\cdot \\ldots \\cdot a_{i_{t}}}$ of an arbitrary subsequence of $a_{1}, \\ldots, a_{n}$. If none of $a_{i_{1}}, \\ldots, a_{i_{t}}$ is divisible by $p$, then they are not greater than $a_{k}$, so\n\n$$\ng=\\sqrt[t]{a_{i_{1}} \\cdot \\ldots \\cdot a_{i_{t}}} \\leqslant a_{k}<\\frac{a_{m}+a_{k}}{2}=b\n$$\n\nand therefore $g \\neq b$.\n\nOtherwise, if at least one of $a_{i_{1}}, \\ldots, a_{i_{t}}$ is divisible by $p$, then $2 g=2 \\sqrt[t]{a_{i_{1}} \\cdot \\ldots \\cdot a_{i_{t}}}$ is either not an integer or is divisible by $p$, while $2 b=a_{m}+a_{k}$ is an integer not divisible by $p$, so $g \\neq b$ again.', ""Like in the previous solution, we assume that the numbers $a_{1}, \\ldots, a_{n}$ have no common divisor greater than 1 . The arithmetic mean of any two numbers in the sequence is half of an integer; on the other hand, it is a (some integer order) root of an integer. This means each pair's mean is an integer, so all terms in the sequence must be of the same parity; hence they all are odd. Let $d=\\min \\left\\{\\operatorname{gcd}\\left(a_{i}, a_{j}\\right): a_{i} \\neq a_{j}\\right\\}$. By reordering the sequence we can assume that $\\operatorname{gcd}\\left(a_{1}, a_{2}\\right)=d$, the sum $a_{1}+a_{2}$ is maximal among such pairs, and $a_{1}>a_{2}$.\n\nWe will show that $\\frac{a_{1}+a_{2}}{2}$ cannot be the geometric mean of any subsequence of $a_{1} \\ldots, a_{n}$.\n\nLet $a_{1}=x d$ and $a_{2}=y d$ where $x, y$ are coprime, and suppose that there exist some $b_{1}, \\ldots, b_{t} \\in\\left\\{a_{1}, \\ldots, a_{n}\\right\\}$ whose geometric mean is $\\frac{a_{1}+a_{2}}{2}$. Let $d_{i}=\\operatorname{gcd}\\left(a_{1}, b_{i}\\right)$ for $i=1,2, \\ldots, t$ and let $D=d_{1} d_{2} \\cdot \\ldots \\cdot d_{t}$. Then\n\n$$\nD=d_{1} d_{2} \\cdot \\ldots \\cdot d_{t} \\mid b_{1} b_{2} \\cdot \\ldots \\cdot b_{t}=\\left(\\frac{a_{1}+a_{2}}{2}\\right)^{t}=\\left(\\frac{x+y}{2}\\right)^{t} d^{t}\n$$\n\nWe claim that $D \\mid d^{t}$. Consider an arbitrary prime divisor $p$ of $D$. Let $\\nu_{p}(x)$ denote the exponent of $p$ in the prime factorization of $x$. If $p \\mid \\frac{x+y}{2}$, then $p \\nmid x, y$, so $p$ is coprime with $x$; hence, $\\nu_{p}\\left(d_{i}\\right) \\leqslant \\nu_{p}\\left(a_{1}\\right)=\\nu_{p}(x d)=\\nu_{p}(d)$ for every $1 \\leqslant i \\leqslant t$, therefore $\\nu_{p}(D)=\\sum_{i} \\nu_{p}\\left(d_{i}\\right) \\leqslant$ $t \\nu_{p}(d)=\\nu_{p}\\left(d^{t}\\right)$. Otherwise, if $p$ is coprime to $\\frac{x+y}{2}$, we have $\\nu_{p}(D) \\leqslant \\nu_{p}\\left(d^{t}\\right)$ trivially. The claim has been proved.\n\n\n\nNotice that $d_{i}=\\operatorname{gcd}\\left(b_{i}, a_{1}\\right) \\geqslant d$ for $1 \\leqslant i \\leqslant t$ : if $b_{i} \\neq a_{1}$ then this follows from the definition of $d$; otherwise we have $b_{i}=a_{1}$, so $d_{i}=a_{1} \\geqslant d$. Hence, $D=d_{1} \\cdot \\ldots \\cdot d_{t} \\geqslant d^{t}$, and the claim forces $d_{1}=\\ldots=d_{t}=d$.\n\nFinally, by $\\frac{a_{1}+a_{2}}{2}>a_{2}$ there must be some $b_{k}$ which is greater than $a_{2}$. From $a_{1}>a_{2} \\geqslant$ $d=\\operatorname{gcd}\\left(a_{1}, b_{k}\\right)$ it follows that $a_{1} \\neq b_{k}$. Now the have a pair $a_{1}, b_{k} \\operatorname{such}$ that $\\operatorname{gcd}\\left(a_{1}, b_{k}\\right)=d$ but $a_{1}+b_{k}>a_{1}+a_{2}$; that contradicts the choice of $a_{1}$ and $a_{2}$.""]",['No such sequence exists'],False,,Need_human_evaluate, 1900,Number Theory,,"For any odd prime $p$ and any integer $n$, let $d_{p}(n) \in\{0,1, \ldots, p-1\}$ denote the remainder when $n$ is divided by $p$. We say that $\left(a_{0}, a_{1}, a_{2}, \ldots\right)$ is a $p$-sequence, if $a_{0}$ is a positive integer coprime to $p$, and $a_{n+1}=a_{n}+d_{p}\left(a_{n}\right)$ for $n \geqslant 0$. (a) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $\left(a_{0}, a_{1}, a_{2}, \ldots\right)$ and $\left(b_{0}, b_{1}, b_{2}, \ldots\right)$ such that $a_{n}>b_{n}$ for infinitely many $n$, and $b_{n}>a_{n}$ for infinitely many $n$ ? (b) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $\left(a_{0}, a_{1}, a_{2}, \ldots\right)$ and $\left(b_{0}, b_{1}, b_{2}, \ldots\right)$ such that $a_{0}b_{n}$ for all $n \geqslant 1$ ?","['Fix some odd prime $p$, and let $T$ be the smallest positive integer such that $p \\mid 2^{T}-1$; in other words, $T$ is the multiplicative order of 2 modulo $p$.\n\nConsider any $p$-sequence $\\left(x_{n}\\right)=\\left(x_{0}, x_{1}, x_{2}, \\ldots\\right)$. Obviously, $x_{n+1} \\equiv 2 x_{n}(\\bmod p)$ and therefore $x_{n} \\equiv 2^{n} x_{0}(\\bmod p)$. This yields $x_{n+T} \\equiv x_{n}(\\bmod p)$ and therefore $d\\left(x_{n+T}\\right)=d\\left(x_{n}\\right)$ for all $n \\geqslant 0$. It follows that the sum $d\\left(x_{n}\\right)+d\\left(x_{n+1}\\right)+\\ldots+d\\left(x_{n+T-1}\\right)$ does not depend on $n$ and is thus a function of $x_{0}$ (and $p$ ) only; we shall denote this sum by $S_{p}\\left(x_{0}\\right)$, and extend the function $S_{p}(\\cdot)$ to all (not necessarily positive) integers. Therefore, we have $x_{n+k T}=x_{n}+k S_{p}\\left(x_{0}\\right)$ for all positive integers $n$ and $k$. Clearly, $S_{p}\\left(x_{0}\\right)=S_{p}\\left(2^{t} x_{0}\\right)$ for every integer $t \\geqslant 0$.\n\nIn both parts, we use the notation\n\n$$\nS_{p}^{+}=S_{p}(1)=\\sum_{i=0}^{T-1} d_{p}\\left(2^{i}\\right) \\quad \\text { and } \\quad S_{p}^{-}=S_{p}(-1)=\\sum_{i=0}^{T-1} d_{p}\\left(p-2^{i}\\right)\n$$\n\n(a) Let $q>3$ be a prime and $p$ a prime divisor of $2^{q}+1$ that is greater than 3 . We will show that $p$ is suitable for part (a). Notice that $9 \\nmid 2^{q}+1$, so that such a $p$ exists. Moreover, for any two odd primes $qb_{0}$ and $a_{1}=p+2b_{0}+k S_{p}^{+}=b_{k \\cdot 2 q} \\quad \\text { and } \\quad a_{k \\cdot 2 q+1}=a_{1}+k S_{p}^{+}S_{p}\\left(y_{0}\\right)$ but $x_{0}y_{n}$ for every $n \\geqslant q+q \\cdot \\max \\left\\{y_{r}-x_{r}: r=0,1, \\ldots, q-1\\right\\}$. Now, since $x_{0}S_{p}^{-}$, then we can put $x_{0}=1$ and $y_{0}=p-1$. Otherwise, if $S_{p}^{+}p$ must be divisible by $p$. Indeed, if $n=p k+r$ is a good number, $k>0,00$. Let $\\mathcal{B}$ be the set of big primes, and let $p_{1}p_{1} p_{2}$, and let $p_{1}^{\\alpha}$ be the largest power of $p_{1}$ smaller than $n$, so that $n \\leqslant p_{1}^{\\alpha+1}0$. If $f(k)=f(n-k)$ for all $k$, it implies that $\\left(\\begin{array}{c}n-1 \\\\ k\\end{array}\\right)$ is not divisible by $p$ for all $k=1,2, \\ldots, n-2$. It is well known that it implies $n=a \\cdot p^{s}, a0, f(q)>0$, there exist only finitely many $n$ which are equal both to $a \\cdot p^{s}, a0$ for at least two primes less than $n$. Let $p_{0}$ be the prime with maximal $g(f, p)$ among all primes $pa, c$, then $b \\nmid a+c$. Otherwise, since $b \\neq a+c$, we would have $b \\leqslant(a+c) / 2<\\max \\{a, c\\}$.\n\nWe prove the following stronger statement.\n\nClaim. Let $\\mathcal{S}$ be a good set consisting of $n \\geqslant 2$ positive integers. Then the elements of $\\mathcal{S}$ may be ordered as $a_{1}, a_{2}, \\ldots, a_{n}$ so that $a_{i} \\nmid a_{i-1}+a_{i+1}$ and $a_{i} \\nmid a_{i-1}-a_{i+1}$, for all $i=2,3, \\ldots, n-1$.\n\nProof. Say that the ordering $a_{1}, \\ldots, a_{n}$ of $\\mathcal{S}$ is nice if it satisfies the required property.\n\nWe proceed by induction on $n$. The base case $n=2$ is trivial, as there are no restrictions on the ordering.\n\nTo perform the step of induction, suppose that $n \\geqslant 3$. Let $a=\\max \\mathcal{S}$, and set $\\mathcal{T}=\\mathcal{S} \\backslash\\{a\\}$. Use the inductive hypothesis to find a nice ordering $b_{1}, \\ldots, b_{n-1}$ of $\\mathcal{T}$. We will show that $a$ may be inserted into this sequence so as to reach a nice ordering of $\\mathcal{S}$. In other words, we will show that there exists a $j \\in\\{1,2, \\ldots, n\\}$ such that the ordering\n\n$$\nN_{j}=\\left(b_{1}, \\ldots, b_{j-1}, a, b_{j}, b_{j+1}, \\ldots, b_{n-1}\\right)\n$$\n\nis nice.\n\nAssume that, for some $j$, the ordering $N_{j}$ is not nice, so that some element $x$ in it divides either the sum or the difference of two adjacent ones. This did not happen in the ordering of $\\mathcal{T}$, hence $x \\in\\left\\{b_{j-1}, a, b_{j}\\right\\}$ (if, say, $b_{j-1}$ does not exist, then $x \\in\\left\\{a, b_{j}\\right\\}$; a similar agreement is applied hereafter). But the case $x=a$ is impossible: $a$ cannot divide $b_{j-1}-b_{j}$, since $0<\\left|b_{j-1}-b_{j}\\right|a, c$, then $b \\nmid a+c$. Otherwise, since $b \\neq a+c$, we would have $b \\leqslant(a+c) / 2<\\max \\{a, c\\}$.\n\n\nWe again prove a stronger statement.\n\nClaim. Let $\\mathcal{S}$ be an arbitrary set of $n \\geqslant 3$ positive integers. Then its elements can be ordered as $a_{1}, \\ldots, a_{n}$ so that, if $a_{i} \\mid a_{i-1}+a_{i+1}$, then $a_{i}=\\max \\mathcal{S}$.\n\nThe claim easily implies what we need to prove, due to Observation A.\n\nTo prove the Claim, introduce the function $f$ which assigns to any two elements $a, b \\in \\mathcal{S}$ with $aa_{j+1}>\\ldots>$ $a_{n} ;$ and\n\n(ii) $f$-avoidance: If $a0$ and $a<0$. Indeed, since $\\sum_{i=1}^{2019} u_{i}^{2}=1$, the variables $u_{i}$ cannot be all zero, and, since $\\sum_{i=1}^{2019} u_{i}=0$, the nonzero elements cannot be all positive or all negative.\n\nLet $P=\\left\\{i: u_{i}>0\\right\\}$ and $N=\\left\\{i: u_{i} \\leqslant 0\\right\\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\\sum_{i=1}^{2019} u_{i}=0$ we have $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i \\in P} u_{i}-\\sum_{i \\in N}\\left|u_{i}\\right|$, so\n\n$$\n\\sum_{i \\in P} u_{i}=\\sum_{i \\in N}\\left|u_{i}\\right|\n\\tag{1}\n$$\n\nAfter this preparation, estimate the sum of squares of the positive and nonpositive elements as follows:\n\n$$\n\\sum_{i \\in P} u_{i}^{2} \\leqslant \\sum_{i \\in P} b u_{i}=b \\sum_{i \\in P} u_{i}=b \\sum_{i \\in N}\\left|u_{i}\\right| \\leqslant b \\sum_{i \\in N}|a|=-n a b\n\\tag{2}\n$$\n$$\n\\sum_{i \\in N} u_{i}^{2} \\leqslant \\sum_{i \\in N}|a| \\cdot\\left|u_{i}\\right|=|a| \\sum_{i \\in N}\\left|u_{i}\\right|=|a| \\sum_{i \\in P} u_{i} \\leqslant|a| \\sum_{i \\in P} b=-p a b .\n\\tag{3}\n$$\n\nThe sum of these estimates is\n\n$$\n1=\\sum_{i=1}^{2019} u_{i}^{2}=\\sum_{i \\in P} u_{i}^{2}+\\sum_{i \\in N} u_{i}^{2} \\leqslant-(p+n) a b=-2019 a b\n$$\n\nthat proves $a b \\leqslant \\frac{-1}{2019}$.', 'As in the previous solution we conclude that $a<0$ and $b>0$.\n\nFor every index $i$, the number $u_{i}$ is a convex combination of $a$ and $b$, so\n\n$$\nu_{i}=x_{i} a+y_{i} b \\quad \\text { with some weights } 0 \\leqslant x_{i}, y_{i} \\leqslant 1 \\text {, with } x_{i}+y_{i}=1 \\text {. }\n$$\n\nLet $X=\\sum_{i=1}^{2019} x_{i}$ and $Y=\\sum_{i=1}^{2019} y_{i}$. From $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i=1}^{2019}\\left(x_{i} a+y_{i} b\\right)=-|a| X+b Y$, we get\n\n$$\n|a| X=b Y\n\\tag{4}\n$$\n\n\n\nFrom $\\sum_{i=1}^{2019}\\left(x_{i}+y_{i}\\right)=2019$ we have\n\n$$\nX+Y=2019\n\\tag{5}\n$$\n\nThe system of linear equations $(4,5)$ has a unique solution:\n\n$$\nX=\\frac{2019 b}{|a|+b}, \\quad Y=\\frac{2019|a|}{|a|+b}\n$$\n\nNow apply the following estimate to every $u_{i}^{2}$ in their sum:\n\n$$\nu_{i}^{2}=x_{i}^{2} a^{2}+2 x_{i} y_{i} a b+y_{i}^{2} b^{2} \\leqslant x_{i} a^{2}+y_{i} b^{2}\n$$\n\nwe obtain that\n\n$$\n1=\\sum_{i=1}^{2019} u_{i}^{2} \\leqslant \\sum_{i=1}^{2019}\\left(x_{i} a^{2}+y_{i} b^{2}\\right)=X a^{2}+Y b^{2}=\\frac{2019 b}{|a|+b}|a|^{2}+\\frac{2019|a|}{|a|+b} b^{2}=2019|a| b=-2019 a b\n$$\n\nHence, $a b \\leqslant \\frac{-1}{2019}$.']",,True,,, 1906,Algebra,,"Let $n \geqslant 3$ be a positive integer and let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2 . Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of $$ \left|1-\sum_{i \in X} a_{i}\right| $$ is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ with sum equal to 2 such that $$ \sum_{i \in X} b_{i}=1 $$","['we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that\n\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|\n$$\n\nso we may exchange $X$ and $X^{c}$ where convenient. Let\n\n$$\n\\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i}\n$$\n\nand note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$.\n\nIf we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that\n\n$$\n\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}\n$$\n\nthen we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 .\n\n\nWithout loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, and we may assume strict inequality as otherwise $b_{i}=a_{i}$ works. Also, $X$ clearly cannot be empty.\n\nIf $n \\in X$, add $\\Delta$ to $a_{n}$, producing a sequence of $c_{i}$ with $\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}$, and then scale as described above to make the sum equal to 2 . Otherwise, there is some $k$ with $k \\in X$ and $k+1 \\in X^{c}$. Let $\\delta=a_{k+1}-a_{k}$.\n\n- If $\\delta>\\Delta$, add $\\Delta$ to $a_{k}$ and then scale.\n- If $\\delta<\\Delta$, then considering $X \\cup\\{k+1\\} \\backslash\\{k\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising.\n- If $\\delta=\\Delta$, choose any $j \\neq k, k+1$ (possible since $n \\geqslant 3$ ), and any $\\epsilon$ less than the least of $a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $j \\in X$ then add $\\Delta-\\epsilon$ to $a_{k}$ and $\\epsilon$ to $a_{j}$, then scale; otherwise, add $\\Delta$ to $a_{k}$ and $\\epsilon / 2$ to $a_{k+1}$, and subtract $\\epsilon / 2$ from $a_{j}$, then scale.', 'we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that\n\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|\n$$\n\nso we may exchange $X$ and $X^{c}$ where convenient. Let\n\n$$\n\\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i}\n$$\n\nand note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$.\n\nIf we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that\n\n$$\n\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}\n$$\n\nthen we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 .\n\nwithout loss of generality, assume $\\sum_{i \\in X} a_{i}<1$.\n\nSuppose there exists $1 \\leqslant j \\leqslant n-1$ such that $j \\in X$ but $j+1 \\in X^{c}$. Then $a_{j+1}-a_{j} \\geqslant \\Delta$, because otherwise considering $X \\cup\\{j+1\\} \\backslash\\{j\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising.\n\nIf $a_{j+1}-a_{j}>\\Delta$, put\n\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{i}, & \\text { otherwise }\\end{cases}\n$$\n\nIf $a_{j+1}-a_{j}=\\Delta$, choose any $\\epsilon$ less than the least of $\\Delta / 2, a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $|X| \\geqslant 2$, choose $k \\in X$ with $k \\neq j$, and put\n\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2-\\epsilon, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text { if } i=j+1 \\\\ a_{k}+\\epsilon, & \\text { if } i=k \\\\ a_{i}, & \\text { otherwise. }\\end{cases}\n$$\n\nOtherwise, $\\left|X^{c}\\right| \\geqslant 2$, so choose $k \\in X^{c}$ with $k \\neq j+1$, and put\n\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text { if } i=j \\\\ a_{j+1}-\\Delta / 2+\\epsilon, & \\text { if } i=j+1 \\\\ a_{k}-\\epsilon, & \\text { if } i=k ; \\\\ a_{i}, & \\text { otherwise }\\end{cases}\n$$\n\nIf there is no $1 \\leqslant j \\leqslant n$ such that $j \\in X$ but $j+1 \\in X^{c}$, there must be some $1\\Delta$, as otherwise considering $X \\cup\\{1\\}$ contradicts $X$ being $\\left(a_{i}\\right)$-minimising. Now put\n\n$$\nb_{i}= \\begin{cases}a_{1}-\\Delta / 2, & \\text { if } i=1 \\\\ a_{n}+\\Delta / 2, & \\text { if } i=n \\\\ a_{i}, & \\text { otherwise }\\end{cases}\n$$', 'we say an index set $X$ is $\\left(a_{i}\\right)$-minimising if it has the property in the problem for the given sequence $\\left(a_{i}\\right)$. Write $X^{c}$ for the complement of $X$, and $[a, b]$ for the interval of integers $k$ such that $a \\leqslant k \\leqslant b$. Note that\n\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|=\\left|1-\\sum_{i \\in X^{c}} a_{i}\\right|\n$$\n\nso we may exchange $X$ and $X^{c}$ where convenient. Let\n\n$$\n\\Delta=\\sum_{i \\in X^{c}} a_{i}-\\sum_{i \\in X} a_{i}\n$$\n\nand note that $X$ is $\\left(a_{i}\\right)$-minimising if and only if it minimises $|\\Delta|$, and that $\\sum_{i \\in X} a_{i}=1$ if and only if $\\Delta=0$.\n\nIf we have a strictly increasing sequence of positive real numbers $c_{i}$ (typically obtained by perturbing the $a_{i}$ in some way) such that\n\n$$\n\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}\n$$\n\nthen we may put $b_{i}=2 c_{i} / \\sum_{j=1}^{n} c_{j}$. So it suffices to construct such a sequence without needing its sum to be 2 .\n\n\nWithout loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, so $\\Delta \\geqslant 0$. If $\\Delta=0$ we can take $b_{i}=a_{i}$, so now assume that $\\Delta>0$.\n\nSuppose that there is some $k \\leqslant n$ such that $|X \\cap[k, n]|>\\left|X^{c} \\cap[k, n]\\right|$. If we choose the largest such $k$ then $|X \\cap[k, n]|-\\left|X^{c} \\cap[k, n]\\right|=1$. We can now find the required sequence $\\left(b_{i}\\right)$ by starting with $c_{i}=a_{i}$ for $i\\frac{n-k+1}{2}$ and $|X \\cap[\\ell, n]|<\\frac{n-\\ell+1}{2}$.\n\nWe now construct our sequence $\\left(b_{i}\\right)$ using this claim. Let $k$ and $\\ell$ be the greatest values satisfying the claim, and without loss of generality suppose $k=n$ and $\\ell0\n$$\n\nPartition the indices into sets $P, Q, R$, and $S$ such that\n\n$$\n\\begin{array}{ll}\nP=\\left\\{i \\mid a_{i} \\leqslant-1\\right\\} & R=\\left\\{i \\mid 00\n$$\n\nPartition the indices into sets $P, Q, R$, and $S$ such that\n\n$$\n\\begin{array}{ll}\nP=\\left\\{i \\mid a_{i} \\leqslant-1\\right\\} & R=\\left\\{i \\mid 01$ ). Therefore,\n\n$$\nt_{+}+t_{-} \\leqslant \\frac{p^{2}+s^{2}}{2}+p q+r s+p r+p s+q s=\\frac{(p+q+r+s)^{2}}{2}-\\frac{(q+r)^{2}}{2}=-\\frac{(q+r)^{2}}{2} \\leqslant 0 .\n$$\n\nIf $A$ is not empty and $p=s=0$, then there must exist $i \\in Q, j \\in R$ with $\\left|a_{i}-a_{j}\\right|>1$, and hence the earlier equality conditions cannot both occur.']",,True,,, 1908,Algebra,,"Let $x_{1}, x_{2}, \ldots, x_{n}$ be different real numbers. Prove that $$ \sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}= \begin{cases}0, & \text { if } n \text { is even } \\ 1, & \text { if } n \text { is odd }\end{cases} $$","['Since both sides of the identity are rational functions, it suffices to prove it when all $x_{i} \\notin\\{ \\pm 1\\}$. Define\n\n$$\nf(t)=\\prod_{i=1}^{n}\\left(1-x_{i} t\\right)\n$$\n\nand note that\n\n$$\nf\\left(x_{i}\\right)=\\left(1-x_{i}^{2}\\right) \\prod_{j \\neq i} 1-x_{i} x_{j}\n$$\n\nUsing the nodes $+1,-1, x_{1}, \\ldots, x_{n}$, the Lagrange interpolation formula gives us the following expression for $f$ :\n\n$$\n\\sum_{i=1}^{n} f\\left(x_{i}\\right) \\frac{(x-1)(x+1)}{\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)} \\prod_{j \\neq i} \\frac{x-x_{j}}{x_{i}-x_{j}}+f(1) \\frac{x+1}{1+1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}}+f(-1) \\frac{x-1}{-1-1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}}\n$$\n\nThe coefficient of $t^{n+1}$ in $f(t)$ is 0 , since $f$ has degree $n$. The coefficient of $t^{n+1}$ in the above expression of $f$ is\n\n$$\n\\begin{aligned}\n0 & =\\sum_{1 \\leqslant i \\leqslant n} \\frac{f\\left(x_{i}\\right)}{\\prod_{j \\neq i}\\left(x_{i}-x_{j}\\right) \\cdot\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)}+\\frac{f(1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(1-x_{j}\\right) \\cdot(1+1)}+\\frac{f(-1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(-1-x_{j}\\right) \\cdot(-1-1)} \\\\\n& =-G\\left(x_{1}, \\ldots, x_{n}\\right)+\\frac{1}{2}+\\frac{(-1)^{n+1}}{2} .\n\\end{aligned}\n$$', 'Observe that $G$ is symmetric in the variables $x_{1}, \\ldots, x_{n}$. Define $V=\\prod_{i0$. By Step 2, each of its monomials $\\mu x^{a} y^{b} z^{c}$ of the highest total degree satisfies $a, b \\geqslant c$. Applying other weak symmetries, we obtain $a, c \\geqslant b$ and $b, c \\geqslant a$; therefore, $P$ has a unique leading monomial of the form $\\mu(x y z)^{c}$. The polynomial $P_{0}(x, y, z)=P(x, y, z)-\\mu\\left(x y z-x^{2}-y^{2}-z^{2}\\right)^{c}$ has smaller total degree. Since $P_{0}$ is symmetric, it is representable as a polynomial function of $x y z-x^{2}-y^{2}-z^{2}$. Then $P$ is also of this form, completing the inductive step.', 'We will rely on the well-known identity\n\n$$\n\\cos ^{2} u+\\cos ^{2} v+\\cos ^{2} w-2 \\cos u \\cos v \\cos w-1=0 \\quad \\text { whenever } u+v+w=0 \\text {. }\n\\tag{2.1}\n$$\n\nClaim 1. The polynomial $P(x, y, z)$ is constant on the surface\n\n$$\n\\mathfrak{S}=\\{(2 \\cos u, 2 \\cos v, 2 \\cos w): u+v+w=0\\}\n$$\n\nProof. Notice that for $x=2 \\cos u, y=2 \\cos v, z=2 \\cos w$, the Vieta jumps $x \\mapsto y z-x$, $y \\mapsto z x-y, z \\mapsto x y-z$ in $(*)$ replace $(u, v, w)$ by $(v-w,-v, w),(u, w-u,-w)$ and $(-u, v, u-v)$, respectively. For example, for the first type of jump we have\n\n$$\ny z-x=4 \\cos v \\cos w-2 \\cos u=2 \\cos (v+w)+2 \\cos (v-w)-2 \\cos u=2 \\cos (v-w) \\text {. }\n$$\n\nDefine $G(u, v, w)=P(2 \\cos u, 2 \\cos v, 2 \\cos w)$. For $u+v+w=0$, the jumps give\n\n$$\n\\begin{aligned}\nG(u, v, w) & =G(v-w,-v, w)=G(w-v,-v,(v-w)-(-v))=G(-u-2 v,-v, 2 v-w) \\\\\n& =G(u+2 v, v, w-2 v)\n\\end{aligned}\n$$\n\nBy induction,\n\n$$\nG(u, v, w)=G(u+2 k v, v, w-2 k v) \\quad(k \\in \\mathbb{Z}) .\n\\tag{2.2}\n$$\n\nSimilarly,\n\n$$\nG(u, v, w)=G(u, v-2 \\ell u, w+2 \\ell u) \\quad(\\ell \\in \\mathbb{Z})\n\\tag{2.3}\n$$\n\nAnd, of course, we have\n\n$$\nG(u, v, w)=G(u+2 p \\pi, v+2 q \\pi, w-2(p+q) \\pi) \\quad(p, q \\in \\mathbb{Z}) .\n\\tag{2.4}\n$$\n\nTake two nonzero real numbers $u, v$ such that $u, v$ and $\\pi$ are linearly independent over $\\mathbb{Q}$. By combining $(2.2-2.4)$, we can see that $G$ is constant on a dense subset of the plane $u+v+w=0$. By continuity, $G$ is constant on the entire plane and therefore $P$ is constant on $\\mathfrak{S}$.\n\nClaim 2. The polynomial $T(x, y, z)=x^{2}+y^{2}+z^{2}-x y z-4$ divides $P(x, y, z)-P(2,2,2)$. Proof. By dividing $P$ by $T$ with remainders, there exist some polynomials $R(x, y, z), A(y, z)$ and $B(y, z)$ such that\n\n$$\nP(x, y, z)-P(2,2,2)=T(x, y, z) \\cdot R(x, y, z)+A(y, z) x+B(y, z)\n\\tag{2.5}\n$$\n\nOn the surface $\\mathfrak{S}$ the LHS of $(2.5)$ is zero by Claim 1 (since $(2,2,2) \\in \\mathfrak{S}$ ) and $T=0$ by (2.1). Hence, $A(y, z) x+B(y, z)$ vanishes on $\\mathfrak{S}$.\n\nNotice that for every $y=2 \\cos v$ and $z=2 \\cos w$ with $\\frac{\\pi}{3}0$ we have\n\n$$\nS_{k}=S_{k-1} \\cup\\left(x_{k}+S_{k-1}\\right)\n$$\n\n(where $\\left(x_{k}+S_{k-1}\\right)$ denotes $\\left\\{x_{k}+s: s \\in S_{k-1}\\right\\}$ ), so it suffices to prove that $x_{k} \\leqslant \\sum_{j\\sum_{j(n+1-k)(k+1)+k-1\n$$\n\nThis rearranges to $n>k(n+1-k)$, which is false for $1 \\leqslant k \\leqslant n$, giving the desired contradiction.']",,True,,, 1914,Combinatorics,,"Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\text {th }}$ coin over; otherwise he stops the process. (For example, the process starting with THT would be THT $\rightarrow H H T \rightarrow H T T \rightarrow T T T$, which takes three steps.) Letting $C$ denote the initial configuration (a sequence of $n H$ 's and $T$ 's), write $\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.","[""We represent the problem using a directed graph $G_{n}$ whose vertices are the length- $n$ strings of $H$ 's and $T$ 's. The graph features an edge from each string to its successor (except for $T T \\cdots T T$, which has no successor). We will also write $\\bar{H}=T$ and $\\bar{T}=H$.\n\nThe graph $G_{0}$ consists of a single vertex: the empty string. The main claim is that $G_{n}$ can be described explicitly in terms of $G_{n-1}$ :\n\n- We take two copies, $X$ and $Y$, of $G_{n-1}$.\n- In $X$, we take each string of $n-1$ coins and just append a $T$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $s_{1} \\cdots s_{n-1} T$.\n- In $Y$, we take each string of $n-1$ coins, flip every coin, reverse the order, and append an $H$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $\\bar{s}_{n-1} \\bar{s}_{n-2} \\cdots \\bar{s}_{1} H$.\n- Finally, we add one new edge from $Y$ to $X$, namely $H H \\cdots H H H \\rightarrow H H \\cdots H H T$.\n\nWe depict $G_{4}$ below, in a way which indicates this recursive construction:\n\n\n\nWe prove the claim inductively. Firstly, $X$ is correct as a subgraph of $G_{n}$, as the operation on coins is unchanged by an extra $T$ at the end: if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $s_{1} \\cdots s_{n-1} T$ is sent to $t_{1} \\cdots t_{n-1} T$.\n\nNext, $Y$ is also correct as a subgraph of $G_{n}$, as if $s_{1} \\cdots s_{n-1}$ has $k$ occurrences of $H$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ has $(n-1-k)+1=n-k$ occurrences of $H$, and thus (provided that $k>0$ ), if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ is sent to $\\bar{t}_{n-1} \\cdots \\bar{t}_{1} H$.\n\nFinally, the one edge from $Y$ to $X$ is correct, as the operation does send $H H \\cdots H H H$ to HH $\\cdots H H T$.\n\n\n\nTo finish, note that the sequences in $X$ take an average of $E(n-1)$ steps to terminate, whereas the sequences in $Y$ take an average of $E(n-1)$ steps to reach $H H \\cdots H$ and then an additional $n$ steps to terminate. Therefore, we have\n\n$$\nE(n)=\\frac{1}{2}(E(n-1)+(E(n-1)+n))=E(n-1)+\\frac{n}{2}\n$$\n\nWe have $E(0)=0$ from our description of $G_{0}$. Thus, by induction, we have $E(n)=\\frac{1}{2}(1+\\cdots+$ $n)=\\frac{1}{4} n(n+1)$, which in particular is finite."", 'We consider what happens with configurations depending on the coins they start and end with.\n\n- If a configuration starts with $H$, the last $n-1$ coins follow the given rules, as if they were all the coins, until they are all $T$, then the first coin is turned over.\n- If a configuration ends with $T$, the last coin will never be turned over, and the first $n-1$ coins follow the given rules, as if they were all the coins.\n- If a configuration starts with $T$ and ends with $H$, the middle $n-2$ coins follow the given rules, as if they were all the coins, until they are all $T$. After that, there are $2 n-1$ more steps: first coins $1,2, \\ldots, n-1$ are turned over in that order, then coins $n, n-1, \\ldots, 1$ are turned over in that order.\n\nAs this covers all configurations, and the number of steps is clearly finite for 0 or 1 coins, it follows by induction on $n$ that the number of steps is always finite.\n\nWe define $E_{A B}(n)$, where $A$ and $B$ are each one of $H, T$ or *, to be the average number of steps over configurations of length $n$ restricted to those that start with $A$, if $A$ is not *, and that end with $B$, if $B$ is not * (so * represents ""either $H$ or $T$ ""). The above observations tell us that, for $n \\geqslant 2$ :\n\n- $E_{H *}(n)=E(n-1)+1$.\n- $E_{* T}(n)=E(n-1)$.\n- $E_{H T}(n)=E(n-2)+1$ (by using both the observations for $H *$ and for $* T$ ).\n- $E_{T H}(n)=E(n-2)+2 n-1$.\n\nNow $E_{H *}(n)=\\frac{1}{2}\\left(E_{H H}(n)+E_{H T}(n)\\right)$, so $E_{H H}(n)=2 E(n-1)-E(n-2)+1$. Similarly, $E_{T T}(n)=2 E(n-1)-E(n-2)-1$. So\n\n$$\nE(n)=\\frac{1}{4}\\left(E_{H T}(n)+E_{H H}(n)+E_{T T}(n)+E_{T H}(n)\\right)=E(n-1)+\\frac{n}{2}\n$$\n\nWe have $E(0)=0$ and $E(1)=\\frac{1}{2}$, so by induction on $n$ we have $E(n)=\\frac{1}{4} n(n+1)$.', 'Let $H_{i}$ be the number of heads in positions 1 to $i$ inclusive (so $H_{n}$ is the total number of heads), and let $I_{i}$ be 1 if the $i^{\\text {th }}$ coin is a head, 0 otherwise. Consider the function\n\n$$\nt(i)=I_{i}+2\\left(\\min \\left\\{i, H_{n}\\right\\}-H_{i}\\right)\n$$\n\nWe claim that $t(i)$ is the total number of times coin $i$ is turned over (which implies that the process terminates). Certainly $t(i)=0$ when all coins are tails, and $t(i)$ is always a nonnegative integer, so it suffices to show that when the $k^{\\text {th }}$ coin is turned over (where $k=H_{n}$ ), $t(k)$ goes down by 1 and all the other $t(i)$ are unchanged. We show this by splitting into cases:\n\n\n\n- If $ik, \\min \\left\\{i, H_{n}\\right\\}=H_{n}$ both before and after the coin flip, and both $H_{n}$ and $H_{i}$ change by the same amount, so $t(i)$ is unchanged.\n- If $i=k$ and the coin is heads, $I_{i}$ goes down by 1 , as do both $\\min \\left\\{i, H_{n}\\right\\}=H_{n}$ and $H_{i}$; so $t(i)$ goes down by 1 .\n- If $i=k$ and the coin is tails, $I_{i}$ goes up by $1, \\min \\left\\{i, H_{n}\\right\\}=i$ is unchanged and $H_{i}$ goes up by 1 ; so $t(i)$ goes down by 1 .\n\nWe now need to compute the average value of\n\n$$\n\\sum_{i=1}^{n} t(i)=\\sum_{i=1}^{n} I_{i}+2 \\sum_{i=1}^{n} \\min \\left\\{i, H_{n}\\right\\}-2 \\sum_{i=1}^{n} H_{i}\n$$\n\nThe average value of the first term is $\\frac{1}{2} n$, and that of the third term is $-\\frac{1}{2} n(n+1)$. To compute the second term, we sum over choices for the total number of heads, and then over the possible values of $i$, getting\n\n$$\n2^{1-n} \\sum_{j=0}^{n}\\left(\\begin{array}{c}\nn \\\\\nj\n\\end{array}\\right) \\sum_{i=1}^{n} \\min \\{i, j\\}=2^{1-n} \\sum_{j=0}^{n}\\left(\\begin{array}{c}\nn \\\\\nj\n\\end{array}\\right)\\left(n j-\\left(\\begin{array}{l}\nj \\\\\n2\n\\end{array}\\right)\\right)\n$$\n\nNow, in terms of trinomial coefficients,\n\n$$\n\\sum_{j=0}^{n} j\\left(\\begin{array}{l}\nn \\\\\nj\n\\end{array}\\right)=\\sum_{j=1}^{n}\\left(\\begin{array}{c}\nn \\\\\nn-j, j-1,1\n\\end{array}\\right)=n \\sum_{j=0}^{n-1}\\left(\\begin{array}{c}\nn-1 \\\\\nj\n\\end{array}\\right)=2^{n-1} n\n$$\n\nand\n\n$$\n\\sum_{j=0}^{n}\\left(\\begin{array}{l}\nj \\\\\n2\n\\end{array}\\right)\\left(\\begin{array}{l}\nn \\\\\nj\n\\end{array}\\right)=\\sum_{j=2}^{n}\\left(\\begin{array}{c}\nn \\\\\nn-j, j-2,2\n\\end{array}\\right)=\\left(\\begin{array}{l}\nn \\\\\n2\n\\end{array}\\right) \\sum_{j=0}^{n-2}\\left(\\begin{array}{c}\nn-2 \\\\\nj\n\\end{array}\\right)=2^{n-2}\\left(\\begin{array}{l}\nn \\\\\n2\n\\end{array}\\right)\n$$\n\nSo the second term above is\n\n$$\n2^{1-n}\\left(2^{n-1} n^{2}-2^{n-2}\\left(\\begin{array}{l}\nn \\\\\n2\n\\end{array}\\right)\\right)=n^{2}-\\frac{n(n-1)}{4}\n$$\n\nand the required average is\n\n$$\nE(n)=\\frac{1}{2} n+n^{2}-\\frac{n(n-1)}{4}-\\frac{1}{2} n(n+1)=\\frac{n(n+1)}{4} .\n$$', 'Harry has built a Turing machine to flip the coins for him. The machine is initially positioned at the $k^{\\text {th }}$ coin, where there are $k$ heads (and the position before the first coin is considered to be the $0^{\\text {th }}$ coin). The machine then moves according to the following rules, stopping when it reaches the position before the first coin: if the coin at its current position is $H$, it flips the coin and moves to the previous coin, while if the coin at its current position is $T$, it flips the coin and moves to the next position.\n\nConsider the maximal sequences of consecutive moves in the same direction. Suppose the machine has $a$ consecutive moves to the next coin, before a move to the previous coin. After those $a$ moves, the $a$ coins flipped in those moves are all heads, as is the coin the machine is now at, so at least the next $a+1$ moves will all be moves to the previous coin. Similarly, $a$ consecutive moves to the previous coin are followed by at least $a+1$ consecutive moves to\n\n\n\nthe next coin. There cannot be more than $n$ consecutive moves in the same direction, so this proves that the process terminates (with a move from the first coin to the position before the first coin).\n\nThus we have a (possibly empty) sequence $a_{1}<\\cdotsj$, meaning that, as we follow the moves backwards, each coin is always in the correct state when flipped to result in a move in the required direction. (Alternatively, since there are $2^{n}$ possible configurations of coins and $2^{n}$ possible such ascending sequences, the fact that the sequence of moves determines at most one configuration of coins, and thus that there is an injection from configurations of coins to such ascending sequences, is sufficient for it to be a bijection, without needing to show that coins are in the right state as we move backwards.)', 'We explicitly describe what happens with an arbitrary sequence $C$ of $n$ coins. Suppose that $C$ contain $k$ heads at positions $1 \\leqslant c_{1}\n\nWe say that a region is ""on the right"" if it has $x$-coordinate unbounded above (note that if we only have one wall, then both regions are on the right). We claim inductively that, after placing $n$ lines, there are $n+1$ connected components in the resulting labyrinth, each of which contains exactly one region on the right. This is certainly true after placing 0 lines, as then there is only one region (and hence one connected component) and it is on the right.\n\nWhen placing the $n^{\\text {th }}$ line, it then cuts every one of the $n-1$ previously placed lines, and since it is to the right of all intersection points, the regions it cuts are exactly the $n$ regions on the right.\n\n\n\nThe addition of this line leaves all previous connected components with exactly one region on the right, and creates a new connected component containing exactly one region, and that region is also on the right. As a result, by induction, this particular labyrinth will have $n+1$ connected components.\n\nHaving built this labyrinth, Merlin then moves the walls one-by-one (by a sequence of continuous translations and rotations of lines) into the proper position of the given labyrinth, in such a way that no two lines ever become parallel.\n\n\n\nThe only time the configuration is changed is when one wall is moved through an intersection point of two others:\n\n\nNote that all moves really do switch between two configurations like this: all sets of three lines have this colour configuration initially, and the rules on rotations mean they are preserved (in particular, we cannot create three lines creating a triangle with three red edges inwards, or three blue edges inwards).\n\nHowever, as can be seen, such a move preserves the number of connected components, so in the painting this provides for Arthur\'s actual labyrinth, Morgana can still only place at most $n+1$ knights.']",['$k=n+1$'],False,,Expression, 1916,Combinatorics,,"On a certain social network, there are 2019 users, some pairs of which are friends, where friendship is a symmetric relation. Initially, there are 1010 people with 1009 friends each and 1009 people with 1010 friends each. However, the friendships are rather unstable, so events of the following kind may happen repeatedly, one at a time: Let $A, B$, and $C$ be people such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends; then $B$ and $C$ become friends, but $A$ is no longer friends with them. Prove that, regardless of the initial friendships, there exists a sequence of such events after which each user is friends with at most one other user.","['The problem has an obvious rephrasing in terms of graph theory. One is given a graph $G$ with 2019 vertices, 1010 of which have degree 1009 and 1009 of which have degree 1010. One is allowed to perform operations on $G$ of the following kind:\n\nSuppose that vertex $A$ is adjacent to two distinct vertices $B$ and $C$ which are not adjacent to each other. Then one may remove the edges $A B$ and $A C$ from $G$ and add the edge $B C$ into $G$.\n\nCall such an operation a refriending. One wants to prove that, via a sequence of such refriendings, one can reach a graph which is a disjoint union of single edges and vertices.\n\n\nNote that the given graph is connected, since the total degree of any two vertices is at least 2018 and hence they are either adjacent or have at least one neighbour in common. Hence the given graph satisfies the following condition:\n$$\n\\text{Every connected component of $G$ with at least three vertices is not complete and has a vertex of odd degree.}\n\\tag{1}\n$$\nWe will show that if a graph $G$ satisfies condition (1) and has a vertex of degree at least 2 , then there is a refriending on $G$ that preserves condition (1). Since refriendings decrease the total number of edges of $G$, by using a sequence of such refriendings, we must reach a graph $G$ with maximal degree at most 1 , so we are done.\n\n\n\n\n\nPick a vertex $A$ of degree at least 2 in a connected component $G^{\\prime}$ of $G$. Since no component of $G$ with at least three vertices is complete we may assume that not all of the neighbours of $A$ are adjacent to one another. (For example, pick a maximal complete subgraph $K$ of $G^{\\prime}$. Some vertex $A$ of $K$ has a neighbour outside $K$, and this neighbour is not adjacent to every vertex of $K$ by maximality.) Removing $A$ from $G$ splits $G^{\\prime}$ into smaller connected components $G_{1}, \\ldots, G_{k}$ (possibly with $k=1$ ), to each of which $A$ is connected by at least one edge. We divide into several cases.\n\nCase 1: $k \\geqslant 2$ and $A$ is connected to some $G_{i}$ by at least two edges.\n\nChoose a vertex $B$ of $G_{i}$ adjacent to $A$, and a vertex $C$ in another component $G_{j}$ adjacent to $A$. The vertices $B$ and $C$ are not adjacent, and hence removing edges $A B$ and $A C$ and adding in edge $B C$ does not disconnect $G^{\\prime}$. It is easy to see that this preserves the condition, since the refriending does not change the parity of the degrees of vertices.\n\nCase 2: $k \\geqslant 2$ and $A$ is connected to each $G_{i}$ by exactly one edge.\n\nConsider the induced subgraph on any $G_{i}$ and the vertex $A$. The vertex $A$ has degree 1 in this subgraph; since the number of odd-degree vertices of a graph is always even, we see that $G_{i}$ has a vertex of odd degree (in $G$ ). Thus if we let $B$ and $C$ be any distinct neighbours of $A$, then removing edges $A B$ and $A C$ and adding in edge $B C$ preserves the above condition: the refriending creates two new components, and if either of these components has at least three vertices, then it cannot be complete and must contain a vertex of odd degree (since each $G_{i}$ does).\n\nCase 3: $k=1$ and $A$ is connected to $G_{1}$ by at least three edges.\n\nBy assumption, $A$ has two neighbours $B$ and $C$ which are not adjacent to one another. Removing edges $A B$ and $A C$ and adding in edge $B C$ does not disconnect $G^{\\prime}$. We are then done as in Case 1.\n\nCase 4: $k=1$ and $A$ is connected to $G_{1}$ by exactly two edges.\n\nLet $B$ and $C$ be the two neighbours of $A$, which are not adjacent. Removing edges $A B$ and $A C$ and adding in edge $B C$ results in two new components: one consisting of a single vertex; and the other containing a vertex of odd degree. We are done unless this second component would be a complete graph on at least 3 vertices. But in this case, $G_{1}$ would be a complete graph minus the single edge $B C$, and hence has at least 4 vertices since $G^{\\prime}$ is not a 4-cycle. If we let $D$ be a third vertex of $G_{1}$, then removing edges $B A$ and $B D$ and adding in edge $A D$ does not disconnect $G^{\\prime}$. We are then done as in Case 1.\n\n', 'The problem has an obvious rephrasing in terms of graph theory. One is given a graph $G$ with 2019 vertices, 1010 of which have degree 1009 and 1009 of which have degree 1010. One is allowed to perform operations on $G$ of the following kind:\n\nSuppose that vertex $A$ is adjacent to two distinct vertices $B$ and $C$ which are not adjacent to each other. Then one may remove the edges $A B$ and $A C$ from $G$ and add the edge $B C$ into $G$.\n\nCall such an operation a refriending. One wants to prove that, via a sequence of such refriendings, one can reach a graph which is a disjoint union of single edges and vertices.\n\n\nnote that a refriending preserves the property that a graph has a vertex of odd degree and (trivially) the property that it is not complete; note also that our initial graph is connected. We describe an algorithm to reduce our initial graph to a graph of maximal degree at most 1 , proceeding in two steps.\n\nStep 1: There exists a sequence of refriendings reducing the graph to a tree.\n\nProof. Since the number of edges decreases with each refriending, it suffices to prove the following: as long as the graph contains a cycle, there exists a refriending such that the resulting graph is still connected. We will show that the graph in fact contains a cycle $Z$ and vertices $A, B, C$ such that $A$ and $B$ are adjacent in the cycle $Z, C$ is not in $Z$, and is adjacent to $A$ but not $B$. Removing edges $A B$ and $A C$ and adding in edge $B C$ keeps the graph connected, so we are done.\n\n\n\nTo find this cycle $Z$ and vertices $A, B, C$, we pursue one of two strategies. If the graph contains a triangle, we consider a largest complete subgraph $K$, which thus contains at least three vertices. Since the graph itself is not complete, there is a vertex $C$ not in $K$ connected to a vertex $A$ of $K$. By maximality of $K$, there is a vertex $B$ of $K$ not connected to $C$, and hence we are done by choosing a cycle $Z$ in $K$ through the edge $A B$.\n\n\n\nIf the graph is triangle-free, we consider instead a smallest cycle $Z$. This cycle cannot be Hamiltonian (i.e. it cannot pass through every vertex of the graph), since otherwise by minimality the graph would then have no other edges, and hence would have even degree at every vertex. We may thus choose a vertex $C$ not in $Z$ adjacent to a vertex $A$ of $Z$. Since the graph is triangle-free, it is not adjacent to any neighbour $B$ of $A$ in $Z$, and we are done.\n\nStep 2: Any tree may be reduced to a disjoint union of single edges and vertices by a sequence of refriendings.\n\nProof. The refriending preserves the property of being acyclic. Hence, after applying a sequence of refriendings, we arrive at an acyclic graph in which it is impossible to perform any further refriendings. The maximal degree of any such graph is 1: if it had a vertex $A$ with two neighbours $B, C$, then $B$ and $C$ would necessarily be nonadjacent since the graph is cycle-free, and so a refriending would be possible. Thus we reach a graph with maximal degree at most 1 as desired.']",,True,,, 1917,Combinatorics,,"Let $n>1$ be an integer. Suppose we are given $2 n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \ldots, A_{2 n}$ in some order. We then consider the $2 n$ angles $\angle A_{1} A_{2} A_{3}, \angle A_{2} A_{3} A_{4}, \ldots, \angle A_{2 n-2} A_{2 n-1} A_{2 n}, \angle A_{2 n-1} A_{2 n} A_{1}$, $\angle A_{2 n} A_{1} A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$ ). Prove that there exists an ordering of the given points such that the resulting $2 n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.","['Let $\\ell$ be a line separating the points into two groups ( $L$ and $R$ ) with $n$ points in each. Label the points $A_{1}, A_{2}, \\ldots, A_{2 n}$ so that $L=\\left\\{A_{1}, A_{3}, \\ldots, A_{2 n-1}\\right\\}$. We claim that this labelling works.\n\nTake a line $s=A_{2 n} A_{1}$.\n\n(a) Rotate $s$ around $A_{1}$ until it passes through $A_{2}$; the rotation is performed in a direction such that $s$ is never parallel to $\\ell$.\n\n(b) Then rotate the new $s$ around $A_{2}$ until it passes through $A_{3}$ in a similar manner.\n\n(c) Perform $2 n-2$ more such steps, after which $s$ returns to its initial position.\n\nThe total (directed) rotation angle $\\Theta$ of $s$ is clearly a multiple of $180^{\\circ}$. On the other hand, $s$ was never parallel to $\\ell$, which is possible only if $\\Theta=0$. Now it remains to partition all the $2 n$ angles into those where $s$ is rotated anticlockwise, and the others.', 'When tracing a cyclic path through the $A_{i}$ in order, with straight line segments between consecutive points, let $\\theta_{i}$ be the exterior angle at $A_{i}$, with a sign convention that it is positive if the path turns left and negative if the path turns right. Then $\\sum_{i=1}^{2 n} \\theta_{i}=360 k^{\\circ}$ for some integer $k$. Let $\\phi_{i}=\\angle A_{i-1} A_{i} A_{i+1}$ (indices $\\bmod 2 n$ ), defined as in the problem; thus $\\phi_{i}=180^{\\circ}-\\left|\\theta_{i}\\right|$.\n\nLet $L$ be the set of $i$ for which the path turns left at $A_{i}$ and let $R$ be the set for which it turns right. Then $S=\\sum_{i \\in L} \\phi_{i}-\\sum_{i \\in R} \\phi_{i}=(180(|L|-|R|)-360 k)^{\\circ}$, which is a multiple of $360^{\\circ}$ since the number of points is even. We will show that the points can be labelled such that $S=0$, in which case $L$ and $R$ satisfy the required condition of the problem.\n\nNote that the value of $S$ is defined for a slightly larger class of configurations: it is OK for two points to coincide, as long as they are not consecutive, and OK for three points to be collinear, as long as $A_{i}, A_{i+1}$ and $A_{i+2}$ do not appear on a line in that order. In what follows it will be convenient, although not strictly necessary, to consider such configurations.\n\nConsider how $S$ changes if a single one of the $A_{i}$ is moved along some straight-line path (not passing through any $A_{j}$ and not lying on any line $A_{j} A_{k}$, but possibly crossing such lines). Because $S$ is a multiple of $360^{\\circ}$, and the angles change continuously, $S$ can only change when a point moves between $R$ and $L$. Furthermore, if $\\phi_{j}=0$ when $A_{j}$ moves between $R$ and $L, S$ is unchanged; it only changes if $\\phi_{j}=180^{\\circ}$ when $A_{j}$ moves between those sets.\n\nFor any starting choice of points, we will now construct a new configuration, with labels such that $S=0$, that can be perturbed into the original one without any $\\phi_{i}$ passing through $180^{\\circ}$, so that $S=0$ for the original configuration with those labels as well.\n\nTake some line such that there are $n$ points on each side of that line. The new configuration has $n$ copies of a single point on each side of the line, and a path that alternates between\n\n\n\nsides of the line; all angles are 0 , so this configuration has $S=0$. Perturbing the points into their original positions, while keeping each point on its side of the line, no angle $\\phi_{i}$ can pass through $180^{\\circ}$, because no straight line can go from one side of the line to the other and back. So the perturbation process leaves $S=0$.', 'First, let $\\ell$ be a line in the plane such that there are $n$ points on one side and the other $n$ points on the other side. For convenience, assume $\\ell$ is horizontal (otherwise, we can rotate the plane). Then we can use the terms ""above"", ""below"", ""left"" and ""right"" in the usual way. We denote the $n$ points above the line in an arbitrary order as $P_{1}, P_{2}, \\ldots, P_{n}$, and the $n$ points below the line as $Q_{1}, Q_{2}, \\ldots, Q_{n}$.\n\nIf we connect $P_{i}$ and $Q_{j}$ with a line segment, the line segment will intersect with the line $\\ell$. Denote the intersection as $I_{i j}$. If $P_{i}$ is connected to $Q_{j}$ and $Q_{k}$, where $jk$, then the sign of $\\angle Q_{j} P_{i} Q_{k}$ is taken to be the same as for $\\angle Q_{k} P_{i} Q_{j}$.\n\nSimilarly, we can define the sign of $\\angle P_{j} Q_{i} P_{k}$ with $j\n\nSimilarly, we have\n\n$$\n\\angle P_{i_{1}} Q_{k} P_{i_{3}}=\\angle P_{i_{1}} Q_{k} P_{i_{2}}+\\angle P_{i_{2}} Q_{k} P_{i_{3}},\n\\tag{2}\n$$\n\nfor all points $Q_{k}, P_{i_{1}}, P_{i_{2}}$ and $P_{i_{3}}$, with $i_{1}\n\nThen $\\angle P_{1}$ and $\\angle Q_{1}$ are positive, $\\angle P_{2}$ and $\\angle Q_{2}$ are negative, and we have\n\n$$\n\\left|\\angle P_{1}\\right|+\\left|\\angle Q_{1}\\right|=\\left|\\angle P_{2}\\right|+\\left|\\angle Q_{2}\\right| .\n$$\n\nWith signed measures, we have\n\n$$\n\\angle P_{1}+\\angle Q_{1}+\\angle P_{2}+\\angle Q_{2}=0\n\\tag{3}\n$$\n\nIf we switch the labels of $P_{1}$ and $P_{2}$, we have the following picture:\n\n\n\nSwitching labels $P_{1}$ and $P_{2}$ has the effect of flipping the sign of all four angles (as well as swapping the magnitudes on the relabelled points); that is, the new values of $\\left(\\angle P_{1}, \\angle P_{2}, \\angle Q_{1}, \\angle Q_{2}\\right.$ ) equal the old values of $\\left(-\\angle P_{2},-\\angle P_{1},-\\angle Q_{1},-\\angle Q_{2}\\right)$. Consequently, equation (3) still holds. Similarly, when switching the labels of $Q_{1}$ and $Q_{2}$, or both the $P$ \'s and the $Q$ \'s, equation (3) still holds.\n\n\n\nThe remaining subcase of $n=2$ is that one point lies inside the triangle formed by the other three. We have the following picture.\n\n\n\nWe have\n\n$$\n\\left|\\angle P_{1}\\right|+\\left|\\angle Q_{1}\\right|+\\left|\\angle Q_{2}\\right|=\\left|\\angle P_{2}\\right|\n$$\n\nand equation (3) holds.\n\nAgain, switching the labels for $P$ \'s or the $Q$ \'s will not affect the validity of equation (3). Also, if the point lying inside the triangle of the other three is one of the $Q$ \'s rather than the $P$ \'s, the result still holds, since our sign convention is preserved when we relabel $Q$ \'s as $P$ \'s and vice-versa and reflect across $\\ell$.\n\nWe have completed the proof of the claim for $n=2$.\n\nAssume the claim holds for $n=k$, and we wish to prove it for $n=k+1$. Suppose we are given our $2(k+1)$ points. First ignore $P_{k+1}$ and $Q_{k+1}$, and form $2 k$ angles from $P_{1}, \\ldots, P_{k}$, $Q_{1}, \\ldots, Q_{k}$ as in the $n=k$ case. By the induction hypothesis we have\n\n$$\n\\sum_{i=1}^{k}\\left(\\angle P_{i}+\\angle Q_{i}\\right)=0\n$$\n\nWhen we add in the two points $P_{k+1}$ and $Q_{k+1}$, this changes our angles as follows:\n\n- the angle at $P_{k}$ changes from $\\angle Q_{k-1} P_{k} Q_{k}$ to $\\angle Q_{k-1} P_{k} Q_{k+1}$;\n- the angle at $Q_{k}$ changes from $\\angle P_{k-1} Q_{k} P_{k}$ to $\\angle P_{k-1} Q_{k} P_{k+1}$;\n- two new angles $\\angle Q_{k} P_{k+1} Q_{k+1}$ and $\\angle P_{k} Q_{k+1} P_{k+1}$ are added.\n\nWe need to prove the changes have no impact on the total sum. In other words, we need to prove\n\n$$\n\\left(\\angle Q_{k-1} P_{k} Q_{k+1}-\\angle Q_{k-1} P_{k} Q_{k}\\right)+\\left(\\angle P_{k-1} Q_{k} P_{k+1}-\\angle P_{k-1} Q_{k} P_{k}\\right)+\\left(\\angle P_{k+1}+\\angle Q_{k+1}\\right)=0\n$$\n\nIn fact, from equations (1) and (2), we have\n\n$$\n\\angle Q_{k-1} P_{k} Q_{k+1}-\\angle Q_{k-1} P_{k} Q_{k}=\\angle Q_{k} P_{k} Q_{k+1}\n$$\n\nand\n\n$$\n\\angle P_{k-1} Q_{k} P_{k+1}-\\angle P_{k-1} Q_{k} P_{k}=\\angle P_{k} Q_{k} P_{k+1} .\n$$\n\nTherefore, the left hand side of equation (4) becomes $\\angle Q_{k} P_{k} Q_{k+1}+\\angle P_{k} Q_{k} P_{k+1}+\\angle Q_{k} P_{k+1} Q_{k+1}+$ $\\angle P_{k} Q_{k+1} P_{k+1}$, which equals 0 , simply by applying the $n=2$ case of the claim. This completes the induction.', 'We shall think instead of the problem as asking us to assign a weight \\pm 1 to each angle, such that the weighted sum of all the angles is zero.\n\nGiven an ordering $A_{1}, \\ldots, A_{2 n}$ of the points, we shall assign weights according to the following recipe: walk in order from point to point, and assign the left turns +1 and the right turns -1 . This is the same weighting as in', '3, and as in that solution, the weighted sum is a multiple of $360^{\\circ}$.\n\nWe now aim to show the following:\n\nLemma. Transposing any two consecutive points in the ordering changes the weighted sum by $\\pm 360^{\\circ}$ or 0 .\n\nKnowing that, we can conclude quickly: if the ordering $A_{1}, \\ldots, A_{2 n}$ has weighted angle sum $360 k^{\\circ}$, then the ordering $A_{2 n}, \\ldots, A_{1}$ has weighted angle sum $-360 k^{\\circ}$ (since the angles are the same, but left turns and right turns are exchanged). We can reverse the ordering of $A_{1}$, $\\ldots, A_{2 n}$ by a sequence of transpositions of consecutive points, and in doing so the weighted angle sum must become zero somewhere along the way.\n\nWe now prove that lemma:\n\nProof. Transposing two points amounts to taking a section $A_{k} A_{k+1} A_{k+2} A_{k+3}$ as depicted, reversing the central line segment $A_{k+1} A_{k+2}$, and replacing its two neighbours with the dotted lines.\n\n\nFigure 1: Transposing two consecutive vertices: before (left) and afterwards (right)\n\nIn each triangle, we alter the sum by $\\pm 180^{\\circ}$. Indeed, using (anticlockwise) directed angles modulo $360^{\\circ}$, we either add or subtract all three angles of each triangle.\n\nHence both triangles together alter the sum by $\\pm 180 \\pm 180^{\\circ}$, which is $\\pm 360^{\\circ}$ or 0 .']",,True,,, 1918,Combinatorics,,"There are 60 empty boxes $B_{1}, \ldots, B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game. In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps: (a) Bob chooses an integer $k$ with $1 \leqslant k \leqslant 59$ and splits the boxes into the two groups $B_{1}, \ldots, B_{k}$ and $B_{k+1}, \ldots, B_{60}$. (b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group. Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning.","[""We present solutions for the general case of $N>1$ boxes, and write $M=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor\\left\\lceil\\frac{N}{2}+1\\right\\rceil-1$ for the claimed answer. For $1 \\leqslant kk$. We may then assume, without loss of generality, that Alice again picks the left group.\n\nProof. Suppose Alice picks the right group in the second round. Then the combined effect of the two rounds is that each of the boxes $B_{k+1}, \\ldots, B_{\\ell}$ lost two pebbles (and the other boxes are unchanged). Hence this configuration is strictly dominated by that before the first round, and it suffices to consider only Alice's other response.\n\nFor Alice. Alice initially distributes pebbles according to $V_{\\left\\lceil\\frac{N}{2}\\right\\rceil}$. Suppose the current configuration of pebbles dominates $V_{i}$. If Bob makes a $k$-move with $k \\geqslant i$ then Alice picks the left group, which results in a configuration that dominates $V_{i+1}$. Likewise, if Bob makes a $k$-move with $k1$ boxes, and write $M=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor\\left\\lceil\\frac{N}{2}+1\\right\\rceil-1$ for the claimed answer. For $1 \\leqslant kk$. We may then assume, without loss of generality, that Alice again picks the left group.\n\nProof. Suppose Alice picks the right group in the second round. Then the combined effect of the two rounds is that each of the boxes $B_{k+1}, \\ldots, B_{\\ell}$ lost two pebbles (and the other boxes are unchanged). Hence this configuration is strictly dominated by that before the first round, and it suffices to consider only Alice's other response.\n\n\nFor Alice. Let $K=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor$. Alice starts with the boxes in the configuration $V_{K}$. For each of Bob's $N-1$ possible choices, consider the subset of rounds in which he makes that choice. In that subset of rounds, Alice alternates between picking the left group and picking the right group; the first time Bob makes that choice, Alice picks the group containing the $K^{\\text {th }}$ box. Thus, at any time during the game, the number of pebbles in each box depends only on which choices Bob has made an odd number of times. This means that the number of pebbles in a box could decrease by at most the number of choices for which Alice would have started by removing a pebble from the group containing that box. These numbers are, for each box,\n\n$$\n\\left\\lfloor\\frac{N}{2}\\right\\rfloor,\\left\\lfloor\\frac{N}{2}-1\\right\\rfloor, \\ldots, 1,0,1, \\ldots,\\left\\lceil\\frac{N}{2}-1\\right\\rceil\n$$\n\nThese are pointwise less than the numbers of pebbles the boxes started with, meaning that no box ever becomes empty with this strategy.\n\nSo the final answer is $n=960$. In general, if there are $N>1$ boxes, the answer is $n=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor\\left\\lceil\\frac{N}{2}+1\\right\\rceil-1$.\n\nCommon remarks. We present solutions for the general case of $N>1$ boxes, and write $M=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor\\left\\lceil\\frac{N}{2}+1\\right\\rceil-1$ for the claimed answer. For $1 \\leqslant k2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase.\n\nAt the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \\leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total.""]",,True,,, 1920,Combinatorics,,"For any two different real numbers $x$ and $y$, we define $D(x, y)$ to be the unique integer $d$ satisfying $2^{d} \leqslant|x-y|<2^{d+1}$. Given a set of reals $\mathcal{F}$, and an element $x \in \mathcal{F}$, we say that the scales of $x$ in $\mathcal{F}$ are the values of $D(x, y)$ for $y \in \mathcal{F}$ with $x \neq y$. Let $k$ be a given positive integer. Suppose that each member $x$ of $\mathcal{F}$ has at most $k$ different scales in $\mathcal{F}$ (note that these scales may depend on $x$ ). What is the maximum possible size of $\mathcal{F}$ ?","['We first construct a set $\\mathcal{F}$ with $2^{k}$ members, each member having at most $k$ different scales in $\\mathcal{F}$. Take $\\mathcal{F}=\\left\\{0,1,2, \\ldots, 2^{k}-1\\right\\}$. The scale between any two members of $\\mathcal{F}$ is in the set $\\{0,1, \\ldots, k-1\\}$.\n\nWe now show that $2^{k}$ is an upper bound on the size of $\\mathcal{F}$. For every finite set $\\mathcal{S}$ of real numbers, and every real $x$, let $r_{\\mathcal{S}}(x)$ denote the number of different scales of $x$ in $\\mathcal{S}$. That is, $r_{\\mathcal{S}}(x)=|\\{D(x, y): x \\neq y \\in \\mathcal{S}\\}|$. Thus, for every element $x$ of the set $\\mathcal{F}$ in the problem statement, we have $r_{\\mathcal{F}}(x) \\leqslant k$. The condition $|\\mathcal{F}| \\leqslant 2^{k}$ is an immediate consequence of the following lemma.\n\nLemma. Let $\\mathcal{S}$ be a finite set of real numbers, and define\n\n$$\nw(\\mathcal{S})=\\sum_{x \\in \\mathcal{S}} 2^{-r_{\\mathcal{S}}(x)}\n$$\n\nThen $w(\\mathcal{S}) \\leqslant 1$.\n\nProof. Induction on $n=|\\mathcal{S}|$. If $\\mathcal{S}=\\{x\\}$, then $r_{\\mathcal{S}}(x)=0$, so $w(\\mathcal{S})=1$.\n\nAssume now $n \\geqslant 2$, and let $x_{1}<\\cdots1$ we have $D\\left(x_{i}, x_{j}\\right)>d$, since\n\n$$\n\\left|x_{i}-x_{j}\\right|=\\left|x_{i+1}-x_{i}\\right|+\\left|x_{j}-x_{i+1}\\right| \\geqslant 2^{d}+2^{d}=2^{d+1}\n$$\n\nNow choose the minimal $i \\leqslant t$ and the maximal $j \\geqslant t+1$ such that $D\\left(x_{i}, x_{i+1}\\right)=$ $D\\left(x_{i+1}, x_{i+2}\\right)=\\cdots=D\\left(x_{j-1}, x_{j}\\right)=d$.\n\nLet $E$ be the set of all the $x_{s}$ with even indices $i \\leqslant s \\leqslant j, O$ be the set of those with odd indices $i \\leqslant s \\leqslant j$, and $R$ be the rest of the elements (so that $\\mathcal{S}$ is the disjoint union of $E, O$ and $R$ ). Set $\\mathcal{S}_{O}=R \\cup O$ and $\\mathcal{S}_{E}=R \\cup E$; we have $\\left|\\mathcal{S}_{O}\\right|<|\\mathcal{S}|$ and $\\left|\\mathcal{S}_{E}\\right|<|\\mathcal{S}|$, so $w\\left(\\mathcal{S}_{O}\\right), w\\left(\\mathcal{S}_{E}\\right) \\leqslant 1$ by the inductive hypothesis.\n\nClearly, $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)$ and $r_{\\mathcal{S}_{E}}(x) \\leqslant r_{\\mathcal{S}}(x)$ for any $x \\in R$, and thus\n\n$$\n\\begin{aligned}\n\\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)} & =\\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}}(x)}+2^{-r_{\\mathcal{S}}(x)}\\right) \\\\\n& \\leqslant \\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}_{O}}(x)}+2^{-r_{\\mathcal{S}_{E}}(x)}\\right)\n\\end{aligned}\n$$\n\nOn the other hand, for every $x \\in O$, there is no $y \\in \\mathcal{S}_{O}$ such that $D_{\\mathcal{S}_{O}}(x, y)=d$ (as all candidates from $\\mathcal{S}$ were in $E$ ). Hence, we have $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)-1$, and thus\n\n$$\n\\sum_{x \\in O} 2^{-r_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in O} 2^{-r_{\\mathcal{S}_{O}}(x)}\n$$\n\n\n\nSimilarly, for every $x \\in E$, we have\n\n$$\n\\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)}\n$$\n\nWe can then combine these to give\n\n$$\n\\begin{aligned}\n& w(S)=\\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)}+\\sum_{x \\in O} 2^{-r_{\\mathcal{S}}(x)}+\\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\\\\n& \\leqslant \\frac{1}{2} \\sum_{x \\in R}\\left(2^{-r_{\\mathcal{S}_{O}}(x)}+2^{-r_{\\mathcal{S}_{E}}(x)}\\right)+\\frac{1}{2} \\sum_{x \\in O} 2^{-r_{\\mathcal{S}_{O}}(x)}+\\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)} \\\\\n& =\\frac{1}{2}\\left(\\sum_{x \\in \\mathcal{S}_{O}} 2^{-{ }^{-} \\mathcal{S}_{O}}(x)+\\sum_{x \\in \\mathcal{S}_{E}} 2^{-r_{\\mathcal{S}_{E}}(x)}\\right) \\quad\\left(\\text { since } \\mathcal{S}_{O}=O \\cup R \\text { and } \\mathcal{S}_{E}=E \\cup R\\right) \\\\\n& \\left.=\\frac{1}{2}\\left(w\\left(\\mathcal{S}_{O}\\right)+w\\left(\\mathcal{S}_{E}\\right)\\right)\\right) \\quad(\\text { by definition of } w(\\cdot)) \\\\\n& \\leqslant 1 \\quad \\text { (by the inductive hypothesis) }\n\\end{aligned}\n$$\n\nwhich completes the induction.']",['$2^{k}$'],False,,Expression, 1921,Geometry,,"Let $A B C$ be a triangle. Circle $\Gamma$ passes through $A$, meets segments $A B$ and $A C$ again at points $D$ and $E$ respectively, and intersects segment $B C$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $B D F$ at $F$ and the tangent to circle $C E G$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $A T$ is parallel to $B C$.","['Notice that $\\angle T F B=\\angle F D A$ because $F T$ is tangent to circle $B D F$, and moreover $\\angle F D A=\\angle C G A$ because quadrilateral $A D F G$ is cyclic. Similarly, $\\angle T G B=\\angle G E C$ because $G T$ is tangent to circle $C E G$, and $\\angle G E C=\\angle C F A$. Hence,\n\n$$\n\\angle T F B=\\angle C G A \\text { and } \\angle T G B=\\angle C F A \\text {. }\n\\tag{1}\n$$\n\n\n\nTriangles $F G A$ and $G F T$ have a common side $F G$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $A T$ is parallel to line $B F G C$.']",,True,,, 1922,Geometry,,"Let $A B C$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $B C, C A$, and $A B$, respectively. Denote by $\omega_{B}$ and $\omega_{C}$ the incircles of triangles $B D F$ and $C D E$, and let these circles be tangent to segments $D F$ and $D E$ at $M$ and $N$, respectively. Let line $M N$ meet circles $\omega_{B}$ and $\omega_{C}$ again at $P \neq M$ and $Q \neq N$, respectively. Prove that $M P=N Q$.","['Denote the centres of $\\omega_{B}$ and $\\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $B C$ be tangent to the two circles at $T$ and $U$, respectively.\n\n\n\nFrom the cyclic quadrilaterals $A F D C$ and $A B D E$ we have\n\n$$\n\\angle M D O_{B}=\\frac{1}{2} \\angle F D B=\\frac{1}{2} \\angle B A C=\\frac{1}{2} \\angle C D E=\\angle O_{C} D N\n$$\n\nso the right-angled triangles $D M O_{B}$ and $D N O_{C}$ are similar. The ratio of similarity between the two triangles is\n\n$$\n\\frac{D N}{D M}=\\frac{O_{C} N}{O_{B} M}=\\frac{r_{C}}{r_{B}}\n$$\n\nLet $\\varphi=\\angle D M N$ and $\\psi=\\angle M N D$. The lines $F M$ and $E N$ are tangent to $\\omega_{B}$ and $\\omega_{C}$, respectively, so\n\n$$\n\\angle M T P=\\angle F M P=\\angle D M N=\\varphi \\quad \\text { and } \\quad \\angle Q U N=\\angle Q N E=\\angle M N D=\\psi .\n$$\n\n(It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $D M T$ or $D U N$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $M T P$ and $Q U N$.)\n\nIn the circles $\\omega_{B}$ and $\\omega_{C}$ the lengths of chords $M P$ and $N Q$ are\n\n$$\nM P=2 r_{B} \\cdot \\sin \\angle M T P=2 r_{B} \\cdot \\sin \\varphi \\quad \\text { and } \\quad N Q=2 r_{C} \\cdot \\sin \\angle Q U N=2 r_{C} \\cdot \\sin \\psi\n$$\n\nBy applying the sine rule to triangle $D N M$ we get\n\n$$\n\\frac{D N}{D M}=\\frac{\\sin \\angle D M N}{\\sin \\angle M N D}=\\frac{\\sin \\varphi}{\\sin \\psi}\n$$\n\nFinally, putting the above observations together, we get\n\n$$\n\\frac{M P}{N Q}=\\frac{2 r_{B} \\sin \\varphi}{2 r_{C} \\sin \\psi}=\\frac{r_{B}}{r_{C}} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{D M}{D N} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=\\frac{\\sin \\psi}{\\sin \\varphi} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi}=1\n$$\n\nso $M P=N Q$ as required.']",,True,,, 1923,Geometry,,"In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\angle P P_{1} C=\angle B A C$. Similarly, on ray $Q A_{1}$, beyond $A_{1}$, let $Q_{1}$ be a point so that $\angle C Q_{1} Q=\angle C B A$. Show that points $P, Q, P_{1}$, and $Q_{1}$ are concyclic.","['Throughout the solution we use oriented angles.\n\nLet rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By\n\n$$\n\\angle Q P A_{2}=\\angle B A A_{2}=\\angle B B_{2} A_{2}=\\angle Q B_{2} A_{2},\n$$\n\npoints $P, Q, A_{2}, B_{2}$ are concyclic; denote the circle passing through these points by $\\omega$. We shall prove that $P_{1}$ and $Q_{1}$ also lie on $\\omega$.\n\n\n\nBy\n\n$$\n\\angle C A_{2} A_{1}=\\angle C A_{2} A=\\angle C B A=\\angle C Q_{1} Q=\\angle C Q_{1} A_{1},\n$$\n\npoints $C, Q_{1}, A_{2}, A_{1}$ are also concyclic. From that we get\n\n$$\n\\angle Q Q_{1} A_{2}=\\angle A_{1} Q_{1} A_{2}=\\angle A_{1} C A_{2}=\\angle B C A_{2}=\\angle B A A_{2}=\\angle Q P A_{2},\n$$\n\nso $Q_{1}$ lies on $\\omega$.\n\nIt follows similarly that $P_{1}$ lies on $\\omega$.', ""First consider the case when lines $P P_{1}$ and $Q Q_{1}$ intersect each other at some point $R$.\n\nLet line $P Q$ meet the sides $A C$ and $B C$ at $E$ and $F$, respectively. Then\n\n$$\n\\angle P P_{1} C=\\angle B A C=\\angle P E C,\n$$\n\nso points $C, E, P, P_{1}$ lie on a circle; denote that circle by $\\omega_{P}$. It follows analogously that points $C, F, Q, Q_{1}$ lie on another circle; denote it by $\\omega_{Q}$.\n\nLet $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem to the lines $A A_{1} P$ and $B B_{1} Q$ provides that points $C=A B_{1} \\cap B A_{1}, R=A_{1} Q \\cap B_{1} P$ and $T=A Q \\cap B P$ are collinear.\n\nLet line $R C T$ meet $P Q$ and $A B$ at $S$ and $U$, respectively. From $A B \\| P Q$ we obtain\n\n$$\n\\frac{S P}{S Q}=\\frac{U B}{U A}=\\frac{S F}{S E}\n$$\n\nSO\n\n$$\nS P \\cdot S E=S Q \\cdot S F\n$$\n\n\n\n\n\nSo, point $S$ has equal powers with respect to $\\omega_{P}$ and $\\omega_{Q}$, hence line $R C S$ is their radical axis; then $R$ also has equal powers to the circles, so $R P \\cdot R P_{1}=R Q \\cdot R Q_{1}$, proving that points $P, P_{1}, Q, Q_{1}$ are indeed concyclic.\n\nNow consider the case when $P P_{1}$ and $Q Q_{1}$ are parallel. Like in the previous case, let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem again to the lines $A A_{1} P$ and $B B_{1} Q$, in this limit case it shows that line $C T$ is parallel to $P P_{1}$ and $Q Q_{1}$.\n\nLet line $C T$ meet $P Q$ and $A B$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $S P \\cdot S E=S Q \\cdot S F$, so $S$ lies on the radical axis between $\\omega_{P}$ and $\\omega_{Q}$.\n\n\n\nLine $C S T$, that is the radical axis between $\\omega_{P}$ and $\\omega_{Q}$, is perpendicular to the line $\\ell$ of centres of $\\omega_{P}$ and $\\omega_{Q}$. Hence, the chords $P P_{1}$ and $Q Q_{1}$ are perpendicular to $\\ell$. So the quadrilateral $P P_{1} Q_{1} Q$ is an isosceles trapezium with symmetry axis $\\ell$, and hence is cyclic.""]",,True,,, 1924,Geometry,,"Let $P$ be a point inside triangle $A B C$. Let $A P$ meet $B C$ at $A_{1}$, let $B P$ meet $C A$ at $B_{1}$, and let $C P$ meet $A B$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $P A_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $P B_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $P C_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $A B C$.","['Since\n\n$$\n\\angle A P B+\\angle B P C+\\angle C P A=2 \\pi=(\\pi-\\angle A C B)+(\\pi-\\angle B A C)+(\\pi-\\angle C B A),\n$$\n\nat least one of the following inequalities holds:\n\n$$\n\\angle A P B \\geqslant \\pi-\\angle A C B, \\quad \\angle B P C \\geqslant \\pi-\\angle B A C, \\quad \\angle C P A \\geqslant \\pi-\\angle C B A .\n$$\n\nWithout loss of generality, we assume that $\\angle B P C \\geqslant \\pi-\\angle B A C$. We have $\\angle B P C>\\angle B A C$ because $P$ is inside $\\triangle A B C$. So $\\angle B P C \\geqslant \\max (\\angle B A C, \\pi-\\angle B A C)$ and hence\n\n$$\n\\sin \\angle B P C \\leqslant \\sin \\angle B A C .\n\\tag{*}\n$$\n\nLet the rays $A P, B P$, and $C P$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\\frac{P B_{1}}{B_{1} B_{3}}$ and $\\frac{P C_{1}}{C_{1} C_{3}}$ is at least 1 , which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\\Omega$.\n\nBecause $A, B, C, B_{3}$ lie on a circle, the triangles $C B_{1} B_{3}$ and $B B_{1} A$ are similar, so\n\n$$\n\\frac{C B_{1}}{B_{1} B_{3}}=\\frac{B B_{1}}{B_{1} A}\n$$\n\nApplying the sine rule we obtain\n\n$$\n\\frac{P B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{C B_{1}}{B_{1} B_{3}}=\\frac{P B_{1}}{C B_{1}} \\cdot \\frac{B B_{1}}{B_{1} A}=\\frac{\\sin \\angle A C P}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle P B A}\n$$\n\nSimilarly,\n\n$$\n\\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin \\angle P B A}{\\sin \\angle B P C} \\cdot \\frac{\\sin \\angle B A C}{\\sin \\angle A C P}\n$$\n\nMultiplying these two equations we get\n\n$$\n\\frac{P B_{1}}{B_{1} B_{3}} \\cdot \\frac{P C_{1}}{C_{1} C_{3}}=\\frac{\\sin ^{2} \\angle B A C}{\\sin ^{2} \\angle B P C} \\geqslant 1\n$$\n\nusing (*), which yields the desired conclusion.', 'Let the rays $A P, B P$, and $C P$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. \nAssume for the sake of contradiction that $A_{2}, B_{2}$, and $C_{2}$ all lie strictly inside circle $A B C$. It follows that $P A_{1}\n\nLet $Y$ be the foot of the bisector of $\\angle B_{3} C P$ in $\\triangle P C B_{3}$. Since $P C_{1}0$ and $\\alpha+\\beta+\\gamma=1$. Then\n\n$$\nA_{1}=\\frac{\\beta B+\\gamma C}{\\beta+\\gamma}=\\frac{1}{1-\\alpha} P-\\frac{\\alpha}{1-\\alpha} A\n$$\n\nso\n\n$$\nA_{2}=2 A_{1}-P=\\frac{1+\\alpha}{1-\\alpha} P-\\frac{2 \\alpha}{1-\\alpha} A\n$$\n\nHence\n\n$$\n\\left|A_{2}\\right|^{2}=\\left(\\frac{1+\\alpha}{1-\\alpha}\\right)^{2}|P|^{2}+\\left(\\frac{2 \\alpha}{1-\\alpha}\\right)^{2}|A|^{2}-\\frac{4 \\alpha(1+\\alpha)}{(1-\\alpha)^{2}} A \\cdot P \\text {. }\n$$\n\nUsing $|A|^{2}=1$ we obtain\n\n$$\n\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}=\\frac{1+\\alpha}{2}|P|^{2}+\\frac{2 \\alpha^{2}}{1+\\alpha}-2 \\alpha A \\cdot P\n\\tag{1}\n$$\n\nLikewise\n\n$$\n\\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}=\\frac{1+\\beta}{2}|P|^{2}+\\frac{2 \\beta^{2}}{1+\\beta}-2 \\beta B \\cdot P\n\\tag{2}\n$$\n\nand\n\n$$\n\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2}=\\frac{1+\\gamma}{2}|P|^{2}+\\frac{2 \\gamma^{2}}{1+\\gamma}-2 \\gamma C \\cdot P\n\\tag{3}\n$$\n\nSumming (1), (2) and (3) we obtain on the LHS the positive linear combination\n\n$$\n\\operatorname{LHS}=\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}\\left|A_{2}\\right|^{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}\\left|B_{2}\\right|^{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\\left|C_{2}\\right|^{2}\n$$\n\nand on the RHS the quantity\n\n$$\n\\left(\\frac{1+\\alpha}{2}+\\frac{1+\\beta}{2}+\\frac{1+\\gamma}{2}\\right)|P|^{2}+\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right)-2(\\alpha A \\cdot P+\\beta B \\cdot P+\\gamma C \\cdot P) .\n$$\n\nThe first term is $2|P|^{2}$ and the last term is $-2 P \\cdot P$, so\n\n$$\n\\begin{aligned}\n\\text { RHS } & =\\left(\\frac{2 \\alpha^{2}}{1+\\alpha}+\\frac{2 \\beta^{2}}{1+\\beta}+\\frac{2 \\gamma^{2}}{1+\\gamma}\\right) \\\\\n& =\\frac{3 \\alpha-1}{2}+\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{3 \\beta-1}{2}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{3 \\gamma-1}{2}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\\\\n& =\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)}+\\frac{(1-\\beta)^{2}}{2(1+\\beta)}+\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} .\n\\end{aligned}\n$$\n\nHere we used the fact that\n\n$$\n\\frac{3 \\alpha-1}{2}+\\frac{3 \\beta-1}{2}+\\frac{3 \\gamma-1}{2}=0 .\n$$\n\nWe have shown that a linear combination of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $\\left|A_{1}\\right|^{2},\\left|B_{1}\\right|^{2}$, and $\\left|C_{1}\\right|^{2}$ must be at least 1 , as required.']",,True,,, 1925,Geometry,,Let $A B C D E$ be a convex pentagon with $C D=D E$ and $\angle E D C \neq 2 \cdot \angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P=A E$ and $B P=B C$. Prove that $P$ lies on the diagonal $C E$ if and only if area $(B C D)+\operatorname{area}(A D E)=$ $\operatorname{area}(A B D)+\operatorname{area}(A B P)$.,"['Let $P^{\\prime}$ be the reflection of $P$ across line $A B$, and let $M$ and $N$ be the midpoints of $P^{\\prime} E$ and $P^{\\prime} C$ respectively. Convexity ensures that $P^{\\prime}$ is distinct from both $E$ and $C$, and hence from both $M$ and $N$. We claim that both the area condition and the collinearity condition in the problem are equivalent to the condition that the (possibly degenerate) right-angled triangles $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar (equivalently, $A P^{\\prime} E$ and $B P^{\\prime} C$ are directly similar).\n\n\n\nFor the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\\angle A F M=\\angle B F N$ of signed angles modulo $\\pi$. By concyclicity of $A P^{\\prime} F M$ and $B F P^{\\prime} N$, this is equivalent to $\\angle A P^{\\prime} M=\\angle B P^{\\prime} N$, which occurs if and only if $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar.\n\n\n\nFor the other equivalence with the area condition, we have the equality of signed areas area $(A B D)+\\operatorname{area}(A B P)=\\operatorname{area}\\left(A P^{\\prime} B D\\right)=\\operatorname{area}\\left(A P^{\\prime} D\\right)+\\operatorname{area}\\left(B D P^{\\prime}\\right)$. Using the identity area $(A D E)-\\operatorname{area}\\left(A P^{\\prime} D\\right)=\\operatorname{area}(A D E)+\\operatorname{area}\\left(A D P^{\\prime}\\right)=2 \\operatorname{area}(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality\n\n$$\n\\operatorname{area}(D A M)=\\operatorname{area}(D B N)\n$$\n\nNow note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\\prime} E$ and $P^{\\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as\n\n$$\nM A \\cdot G D=N B \\cdot H D \\text {. }\n$$\n\n(In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\\prime} E$ from $P^{\\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\\prime} E$ at an angle $\\pi / 2$ clockwise from the oriented segment $P^{\\prime} E$ - we adopt the analogous conventions at $B$.)\n\n\n\n\n\nTo relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\\prime} M$ and $B P^{\\prime} N$, we use the following calculation.\n\nClaim. Let $\\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\\prime}$ with respect to $\\Gamma$. Then we have the equality\n\n$$\nG D \\cdot P^{\\prime} M=H D \\cdot P^{\\prime} N=\\frac{1}{4} h \\neq 0\n$$\n\nProof. Firstly, we have $h \\neq 0$, since otherwise $P^{\\prime}$ would lie on $\\Gamma$, and hence the internal angle bisectors of $\\angle E D P^{\\prime}$ and $\\angle P^{\\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\\angle E D C \\neq 2 \\cdot \\angle A D B$ given in the question.\n\nNext, let $E^{\\prime}$ denote the second point of intersection of $P^{\\prime} E$ with $\\Gamma$, and let $E^{\\prime \\prime}$ denote the point on $\\Gamma$ diametrically opposite $E^{\\prime}$, so that $E^{\\prime \\prime} E$ is perpendicular to $P^{\\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\\prime} E$ and $E E^{\\prime \\prime}$ of the right-angled triangle $P^{\\prime} E E^{\\prime \\prime}$; it follows that $G$ is the midpoint of $P^{\\prime} E^{\\prime \\prime}$. Since $D$ is the midpoint of $E^{\\prime} E^{\\prime \\prime}$, we have that $G D=\\frac{1}{2} P^{\\prime} E^{\\prime}$. Since $P^{\\prime} M=\\frac{1}{2} P^{\\prime} E$, we have $G D \\cdot P^{\\prime} M=\\frac{1}{4} P^{\\prime} E^{\\prime} \\cdot P^{\\prime} E=\\frac{1}{4} h$. The other equality $H D \\cdot P^{\\prime} N$ follows by exactly the same argument.\n\n\n\nFrom this claim, we see that the area condition is equivalent to the equality\n\n$$\n\\left(M A: P^{\\prime} M\\right)=\\left(N B: P^{\\prime} N\\right)\n$$\n\nof ratios of signed lengths, which is equivalent to direct similarity of $A P^{\\prime} M$ and $B P^{\\prime} N$, as desired.', 'Along the perpendicular bisector of $C E$, define the linear function\n\n$$\nf(X)=\\operatorname{area}(B C X)+\\operatorname{area}(A X E)-\\operatorname{area}(A B X)-\\operatorname{area}(A B P)\n$$\n\nwhere, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D)=0$.\n\n\n\nLet $P^{\\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\\prime}$ does not lie on the line $C E$. To see this, we let $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\\prime}$, so that $P$ is the orthogonal projection of $P^{\\prime}$ onto $A^{\\prime \\prime} B^{\\prime \\prime}$. Since $A$ lies on the perpendicular bisector of $P^{\\prime} E$, the triangle $A^{\\prime \\prime} E P^{\\prime}$ is right-angled at $E$ (and $B^{\\prime \\prime} C P^{\\prime}$ similarly). If $P^{\\prime}$ were to lie on $C E$, then the lines $A^{\\prime \\prime} E$ and $B^{\\prime \\prime} C$ would be perpendicular to $C E$ and $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\\prime \\prime} B^{\\prime \\prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\\prime}$ on $C E$.\n\nWe thus let $O$ be the centre of the circle $C E P^{\\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\\prime}$ and $C P^{\\prime}$, respectively, so\n\n$$\n\\begin{aligned}\n\\operatorname{area}(B C O)+\\operatorname{area}(A O E) & =\\operatorname{area}\\left(O P^{\\prime} B\\right)+\\operatorname{area}\\left(P^{\\prime} O A\\right)=\\operatorname{area}\\left(P^{\\prime} B O A\\right) \\\\\n& =\\operatorname{area}(A B O)+\\operatorname{area}\\left(B A P^{\\prime}\\right)=\\operatorname{area}(A B O)+\\operatorname{area}(A B P)\n\\end{aligned}\n$$\n\nand hence $f(O)=0$.\n\nNotice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\\angle E O C=2 \\cdot \\angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D)=0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear.\n\n\nIn the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides\n\n$$\n\\operatorname{area}(B C T)+\\operatorname{area}(A T E)=\\operatorname{area}(P B T A)=\\operatorname{area}(A B T)-\\operatorname{area}(A B P)\n$$\n\n$$\nf(T)=-2 \\operatorname{area}(A B P) \\neq 0 .\n$$\n\n\n\nHence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \\neq 0$.\n\nIn the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes.\n\n\n\nLet $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\\prime}$ and $B^{\\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2 , then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\\prime}$ and $B^{\\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so\n\n$$\nf(Q)=(\\operatorname{area}(B C Q)+\\operatorname{area}(A Q E))-(\\operatorname{area}(A B Q)-\\operatorname{area}(B A P))=0-0=0\n$$\n\nThe points $O$ and $Q$ are distinct. To see this, consider the circle $\\omega$ centred on $Q$ with $P^{\\prime}$ on the circumference; since triangle $P P^{\\prime} Q$ is right-angled at $P^{\\prime}$, it follows that $P$ lies outside $\\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\\omega$, so that $\\omega$ is not the circle $C E P^{\\prime}$ and $Q \\neq O$.\n\nSince $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D)=0$ as desired.']",,True,,, 1926,Geometry,,"Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C, C A$, and $A B$ at $D, E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle D P A+\angle A Q D=\angle Q I P$.","[""Let $N$ and $M$ be the midpoints of the $\\operatorname{arcs} \\overparen{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\\angle F A E=\\angle B A C=\\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\\varphi$ (dilation in case of $A B=A C$ ) that moves $A F I E$ to $N B M C$. The directed angle in which $\\varphi$ changes directions is $\\angle(A F, N B)$, same as $\\angle(A P, N P)$ and $\\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P=E F \\cap A P$ and $Q=E F \\cap A Q$ are mapped to points $B C \\cap N P$ and $B C \\cap N Q$, respectively. Denote these points by $P^{\\prime}$ and $Q^{\\prime}$, respectively.\n\n\n\nLet $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D=L$ then line $B C$ is tangent to circle $P Q D$ ). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have\n\n$$\n\\frac{B D}{D C}=\\frac{B F}{F A} \\cdot \\frac{A E}{E C}=-\\frac{B Z}{Z C}\n$$\n\nso the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \\cdot Z C=Z D \\cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \\cdot Z L=Z B \\cdot Z C=Z P \\cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim.\n\nBy $\\angle M P P^{\\prime}=\\angle M Q Q^{\\prime}=\\angle M L P^{\\prime}=\\angle M L Q^{\\prime}=90^{\\circ}$, the quadrilaterals $M L P P^{\\prime}$ and $M L Q Q^{\\prime}$ are cyclic. Then the problem statement follows by\n\n$$\n\\begin{aligned}\n\\angle D P A+\\angle A Q D & =360^{\\circ}-\\angle P A Q-\\angle Q D P=360^{\\circ}-\\angle P N Q-\\angle Q L P \\\\\n& =\\angle L P N+\\angle N Q L=\\angle P^{\\prime} M L+\\angle L M Q^{\\prime}=\\angle P^{\\prime} M Q^{\\prime}=\\angle P I Q .\n\\end{aligned}\n$$"", 'Define the point $M$ and the same spiral similarity $\\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$.\n\nBy $\\varphi(F)=B$ and $\\varphi(E)=C$ the triangles $S B F$ and $S C E$ are similar, so we have\n\n$$\n\\frac{S B}{S C}=\\frac{B F}{C E}=\\frac{B D}{C D}\n$$\n\nBy the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\\angle B S C$ and hence passes through $M$.\n\nLet $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\\varphi(K)=\\varphi(F E \\cap S I)=B C \\cap S M=D$. By $\\varphi(I)=M$, we have $K D \\| I M$.\n\n\n\nWe claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\\angle A M L=\\angle A S L=\\angle A S I=90^{\\circ}$, we have $M L \\| P Q$. Hence, $\\overparen{P L}=\\overparen{M Q}$ and therefore $\\angle P S L=\\angle M S Q=\\angle D S Q$. By $\\angle Q P S=\\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally,\n\n$$\n\\frac{S P}{S I}=\\frac{S P}{S K} \\cdot \\frac{S K}{S I}=\\frac{S M}{S Q} \\cdot \\frac{S D}{S M}=\\frac{S D}{S Q}\n$$\n\nshows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously.\n\nNow the problem statement can be proved by\n\n$$\n\\angle D P A+\\angle A Q D=\\angle D P S+\\angle S Q D=\\angle Q I S+\\angle S I P=\\angle Q I P\n$$', 'Denote the circumcircle of triangle $A B C$ by $\\Gamma$, and let rays $P D$ and $Q D$ meet $\\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \\perp I P$ and $A V \\perp I Q$. Then the problem statement will follow as\n\n$$\n\\angle D P A+\\angle A Q D=\\angle V U A+\\angle A V U=180^{\\circ}-\\angle U A V=\\angle Q I P .\n$$\n\nLet $M$ be the midpoint of arc $\\overparen{B U V C}$ and let $N$ be the midpoint of $\\operatorname{arc} \\overparen{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively.\n\nwe observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively.\n\nCombine the inversion with projecting $\\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively.\n\n\n\nThe combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \\perp I F$, $A C \\perp I E$ and $A N \\perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \\perp I P$ and $A V \\perp I Q$, and hence completes the solution.']",,True,,, 1927,Geometry,,"The incircle $\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\omega$ again at $R$. Line $A R$ meets $\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$. $\angle(a, b)$ denotes the directed angle between lines $a$ and $b$, measured modulo $\pi$.","['Step 1. The external bisector of $\\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\\omega$. Since $A N \\cdot A I=A E^{2}=A R \\cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R=I P$, the line $N I$ is the external bisector of $\\angle P N R$, so $P N$ meets $\\omega$ again at the point symmetric to $R$ with respect to $A N$ - i.e. at $K$.\n\nLet $D N$ cross $\\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\\omega$ meet on the polar line of the intersection of the diagonals with respect to $\\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear.\n\n\nStep 2. Let $\\Gamma$ be the circumcircle of $\\triangle B I C$. Notice that\n\n$$\n\\begin{aligned}\n\\angle(B Q, Q C)=\\angle & (B Q, Q P)+\\angle(P Q, Q C)=\\angle(B F, F P)+\\angle(P E, E C) \\\\\n& =\\angle(E F, E P)+\\angle(F P, F E)=\\angle(F P, E P)=\\angle(D F, D E)=\\angle(B I, I C),\n\\end{aligned}\n$$\n\nso $Q$ lies on $\\Gamma$. Let $Q P$ meet $\\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\\angle(B I, I T)=\\angle(B Q, Q T)=\\angle(B F, F P)=\\angle(F K, K P)$. Note $F D \\perp F K$ and $F D \\perp B I$ so $F K \\| B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I=I K$, the line $I T$ crosses $D N$ at its midpoint $M$.\n\nStep 3. Let $F^{\\prime}$ and $E^{\\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\\prime} \\cdot E^{\\prime} F=D E^{\\prime 2}=$ $B E^{\\prime} \\cdot E^{\\prime} I$, the point $E^{\\prime}$ lies on the radical axis of $\\omega$ and $\\Gamma$; the same holds for $F^{\\prime}$. Therefore, this radical axis is $E^{\\prime} F^{\\prime}$, and it passes through $M$. Thus $I M \\cdot M T=D M \\cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\\angle(D S, S T)=\\angle(D I, I T)=\\angle(D K, K P)=\\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired.\n\n\n\n', 'Namely, we introduce the points $K, L, N$, and $S$, and show that the triples $(P, N, K)$ and $(P, S, L)$ are collinear. We conclude that $K$ and $R$ are symmetric in $A I$, and reduce the problem statement to showing that $P, Q$, and $S$ are collinear.\n\nStep 1. Let $A R$ meet the circumcircle $\\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\\Omega$ again at the midpoint $M^{\\prime}$ of arc $B A C$.\n\nStep 2. Denote the circles $B F P$ and $C E P$ by $\\Omega_{B}$ and $\\Omega_{C}$, respectively. Let $\\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have\n\n$$\n\\angle(U B, B F)=\\angle(U P, P F)=\\angle(R P, P F)=\\angle(R F, F A),\n$$\n\nso $U B \\| R F$.\n\n\n\nNext, we show that the points $B, I, U$, and $X$ are concyclic. Since\n\n$$\n\\angle(U B, U X)=\\angle(R F, R X)=\\angle(A F, A R)+\\angle(F R, F A)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F),\n$$\n\nit suffices to prove $\\angle(I B, I X)=\\angle\\left(M^{\\prime} B, M^{\\prime} X\\right)+\\angle(D R, D F)$, or $\\angle\\left(I B, M^{\\prime} B\\right)=\\angle(D R, D F)$. But both angles equal $\\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\\prime}$ is the midpoint of $\\operatorname{arc} B A C$ of $\\Omega$.)\n\nIt follows now from circles $B U I X$ and $B P U F Y$ that\n\n$$\n\\begin{aligned}\n\\angle(I U, U B)=\\angle(I X, B X)=\\angle\\left(M^{\\prime} X, B X\\right)= & \\frac{\\pi-\\angle A}{2} \\\\\n& =\\angle(E F, A F)=\\angle(Y F, B F)=\\angle(Y U, B U),\n\\end{aligned}\n$$\n\nso the points $Y, U$, and $I$ are collinear.\n\nLet $E F$ meet $B C$ at $W$. We have\n\n$$\n\\angle(I Y, Y W)=\\angle(U Y, F Y)=\\angle(U B, F B)=\\angle(R F, A F)=\\angle(C I, C W)\n$$\n\nso the points $W, Y, I$, and $C$ are concyclic.\n\n\n\nSimilarly, if $V$ and $Z$ are the second meeting points of $\\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples $(C, V, I, X)$ and $(B, I, Z, W)$ are both concyclic.\n\nStep 3. Let $Q^{\\prime}=C Y \\cap B Z$. We will show that $Q^{\\prime}=Q$.\n\nFirst of all, we have\n\n$$\n\\begin{aligned}\n& \\angle\\left(Q^{\\prime} Y, Q^{\\prime} B\\right)=\\angle(C Y, Z B)=\\angle(C Y, Z Y)+\\angle(Z Y, B Z) \\\\\n& \\quad=\\angle(C I, I W)+\\angle(I W, I B)=\\angle(C I, I B)=\\frac{\\pi-\\angle A}{2}=\\angle(F Y, F B)\n\\end{aligned}\n$$\n\nso $Q^{\\prime} \\in \\Omega_{B}$. Similarly, $Q^{\\prime} \\in \\Omega_{C}$. Thus $Q^{\\prime} \\in \\Omega_{B} \\cap \\Omega_{C}=\\{P, Q\\}$ and it remains to prove that $Q^{\\prime} \\neq P$. If we had $Q^{\\prime}=P$, we would have $\\angle(P Y, P Z)=\\angle\\left(Q^{\\prime} Y, Q^{\\prime} Z\\right)=\\angle(I C, I B)$. This would imply\n\n$$\n\\angle(P Y, Y F)+\\angle(E Z, Z P)=\\angle(P Y, P Z)=\\angle(I C, I B)=\\angle(P E, P F)\n$$\n\nso circles $\\Omega_{B}$ and $\\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\\prime}=Q$.\n\n\n\nStep 4. Now we are ready to show that $P, Q$, and $S$ are collinear.\n\nNotice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\\omega$, so $W$ is the pole of $A D$. Hence, $W I \\perp A D$. Since $C I \\perp D E$, this yields $\\angle(I C, W I)=\\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\\triangle D E F$, so $\\angle(D E, D A)=\\angle(D N, D F)=\\angle(D S, D F)$. Therefore,\n\n$$\n\\begin{aligned}\n\\angle(P S, P F)=\\angle(D S, D F)=\\angle(D E, D A)= & \\angle(I C, I W) \\\\\n& =\\angle(Y C, Y W)=\\angle(Y Q, Y F)=\\angle(P Q, P F)\n\\end{aligned}\n$$\n\nwhich yields the desired collinearity.']",,True,,, 1928,Geometry,,"Let $\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$. Suppose that for any point $X$, and for any three lines $\ell_{1}, \ell_{2}, \ell_{3}$ passing through $X$, the points $f\left(\ell_{1}\right), f\left(\ell_{2}\right), f\left(\ell_{3}\right)$ and $X$ lie on a circle. Prove that there is a unique point $P$ such that $f(\ell)=P$ for any line $\ell$ passing through $P$.","[""We provide a complete characterisation of the functions satisfying the given condition.\n\nWrite $\\angle\\left(\\ell_{1}, \\ell_{2}\\right)$ for the directed angle modulo $180^{\\circ}$ between the lines $\\ell_{1}$ and $\\ell_{2}$. Given a point $P$ and an angle $\\alpha \\in\\left(0,180^{\\circ}\\right)$, for each line $\\ell$, let $\\ell^{\\prime}$ be the line through $P$ satisfying $\\angle\\left(\\ell^{\\prime}, \\ell\\right)=\\alpha$, and let $h_{P, \\alpha}(\\ell)$ be the intersection point of $\\ell$ and $\\ell^{\\prime}$. We will prove that there is some pair $(P, \\alpha)$ such that $f$ and $h_{P, \\alpha}$ are the same function. Then $P$ is the unique point in the problem statement.\n\nGiven an angle $\\alpha$ and a point $P$, let a line $\\ell$ be called $(P, \\alpha)$-good if $f(\\ell)=h_{P, \\alpha}(\\ell)$. Let a point $X \\neq P$ be called $(P, \\alpha)$-good if the circle $g(X)$ passes through $P$ and some point $Y \\neq P, X$ on $g(X)$ satisfies $\\angle(P Y, Y X)=\\alpha$. It follows from this definition that if $X$ is $(P, \\alpha)$ good then every point $Y \\neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \\alpha}(X Y)=Y$ for every $Y \\in g(X)$. Equivalently, $f(\\ell) \\in\\left\\{X, h_{P, \\alpha}(\\ell)\\right\\}$ for each line $\\ell$ passing through $X$. This shows the following lemma.\n\nLemma 1. If $X$ is $(P, \\alpha)$-good and $\\ell$ is a line passing through $X$ then either $f(\\ell)=X$ or $\\ell$ is $(P, \\alpha)$-good.\n\nLemma 2. If $X$ and $Y$ are different $(P, \\alpha)$-good points, then line $X Y$ is $(P, \\alpha)$-good.\n\nProof. If $X Y$ is not $(P, \\alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \\neq Y$.\n\nLemma 3. If $\\ell_{1}$ and $\\ell_{2}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then either $f\\left(\\ell_{1}\\right)=X$ or $f\\left(\\ell_{2}\\right)=X$ or $X$ is $(P, \\alpha)$-good.\n\nProof. If $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right) \\neq X$, then $g(X)$ is the circumcircle of $X, f\\left(\\ell_{1}\\right)$ and $f\\left(\\ell_{2}\\right)$. Since $\\ell_{1}$ and $\\ell_{2}$ are $(P, \\alpha)$-good lines, the angles\n\n$$\n\\angle\\left(\\operatorname{Pf}\\left(\\ell_{1}\\right), f\\left(\\ell_{1}\\right) X\\right)=\\angle\\left(\\operatorname{Pf}\\left(\\ell_{2}\\right), f\\left(\\ell_{2}\\right) X\\right)=\\alpha\n$$\n\nso $P$ lies on $g(X)$. Hence, $X$ is $(P, \\alpha)$-good.\n\nLemma 4. If $\\ell_{1}, \\ell_{2}$ and $\\ell_{3}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then $X$ is $(P, \\alpha)$-good.\n\nProof. This follows from the previous Lemma since at most one of the three lines $\\ell_{i}$ can satisfy $f\\left(\\ell_{i}\\right)=X$ as the three lines are all $(P, \\alpha)$-good.\n\nLemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\\alpha$ such that $A, B$ and $C$ are $(P, \\alpha) \\operatorname{good}$ points and $A B, B C$ and $C A$ are $(P, \\alpha)$-good lines.\n\n\n\n\n\nProof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \\neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles\n\n$$\n\\angle(P D, D C)=\\angle(P F, F B)=\\angle(P E, E A)=\\alpha\n$$\n\nfor some angle $\\alpha$. Hence, lines $A B, B C$ and $C A$ are all $(P, \\alpha)$-good, so by Lemma $3, A, B$ and $C$ are $(P, \\alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.)\n\nConsider the set $\\Omega$ of all points $(x, y)$ with integer coordinates $1 \\leqslant x, y \\leqslant 1000$, and consider the set $L_{\\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\\Omega}$. For each line $\\ell$ in $L_{\\Omega}$ we colour the point $f(\\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5 , so these three vertices are all $(P, \\alpha)$-good for some point $P$ and angle $\\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \\alpha)$-good for this particular pair $(P, \\alpha)$. Now by Lemma 1 , the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that $(m+1, n+1)$ is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points $(m, n+2)$ and $(m+2, n)$ then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\\Omega_{10}$ are good.\n\nNow we will use this to prove that every point $S \\neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\\ell$ through $S$ intersects at most 10 points in $\\Omega_{10}$. Hence, there are at least eight lines $\\ell$ through $S$ which contain a point $Q$ in $\\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\\ell) \\neq Q$. Hence, by Lemma 1, the line $\\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4 , the point $S$ is good. Hence, every point $S \\neq P$ is good, so by Lemma 2 , every line is good. In particular, every line $\\ell$ passing through $P$ is good, and therefore satisfies $f(\\ell)=P$, as required."", ""Note that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \\in g(X) \\cap g(Y) \\cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$.\n\nLemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \\notin\\{X, Y\\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$.\n\nProof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram below)\n\n\n\n\n\nNow pick any line $\\ell$ and any six different points $Y_{1}, \\ldots, Y_{6}$ on $\\ell \\backslash\\{f(\\ell)\\}$. Pick a point $X$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \\ldots, Y_{4}$ do not lie on $g(X)$, so that $f\\left(X Y_{i}\\right) \\notin\\left\\{X, Y_{i}\\right\\}$ for $1 \\leqslant i \\leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \\leqslant i0$, there is a point $P_{\\epsilon}$ with $g\\left(P_{\\epsilon}\\right)$ of radius at most $\\epsilon$. Then there is a point $P$ with the given property.\n\nProof. Consider a sequence $\\epsilon_{i}=2^{-i}$ and corresponding points $P_{\\epsilon_{i}}$. Because the two circles $g\\left(P_{\\epsilon_{i}}\\right)$ and $g\\left(P_{\\epsilon_{j}}\\right)$ meet, the distance between $P_{\\epsilon_{i}}$ and $P_{\\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\\sum_{i} \\epsilon_{i}$ converges, these points converge to some point $P$. For all $\\epsilon>0$, the point $P$ has distance at most $2 \\epsilon$ from $P_{\\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \\epsilon$ from $P_{\\epsilon}$, so distance at most $4 \\epsilon$ from $P$. A circle that passes distance at most $4 \\epsilon$ from $P$ for all $\\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property.\n\nLemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property.\n\nProof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2 .\n\nLemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\\left(B_{i}\\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property.\n\nProof. Consider some line $\\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\\left(B_{i}\\right)$. Fix some direction along that line, and let $X_{\\epsilon}$ be the point on $\\ell$ that has distance $\\epsilon$ from $A$ in that direction. In what follows we consider only those $\\epsilon$ for which $X_{\\epsilon}$ does not lie on any $g\\left(B_{i}\\right)$ (this restriction excludes only finitely many possible values of $\\epsilon$ ).\n\nConsider the circle $g\\left(X_{\\epsilon}\\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\\epsilon \\rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\\left(X_{\\epsilon}\\right)$ for any $\\epsilon$ in that sequence.\n\n\n\nThen $f\\left(X_{\\epsilon} B_{1}\\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\\epsilon} B_{1}$ with $g\\left(B_{1}\\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\\epsilon}, f\\left(X_{\\epsilon} B_{1}\\right)$ and $f\\left(X_{\\epsilon} B_{2}\\right)$. As $\\epsilon \\rightarrow 0$, the angle at $X_{\\epsilon}$ tends to $\\angle B_{1} A B_{2}$ or $180^{\\circ}-\\angle B_{1} A B_{2}$, which is not 0 or $180^{\\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \\epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\\left(X_{\\epsilon}\\right)$, is also bounded above by a constant multiple of $\\epsilon$, and so the result follows by Lemma 2 .\n\nLemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property.\n\nProof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.\n\nCase 1: D does not lie on that line:\n\nIn this case, consider a sequence of $X_{\\epsilon}$ at distance $\\epsilon$ from $D$, tending to $D$ along some line that is not a tangent to either circle, but perturbed slightly (by at most $\\epsilon^{2}$ ) to ensure that no three of the points $A, B$ and $X_{\\epsilon}$ are collinear and no four are concyclic.\n\nConsider the points $f\\left(X_{\\epsilon} A\\right)$ and $f\\left(X_{\\epsilon} B\\right)$, and the circles $g\\left(X_{\\epsilon}\\right)$ on which they lie. The point $f\\left(X_{\\epsilon} A\\right)$ might be either $A$ or the other intersection of $X_{\\epsilon} A$ with the circle $g(A)$, and the same applies for $B$. If, for some sequence of $\\epsilon \\rightarrow 0$, both those points are the other point of intersection, the same argument as in the proof of Lemma 4 applies to find arbitrarily small circles. Otherwise, we have either infinitely many of those circles passing through $A$, or infinitely many passing through $B$; without loss of generality, suppose infinitely many through $A$.\n\nWe now show we can find five points $B_{i}$ satisfying the conditions of Lemma 4 (together with $A$ ). Let $B_{1}$ be any of the $X_{\\epsilon}$ for which $g\\left(X_{\\epsilon}\\right)$ passes through $A$. Then repeat the following four times, for $2 \\leqslant i \\leqslant 5$.\n\nConsider some line $\\ell=X_{\\epsilon} A$ (different from those considered for previous $i$ ) that is not tangent to any of the $g\\left(B_{j}\\right)$ for $j\n\nSo, Case 2.1 can occur for only finitely many points $X$.\n\nCase 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.\n\nClearly, $R_{1} \\neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\\ell$ only at $X$ and $Q$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P X R_{i}\\right)$. Let $R_{i} Q$ meet $g\\left(R_{i}\\right)$ again at $S_{i}$; then $S_{i} \\neq Q$, as $g\\left(R_{i}\\right)$ meets $\\ell$ only at $P$ and $X$. Then $f\\left(R_{i} Q\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases.\n\n\n\nSubcase 2.2.1: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=S_{2}$; so $S_{1}, S_{2} \\in g(Q)$.\n\nIn this case we have $0=\\angle\\left(R_{1} X, X P\\right)+\\angle\\left(X P, R_{2} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} R_{2}\\right)=$ $\\angle\\left(Q S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} Q\\right)$, which shows $P \\in g(Q)$.\n\nSubcase 2.2.2: $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=R_{2}$; so $R_{1}, R_{2} \\in g(Q)$.\n\nThis can happen for at most four positions of $X$ - namely, at the intersections of $\\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \\in g(Q) \\cap p_{i}$.\n\n\n\nSubcase 2.2.3: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=R_{2}$ (the case $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=S_{2}$ is similar).\n\nIn this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\\angle\\left(Q S_{1}, S_{1} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} X\\right)=$ $\\angle\\left(R_{1} P, P X\\right)=\\angle\\left(p_{1}, \\ell\\right) \\neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \\rightarrow \\ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction.\n\nThus Case 2.2 also occurs for only finitely many points $X$, as desired.\n\nStep 3: We show that $f(P Q)=P$, as desired.\n\nThe argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \\in \\ell \\backslash\\{P, Q\\}$, which is absurd.\n\nCase 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$.\n\nLet $t(Q) \\operatorname{cross} p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X$ cross $g(R)$ again at $S$. Then $f(R X) \\in\\{R, S\\} \\cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$.\n\n\n\nSubcase 3.1.1: $S=f(R X) \\in g(X)$.\n\nWe have $\\angle(t(X), Q X)=\\angle(S X, S Q)=\\angle(S R, S Q)=\\angle(P R, P Q)=\\angle\\left(p_{2}, \\ell\\right)$. Hence $t(X) \\| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\\prime}$, so $g\\left(R^{\\prime}\\right)=\\left(P R^{\\prime} X\\right)$, and let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$; notice that $S^{\\prime} \\neq Q$. Excluding one position of $X$, we may assume that $R^{\\prime} \\notin g(Q)$, so $R^{\\prime} \\neq f\\left(R^{\\prime} Q\\right)$. Therefore, $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then, as in Case 2.1, we get $\\angle(t(Q), P Q)=\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(p_{2}, \\ell\\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible.\n\nSubcase 3.1.2: $R=f(R X) \\in g(X)$.\n\nIn this case, we have $\\angle(t(X), \\ell)=\\angle(R X, R Q)=\\angle\\left(R X, p_{1}\\right)$. Again, let $R^{\\prime}=t(X) \\cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\\left(R^{\\prime}\\right)=\\left(R^{\\prime} X P\\right)$; let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$. Due to $\\angle\\left(R^{\\prime} X, X R\\right)=\\angle(Q X, Q R)=\\angle\\left(\\ell, p_{1}\\right), R^{\\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\\prime} \\notin g(Q)$. Therefore, for those positions we have $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then $\\angle\\left(R X, p_{1}\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(R^{\\prime} S^{\\prime}, S^{\\prime} P\\right)=$ $\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well.\n\nThus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction.\n\n\n\nCase 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.\n\nBy Step $2, R_{1} \\neq R_{2}$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P Q R_{i}\\right)$. Let $R_{i} X$ meet $g\\left(R_{i}\\right)$ at $S_{i}$; then $S_{i} \\neq X$. Then $f\\left(R_{i} X\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases.\n\n\n\nSubcase 3.2.1: $f\\left(R_{1} X\\right)=S_{1}$ and $f\\left(R_{2} X\\right)=S_{2}$, so $S_{1}, S_{2} \\in g(X)$.\n\nAs in Subcase 2.2.1, we have $0=\\angle\\left(R_{1} Q, Q P\\right)+\\angle\\left(Q P, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} X\\right)$, which shows $P \\in g(X)$. But $X, Q \\in g(X)$ as well, so $g(X)$ meets $\\ell$ at three distinct points, which is absurd.\n\nSubcase 3.2.2: $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=R_{2}$, so $R_{1}, R_{2} \\in g(X)$.\n\nNow three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible.\n\nSubcase 3.2.3: $f\\left(R_{1} X\\right)=S_{1}, f\\left(R_{2} X\\right)=R_{2}$ (the case $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=S_{2}$ is similar).\n\nWe have $\\angle\\left(X R_{2}, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} P, P Q\\right)=\\angle\\left(p_{1}, \\ell\\right)$, so this case can occur for a unique position of $X$.\n\nThus, in Case 3.2, there is only a unique position of $X$, again yielding the required contradiction.\n\n\n### 分析\nThe condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\\ell$ passing through $X$, the point $f(\\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\\ell)$ other than $X$; for such points, an arbitrary choice is made.)\n\nIf there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.']",,True,,, 1929,Number Theory,,"Find all pairs $(m, n)$ of positive integers satisfying the equation $$ \left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right)=m ! \tag{1} $$","['For any prime $p$ and positive integer $N$, we will denote by $v_{p}(N)$ the exponent of the largest power of $p$ that divides $N$. The left-hand side of (1) will be denoted by $L_{n}$; that is, $L_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right)\\left(2^{n}-4\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)$.\n\nWe will get an upper bound on $n$ from the speed at which $v_{2}\\left(L_{n}\\right)$ grows.\n\nFrom\n\n$$\nL_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)=2^{1+2+\\cdots+(n-1)}\\left(2^{n}-1\\right)\\left(2^{n-1}-1\\right) \\cdots\\left(2^{1}-1\\right)\n$$\n\nwe read\n\n$$\nv_{2}\\left(L_{n}\\right)=1+2+\\cdots+(n-1)=\\frac{n(n-1)}{2}\n$$\n\nOn the other hand, $v_{2}(m !)$ is expressed by the Legendre formula as\n\n$$\nv_{2}(m !)=\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{m}{2^{i}}\\right\\rfloor\n$$\n\nAs usual, by omitting the floor functions,\n\n$$\nv_{2}(m !)<\\sum_{i=1}^{\\infty} \\frac{m}{2^{i}}=m\n$$\n\nThus, $L_{n}=m$ ! implies the inequality\n\n$$\n\\frac{n(n-1)}{2}1.3 \\cdot 10^{12}$.\n\nFor $n \\geqslant 7$ we prove (3) by the following inequalities:\n\n$$\n\\begin{aligned}\n\\left(\\frac{n(n-1)}{2}\\right) ! & =15 ! \\cdot 16 \\cdot 17 \\cdots \\frac{n(n-1)}{2}>2^{36} \\cdot 16^{\\frac{n(n-1)}{2}-15} \\\\\n& =2^{2 n(n-1)-24}=2^{n^{2}} \\cdot 2^{n(n-2)-24}>2^{n^{2}}\n\\end{aligned}\n$$\n\n\n\nPutting together (2) and (3), for $n \\geqslant 6$ we get a contradiction, since\n\n$$\nL_{n}<2^{n^{2}}<\\left(\\frac{n(n-1)}{2}\\right) !a^{3} .\n$$\n\nSo $3 a^{3} \\geqslant(a b c)^{2}>a^{3}$ and hence $3 a \\geqslant b^{2} c^{2}>a$. Now $b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right) \\geqslant a^{2}$, and so\n\n$$\n18 b^{3} \\geqslant 9\\left(b^{3}+c^{3}\\right) \\geqslant 9 a^{2} \\geqslant b^{4} c^{4} \\geqslant b^{3} c^{5},\n$$\n\nso $18 \\geqslant c^{5}$ which yields $c=1$.\n\nNow, note that we must have $a>b$, as otherwise we would have $2 b^{3}+1=b^{4}$ which has no positive integer solutions. So\n\n$$\na^{3}-b^{3} \\geqslant(b+1)^{3}-b^{3}>1\n$$\n\nand\n\n$$\n2 a^{3}>1+a^{3}+b^{3}>a^{3},\n$$\n\nwhich implies $2 a^{3}>a^{2} b^{2}>a^{3}$ and so $2 a>b^{2}>a$. Therefore\n\n$$\n4\\left(1+b^{3}\\right)=4 a^{2}\\left(b^{2}-a\\right) \\geqslant 4 a^{2}>b^{4},\n$$\n\nso $4>b^{3}(b-4)$; that is, $b \\leqslant 4$.\n\nNow, for each possible value of $b$ with $2 \\leqslant b \\leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows:\n\n$$\n\\begin{aligned}\n& b=2: \\quad a^{3}-4 a^{2}+9=0 \\\\\n& b=3: \\quad a^{3}-9 a^{2}+28=0 \\\\\n& b=4: \\quad a^{3}-16 a^{2}+65=0\n\\end{aligned}\n$$\n\nThe only case with an integer solution for $a$ with $b \\leqslant a$ is $b=2$, leading to $(a, b, c)=(3,2,1)$.', 'Note that the equation is symmetric. We will assume without loss of generality that $a \\geqslant b \\geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.\n\nAgain, we will start by proving that $c=1$. Suppose otherwise that $c \\geqslant 2$. We have $a^{3}+b^{3}+c^{3} \\leqslant 3 a^{3}$, so $b^{2} c^{2} \\leqslant 3 a$. Since $c \\geqslant 2$, this tells us that $b \\leqslant \\sqrt{3 a / 4}$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \\leqslant 2 b^{3} \\leqslant 2(3 a / 4)^{3 / 2}$. In other words, $a \\leqslant \\frac{27}{16}<2$, which contradicts the assertion that $a \\geqslant c \\geqslant 2$. So there are no solutions in this case, and so we must have $c=1$.\n\nNow, the original equation becomes $a^{3}+b^{3}+1=a^{2} b^{2}$. Observe that $a \\geqslant 2$, since otherwise $a=b=1$ as $a \\geqslant b$.\n\nThe right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3}+1 \\geqslant$ $a^{2}$. Since $a \\geqslant b$, we also have\n\n$$\nb^{2}=a+\\frac{b^{3}+1}{a^{2}} \\leqslant 2 a+\\frac{1}{a^{2}}\n$$\n\nand so $b^{2} \\leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2}+1 \\geqslant b^{3}+1 \\geqslant a^{2}$, from which we deduce $a \\leqslant 8$.\n\nNow, for each possible value of $a$ with $2 \\leqslant a \\leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows:\n\n$$\n\\begin{array}{ll}\na=2: & b^{3}-4 b^{2}+9=0 \\\\\na=3: & b^{3}-9 b^{2}+28=0 \\\\\na=4: & b^{3}-16 b^{2}+65=0 \\\\\na=5: & b^{3}-25 b^{2}+126=0 \\\\\na=6: & b^{3}-36 b^{2}+217=0 \\\\\na=7: & b^{3}-49 b^{2}+344=0 \\\\\na=8: & b^{3}-64 b^{2}+513=0 .\n\\end{array}\n$$\n\nThe only case with an integer solution for $b$ with $a \\geqslant b$ is $a=3$, leading to $(a, b, c)=(3,2,1)$.', 'Note that the equation is symmetric. We will assume without loss of generality that $a \\geqslant b \\geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.\n\nOne approach to finish the problem after establishing that $c \\leqslant 1$ is to set $k=b^{2} c^{2}-a$, which is clearly an integer and must be positive as it is equal to $\\left(b^{3}+c^{3}\\right) / a^{2}$. Then we divide into cases based on whether $k=1$ or $k \\geqslant 2$; in the first case, we have $b^{3}+1=a^{2}=\\left(b^{2}-1\\right)^{2}$ whose only positive root is $b=2$, and in the second case we have $b^{2} \\leqslant 3 a$, and so\n\n$$\nb^{4} \\leqslant(3 a)^{2} \\leqslant \\frac{9}{2}\\left(k a^{2}\\right)=\\frac{9}{2}\\left(b^{3}+1\\right),\n$$\n\nwhich implies that $b \\leqslant 4$.\n\nSet $k=\\left(b^{3}+c^{3}\\right) / a^{2} \\leqslant 2 a$, and rewrite the original equation as $a+k=(b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \\geqslant b^{3}+c^{3}-1=k a^{2}-1$, so\n\n$$\na+k \\geqslant\\left(k a^{2}-1\\right)^{2 / 3}\n$$\n\nAs proved before, $k$ is a positive integer; for each value of $k \\geqslant 1$, this gives us a polynomial inequality satisfied by $a$ :\n\n$$\nk^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\\left(k^{3}-1\\right) \\leqslant 0\n$$\n\nWe now prove that $a \\leqslant 3$. Indeed,\n\n$$\n0 \\geqslant \\frac{k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\\left(k^{3}-1\\right)}{k^{2}} \\geqslant a^{4}-a^{3}-5 a^{2}-3 a-k \\geqslant a^{4}-a^{3}-5 a^{2}-5 a,\n$$\n\nwhich fails when $a \\geqslant 4$.\n\nThis leaves ten triples with $3 \\geqslant a \\geqslant b \\geqslant c \\geqslant 1$, which may be checked manually to give $(a, b, c)=(3,2,1)$.', 'Note that the equation is symmetric. We will assume without loss of generality that $a \\geqslant b \\geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.\n\nAgain, observe that $b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right)$, so $b \\leqslant a \\leqslant b^{2} c^{2}-1$.\n\nWe consider the function $f(x)=x^{2}\\left(b^{2} c^{2}-x\\right)$. It can be seen that that on the interval $\\left[0, b^{2} c^{2}-1\\right]$ the function $f$ is increasing if $x<\\frac{2}{3} b^{2} c^{2}$ and decreasing if $x>\\frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that\n\n$$\nb^{3}+c^{3}=f(a) \\geqslant \\min \\left(f(b), f\\left(b^{2} c^{2}-1\\right)\\right)\n$$\n\nFirst, suppose that $b^{3}+c^{3} \\geqslant f\\left(b^{2} c^{2}-1\\right)$. This may be written $b^{3}+c^{3} \\geqslant\\left(b^{2} c^{2}-1\\right)^{2}$, and so\n\n$$\n2 b^{3} \\geqslant b^{3}+c^{3} \\geqslant\\left(b^{2} c^{2}-1\\right)^{2}>b^{4} c^{4}-2 b^{2} c^{2} \\geqslant b^{4} c^{4}-2 b^{3} c^{4}\n$$\n\nThus, $(b-2) c^{4}<2$, and the only solutions to this inequality have $(b, c)=(2,2)$ or $b \\leqslant 3$ and $c=1$. It is easy to verify that the only case giving a solution for $a \\geqslant b$ is $(a, b, c)=(3,2,1)$.\n\nOtherwise, suppose that $b^{3}+c^{3}=f(a) \\geqslant f(b)$. Then, we have\n\n$$\n2 b^{3} \\geqslant b^{3}+c^{3}=a^{2}\\left(b^{2} c^{2}-a\\right) \\geqslant b^{2}\\left(b^{2} c^{2}-b\\right) .\n$$\n\nConsequently $b c^{2} \\leqslant 3$, with strict inequality in the case that $b \\neq c$. Hence $c=1$ and $b \\leqslant 2$. Both of these cases have been considered already, so we are done.']","['$(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$']",True,,Tuple, 1931,Number Theory,,"We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \ldots, a_{n} \in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers $a$ and $b$.","['The set $\\mathbb{Z}$ of all integers is clearly rootiful. We shall prove that any rootiful set $S$ containing all the numbers of the form $2^{a}-2^{b}$ for $a, b \\in \\mathbb{Z}_{>0}$ must be all of $\\mathbb{Z}$.\n\nFirst, note that $0=2^{1}-2^{1} \\in S$ and $2=2^{2}-2^{1} \\in S$. Now, $-1 \\in S$, since it is a root of $2 x+2$, and $1 \\in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \\in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$.\n\nNow, we claim that any positive integer $n$ has a multiple in $S$. Indeed, suppose that $n=2^{\\alpha} \\cdot t$ for $\\alpha \\in \\mathbb{Z}_{\\geqslant 0}$ and $t$ odd. Then $t \\mid 2^{\\phi(t)}-1$, so $n \\mid 2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1}$. Moreover, $2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1} \\in S$, and so $S$ contains a multiple of every positive integer $n$.\n\nWe will now prove by induction that all positive integers are in $S$. Suppose that $0,1, \\ldots, n-$ $1 \\in S$; furthermore, let $N$ be a multiple of $n$ in $S$. Consider the base- $n$ expansion of $N$, say $N=a_{k} n^{k}+a_{k-1} n^{k-1}+\\cdots+a_{1} n+a_{0}$. Since $0 \\leqslant a_{i}0}$.\n\nWe show that, in fact, every integer $k$ with $|k|>2$ can be expressed as a root of a polynomial whose coefficients are of the form $2^{a}-2^{b}$. Observe that it suffices to consider the case where $k$ is positive, as if $k$ is a root of $a_{n} x^{n}+\\cdots+a_{1} x+a_{0}=0$, then $-k$ is a root of $(-1)^{n} a_{n} x^{n}+\\cdots-$ $a_{1} x+a_{0}=0$.\n\nNote that\n\n$$\n\\left(2^{a_{n}}-2^{b_{n}}\\right) k^{n}+\\cdots+\\left(2^{a_{0}}-2^{b_{0}}\\right)=0\n$$\n\nis equivalent to\n\n$$\n2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}=2^{b_{n}} k^{n}+\\cdots+2^{b_{0}}\n$$\n\nHence our aim is to show that two numbers of the form $2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}$ are equal, for a fixed value of $n$. We consider such polynomials where every term $2^{a_{i}} k^{i}$ is at most $2 k^{n}$; in other words, where $2 \\leqslant 2^{a_{i}} \\leqslant 2 k^{n-i}$, or, equivalently, $1 \\leqslant a_{i} \\leqslant 1+(n-i) \\log _{2} k$. Therefore, there must be $1+\\left\\lfloor(n-i) \\log _{2} k\\right\\rfloor$ possible choices for $a_{i}$ satisfying these constraints.\n\nThe number of possible polynomials is then\n\n$$\n\\prod_{i=0}^{n}\\left(1+\\left\\lfloor(n-i) \\log _{2} k\\right\\rfloor\\right) \\geqslant \\prod_{i=0}^{n-1}(n-i) \\log _{2} k=n !\\left(\\log _{2} k\\right)^{n}\n$$\n\nwhere the inequality holds as $1+\\lfloor x\\rfloor \\geqslant x$.\n\nAs there are $(n+1)$ such terms in the polynomial, each at most $2 k^{n}$, such a polynomial must have value at most $2 k^{n}(n+1)$. However, for large $n$, we have $n !\\left(\\log _{2} k\\right)^{n}>2 k^{n}(n+1)$. Therefore there are more polynomials than possible values, so some two must be equal, as required.']",['The set $\\mathbb{Z}$ of all integers is the only such rootiful set'],False,,Need_human_evaluate, 1932,Number Theory,,"Let $\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$, $$ a+f(b) \mid a^{2}+b f(a) \tag{*} $$","['It is easy to verify that the functions $f(a)=k a$ satisfy $(*)$. Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$.\n\nA common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality\n\n$$\nf(b) \\leqslant b \\cdot f(1)\n$$\n\nfor all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that\n\n$$\na+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a)\n$$\n\nfor all sufficiently large $a$. It follows from this that one has the lower bound\n\n$$\nf(a) \\geqslant a+f(1) \\cdot(1-f(1))\n$$\n\nagain for all sufficiently large $a$.\n\nEach of the following proofs makes use of at least one of these bounds.\n\n\nFirst, we show that $b \\mid f(b)^{2}$ for all $b$. To do this, we choose a large positive integer $n$ so that $n b-f(b) \\geqslant C$. Setting $a=n b-f(b)$ in $(*)$ then shows that\n\n$$\nn b \\mid(n b-f(b))^{2}+b f(n b-f(b))\n$$\n\nso that $b \\mid f(b)^{2}$ as claimed.\n\nNow in particular we have that $p \\mid f(p)$ for every prime $p$. If we write $f(p)=k(p) \\cdot p$, then the bound $f(p) \\leqslant f(1) \\cdot p$ (valid for $p$ sufficiently large) shows that some value $k$ of $k(p)$ must be attained for infinitely many $p$. We will show that $f(a)=k a$ for all positive integers $a$. To do this, we substitute $b=p$ in $(*)$, where $p$ is any sufficiently large prime for which $k(p)=k$, obtaining\n\n$$\na+k p \\mid\\left(a^{2}+p f(a)\\right)-a(a+k p)=p f(a)-p k a .\n$$\n\nFor suitably large $p$ we have $\\operatorname{gcd}(a+k p, p)=1$, and hence we have\n\n$$\na+k p \\mid f(a)-k a\n$$\n\nBut the only way this can hold for arbitrarily large $p$ is if $f(a)-k a=0$. This concludes the proof.', 'It is easy to verify that the functions $f(a)=k a$ satisfy $(*)$. Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$.\n\nA common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality\n\n$$\nf(b) \\leqslant b \\cdot f(1)\n$$\n\nfor all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that\n\n$$\na+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a)\n$$\n\nfor all sufficiently large $a$. It follows from this that one has the lower bound\n\n$$\nf(a) \\geqslant a+f(1) \\cdot(1-f(1))\n$$\n\nagain for all sufficiently large $a$.\n\nEach of the following proofs makes use of at least one of these bounds.\n\nFirst, we substitute $b=1$ in $(*)$ and rearrange to find that\n\n$$\n\\frac{f(a)+f(1)^{2}}{a+f(1)}=f(1)-a+\\frac{a^{2}+f(a)}{a+f(1)}\n$$\n\nis a positive integer for sufficiently large $a$. Since $f(a) \\leqslant a f(1)$, for all sufficiently large $a$, it follows that $\\frac{f(a)+f(1)^{2}}{a+f(1)} \\leqslant f(1)$ also and hence there is a positive integer $k$ such that $\\frac{f(a)+f(1)^{2}}{a+f(1)}=k$ for infinitely many values of $a$. In other words,\n\n$$\nf(a)=k a+f(1) \\cdot(k-f(1))\n$$\n\nfor infinitely many $a$.\n\nFixing an arbitrary choice of $a$ in (*), we have that\n\n$$\n\\frac{a^{2}+b f(a)}{a+k b+f(1) \\cdot(k-f(1))}\n$$\n\nis an integer for infinitely many $b$ (the same $b$ as above, maybe with finitely many exceptions). On the other hand, for $b$ taken sufficiently large, this quantity becomes arbitrarily close to $\\frac{f(a)}{k}$; this is only possible if $\\frac{f(a)}{k}$ is an integer and\n\n$$\n\\frac{a^{2}+b f(a)}{a+k b+f(1) \\cdot(k-f(1))}=\\frac{f(a)}{k}\n$$\n\nfor infinitely many $b$. This rearranges to\n\n$$\n\\frac{f(a)}{k} \\cdot(a+f(1) \\cdot(k-f(1)))=a^{2}\n\\tag{**}\n$$\n\nHence $a^{2}$ is divisible by $a+f(1) \\cdot(k-f(1))$, and hence so is $f(1)^{2}(k-f(1))^{2}$. The only way this can occur for all $a$ is if $k=f(1)$, in which case $(* *)$ provides that $f(a)=k a$ for all $a$, as desired.', 'It is easy to verify that the functions $f(a)=k a$ satisfy $(*)$. Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$.\n\nA common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality\n\n$$\nf(b) \\leqslant b \\cdot f(1)\n$$\n\nfor all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that\n\n$$\na+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a)\n$$\n\nfor all sufficiently large $a$. It follows from this that one has the lower bound\n\n$$\nf(a) \\geqslant a+f(1) \\cdot(1-f(1))\n$$\n\nagain for all sufficiently large $a$.\n\nEach of the following proofs makes use of at least one of these bounds.\n\nFix any two distinct positive integers $a$ and $b$. From $(*)$ it follows that the two integers\n\n$$\n\\left(a^{2}+c f(a)\\right) \\cdot(b+f(c)) \\text { and }\\left(b^{2}+c f(b)\\right) \\cdot(a+f(c))\n$$\n\nare both multiples of $(a+f(c)) \\cdot(b+f(c))$ for all sufficiently large $c$. Taking an appropriate linear combination to eliminate the $c f(c)$ term, we find after expanding out that the integer\n\n$$\n\\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot f(c)+[(b-a) f(a) f(b)] \\cdot c+[a b(a f(b)-b f(a))]\n\\tag{1}\n$$\n\nis also a multiple of $(a+f(c)) \\cdot(b+f(c))$.\n\nBut as $c$ varies, $(1)$ is bounded above by a positive multiple of $c$ while $(a+f(c)) \\cdot(b+f(c))$ is bounded below by a positive multiple of $c^{2}$. The only way that such a divisibility can hold is if in fact\n\n$$\n\\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot f(c)+[(b-a) f(a) f(b)] \\cdot c+[a b(a f(b)-b f(a))]=0\n\\tag{2}\n$$\n\nfor sufficiently large $c$. Since the coefficient of $c$ in this linear relation is nonzero, it follows that there are constants $k, \\ell$ such that $f(c)=k c+\\ell$ for all sufficiently large $c$; the constants $k$ and $\\ell$ are necessarily integers.\n\nThe value of $\\ell$ satisfies\n\n$$\n\\left[a^{2} f(b)-b^{2} f(a)\\right] \\cdot \\ell+[a b(a f(b)-b f(a))]=0\n\\tag{3}\n$$\n\nand hence $b \\mid \\ell a^{2} f(b)$ for all $a$ and $b$. Taking $b$ sufficiently large so that $f(b)=k b+\\ell$, we thus have that $b \\mid \\ell^{2} a^{2}$ for all sufficiently large $b$; this implies that $\\ell=0$. From (3) it then follows that $\\frac{f(a)}{a}=\\frac{f(b)}{b}$ for all $a \\neq b$, so that there is a constant $k$ such that $f(a)=k a$ for all $a$ ( $k$ is equal to the constant defined earlier).', 'It is easy to verify that the functions $f(a)=k a$ satisfy $(*)$. Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$.\n\nA common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality\n\n$$\nf(b) \\leqslant b \\cdot f(1)\n$$\n\nfor all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that\n\n$$\na+f(1) \\mid\\left(a^{2}+f(a)\\right)-(a-f(1)) \\cdot(a+f(1))=f(1)^{2}+f(a)\n$$\n\nfor all sufficiently large $a$. It follows from this that one has the lower bound\n\n$$\nf(a) \\geqslant a+f(1) \\cdot(1-f(1))\n$$\n\nagain for all sufficiently large $a$.\n\nEach of the following proofs makes use of at least one of these bounds.\n\n\nLet $\\Gamma$ denote the set of all points $(a, f(a))$, so that $\\Gamma$ is an infinite subset of the upper-right quadrant of the plane. For a point $A=(a, f(a))$ in $\\Gamma$, we define a point $A^{\\prime}=\\left(-f(a),-f(a)^{2} / a\\right)$ in the lower-left quadrant of the plane, and let $\\Gamma^{\\prime}$ denote the set of all such points $A^{\\prime}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_5db419644d3b656627a9g-1.jpg?height=551&width=645&top_left_y=450&top_left_x=708)\n\nClaim. For any point $A \\in \\Gamma$, the set $\\Gamma$ is contained in finitely many lines through the point $A^{\\prime}$. Proof. Let $A=(a, f(a))$. The functional equation (with $a$ and $b$ interchanged) can be rewritten as $b+f(a) \\mid a f(b)-b f(a)$, so that all but finitely many points in $\\Gamma$ are contained in one of the lines with equation\n\n$$\na y-f(a) x=m(x+f(a))\n$$\n\nfor $m$ an integer. Geometrically, these are the lines through $A^{\\prime}=\\left(-f(a),-f(a)^{2} / a\\right)$ with gradient $\\frac{f(a)+m}{a}$. Since $\\Gamma$ is contained, with finitely many exceptions, in the region $0 \\leqslant y \\leqslant$ $f(1) \\cdot x$ and the point $A^{\\prime}$ lies strictly in the lower-left quadrant of the plane, there are only finitely many values of $m$ for which this line meets $\\Gamma$. This concludes the proof of the claim.\n\nNow consider any distinct points $A, B \\in \\Gamma$. It is clear that $A^{\\prime}$ and $B^{\\prime}$ are distinct. A line through $A^{\\prime}$ and a line through $B^{\\prime}$ only meet in more than one point if these two lines are equal to the line $A^{\\prime} B^{\\prime}$. It then follows from the above claim that the line $A^{\\prime} B^{\\prime}$ must contain all but finitely many points of $\\Gamma$. If $C$ is another point of $\\Gamma$, then the line $A^{\\prime} C^{\\prime}$ also passes through all but finitely many points of $\\Gamma$, which is only possible if $A^{\\prime} C^{\\prime}=A^{\\prime} B^{\\prime}$.\n\nWe have thus seen that there is a line $\\ell$ passing through all points of $\\Gamma^{\\prime}$ and through all but finitely many points of $\\Gamma$. We claim that this line passes through the origin $O$ and passes through every point of $\\Gamma$. To see this, note that by construction $A, O, A^{\\prime}$ are collinear for every point $A \\in \\Gamma$. Since $\\ell=A A^{\\prime}$ for all but finitely many points $A \\in \\Gamma$, it thus follows that $O \\in \\ell$. Thus any $A \\in \\Gamma$ lies on the line $\\ell=A^{\\prime} O$.\n\nSince $\\Gamma$ is contained in a line through $O$, it follows that there is a real constant $k$ (the gradient of $\\ell$ ) such that $f(a)=k a$ for all $a$. The number $k$ is, of course, a positive integer.']",['$f(a)=k a$ for some constant $k \\in \\mathbb{Z}_{>0}$ (irrespective of the value of $C$ )'],False,,Need_human_evaluate, 1933,Number Theory,,"Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\left(\begin{array}{c}a n \\ b\end{array}\right)-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.","['For $p$ a prime and $n$ a nonzero integer, we write $v_{p}(n)$ for the $p$-adic valuation of $n$ : the largest integer $t$ such that $p^{t} \\mid n$.\n\nWe first show that $b$ is $a$-good if and only if $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$.\n\nTo start with, the condition that $a n+1 \\mid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)-1$ can be rewritten as saying that\n\n$$\n\\frac{a n(a n-1) \\cdots(a n-b+1)}{b !} \\equiv 1 \\quad(\\bmod a n+1)\n\\tag{1}\n$$\n\nSuppose, on the one hand, there is a prime $p \\leqslant b$ with $p \\nmid a$. Take $t=v_{p}(b !)$. Then there exist positive integers $c$ such that $a c \\equiv 1\\left(\\bmod p^{t+1}\\right)$. If we take $c$ big enough, and then take $n=(p-1) c$, then $a n=a(p-1) c \\equiv p-1\\left(\\bmod p^{t+1}\\right)$ and $a n \\geqslant b$. Since $p \\leqslant b$, one of the terms of the numerator $a n(a n-1) \\cdots(a n-b+1)$ is $a n-p+1$, which is divisible by $p^{t+1}$. Hence the $p$-adic valuation of the numerator is at least $t+1$, but that of the denominator is exactly $t$. This means that $p \\mid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$, so $p \\nmid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)-1$. As $p \\mid a n+1$, we get that $a n+1 \\nmid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$, so $b$ is not $a$-good.\n\nOn the other hand, if for all primes $p \\leqslant b$ we have $p \\mid a$, then every factor of $b$ ! is coprime to $a n+1$, and hence invertible modulo $a n+1$ : hence $b$ ! is also invertible modulo $a n+1$. Then equation (1) reduces to:\n\n$$\na n(a n-1) \\cdots(a n-b+1) \\equiv b ! \\quad(\\bmod a n+1)\n$$\n\nHowever, we can rewrite the left-hand side as follows:\n\n$$\na n(a n-1) \\cdots(a n-b+1) \\equiv(-1)(-2) \\cdots(-b) \\equiv(-1)^{b} b ! \\quad(\\bmod a n+1)\n$$\n\nProvided that $a n>1$, if $b$ is even we deduce $(-1)^{b} b ! \\equiv b$ ! as needed. On the other hand, if $b$ is odd, and we take $a n+1>2(b !)$, then we will not have $(-1)^{b} b ! \\equiv b$ !, so $b$ is not $a$-good. This completes the claim.\n\nTo conclude from here, suppose that $b$ is $a$-good, but $b+2$ is not. Then $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$, but there is a prime $q \\leqslant b+2$ for which $q \\nmid a$ : so $q=b+1$ or $q=b+2$. We cannot have $q=b+2$, as that is even too, so we have $q=b+1$ : in other words, $b+1$ is prime.', ""We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p \\mid a$ for all primes $p \\leqslant b$. We do this with Lucas' theorem.\n\nSuppose that we have $p \\leqslant b$ with $p \\nmid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p \\leqslant b$. Suppose it is the $p^{t}$ digit for $t \\geqslant 1$.\n\nNow, as $n$ varies over the integers, $a n+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $a n>b$ ) such that the $p^{0}$ digit of an is $p-1$ (so $p \\mid a n+1)$ and the $p^{t}$ digit of an is 0 . Consequently, $p \\mid a n+1$ but $p \\mid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$ (by Lucas' theorem) so $p \\nmid\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)-1$. Thus $b$ is not $a$-good.\n\nNow we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $a n+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\\left(\\begin{array}{c}a n \\\\ b+2\\end{array}\\right)$ and $\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right)$ is $(b+1)(b+2) /(a n-b)(a n-b-1)$. We know that there must be a choice of $a n+1$ such that the former binomial coefficient is 1 modulo $a n+1$ but the latter is not, which means that the given ratio must not be $1 \\bmod a n+1$. If $b+1$ and $b+2$ are both coprime to $a n+1$ then\n\n\n\nthe ratio is 1 , so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.\n\nHowever, we can observe that $b$ must be even by insisting that $a n+1$ is prime (which is possible by Dirichlet's theorem) and hence $\\left(\\begin{array}{c}a n \\\\ b\\end{array}\\right) \\equiv(-1)^{b}=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.""]",,True,,, 1934,Number Theory,,"Let $H=\left\{\lfloor i \sqrt{2}\rfloor: i \in \mathbb{Z}_{>0}\right\}=\{1,2,4,5,7, \ldots\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \subset\{1,2, \ldots, n\}$ satisfies $|A| \geqslant C \sqrt{n}$, then there exist $a, b \in A$ such that $a-b \in H$. (Here $\mathbb{Z}_{>0}$ is the set of positive integers, and $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z$.)","['we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|1-\\frac{1}{\\sqrt{2}}\n\\tag{1}\n$$\n\nTo see why, observe that $n \\in H$ if and only if $00}$. In other words, $01$ since $a_{i}-a_{1} \\notin H$. Furthermore, we must have $\\left\\{a_{i} / \\sqrt{2}\\right\\}<\\left\\{a_{j} / \\sqrt{2}\\right\\}$ whenever $i1 / \\sqrt{2}>$ $1-1 / \\sqrt{2}$, contradicting (1).\n\nNow, we have a sequence $0=a_{1}\\frac{1}{2 d \\sqrt{2}}\n\\tag{2}\n$$\n\nTo see why this is the case, let $h=\\lfloor d / \\sqrt{2}\\rfloor$, so $\\{d / \\sqrt{2}\\}=d / \\sqrt{2}-h$. Then\n\n$$\n\\left\\{\\frac{d}{\\sqrt{2}}\\right\\}\\left(\\frac{d}{\\sqrt{2}}+h\\right)=\\frac{d^{2}-2 h^{2}}{2} \\geqslant \\frac{1}{2}\n$$\n\nsince the numerator is a positive integer. Because $d / \\sqrt{2}+h<2 d / \\sqrt{2}$, inequality (2) follows.\n\nLet $d_{i}=a_{i+1}-a_{i}$, for $1 \\leqslant i\\sum_{i}\\left\\{\\frac{d_{i}}{\\sqrt{2}}\\right\\}>\\frac{1}{2 \\sqrt{2}} \\sum_{i} \\frac{1}{d_{i}} \\geqslant \\frac{(k-1)^{2}}{2 \\sqrt{2}} \\frac{1}{\\sum_{i} d_{i}}>\\frac{(k-1)^{2}}{2 \\sqrt{2}} \\cdot \\frac{1}{n}\n\\tag{3}\n$$\n\nHere, the first inequality holds because $\\left\\{a_{k} / \\sqrt{2}\\right\\}<1-1 / \\sqrt{2}$, the second follows from (2), the third follows from an easy application of the AM-HM inequality (or Cauchy-Schwarz), and the fourth follows from the fact that $\\sum_{i} d_{i}=a_{k}k-1\n$$\n\nwhich provides the required bound on $k$.', 'we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|0}\\right\\}$ is the complementary Beatty sequence to $H$ (in other words, $H$ and $J$ are disjoint with $H \\cup J=\\mathbb{Z}_{>0}$ ). Write $A=\\left\\{a_{1}0}$.\n\nFor any $j>i$, we have $a_{j}-a_{i}=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. Because $a_{j}-a_{i} \\in J$, we also have $a_{j}-a_{i}=\\lfloor\\alpha t\\rfloor$ for some positive integer $t$. Thus, $\\lfloor\\alpha t\\rfloor=\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor$. The right hand side must equal either $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$ or $\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor-1$, the latter of which is not a member of $J$ as $\\alpha>2$. Therefore, $t=b_{j}-b_{i}$ and so we have $\\left\\lfloor\\alpha b_{j}\\right\\rfloor-\\left\\lfloor\\alpha b_{i}\\right\\rfloor=\\left\\lfloor\\alpha\\left(b_{j}-b_{i}\\right)\\right\\rfloor$.\n\nFor $1 \\leqslant i1 /(2 d \\sqrt{2})$ for positive integers $d)$ proves that $1>\\left((k-1)^{2} /(2 \\sqrt{2})\\right)(\\alpha / n)$, which again rearranges to give\n\n$$\n\\sqrt{2 \\sqrt{2}-2} \\cdot \\sqrt{n}>k-1\n$$', 'we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|0} \\backslash H$, so all differences between elements of $A$ are in $J$. We start by making the following observation. Suppose we have a set $B \\subseteq\\{1,2, \\ldots, n\\}$ such that all of the differences between elements of $B$ are in $H$. Then $|A| \\cdot|B| \\leqslant 2 n$.\n\nTo see why, observe that any two sums of the form $a+b$ with $a \\in A, b \\in B$ are different; otherwise, we would have $a_{1}+b_{1}=a_{2}+b_{2}$, and so $\\left|a_{1}-a_{2}\\right|=\\left|b_{2}-b_{1}\\right|$. However, then the left hand side is in $J$ whereas the right hand side is in $H$. Thus, $\\{a+b: a \\in A, b \\in B\\}$ is a set of size $|A| \\cdot|B|$ all of whose elements are no greater than $2 n$, yielding the claimed inequality.\n\nWith this in mind, it suffices to construct a set $B$, all of whose differences are in $H$ and whose size is at least $C^{\\prime} \\sqrt{n}$ for some constant $C^{\\prime}>0$.\n\nTo do so, we will use well-known facts about the negative Pell equation $X^{2}-2 Y^{2}=-1$; in particular, that there are infinitely many solutions and the values of $X$ are given by the recurrence $X_{1}=1, X_{2}=7$ and $X_{m}=6 X_{m-1}-X_{m-2}$. Therefore, we may choose $X$ to be a solution with $\\sqrt{n} / 6\\left(\\frac{X}{\\sqrt{2}}-Y\\right) \\geqslant \\frac{-3}{\\sqrt{2 n}}\n$$\n\nfrom which it follows that $\\{X / \\sqrt{2}\\}>1-(3 / \\sqrt{2 n})$. Combined with (1), this shows that all differences between elements of $B$ are in $H$.', 'we will assume that $A$ is a set such that $\\{a-b: a, b \\in A\\}$ is disjoint from $H$, and prove that $|A|\\sqrt{Y}$ elements; if that subsequence is decreasing, we may reflect (using (4) or (5)) to ensure that it is increasing. Call the subsequence $\\left\\{y_{1} \\sqrt{2}\\right\\},\\left\\{y_{2} \\sqrt{2}\\right\\}, \\ldots,\\left\\{y_{t} \\sqrt{2}\\right\\}$ for $t>\\sqrt{Y}$.\n\nNow, set $B=\\left\\{\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor: 1 \\leqslant i \\leqslant t\\right\\}$. We have $\\left\\lfloor y_{j} \\sqrt{2}\\right\\rfloor-\\left\\lfloor y_{i} \\sqrt{2}\\right\\rfloor=\\left\\lfloor\\left(y_{j}-y_{i}\\right) \\sqrt{2}\\right\\rfloor$ for $i\\sqrt{Y}>\\sqrt{n} / \\sqrt{3 \\sqrt{2}}$, this is the required set.']",,True,,, 1935,Number Theory,,Prove that there is a constant $c>0$ and infinitely many positive integers $n$ with the following property: there are infinitely many positive integers that cannot be expressed as the sum of fewer than $c n \log (n)$ pairwise coprime $n^{\text {th }}$ powers.,"['Suppose, for an integer $n$, that we can find another integer $N$ satisfying the following property:\n\n$$\nn\\text{ is divisible by }\\varphi\\left(p^{e}\\right)\\text{ for every prime power }p^{e} \\text{exactly dividing }N.\n\\tag{1}\n$$\n\nThis property ensures that all $n^{\\text {th }}$ powers are congruent to 0 or 1 modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\\text {th }}$ powers is congruent to $m$ or $m-1$ modulo $p^{e}$, since at most one of the $n^{\\text {th }}$ powers is divisible by $p$. Thus, if $k$ denotes the number of distinct prime factors of $N$, we find by the Chinese Remainder Theorem at most $2^{k} m$ residue classes modulo $N$ which are sums of at most $m$ pairwise coprime $n^{\\text {th }}$ powers. In particular, if $N>2^{k} m$ then there are infinitely many positive integers not expressible as a sum of at most $m$ pairwise coprime $n^{\\text {th }}$ powers.\n\nIt thus suffices to prove that there are arbitrarily large pairs $(n, N)$ of integers satisfying $(1)$ such that\n\n$$\nN>c \\cdot 2^{k} n \\log (n)\n$$\n\nfor some positive constant $c$.\n\nWe construct such pairs as follows. Fix a positive integer $t$ and choose (distinct) prime numbers $p \\mid 2^{2^{t-1}}+1$ and $q \\mid 2^{2^{t}}+1$; we set $N=p q$. It is well-known that $2^{t} \\mid p-1$ and $2^{t+1} \\mid q-1$, hence\n\n$$\nn=\\frac{(p-1)(q-1)}{2^{t}}\n$$\n\nis an integer and the pair $(n, N)$ satisfies $(1)$.\n\nEstimating the size of $N$ and $n$ is now straightforward. We have\n\n$$\n\\log _{2}(n) \\leqslant 2^{t-1}+2^{t}-t<2^{t+1}<2 \\cdot \\frac{N}{n}\n$$\n\nwhich rearranges to\n\n$$\nN>\\frac{1}{8} \\cdot 2^{2} n \\log _{2}(n)\n$$\n\nand so we are done if we choose $c<\\frac{1}{8 \\log (2)} \\approx 0.18$.', 'we seek arbitrarily large pairs of integers $n$ and $N$ satisfying ( $\\dagger$ ) such that $N>c 2^{k} n \\log (n)$.\n\nThis time, to construct such pairs, we fix an integer $x \\geqslant 4$, set $N$ to be the lowest common multiple of $1,2, \\ldots, 2 x$, and set $n$ to be twice the lowest common multiple of $1,2, \\ldots, x$. The pair $(n, N)$ does indeed satisfy the condition, since if $p^{e}$ is a prime power divisor of $N$ then $\\frac{\\varphi\\left(p^{e}\\right)}{2} \\leqslant x$ is a factor of $\\frac{n}{2}=\\operatorname{lcm}_{r \\leqslant x}(r)$.\n\nNow $2 N / n$ is the product of all primes having a power lying in the interval $(x, 2 x]$, and hence $2 N / n>x^{\\pi(2 x)-\\pi(x)}$. Thus for sufficiently large $x$ we have\n\n$$\n\\log \\left(\\frac{2 N}{2^{\\pi(2 x)} n}\\right)>(\\pi(2 x)-\\pi(x)) \\log (x)-\\log (2) \\pi(2 x) \\sim x\n$$\n\nusing the Prime Number Theorem $\\pi(t) \\sim t / \\log (t)$.\n\nOn the other hand, $n$ is a product of at most $\\pi(x)$ prime powers less than or equal to $x$, and so we have the upper bound\n\n$$\n\\log (n) \\leqslant \\pi(x) \\log (x) \\sim x\n$$\n\nagain by the Prime Number Theorem. Combined with the above inequality, we find that for any $\\epsilon>0$, the inequality\n\n$$\n\\log \\left(\\frac{N}{2^{\\pi(2 x)} n}\\right)>(1-\\epsilon) \\log (n)\n$$\n\nholds for sufficiently large $x$. Rearranging this shows that\n\n$$\nN>2^{\\pi(2 x)} n^{2-\\epsilon}>2^{\\pi(2 x)} n \\log (n)\n$$\n\nfor all sufficiently large $x$ and we are done.']",,True,,, 1936,Number Theory,,"Let $a$ and $b$ be two positive integers. Prove that the integer $$ a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil $$ is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to $z$.)","['Arguing indirectly, assume that\n\n$$\na^{2}+\\left\\lceil\\frac{4 a^{2}}{b}\\right\\rceil=(a+k)^{2}, \\quad \\text { or } \\quad\\left\\lceil\\frac{(2 a)^{2}}{b}\\right\\rceil=(2 a+k) k\n$$\n\nClearly, $k \\geqslant 1$. In other words, the equation\n\n$$\n\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k\n\\tag{1}\n$$\n\nhas a positive integer solution $(c, k)$, with an even value of $c$.\n\nChoose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From\n\n$$\n\\frac{c^{2}}{b}>\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil-1=c k+k^{2}-1 \\geqslant c k\n$$\n\nand\n\n$$\n\\frac{(c-k)(c+k)}{b}<\\frac{c^{2}}{b} \\leqslant\\left\\lceil\\frac{c^{2}}{b}\\right\\rceil=(c+k) k\n$$\n\nit can be seen that $c>b k>c-k$, so\n\n$$\nc=k b+r \\quad \\text { with some } 00$ and $0a$, so\n\n$$\n\\begin{aligned}\n& c^{2}-1(n-1)+2 .\n$$\n\nThat completes the solution.']",,True,,, 1938,Algebra,,"Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with the property that $$ f(x-f(y))=f(f(x))-f(y)-1 \tag{1} $$ holds for all $x, y \in \mathbb{Z}$.","['It is immediately checked that both functions mentioned in the answer are as desired.\n\nNow let $f$ denote any function satisfying (1) for all $x, y \\in \\mathbb{Z}$. Substituting $x=0$ and $y=f(0)$ into (1) we learn that the number $z=-f(f(0))$ satisfies $f(z)=-1$. So by plugging $y=z$ into (1) we deduce that\n\n$$\nf(x+1)=f(f(x))\n\\tag{2}\n$$\n\nholds for all $x \\in \\mathbb{Z}$. Thereby (1) simplifies to\n\n$$\nf(x-f(y))=f(x+1)-f(y)-1 .\n\\tag{3}\n$$\n\nWe now work towards showing that $f$ is linear by contemplating the difference $f(x+1)-f(x)$ for any $x \\in \\mathbb{Z}$. By applying (3) with $y=x$ and (2) in this order, we obtain\n\n$$\nf(x+1)-f(x)=f(x-f(x))+1=f(f(x-1-f(x)))+1 .\n$$\n\nSince (3) shows $f(x-1-f(x))=f(x)-f(x)-1=-1$, this simplifies to\n\n$$\nf(x+1)=f(x)+A\n$$\n\nwhere $A=f(-1)+1$ is some absolute constant.\n\nNow a standard induction in both directions reveals that $f$ is indeed linear and that in fact we have $f(x)=A x+B$ for all $x \\in \\mathbb{Z}$, where $B=f(0)$. Substituting this into (2) we obtain that\n\n$$\nA x+(A+B)=A^{2} x+(A B+B)\n$$\n\nholds for all $x \\in \\mathbb{Z}$; applying this to $x=0$ and $x=1$ we infer $A+B=A B+B$ and $A^{2}=A$. The second equation leads to $A=0$ or $A=1$. In case $A=1$, the first equation gives $B=1$, meaning that $f$ has to be the successor function. If $A=0$, then $f$ is constant and (1) shows that its constant value has to be -1 . Thereby the solution is complete.', 'We commence by deriving (2) and (3) as in the first solution. Now provided that $f$ is injective, (2) tells us that $f$ is the successor function. Thus we may assume from now on that $f$ is not injective, i.e., that there are two integers $a>b$ with $f(a)=f(b)$. A straightforward induction using (2) in the induction step reveals that we have $f(a+n)=f(b+n)$ for all nonnegative integers $n$. Consequently, the sequence $\\gamma_{n}=f(b+n)$ is periodic and thus in particular bounded, which means that the numbers\n\n$$\n\\varphi=\\min _{n \\geqslant 0} \\gamma_{n} \\quad \\text { and } \\quad \\psi=\\max _{n \\geqslant 0} \\gamma_{n}\n$$\n\nexist.\n\nLet us pick any integer $y$ with $f(y)=\\varphi$ and then an integer $x \\geqslant a$ with $f(x-f(y))=\\varphi$. Due to the definition of $\\varphi$ and (3) we have\n\n$$\n\\varphi \\leqslant f(x+1)=f(x-f(y))+f(y)+1=2 \\varphi+1\n$$\n\nwhence $\\varphi \\geqslant-1$. The same reasoning applied to $\\psi$ yields $\\psi \\leqslant-1$. Since $\\varphi \\leqslant \\psi$ holds trivially, it follows that $\\varphi=\\psi=-1$, or in other words that we have $f(t)=-1$ for all integers $t \\geqslant a$.\n\nFinally, if any integer $y$ is given, we may find an integer $x$ which is so large that $x+1 \\geqslant a$ and $x-f(y) \\geqslant a$ hold. Due to (3) and the result from the previous paragraph we get\n\n$$\nf(y)=f(x+1)-f(x-f(y))-1=(-1)-(-1)-1=-1 .\n$$\n\nThereby the problem is solved.', 'Set $d=f(0)$. By plugging $x=f(y)$ into (1) we obtain\n\n$$\nf^{3}(y)=f(y)+d+1\n\\tag{4}\n$$\n\nfor all $y \\in \\mathbb{Z}$, where the left-hand side abbreviates $f(f(f(y)))$. When we replace $x$ in (1) by $f(x)$ we obtain $f(f(x)-f(y))=f^{3}(x)-f(y)-1$ and as a consequence of (4) this simplifies to\n\n$$\nf(f(x)-f(y))=f(x)-f(y)+d\n\\tag{5}\n$$\n\nNow we consider the set\n\n$$\nE=\\{f(x)-d \\mid x \\in \\mathbb{Z}\\}\n\\tag{6}\n$$\n\nGiven two integers $a$ and $b$ from $E$, we may pick some integers $x$ and $y$ with $f(x)=a+d$ and $f(y)=b+d$; now (5) tells us that $f(a-b)=(a-b)+d$, which means that $a-b$ itself exemplifies $a-b \\in E$. Thus,\n\n$$\nE \\text { is closed under taking differences. }\n$$\n\nAlso, the definitions of $d$ and $E$ yield $0 \\in E$. If $E=\\{0\\}$, then $f$ is a constant function and (1) implies that the only value attained by $f$ is indeed -1 .\n\nSo let us henceforth suppose that $E$ contains some number besides zero. It is known that in this case (6) entails $E$ to be the set of all integer multiples of some positive integer $k$. Indeed, this holds for\n\n$$\nk=\\min \\{|x| \\mid x \\in E \\text { and } x \\neq 0\\}\n$$\n\nas one may verify by an argument based on division with remainder.\n\nThus we have\n\n$$\n\\{f(x) \\mid x \\in \\mathbb{Z}\\}=\\{k \\cdot t+d \\mid t \\in \\mathbb{Z}\\}\n\\tag{7}\n$$\n\nDue to (5) and (7) we get\n\n$$\nf(k \\cdot t)=k \\cdot t+d\n$$\n\n\n\nfor all $t \\in \\mathbb{Z}$, whence in particular $f(k)=k+d$. So by comparing the results of substituting $y=0$ and $y=k$ into (1) we learn that\n\n$$\nf(z+k)=f(z)+k\n\\tag{8}\n$$\n\nholds for all integers $z$. In plain English, this means that on any residue class modulo $k$ the function $f$ is linear with slope 1 .\n\nNow by (7) the set of all values attained by $f$ is such a residue class. Hence, there exists an absolute constant $c$ such that $f(f(x))=f(x)+c$ holds for all $x \\in \\mathbb{Z}$. Thereby (1) simplifies to\n\n$$\nf(x-f(y))=f(x)-f(y)+c-1 .\n\\tag{9}\n$$\n\nOn the other hand, considering (1) modulo $k$ we obtain $d \\equiv-1(\\bmod k)$ because of $(7)$. So by (7) again, $f$ attains the value -1 .\n\nThus we may apply (9) to some integer $y$ with $f(y)=-1$, which gives $f(x+1)=f(x)+c$. So $f$ is a linear function with slope $c$. Hence, (8) leads to $c=1$, wherefore there is an absolute constant $d^{\\prime}$ with $f(x)=x+d^{\\prime}$ for all $x \\in \\mathbb{Z}$. Using this for $x=0$ we obtain $d^{\\prime}=d$ and finally (4) discloses $d=1$, meaning that $f$ is indeed the successor function.']","['$f(x)=-1$,$f(x)=x+1$']",True,,Expression, 1939,Algebra,,"Let $n$ be a fixed positive integer. Find the maximum possible value of $$ \sum_{1 \leqslant rp^{+}$, then $S\\left(p^{-}\\right)=P\\left(p^{-}\\right)+Q\\left(p^{-}\\right) \\leqslant Q\\left(p^{+}\\right)+P\\left(p^{+}\\right)=S\\left(p^{+}\\right)$, so our claim holds.\n\nWe now show that the remaining case $p^{-}$ $Q\\left(p^{+}\\right)$. Then, like in Step 1 , we have $R\\left(p^{-}\\right) \\leqslant 0, R\\left(p^{+}\\right)>0$, and $R(i k) \\leqslant 0$, so $R(x)$ has a root in each of the intervals $\\left[p^{-}, p^{+}\\right)$and $\\left(p^{+}, i k\\right]$. This contradicts the result of Step 1.\n\nWe are left only with the case $p^{-}\\ell_{j}>r_{i}$.\n\nClearly, there is no town which can sweep $T_{n}$ away from the right. Then we may choose the leftmost town $T_{k}$ which cannot be swept away from the right. One can observe now that no town $T_{i}$ with $i>k$ may sweep away some town $T_{j}$ with $jm$. As we have already observed, $p$ cannot be greater than $k$. On the other hand, $T_{m}$ cannot sweep $T_{p}$ away, so a fortiori it cannot sweep $T_{k}$ away.\n\nClaim 2. Any town $T_{m}$ with $m \\neq k$ can be swept away by some other town.\n\n\n\nProof. If $mk$.\n\nLet $T_{p}$ be a town among $T_{k}, T_{k+1}, \\ldots, T_{m-1}$ having the largest right bulldozer. We claim that $T_{p}$ can sweep $T_{m}$ away. If this is not the case, then $r_{p}<\\ell_{q}$ for some $q$ with $pm \geqslant N$.","['We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \\neq n+a_{n}$ for $m \\neq n$, each positive integer receives at most one arrow. There are some positive integers, such as 1 , that receive no arrows; these will be referred to as starting points in the sequel. When one starts at any of the starting points and keeps following the arrows, one is led to an infinite path, called its ray, that visits a strictly increasing sequence of positive integers. Since the length of any arrow is at most 2015, such a ray, say with starting point $s$, meets every interval of the form $[n, n+2014]$ with $n \\geqslant s$ at least once.\n\nSuppose for the sake of contradiction that there would be at least 2016 starting points. Then we could take an integer $n$ that is larger than the first 2016 starting points. But now the interval $[n, n+2014]$ must be met by at least 2016 rays in distinct points, which is absurd. We have thereby shown that the number $b$ of starting points satisfies $1 \\leqslant b \\leqslant 2015$. Let $N$ denote any integer that is larger than all starting points. We contend that $b$ and $N$ are as required.\n\nTo see this, let any two integers $m$ and $n$ with $n>m \\geqslant N$ be given. The sum $\\sum_{i=m+1}^{n} a_{i}$ gives the total length of the arrows emerging from $m+1, \\ldots, n$. Taken together, these arrows form $b$ subpaths of our rays, some of which may be empty. Now on each ray we look at the first number that is larger than $m$; let $x_{1}, \\ldots, x_{b}$ denote these numbers, and let $y_{1}, \\ldots, y_{b}$ enumerate in corresponding order the numbers defined similarly with respect to $n$. Then the list of differences $y_{1}-x_{1}, \\ldots, y_{b}-x_{b}$ consists of the lengths of these paths and possibly some zeros corresponding to empty paths. Consequently, we obtain\n\n$$\n\\sum_{i=m+1}^{n} a_{i}=\\sum_{j=1}^{b}\\left(y_{j}-x_{j}\\right)\n$$\n\nwhence\n\n$$\n\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)=\\sum_{j=1}^{b}\\left(y_{j}-n\\right)-\\sum_{j=1}^{b}\\left(x_{j}-m\\right)\n$$\n\nNow each of the $b$ rays meets the interval $[m+1, m+2015]$ at some point and thus $x_{1}-$ $m, \\ldots, x_{b}-m$ are $b$ distinct members of the set $\\{1,2, \\ldots, 2015\\}$. Moreover, since $m+1$ is not a starting point, it must belong to some ray; so 1 has to appear among these numbers, wherefore\n\n$$\n1+\\sum_{j=1}^{b-1}(j+1) \\leqslant \\sum_{j=1}^{b}\\left(x_{j}-m\\right) \\leqslant 1+\\sum_{j=1}^{b-1}(2016-b+j)\n$$\n\nThe same argument applied to $n$ and $y_{1}, \\ldots, y_{b}$ yields\n\n$$\n1+\\sum_{j=1}^{b-1}(j+1) \\leqslant \\sum_{j=1}^{b}\\left(y_{j}-n\\right) \\leqslant 1+\\sum_{j=1}^{b-1}(2016-b+j)\n$$\n\n\n\nSo altogether we get\n\n$$\n\\begin{aligned}\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| & \\leqslant \\sum_{j=1}^{b-1}((2016-b+j)-(j+1))=(b-1)(2015-b) \\\\\n& \\leqslant\\left(\\frac{(b-1)+(2015-b)}{2}\\right)^{2}=1007^{2},\n\\end{aligned}\n$$\n\nas desired.', 'Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have\n\n$$\nn+1 \\leqslant s_{n} \\leqslant n+2015\n$$\n\nfor all $n \\in \\mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \\ldots$ are distinct. We shall investigate the set\n\n$$\nM=\\mathbb{Z}_{>0} \\backslash\\left\\{s_{1}, s_{2}, \\ldots\\right\\}\n$$\n\nClaim. At most 2015 numbers belong to $M$.\n\nProof. Otherwise let $m_{1}m \\geqslant N$. Plugging $r=n$ and $r=m$ into (3) and subtracting the estimates that result, we deduce\n\n$$\n\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)=\\sum C_{n}-\\sum C_{m}\n$$\n\n\n\nSince $C_{n}$ and $C_{m}$ are subsets of $\\{1,2, \\ldots, 2014\\}$ with $\\left|C_{n}\\right|=\\left|C_{m}\\right|=b-1$, it is clear that the absolute value of the right-hand side of the above inequality attains its largest possible value if either $C_{m}=\\{1,2, \\ldots, b-1\\}$ and $C_{n}=\\{2016-b, \\ldots, 2014\\}$, or the other way around. In these two cases we have\n\n$$\n\\left|\\sum C_{n}-\\sum C_{m}\\right|=(b-1)(2015-b)\n$$\n\nso in the general case we find\n\n$$\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| \\leqslant(b-1)(2015-b) \\leqslant\\left(\\frac{(b-1)+(2015-b)}{2}\\right)^{2}=1007^{2}\n$$\n\nas desired.']",,True,,, 1948,Combinatorics,,Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.,"['Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\\mathbb{Z}_{>0}$ denote the set of all positive integers.\n\nSuppose, to the contrary, that there exist only finitely many positive integers that are not clean. Therefore, there exists a positive integer $N$ such that every integer $n>N$ has exactly one odd representation.\n\nClearly, $S$ is infinite. We now claim the following properties of odd and even representations.\n\nProperty 1. Any positive integer $n$ has at most one odd and at most one even representation. Proof. We first show that every integer $n$ has at most one even representation. Since $S$ is infinite, there exists $x \\in S$ such that $x>\\max \\{n, N\\}$. Then, the number $n+x$ must be clean, and $x$ does not appear in any even representation of $n$. If $n$ has more than one even representation, then we obtain two distinct odd representations of $n+x$ by adding $x$ to the even representations of $n$, which is impossible. Therefore, $n$ can have at most one even representation.\n\nSimilarly, there exist two distinct elements $y, z \\in S$ such that $y, z>\\max \\{n, N\\}$. If $n$ has more than one odd representation, then we obtain two distinct odd representations of $n+y+z$ by adding $y$ and $z$ to the odd representations of $n$. This is again a contradiction.\n\nProperty 2. Fix $s \\in S$. Suppose that a number $n>N$ has no even representation. Then $n+2$ as has an even representation containing $s$ for all integers $a \\geqslant 1$.\n\nProof. It is sufficient to prove the following statement: If $n$ has no even representation without $s$, then $n+2 s$ has an even representation containing $s$ (and hence no even representation without $s$ by Property 1).\n\nNotice that the odd representation of $n+s$ does not contain $s$; otherwise, we have an even representation of $n$ without $s$. Then, adding $s$ to this odd representation of $n+s$, we get that $n+2 s$ has an even representation containing $s$, as desired.\n\nProperty 3. Every sufficiently large integer has an even representation.\n\nProof. Fix any $s \\in S$, and let $r$ be an arbitrary element in $\\{1,2, \\ldots, 2 s\\}$. Then, Property 2 implies that the set $Z_{r}=\\{r+2 a s: a \\geqslant 0\\}$ contains at most one number exceeding $N$ with no even representation. Therefore, $Z_{r}$ contains finitely many positive integers with no even representation, and so does $\\mathbb{Z}_{>0}=\\bigcup_{r=1}^{2 s} Z_{r}$.\n\nIn view of Properties 1 and 3, we may assume that $N$ is chosen such that every $n>N$ has exactly one odd and exactly one even representation. In particular, each element $s>N$ of $S$ has an even representation.\n\nProperty 4. For any $s, t \\in S$ with $NN$. Then, Property 4 implies that for every $i>k$ the even representation of $s_{i}$ contains all the numbers $s_{k}, s_{k+1}, \\ldots, s_{i-1}$. Therefore,\n\n$$\ns_{i}=s_{k}+s_{k+1}+\\cdots+s_{i-1}+R_{i}=\\sigma_{i-1}-\\sigma_{k-1}+R_{i},\\tag{1}\n$$\n\nwhere $R_{i}$ is a sum of some of $s_{1}, \\ldots, s_{k-1}$. In particular, $0 \\leqslant R_{i} \\leqslant s_{1}+\\cdots+s_{k-1}=\\sigma_{k-1}$.\n\n\n\nLet $j_{0}$ be an integer satisfying $j_{0}>k$ and $\\sigma_{j_{0}}>2 \\sigma_{k-1}$. Then (1) shows that, for every $j>j_{0}$,\n\n$$\ns_{j+1} \\geqslant \\sigma_{j}-\\sigma_{k-1}>\\sigma_{j} / 2\\tag{2}\n$$\n\nNext, let $p>j_{0}$ be an index such that $R_{p}=\\min _{i>j_{0}} R_{i}$. Then,\n\n$$\ns_{p+1}=s_{k}+s_{k+1}+\\cdots+s_{p}+R_{p+1}=\\left(s_{p}-R_{p}\\right)+s_{p}+R_{p+1} \\geqslant 2 s_{p}\n$$\n\nTherefore, there is no element of $S$ larger than $s_{p}$ but smaller than $2 s_{p}$. It follows that the even representation $\\tau$ of $2 s_{p}$ does not contain any element larger than $s_{p}$. On the other hand, inequality (2) yields $2 s_{p}>s_{1}+\\cdots+s_{p-1}$, so $\\tau$ must contain a term larger than $s_{p-1}$. Thus, it must contain $s_{p}$. After removing $s_{p}$ from $\\tau$, we have that $s_{p}$ has an odd representation not containing $s_{p}$, which contradicts Property 1 since $s_{p}$ itself also forms an odd representation of $s_{p}$.']",,True,,, 1949,Combinatorics,,"In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3 , and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that it is possible to partition the company into 11 parts so that no two enemies are in the same part.","['Let $G=(V, E)$ be a graph where $V$ is the set of people in the company and $E$ is the set of the enemy pairs - the edges of the graph. In this language, partitioning into 11 disjoint enemy-free subsets means properly coloring the vertices of this graph with 11 colors.\n\nWe will prove the following more general statement.\n\nClaim. Let $G$ be a graph with chromatic number $k \\geqslant 3$. Then $G$ contains at least $2^{k-1}-k$ unsociable groups.\n\nRecall that the chromatic number of $G$ is the least $k$ such that a proper coloring\n\n$$\nV=V_{1} \\sqcup \\cdots \\sqcup V_{k}\n\\tag{1}\n$$\n\nexists. In view of $2^{11}-12>2015$, the claim implies the problem statement.\n\nLet $G$ be a graph with chromatic number $k$. We say that a proper coloring (1) of $G$ is leximinimal, if the $k$-tuple $\\left(\\left|V_{1}\\right|,\\left|V_{2}\\right|, \\ldots,\\left|V_{k}\\right|\\right)$ is lexicographically minimal; in other words, the following conditions are satisfied: the number $n_{1}=\\left|V_{1}\\right|$ is minimal; the number $n_{2}=\\left|V_{2}\\right|$ is minimal, subject to the previously chosen value of $n_{1} ; \\ldots$; the number $n_{k-1}=\\left|V_{k-1}\\right|$ is minimal, subject to the previously chosen values of $n_{1}, \\ldots, n_{k-2}$.\n\nThe following lemma is the core of the proof.\n\nLemma 1. Suppose that $G=(V, E)$ is a graph with odd chromatic number $k \\geqslant 3$, and let (1) be one of its leximinimal colorings. Then $G$ contains an odd cycle which visits all color classes $V_{1}, V_{2}, \\ldots, V_{k}$.\n\nProof of Lemma 1. Let us call a cycle colorful if it visits all color classes.\n\nDue to the definition of the chromatic number, $V_{1}$ is nonempty. Choose an arbitrary vertex $v \\in V_{1}$. We construct a colorful odd cycle that has only one vertex in $V_{1}$, and this vertex is $v$.\n\nWe draw a subgraph of $G$ as follows. Place $v$ in the center, and arrange the sets $V_{2}, V_{3}, \\ldots, V_{k}$ in counterclockwise circular order around it. For convenience, let $V_{k+1}=V_{2}$. We will draw arrows to add direction to some edges of $G$, and mark the vertices these arrows point to. First we draw arrows from $v$ to all its neighbors in $V_{2}$, and mark all those neighbors. If some vertex $u \\in V_{i}$ with $i \\in\\{2,3, \\ldots, k\\}$ is already marked, we draw arrows from $u$ to all its neighbors in $V_{i+1}$ which are not marked yet, and we mark all of them. We proceed doing this as long as it is possible. The process of marking is exemplified in Figure 1.\n\nNotice that by the rules of our process, in the final state, marked vertices in $V_{i}$ cannot have unmarked neighbors in $V_{i+1}$. Moreover, $v$ is connected to all marked vertices by directed paths.\n\nNow move each marked vertex to the next color class in circular order (see an example in Figure 3). In view of the arguments above, the obtained coloring $V_{1} \\sqcup W_{2} \\sqcup \\cdots \\sqcup W_{k}$ is proper. Notice that $v$ has a neighbor $w \\in W_{2}$, because otherwise\n\n$$\n\\left(V_{1} \\backslash\\{v\\}\\right) \\sqcup\\left(W_{2} \\cup\\{v\\}\\right) \\sqcup W_{3} \\sqcup \\cdots \\sqcup W_{k}\n$$\n\nwould be a proper coloring lexicographically smaller than (1). If $w$ was unmarked, i.e., $w$ was an element of $V_{2}$, then it would be marked at the beginning of the process and thus moved to $V_{3}$, which did not happen. Therefore, $w$ is marked and $w \\in V_{k}$.\n\n\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nSince $w$ is marked, there exists a directed path from $v$ to $w$. This path moves through the sets $V_{2}, \\ldots, V_{k}$ in circular order, so the number of edges in it is divisible by $k-1$ and thus even. Closing this path by the edge $w \\rightarrow v$, we get a colorful odd cycle, as required.\n\nProof of the claim. Let us choose a leximinimal coloring (1) of $G$. For every set $C \\subseteq\\{1,2, \\ldots, k\\}$ such that $|C|$ is odd and greater than 1 , we will provide an odd cycle visiting exactly those color classes whose indices are listed in the set $C$. This property ensures that we have different cycles for different choices of $C$, and it proves the claim because there are $2^{k-1}-k$ choices for the set $C$.\n\nLet $V_{C}=\\bigcup_{c \\in C} V_{c}$, and let $G_{C}$ be the induced subgraph of $G$ on the vertex set $V_{C}$. We also have the induced coloring of $V_{C}$ with $|C|$ colors; this coloring is of course proper. Notice further that the induced coloring is leximinimal: if we had a lexicographically smaller coloring $\\left(W_{c}\\right)_{c \\in C}$ of $G_{C}$, then these classes, together the original color classes $V_{i}$ for $i \\notin C$, would provide a proper coloring which is lexicographically smaller than (1). Hence Lemma 1, applied to the subgraph $G_{C}$ and its leximinimal coloring $\\left(V_{c}\\right)_{c \\in C}$, provides an odd cycle that visits exactly those color classes that are listed in the set $C$.', ""We say that a graph is critical if deleting any vertex from the graph decreases the graph's chromatic number. Obviously every graph contains a critical induced subgraph with the same chromatic number.\n\nLemma 2. Suppose that $G=(V, E)$ is a critical graph with chromatic number $k \\geqslant 3$. Then every vertex $v$ of $G$ is contained in at least $2^{k-2}-1$ unsociable groups.\n\nProof. For every set $X \\subseteq V$, denote by $n(X)$ the number of neighbors of $v$ in the set $X$.\n\nSince $G$ is critical, there exists a proper coloring of $G \\backslash\\{v\\}$ with $k-1$ colors, so there exists a proper coloring $V=V_{1} \\sqcup V_{2} \\sqcup \\cdots \\sqcup V_{k}$ of $G$ such that $V_{1}=\\{v\\}$. Among such colorings, take one for which the sequence $\\left(n\\left(V_{2}\\right), n\\left(V_{3}\\right), \\ldots, n\\left(V_{k}\\right)\\right)$ is lexicographically minimal. Clearly, $n\\left(V_{i}\\right)>0$ for every $i=2,3, \\ldots, k$; otherwise $V_{2} \\sqcup \\ldots \\sqcup V_{i-1} \\sqcup\\left(V_{i} \\cup V_{1}\\right) \\sqcup V_{i+1} \\sqcup \\ldots V_{k}$ would be a proper coloring of $G$ with $k-1$ colors.\n\nWe claim that for every $C \\subseteq\\{2,3, \\ldots, k\\}$ with $|C| \\geqslant 2$ being even, $G$ contains an unsociable group so that the set of its members' colors is precisely $C \\cup\\{1\\}$. Since the number of such sets $C$ is $2^{k-2}-1$, this proves the lemma. Denote the elements of $C$ by $c_{1}, \\ldots, c_{2 \\ell}$ in increasing order. For brevity, let $U_{i}=V_{c_{i}}$. Denote by $N_{i}$ the set of neighbors of $v$ in $U_{i}$.\n\n\n\nWe show that for every $i=1, \\ldots, 2 \\ell-1$ and $x \\in N_{i}$, the subgraph induced by $U_{i} \\cup U_{i+1}$ contains a path that connects $x$ with another point in $N_{i+1}$. For the sake of contradiction, suppose that no such path exists. Let $S$ be the set of vertices that lie in the connected component of $x$ in the subgraph induced by $U_{i} \\cup U_{i+1}$, and let $P=U_{i} \\cap S$, and $Q=U_{i+1} \\cap S$ (see Figure 3). Since $x$ is separated from $N_{i+1}$, the sets $Q$ and $N_{i+1}$ are disjoint. So, if we re-color $G$ by replacing $U_{i}$ and $U_{i+1}$ by $\\left(U_{i} \\cup Q\\right) \\backslash P$ and $\\left(U_{i+1} \\cup P\\right) \\backslash Q$, respectively, we obtain a proper coloring such that $n\\left(U_{i}\\right)=n\\left(V_{c_{i}}\\right)$ is decreased and only $n\\left(U_{i+1}\\right)=n\\left(V_{c_{i+1}}\\right)$ is increased. That contradicts the lexicographical minimality of $\\left(n\\left(V_{2}\\right), n\\left(V_{3}\\right), \\ldots, n\\left(V_{k}\\right)\\right)$.\n\n\n\nFigure 3\n\nNext, we build a path through $U_{1}, U_{2}, \\ldots, U_{2 \\ell}$ as follows. Let the starting point of the path be an arbitrary vertex $v_{1}$ in the set $N_{1}$. For $i \\leqslant 2 \\ell-1$, if the vertex $v_{i} \\in N_{i}$ is already defined, connect $v_{i}$ to some vertex in $N_{i+1}$ in the subgraph induced by $U_{i} \\cup U_{i+1}$, and add these edges to the path. Denote the new endpoint of the path by $v_{i+1}$; by the construction we have $v_{i+1} \\in N_{i+1}$ again, so the process can be continued. At the end we have a path that starts at $v_{1} \\in N_{1}$ and ends at some $v_{2 \\ell} \\in N_{2 \\ell}$. Moreover, all edges in this path connect vertices in neighboring classes: if a vertex of the path lies in $U_{i}$, then the next vertex lies in $U_{i+1}$ or $U_{i-1}$. Notice that the path is not necessary simple, so take a minimal subpath of it. The minimal subpath is simple and connects the same endpoints $v_{1}$ and $v_{2 \\ell}$. The property that every edge steps to a neighboring color class (i.e., from $U_{i}$ to $U_{i+1}$ or $U_{i-1}$ ) is preserved. So the resulting path also visits all of $U_{1}, \\ldots, U_{2 \\ell}$, and its length must be odd. Closing the path with the edges $v v_{1}$ and $v_{2 \\ell} v$ we obtain the desired odd cycle (see Figure 4).\n\n\n\nFigure 4\n\nNow we prove the claim by induction on $k \\geqslant 3$. The base case $k=3$ holds by applying Lemma 2 to a critical subgraph. For the induction step, let $G_{0}$ be a critical $k$-chromatic subgraph of $G$, and let $v$ be an arbitrary vertex of $G_{0}$. By Lemma $2, G_{0}$ has at least $2^{k-2}-1$ unsociable groups containing $v$. On the other hand, the graph $G_{0} \\backslash\\{v\\}$ has chromatic number $k-1$, so it contains at least $2^{k-2}-(k-1)$ unsociable groups by the induction hypothesis. Altogether, this gives $2^{k-2}-1+2^{k-2}-(k-1)=2^{k-1}-k$ distinct unsociable groups in $G_{0}$ (and thus in $G)$.""]",,True,,, 1950,Geometry,,Let $A B C$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $A B G H$ is a parallelogram. Let $I$ be the point on the line $G H$ such that $A C$ bisects $H I$. Suppose that the line $A C$ intersects the circumcircle of the triangle $G C I$ at $C$ and $J$. Prove that $I J=A H$.,"['Since $H G \\| A B$ and $B G \\| A H$, we have $B G \\perp B C$ and $C H \\perp G H$. Therefore, the quadrilateral $B G C H$ is cyclic. Since $H$ is the orthocenter of the triangle $A B C$, we have $\\angle H A C=90^{\\circ}-\\angle A C B=\\angle C B H$. Using that $B G C H$ and $C G J I$ are cyclic quadrilaterals, we get\n\n$$\n\\angle C J I=\\angle C G H=\\angle C B H=\\angle H A C .\n$$\n\nLet $M$ be the intersection of $A C$ and $G H$, and let $D \\neq A$ be the point on the line $A C$ such that $A H=H D$. Then $\\angle M J I=\\angle H A C=\\angle M D H$.\n\nSince $\\angle M J I=\\angle M D H, \\angle I M J=\\angle H M D$, and $I M=M H$, the triangles $I M J$ and $H M D$ are congruent, and thus $I J=H D=A H$.\n\n', 'Obtain $\\angle C G H=\\angle H A C$ as in the previous solution. In the parallelogram $A B G H$ we have $\\angle B A H=\\angle H G B$. It follows that\n\n$$\n\\angle H M C=\\angle B A C=\\angle B A H+\\angle H A C=\\angle H G B+\\angle C G H=\\angle C G B .\n$$\n\nSo the right triangles $C M H$ and $C G B$ are similar. Also, in the circumcircle of triangle $G C I$ we have similar triangles $M I J$ and $M C G$. Therefore,\n\n$$\n\\frac{I J}{C G}=\\frac{M I}{M C}=\\frac{M H}{M C}=\\frac{G B}{G C}=\\frac{A H}{C G}\n$$\n\nHence $I J=A H$.']",,True,,, 1951,Geometry,,"Let $A B C$ be a triangle inscribed into a circle $\Omega$ with center $O$. A circle $\Gamma$ with center $A$ meets the side $B C$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\Gamma$ and $\Omega$. We assume that $F$ lies on the arc $A B$ of $\Omega$ not containing $C$, and $G$ lies on the arc $A C$ of $\Omega$ not containing $B$. The circumcircles of the triangles $B D F$ and $C E G$ meet the sides $A B$ and $A C$ again at $K$ and $L$, respectively. Suppose that the lines $F K$ and $G L$ are distinct and intersect at $X$. Prove that the points $A, X$, and $O$ are collinear.","['It suffices to prove that the lines $F K$ and $G L$ are symmetric about $A O$. Now the segments $A F$ and $A G$, being chords of $\\Omega$ with the same length, are clearly symmetric with respect to $A O$. Hence it is enough to show\n\n$$\n\\angle K F A=\\angle A G L \\text {. }\n\\tag{1}\n$$\n\nLet us denote the circumcircles of $B D F$ and $C E G$ by $\\omega_{B}$ and $\\omega_{C}$, respectively. To prove (1), we start from\n\n$$\n\\angle K F A=\\angle D F G+\\angle G F A-\\angle D F K .\n$$\n\nIn view of the circles $\\omega_{B}, \\Gamma$, and $\\Omega$, this may be rewritten as\n\n$$\n\\angle K F A=\\angle C E G+\\angle G B A-\\angle D B K=\\angle C E G-\\angle C B G .\n$$\n\nDue to the circles $\\omega_{C}$ and $\\Omega$, we obtain $\\angle K F A=\\angle C L G-\\angle C A G=\\angle A G L$. Thereby the problem is solved.\n\n\n\nFigure 1', 'Again, we denote the circumcircle of $B D K F$ by $\\omega_{B}$. In addition, we set $\\alpha=$ $\\angle B A C, \\varphi=\\angle A B F$, and $\\psi=\\angle E D A=\\angle A E D$ (see Figure 2). Notice that $A F=A G$ entails $\\varphi=\\angle G C A$, so all three of $\\alpha, \\varphi$, and $\\psi$ respect the ""symmetry"" between $B$ and $C$ of our configuration. Again, we reduce our task to proving (1).\n\nThis time, we start from\n\n$$\n2 \\angle K F A=2(\\angle D F A-\\angle D F K) .\n$$\n\nSince the triangle $A F D$ is isosceles, we have\n\n$$\n\\angle D F A=\\angle A D F=\\angle E D F-\\psi=\\angle B F D+\\angle E B F-\\psi .\n$$\n\n\n\nMoreover, because of the circle $\\omega_{B}$ we have $\\angle D F K=\\angle C B A$. Altogether, this yields\n\n$$\n2 \\angle K F A=\\angle D F A+(\\angle B F D+\\angle E B F-\\psi)-2 \\angle C B A,\n$$\n\nwhich simplifies to\n\n$$\n2 \\angle K F A=\\angle B F A+\\varphi-\\psi-\\angle C B A .\n$$\n\nNow the quadrilateral $A F B C$ is cyclic, so this entails $2 \\angle K F A=\\alpha+\\varphi-\\psi$.\n\nDue to the ""symmetry"" between $B$ and $C$ alluded to above, this argument also shows that $2 \\angle A G L=\\alpha+\\varphi-\\psi$. This concludes the proof of (1).\n\n\n\nFigure 2']",,True,,, 1952,Geometry,,"Let $A B C$ be a triangle with $\angle C=90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $C B H$ so that $C H$ bisects $A D$. Let $P$ be the intersection point of the lines $B D$ and $C H$. Let $\omega$ be the semicircle with diameter $B D$ that meets the segment $C B$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $C Q$ and $A D$ meet on $\omega$.","['Let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 1). Since $P H \\| D K$, we have\n\n$$\n\\frac{P D}{P B}=\\frac{H K}{H B}=\\frac{A H}{H B}\n\\tag{1}\n$$\n\nLet $L$ be the projection of $Q$ onto $D B$. Since $P Q$ is tangent to $\\omega$ and $\\angle D Q B=\\angle B L Q=$ $90^{\\circ}$, we have $\\angle P Q D=\\angle Q B P=\\angle D Q L$. Therefore, $Q D$ and $Q B$ are respectively the internal and the external bisectors of $\\angle P Q L$. By the angle bisector theorem, we obtain\n\n$$\n\\frac{P D}{D L}=\\frac{P Q}{Q L}=\\frac{P B}{B L}\n\\tag{2}\n$$\n\nThe relations (1) and (2) yield $\\frac{A H}{H B}=\\frac{P D}{P B}=\\frac{D L}{L B}$. So, the spiral similarity $\\tau$ centered at $B$ and sending $A$ to $D$ maps $H$ to $L$. Moreover, $\\tau$ sends the semicircle with diameter $A B$ passing through $C$ to $\\omega$. Due to $C H \\perp A B$ and $Q L \\perp D B$, it follows that $\\tau(C)=Q$.\n\nHence, the triangles $A B D$ and $C B Q$ are similar, so $\\angle A D B=\\angle C Q B$. This means that the lines $A D$ and $C Q$ meet at some point $T$, and this point satisfies $\\angle B D T=\\angle B Q T$. Therefore, $T$ lies on $\\omega$, as needed.\n\n\n\nFigure 1\n\n\n\nFigure 2', 'Let $\\Gamma$ be the circumcircle of $A B C$, and let $A D$ meet $\\omega$ at $T$. Then $\\angle A T B=$ $\\angle A C B=90^{\\circ}$, so $T$ lies on $\\Gamma$ as well. As in the previous solution, let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 2).\n\nOur goal now is to prove that the points $C, Q$, and $T$ are collinear. Let $C T$ meet $\\omega$ again at $Q^{\\prime}$. Then, it suffices to show that $P Q^{\\prime}$ is tangent to $\\omega$, or that $\\angle P Q^{\\prime} D=\\angle Q^{\\prime} B D$.\n\nSince the quadrilateral $B D Q^{\\prime} T$ is cyclic and the triangles $A H C$ and $K H C$ are congruent, we have $\\angle Q^{\\prime} B D=\\angle Q^{\\prime} T D=\\angle C T A=\\angle C B A=\\angle A C H=\\angle H C K$. Hence, the right triangles $C H K$ and $B Q^{\\prime} D$ are similar. This implies that $\\frac{H K}{C K}=\\frac{Q^{\\prime} D}{B D}$, and thus $H K \\cdot B D=C K \\cdot Q^{\\prime} D$.\n\nNotice that $P H \\| D K$; therefore, we have $\\frac{P D}{B D}=\\frac{H K}{B K}$, and so $P D \\cdot B K=H K \\cdot B D$. Consequently, $P D \\cdot B K=H K \\cdot B D=C K \\cdot Q^{\\prime} D$, which yields $\\frac{P D}{Q^{\\prime} D}=\\frac{C K}{B K}$.\n\nSince $\\angle C K A=\\angle K A C=\\angle B D Q^{\\prime}$, the triangles $C K B$ and $P D Q^{\\prime}$ are similar, so $\\angle P Q^{\\prime} D=$ $\\angle C B A=\\angle Q^{\\prime} B D$, as required.', ""Introduce the points $T$ and $Q^{\\prime}$ as in the previous solution. Note that $T$ lies on the circumcircle of $A B C$. Here we present yet another proof that $P Q^{\\prime}$ is tangent to $\\omega$.\n\nLet $\\Omega$ be the circle completing the semicircle $\\omega$. Construct a point $F$ symmetric to $C$ with respect to $A B$. Let $S \\neq T$ be the second intersection point of $F T$ and $\\Omega$ (see Figure 4).\n\n\n\nFigure 4\n\nSince $A C=A F$, we have $\\angle D K C=\\angle H C K=\\angle C B A=\\angle C T A=\\angle D T S=180^{\\circ}-$ $\\angle S K D$. Thus, the points $C, K$, and $S$ are collinear. Notice also that $\\angle Q^{\\prime} K D=\\angle Q^{\\prime} T D=$ $\\angle H C K=\\angle K F H=180^{\\circ}-\\angle D K F$. This implies that the points $F, K$, and $Q^{\\prime}$ are collinear.\n\nApplying PASCAL's theorem to the degenerate hexagon $K Q^{\\prime} Q^{\\prime} T S S$, we get that the tangents to $\\Omega$ passing through $Q^{\\prime}$ and $S$ intersect on $C F$. The relation $\\angle Q^{\\prime} T D=\\angle D T S$ yields that $Q^{\\prime}$ and $S$ are symmetric with respect to $B D$. Therefore, the two tangents also intersect on $B D$. Thus, the two tangents pass through $P$. Hence, $P Q^{\\prime}$ is tangent to $\\omega$, as needed.""]",,True,,, 1953,Geometry,,"Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triangle $A B C$. Determine all possible values of $B T / B M$.","['Let $S$ be the center of the parallelogram $B P T Q$, and let $B^{\\prime} \\neq B$ be the point on the ray $B M$ such that $B M=M B^{\\prime}$ (see Figure 1). It follows that $A B C B^{\\prime}$ is a parallelogram. Then, $\\angle A B B^{\\prime}=\\angle P Q M$ and $\\angle B B^{\\prime} A=\\angle B^{\\prime} B C=\\angle M P Q$, and so the triangles $A B B^{\\prime}$ and $M Q P$ are similar. It follows that $A M$ and $M S$ are corresponding medians in these triangles. Hence,\n\n$$\n\\angle S M P=\\angle B^{\\prime} A M=\\angle B C A=\\angle B T A .\n\\tag{1}\n$$\n\nSince $\\angle A C T=\\angle P B T$ and $\\angle T A C=\\angle T B C=\\angle B T P$, the triangles $T C A$ and $P B T$ are similar. Again, as $T M$ and $P S$ are corresponding medians in these triangles, we have\n\n$$\n\\angle M T A=\\angle T P S=\\angle B Q P=\\angle B M P .\n\\tag{2}\n$$\n\nNow we deal separately with two cases.\n\nCase 1. $S$ does not lie on $B M$. Since the configuration is symmetric between $A$ and $C$, we may assume that $S$ and $A$ lie on the same side with respect to the line $B M$.\n\nApplying (1) and (2), we get\n\n$$\n\\angle B M S=\\angle B M P-\\angle S M P=\\angle M T A-\\angle B T A=\\angle M T B\n$$\n\nand so the triangles $B S M$ and $B M T$ are similar. We now have $B M^{2}=B S \\cdot B T=B T^{2} / 2$, so $B T=\\sqrt{2} B M$.\n\nCase 2. S lies on $B M$. It follows from (2) that $\\angle B C A=\\angle M T A=\\angle B Q P=\\angle B M P$ (see Figure 2). Thus, $P Q \\| A C$ and $P M \\| A T$. Hence, $B S / B M=B P / B A=B M / B T$, so $B T^{2}=2 B M^{2}$ and $B T=\\sqrt{2} B M$.\n\n\n\nFigure 1\n\n\n\nFigure 2', 'Again, we denote by $\\Omega$ the circumcircle of the triangle $A B C$.\n\nChoose the points $X$ and $Y$ on the rays $B A$ and $B C$ respectively, so that $\\angle M X B=\\angle M B C$ and $\\angle B Y M=\\angle A B M$ (see Figure 4). Then the triangles $B M X$ and $Y M B$ are similar. Since $\\angle X P M=\\angle B Q M$, the points $P$ and $Q$ correspond to each other in these triangles. So, if $\\overrightarrow{B P}=\\mu \\cdot \\overrightarrow{B X}$, then $\\overrightarrow{B Q}=(1-\\mu) \\cdot \\overrightarrow{B Y}$. Thus\n\n$$\n\\overrightarrow{B T}=\\overrightarrow{B P}+\\overrightarrow{B Q}=\\overrightarrow{B Y}+\\mu \\cdot(\\overrightarrow{B X}-\\overrightarrow{B Y})=\\overrightarrow{B Y}+\\mu \\cdot \\overrightarrow{Y X}\n$$\n\nwhich means that $T$ lies on the line $X Y$.\n\nLet $B^{\\prime} \\neq B$ be the point on the ray $B M$ such that $B M=M B^{\\prime}$. Then $\\angle M B^{\\prime} A=$ $\\angle M B C=\\angle M X B$ and $\\angle C B^{\\prime} M=\\angle A B M=\\angle B Y M$. This means that the triangles $B M X$, $B A B^{\\prime}, Y M B$, and $B^{\\prime} C B$ are all similar; hence $B A \\cdot B X=B M \\cdot B B^{\\prime}=B C \\cdot B Y$. Thus there exists an inversion centered at $B$ which swaps $A$ with $X, M$ with $B^{\\prime}$, and $C$ with $Y$. This inversion then swaps $\\Omega$ with the line $X Y$, and hence it preserves $T$. Therefore, we have $B T^{2}=B M \\cdot B B^{\\prime}=2 B M^{2}$, and $B T=\\sqrt{2} B M$.', 'We begin with the following lemma.\n\nLemma. Let $A B C T$ be a cyclic quadrilateral. Let $P$ and $Q$ be points on the sides $B A$ and $B C$ respectively, such that $B P T Q$ is a parallelogram. Then $B P \\cdot B A+B Q \\cdot B C=B T^{2}$.\n\nProof. Let the circumcircle of the triangle $Q T C$ meet the line $B T$ again at $J$ (see Figure 5). The power of $B$ with respect to this circle yields\n\n$$\nB Q \\cdot B C=B J \\cdot B T \\text {. }\\tag{3}\n$$\n\n\n\nWe also have $\\angle T J Q=180^{\\circ}-\\angle Q C T=\\angle T A B$ and $\\angle Q T J=\\angle A B T$, and so the triangles $T J Q$ and $B A T$ are similar. We now have $T J / T Q=B A / B T$. Therefore,\n\n$$\nT J \\cdot B T=T Q \\cdot B A=B P \\cdot B A \\text {. }\n\\tag{4}\n$$\n\nCombining (3) and (4) now yields the desired result.\n\nLet $X$ and $Y$ be the midpoints of $B A$ and $B C$ respectively (see Figure 6). Applying the lemma to the cyclic quadrilaterals $P B Q M$ and $A B C T$, we obtain\n\n$$\nB X \\cdot B P+B Y \\cdot B Q=B M^{2}\n$$\n\nand\n\n$$\nB P \\cdot B A+B Q \\cdot B C=B T^{2} \\text {. }\n$$\n\nSince $B A=2 B X$ and $B C=2 B Y$, we have $B T^{2}=2 B M^{2}$, and so $B T=\\sqrt{2} B M$.\n\n\n\nFigure 5\n\n\n\nFigure 6']",['$\\sqrt{2}$'],False,,Numerical, 1954,Geometry,,"Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$.","['We may assume that $C A>C B$. Observe that $H^{\\prime}$ and $I^{\\prime}$ lie inside the segments $C F$ and $C G$, respectively. Therefore, $M$ lies outside $\\triangle A B C$ (see Figure 1).\n\nDue to the powers of points $A$ and $B$ with respect to the circle $\\Gamma$, we have\n\n$$\nC H^{\\prime} \\cdot C A=A H \\cdot A C=A D^{2}=B D^{2}=B I \\cdot B C=C I^{\\prime} \\cdot C B .\n$$\n\nTherefore, $C H^{\\prime} \\cdot C F=C I^{\\prime} \\cdot C G$. Hence, the quadrilateral $H^{\\prime} I^{\\prime} G F$ is cyclic, and so $\\angle I^{\\prime} H^{\\prime} C=$ $\\angle C G F$.\n\nLet $D F$ and $D G$ meet $\\Gamma$ again at $R$ and $S$, respectively. We claim that the points $R$ and $S$ lie on the line $H^{\\prime} I^{\\prime}$.\n\nObserve that $F H^{\\prime} \\cdot F A=F H \\cdot F C=F R \\cdot F D$. Thus, the quadrilateral $A D H^{\\prime} R$ is cyclic, and hence $\\angle R H^{\\prime} F=\\angle F D A=\\angle C G F=\\angle I^{\\prime} H^{\\prime} C$. Therefore, the points $R, H^{\\prime}$, and $I^{\\prime}$ are collinear. Similarly, the points $S, H^{\\prime}$, and $I^{\\prime}$ are also collinear, and so all the points $R, H^{\\prime}, Q, I^{\\prime}, S$, and $M$ are all collinear.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nThen, $\\angle R S D=\\angle R D A=\\angle D F G$. Hence, the quadrilateral $R S G F$ is cyclic (see Figure 2). Therefore, $M H^{\\prime} \\cdot M I^{\\prime}=M F \\cdot M G=M R \\cdot M S=M P \\cdot M C$. Thus, the quadrilateral $C P I^{\\prime} H^{\\prime}$ is also cyclic. Let $\\omega$ be its circumcircle.\n\nNotice that $\\angle H^{\\prime} C Q=\\angle S D C=\\angle S R C$ and $\\angle Q C I^{\\prime}=\\angle C D R=\\angle C S R$. Hence, $\\triangle C H^{\\prime} Q \\sim \\triangle R C Q$ and $\\triangle C I^{\\prime} Q \\sim \\triangle S C Q$, and therefore $Q H^{\\prime} \\cdot Q R=Q C^{2}=Q I^{\\prime} \\cdot Q S$.\n\nWe apply the inversion with center $Q$ and radius $Q C$. Observe that the points $R, C$, and $S$ are mapped to $H^{\\prime}, C$, and $I^{\\prime}$, respectively. Therefore, the circumcircle $\\Gamma$ of $\\triangle R C S$ is mapped to the circumcircle $\\omega$ of $\\triangle H^{\\prime} C I^{\\prime}$. Since $P$ and $C$ belong to both circles and the point $C$ is preserved by the inversion, we have that $P$ is also mapped to itself. We then get $Q P^{2}=Q C^{2}$. Hence, $Q P=Q C$.', ""Let $X=H I \\cap A B$, and let the tangent to $\\Gamma$ at $C$ meet $A B$ at $Y$. Let $X C$ meet $\\Gamma$ again at $X^{\\prime}$ (see Figure 3). Projecting from $C, X$, and $C$ again, we have $(X, A ; D, B)=$ $\\left(X^{\\prime}, H ; D, I\\right)=(C, I ; D, H)=(Y, B ; D, A)$. Since $A$ and $B$ are symmetric about $D$, it follows that $X$ and $Y$ are also symmetric about $D$.\n\nNow, Menelaus' theorem applied to $\\triangle A B C$ with the line $H I X$ yields\n\n$$\n1=\\frac{C H}{H A} \\cdot \\frac{B I}{I C} \\cdot \\frac{A X}{X B}=\\frac{A H^{\\prime}}{H^{\\prime} C} \\cdot \\frac{C I^{\\prime}}{I^{\\prime} B} \\cdot \\frac{B Y}{Y A}\n$$\n\nBy the converse of Menelaus' theorem applied to $\\triangle A B C$ with points $H^{\\prime}, I^{\\prime}, Y$, we get that the points $H^{\\prime}, I^{\\prime}, Y$ are collinear.\n\n\n\nFigure 3\n\nLet $T$ be the midpoint of $C D$, and let $O$ be the center of $\\Gamma$. Let $C M$ meet $T Y$ at $N$. To avoid confusion, we clean some superfluous details out of the picture (see Figure 4).\n\nLet $V=M T \\cap C Y$. Since $M T \\| Y D$ and $D T=T C$, we get $C V=V Y$. Then CEvA's theorem applied to $\\triangle C T Y$ with the point $M$ yields\n\n$$\n1=\\frac{T Q}{Q C} \\cdot \\frac{C V}{V Y} \\cdot \\frac{Y N}{N T}=\\frac{T Q}{Q C} \\cdot \\frac{Y N}{N T}\n$$\n\nTherefore, $\\frac{T Q}{Q C}=\\frac{T N}{N Y}$. So, $N Q \\| C Y$, and thus $N Q \\perp O C$.\n\nNote that the points $O, N, T$, and $Y$ are collinear. Therefore, $C Q \\perp O N$. So, $Q$ is the orthocenter of $\\triangle O C N$, and hence $O Q \\perp C P$. Thus, $Q$ lies on the perpendicular bisector of $C P$, and therefore $C Q=Q P$, as required.\n\n\n\nFigure 4""]",,True,,, 1955,Geometry,,"Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other.","['Let $A^{\\prime}$ be the point diametrically opposite to $A$ on $\\Gamma$. Since $\\angle A Q A^{\\prime}=90^{\\circ}$ and $\\angle A Q H=90^{\\circ}$, the points $Q, H$, and $A^{\\prime}$ are collinear. Similarly, if $Q^{\\prime}$ denotes the point on $\\Gamma$ diametrically opposite to $Q$, then $K, H$, and $Q^{\\prime}$ are collinear. Let the line $A H F$ intersect $\\Gamma$ again at $E$; it is known that $M$ is the midpoint of the segment $H A^{\\prime}$ and that $F$ is the midpoint of $H E$. Let $J$ be the midpoint of $H Q^{\\prime}$.\n\nConsider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\\angle H K T=\\angle H Q K$ and we are to prove that $\\angle M K T=\\angle C F K$. Thus it remains to show that $\\angle H Q K=\\angle C F K+\\angle H K M$. Due to $\\angle H Q K=90^{\\circ}-\\angle Q^{\\prime} H A^{\\prime}$ and $\\angle C F K=90^{\\circ}-\\angle K F A$, this means the same as $\\angle Q^{\\prime} H A^{\\prime}=$ $\\angle K F A-\\angle H K M$. Now, since the triangles $K H E$ and $A H Q^{\\prime}$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\\angle K F A=\\angle H J A$, and analogously one may obtain $\\angle H K M=\\angle J Q H$. Thereby our task is reduced to verifying\n\n$$\n\\angle Q^{\\prime} H A^{\\prime}=\\angle H J A-\\angle J Q H .\n$$\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nTo avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\\angle Q^{\\prime} H A^{\\prime}=\\angle J Q H+\\angle H J Q$ and $\\angle H J A=\\angle Q J A+\\angle H J Q$, we just have to show that $2 \\angle J Q H=\\angle Q J A$. To this end, it suffices to remark that $A Q A^{\\prime} Q^{\\prime}$ is a rectangle and that $J$, being defined to be the midpoint of $H Q^{\\prime}$, has to lie on the mid parallel of $Q A^{\\prime}$ and $Q^{\\prime} A$.', 'We define the points $A^{\\prime}$ and $E$ and prove that the ray $M H$ passes through $Q$ in the same way as in the first solution. Notice that the points $A^{\\prime}$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A^{\\prime}$ is the second intersection of the line $M H$ with $\\Gamma$, and $E$ is the point on $\\Gamma$ with the property $\\angle H E A^{\\prime}=90^{\\circ}$ (see Figure 3).\n\nIn the circles $K Q H$ and $E A^{\\prime} H$, the line segments $H Q$ and $H A^{\\prime}$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A^{\\prime} H$. Their pairwise radical axes are the lines $Q K$, $A^{\\prime} E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\\angle Q K H=$\n\n\n\n\n\nFigure 3\n\n$\\angle H E A^{\\prime}=90^{\\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K=S E=S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$.\n\nThe circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have\n\n$$\nS M \\cdot S F=S H^{2}=S K^{2}\n$$\n\nSo, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.']",,True,,, 1956,Geometry,,"Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other.","[""Denote by $\\gamma_{A}, \\gamma_{B}, \\gamma_{C}$, and $\\gamma_{D}$ the incircles of the quadrilaterals $A P O S, B Q O P$, $C R O Q$, and $D S O R$, respectively.\n\nWe start with proving that the quadrilateral $A B C D$ also has an incircle which will be referred to as $\\Omega$. Denote the points of tangency as in Figure 1. It is well-known that $Q Q_{1}=O O_{1}$ (if $B C \\| P R$, this is obvious; otherwise, one may regard the two circles involved as the incircle and an excircle of the triangle formed by the lines $O Q, P R$, and $B C$ ). Similarly, $O O_{1}=P P_{1}$. Hence we have $Q Q_{1}=P P_{1}$. The other equalities of segment lengths marked in Figure 1 can be proved analogously. These equalities, together with $A P_{1}=A S_{1}$ and similar ones, yield $A B+C D=A D+B C$, as required.\n\n\n\nFigure 1\n\nNext, let us draw the lines parallel to $Q S$ through $P$ and $R$, and also draw the lines parallel to $P R$ through $Q$ and $S$. These lines form a parallelogram; denote its vertices by $A^{\\prime}, B^{\\prime}, C^{\\prime}$, and $D^{\\prime}$ as shown in Figure 2.\n\nSince the quadrilateral $A P O S$ has an incircle, we have $A P-A S=O P-O S=A^{\\prime} S-A^{\\prime} P$. It is well-known that in this case there also exists a circle $\\omega_{A}$ tangent to the four rays $A P$, $A S, A^{\\prime} P$, and $A^{\\prime} S$. It is worth mentioning here that in case when, say, the lines $A B$ and $A^{\\prime} B^{\\prime}$ coincide, the circle $\\omega_{A}$ is just tangent to $A B$ at $P$. We introduce the circles $\\omega_{B}, \\omega_{C}$, and $\\omega_{D}$ in a similar manner.\n\nAssume that the radii of the circles $\\omega_{A}$ and $\\omega_{C}$ are different. Let $X$ be the center of the homothety having a positive scale factor and mapping $\\omega_{A}$ to $\\omega_{C}$.\n\nNow, Monge's theorem applied to the circles $\\omega_{A}, \\Omega$, and $\\omega_{C}$ shows that the points $A, C$, and $X$ are collinear. Applying the same theorem to the circles $\\omega_{A}, \\omega_{B}$, and $\\omega_{C}$, we see that the points $P, Q$, and $X$ are also collinear. Similarly, the points $R, S$, and $X$ are collinear, as required.\n\nIf the radii of $\\omega_{A}$ and $\\omega_{C}$ are equal but these circles do not coincide, then the degenerate version of the same theorem yields that the three lines $A C, P Q$, and $R S$ are parallel to the line of centers of $\\omega_{A}$ and $\\omega_{C}$.\n\nFinally, we need to say a few words about the case when $\\omega_{A}$ and $\\omega_{C}$ coincide (and thus they also coincide with $\\Omega, \\omega_{B}$, and $\\omega_{D}$ ). It may be regarded as the limit case in the following manner.\n\n\n\n\n\nFigure 2\n\nLet us fix the positions of $A, P, O$, and $S$ (thus we also fix the circles $\\omega_{A}, \\gamma_{A}, \\gamma_{B}$, and $\\gamma_{D}$ ). Now we vary the circle $\\gamma_{C}$ inscribed into $\\angle Q O R$; for each of its positions, one may reconstruct the lines $B C$ and $C D$ as the external common tangents to $\\gamma_{B}, \\gamma_{C}$ and $\\gamma_{C}, \\gamma_{D}$ different from $P R$ and $Q S$, respectively. After such variation, the circle $\\Omega$ changes, so the result obtained above may be applied."", ""Applying Menelaus' theorem to $\\triangle A B C$ with the line $P Q$ and to $\\triangle A C D$ with the line $R S$, we see that the line $A C$ meets $P Q$ and $R S$ at the same point (possibly at infinity) if and only if\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=1\n\\tag{1}\n$$\n\nSo, it suffices to prove (1).\n\nWe start with the following result.\n\nLemma 1. Let $E F G H$ be a circumscribed quadrilateral, and let $M$ be its incenter. Then\n\n$$\n\\frac{E F \\cdot F G}{G H \\cdot H E}=\\frac{F M^{2}}{H M^{2}}\n$$\n\nProof. Notice that $\\angle E M H+\\angle G M F=\\angle F M E+\\angle H M G=180^{\\circ}, \\angle F G M=\\angle M G H$, and $\\angle H E M=\\angle M E F$ (see Figure 3). By the law of sines, we get\n\n$$\n\\frac{E F}{F M} \\cdot \\frac{F G}{F M}=\\frac{\\sin \\angle F M E \\cdot \\sin \\angle G M F}{\\sin \\angle M E F \\cdot \\sin \\angle F G M}=\\frac{\\sin \\angle H M G \\cdot \\sin \\angle E M H}{\\sin \\angle M G H \\cdot \\sin \\angle H E M}=\\frac{G H}{H M} \\cdot \\frac{H E}{H M} .\n$$\n\n\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nWe denote by $I, J, K$, and $L$ the incenters of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$, respectively. Applying Lemma 1 to these four quadrilaterals we get\n\n$$\n\\frac{A P \\cdot P O}{O S \\cdot S A} \\cdot \\frac{B Q \\cdot Q O}{O P \\cdot P B} \\cdot \\frac{C R \\cdot R O}{O Q \\cdot Q C} \\cdot \\frac{D S \\cdot S O}{O R \\cdot R D}=\\frac{P I^{2}}{S I^{2}} \\cdot \\frac{Q J^{2}}{P J^{2}} \\cdot \\frac{R K^{2}}{Q K^{2}} \\cdot \\frac{S L^{2}}{R L^{2}}\n$$\n\nwhich reduces to\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=\\frac{P I^{2}}{P J^{2}} \\cdot \\frac{Q J^{2}}{Q K^{2}} \\cdot \\frac{R K^{2}}{R L^{2}} \\cdot \\frac{S L^{2}}{S I^{2}}\n\\tag{2}\n$$\n\nNext, we have $\\angle I P J=\\angle J O I=90^{\\circ}$, and the line $O P$ separates $I$ and $J$ (see Figure 4). This means that the quadrilateral $I P J O$ is cyclic. Similarly, we get that the quadrilateral $J Q K O$ is cyclic with $\\angle J Q K=90^{\\circ}$. Thus, $\\angle Q K J=\\angle Q O J=\\angle J O P=\\angle J I P$. Hence, the right triangles $I P J$ and $K Q J$ are similar. Therefore, $\\frac{P I}{P J}=\\frac{Q K}{Q J}$. Likewise, we obtain $\\frac{R K}{R L}=\\frac{S I}{S L}$. These two equations together with (2) yield (1)."", ""We present another approach for showing this equation:\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=1 \\tag{1}\n$$\n\nLemma 2. Let $E F G H$ and $E^{\\prime} F^{\\prime} G^{\\prime} H^{\\prime}$ be circumscribed quadrilaterals such that $\\angle E+\\angle E^{\\prime}=$ $\\angle F+\\angle F^{\\prime}=\\angle G+\\angle G^{\\prime}=\\angle H+\\angle H^{\\prime}=180^{\\circ}$. Then\n\n$$\n\\frac{E F \\cdot G H}{F G \\cdot H E}=\\frac{E^{\\prime} F^{\\prime} \\cdot G^{\\prime} H^{\\prime}}{F^{\\prime} G^{\\prime} \\cdot H^{\\prime} E^{\\prime}}\n$$\n\nProof. Let $M$ and $M^{\\prime}$ be the incenters of $E F G H$ and $E^{\\prime} F^{\\prime} G^{\\prime} H^{\\prime}$, respectively. We use the notation $[X Y Z]$ for the area of a triangle $X Y Z$.\n\nTaking into account the relation $\\angle F M E+\\angle F^{\\prime} M^{\\prime} E^{\\prime}=180^{\\circ}$ together with the analogous ones, we get\n\n$$\n\\begin{aligned}\n\\frac{E F \\cdot G H}{F G \\cdot H E} & =\\frac{[M E F] \\cdot[M G H]}{[M F G] \\cdot[M H E]}=\\frac{M E \\cdot M F \\cdot \\sin \\angle F M E \\cdot M G \\cdot M H \\cdot \\sin \\angle H M G}{M F \\cdot M G \\cdot \\sin \\angle G M F \\cdot M H \\cdot M E \\cdot \\sin \\angle E M H} \\\\\n& =\\frac{M^{\\prime} E^{\\prime} \\cdot M^{\\prime} F^{\\prime} \\cdot \\sin \\angle F^{\\prime} M^{\\prime} E^{\\prime} \\cdot M^{\\prime} G^{\\prime} \\cdot M^{\\prime} H^{\\prime} \\cdot \\sin \\angle H^{\\prime} M^{\\prime} G^{\\prime}}{M^{\\prime} F^{\\prime} \\cdot M^{\\prime} G^{\\prime} \\cdot \\sin \\angle G^{\\prime} M^{\\prime} F^{\\prime} \\cdot M^{\\prime} H^{\\prime} \\cdot M^{\\prime} E^{\\prime} \\cdot \\sin \\angle E^{\\prime} M^{\\prime} H^{\\prime}}=\\frac{E^{\\prime} F^{\\prime} \\cdot G^{\\prime} H^{\\prime}}{F^{\\prime} G^{\\prime} \\cdot H^{\\prime} E^{\\prime}}\n\\end{aligned}\n$$\n\n\n\nFigure 6\n\nDenote by $h$ the homothety centered at $O$ that maps the incircle of $C R O Q$ to the incircle of $A P O S$. Let $Q^{\\prime}=h(Q), C^{\\prime}=h(C), R^{\\prime}=h(R), O^{\\prime}=O, S^{\\prime}=S, A^{\\prime}=A$, and $P^{\\prime}=P$. Furthermore, define $B^{\\prime}=A^{\\prime} P^{\\prime} \\cap C^{\\prime} Q^{\\prime}$ and $D^{\\prime}=A^{\\prime} S^{\\prime} \\cap C^{\\prime} R^{\\prime}$ as shown in Figure 6. Then\n\n$$\n\\frac{A P \\cdot O S}{P O \\cdot S A}=\\frac{A^{\\prime} P^{\\prime} \\cdot O^{\\prime} S^{\\prime}}{P^{\\prime} O^{\\prime} \\cdot S^{\\prime} A^{\\prime}}\n$$\n\nholds trivially. We also have\n\n$$\n\\frac{C R \\cdot O Q}{R O \\cdot Q C}=\\frac{C^{\\prime} R^{\\prime} \\cdot O^{\\prime} Q^{\\prime}}{R^{\\prime} O^{\\prime} \\cdot Q^{\\prime} C^{\\prime}}\n$$\n\nby the similarity of the quadrilaterals $C R O Q$ and $C^{\\prime} R^{\\prime} O^{\\prime} Q^{\\prime}$.\n\n\n\nNext, consider the circumscribed quadrilaterals $B Q O P$ and $B^{\\prime} Q^{\\prime} O^{\\prime} P^{\\prime}$ whose incenters lie on different sides of the quadrilaterals' shared side line $O P=O^{\\prime} P^{\\prime}$. Observe that $B Q \\| B^{\\prime} Q^{\\prime}$ and that $B^{\\prime}$ and $Q^{\\prime}$ lie on the lines $B P$ and $Q O$, respectively. It is now easy to see that the two quadrilaterals satisfy the hypotheses of Lemma 2 . Thus, we deduce\n\n$$\n\\frac{B Q \\cdot O P}{Q O \\cdot P B}=\\frac{B^{\\prime} Q^{\\prime} \\cdot O^{\\prime} P^{\\prime}}{Q^{\\prime} O^{\\prime} \\cdot P^{\\prime} B^{\\prime}}\n$$\n\nSimilarly, we get\n\n$$\n\\frac{D S \\cdot O R}{S O \\cdot R D}=\\frac{D^{\\prime} S^{\\prime} \\cdot O^{\\prime} R^{\\prime}}{S^{\\prime} O^{\\prime} \\cdot R^{\\prime} D^{\\prime}}\n$$\n\nMultiplying these four equations, we obtain\n\n$$\n\\frac{A P}{P B} \\cdot \\frac{B Q}{Q C} \\cdot \\frac{C R}{R D} \\cdot \\frac{D S}{S A}=\\frac{A^{\\prime} P^{\\prime}}{P^{\\prime} B^{\\prime}} \\cdot \\frac{B^{\\prime} Q^{\\prime}}{Q^{\\prime} C^{\\prime}} \\cdot \\frac{C^{\\prime} R^{\\prime}}{R^{\\prime} D^{\\prime}} \\cdot \n\\frac{D^{\\prime} S^{\\prime}}{S^{\\prime} A^{\\prime}}\n\\tag{3}\n$$\n\nFinally, we apply BRianchon's theorem to the circumscribed hexagon $A^{\\prime} P^{\\prime} R^{\\prime} C^{\\prime} Q^{\\prime} S^{\\prime}$ and deduce that the lines $A^{\\prime} C^{\\prime}, P^{\\prime} Q^{\\prime}$, and $R^{\\prime} S^{\\prime}$ are either concurrent or parallel to each other. So, by Menelaus' theorem, we obtain\n\n$$\n\\frac{A^{\\prime} P^{\\prime}}{P^{\\prime} B^{\\prime}} \\cdot \\frac{B^{\\prime} Q^{\\prime}}{Q^{\\prime} C^{\\prime}} \\cdot \\frac{C^{\\prime} R^{\\prime}}{R^{\\prime} D^{\\prime}} \\cdot \\frac{D^{\\prime} S^{\\prime}}{S^{\\prime} A^{\\prime}}=1\n$$\n\nThis equation together with (3) yield (1).""]",,True,,, 1957,Geometry,,"A triangulation of a convex polygon $\Pi$ is a partitioning of $\Pi$ into triangles by diagonals having no common points other than the vertices of the polygon. We say that a triangulation is a Thaiangulation if all triangles in it have the same area. Prove that any two different Thaiangulations of a convex polygon $\Pi$ differ by exactly two triangles. (In other words, prove that it is possible to replace one pair of triangles in the first Thaiangulation with a different pair of triangles so as to obtain the second Thaiangulation.)","['We denote by $[S]$ the area of a polygon $S$.\n\nRecall that each triangulation of a convex $n$-gon has exactly $n-2$ triangles. This means that all triangles in any two Thaiangulations of a convex polygon $\\Pi$ have the same area.\n\nLet $\\mathcal{T}$ be a triangulation of a convex polygon $\\Pi$. If four vertices $A, B, C$, and $D$ of $\\Pi$ form a parallelogram, and $\\mathcal{T}$ contains two triangles whose union is this parallelogram, then we say that $\\mathcal{T}$ contains parallelogram $A B C D$. Notice here that if two Thaiangulations $\\mathcal{T}_{1}$ and $\\mathcal{T}_{2}$ of $\\Pi$ differ by two triangles, then the union of these triangles is a quadrilateral each of whose diagonals bisects its area, i.e., a parallelogram.\n\nWe start with proving two properties of triangulations.\n\nLemma 1. A triangulation of a convex polygon $\\Pi$ cannot contain two parallelograms.\n\nProof. Arguing indirectly, assume that $P_{1}$ and $P_{2}$ are two parallelograms contained in some triangulation $\\mathcal{T}$. If they have a common triangle in $\\mathcal{T}$, then we may assume that $P_{1}$ consists of triangles $A B C$ and $A D C$ of $\\mathcal{T}$, while $P_{2}$ consists of triangles $A D C$ and $C D E$ (see Figure 1). But then $B C\\|A D\\| C E$, so the three vertices $B, C$, and $E$ of $\\Pi$ are collinear, which is absurd.\n\nAssume now that $P_{1}$ and $P_{2}$ contain no common triangle. Let $P_{1}=A B C D$. The sides $A B$, $B C, C D$, and $D A$ partition $\\Pi$ into several parts, and $P_{2}$ is contained in one of them; we may assume that this part is cut off from $P_{1}$ by $A D$. Then one may label the vertices of $P_{2}$ by $X$, $Y, Z$, and $T$ so that the polygon $A B C D X Y Z T$ is convex (see Figure 2; it may happen that $D=X$ and/or $T=A$, but still this polygon has at least six vertices). But the sum of the external angles of this polygon at $B, C, Y$, and $Z$ is already $360^{\\circ}$, which is impossible. A final contradiction.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nLemma 2. Every triangle in a Thaiangulation $\\mathcal{T}$ of $\\Pi$ contains a side of $\\Pi$.\n\nProof. Let $A B C$ be a triangle in $\\mathcal{T}$. Apply an affine transform such that $A B C$ maps to an equilateral triangle; let $A^{\\prime} B^{\\prime} C^{\\prime}$ be the image of this triangle, and $\\Pi^{\\prime}$ be the image of $\\Pi$. Clearly, $\\mathcal{T}$ maps into a Thaiangulation $\\mathcal{T}^{\\prime}$ of $\\Pi^{\\prime}$.\n\nAssume that none of the sides of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ is a side of $\\Pi^{\\prime}$. Then $\\mathcal{T}^{\\prime}$ contains some other triangles with these sides, say, $A^{\\prime} B^{\\prime} Z, C^{\\prime} A^{\\prime} Y$, and $B^{\\prime} C^{\\prime} X$; notice that $A^{\\prime} Z B^{\\prime} X C^{\\prime} Y$ is a convex hexagon (see Figure 3). The sum of its external angles at $X, Y$, and $Z$ is less than $360^{\\circ}$. So one of these angles (say, at $Z$ ) is less than $120^{\\circ}$, hence $\\angle A^{\\prime} Z B^{\\prime}>60^{\\circ}$. Then $Z$ lies on a circular arc subtended by $A^{\\prime} B^{\\prime}$ and having angular measure less than $240^{\\circ}$; consequently, the altitude $Z H$ of $\\triangle A^{\\prime} B^{\\prime} Z$ is less than $\\sqrt{3} A^{\\prime} B^{\\prime} / 2$. Thus $\\left[A^{\\prime} B^{\\prime} Z\\right]<\\left[A^{\\prime} B^{\\prime} C^{\\prime}\\right]$, and $\\mathcal{T}^{\\prime}$ is not a Thaiangulation. A contradiction.\n\n\n\nNow we pass to the solution. We say that a triangle in a triangulation of $\\Pi$ is an ear if it contains two sides of $\\Pi$. Note that each triangulation of a polygon contains some ear.\n\nArguing indirectly, we choose a convex polygon $\\Pi$ with the least possible number of sides such that some two Thaiangulations $\\mathcal{T}_{1}$ and $\\mathcal{T}_{2}$ of $\\Pi$ violate the problem statement (thus $\\Pi$ has at least five sides). Consider now any ear $A B C$ in $\\mathcal{T}_{1}$, with $A C$ being a diagonal of $\\Pi$. If $\\mathcal{T}_{2}$ also contains $\\triangle A B C$, then one may cut $\\triangle A B C$ off from $\\Pi$, getting a polygon with a smaller number of sides which also violates the problem statement. This is impossible; thus $\\mathcal{T}_{2}$ does not contain $\\triangle A B C$.\n\nNext, $\\mathcal{T}_{1}$ contains also another triangle with side $A C$, say $\\triangle A C D$. By Lemma 2, this triangle contains a side of $\\Pi$, so $D$ is adjacent to either $A$ or $C$ on the boundary of $\\Pi$. We may assume that $D$ is adjacent to $C$.\n\nAssume that $\\mathcal{T}_{2}$ does not contain the triangle $B C D$. Then it contains two different triangles $B C X$ and $C D Y$ (possibly, with $X=Y$ ); since these triangles have no common interior points, the polygon $A B C D Y X$ is convex (see Figure 4). But, since $[A B C]=[B C X]=$ $[A C D]=[C D Y]$, we get $A X \\| B C$ and $A Y \\| C D$ which is impossible. Thus $\\mathcal{T}_{2}$ contains $\\triangle B C D$.\n\nTherefore, $[A B D]=[A B C]+[A C D]-[B C D]=[A B C]$, and $A B C D$ is a parallelogram contained in $\\mathcal{T}_{1}$. Let $\\mathcal{T}^{\\prime}$ be the Thaiangulation of $\\Pi$ obtained from $\\mathcal{T}_{1}$ by replacing the diagonal $A C$ with $B D$; then $\\mathcal{T}^{\\prime}$ is distinct from $\\mathcal{T}_{2}$ (otherwise $\\mathcal{T}_{1}$ and $\\mathcal{T}_{2}$ would differ by two triangles). Moreover, $\\mathcal{T}^{\\prime}$ shares a common ear $B C D$ with $\\mathcal{T}_{2}$. As above, cutting this ear away we obtain that $\\mathcal{T}_{2}$ and $\\mathcal{T}^{\\prime}$ differ by two triangles forming a parallelogram different from $A B C D$. Thus $\\mathcal{T}^{\\prime}$ contains two parallelograms, which contradicts Lemma 1.\n\n\n\nFigure 4\n\n\n\nFigure 5']",,True,,, 1958,Number Theory,,"Determine all positive integers $M$ for which the sequence $a_{0}, a_{1}, a_{2}, \ldots$, defined by $a_{0}=\frac{2 M+1}{2}$ and $a_{k+1}=a_{k}\left\lfloor a_{k}\right\rfloor$ for $k=0,1,2, \ldots$, contains at least one integer term.","['Define $b_{k}=2 a_{k}$ for all $k \\geqslant 0$. Then\n\n$$\nb_{k+1}=2 a_{k+1}=2 a_{k}\\left\\lfloor a_{k}\\right\\rfloor=b_{k}\\left\\lfloor\\frac{b_{k}}{2}\\right\\rfloor\n$$\n\nSince $b_{0}$ is an integer, it follows that $b_{k}$ is an integer for all $k \\geqslant 0$.\n\nSuppose that the sequence $a_{0}, a_{1}, a_{2}, \\ldots$ does not contain any integer term. Then $b_{k}$ must be an odd integer for all $k \\geqslant 0$, so that\n\n$$\nb_{k+1}=b_{k}\\left\\lfloor\\frac{b_{k}}{2}\\right\\rfloor=\\frac{b_{k}\\left(b_{k}-1\\right)}{2}\\tag{1}\n$$\n\nHence\n\n$$\nb_{k+1}-3=\\frac{b_{k}\\left(b_{k}-1\\right)}{2}-3=\\frac{\\left(b_{k}-3\\right)\\left(b_{k}+2\\right)}{2}\\tag{2}\n$$\n\nfor all $k \\geqslant 0$.\n\nSuppose that $b_{0}-3>0$. Then equation (2) yields $b_{k}-3>0$ for all $k \\geqslant 0$. For each $k \\geqslant 0$, define $c_{k}$ to be the highest power of 2 that divides $b_{k}-3$. Since $b_{k}-3$ is even for all $k \\geqslant 0$, the number $c_{k}$ is positive for every $k \\geqslant 0$.\n\nNote that $b_{k}+2$ is an odd integer. Therefore, from equation (2), we have that $c_{k+1}=c_{k}-1$. Thus, the sequence $c_{0}, c_{1}, c_{2}, \\ldots$ of positive integers is strictly decreasing, a contradiction. So, $b_{0}-3 \\leqslant 0$, which implies $M=1$.\n\nFor $M=1$, we can check that the sequence is constant with $a_{k}=\\frac{3}{2}$ for all $k \\geqslant 0$. Therefore, the answer is $M \\geqslant 2$.', 'We provide an alternative way to show $M=1$ once equation (1) has been reached. We claim that $b_{k} \\equiv 3\\left(\\bmod 2^{m}\\right)$ for all $k \\geqslant 0$ and $m \\geqslant 1$. If this is true, then we would have $b_{k}=3$ for all $k \\geqslant 0$ and hence $M=1$.\n\nTo establish our claim, we proceed by induction on $m$. The base case $b_{k} \\equiv 3(\\bmod 2)$ is true for all $k \\geqslant 0$ since $b_{k}$ is odd. Now suppose that $b_{k} \\equiv 3\\left(\\bmod 2^{m}\\right)$ for all $k \\geqslant 0$. Hence $b_{k}=2^{m} d_{k}+3$ for some integer $d_{k}$. We have\n\n$$\n3 \\equiv b_{k+1} \\equiv\\left(2^{m} d_{k}+3\\right)\\left(2^{m-1} d_{k}+1\\right) \\equiv 3 \\cdot 2^{m-1} d_{k}+3 \\quad\\left(\\bmod 2^{m}\\right)\n$$\n\nso that $d_{k}$ must be even. This implies that $b_{k} \\equiv 3\\left(\\bmod 2^{m+1}\\right)$, as required.']",['All integers $M \\geqslant 2$'],False,,Need_human_evaluate, 1959,Number Theory,,Let $a$ and $b$ be positive integers such that $a ! b !$ is a multiple of $a !+b !$. Prove that $3 a \geqslant 2 b+2$.,"['If $a>b$, we immediately get $3 a \\geqslant 2 b+2$. In the case $a=b$, the required inequality is equivalent to $a \\geqslant 2$, which can be checked easily since $(a, b)=(1,1)$ does not satisfy $a !+b ! \\mid a ! b !$. We now assume $aa !$, which is impossible. We observe that $c ! \\mid M$ since $M$ is a product of $c$ consecutive integers. Thus $\\operatorname{gcd}(1+M, c !)=1$, which implies\n\n$$\n1+M \\mid \\frac{a !}{c !}=(c+1)(c+2) \\cdots a\n\\tag{1}\n$$\n\nIf $a \\leqslant 2 c$, then $\\frac{a !}{c !}$ is a product of $a-c \\leqslant c$ integers not exceeding $a$ whereas $M$ is a product of $c$ integers exceeding $a$. Therefore, $1+M>\\frac{a !}{c !}$, which is a contradiction.\n\nIt remains to exclude the case $a=2 c+1$. Since $a+1=2(c+1)$, we have $c+1 \\mid M$. Hence, we can deduce from (1) that $1+M \\mid(c+2)(c+3) \\cdots a$. Now $(c+2)(c+3) \\cdots a$ is a product of $a-c-1=c$ integers not exceeding $a$; thus it is smaller than $1+M$. Again, we arrive at a contradiction.', 'we may assume that $aa+c$; otherwise, $a+1 \\leqslant 2 c+2 \\leqslant 2 p \\leqslant a+c$ so $p \\mid N-1$, again impossible. Thus, we have $p \\in\\left(\\frac{a+c}{2}, a\\right]$, and $p^{2} \\nmid(a+c)$ ! since $2 p>a+c$. Therefore, $p^{2} \\nmid N$ as well.\n\nIf $a \\leqslant c+2$, then the interval $\\left(\\frac{a+c}{2}, a\\right]$ contains at most one integer and hence at most one prime number, which has to be $a$. Since $p^{2} \\nmid N$, we must have $N=p=a$ or $N=1$, which is absurd since $N>a \\geqslant 1$. Thus, we have $a \\geqslant c+3$, and so $\\frac{a+c+1}{2} \\geqslant c+2$. It follows that $p$ lies in the interval $[c+2, a]$.\n\nThus, every prime appearing in the prime factorization of $N$ lies in the interval $[c+2, a]$, and its exponent is exactly 1. So we must have $N \\mid(c+2)(c+3) \\cdots a$. However, $(c+2)(c+3) \\cdots a$ is a product of $a-c-1 \\leqslant c$ numbers not exceeding $a$, so it is less than $N$. This is a contradiction.']",,True,,, 1960,Number Theory,,"Let $m$ and $n$ be positive integers such that $m>n$. Define $x_{k}=(m+k) /(n+k)$ for $k=$ $1,2, \ldots, n+1$. Prove that if all the numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers, then $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime.","['Assume that $x_{1}, x_{2}, \\ldots, x_{n+1}$ are integers. Define the integers\n\n$$\na_{k}=x_{k}-1=\\frac{m+k}{n+k}-1=\\frac{m-n}{n+k}>0\n$$\n\nfor $k=1,2, \\ldots, n+1$.\n\nLet $P=x_{1} x_{2} \\cdots x_{n+1}-1$. We need to prove that $P$ is divisible by an odd prime, or in other words, that $P$ is not a power of 2 . To this end, we investigate the powers of 2 dividing the numbers $a_{k}$.\n\nLet $2^{d}$ be the largest power of 2 dividing $m-n$, and let $2^{c}$ be the largest power of 2 not exceeding $2 n+1$. Then $2 n+1 \\leqslant 2^{c+1}-1$, and so $n+1 \\leqslant 2^{c}$. We conclude that $2^{c}$ is one of the numbers $n+1, n+2, \\ldots, 2 n+1$, and that it is the only multiple of $2^{c}$ appearing among these numbers. Let $\\ell$ be such that $n+\\ell=2^{c}$. Since $\\frac{m-n}{n+\\ell}$ is an integer, we have $d \\geqslant c$. Therefore, $2^{d-c+1} \\nmid a_{\\ell}=\\frac{m-n}{n+\\ell}$, while $2^{d-c+1} \\mid a_{k}$ for all $k \\in\\{1, \\ldots, n+1\\} \\backslash\\{\\ell\\}$.\n\nComputing modulo $2^{d-c+1}$, we get\n\n$$\nP=\\left(a_{1}+1\\right)\\left(a_{2}+1\\right) \\cdots\\left(a_{n+1}+1\\right)-1 \\equiv\\left(a_{\\ell}+1\\right) \\cdot 1^{n}-1 \\equiv a_{\\ell} \\not \\equiv 0 \\quad\\left(\\bmod 2^{d-c+1}\\right) .\n$$\n\nTherefore, $2^{d-c+1} \\nmid P$.\n\nOn the other hand, for any $k \\in\\{1, \\ldots, n+1\\} \\backslash\\{\\ell\\}$, we have $2^{d-c+1} \\mid a_{k}$. So $P \\geqslant a_{k} \\geqslant 2^{d-c+1}$, and it follows that $P$ is not a power of 2 .']",,True,,, 1961,Number Theory,,"Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\left(a_{n}\right)$ is eventually periodic; in other words, there exist integers $N \geqslant 0$ and $t>0$ such that $a_{n+t}=a_{n}$ for all $n \geqslant N$.","['Let $s_{n}=a_{n}+b_{n}$. Notice that if $a_{n} \\mid b_{n}$, then $a_{n+1}=a_{n}+1, b_{n+1}=b_{n}-1$ and $s_{n+1}=s_{n}$. So, $a_{n}$ increases by 1 and $s_{n}$ does not change until the first index is reached with $a_{n} \\nmid s_{n}$. Define\n\n$$\nW_{n}=\\left\\{m \\in \\mathbb{Z}_{>0}: m \\geqslant a_{n} \\text { and } m \\nmid s_{n}\\right\\} \\quad \\text { and } \\quad w_{n}=\\min W_{n}\n$$\n\nClaim 1. The sequence $\\left(w_{n}\\right)$ is non-increasing.\n\nProof. If $a_{n} \\mid b_{n}$ then $a_{n+1}=a_{n}+1$. Due to $a_{n} \\mid s_{n}$, we have $a_{n} \\notin W_{n}$. Moreover $s_{n+1}=s_{n}$; therefore, $W_{n+1}=W_{n}$ and $w_{n+1}=w_{n}$.\n\nOtherwise, if $a_{n} \\nmid b_{n}$, then $a_{n} \\nmid s_{n}$, so $a_{n} \\in W_{n}$ and thus $w_{n}=a_{n}$. We show that $a_{n} \\in W_{n+1}$; this implies $w_{n+1} \\leqslant a_{n}=w_{n}$. By the definition of $W_{n+1}$, we need that $a_{n} \\geqslant a_{n+1}$ and $a_{n} \\nmid s_{n+1}$. The first relation holds because of $\\operatorname{gcd}\\left(a_{n}, b_{n}\\right)\\beta>\\gamma\\tag{5}\n$$\n\nDepending on how large $a$ is, we divide the argument into two further cases.\n\nCase 2.1. $\\quad a=2$.\n\nWe first prove that $\\gamma=0$. Assume for the sake of contradiction that $\\gamma>0$. Then $c$ is even by (4) and, similarly, $b$ is even by (5) and (3). So the left-hand side of (2) is congruent to 2 modulo 4 , which is only possible if $b c=4$. As this contradicts (1), we have thereby shown that $\\gamma=0$, i.e., that $c=2 b-1$.\n\nNow (3) yields $3 b-2=2^{\\beta}$. Due to $b>2$ this is only possible if $\\beta \\geqslant 4$. If $\\beta=4$, then we get $b=6$ and $c=2 \\cdot 6-1=11$, which is a solution. It remains to deal with the case $\\beta \\geqslant 5$. Now (2) implies\n\n$$\n9 \\cdot 2^{\\alpha}=9 b(2 b-1)-18=(3 b-2)(6 b+1)-16=2^{\\beta}\\left(2^{\\beta+1}+5\\right)-16,\n$$\n\nand by $\\beta \\geqslant 5$ the right-hand side is not divisible by 32 . Thus $\\alpha \\leqslant 4$ and we get a contradiction to (5).\n\n\n\nCase 2.2. $a \\geqslant 3$.\n\nPick an integer $\\vartheta \\in\\{-1,+1\\}$ such that $c-\\vartheta$ is not divisible by 4 . Now\n\n$$\n2^{\\alpha}+\\vartheta \\cdot 2^{\\beta}=\\left(b c-a \\vartheta^{2}\\right)+\\vartheta(c a-b)=(b+a \\vartheta)(c-\\vartheta)\n$$\n\nis divisible by $2^{\\beta}$ and, consequently, $b+a \\vartheta$ is divisible by $2^{\\beta-1}$. On the other hand, $2^{\\beta}=a c-b>$ $(a-1) c \\geqslant 2 c$ implies in view of (1) that $a$ and $b$ are smaller than $2^{\\beta-1}$. All this is only possible if $\\vartheta=1$ and $a+b=2^{\\beta-1}$. Now (3) yields\n\n$$\na c-b=2(a+b),\n\\tag{6}\n$$\n\nwhence $4 b>a+3 b=a(c-1) \\geqslant a b$, which in turn yields $a=3$.\n\nSo (6) simplifies to $c=b+2$ and (2) tells us that $b(b+2)-3=(b-1)(b+3)$ is a power of 2. Consequently, the factors $b-1$ and $b+3$ are powers of 2 themselves. Since their difference is 4 , this is only possible if $b=5$ and thus $c=7$. Thereby the solution is complete.', 'As in the beginning of the first solution, we observe that $a, b, c \\geqslant 2$. Depending on the parities of $a, b$, and $c$ we distinguish three cases.\n\nCase 1. The numbers $a, b$, and $c$ are even.\n\nLet $2^{A}, 2^{B}$, and $2^{C}$ be the largest powers of 2 dividing $a, b$, and $c$ respectively. We may assume without loss of generality that $1 \\leqslant A \\leqslant B \\leqslant C$. Now $2^{B}$ is the highest power of 2 dividing $a c-b$, whence $a c-b=2^{B} \\leqslant b$. Similarly, we deduce $b c-a=2^{A} \\leqslant a$. Adding both estimates we get $(a+b) c \\leqslant 2(a+b)$, whence $c \\leqslant 2$. So $c=2$ and thus $A=B=C=1$; moreover, we must have had equality throughout, i.e., $a=2^{A}=2$ and $b=2^{B}=2$. We have thereby found the solution $(a, b, c)=(2,2,2)$.\n\nCase 2. The numbers $a, b$, and $c$ are odd.\n\nIf any two of these numbers are equal, say $a=b$, then $a c-b=a(c-1)$ has a nontrivial odd divisor and cannot be a power of 2 . Hence $a, b$, and $c$ are distinct. So we may assume without loss of generality that $a\\beta$, and thus $2^{\\beta}$ divides\n\n$$\na \\cdot 2^{\\alpha}-b \\cdot 2^{\\beta}=a(b c-a)-b(a c-b)=b^{2}-a^{2}=(b+a)(b-a) .\n$$\n\nSince $a$ is odd, it is not possible that both factors $b+a$ and $b-a$ are divisible by 4 . Consequently, one of them has to be a multiple of $2^{\\beta-1}$. Hence one of the numbers $2(b+a)$ and $2(b-a)$ is divisible by $2^{\\beta}$ and in either case we have\n\n$$\na c-b=2^{\\beta} \\leqslant 2(a+b) .\n\\tag{7}\n$$\n\nThis in turn yields $(a-1) b\\beta$ denote the integers satisfying\n\n$$\n2^{\\alpha}=b c-a \\quad \\text { and } \\quad 2^{\\beta}=a c-b\n\\tag{9}\n$$\n\nIf $\\beta=0$ it would follow that $a c-b=a b-c=1$ and hence that $b=c=1$, which is absurd. So $\\beta$ and $\\alpha$ are positive and consequently $a$ and $b$ are even. Substituting $c=a b-1$ into (9) we obtain\n\n$$\n2^{\\alpha} =a b^{2}-(a+b), \\tag{10}\n$$\n$$\n\\text { and } \\quad 2^{\\beta} =a^{2} b-(a+b) .\n\\tag{11}\n$$\n\nThe addition of both equation yields $2^{\\alpha}+2^{\\beta}=(a b-2)(a+b)$. Now $a b-2$ is even but not divisible by 4 , so the highest power of 2 dividing $a+b$ is $2^{\\beta-1}$. For this reason, the equations (10) and (11) show that the highest powers of 2 dividing either of the numbers $a b^{2}$ and $a^{2} b$ is likewise $2^{\\beta-1}$. Thus there is an integer $\\tau \\geqslant 1$ together with odd integers $A, B$, and $C$ such that $a=2^{\\tau} A, b=2^{\\tau} B, a+b=2^{3 \\tau} C$, and $\\beta=1+3 \\tau$.\n\nNotice that $A+B=2^{2 \\tau} C \\geqslant 4 C$. Moreover, (11) entails $A^{2} B-C=2$. Thus $8=$ $4 A^{2} B-4 C \\geqslant 4 A^{2} B-A-B \\geqslant A^{2}(3 B-1)$. Since $A$ and $B$ are odd with $A0}$ denote the set of positive integers. Consider a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$. For any $m, n \in \mathbb{Z}_{>0}$ we write $f^{n}(m)=\underbrace{f(f(\ldots f}_{n}(m) \ldots))$. Suppose that $f$ has the following two properties: (i) If $m, n \in \mathbb{Z}_{>0}$, then $\frac{f^{n}(m)-m}{n} \in \mathbb{Z}_{>0}$; (ii) The set $\mathbb{Z}_{>0} \backslash\left\{f(n) \mid n \in \mathbb{Z}_{>0}\right\}$ is finite. Prove that the sequence $f(1)-1, f(2)-2, f(3)-3, \ldots$ is periodic.","['We split the solution into three steps. In the first of them, we show that the function $f$ is injective and explain how this leads to a useful visualization of $f$. Then comes the second step, in which most of the work happens: its goal is to show that for any $n \\in \\mathbb{Z}_{>0}$ the sequence $n, f(n), f^{2}(n), \\ldots$ is an arithmetic progression. Finally, in the third step we put everything together, thus solving the problem.\n\nStep 1. We commence by checking that $f$ is injective. For this purpose, we consider any $m, k \\in \\mathbb{Z}_{>0}$ with $f(m)=f(k)$. By $(i)$, every positive integer $n$ has the property that\n\n$$\n\\frac{k-m}{n}=\\frac{f^{n}(m)-m}{n}-\\frac{f^{n}(k)-k}{n}\n$$\n\nis a difference of two integers and thus integral as well. But for $n=|k-m|+1$ this is only possible if $k=m$. Thereby, the injectivity of $f$ is established.\n\nNow recall that due to condition $(i i)$ there are finitely many positive integers $a_{1}, \\ldots, a_{k}$ such that $\\mathbb{Z}_{>0}$ is the disjoint union of $\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$ and $\\left\\{f(n) \\mid n \\in \\mathbb{Z}_{>0}\\right\\}$. Notice that by plugging $n=1$ into condition $(i)$ we get $f(m)>m$ for all $m \\in \\mathbb{Z}_{>0}$.\n\nWe contend that every positive integer $n$ may be expressed uniquely in the form $n=f^{j}\\left(a_{i}\\right)$ for some $j \\geqslant 0$ and $i \\in\\{1, \\ldots, k\\}$. The uniqueness follows from the injectivity of $f$. The existence can be proved by induction on $n$ in the following way. If $n \\in\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$, then we may take $j=0$; otherwise there is some $n^{\\prime}0$; and $T=1$ and $A=0$ if $t=0$. For every integer $n \\geqslant A$, the interval $\\Delta_{n}=[n+1, n+T]$ contains exactly $T / T_{i}$\n\n\n\nelements of the $i^{\\text {th }}$ row $(1 \\leqslant i \\leqslant t)$. Therefore, the number of elements from the last $(k-t)$ rows of the Table contained in $\\Delta_{n}$ does not depend on $n \\geqslant A$. It is not possible that none of these intervals $\\Delta_{n}$ contains an element from the $k-t$ last rows, because infinitely many numbers appear in these rows. It follows that for each $n \\geqslant A$ the interval $\\Delta_{n}$ contains at least one member from these rows.\n\nThis yields that for every positive integer $d$, the interval $[A+1, A+(d+1)(k-t) T]$ contains at least $(d+1)(k-t)$ elements from the last $k-t$ rows; therefore, there exists an index $x$ with $t+1 \\leqslant x \\leqslant k$, possibly depending on $d$, such that our interval contains at least $d+1$ elements from the $x^{\\text {th }}$ row. In this situation we have\n\n$$\nf^{d}\\left(a_{x}\\right) \\leqslant A+(d+1)(k-t) T .\n$$\n\nFinally, since there are finitely many possibilities for $x$, there exists an index $x \\geqslant t+1$ such that the set\n\n$$\nX=\\left\\{d \\in \\mathbb{Z}_{>0} \\mid f^{d}\\left(a_{x}\\right) \\leqslant A+(d+1)(k-t) T\\right\\}\n$$\n\nis infinite. Thereby we have found the ""dense row"" promised above.\n\nBy assumption $(i)$, for every $d \\in X$ the number\n\n$$\n\\beta_{d}=\\frac{f^{d}\\left(a_{x}\\right)-a_{x}}{d}\n$$\n\nis a positive integer not exceeding\n\n$$\n\\frac{A+(d+1)(k-t) T}{d} \\leqslant \\frac{A d+2 d(k-t) T}{d}=A+2(k-t) T\n$$\n\nThis leaves us with finitely many choices for $\\beta_{d}$, which means that there exists a number $T_{x}$ such that the set\n\n$$\nY=\\left\\{d \\in X \\mid \\beta_{d}=T_{x}\\right\\}\n$$\n\nis infinite. Notice that we have $f^{d}\\left(a_{x}\\right)=a_{x}+d \\cdot T_{x}$ for all $d \\in Y$.\n\nNow we are prepared to prove that the numbers in the $x^{\\text {th }}$ row form an arithmetic progression, thus coming to a contradiction with our assumption. Let us fix any positive integer $j$. Since the set $Y$ is infinite, we can choose a number $y \\in Y$ such that $y-j>\\left|f^{j}\\left(a_{x}\\right)-\\left(a_{x}+j T_{x}\\right)\\right|$. Notice that both numbers\n\n$$\nf^{y}\\left(a_{x}\\right)-f^{j}\\left(a_{x}\\right)=f^{y-j}\\left(f^{j}\\left(a_{x}\\right)\\right)-f^{j}\\left(a_{x}\\right) \\quad \\text { and } \\quad f^{y}\\left(a_{x}\\right)-\\left(a_{x}+j T_{x}\\right)=(y-j) T_{x}\n$$\n\nare divisible by $y-j$. Thus, the difference between these numbers is also divisible by $y-j$. Since the absolute value of this difference is less than $y-j$, it has to vanish, so we get $f^{j}\\left(a_{x}\\right)=$ $a_{x}+j \\cdot T_{x}$.\n\nHence, it is indeed true that all rows of the Table are arithmetic progressions.\n\nStep 3. Keeping the above notation in force, we denote the step of the $i^{\\text {th }}$ row of the table by $T_{i}$.\n\n\n\n$$\nT=\\operatorname{lcm}\\left(T_{1}, \\ldots, T_{k}\\right)\n$$\n\nTo see this, let any $n \\in \\mathbb{Z}_{>0}$ be given and denote the index of the row in which it appears in the Table by $i$. Then we have $f^{j}(n)=n+j \\cdot T_{i}$ for all $j \\in \\mathbb{Z}_{>0}$, and thus indeed\n\n$$\nf(n+T)-f(n)=f^{1+T / T_{i}}(n)-f(n)=\\left(n+T+T_{i}\\right)-\\left(n+T_{i}\\right)=T .\n$$\n\nThis concludes the solution.']",,True,,, 1964,Number Theory,,"Let $\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ is called $k$-good if $\operatorname{gcd}(f(m)+n, f(n)+m) \leqslant k$ for all $m \neq n$. Find all $k$ such that there exists a $k$-good function.","['For any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$, let $G_{f}(m, n)=\\operatorname{gcd}(f(m)+n, f(n)+m)$. Note that a $k$-good function is also $(k+1)$-good for any positive integer $k$. Hence, it suffices to show that there does not exist a 1-good function and that there exists a 2-good function.\n\nWe first show that there is no 1-good function. Suppose that there exists a function $f$ such that $G_{f}(m, n)=1$ for all $m \\neq n$. Now, if there are two distinct even numbers $m$ and $n$ such that $f(m)$ and $f(n)$ are both even, then $2 \\mid G_{f}(m, n)$, a contradiction. A similar argument holds if there are two distinct odd numbers $m$ and $n$ such that $f(m)$ and $f(n)$ are both odd. Hence we can choose an even $m$ and an odd $n$ such that $f(m)$ is odd and $f(n)$ is even. This also implies that $2 \\mid G_{f}(m, n)$, a contradiction.\n\nWe now construct a 2 -good function. Define $f(n)=2^{g(n)+1}-n-1$, where $g$ is defined recursively by $g(1)=1$ and $g(n+1)=\\left(2^{g(n)+1}\\right) !$.\n\nFor any positive integers $m>n$, set\n\n$$\nA=f(m)+n=2^{g(m)+1}-m+n-1, \\quad B=f(n)+m=2^{g(n)+1}-n+m-1 .\n$$\n\nWe need to show that $\\operatorname{gcd}(A, B) \\leqslant 2$. First, note that $A+B=2^{g(m)+1}+2^{g(n)+1}-2$ is not divisible by 4 , so that $4 \\nmid \\operatorname{gcd}(A, B)$. Now we suppose that there is an odd prime $p$ for which $p \\mid \\operatorname{gcd}(A, B)$ and derive a contradiction.\n\nWe first claim that $2^{g(m-1)+1} \\geqslant B$. This is a rather weak bound; one way to prove it is as follows. Observe that $g(k+1)>g(k)$ and hence $2^{g(k+1)+1} \\geqslant 2^{g(k)+1}+1$ for every positive integer $k$. By repeatedly applying this inequality, we obtain $2^{g(m-1)+1} \\geqslant 2^{g(n)+1}+(m-1)-n=B$.\n\nNow, since $p \\mid B$, we have $p-110^{100}} \alpha_{i}\tag{1} $$ That is, $\mho(n)$ is the number of prime factors of $n$ greater than $10^{100}$, counted with multiplicity. Find all strictly increasing functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ \mho(f(a)-f(b)) \leqslant \mho(a-b) \quad \text { for all integers } a \text { and } b \text { with } a>b \text {. } $$","['A straightforward check shows that all the functions listed in the answer satisfy the problem condition. It remains to show the converse.\n\nAssume that $f$ is a function satisfying the problem condition. Notice that the function $g(x)=f(x)-f(0)$ also satisfies this condition. Replacing $f$ by $g$, we assume from now on that $f(0)=0$; then $f(n)>0$ for any positive integer $n$. Thus, we aim to prove that there exists a positive integer $a$ with $\\mho(a)=0$ such that $f(n)=a n$ for all $n \\in \\mathbb{Z}$.\n\nWe start by introducing some notation. Set $N=10^{100}$. We say that a prime $p$ is large if $p>N$, and $p$ is small otherwise; let $\\mathcal{S}$ be the set of all small primes. Next, we say that a positive integer is large or small if all its prime factors are such (thus, the number 1 is the unique number which is both large and small). For a positive integer $k$, we denote the greatest large divisor of $k$ and the greatest small divisor of $k$ by $L(k)$ and $S(k)$, respectively; thus, $k=L(k) S(k)$.\n\nWe split the proof into three steps.\n\nStep 1. We prove that for every large $k$, we have $k|f(a)-f(b) \\Longleftrightarrow k| a-b$. In other $\\overline{\\text { words, }} L(f(a)-f(b))=L(a-b)$ for all integers $a$ and $b$ with $a>b$.\n\nWe use induction on $k$. The base case $k=1$ is trivial. For the induction step, assume that $k_{0}$ is a large number, and that the statement holds for all large numbers $k$ with $k\\ell$. But then\n\n$$\n\\mho(f(x)-f(y)) \\geqslant \\mho\\left(\\operatorname{lcm}\\left(k_{0}, \\ell\\right)\\right)>\\mho(\\ell)=\\mho(x-y),\n$$\n\nwhich is impossible.\n\nNow we complete the induction step. By Claim 1, for every integer $a$ each of the sequences\n\n$$\nf(a), f(a+1), \\ldots, f\\left(a+k_{0}-1\\right) \\quad \\text { and } \\quad f(a+1), f(a+2), \\ldots, f\\left(a+k_{0}\\right)\n$$\n\nforms a complete residue system modulo $k_{0}$. This yields $f(a) \\equiv f\\left(a+k_{0}\\right)\\left(\\bmod k_{0}\\right)$. Thus, $f(a) \\equiv f(b)\\left(\\bmod k_{0}\\right)$ whenever $a \\equiv b\\left(\\bmod k_{0}\\right)$.\n\nFinally, if $a \\not \\equiv b\\left(\\bmod k_{0}\\right)$ then there exists an integer $b^{\\prime}$ such that $b^{\\prime} \\equiv b\\left(\\bmod k_{0}\\right)$ and $\\left|a-b^{\\prime}\\right|N$.\n\nProof. Let $d$ be the product of all small primes, and let $\\alpha$ be a positive integer such that $2^{\\alpha}>f(N)$. Then, for every $p \\in \\mathcal{S}$ the numbers $f(0), f(1), \\ldots, f(N)$ are distinct modulo $p^{\\alpha}$. Set $P=d^{\\alpha}$ and $c=P+f(N)$.\n\nChoose any integer $t>N$. Due to the choice of $\\alpha$, for every $p \\in \\mathcal{S}$ there exists at most one nonnegative integer $i \\leqslant N$ with $p^{\\alpha} \\mid f(t)-f(i)$. Since $|\\mathcal{S}||f(x)-a x|$. Since $n-x \\equiv r-(r+1)=-1(\\bmod N !)$, the number $|n-x|$ is large. Therefore, by Step 1 we have $f(x) \\equiv f(n)=a n \\equiv a x(\\bmod n-x)$, so $n-x \\mid f(x)-a x$. Due to the choice of $n$, this yields $f(x)=a x$.\n\nTo complete Step 3, notice that the set $\\mathcal{T}^{\\prime}$ found in Step 2 contains infinitely many elements of some residue class $R_{i}$. Applying Claim 3, we successively obtain that $f(x)=a x$ for all $x \\in R_{i+1}, R_{i+2}, \\ldots, R_{i+N !}=R_{i}$. This finishes the solution.']","['$f(x)=a x+b$, where $b$ is an arbitrary integer, and $a$ is an arbitrary positive integer with $\\mho(a)=0$']",True,,Expression, 1966,Algebra,,"Let $z_{0}0$.\n\nNotice that\n\n$$\nn z_{n+1}-\\left(z_{0}+z_{1}+\\cdots+z_{n}\\right)=(n+1) z_{n+1}-\\left(z_{0}+z_{1}+\\cdots+z_{n}+z_{n+1}\\right)=-d_{n+1}\n$$\n\nso the second inequality in (1) is equivalent to $d_{n+1} \\leqslant 0$. Therefore, we have to prove that there is a unique index $n \\geqslant 1$ that satisfies $d_{n}>0 \\geqslant d_{n+1}$.\n\nBy its definition the sequence $d_{1}, d_{2}, \\ldots$ consists of integers and we have\n\n$$\nd_{1}=\\left(z_{0}+z_{1}\\right)-1 \\cdot z_{1}=z_{0}>0\n$$\n\nFrom\n\n$d_{n+1}-d_{n}=\\left(\\left(z_{0}+\\cdots+z_{n}+z_{n+1}\\right)-(n+1) z_{n+1}\\right)-\\left(\\left(z_{0}+\\cdots+z_{n}\\right)-n z_{n}\\right)=n\\left(z_{n}-z_{n+1}\\right)<0$\n\nwe can see that $d_{n+1}d_{2}>\\ldots$ of integers such that its first element $d_{1}$ is positive. The sequence must drop below 0 at some point, and thus there is a unique index $n$, that is the index of the last positive term, satisfying $d_{n}>0 \\geqslant d_{n+1}$.']",,True,,, 1967,Algebra,,"Define the function $f:(0,1) \rightarrow(0,1)$ by $$ f(x)= \begin{cases}x+\frac{1}{2} & \text { if } x<\frac{1}{2} \\ x^{2} & \text { if } x \geqslant \frac{1}{2}\end{cases} $$ Let $a$ and $b$ be two real numbers such that $00$. Show that there exists a positive integer $n$ such that $$ \left(a_{n}-a_{n-1}\right)\left(b_{n}-b_{n-1}\right)<0 . $$","['Note that\n\n$$\nf(x)-x=\\frac{1}{2}>0\n$$\n\nif $x<\\frac{1}{2}$ and\n\n$$\nf(x)-x=x^{2}-x<0\n$$\n\nif $x \\geqslant \\frac{1}{2}$. So if we consider $(0,1)$ as being divided into the two subintervals $I_{1}=\\left(0, \\frac{1}{2}\\right)$ and $I_{2}=\\left[\\frac{1}{2}, 1\\right)$, the inequality\n\n$$\n\\left(a_{n}-a_{n-1}\\right)\\left(b_{n}-b_{n-1}\\right)=\\left(f\\left(a_{n-1}\\right)-a_{n-1}\\right)\\left(f\\left(b_{n-1}\\right)-b_{n-1}\\right)<0\n$$\n\nholds if and only if $a_{n-1}$ and $b_{n-1}$ lie in distinct subintervals.\n\nLet us now assume, to the contrary, that $a_{k}$ and $b_{k}$ always lie in the same subinterval. Consider the distance $d_{k}=\\left|a_{k}-b_{k}\\right|$. If both $a_{k}$ and $b_{k}$ lie in $I_{1}$, then\n\n$$\nd_{k+1}=\\left|a_{k+1}-b_{k+1}\\right|=\\left|a_{k}+\\frac{1}{2}-b_{k}-\\frac{1}{2}\\right|=d_{k}\n$$\n\nIf, on the other hand, $a_{k}$ and $b_{k}$ both lie in $I_{2}$, then $\\min \\left(a_{k}, b_{k}\\right) \\geqslant \\frac{1}{2}$ and $\\max \\left(a_{k}, b_{k}\\right)=$ $\\min \\left(a_{k}, b_{k}\\right)+d_{k} \\geqslant \\frac{1}{2}+d_{k}$, which implies\n\n$$\nd_{k+1}=\\left|a_{k+1}-b_{k+1}\\right|=\\left|a_{k}^{2}-b_{k}^{2}\\right|=\\left|\\left(a_{k}-b_{k}\\right)\\left(a_{k}+b_{k}\\right)\\right| \\geqslant\\left|a_{k}-b_{k}\\right|\\left(\\frac{1}{2}+\\frac{1}{2}+d_{k}\\right)=d_{k}\\left(1+d_{k}\\right) \\geqslant d_{k}\n$$\n\nThis means that the difference $d_{k}$ is non-decreasing, and in particular $d_{k} \\geqslant d_{0}>0$ for all $k$.\n\nWe can even say more. If $a_{k}$ and $b_{k}$ lie in $I_{2}$, then\n\n$$\nd_{k+2} \\geqslant d_{k+1} \\geqslant d_{k}\\left(1+d_{k}\\right) \\geqslant d_{k}\\left(1+d_{0}\\right)\n$$\n\nIf $a_{k}$ and $b_{k}$ both lie in $I_{1}$, then $a_{k+1}$ and $b_{k+1}$ both lie in $I_{2}$, and so we have\n\n$$\nd_{k+2} \\geqslant d_{k+1}\\left(1+d_{k+1}\\right) \\geqslant d_{k+1}\\left(1+d_{0}\\right)=d_{k}\\left(1+d_{0}\\right)\n$$\n\nIn either case, $d_{k+2} \\geqslant d_{k}\\left(1+d_{0}\\right)$, and inductively we get\n\n$$\nd_{2 m} \\geqslant d_{0}\\left(1+d_{0}\\right)^{m}\n$$\n\nFor sufficiently large $m$, the right-hand side is greater than 1 , but since $a_{2 m}, b_{2 m}$ both lie in $(0,1)$, we must have $d_{2 m}<1$, a contradiction.\n\nThus there must be a positive integer $n$ such that $a_{n-1}$ and $b_{n-1}$ do not lie in the same subinterval, which proves the desired statement.']",,True,,, 1968,Algebra,,"For a sequence $x_{1}, x_{2}, \ldots, x_{n}$ of real numbers, we define its price as $$ \max _{1 \leqslant i \leqslant n}\left|x_{1}+\cdots+x_{i}\right| $$ Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_{1}$ such that $\left|x_{1}\right|$ is as small as possible; among the remaining numbers, he chooses $x_{2}$ such that $\left|x_{1}+x_{2}\right|$ is as small as possible, and so on. Thus, in the $i^{\text {th }}$ step he chooses $x_{i}$ among the remaining numbers so as to minimise the value of $\left|x_{1}+x_{2}+\cdots+x_{i}\right|$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$. Find the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G \leqslant c D$.","[""If the initial numbers are $1,-1,2$, and -2 , then Dave may arrange them as $1,-2,2,-1$, while George may get the sequence $1,-1,2,-2$, resulting in $D=1$ and $G=2$. So we obtain $c \\geqslant 2$.\n\nTherefore, it remains to prove that $G \\leqslant 2 D$. Let $x_{1}, x_{2}, \\ldots, x_{n}$ be the numbers Dave and George have at their disposal. Assume that Dave and George arrange them into sequences $d_{1}, d_{2}, \\ldots, d_{n}$ and $g_{1}, g_{2}, \\ldots, g_{n}$, respectively. Put\n\n$$\nM=\\max _{1 \\leqslant i \\leqslant n}\\left|x_{i}\\right|, \\quad S=\\left|x_{1}+\\cdots+x_{n}\\right|, \\quad \\text { and } \\quad N=\\max \\{M, S\\}\n$$\n\nWe claim that\n\n$$\nD \\geqslant S,\n\\tag{1}\n$$\n$$\nD \\geqslant \\frac{M}{2}, \\quad \\text { and } \n\\tag{2}\n$$\n$$\nG \\leqslant N=\\max \\{M, S\\} \n\\tag{3}\n$$\n\nThese inequalities yield the desired estimate, as $G \\leqslant \\max \\{M, S\\} \\leqslant \\max \\{M, 2 S\\} \\leqslant 2 D$.\n\nThe inequality (1) is a direct consequence of the definition of the price.\n\nTo prove (2), consider an index $i$ with $\\left|d_{i}\\right|=M$. Then we have\n\n$$\nM=\\left|d_{i}\\right|=\\left|\\left(d_{1}+\\cdots+d_{i}\\right)-\\left(d_{1}+\\cdots+d_{i-1}\\right)\\right| \\leqslant\\left|d_{1}+\\cdots+d_{i}\\right|+\\left|d_{1}+\\cdots+d_{i-1}\\right| \\leqslant 2 D\n$$\n\nas required.\n\nIt remains to establish (3). Put $h_{i}=g_{1}+g_{2}+\\cdots+g_{i}$. We will prove by induction on $i$ that $\\left|h_{i}\\right| \\leqslant N$. The base case $i=1$ holds, since $\\left|h_{1}\\right|=\\left|g_{1}\\right| \\leqslant M \\leqslant N$. Notice also that $\\left|h_{n}\\right|=S \\leqslant N$.\n\nFor the induction step, assume that $\\left|h_{i-1}\\right| \\leqslant N$. We distinguish two cases.\n\nCase 1. Assume that no two of the numbers $g_{i}, g_{i+1}, \\ldots, g_{n}$ have opposite signs.\n\nWithout loss of generality, we may assume that they are all nonnegative. Then one has $h_{i-1} \\leqslant h_{i} \\leqslant \\cdots \\leqslant h_{n}$, thus\n\n$$\n\\left|h_{i}\\right| \\leqslant \\max \\left\\{\\left|h_{i-1}\\right|,\\left|h_{n}\\right|\\right\\} \\leqslant N\n$$\n\nCase 2. Among the numbers $g_{i}, g_{i+1}, \\ldots, g_{n}$ there are positive and negative ones.\n\n\n\nThen there exists some index $j \\geqslant i$ such that $h_{i-1} g_{j} \\leqslant 0$. By the definition of George's sequence we have\n\n$$\n\\left|h_{i}\\right|=\\left|h_{i-1}+g_{i}\\right| \\leqslant\\left|h_{i-1}+g_{j}\\right| \\leqslant \\max \\left\\{\\left|h_{i-1}\\right|,\\left|g_{j}\\right|\\right\\} \\leqslant N\n$$\n\nThus, the induction step is established.""]",['2'],False,,Numerical, 1969,Algebra,,"Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying $$ f(f(m)+n)+f(m)=f(n)+f(3 m)+2014 \tag{1} $$ for all integers $m$ and $n$.","['Let $f$ be a function satisfying (1). Set $C=1007$ and define the function $g: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ by $g(m)=f(3 m)-f(m)+2 C$ for all $m \\in \\mathbb{Z}$; in particular, $g(0)=2 C$. Now (1) rewrites as\n\n$$\nf(f(m)+n)=g(m)+f(n)\n$$\n\nfor all $m, n \\in \\mathbb{Z}$. By induction in both directions it follows that\n\n$$\nf(t f(m)+n)=\\operatorname{tg}(m)+f(n)\n\\tag{2}\n$$\n\nholds for all $m, n, t \\in \\mathbb{Z}$. Applying this, for any $r \\in \\mathbb{Z}$, to the triples $(r, 0, f(0))$ and $(0,0, f(r))$ in place of $(m, n, t)$ we obtain\n\n$$\nf(0) g(r)=f(f(r) f(0))-f(0)=f(r) g(0) \\text {. }\n$$\n\nNow if $f(0)$ vanished, then $g(0)=2 C>0$ would entail that $f$ vanishes identically, contrary to (1). Thus $f(0) \\neq 0$ and the previous equation yields $g(r)=\\alpha f(r)$, where $\\alpha=\\frac{g(0)}{f(0)}$ is some nonzero constant.\n\nSo the definition of $g$ reveals $f(3 m)=(1+\\alpha) f(m)-2 C$, i.e.,\n\n$$\nf(3 m)-\\beta=(1+\\alpha)(f(m)-\\beta)\n\\tag{3}\n$$\n\nfor all $m \\in \\mathbb{Z}$, where $\\beta=\\frac{2 C}{\\alpha}$. By induction on $k$ this implies\n\n$$\nf\\left(3^{k} m\\right)-\\beta=(1+\\alpha)^{k}(f(m)-\\beta)\n\\tag{4}\n$$\n\nfor all integers $k \\geqslant 0$ and $m$.\n\nSince $3 \\nmid 2014$, there exists by (1) some value $d=f(a)$ attained by $f$ that is not divisible by 3 . Now by (2) we have $f(n+t d)=f(n)+t g(a)=f(n)+\\alpha \\cdot t f(a)$, i.e.,\n\n$$\nf(n+t d)=f(n)+\\alpha \\cdot t d\n\\tag{5}\n$$\n\nfor all $n, t \\in \\mathbb{Z}$.\n\nLet us fix any positive integer $k$ with $d \\mid\\left(3^{k}-1\\right)$, which is possible, since $\\operatorname{gcd}(3, d)=1$. E.g., by the Euler-Fermat theorem, we may take $k=\\varphi(|d|)$. Now for each $m \\in \\mathbb{Z}$ we get\n\n$$\nf\\left(3^{k} m\\right)=f(m)+\\alpha\\left(3^{k}-1\\right) m\n$$\n\nfrom (5), which in view of (4) yields $\\left((1+\\alpha)^{k}-1\\right)(f(m)-\\beta)=\\alpha\\left(3^{k}-1\\right) m$. Since $\\alpha \\neq 0$, the right hand side does not vanish for $m \\neq 0$, wherefore the first factor on the left hand side cannot vanish either. It follows that\n\n$$\nf(m)=\\frac{\\alpha\\left(3^{k}-1\\right)}{(1+\\alpha)^{k}-1} \\cdot m+\\beta\n$$\n\n\n\nSo $f$ is a linear function, say $f(m)=A m+\\beta$ for all $m \\in \\mathbb{Z}$ with some constant $A \\in \\mathbb{Q}$. Plugging this into (1) one obtains $\\left(A^{2}-2 A\\right) m+(A \\beta-2 C)=0$ for all $m$, which is equivalent to the conjunction of\n\n$$\nA^{2}=2 A \\quad \\text { and } \\quad A \\beta=2 C .\n\\tag{6}\n$$\n\nThe first equation is equivalent to $A \\in\\{0,2\\}$, and as $C \\neq 0$ the second one gives\n\n$$\nA=2 \\quad \\text { and } \\quad \\beta=C .\n\\tag{7}\n$$\n\nThis shows that $f$ is indeed the function mentioned in the answer and as the numbers found in (7) do indeed satisfy the equations (6) this function is indeed as desired.']",['$f(n) = 2 n+1007$'],False,,Expression, 1970,Algebra,,"Consider all polynomials $P(x)$ with real coefficients that have the following property: for any two real numbers $x$ and $y$ one has $$ \left|y^{2}-P(x)\right| \leqslant 2|x| \text { if and only if }\left|x^{2}-P(y)\right| \leqslant 2|y| \tag{1} $$ Determine all possible values of $P(0)$.","['Part I. We begin by verifying that these numbers are indeed possible values of $P(0)$. To see that each negative real number $-C$ can be $P(0)$, it suffices to check that for every $C>0$ the polynomial $P(x)=-\\left(\\frac{2 x^{2}}{C}+C\\right)$ has the property described in the statement of the problem. Due to symmetry it is enough for this purpose to prove $\\left|y^{2}-P(x)\\right|>2|x|$ for any two real numbers $x$ and $y$. In fact we have\n\n$$\n\\left|y^{2}-P(x)\\right|=y^{2}+\\frac{x^{2}}{C}+\\frac{(|x|-C)^{2}}{C}+2|x| \\geqslant \\frac{x^{2}}{C}+2|x| \\geqslant 2|x|\n$$\n\nwhere in the first estimate equality can only hold if $|x|=C$, whilst in the second one it can only hold if $x=0$. As these two conditions cannot be met at the same time, we have indeed $\\left|y^{2}-P(x)\\right|>2|x|$.\n\nTo show that $P(0)=1$ is possible as well, we verify that the polynomial $P(x)=x^{2}+1$ satisfies (1). Notice that for all real numbers $x$ and $y$ we have\n\n$$\n\\begin{aligned}\n\\left|y^{2}-P(x)\\right| \\leqslant 2|x| & \\Longleftrightarrow\\left(y^{2}-x^{2}-1\\right)^{2} \\leqslant 4 x^{2} \\\\\n& \\Longleftrightarrow 0 \\leqslant\\left(\\left(y^{2}-(x-1)^{2}\\right)\\left((x+1)^{2}-y^{2}\\right)\\right. \\\\\n& \\Longleftrightarrow 0 \\leqslant(y-x+1)(y+x-1)(x+1-y)(x+1+y) \\\\\n& \\Longleftrightarrow 0 \\leqslant\\left((x+y)^{2}-1\\right)\\left(1-(x-y)^{2}\\right) .\n\\end{aligned}\n$$\n\nSince this inequality is symmetric in $x$ and $y$, we are done.\n\nPart II. Now we show that no values other than those mentioned in the answer are possible for $P(0)$. To reach this we let $P$ denote any polynomial satisfying (1) and $P(0) \\geqslant 0$; as we shall see, this implies $P(x)=x^{2}+1$ for all real $x$, which is actually more than what we want.\n\nFirst step: We prove that $P$ is even.\n\nBy (1) we have\n\n$$\n\\left|y^{2}-P(x)\\right| \\leqslant 2|x| \\Longleftrightarrow\\left|x^{2}-P(y)\\right| \\leqslant 2|y| \\Longleftrightarrow\\left|y^{2}-P(-x)\\right| \\leqslant 2|x|\n$$\n\nfor all real numbers $x$ and $y$. Considering just the equivalence of the first and third statement and taking into account that $y^{2}$ may vary through $\\mathbb{R}_{\\geqslant 0}$ we infer that\n\n$$\n[P(x)-2|x|, P(x)+2|x|] \\cap \\mathbb{R}_{\\geqslant 0}=[P(-x)-2|x|, P(-x)+2|x|] \\cap \\mathbb{R}_{\\geqslant 0}\n$$\n\nholds for all $x \\in \\mathbb{R}$. We claim that there are infinitely many real numbers $x$ such that $P(x)+2|x| \\geqslant 0$. This holds in fact for any real polynomial with $P(0) \\geqslant 0$; in order to see this, we may assume that the coefficient of $P$ appearing in front of $x$ is nonnegative. In this case the desired inequality holds for all sufficiently small positive real numbers.\n\nFor such numbers $x$ satisfying $P(x)+2|x| \\geqslant 0$ we have $P(x)+2|x|=P(-x)+2|x|$ by the previous displayed formula, and hence also $P(x)=P(-x)$. Consequently the polynomial $P(x)-P(-x)$ has infinitely many zeros, wherefore it has to vanish identically. Thus $P$ is indeed even.\n\n\n\nSecond step: We prove that $P(t)>0$ for all $t \\in \\mathbb{R}$.\n\nLet us assume for a moment that there exists a real number $t \\neq 0$ with $P(t)=0$. Then there is some open interval $I$ around $t$ such that $|P(y)| \\leqslant 2|y|$ holds for all $y \\in I$. Plugging $x=0$ into (1) we learn that $y^{2}=P(0)$ holds for all $y \\in I$, which is clearly absurd. We have thus shown $P(t) \\neq 0$ for all $t \\neq 0$.\n\nIn combination with $P(0) \\geqslant 0$ this informs us that our claim could only fail if $P(0)=0$. In this case there is by our first step a polynomial $Q(x)$ such that $P(x)=x^{2} Q(x)$. Applying (1) to $x=0$ and an arbitrary $y \\neq 0$ we get $|y Q(y)|>2$, which is surely false when $y$ is sufficiently small.\n\nThird step: We prove that $P$ is a quadratic polynomial.\n\nNotice that $P$ cannot be constant, for otherwise if $x=\\sqrt{P(0)}$ and $y$ is sufficiently large, the first part of (1) is false whilst the second part is true. So the degree $n$ of $P$ has to be at least 1 . By our first step $n$ has to be even as well, whence in particular $n \\geqslant 2$.\n\nNow assume that $n \\geqslant 4$. Plugging $y=\\sqrt{P(x)}$ into (1) we get $\\left|x^{2}-P(\\sqrt{P(x)})\\right| \\leqslant 2 \\sqrt{P(x)}$ and hence\n\n$$\nP(\\sqrt{P(x)}) \\leqslant x^{2}+2 \\sqrt{P(x)}\n$$\n\nfor all real $x$. Choose positive real numbers $x_{0}, a$, and $b$ such that if $x \\in\\left(x_{0}, \\infty\\right)$, then $a x^{n}<$ $P(x)0$ denotes the leading coefficient of $P$, then $\\lim _{x \\rightarrow \\infty} \\frac{P(x)}{x^{n}}=d$, whence for instance the numbers $a=\\frac{d}{2}$ and $b=2 d$ work provided that $x_{0}$ is chosen large enough.\n\nNow for all sufficiently large real numbers $x$ we have\n\n$$\na^{n / 2+1} x^{n^{2} / 2}0$ and $b$ such that $P(x)=$ $a x^{2}+b$. Now if $x$ is large enough and $y=\\sqrt{a} x$, the left part of (1) holds and the right part reads $\\left|\\left(1-a^{2}\\right) x^{2}-b\\right| \\leqslant 2 \\sqrt{a} x$. In view of the fact that $a>0$ this is only possible if $a=1$. Finally, substituting $y=x+1$ with $x>0$ into (1) we get\n\n$$\n|2 x+1-b| \\leqslant 2 x \\Longleftrightarrow|2 x+1+b| \\leqslant 2 x+2,\n$$\n\ni.e.,\n\n$$\nb \\in[1,4 x+1] \\Longleftrightarrow b \\in[-4 x-3,1]\n$$\n\nfor all $x>0$. Choosing $x$ large enough, we can achieve that at least one of these two statements holds; then both hold, which is only possible if $b=1$, as desired.']","['$(-\\infty, 0) \\cup\\{1\\}$.']",False,,Interval, 1971,Algebra,,"Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ n^{2}+4 f(n)=f(f(n))^{2} \tag{1} $$ for all $n \in \mathbb{Z}$.","['Part I. Let us first check that each of the functions above really satisfies the given functional equation. If $f(n)=n+1$ for all $n$, then we have\n\n$$\nn^{2}+4 f(n)=n^{2}+4 n+4=(n+2)^{2}=f(n+1)^{2}=f(f(n))^{2} .\n$$\n\nIf $f(n)=n+1$ for $n>-a$ and $f(n)=-n+1$ otherwise, then we have the same identity for $n>-a$ and\n\n$$\nn^{2}+4 f(n)=n^{2}-4 n+4=(2-n)^{2}=f(1-n)^{2}=f(f(n))^{2}\n$$\n\notherwise. The same applies to the third solution (with $a=0$ ), where in addition one has\n\n$$\n0^{2}+4 f(0)=0=f(f(0))^{2}\n$$\n\nPart II. It remains to prove that these are really the only functions that satisfy our functional equation. We do so in three steps:\n\nStep 1: We prove that $f(n)=n+1$ for $n>0$.\n\nConsider the sequence $\\left(a_{k}\\right)$ given by $a_{k}=f^{k}(1)$ for $k \\geqslant 0$. Setting $n=a_{k}$ in (1), we get\n\n$$\na_{k}^{2}+4 a_{k+1}=a_{k+2}^{2}\n$$\n\nOf course, $a_{0}=1$ by definition. Since $a_{2}^{2}=1+4 a_{1}$ is odd, $a_{2}$ has to be odd as well, so we set $a_{2}=2 r+1$ for some $r \\in \\mathbb{Z}$. Then $a_{1}=r^{2}+r$ and consequently\n\n$$\na_{3}^{2}=a_{1}^{2}+4 a_{2}=\\left(r^{2}+r\\right)^{2}+8 r+4\n$$\n\nSince $8 r+4 \\neq 0, a_{3}^{2} \\neq\\left(r^{2}+r\\right)^{2}$, so the difference between $a_{3}^{2}$ and $\\left(r^{2}+r\\right)^{2}$ is at least the distance from $\\left(r^{2}+r\\right)^{2}$ to the nearest even square (since $8 r+4$ and $r^{2}+r$ are both even). This implies that\n\n$$\n|8 r+4|=\\left|a_{3}^{2}-\\left(r^{2}+r\\right)^{2}\\right| \\geqslant\\left(r^{2}+r\\right)^{2}-\\left(r^{2}+r-2\\right)^{2}=4\\left(r^{2}+r-1\\right),\n$$\n\n(for $r=0$ and $r=-1$, the estimate is trivial, but this does not matter). Therefore, we ave\n\n$$\n4 r^{2} \\leqslant|8 r+4|-4 r+4\n$$\n\n\n\nIf $|r| \\geqslant 4$, then\n\n$$\n4 r^{2} \\geqslant 16|r| \\geqslant 12|r|+16>8|r|+4+4|r|+4 \\geqslant|8 r+4|-4 r+4\n$$\n\na contradiction. Thus $|r|<4$. Checking all possible remaining values of $r$, we find that $\\left(r^{2}+r\\right)^{2}+8 r+4$ is only a square in three cases: $r=-3, r=0$ and $r=1$. Let us now distinguish these three cases:\n\n- $r=-3$, thus $a_{1}=6$ and $a_{2}=-5$. For each $k \\geqslant 1$, we have\n\n$$\na_{k+2}= \\pm \\sqrt{a_{k}^{2}+4 a_{k+1}}\n$$\n\nand the sign needs to be chosen in such a way that $a_{k+1}^{2}+4 a_{k+2}$ is again a square. This yields $a_{3}=-4, a_{4}=-3, a_{5}=-2, a_{6}=-1, a_{7}=0, a_{8}=1, a_{9}=2$. At this point we have reached a contradiction, since $f(1)=f\\left(a_{0}\\right)=a_{1}=6$ and at the same time $f(1)=f\\left(a_{8}\\right)=a_{9}=2$.\n\n- $r=0$, thus $a_{1}=0$ and $a_{2}=1$. Then $a_{3}^{2}=a_{1}^{2}+4 a_{2}=4$, so $a_{3}= \\pm 2$. This, however, is a contradiction again, since it gives us $f(1)=f\\left(a_{0}\\right)=a_{1}=0$ and at the same time $f(1)=f\\left(a_{2}\\right)=a_{3}= \\pm 2$.\n- $r=1$, thus $a_{1}=2$ and $a_{2}=3$. We prove by induction that $a_{k}=k+1$ for all $k \\geqslant 0$ in this case, which we already know for $k \\leqslant 2$ now. For the induction step, assume that $a_{k-1}=k$ and $a_{k}=k+1$. Then\n\n$$\na_{k+1}^{2}=a_{k-1}^{2}+4 a_{k}=k^{2}+4 k+4=(k+2)^{2}\n$$\n\nso $a_{k+1}= \\pm(k+2)$. If $a_{k+1}=-(k+2)$, then\n\n$$\na_{k+2}^{2}=a_{k}^{2}+4 a_{k+1}=(k+1)^{2}-4 k-8=k^{2}-2 k-7=(k-1)^{2}-8\n$$\n\nThe latter can only be a square if $k=4$ (since 1 and 9 are the only two squares whose difference is 8). Then, however, $a_{4}=5, a_{5}=-6$ and $a_{6}= \\pm 1$, so\n\n$$\na_{7}^{2}=a_{5}^{2}+4 a_{6}=36 \\pm 4\n$$\n\nbut neither 32 nor 40 is a perfect square. Thus $a_{k+1}=k+2$, which completes our induction. This also means that $f(n)=f\\left(a_{n-1}\\right)=a_{n}=n+1$ for all $n \\geqslant 1$.\n\nStep 2: We prove that either $f(0)=1$, or $f(0)=0$ and $f(n) \\neq 0$ for $n \\neq 0$.\n\nSet $n=0$ in (1) to get\n\n$$\n4 f(0)=f(f(0))^{2}\n$$\n\nThis means that $f(0) \\geqslant 0$. If $f(0)=0$, then $f(n) \\neq 0$ for all $n \\neq 0$, since we would otherwise have\n\n$$\nn^{2}=n^{2}+4 f(n)=f(f(n))^{2}=f(0)^{2}=0 \\text {. }\n$$\n\nIf $f(0)>0$, then we know that $f(f(0))=f(0)+1$ from the first step, so\n\n$$\n4 f(0)=(f(0)+1)^{2}\n$$\n\nwhich yields $f(0)=1$.\n\n\n\nStep 3: We discuss the values of $f(n)$ for $n<0$.\n\nLemma. For every $n \\geqslant 1$, we have $f(-n)=-n+1$ or $f(-n)=n+1$. Moreover, if $f(-n)=$ $-n+1$ for some $n \\geqslant 1$, then also $f(-n+1)=-n+2$.\n\nProof. We prove this statement by strong induction on $n$. For $n=1$, we get\n\n$$\n1+4 f(-1)=f(f(-1))^{2}\n$$\n\nThus $f(-1)$ needs to be nonnegative. If $f(-1)=0$, then $f(f(-1))=f(0)= \\pm 1$, so $f(0)=1$ (by our second step). Otherwise, we know that $f(f(-1))=f(-1)+1$, so\n\n$$\n1+4 f(-1)=(f(-1)+1)^{2}\n$$\n\nwhich yields $f(-1)=2$ and thus establishes the base case. For the induction step, we consider two cases:\n\n- If $f(-n) \\leqslant-n$, then\n\n$$\nf(f(-n))^{2}=(-n)^{2}+4 f(-n) \\leqslant n^{2}-4 n<(n-2)^{2},\n$$\n\nso $|f(f(-n))| \\leqslant n-3$ (for $n=2$, this case cannot even occur). If $f(f(-n)) \\geqslant 0$, then we already know from the first two steps that $f(f(f(-n)))=f(f(-n))+1$, unless perhaps if $f(0)=0$ and $f(f(-n))=0$. However, the latter would imply $f(-n)=0$ (as shown in Step 2 ) and thus $n=0$, which is impossible. If $f(f(-n))<0$, we can apply the induction hypothesis to $f(f(-n))$. In either case, $f(f(f(-n)))= \\pm f(f(-n))+1$. Therefore,\n\n$$\nf(-n)^{2}+4 f(f(-n))=f(f(f(-n)))^{2}=( \\pm f(f(-n))+1)^{2}\n$$\n\nwhich gives us\n\n$$\n\\begin{aligned}\nn^{2} & \\leqslant f(-n)^{2}=( \\pm f(f(-n))+1)^{2}-4 f(f(-n)) \\leqslant f(f(-n))^{2}+6|f(f(-n))|+1 \\\\\n& \\leqslant(n-3)^{2}+6(n-3)+1=n^{2}-8,\n\\end{aligned}\n$$\n\na contradiction.\n\n- Thus, we are left with the case that $f(-n)>-n$. Now we argue as in the previous case: if $f(-n) \\geqslant 0$, then $f(f(-n))=f(-n)+1$ by the first two steps, since $f(0)=0$ and $f(-n)=0$ would imply $n=0$ (as seen in Step 2) and is thus impossible. If $f(-n)<0$, we can apply the induction hypothesis, so in any case we can infer that $f(f(-n))= \\pm f(-n)+1$. We obtain\n\n$$\n(-n)^{2}+4 f(-n)=( \\pm f(-n)+1)^{2}\n$$\n\nso either\n\n$$\nn^{2}=f(-n)^{2}-2 f(-n)+1=(f(-n)-1)^{2}\n$$\n\nwhich gives us $f(-n)= \\pm n+1$, or\n\n$$\nn^{2}=f(-n)^{2}-6 f(-n)+1=(f(-n)-3)^{2}-8\n$$\n\nSince 1 and 9 are the only perfect squares whose difference is 8 , we must have $n=1$, which we have already considered.\n\nFinally, suppose that $f(-n)=-n+1$ for some $n \\geqslant 2$. Then\n\n$$\nf(-n+1)^{2}=f(f(-n))^{2}=(-n)^{2}+4 f(-n)=(n-2)^{2}\n$$\n\nso $f(-n+1)= \\pm(n-2)$. However, we already know that $f(-n+1)=-n+2$ or $f(-n+1)=n$, so $f(-n+1)=-n+2$.\n\n\n\nCombining everything we know, we find the solutions as stated in the answer:\n\n- One solution is given by $f(n)=n+1$ for all $n$.\n- If $f(n)$ is not always equal to $n+1$, then there is a largest integer $m$ (which cannot be positive) for which this is not the case. In view of the lemma that we proved, we must then have $f(n)=-n+1$ for any integer $n(|b|-4)^{2}\n$$\n\nbecause $|b| \\geqslant|a|-1 \\geqslant 9$. Thus (3) can be refined to\n\n$$\n|a|+3 \\geqslant|f(a)| \\geqslant|a|-1 \\quad \\text { for }|a| \\geqslant E\n$$\n\nNow, from $c^{2}=a^{2}+4 b$ with $|b| \\in[|a|-1,|a|+3]$ we get $c^{2}=(a \\pm 2)^{2}+d$, where $d \\in\\{-16,-12,-8,-4,0,4,8\\}$. Since $|a \\pm 2| \\geqslant 8$, this can happen only if $c^{2}=(a \\pm 2)^{2}$, which in turn yields $b= \\pm a+1$. To summarise,\n\n$$\nf(a)=1 \\pm a \\quad \\text { for }|a| \\geqslant E\n\\tag{4}\n$$\n\nWe have shown that, with at most finitely many exceptions, $f(a)=1 \\pm a$. Thus it will be convenient for our second step to introduce the sets\n\n$$\nZ_{+}=\\{a \\in \\mathbb{Z}: f(a)=a+1\\}, \\quad Z_{-}=\\{a \\in \\mathbb{Z}: f(a)=1-a\\}, \\quad \\text { and } \\quad Z_{0}=\\mathbb{Z} \\backslash\\left(Z_{+} \\cup Z_{-}\\right)\n$$\n\n\n\nStep 2. Now we investigate the structure of the sets $Z_{+}, Z_{-}$, and $Z_{0}$.\n\n4. Note that $f(E+1)=1 \\pm(E+1)$. If $f(E+1)=E+2$, then $E+1 \\in Z_{+}$. Otherwise we have $f(1+E)=-E$; then the original equation (1) with $n=E+1$ gives us $(E-1)^{2}=f(-E)^{2}$, so $f(-E)= \\pm(E-1)$. By (4) this may happen only if $f(-E)=1-E$, so in this case $-E \\in Z_{+}$. In any case we find that $Z_{+} \\neq \\varnothing$.\n5. Now take any $a \\in Z_{+}$. We claim that every integer $x \\geqslant a$ also lies in $Z_{+}$. We proceed by induction on $x$, the base case $x=a$ being covered by our assumption. For the induction step, assume that $f(x-1)=x$ and plug $n=x-1$ into (1). We get $f(x)^{2}=(x+1)^{2}$, so either $f(x)=x+1$ or $f(x)=-(x+1)$.\n\nAssume that $f(x)=-(x+1)$ and $x \\neq-1$, since otherwise we already have $f(x)=x+1$. Plugging $n=x$ into (1), we obtain $f(-x-1)^{2}=(x-2)^{2}-8$, which may happen only if $x-2= \\pm 3$ and $f(-x-1)= \\pm 1$. Plugging $n=-x-1$ into (1), we get $f( \\pm 1)^{2}=(x+1)^{2} \\pm 4$, which in turn may happen only if $x+1 \\in\\{-2,0,2\\}$.\n\nThus $x \\in\\{-1,5\\}$ and at the same time $x \\in\\{-3,-1,1\\}$, which gives us $x=-1$. Since this has already been excluded, we must have $f(x)=x+1$, which completes our induction.\n\n6. Now we know that either $Z_{+}=\\mathbb{Z}$ (if $Z_{+}$is not bounded below), or $Z_{+}=\\left\\{a \\in \\mathbb{Z}: a \\geqslant a_{0}\\right\\}$, where $a_{0}$ is the smallest element of $Z_{+}$. In the former case, $f(n)=n+1$ for all $n \\in \\mathbb{Z}$, which is our first solution. So we assume in the following that $Z_{+}$is bounded below and has a smallest element $a_{0}$.\n\nIf $Z_{0}=\\varnothing$, then we have $f(x)=x+1$ for $x \\geqslant a_{0}$ and $f(x)=1-x$ for $x-a, \\\\ -n+1, & n \\leqslant-a\\end{cases}$\n- or $f(n)= \\begin{cases}n+1, & n>0 \\\\ 0, & n=0 \\\\ -n+1, & n<0\\end{cases}$']",True,,Need_human_evaluate, 1972,Combinatorics,,Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect $R$ into at least $n+1$ smaller rectangles.,"['Let $k$ be the number of rectangles in the dissection. The set of all points that are corners of one of the rectangles can be divided into three disjoint subsets:\n\n- $A$, which consists of the four corners of the original rectangle $R$, each of which is the corner of exactly one of the smaller rectangles,\n- $B$, which contains points where exactly two of the rectangles have a common corner (T-junctions, see the figure below),\n- $C$, which contains points where four of the rectangles have a common corner (crossings, see the figure below).\n\n\nFigure 1: A T-junction and a crossing\n\nWe denote the number of points in $B$ by $b$ and the number of points in $C$ by $c$. Since each of the $k$ rectangles has exactly four corners, we get\n\n$$\n4 k=4+2 b+4 c\n$$\n\nIt follows that $2 b \\leqslant 4 k-4$, so $b \\leqslant 2 k-2$.\n\nEach of the $n$ given points has to lie on a side of one of the smaller rectangles (but not of the original rectangle $R$ ). If we extend this side as far as possible along borders between rectangles, we obtain a line segment whose ends are T-junctions. Note that every point in $B$ can only be an endpoint of at most one such segment containing one of the given points, since it is stated that no two of them lie on a common line parallel to the sides of $R$. This means that\n\n$$\nb \\geqslant 2 n \\text {. }\n$$\n\nCombining our two inequalities for $b$, we get\n\n$$\n2 k-2 \\geqslant b \\geqslant 2 n\n$$\n\nthus $k \\geqslant n+1$, which is what we wanted to prove.', ""Let $k$ denote the number of rectangles. In the following, we refer to the directions of the sides of $R$ as 'horizontal' and 'vertical' respectively. Our goal is to prove the inequality $k \\geqslant n+1$ for fixed $n$. Equivalently, we can prove the inequality $n \\leqslant k-1$ for each $k$, which will be done by induction on $k$. For $k=1$, the statement is trivial.\n\nNow assume that $k>1$. If none of the line segments that form the borders between the rectangles is horizontal, then we have $k-1$ vertical segments dividing $R$ into $k$ rectangles. On each of them, there can only be one of the $n$ points, so $n \\leqslant k-1$, which is exactly what we want to prove.\n\nOtherwise, consider the lowest horizontal line $h$ that contains one or more of these line segments. Let $R^{\\prime}$ be the rectangle that results when everything that lies below $h$ is removed from $R$ (see the example in the figure below).\n\nThe rectangles that lie entirely below $h$ form blocks of rectangles separated by vertical line segments. Suppose there are $r$ blocks and $k_{i}$ rectangles in the $i^{\\text {th }}$ block. The left and right border of each block has to extend further upwards beyond $h$. Thus we can move any points that lie on these borders upwards, so that they now lie inside $R^{\\prime}$. This can be done without violating the conditions, one only needs to make sure that they do not get to lie on a common horizontal line with one of the other given points.\n\nAll other borders between rectangles in the $i^{\\text {th }}$ block have to lie entirely below $h$. There are $k_{i}-1$ such line segments, each of which can contain at most one of the given points. Finally, there can be one point that lies on $h$. All other points have to lie in $R^{\\prime}$ (after moving some of them as explained in the previous paragraph).\n\n\n\nFigure 2: Illustration of the inductive argument\n\nWe see that $R^{\\prime}$ is divided into $k-\\sum_{i=1}^{r} k_{i}$ rectangles. Applying the induction hypothesis to $R^{\\prime}$, we find that there are at most\n\n$$\n\\left(k-\\sum_{i=1}^{r} k_{i}\\right)-1+\\sum_{i=1}^{r}\\left(k_{i}-1\\right)+1=k-r\n$$\n\npoints. Since $r \\geqslant 1$, this means that $n \\leqslant k-1$, which completes our induction.""]",,True,,, 1973,Combinatorics,,"We have $2^{m}$ sheets of paper, with the number 1 written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a+b$ on both sheets. Prove that after $m 2^{m-1}$ steps, the sum of the numbers on all the sheets is at least $4^{m}$.","['Let $P_{k}$ be the product of the numbers on the sheets after $k$ steps.\n\nSuppose that in the $(k+1)^{\\text {th }}$ step the numbers $a$ and $b$ are replaced by $a+b$. In the product, the number $a b$ is replaced by $(a+b)^{2}$, and the other factors do not change. Since $(a+b)^{2} \\geqslant 4 a b$, we see that $P_{k+1} \\geqslant 4 P_{k}$. Starting with $P_{0}=1$, a straightforward induction yields\n\n$$\nP_{k} \\geqslant 4^{k}\n$$\n\nfor all integers $k \\geqslant 0$; in particular\n\n$$\nP_{m \\cdot 2^{m-1}} \\geqslant 4^{m \\cdot 2^{m-1}}=\\left(2^{m}\\right)^{2^{m}}\n$$\n\nso by the AM-GM inequality, the sum of the numbers written on the sheets after $m 2^{m-1}$ steps is at least\n\n$$\n2^{m} \\cdot \\sqrt[2^{m}]{P_{m \\cdot 2^{m-1}}} \\geqslant 2^{m} \\cdot 2^{m}=4^{m}\n$$']",,True,,, 1974,Combinatorics,,"Let $n \geqslant 2$ be an integer. Consider an $n \times n$ chessboard divided into $n^{2}$ unit squares. We call a configuration of $n$ rooks on this board happy if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that for every happy configuration of rooks, we can find a $k \times k$ square without a rook on any of its $k^{2}$ unit squares.","['Let $\\ell$ be a positive integer. We will show that (i) if $n>\\ell^{2}$ then each happy configuration contains an empty $\\ell \\times \\ell$ square, but (ii) if $n \\leqslant \\ell^{2}$ then there exists a happy configuration not containing such a square. These two statements together yield the answer.\n\n(i). Assume that $n>\\ell^{2}$. Consider any happy configuration. There exists a row $R$ containing a rook in its leftmost square. Take $\\ell$ consecutive rows with $R$ being one of them. Their union $U$ contains exactly $\\ell$ rooks. Now remove the $n-\\ell^{2} \\geqslant 1$ leftmost columns from $U$ (thus at least one rook is also removed). The remaining part is an $\\ell^{2} \\times \\ell$ rectangle, so it can be split into $\\ell$ squares of size $\\ell \\times \\ell$, and this part contains at most $\\ell-1$ rooks. Thus one of these squares is empty.\n\n(ii). Now we assume that $n \\leqslant \\ell^{2}$. Firstly, we will construct a happy configuration with no empty $\\ell \\times \\ell$ square for the case $n=\\ell^{2}$. After that we will modify it to work for smaller values of $n$.\n\nLet us enumerate the rows from bottom to top as well as the columns from left to right by the numbers $0,1, \\ldots, \\ell^{2}-1$. Every square will be denoted, as usual, by the pair $(r, c)$ of its row and column numbers. Now we put the rooks on all squares of the form $(i \\ell+j, j \\ell+i)$ with $i, j=0,1, \\ldots, \\ell-1$ (the picture below represents this arrangement for $\\ell=3$ ). Since each number from 0 to $\\ell^{2}-1$ has a unique representation of the form $i \\ell+j(0 \\leqslant i, j \\leqslant \\ell-1)$, each row and each column contains exactly one rook.\n\n\n\nNext, we show that each $\\ell \\times \\ell$ square $A$ on the board contains a rook. Consider such a square $A$, and consider $\\ell$ consecutive rows the union of which contains $A$. Let the lowest of these rows have number $p \\ell+q$ with $0 \\leqslant p, q \\leqslant \\ell-1$ (notice that $p \\ell+q \\leqslant \\ell^{2}-\\ell$ ). Then the rooks in this union are placed in the columns with numbers $q \\ell+p,(q+1) \\ell+p, \\ldots,(\\ell-1) \\ell+p$, $p+1, \\ell+(p+1), \\ldots,(q-1) \\ell+p+1$, or, putting these numbers in increasing order,\n\n$$\np+1, \\ell+(p+1), \\ldots,(q-1) \\ell+(p+1), q \\ell+p,(q+1) \\ell+p, \\ldots,(\\ell-1) \\ell+p\n$$\n\nOne readily checks that the first number in this list is at most $\\ell-1$ (if $p=\\ell-1$, then $q=0$, and the first listed number is $q \\ell+p=\\ell-1)$, the last one is at least $(\\ell-1) \\ell$, and the difference between any two consecutive numbers is at most $\\ell$. Thus, one of the $\\ell$ consecutive columns intersecting $A$ contains a number listed above, and the rook in this column is inside $A$, as required. The construction for $n=\\ell^{2}$ is established.\n\n\n\nIt remains to construct a happy configuration of rooks not containing an empty $\\ell \\times \\ell$ square for $n<\\ell^{2}$. In order to achieve this, take the construction for an $\\ell^{2} \\times \\ell^{2}$ square described above and remove the $\\ell^{2}-n$ bottom rows together with the $\\ell^{2}-n$ rightmost columns. We will have a rook arrangement with no empty $\\ell \\times \\ell$ square, but several rows and columns may happen to be empty. Clearly, the number of empty rows is equal to the number of empty columns, so one can find a bijection between them, and put a rook on any crossing of an empty row and an empty column corresponding to each other.']",['$\\lfloor\\sqrt{n-1}\\rfloor$'],False,,Expression, 1975,Combinatorics,,"Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively. S-tetrominoes Z-tetrominoes Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove than no matter how we tile $P$ using only $\mathrm{S}$ - and Z-tetrominoes, we always use an even number of Z-tetrominoes.","['We may assume that polygon $P$ is the union of some squares of an infinite chessboard. Colour the squares of the chessboard with two colours as the figure below illustrates.\n\n\n\nObserve that no matter how we tile $P$, any S-tetromino covers an even number of black squares, whereas any Z-tetromino covers an odd number of them. As $P$ can be tiled exclusively by S-tetrominoes, it contains an even number of black squares. But if some S-tetrominoes and some Z-tetrominoes cover an even number of black squares, then the number of Z-tetrominoes must be even.', 'Let us assign coordinates to the squares of the infinite chessboard in such a way that the squares of $P$ have nonnegative coordinates only, and that the first coordinate increases as one moves to the right, while the second coordinate increases as one moves upwards. Write the integer $3^{i} \\cdot(-3)^{j}$ into the square with coordinates $(i, j)$, as in the following figure:\n\n| $\\vdots$ | | | | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 81 | $\\vdots$ | | | | |\n| -27 | -81 | $\\vdots$ | | | |\n| 9 | 27 | 81 | $\\cdots$ | | |\n| -3 | -9 | -27 | -81 | $\\ldots$ | |\n| 1 | 3 | 9 | 27 | 81 | $\\ldots$ |\n\nThe sum of the numbers written into four squares that can be covered by an $S$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+3+3 \\cdot(-3)+3^{2} \\cdot(-3)\\right)=-32 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\n(for the first type of $S$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3 \\cdot(-3)+(-3)+(-3)^{2}\\right)=0\n$$\n\nand thus divisible by 32. For this reason, the sum of the numbers written into the squares of $P$, and thus also the sum of the numbers covered by $Z$-tetrominoes in the second covering, is likewise divisible by 32 . Now the sum of the entries of a $Z$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3^{2}+(-3)+3 \\cdot(-3)\\right)=0\n$$\n\n(for the first type of $Z$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+(-3)+3 \\cdot(-3)+3 \\cdot(-3)^{2}\\right)=16 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\ni.e., 16 times an odd number. Thus in order to obtain a total that is divisible by 32, an even number of the latter kind of $Z$-tetrominoes needs to be used. Rotating everything by $90^{\\circ}$, we find that the number of $Z$-tetrominoes of the first kind is even as well. So we have even proven slightly more than necessary.']",,True,,, 1975,Combinatorics,,"Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively. ![](https://cdn.mathpix.com/cropped/2023_12_21_f538cb2117a5b64c3bddg-1.jpg?height=163&width=312&top_left_y=455&top_left_x=632) S-tetrominoes ![](https://cdn.mathpix.com/cropped/2023_12_21_f538cb2117a5b64c3bddg-1.jpg?height=163&width=323&top_left_y=455&top_left_x=1112) Z-tetrominoes Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove than no matter how we tile $P$ using only $\mathrm{S}$ - and Z-tetrominoes, we always use an even number of Z-tetrominoes.","['We may assume that polygon $P$ is the union of some squares of an infinite chessboard. Colour the squares of the chessboard with two colours as the figure below illustrates.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_f538cb2117a5b64c3bddg-1.jpg?height=665&width=874&top_left_y=978&top_left_x=591)\n\nObserve that no matter how we tile $P$, any S-tetromino covers an even number of black squares, whereas any Z-tetromino covers an odd number of them. As $P$ can be tiled exclusively by S-tetrominoes, it contains an even number of black squares. But if some S-tetrominoes and some Z-tetrominoes cover an even number of black squares, then the number of Z-tetrominoes must be even.', 'Let us assign coordinates to the squares of the infinite chessboard in such a way that the squares of $P$ have nonnegative coordinates only, and that the first coordinate increases as one moves to the right, while the second coordinate increases as one moves upwards. Write the integer $3^{i} \\cdot(-3)^{j}$ into the square with coordinates $(i, j)$, as in the following figure:\n\n| $\\vdots$ | | | | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 81 | $\\vdots$ | | | | |\n| -27 | -81 | $\\vdots$ | | | |\n| 9 | 27 | 81 | $\\cdots$ | | |\n| -3 | -9 | -27 | -81 | $\\ldots$ | |\n| 1 | 3 | 9 | 27 | 81 | $\\ldots$ |\n\nThe sum of the numbers written into four squares that can be covered by an $S$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+3+3 \\cdot(-3)+3^{2} \\cdot(-3)\\right)=-32 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\n(for the first type of $S$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3 \\cdot(-3)+(-3)+(-3)^{2}\\right)=0\n$$\n\nand thus divisible by 32. For this reason, the sum of the numbers written into the squares of $P$, and thus also the sum of the numbers covered by $Z$-tetrominoes in the second covering, is likewise divisible by 32 . Now the sum of the entries of a $Z$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3^{2}+(-3)+3 \\cdot(-3)\\right)=0\n$$\n\n(for the first type of $Z$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+(-3)+3 \\cdot(-3)+3 \\cdot(-3)^{2}\\right)=16 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\ni.e., 16 times an odd number. Thus in order to obtain a total that is divisible by 32, an even number of the latter kind of $Z$-tetrominoes needs to be used. Rotating everything by $90^{\\circ}$, we find that the number of $Z$-tetrominoes of the first kind is even as well. So we have even proven slightly more than necessary.']",['证明题，略'],True,,Need_human_evaluate, 1976,Combinatorics,,"Consider $n \geqslant 3$ lines in the plane such that no two lines are parallel and no three have a common point. These lines divide the plane into polygonal regions; let $\mathcal{F}$ be the set of regions having finite area. Prove that it is possible to colour $\lceil\sqrt{n / 2}\rceil$ of the lines blue in such a way that no region in $\mathcal{F}$ has a completely blue boundary. (For a real number $x,\lceil x\rceil$ denotes the least integer which is not smaller than $x$.)","['Let $L$ be the given set of lines. Choose a maximal (by inclusion) subset $B \\subseteq L$ such that when we colour the lines of $B$ blue, no region in $\\mathcal{F}$ has a completely blue boundary. Let $|B|=k$. We claim that $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$.\n\nLet us colour all the lines of $L \\backslash B$ red. Call a point blue if it is the intersection of two blue lines. Then there are $\\left(\\begin{array}{l}k \\\\ 2\\end{array}\\right)$ blue points.\n\nNow consider any red line $\\ell$. By the maximality of $B$, there exists at least one region $A \\in \\mathcal{F}$ whose only red side lies on $\\ell$. Since $A$ has at least three sides, it must have at least one blue vertex. Let us take one such vertex and associate it to $\\ell$.\n\nSince each blue point belongs to four regions (some of which may be unbounded), it is associated to at most four red lines. Thus the total number of red lines is at most $4\\left(\\begin{array}{l}k \\\\ 2\\end{array}\\right)$. On the other hand, this number is $n-k$, so\n\n$$\nn-k \\leqslant 2 k(k-1), \\quad \\text { thus } n \\leqslant 2 k^{2}-k \\leqslant 2 k^{2} \\text {, }\n$$\n\nand finally $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$, which gives the desired result.']",,True,,, 1977,Combinatorics,,"We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of 100 cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions: 1. The winner only depends on the relative order of the 200 cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner. 2. If we write the elements of both sets in increasing order as $A=\left\{a_{1}, a_{2}, \ldots, a_{100}\right\}$ and $B=\left\{b_{1}, b_{2}, \ldots, b_{100}\right\}$, and $a_{i}>b_{i}$ for all $i$, then $A$ beats $B$. 3. If three players draw three disjoint sets $A, B, C$ from the deck, $A$ beats $B$ and $B$ beats $C$, then $A$ also beats $C$. How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets $A$ and $B$ such that $A$ beats $B$ according to one rule, but $B$ beats $A$ according to the other.","['We prove a more general statement for sets of cardinality $n$ (the problem being the special case $n=100$, then the answer is $n$ ). In the following, we write $A>B$ or $Bb_{k}$. This rule clearly satisfies all three conditions, and the rules corresponding to different $k$ are all different. Thus there are at least $n$ different rules.\n\nPart II. Now we have to prove that there is no other way to define such a rule. Suppose that our rule satisfies the conditions, and let $k \\in\\{1,2, \\ldots, n\\}$ be minimal with the property that\n\n$$\nA_{k}=\\{1,2, \\ldots, k, n+k+1, n+k+2, \\ldots, 2 n\\} \\prec B_{k}=\\{k+1, k+2, \\ldots, n+k\\} .\n$$\n\nClearly, such a $k$ exists, since this holds for $k=n$ by assumption. Now consider two disjoint sets $X=\\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}$ and $Y=\\left\\{y_{1}, y_{2}, \\ldots, y_{n}\\right\\}$, both in increasing order (i.e., $x_{1}B_{k-1}$ by our choice of $k$, we also have $U>W$ (if $k=1$, this is trivial).\n- The elements of $V \\cup W$ are ordered in the same way as those of $A_{k} \\cup B_{k}$, and since $A_{k} \\prec B_{k}$ by our choice of $k$, we also have $V(\\{1\\} \\cup\\{3 i-1 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup\\{3 n\\}) .\n$$\n\nLikewise, if the second relation does not hold, then we must also have\n\n$$\n(\\{1\\} \\cup\\{3 i-1 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup\\{3 n\\})>(\\{3\\} \\cup\\{3 i \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup\\{3 n-1\\}) .\n$$\n\nNow condition 3 implies that\n\n$$\n(\\{2\\} \\cup\\{3 i-2 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup\\{3 n-2\\})>(\\{3\\} \\cup\\{3 i \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup\\{3 n-1\\}),\n$$\n\nwhich contradicts the second condition.\n\nNow we distinguish two cases, depending on which of the two relations actually holds:\n\nFirst case: $(\\{2\\} \\cup\\{2 i-1 \\mid 2 \\leqslant i \\leqslant n\\})<(\\{1\\} \\cup\\{2 i \\mid 2 \\leqslant i \\leqslant n\\})$.\n\nLet $A=\\left\\{a_{1}, a_{2}, \\ldots, a_{n}\\right\\}$ and $B=\\left\\{b_{1}, b_{2}, \\ldots, b_{n}\\right\\}$ be two disjoint sets, both in increasing order. We claim that the winner can be decided only from the values of $a_{2}, \\ldots, a_{n}$ and $b_{2}, \\ldots, b_{n}$, while $a_{1}$ and $b_{1}$ are actually irrelevant. Suppose that this was not the case, and assume without loss of generality that $a_{2}0$ be smaller than half the distance between any two of the numbers in $B_{x} \\cup B_{y} \\cup A$. For any set $M$, let $M \\pm \\varepsilon$ be the set obtained by adding/subtracting $\\varepsilon$ to all elements of $M$. By our choice of $\\varepsilon$, the relative order of the elements of $\\left(B_{y}+\\varepsilon\\right) \\cup A$ is still the same as for $B_{y} \\cup A$, while the relative order of the elements of $\\left(B_{x}-\\varepsilon\\right) \\cup A$ is still the same as for $B_{x} \\cup A$. Thus $AB_{y}+\\varepsilon$. Moreover, if $y>x$, then $B_{x}-\\varepsilon \\prec B_{y}+\\varepsilon$ by condition 2, while otherwise the relative order of\n\n\n\nthe elements in $\\left(B_{x}-\\varepsilon\\right) \\cup\\left(B_{y}+\\varepsilon\\right)$ is the same as for the two sets $\\{2\\} \\cup\\{2 i-1 \\mid 2 \\leqslant i \\leqslant n\\}$ and $\\{1\\} \\cup\\{2 i \\mid 2 \\leqslant i \\leqslant n\\}$, so that $B_{x}-\\varepsilon\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nSimilarly but more easily one obtains\n\n$$\n\\left|X_{A C} \\cap X_{B D}\\right| \\leqslant\\left|X_{A B} \\cap X_{C D}\\right|\n\\tag{2}\n$$\n\nIndeed, a red line $h$ appearing in $X_{A C} \\cap X_{B D}$ belongs, for similar reasons as above, also to $X_{A B} \\cap X_{C D}$. To make the argument precise, one may just distinguish the cases $S \\in h$ (see Figure 2) and $S \\notin h$ (see Figure 3). Thereby (2) is proved.\n\nAdding (1) and (2) we obtain the desired conclusion, thus completing the solution of this problem.']",,True,,, 1979,Combinatorics,,"A card deck consists of 1024 cards. On each card, a set of distinct decimal digits is written in such a way that no two of these sets coincide (thus, one of the cards is empty). Two players alternately take cards from the deck, one card per turn. After the deck is empty, each player checks if he can throw out one of his cards so that each of the ten digits occurs on an even number of his remaining cards. If one player can do this but the other one cannot, the one who can is the winner; otherwise a draw is declared. Determine all possible first moves of the first player after which he has a winning strategy.","[""Let us identify each card with the set of digits written on it. For any collection of cards $C_{1}, C_{2}, \\ldots, C_{k}$ denote by their sum the set $C_{1} \\Delta C_{2} \\Delta \\cdots \\Delta C_{k}$ consisting of all elements belonging to an odd number of the $C_{i}$ 's. Denote the first and the second player by $\\mathcal{F}$ and $\\mathcal{S}$, respectively.\n\nSince each digit is written on exactly 512 cards, the sum of all the cards is $\\varnothing$. Therefore, at the end of the game the sum of all the cards of $\\mathcal{F}$ will be the same as that of $\\mathcal{S}$; denote this sum by $C$. Then the player who took $C$ can throw it out and get the desired situation, while the other one cannot. Thus, the player getting card $C$ wins, and no draw is possible.\n\nNow, given a nonempty card $B$, one can easily see that all the cards can be split into 512 pairs of the form $(X, X \\triangle B)$ because $(X \\triangle B) \\triangle B=X$. The following lemma shows a property of such a partition that is important for the solution.\n\nLemma. Let $B \\neq \\varnothing$ be some card. Let us choose 512 cards so that exactly one card is chosen from every pair $(X, X \\triangle B)$. Then the sum of all chosen cards is either $\\varnothing$ or $B$.\n\nProof. Let $b$ be some element of $B$. Enumerate the pairs; let $X_{i}$ be the card not containing $b$ in the $i^{\\text {th }}$ pair, and let $Y_{i}$ be the other card in this pair. Then the sets $X_{i}$ are exactly all the sets not containing $b$, therefore each digit $a \\neq b$ is written on exactly 256 of these cards, so $X_{1} \\Delta X_{2} \\Delta \\cdots \\Delta X_{512}=\\varnothing$. Now, if we replace some summands in this sum by the other elements from their pairs, we will simply add $B$ several times to this sum, thus the sum will either remain unchanged or change by $B$, as required.\n\nNow we consider two cases.\n\nCase 1. Assume that $\\mathcal{F}$ takes the card $\\varnothing$ on his first move. In this case, we present a winning strategy for $\\mathcal{S}$.\n\nLet $\\mathcal{S}$ take an arbitrary card $A$. Assume that $\\mathcal{F}$ takes card $B$ after that; then $\\mathcal{S}$ takes $A \\triangle B$. Split all 1024 cards into 512 pairs of the form $(X, X \\triangle B)$; we call two cards in one pair partners. Then the four cards taken so far form two pairs $(\\varnothing, B)$ and $(A, A \\triangle B)$ belonging to $\\mathcal{F}$ and $\\mathcal{S}$, respectively. On each of the subsequent moves, when $\\mathcal{F}$ takes some card, $\\mathcal{S}$ should take the partner of this card in response.\n\nConsider the situation at the end of the game. Let us for a moment replace card $A$ belonging to $\\mathcal{S}$ by $\\varnothing$. Then he would have one card from each pair; by our lemma, the sum of all these cards would be either $\\varnothing$ or $B$. Now, replacing $\\varnothing$ back by $A$ we get that the actual sum of the cards of $\\mathcal{S}$ is either $A$ or $A \\triangle B$, and he has both these cards. Thus $\\mathcal{S}$ wins.\n\nCase 2. Now assume that $\\mathcal{F}$ takes some card $A \\neq \\varnothing$ on his first move. Let us present a winning strategy for $\\mathcal{F}$ in this case.\n\nAssume that $\\mathcal{S}$ takes some card $B \\neq \\varnothing$ on his first move; then $\\mathcal{F}$ takes $A \\triangle B$. Again, let us split all the cards into pairs of the form $(X, X \\triangle B)$; then the cards which have not been taken yet form several complete pairs and one extra element (card $\\varnothing$ has not been taken while its partner $B$ has). Now, on each of the subsequent moves, if $\\mathcal{S}$ takes some element from a\n\n\n\ncomplete pair, then $\\mathcal{F}$ takes its partner. If $\\mathcal{S}$ takes the extra element, then $\\mathcal{F}$ takes an arbitrary card $Y$, and the partner of $Y$ becomes the new extra element.\n\nThus, on his last move $\\mathcal{S}$ is forced to take the extra element. After that player $\\mathcal{F}$ has cards $A$ and $A \\triangle B$, player $\\mathcal{S}$ has cards $B$ and $\\varnothing$, and $\\mathcal{F}$ has exactly one element from every other pair. Thus the situation is the same as in the previous case with roles reversed, and $\\mathcal{F}$ wins.\n\nFinally, if $\\mathcal{S}$ takes $\\varnothing$ on his first move then $\\mathcal{F}$ denotes any card which has not been taken yet by $B$ and takes $A \\triangle B$. After that, the same strategy as above is applicable.""]",['All the moves except for taking the empty card'],False,,Need_human_evaluate, 1980,Combinatorics,,"There are $n$ circles drawn on a piece of paper in such a way that any two circles intersect in two points, and no three circles pass through the same point. Turbo the snail slides along the circles in the following fashion. Initially he moves on one of the circles in clockwise direction. Turbo always keeps sliding along the current circle until he reaches an intersection with another circle. Then he continues his journey on this new circle and also changes the direction of moving, i.e. from clockwise to anticlockwise or vice versa. Suppose that Turbo's path entirely covers all circles. Prove that $n$ must be odd.","['Replace every cross (i.e. intersection of two circles) by two small circle arcs that indicate the direction in which the snail should leave the cross (see Figure 1.1). Notice that the placement of the small arcs does not depend on the direction of moving on the curves; no matter which direction the snail is moving on the circle arcs, he will follow the same curves (see Figure 1.2). In this way we have a set of curves, that are the possible paths of the snail. Call these curves snail orbits or just orbits. Every snail orbit is a simple closed curve that has no intersection with any other orbit.\n\n\n\nFigure 1.1\n\n\n\nFigure 1.2\n\nWe prove the following, more general statement.\n\n(*) In any configuration of $n$ circles such that no two of them are tangent, the number of snail orbits has the same parity as the number $n$. (Note that it is not assumed that all circle pairs intersect.)\n\nThis immediately solves the problem.\n\nLet us introduce the following operation that will be called flipping a cross. At a cross, remove the two small arcs of the orbits, and replace them by the other two arcs. Hence, when the snail arrives at a flipped cross, he will continue on the other circle as before, but he will preserve the orientation in which he goes along the circle arcs (see Figure 2).\n\n\nFigure 2\n\nConsider what happens to the number of orbits when a cross is flipped. Denote by $a, b, c$, and $d$ the four arcs that meet at the cross such that $a$ and $b$ belong to the same circle. Before the flipping $a$ and $b$ were connected to $c$ and $d$, respectively, and after the flipping $a$ and $b$ are connected to $d$ and $c$, respectively.\n\nThe orbits passing through the cross are closed curves, so each of the arcs $a, b, c$, and $d$ is connected to another one by orbits outside the cross. We distinguish three cases.\n\nCase 1: $a$ is connected to $b$ and $c$ is connected to $d$ by the orbits outside the cross (see Figure 3.1).\n\n\n\nWe show that this case is impossible. Remove the two small arcs at the cross, connect $a$ to $b$, and connect $c$ to $d$ at the cross. Let $\\gamma$ be the new closed curve containing $a$ and $b$, and let $\\delta$ be the new curve that connects $c$ and $d$. These two curves intersect at the cross. So one of $c$ and $d$ is inside $\\gamma$ and the other one is outside $\\gamma$. Then the two closed curves have to meet at least one more time, but this is a contradiction, since no orbit can intersect itself.\n\n\n\nFigure 3.1\n\n\n\nFigure 3.2\n\n\nFigure 3.3\n\nCase 2: $a$ is connected to $c$ and $b$ is connected to $d$ (see Figure 3.2).\n\nBefore the flipping $a$ and $c$ belong to one orbit and $b$ and $d$ belong to another orbit. Flipping the cross merges the two orbits into a single orbit. Hence, the number of orbits decreases by 1 .\n\nCase 3: $a$ is connected to $d$ and $b$ is connected to $c$ (see Figure 3.3).\n\nBefore the flipping the $\\operatorname{arcs} a, b, c$, and $d$ belong to a single orbit. Flipping the cross splits that orbit in two. The number of orbits increases by 1 .\n\nAs can be seen, every flipping decreases or increases the number of orbits by one, thus changes its parity.\n\nNow flip every cross, one by one. Since every pair of circles has 0 or 2 intersections, the number of crosses is even. Therefore, when all crosses have been flipped, the original parity of the number of orbits is restored. So it is sufficient to prove $(*)$ for the new configuration, where all crosses are flipped. Of course also in this new configuration the (modified) orbits are simple closed curves not intersecting each other.\n\nOrient the orbits in such a way that the snail always moves anticlockwise along the circle arcs. Figure 4 shows the same circles as in Figure 1 after flipping all crosses and adding orientation. (Note that this orientation may be different from the orientation of the orbit as a planar curve; the orientation of every orbit may be negative as well as positive, like the middle orbit in Figure 4.) If the snail moves around an orbit, the total angle change in his moving direction, the total curvature, is either $+2 \\pi$ or $-2 \\pi$, depending on the orientation of the orbit. Let $P$ and $N$ be the number of orbits with positive and negative orientation, respectively. Then the total curvature of all orbits is $(P-N) \\cdot 2 \\pi$.\n\n\n\nFigure 4\n\n\n\nFigure 5\n\nDouble-count the total curvature of all orbits. Along every circle the total curvature is $2 \\pi$. At every cross, the two turnings make two changes with some angles having the same absolute value but opposite signs, as depicted in Figure 5. So the changes in the direction at the crosses cancel out. Hence, the total curvature is $n \\cdot 2 \\pi$.\n\nNow we have $(P-N) \\cdot 2 \\pi=n \\cdot 2 \\pi$, so $P-N=n$. The number of (modified) orbits is $P+N$, that has a same parity as $P-N=n$.', 'We present a different proof of $(*)$.\n\nWe perform a sequence of small modification steps on the configuration of the circles in such a way that at the end they have no intersection at all (see Figure 6.1). We use two kinds of local changes to the structure of the orbits (see Figure 6.2):\n\n- Type-1 step: An arc of a circle is moved over an arc of another circle; such a step creates or removes two intersections.\n- Type-2 step: An arc of a circle is moved through the intersection of two other circles.\n\n\n\nFigure 6.1\n\n\n\nype-1\n\n\n\nFigure 6.2\n\nWe assume that in every step only one circle is moved, and that this circle is moved over at most one arc or intersection point of other circles.\n\nWe will show that the parity of the number of orbits does not change in any step. As every circle becomes a separate orbit at the end of the procedure, this fact proves (*).\n\nConsider what happens to the number of orbits when a Type- 1 step is performed. The two intersection points are created or removed in a small neighbourhood. Denote some points of the two circles where they enter or leave this neighbourhood by $a, b, c$, and $d$ in this order around the neighbourhood; let $a$ and $b$ belong to one circle and let $c$ and $d$ belong to the other circle. The two circle arcs may have the same or opposite orientations. Moreover, the four end-points of the two arcs are connected by the other parts of the orbits. This can happen in two ways without intersection: either $a$ is connected to $d$ and $b$ is connected to $c$, or $a$ is connected to $b$ and $c$ is connected to $d$. Altogether we have four cases, as shown in Figure 7.\n\n\nFigure 7\n\nWe can see that the number of orbits is changed by -2 or +2 in the leftmost case when the arcs have the same orientation, $a$ is connected to $d$, and $b$ is connected to $c$. In the other three cases the number of orbits is not changed. Hence, Type-1 steps do not change the parity of the number of orbits.\n\nNow consider a Type-2 step. The three circles enclose a small, triangular region; by the step, this triangle is replaced by another triangle. Again, the modification of the orbits is done in some small neighbourhood; the structure does not change outside. Each side of the triangle shaped region can be convex or concave; the number of concave sides can be $0,1,2$ or 3 , so there are 4 possible arrangements of the orbits inside the neighbourhood, as shown in Figure 8.\n\n\n\n\n\nFigure 8\n\nDenote the points where the three circles enter or leave the neighbourhood by $a, b, c, d$, $e$, and $f$ in this order around the neighbourhood. As can be seen in Figure 8, there are only two essentially different cases; either $a, c, e$ are connected to $b, d, f$, respectively, or $a, c, e$ are connected to $f, b, d$, respectively. The step either preserves the set of connections or switches to the other arrangement. Obviously, in the earlier case the number of orbits is not changed; therefore we have to consider only the latter case.\n\nThe points $a, b, c, d, e$, and $f$ are connected by the orbits outside, without intersection. If $a$ was connected to $c$, say, then this orbit would isolate $b$, so this is impossible. Hence, each of $a, b, c, d, e$ and $f$ must be connected either to one of its neighbours or to the opposite point. If say $a$ is connected to $d$, then this orbit separates $b$ and $c$ from $e$ and $f$, therefore $b$ must be connected to $c$ and $e$ must be connected to $f$. Altogether there are only two cases and their reverses: either each point is connected to one of its neighbours or two opposite points are connected and the the remaining neigh boring pairs are connected to each other. See Figure 9.\n\n\nFigure 9\n\nWe can see that if only neighbouring points are connected, then the number of orbits is changed by +2 or -2 . If two opposite points are connected ( $a$ and $d$ in the figure), then the orbits are re-arranged, but their number is unchanged. Hence, Type-2 steps also preserve the parity. This completes the proof of $(*)$.', ""Like in the previous solutions, we do not need all circle pairs to intersect but we assume that the circles form a connected set. Denote by $\\mathcal{C}$ and $\\mathcal{P}$ the sets of circles and their intersection points, respectively.\n\nThe circles divide the plane into several simply connected, bounded regions and one unbounded region. Denote the set of these regions by $\\mathcal{R}$. We say that an intersection point or a region is odd or even if it is contained inside an odd or even number of circles, respectively. Let $\\mathcal{P}_{\\text {odd }}$ and $\\mathcal{R}_{\\text {odd }}$ be the sets of odd intersection points and odd regions, respectively.\n\nClaim.\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right|-\\left|\\mathcal{P}_{\\text {odd }}\\right| \\equiv n \\quad(\\bmod 2)\n\\tag{1}\n$$\n\nProof. For each circle $c \\in \\mathcal{C}$, denote by $R_{c}, P_{c}$, and $X_{c}$ the number of regions inside $c$, the number of intersection points inside $c$, and the number of circles intersecting $c$, respectively. The circles divide each other into several arcs; denote by $A_{c}$ the number of such arcs inside $c$. By double counting the regions and intersection points inside the circles we get\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right| \\equiv \\sum_{c \\in \\mathcal{C}} R_{c} \\quad(\\bmod 2) \\quad \\text { and } \\quad\\left|\\mathcal{P}_{\\text {odd }}\\right| \\equiv \\sum_{c \\in \\mathcal{C}} P_{c} \\quad(\\bmod 2)\n$$\n\n\n\nFor each circle $c$, apply EULER's polyhedron theorem to the (simply connected) regions in $c$. There are $2 X_{c}$ intersection points on $c$; they divide the circle into $2 X_{c}$ arcs. The polyhedron theorem yields $\\left(R_{c}+1\\right)+\\left(P_{c}+2 X_{c}\\right)=\\left(A_{c}+2 X_{c}\\right)+2$, considering the exterior of $c$ as a single region. Therefore,\n\n$$\nR_{c}+P_{c}=A_{c}+1 .\n\\tag{2}\n$$\n\nMoreover, we have four arcs starting from every interior points inside $c$ and a single arc starting into the interior from each intersection point on the circle. By double-counting the end-points of the interior arcs we get $2 A_{c}=4 P_{c}+2 X_{c}$, so\n\n$$\nA_{c}=2 P_{c}+X_{c} .\n\\tag{3}\n$$\n\nThe relations (2) and (3) together yield\n\n$$\nR_{c}-P_{c}=X_{c}+1\n\\tag{4}\n$$\n\nBy summing up (4) for all circles we obtain\n\n$$\n\\sum_{c \\in \\mathcal{C}} R_{c}-\\sum_{c \\in \\mathcal{C}} P_{c}=\\sum_{c \\in \\mathcal{C}} X_{c}+|\\mathcal{C}|\n$$\n\nwhich yields\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right|-\\left|\\mathcal{P}_{\\text {odd }}\\right| \\equiv \\sum_{c \\in \\mathcal{C}} X_{c}+n \\quad(\\bmod 2)\n\\tag{5}\n$$\n\nNotice that in $\\sum_{c \\in \\mathcal{C}} X_{c}$ each intersecting circle pair is counted twice, i.e., for both circles in the pair, so\n\n$$\n\\sum_{c \\in \\mathcal{C}} X_{c} \\equiv 0 \\quad(\\bmod 2)\n$$\n\nwhich finishes the proof of the Claim.\n\nNow insert the same small arcs at the intersections as in the first solution, and suppose that there is a single snail orbit $b$.\n\nFirst we show that the odd regions are inside the curve $b$, while the even regions are outside. Take a region $r \\in \\mathcal{R}$ and a point $x$ in its interior, and draw a ray $y, \\operatorname{starting}$ from $x$, that does not pass through any intersection point of the circles and is neither tangent to any of the circles. As is well-known, $x$ is inside the curve $b$ if and only if $y$ intersects $b$ an odd number of times (see Figure 10). Notice that if an arbitrary circle $c$ contains $x$ in its interior, then $c$ intersects $y$ at a single point; otherwise, if $x$ is outside $c$, then $c$ has 2 or 0 intersections with $y$. Therefore, $y$ intersects $b$ an odd number of times if and only if $x$ is contained in an odd number of circles, so if and only if $r$ is odd.\n\n\n\nFigure 10\n\nNow consider an intersection point $p$ of two circles $c_{1}$ and $c_{2}$ and a small neighbourhood around $p$. Suppose that $p$ is contained inside $k$ circles.\n\n\n\nWe have four regions that meet at $p$. Let $r_{1}$ be the region that lies outside both $c_{1}$ and $c_{2}$, let $r_{2}$ be the region that lies inside both $c_{1}$ and $c_{2}$, and let $r_{3}$ and $r_{4}$ be the two remaining regions, each lying inside exactly one of $c_{1}$ and $c_{2}$. The region $r_{1}$ is contained inside the same $k$ circles as $p$; the region $r_{2}$ is contained also by $c_{1}$ and $c_{2}$, so by $k+2$ circles in total; each of the regions $r_{3}$ and $r_{4}$ is contained inside $k+1$ circles. After the small arcs have been inserted at $p$, the regions $r_{1}$ and $r_{2}$ get connected, and the regions $r_{3}$ and $r_{4}$ remain separated at $p$ (see Figure 11). If $p$ is an odd point, then $r_{1}$ and $r_{2}$ are odd, so two odd regions are connected at $p$. Otherwise, if $p$ is even, then we have two even regions connected at $p$.\n\n\n\nFigure 11\n\n\n\nFigure 12\n\nConsider the system of odd regions and their connections at the odd points as a graph. In this graph the odd regions are the vertices, and each odd point establishes an edge that connects two vertices (see Figure 12). As $b$ is a single closed curve, this graph is connected and contains no cycle, so the graph is a tree. Then the number of vertices must be by one greater than the number of edges, so\n\n$$\n\\left|\\mathcal{R}_{\\text {odd }}\\right|-\\left|\\mathcal{P}_{\\text {odd }}\\right|=1\n\\tag{6}\n$$\n\nThe relations (1) and (6) together prove that $n$ must be odd.""]",,True,,, 1981,Geometry,,"The points $P$ and $Q$ are chosen on the side $B C$ of an acute-angled triangle $A B C$ so that $\angle P A B=\angle A C B$ and $\angle Q A C=\angle C B A$. The points $M$ and $N$ are taken on the rays $A P$ and $A Q$, respectively, so that $A P=P M$ and $A Q=Q N$. Prove that the lines $B M$ and $C N$ intersect on the circumcircle of the triangle $A B C$.","['Denote by $S$ the intersection point of the lines $B M$ and $C N$. Let moreover $\\beta=\\angle Q A C=\\angle C B A$ and $\\gamma=\\angle P A B=\\angle A C B$. From these equalities it follows that the triangles $A B P$ and $C A Q$ are similar (see Figure 1). Therefore we obtain\n\n$$\n\\frac{B P}{P M}=\\frac{B P}{P A}=\\frac{A Q}{Q C}=\\frac{N Q}{Q C}\n$$\n\nMoreover,\n\n$$\n\\angle B P M=\\beta+\\gamma=\\angle C Q N\n$$\n\nHence the triangles $B P M$ and $N Q C$ are similar. This gives $\\angle B M P=\\angle N C Q$, so the triangles $B P M$ and $B S C$ are also similar. Thus we get\n\n$$\n\\angle C S B=\\angle B P M=\\beta+\\gamma=180^{\\circ}-\\angle B A C\n$$\n\nwhich completes the solution.\n\n\n\nFigure 1\n\n\n\nFigure 2', ""As in the previous solution, denote by $S$ the intersection point of the lines $B M$ and $N C$. Let moreover the circumcircle of the triangle $A B C$ intersect the lines $A P$ and $A Q$ again at $K$ and $L$, respectively (see Figure 2).\n\nNote that $\\angle L B C=\\angle L A C=\\angle C B A$ and similarly $\\angle K C B=\\angle K A B=\\angle B C A$. It implies that the lines $B L$ and $C K$ meet at a point $X$, being symmetric to the point $A$ with respect to the line $B C$. Since $A P=P M$ and $A Q=Q N$, it follows that $X$ lies on the line $M N$. Therefore, using PASCAL's theorem for the hexagon $A L B S C K$, we infer that $S$ lies on the circumcircle of the triangle $A B C$, which finishes the proof.""]",,True,,, 1982,Geometry,,"Let $A B C$ be a triangle. The points $K, L$, and $M$ lie on the segments $B C, C A$, and $A B$, respectively, such that the lines $A K, B L$, and $C M$ intersect in a common point. Prove that it is possible to choose two of the triangles $A L M, B M K$, and $C K L$ whose inradii sum up to at least the inradius of the triangle $A B C$.","[""Denote\n\n$$\na=\\frac{B K}{K C}, \\quad b=\\frac{C L}{L A}, \\quad c=\\frac{A M}{M B}\n$$\n\nBy CEvA's theorem, $a b c=1$, so we may, without loss of generality, assume that $a \\geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than 1. Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first component not less than 1 and the second one not greater than 1 . Without loss of generality, assume that $1 \\leqslant a$ and $b \\leqslant 1$.\n\nTherefore, we obtain $b c \\leqslant 1$ and $1 \\leqslant c a$, or equivalently\n\n$$\n\\frac{A M}{M B} \\leqslant \\frac{L A}{C L} \\quad \\text { and } \\quad \\frac{M B}{A M} \\leqslant \\frac{B K}{K C}\n$$\n\nThe first inequality implies that the line passing through $M$ and parallel to $B C$ intersects the segment $A L$ at a point $X$ (see Figure 1). Therefore the inradius of the triangle $A L M$ is not less than the inradius $r_{1}$ of triangle $A M X$.\n\nSimilarly, the line passing through $M$ and parallel to $A C$ intersects the segment $B K$ at a point $Y$, so the inradius of the triangle $B M K$ is not less than the inradius $r_{2}$ of the triangle $B M Y$. Thus, to complete our solution, it is enough to show that $r_{1}+r_{2} \\geqslant r$, where $r$ is the inradius of the triangle $A B C$. We prove that in fact $r_{1}+r_{2}=r$.\n\n\n\nFigure 1\n\nSince $M X \\| B C$, the dilation with centre $A$ that takes $M$ to $B$ takes the incircle of the triangle $A M X$ to the incircle of the triangle $A B C$. Therefore\n\n$$\n\\frac{r_{1}}{r}=\\frac{A M}{A B}, \\quad \\text { and similarly } \\quad \\frac{r_{2}}{r}=\\frac{M B}{A B}\n$$\n\nAdding these equalities gives $r_{1}+r_{2}=r$, as required.""]",,True,,, 1983,Geometry,,"Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $A B C$ with $A B>B C$. The angle bisector of $\angle A B C$ intersects $\Omega$ at $M \neq B$. Let $\Gamma$ be the circle with diameter $B M$. The angle bisectors of $\angle A O B$ and $\angle B O C$ intersect $\Gamma$ at points $P$ and $Q$, respectively. The point $R$ is chosen on the line $P Q$ so that $B R=M R$. Prove that $B R \| A C$. (Here we always assume that an angle bisector is a ray.)","['Let $K$ be the midpoint of $B M$, i.e., the centre of $\\Gamma$. Notice that $A B \\neq B C$ implies $K \\neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.\n\nLet $N$ be the second point of intersection of $\\Gamma$ with the line $O M$. Since $B M$ is a diameter of $\\Gamma$, the lines $B N$ and $A C$ are both perpendicular to $O M$. Hence $B N \\| A C$, and it suffices to prove that $B N$ passes through $R$. Our plan for doing this is to interpret the lines $B N, O K$, and $P Q$ as the radical axes of three appropriate circles.\n\nLet $\\omega$ be the circle with diameter $B O$. Since $\\angle B N O=\\angle B K O=90^{\\circ}$, the points $N$ and $K$ lie on $\\omega$.\n\nNext we show that the points $O, K, P$, and $Q$ are concyclic. To this end, let $D$ and $E$ be the midpoints of $B C$ and $A B$, respectively. Clearly, $D$ and $E$ lie on the rays $O Q$ and $O P$, respectively. By our assumptions about the triangle $A B C$, the points $B, E, O, K$, and $D$ lie in this order on $\\omega$. It follows that $\\angle E O R=\\angle E B K=\\angle K B D=\\angle K O D$, so the line $K O$ externally bisects the angle $P O Q$. Since the point $K$ is the centre of $\\Gamma$, it also lies on the perpendicular bisector of $P Q$. So $K$ coincides with the midpoint of the arc $P O Q$ of the circumcircle $\\gamma$ of triangle $P O Q$.\n\nThus the lines $O K, B N$, and $P Q$ are pairwise radical axes of the circles $\\omega, \\gamma$, and $\\Gamma$. Hence they are concurrent at $R$, as required.\n\n']",,True,,, 1984,Geometry,,"Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circumcircles of the triangles $A M P$ and $B M C$. Prove that as $P$ varies, the point $Q$ lies on a fixed circle.","[""Throughout the solution, we denote by $\\sphericalangle(a, b)$ the directed angle between the lines $a$ and $b$.\n\nLet $D$ be the point on the segment $A B$ such that $B D=\\lambda \\cdot B A$. We will show that either $Q=D$, or $\\sphericalangle(D Q, Q B)=\\sphericalangle(A B, B C)$; this would mean that the point $Q$ varies over the constant circle through $D$ tangent to $B C$ at $B$, as required.\n\nDenote the circumcircles of the triangles $A M P$ and $B M C$ by $\\omega_{A}$ and $\\omega_{B}$, respectively. The lines $A P, B C$, and $M Q$ are pairwise radical axes of the circles $\\Gamma, \\omega_{A}$, and $\\omega_{B}$, thus either they are parallel, or they share a common point $X$.\n\nAssume that these lines are parallel (see Figure 1). Then the segments $A P, Q M$, and $B C$ have a common perpendicular bisector; the reflection in this bisector maps the segment $C P$ to $B A$, and maps $M$ to $Q$. Therefore, in this case $Q$ lies on $A B$, and $B Q / A B=C M / C P=$ $B D / A B$; so we have $Q=D$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nNow assume that the lines $A P, Q M$, and $B C$ are concurrent at some point $X$ (see Figure 2). Notice that the points $A, B, Q$, and $X$ lie on a common circle $\\Omega$ by MiQUEL's theorem applied to the triangle $X P C$. Let us denote by $Y$ the symmetric image of $X$ about the perpendicular bisector of $A B$. Clearly, $Y$ lies on $\\Omega$, and the triangles $Y A B$ and $\\triangle X B A$ are congruent. Moreover, the triangle $X P C$ is similar to the triangle $X B A$, so it is also similar to the triangle $Y A B$.\n\nNext, the points $D$ and $M$ correspond to each other in similar triangles $Y A B$ and $X P C$, since $B D / B A=C M / C P=\\lambda$. Moreover, the triangles $Y A B$ and $X P C$ are equi-oriented, so $\\sphericalangle(M X, X P)=\\sphericalangle(D Y, Y A)$. On the other hand, since the points $A, Q, X$, and $Y$ lie on $\\Omega$, we have $\\sphericalangle(Q Y, Y A)=\\sphericalangle(M X, X P)$. Therefore, $\\sphericalangle(Q Y, Y A)=\\sphericalangle(D Y, Y A)$, so the points $Y, D$, and $Q$ are collinear.\n\nFinally, we have $\\sphericalangle(D Q, Q B)=\\sphericalangle(Y Q, Q B)=\\sphericalangle(Y A, A B)=\\sphericalangle(A B, B X)=\\sphericalangle(A B, B C)$, as desired."", 'As in the previous solution, we introduce the radical centre $X=A P \\cap B C \\cap M Q$ of the circles $\\omega_{A}, \\omega_{B}$, and $\\Gamma$. Next, we also notice that the points $A, Q, B$, and $X$ lie on a common circle $\\Omega$.\n\nIf the point $P$ lies on the $\\operatorname{arc} B A C$ of $\\Gamma$, then the point $X$ is outside $\\Gamma$, thus the point $Q$ belongs to the ray $X M$, and therefore the points $P, A$, and $Q$ lie on the same side of $B C$. Otherwise, if $P$ lies on the $\\operatorname{arc} B C$ not containing $A$, then $X$ lies inside $\\Gamma$, so $M$ and $Q$ lie on different sides of $B C$; thus again $Q$ and $A$ lie on the same side of $B C$. So, in each case the points $Q$ and $A$ lie on the same side of $B C$.\n\n\n\nFigure 3\n\nNow we prove that the ratio\n\n$$\n\\frac{Q B}{\\sin \\angle Q B C}=\\frac{Q B}{Q X} \\cdot \\frac{Q X}{\\sin \\angle Q B X}\n$$\n\nis constant. Since the points $A, Q, B$, and $X$ are concyclic, we have\n\n$$\n\\frac{Q X}{\\sin \\angle Q B X}=\\frac{A X}{\\sin \\angle A B C}\n$$\n\nNext, since the points $B, Q, M$, and $C$ are concyclic, the triangles $X B Q$ and $X M C$ are similar, so\n\n$$\n\\frac{Q B}{Q X}=\\frac{C M}{C X}=\\lambda \\cdot \\frac{C P}{C X}\n$$\n\nAnalogously, the triangles $X C P$ and $X A B$ are also similar, so\n\n$$\n\\frac{C P}{C X}=\\frac{A B}{A X}\n$$\n\nTherefore, we obtain\n\n$$\n\\frac{Q B}{\\sin \\angle Q B C}=\\lambda \\cdot \\frac{A B}{A X} \\cdot \\frac{A X}{\\sin \\angle A B C}=\\lambda \\cdot \\frac{A B}{\\sin \\angle A B C}\n$$\n\nso this ratio is indeed constant. Thus the circle passing through $Q$ and tangent to $B C$ at $B$ is also constant, and $Q$ varies over this fixed circle.', ""Let us perform an inversion centred at $C$. Denote by $X^{\\prime}$ the image of a point $X$ under this inversion.\n\nThe circle $\\Gamma$ maps to the line $\\Gamma^{\\prime}$ passing through the constant points $A^{\\prime}$ and $B^{\\prime}$, and containing the variable point $P^{\\prime}$. By the problem condition, the point $M$ varies over the circle $\\gamma$ which is the homothetic image of $\\Gamma$ with centre $C$ and coefficient $\\lambda$. Thus $M^{\\prime}$ varies over the constant line $\\gamma^{\\prime} \\| A^{\\prime} B^{\\prime}$ which is the homothetic image of $A^{\\prime} B^{\\prime}$ with centre $C$ and coefficient $1 / \\lambda$, and $M=\\gamma^{\\prime} \\cap C P^{\\prime}$. Next, the circumcircles $\\omega_{A}$ and $\\omega_{B}$ of the triangles $A M P$ and $B M C$ map to the circumcircle $\\omega_{A}^{\\prime}$ of the triangle $A^{\\prime} M^{\\prime} P^{\\prime}$ and to the line $B^{\\prime} M^{\\prime}$, respectively; the point $Q$ thus maps to the second point of intersection of $B^{\\prime} M^{\\prime}$ with $\\omega_{A}^{\\prime}$ (see Figure 4).\n\n\n\nFigure 4\n\nLet $J$ be the (constant) common point of the lines $\\gamma^{\\prime}$ and $C A^{\\prime}$, and let $\\ell$ be the (constant) line through $J$ parallel to $C B^{\\prime}$. Let $V$ be the common point of the lines $\\ell$ and $B^{\\prime} M^{\\prime}$. Applying PAppus' theorem to the triples $\\left(C, J, A^{\\prime}\\right)$ and $\\left(V, B^{\\prime}, M^{\\prime}\\right)$ we get that the points $C B^{\\prime} \\cap J V$, $J M^{\\prime} \\cap A^{\\prime} B^{\\prime}$, and $C M^{\\prime} \\cap A^{\\prime} V$ are collinear. The first two of these points are ideal, hence so is the third, which means that $C M^{\\prime} \\| A^{\\prime} V$.\n\nNow we have $\\sphericalangle\\left(Q^{\\prime} A^{\\prime}, A^{\\prime} P^{\\prime}\\right)=\\sphericalangle\\left(Q^{\\prime} M^{\\prime}, M^{\\prime} P^{\\prime}\\right)=\\angle\\left(V M^{\\prime}, A^{\\prime} V\\right)$, which means that the triangles $B^{\\prime} Q^{\\prime} A^{\\prime}$ and $B^{\\prime} A^{\\prime} V$ are similar, and $\\left(B^{\\prime} A^{\\prime}\\right)^{2}=B^{\\prime} Q^{\\prime} \\cdot B^{\\prime} V$. Thus $Q^{\\prime}$ is the image of $V$ under the second (fixed) inversion with centre $B^{\\prime}$ and radius $B^{\\prime} A^{\\prime}$. Since $V$ varies over the constant line $\\ell, Q^{\\prime}$ varies over some constant circle $\\Theta$. Thus, applying the first inversion back we get that $Q$ also varies over some fixed circle.\n\nOne should notice that this last circle is not a line; otherwise $\\Theta$ would contain $C$, and thus $\\ell$ would contain the image of $C$ under the second inversion. This is impossible, since $C B^{\\prime} \\| \\ell$.""]",,True,,, 1985,Geometry,,"Let $A B C D$ be a convex quadrilateral with $\angle B=\angle D=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $B D$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle $S C T$ and $$ \angle S H C-\angle B S C=90^{\circ}, \quad \angle T H C-\angle D T C=90^{\circ} . $$ Prove that the circumcircle of triangle $S H T$ is tangent to the line $B D$.","['Let the line passing through $C$ and perpendicular to the line $S C$ intersect the line $A B$ at $Q$ (see Figure 1). Then\n\n$$\n\\angle S Q C=90^{\\circ}-\\angle B S C=180^{\\circ}-\\angle S H C\n$$\n\nwhich implies that the points $C, H, S$, and $Q$ lie on a common circle. Moreover, since $S Q$ is a diameter of this circle, we infer that the circumcentre $K$ of triangle $S H C$ lies on the line $A B$. Similarly, we prove that the circumcentre $L$ of triangle $C H T$ lies on the line $A D$.\n\n\n\nFigure 1\n\nIn order to prove that the circumcircle of triangle $S H T$ is tangent to $B D$, it suffices to show that the perpendicular bisectors of $H S$ and $H T$ intersect on the line $A H$. However, these two perpendicular bisectors coincide with the angle bisectors of angles $A K H$ and $A L H$. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that\n\n$$\n\\frac{A K}{K H}=\\frac{A L}{L H}\\tag{1}\n$$\n\nWe present two proofs of this equality.\n\nFirst proof. Let the lines $K L$ and $H C$ intersect at $M$ (see Figure 2). Since $K H=K C$ and $L H=L C$, the points $H$ and $C$ are symmetric to each other with respect to the line $K L$. Therefore $M$ is the midpoint of $H C$. Denote by $O$ the circumcentre of quadrilateral $A B C D$. Then $O$ is the midpoint of $A C$. Therefore we have $O M \\| A H$ and hence $O M \\perp B D$. This together with the equality $O B=O D$ implies that $O M$ is the perpendicular bisector of $B D$ and therefore $B M=D M$.\n\nSince $C M \\perp K L$, the points $B, C, M$, and $K$ lie on a common circle with diameter $K C$. Similarly, the points $L, C, M$, and $D$ lie on a circle with diameter $L C$. Thus, using the sine law, we obtain\n\n$$\n\\frac{A K}{A L}=\\frac{\\sin \\angle A L K}{\\sin \\angle A K L}=\\frac{D M}{C L} \\cdot \\frac{C K}{B M}=\\frac{C K}{C L}=\\frac{K H}{L H}\n$$\n\n\n\nwhich finishes the proof of (1).\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nSecond proof. If the points $A, H$, and $C$ are collinear, then $A K=A L$ and $K H=L H$, so the equality (1) follows. Assume therefore that the points $A, H$, and $C$ do not lie in a line and consider the circle $\\omega$ passing through them (see Figure 3). Since the quadrilateral $A B C D$ is cyclic,\n\n$$\n\\angle B A C=\\angle B D C=90^{\\circ}-\\angle A D H=\\angle H A D .\n$$\n\nLet $N \\neq A$ be the intersection point of the circle $\\omega$ and the angle bisector of $\\angle C A H$. Then $A N$ is also the angle bisector of $\\angle B A D$. Since $H$ and $C$ are symmetric to each other with respect to the line $K L$ and $H N=N C$, it follows that both $N$ and the centre of $\\omega$ lie on the line $K L$. This means that the circle $\\omega$ is an Apollonius circle of the points $K$ and $L$. This immediately yields (1).']",,True,,, 1986,Geometry,,"Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ at $S$ and $T$, respectively. We call the pair $(E, F)$ interesting, if the quadrilateral $K S A T$ is cyclic. Suppose that the pairs $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ are interesting. Prove that $$ \frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C} $$","[""For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting.\n\nLet $E F K$ be an interesting triangle. Firstly, we prove that $\\angle K E F=\\angle K F E=\\angle A$, which also means that the circumcircle $\\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $K F$.\n\nDenote by $\\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $A M$ intersect the line $S T$ and the circle $\\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1).\n\nSince $E F \\| T S$ and $M$ is the midpoint of $E F, N$ is the midpoint of $S T$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $S T$, we have $\\angle K N S=\\angle M N S=$ $\\angle L N T$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $S T$. Therefore $K L \\| S T$.\n\nLet $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $E F$, and we may assume that it lies on the ray $M F$. One has\n\n$$\n\\angle K G E=\\angle K N S=\\angle S N M=\\angle K L A=180^{\\circ}-\\angle K S A\n$$\n\n(if $K=L$, then the angle $K L A$ is understood to be the angle between $A L$ and the tangent to $\\omega$ at $L$ ). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $K S G T$ is a parallelogram, we obtain $\\angle K E F=\\angle K S G=180^{\\circ}-\\angle T K S=\\angle A$. Since $K E=K F$, we also have $\\angle K F E=\\angle K E F=\\angle A$.\n\nAfter having proved this fact, one may finish the solution by different methods.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nFirst method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.\n\n\n\nLemma. Let $A B C$ be an arbitrary triangle. Choose two points $E_{1}$ and $E_{2}$ on the side $A C$, two points $F_{1}$ and $F_{2}$ on the side $A B$, and two points $K_{1}$ and $K_{2}$ on the side $B C$, in a way that the triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ are similar. Then the six circumcircles of the triangles $A E_{i} F_{i}$, $B F_{i} K_{i}$, and $C E_{i} K_{i}(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_{1} F_{1} K_{1}$ to the triangle $E_{2} F_{2} K_{2}$.\n\nProof. Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $A E_{i} F_{i}, B F_{i} K_{i}$, and $C K_{i} E_{i}$ have a common point $Z_{i}$ by MiqueL's theorem. Moreover, we have\n\n$\\sphericalangle\\left(Z_{i} F_{i}, Z_{i} E_{i}\\right)=\\sphericalangle(A B, C A), \\quad \\sphericalangle\\left(Z_{i} K_{i}, Z_{i} F_{i}\\right)=\\sphericalangle(B C, A B), \\quad \\sphericalangle\\left(Z_{i} E_{i}, Z_{i} K_{i}\\right)=\\sphericalangle(C A, B C)$.\n\nThis yields that the points $Z_{1}$ and $Z_{2}$ correspond to each other in similar triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.\n\nFinally, in order to show that $Z_{1}=Z_{2}$, one may notice that $\\sphericalangle\\left(A B, A Z_{1}\\right)=\\sphericalangle\\left(E_{1} F_{1}, E_{1} Z_{1}\\right)=$ $\\sphericalangle\\left(E_{2} F_{2}, E_{2} Z_{2}\\right)=\\sphericalangle\\left(A B, A Z_{2}\\right)$ (see Figure 2$)$. Similarly, one has $\\sphericalangle\\left(B C, B Z_{1}\\right)=\\sphericalangle\\left(B C, B Z_{2}\\right)$ and $\\sphericalangle\\left(C A, C Z_{1}\\right)=\\sphericalangle\\left(C A, C Z_{2}\\right)$. This yields $Z_{1}=Z_{2}$.\n\nNow, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively, and let $R$ be the midpoint of $B C$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $B C P Q$. Thus we obtain $\\angle A P Q=\\angle B$ and $\\angle R P C=\\angle C$, which yields $\\angle Q P R=\\angle A$. Similarly, we show that $\\angle P Q R=\\angle A$. Thus, all interesting triangles are similar to the triangle $P Q R$.\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nDenote now by $Z$ the common point of the circumcircles of $A P Q, B Q R$, and $C P R$. Let $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ be two interesting triangles. By the lemma, $Z$ is the centre of any spiral similarity taking one of the triangles $E_{1} F_{1} K_{1}, E_{2} F_{2} K_{2}$, and $P Q R$ to some other of them. Therefore the triangles $Z E_{1} E_{2}$ and $Z F_{1} F_{2}$ are similar, as well as the triangles $Z E_{1} F_{1}$ and $Z P Q$. Hence\n\n$$\n\\frac{E_{1} E_{2}}{F_{1} F_{2}}=\\frac{Z E_{1}}{Z F_{1}}=\\frac{Z P}{Z Q}\n$$\n\nMoreover, the equalities $\\angle A Z Q=\\angle A P Q=\\angle A B C=180^{\\circ}-\\angle Q Z R$ show that the point $Z$ lies on the line $A R$ (see Figure 4). Therefore the triangles $A Z P$ and $A C R$ are similar, as well as the triangles $A Z Q$ and $A B R$. This yields\n\n$$\n\\frac{Z P}{Z Q}=\\frac{Z P}{R C} \\cdot \\frac{R B}{Z Q}=\\frac{A Z}{A C} \\cdot \\frac{A B}{A Z}=\\frac{A B}{A C},\n$$\n\nwhich completes the solution.\n\n\n\nSecond method. Now we will start from the fact that $\\omega_{1}$ is tangent to the lines $K E$ and $K F$ (see Figure 5$)$. We prove that if $(E, F)$ is an interesting pair, then\n\n$$\n\\frac{A E}{A B}+\\frac{A F}{A C}=2 \\cos \\angle A\n\\tag{1}\n$$\n\nLet $Y$ be the intersection point of the segments $B E$ and $C F$. The points $B, K$, and $C$ are collinear, hence applying PASCAL's theorem to the degenerated hexagon $A F F Y E E$, we infer that $Y$ lies on the circle $\\omega_{1}$.\n\nDenote by $Z$ the second intersection point of the circumcircle of the triangle $B F Y$ with the line $B C$ (see Figure 6). By Miquel's theorem, the points $C, Z, Y$, and $E$ are concyclic. Therefore we obtain\n\n$$\nB F \\cdot A B+C E \\cdot A C=B Y \\cdot B E+C Y \\cdot C F=B Z \\cdot B C+C Z \\cdot B C=B C^{2}\n$$\n\nOn the other hand, $B C^{2}=A B^{2}+A C^{2}-2 A B \\cdot A C \\cos \\angle A$, by the cosine law. Hence\n\n$$\n(A B-A F) \\cdot A B+(A C-A E) \\cdot A C=A B^{2}+A C^{2}-2 A B \\cdot A C \\cos \\angle A\n$$\n\nwhich simplifies to the desired equality (1).\n\nLet now $\\left(E_{1}, F_{1}\\right)$ and $\\left(E_{2}, F_{2}\\right)$ be two interesting pairs of points. Then we get\n\n$$\n\\frac{A E_{1}}{A B}+\\frac{A F_{1}}{A C}=\\frac{A E_{2}}{A B}+\\frac{A F_{2}}{A C}\n$$\n\nwhich gives the desired result.\n\n\n\nFigure 5\n\n\n\nFigure 6\n\nThird method. Again, we make use of the fact that all interesting triangles are similar (and equi-oriented). Let us put the picture onto a complex plane such that $A$ is at the origin, and identify each point with the corresponding complex number.\n\nLet $E F K$ be any interesting triangle. The equalities $\\angle K E F=\\angle K F E=\\angle A$ yield that the ratio $\\nu=\\frac{K-E}{F-E}$ is the same for all interesting triangles. This in turn means that the numbers $E$, $F$, and $K$ satisfy the linear equation\n\n$$\nK=\\mu E+\\nu F, \\quad \\text { where } \\quad \\mu=1-\\nu\n\\tag{2}\n$$\n\n\n\nNow let us choose the points $X$ and $Y$ on the rays $A B$ and $A C$, respectively, so that $\\angle C X A=\\angle A Y B=\\angle A=\\angle K E F$ (see Figure 7). Then each of the triangles $A X C$ and $Y A B$ is similar to any interesting triangle, which also means that\n\n$$\nC=\\mu A+\\nu X=\\nu X \\quad \\text { and } \\quad B=\\mu Y+\\nu A=\\mu Y\n\\tag{3}\n$$\n\nMoreover, one has $X / Y=\\overline{C / B}$.\n\nSince the points $E, F$, and $K$ lie on $A C, A B$, and $B C$, respectively, one gets\n\n$$\nE=\\rho Y, \\quad F=\\sigma X, \\quad \\text { and } \\quad K=\\lambda B+(1-\\lambda) C\n$$\n\nfor some real $\\rho, \\sigma$, and $\\lambda$. In view of (3), the equation (2) now reads $\\lambda B+(1-\\lambda) C=K=$ $\\mu E+\\nu F=\\rho B+\\sigma C$, or\n\n$$\n(\\lambda-\\rho) B=(\\sigma+\\lambda-1) C .\n$$\n\nSince the nonzero complex numbers $B$ and $C$ have different arguments, the coefficients in the brackets vanish, so $\\rho=\\lambda$ and $\\sigma=1-\\lambda$. Therefore,\n\n$$\n\\frac{E}{Y}+\\frac{F}{X}=\\rho+\\sigma=1\n\\tag{4}\n$$\n\nNow, if $\\left(E_{1}, F_{1}\\right)$ and $\\left(E_{2}, F_{2}\\right)$ are two distinct interesting pairs, one may apply (4) to both pairs. Subtracting, we get\n\n$$\n\\frac{E_{1}-E_{2}}{Y}=\\frac{F_{2}-F_{1}}{X}, \\quad \\text { so } \\quad \\frac{E_{1}-E_{2}}{F_{2}-F_{1}}=\\frac{Y}{X}=\\frac{\\bar{B}}{\\bar{C}}\n$$\n\nTaking absolute values provides the required result.\n\n\n\nFigure 7"", ""Let $(E, F)$ be an interesting pair. This time we prove that\n\n$$\n\\frac{A M}{A K}=\\cos \\angle A\n\\tag{5}\n$$\n\nWe introduce the circle $\\omega$ passing through the points $K, S$, $A$, and $T$, together with the points $N$ and $L$ at which the line $A M$ intersect the line $S T$ and the circle $\\omega$ for the second time, respectively. Let moreover $O$ be the centre of $\\omega$ (see Figures 8 and 9). we note that $N$ is the midpoint of $S T$ and show that $K L \\| S T$, which implies $\\angle F A M=\\angle E A K$.\n\n\n\nFigure 8\n\n\n\nFigure 9\n\nSuppose now that $K \\neq L$ (see Figure 8). Then $K L \\| S T$, and consequently the lines $K M$ and $K L$ are perpendicular. It implies that the lines $L O$ and $K M$ meet at a point $X$ lying on the circle $\\omega$. Since the lines $O N$ and $X M$ are both perpendicular to the line $S T$, they are parallel to each other, and hence $\\angle L O N=\\angle L X K=\\angle M A K$. On the other hand, $\\angle O L N=\\angle M K A$, so we infer that triangles $N O L$ and $M A K$ are similar. This yields\n\n$$\n\\frac{A M}{A K}=\\frac{O N}{O L}=\\frac{O N}{O T}=\\cos \\angle T O N=\\cos \\angle A\n$$\n\nIf, on the other hand, $K=L$, then the points $A, M, N$, and $K$ lie on a common line, and this line is the perpendicular bisector of $S T$ (see Figure 9). This implies that $A K$ is a diameter of $\\omega$, which yields $A M=2 O K-2 N K=2 O N$. So also in this case we obtain\n\n$$\n\\frac{A M}{A K}=\\frac{2 O N}{2 O T}=\\cos \\angle T O N=\\cos \\angle A\n$$\n\nThus (5) is proved.\n\nLet $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively (see Figure 10). We claim that the point $M$ lies on the line $P Q$. Consider now the composition of the dilatation with factor $\\cos \\angle A$ and centre $A$, and the reflection with respect to the angle bisector of $\\angle B A C$. This transformation is a similarity that takes $B, C$, and $K$ to $P, Q$, and $M$, respectively. Since $K$ lies on the line $B C$, the point $M$ lies on the line $P Q$.\n\n\n\n\n\nFigure 10\n\nSuppose that $E \\neq P$. Then also $F \\neq Q$, and by Menelaus' theorem, we obtain\n\n$$\n\\frac{A Q}{F Q} \\cdot \\frac{F M}{E M} \\cdot \\frac{E P}{A P}=1\n$$\n\nUsing the similarity of the triangles $A P Q$ and $A B C$, we infer that\n\n$$\n\\frac{E P}{F Q}=\\frac{A P}{A Q}=\\frac{A B}{A C}, \\quad \\text { and hence } \\quad \\frac{E P}{A B}=\\frac{F Q}{A C}\n$$\n\nThe last equality holds obviously also in case $E=P$, because then $F=Q$. Moreover, since the line $P Q$ intersects the segment $E F$, we infer that the point $E$ lies on the segment $A P$ if and only if the point $F$ lies outside of the segment $A Q$.\n\nLet now $\\left(E_{1}, F_{1}\\right)$ and $\\left(E_{2}, F_{2}\\right)$ be two interesting pairs. Then we obtain\n\n$$\n\\frac{E_{1} P}{A B}=\\frac{F_{1} Q}{A C} \\quad \\text { and } \\quad \\frac{E_{2} P}{A B}=\\frac{F_{2} Q}{A C}\n$$\n\nIf $P$ lies between the points $E_{1}$ and $E_{2}$, we add the equalities above, otherwise we subtract them. In any case we obtain\n\n$$\n\\frac{E_{1} E_{2}}{A B}=\\frac{F_{1} F_{2}}{A C}\n$$\n\nwhich completes the solution.""]",,True,,, 1987,Geometry,,"Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C$ (not containing $A$ ) of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ intersect $A V$ at $X$, and let the line passing through $V$ and parallel to $A I$ intersect $A B$ at $Y$. Let $W$ and $Z$ be the midpoints of $A X$ and $B C$, respectively. Prove that if the points $I, X$, and $Y$ are collinear, then the points $I, W$, and $Z$ are also collinear.","['We start with some general observations. Set $\\alpha=\\angle A / 2, \\beta=\\angle B / 2, \\gamma=\\angle C / 2$. Then obviously $\\alpha+\\beta+\\gamma=90^{\\circ}$. Since $\\angle U I C=90^{\\circ}$, we obtain $\\angle I U C=\\alpha+\\beta$. Therefore $\\angle B I V=\\angle I U C-\\angle I B C=\\alpha=\\angle B A I=\\angle B Y V$, which implies that the points $B, Y, I$, and $V$ lie on a common circle (see Figure 1).\n\nAssume now that the points $I, X$ and $Y$ are collinear. We prove that $\\angle Y I A=90^{\\circ}$.\n\nLet the line $X U$ intersect $A B$ at $N$. Since the lines $A I, U X$, and $V Y$ are parallel, we get\n\n$$\n\\frac{N X}{A I}=\\frac{Y N}{Y A}=\\frac{V U}{V I}=\\frac{X U}{A I}\n$$\n\nimplying $N X=X U$. Moreover, $\\angle B I U=\\alpha=\\angle B N U$. This implies that the quadrilateral $B U I N$ is cyclic, and since $B I$ is the angle bisector of $\\angle U B N$, we infer that $N I=U I$. Thus in the isosceles triangle $N I U$, the point $X$ is the midpoint of the base $N U$. This gives $\\angle I X N=90^{\\circ}$, i.e., $\\angle Y I A=90^{\\circ}$.\n\n\n\nFigure 1\n\nLet $S$ be the midpoint of the segment $V C$. Let moreover $T$ be the intersection point of the lines $A X$ and $S I$, and set $x=\\angle B A V=\\angle B C V$. Since $\\angle C I A=90^{\\circ}+\\beta$ and $S I=S C$, we obtain\n\n$$\n\\angle T I A=180^{\\circ}-\\angle A I S=90^{\\circ}-\\beta-\\angle C I S=90^{\\circ}-\\beta-\\gamma-x=\\alpha-x=\\angle T A I,\n$$\n\nwhich implies that $T I=T A$. Therefore, since $\\angle X I A=90^{\\circ}$, the point $T$ is the midpoint of $A X$, i.e., $T=W$.\n\nTo complete our solution, it remains to show that the intersection point of the lines $I S$ and $B C$ coincide with the midpoint of the segment $B C$. But since $S$ is the midpoint of the segment $V C$, it suffices to show that the lines $B V$ and $I S$ are parallel.\n\n\n\nSince the quadrilateral $B Y I V$ is cyclic, $\\angle V B I=\\angle V Y I=\\angle Y I A=90^{\\circ}$. This implies that $B V$ is the external angle bisector of the angle $A B C$, which yields $\\angle V A C=\\angle V C A$. Therefore $2 \\alpha-x=2 \\gamma+x$, which gives $\\alpha=\\gamma+x$. Hence $\\angle S C I=\\alpha$, so $\\angle V S I=2 \\alpha$.\n\nOn the other hand, $\\angle B V C=180^{\\circ}-\\angle B A C=180^{\\circ}-2 \\alpha$, which implies that the lines $B V$ and $I S$ are parallel. This completes the solution.', 'As in', '1, we first prove that the points $B, Y, I, V$ lie on a common circle and $\\angle Y I A=90^{\\circ}$. The remaining part of the solution is based on the following lemma, which holds true for any triangle $A B C$, not necessarily with the property that $I, X, Y$ are collinear.\n\nLemma. Let $A B C$ be the triangle inscribed in a circle $\\Gamma$ and let $I$ be its incentre. Assume that the line passing through $I$ and perpendicular to the line $A I$ intersects the side $A B$ at the point $Y$. Let the circumcircle of the triangle $B Y I$ intersect the circle $\\Gamma$ for the second time at $V$, and let the excircle of the triangle $A B C$ opposite to the vertex $A$ be tangent to the side $B C$ at $E$. Then\n\n$$\n\\angle B A V=\\angle C A E .\n$$\n\nProof. Let $\\rho$ be the composition of the inversion with centre $A$ and radius $\\sqrt{A B \\cdot A C}$, and the symmetry with respect to $A I$. Clearly, $\\rho$ interchanges $B$ and $C$.\n\nLet $J$ be the excentre of the triangle $A B C$ opposite to $A$ (see Figure 2). Then we have $\\angle J A C=\\angle B A I$ and $\\angle J C A=90^{\\circ}+\\gamma=\\angle B I A$, so the triangles $A C J$ and $A I B$ are similar, and therefore $A B \\cdot A C=A I \\cdot A J$. This means that $\\rho$ interchanges $I$ and $J$. Moreover, since $Y$ lies on $A B$ and $\\angle A I Y=90^{\\circ}$, the point $Y^{\\prime}=\\rho(Y)$ lies on $A C$, and $\\angle J Y^{\\prime} A=90^{\\circ}$. Thus $\\rho$ maps the circumcircle $\\gamma$ of the triangle $B Y I$ to a circle $\\gamma^{\\prime}$ with diameter $J C$.\n\nFinally, since $V$ lies on both $\\Gamma$ and $\\gamma$, the point $V^{\\prime}=\\rho(V)$ lies on the line $\\rho(\\Gamma)=A B$ as well as on $\\gamma^{\\prime}$, which in turn means that $V^{\\prime}=E$. This implies the desired result.\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nNow we turn to the solution of the problem.\n\nAssume that the incircle $\\omega_{1}$ of the triangle $A B C$ is tangent to $B C$ at $D$, and let the excircle $\\omega_{2}$ of the triangle $A B C$ opposite to the vertex $A$ touch the side $B C$ at $E$ (see Figure 3 ). The homothety with centre $A$ that takes $\\omega_{2}$ to $\\omega_{1}$ takes the point $E$ to some point $F$, and the\n\n\n\ntangent to $\\omega_{1}$ at $F$ is parallel to $B C$. Therefore $D F$ is a diameter of $\\omega_{1}$. Moreover, $Z$ is the midpoint of $D E$. This implies that the lines $I Z$ and $F E$ are parallel.\n\nLet $K=Y I \\cap A E$. Since $\\angle Y I A=90^{\\circ}$, the lemma yields that $I$ is the midpoint of $X K$. This implies that the segments $I W$ and $A K$ are parallel. Therefore, the points $W, I$ and $Z$ are collinear.']",,True,,, 1988,Number Theory,,"Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k2$, and take an integer $m>(n-2) 2^{n}+1$. If $m$ is even, then consider\n\n$$\n\\frac{m}{2} \\geqslant \\frac{(n-2) 2^{n}+2}{2}=(n-2) 2^{n-1}+1>(n-3) 2^{n-1}+1\n$$\n\nBy the induction hypothesis, there is a representation of the form\n\n$$\n\\frac{m}{2}=\\left(2^{n-1}-2^{k_{1}}\\right)+\\left(2^{n-1}-2^{k_{2}}\\right)+\\cdots+\\left(2^{n-1}-2^{k_{r}}\\right)\n$$\n\nfor some $k_{i}$ with $0 \\leqslant k_{i}\\frac{(n-2) 2^{n}+1-\\left(2^{n}-1\\right)}{2}=(n-3) 2^{n-1}+1\n$$\n\nBy the induction hypothesis, there is a representation of the form\n\n$$\n\\frac{m-\\left(2^{n}-1\\right)}{2}=\\left(2^{n-1}-2^{k_{1}}\\right)+\\left(2^{n-1}-2^{k_{2}}\\right)+\\cdots+\\left(2^{n-1}-2^{k_{r}}\\right)\n$$\n\nfor some $k_{i}$ with $0 \\leqslant k_{i}2$, assume that there exist integers $a, b$ with $a \\geqslant 0, b \\geqslant 1$ and $a+by$ due to symmetry. Then the integer $n=x-y$ is positive and (1) may be rewritten as\n\n$$\n\\sqrt[3]{7(y+n)^{2}-13(y+n) y+7 y^{2}}=n+1\n$$\n\nRaising this to the third power and simplifying the result one obtains\n\n$$\ny^{2}+y n=n^{3}-4 n^{2}+3 n+1\n$$\n\nTo complete the square on the left hand side, we multiply by 4 and add $n^{2}$, thus getting\n\n$$\n(2 y+n)^{2}=4 n^{3}-15 n^{2}+12 n+4=(n-2)^{2}(4 n+1) .\n$$\n\nThis shows that the cases $n=1$ and $n=2$ are impossible, whence $n>2$, and $4 n+1$ is the square of the rational number $\\frac{2 y+n}{n-2}$. Consequently, it has to be a perfect square, and, since it is odd as well, there has to exist some nonnegative integer $m$ such that $4 n+1=(2 m+1)^{2}$, i.e.\n\n$$\nn=m^{2}+m \\text {. }\n$$\n\nNotice that $n>2$ entails $m \\geqslant 2$. Substituting the value of $n$ just found into the previous displayed equation we arrive at\n\n$$\n\\left(2 y+m^{2}+m\\right)^{2}=\\left(m^{2}+m-2\\right)^{2}(2 m+1)^{2}=\\left(2 m^{3}+3 m^{2}-3 m-2\\right)^{2} .\n$$\n\nExtracting square roots and taking $2 m^{3}+3 m^{2}-3 m-2=(m-1)\\left(2 m^{2}+5 m+2\\right)>0$ into account we derive $2 y+m^{2}+m=2 m^{3}+3 m^{2}-3 m-2$, which in turn yields\n\n$$\ny=m^{3}+m^{2}-2 m-1 \\text {. }\n$$\n\nNotice that $m \\geqslant 2$ implies that $y=\\left(m^{3}-1\\right)+(m-2) m$ is indeed positive, as it should be. In view of $x=y+n=y+m^{2}+m$ it also follows that\n\n$$\nx=m^{3}+2 m^{2}-m-1,\n$$\n\nand that this integer is positive as well.']","['Either $(x, y)=(1,1)$ or $\\{x, y\\}=\\left\\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\\right\\}$ for some positive integer $m \\geqslant 2$']",True,,Need_human_evaluate, 1990,Number Theory,,"A coin is called a Cape Town coin if its value is $1 / n$ for some positive integer $n$. Given a collection of Cape Town coins of total value at most $99+\frac{1}{2}$, prove that it is possible to split this collection into at most 100 groups each of total value at most 1 .","['We will show that for every positive integer $N$ any collection of Cape Town coins of total value at most $N-\\frac{1}{2}$ can be split into $N$ groups each of total value at most 1 . The problem statement is a particular case for $N=100$.\n\nWe start with some preparations. If several given coins together have a total value also of the form $\\frac{1}{k}$ for a positive integer $k$, then we may merge them into one new coin. Clearly, if the resulting collection can be split in the required way then the initial collection can also be split.\n\nAfter each such merging, the total number of coins decreases, thus at some moment we come to a situation when no more merging is possible. At this moment, for every even $k$ there is at most one coin of value $\\frac{1}{k}$ (otherwise two such coins may be merged), and for every odd $k>1$ there are at most $k-1$ coins of value $\\frac{1}{k}$ (otherwise $k$ such coins may also be merged).\n\nNow, clearly, each coin of value 1 should form a single group; if there are $d$ such coins then we may remove them from the collection and replace $N$ by $N-d$. So from now on we may assume that there are no coins of value 1 .\n\nFinally, we may split all the coins in the following way. For each $k=1,2, \\ldots, N$ we put all the coins of values $\\frac{1}{2 k-1}$ and $\\frac{1}{2 k}$ into a group $G_{k}$; the total value of $G_{k}$ does not exceed\n\n$$\n(2 k-2) \\cdot \\frac{1}{2 k-1}+\\frac{1}{2 k}<1\n$$\n\nIt remains to distribute the ""small"" coins of values which are less than $\\frac{1}{2 N}$; we will add them one by one. In each step, take any remaining small coin. The total value of coins in the groups at this moment is at most $N-\\frac{1}{2}$, so there exists a group of total value at most $\\frac{1}{N}\\left(N-\\frac{1}{2}\\right)=1-\\frac{1}{2 N}$; thus it is possible to put our small coin into this group. Acting so, we will finally distribute all the coins.']",,True,,, 1991,Number Theory,,"Let $n>1$ be a given integer. Prove that infinitely many terms of the sequence $\left(a_{k}\right)_{k \geqslant 1}$, defined by $$ a_{k}=\left\lfloor\frac{n^{k}}{k}\right\rfloor $$ are odd. (For a real number $x,\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.)","[""If $n$ is odd, let $k=n^{m}$ for $m=1,2, \\ldots$ Then $a_{k}=n^{n^{m}-m}$, which is odd for each $m$.\n\nHenceforth, assume that $n$ is even, say $n=2 t$ for some integer $t \\geqslant 1$. Then, for any $m \\geqslant 2$, the integer $n^{2^{m}}-2^{m}=2^{m}\\left(2^{2^{m}-m} \\cdot t^{2^{m}}-1\\right)$ has an odd prime divisor $p$, since $2^{m}-m>1$. Then, for $k=p \\cdot 2^{m}$, we have\n\n$$\nn^{k}=\\left(n^{2^{m}}\\right)^{p} \\equiv\\left(2^{m}\\right)^{p}=\\left(2^{p}\\right)^{m} \\equiv 2^{m}\n$$\n\nwhere the congruences are taken modulo $p\\left(\\right.$ recall that $2^{p} \\equiv 2(\\bmod p)$, by FERMAT's little theorem). Also, from $n^{k}-2^{m}m$ ). Note that for different values of $m$, we get different values of $k$, due to the different powers of 2 in the prime factorisation of $k$."", 'Treat the (trivial) case when $n$ is odd.\nNow assume that $n$ is even and $n>2$. Let $p$ be a prime divisor of $n-1$.\n\nProceed by induction on $i$ to prove that $p^{i+1}$ is a divisor of $n^{p^{i}}-1$ for every $i \\geqslant 0$. The case $i=0$ is true by the way in which $p$ is chosen. Suppose the result is true for some $i \\geqslant 0$. The factorisation\n\n$$\nn^{p^{i+1}}-1=\\left(n^{p^{i}}-1\\right)\\left[n^{p^{i}(p-1)}+n^{p^{i}(p-2)}+\\cdots+n^{p^{i}}+1\\right]\n$$\n\ntogether with the fact that each of the $p$ terms between the square brackets is congruent to 1 modulo $p$, implies that the result is also true for $i+1$.\n\nHence $\\left\\lfloor\\frac{n^{p^{i}}}{p^{i}}\\right\\rfloor=\\frac{n^{p^{i}}-1}{p^{i}}$, an odd integer for each $i \\geqslant 1$.\n\nFinally, we consider the case $n=2$. We observe that $3 \\cdot 4^{i}$ is a divisor of $2^{3 \\cdot 4^{i}}-4^{i}$ for every $i \\geqslant 1$ : Trivially, $4^{i}$ is a divisor of $2^{3 \\cdot 4^{i}}-4^{i}$, since $3 \\cdot 4^{i}>2 i$. Furthermore, since $2^{3 \\cdot 4^{i}}$ and $4^{i}$ are both congruent to 1 modulo 3 , we have $3 \\mid 2^{3 \\cdot 4^{i}}-4^{i}$. Hence, $\\left\\lfloor\\frac{2^{3 \\cdot 4^{i}}}{3 \\cdot 4^{i}}\\right\\rfloor=\\frac{2^{3 \\cdot 4^{i}}-4^{i}}{3 \\cdot 4^{i}}=\\frac{2^{3 \\cdot 4^{i}-2 i}-1}{3}$, which is odd for every $i \\geqslant 1$.', 'Treat the (trivial) case when $n$ is odd.\nLet $n$ be even, and let $p$ be a prime divisor of $n+1$. Define the sequence $\\left(a_{i}\\right)_{i \\geqslant 1}$ by\n\n$$\na_{i}=\\min \\left\\{a \\in \\mathbb{Z}_{>0}: 2^{i} \\text { divides } a p+1\\right\\}\n$$\n\nRecall that there exists $a$ with $1 \\leqslant a<2^{i}$ such that $a p \\equiv-1\\left(\\bmod 2^{i}\\right)$, so each $a_{i}$ satisfies $1 \\leqslant a_{i}<2^{i}$. This implies that $a_{i} p+1

2$, and we let $a$ and $b$ be positive integers such that $x^{p-1}+y=p^{a}$ and $x+y^{p-1}=p^{b}$. Assume further, without loss of generality, that $x \\leqslant y$, so that $p^{a}=x^{p-1}+y \\leqslant x+y^{p-1}=p^{b}$, which means that $a \\leqslant b$ (and thus $\\left.p^{a} \\mid p^{b}\\right)$.\n\nNow we have\n\n$$\np^{b}=y^{p-1}+x=\\left(p^{a}-x^{p-1}\\right)^{p-1}+x .\n$$\n\nWe take this equation modulo $p^{a}$ and take into account that $p-1$ is even, which gives us\n\n$$\n0 \\equiv x^{(p-1)^{2}}+x \\quad\\left(\\bmod p^{a}\\right)\n$$\n\nIf $p \\mid x$, then $p^{a} \\mid x$, since $x^{(p-1)^{2}-1}+1$ is not divisible by $p$ in this case. However, this is impossible, since $x \\leqslant x^{p-1}2$. Thus $a=r+1$. Now since $p^{r} \\leqslant x+1$, we get\n\n$$\nx=\\frac{x^{2}+x}{x+1} \\leqslant \\frac{x^{p-1}+y}{x+1}=\\frac{p^{a}}{x+1} \\leqslant \\frac{p^{a}}{p^{r}}=p\n$$\n\nso we must have $x=p-1$ for $p$ to divide $x+1$.\n\nIt follows that $r=1$ and $a=2$. If $p \\geqslant 5$, we obtain\n\n$$\np^{a}=x^{p-1}+y>(p-1)^{4}=\\left(p^{2}-2 p+1\\right)^{2}>(3 p)^{2}>p^{2}=p^{a}\n$$\n\na contradiction. So the only case that remains is $p=3$, and indeed $x=2$ and $y=p^{a}-x^{p-1}=5$ satisfy the conditions."", ""Again, we can focus on the case that $p>2$. If $p \\mid x$, then also $p \\mid y$. In this case, let $p^{k}$ and $p^{\\ell}$ be the highest powers of $p$ that divide $x$ and $y$ respectively, and assume without loss of generality that $k \\leqslant \\ell$. Then $p^{k}$ divides $x+y^{p-1}$ while $p^{k+1}$ does not, but $p^{k}p$, so $x^{p-1}+y$ and $y^{p-1}+x$ are both at least equal to $p^{2}$. Now we have\n\n$$\nx^{p-1} \\equiv-y \\quad\\left(\\bmod p^{2}\\right) \\quad \\text { and } \\quad y^{p-1} \\equiv-x \\quad\\left(\\bmod p^{2}\\right)\n$$\n\nThese two congruences, together with the EulER-FERMAT theorem, give us\n\n$$\n1 \\equiv x^{p(p-1)} \\equiv(-y)^{p} \\equiv-y^{p} \\equiv x y \\quad\\left(\\bmod p^{2}\\right)\n$$\n\nSince $x \\equiv y \\equiv-1(\\bmod p), x-y$ is divisible by $p$, so $(x-y)^{2}$ is divisible by $p^{2}$. This means that\n\n$$\n(x+y)^{2}=(x-y)^{2}+4 x y \\equiv 4 \\quad\\left(\\bmod p^{2}\\right)\n$$\n\nso $p^{2}$ divides $(x+y-2)(x+y+2)$. We already know that $x+y \\equiv-2(\\bmod p)$, so $x+y-2 \\equiv$ $-4 \\not \\equiv 0(\\bmod p)$. This means that $p^{2}$ divides $x+y+2$.\n\nUsing the same notation as in the first solution, we subtract the two original equations to obtain\n\n$$\np^{b}-p^{a}=y^{p-1}-x^{p-1}+x-y=(y-x)\\left(y^{p-2}+y^{p-3} x+\\cdots+x^{p-2}-1\\right) .\\tag{1}\n$$\n\nThe second factor is symmetric in $x$ and $y$, so it can be written as a polynomial of the elementary symmetric polynomials $x+y$ and $x y$ with integer coefficients. In particular, its value modulo\n\n\n\n$p^{2}$ is characterised by the two congruences $x y \\equiv 1\\left(\\bmod p^{2}\\right)$ and $x+y \\equiv-2\\left(\\bmod p^{2}\\right)$. Since both congruences are satisfied when $x=y=-1$, we must have\n\n$$\ny^{p-2}+y^{p-3} x+\\cdots+x^{p-2}-1 \\equiv(-1)^{p-2}+(-1)^{p-3}(-1)+\\cdots+(-1)^{p-2}-1 \\quad\\left(\\bmod p^{2}\\right)\n$$\n\nwhich simplifies to $y^{p-2}+y^{p-3} x+\\cdots+x^{p-2}-1 \\equiv-p\\left(\\bmod p^{2}\\right)$. Thus the second factor in (1) is divisible by $p$, but not $p^{2}$.\n\nThis means that $p^{a-1}$ has to divide the other factor $y-x$. It follows that\n\n$$\n0 \\equiv x^{p-1}+y \\equiv x^{p-1}+x \\equiv x(x+1)\\left(x^{p-3}-x^{p-4}+\\cdots+1\\right) \\quad\\left(\\bmod p^{a-1}\\right) .\n$$\n\nSince $x \\equiv-1(\\bmod p)$, the last factor is $x^{p-3}-x^{p-4}+\\cdots+1 \\equiv p-2(\\bmod p)$ and in particular not divisible by $p$. We infer that $p^{a-1} \\mid x+1$ and continue as in the first solution.""]","['$(p, x, y) \\in\\{(3,2,5),(3,5,2)\\} \\cup\\left\\{\\left(2, n, 2^{k}-n\\right) \\mid 0x_{n}\\tag{2}\n$$\n\nholds for all integers $n \\geqslant 0$, it is also strictly increasing. Since $x_{n+1}$ is by (1) coprime to $c$ for any $n \\geqslant 0$, it suffices to prove that for each $n \\geqslant 2$ there exists a prime number $p$ dividing $x_{n}$ but none of the numbers $x_{1}, \\ldots, x_{n-1}$. Let us begin by establishing three preliminary claims.\n\nClaim 1. If $i \\equiv j(\\bmod m)$ holds for some integers $i, j \\geqslant 0$ and $m \\geqslant 1$, then $x_{i} \\equiv x_{j}\\left(\\bmod x_{m}\\right)$ holds as well.\n\nProof. Evidently, it suffices to show $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}\\right)$ for all integers $i \\geqslant 0$ and $m \\geqslant 1$. For this purpose we may argue for fixed $m$ by induction on $i$ using $x_{0}=0$ in the base case $i=0$. Now, if we have $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}\\right)$ for some integer $i$, then the recursive equation (1) yields\n\n$$\nx_{i+m+1} \\equiv c^{2}\\left(x_{i+m}^{3}-4 x_{i+m}^{2}+5 x_{i+m}\\right)+1 \\equiv c^{2}\\left(x_{i}^{3}-4 x_{i}^{2}+5 x_{i}\\right)+1 \\equiv x_{i+1} \\quad\\left(\\bmod x_{m}\\right),\n$$\n\nwhich completes the induction.\n\nClaim 2. If the integers $i, j \\geqslant 2$ and $m \\geqslant 1$ satisfy $i \\equiv j(\\bmod m)$, then $x_{i} \\equiv x_{j}\\left(\\bmod x_{m}^{2}\\right)$ holds as well.\n\nProof. Again it suffices to prove $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}^{2}\\right)$ for all integers $i \\geqslant 2$ and $m \\geqslant 1$. As above, we proceed for fixed $m$ by induction on $i$. The induction step is again easy using (1), but this time the base case $i=2$ requires some calculation. Set $L=5 c^{2}$. By (1) we have $x_{m+1} \\equiv L x_{m}+1\\left(\\bmod x_{m}^{2}\\right)$, and hence\n\n$$\n\\begin{aligned}\nx_{m+1}^{3}-4 x_{m+1}^{2}+5 x_{m+1} & \\equiv\\left(L x_{m}+1\\right)^{3}-4\\left(L x_{m}+1\\right)^{2}+5\\left(L x_{m}+1\\right) \\\\\n& \\equiv\\left(3 L x_{m}+1\\right)-4\\left(2 L x_{m}+1\\right)+5\\left(L x_{m}+1\\right) \\equiv 2 \\quad\\left(\\bmod x_{m}^{2}\\right),\n\\end{aligned}\n$$\n\nwhich in turn gives indeed $x_{m+2} \\equiv 2 c^{2}+1 \\equiv x_{2}\\left(\\bmod x_{m}^{2}\\right)$.\n\nClaim 3. For each integer $n \\geqslant 2$, we have $x_{n}>x_{1} \\cdot x_{2} \\cdots x_{n-2}$.\n\nProof. The cases $n=2$ and $n=3$ are clear. Arguing inductively, we assume now that the claim holds for some $n \\geqslant 3$. Recall that $x_{2} \\geqslant 3$, so by monotonicity and (2) we get $x_{n} \\geqslant x_{3} \\geqslant x_{2}\\left(x_{2}-2\\right)^{2}+x_{2}+1 \\geqslant 7$. It follows that\n\n$$\nx_{n+1}>x_{n}^{3}-4 x_{n}^{2}+5 x_{n}>7 x_{n}^{2}-4 x_{n}^{2}>x_{n}^{2}>x_{n} x_{n-1},\n$$\n\nwhich by the induction hypothesis yields $x_{n+1}>x_{1} \\cdot x_{2} \\cdots x_{n-1}$, as desired.\n\n\n\nNow we direct our attention to the problem itself: let any integer $n \\geqslant 2$ be given. By Claim 3 there exists a prime number $p$ appearing with a higher exponent in the prime factorisation of $x_{n}$ than in the prime factorisation of $x_{1} \\cdots x_{n-2}$. In particular, $p \\mid x_{n}$, and it suffices to prove that $p$ divides none of $x_{1}, \\ldots, x_{n-1}$.\n\nOtherwise let $k \\in\\{1, \\ldots, n-1\\}$ be minimal such that $p$ divides $x_{k}$. Since $x_{n-1}$ and $x_{n}$ are coprime by (1) and $x_{1}=1$, we actually have $2 \\leqslant k \\leqslant n-2$. Write $n=q k+r$ with some integers $q \\geqslant 0$ and $0 \\leqslant r1$, so there exists a prime $p$ with $v_{p}(N)>0$. Since $N$ is a fraction of two odd numbers, $p$ is odd.\n\nBy our lemma,\n\n$$\n01$ and $a_{n+2}>1$. Since $\\left(a_{n+1}\\right)^{2}-1 \\leqslant\\left(1-a_{n}\\right)\\left(a_{n+2}-1\\right)$, we deduce that $0<1-a_{n}<1<1+a_{n+2}$, thus $\\left(a_{n+1}\\right)^{2}-1<\\left(a_{n+2}+1\\right)\\left(a_{n+2}-1\\right)=\\left(a_{n+2}\\right)^{2}-1$.\n\nOn the other hand, $\\left(a_{n+2}\\right)^{2}-1 \\leqslant\\left(1-a_{n+3}\\right)\\left(a_{n+1}-1\\right)<\\left(1+a_{n+1}\\right)\\left(a_{n+1}-1\\right)=\\left(a_{n+1}\\right)^{2}-1$, a contradiction. We have shown that we cannot have two consecutive terms, except maybe $a_{1}$ and $a_{2}$, strictly greater than 1 .\n\nFinally, suppose $a_{2022}>1$. This implies that $a_{2021} \\leqslant 1$ and $a_{2023} \\leqslant 1$. Therefore $0<$ $\\left(a_{2022}\\right)^{2}-1 \\leqslant\\left(1-a_{2021}\\right)\\left(a_{2023}-1\\right) \\leqslant 0$, a contradiction. We conclude that $a_{2022} \\leqslant 1$.']",,True,,, 1997,Algebra,,Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set.,"['First we show that $n \\geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1}a_{1}+\\cdots+a_{k} \\geqslant a_{k+1}$, which gives a contradiction.\n\nIf $n=k+2$ then we have $a_{1} \\geqslant a_{2}+\\cdots+a_{k+1} \\geqslant a_{k+2}$, that again gives a contradiction.\n\nIf $n=k+3$ then we have $a_{1} \\geqslant a_{2}+\\cdots+a_{k+1}$ and $a_{3}+\\cdots+a_{k+2} \\geqslant a_{k+3}$. Adding the two inequalities we get $a_{1}+a_{k+2} \\geqslant a_{2}+a_{k+3}$, again a contradiction.\n\nIt remains to give an example of a set with $k+4$ elements satisfying the condition of the problem. We start with the case when $k=2 l$ and $l \\geqslant 1$. In that case, denote by $A_{i}=\\{-i, i\\}$ and take the set $A_{1} \\cup \\cdots \\cup A_{l+2}$, which has exactly $k+4=2 l+4$ elements. We are left to show that this set satisfies the required condition.\n\nNote that if a number $i$ can be expressed in the desired way, then so can $-i$ by negating the expression. Therefore, we consider only $1 \\leqslant i \\leqslant l+2$.\n\nIf $i0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying $$ x f(y)+y f(x) \leqslant 2 . $$","['First we prove that the function $f(x)=1 / x$ satisfies the condition of the problem statement. The AM-GM inequality gives\n\n$$\n\\frac{x}{y}+\\frac{y}{x} \\geqslant 2\n$$\n\nfor every $x, y>0$, with equality if and only if $x=y$. This means that, for every $x>0$, there exists a unique $y>0$ such that\n\n$$\n\\frac{x}{y}+\\frac{y}{x} \\leqslant 2\n$$\n\nnamely $y=x$.\n\nLet now $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ be a function that satisfies the condition of the problem statement. We say that a pair of positive real numbers $(x, y)$ is $\\operatorname{good}$ if $x f(y)+y f(x) \\leqslant 2$. Observe that if $(x, y)$ is good, then so is $(y, x)$.\n\nLemma 1.0. If $(x, y)$ is good, then $x=y$.\n\nProof. Assume that there exist positive real numbers $x \\neq y$ such that $(x, y)$ is good. The uniqueness assumption says that $y$ is the unique positive real number such that $(x, y)$ is good. In particular, $(x, x)$ is not a good pair. This means that\n\n$$\nx f(x)+x f(x)>2\n$$\n\nand thus $x f(x)>1$. Similarly, $(y, x)$ is a good pair, so $(y, y)$ is not a good pair, which implies $y f(y)>1$. We apply the AM-GM inequality to obtain\n\n$$\nx f(y)+y f(x) \\geqslant 2 \\sqrt{x f(y) \\cdot y f(x)}=2 \\sqrt{x f(x) \\cdot y f(y)}>2\n$$\n\nThis is a contradiction, since $(x, y)$ is a good pair.\n\nBy assumption, for any $x>0$, there always exists a good pair containing $x$, however Lemma 1 implies that the only good pair that can contain $x$ is $(x, x)$, so\n\n$$\nx f(x) \\leqslant 1 \\quad \\Longleftrightarrow \\quad f(x) \\leqslant \\frac{1}{x},\n$$\n\nfor every $x>0$.\n\nIn particular, with $x=1 / f(t)$ for $t>0$, we obtain\n\n$$\n\\frac{1}{f(t)} \\cdot f\\left(\\frac{1}{f(t)}\\right) \\leqslant 1\n$$\n\nHence\n\n$$\nt \\cdot f\\left(\\frac{1}{f(t)}\\right) \\leqslant t f(t) \\leqslant 1\n$$\n\nWe claim that $(t, 1 / f(t))$ is a good pair for every $t>0$. Indeed,\n\n$$\nt \\cdot f\\left(\\frac{1}{f(t)}\\right)+\\frac{1}{f(t)} f(t)=t \\cdot f\\left(\\frac{1}{f(t)}\\right)+1 \\leqslant 2\n$$\n\nLemma 1 implies that $t=1 / f(t) \\Longleftrightarrow f(t)=1 / t$ for every $t>0$.\n\n\n1. We give an alternative way to prove that $f(x)=1 / x$ assuming $f(x) \\leqslant 1 / x$ for every $x>0$.\n\nIndeed, if $f(x)<1 / x$ then for every $a>0$ with $f(x)<1 / a<1 / x$ (and there are at least two of them), we have\n\n$$\na f(x)+x f(a)<1+\\frac{x}{a}<2 .\n$$\n\nHence $(x, a)$ is a good pair for every such $a$, a contradiction. We conclude that $f(x)=1 / x$.\n\n\n2. We can also conclude from Lemma 1 and $f(x) \\leqslant 1 / x$ as follows.\n\nLemma 2. The function $f$ is decreasing.\n\nProof. Let $y>x>0$. Lemma 1 says that $(x, y)$ is not a good pair, but $(y, y)$ is. Hence\n\n$$\nx f(y)+y f(x)>2 \\geqslant 2 y f(y)>y f(y)+x f(y),\n$$\n\nwhere we used $y>x$ (and $f(y)>0$ ) in the last inequality. This implies that $f(x)>f(y)$, showing that $f$ is decreasing.\n\nWe now prove that $f(x)=1 / x$ for all $x$. Fix a value of $x$ and note that for $y>x$ we must have $x f(x)+y f(x)>x f(y)+y f(x)>2$ (using that $f$ is decreasing for the first step), hence $f(x)>\\frac{2}{x+y}$. The last inequality is true for every $y>x>0$. If we fix $x$ and look for the supremum of the expression $\\frac{2}{x+y}$ over all $y>x$, we get\n\n$$\nf(x) \\geqslant \\frac{2}{x+x}=\\frac{1}{x}\n$$\n\nSince we already know that $f(x) \\leqslant 1 / x$, we conclude that $f(x)=1 / x$.', 'As in the first solution, we note that $f(x)=1 / x$ is a solution, and we set out to prove that it is the only one. We write $g(x)$ for the unique positive real number such that $(x, g(x))$ is a good pair. In this solution, we prove Lemma 2 without assuming Lemma 1.\n\nLemma 2. The function $f$ is decreasing.\n\nProof. Consider $x$ 2. Combining these two inequalities yields\n\n$$\nx f(g(y))+g(y) f(x)>2 \\geqslant y f(g(y))+g(y) f(y)\n$$\n\nor $f(g(y))(x-y)>g(y)(f(y)-f(x))$. Because $g(y)$ and $f(g(y))$ are both positive while $x-y$ is negative, it follows that $f(y)2$ and $y f(y)+y f(y)>2$, which implies $x f(x)+y f(y)>2$. Now\n\n$$\nx f(x)+y f(y)>2 \\geqslant x f(y)+y f(x)\n$$\n\nimplies $(x-y)(f(x)-f(y))>0$, which contradicts the fact that $f$ is decreasing. So $y=x$ is the unique $y$ such that $(x, y)$ is a good pair, and in particular we have $f(x) \\leqslant 1 / x$.\n\nWe can now conclude the proof', 'As in the other solutions we verify that the function $f(x)=1 / x$ is a solution. We first want to prove the following lemma:\n\nLemma 3. For all $x \\in \\mathbb{R}_{>0}$ we actually have $x f(g(x))+g(x) f(x)=2$ (that is: the inequality is actually an equality).\n\n\n\nProof. We proceed by contradiction: Assume there exists some number $x>0$ such that for $y=g(x)$ we have $x f(y)+y f(x)<2$. Then for any $0<\\epsilon<\\frac{2-x f(y)-y f(x)}{2 f(x)}$ we have, by uniqueness of $y$, that $x f(y+\\epsilon)+(y+\\epsilon) f(x)>2$. Therefore\n\n$$\n\\begin{aligned}\nf(y+\\epsilon) & >\\frac{2-(y+\\epsilon) f(x)}{x}=\\frac{2-y f(x)-\\epsilon f(x)}{x} \\\\\n& >\\frac{2-y f(x)-\\frac{2-x f(y)-y f(x)}{2}}{x} \\\\\n& =\\frac{2-x f(y)-y f(x)}{2 x}+f(y)>f(y) .\n\\end{aligned}\n\\tag{1}\n$$\n\nFurthermore, for every such $\\epsilon$ we have $g(y+\\epsilon) f(y+\\epsilon)+(y+\\epsilon) f(g(y+\\epsilon)) \\leqslant 2$ and $g(y+\\epsilon) f(y)+y f(g(y+\\epsilon))>2($ since $y \\neq y+\\epsilon=g(g(y+\\epsilon)))$. This gives us the two inequalities\n\n$$\nf(g(y+\\epsilon)) \\leqslant \\frac{2-g(y+\\epsilon) f(y+\\epsilon)}{y+\\epsilon} \\quad \\text { and } \\quad f(g(y+\\epsilon))>\\frac{2-g(y+\\epsilon) f(y)}{y} \\text {. }\n$$\n\nCombining these two inequalities and rearranging the terms leads to the inequality\n\n$$\n2 \\epsilon0}$ we have\n\n$$\nx f(y)+y f(x) \\geqslant 2\n$$\n\nsince for $y=g(x)$ we have equality and by uniqueness for $y \\neq g(x)$ the inequality is strict.\n\nIn particular for every $x \\in \\mathbb{R}_{>0}$ and for $y=x$ we have $2 x f(x) \\geqslant 2$, or equivalently $f(x) \\geqslant 1 / x$ for all $x \\in \\mathbb{R}_{>0}$. With this inequality we obtain for all $x \\in \\mathbb{R}_{>0}$\n\n$$\n2 \\geqslant x f(g(x))+g(x) f(x) \\geqslant \\frac{x}{g(x)}+\\frac{g(x)}{x} \\geqslant 2\n$$\n\nwhere the first inequality comes from the problem statement. Consequently each of these inequalities must actually be an equality, and in particular we obtain $f(x)=1 / x$ for all $x \\in \\mathbb{R}_{>0}$.', 'Again, let us prove that $f(x)=1 / x$ is the only solution. Let again $g(x)$ be the unique positive real number such that $(x, g(x))$ is a good pair.\n\nLemma 4. The function $f$ is strictly convex.\n\nProof. Consider the function $q_{s}(x)=f(x)+s x$ for some real number $s$. If $f$ is not strictly convex, then there exist $u2^{s-3} $$","[""Let $1 \\leqslant a0$ such that the $\frac{1}{2} n(n-1)$ differences $a_{j}-a_{i}$ for $1 \leqslant i1$ of $x^{2}-x-1=0$ (the golden ratio) and set $\\left(a_{1}, a_{2}, a_{3}\\right)=$ $\\left(0, r, r+r^{2}\\right)$. Then\n\n$$\n\\left(a_{2}-a_{1}, a_{3}-a_{2}, a_{3}-a_{1}\\right)=\\left(r, r^{2}, r+r^{2}=r^{3}\\right)\n$$\n\nFor $n=4$, take the root $r \\in(1,2)$ of $x^{3}-x-1=0$ (such a root exists because $1^{3}-1-1<0$ and $\\left.2^{3}-2-1>0\\right)$ and set $\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)=\\left(0, r, r+r^{2}, r+r^{2}+r^{3}\\right)$. Then\n\n$$\n\\left(a_{2}-a_{1}, a_{3}-a_{2}, a_{4}-a_{3}, a_{3}-a_{1}, a_{4}-a_{2}, a_{4}-a_{1}\\right)=\\left(r, r^{2}, r^{3}, r^{4}, r^{5}, r^{6}\\right)\n$$\n\nFor $n \\geqslant 5$, we will proceed by contradiction. Suppose there exist numbers $a_{1}<\\cdots1$ satisfying the conditions of the problem. We start with a lemma:\n\nLemma. We have $r^{n-1}>2$.\n\nProof. There are only $n-1$ differences $a_{j}-a_{i}$ with $j=i+1$, so there exists an exponent $e \\leqslant n$ and a difference $a_{j}-a_{i}$ with $j \\geqslant i+2$ such that $a_{j}-a_{i}=r^{e}$. This implies\n\n$$\nr^{n} \\geqslant r^{e}=a_{j}-a_{i}=\\left(a_{j}-a_{j-1}\\right)+\\left(a_{j-1}-a_{i}\\right)>r+r=2 r\n$$\n\nthus $r^{n-1}>2$ as desired.\n\nTo illustrate the general approach, we first briefly sketch the idea behind the argument in the special case $n=5$. In this case, we clearly have $a_{5}-a_{1}=r^{10}$. Note that there are 3 ways to rewrite $a_{5}-a_{1}$ as a sum of two differences, namely\n\n$$\n\\left(a_{5}-a_{4}\\right)+\\left(a_{4}-a_{1}\\right),\\left(a_{5}-a_{3}\\right)+\\left(a_{3}-a_{1}\\right),\\left(a_{5}-a_{2}\\right)+\\left(a_{2}-a_{1}\\right) .\n$$\n\nUsing the lemma above and convexity of the function $f(n)=r^{n}$, we argue that those three ways must be $r^{10}=r^{9}+r^{1}=r^{8}+r^{4}=r^{7}+r^{6}$. That is, the ""large"" exponents keep dropping by 1 , while the ""small"" exponents keep increasing by $n-2, n-3, \\ldots, 2$. Comparing any two such equations, we then get a contradiction unless $n \\leqslant 4$.\n\nNow we go back to the full proof for any $n \\geqslant 5$. Denote $b=\\frac{1}{2} n(n-1)$. Clearly, we have $a_{n}-a_{1}=r^{b}$. Consider the $n-2$ equations of the form:\n\n$$\na_{n}-a_{1}=\\left(a_{n}-a_{i}\\right)+\\left(a_{i}-a_{1}\\right) \\text { for } i \\in\\{2, \\ldots, n-1\\}\n$$\n\nIn each equation, one of the two terms on the right-hand side must be at least $\\frac{1}{2}\\left(a_{n}-a_{1}\\right)$. But from the lemma we have $r^{b-(n-1)}=r^{b} / r^{n-1}<\\frac{1}{2}\\left(a_{n}-a_{1}\\right)$, so there are at most $n-2$ sufficiently large elements in $\\left\\{r^{k} \\mid 1 \\leqslant k1$ and $f(r)=r^{n}$ is convex, we have\n\n$$\nr^{b-1}-r^{b-2}>r^{b-2}-r^{b-3}>\\ldots>r^{b-(n-3)}-r^{b-(n-2)},\n$$\n\nimplying\n\n$$\nr^{\\alpha_{2}}-r^{\\alpha_{1}}>r^{\\alpha_{3}}-r^{\\alpha_{2}}>\\ldots>r^{\\alpha_{n-2}}-r^{\\alpha_{n-3}} .\n$$\n\nConvexity of $f(r)=r^{n}$ further implies\n\n$$\n\\alpha_{2}-\\alpha_{1}>\\alpha_{3}-\\alpha_{2}>\\ldots>\\alpha_{n-2}-\\alpha_{n-3}\n$$\n\nNote that $\\alpha_{n-2}-\\alpha_{n-3} \\geqslant 2$ : Otherwise we would have $\\alpha_{n-2}-\\alpha_{n-3}=1$ and thus\n\n$$\nr^{\\alpha_{n-3}} \\cdot(r-1)=r^{\\alpha_{n-2}}-r^{\\alpha_{n-3}}=r^{b-(n-3)}-r^{b-(n-2)}=r^{b-(n-2)} \\cdot(r-1)\n$$\n\nimplying that $\\alpha_{n-3}=b-(n-2)$, a contradiction. Therefore, we have\n\n$$\n\\begin{aligned}\n\\alpha_{n-2}-\\alpha_{1} & =\\left(\\alpha_{n-2}-\\alpha_{n-3}\\right)+\\cdots+\\left(\\alpha_{2}-\\alpha_{1}\\right) \\\\\n& \\geqslant 2+3+\\cdots+(n-2) \\\\\n& =\\frac{1}{2}(n-2)(n-1)-1=\\frac{1}{2} n(n-3) .\n\\end{aligned}\n$$\n\nOn the other hand, from $\\alpha_{n-2} \\leqslant b-(n-1)$ and $\\alpha_{1} \\geqslant 1$ we get\n\n$$\n\\alpha_{n-2}-\\alpha_{1} \\leqslant b-n=\\frac{1}{2} n(n-1)-n=\\frac{1}{2} n(n-3),\n$$\n\nimplying that equalities must occur everywhere and the claim about the small terms follows.\n\nNow, assuming $n-2 \\geqslant 2$, we have the two different equations:\n\n$$\nr^{b}=r^{b-(n-2)}+r^{b-(n-2)-1} \\text { and } r^{b}=r^{b-(n-3)}+r^{b-(n-2)-3}\n$$\n\nwhich can be rewritten as\n\n$$\nr^{n-1}=r+1 \\quad \\text { and } \\quad r^{n+1}=r^{4}+1\n\\tag{1}\n$$\n\nSimple algebra now gives\n\n$$\nr^{4}+1=r^{n+1}=r^{n-1} \\cdot r^{2}=r^{3}+r^{2} \\Longrightarrow(r-1)\\left(r^{3}-r-1\\right)=0 .\n$$\n\nSince $r \\neq 1$, using Equation (1) we conclude $r^{3}=r+1=r^{n-1}$, thus $n=4$, which gives a contradiction.']","['2,3,4']",True,,Numerical, 2001,Algebra,,"Let $\mathbb{R}$ be the set of real numbers. We denote by $\mathcal{F}$ the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(x+f(y))=f(x)+f(y) $$ for every $x, y \in \mathbb{R}$. Find all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{R}$ satisfying $f(z)=q z$.","['Let $Z$ be the set of all rational numbers $q$ such that for every function $f \\in \\mathcal{F}$, there exists some $z \\in \\mathbb{R}$ satisfying $f(z)=q z$. Let further\n\n$$\nS=\\left\\{\\frac{n+1}{n}: n \\in \\mathbb{Z}, n \\neq 0\\right\\}\n$$\n\nWe prove that $Z=S$ by showing the two inclusions: $S \\subseteq Z$ and $Z \\subseteq S$.\n\nWe first prove that $S \\subseteq Z$. Let $f \\in \\mathcal{F}$ and let $P(x, y)$ be the relation $f(x+f(y))=f(x)+$ $f(y)$. First note that $P(0,0)$ gives $f(f(0))=2 f(0)$. Then, $P(0, f(0))$ gives $f(2 f(0))=3 f(0)$. We claim that\n\n$$\nf(k f(0))=(k+1) f(0)\n$$\n\nfor every integer $k \\geqslant 1$. The claim can be proved by induction. The cases $k=1$ and $k=2$ have already been established. Assume that $f(k f(0))=(k+1) f(0)$ and consider $P(0, k f(0))$ which gives\n\n$$\nf((k+1) f(0))=f(0)+f(k f(0))=(k+2) f(0) \\text {. }\n$$\n\nThis proves the claim. We conclude that $\\frac{k+1}{k} \\in Z$ for every integer $k \\geqslant 1$. Note that $P(-f(0), 0)$ gives $f(-f(0))=0$. We now claim that\n\n$$\nf(-k f(0))=(-k+1) f(0)\n$$\n\nfor every integer $k \\geqslant 1$. The proof by induction is similar to the one above. We conclude that $\\frac{-k+1}{-k} \\in Z$ for every integer $k \\geqslant 1$. This shows that $S \\subseteq Z$.\n\nWe now prove that $Z \\subseteq S$. Let $p$ be a rational number outside the set $S$. We want to prove that $p$ does not belong to $Z$. To that end, we construct a function $f \\in \\mathcal{F}$ such that $f(z) \\neq p z$ for every $z \\in \\mathbb{R}$. The strategy is to first construct a function\n\n$$\ng:[0,1) \\rightarrow \\mathbb{Z}\n$$\n\nand then define $f$ as $f(x)=g(\\{x\\})+\\lfloor x\\rfloor$. This function $f$ belongs to $\\mathcal{F}$. Indeed,\n\n$$\n\\begin{aligned}\nf(x+f(y)) & =g(\\{x+f(y)\\})+\\lfloor x+f(y)\\rfloor \\\\\n& =g(\\{x+g(\\{y\\})+\\lfloor y\\rfloor\\})+\\lfloor x+g(\\{y\\})+\\lfloor y\\rfloor\\rfloor \\\\\n& =g(\\{x\\})+\\lfloor x\\rfloor+g(\\{y\\})+\\lfloor y\\rfloor \\\\\n& =f(x)+f(y),\n\\end{aligned}\n$$\n\nwhere we used that $g$ only takes integer values.\n\nLemma 1. For every $\\alpha \\in[0,1)$, there exists $m \\in \\mathbb{Z}$ such that\n\n$$\nm+n \\neq p(\\alpha+n)\n$$\n\nfor every $n \\in \\mathbb{Z}$.\n\n\n\nProof. Note that if $p=1$ the claim is trivial. If $p \\neq 1$, then the claim is equivalent to the existence of an integer $m$ such that\n\n$$\n\\frac{m-p \\alpha}{p-1}\n$$\n\nis never an integer. Assume the contrary. That would mean that both\n\n$$\n\\frac{m-p \\alpha}{p-1} \\text { and } \\quad \\frac{(m+1)-p \\alpha}{p-1}\n$$\n\nare integers, and so is their difference. The latter is equal to\n\n$$\n\\frac{1}{p-1}\n$$\n\nSince we assumed $p \\notin S, 1 /(p-1)$ is never an integer. This is a contradiction.\n\nDefine $g:[0,1) \\rightarrow \\mathbb{Z}$ by $g(\\alpha)=m$ for any integer $m$ that satisfies the conclusion of Lemma 1. Note that $f(z) \\neq p z$ if and and only if\n\n$$\ng(\\{z\\})+\\lfloor z\\rfloor \\neq p(\\{z\\}+\\lfloor z\\rfloor)\n$$\n\nThe latter is guaranteed by the construction of the function $g$. We conclude that $p \\notin Z$ as desired. This shows that $Z \\subset S$.']","['$\\left\\{\\frac{n+1}{n}: n \\in \\mathbb{Z}, n \\neq 0\\right\\}$']",True,,Need_human_evaluate, 2002,Algebra,,"For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)=$ $x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ be a polynomial, where $n \geqslant 2$ and $a_{i}$ is a positive integer for all $0 \leqslant i \leqslant n-1$. Could it be the case that, for all positive integers $k, s(k)$ and $s(P(k))$ have the same parity?","[""With the notation above, we begin by choosing a positive integer $t$ such that\n\n$$\n10^{t}>\\max \\left\\{\\frac{100^{n-1} a_{n-1}}{\\left(10^{\\frac{1}{n-1}}-9^{\\frac{1}{n-1}}\\right)^{n-1}}, \\frac{a_{n-1}}{9} 10^{n-1}, \\frac{a_{n-1}}{9}\\left(10 a_{n-1}\\right)^{n-1}, \\ldots, \\frac{a_{n-1}}{9}\\left(10 a_{0}\\right)^{n-1}\\right\\} .\n$$\n\nAs a direct consequence of $10^{t}$ being bigger than the first quantity listed in the above set, we get that the interval\n\n$$\nI=\\left[\\left(\\frac{9}{a_{n-1}} 10^{t}\\right)^{\\frac{1}{n-1}},\\left(\\frac{1}{a_{n-1}} 10^{t+1}\\right)^{\\frac{1}{n-1}}\\right)\n$$\n\ncontains at least 100 consecutive positive integers.\n\nLet $X$ be a positive integer in $I$ such that $X$ is congruent to $1 \\bmod 100$. Since $X \\in I$ we have\n\n$$\n9 \\cdot 10^{t} \\leqslant a_{n-1} X^{n-1}<10^{t+1}\n$$\n\nthus the first digit (from the left) of $a_{n-1} X^{n-1}$ must be 9 .\n\nNext, we observe that $a_{n-1}\\left(10 a_{i}\\right)^{n-1}<9 \\cdot 10^{t} \\leqslant a_{n-1} X^{n-1}$, thus $10 a_{i}10^{\\alpha i} a_{n-1} X^{n-1}>$ $10^{\\alpha i} a_{i} X^{i}$, the terms $a_{i} 10^{\\alpha i} X^{i}$ do not interact when added; in particular, there is no carryover caused by addition. Thus we have $s\\left(P\\left(10^{\\alpha} X\\right)\\right)=s\\left(X^{n}\\right)+s\\left(a_{n-1} X^{n-1}\\right)+\\cdots+s\\left(a_{0}\\right)$.\n\nWe now look at $P\\left(10^{\\alpha-1} X\\right)=10^{(\\alpha-1) n} X^{n}+a_{n-1} 10^{(\\alpha-1)(n-1)} X^{n-1}+\\cdots+a_{0}$. Firstly, if $i10^{(\\alpha-1) i} a_{n-1} X^{n-1} \\geqslant 10^{(\\alpha-1) i+1} a_{i} X^{i}$, thus all terms $10^{(\\alpha-1) i} a_{i} X^{i}$ for $0 \\leqslant i \\leqslant n-1$ come in 'blocks', exactly as in the previous case.\n\nFinally, $10^{(\\alpha-1) n+1}>10^{(\\alpha-1)(n-1)} a_{n-1} X^{n-1} \\geqslant 10^{(\\alpha-1) n}$, thus $10^{(\\alpha-1)(n-1)} a_{n-1} X^{n-1}$ has exactly $(\\alpha-1) n+1$ digits, and its first digit is 9 , as established above. On the other hand, $10^{(\\alpha-1) n} X^{n}$ has exactly $(\\alpha-1) n$ zeros, followed by $01($ as $X$ is $1 \\bmod 100)$. Therefore, when we add the terms, the 9 and 1 turn into 0 , the 0 turns into 1 , and nothing else is affected.\n\nPutting everything together, we obtain\n\n$$\ns\\left(P\\left(10^{\\alpha-1} X\\right)\\right)=s\\left(X^{n}\\right)+s\\left(a_{n-1} X^{n-1}\\right)+\\cdots+s\\left(a_{0}\\right)-9=s\\left(P\\left(10^{\\alpha} X\\right)\\right)-9\n$$\n\nthus $s\\left(P\\left(10^{\\alpha} X\\right)\\right)$ and $s\\left(P\\left(10^{\\alpha-1} X\\right)\\right)$ have different parities, as claimed.""]",,True,,, 2003,Algebra,,"For a positive integer $n$, an $n$-sequence is a sequence $\left(a_{0}, \ldots, a_{n}\right)$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \leqslant n$, then $a_{i}+a_{j} \leqslant n$ and $a_{a_{i}+a_{j}}=a_{i+j}$. Let $f(n)$ be the number of $n$-sequences. Prove that there exist positive real numbers $c_{1}, c_{2}$ and $\lambda$ such that $$ c_{1} \lambda^{n}k$ for some $i$, and small if no such $i$ exists. For now we will assume that $\\left(a_{i}\\right)$ is not the identity sequence (in other words, that $a_{i} \\neq i$ for some $i$ ).\n\nLemma 1. If $a_{r}=a_{s}$ and $r, sr$, and let $d$ be the minimum positive integer such that $a_{r+d}=a_{r}$. Then\n\n1. The subsequence $\\left(a_{r}, a_{r+1}, \\ldots, a_{n}\\right)$ is periodic with minimal period $d$. That is, for $uv_{0}$. Thus $u_{0}=v_{0}$, so $d \\mid u-v$.\n2. If $r=0$ there is nothing to prove. Otherwise $a_{0}=a_{2 a_{0}}$ so $2 a_{0}=0$. Then we have $a_{a_{i}}=a_{i}$ for all $i$, so $a_{i}=i$ for $ik+1$. We show that $a_{i} \\leqslant k$ for all $i$ by induction. Note that Lemma 2 already establishes this for $i \\leqslant k$. We must have $d \\mid a_{d / 2}$ and $a_{d / 2} \\leqslant kk$, if $a_{j} \\leqslant k$ for $jk+1$ and $a_{i}=i$ for all $0 \\leqslant ik$.\n\nWe already have $a_{i}=i$ for $id$, this means that $a_{i}=i$ for $i \\leqslant k$.\n\nFinally, one can show inductively that $a_{i}=i$ for $kc_{1} \\lambda^{n}$ for some $c_{1}$, we note that\n\n$$\nf(n)>g\\left(k+1,\\left\\lfloor\\frac{k+1}{3}\\right\\rfloor\\right) \\geqslant 3^{\\lfloor(k+1) / 3\\rfloor}>3^{n / 6}-1\n$$\n\nTo show that $f(n)\\frac{3 n+1}{2}$, let $a=k-n-1, b=2 n-k+1$. Then $k>2 a+b, k>2 b+a$, so the configuration $A^{a} C^{b} A^{b} C^{a}$ will always have four blocks:\n\n$$\nA^{a} C^{b} A^{b} C^{a} \\rightarrow C^{a} A^{a} C^{b} A^{b} \\rightarrow A^{b} C^{a} A^{a} C^{b} \\rightarrow C^{b} A^{b} C^{a} A^{a} \\rightarrow A^{a} C^{b} A^{b} C^{a} \\rightarrow \\ldots\n$$\n\nTherefore a pair $(n, k)$ can have the desired property only if $n \\leqslant k \\leqslant \\frac{3 n+1}{2}$. We claim that all such pairs in fact do have the desired property. Clearly, the number of blocks in a configuration cannot increase, so whenever the operation is applied, it either decreases or remains constant. We show that unless there are only two blocks, after a finite amount of steps the number of blocks will decrease.\n\nConsider an arbitrary configuration with $c \\geqslant 3$ blocks. We note that as $k \\geqslant n$, the leftmost block cannot be moved, because in this case all $n$ coins of one type are in the leftmost block, meaning there are only two blocks. If a block which is not the leftmost or rightmost block is moved, its neighbor blocks will be merged, causing the number of blocks to decrease.\n\nHence the only case in which the number of blocks does not decrease in the next step is if the rightmost block is moved. If $c$ is odd, the leftmost and the rightmost blocks are made of the same metal, so this would merge two blocks. Hence $c \\geqslant 4$ must be even. Suppose there is a configuration of $c$ blocks with the $i$-th block having size $a_{i}$ so that the operation always moves the rightmost block:\n\n$$\nA^{a_{1}} \\ldots A^{a_{c-1}} C^{a_{c}} \\rightarrow C^{a_{c}} A^{a_{1}} \\ldots A^{a_{c-1}} \\rightarrow A^{a_{c-1}} C^{a_{c}} A^{a_{1}} \\ldots C^{a_{c-2}} \\rightarrow \\ldots\n$$\n\nBecause the rightmost block is always moved, $k \\geqslant 2 n+1-a_{i}$ for all $i$. Because $\\sum a_{i}=2 n$, summing this over all $i$ we get $c k \\geqslant 2 c n+c-\\sum a_{i}=2 c n+c-2 n$, so $k \\geqslant 2 n+1-\\frac{2 n}{c} \\geqslant \\frac{3 n}{2}+1$. But this contradicts $k \\leqslant \\frac{3 n+1}{2}$. Hence at some point the operation will not move the rightmost block, meaning that the number of blocks will decrease, as desired.']","['All pairs $(n, k)$ such that $n \\leqslant k \\leqslant \\frac{3 n+1}{2}$']",False,,Need_human_evaluate, 2006,Combinatorics,,"In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height 0 . A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn: - The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller. - The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter. We say that a tree is majestic if its height is at least $10^{6}$. Determine the largest number $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.","[""We solve the problem for a general $3 N \\times 3 N$ board. First, we prove that the lumberjack has a strategy to ensure there are never more than $5 N^{2}$ majestic trees. Giving the squares of the board coordinates in the natural manner, colour each square where at least one of its coordinates are divisible by 3 , shown below for a $9 \\times 9$ board:\n\n\n\nThen, as each $3 \\times 3$ square on the board contains exactly 5 coloured squares, each move of the gardener will cause at most 4 trees on non-coloured squares to grow. The lumberjack may therefore cut those trees, ensuring no tree on a non-coloured square has positive height after his turn. Hence there cannot ever be more majestic trees than coloured squares, which is $5 N^{2}$.\n\nNext, we prove the gardener may ensure there are $5 N^{2}$ majestic trees. In fact, we prove this statement in a modified game which is more difficult for the gardener: on the lumberjack's turn in the modified game, he may decrement the height of all trees on the board except those the gardener did not just grow, in addition to four of the trees the gardener just grew. Clearly, a sequence of moves for the gardener which ensures that there are $K$ majestic trees in the modified game also ensures this in the original game.\n\n\n\nLet $M=\\left(\\begin{array}{l}9 \\\\ 5\\end{array}\\right)$; we say that a $m a p$ is one of the $M$ possible ways to mark 5 squares on a $3 \\times 3$ board. In the modified game, after the gardener chooses a $3 \\times 3$ subboard on the board, the lumberjack chooses a map in this subboard, and the total result of the two moves is that each tree marked on the map increases its height by 1, each tree in the subboard which is not in the map remains unchanged, and each tree outside the subboard decreases its height by 1 . Also note that if the gardener chooses a $3 \\times 3$ subboard $M l$ times, the lumberjack will have to choose some map at least $l$ times, so there will be at least 5 trees which each have height $\\geqslant l$.\n\nThe strategy for the gardener will be to divide the board into $N^{2}$ disjoint $3 \\times 3$ subboards, number them $0, \\ldots, N^{2}-1$ in some order. Then, for $b=N^{2}-1, \\ldots, 0$ in order, he plays $10^{6} M(M+1)^{b}$ times on subboard number $b$. Hence, on subboard number $b$, the moves on that subboard will first ensure 5 of its trees grows by at least $10^{6}(M+1)^{b}$, and then each move after that will decrease their heights by 1 . (As the trees on subboard $b$ had height 0 before the gardener started playing there, no move made on subboards $\\geqslant b$ decreased their heights.) As the gardener makes $10^{6} M(M+1)^{b-1}+\\ldots=10^{6}\\left((M+1)^{b}-1\\right)$ moves after he finishes playing on subboard $b$, this means that on subboard $b$, there will be 5 trees of height at least $10^{6}(M+1)^{b}-10^{6}\\left((M+1)^{b}-1\\right)=10^{6}$, hence each of the subboard has 5 majestic trees, which was what we wanted.""]",['2271380'],False,,Numerical, 2007,Combinatorics,,"Let $n>3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their immediate neighbours on the right and left. Determine all initial distributions of coins from which it is possible that, after a finite number of steps, each child has exactly one coin.","['Number the children $1, \\ldots, n$, and denote the number of coins the $i$-th child has by $c_{i}$. A step of this process consists of reducing some $c_{i}$ by 2 , and increasing $c_{i-1}, c_{i+1}$ by 1 . (Indices are considered $(\\bmod n)$.$) Because (i-1)-2 i+(i+1)=0$, the quantity $\\sum_{i=1}^{n} i c_{i}(\\bmod n)$ will be invariant under this process. Hence a necessary condition for the children to end up with an uniform distribution of coins is that\n\n$$\n\\sum_{i=1}^{n} i c_{i}=\\frac{n(n+1)}{2}(\\bmod n)\n$$\n\nWe will show that this condition is also sufficient. Consider an arbitrary initial distribution of coins. First, whenever child $i$ has more than one coin and $i \\neq n$, have child $i$ pass coins to its neighbors. (Child $i$ does nothing.) Then, after some amount of such steps, it must eventually become impossible to do any more steps because no child except perhaps child $i$ has more than 1 coin. (To see this, consider e.g. the quantity $\\sum_{i=1}^{n-1} i^{2} c_{i}$, which $\\left(\\right.$ as $(i-1)^{2}+(i+1)^{2}>2 i^{2}$ ) increases at each step.)\n\nHence we can reach a state of the form $\\left(z_{1}, \\ldots, z_{n-1}, M\\right)$, where $z_{i}=0$ or 1 . Call such states semi-uniform states of irregularity $M$.\n\nLemma. If there is a string of children having coins $a, \\overbrace{1, \\ldots, 1}^{k \\text { ones }}, b, \\overbrace{1, \\ldots, 1}^{k \\text { ones }}, c$, with $b \\geqslant 2$, after some sequence of steps we may reach the state $a+1, \\overbrace{1, \\ldots, 1}^{k \\text { ones }}, b-2, \\overbrace{1, \\ldots, 1}^{k \\text { ones }}, c+1$. We call performing this sequence of steps long-passing coins.\n\nProof. This is simply repeated application of the operation. We prove the lemma by induction on $k$. For $k=0$, this is just the operation of the problem. If $k=1$, have the child with $b$ coins pass coins, then both of their neighbors pass coins, then the child with $b$ coins pass coins again. For $k \\geqslant 2$, first, have the child with $b$ coins pass coins, then have both their neigbors send coins, giving the state\n\n$$\na, \\overbrace{1, \\ldots, 1}^{k-2 \\text { ones }}, 2,0, b, 0,2, \\overbrace{1, \\ldots, 1}^{k-2 \\text { ones }}, c\n$$\n\nNow set aside the children with $a, b$ and $c$ coins, and have each child with 2 coins give them to their neighbors until there are no such children remaining. This results in the state\n\n$$\na+1,0, \\overbrace{1, \\ldots, 1}^{k-2 \\text { ones }}, b, \\overbrace{1, \\ldots, 1}^{k-2 \\text { ones }}, 0, c+1\n$$\n\nBy the induction hypothesis, we can have the child with $b$ coins may pass a coin to each of the children with 0 coins, proving the lemma.\n\n\n\nClaim. We can reach a semi-uniform state of irregularity $M \\leqslant 2$.\n\nProof. If $M>3$, because there are only $n$ coins in total, there must be at least two children with 0 coins. Consider the arc of the circle spanned by the two such children closest to the child with $M$ coins. It has the form\n\n$$\n0, \\overbrace{1, \\ldots, 1}^{a \\text { ones }}, M, \\overbrace{1, \\ldots, 1}^{b \\text { ones }}, 0\n$$\n\nIf $a=b$, applying the previous lemma we can have the child with $M$ coins long-pass a coin to each of the children with 0 coins, which yields a semi-uniform state with lower $M$. Otherwise, WLOG $a>b$, so we can have the child with $M$ coins long-pass a coin to each of the children at distance $b$ from it, reaching a state of the form $(\\alpha:=a-b-1, \\beta:=b)$\n\n$$\n0, \\overbrace{1, \\ldots, 1}^{\\alpha \\text { ones }}, 2, \\overbrace{1, \\ldots, 1}^{\\beta \\text { ones }}, M-2, \\overbrace{1, \\ldots, 1}^{c \\text { ones }}\n$$\n\nThe children in the rightmost string of ones need make no further moves, so consider only the leftmost string. If $\\alpha<\\beta$, have the child with 2 coins long-pass coins to the child with 0 coins to its left and some child with 1 coin to its right, reaching a new state of the form $0, \\overbrace{1, \\ldots, 1}^{\\alpha \\text { ones }}, 2, \\overbrace{1, \\ldots, 1}^{\\beta \\text { ones }}, M-2$ with a smaller $\\beta$. As $\\beta$ cannot decrease indefinitely, eventually $\\alpha \\geqslant \\beta$. If $\\alpha=\\beta$, have the child with 2 coins long-pass to the child with $M$ coins and the child with 0 coins, reaching a semi-uniform state of irregularity $M-1$ as desired. Otherwise, $\\alpha<\\beta$, so have the child with 2 coins long-pass to the child with $M$ coins and a child with 1 coin, reaching a state of the form\n\n$$\n0, \\overbrace{1, \\ldots, 1}^{x \\text { ones }}, 2, \\overbrace{1, \\ldots, 1}^{y \\text { ones }}, 0, \\overbrace{1, \\ldots, 1}^{z \\text { ones }}, M-1\n$$\n\nNow, consider only the substring between the two children with 0 coins, which has the form $0, \\overbrace{1, \\ldots, 1}^{x \\text { ones }}, 2, \\overbrace{1, \\ldots, 1}^{y \\text { ones }}, 0$ Repeatedly have the child in this substring with 2 coins long-pass to the closest child with 0 coins and some other child. If the other child has 1 coin, we have a new strictly shorter substring of the form $0, \\overbrace{1, \\ldots, 1}^{x \\text { ones }}, 2, \\overbrace{1, \\ldots, 1}^{y \\text { ones }}, 0$. Hence eventually it must happen that the other child also has 0 coins, at which point we reach a semi-uniform state of irregularity $M-1$, proving the claim.\n\nWe have now shown that we can reach a semi-regular state of irregularity $M \\leqslant 2$, If $M=1$, each child must have one coin, as desired. Otherwise, we must have $M=2$, so there is one child with 0 coins, one child with 2 coins, and the remaining children all have 1 coin. Recall that the state we started with satisfied the invariant\n\n$$\n\\sum_{i=1}^{n} i c_{i}=\\frac{n(n+1)}{2} \\quad(\\bmod n)\n$$\n\nBecause each step preserves this invariant, this must also be true of the current state. Number the children so that the child with $M$ coins is child number $n$, and suppose the child with 0 coins is child $k$. Then $\\frac{n(n+1)}{2}=\\sum_{i=1}^{n} i c_{i}=\\left(\\sum_{i=1}^{n} 1 \\cdot c_{i}\\right)-k=\\frac{n(n+1)}{2}-k(\\bmod n)$, so $k=0$ $(\\bmod n)$. But this is impossible, as no child except the child with $M$ coins has an index divisible by $n$. Hence we cannot end up in a semi-regular state of irregularity 2 , so we are done.']","['All distributions where $\\sum_{i=1}^{n} i c_{i}=\\frac{n(n+1)}{2}(\\bmod n)$, where $c_{i}$ denotes the number of coins the $i$-th child starts with']",True,,Need_human_evaluate, 2008,Combinatorics,,"Let $m, n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_{1}, X_{2}, \ldots, X_{m}$ be pairwise distinct non-empty, not necessary disjoint subsets of $X$. A function $f: X \rightarrow$ $\{1,2, \ldots, n+1\}$ is called nice if there exists an index $k$ such that $$ \sum_{x \in X_{k}} f(x)>\sum_{x \in X_{i}} f(x) \text { for all } i \neq k $$ Prove that the number of nice functions is at least $n^{n}$.","['For a subset $Y \\subseteq X$, we write $f(Y)$ for $\\sum_{y \\in Y} f(y)$. Note that a function $f: X \\rightarrow$ $\\{1, \\ldots, n+1\\}$ is nice, if and only if $f\\left(X_{i}\\right)$ is maximized by a unique index $i \\in\\{1, \\ldots, m\\}$.\n\nWe will first investigate the set $\\mathcal{F}$ of functions $f: X \\rightarrow\\{1, \\ldots, n\\}$; note that $|\\mathcal{F}|=n^{n}$. For every function $f \\in \\mathcal{F}$, define a corresponding function $f^{+}: X \\rightarrow\\{1,2, \\ldots, n+1\\}$ in the following way: Pick some set $X_{l}$ that maximizes the value $f\\left(X_{l}\\right)$.\n\n- For all $x \\in X_{l}$, define $f^{+}(x)=f(x)+1$.\n- For all $x \\in X \\backslash X_{l}$, define $f^{+}(x)=f(x)$.\n\nClaim. The resulting function $f^{+}$is nice.\n\nProof. Note that $f^{+}\\left(X_{i}\\right)=f\\left(X_{i}\\right)+\\left|X_{i} \\cap X_{l}\\right|$ holds for all $X_{i}$. We show that $f^{+}\\left(X_{i}\\right)$ is maximized at the unique index $i=l$. Hence consider some arbitrary index $j \\neq l$. Then $X_{l} \\subset X_{j}$ is impossible, as this would imply $f\\left(X_{j}\\right)>f\\left(X_{l}\\right)$ and thereby contradict the choice of set $X_{l}$; this in particular yields $\\left|X_{l}\\right|>\\left|X_{j} \\cap X_{l}\\right|$.\n\n$$\nf^{+}\\left(X_{l}\\right)=f\\left(X_{l}\\right)+\\left|X_{l}\\right| \\geqslant f\\left(X_{j}\\right)+\\left|X_{l}\\right|>f\\left(X_{j}\\right)+\\left|X_{j} \\cap X_{l}\\right|=f^{+}\\left(X_{j}\\right)\n$$\n\nThe first inequality follows since $X_{l}$ was chosen to maximize the value $f\\left(X_{l}\\right)$. The second (strict) inequality follows from $\\left|X_{l}\\right|>\\left|X_{j} \\cap X_{l}\\right|$ as observed above. This completes the proof of the claim.\n\nNext observe that function $f$ can be uniquely reconstructed from $f^{+}$: the claim yields that $f^{+}$has a unique maximizer $X_{l}$, and by decreasing the value of $f^{+}$on $X_{l}$ by 1 , we get we can fully determine the values of $f$. As each of the $n^{n}$ functions $f \\in \\mathcal{F}$ yields a (unique) corresponding nice function $f^{+}: X \\rightarrow\\{1,2, \\ldots, n+1\\}$, the proof is complete.']",,True,,, 2009,Combinatorics,,"Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of non-empty piles that one can obtain by performing a finite sequence of moves of this form.","['The solution we describe is simple, but not the most effective one.\n\nWe can combine two piles of $2^{k-1}$ pebbles to make one pile of $2^{k}$ pebbles. In particular, given $2^{k}$ piles of one pebble, we may combine them as follows:\n\n$$\n\\begin{array}{ll}\n2^{k} \\text { piles of } 1 \\text { pebble } & \\rightarrow 2^{k-1} \\text { piles of } 2 \\text { pebbles } \\\\\n2^{k-1} \\text { piles of } 2 \\text { pebbles } & \\rightarrow 2^{k-2} \\text { piles of } 2 \\text { pebbles } \\\\\n2^{k-2} \\text { piles of } 4 \\text { pebbles } & \\rightarrow 2^{k-3} \\text { piles of } 4 \\text { pebbles } \\\\\n& \\vdots \\\\\n2 \\text { piles of } 2^{k-1} \\text { pebbles } & \\rightarrow 1 \\text { pile of } 2^{k} \\text { pebbles }\n\\end{array}\n$$\n\nThis proves the desired result in the case when $n$ is a power of 2 .\n\nIf $n$ is not a power of 2 , choose $N$ such that $2^{N}1$. In order to make a single pile of $n$ pebbles, we would have to start with a distribution in which the number of pebbles in each pile is divisible by the integer $m$. This is impossible when we start with all piles containing a single pebble.', 'We show an alternative strategy if $n$ is not a power of 2 . Write $n$ in binary form: $n=2^{i_{1}}+2^{i_{2}}+\\cdots+2^{i_{k}}$, where $i_{1}>i_{2}>\\cdots>i_{k}$. Now we make piles of sizes $2^{i_{1}}, 2^{i_{2}}, \\ldots, 2^{i_{k}}$. We call the pile with $2^{i_{1}}$ the large pile, all the others are the small piles.\n\nThe strategy is the following: take the two smallest small pile. If they have the same size of $2^{a}$, we make a pile of size $2^{a+1}$. If they have different sizes, we double the smaller pile using the large pile (we allow the large pile to have a negative number of pebbles: later we prove that it is not possible). We call all the new piles small. When we have only one small pile, we terminate the process: we have at most 2 piles.\n\nAfter each move we have one less number of piles, and all the piles have cardinality power of 2. The number of pebbles is decreasing, and at the end of the process, it has a size of $n-2^{i_{2}+1} \\geqslant n-2^{i_{1}}>0$, thus we can manage to have two piles.', ""Throughout the solution, we will consider the moves in reverse order. Namely, imagine we have some piles of pebbles, and we are allowed to perform moves as follows: take a pile with an even number of pebbles, split it into two equal halves and add the pebbles from each half to a different pile, possibly forming new piles (we may assume for convenience that there are infinitely many empty piles at any given moment). Given a configuration of piles $\\mathcal{C}$, we will use $|\\mathcal{C}|$ to denote the number of non-empty piles in $\\mathcal{C}$. Given two configurations $\\mathcal{C}_{1}, \\mathcal{C}_{2}$, we will say that $\\mathcal{C}_{2}$ is reachable from $\\mathcal{C}_{1}$ if $\\mathcal{C}_{2}$ can be obtained by performing a finite sequence of moves starting from $\\mathcal{C}_{1}$. Call a configuration of piles $\\mathcal{C}$\n\n- simple if each (non-empty) pile in $\\mathcal{C}$ consists of a single pebble;\n- good if at least one (non-empty) pile in $\\mathcal{C}$ has an even number of pebbles and the numbers of pebbles on the piles in $\\mathcal{C}$ have no non-trivial odd common divisor $(\\mathcal{C}$ has the odd divisor property);\n- solvable if there exists a simple configuration which is reachable from $\\mathcal{C}$.\n\nThe problem asks to find the smallest number of non-empty piles in a solvable configuration consisting of $n$ pebbles. We begin the process of answering this question by making the following observation:\n\nLemma 1. Let $\\mathcal{C}$ be a configuration of piles. Let $\\mathcal{C}^{\\prime}$ be a configuration obtained by applying a single move to $\\mathcal{C}$. Then\n\n(i) if $\\mathcal{C}^{\\prime}$ has the odd divisor property, then so does $\\mathcal{C}$;\n\n(ii) the converse to (i) holds if $\\left|\\mathcal{C}^{\\prime}\\right| \\geqslant|\\mathcal{C}|$.\n\n\n\nProof. Suppose that the move consists of splitting a pile of size $2 a$ and adding $a$ pebbles to each of two piles of sizes $b$ and $c$. Here, $a$ is a positive integer and $b, c$ are non-negative integers. Thus, $\\mathcal{C}^{\\prime}$ can be obtained from $\\mathcal{C}$ by replacing the piles of sizes $2 a, b, c$ by two piles of sizes $a+b$ and $a+c$. Note that the extra assumption $\\left|\\mathcal{C}^{\\prime}\\right| \\geqslant|\\mathcal{C}|$ of part (ii) holds if and only if at least one of $b, c$ is zero.\n\n(i) Suppose $\\mathcal{C}$ doesn't have the odd divisor property, i.e. there exists an odd integer $d>1$ such that the size of each pile in $\\mathcal{C}$ is divisible by $d$. In particular, $2 a, b, c$ are multiples of $d$, so since $d$ is odd, it follows that $a, b, c$ are all divisible by $d$. Thus, $a+b$ and $a+c$ are also divisible by $d$, so $d$ divides the size of each pile in $\\mathcal{C}^{\\prime}$. In conclusion, $\\mathcal{C}^{\\prime}$ doesn't have the odd divisor property.\n\n(ii) If $\\mathcal{C}^{\\prime}$ doesn't have the odd divisor property and at least one of $b, c$ is zero, then there exists an odd integer $d>1$ such that the size of each pile in $\\mathcal{C}^{\\prime}$ is divisible by $d$. In particular, $d$ divides $a+b$ and $a+c$, so since at least one of these numbers is equal to $a$, it follows that $d$ divides $a$. But then $d$ must divide all three of $a, b$ and $c$, and hence it certainly divides $2 a, b$ and $c$. Thus, $\\mathcal{C}$ doesn't have the odd divisor property, as desired.\n\nLemma 2. If $\\mathcal{C}_{2}$ is reachable from $\\mathcal{C}_{1}$ and $\\mathcal{C}_{2}$ has the odd divisor property, then so does $\\mathcal{C}_{1}$. In particular, any solvable configuration has the odd divisor property.\n\nProof. The first statement follows by inductively applying part (i) of Lemma 1. The second statement follows from the first because every simple configuration has the odd divisor property.\n\nThe main claim is the following:\n\nLemma 3. Let $\\mathcal{C}$ be a good configuration. Then there exists a configuration $\\mathcal{C}^{\\prime}$ with the following properties:\n\n- $\\mathcal{C}^{\\prime}$ is reachable from $\\mathcal{C}$ and $\\left|\\mathcal{C}^{\\prime}\\right|>|\\mathcal{C}| ;$\n- $\\mathcal{C}^{\\prime}$ is either simple or good.\n\nProof. Call a configuration terminal if it is a counterexample to the claim. The following claim is at the heart of the proof:\n\nClaim. Let $a_{1}, \\ldots, a_{k}$ be the numbers of pebbles on the non-empty piles of a terminal configuration $\\mathcal{C}$. Then there exists a unique $i \\in[k]$ such that $a_{i}$ is even. Moreover, for all $t \\geqslant 1$ we have $a_{j} \\equiv \\frac{a_{i}}{2}\\left(\\bmod 2^{t}\\right)$ for all $j \\neq i$.\n\nProof of Claim. Since the configuration is good, there must exist $i \\in[k]$ such that $a_{i}$ is even. Moreover, by assumption, if we split the pile with $a_{i}$ pebbles into two equal halves, the resulting configuration will not be good. By part (ii) of Lemma 2,the only way this can happen is that $\\frac{a_{i}}{2}$ and $a_{j}$ for all $j \\neq i$ are odd. To prove the second assertion, we proceed by induction on $t$, with the case $t=1$ already being established. If $t \\geqslant 2$, then split the pile with $a_{i}$ pebbles into two equal halves and move one half to the pile with $a_{j}$ pebbles. Since $\\frac{a_{i}}{2}$ and $a_{j}$ are both odd, $a_{j}+\\frac{a_{i}}{2}$ is even, so by part (ii) of Lemma 2, the resulting configuration $\\mathcal{C}^{\\prime}$ is good. Thus, $\\mathcal{C}^{\\prime}$ is terminal, so by the induction hypothesis, we have $\\frac{a_{i}}{2} \\equiv \\frac{1}{2}\\left(a_{j}+\\frac{a_{i}}{2}\\right)\\left(\\bmod 2^{t-1}\\right)$, whence $a_{j} \\equiv \\frac{a_{i}}{2}($ $\\bmod 2^{t}$ ), as desired.\n\nSuppose for contradiction that there exists a configuration as in the Claim. It follows that there exists $i \\in[k]$ and an odd integer $x$ such that $a_{i}=2 x$ and $a_{j}=x$ for all $j \\neq i$. Thus, $x$ is an odd common divisor of $a_{1}, \\ldots, a_{k}$, so by the odd divisor property, we must have $x=1$. But then we can obtain a simple configuration by splitting the only pile with two pebbles into two piles each consisting of a single pebble, which is a contradiction.\n\nWith the aid of Lemmas 2 and 3, it is not hard to characterise all solvable configurations: Lemma 4. A configuration of piles $\\mathcal{C}$ is solvable if and only if it is simple or good.\n\n\n\nProof. $(\\Longrightarrow)$ Suppose $\\mathcal{C}$ is not simple. Then since we have to be able to perform at least one move, there must be at least one non-empty pile in $\\mathcal{C}$ with an even number of pebbles. Moreover, by Lemma $2, \\mathcal{C}$ has the odd divisor property, so it must be good.\n\n$(\\Longleftarrow)$ This follows by repeatedly applying Lemma 3 until we reach a simple configuration. Note that the process must stop eventually since the number of non-empty piles is bounded from above.\n\nFinally, the answer to the problem is implied by the following corollary of Lemma 4:\n\nLemma 5. Let $n$ be a positive integer. Then\n\n(i) a configuration consisting of a single pile of $n$ pebbles is solvable if and only if $n$ is a power of two;\n\n(ii) if $n \\geqslant 2$, then the configuration consisting of piles of sizes 2 and $n-2$ is good.\n\nProof. (i) By Lemma 4, this configuration is solvable if and only if either $n=1$ or $n$ is even and has no non-trivial odd divisor, so the conclusion follows.\n\n(ii) Since 2 is even and has no non-trivial odd divisor, this configuration is certainly good, so the conclusion follows by Lemma 4.\n\nCommon remarks. Instead of choosing pebbles from two piles, one could allow choosing an equal number of pebbles from each of $k$ piles, where $k \\geqslant 2$ is a fixed (prime) integer. However, this seems to yield a much harder problem - if $k=3$, numerical evidence suggests the same answer as in the case $k=2$ (with powers of two replaced by powers of three), but the case $k=5$ is already unclear.""]","['1 if $n$ is a power of two, and 2 otherwise']",True,,Need_human_evaluate, 2010,Combinatorics,,"Lucy starts by writing $s$ integer-valued 2022-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\mathbf{v}=\left(v_{1}, \ldots, v_{2022}\right)$ and $\mathbf{w}=\left(w_{1}, \ldots, w_{2022}\right)$ that she has already written, and apply one of the following operations to obtain a new tuple: $$ \begin{aligned} & \mathbf{v}+\mathbf{w}=\left(v_{1}+w_{1}, \ldots, v_{2022}+w_{2022}\right) \\ & \mathbf{v} \vee \mathbf{w}=\left(\max \left(v_{1}, w_{1}\right), \ldots, \max \left(v_{2022}, w_{2022}\right)\right) \end{aligned} $$ and then write this tuple on the blackboard. It turns out that, in this way, Lucy can write any integer-valued 2022-tuple on the blackboard after finitely many steps. What is the smallest possible number $s$ of tuples that she initially wrote?","['We solve the problem for $n$-tuples for any $n \\geqslant 3$ : we will show that the answer is $s=3$, regardless of the value of $n$.\n\nFirst, let us briefly introduce some notation. For an $n$-tuple $\\mathbf{v}$, we will write $\\mathbf{v}_{i}$ for its $i$-th coordinate (where $1 \\leqslant i \\leqslant n$ ). For a positive integer $n$ and a tuple $\\mathbf{v}$ we will denote by $n \\cdot \\mathbf{v}$ the tuple obtained by applying addition on $\\mathbf{v}$ with itself $n$ times. Furthermore, we denote by $\\mathbf{e}(i)$ the tuple which has $i$-th coordinate equal to one and all the other coordinates equal to zero. We say that a tuple is positive if all its coordinates are positive, and negative if all its coordinates are negative.\n\nWe will show that three tuples suffice, and then that two tuples do not suffice.\n\n**Three tuples suffice.** Write $\\mathbf{c}$ for the constant-valued tuple $\\mathbf{c}=(-1, \\ldots,-1)$.\n\nWe note that it is enough for Lucy to be able to make the tuples $\\mathbf{e}(1), \\ldots, \\mathbf{e}(n)$, $\\mathbf{c}$; from those any other tuple $\\mathbf{v}$ can be made as follows. First we choose some positive integer $k$ such that $k+\\mathbf{v}_{i}>0$ for all $i$. Then, by adding a positive number of copies of $\\mathbf{c}, \\mathbf{e}(1), \\ldots, \\mathbf{e}(n)$, she can make\n\n$$\nk \\mathbf{c}+\\left(k+\\mathbf{v}_{1}\\right) \\cdot \\mathbf{e}(1)+\\cdots+\\left(k+\\mathbf{v}_{n}\\right) \\cdot \\mathbf{e}(n)\n$$\n\nwhich we claim is equal to $\\mathbf{v}$. Indeed, this can be checked by comparing coordinates: the $i$-th coordinate of the right-hand side is $-k+\\left(k+\\mathbf{v}_{i}\\right)=\\mathbf{v}_{i}$ as needed.\n\nLucy can take her three starting tuples to be $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$, such that $\\mathbf{a}_{i}=-i^{2}, \\mathbf{b}_{i}=i$ and $\\mathbf{c}=-1$ (as above).\n\nFor any $1 \\leqslant j \\leqslant n$, write $\\mathbf{d}(j)$ for the tuple $2 \\cdot \\mathbf{a}+4 j \\cdot \\mathbf{b}+\\left(2 j^{2}-1\\right) \\cdot \\mathbf{c}$, which Lucy can make by adding together $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ repeatedly. This has $i$ th term\n\n$$\n\\begin{aligned}\n\\mathbf{d}(j)_{i} & =2 \\mathbf{a}_{i}+4 j \\mathbf{b}_{i}+\\left(2 j^{2}-1\\right) \\mathbf{c}_{i} \\\\\n& =-2 i^{2}+4 i j-\\left(2 j^{2}-1\\right) \\\\\n& =1-2(i-j)^{2}\n\\end{aligned}\n$$\n\nThis is 1 if $j=i$, and at most -1 otherwise. Hence Lucy can produce the tuple $\\mathbf{1}=(1, \\ldots, 1)$ as $\\mathbf{d}(1) \\vee \\cdots \\vee \\mathbf{d}(n)$.\n\nShe can then produce the constant tuple $\\mathbf{0}=(0, \\ldots, 0)$ as $\\mathbf{1}+\\mathbf{c}$, and for any $1 \\leqslant j \\leqslant n$ she can then produce the tuple $\\mathbf{e}(j)$ as $\\mathbf{d}(j) \\vee \\mathbf{0}$. Since she can now produce $\\mathbf{e}(1), \\ldots, \\mathbf{e}(n)$ and already had $\\mathbf{c}$, she can (as we argued earlier) produce any integer-valued tuple.\n\n\n\n**Two tuples do not suffice.** We start with an observation: Let $a$ be a non-negative real number and suppose that two tuples $\\mathbf{v}$ and $\\mathbf{w}$ satisfy $\\mathbf{v}_{j} \\geqslant a \\mathbf{v}_{k}$ and $\\mathbf{w}_{j} \\geqslant a \\mathbf{w}_{k}$ for some $1 \\leqslant j, k \\leqslant n$. Then we claim that the same inequality holds for $\\mathbf{v}+\\mathbf{w}$ and $\\mathbf{v} \\vee \\mathbf{w}$ : Indeed, the property for the sum is verified by an easy computation:\n\n$$\n(\\mathbf{v}+\\mathbf{w})_{j}=\\mathbf{v}_{j}+\\mathbf{w}_{j} \\geqslant a \\mathbf{v}_{k}+a \\mathbf{w}_{k}=a(\\mathbf{v}+\\mathbf{w})_{k}\n$$\n\nFor the second operation, we denote by $\\mathbf{m}$ the tuple $\\mathbf{v} \\vee \\mathbf{w}$. Then $\\mathbf{m}_{j} \\geqslant \\mathbf{v}_{j} \\geqslant a \\mathbf{v}_{k}$ and $\\mathbf{m}_{j} \\geqslant \\mathbf{w}_{j} \\geqslant a \\mathbf{w}_{k}$. Since $\\mathbf{m}_{k}=\\mathbf{v}_{k}$ or $\\mathbf{m}_{k}=\\mathbf{w}_{k}$, the observation follows.\n\nAs a consequence of this observation we have that if all starting tuples satisfy such an inequality, then all generated tuples will also satisfy it, and so we would not be able to obtain every integer-valued tuple.\n\nLet us now prove that Lucy needs at least three starting tuples. For contradiction, let us suppose that Lucy started with only two tuples $\\mathbf{v}$ and $\\mathbf{w}$. We are going to distinguish two cases. In the first case, suppose we can find a coordinate $i$ such that $\\mathbf{v}_{i}, \\mathbf{w}_{i} \\geqslant 0$. Both operations preserve the sign, thus we can not generate any tuple that has negative $i$-th coordinate. Similarly for $\\mathbf{v}_{i}, \\mathbf{w}_{i} \\leqslant 0$.\n\nSuppose the opposite, i.e., for every $i$ we have either $\\mathbf{v}_{i}>0>\\mathbf{w}_{i}$, or $\\mathbf{v}_{i}<0<\\mathbf{w}_{i}$. Since we assumed that our tuples have at least three coordinates, by pigeonhole principle there exist two coordinates $j \\neq k$ such that $\\mathbf{v}_{j}$ has the same sign as $\\mathbf{v}_{k}$ and $\\mathbf{w}_{j}$ has the same sign as $\\mathbf{w}_{k}$ (because there are only two possible combinations of signs).\n\nWithout loss of generality assume that $\\mathbf{v}_{j}, \\mathbf{v}_{k}>0$ and $\\mathbf{w}_{j}, \\mathbf{w}_{k}<0$. Let us denote the positive real number $\\mathbf{v}_{j} / \\mathbf{v}_{k}$ by $a$. If $\\mathbf{w}_{j} / \\mathbf{w}_{k} \\leqslant a$, then both inequalities $\\mathbf{v}_{j} \\geqslant a \\mathbf{v}_{k}$ and $\\mathbf{w}_{j} \\geqslant a \\mathbf{w}_{k}$ are satisfied. On the other hand, if $\\mathbf{w}_{j} / \\mathbf{w}_{k} \\leqslant a$, then both $\\mathbf{v}_{k} \\geqslant(1 / a) \\mathbf{v}_{j}$ and $\\mathbf{w}_{k} \\geqslant(1 / a) \\mathbf{w}_{j}$ are satisfied. In either case, we have found the desired inequality satisfied by both starting tuples, a contradiction with the observation above.']",['3'],False,,Numerical, 2011,Combinatorics,,"Alice fills the fields of an $n \times n$ board with numbers from 1 to $n^{2}$, each number being used exactly once. She then counts the total number of good paths on the board. A good path is a sequence of fields of arbitrary length (including 1) such that: (i) The first field in the sequence is one that is only adjacent to fields with larger numbers, (ii) Each subsequent field in the sequence is adjacent to the previous field, (iii) The numbers written on the fields in the sequence are in increasing order. Two fields are considered adjacent if they share a common side. Find the smallest possible number of good paths Alice can obtain, as a function of $n$.","['We will call any field that is only adjacent to fields with larger numbers a well. Other fields will be called non-wells. Let us make a second $n \\times n$ board $B$ where in each field we will write the number of good sequences which end on the corresponding field in the original board $A$. We will thus look for the minimal possible value of the sum of all entries in $B$.\n\nWe note that any well has just one good path ending in it, consisting of just the well, and that any other field has the number of good paths ending in it equal to the sum of this quantity for all the adjacent fields with smaller values, since a good path can only come into the field from a field of lower value. Therefore, if we fill in the fields in $B$ in increasing order with respect to their values in $A$, it follows that each field not adjacent to any already filled field will receive a 1, while each field adjacent to already filled fields will receive the sum of the numbers already written on these adjacent fields.\n\nWe note that there is at least one well in $A$, that corresponding with the field with the entry 1 in $A$. Hence, the sum of values of fields in $B$ corresponding to wells in $A$ is at least 1 . We will now try to minimize the sum of the non-well entries, i.e. of the entries in $B$ corresponding to the non-wells in $A$. We note that we can ascribe to each pair of adjacent fields the value of the lower assigned number and that the sum of non-well entries will then equal to the sum of the ascribed numbers. Since the lower number is still at least 1, the sum of non-well entries will at least equal the number of pairs of adjacent fields, which is $2 n(n-1)$. Hence, the total minimum sum of entries in $B$ is at least $2 n(n-1)+1=2 n^{2}-2 n+1$. The necessary conditions for the minimum to be achieved is for there to be only one well and for no two entries in $B$ larger than 1 to be adjacent to each other.\n\nWe will now prove that the lower limit of $2 n^{2}-2 n+1$ entries can be achieved. This amounts to finding a way of marking a certain set of squares, those that have a value of 1 in $B$, such that no two unmarked squares are adjacent and that the marked squares form a connected tree with respect to adjacency.\n\nFor $n=1$ and $n=2$ the markings are respectively the lone field and the L-trimino. Now, for $n>2$, let $s=2$ for $n \\equiv 0,2 \\bmod 3$ and $s=1$ for $n \\equiv 1 \\bmod 3$. We will take indices $k$ and $l$ to be arbitrary non-negative integers. For $n \\geqslant 3$ we will construct a path of marked squares in the first two columns consisting of all squares of the form $(1, i)$ where $i$ is not of the form $6 k+s$ and $(2, j)$ where $j$ is of the form $6 k+s-1,6 k+s$ or $6+s+1$. Obviously, this path is connected. Now, let us consider the fields $(2,6 k+s)$ and $(1,6 k+s+3)$. For each considered field $(i, j)$ we will mark all squares of the form $(l, j)$ for $l>i$ and $(i+2 k, j \\pm 1)$. One can easily see that no set of marked fields will produce a cycle, that the only fields of the unmarked form $(1,6 k+s),(2+2 l+1,6 k+s \\pm 1)$ and $(2+2 l, 6 k+s+3 \\pm 1)$ and that no two are adjacent, since\n\n\n\nthe consecutive considered fields are in columns of opposite parity. Examples of markings are given for $n=3,4,5,6,7$, and the corresponding constructions for $A$ and $B$ are given for $n=5$.\n']",['$2 n^{2}-2 n+1$'],False,,Expression, 2012,Combinatorics,,"Let $\mathbb{Z}_{\geqslant 0}$ be the set of non-negative integers, and let $f: \mathbb{Z}_{\geqslant 0} \times \mathbb{Z}_{\geqslant 0} \rightarrow \mathbb{Z}_{\geqslant 0}$ be a bijection such that whenever $f\left(x_{1}, y_{1}\right)>f\left(x_{2}, y_{2}\right)$, we have $f\left(x_{1}+1, y_{1}\right)>f\left(x_{2}+1, y_{2}\right)$ and $f\left(x_{1}, y_{1}+1\right)>f\left(x_{2}, y_{2}+1\right)$. Let $N$ be the number of pairs of integers $(x, y)$, with $0 \leqslant x, y<100$, such that $f(x, y)$ is odd. Find the smallest and largest possible value of $N$.","['We defer the constructions to the end of the solution. Instead, we begin by characterizing all such functions $f$, prove a formula and key property for such functions, and then solve the problem, providing constructions.\n\n**Characterization** Suppose $f$ satisfies the given relation. The condition can be written more strongly as\n\n$$\n\\begin{aligned}\nf\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right) & \\Longleftrightarrow f\\left(x_{1}+1, y_{1}\\right)>f\\left(x_{2}+1, y_{2}\\right) \\\\\n& \\Longleftrightarrow f\\left(x_{1}, y_{1}+1\\right)>f\\left(x_{2}, y_{2}+1\\right)\n\\end{aligned}\n$$\n\nIn particular, this means for any $(k, l) \\in \\mathbb{Z}^{2}, f(x+k, y+l)-f(x, y)$ has the same sign for all $x$ and $y$.\n\nCall a non-zero vector $(k, l) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0}$ a needle if $f(x+k, y)-f(x, y+l)>0$ for all $x$ and $y$. It is not hard to see that needles and non-needles are both closed under addition, and thus under scalar division (whenever the quotient lives in $\\mathbb{Z}^{2}$ ).\n\nIn addition, call a positive rational number $\\frac{k}{l}$ a grade if the vector $(k, l)$ is a needle. (Since needles are closed under scalar multiples and quotients, this is well-defined.)\n\nClaim. Grades are closed upwards.\n\nProof. Consider positive rationals $k_{1} / l_{1}0$ and $(1,0)$ is a needle).\n\nThus $\\left(k_{2}, l_{2}\\right)$ is a needle, as wanted.\n\nClaim. A grade exists.\n\nProof. If no positive integer $n$ is a grade, then $f(1,0)>f(0, n)$ for all $n$ which is impossible.\n\nSimilarly, there is an $n$ such that $f(0,1)x_{2}+y_{2} \\alpha$ then $f\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right)$.\n\nProof. If both $x_{1} \\geqslant x_{2}$ and $y_{1} \\geqslant y_{2}$ this is clear.\n\nSuppose $x_{1} \\geqslant x_{2}$ and $y_{1}\\alpha$ is a grade. This gives $f\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right)$. Suppose $x_{1}0} \\mid x+(y+1) \\alpha \\leqslant a+b \\alpha<(x+1)+(y+1) \\alpha\\right\\}+ \\\\\n\\#\\left\\{(a, 0) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid(x+1)+y \\alpha \\leqslant a<(x+1)+(y+1) \\alpha\\right\\}= \\\\\n\\#\\left\\{(a, b) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid x+y \\alpha \\leqslant a+b \\alpha<(x+1)+y \\alpha\\right\\}+1=f(x+1, y)-f(x, y) .\n\\end{gathered}\n$$\n\nFrom this claim we immediately get that $2500 \\leqslant N \\leqslant 7500$; now we show that those bounds are indeed sharp.\n\nRemember that if $\\alpha$ is irrational then\n\n$$\nf(a, b)=\\#\\left\\{(x, y) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid x+y \\alpha\n\nThe numbers mod 2 in the construction for 7500.\n\nConstruction for 2500 Select $\\alpha \\approx 200.001$.\n\nClaim. \n\n1. $f(n, 0)=n$ for $0 \\leqslant n \\leqslant 100$.\n2. $f(0, k) \\equiv 0 \\bmod 2$ for $0 \\leqslant k \\leqslant 100$.\n\nProof.\n\n1. As above.\n2. Similarly to the above:\n\n$$\n\\begin{aligned}\nf(0, k) & =\\#\\{(x, y) \\mid x+y \\alpha\n\nThe numbers mod 2 in the construction for 7500.']","['2500,7500']",True,,Numerical, 2013,Geometry,,"Let $A B C D E$ be a convex pentagon such that $B C=D E$. Assume there is a point $T$ inside $A B C D E$ with $T B=T D, T C=T E$ and $\angle T B A=\angle A E T$. Let lines $C D$ and $C T$ intersect line $A B$ at points $P$ and $Q$, respectively, and let lines $C D$ and $D T$ intersect line $A E$ at points $R$ and $S$, respectively. Assume that points $P, B, A, Q$ and $R, E, A, S$ respectively, are collinear and occur on their lines in this order. Prove that the points $P, S, Q, R$ are concyclic.","['By the conditions we have $B C=D E, C T=E T$ and $T B=T D$, so the triangles $T B C$ and $T D E$ are congruent, in particular $\\angle B T C=\\angle D T E$.\n\nIn triangles $T B Q$ and $T E S$ we have $\\angle T B Q=\\angle S E T$ and $\\angle Q T B=180^{\\circ}-\\angle B T C=180^{\\circ}-$ $\\angle D T E=\\angle E T S$, so these triangles are similar to each other. It follows that $\\angle T S E=\\angle B Q T$ and\n\n$$\n\\frac{T D}{T Q}=\\frac{T B}{T Q}=\\frac{T E}{T S}=\\frac{T C}{T S}\n$$\n\nBy rearranging this relation we get $T D \\cdot T S=T C \\cdot T Q$, so $C, D, Q$ and $S$ are concyclic. (Alternatively, we can get $\\angle C Q D=\\angle C S D$ from the similar triangles $T C S$ and $T D Q$.) Hence, $\\angle D C Q=\\angle D S Q$.\n\nFinally, from the angles of triangle $C Q P$ we get\n\n$$\n\\angle R P Q=\\angle R C Q-\\angle P Q C=\\angle D S Q-\\angle D S R=\\angle R S Q\n$$\n\nwhich proves that $P, Q, R$ and $S$ are concyclic.\n\n', 'As in the previous solution, we note that triangles $T B C$ and $T D E$ are congruent. Denote the intersection point of $D T$ and $B A$ by $V$, and the intersection point of $C T$ and $E A$ by $W$. From triangles $B C Q$ and $D E S$ we then have\n\n$$\n\\begin{aligned}\n\\angle V S W & =\\angle D S E=180^{\\circ}-\\angle S E D-\\angle E D S=180^{\\circ}-\\angle A E T-\\angle T E D-\\angle E D T \\\\\n& =180^{\\circ}-\\angle T B A-\\angle T C B-\\angle C B T=180^{\\circ}-\\angle Q C B-\\angle C B Q=\\angle B Q C=\\angle V Q W,\n\\end{aligned}\n$$\n\nmeaning that $V S Q W$ is cyclic, and in particular $\\angle W V Q=\\angle W S Q$. Since\n\n$$\n\\angle V T B=180^{\\circ}-\\angle B T C-\\angle C T D=180^{\\circ}-\\angle C T D-\\angle D T E=\\angle E T W\n$$\n\nand $\\angle T B V=\\angle W E T$ by assumption, we have that the triangles $V T B$ and $W T E$ are similar, hence\n\n$$\n\\frac{V T}{W T}=\\frac{B T}{E T}=\\frac{D T}{C T}\n$$\n\n\n\nThus $C D \\| V W$, and angle chasing yields\n\n$$\n\\angle R P Q=\\angle W V Q=\\angle W S Q=\\angle R S Q\n$$\n\nconcluding the proof.']",,True,,, 2014,Geometry,,"In the acute-angled triangle $A B C$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $A F$. The lines through $P$ parallel to $A C$ and $A B$ meet $B C$ at $D$ and $E$, respectively. Points $X \neq A$ and $Y \neq A$ lie on the circles $A B D$ and $A C E$, respectively, such that $D A=D X$ and $E A=E Y$. Prove that $B, C, X$ and $Y$ are concyclic.","['Let $A^{\\prime}$ be the intersection of lines $B X$ and $C Y$. By power of a point, it suffices to prove that $A^{\\prime} B \\cdot A^{\\prime} X=A^{\\prime} C \\cdot A^{\\prime} Y$, or, equivalently, that $A^{\\prime}$ lies on the radical axis of the circles $A B D X$ and $A C E Y$.\n\nFrom $D A=D X$ it follows that in circle $A B D X$, point $D$ bisects of one of the arcs $A X$. Therefore, depending on the order of points, the line $B C$ is either the internal or external bisector of $\\angle A B X$. In both cases, line $B X$ is the reflection of $B A$ in line $B D C$. Analogously, line $C Y$ is the reflection of $C A$ in line $B C$; we can see that $A^{\\prime}$ is the reflection of $A$ in line $B C$, so $A, F$ and $A^{\\prime}$ are collinear.\n\nBy $P D \\| A C$ and $P E \\| A B$ we have $\\frac{F D}{F C}=\\frac{F P}{F A}=\\frac{F E}{F B}$, hence $F D \\cdot F B=F E \\cdot F C$. So, point $F$ has equal powers with respect to circles $A B D X$ and $A C E Y$.\n\nPoint $A$, being a common point of the two circles, is another point with equal powers, so the radical axis of circles $A B D X$ and $A C E Y$ is the altitude $A F$ that passes through $A^{\\prime}$.\n\n']",,True,,, 2015,Geometry,,"Let $A B C D$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $A C$ is tangent to the circle $A D Q$, and the line $B D$ is tangent to the circle $B C P$. Let $M$ and $N$ be the midpoints of $B C$ and $A D$, respectively. Prove that the following three lines are concurrent: line $C D$, the tangent of circle $A N Q$ at point $A$, and the tangent to circle $B M P$ at point $B$.","['We first prove that triangles $A D Q$ and $C D B$ are similar. Since $A B C D$ is cyclic, we have $\\angle D A Q=\\angle D C B$. By the tangency of $A C$ to the circle $A Q D$ we also have $\\angle C B D=\\angle C A D=\\angle A Q D$. The claimed similarity is proven.\n\nLet $R$ be the midpoint of $C D$. Points $N$ and $R$ correspond in the proven similarity, and so $\\angle Q N A=\\angle B R C$.\n\n\n\nLet $K$ be the second common point of line $C D$ with circle $A B R$ (i.e., if $C D$ intersects circle $A B R$, then $K \\neq R$ is the other intersection; otherwise, if $C D$ is tangent to $C D$, then $K=R$ ). In both cases, we have $\\angle B A K=\\angle B R C=\\angle Q N A$; that indicates that $A K$ is tangent to circle $A N Q$. It can be showed analogously that $B K$ is tangent to circle $B M P$.', 'We present a second solution, without using the condition that $A B C D$ is cyclic. Again, $M$ and $N$ can be any points on lines $B C$ and $A D$ such that $B M: M C=D N: N A$.\n\nLet $A B$ and $C D$ meet at $T$ (if $A B \\| C D$ then $T$ is their common ideal point). Let $C D$ meet the tangent to the circle $A N Q$ at $A$, and the tangent to the circle $B M P$ at $B$ at points $K_{1}$ and $K_{2}$, respectively.\n\n\n\nLet $I$ and $J$ be the ideal points of $A D$ and $B C$, respectively. Notice that the pencils $\\left(A D, A C, A T, A K_{1}\\right)$ and $(Q A, Q D, Q I, Q N)$ of lines are congruent, because $\\angle K_{1} A D=\\angle A Q N$, $\\angle C A D=\\angle A Q D$ and $\\angle I A T=\\angle I Q T$. Hence,\n\n$$\n\\left(D, C ; T, K_{1}\\right)=\\left(A D, A C ; A T, A K_{1}\\right)=(Q A, Q D ; Q I, Q N)=(A, D ; I, N)=\\frac{D N}{N A}\n$$\n\nIt can be obtained analogously that\n\n$$\n\\left(D, C ; T, K_{2}\\right)=\\left(B D, B C ; B T, B K_{2}\\right)=(P C, P B ; P J, P M)=(C, B ; I, N)=\\frac{B M}{M C}\n$$\n\nFrom $B M: M C=D N: D A$ we get $\\left(D, C ; T, K_{1}\\right)=\\left(D, C ; T, K_{2}\\right)$ and hence $K_{1}=K_{2}$.']",,True,,, 2016,Geometry,,"Let $A B C$ be an acute-angled triangle with $A C>A B$, let $O$ be its circumcentre, and let $D$ be a point on the segment $B C$. The line through $D$ perpendicular to $B C$ intersects the lines $A O, A C$ and $A B$ at $W, X$ and $Y$, respectively. The circumcircles of triangles $A X Y$ and $A B C$ intersect again at $Z \neq A$. Prove that if $O W=O D$, then $D Z$ is tangent to the circle $A X Y$.","['Let $A O$ intersect $B C$ at $E$. As $E D W$ is a right-angled triangle and $O$ is on $W E$, the condition $O W=O D$ means $O$ is the circumcentre of this triangle. So $O D=O E$ which establishes that $D, E$ are reflections in the perpendicular bisector of $B C$.\n\nNow observe:\n\n$$\n180^{\\circ}-\\angle D X Z=\\angle Z X Y=\\angle Z A Y=\\angle Z C D \\text {, }\n$$\n\nwhich shows $C D X Z$ is cyclic.\n\n\n\nWe next show that $A Z \\| B C$. To do this, introduce point $Z^{\\prime}$ on circle $A B C$ such that $A Z^{\\prime} \\| B C$. By the previous result, it suffices to prove that $C D X Z^{\\prime}$ is cyclic. Notice that triangles $B A E$ and $C Z^{\\prime} D$ are reflections in the perpendicular bisector of $B C$. Using this and that $A, O, E$ are collinear:\n\n$$\n\\angle D Z^{\\prime} C=\\angle B A E=\\angle B A O=90^{\\circ}-\\frac{1}{2} \\angle A O B=90^{\\circ}-\\angle C=\\angle D X C,\n$$\n\nso $D X Z^{\\prime} C$ is cyclic, giving $Z \\equiv Z^{\\prime}$ as desired.\n\nUsing $A Z \\| B C$ and $C D X Z$ cyclic we get:\n\n$$\n\\angle A Z D=\\angle C D Z=\\angle C X Z=\\angle A Y Z\n$$\n\nwhich by the converse of alternate segment theorem shows $D Z$ is tangent to circle $A X Y$.', 'Notice that point $Z$ is the Miquel-point of lines $A C, B C, B A$ and $D Y$; then $B, D, Z, Y$ and $C, D, X, Y$ are concyclic. Moreover, $Z$ is the centre of the spiral similarity that maps $B C$ to $Y X$.\n\nBy $B C \\perp Y X$, the angle of that similarity is $90^{\\circ}$; hence the circles $A B C Z$ and $A X Y Z$ are perpendicular, therefore the radius $O Z$ in circle $A B C Z$ is tangent to circle $A X Y Z$.\n\n\n\nBy $O W=O D$, the triangle $O W D$ is isosceles, and\n\n$$\n\\angle Z O A=2 \\angle Z B A=2 \\angle Z B Y=2 \\angle Z D Y=\\angle O D W+\\angle D W O\n$$\n\nso $D$ lies on line $Z O$ that is tangent to circle $A X Y$.']",,True,,, 2017,Geometry,,"Let $A B C$ be a triangle, and let $\ell_{1}$ and $\ell_{2}$ be two parallel lines. For $i=1,2$, let $\ell_{i}$ meet the lines $B C, C A$, and $A B$ at $X_{i}, Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $B C$, the line through $Y_{i}$ perpendicular to $C A$, and finally the line through $Z_{i}$ perpendicular to $A B$, determine a non-degenerate triangle $\Delta_{i}$. Show that the circumcircles of $\Delta_{1}$ and $\Delta_{2}$ are tangent to each other.","['Throughout the solutions, $\\sphericalangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\\circ}$.\n\nLet the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, and denote the circumcircle of $\\Delta_{i}$ by $\\omega_{i}$.\n\nIn triangles $D_{1} Y_{1} Z_{1}$ and $D_{2} Y_{2} Z_{2}$ we have $Y_{1} Z_{1} \\| Y_{2} Z_{2}$ because they are parts of $\\ell_{1}$ and $\\ell_{2}$. Moreover, $D_{1} Y_{1} \\| D_{2} Y_{2}$ are perpendicular to $A C$ and $D_{1} Z_{1} \\| D_{2} Z_{2}$ are perpendicular to $A B$, so the two triangles are homothetic and their homothetic centre is $Y_{1} Y_{2} \\cap Z_{1} Z_{2}=A$. Hence, line $D_{1} D_{2}$ passes through $A$. Analogously, line $E_{1} E_{2}$ passes through $B$ and $F_{1} F_{2}$ passes through $C$.\n\n\n\nThe corresponding sides of $\\Delta_{1}$ and $\\Delta_{2}$ are parallel, because they are perpendicular to the respective sides of triangle $A B C$. Hence, $\\Delta_{1}$ and $\\Delta_{2}$ are either homothetic, or they can be translated to each other. Using that $B, X_{2}, Z_{2}$ and $E_{2}$ are concyclic, $C, X_{2}, Y_{2}$ and $F_{2}$ are concyclic, $Z_{2} E_{2} \\perp A B$ and $Y_{2}, F_{2} \\perp A C$ we can calculate\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(E_{1} E_{2}, F_{1} F_{2}\\right) & =\\sphericalangle\\left(E_{1} E_{2}, X_{1} X_{2}\\right)+\\sphericalangle\\left(X_{1} X_{2}, F_{1} F_{2}\\right)=\\sphericalangle\\left(B E_{2}, B X_{2}\\right)+\\sphericalangle\\left(C X_{2}, C F_{2}\\right) \\\\\n& =\\sphericalangle\\left(Z_{2} E_{2}, Z_{2} X_{2}\\right)+\\sphericalangle\\left(Y_{2} X_{2}, Y_{2} F_{2}\\right)=\\sphericalangle\\left(Z_{2} E_{2}, \\ell_{2}\\right)+\\sphericalangle\\left(\\ell_{2}, Y_{2} F_{2}\\right) \\\\\n& =\\sphericalangle\\left(Z_{2} E_{2}, Y_{2} F_{2}\\right)=\\sphericalangle(A B, A C) \\not \\equiv 0,\n\\end{aligned}\n\\tag{1}\n$$\n\nand conclude that lines $E_{1} E_{2}$ and $F_{1} F_{2}$ are not parallel. Hence, $\\Delta_{1}$ and $\\Delta_{2}$ are homothetic; the lines $D_{1} D_{2}, E_{1} E_{2}$, and $F_{1} F_{2}$ are concurrent at the homothetic centre of the two triangles. Denote this homothetic centre by $H$.\n\nFor $i=1,2$, using (1), and that $A, Y_{i}, Z_{i}$ and $D_{i}$ are concyclic,\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(H E_{i}, H F_{i}\\right) & =\\sphericalangle\\left(E_{1} E_{2}, F_{1} F_{2}\\right)=\\sphericalangle(A B, A C) \\\\\n& =\\sphericalangle\\left(A Z_{i}, A Y_{i}\\right)=\\sphericalangle\\left(D_{i} Z_{i}, D_{i} Y_{i}\\right)=\\sphericalangle\\left(D_{i} E_{i}, D_{i} F_{i}\\right),\n\\end{aligned}\n$$\n\n\n\nso $H$ lies on circle $\\omega_{i}$.\n\nThe same homothety that maps $\\Delta_{1}$ to $\\Delta_{2}$, sends $\\omega_{1}$ to $\\omega_{2}$ as well. Point $H$, that is the centre of the homothety, is a common point of the two circles, That finishes proving that $\\omega_{1}$ and $\\omega_{2}$ are tangent to each other.', ""As in the first solution, let the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that $E_{i} F_{i}, F_{i} D_{i}$ and $D_{i} E_{i}$ are the perpendiculars through $X_{i}, Y_{i}$ and $Z_{i}$, respectively. In the same way we conclude that $\\left(A, D_{1}, D_{2}\\right),\\left(B, E_{1}, E_{2}\\right)$ and $\\left(C, F_{1}, F_{2}\\right)$ are collinear.\n\nThe corresponding sides of triangles $A B C$ and $D_{i} E_{i} F_{i}$ are perpendicular to each other. Hence, there is a spiral similarity with rotation $\\pm 90^{\\circ}$ that maps $A B C$ to $D_{i} E_{i} F_{i}$; let $M_{i}$ be the centre of that similarity. Hence, $\\sphericalangle\\left(M_{i} A, M_{i} D_{i}\\right)=\\sphericalangle\\left(M_{i} B, M_{i} E_{i}\\right)=\\sphericalangle\\left(M_{i} C, M_{i} F_{i}\\right)=90^{\\circ}$. The circle with diameter $A D_{i}$ passes through $M_{i}, Y_{i}, Z_{i}$, so $M_{i}, A, Y_{i}, Z_{i}, D_{i}$ are concyclic; analogously $\\left(M_{i}, B, X_{i}, Z_{i}, E_{i}\\right)$ and $\\left(M_{i}, C, X_{i}, Y_{i}, F_{i}\\right)$ are concyclic.\n\nBy applying Desargues' theorem to triangles $A B C$ and $D_{i} E_{i} F_{i}$ we conclude that the lines $A D_{i}, B E_{i}$ and $B F_{i}$ are concurrent; let their intersection be $H$. Since $\\left(A, D_{1} . D_{2}\\right),\\left(B, E_{1} \\cdot E_{2}\\right)$ and $\\left(C, F_{1} . F_{2}\\right)$ are collinear, we obtain the same point $H$ for $i=1$ and $i=2$.\n\n\n\nBy $\\sphericalangle(C B, C H)=\\sphericalangle\\left(C X_{i}, C F_{i}\\right)=\\sphericalangle\\left(Y_{i} X_{i}, Y_{i} F_{i}\\right)=\\sphericalangle\\left(Y_{i} Z_{i}, Y_{i} D_{i}\\right)=\\sphericalangle\\left(A Z_{i}, A D_{i}\\right)=$ $\\sphericalangle(A B, A H)$, point $H$ lies on circle $A B C$.\n\nAnalogously, from $\\sphericalangle\\left(F_{i} D_{i}, F_{i} H\\right)=\\sphericalangle\\left(F_{i} Y_{i}, F_{i} C\\right)=\\sphericalangle\\left(X_{i} Y_{i}, X_{i} C\\right)=\\sphericalangle\\left(X_{i} Z_{i}, X_{i} B\\right)=$ $\\sphericalangle\\left(E_{i} Z_{i}, E_{i} B\\right)=\\sphericalangle\\left(E_{i} D_{i}, E_{i} H\\right)$, we can see that point $H$ lies on circle $D_{i} E_{i} F_{i}$ as well. Therefore, circles $A B C$ and $D_{i} E_{i} F_{i}$ intersect at point $H$.\n\nThe spiral similarity moves the circle $A B C$ to circle $D_{i} E_{i} F_{i}$, so the two circles are perpendicular. Hence, both circles $D_{1} E_{1} F_{1}$ and $D_{2} E_{2} F_{2}$ are tangent to the radius of circle $A B C$ at $H$.""]",,True,,, 2018,Geometry,,"In an acute-angled triangle $A B C$, point $H$ is the foot of the altitude from $A$. Let $P$ be a moving point such that the bisectors $k$ and $\ell$ of angles $P B C$ and $P C B$, respectively, intersect each other on the line segment $A H$. Let $k$ and $A C$ meet at $E$, let $\ell$ and $A B$ meet at $F$, and let $E F$ and $A H$ meet at $Q$. Prove that, as $P$ varies, the line $P Q$ passes through a fixed point.","[""Let the reflections of the line $B C$ with respect to the lines $A B$ and $A C$ intersect at point $K$. We will prove that $P, Q$ and $K$ are collinear, so $K$ is the common point of the varying line $P Q$.\n\nLet lines $B E$ and $C F$ intersect at $I$. For every point $O$ and $d>0$, denote by $(O, d)$ the circle centred at $O$ with radius $d$, and define $\\omega_{I}=(I, I H)$ and $\\omega_{A}=(A, A H)$. Let $\\omega_{K}$ and $\\omega_{P}$ be the incircle of triangle $K B C$ and the $P$-excircle of triangle $P B C$, respectively.\n\nSince $I H \\perp B C$ and $A H \\perp B C$, the circles $\\omega_{A}$ and $\\omega_{I}$ are tangent to each other at $H$. So, $H$ is the external homothetic centre of $\\omega_{A}$ and $\\omega_{I}$. From the complete quadrangle $B C E F$ we have $(A, I ; Q, H)=-1$, therefore $Q$ is the internal homothetic centre of $\\omega_{A}$ and $\\omega_{I}$. Since $B A$ and $C A$ are the external bisectors of angles $\\angle K B C$ and $\\angle K C B$, circle $\\omega_{A}$ is the $K$-excircle in triangle $B K C$. Hence, $K$ is the external homothetic centre of $\\omega_{A}$ and $\\omega_{K}$. Also it is clear that $P$ is the external homothetic centre of $\\omega_{I}$ and $\\omega_{P}$. Let point $T$ be the tangency point of $\\omega_{P}$ and $B C$, and let $T^{\\prime}$ be the tangency point of $\\omega_{K}$ and $B C$. Since $\\omega_{I}$ is the incircle and $\\omega_{P}$ is the $P$-excircle of $P B C, T C=B H$ and since $\\omega_{K}$ is the incircle and $\\omega_{A}$ is the $K$-excircle of $K B C, T^{\\prime} C=B H$. Therefore $T C=T^{\\prime} C$ and $T \\equiv T^{\\prime}$. It yields that $\\omega_{K}$ and $\\omega_{P}$ are tangent to each other at $T$.\n\n\n\nLet point $S$ be the internal homothetic centre of $\\omega_{A}$ and $\\omega_{P}$, and let $S^{\\prime}$ be the internal homothetic centre of $\\omega_{I}$ and $\\omega_{K}$. It's obvious that $S$ and $S^{\\prime}$ lie on $B C$. We claim that $S \\equiv S^{\\prime}$. To prove our claim, let $r_{A}, r_{I}, r_{P}$, and $r_{K}$ be the radii of $\\omega_{A}, \\omega_{I}, \\omega_{P}$ and $\\omega_{k}$, respectively.\n\nIt is well known that if the sides of a triangle are $a, b, c$, its semiperimeter is $s=(a+b+c) / 2$, and the radii of the incircle and the $a$-excircle are $r$ and $r_{a}$, respectively, then $r \\cdot r_{a}=(s-b)(s-c)$. Applying this fact to triangle $P B C$ we get $r_{I} \\cdot r_{P}=B H \\cdot C H$. The same fact in triangle $K C B$\n\n\n\nyields $r_{K} \\cdot r_{A}=C T \\cdot B T$. Since $B H=C T$ and $B T=C H$, from these two we get\n\n$$\n\\frac{H S}{S T}=\\frac{r_{A}}{r_{P}}=\\frac{r_{I}}{r_{K}}=\\frac{H S^{\\prime}}{S^{\\prime} T}\n$$\n\nso $S=S^{\\prime}$ indeed.\n\nFinally, by applying the generalised Monge's theorem to the circles $\\omega_{A}, \\omega_{I}$, and $\\omega_{K}$ (with two pairs of internal and one pair of external common tangents), we can see that points $Q$, $S$, and $K$ are collinear. Similarly one can show that $Q, S$ and $P$ are collinear, and the result follows."", ""Again, let $B E$ and $C F$ meet at $I$, that is the incentre in triangle $B C P$; then $P I$ is the third angle bisector. From the tangent segments of the incircle we have $B P-C P=$ $B H-C H$; hence, the possible points $P$ lie on a branch of a hyperbola $\\mathcal{H}$ with foci $B, C$, and $H$ is a vertex of $\\mathcal{H}$. Since $P I$ bisects the angle between the radii $B P$ and $C P$ of the hyperbola, line $P I$ is tangent to $\\mathcal{H}$.\n\n\n\nLet $K$ be the second intersection of $P Q$ and $\\mathcal{H}$, we will show that $A K$ is tangent to $\\mathcal{H}$ at $K$; this property determines $K$.\n\nLet $G=K I \\cap A P$ and $M=P I \\cap A K$. From the complete quadrangle $B C E F$ we can see that $(H, Q ; I, A)$ is harmonic, so in the complete quadrangle $A P I K$, point $H$ lies on line $G M$.\n\nConsider triangle $A I M$. Its side $A I$ is tangent to $\\mathcal{H}$ at $H$, the side $I M$ is tangent to $\\mathcal{H}$ at $P$, and $K$ is a common point of the third side $A M$ and the hyperbola such that the lines $A P$, $I K$ and $M H$ are concurrent at the generalised Gergonne-point $G$. It follows that the third side, $A M$ is also tangent to $\\mathcal{H}$ at $K$.\n\n(Alternatively, in the last step we can apply the converse of Brianchon's theorem to the degenerate hexagon $A H I P M K$. By the theorem there is a conic section $\\mathcal{H}^{\\prime}$ such that lines $A I, I M$ and $M A$ are tangent to $\\mathcal{H}^{\\prime}$ at $H, P$ and $K$, respectively. But the three points $H, K$ and $P$, together with the tangents at $H$ and $P$ uniquely determine $\\mathcal{H}^{\\prime}$, so indeed $\\mathcal{H}^{\\prime}=\\mathcal{H}$.)""]",,True,,, 2019,Geometry,,"Let $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ be two triangles having the same circumcircle $\omega$, and the same orthocentre $H$. Let $\Omega$ be the circumcircle of the triangle determined by the lines $A A^{\prime}, B B^{\prime}$ and $C C^{\prime}$. Prove that $H$, the centre of $\omega$, and the centre of $\Omega$ are collinear.","[""In what follows, $\\sphericalangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\\circ}$. Denote by $O$ the centre of $\\omega$. In any triangle, the homothety with ratio $-\\frac{1}{2}$ centred at the centroid of the triangle takes the vertices to the midpoints of the opposite sides and it takes the orthocentre to the circumcentre. Therefore the triangles $A B C$ and $A^{\\prime} B^{\\prime} C^{\\prime}$ share the same centroid $G$ and the midpoints of their sides lie on a circle $\\rho$ with centre on $O H$. We will prove that $\\omega, \\Omega$, and $\\rho$ are coaxial, so in particular it follows that their centres are collinear on $O H$.\n\nLet $D=B B^{\\prime} \\cap C C^{\\prime}, E=C C^{\\prime} \\cap A A^{\\prime}, F=A A^{\\prime} \\cap B B^{\\prime}, S=B C^{\\prime} \\cap B^{\\prime} C$, and $T=B C \\cap B^{\\prime} C^{\\prime}$. Since $D, S$, and $T$ are the intersections of opposite sides and of the diagonals in the quadrilateral $B B^{\\prime} C C^{\\prime}$ inscribed in $\\omega$, by Brocard's theorem triangle $D S T$ is self-polar with respect to $\\omega$, i.e. each vertex is the pole of the opposite side. We apply this in two ways.\n\n\n\nFirst, from $D$ being the pole of $S T$ it follows that the inverse $D^{*}$ of $D$ with respect to $\\omega$ is the projection of $D$ onto $S T$. In particular, $D^{*}$ lies on the circle with diameter $S D$. If $N$ denotes the midpoint of $S D$ and $R$ the radius of $\\omega$, then the power of $O$ with respect to this circle is $O N^{2}-N D^{2}=O D \\cdot O D^{*}=R^{2}$. By rearranging, we see that $N D^{2}$ is the power of $N$ with respect to $\\omega$.\n\nSecond, from $T$ being the pole of $S D$ it follows that $O T$ is perpendicular to $S D$. Let $M$ and $M^{\\prime}$ denote the midpoints of $B C$ and $B^{\\prime} C^{\\prime}$. Then since $O M \\perp B C$ and $O M^{\\prime} \\perp B^{\\prime} C^{\\prime}$ it follows that $O M M^{\\prime} T$ is cyclic and\n\n$$\n\\sphericalangle(S D, B C)=\\sphericalangle(O T, O M)=\\sphericalangle\\left(B^{\\prime} C^{\\prime}, M M^{\\prime}\\right) .\n$$\n\nFrom $B B^{\\prime} C C^{\\prime}$ being cyclic we also have $\\sphericalangle\\left(B C, B B^{\\prime}\\right)=\\sphericalangle\\left(C C^{\\prime}, B^{\\prime} C^{\\prime}\\right)$, hence we obtain\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(S D, B B^{\\prime}\\right) & =\\sphericalangle(S D, B C)+\\sphericalangle\\left(B C, B B^{\\prime}\\right) \\\\\n& =\\sphericalangle\\left(B^{\\prime} C^{\\prime}, M M^{\\prime}\\right)+\\sphericalangle\\left(C C^{\\prime}, B^{\\prime} C^{\\prime}\\right)=\\sphericalangle\\left(C C^{\\prime}, M M^{\\prime}\\right) .\n\\end{aligned}\n$$\n\nNow from the homothety mentioned in the beginning, we know that $M M^{\\prime}$ is parallel to $A A^{\\prime}$, hence the above implies that $\\sphericalangle\\left(S D, B B^{\\prime}\\right)=\\sphericalangle\\left(C C^{\\prime}, A A^{\\prime}\\right)$, which shows that $\\Omega$ is tangent to $S D$ at $D$. In particular, $N D^{2}$ is also the power of $N$ with respect to $\\Omega$.\n\n\n\nAdditionally, from $B B^{\\prime} C C^{\\prime}$ being cyclic it follows that triangles $D B C$ and $D C^{\\prime} B^{\\prime}$ are inversely similar, so $\\sphericalangle\\left(B B^{\\prime}, D M^{\\prime}\\right)=\\sphericalangle\\left(D M, C C^{\\prime}\\right)$. This yields\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(S D, D M^{\\prime}\\right) & =\\sphericalangle\\left(S D, B B^{\\prime}\\right)+\\sphericalangle\\left(B B^{\\prime}, D M^{\\prime}\\right) \\\\\n& =\\sphericalangle\\left(C C^{\\prime}, M M^{\\prime}\\right)+\\sphericalangle\\left(D M, C C^{\\prime}\\right)=\\sphericalangle\\left(D M, M M^{\\prime}\\right),\n\\end{aligned}\n$$\n\nwhich shows that the circle $D M M^{\\prime}$ is also tangent to $S D$. Since $N, M$, and $M^{\\prime}$ are collinear on the Newton-Gauss line of the complete quadrilateral determined by the lines $B B^{\\prime}, C C^{\\prime}, B C^{\\prime}$, and $B^{\\prime} C$, it follows that $N D^{2}=N M \\cdot N M^{\\prime}$. Hence $N$ has the same power with respect to $\\omega$, $\\Omega$, and $\\rho$.\n\nBy the same arguments there exist points on the tangents to $\\Omega$ at $E$ and $F$ which have the same power with respect to $\\omega, \\Omega$, and $\\rho$. The tangents to a given circle at three distinct points cannot be concurrent, hence we obtain at least two distinct points with the same power with respect to $\\omega, \\Omega$, and $\\rho$. Hence the three circles are coaxial, as desired.""]",,True,,, 2020,Geometry,,"Let $A A^{\prime} B C C^{\prime} B^{\prime}$ be a convex cyclic hexagon such that $A C$ is tangent to the incircle of the triangle $A^{\prime} B^{\prime} C^{\prime}$, and $A^{\prime} C^{\prime}$ is tangent to the incircle of the triangle $A B C$. Let the lines $A B$ and $A^{\prime} B^{\prime}$ meet at $X$ and let the lines $B C$ and $B^{\prime} C^{\prime}$ meet at $Y$. Prove that if $X B Y B^{\prime}$ is a convex quadrilateral, then it has an incircle.","[""Denote by $\\omega$ and $\\omega^{\\prime}$ the incircles of $\\triangle A B C$ and $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ and let $I$ and $I^{\\prime}$ be the centres of these circles. Let $N$ and $N^{\\prime}$ be the second intersections of $B I$ and $B^{\\prime} I^{\\prime}$ with $\\Omega$, the circumcircle of $A^{\\prime} B C C^{\\prime} B^{\\prime} A$, and let $O$ be the centre of $\\Omega$. Note that $O N \\perp A C, O N^{\\prime} \\perp A^{\\prime} C^{\\prime}$ and $O N=O N^{\\prime}$ so $N N^{\\prime}$ is parallel to the angle bisector $I I^{\\prime}$ of $A C$ and $A^{\\prime} C^{\\prime}$. Thus $I I^{\\prime} \\| N N^{\\prime}$ which is antiparallel to $B B^{\\prime}$ with respect to $B I$ and $B^{\\prime} I^{\\prime}$. Therefore $B, I, I^{\\prime}, B^{\\prime}$ are concyclic.\n\n\n\nFurther define $P$ as the intersection of $A C$ and $A^{\\prime} C^{\\prime}$ and $M$ as the antipode of $N^{\\prime}$ in $\\Omega$. Consider the circle $\\Gamma_{1}$ with centre $N$ and radius $N A=N C$ and the circle $\\Gamma_{2}$ with centre $M$ and radius $M A^{\\prime}=M C^{\\prime}$. Their radical axis passes through $P$ and is perpendicular to $M N \\perp N N^{\\prime} \\| I P$, so $I$ lies on their radical axis. Therefore, since $I$ lies on $\\Gamma_{1}$, it must also lie on $\\Gamma_{2}$. Thus, if we define $Z$ as the second intersection of $M I$ with $\\Omega$, we have that $I$ is the incentre of triangle $Z A^{\\prime} C^{\\prime}$. (Note that the point $Z$ can also be constructed directly via Poncelet's porism.)\n\nConsider the incircle $\\omega_{c}$ with centre $I_{c}$ of triangle $C^{\\prime} B^{\\prime} Z$. Note that $\\angle Z I C^{\\prime}=90^{\\circ}+$ $\\frac{1}{2} \\angle Z A^{\\prime} C^{\\prime}=90^{\\circ}+\\frac{1}{2} \\angle Z B^{\\prime} C^{\\prime}=\\angle Z I_{c} C^{\\prime}$, so $Z, I, I_{c}, C^{\\prime}$ are concyclic. Similarly $B^{\\prime}, I^{\\prime}, I_{c}, C^{\\prime}$ are concyclic.\n\nThe external centre of dilation from $\\omega$ to $\\omega_{c}$ is the intersection of $I I_{c}$ and $C^{\\prime} Z$ ( $D$ in the picture), that is the radical centre of circles $\\Omega, C^{\\prime} I_{c} I Z$ and $I I^{\\prime} I_{c}$. Similarly, the external centre of dilation from $\\omega^{\\prime}$ to $\\omega_{c}$ is the intersection of $I^{\\prime} I_{c}$ and $B^{\\prime} C^{\\prime}$ ( $D^{\\prime}$ in the picture), that is the radical centre of circles $\\Omega, B^{\\prime} I^{\\prime} I_{c} C^{\\prime}$ and $I I^{\\prime} I_{c}$. Therefore the Monge line of $\\omega, \\omega^{\\prime}$ and $\\omega_{c}$ is line $D D^{\\prime}$, and the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$ coincide. Hence the external centre $T$ of dilation from $\\omega$ to $\\omega^{\\prime}$ is also on the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$.\n\n\n\n\n\nNow since $B, I, I^{\\prime}, B^{\\prime}$ are concyclic, the intersection $T^{\\prime}$ of $B B^{\\prime}$ and $I I^{\\prime}$ is on the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$. Thus $T^{\\prime}=T$ and $T$ lies on line $B B^{\\prime}$. Finally, construct a circle $\\Omega_{0}$ tangent to $A^{\\prime} B^{\\prime}, B^{\\prime} C^{\\prime}, A B$ on the same side of these lines as $\\omega^{\\prime}$. The centre of dilation from $\\omega^{\\prime}$ to $\\Omega_{0}$ is $B^{\\prime}$, so by Monge's theorem the external centre of dilation from $\\Omega_{0}$ to $\\omega$ must be on the line $T B B^{\\prime}$. However, it is on line $A B$, so it must be $B$ and $B C$ must be tangent to $\\Omega_{0}$ as desired.\n\n""]",,True,,, 2021,Number Theory,,"A number is called Norwegian if it has three distinct positive divisors whose sum is equal to 2022. Determine the smallest Norwegian number. (Note: The total number of positive divisors of a Norwegian number is allowed to be larger than 3.)","['Observe that 1344 is a Norwegian number as 6, 672 and 1344 are three distinct divisors of 1344 and $6+672+1344=2022$. It remains to show that this is the smallest such number.\n\nAssume for contradiction that $N<1344$ is Norwegian and let $N / a, N / b$ and $N / c$ be the three distinct divisors of $N$, with $a\\frac{2022}{1344}=\\frac{337}{224}=\\frac{3}{2}+\\frac{1}{224} .\n$$\n\nIf $a>1$ then\n\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\leqslant \\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}=\\frac{13}{12}<\\frac{3}{2}\n$$\n\nso it must be the case that $a=1$. Similarly, it must hold that $b<4$ since otherwise\n\n$$\n1+\\frac{1}{b}+\\frac{1}{c} \\leqslant 1+\\frac{1}{4}+\\frac{1}{5}<\\frac{3}{2}\n$$\n\nThis leaves two cases to check, $b=2$ and $b=3$.\n\nCase $b=3$. Then\n\n$$\n\\frac{1}{c}>\\frac{3}{2}+\\frac{1}{224}-1-\\frac{1}{3}>\\frac{1}{6}\n$$\n\nso $c=4$ or $c=5$. If $c=4$ then\n\n$$\n2022=N\\left(1+\\frac{1}{3}+\\frac{1}{4}\\right)=\\frac{19}{12} N\n$$\n\nbut this is impossible as $19 \\nmid 2022$. If $c=5$ then\n\n$$\n2022=N\\left(1+\\frac{1}{3}+\\frac{1}{5}\\right)=\\frac{23}{15} N\n$$\n\nwhich again is impossible, as $23 \\nmid 2022$.\n\nCase $b=2$. Note that $c<224$ since\n\n$$\n\\frac{1}{c}>\\frac{3}{2}+\\frac{1}{224}-1-\\frac{1}{2}=\\frac{1}{224}\n$$\n\nIt holds that\n\n$$\n2022=N\\left(1+\\frac{1}{2}+\\frac{1}{c}\\right)=\\frac{3 c+2}{2 c} N \\Rightarrow(3 c+2) N=4044 c\n$$\n\nSince $(c, 3 c-2)=(c, 2) \\in\\{1,2\\}$, then $3 c+2 \\mid 8088=2^{3} \\cdot 3 \\cdot 337$ which implies that $3 c+2 \\mid 2^{3} \\cdot 337$. But since $3 c+2 \\geqslant 3 \\cdot 3+2>8=2^{3}$ and $3 c+2 \\neq 337$, then it must hold that $3 c+2 \\geqslant 2 \\cdot 337$, contradicting $c<224$.']",['1344'],False,,Numerical, 2022,Number Theory,,"Find all positive integers $n>2$ such that $$ n ! \mid \prod_{\substack{p2$ and 3 divides one of $p_{m-2}, p_{m-1}=p_{m-2}+2$ and $p_{m}=p_{m-2}+4$. This implies $p_{m-2}=3$ and thus $p_{m}=7$, giving $7 \\leqslant n<11$.\n\nFinally, a quick computation shows that $7 ! \\mid \\prod_{p1$ be a positive integer, and let $d>1$ be a positive integer coprime to $a$. Let $x_{1}=1$ and, for $k \geqslant 1$, define $$ x_{k+1}= \begin{cases}x_{k}+d & \text { if } a \text { doesn't divide } x_{k} \\ x_{k} / a & \text { if } a \text { divides } x_{k}\end{cases} $$ Find the greatest positive integer $n$ for which there exists an index $k$ such that $x_{k}$ is divisible by $a^{n}$.","['By trivial induction, $x_{k}$ is coprime to $d$.\n\nBy induction and the fact that there can be at most $a-1$ consecutive increasing terms in the sequence, it also holds that $x_{k}0}: 0d$ then $y-d \\in S$ but $a \\cdot y \\notin S$; otherwise, if $yd, \\\\\na \\cdot y & \\text { if } y1$ be an index such that $x_{k_{1}}=1$. Then\n\n$$\nx_{k_{1}}=1, \\quad x_{k_{1}-1}=f^{-1}(1)=a, x_{k_{1}-2}=f^{-1}(a)=a^{2}, \\ldots, \\quad x_{k_{1}-n}=a^{n}\n$$', 'By trivial induction, $x_{k}$ is coprime to $d$.\n\nBy induction and the fact that there can be at most $a-1$ consecutive increasing terms in the sequence, it also holds that $x_{k}a^{n} / a=a^{n-1}$ the LHS is strictly less than $a^{v-n}$. This implies that on the RHS, the coefficients of $a^{v-n}, a^{v-n+1}, \\ldots$ must all be zero, i.e. $z_{v-n}=z_{v-n+1}=\\cdots=z_{v-1}=0$. This implies that there are $n$ consecutive decreasing indices in the original sequence.']",['$n$ is the exponent with $d1$. We consider three cases.\n\nCase 1: We have $ab$ which is also impossible since in this case we have $b ! \\leqslant a !a>1$.\n\nCase 2: We have $a>p$. In this case $b !=a^{p}-p>p^{p}-p \\geqslant p$ ! so $b>p$ which means that $a^{p}=b !+p$ is divisible by $p$. Hence, $a$ is divisible by $p$ and $b !=a^{p}-p$ is not divisible by $p^{2}$. This means that $b<2 p$. If $a(2 p-1) !+p \\geqslant b !+p$.\n\nCase 3: We have $a=p$. In this case $b !=p^{p}-p$. One can check that the values $p=2,3$ lead to the claimed solutions and $p=5$ does not lead to a solution. So we now assume that $p \\geqslant 7$. We have $b !=p^{p}-p>p !$ and so $b \\geqslant p+1$ which implies that\n\n$$\nv_{2}((p+1) !) \\leqslant v_{2}(b !)=v_{2}\\left(p^{p-1}-1\\right) \\stackrel{L T E}{=} 2 v_{2}(p-1)+v_{2}(p+1)-1=v_{2}\\left(\\frac{p-1}{2} \\cdot(p-1) \\cdot(p+1)\\right)\n$$\n\nwhere in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of $(p+1)$ !. But, due to $p+1 \\geqslant 8$, there are at least 4 even numbers among $1,2, \\ldots, p+1$, so this case is not possible.', ""Clearly, $a>1$. We consider three cases.\n\nCase 1: We have $ab$ which is also impossible since in this case we have $b ! \\leqslant a !a>1$.\n\nCase 2: We have $a>p$. In this case $b !=a^{p}-p>p^{p}-p \\geqslant p$ ! so $b>p$ which means that $a^{p}=b !+p$ is divisible by $p$. Hence, $a$ is divisible by $p$ and $b !=a^{p}-p$ is not divisible by $p^{2}$. This means that $b<2 p$. If $a(2 p-1) !+p \\geqslant b !+p$.\n\nCase 3: We have $a=p$. In this case $b !=p^{p}-p$. One can check that the values $p=2,3$ lead to the claimed solutions and $p=5$ does not lead to a solution. For $p \\geqslant 5$ we have $b !=p\\left(p^{p-1}-1\\right)$. By Zsigmondy's Theorem there exists some prime $q$ that divides $p^{p-1}-1$ but does not divide $p^{k}-1$ for $k(2 p-1)^{p-1} p>p^{p}>p^{p}-p$, a contradiction.""]","['$(2,2,2),(3,4,3)$']",True,,Tuple, 2025,Number Theory,,"For each $1 \leqslant i \leqslant 9$ and $T \in \mathbb{N}$, define $d_{i}(T)$ to be the total number of times the digit $i$ appears when all the multiples of 1829 between 1 and $T$ inclusive are written out in base 10 . Show that there are infinitely many $T \in \mathbb{N}$ such that there are precisely two distinct values among $d_{1}(T), d_{2}(T), \ldots, d_{9}(T)$.","['Let $n:=1829$. First, we choose some $k$ such that $n \\mid 10^{k}-1$. For instance, any multiple of $\\varphi(n)$ would work since $n$ is coprime to 10 . We will show that either $T=10^{k}-1$ or $T=10^{k}-2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrary large.\n\nFor this it suffices to show that $\\#\\left\\{d_{i}\\left(10^{k}-1\\right): 1 \\leqslant i \\leqslant 9\\right\\} \\leqslant 2$. Indeed, if\n$$\n\\#\\left\\{d_{i}\\left(10^{k}-1\\right): 1 \\leqslant i \\leqslant 9\\right\\}=1\n$$\nthen, since $10^{k}-1$ which consists of all nines is a multiple of $n$, we have\n$$\nd_{i}\\left(10^{k}-2\\right)=d_{i}\\left(10^{k}-1\\right) \\text { for } i \\in\\{1, \\ldots, 8\\} \\text {, and } d_{9}\\left(10^{k}-2\\right)0$ independent of the set $Q$ such that for any positive integer $N>100$, the number of special integers in $[1, N]$ is at least $c N$. (For example, if $Q=\{3,7\}$, then $p(42)=3, q(42)=2, p(63)=3, q(63)=3, p(2022)=3$, $q(2022)=1$.)","['Let us call two positive integers $m, n$ friends if $p(m)+p(n)$ and $q(m)+q(n)$ are both even integers. We start by noting that the pairs $(p(k), q(k))$ modulo 2 can take at most 4 different values; thus, among any five different positive integers there are two which are friends.\n\nIn addition, both functions $p$ and $q$ satisfy $f(a b)=f(a)+f(b)$ for any $a, b$. Therefore, if $m$ and $n$ are divisible by $d$, then both $p$ and $q$ satisfy the equality $f(m)+f(n)=f(m / d)+$ $f(n / d)+2 f(d)$. This implies that $m, n$ are friends if and only if $m / d, n / d$ are friends.\n\nLet us call a set of integers $\\left\\{n_{1}, n_{2}, \\ldots, n_{5}\\right\\}$ an interesting set if for any indexes $i, j$, the difference $d_{i j}=\\left|n_{i}-n_{j}\\right|$ divides both $n_{i}$ and $n_{j}$. We claim that if elements of an interesting set are all positive, then we can obtain a special integer. Indeed, if we were able to construct such a set, then there would be a pair of integers $\\left\\{n_{i}, n_{j}\\right\\}$ which are friends, according to the first observation. Additionally, the second observation yields that the quotients $n_{i} / d_{i j}, n_{j} / d_{i j}$ form a pair of friends, which happen to be consecutive integers, thus giving a special integer as desired.\n\nIn order to construct a family of interesting sets, we can start by observing that the set $\\{0,6,8,9,12\\}$ is an interesting set. Using that $72=2^{3} \\cdot 3^{2}$ is the least common multiple of all pairwise differences in this set, we obtain a family of interesting sets by considering\n$$\n\\{72 k, 72 k+6,72 k+8,72 k+9,72 k+12\\}\n$$\nfor any $k \\geqslant 1$. If we consider the quotients (of these numbers by the appropriate differences), then we obtain that the set\n$$\nS_{k}=\\{6 k, 8 k, 9 k, 12 k, 12 k+1,18 k+2,24 k+2,24 k+3,36 k+3,72 k+8\\}\n$$\nhas at least one special integer. In particular, the interval $[1,100 k]$ contains the sets $S_{1}, S_{2}, \\ldots, S_{k}$, each of which has a special number. Any special number can be contained in at most ten sets $S_{k}$, from where we conclude that the number of special integers in $[1,100 k]$ is at least $k / 10$.\n\nFinally, let $N=100 k+r$, with $k \\geqslant 1$ and $0 \\leqslant r<100$, so that we have $N<100(k+1) \\leqslant$ $200 k$. Then the number of special integers in $[1, N]$ is at least $k / 10>N / 2000$, as we wanted to prove.']",,True,,, 2027,Number Theory,,Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers. Prove that there is at most one way (modulo rotation and reflection) to place the elements of $S$ around a circle such that the product of any two neighbors is of the form $x^{2}+x+k$ for some positive integer $x$.,"[""Let us allow the value $x=0$ as well; we prove the same statement under this more general constraint. Obviously that implies the statement with the original conditions.\n\nCall a pair $\\{p, q\\}$ of primes with $p \\neq q$ special if $p q=x^{2}+x+k$ for some nonnegative integer $x$. The following claim is the key mechanism of the problem:\n\nClaim.\n\n(a) For every prime $r$, there are at most two primes less than $r$ forming a special pair with $r$.\n\n(b) If such $p$ and $q$ exist, then $\\{p, q\\}$ is itself special.\n\nWe present two proofs of the claim.\n\nProof 1. We are interested in integers $1 \\leqslant x\\sqrt{m}$ choices for $x$ and $y$, so there are more than $m$ possible pairs $(x, y)$. Hence, two of these sums are congruent modulo $m$ : $5^{k+1} x_{1}+3^{k+1} y_{1} \\equiv 5^{k+1} x_{2}+3^{k+1} y_{2}(\\bmod m)$.\n\nNow choose $a=x_{1}-x_{2}$ and $b=y_{1}-y_{2}$; at least one of $a, b$ is nonzero, and\n$$\n5^{k+1} a+3^{k+1} b \\equiv 0 \\quad(\\bmod m), \\quad|a|,|b| \\leqslant \\sqrt{m}\n$$\nFrom\n$$\n0 \\equiv\\left(5^{k+1} a\\right)^{2}-\\left(3^{k+1} b\\right)^{2}=5^{n+1} a^{2}-3^{n+1} b^{2} \\equiv 5 \\cdot 3^{n} a^{2}-3^{n+1} b^{2}=3^{n}\\left(5 a^{2}-3 b^{2}\\right) \\quad(\\bmod m)\n$$\nwe can see that $\\left|5 a^{2}-3 b^{2}\\right|$ is a multiple of $m$. Since at least one of $a$ and $b$ is nonzero, $5 a^{2} \\neq 3 b^{2}$. Hence, by the choice of $a, b$, we have $0<\\left|5 a^{2}-3 b^{2}\\right| \\leqslant \\max \\left(5 a^{2}, 3 b^{2}\\right) \\leqslant 5 m$. That shows that $m_{1} \\leqslant 5 m$\n\nII. Next, we show that $m_{1}$ cannot be divisible by 2,3 and 5 . Since $m_{1}$ equals either $\\left|5 a^{2}-3 b^{2}\\right|$ or $\\left|a^{2}-15 b^{2}\\right|$ with some integers $a, b$, we have six cases to check. In all six cases, we will get a contradiction by presenting another multiple of $m$, smaller than $m_{1}$.\n\n- If $5 \\mid m_{1}$ and $m_{1}=\\left|5 a^{2}-3 b^{2}\\right|$, then $5 \\mid b$ and $\\left|a^{2}-15\\left(\\frac{b}{5}\\right)^{2}\\right|=\\frac{m_{1}}{5}0}$ denote the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Q}_{>0}$ satisfying $$ f\left(x^{2} f(y)^{2}\right)=f(x)^{2} f(y) \tag{*} $$ for all $x, y \in \mathbb{Q}_{>0}$.","['Take any $a, b \\in \\mathbb{Q}_{>0}$. By substituting $x=f(a), y=b$ and $x=f(b), y=a$ into $(*)$ we get\n\n$$\nf(f(a))^{2} f(b)=f\\left(f(a)^{2} f(b)^{2}\\right)=f(f(b))^{2} f(a)\n$$\n\nwhich yields\n\n$$\n\\frac{f(f(a))^{2}}{f(a)}=\\frac{f(f(b))^{2}}{f(b)} \\quad \\text { for all } a, b \\in \\mathbb{Q}>0\n$$\n\nIn other words, this shows that there exists a constant $C \\in \\mathbb{Q}_{>0}$ such that $f(f(a))^{2}=C f(a)$, or\n\n$$\n\\left(\\frac{f(f(a))}{C}\\right)^{2}=\\frac{f(a)}{C} \\quad \\text { for all } a \\in \\mathbb{Q}_{>0}\n\\tag{1}\n$$\n\nDenote by $f^{n}(x)=\\underbrace{f(f(\\ldots(f}_{n}(x)) \\ldots))$ the $n^{\\text {th }}$ iteration of $f$. Equality (1) yields\n\n$$\n\\frac{f(a)}{C}=\\left(\\frac{f^{2}(a)}{C}\\right)^{2}=\\left(\\frac{f^{3}(a)}{C}\\right)^{4}=\\cdots=\\left(\\frac{f^{n+1}(a)}{C}\\right)^{2^{n}}\n$$\n\nfor all positive integer $n$. So, $f(a) / C$ is the $2^{n}$-th power of a rational number for all positive integer $n$. This is impossible unless $f(a) / C=1$, since otherwise the exponent of some prime in the prime decomposition of $f(a) / C$ is not divisible by sufficiently large powers of 2 . Therefore, $f(a)=C$ for all $a \\in \\mathbb{Q}_{>0}$.\n\nFinally, after substituting $f \\equiv C$ into $(*)$ we get $C=C^{3}$, whence $C=1$. So $f(x) \\equiv 1$ is the unique function satisfying $(*)$.']",['$f(x)=1$'],False,,Expression, 2030,Algebra,,"Find all positive integers $n \geqslant 3$ for which there exist real numbers $a_{1}, a_{2}, \ldots, a_{n}$, $a_{n+1}=a_{1}, a_{n+2}=a_{2}$ such that $$ a_{i} a_{i+1}+1=a_{i+2} $$ for all $i=1,2, \ldots, n$.","['For the sake of convenience, extend the sequence $a_{1}, \\ldots, a_{n+2}$ to an infinite periodic sequence with period $n$. ( $n$ is not necessarily the shortest period.)\n\nIf $n$ is divisible by 3 , then $\\left(a_{1}, a_{2}, \\ldots\\right)=(-1,-1,2,-1,-1,2, \\ldots)$ is an obvious solution.\n\nWe will show that in every periodic sequence satisfying the recurrence, each positive term is followed by two negative values, and after them the next number is positive again. From this, it follows that $n$ is divisible by 3 .\n\nIf the sequence contains two consecutive positive numbers $a_{i}, a_{i+1}$, then $a_{i+2}=a_{i} a_{i+1}+1>1$, so the next value is positive as well; by induction, all numbers are positive and greater than 1 . But then $a_{i+2}=a_{i} a_{i+1}+1 \\geqslant 1 \\cdot a_{i+1}+1>a_{i+1}$ for every index $i$, which is impossible: our sequence is periodic, so it cannot increase everywhere.\n\nIf the number 0 occurs in the sequence, $a_{i}=0$ for some index $i$, then it follows that $a_{i+1}=a_{i-1} a_{i}+1$ and $a_{i+2}=a_{i} a_{i+1}+1$ are two consecutive positive elements in the sequences and we get the same contradiction again.\n\nNotice that after any two consecutive negative numbers the next one must be positive: if $a_{i}<0$ and $a_{i+1}<0$, then $a_{i+2}=a_{1} a_{i+1}+1>1>0$. Hence, the positive and negative numbers follow each other in such a way that each positive term is followed by one or two negative values and then comes the next positive term.\n\nConsider the case when the positive and negative values alternate. So, if $a_{i}$ is a negative value then $a_{i+1}$ is positive, $a_{i+2}$ is negative and $a_{i+3}$ is positive again.\n\nNotice that $a_{i} a_{i+1}+1=a_{i+2}<00$ we conclude $a_{i}1$. The number $a_{i+3}$ must be negative. We show that $a_{i+4}$ also must be negative.\n\nNotice that $a_{i+3}$ is negative and $a_{i+4}=a_{i+2} a_{i+3}+1<10\n$$\n\ntherefore $a_{i+5}>a_{i+4}$. Since at most one of $a_{i+4}$ and $a_{i+5}$ can be positive, that means that $a_{i+4}$ must be negative.\n\nNow $a_{i+3}$ and $a_{i+4}$ are negative and $a_{i+5}$ is positive; so after two negative and a positive terms, the next three terms repeat the same pattern. That completes the solution.', 'We prove that the shortest period of the sequence must be 3. Then it follows that $n$ must be divisible by 3 .\n\nNotice that the equation $x^{2}+1=x$ has no real root, so the numbers $a_{1}, \\ldots, a_{n}$ cannot be all equal, hence the shortest period of the sequence cannot be 1 .\n\nBy applying the recurrence relation for $i$ and $i+1$,\n\n$$\n\\begin{gathered}\n\\left(a_{i+2}-1\\right) a_{i+2}=a_{i} a_{i+1} a_{i+2}=a_{i}\\left(a_{i+3}-1\\right), \\quad \\text { so } \\\\\na_{i+2}^{2}-a_{i} a_{i+3}=a_{i+2}-a_{i} .\n\\end{gathered}\n$$\n\nBy summing over $i=1,2, \\ldots, n$, we get\n\n$$\n\\sum_{i=1}^{n}\\left(a_{i}-a_{i+3}\\right)^{2}=0\n$$\n\nThat proves that $a_{i}=a_{i+3}$ for every index $i$, so the sequence $a_{1}, a_{2}, \\ldots$ is indeed periodic with period 3. The shortest period cannot be 1 , so it must be 3 ; therefore, $n$ is divisible by 3 .']",['$n$ can be any multiple of 3'],False,,Need_human_evaluate, 2031,Algebra,,"Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.","['Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\\sum_{x \\in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\\sum_{x \\in F_{r}} 1 / x=r$. The argument hinges on the lemma below.\n\nLemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$.\n\nProof. If $x$ is a member of $F_{q}$, then\n\n$$\n\\sum_{y \\in F_{q} \\backslash\\{x\\}} \\frac{1}{y}=\\sum_{y \\in F_{q}} \\frac{1}{y}-\\frac{1}{x}=q-\\frac{1}{x}=r=\\sum_{y \\in F_{r}} \\frac{1}{y}\n$$\n\nso $F_{r}=F_{q} \\backslash\\{x\\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then\n\n$$\n\\sum_{y \\in F_{r} \\cup\\{x\\}} \\frac{1}{y}=\\sum_{y \\in F_{r}} \\frac{1}{y}+\\frac{1}{x}=r+\\frac{1}{x}=q=\\sum_{y \\in F_{q}} \\frac{1}{y}\n$$\n\nso $F_{q}=F_{r} \\cup\\{x\\}$, and $x$ is a member of $F_{q}$.\n\nConsider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=|r x|$ and consider the sets $F_{r-k / x}, k=0, \\ldots, n$. Since $0 \\leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd.\n\nFinally, consider $F_{2 / 3}$. By the preceding, $\\lfloor 2 x / 3\\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\\varepsilon$ such that $\\lfloor(2 / 3-\\varepsilon) x\\rfloor=\\lfloor 2 x / 3\\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\\varepsilon}$ which is impossible.', 'A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \\backslash\\{1\\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}2 x_{n}$ for some $n$, then $\\sum_{x \\in F} 1 / x2$, we have $\\Delta_{n} \\leqslant \\frac{n-1}{n} \\Delta_{n-1}$.\n\nProof. Choose positive integers $k, \\ell \\leqslant n$ such that $M_{n}=S(n, k) / k$ and $m_{n}=S(n, \\ell) / \\ell$. We have $S(n, k)=a_{n-1}+S(n-1, k-1)$, so\n\n$$\nk\\left(M_{n}-a_{n-1}\\right)=S(n, k)-k a_{n-1}=S(n-1, k-1)-(k-1) a_{n-1} \\leqslant(k-1)\\left(M_{n-1}-a_{n-1}\\right),\n$$\n\nsince $S(n-1, k-1) \\leqslant(k-1) M_{n-1}$. Similarly, we get\n\n$$\n\\ell\\left(a_{n-1}-m_{n}\\right)=(\\ell-1) a_{n-1}-S(n-1, \\ell-1) \\leqslant(\\ell-1)\\left(a_{n-1}-m_{n-1}\\right) .\n$$\n\nSince $m_{n-1} \\leqslant a_{n-1} \\leqslant M_{n-1}$ and $k, \\ell \\leqslant n$, the obtained inequalities yield\n\n$$\n\\begin{array}{ll}\nM_{n}-a_{n-1} \\leqslant \\frac{k-1}{k}\\left(M_{n-1}-a_{n-1}\\right) \\leqslant \\frac{n-1}{n}\\left(M_{n-1}-a_{n-1}\\right) \\text { and } \\\\\na_{n-1}-m_{n} \\leqslant \\frac{\\ell-1}{\\ell}\\left(a_{n-1}-m_{n-1}\\right) \\leqslant \\frac{n-1}{n}\\left(a_{n-1}-m_{n-1}\\right) .\n\\end{array}\n$$\n\nTherefore,\n\n$$\n\\Delta_{n}=\\left(M_{n}-a_{n-1}\\right)+\\left(a_{n-1}-m_{n}\\right) \\leqslant \\frac{n-1}{n}\\left(\\left(M_{n-1}-a_{n-1}\\right)+\\left(a_{n-1}-m_{n-1}\\right)\\right)=\\frac{n-1}{n} \\Delta_{n-1}\n$$\n\nBack to the problem, if $a_{n}=1$ for all $n \\leqslant 2017$, then $a_{2018} \\leqslant 1$ and hence $a_{2018}-a_{2017} \\leqslant 0$. Otherwise, let $2 \\leqslant q \\leqslant 2017$ be the minimal index with $a_{q}<1$. We have $S(q, i)=i$ for all $i=1,2, \\ldots, q-1$, while $S(q, q)=q-1$. Therefore, $a_{q}<1$ yields $a_{q}=S(q, q) / q=1-\\frac{1}{q}$.\n\nNow we have $S(q+1, i)=i-\\frac{1}{q}$ for $i=1,2, \\ldots, q$, and $S(q+1, q+1)=q-\\frac{1}{q}$. This gives us\n\n$$\nm_{q+1}=\\frac{S(q+1,1)}{1}=\\frac{S(q+1, q+1)}{q+1}=\\frac{q-1}{q} \\quad \\text { and } \\quad M_{q+1}=\\frac{S(q+1, q)}{q}=\\frac{q^{2}-1}{q^{2}}\n$$\n\nso $\\Delta_{q+1}=M_{q+1}-m_{q+1}=(q-1) / q^{2}$. Denoting $N=2017 \\geqslant q$ and using Claim 1 for $n=q+2, q+3, \\ldots, N+1$ we finally obtain\n\n$$\n\\Delta_{N+1} \\leqslant \\frac{q-1}{q^{2}} \\cdot \\frac{q+1}{q+2} \\cdot \\frac{q+2}{q+3} \\cdots \\frac{N}{N+1}=\\frac{1}{N+1}\\left(1-\\frac{1}{q^{2}}\\right) \\leqslant \\frac{1}{N+1}\\left(1-\\frac{1}{N^{2}}\\right)=\\frac{N-1}{N^{2}},\n$$\n\nas required.', 'We present a different proof of the estimate $a_{2018}-a_{2017} \\leqslant \\frac{2016}{2017^{2}}$. We keep the same notations of $S(n, k), m_{n}$ and $M_{n}$ from the previous solution.\n\nNotice that $S(n, n)=S(n, n-1)$, as $a_{0}=0$. Also notice that for $0 \\leqslant k \\leqslant \\ell \\leqslant n$ we have $S(n, \\ell)=S(n, k)+S(n-k, \\ell-k)$.\n\nClaim 2. For every positive integer $n$, we have $m_{n} \\leqslant m_{n+1}$ and $M_{n+1} \\leqslant M_{n}$, so the segment $\\left[m_{n+1}, M_{n+1}\\right]$ is contained in $\\left[m_{n}, M_{n}\\right]$.\n\nProof. Choose a positive integer $k \\leqslant n+1$ such that $m_{n+1}=S(n+1, k) / k$. Then we have\n\n$$\nk m_{n+1}=S(n+1, k)=a_{n}+S(n, k-1) \\geqslant m_{n}+(k-1) m_{n}=k m_{n},\n$$\n\nwhich establishes the first inequality in the Claim. The proof of the second inequality is similar.\n\nClaim 3. For every positive integers $k \\geqslant n$, we have $m_{n} \\leqslant a_{k} \\leqslant M_{n}$.\n\nProof. By Claim 2, we have $\\left[m_{k}, M_{k}\\right] \\subseteq\\left[m_{k-1}, M_{k-1}\\right] \\subseteq \\cdots \\subseteq\\left[m_{n}, M_{n}\\right]$. Since $a_{k} \\in\\left[m_{k}, M_{k}\\right]$, the claim follows.\n\nClaim 4. For every integer $n \\geqslant 2$, we have $M_{n}=S(n, n-1) /(n-1)$ and $m_{n}=S(n, n) / n$.\n\nProof. We use induction on $n$. The base case $n=2$ is routine. To perform the induction step, we need to prove the inequalities\n\n$$\n\\frac{S(n, n)}{n} \\leqslant \\frac{S(n, k)}{k} \\quad \\text { and } \\quad \\frac{S(n, k)}{k} \\leqslant \\frac{S(n, n-1)}{n-1}\n\\tag{1}\n$$\n\nfor every positive integer $k \\leqslant n$. Clearly, these inequalities hold for $k=n$ and $k=n-1$, as $S(n, n)=S(n, n-1)>0$. In the sequel, we assume that $k0$.","['Fix a real number $a>1$, and take a new variable $t$. For the values $f(t), f\\left(t^{2}\\right)$, $f(a t)$ and $f\\left(a^{2} t^{2}\\right)$, the relation (1) provides a system of linear equations:\n\n$$\nx=y=t: \\quad\\left(t+\\frac{1}{t}\\right) f(t) \\quad=f\\left(t^{2}\\right)+f(1)\n\\tag{2a}\n$$\n\n$$\nx=\\frac{t}{a}, y=a t: \\quad\\left(\\frac{t}{a}+\\frac{a}{t}\\right) f(a t) \\quad=f\\left(t^{2}\\right)+f\\left(a^{2}\\right)\n\\tag{2b}\n$$\n\n$$\nx=a^{2} t, y=t: \\quad\\left(a^{2} t+\\frac{1}{a^{2} t}\\right) f(t)=f\\left(a^{2} t^{2}\\right)+f\\left(\\frac{1}{a^{2}}\\right)\n\\tag{2c}\n$$\n\n$$\nx=y=a t: \\quad\\left(a t+\\frac{1}{a t}\\right) f(a t)=f\\left(a^{2} t^{2}\\right)+f(1)\n\\tag{2d}\n$$\n\nIn order to eliminate $f\\left(t^{2}\\right)$, take the difference of (2a) and (2b); from (2c) and (2d) eliminate $f\\left(a^{2} t^{2}\\right)$; then by taking a linear combination, eliminate $f(a t)$ as well:\n\n$$\n\\begin{aligned}\n\\left(t+\\frac{1}{t}\\right) f(t)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right) f(a t) & =f(1)-f\\left(a^{2}\\right) \\text { and } \\\\\n\\left(a^{2} t+\\frac{1}{a^{2} t}\\right) f(t)-\\left(a t+\\frac{1}{a t}\\right) f(a t) & =f\\left(1 / a^{2}\\right)-f(1), \\text { so } \\\\\n\\left(\\left(a t+\\frac{1}{a t}\\right)\\left(t+\\frac{1}{t}\\right)\\right. & \\left.-\\left(\\frac{t}{a}+\\frac{a}{t}\\right)\\left(a^{2} t+\\frac{1}{a^{2} t}\\right)\\right) f(t) \\\\\n& =\\left(a t+\\frac{1}{a t}\\right)\\left(f(1)-f\\left(a^{2}\\right)\\right)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right)\\left(f\\left(1 / a^{2}\\right)-f(1)\\right) .\n\\end{aligned}\n$$\n\nNotice that on the left-hand side, the coefficient of $f(t)$ is nonzero and does not depend on $t$ :\n\n$$\n\\left(a t+\\frac{1}{a t}\\right)\\left(t+\\frac{1}{t}\\right)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right)\\left(a^{2} t+\\frac{1}{a^{2} t}\\right)=a+\\frac{1}{a}-\\left(a^{3}+\\frac{1}{a^{3}}\\right)<0 .\n$$\n\nAfter dividing by this fixed number, we get\n\n$$\nf(t)=C_{1} t+\\frac{C_{2}}{t}\n\\tag{3}\n$$\n\nwhere the numbers $C_{1}$ and $C_{2}$ are expressed in terms of $a, f(1), f\\left(a^{2}\\right)$ and $f\\left(1 / a^{2}\\right)$, and they do not depend on $t$.\n\nThe functions of the form (3) satisfy the equation:\n\n$$\n\\left(x+\\frac{1}{x}\\right) f(y)=\\left(x+\\frac{1}{x}\\right)\\left(C_{1} y+\\frac{C_{2}}{y}\\right)=\\left(C_{1} x y+\\frac{C_{2}}{x y}\\right)+\\left(C_{1} \\frac{y}{x}+C_{2} \\frac{x}{y}\\right)=f(x y)+f\\left(\\frac{y}{x}\\right) .\n$$', 'We start with an observation. If we substitute $x=a \\neq 1$ and $y=a^{n}$ in (1), we obtain\n\n$$\nf\\left(a^{n+1}\\right)-\\left(a+\\frac{1}{a}\\right) f\\left(a^{n}\\right)+f\\left(a^{n-1}\\right)=0\n$$\n\nFor the sequence $z_{n}=a^{n}$, this is a homogeneous linear recurrence of the second order, and its characteristic polynomial is $t^{2}-\\left(a+\\frac{1}{a}\\right) t+1=(t-a)\\left(t-\\frac{1}{a}\\right)$ with two distinct nonzero roots, namely $a$ and $1 / a$. As is well-known, the general solution is $z_{n}=C_{1} a^{n}+C_{2}(1 / a)^{n}$ where the index $n$ can be as well positive as negative. Of course, the numbers $C_{1}$ and $C_{2}$ may depend of the choice of $a$, so in fact we have two functions, $C_{1}$ and $C_{2}$, such that\n\n$$\nf\\left(a^{n}\\right)=C_{1}(a) \\cdot a^{n}+\\frac{C_{2}(a)}{a^{n}} \\quad \\text { for every } a \\neq 1 \\text { and every integer } n\n\\tag{4}\n$$\n\nThe relation (4) can be easily extended to rational values of $n$, so we may conjecture that $C_{1}$ and $C_{2}$ are constants, and whence $f(t)=C_{1} t+\\frac{C_{2}}{t}$. As it was seen in the previous solution, such functions indeed satisfy (1).\n\nThe equation (1) is linear in $f$; so if some functions $f_{1}$ and $f_{2}$ satisfy (1) and $c_{1}, c_{2}$ are real numbers, then $c_{1} f_{1}(x)+c_{2} f_{2}(x)$ is also a solution of (1). In order to make our formulas simpler, define\n\n$$\nf_{0}(x)=f(x)-f(1) \\cdot x .\n$$\n\nThis function is another one satisfying (1) and the extra constraint $f_{0}(1)=0$. Repeating the same argument on linear recurrences, we can write $f_{0}(a)=K(a) a^{n}+\\frac{L(a)}{a^{n}}$ with some functions $K$ and $L$. By substituting $n=0$, we can see that $K(a)+L(a)=f_{0}(1)=0$ for every $a$. Hence,\n\n$$\nf_{0}\\left(a^{n}\\right)=K(a)\\left(a^{n}-\\frac{1}{a^{n}}\\right)\n$$\n\nNow take two numbers $a>b>1$ arbitrarily and substitute $x=(a / b)^{n}$ and $y=(a b)^{n}$ in (1):\n\n$$\n\\begin{aligned}\n\\left(\\frac{a^{n}}{b^{n}}+\\frac{b^{n}}{a^{n}}\\right) f_{0}\\left((a b)^{n}\\right) & =f_{0}\\left(a^{2 n}\\right)+f_{0}\\left(b^{2 n}\\right), \\quad \\text { so } \\\\\n\\left(\\frac{a^{n}}{b^{n}}+\\frac{b^{n}}{a^{n}}\\right) K(a b)\\left((a b)^{n}-\\frac{1}{(a b)^{n}}\\right) & =K(a)\\left(a^{2 n}-\\frac{1}{a^{2 n}}\\right)+K(b)\\left(b^{2 n}-\\frac{1}{b^{2 n}}\\right), \\quad \\text { or equivalently } \\\\\nK(a b)\\left(a^{2 n}-\\frac{1}{a^{2 n}}+b^{2 n}-\\frac{1}{b^{2 n}}\\right) & =K(a)\\left(a^{2 n}-\\frac{1}{a^{2 n}}\\right)+K(b)\\left(b^{2 n}-\\frac{1}{b^{2 n}}\\right) .\n\\end{aligned}\n\\tag{5}\n$$\n\nBy dividing (5) by $a^{2 n}$ and then taking limit with $n \\rightarrow+\\infty$ we get $K(a b)=K(a)$. Then (5) reduces to $K(a)=K(b)$. Hence, $K(a)=K(b)$ for all $a>b>1$.\n\nFix $a>1$. For every $x>0$ there is some $b$ and an integer $n$ such that $10$, at least one of $a_{1}, \\ldots, a_{n}$ is positive; without loss of generality suppose $a_{1} \\geqslant 1$.\n\nConsider the polynomials $F_{1}=\\Delta_{1} F$ and $G_{1}=\\Delta G$. On the $\\operatorname{grid}\\left\\{0, \\ldots, a_{1}-1\\right\\} \\times\\left\\{0, \\ldots, a_{2}\\right\\} \\times$ $\\ldots \\times\\left\\{0, \\ldots, a_{n}\\right\\}$ we have\n\n$$\n\\begin{aligned}\nF_{1}\\left(x_{1}, \\ldots, x_{n}\\right) & =F\\left(x_{1}+1, x_{2}, \\ldots, x_{n}\\right)-F\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)= \\\\\n& =G\\left(x_{1}+\\ldots+x_{n}+1\\right)-G\\left(x_{1}+\\ldots+x_{n}\\right)=G_{1}\\left(x_{1}+\\ldots+x_{n}\\right) .\n\\end{aligned}\n$$\n\nSince $G$ is nonconstant, we have $\\operatorname{deg} G_{1}=\\operatorname{deg} G-1 \\leqslant\\left(a_{1}-1\\right)+a_{2}+\\ldots+a_{n}$. Therefore we can apply the induction hypothesis to $F_{1}$ and $G_{1}$ and conclude that $F_{1}$ is not the zero polynomial and $\\operatorname{deg} F_{1} \\geqslant \\operatorname{deg} G_{1}$. Hence, $\\operatorname{deg} F \\geqslant \\operatorname{deg} F_{1}+1 \\geqslant \\operatorname{deg} G_{1}+1=\\operatorname{deg} G$. That finishes the proof.\n\nTo prove the problem statement, take the unique polynomial $g(x)$ so that $g(x)=\\left\\lfloor\\frac{x}{m}\\right\\rfloor$ for $x \\in\\{0,1, \\ldots, n(m-1)\\}$ and $\\operatorname{deg} g \\leqslant n(m-1)$. Notice that precisely $n(m-1)+1$ values of $g$ are prescribed, so $g(x)$ indeed exists and is unique. Notice further that the constraints $g(0)=g(1)=0$ and $g(m)=1$ together enforce $\\operatorname{deg} g \\geqslant 2$.\n\nBy applying the lemma to $a_{1}=\\ldots=a_{n}=m-1$ and the polynomials $f$ and $g$, we achieve $\\operatorname{deg} f \\geqslant \\operatorname{deg} g$. Hence we just need a suitable lower bound on $\\operatorname{deg} g$.\n\nConsider the polynomial $h(x)=g(x+m)-g(x)-1$. The degree of $g(x+m)-g(x)$ is $\\operatorname{deg} g-1 \\geqslant 1$, so $\\operatorname{deg} h=\\operatorname{deg} g-1 \\geqslant 1$, and therefore $h$ cannot be the zero polynomial. On the other hand, $h$ vanishes at the points $0,1, \\ldots, n(m-1)-m$, so $h$ has at least $(n-1)(m-1)$ roots. Hence,\n\n$$\n\\operatorname{deg} f \\geqslant \\operatorname{deg} g=\\operatorname{deg} h+1 \\geqslant(n-1)(m-1)+1 \\geqslant n\n$$']",,True,,, 2035,Algebra,,"Find the maximal value of $$ S=\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}}+\sqrt[3]{\frac{d}{a+7}} $$ where $a, b, c, d$ are nonnegative real numbers which satisfy $a+b+c+d=100$.","['Since the value $8 / \\sqrt[3]{7}$ is reached, it suffices to prove that $S \\leqslant 8 / \\sqrt[3]{7}$.\n\nAssume that $x, y, z, t$ is a permutation of the variables, with $x \\leqslant y \\leqslant z \\leqslant t$. Then, by the rearrangement inequality,\n\n$$\nS \\leqslant\\left(\\sqrt[3]{\\frac{x}{t+7}}+\\sqrt[3]{\\frac{t}{x+7}}\\right)+\\left(\\sqrt[3]{\\frac{y}{z+7}}+\\sqrt[3]{\\frac{z}{y+7}}\\right)\n$$\n\nClaim. The first bracket above does not exceed $\\sqrt[3]{\\frac{x+t+14}{7}}$.\n\nProof. Since\n\n$$\nX^{3}+Y^{3}+3 X Y Z-Z^{3}=\\frac{1}{2}(X+Y-Z)\\left((X-Y)^{2}+(X+Z)^{2}+(Y+Z)^{2}\\right)\n$$\n\nthe inequality $X+Y \\leqslant Z$ is equivalent (when $X, Y, Z \\geqslant 0$ ) to $X^{3}+Y^{3}+3 X Y Z \\leqslant Z^{3}$. Therefore, the claim is equivalent to\n\n$$\n\\frac{x}{t+7}+\\frac{t}{x+7}+3 \\sqrt[3]{\\frac{x t(x+t+14)}{7(x+7)(t+7)}} \\leqslant \\frac{x+t+14}{7}\n$$\n\nNotice that\n\n$$\n\\begin{array}{r}\n3 \\sqrt[3]{\\frac{x t(x+t+14)}{7(x+7)(t+7)}}=3 \\sqrt[3]{\\frac{t(x+7)}{7(t+7)} \\cdot \\frac{x(t+7)}{7(x+7)} \\cdot \\frac{7(x+t+14)}{(t+7)(x+7)}} \\\\\n\\leqslant \\frac{t(x+7)}{7(t+7)}+\\frac{x(t+7)}{7(x+7)}+\\frac{7(x+t+14)}{(t+7)(x+7)}\n\\end{array}\n$$\n\nby the AM-GM inequality, so it suffices to prove\n\n$$\n\\frac{x}{t+7}+\\frac{t}{x+7}+\\frac{t(x+7)}{7(t+7)}+\\frac{x(t+7)}{7(x+7)}+\\frac{7(x+t+14)}{(t+7)(x+7)} \\leqslant \\frac{x+t+14}{7}\n$$\n\nA straightforward check verifies that the last inequality is in fact an equality.\n\nThe claim leads now to\n\n$$\nS \\leqslant \\sqrt[3]{\\frac{x+t+14}{7}}+\\sqrt[3]{\\frac{y+z+14}{7}} \\leqslant 2 \\sqrt[3]{\\frac{x+y+z+t+28}{14}}=\\frac{8}{\\sqrt[3]{7}}\n$$\n\nthe last inequality being due to the AM-CM inequality (or to the fact that $\\sqrt[3]{ }$ is concave on $[0, \\infty))$.', ""We present a different proof for the estimate $S \\leqslant 8 / \\sqrt[3]{7}$.\n\nStart by using Hölder's inequality:\n\n$$\nS^{3}=\\left(\\sum_{\\mathrm{cyc}} \\frac{\\sqrt[6]{a} \\cdot \\sqrt[6]{a}}{\\sqrt[3]{b+7}}\\right)^{3} \\leqslant \\sum_{\\mathrm{cyc}}(\\sqrt[6]{a})^{3} \\cdot \\sum_{\\mathrm{cyc}}(\\sqrt[6]{a})^{3} \\cdot \\sum_{\\mathrm{cyc}}\\left(\\frac{1}{\\sqrt[3]{b+7}}\\right)^{3}=\\left(\\sum_{\\text {cyc }} \\sqrt{a}\\right)^{2} \\sum_{\\mathrm{cyc}} \\frac{1}{b+7}\n$$\n\nNotice that\n\n$$\n\\frac{(x-1)^{2}(x-7)^{2}}{x^{2}+7} \\geqslant 0 \\Longleftrightarrow x^{2}-16 x+71 \\geqslant \\frac{448}{x^{2}+7}\n$$\n\nyields\n\n$$\n\\sum \\frac{1}{b+7} \\leqslant \\frac{1}{448} \\sum(b-16 \\sqrt{b}+71)=\\frac{1}{448}\\left(384-16 \\sum \\sqrt{b}\\right)=\\frac{48-2 \\sum \\sqrt{b}}{56} .\n$$\n\nFinally,\n\n$$\nS^{3} \\leqslant \\frac{1}{56}\\left(\\sum \\sqrt{a}\\right)^{2}\\left(48-2 \\sum \\sqrt{a}\\right) \\leqslant \\frac{1}{56}\\left(\\frac{\\sum \\sqrt{a}+\\sum \\sqrt{a}+\\left(48-2 \\sum \\sqrt{a}\\right)}{3}\\right)^{3}=\\frac{512}{7}\n$$\n\nby the AM-GM inequality. The conclusion follows.""]",['$\\frac{8}{\\sqrt[3]{7}}$'],False,,Numerical, 2036,Combinatorics,,"Let $n \geqslant 3$ be an integer. Prove that there exists a set $S$ of $2 n$ positive integers satisfying the following property: For every $m=2,3, \ldots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$.","['We show that one of possible examples is the set\n\n$$\nS=\\left\\{1 \\cdot 3^{k}, 2 \\cdot 3^{k}: k=1,2, \\ldots, n-1\\right\\} \\cup\\left\\{1, \\frac{3^{n}+9}{2}-1\\right\\}\n$$\n\nIt is readily verified that all the numbers listed above are distinct (notice that the last two are not divisible by 3 ).\n\nThe sum of elements in $S$ is\n\n$$\n\\Sigma=1+\\left(\\frac{3^{n}+9}{2}-1\\right)+\\sum_{k=1}^{n-1}\\left(1 \\cdot 3^{k}+2 \\cdot 3^{k}\\right)=\\frac{3^{n}+9}{2}+\\sum_{k=1}^{n-1} 3^{k+1}=\\frac{3^{n}+9}{2}+\\frac{3^{n+1}-9}{2}=2 \\cdot 3^{n}\n$$\n\nHence, in order to show that this set satisfies the problem requirements, it suffices to present, for every $m=2,3, \\ldots, n$, an $m$-element subset $A_{m} \\subset S$ whose sum of elements equals $3^{n}$.\n\nSuch a subset is\n\n$$\nA_{m}=\\left\\{2 \\cdot 3^{k}: k=n-m+1, n-m+2, \\ldots, n-1\\right\\} \\cup\\left\\{1 \\cdot 3^{n-m+1}\\right\\} .\n$$\n\nClearly, $\\left|A_{m}\\right|=m$. The sum of elements in $A_{m}$ is\n\n$$\n3^{n-m+1}+\\sum_{k=n-m+1}^{n-1} 2 \\cdot 3^{k}=3^{n-m+1}+\\frac{2 \\cdot 3^{n}-2 \\cdot 3^{n-m+1}}{2}=3^{n}\n$$\n\nas required.']",,True,,, 2037,Combinatorics,,"Queenie and Horst play a game on a $20 \times 20$ chessboard. In the beginning the board is empty. In every turn, Horst places a black knight on an empty square in such a way that his new knight does not attack any previous knights. Then Queenie places a white queen on an empty square. The game gets finished when somebody cannot move. Find the maximal positive $K$ such that, regardless of the strategy of Queenie, Horst can put at least $K$ knights on the board.","['We show two strategies, one for Horst to place at least 100 knights, and another strategy for Queenie that prevents Horst from putting more than 100 knights on the board.\n\nA strategy for Horst: Put knights only on black squares, until all black squares get occupied.\n\nColour the squares of the board black and white in the usual way, such that the white and black squares alternate, and let Horst put his knights on black squares as long as it is possible. Two knights on squares of the same colour never attack each other. The number of black squares is $20^{2} / 2=200$. The two players occupy the squares in turn, so Horst will surely find empty black squares in his first 100 steps.\n\nA strategy for Queenie: Group the squares into cycles of length 4, and after each step of Horst, occupy the opposite square in the same cycle.\n\nConsider the squares of the board as vertices of a graph; let two squares be connected if two knights on those squares would attack each other. Notice that in a $4 \\times 4$ board the squares can be grouped into 4 cycles of length 4 , as shown in Figure 1. Divide the board into parts of size $4 \\times 4$, and perform the same grouping in every part; this way we arrange the 400 squares of the board into 100 cycles (Figure 2).\n\n\n\nFigure 1\n\n\n\nFigure 2\n\n\n\nFigure 3\n\nThe strategy of Queenie can be as follows: Whenever Horst puts a new knight to a certain square $A$, which is part of some cycle $A-B-C-D-A$, let Queenie put her queen on the opposite square $C$ in that cycle (Figure 3). From this point, Horst cannot put any knight on $A$ or $C$ because those squares are already occupied, neither on $B$ or $D$ because those squares are attacked by the knight standing on $A$. Hence, Horst can put at most one knight on each cycle, that is at most 100 knights in total.']",['100'],False,,Numerical, 2038,Combinatorics,,"Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n+1$ squares in a row, numbered 0 to $n$ from left to right. Initially, $n$ stones are put into square 0, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of those stones and moves it to the right by at most $k$ squares (the stone should stay within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than $$ \left\lceil\frac{n}{1}\right\rceil+\left\lceil\frac{n}{2}\right\rceil+\left\lceil\frac{n}{3}\right\rceil+\cdots+\left\lceil\frac{n}{n}\right\rceil $$ turns. (As usual, $\lceil x\rceil$ stands for the least integer not smaller than $x$.)","['The stones are indistinguishable, and all have the same origin and the same final position. So, at any turn we can prescribe which stone from the chosen square to move. We do it in the following manner. Number the stones from 1 to $n$. At any turn, after choosing a square, Sisyphus moves the stone with the largest number from this square.\n\nThis way, when stone $k$ is moved from some square, that square contains not more than $k$ stones (since all their numbers are at most $k$ ). Therefore, stone $k$ is moved by at most $k$ squares at each turn. Since the total shift of the stone is exactly $n$, at least $\\lceil n / k\\rceil$ moves of stone $k$ should have been made, for every $k=1,2, \\ldots, n$.\n\nBy summing up over all $k=1,2, \\ldots, n$, we get the required estimate.']",,True,,, 2039,Combinatorics,,"An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once: Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once.","[""Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\\cdots+n$, it follows that they form a permutation of $1,2, \\ldots, n$.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nConsider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\\prime}$, has side length $\\ell \\geqslant\\lceil(n-2) / 2\\rceil$. Since $T^{\\prime}$ obeys the anti-Pascal rule, it contains $\\ell$ pairwise distinct positive integers $a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{\\ell}^{\\prime}$, where $a_{1}^{\\prime}$ is at the apex, and $a_{k}^{\\prime}$ and $b_{k}^{\\prime}=a_{1}^{\\prime}+a_{2}^{\\prime}+\\cdots+a_{k}^{\\prime}$ are the two neighbours below $b_{k-1}^{\\prime}$ for each $k=2,3 \\ldots, \\ell$. Since the $a_{k}$ all lie outside $T^{\\prime}$, and they form a permutation of $1,2, \\ldots, n$, the $a_{k}^{\\prime}$ are all greater than $n$. Consequently,\n\n$$\n\\begin{array}{r}\nb_{\\ell}^{\\prime} \\geqslant(n+1)+(n+2)+\\cdots+(n+\\ell)=\\frac{\\ell(2 n+\\ell+1)}{2} \\\\\n\\geqslant \\frac{1}{2} \\cdot \\frac{n-2}{2}\\left(2 n+\\frac{n-2}{2}+1\\right)=\\frac{5 n(n-2)}{8},\n\\end{array}\n$$\n\nwhich is greater than $1+2+\\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.\n\nSo it is not possible.""]",,True,,, 2039,Combinatorics,,"An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once: ![](https://cdn.mathpix.com/cropped/2023_12_21_9a7a2c8e401502a85653g-1.jpg?height=208&width=274&top_left_y=553&top_left_x=911) Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once.","[""Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\\cdots+n$, it follows that they form a permutation of $1,2, \\ldots, n$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9a7a2c8e401502a85653g-1.jpg?height=389&width=442&top_left_y=1413&top_left_x=450)\n\nFigure 1\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9a7a2c8e401502a85653g-1.jpg?height=389&width=459&top_left_y=1413&top_left_x=1164)\n\nFigure 2\n\nConsider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\\prime}$, has side length $\\ell \\geqslant\\lceil(n-2) / 2\\rceil$. Since $T^{\\prime}$ obeys the anti-Pascal rule, it contains $\\ell$ pairwise distinct positive integers $a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{\\ell}^{\\prime}$, where $a_{1}^{\\prime}$ is at the apex, and $a_{k}^{\\prime}$ and $b_{k}^{\\prime}=a_{1}^{\\prime}+a_{2}^{\\prime}+\\cdots+a_{k}^{\\prime}$ are the two neighbours below $b_{k-1}^{\\prime}$ for each $k=2,3 \\ldots, \\ell$. Since the $a_{k}$ all lie outside $T^{\\prime}$, and they form a permutation of $1,2, \\ldots, n$, the $a_{k}^{\\prime}$ are all greater than $n$. Consequently,\n\n$$\n\\begin{array}{r}\nb_{\\ell}^{\\prime} \\geqslant(n+1)+(n+2)+\\cdots+(n+\\ell)=\\frac{\\ell(2 n+\\ell+1)}{2} \\\\\n\\geqslant \\frac{1}{2} \\cdot \\frac{n-2}{2}\\left(2 n+\\frac{n-2}{2}+1\\right)=\\frac{5 n(n-2)}{8},\n\\end{array}\n$$\n\nwhich is greater than $1+2+\\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.\n\nSo it is not possible.""]",['证明题，略'],True,,Need_human_evaluate, 2040,Combinatorics,,"Let $k$ be a positive integer. The organising committee of a tennis tournament is to schedule the matches for $2 k$ players so that every two players play once, each day exactly one match is played, and each player arrives to the tournament site the day of his first match, and departs the day of his last match. For every day a player is present on the tournament, the committee has to pay 1 coin to the hotel. The organisers want to design the schedule so as to minimise the total cost of all players' stays. Determine this minimum cost.","[""Enumerate the days of the tournament $1,2, \\ldots,\\left(\\begin{array}{c}2 k \\\\ 2\\end{array}\\right)$. Let $b_{1} \\leqslant b_{2} \\leqslant \\cdots \\leqslant b_{2 k}$ be the days the players arrive to the tournament, arranged in nondecreasing order; similarly, let $e_{1} \\geqslant \\cdots \\geqslant e_{2 k}$ be the days they depart arranged in nonincreasing order (it may happen that a player arrives on day $b_{i}$ and departs on day $e_{j}$, where $i \\neq j$ ). If a player arrives on day $b$ and departs on day $e$, then his stay cost is $e-b+1$. Therefore, the total stay cost is\n\n$$\n\\Sigma=\\sum_{i=1}^{2 k} e_{i}-\\sum_{i=1}^{2 k} b_{i}+n=\\sum_{i=1}^{2 k}\\left(e_{i}-b_{i}+1\\right)\n$$\n\nBounding the total cost from below. To this end, estimate $e_{i+1}-b_{i+1}+1$. Before day $b_{i+1}$, only $i$ players were present, so at most $\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)$ matches could be played. Therefore, $b_{i+1} \\leqslant\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)+1$. Similarly, at most $\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)$ matches could be played after day $e_{i+1}$, so $e_{i} \\geqslant\\left(\\begin{array}{c}2 k \\\\ 2\\end{array}\\right)-\\left(\\begin{array}{c}i \\\\ 2\\end{array}\\right)$. Thus,\n\n$$\ne_{i+1}-b_{i+1}+1 \\geqslant\\left(\\begin{array}{c}\n2 k \\\\\n2\n\\end{array}\\right)-2\\left(\\begin{array}{c}\ni \\\\\n2\n\\end{array}\\right)=k(2 k-1)-i(i-1)\n$$\n\nThis lower bound can be improved for $i>k$ : List the $i$ players who arrived first, and the $i$ players who departed last; at least $2 i-2 k$ players appear in both lists. The matches between these players were counted twice, though the players in each pair have played only once. Therefore, if $i>k$, then\n\n$$\ne_{i+1}-b_{i+1}+1 \\geqslant\\left(\\begin{array}{c}\n2 k \\\\\n2\n\\end{array}\\right)-2\\left(\\begin{array}{l}\ni \\\\\n2\n\\end{array}\\right)+\\left(\\begin{array}{c}\n2 i-2 k \\\\\n2\n\\end{array}\\right)=(2 k-i)^{2}\n$$\n\nAn optimal tournament, We now describe a schedule in which the lower bounds above are all achieved simultaneously. Split players into two groups $X$ and $Y$, each of cardinality $k$. Next, partition the schedule into three parts. During the first part, the players from $X$ arrive one by one, and each newly arrived player immediately plays with everyone already present. During the third part (after all players from $X$ have already departed) the players from $Y$ depart one by one, each playing with everyone still present just before departing.\n\nIn the middle part, everyone from $X$ should play with everyone from $Y$. Let $S_{1}, S_{2}, \\ldots, S_{k}$ be the players in $X$, and let $T_{1}, T_{2}, \\ldots, T_{k}$ be the players in $Y$. Let $T_{1}, T_{2}, \\ldots, T_{k}$ arrive in this order; after $T_{j}$ arrives, he immediately plays with all the $S_{i}, i>j$. Afterwards, players $S_{k}$, $S_{k-1}, \\ldots, S_{1}$ depart in this order; each $S_{i}$ plays with all the $T_{j}, i \\leqslant j$, just before his departure, and $S_{k}$ departs the day $T_{k}$ arrives. For $0 \\leqslant s \\leqslant k-1$, the number of matches played between $T_{k-s}$ 's arrival and $S_{k-s}$ 's departure is\n\n$$\n\\sum_{j=k-s}^{k-1}(k-j)+1+\\sum_{j=k-s}^{k-1}(k-j+1)=\\frac{1}{2} s(s+1)+1+\\frac{1}{2} s(s+3)=(s+1)^{2}\n$$\n\nThus, if $i>k$, then the number of matches that have been played between $T_{i-k+1}$ 's arrival, which is $b_{i+1}$, and $S_{i-k+1}$ 's departure, which is $e_{i+1}$, is $(2 k-i)^{2}$; that is, $e_{i+1}-b_{i+1}+1=(2 k-i)^{2}$, showing the second lower bound achieved for all $i>k$.\n\nIf $i \\leqslant k$, then the matches between the $i$ players present before $b_{i+1}$ all fall in the first part of the schedule, so there are $\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)$ such, and $b_{i+1}=\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)+1$. Similarly, after $e_{i+1}$, there are $i$ players left, all $\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)$ matches now fall in the third part of the schedule, and $e_{i+1}=\\left(\\begin{array}{c}2 k \\\\ 2\\end{array}\\right)-\\left(\\begin{array}{l}i \\\\ 2\\end{array}\\right)$. The first lower bound is therefore also achieved for all $i \\leqslant k$.\n\nConsequently, all lower bounds are achieved simultaneously, and the schedule is indeed optimal.\n\nEvaluation. Finally, evaluate the total cost for the optimal schedule:\n\n$$\n\\begin{aligned}\n\\Sigma & =\\sum_{i=0}^{k}(k(2 k-1)-i(i-1))+\\sum_{i=k+1}^{2 k-1}(2 k-i)^{2}=(k+1) k(2 k-1)-\\sum_{i=0}^{k} i(i-1)+\\sum_{j=1}^{k-1} j^{2} \\\\\n& =k(k+1)(2 k-1)-k^{2}+\\frac{1}{2} k(k+1)=\\frac{1}{2} k\\left(4 k^{2}+k-1\\right)\n\\end{aligned}\n$$"", ""Consider any tournament schedule. Label players $P_{1}, P_{2}, \\ldots, P_{2 k}$ in order of their arrival, and label them again $Q_{2 k}, Q_{2 k-1}, \\ldots, Q_{1}$ in order of their departure, to define a permutation $a_{1}, a_{2}, \\ldots, a_{2 k}$ of $1,2, \\ldots, 2 k$ by $P_{i}=Q_{a_{i}}$.\n\nWe first describe an optimal tournament for any given permutation $a_{1}, a_{2}, \\ldots, a_{2 k}$ of the indices $1,2, \\ldots, 2 k$. Next, we find an optimal permutation and an optimal tournament.\n\nOptimisation for a fixed $a_{1}, \\ldots, a_{2 k}$. We say that the cost of the match between $P_{i}$ and $P_{j}$ is the number of players present at the tournament when this match is played. Clearly, the Committee pays for each day the cost of the match of that day. Hence, we are to minimise the total cost of all matches.\n\nNotice that $Q_{2 k}$ 's departure does not precede $P_{2 k}$ 's arrival. Hence, the number of players at the tournament monotonically increases (non-strictly) until it reaches $2 k$, and then monotonically decreases (non-strictly). So, the best time to schedule the match between $P_{i}$ and $P_{j}$ is either when $P_{\\max (i, j)}$ arrives, or when $Q_{\\max \\left(a_{i}, a_{j}\\right)}$ departs, in which case the cost is $\\min \\left(\\max (i, j), \\max \\left(a_{i}, a_{j}\\right)\\right)$.\n\nConversely, assuming that $i>j$, if this match is scheduled between the arrivals of $P_{i}$ and $P_{i+1}$, then its cost will be exactly $i=\\max (i, j)$. Similarly, one can make it cost $\\max \\left(a_{i}, a_{j}\\right)$. Obviously, these conditions can all be simultaneously satisfied, so the minimal cost for a fixed sequence $a_{1}, a_{2}, \\ldots, a_{2 k}$ is\n\n$$\n\\Sigma\\left(a_{1}, \\ldots, a_{2 k}\\right)=\\sum_{1 \\leqslant ia b-a-b$ is expressible in the form $x=s a+t b$, with integer $s, t \\geqslant 0$.\n\nWe will prove by induction on $s+t$ that if $x=s a+b t$, with $s, t$ nonnegative integers, then\n\n$$\nf(x)>\\frac{f(0)}{2^{s+t}}-2\n\\tag{3}\n$$\n\nThe base case $s+t=0$ is trivial. Assume now that (3) is true for $s+t=v$. Then, if $s+t=v+1$ and $x=s a+t b$, at least one of the numbers $s$ and $t$ - say $s-$ is positive, hence by (2),\n\n$$\nf(x)=f(s a+t b) \\geqslant \\frac{f((s-1) a+t b)}{2}-1>\\frac{1}{2}\\left(\\frac{f(0)}{2^{s+t-1}}-2\\right)-1=\\frac{f(0)}{2^{s+t}}-2 .\n$$\n\nAssume now that we must perform moves of type (ii) ad infinitum. Take $n=a b-a-b$ and suppose $b>a$. Since each of the numbers $n+1, n+2, \\ldots, n+b$ can be expressed in the form $s a+t b$, with $0 \\leqslant s \\leqslant b$ and $0 \\leqslant t \\leqslant a$, after moves of type (ii) have been performed $2^{a+b+1}$ times and we have to add a new pair of zeros, each $f(n+k), k=1,2, \\ldots, b$, is at least 2 . In this case (1) yields inductively $f(n+k) \\geqslant 2$ for all $k \\geqslant 1$. But this is absurd: after a finite number of moves, $f$ cannot attain nonzero values at infinitely many points.', 'We start by showing that the result of the process in the problem does not depend on the way the operations are performed. For that purpose, it is convenient to modify the process a bit.\n\nClaim 1. Suppose that the board initially contains a finite number of nonnegative integers, and one starts performing type (i) moves only. Assume that one had applied $k$ moves which led to a final arrangement where no more type $(i)$ moves are possible. Then, if one starts from the same initial arrangement, performing type (i) moves in an arbitrary fashion, then the process will necessarily stop at the same final arrangement\n\nProof. Throughout this proof, all moves are supposed to be of type $(i)$.\n\nInduct on $k$; the base case $k=0$ is trivial, since no moves are possible. Assume now that $k \\geqslant 1$. Fix some canonical process, consisting of $k$ moves $M_{1}, M_{2}, \\ldots, M_{k}$, and reaching the final arrangement $A$. Consider any sample process $m_{1}, m_{2}, \\ldots$ starting with the same initial arrangement and proceeding as long as possible; clearly, it contains at least one move. We need to show that this process stops at $A$.\n\nLet move $m_{1}$ consist in replacing two copies of $x$ with $x+a$ and $x+b$. If move $M_{1}$ does the same, we may apply the induction hypothesis to the arrangement appearing after $m_{1}$. Otherwise, the canonical process should still contain at least one move consisting in replacing $(x, x) \\mapsto(x+a, x+b)$, because the initial arrangement contains at least two copies of $x$, while the final one contains at most one such.\n\nLet $M_{i}$ be the first such move. Since the copies of $x$ are indistinguishable and no other copy of $x$ disappeared before $M_{i}$ in the canonical process, the moves in this process can be permuted as $M_{i}, M_{1}, \\ldots, M_{i-1}, M_{i+1}, \\ldots, M_{k}$, without affecting the final arrangement. Now it suffices to perform the move $m_{1}=M_{i}$ and apply the induction hypothesis as above.\n\nClaim 2. Consider any process starting from the empty board, which involved exactly $n$ moves of type (ii) and led to a final arrangement where all the numbers are distinct. Assume that one starts with the board containing $2 n$ zeroes (as if $n$ moves of type (ii) were made in the beginning), applying type (i) moves in an arbitrary way. Then this process will reach the same final arrangement.\n\nProof. Starting with the board with $2 n$ zeros, one may indeed model the first process mentioned in the statement of the claim, omitting the type (ii) moves. This way, one reaches the same final arrangement. Now, Claim 1 yields that this final arrangement will be obtained when type (i) moves are applied arbitrarily.\n\nClaim 2 allows now to reformulate the problem statement as follows: There exists an integer $n$ such that, starting from $2 n$ zeroes, one may apply type (i) moves indefinitely.\n\nIn order to prove this, we start with an obvious induction on $s+t=k \\geqslant 1$ to show that if we start with $2^{s+t}$ zeros, then we can get simultaneously on the board, at some point, each of the numbers $s a+t b$, with $s+t=k$.\n\nSuppose now that $a\n\nNow $\\angle F B D=\\angle F D B$ gives $\\overparen{A F}=\\overparen{B F}+\\overparen{A K}=\\overparen{B F}+\\overparen{A L}$, hence $\\overparen{B F}=\\overparen{L F}$. In a similar way, we get $\\overparen{C G}=\\overparen{G K}$. This yields\n\n$$\n\\angle(A P, F G)=\\frac{\\overparen{A F}+\\overparen{P G}}{2}=\\frac{\\overparen{A L}+\\overparen{L F}+\\overparen{P C}+\\overparen{C G}}{2}=\\frac{\\overparen{K L}+\\overparen{L B}+\\overparen{B C}+\\overparen{C K}}{4}=90^{\\circ}\n$$', 'Let $Z=A B \\cap F G, T=A C \\cap F G$. It suffices to prove that $\\angle A T Z=\\angle A Z T$.\n\nLet $X$ be the point for which $F X A D$ is a parallelogram. Then\n\n$$\n\\angle F X A=\\angle F D A=180^{\\circ}-\\angle F D B=180^{\\circ}-\\angle F B D \\text {, }\n$$\n\nwhere in the last equality we used that $F D=F B$. It follows that the quadrilateral $B F X A$ is cyclic, so $X$ lies on $\\Gamma$.\n\n\n\n\n\nAnalogously, if $Y$ is the point for which $G Y A E$ is a parallelogram, then $Y$ lies on $\\Gamma$. So the quadrilateral $X F G Y$ is cyclic and $F X=A D=A E=G Y$, hence $X F G Y$ is an isosceles trapezoid.\n\nNow, by $X F \\| A Z$ and $Y G \\| A T$, it follows that $\\angle A T Z=\\angle Y G F=\\angle X F G=\\angle A Z T$.', 'As in the first solution, we prove that $F G \\perp A P$, where $P$ is the midpoint of the small arc $\\overparen{B C}$.\n\nLet $O$ be the circumcentre of the triangle $A B C$, and let $M$ and $N$ be the midpoints of the small arcs $\\overparen{A B}$ and $\\overparen{A C}$, respectively. Then $O M$ and $O N$ are the perpendicular bisectors of $A B$ and $A C$, respectively.\n\n\n\nThe distance $d$ between $O M$ and the perpendicular bisector of $B D$ is $\\frac{1}{2} A B-\\frac{1}{2} B D=\\frac{1}{2} A D$, hence it is equal to the distance between $O N$ and the perpendicular bisector of $C E$.\n\nThis shows that the isosceles trapezoid determined by the diameter $\\delta$ of $\\Gamma$ through $M$ and the chord parallel to $\\delta$ through $F$ is congruent to the isosceles trapezoid determined by the diameter $\\delta^{\\prime}$ of $\\Gamma$ through $N$ and the chord parallel to $\\delta^{\\prime}$ through $G$. Therefore $M F=N G$, yielding $M N \\| F G$.\n\nNow\n\n$$\n\\angle(M N, A P)=\\frac{1}{2}(\\overparen{A M}+\\overparen{P C}+\\overparen{C N})=\\frac{1}{4}(\\overparen{A B}+\\overparen{B C}+\\overparen{C A})=90^{\\circ}\n$$\n\nhence $M N \\perp A P$, and the conclusion follows.']",,True,,, 2044,Geometry,,"Let $A B C$ be a triangle with $A B=A C$, and let $M$ be the midpoint of $B C$. Let $P$ be a point such that $P B

\n\nNow let $Z$ be the common point of $A M$ and the perpendicular through $Y$ to $P C$ (notice that $Z$ lies on to the ray $A M$ beyond $M$ ). We have $\\angle P A Z=\\angle P Y Z=90^{\\circ}$. Thus the points $P, A, Y$, and $Z$ are concyclic.\n\nSince $\\angle C M Z=\\angle C Y Z=90^{\\circ}$, the quadrilateral $C Y Z M$ is cyclic, hence $\\angle C Z M=$ $\\angle C Y M$. By the condition in the statement, $\\angle C Y M=\\angle B X M$, and, by symmetry in $Z M$, $\\angle C Z M=\\angle B Z M$. Therefore, $\\angle B X M=\\angle B Z M$. It follows that the points $B, X, Z$, and $M$ are concyclic, hence $\\angle B X Z=180^{\\circ}-\\angle B M Z=90^{\\circ}$.\n\nFinally, we have $\\angle P X Z=\\angle P Y Z=\\angle P A Z=90^{\\circ}$, hence the five points $P, A, X, Y, Z$ are concyclic. In particular, the quadrilateral $A P X Y$ is cyclic, as required.']",,True,,, 2045,Geometry,,"A circle $\omega$ of radius 1 is given. A collection $T$ of triangles is called good, if the following conditions hold: (i) each triangle from $T$ is inscribed in $\omega$; (ii) no two triangles from $T$ have a common interior point. Determine all positive real numbers $t$ such that, for each positive integer $n$, there exists a good collection of $n$ triangles, each of perimeter greater than $t$.","[""First, we show how to construct a good collection of $n$ triangles, each of perimeter greater than 4 . This will show that all $t \\leqslant 4$ satisfy the required conditions.\n\nConstruct inductively an $(n+2)$-gon $B A_{1} A_{2} \\ldots A_{n} C$ inscribed in $\\omega$ such that $B C$ is a diameter, and $B A_{1} A_{2}, B A_{2} A_{3}, \\ldots, B A_{n-1} A_{n}, B A_{n} C$ is a good collection of $n$ triangles. For $n=1$, take any triangle $B A_{1} C$ inscribed in $\\omega$ such that $B C$ is a diameter; its perimeter is greater than $2 B C=4$. To perform the inductive step, assume that the $(n+2)$-gon $B A_{1} A_{2} \\ldots A_{n} C$ is already constructed. Since $A_{n} B+A_{n} C+B C>4$, one can choose a point $A_{n+1}$ on the small arc $\\widehat{C A_{n}}$, close enough to $C$, so that $A_{n} B+A_{n} A_{n+1}+B A_{n+1}$ is still greater than 4 . Thus each of these new triangles $B A_{n} A_{n+1}$ and $B A_{n+1} C$ has perimeter greater than 4 , which completes the induction step.\n\n\n\nWe proceed by showing that no $t>4$ satisfies the conditions of the problem. To this end, we assume that there exists a good collection $T$ of $n$ triangles, each of perimeter greater than $t$, and then bound $n$ from above.\n\nTake $\\varepsilon>0$ such that $t=4+2 \\varepsilon$.\n\nClaim. There exists a positive constant $\\sigma=\\sigma(\\varepsilon)$ such that any triangle $\\Delta$ with perimeter $2 s \\geqslant 4+2 \\varepsilon$, inscribed in $\\omega$, has area $S(\\Delta)$ at least $\\sigma$.\n\nProof. Let $a, b, c$ be the side lengths of $\\Delta$. Since $\\Delta$ is inscribed in $\\omega$, each side has length at most 2. Therefore, $s-a \\geqslant(2+\\varepsilon)-2=\\varepsilon$. Similarly, $s-b \\geqslant \\varepsilon$ and $s-c \\geqslant \\varepsilon$. By Heron's formula, $S(\\Delta)=\\sqrt{s(s-a)(s-b)(s-c)} \\geqslant \\sqrt{(2+\\varepsilon) \\varepsilon^{3}}$. Thus we can set $\\sigma(\\varepsilon)=\\sqrt{(2+\\varepsilon) \\varepsilon^{3}}$.\n\nNow we see that the total area $S$ of all triangles from $T$ is at least $n \\sigma(\\varepsilon)$. On the other hand, $S$ does not exceed the area of the disk bounded by $\\omega$. Thus $n \\sigma(\\varepsilon) \\leqslant \\pi$, which means that $n$ is bounded from above.""]","['$t(0,4]$']",False,,Interval, 2046,Geometry,,"A point $T$ is chosen inside a triangle $A B C$. Let $A_{1}, B_{1}$, and $C_{1}$ be the reflections of $T$ in $B C, C A$, and $A B$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_{1} B_{1} C_{1}$. The lines $A_{1} T, B_{1} T$, and $C_{1} T$ meet $\Omega$ again at $A_{2}, B_{2}$, and $C_{2}$, respectively. Prove that the lines $A A_{2}, B B_{2}$, and $C C_{2}$ are concurrent on $\Omega$.","['By $\\sphericalangle(\\ell, n)$ we always mean the directed angle of the lines $\\ell$ and $n$, taken modulo $180^{\\circ}$.\n\nLet $C C_{2}$ meet $\\Omega$ again at $K$ (as usual, if $C C_{2}$ is tangent to $\\Omega$, we set $T=C_{2}$ ). We show that the line $B B_{2}$ contains $K$; similarly, $A A_{2}$ will also pass through $K$. For this purpose, it suffices to prove that\n\n$$\n\\sphericalangle\\left(C_{2} C, C_{2} A_{1}\\right)=\\sphericalangle\\left(B_{2} B, B_{2} A_{1}\\right) .\n\\tag{1}\n$$\n\nBy the problem condition, $C B$ and $C A$ are the perpendicular bisectors of $T A_{1}$ and $T B_{1}$, respectively. Hence, $C$ is the circumcentre of the triangle $A_{1} T B_{1}$. Therefore,\n\n$$\n\\sphericalangle\\left(C A_{1}, C B\\right)=\\sphericalangle(C B, C T)=\\sphericalangle\\left(B_{1} A_{1}, B_{1} T\\right)=\\sphericalangle\\left(B_{1} A_{1}, B_{1} B_{2}\\right) .\n$$\n\nIn circle $\\Omega$ we have $\\sphericalangle\\left(B_{1} A_{1}, B_{1} B_{2}\\right)=\\sphericalangle\\left(C_{2} A_{1}, C_{2} B_{2}\\right)$. Thus,\n\n$$\n\\sphericalangle\\left(C A_{1}, C B\\right)=\\sphericalangle\\left(B_{1} A_{1}, B_{1} B_{2}\\right)=\\sphericalangle\\left(C_{2} A_{1}, C_{2} B_{2}\\right) .\n\\tag{2}\n$$\n\nSimilarly, we get\n\n$$\n\\sphericalangle\\left(B A_{1}, B C\\right)=\\sphericalangle\\left(C_{1} A_{1}, C_{1} C_{2}\\right)=\\sphericalangle\\left(B_{2} A_{1}, B_{2} C_{2}\\right) .\n\\tag{3}\n$$\n\nThe two obtained relations yield that the triangles $A_{1} B C$ and $A_{1} B_{2} C_{2}$ are similar and equioriented, hence\n\n$$\n\\frac{A_{1} B_{2}}{A_{1} B}=\\frac{A_{1} C_{2}}{A_{1} C} \\quad \\text { and } \\quad \\sphericalangle\\left(A_{1} B, A_{1} C\\right)=\\sphericalangle\\left(A_{1} B_{2}, A_{1} C_{2}\\right) \\text {. }\n$$\n\nThe second equality may be rewritten as $\\sphericalangle\\left(A_{1} B, A_{1} B_{2}\\right)=\\sphericalangle\\left(A_{1} C, A_{1} C_{2}\\right)$, so the triangles $A_{1} B B_{2}$ and $A_{1} C C_{2}$ are also similar and equioriented. This establishes (1).\n\n']",,True,,, 2047,Geometry,,"Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark) Preamble. Let $X=y \cap z, Y=x \cap z, Z=x \cap y$ and let $\Omega$ denote the circumcircle of the triangle $X Y Z$. Denote by $X_{0}, Y_{0}$, and $Z_{0}$ the second intersection points of $A I, B I$ and $C I$, respectively, with $\omega$. It is known that $Y_{0} Z_{0}$ is the perpendicular bisector of $A I, Z_{0} X_{0}$ is the perpendicular bisector of $B I$, and $X_{0} Y_{0}$ is the perpendicular bisector of $C I$. In particular, the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$ are homothetic, because their corresponding sides are parallel. The solutions below mostly exploit the following approach. Consider the triangles $X Y Z$ and $X_{0} Y_{0} Z_{0}$, or some other pair of homothetic triangles $\Delta$ and $\delta$ inscribed into $\Omega$ and $\omega$, respectively. In order to prove that $\Omega$ and $\omega$ are tangent, it suffices to show that the centre $T$ of the homothety taking $\Delta$ to $\delta$ lies on $\omega$ (or $\Omega$ ), or, in other words, to show that $\Delta$ and $\delta$ are perspective (i.e., the lines joining corresponding vertices are concurrent), with their perspector lying on $\omega$ (or $\Omega$ ). We use directed angles throughout all the solutions.","['Claim 1. The reflections $\\ell_{a}, \\ell_{b}$ and $\\ell_{c}$ of the line $\\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\\omega$.\n\n\n\nProof. Notice that $\\sphericalangle\\left(\\ell_{b}, \\ell_{c}\\right)=\\sphericalangle\\left(\\ell_{b}, \\ell\\right)+\\sphericalangle\\left(\\ell, \\ell_{c}\\right)=2 \\sphericalangle(y, \\ell)+2 \\sphericalangle(\\ell, z)=2 \\sphericalangle(y, z)$. But $y \\perp B I$ and $z \\perp C I$ implies $\\sphericalangle(y, z)=\\sphericalangle(B I, I C)$, so, since $2 \\sphericalangle(B I, I C)=\\sphericalangle(B A, A C)$, we obtain\n\n$$\n\\sphericalangle\\left(\\ell_{b}, \\ell_{c}\\right)=\\sphericalangle(B A, A C) .\n\\tag{1}\n$$\n\nSince $A$ is the reflection of $D$ in $x, A$ belongs to $\\ell_{a}$; similarly, $B$ belongs to $\\ell_{b}$. Then (1) shows that the common point $T^{\\prime}$ of $\\ell_{a}$ and $\\ell_{b}$ lies on $\\omega$; similarly, the common point $T^{\\prime \\prime}$ of $\\ell_{c}$ and $\\ell_{b}$ lies on $\\omega$.\n\nIf $B \\notin \\ell_{a}$ and $B \\notin \\ell_{c}$, then $T^{\\prime}$ and $T^{\\prime \\prime}$ are the second point of intersection of $\\ell_{b}$ and $\\omega$, hence they coincide. Otherwise, if, say, $B \\in \\ell_{c}$, then $\\ell_{c}=B C$, so $\\sphericalangle(B A, A C)=\\sphericalangle\\left(\\ell_{b}, \\ell_{c}\\right)=\\sphericalangle\\left(\\ell_{b}, B C\\right)$, which shows that $\\ell_{b}$ is tangent at $B$ to $\\omega$ and $T^{\\prime}=T^{\\prime \\prime}=B$. So $T^{\\prime}$ and $T^{\\prime \\prime}$ coincide in all the cases, and the conclusion of the claim follows.\n\n\n\nNow we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\\ell_{b}, D_{c}$ lies on $\\ell_{c}$, and\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(D_{b} X, X D_{c}\\right) & =\\sphericalangle\\left(D_{b} X, D X\\right)+\\sphericalangle\\left(D X, X D_{c}\\right)=2 \\sphericalangle(y, D X)+2 \\sphericalangle(D X, z)=2 \\sphericalangle(y, z) \\\\\n& =\\sphericalangle(B A, A C)=\\sphericalangle(B T, T C),\n\\end{aligned}\n$$\n\nhence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\\prime}$ yields $\\sphericalangle\\left(D_{b} D_{c}, D_{c} X\\right)=90^{\\circ}+\\sphericalangle\\left(D_{b} D_{c}^{\\prime}, D_{c}^{\\prime} X\\right)=90^{\\circ}+\\sphericalangle\\left(D_{b} D, D D_{c}\\right)$. Therefore,\n\n$$\n\\begin{gathered}\n\\sphericalangle\\left(\\ell_{b}, X T\\right)=\\sphericalangle\\left(D_{b} T, X T\\right)=\\sphericalangle\\left(D_{b} D_{c}, D_{c} X\\right)=90^{\\circ}+\\sphericalangle\\left(D_{b} D, D D_{c}\\right) \\\\\n=90^{\\circ}+\\sphericalangle(B I, I C)=\\sphericalangle(B A, A I)=\\sphericalangle\\left(B A, A X_{0}\\right)=\\sphericalangle\\left(B T, T X_{0}\\right)=\\sphericalangle\\left(\\ell_{b}, X_{0} T\\right),\n\\end{gathered}\n$$\n\nso the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows.', 'As mentioned in the preamble, it is sufficient to prove that the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\\omega$. Thus, it suffices to prove that $\\sphericalangle\\left(T X_{0}, T Y_{0}\\right)=$ $\\sphericalangle\\left(Z_{0} X_{0}, Z_{0} Y_{0}\\right)$, or, equivalently, $\\sphericalangle\\left(X X_{0}, Y Y_{0}\\right)=\\sphericalangle\\left(Z_{0} X_{0}, Z_{0} Y_{0}\\right)$.\n\nRecall that $Y Z$ and $Y_{0} Z_{0}$ are the perpendicular bisectors of $A D$ and $A I$, respectively. Then, the vector $\\vec{x}$ perpendicular to $Y Z$ and shifting the line $Y_{0} Z_{0}$ to $Y Z$ is equal to $\\frac{1}{2} \\overrightarrow{I D}$. Define the shifting vectors $\\vec{y}=\\frac{1}{2} \\overrightarrow{I E}, \\vec{z}=\\frac{1}{2} \\overrightarrow{I F}$ similarly. Consider now the triangle $U V W$ formed by the perpendiculars to $A I, B I$, and $C I$ through $D, E$, and $F$, respectively (see figure below). This is another triangle whose sides are parallel to the corresponding sides of $X Y Z$.\n\nClaim 2. $\\overrightarrow{I U}=2 \\overrightarrow{X_{0} X}, \\overrightarrow{I V}=2 \\overrightarrow{Y_{0} Y}, \\overrightarrow{I W}=2 \\overrightarrow{Z_{0} Z}$.\n\nProof. We prove one of the relations, the other proofs being similar. To prove the equality of two vectors it suffices to project them onto two non-parallel axes and check that their projections are equal.\n\nThe projection of $\\overrightarrow{X_{0} X}$ onto $I B$ equals $\\vec{y}$, while the projection of $\\overrightarrow{I U}$ onto $I B$ is $\\overrightarrow{I E}=2 \\vec{y}$. The projections onto the other axis $I C$ are $\\vec{z}$ and $\\overrightarrow{I F}=2 \\vec{z}$. Then $\\overrightarrow{I U}=2 \\overrightarrow{X_{0} X}$ follows.\n\nNotice that the line $\\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$; thus $U, V, W$, and $I$ are concyclic. It follows from Claim 2 that $\\sphericalangle\\left(X X_{0}, Y Y_{0}\\right)=\\sphericalangle(I U, I V)=$ $\\sphericalangle(W U, W V)=\\sphericalangle\\left(Z_{0} X_{0}, Z_{0} Y_{0}\\right)$, and we are done.\n\n', 'Let $I_{a}, I_{b}$, and $I_{c}$ be the excentres of triangle $A B C$ corresponding to $A, B$, and $C$, respectively. Also, let $u, v$, and $w$ be the lines through $D, E$, and $F$ which are perpendicular to $A I, B I$, and $C I$, respectively, and let $U V W$ be the triangle determined by these lines, where $u=V W, v=U W$ and $w=U V$ (see figure above).\n\nNotice that the line $u$ is the reflection of $I_{b} I_{c}$ in the line $x$, because $u, x$, and $I_{b} I_{c}$ are perpendicular to $A D$ and $x$ is the perpendicular bisector of $A D$. Likewise, $v$ and $I_{a} I_{c}$ are reflections of each other in $y$, while $w$ and $I_{a} I_{b}$ are reflections of each other in $z$. It follows that $X, Y$, and $Z$ are the midpoints of $U I_{a}, V I_{b}$ and $W I_{c}$, respectively, and that the triangles $U V W$, $X Y Z$ and $I_{a} I_{b} I_{c}$ are either translates of each other or homothetic with a common homothety centre.\n\nConstruct the points $T$ and $S$ such that the quadrilaterals $U V I W, X Y T Z$ and $I_{a} I_{b} S I_{c}$ are homothetic. Then $T$ is the midpoint of $I S$. Moreover, note that $\\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$, hence $I$ belongs to the circumcircle of the triangle $U V W$, therefore $T$ belongs to $\\Omega$.\n\n\n\nConsider now the homothety or translation $h_{1}$ that maps $X Y Z T$ to $I_{a} I_{b} I_{c} S$ and the homothety $h_{2}$ with centre $I$ and factor $\\frac{1}{2}$. Furthermore, let $h=h_{2} \\circ h_{1}$. The transform $h$ can be a homothety or a translation, and\n\n$$\nh(T)=h_{2}\\left(h_{1}(T)\\right)=h_{2}(S)=T\n$$\n\nhence $T$ is a fixed point of $h$. So, $h$ is a homothety with centre $T$. Note that $h_{2}$ maps the excentres $I_{a}, I_{b}, I_{c}$ to $X_{0}, Y_{0}, Z_{0}$ defined in the preamble. Thus the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\\Omega$, and this completes the proof.']",,True,,, 2048,Geometry,,A convex quadrilateral $A B C D$ satisfies $A B \cdot C D=B C \cdot D A$. A point $X$ is chosen inside the quadrilateral so that $\angle X A B=\angle X C D$ and $\angle X B C=\angle X D A$. Prove that $\angle A X B+$ $\angle C X D=180^{\circ}$.,"['Let $B^{\\prime}$ be the reflection of $B$ in the internal angle bisector of $\\angle A X C$, so that $\\angle A X B^{\\prime}=\\angle C X B$ and $\\angle C X B^{\\prime}=\\angle A X B$. If $X, D$, and $B^{\\prime}$ are collinear, then we are done. Now assume the contrary.\n\nOn the ray $X B^{\\prime}$ take a point $E$ such that $X E \\cdot X B=X A \\cdot X C$, so that $\\triangle A X E \\sim$ $\\triangle B X C$ and $\\triangle C X E \\sim \\triangle B X A$. We have $\\angle X C E+\\angle X C D=\\angle X B A+\\angle X A B<180^{\\circ}$ and $\\angle X A E+\\angle X A D=\\angle X D A+\\angle X A D<180^{\\circ}$, which proves that $X$ lies inside the angles $\\angle E C D$ and $\\angle E A D$ of the quadrilateral $E A D C$. Moreover, $X$ lies in the interior of exactly one of the two triangles $E A D, E C D$ (and in the exterior of the other).\n\n\n\nThe similarities mentioned above imply $X A \\cdot B C=X B \\cdot A E$ and $X B \\cdot C E=X C \\cdot A B$. Multiplying these equalities with the given equality $A B \\cdot C D=B C \\cdot D A$, we obtain $X A \\cdot C D$. $C E=X C \\cdot A D \\cdot A E$, or, equivalently,\n\n$$\n\\frac{X A \\cdot D E}{A D \\cdot A E}=\\frac{X C \\cdot D E}{C D \\cdot C E}\n\\tag{*}\n$$\n\nLemma. Let $P Q R$ be a triangle, and let $X$ be a point in the interior of the angle $Q P R$ such that $\\angle Q P X=\\angle P R X$. Then $\\frac{P X \\cdot Q R}{P Q \\cdot P R}<1$ if and only if $X$ lies in the interior of the triangle $P Q R$. Proof. The locus of points $X$ with $\\angle Q P X=\\angle P R X$ lying inside the angle $Q P R$ is an arc $\\alpha$ of the circle $\\gamma$ through $R$ tangent to $P Q$ at $P$. Let $\\gamma$ intersect the line $Q R$ again at $Y$ (if $\\gamma$ is tangent to $Q R$, then set $Y=R$ ). The similarity $\\triangle Q P Y \\sim \\triangle Q R P$ yields $P Y=\\frac{P Q \\cdot P R}{Q R}$. Now it suffices to show that $P X

P Y$ for $X \\in \\overparen{Y R}$.\n\n\n\nCase 2: $Y$ lies on the ray $Q R$ beyond $R$ (see the right figure below).\n\nIn this case the whole arc $\\alpha$ lies inside triangle $P Q R$, and between $m$ and $P Q$, thus $P X<$ $P Y$ for all $X \\in \\alpha$.\n\n\nApplying the Lemma (to $\\triangle E A D$ with the point $X$, and to $\\triangle E C D$ with the point $X$ ), we obtain that exactly one of two expressions $\\frac{X A \\cdot D E}{A D \\cdot A E}$ and $\\frac{X C \\cdot D E}{C D \\cdot C E}$ is less than 1, which contradicts $(*)$.', 'The solution consists of two parts. In Part 1 we show that it suffices to prove that\n\n$$\n\\frac{X B}{X D}=\\frac{A B}{C D}\n$$\n\nand\n\n$$\n\\frac{X A}{X C}=\\frac{D A}{B C}\n$$\n\nIn Part 2 we establish these equalities.\n\nPart 1. Using the sine law and applying (1) we obtain\n\n$$\n\\frac{\\sin \\angle A X B}{\\sin \\angle X A B}=\\frac{A B}{X B}=\\frac{C D}{X D}=\\frac{\\sin \\angle C X D}{\\sin \\angle X C D}\n$$\n\nso $\\sin \\angle A X B=\\sin \\angle C X D$ by the problem conditions. Similarly, (2) yields $\\sin \\angle D X A=$ $\\sin \\angle B X C$. If at least one of the pairs $(\\angle A X B, \\angle C X D)$ and $(\\angle B X C, \\angle D X A)$ consists of supplementary angles, then we are done. Otherwise, $\\angle A X B=\\angle C X D$ and $\\angle D X A=\\angle B X C$. In this case $X=A C \\cap B D$, and the problem conditions yield that $A B C D$ is a parallelogram and hence a rhombus. In this last case the claim also holds.\n\nPart 2. To prove the desired equality (1), invert $A B C D$ at centre $X$ with unit radius; the images of points are denoted by primes.\n\nWe have\n\n$$\n\\angle A^{\\prime} B^{\\prime} C^{\\prime}=\\angle X B^{\\prime} A^{\\prime}+\\angle X B^{\\prime} C^{\\prime}=\\angle X A B+\\angle X C B=\\angle X C D+\\angle X C B=\\angle B C D .\n$$\n\n\n\nSimilarly, the corresponding angles of quadrilaterals $A B C D$ and $D^{\\prime} A^{\\prime} B^{\\prime} C^{\\prime}$ are equal.\n\nMoreover, we have\n\n$$\nA^{\\prime} B^{\\prime} \\cdot C^{\\prime} D^{\\prime}=\\frac{A B}{X A \\cdot X B} \\cdot \\frac{C D}{X C \\cdot X D}=\\frac{B C}{X B \\cdot X C} \\cdot \\frac{D A}{X D \\cdot D A}=B^{\\prime} C^{\\prime} \\cdot D^{\\prime} A^{\\prime}\n$$\n\n\n\nNow we need the following Lemma.\n\nLemma. Assume that the corresponding angles of convex quadrilaterals $X Y Z T$ and $X^{\\prime} Y^{\\prime} Z^{\\prime} T^{\\prime}$ are equal, and that $X Y \\cdot Z T=Y Z \\cdot T X$ and $X^{\\prime} Y^{\\prime} \\cdot Z^{\\prime} T^{\\prime}=Y^{\\prime} Z^{\\prime} \\cdot T^{\\prime} X^{\\prime}$. Then the two quadrilaterals are similar.\n\nProof. Take the quadrilateral $X Y Z_{1} T_{1}$ similar to $X^{\\prime} Y^{\\prime} Z^{\\prime} T^{\\prime}$ and sharing the side $X Y$ with $X Y Z T$, such that $Z_{1}$ and $T_{1}$ lie on the rays $Y Z$ and $X T$, respectively, and $Z_{1} T_{1} \\| Z T$. We need to prove that $Z_{1}=Z$ and $T_{1}=T$. Assume the contrary. Without loss of generality, $T X>X T_{1}$. Let segments $X Z$ and $Z_{1} T_{1}$ intersect at $U$. We have\n\n$$\n\\frac{T_{1} X}{T_{1} Z_{1}}<\\frac{T_{1} X}{T_{1} U}=\\frac{T X}{Z T}=\\frac{X Y}{Y Z}<\\frac{X Y}{Y Z_{1}}\n$$\n\nthus $T_{1} X \\cdot Y Z_{1}\n\nIt follows from the Lemma that the quadrilaterals $A B C D$ and $D^{\\prime} A^{\\prime} B^{\\prime} C^{\\prime}$ are similar, hence\n\n$$\n\\frac{B C}{A B}=\\frac{A^{\\prime} B^{\\prime}}{D^{\\prime} A^{\\prime}}=\\frac{A B}{X A \\cdot X B} \\cdot \\frac{X D \\cdot X A}{D A}=\\frac{A B}{A D} \\cdot \\frac{X D}{X B}\n$$\n\nand therefore\n\n$$\n\\frac{X B}{X D}=\\frac{A B^{2}}{B C \\cdot A D}=\\frac{A B^{2}}{A B \\cdot C D}=\\frac{A B}{C D}\n$$\n\nWe obtain (1), as desired; (2) is proved similarly.']",,True,,, 2049,Geometry,,"Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\ell_{A}, \ell_{B}$, and $\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$ is tangent to the line $O P$.","['As usual, we denote the directed angle between the lines $a$ and $b$ by $\\sphericalangle(a, b)$. We frequently use the fact that $a_{1} \\perp a_{2}$ and $b_{1} \\perp b_{2}$ yield $\\sphericalangle\\left(a_{1}, b_{1}\\right)=\\sphericalangle\\left(a_{2}, b_{2}\\right)$.\n\nLet the lines $\\ell_{B}$ and $\\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.\n\n\n\nClaim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic.\n\nProof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have\n\n$$\n\\sphericalangle\\left(O_{A} P, O_{C} P\\right)=\\sphericalangle\\left(O_{C} O, O_{A} O\\right)=\\sphericalangle(C P, A P)=\\sphericalangle(C B, A B)=\\sphericalangle\\left(O_{A} L_{B}, O_{C} L_{B}\\right) .\n$$\n\nDenote the circle through $L_{B}, P, O_{A}$, and $O_{C}$ by $\\omega_{B}$. Define the circles $\\omega_{A}$ and $\\omega_{C}$ similarly. Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$.\n\nProof. From cyclic quadruples of points in the circles $\\omega_{B}$ and $\\omega_{C}$, we have\n\n$$\n\\begin{aligned}\n\\sphericalangle\\left(L_{C} L_{A}, L_{C} P\\right) & =\\sphericalangle\\left(L_{C} O_{B}, L_{C} P\\right)=\\sphericalangle\\left(O_{A} O_{B}, O_{A} P\\right) \\\\\n& =\\sphericalangle\\left(O_{A} O_{C}, O_{A} P\\right)=\\sphericalangle\\left(L_{B} O_{C}, L_{B} P\\right)=\\sphericalangle\\left(L_{B} L_{A}, L_{B} P\\right) .\n\\end{aligned}\n$$\n\nClaim 3. The points $P, L_{C}$, and $C$ are collinear.\n\nProof. We have $\\sphericalangle\\left(P L_{C}, L_{C} L_{A}\\right)=\\sphericalangle\\left(P L_{C}, L_{C} O_{B}\\right)=\\sphericalangle\\left(P O_{A}, O_{A} O_{B}\\right)$. Further, since $O_{A}$ is the centre of the circle $A O P, \\sphericalangle\\left(P O_{A}, O_{A} O_{B}\\right)=\\sphericalangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \\sphericalangle(P A, A O)=\\pi / 2-\\sphericalangle(C A, C P)=\\sphericalangle\\left(C P, L_{C} L_{A}\\right)$. We obtain $\\sphericalangle\\left(P L_{C}, L_{C} L_{A}\\right)=$ $\\sphericalangle\\left(C P, L_{C} L_{A}\\right)$, which shows that $P \\in C L_{C}$.\n\n\n\nSimilarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that\n\n$$\n\\sphericalangle\\left(O P, P L_{A}\\right)=\\sphericalangle(P A, A O)=\\sphericalangle\\left(P L_{C}, L_{C} L_{A}\\right),\n$$\n\nwhich means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$.']",,True,,, 2050,Number Theory,,"Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the numbers of divisors of $s n$ and of $s k$ are equal.","['As usual, the number of divisors of a positive integer $n$ is denoted by $d(n)$. If $n=\\prod_{i} p_{i}^{\\alpha_{i}}$ is the prime factorisation of $n$, then $d(n)=\\prod_{i}\\left(\\alpha_{i}+1\\right)$.\n\nWe start by showing that one cannot find any suitable number $s$ if $k \\mid n$ or $n \\mid k$ (and $k \\neq n$ ). Suppose that $n \\mid k$, and choose any positive integer $s$. Then the set of divisors of $s n$ is a proper subset of that of $s k$, hence $d(s n)\\beta$ be nonnegative integers. Then, for every integer $M \\geqslant \\beta+1$, there exists a nonnegative integer $\\gamma$ such that\n\n$$\n\\frac{\\alpha+\\gamma+1}{\\beta+\\gamma+1}=1+\\frac{1}{M}=\\frac{M+1}{M}\n$$\n\nProof.\n\n$$\n\\frac{\\alpha+\\gamma+1}{\\beta+\\gamma+1}=1+\\frac{1}{M} \\Longleftrightarrow \\frac{\\alpha-\\beta}{\\beta+\\gamma+1}=\\frac{1}{M} \\Longleftrightarrow \\gamma=M(\\alpha-\\beta)-(\\beta+1) \\geqslant 0 .\n$$\n\nNow we can finish the solution. Without loss of generality, there exists an index $u$ such that $\\alpha_{i}>\\beta_{i}$ for $i=1,2, \\ldots, u$, and $\\alpha_{i}<\\beta_{i}$ for $i=u+1, \\ldots, t$. The conditions $n \\nmid k$ and $k \\nmid n$ mean that $1 \\leqslant u \\leqslant t-1$.\n\nChoose an integer $X$ greater than all the $\\alpha_{i}$ and $\\beta_{i}$. By the lemma, we can define the numbers $\\gamma_{i}$ so as to satisfy\n\n$$\n\\begin{array}{lr}\n\\frac{\\alpha_{i}+\\gamma_{i}+1}{\\beta_{i}+\\gamma_{i}+1}=\\frac{u X+i}{u X+i-1} & \\text { for } i=1,2, \\ldots, u, \\text { and } \\\\\n\\frac{\\beta_{u+i}+\\gamma_{u+i}+1}{\\alpha_{u+i}+\\gamma_{u+i}+1}=\\frac{(t-u) X+i}{(t-u) X+i-1} & \\text { for } i=1,2, \\ldots, t-u .\n\\end{array}\n$$\n\nThen we will have\n\n$$\n\\frac{d(s n)}{d(s k)}=\\prod_{i=1}^{u} \\frac{u X+i}{u X+i-1} \\cdot \\prod_{i=1}^{t-u} \\frac{(t-u) X+i-1}{(t-u) X+i}=\\frac{u(X+1)}{u X} \\cdot \\frac{(t-u) X}{(t-u)(X+1)}=1,\n$$\n\nas required.']","['All pairs $(n, k)$ such that $n \\nmid k$ and $k \\nmid n$']",False,,Need_human_evaluate, 2051,Number Theory,,"Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied: (i) Each number in the table is congruent to 1 modulo $n$; (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$. Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$.","['Let $A_{i, j}$ be the entry in the $i^{\\text {th }}$ row and the $j^{\\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that\n\n$$\n\\sum_{i=1}^{n} R_{i} \\equiv(n-1)+P \\quad\\left(\\bmod n^{4}\\right)\n\\tag{1}\n$$\n\nDue to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion.\n\nBy condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence\n\n$R_{i}=\\prod_{j=1}^{n}\\left(1+a_{i, j}\\right)=1+\\sum_{j=1}^{n} a_{i, j}+\\sum_{1 \\leqslant j_{1}3$, so\n\n$$\nS_{n-1}=2^{n}+2^{\\lceil n / 2\\rceil}+2^{\\lfloor n / 2\\rfloor}-3>2^{n}+2^{\\lfloor n / 2\\rfloor}=a_{n} .\n$$\n\nAlso notice that $S_{n-1}-a_{n}=2^{[n / 2]}-3a_{n}$. Denote $c=S_{n-1}-b$; then $S_{n-1}-a_{n}2^{t}-3$.\n\nProof. The inequality follows from $t \\geqslant 3$. In order to prove the equivalence, we apply Claim 1 twice in the following manner.\n\nFirst, since $S_{2 t-3}-a_{2 t-2}=2^{t-1}-3<2^{t}-3<2^{2 t-2}+2^{t-1}=a_{2 t-2}$, by Claim 1 we have $2^{t}-3 \\sim S_{2 t-3}-\\left(2^{t}-3\\right)=2^{2 t-2}$.\n\nSecond, since $S_{4 t-7}-a_{4 t-6}=2^{2 t-3}-3<2^{2 t-2}<2^{4 t-6}+2^{2 t-3}=a_{4 t-6}$, by Claim 1 we have $2^{2 t-2} \\sim S_{4 t-7}-2^{2 t-2}=2^{4 t-6}-3$.\n\nTherefore, $2^{t}-3 \\sim 2^{2 t-2} \\sim 2^{4 t-6}-3$, as required.\n\nNow it is easy to find the required numbers. Indeed, the number $2^{3}-3=5=a_{0}+a_{1}$ is representable, so Claim 3 provides an infinite sequence of representable numbers\n\n$$\n2^{3}-3 \\sim 2^{6}-3 \\sim 2^{18}-3 \\sim \\cdots \\sim 2^{t}-3 \\sim 2^{4 t-6}-3 \\sim \\cdots\n$$\n\nOn the other hand, the number $2^{7}-3=125$ is non-representable (since by Claim 1 we have $125 \\sim S_{6}-125=24 \\sim S_{4}-24=17 \\sim S_{3}-17=4$ which is clearly non-representable). So Claim 3 provides an infinite sequence of non-representable numbers\n\n$$\n2^{7}-3 \\sim 2^{22}-3 \\sim 2^{82}-3 \\sim \\cdots \\sim 2^{t}-3 \\sim 2^{4 t-6}-3 \\sim \\cdots\n$$', 'We keep the notion of representability and the notation $S_{n}$ from the previous solution. We say that an index $n$ is good if $a_{n}$ writes as a sum of smaller terms from the sequence $a_{0}, a_{1}, \\ldots$ Otherwise we say it is bad. We must prove that there are infinitely many good indices, as well as infinitely many bad ones.\n\nLemma 1. If $m \\geqslant 0$ is an integer, then $4^{m}$ is representable if and only if either of $2 m+1$ and $2 m+2$ is good.\n\nProof. The case $m=0$ is obvious, so we may assume that $m \\geqslant 1$. Let $n=2 m+1$ or $2 m+2$. Then $n \\geqslant 3$. We notice that\n\n$$\nS_{n-1}4^{s}$.\n\nProof. We have $2^{4 k-2}s$.\n\nNow $4^{2}=a_{2}+a_{3}$ is representable, whereas $4^{6}=4096$ is not. Indeed, note that $4^{6}=2^{12}v_{p}\\left(a_{n}\\right)$, then $v_{p}\\left(a_{n} / a_{n+1}\\right)<0$, while $v_{p}\\left(\\left(a_{n+1}-a_{n}\\right) / a_{1}\\right) \\geqslant 0$, so $(*)$ is not integer again. Thus, $v_{p}\\left(a_{1}\\right) \\leqslant v_{p}\\left(a_{n+1}\\right) \\leqslant v_{p}\\left(a_{n}\\right)$.\n\nThe above arguments can now be applied successively to indices $n+1, n+2, \\ldots$, showing that all the indices greater than $n$ are large, and the sequence $v_{p}\\left(a_{n}\\right), v_{p}\\left(a_{n+1}\\right), v_{p}\\left(a_{n+2}\\right), \\ldots$ is nonincreasing - hence eventually constant.\n\nCase 2: There is no large index.\n\nWe have $v_{p}\\left(a_{1}\\right)>v_{p}\\left(a_{n}\\right)$ for all $n \\geqslant k$. If we had $v_{p}\\left(a_{n+1}\\right)1$ which divides all values of $f$.","['For every positive integer $m$, define $S_{m}=\\{n: m \\mid f(n)\\}$.\n\nLemma. If the set $S_{m}$ is infinite, then $S_{m}=\\{d, 2 d, 3 d, \\ldots\\}=d \\cdot \\mathbb{Z}_{>0}$ for some positive integer $d$. Proof. Let $d=\\min S_{m}$; the definition of $S_{m}$ yields $m \\mid f(d)$.\n\nWhenever $n \\in S_{m}$ and $n>d$, we have $m|f(n)| f(n-d)+f(d)$, so $m \\mid f(n-d)$ and therefore $n-d \\in S_{m}$. Let $r \\leqslant d$ be the least positive integer with $n \\equiv r(\\bmod d)$; repeating the same step, we can see that $n-d, n-2 d, \\ldots, r \\in S_{m}$. By the minimality of $d$, this shows $r=d$ and therefore $d \\mid n$.\n\nStarting from an arbitrarily large element of $S_{m}$, the process above reaches all multiples of $d$; so they all are elements of $S_{m}$.\n\nThe solution for the problem will be split into two cases.\n\nCase 1: The function $f$ is bounded.\n\nCall a prime $p$ frequent if the set $S_{p}$ is infinite, i.e., if $p$ divides $f(n)$ for infinitely many positive integers $n$; otherwise call $p$ sporadic. Since the function $f$ is bounded, there are only a finite number of primes that divide at least one $f(n)$; so altogether there are finitely many numbers $n$ such that $f(n)$ has a sporadic prime divisor. Let $N$ be a positive integer, greater than all those numbers $n$.\n\nLet $p_{1}, \\ldots, p_{k}$ be the frequent primes. By the lemma we have $S_{p_{i}}=d_{i} \\cdot \\mathbb{Z}_{>0}$ for some $d_{i}$. Consider the number\n\n$$\nn=N d_{1} d_{2} \\cdots d_{k}+1\n$$\n\nDue to $n>N$, all prime divisors of $f(n)$ are frequent primes. Let $p_{i}$ be any frequent prime divisor of $f(n)$. Then $n \\in S_{p_{i}}$, and therefore $d_{i} \\mid n$. But $n \\equiv 1\\left(\\bmod d_{i}\\right)$, which means $d_{i}=1$. Hence $S_{p_{i}}=1 \\cdot \\mathbb{Z}_{>0}=\\mathbb{Z}_{>0}$ and therefore $p_{i}$ is a common divisor of all values $f(n)$.\n\nCase 2: $f$ is unbounded.\n\nWe prove that $f(1)$ divides all $f(n)$.\n\nLet $a=f(1)$. Since $1 \\in S_{a}$, by the lemma it suffices to prove that $S_{a}$ is an infinite set.\n\nCall a positive integer $p$ a peak if $f(p)>\\max (f(1), \\ldots, f(p-1))$. Since $f$ is not bounded, there are infinitely many peaks. Let $1=p_{1}0}$ are coprime then $\\operatorname{gcd}(f(a), f(b)) \\mid f(1)$. In particular, if $a, b \\geqslant n_{0}$ are coprime then $f(a)$ and $f(b)$ are coprime.\n\nProof. Let $d=\\operatorname{gcd}(f(a), f(b))$. We can replicate Euclid's algorithm. Formally, apply induction on $a+b$. If $a=1$ or $b=1$ then we already have $d \\mid f(1)$.\n\nWithout loss of generality, suppose $1C$ (that is possible, because there are arbitrarily long gaps between the primes). Then we establish a contradiction\n\n$$\np_{N+1} \\leqslant \\max \\left(f(1), f\\left(q_{1}\\right), \\ldots, f\\left(q_{N}\\right)\\right)<\\max \\left(1+C, q_{1}+C, \\ldots, q_{N}+C\\right)=p_{N}+C0$ and $c, d$ are coprime.\n\nWe will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \\leqslant i \\leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\\left(b_{i}\\right)5 n\n$$\n\na contradiction.\n\nClaim 3. For every $0 \\leqslant k \\leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \\ldots, b_{k+30}$, at least $\\varphi(30)=8$ are divisible by $d$.\n\nProof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \\cdot(3-1) \\cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \\mid b_{i}$ for every uncovered index $i$.\n\nClaim 4. $|\\Delta|<\\frac{20}{n-2}$ and $d>\\frac{n-2}{20}$.\n\nProof. From the sequence (1), remove all fractions with $b_{n}<\\frac{n}{2}$, There remain at least $\\frac{n}{2}$ fractions, and they cannot exceed $\\frac{5 n}{n / 2}=10$. So we have at least $\\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\\frac{10}{n / 2-1}=\\frac{20}{n-2}$.\n\nThe second inequality follows from $\\frac{1}{d} \\leqslant \\frac{|c|}{d}=|\\Delta|$.\n\nNow we have everything to get the final contradiction. By Claim 3, we have $d \\mid b_{i}$ for at least $\\left\\lfloor\\frac{n}{30}\\right\\rfloor \\cdot 8$ indices $i$. By Claim 4, we have $d \\geqslant \\frac{n-2}{20}$. Therefore,\n\n$$\n5 n \\geqslant \\max \\left\\{b_{i}: d \\mid b_{i}\\right\\} \\geqslant\\left(\\left\\lfloor\\frac{n}{30}\\right\\rfloor \\cdot 8\\right) \\cdot d>\\left(\\frac{n}{30}-1\\right) \\cdot 8 \\cdot \\frac{n-2}{20}>5 n .\n$$']",,True,,, 2057,Algebra,,"Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=v_{1}=1$, and $$ u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text { for } k=1, \ldots, n-1 $$ Prove that $u_{n}=v_{n}$.","['We prove by induction on $k$ that\n\n$$\nu_{k}=\\sum_{\\substack{0i_{1}>\\ldots>i_{t}>n-k \\\\ i_{j}-i_{j+1} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}}\n\\tag{2}\n$$\n\nFor $k=n$ the expressions (1) and (2) coincide, so indeed $u_{n}=v_{n}$.', 'Define recursively a sequence of multivariate polynomials by\n\n$$\nP_{0}=P_{1}=1, \\quad P_{k+1}\\left(x_{1}, \\ldots, x_{k}\\right)=P_{k}\\left(x_{1}, \\ldots, x_{k-1}\\right)+x_{k} P_{k-1}\\left(x_{1}, \\ldots, x_{k-2}\\right),\n$$\n\nso $P_{n}$ is a polynomial in $n-1$ variables for each $n \\geqslant 1$. Two easy inductive arguments show that\n\n$$\nu_{n}=P_{n}\\left(a_{1}, \\ldots, a_{n-1}\\right), \\quad v_{n}=P_{n}\\left(a_{n-1}, \\ldots, a_{1}\\right)\n$$\n\n\n\nso we need to prove $P_{n}\\left(x_{1}, \\ldots, x_{n-1}\\right)=P_{n}\\left(x_{n-1}, \\ldots, x_{1}\\right)$ for every positive integer $n$. The cases $n=1,2$ are trivial, and the cases $n=3,4$ follow from $P_{3}(x, y)=1+x+y$ and $P_{4}(x, y, z)=$ $1+x+y+z+x z$.\n\nNow we proceed by induction, assuming that $n \\geqslant 5$ and the claim hold for all smaller cases. Using $F(a, b)$ as an abbreviation for $P_{|a-b|+1}\\left(x_{a}, \\ldots, x_{b}\\right)$ (where the indices $a, \\ldots, b$ can be either in increasing or decreasing order),\n\n$$\n\\begin{aligned}\nF(n, 1) & =F(n, 2)+x_{1} F(n, 3)=F(2, n)+x_{1} F(3, n) \\\\\n& =\\left(F(2, n-1)+x_{n} F(2, n-2)\\right)+x_{1}\\left(F(3, n-1)+x_{n} F(3, n-2)\\right) \\\\\n& =\\left(F(n-1,2)+x_{1} F(n-1,3)\\right)+x_{n}\\left(F(n-2,2)+x_{1} F(n-2,3)\\right) \\\\\n& =F(n-1,1)+x_{n} F(n-2,1)=F(1, n-1)+x_{n} F(1, n-2) \\\\\n& =F(1, n),\n\\end{aligned}\n$$\n\nas we wished to show.', 'Using matrix notation, we can rewrite the recurrence relation as\n\n$$\n\\left(\\begin{array}{c}\nu_{k+1} \\\\\nu_{k+1}-u_{k}\n\\end{array}\\right)=\\left(\\begin{array}{c}\nu_{k}+a_{k} u_{k-1} \\\\\na_{k} u_{k-1}\n\\end{array}\\right)=\\left(\\begin{array}{cc}\n1+a_{k} & -a_{k} \\\\\na_{k} & -a_{k}\n\\end{array}\\right)\\left(\\begin{array}{c}\nu_{k} \\\\\nu_{k}-u_{k-1}\n\\end{array}\\right)\n$$\n\nfor $1 \\leqslant k \\leqslant n-1$, and similarly\n\n$$\n\\left(v_{k+1} ; v_{k}-v_{k+1}\\right)=\\left(v_{k}+a_{n-k} v_{k-1} ;-a_{n-k} v_{k-1}\\right)=\\left(v_{k} ; v_{k-1}-v_{k}\\right)\\left(\\begin{array}{cc}\n1+a_{n-k} & -a_{n-k} \\\\\na_{n-k} & -a_{n-k}\n\\end{array}\\right)\n$$\n\nfor $1 \\leqslant k \\leqslant n-1$. Hence, introducing the $2 \\times 2$ matrices $A_{k}=\\left(\\begin{array}{cc}1+a_{k} & -a_{k} \\\\ a_{k} & -a_{k}\\end{array}\\right)$ we have\n\n$$\n\\left(\\begin{array}{c}\nu_{k+1} \\\\\nu_{k+1}-u_{k}\n\\end{array}\\right)=A_{k}\\left(\\begin{array}{c}\nu_{k} \\\\\nu_{k}-u_{k-1}\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(v_{k+1} ; v_{k}-v_{k+1}\\right)=\\left(v_{k} ; v_{k-1}-v_{k}\\right) A_{n-k} .\n$$\n\nfor $1 \\leqslant k \\leqslant n-1$. Since $\\left(\\begin{array}{c}u_{1} \\\\ u_{1}-u_{0}\\end{array}\\right)=\\left(\\begin{array}{l}1 \\\\ 0\\end{array}\\right)$ and $\\left(v_{1} ; v_{0}-v_{1}\\right)=(1 ; 0)$, we get\n\n$$\n\\left(\\begin{array}{c}\nu_{n} \\\\\nu_{n}-u_{n-1}\n\\end{array}\\right)=A_{n-1} A_{n-2} \\cdots A_{1} \\cdot\\left(\\begin{array}{l}\n1 \\\\\n0\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(v_{n} ; v_{n-1}-v_{n}\\right)=(1 ; 0) \\cdot A_{n-1} A_{n-2} \\cdots A_{1} \\text {. }\n$$\n\nIt follows that\n\n$$\n\\left(u_{n}\\right)=(1 ; 0)\\left(\\begin{array}{c}\nu_{n} \\\\\nu_{n}-u_{n-1}\n\\end{array}\\right)=(1 ; 0) \\cdot A_{n-1} A_{n-2} \\cdots A_{1} \\cdot\\left(\\begin{array}{l}\n1 \\\\\n0\n\\end{array}\\right)=\\left(v_{n} ; v_{n-1}-v_{n}\\right)\\left(\\begin{array}{l}\n1 \\\\\n0\n\\end{array}\\right)=\\left(v_{n}\\right) .\n$$']",,True,,, 2058,Algebra,,"Prove that in any set of 2000 distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d$, such that $$ \left|\frac{a-b}{c-d}-1\right|<\frac{1}{100000} $$","['For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \\leqslant D_{2} \\leqslant \\cdots \\leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m=n(n-1) / 2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{1}=1$. Let $D_{1}=1=y-x$ for $x, y \\in S$. Evidently $D_{m}=v-u$ is the difference between the largest element $v$ and the smallest element $u$ of $S$.\n\nIf $D_{i+1} / D_{i}<1+10^{-5}$ for some $i=1,2, \\ldots, m-1$ then the required inequality holds, because $0 \\leqslant D_{i+1} / D_{i}-1<10^{-5}$. Otherwise, the reverse inequality\n\n$$\n\\frac{D_{i+1}}{D_{i}} \\geqslant 1+\\frac{1}{10^{5}}\n$$\n\nholds for each $i=1,2, \\ldots, m-1$, and therefore\n\n$$\nv-u=D_{m}=\\frac{D_{m}}{D_{1}}=\\frac{D_{m}}{D_{m-1}} \\cdots \\frac{D_{3}}{D_{2}} \\cdot \\frac{D_{2}}{D_{1}} \\geqslant\\left(1+\\frac{1}{10^{5}}\\right)^{m-1} .\n$$\n\nFrom $m-1=n(n-1) / 2-1=1000 \\cdot 1999-1>19 \\cdot 10^{5}$, together with the fact that for all $n \\geqslant 1$, $\\left(1+\\frac{1}{n}\\right)^{n} \\geqslant 1+\\left(\\begin{array}{l}n \\\\ 1\\end{array}\\right) \\cdot \\frac{1}{n}=2$, we get\n\n$$\n\\left(1+\\frac{1}{10^{5}}\\right)^{19 \\cdot 10^{5}}=\\left(\\left(1+\\frac{1}{10^{5}}\\right)^{10^{5}}\\right)^{19} \\geqslant 2^{19}=2^{9} \\cdot 2^{10}>500 \\cdot 1000>2 \\cdot 10^{5},\n$$\n\nand so $v-u=D_{m}>2 \\cdot 10^{5}$.\n\nSince the distance of $x$ to at least one of the numbers $u, v$ is at least $(u-v) / 2>10^{5}$, we have\n\n$$\n|x-z|>10^{5} .\n$$\n\nfor some $z \\in\\{u, v\\}$. Since $y-x=1$, we have either $z>y>x$ (if $z=v$ ) or $y>x>z$ (if $z=u$ ). If $z>y>x$, selecting $a=z, b=y, c=z$ and $d=x$ (so that $b \\neq d$ ), we obtain\n\n$$\n\\left|\\frac{a-b}{c-d}-1\\right|=\\left|\\frac{z-y}{z-x}-1\\right|=\\left|\\frac{x-y}{z-x}\\right|=\\frac{1}{z-x}<10^{-5}\n$$\n\nOtherwise, if $y>x>z$, we may choose $a=y, b=z, c=x$ and $d=z$ (so that $a \\neq c$ ), and obtain\n\n$$\n\\left|\\frac{a-b}{c-d}-1\\right|=\\left|\\frac{y-z}{x-z}-1\\right|=\\left|\\frac{y-x}{x-z}\\right|=\\frac{1}{x-z}<10^{-5}\n$$\n\nThe desired result follows.']",,True,,, 2059,Algebra,,"Let $\mathbb{Q}_{>0}$ be the set of positive rational numbers. Let $f: \mathbb{Q}_{>0} \rightarrow \mathbb{R}$ be a function satisfying the conditions $$ f(x) f(y) \geqslant f(x y) \tag{1} $$ $$ f(x+y) \geqslant f(x)+f(y) \tag{2} $$ for all $x, y \in \mathbb{Q}_{>0}$. Given that $f(a)=a$ for some rational $a>1$, prove that $f(x)=x$ for all $x \in \mathbb{Q}_{>0}$.","['Denote by $\\mathbb{Z}_{>0}$ the set of positive integers.\n\nPlugging $x=1, y=a$ into (1) we get $f(1) \\geqslant 1$. Next, by an easy induction on $n$ we get from (2) that\n\n$$\nf(n x) \\geqslant n f(x) \\text { for all } n \\in \\mathbb{Z}_{>0} \\text { and } x \\in \\mathbb{Q}_{>0} \\text {. }\n\\tag{3}\n$$\n\nIn particular, we have\n\n$$\nf(n) \\geqslant n f(1) \\geqslant n \\text { for all } n \\in \\mathbb{Z}_{>0}\n\\tag{4}\n$$\n\nFrom (1) again we have $f(m / n) f(n) \\geqslant f(m)$, so $f(q)>0$ for all $q \\in \\mathbb{Q}_{>0}$.\n\nNow, (2) implies that $f$ is strictly increasing; this fact together with (4) yields\n\n$$\nf(x) \\geqslant f(\\lfloor x\\rfloor) \\geqslant\\lfloor x\\rfloor>x-1 \\quad \\text { for all } x \\geqslant 1\n$$\n\nBy an easy induction we get from (1) that $f(x)^{n} \\geqslant f\\left(x^{n}\\right)$, so\n\n$$\nf(x)^{n} \\geqslant f\\left(x^{n}\\right)>x^{n}-1 \\quad \\Longrightarrow \\quad f(x) \\geqslant \\sqrt[n]{x^{n}-1} \\quad \\text { for all } x>1 \\text { and } n \\in \\mathbb{Z}_{>0}\n$$\n\nThis yields\n\n$$\nf(x) \\geqslant x \\text { for every } x>1 \\text {. }\\tag{5}\n$$\n\n(Indeed, if $x>y>1$ then $x^{n}-y^{n}=(x-y)\\left(x^{n-1}+x^{n-2} y+\\cdots+y^{n}\\right)>n(x-y)$, so for a large $n$ we have $x^{n}-1>y^{n}$ and thus $f(x)>y$.)\n\nNow, (1) and (5) give $a^{n}=f(a)^{n} \\geqslant f\\left(a^{n}\\right) \\geqslant a^{n}$, so $f\\left(a^{n}\\right)=a^{n}$. Now, for $x>1$ let us choose $n \\in \\mathbb{Z}_{>0}$ such that $a^{n}-x>1$. Then by (2) and (5) we get\n\n$$\na^{n}=f\\left(a^{n}\\right) \\geqslant f(x)+f\\left(a^{n}-x\\right) \\geqslant x+\\left(a^{n}-x\\right)=a^{n}\n$$\n\nand therefore $f(x)=x$ for $x>1$. Finally, for every $x \\in \\mathbb{Q}_{>0}$ and every $n \\in \\mathbb{Z}_{>0}$, from (1) and (3) we get\n\n$$\nn f(x)=f(n) f(x) \\geqslant f(n x) \\geqslant n f(x)\n$$\n\nwhich gives $f(n x)=n f(x)$. Therefore $f(m / n)=f(m) / n=m / n$ for all $m, n \\in \\mathbb{Z}_{>0}$.']",,True,,, 2060,Algebra,,"Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \ldots$ by defining $a_{n+i}=a_{i}$ for all $i \geqslant 1$. If $$ a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n \tag{1} $$ and $$ a_{a_{i}} \leqslant n+i-1 \quad \text { for } i=1,2, \ldots, n \tag{2} $$ prove that $$ a_{1}+\cdots+a_{n} \leqslant n^{2} $$","['First, we claim that\n\n$$\na_{i} \\leqslant n+i-1 \\quad \\text { for } i=1,2, \\ldots, n\n\\tag{3}\n$$\n\nAssume contrariwise that $i$ is the smallest counterexample. From $a_{n} \\geqslant a_{n-1} \\geqslant \\cdots \\geqslant a_{i} \\geqslant n+i$ and $a_{a_{i}} \\leqslant n+i-1$, taking into account the periodicity of our sequence, it follows that\n\n$$\na_{i} \\text { cannot be congruent to } i, i+1, \\ldots, n-1 \\text {, or } n(\\bmod n) \\text {. }\n\\tag{4}\n$$\n\nThus our assumption that $a_{i} \\geqslant n+i$ implies the stronger statement that $a_{i} \\geqslant 2 n+1$, which by $a_{1}+n \\geqslant a_{n} \\geqslant a_{i}$ gives $a_{1} \\geqslant n+1$. The minimality of $i$ then yields $i=1$, and (4) becomes contradictory. This establishes our first claim.\n\nIn particular we now know that $a_{1} \\leqslant n$. If $a_{n} \\leqslant n$, then $a_{1} \\leqslant \\cdots \\leqslant \\cdots a_{n} \\leqslant n$ and the desired inequality holds trivially. Otherwise, consider the number $t$ with $1 \\leqslant t \\leqslant n-1$ such that\n\n$$\na_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{t} \\leqslant na_{a_{i}}$ ) belongs to $\\left\\{a_{i}+1, \\ldots, n\\right\\}$, and for this reason $b_{i} \\leqslant n-a_{i}$.\n\nIt follows from the definition of the $b_{i}$ and (5) that\n\n$$\na_{t+1}+\\ldots+a_{n} \\leqslant n(n-t)+b_{1}+\\ldots+b_{t} .\n$$\n\nAdding $a_{1}+\\ldots+a_{t}$ to both sides and using that $a_{i}+b_{i} \\leqslant n$ for $1 \\leqslant i \\leqslant t$, we get\n\n$$\na_{1}+a_{2}+\\cdots+a_{n} \\leqslant n(n-t)+n t=n^{2}\n$$\n\nas we wished to prove.', 'In the first quadrant of an infinite grid, consider the increasing ""staircase"" obtained by shading in dark the bottom $a_{i}$ cells of the $i$ th column for $1 \\leqslant i \\leqslant n$. We will prove that there are at most $n^{2}$ dark cells.\n\nTo do it, consider the $n \\times n$ square $S$ in the first quadrant with a vertex at the origin. Also consider the $n \\times n$ square directly to the left of $S$. Starting from its lower left corner, shade in light the leftmost $a_{j}$ cells of the $j$ th row for $1 \\leqslant j \\leqslant n$. Equivalently, the light shading is obtained by reflecting the dark shading across the line $x=y$ and translating it $n$ units to the left. The figure below illustrates this construction for the sequence $6,6,6,7,7,7,8,12,12,14$.\n\n\n\nWe claim that there is no cell in $S$ which is both dark and light. Assume, contrariwise, that there is such a cell in column $i$. Consider the highest dark cell in column $i$ which is inside $S$. Since it is above a light cell and inside $S$, it must be light as well. There are two cases:\n\nCase 1. $a_{i} \\leqslant n$\n\nIf $a_{i} \\leqslant n$ then this dark and light cell is $\\left(i, a_{i}\\right)$, as highlighted in the figure. However, this is the $(n+i)$-th cell in row $a_{i}$, and we only shaded $a_{a_{i}}0$ such that $f(n)=b-1$; but then $f^{3}(n-1)=f(n)+1=b$, so $b \\in R_{3}$.\n\nThis yields\n\n$$\n3 k=\\left|S_{1} \\cup S_{2} \\cup S_{3}\\right| \\leqslant 1+1+\\left|S_{1}\\right|=k+2\n$$\n\nor $k \\leqslant 1$. Therefore $k=1$, and the inequality above comes to equality. So we have $S_{1}=\\{a\\}$, $S_{2}=\\{f(a)\\}$, and $S_{3}=\\left\\{f^{2}(a)\\right\\}$ for some $a \\in \\mathbb{Z}_{\\geqslant 0}$, and each one of the three options $(i)$, (ii), and (iii) should be realized exactly once, which means that\n\n$$\n\\left\\{a, f(a), f^{2}(a)\\right\\}=\\{0, a+1, f(0)+1\\}\n\\tag{3}\n$$\n\n\n\nIII. From (3), we get $a+1 \\in\\left\\{f(a), f^{2}(a)\\right\\}$ (the case $a+1=a$ is impossible). If $a+1=f^{2}(a)$ then we have $f(a+1)=f^{3}(a)=f(a+1)+1$ which is absurd. Therefore\n\n$$\nf(a)=a+1\n\\tag{4}\n$$\n\nNext, again from (3) we have $0 \\in\\left\\{a, f^{2}(a)\\right\\}$. Let us consider these two cases separately.\n\nCase 1. Assume that $a=0$, then $f(0)=f(a)=a+1=1$. Also from (3) we get $f(1)=f^{2}(a)=$ $f(0)+1=2$. Now, let us show that $f(n)=n+1$ by induction on $n$; the base cases $n \\leqslant 1$ are established. Next, if $n \\geqslant 2$ then the induction hypothesis implies\n\n$$\nn+1=f(n-1)+1=f^{3}(n-2)=f^{2}(n-1)=f(n),\n$$\n\nestablishing the step. In this case we have obtained the first of two answers; checking that is satisfies (*) is straightforward.\n\nCase 2. Assume now that $f^{2}(a)=0$; then by (3) we get $a=f(0)+1$. By (4) we get $f(a+1)=$ $f^{2}(a)=0$, then $f(0)=f^{3}(a)=f(a+1)+1=1$, hence $a=f(0)+1=2$ and $f(2)=3$ by (4). To summarize,\n\n$$\nf(0)=1, \\quad f(2)=3, \\quad f(3)=0\n$$\n\nNow let us prove by induction on $m$ that (1) holds for all $n=4 k, 4 k+2,4 k+3$ with $k \\leqslant m$ and for all $n=4 k+1$ with $k0$ and $B(1)=m+2>0$ since $n=2 m$.\n\nTherefore $B(x)=A(x+a+b)$. Writing $c=a+b \\geqslant 1$ we compute\n\n$$\n0=A(x+c)-B(x)=(3 c-2 m) x^{2}+c(3 c-2 m) x+c^{2}(c-m) .\n$$\n\nThen we must have $3 c-2 m=c-m=0$, which gives $m=0$, a contradiction. We conclude that $f(x)=t x$ is the only solution.', 'Multiplying (1) by $x$, we rewrite it as\n\n$$\nx\\left(x^{3}-m x^{2}+1\\right) P(x+1)+x\\left(x^{3}+m x^{2}+1\\right) P(x-1)=[(x+1)+(x-1)]\\left(x^{3}-m x+1\\right) P(x) .\n$$\n\nAfter regrouping, it becomes\n\n$$\n\\left(x^{3}-m x^{2}+1\\right) Q(x)=\\left(x^{3}+m x^{2}+1\\right) Q(x-1)\n\\tag{2}\n$$\n\nwhere $Q(x)=x P(x+1)-(x+1) P(x)$. If $\\operatorname{deg} P \\geqslant 2$ then $\\operatorname{deg} Q=\\operatorname{deg} P$, so $Q(x)$ has a finite multiset of complex roots, which we denote $R_{Q}$. Each root is taken with its multiplicity. Then the multiset of complex roots of $Q(x-1)$ is $R_{Q}+1=\\left\\{z+1: z \\in R_{Q}\\right\\}$.\n\n\n\nLet $\\left\\{x_{1}, x_{2}, x_{3}\\right\\}$ and $\\left\\{y_{1}, y_{2}, y_{3}\\right\\}$ be the multisets of roots of the polynomials $A(x)=x^{3}-m x^{2}+1$ and $B(x)=x^{3}+m x^{2}+1$, respectively. From (2) we get the equality of multisets\n\n$$\n\\left\\{x_{1}, x_{2}, x_{3}\\right\\} \\cup R_{Q}=\\left\\{y_{1}, y_{2}, y_{3}\\right\\} \\cup\\left(R_{Q}+1\\right) .\n$$\n\nFor every $r \\in R_{Q}$, since $r+1$ is in the set of the right hand side, we must have $r+1 \\in R_{Q}$ or $r+1=x_{i}$ for some $i$. Similarly, since $r$ is in the set of the left hand side, either $r-1 \\in R_{Q}$ or $r=y_{i}$ for some $i$. This implies that, possibly after relabelling $y_{1}, y_{2}, y_{3}$, all the roots of (2) may be partitioned into three chains of the form $\\left\\{y_{i}, y_{i}+1, \\ldots, y_{i}+k_{i}=x_{i}\\right\\}$ for $i=1,2,3$ and some integers $k_{1}, k_{2}, k_{3} \\geqslant 0$.\n\nNow we analyze the roots of the polynomial $A_{a}(x)=x^{3}+a x^{2}+1$. Using calculus or elementary methods, we find that the local extrema of $A_{a}(x)$ occur at $x=0$ and $x=-2 a / 3$; their values are $A_{a}(0)=1>0$ and $A_{a}(-2 a / 3)=1+4 a^{3} / 27$, which is positive for integers $a \\geqslant-1$ and negative for integers $a \\leqslant-2$. So when $a \\in \\mathbb{Z}, A_{a}$ has three real roots if $a \\leqslant-2$ and one if $a \\geqslant-1$.\n\nNow, since $y_{i}-x_{i} \\in \\mathbb{Z}$ for $i=1,2,3$, the cubics $A_{m}$ and $A_{-m}$ must have the same number of real roots. The previous analysis then implies that $m=1$ or $m=-1$. Therefore the real root $\\alpha$ of $A_{1}(x)=x^{3}+x^{2}+1$ and the real root $\\beta$ of $A_{-1}(x)=x^{3}-x^{2}+1$ must differ by an integer. But this is impossible, because $A_{1}\\left(-\\frac{3}{2}\\right)=-\\frac{1}{8}$ and $A_{1}(-1)=1$ so $-1.5<\\alpha<-1$, while $A_{-1}(-1)=-1$ and $A_{-1}\\left(-\\frac{1}{2}\\right)=\\frac{5}{8}$, so $-1<\\beta<-0.5$.\n\nIt follows that $\\operatorname{deg} P \\leqslant 1$. Then, we conclude that the solutions are $P(x)=t x$ for all real numbers $t$.']",['$P(x)=t x$ for any real number $t$'],False,,Need_human_evaluate, 2063,Algebra,,"Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty) such that the sum of the numbers in each group is at most 1 .","['If $d=2 n-1$ and $a_{1}=\\cdots=a_{2 n-1}=n /(2 n-1)$, then each group in such a partition can contain at most one number, since $2 n /(2 n-1)>1$. Therefore $k \\geqslant 2 n-1$. It remains to show that a suitable partition into $2 n-1$ groups always exists.\n\nWe proceed by induction on $d$. For $d \\leqslant 2 n-1$ the result is trivial. If $d \\geqslant 2 n$, then since\n\n$$\n\\left(a_{1}+a_{2}\\right)+\\ldots+\\left(a_{2 n-1}+a_{2 n}\\right) \\leqslant n\n$$\n\nwe may find two numbers $a_{i}, a_{i+1}$ such that $a_{i}+a_{i+1} \\leqslant 1$. We ""merge"" these two numbers into one new number $a_{i}+a_{i+1}$. By the induction hypothesis, a suitable partition exists for the $d-1$ numbers $a_{1}, \\ldots, a_{i-1}, a_{i}+a_{i+1}, a_{i+2}, \\ldots, a_{d}$. This induces a suitable partition for $a_{1}, \\ldots, a_{d}$.', 'We will show that it is even possible to split the sequence $a_{1}, \\ldots, a_{d}$ into $2 n-1$ contiguous groups so that the sum of the numbers in each groups does not exceed 1. Consider a segment $S$ of length $n$, and partition it into segments $S_{1}, \\ldots, S_{d}$ of lengths $a_{1}, \\ldots, a_{d}$, respectively, as shown below. Consider a second partition of $S$ into $n$ equal parts by $n-1$ ""empty dots"".\n\n\n\nAssume that the $n-1$ empty dots are in segments $S_{i_{1}}, \\ldots, S_{i_{n-1}}$. (If a dot is on the boundary of two segments, we choose the right segment). These $n-1$ segments are distinct because they have length at most 1 . Consider the partition:\n\n$$\n\\left\\{a_{1}, \\ldots, a_{i_{1}-1}\\right\\},\\left\\{a_{i_{1}}\\right\\},\\left\\{a_{i_{1}+1}, \\ldots, a_{i_{2}-1}\\right\\},\\left\\{a_{i_{2}}\\right\\}, \\ldots\\left\\{a_{i_{n-1}}\\right\\},\\left\\{a_{i_{n-1}+1}, \\ldots, a_{d}\\right\\}\n$$\n\nIn the example above, this partition is $\\left\\{a_{1}, a_{2}\\right\\},\\left\\{a_{3}\\right\\},\\left\\{a_{4}, a_{5}\\right\\},\\left\\{a_{6}\\right\\}, \\varnothing,\\left\\{a_{7}\\right\\},\\left\\{a_{8}, a_{9}, a_{10}\\right\\}$. We claim that in this partition, the sum of the numbers in this group is at most 1.\n\nFor the sets $\\left\\{a_{i_{t}}\\right\\}$ this is obvious since $a_{i_{t}} \\leqslant 1$. For the sets $\\left\\{a_{i_{t}}+1, \\ldots, a_{i_{t+1}-1}\\right\\}$ this follows from the fact that the corresponding segments lie between two neighboring empty dots, or between an endpoint of $S$ and its nearest empty dot. Therefore the sum of their lengths cannot exceed 1.', 'First put all numbers greater than $\\frac{1}{2}$ in their own groups. Then, form the remaining groups as follows: For each group, add new $a_{i} \\mathrm{~s}$ one at a time until their sum exceeds $\\frac{1}{2}$. Since the last summand is at most $\\frac{1}{2}$, this group has sum at most 1 . Continue this procedure until we have used all the $a_{i}$ s. Notice that the last group may have sum less than $\\frac{1}{2}$. If the sum of the numbers in the last two groups is less than or equal to 1, we merge them into one group. In the end we are left with $m$ groups. If $m=1$ we are done. Otherwise the first $m-2$ have sums greater than $\\frac{1}{2}$ and the last two have total sum greater than 1 . Therefore $n>(m-2) / 2+1$ so $m \\leqslant 2 n-1$ as desired.']",['$k=2 n-1$'],False,,Expression, 2064,Combinatorics,,"In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors. Find the minimal value of $k$ such that the goal is attainable for every possible configuration of 4027 points.","['Firstly, let us present an example showing that $k \\geqslant 2013$. Mark 2013 red and 2013 blue points on some circle alternately, and mark one more blue point somewhere in the plane. The circle is thus split into 4026 arcs, each arc having endpoints of different colors. Thus, if the goal is reached, then each arc should intersect some of the drawn lines. Since any line contains at most two points of the circle, one needs at least 4026/2 $=2013$ lines.\n\nIt remains to prove that one can reach the goal using 2013 lines. First of all, let us mention that for every two points $A$ and $B$ having the same color, one can draw two lines separating these points from all other ones. Namely, it suffices to take two lines parallel to $A B$ and lying on different sides of $A B$ sufficiently close to it: the only two points between these lines will be $A$ and $B$.\n\nNow, let $P$ be the convex hull of all marked points. Two cases are possible.\n\nCase 1. Assume that $P$ has a red vertex $A$. Then one may draw a line separating $A$ from all the other points, pair up the other 2012 red points into 1006 pairs, and separate each pair from the other points by two lines. Thus, 2013 lines will be used.\n\nCase 2. Assume now that all the vertices of $P$ are blue. Consider any two consecutive vertices of $P$, say $A$ and $B$. One may separate these two points from the others by a line parallel to $A B$. Then, as in the previous case, one pairs up all the other 2012 blue points into 1006 pairs, and separates each pair from the other points by two lines. Again, 2013 lines will be used.', 'Let us present a different proof of the fact that $k=2013$ suffices. In fact, we will prove a more general statement:\n\nIf $n$ points in the plane, no three of which are collinear, are colored in red and blue arbitrarily, then it suffices to draw $\\lfloor n / 2\\rfloor$ lines to reach the goal.\n\nWe proceed by induction on $n$. If $n \\leqslant 2$ then the statement is obvious. Now assume that $n \\geqslant 3$, and consider a line $\\ell$ containing two marked points $A$ and $B$ such that all the other marked points are on one side of $\\ell$; for instance, any line containing a side of the convex hull works.\n\nRemove for a moment the points $A$ and $B$. By the induction hypothesis, for the remaining configuration it suffices to draw $\\lfloor n / 2\\rfloor-1$ lines to reach the goal. Now return the points $A$ and $B$ back. Three cases are possible.\n\nCase 1. If $A$ and $B$ have the same color, then one may draw a line parallel to $\\ell$ and separating $A$ and $B$ from the other points. Obviously, the obtained configuration of $\\lfloor n / 2\\rfloor$ lines works.\n\nCase 2. If $A$ and $B$ have different colors, but they are separated by some drawn line, then again the same line parallel to $\\ell$ works.\n\n\n\nCase 3. Finally, assume that $A$ and $B$ have different colors and lie in one of the regions defined by the drawn lines. By the induction assumption, this region contains no other points of one of the colors - without loss of generality, the only blue point it contains is $A$. Then it suffices to draw a line separating $A$ from all other points.\n\nThus the step of the induction is proved.']",['2013'],False,,Numerical, 2065,Combinatorics,,"A crazy physicist discovered a new kind of particle which he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time. (i) If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it. (ii) At any moment, he may double the whole family of imons in his lab by creating a copy $I^{\prime}$ of each imon $I$. During this procedure, the two copies $I^{\prime}$ and $J^{\prime}$ become entangled if and only if the original imons $I$ and $J$ are entangled, and each copy $I^{\prime}$ becomes entangled with its original imon $I$; no other entanglements occur or disappear at this moment. Prove that the physicist may apply a sequence of such operations resulting in a family of imons, no two of which are entangled.","['Let us consider a graph with the imons as vertices, and two imons being connected if and only if they are entangled. Recall that a proper coloring of a graph $G$ is a coloring of its vertices in several colors so that every two connected vertices have different colors.\n\nLemma. Assume that a graph $G$ admits a proper coloring in $n$ colors $(n>1)$. Then one may perform a sequence of operations resulting in a graph which admits a proper coloring in $n-1$ colors.\n\nProof. Let us apply repeatedly operation (i) to any appropriate vertices while it is possible. Since the number of vertices decreases, this process finally results in a graph where all the degrees are even. Surely this graph also admits a proper coloring in $n$ colors $1, \\ldots, n$; let us fix this coloring.\n\nNow apply the operation (ii) to this graph. A proper coloring of the resulting graph in $n$ colors still exists: one may preserve the colors of the original vertices and color the vertex $I^{\\prime}$ in a color $k+1(\\bmod n)$ if the vertex $I$ has color $k$. Then two connected original vertices still have different colors, and so do their two connected copies. On the other hand, the vertices $I$ and $I^{\\prime}$ have different colors since $n>1$.\n\nAll the degrees of the vertices in the resulting graph are odd, so one may apply operation $(i)$ to delete consecutively all the vertices of color $n$ one by one; no two of them are connected by an edge, so their degrees do not change during the process. Thus, we obtain a graph admitting a proper coloring in $n-1$ colors, as required. The lemma is proved.\n\nNow, assume that a graph $G$ has $n$ vertices; then it admits a proper coloring in $n$ colors. Applying repeatedly the lemma we finally obtain a graph admitting a proper coloring in one color, that is - a graph with no edges, as required.', 'Again, we will use the graph language.\n\nI. We start with the following observation.\n\nLemma. Assume that a graph $G$ contains an isolated vertex $A$, and a graph $G^{\\circ}$ is obtained from $G$ by deleting this vertex. Then, if one can apply a sequence of operations which makes a graph with no edges from $G^{\\circ}$, then such a sequence also exists for $G$.\n\nProof. Consider any operation applicable to $G^{\\circ}$ resulting in a graph $G_{1}^{\\circ}$; then there exists a sequence of operations applicable to $G$ and resulting in a graph $G_{1}$ differing from $G_{1}^{\\circ}$ by an addition of an isolated vertex $A$. Indeed, if this operation is of type $(i)$, then one may simply repeat it in $G$.\n\n\n\nOtherwise, the operation is of type (ii), and one may apply it to $G$ and then delete the vertex $A^{\\prime}$ (it will have degree 1).\n\nThus one may change the process for $G^{\\circ}$ into a corresponding process for $G$ step by step.\n\nIn view of this lemma, if at some moment a graph contains some isolated vertex, then we may simply delete it; let us call this operation (iii).\n\nII. Let $V=\\left\\{A_{1}^{0}, \\ldots, A_{n}^{0}\\right\\}$ be the vertices of the initial graph. Let us describe which graphs can appear during our operations. Assume that operation (ii) was applied $m$ times. If these were the only operations applied, then the resulting graph $G_{n}^{m}$ has the set of vertices which can be enumerated as\n\n$$\nV_{n}^{m}=\\left\\{A_{i}^{j}: 1 \\leqslant i \\leqslant n, 0 \\leqslant j \\leqslant 2^{m}-1\\right\\},\n$$\n\nwhere $A_{i}^{0}$ is the common ""ancestor"" of all the vertices $A_{i}^{j}$, and the binary expansion of $j$ (adjoined with some zeroes at the left to have $m$ digits) ""keeps the history"" of this vertex: the $d$ th digit from the right is 0 if at the $d$ th doubling the ancestor of $A_{i}^{j}$ was in the original part, and this digit is 1 if it was in the copy.\n\nNext, the two vertices $A_{i}^{j}$ and $A_{k}^{\\ell}$ in $G_{n}^{m}$ are connected with an edge exactly if either (1) $j=\\ell$ and there was an edge between $A_{i}^{0}$ and $A_{k}^{0}$ (so these vertices appeared at the same application of operation (ii)); or (2) $i=k$ and the binary expansions of $j$ and $\\ell$ differ in exactly one digit (so their ancestors became connected as a copy and the original vertex at some application of $(i i))$.\n\nNow, if some operations $(i)$ were applied during the process, then simply some vertices in $G_{n}^{m}$ disappeared. So, in any case the resulting graph is some induced subgraph of $G_{n}^{m}$.\n\nIII. Finally, we will show that from each (not necessarily induced) subgraph of $G_{n}^{m}$ one can obtain a graph with no vertices by applying operations $(i),($ ii) and (iii). We proceed by induction on $n$; the base case $n=0$ is trivial.\n\nFor the induction step, let us show how to apply several operations so as to obtain a graph containing no vertices of the form $A_{n}^{j}$ for $j \\in \\mathbb{Z}$. We will do this in three steps.\n\nStep 1. We apply repeatedly operation $(i)$ to any appropriate vertices while it is possible. In the resulting graph, all vertices have even degrees.\n\nStep 2. Apply operation (ii) obtaining a subgraph of $G_{n}^{m+1}$ with all degrees being odd. In this graph, we delete one by one all the vertices $A_{n}^{j}$ where the sum of the binary digits of $j$ is even; it is possible since there are no edges between such vertices, so all their degrees remain odd. After that, we delete all isolated vertices.\n\nStep 3. Finally, consider any remaining vertex $A_{n}^{j}$ (then the sum of digits of $j$ is odd). If its degree is odd, then we simply delete it. Otherwise, since $A_{n}^{j}$ is not isolated, we consider any vertex adjacent to it. It has the form $A_{k}^{j}$ for some $k3$. Finally, in the sequence $d(a, y), d\\left(a, b_{y}\\right), d\\left(a, c_{y}\\right), d\\left(a, d_{y}\\right)$ the neighboring terms differ by at most 1 , the first term is less than 3 , and the last one is greater than 3 ; thus there exists one which is equal to 3 , as required.']",,True,,, 2068,Combinatorics,,"Let $n \geqslant 2$ be an integer. Consider all circular arrangements of the numbers $0,1, \ldots, n$; the $n+1$ rotations of an arrangement are considered to be equal. A circular arrangement is called beautiful if, for any four distinct numbers $0 \leqslant a, b, c, d \leqslant n$ with $a+c=b+d$, the chord joining numbers $a$ and $c$ does not intersect the chord joining numbers $b$ and $d$. Let $M$ be the number of beautiful arrangements of $0,1, \ldots, n$. Let $N$ be the number of pairs $(x, y)$ of positive integers such that $x+y \leqslant n$ and $\operatorname{gcd}(x, y)=1$. Prove that $$ M=N+1 $$","['Given a circular arrangement of $[0, n]=\\{0,1, \\ldots, n\\}$, we define a $k$-chord to be a (possibly degenerate) chord whose (possibly equal) endpoints add up to $k$. We say that three chords of a circle are aligned if one of them separates the other two. Say that $m \\geqslant 3$ chords are aligned if any three of them are aligned. For instance, in Figure 1, $A, B$, and $C$ are aligned, while $B, C$, and $D$ are not.\n\n\n\nFigure 1\n\n\n\nFigure 2\n\nClaim. In a beautiful arrangement, the $k$-chords are aligned for any integer $k$.\n\nProof. We proceed by induction. For $n \\leqslant 3$ the statement is trivial. Now let $n \\geqslant 4$, and proceed by contradiction. Consider a beautiful arrangement $S$ where the three $k$-chords $A, B, C$ are not aligned. If $n$ is not among the endpoints of $A, B$, and $C$, then by deleting $n$ from $S$ we obtain a beautiful arrangement $S \\backslash\\{n\\}$ of $[0, n-1]$, where $A, B$, and $C$ are aligned by the induction hypothesis. Similarly, if 0 is not among these endpoints, then deleting 0 and decreasing all the numbers by 1 gives a beautiful arrangement $S \\backslash\\{0\\}$ where $A, B$, and $C$ are aligned. Therefore both 0 and $n$ are among the endpoints of these segments. If $x$ and $y$ are their respective partners, we have $n \\geqslant 0+x=k=n+y \\geqslant n$. Thus 0 and $n$ are the endpoints of one of the chords; say it is $C$.\n\nLet $D$ be the chord formed by the numbers $u$ and $v$ which are adjacent to 0 and $n$ and on the same side of $C$ as $A$ and $B$, as shown in Figure 2. Set $t=u+v$. If we had $t=n$, the $n-$ chords $A$, $B$, and $D$ would not be aligned in the beautiful arrangement $S \\backslash\\{0, n\\}$, contradicting the induction hypothesis. If $tn$ is equivalent to the case $t0$ is very small.\n\n\n\nFigure 3\n\nIf $0 \\leqslant a, b, c, d \\leqslant n$ satisfy $a+c=b+d$, then $a \\alpha+c \\alpha=b \\alpha+d \\alpha$, so the chord from $a$ to $c$ is parallel to the chord from $b$ to $d$ in $A(\\alpha)$. Hence in a cyclic arrangement all $k$-chords are parallel. In particular every cyclic arrangement is beautiful.\n\nNext we show that there are exactly $N+1$ distinct cyclic arrangements. To see this, let us see how $A(\\alpha)$ changes as we increase $\\alpha$ from 0 to 1 . The order of points $p$ and $q$ changes precisely when we cross a value $\\alpha=f$ such that $\\{p f\\}=\\{q f\\}$; this can only happen if $f$ is one of the $N$ fractions $f_{1}, \\ldots, f_{N}$. Therefore there are at most $N+1$ different cyclic arrangements.\n\nTo show they are all distinct, recall that $f_{i}=a_{i} / b_{i}$ and let $\\epsilon>0$ be a very small number. In the arrangement $A\\left(f_{i}+\\epsilon\\right)$, point $k$ lands at $\\frac{k a_{i}\\left(\\bmod b_{i}\\right)}{b_{i}}+k \\epsilon$. Therefore the points are grouped into $b_{i}$ clusters next to the points $0, \\frac{1}{b_{i}}, \\ldots, \\frac{b_{i}-1}{b_{i}}$ of the circle. The cluster following $\\frac{k}{b_{i}}$ contains the numbers congruent to $k a_{i}^{-1}$ modulo $b_{i}$, listed clockwise in increasing order. It follows that the first number after 0 in $A\\left(f_{i}+\\epsilon\\right)$ is $b_{i}$, and the first number after 0 which is less than $b_{i}$ is $a_{i}^{-1}\\left(\\bmod b_{i}\\right)$, which uniquely determines $a_{i}$. In this way we can recover $f_{i}$ from the cyclic arrangement. Note also that $A\\left(f_{i}+\\epsilon\\right)$ is not the trivial arrangement where we list $0,1, \\ldots, n$ in order clockwise. It follows that the $N+1$ cyclic arrangements $A(\\epsilon), A\\left(f_{1}+\\epsilon\\right), \\ldots, A\\left(f_{N}+\\epsilon\\right)$ are distinct.\n\nLet us record an observation which will be useful later:\n\n$$\n\\text { if } f_{i}<\\alpha\n\nFigure 4\n\nIn $A_{n}(\\alpha)$ the chord from $n-1$ to $x$ is parallel and adjacent to the chord from $n$ to $x-1$, so $n-1$ is between $x-1$ and $x$ in clockwise order, as shown in Figure 4. Similarly, $n-1$ is between $y$ and $y-1$. Therefore $x, y, x-1, n-1$, and $y-1$ occur in this order in $A_{n}(\\alpha)$ and hence in $A$ (possibly with $y=x-1$ or $x=y-1$ ).\n\nNow, $A$ may only differ from $A_{n}(\\alpha)$ in the location of $n$. In $A$, since the chord from $n-1$ to $x$ and the chord from $n$ to $x-1$ do not intersect, $n$ is between $x$ and $n-1$. Similarly, $n$ is between $n-1$ and $y$. Then $n$ must be between $x$ and $y$ and $A=A_{n}(\\alpha)$. Therefore $A$ is cyclic as desired.\n\nCase 2. There is exactly one $i$ with $\\frac{p_{1}}{q_{1}}<\\frac{i}{n}<\\frac{p_{2}}{q_{2}}$.\n\nIn this case there exist two cyclic arrangements $A_{n}\\left(\\alpha_{1}\\right)$ and $A_{n}\\left(\\alpha_{2}\\right)$ of the numbers $0, \\ldots, n$ extending $A_{n-1}(\\alpha)$, where $\\frac{p_{1}}{q_{1}}<\\alpha_{1}<\\frac{i}{n}$ and $\\frac{i}{n}<\\alpha_{2}<\\frac{p_{2}}{q_{2}}$. In $A_{n-1}(\\alpha), 0$ is the only number between $q_{2}$ and $q_{1}$ by (2). For the same reason, $n$ is between $q_{2}$ and 0 in $A_{n}\\left(\\alpha_{1}\\right)$, and between 0 and $q_{1}$ in $A_{n}\\left(\\alpha_{2}\\right)$.\n\nLetting $x=q_{2}$ and $y=q_{1}$, the argument of Case 1 tells us that $n$ must be between $x$ and $y$ in $A$. Therefore $A$ must equal $A_{n}\\left(\\alpha_{1}\\right)$ or $A_{n}\\left(\\alpha_{2}\\right)$, and therefore it is cyclic.\n\nThis concludes the proof that every beautiful arrangement is cyclic. It follows that there are exactly $N+1$ beautiful arrangements of $[0, n]$ as we wished to show.']",,True,,, 2069,Combinatorics,,"Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Player $B$ then picks an integer $k$ and blackens the interval from $k / 2^{m}$ to $(k+1) / 2^{m}$ (some parts of this interval may have been blackened before). The goal of player $A$ is to reach a situation where the pot is empty and the interval $[0,1]$ is not completely blackened. Prove that there doesn't exist a strategy for player $A$ to win in a finite number of moves.","['We will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty.\n\nAt the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between 0 and $x_{r}$ has already been blackened; for completeness we define $x_{1}=0$. Let $m$ be the integer picked by player $A$ in this round; we define an integer $y_{r}$ by\n\n$$\n\\frac{y_{r}}{2^{m}} \\leqslant x_{r}<\\frac{y_{r}+1}{2^{m}}\n$$\n\nNote that $I_{0}^{r}=\\left[y_{r} / 2^{m},\\left(y_{r}+1\\right) / 2^{m}\\right]$ is the leftmost interval that may be painted in round $r$ and that still contains some uncolored point.\n\nPlayer $B$ now looks at the next interval $I_{1}^{r}=\\left[\\left(y_{r}+1\\right) / 2^{m},\\left(y_{r}+2\\right) / 2^{m}\\right]$. If $I_{1}^{r}$ still contains an uncolored point, then player $B$ blackens the interval $I_{1}^{r}$; otherwise he blackens the interval $I_{0}^{r}$. We make the convention that, at the beginning of the game, the interval $[1,2]$ is already blackened; thus, if $y_{r}+1=2^{m}$, then $B$ blackens $I_{0}^{r}$.\n\nOur aim is to estimate the amount of ink used after each round. Firstly, we will prove by induction that, if before $r$ th round the segment $[0,1]$ is not completely colored, then, before this move,\n\n(i) the amount of ink used for the segment $\\left[0, x_{r}\\right]$ is at most $3 x_{r}$; and\n\n(ii) for every $m, B$ has blackened at most one interval of length $1 / 2^{m}$ to the right of $x_{r}$.\n\nObviously, these conditions are satisfied for $r=0$. Now assume that they were satisfied before the $r$ th move, and consider the situation after this move; let $m$ be the number $A$ has picked at this move.\n\nIf $B$ has blackened the interval $I_{1}^{r}$ at this move, then $x_{r+1}=x_{r}$, and $(i)$ holds by the induction hypothesis. Next, had $B$ blackened before the $r$ th move any interval of length $1 / 2^{m}$ to the right of $x_{r}$, this interval would necessarily coincide with $I_{1}^{r}$. By our strategy, this cannot happen. So, condition (ii) also remains valid.\n\nAssume now that $B$ has blackened the interval $I_{0}^{r}$ at the $r$ th move, but the interval $[0,1]$ still contains uncolored parts (which means that $I_{1}^{r}$ is contained in $[0,1]$ ). Then condition (ii) clearly remains true, and we need to check $(i)$ only. In our case, the intervals $I_{0}^{r}$ and $I_{1}^{r}$ are completely colored after the $r$ th move, so $x_{r+1}$ either reaches the right endpoint of $I_{1}$ or moves even further to the right. So, $x_{r+1}=x_{r}+\\alpha$ for some $\\alpha>1 / 2^{m}$.\n\nNext, any interval blackened by $B$ before the $r$ th move which intersects $\\left(x_{r}, x_{r+1}\\right)$ should be contained in $\\left[x_{r}, x_{r+1}\\right]$; by (ii), all such intervals have different lengths not exceeding $1 / 2^{m}$, so the total amount of ink used for them is less than $2 / 2^{m}$. Thus, the amount of ink used for the segment $\\left[0, x_{r+1}\\right]$ does not exceed the sum of $2 / 2^{m}, 3 x_{r}$ (used for $\\left[0, x_{r}\\right]$ ), and $1 / 2^{m}$ used for the\n\n\n\nsegment $I_{0}^{r}$. In total it gives at most $3\\left(x_{r}+1 / 2^{m}\\right)<3\\left(x_{r}+\\alpha\\right)=3 x_{r+1}$. Thus condition $(i)$ is also verified in this case. The claim is proved.\n\nFinally, we can perform the desired estimation. Consider any situation in the game, say after the $(r-1)$ st move; assume that the segment $[0,1]$ is not completely black. By $(i i)$, in the segment $\\left[x_{r}, 1\\right]$ player $B$ has colored several segments of different lengths; all these lengths are negative powers of 2 not exceeding $1-x_{r}$; thus the total amount of ink used for this interval is at most $2\\left(1-x_{r}\\right)$. Using $(i)$, we obtain that the total amount of ink used is at most $3 x_{r}+2\\left(1-x_{r}\\right)<3$. Thus the pot is not empty, and therefore $A$ never wins.']",,True,,, 2070,Geometry,,"Let $A B C$ be an acute-angled triangle with orthocenter $H$, and let $W$ be a point on side $B C$. Denote by $M$ and $N$ the feet of the altitudes from $B$ and $C$, respectively. Denote by $\omega_{1}$ the circumcircle of $B W N$, and let $X$ be the point on $\omega_{1}$ which is diametrically opposite to $W$. Analogously, denote by $\omega_{2}$ the circumcircle of $C W M$, and let $Y$ be the point on $\omega_{2}$ which is diametrically opposite to $W$. Prove that $X, Y$ and $H$ are collinear.","['Let $L$ be the foot of the altitude from $A$, and let $Z$ be the second intersection point of circles $\\omega_{1}$ and $\\omega_{2}$, other than $W$. We show that $X, Y, Z$ and $H$ lie on the same line.\n\nDue to $\\angle B N C=\\angle B M C=90^{\\circ}$, the points $B, C, N$ and $M$ are concyclic; denote their circle by $\\omega_{3}$. Observe that the line $W Z$ is the radical axis of $\\omega_{1}$ and $\\omega_{2}$; similarly, $B N$ is the radical axis of $\\omega_{1}$ and $\\omega_{3}$, and $C M$ is the radical axis of $\\omega_{2}$ and $\\omega_{3}$. Hence $A=B N \\cap C M$ is the radical center of the three circles, and therefore $W Z$ passes through $A$.\n\nSince $W X$ and $W Y$ are diameters in $\\omega_{1}$ and $\\omega_{2}$, respectively, we have $\\angle W Z X=\\angle W Z Y=90^{\\circ}$, so the points $X$ and $Y$ lie on the line through $Z$, perpendicular to $W Z$.\n\n\n\nThe quadrilateral $B L H N$ is cyclic, because it has two opposite right angles. From the power of $A$ with respect to the circles $\\omega_{1}$ and $B L H N$ we find $A L \\cdot A H=A B \\cdot A N=A W \\cdot A Z$. If $H$ lies on the line $A W$ then this implies $H=Z$ immediately. Otherwise, by $\\frac{A Z}{A H}=\\frac{A L}{A W}$ the triangles $A H Z$ and $A W L$ are similar. Then $\\angle H Z A=\\angle W L A=90^{\\circ}$, so the point $H$ also lies on the line $X Y Z$.']",,True,,, 2071,Geometry,,"Let $\omega$ be the circumcircle of a triangle $A B C$. Denote by $M$ and $N$ the midpoints of the sides $A B$ and $A C$, respectively, and denote by $T$ the midpoint of the $\operatorname{arc} B C$ of $\omega$ not containing $A$. The circumcircles of the triangles $A M T$ and $A N T$ intersect the perpendicular bisectors of $A C$ and $A B$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $A B C$. The lines $M N$ and $X Y$ intersect at $K$. Prove that $K A=K T$.","['Let $O$ be the center of $\\omega$, thus $O=M Y \\cap N X$. Let $\\ell$ be the perpendicular bisector of $A T$ (it also passes through $O$ ). Denote by $r$ the operation of reflection about $\\ell$. Since $A T$ is the angle bisector of $\\angle B A C$, the line $r(A B)$ is parallel to $A C$. Since $O M \\perp A B$ and $O N \\perp A C$, this means that the line $r(O M)$ is parallel to the line $O N$ and passes through $O$, so $r(O M)=O N$. Finally, the circumcircle $\\gamma$ of the triangle $A M T$ is symmetric about $\\ell$, so $r(\\gamma)=\\gamma$. Thus the point $M$ maps to the common point of $O N$ with the $\\operatorname{arc} A M T$ of $\\gamma$ - that is, $r(M)=X$.\n\nSimilarly, $r(N)=Y$. Thus, we get $r(M N)=X Y$, and the common point $K$ of $M N$ nd $X Y$ lies on $\\ell$. This means exactly that $K A=K T$.\n\n', 'Let $L$ be the second common point of the line $A C$ with the circumcircle $\\gamma$ of the triangle $A M T$. From the cyclic quadrilaterals $A B T C$ and $A M T L$ we get $\\angle B T C=180^{\\circ}-$ $\\angle B A C=\\angle M T L$, which implies $\\angle B T M=\\angle C T L$. Since $A T$ is an angle bisector in these quadrilaterals, we have $B T=T C$ and $M T=T L$. Thus the triangles $B T M$ and $C T L$ are congruent, so $C L=B M=A M$.\n\nLet $X^{\\prime}$ be the common point of the line $N X$ with the external bisector of $\\angle B A C$; notice that it lies outside the triangle $A B C$. Then we have $\\angle T A X^{\\prime}=90^{\\circ}$ and $X^{\\prime} A=X^{\\prime} C$, so we get $\\angle X^{\\prime} A M=90^{\\circ}+\\angle B A C / 2=180^{\\circ}-\\angle X^{\\prime} A C=180^{\\circ}-\\angle X^{\\prime} C A=\\angle X^{\\prime} C L$. Thus the triangles $X^{\\prime} A M$ and $X^{\\prime} C L$ are congruent, and therefore\n\n$$\n\\angle M X^{\\prime} L=\\angle A X^{\\prime} C+\\left(\\angle C X^{\\prime} L-\\angle A X^{\\prime} M\\right)=\\angle A X^{\\prime} C=180^{\\circ}-2 \\angle X^{\\prime} A C=\\angle B A C=\\angle M A L .\n$$\n\nThis means that $X^{\\prime}$ lies on $\\gamma$.\n\nThus we have $\\angle T X N=\\angle T X X^{\\prime}=\\angle T A X^{\\prime}=90^{\\circ}$, so $T X \\| A C$. Then $\\angle X T A=\\angle T A C=$ $\\angle T A M$, so the cyclic quadrilateral $M A T X$ is an isosceles trapezoid. Similarly, $N A T Y$ is an isosceles trapezoid, so again the lines $M N$ and $X Y$ are the reflections of each other about the perpendicular bisector of $A T$. Thus $K$ belongs to this perpendicular bisector.\n\n\n\n']",,True,,, 2072,Geometry,,Let $A B C$ be a triangle with $\angle B>\angle C$. Let $P$ and $Q$ be two different points on line $A C$ such that $\angle P B A=\angle Q B A=\angle A C B$ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $B Q$ for which $P D=P B$. Let the ray $A D$ intersect the circle $A B C$ at $R \neq A$. Prove that $Q B=Q R$.,"['Denote by $\\omega$ the circumcircle of the triangle $A B C$, and let $\\angle A C B=\\gamma$. Note that the condition $\\gamma<\\angle C B A$ implies $\\gamma<90^{\\circ}$. Since $\\angle P B A=\\gamma$, the line $P B$ is tangent to $\\omega$, so $P A \\cdot P C=P B^{2}=P D^{2}$. By $\\frac{P A}{P D}=\\frac{P D}{P C}$ the triangles $P A D$ and $P D C$ are similar, and $\\angle A D P=\\angle D C P$.\n\nNext, since $\\angle A B Q=\\angle A C B$, the triangles $A B C$ and $A Q B$ are also similar. Then $\\angle A Q B=$ $\\angle A B C=\\angle A R C$, which means that the points $D, R, C$, and $Q$ are concyclic. Therefore $\\angle D R Q=$ $\\angle D C Q=\\angle A D P$.\n\n\n\nFigure 1\n\nNow from $\\angle A R B=\\angle A C B=\\gamma$ and $\\angle P D B=\\angle P B D=2 \\gamma$ we get\n\n$$\n\\angle Q B R=\\angle A D B-\\angle A R B=\\angle A D P+\\angle P D B-\\angle A R B=\\angle D R Q+\\gamma=\\angle Q R B\n$$\n\nso the triangle $Q R B$ is isosceles, which yields $Q B=Q R$.', 'Again, denote by $\\omega$ the circumcircle of the triangle $A B C$. Denote $\\angle A C B=\\gamma$. Since $\\angle P B A=\\gamma$, the line $P B$ is tangent to $\\omega$.\n\nLet $E$ be the second intersection point of $B Q$ with $\\omega$. If $V^{\\prime}$ is any point on the ray $C E$ beyond $E$, then $\\angle B E V^{\\prime}=180^{\\circ}-\\angle B E C=180^{\\circ}-\\angle B A C=\\angle P A B$; together with $\\angle A B Q=$ $\\angle P B A$ this shows firstly, that the rays $B A$ and $C E$ intersect at some point $V$, and secondly that the triangle $V E B$ is similar to the triangle $P A B$. Thus we have $\\angle B V E=\\angle B P A$. Next, $\\angle A E V=\\angle B E V-\\gamma=\\angle P A B-\\angle A B Q=\\angle A Q B$; so the triangles $P B Q$ and $V A E$ are also similar.\n\nLet $P H$ be an altitude in the isosceles triangle $P B D$; then $B H=H D$. Let $G$ be the intersection point of $P H$ and $A B$. By the symmetry with respect to $P H$, we have $\\angle B D G=\\angle D B G=\\gamma=$ $\\angle B E A$; thus $D G \\| A E$ and hence $\\frac{B G}{G A}=\\frac{B D}{D E}$. Thus the points $G$ and $D$ correspond to each other in the similar triangles $P A B$ and $V E B$, so $\\angle D V B=\\angle G P B=90^{\\circ}-\\angle P B Q=90^{\\circ}-\\angle V A E$. Thus $V D \\perp A E$.\n\n\n\nLet $T$ be the common point of $V D$ and $A E$, and let $D S$ be an altitude in the triangle $B D R$. The points $S$ and $T$ are the feet of corresponding altitudes in the similar triangles $A D E$ and $B D R$, so $\\frac{B S}{S R}=\\frac{A T}{T E}$. On the other hand, the points $T$ and $H$ are feet of corresponding altitudes in the similar triangles $V A E$ and $P B Q$, so $\\frac{A T}{T E}=\\frac{B H}{H Q}$. Thus $\\frac{B S}{S R}=\\frac{A T}{T E}=\\frac{B H}{H Q}$, and the triangles $B H S$ and $B Q R$ are similar.\n\nFinally, $S H$ is a median in the right-angled triangle $S B D$; so $B H=H S$, and hence $B Q=Q R$.\n\n\n\nFigure 2', 'Denote by $\\omega$ and $O$ the circumcircle of the triangle $A B C$ and its center, respectively. From the condition $\\angle P B A=\\angle B C A$ we know that $B P$ is tangent to $\\omega$.\n\nLet $E$ be the second point of intersection of $\\omega$ and $B D$. Due to the isosceles triangle $B D P$, the tangent of $\\omega$ at $E$ is parallel to $D P$ and consequently it intersects $B P$ at some point $L$. Of course, $P D \\| L E$. Let $M$ be the midpoint of $B E$, and let $H$ be the midpoint of $B R$. Notice that $\\angle A E B=\\angle A C B=\\angle A B Q=\\angle A B E$, so $A$ lies on the perpendicular bisector of $B E$; thus the points $L, A, M$, and $O$ are collinear. Let $\\omega_{1}$ be the circle with diameter $B O$. Let $Q^{\\prime}=H O \\cap B E$; since $H O$ is the perpendicular bisector of $B R$, the statement of the problem is equivalent to $Q^{\\prime}=Q$.\n\nConsider the following sequence of projections (see Fig. 3).\n\n1. Project the line $B E$ to the line $L B$ through the center $A$. (This maps $Q$ to $P$.)\n2. Project the line $L B$ to $B E$ in parallel direction with $L E .(P \\mapsto D$.\n3. Project the line $B E$ to the circle $\\omega$ through its point $A$. $(D \\mapsto R$.)\n4. Scale $\\omega$ by the ratio $\\frac{1}{2}$ from the point $B$ to the circle $\\omega_{1} .(R \\mapsto H$.\n5. Project $\\omega_{1}$ to the line $B E$ through its point $O$. $\\left(H \\mapsto Q^{\\prime}\\right.$.)\n\nWe prove that the composition of these transforms, which maps the line $B E$ to itself, is the identity. To achieve this, it suffices to show three fixed points. An obvious fixed point is $B$ which is fixed by all the transformations above. Another fixed point is $M$, its path being $M \\mapsto L \\mapsto$ $E \\mapsto E \\mapsto M \\mapsto M$.\n\n\n\n\n\nFigure 3\n\n\n\nFigure 4\n\nIn order to show a third fixed point, draw a line parallel with $L E$ through $A$; let that line intersect $B E, L B$ and $\\omega$ at $X, Y$ and $Z \\neq A$, respectively (see Fig. 4). We show that $X$ is a fixed point. The images of $X$ at the first three transformations are $X \\mapsto Y \\mapsto X \\mapsto Z$. From $\\angle X B Z=\\angle E A Z=\\angle A E L=\\angle L B A=\\angle B Z X$ we can see that the triangle $X B Z$ is isosceles. Let $U$ be the midpoint of $B Z$; then the last two transformations do $Z \\mapsto U \\mapsto X$, and the point $X$ is fixed.']",,True,,, 2073,Geometry,,"Let $A B C D E F$ be a convex hexagon with $A B=D E, B C=E F, C D=F A$, and $\angle A-\angle D=\angle C-\angle F=\angle E-\angle B$. Prove that the diagonals $A D, B E$, and $C F$ are concurrent.","['we denote $\\theta=\\angle A-\\angle D=\\angle C-\\angle F=\\angle E-\\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nLet $x=A B=D E, y=C D=F A, z=E F=B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P, E F A Q$, and $A B C R$ are parallelograms. We compute\n\n$$\n\\begin{aligned}\n\\angle P E Q & =\\angle F E Q+\\angle D E P-\\angle E=\\left(180^{\\circ}-\\angle F\\right)+\\left(180^{\\circ}-\\angle D\\right)-\\angle E \\\\\n& =360^{\\circ}-\\angle D-\\angle E-\\angle F=\\frac{1}{2}(\\angle A+\\angle B+\\angle C-\\angle D-\\angle E-\\angle F)=\\theta / 2\n\\end{aligned}\n$$\n\nSimilarly, $\\angle Q A R=\\angle R C P=\\theta / 2$.\n\n\n\nIf $\\theta=0$, since $\\triangle R C P$ is isosceles, $R=P$. Therefore $A B\\|R C=P C\\| E D$, so $A B D E$ is a parallelogram. Similarly, $B C E F$ and $C D F A$ are parallelograms. It follows that $A D, B E$ and $C F$ meet at their common midpoint.\n\nNow assume $\\theta>0$. Since $\\triangle P E Q, \\triangle Q A R$, and $\\triangle R C P$ are isosceles and have the same angle at the apex, we have $\\triangle P E Q \\sim \\triangle Q A R \\sim \\triangle R C P$ with ratios of similarity $y: z: x$. Thus\n\n$$\n\\triangle P Q R \\text { is similar to the triangle with sidelengths } y, z, \\text { and } x \\text {. }\n\\tag{1}\n$$\n\nNext, notice that\n\n$$\n\\frac{R Q}{Q P}=\\frac{z}{y}=\\frac{R A}{A F}\n$$\n\nand, using directed angles between rays,\n\n$$\n\\begin{aligned}\n\\sphericalangle(R Q, Q P) & =\\sphericalangle(R Q, Q E)+\\sphericalangle(Q E, Q P) \\\\\n& =\\sphericalangle(R Q, Q E)+\\sphericalangle(R A, R Q)=\\sphericalangle(R A, Q E)=\\sphericalangle(R A, A F) .\n\\end{aligned}\n$$\n\nThus $\\triangle P Q R \\sim \\triangle F A R$. Since $F A=y$ and $A R=z,(1)$ then implies that $F R=x$. Similarly $F P=x$. Therefore $C R F P$ is a rhombus.\n\nWe conclude that $C F$ is the perpendicular bisector of $P R$. Similarly, $B E$ is the perpendicular bisector of $P Q$ and $A D$ is the perpendicular bisector of $Q R$. It follows that $A D, B E$, and $C F$ are concurrent at the circumcenter of $P Q R$.', 'we denote $\\theta=\\angle A-\\angle D=\\angle C-\\angle F=\\angle E-\\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nLet $X=C D \\cap E F, Y=E F \\cap A B, Z=A B \\cap C D, X^{\\prime}=F A \\cap B C, Y^{\\prime}=$ $B C \\cap D E$, and $Z^{\\prime}=D E \\cap F A$. From $\\angle A+\\angle B+\\angle C=360^{\\circ}+\\theta / 2$ we get $\\angle A+\\angle B>180^{\\circ}$ and $\\angle B+\\angle C>180^{\\circ}$, so $Z$ and $X^{\\prime}$ are respectively on the opposite sides of $B C$ and $A B$ from the hexagon. Similar conclusions hold for $X, Y, Y^{\\prime}$, and $Z^{\\prime}$. Then\n\n$$\n\\angle Y Z X=\\angle B+\\angle C-180^{\\circ}=\\angle E+\\angle F-180^{\\circ}=\\angle Y^{\\prime} Z^{\\prime} X^{\\prime},\n$$\n\nand similarly $\\angle Z X Y=\\angle Z^{\\prime} X^{\\prime} Y^{\\prime}$ and $\\angle X Y Z=\\angle X^{\\prime} Y^{\\prime} Z^{\\prime}$, so $\\triangle X Y Z \\sim \\triangle X^{\\prime} Y^{\\prime} Z^{\\prime}$. Thus there is a rotation $R$ which sends $\\triangle X Y Z$ to a triangle with sides parallel to $\\triangle X^{\\prime} Y^{\\prime} Z^{\\prime}$. Since $A B=D E$ we have $R(\\overrightarrow{A B})=\\overrightarrow{D E}$. Similarly, $R(\\overrightarrow{C D})=\\overrightarrow{F A}$ and $R(\\overrightarrow{E F})=\\overrightarrow{B C}$. Therefore\n\n$$\n\\overrightarrow{0}=\\overrightarrow{A B}+\\overrightarrow{B C}+\\overrightarrow{C D}+\\overrightarrow{D E}+\\overrightarrow{E F}+\\overrightarrow{F A}=(\\overrightarrow{A B}+\\overrightarrow{C D}+\\overrightarrow{E F})+R(\\overrightarrow{A B}+\\overrightarrow{C D}+\\overrightarrow{E F})\n$$\n\nIf $R$ is a rotation by $180^{\\circ}$, then any two opposite sides of our hexagon are equal and parallel, so the three diagonals meet at their common midpoint. Otherwise, we must have\n\n$$\n\\overrightarrow{A B}+\\overrightarrow{C D}+\\overrightarrow{E F}=\\overrightarrow{0}\n$$\n\nor else we would have two vectors with different directions whose sum is $\\overrightarrow{0}$.\n\n\n\nThis allows us to consider a triangle $L M N$ with $\\overrightarrow{L M}=\\overrightarrow{E F}, \\overrightarrow{M N}=\\overrightarrow{A B}$, and $\\overrightarrow{N L}=\\overrightarrow{C D}$. Let $O$ be the circumcenter of $\\triangle L M N$ and consider the points $O_{1}, O_{2}, O_{3}$ such that $\\triangle A O_{1} B, \\triangle C O_{2} D$, and $\\triangle E O_{3} F$ are translations of $\\triangle M O N, \\triangle N O L$, and $\\triangle L O M$, respectively. Since $F O_{3}$ and $A O_{1}$ are translations of $M O$, quadrilateral $A F O_{3} O_{1}$ is a parallelogram and $O_{3} O_{1}=F A=C D=N L$. Similarly, $O_{1} O_{2}=L M$ and $O_{2} O_{3}=M N$. Therefore $\\triangle O_{1} O_{2} O_{3} \\cong \\triangle L M N$. Moreover, by means of the rotation $R$ one may check that these triangles have the same orientation.\n\nLet $T$ be the circumcenter of $\\triangle O_{1} O_{2} O_{3}$. We claim that $A D, B E$, and $C F$ meet at $T$. Let us show that $C, T$, and $F$ are collinear. Notice that $C O_{2}=O_{2} T=T O_{3}=O_{3} F$ since they are all equal to the circumradius of $\\triangle L M N$. Therefore $\\triangle T O_{3} F$ and $\\triangle C_{2} T$ are isosceles. Using directed angles between rays again, we get\n\n$$\n\\sphericalangle\\left(T F, T O_{3}\\right)=\\sphericalangle\\left(F O_{3}, F T\\right) \\quad \\text { and } \\quad \\sphericalangle\\left(T O_{2}, T C\\right)=\\sphericalangle\\left(C T, C O_{2}\\right) \\text {. }\n\\tag{2}\n$$\n\nAlso, $T$ and $O$ are the circumcenters of the congruent triangles $\\triangle O_{1} O_{2} O_{3}$ and $\\triangle L M N$ so we have $\\sphericalangle\\left(T O_{3}, T O_{2}\\right)=\\sphericalangle(O N, O M)$. Since $C O_{2}$ and $F_{3}$ are translations of $N O$ and $M O$ respectively, this implies\n\n$$\n\\sphericalangle\\left(T O_{3}, T O_{2}\\right)=\\sphericalangle\\left(C O_{2}, F O_{3}\\right) .\n\\tag{3}\n$$\n\n\n\nAdding the three equations in (2) and (3) gives\n\n$$\n\\sphericalangle(T F, T C)=\\sphericalangle(C T, F T)=-\\sphericalangle(T F, T C)\n$$\n\nwhich implies that $T$ is on $C F$. Analogous arguments show that it is on $A D$ and $B E$ also. The desired result follows.', 'we denote $\\theta=\\angle A-\\angle D=\\angle C-\\angle F=\\angle E-\\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nPlace the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A, B, C, D, E, F$ represent the corresponding complex numbers. Also consider the complex numbers $a, b, c, a^{\\prime}, b^{\\prime}, c^{\\prime}$ given by $B-A=a, D-C=b, F-E=c, E-D=a^{\\prime}$, $A-F=b^{\\prime}$, and $C-B=c^{\\prime}$. Let $k=|a| /|b|$. From $a / b^{\\prime}=-k e^{i \\angle A}$ and $a^{\\prime} / b=-k e^{i \\angle D}$ we get that $\\left(a^{\\prime} / a\\right)\\left(b^{\\prime} / b\\right)=e^{-i \\theta}$ and similarly $\\left(b^{\\prime} / b\\right)\\left(c^{\\prime} / c\\right)=e^{-i \\theta}$ and $\\left(c^{\\prime} / c\\right)\\left(a^{\\prime} / a\\right)=e^{-i \\theta}$. It follows that $a^{\\prime}=a r$, $b^{\\prime}=b r$, and $c^{\\prime}=c r$ for a complex number $r$ with $|r|=1$, as shown below.\n\n\n\nWe have\n\n$$\n0=a+c r+b+a r+c+b r=(a+b+c)(1+r) \\text {. }\n$$\n\nIf $r=-1$, then the hexagon is centrally symmetric and its diagonals intersect at its center of symmetry. Otherwise\n\n$$\na+b+c=0\n$$\n\nTherefore\n\n$$\nA=0, \\quad B=a, \\quad C=a+c r, \\quad D=c(r-1), \\quad E=-b r-c, \\quad F=-b r\n$$\n\nNow consider a point $W$ on $A D$ given by the complex number $c(r-1) \\lambda$, where $\\lambda$ is a real number with $0<\\lambda<1$. Since $D \\neq A$, we have $r \\neq 1$, so we can define $s=1 /(r-1)$. From $r \\bar{r}=|r|^{2}=1$ we get\n\n$$\n1+s=\\frac{r}{r-1}=\\frac{r}{r-r \\bar{r}}=\\frac{1}{1-\\bar{r}}=-\\bar{s}\n$$\n\nNow,\n\n$$\n\\begin{aligned}\nW \\text { is on } B E & \\Longleftrightarrow c(r-1) \\lambda-a\\|a-(-b r-c)=b(r-1) \\Longleftrightarrow c \\lambda-a s\\| b \\\\\n& \\Longleftrightarrow-a \\lambda-b \\lambda-a s\\|b \\Longleftrightarrow a(\\lambda+s)\\| b .\n\\end{aligned}\n$$\n\nOne easily checks that $r \\neq \\pm 1$ implies that $\\lambda+s \\neq 0$ since $s$ is not real. On the other hand,\n\n$$\n\\begin{aligned}\nW \\text { on } C F & \\Longleftrightarrow c(r-1) \\lambda+b r\\|-b r-(a+c r)=a(r-1) \\Longleftrightarrow c \\lambda+b(1+s)\\| a \\\\\n& \\Longleftrightarrow-a \\lambda-b \\lambda-b \\bar{s}\\|a \\Longleftrightarrow b(\\lambda+\\bar{s})\\| a \\Longleftrightarrow b \\| a(\\lambda+s),\n\\end{aligned}\n$$\n\nwhere in the last step we use that $(\\lambda+s)(\\lambda+\\bar{s})=|\\lambda+s|^{2} \\in \\mathbb{R}_{>0}$. We conclude that $A D \\cap B E=$ $C F \\cap B E$, and the desired result follows.']",,True,,, 2074,Geometry,,Let the excircle of the triangle $A B C$ lying opposite to $A$ touch its side $B C$ at the point $A_{1}$. Define the points $B_{1}$ and $C_{1}$ analogously. Suppose that the circumcentre of the triangle $A_{1} B_{1} C_{1}$ lies on the circumcircle of the triangle $A B C$. Prove that the triangle $A B C$ is right-angled.,"['Denote the circumcircles of the triangles $A B C$ and $A_{1} B_{1} C_{1}$ by $\\Omega$ and $\\Gamma$, respectively. Denote the midpoint of the $\\operatorname{arc} C B$ of $\\Omega$ containing $A$ by $A_{0}$, and define $B_{0}$ as well as $C_{0}$ analogously. By our hypothesis the centre $Q$ of $\\Gamma$ lies on $\\Omega$.\n\nLemma. One has $A_{0} B_{1}=A_{0} C_{1}$. Moreover, the points $A, A_{0}, B_{1}$, and $C_{1}$ are concyclic. Finally, the points $A$ and $A_{0}$ lie on the same side of $B_{1} C_{1}$. Similar statements hold for $B$ and $C$.\n\nProof. Let us consider the case $A=A_{0}$ first. Then the triangle $A B C$ is isosceles at $A$, which implies $A B_{1}=A C_{1}$ while the remaining assertions of the Lemma are obvious. So let us suppose $A \\neq A_{0}$ from now on.\n\nBy the definition of $A_{0}$, we have $A_{0} B=A_{0} C$. It is also well known and easy to show that $B C_{1}=$ $C B_{1}$. Next, we have $\\angle C_{1} B A_{0}=\\angle A B A_{0}=\\angle A C A_{0}=\\angle B_{1} C A_{0}$. Hence the triangles $A_{0} B C_{1}$ and $A_{0} C B_{1}$ are congruent. This implies $A_{0} C_{1}=A_{0} B_{1}$, establishing the first part of the Lemma. It also follows that $\\angle A_{0} C_{1} A=\\angle A_{0} B_{1} A$, as these are exterior angles at the corresponding vertices $C_{1}$ and $B_{1}$ of the congruent triangles $A_{0} B C_{1}$ and $A_{0} C B_{1}$. For that reason the points $A, A_{0}, B_{1}$, and $C_{1}$ are indeed the vertices of some cyclic quadrilateral two opposite sides of which are $A A_{0}$ and $B_{1} C_{1}$.\n\nNow we turn to the solution. Evidently the points $A_{1}, B_{1}$, and $C_{1}$ lie interior to some semicircle arc of $\\Gamma$, so the triangle $A_{1} B_{1} C_{1}$ is obtuse-angled. Without loss of generality, we will assume that its angle at $B_{1}$ is obtuse. Thus $Q$ and $B_{1}$ lie on different sides of $A_{1} C_{1}$; obviously, the same holds for the points $B$ and $B_{1}$. So, the points $Q$ and $B$ are on the same side of $A_{1} C_{1}$.\n\nNotice that the perpendicular bisector of $A_{1} C_{1}$ intersects $\\Omega$ at two points lying on different sides of $A_{1} C_{1}$. By the first statement from the Lemma, both points $B_{0}$ and $Q$ are among these points of intersection; since they share the same side of $A_{1} C_{1}$, they coincide (see Figure 1).\n\n\n\nFigure 1\n\n\n\nNow, by the first part of the Lemma again, the lines $Q A_{0}$ and $Q C_{0}$ are the perpendicular bisectors of $B_{1} C_{1}$ and $A_{1} B_{1}$, respectively. Thus\n\n$$\n\\angle C_{1} B_{0} A_{1}=\\angle C_{1} B_{0} B_{1}+\\angle B_{1} B_{0} A_{1}=2 \\angle A_{0} B_{0} B_{1}+2 \\angle B_{1} B_{0} C_{0}=2 \\angle A_{0} B_{0} C_{0}=180^{\\circ}-\\angle A B C,\n$$\n\nrecalling that $A_{0}$ and $C_{0}$ are the midpoints of the $\\operatorname{arcs} C B$ and $B A$, respectively.\n\nOn the other hand, by the second part of the Lemma we have\n\n$$\n\\angle C_{1} B_{0} A_{1}=\\angle C_{1} B A_{1}=\\angle A B C .\n$$\n\nFrom the last two equalities, we get $\\angle A B C=90^{\\circ}$, whereby the problem is solved.', 'Let $Q$ again denote the centre of the circumcircle of the triangle $A_{1} B_{1} C_{1}$, that lies on the circumcircle $\\Omega$ of the triangle $A B C$. We first consider the case where $Q$ coincides with one of the vertices of $A B C$, say $Q=B$. Then $B C_{1}=B A_{1}$ and consequently the triangle $A B C$ is isosceles at $B$. Moreover we have $B C_{1}=B_{1} C$ in any triangle, and hence $B B_{1}=B C_{1}=B_{1} C$; similarly, $B B_{1}=B_{1} A$. It follows that $B_{1}$ is the centre of $\\Omega$ and that the triangle $A B C$ has a right angle at $B$.\n\nSo from now on we may suppose $Q \\notin\\{A, B, C\\}$. We start with the following well known fact. Lemma. Let $X Y Z$ and $X^{\\prime} Y^{\\prime} Z^{\\prime}$ be two triangles with $X Y=X^{\\prime} Y^{\\prime}$ and $Y Z=Y^{\\prime} Z^{\\prime}$.\n\n(i) If $X Z \\neq X^{\\prime} Z^{\\prime}$ and $\\angle Y Z X=\\angle Y^{\\prime} Z^{\\prime} X^{\\prime}$, then $\\angle Z X Y+\\angle Z^{\\prime} X^{\\prime} Y^{\\prime}=180^{\\circ}$.\n\n(ii) If $\\angle Y Z X+\\angle X^{\\prime} Z^{\\prime} Y^{\\prime}=180^{\\circ}$, then $\\angle Z X Y=\\angle Y^{\\prime} X^{\\prime} Z^{\\prime}$.\n\nProof. For both parts, we may move the triangle $X Y Z$ through the plane until $Y=Y^{\\prime}$ and $Z=Z^{\\prime}$. Possibly after reflecting one of the two triangles about $Y Z$, we may also suppose that $X$ and $X^{\\prime}$ lie on the same side of $Y Z$ if we are in case $(i)$ and on different sides if we are in case (ii). In both cases, the points $X, Z$, and $X^{\\prime}$ are collinear due to the angle condition (see Fig. 2). Moreover we have $X \\neq X^{\\prime}$, because in case $(i)$ we assumed $X Z \\neq X^{\\prime} Z^{\\prime}$ and in case $(i i)$ these points even lie on different sides of $Y Z$. Thus the triangle $X X^{\\prime} Y$ is isosceles at $Y$. The claim now follows by considering the equal angles at its base.\n\n\n\nFigure $2(i)$\n\n\n\nFigure $2($ ii $)$\n\nRelabeling the vertices of the triangle $A B C$ if necessary we may suppose that $Q$ lies in the interior of the arc $A B$ of $\\Omega$ not containing $C$. We will sometimes use tacitly that the six triangles $Q B A_{1}, Q A_{1} C, Q C B_{1}, Q B_{1} A, Q C_{1} A$, and $Q B C_{1}$ have the same orientation.\n\nAs $Q$ cannot be the circumcentre of the triangle $A B C$, it is impossible that $Q A=Q B=Q C$ and thus we may also suppose that $Q C \\neq Q B$. Now the above Lemma $(i)$ is applicable to the triangles $Q B_{1} C$ and $Q C_{1} B$, since $Q B_{1}=Q C_{1}$ and $B_{1} C=C_{1} B$, while $\\angle B_{1} C Q=\\angle C_{1} B Q$ holds as both angles appear over the same side of the chord $Q A$ in $\\Omega$ (see Fig. 3). So we get\n\n$$\n\\angle C Q B_{1}+\\angle B Q C_{1}=180^{\\circ}\n\\tag{1}\n$$\n\n\n\nWe claim that $Q C=Q A$. To see this, let us assume for the sake of a contradiction that $Q C \\neq Q A$. Then arguing similarly as before but now with the triangles $Q A_{1} C$ and $Q C_{1} A$ we get\n\n$$\n\\angle A_{1} Q C+\\angle C_{1} Q A=180^{\\circ} .\n$$\n\nAdding this equation to (1), we get $\\angle A_{1} Q B_{1}+\\angle B Q A=360^{\\circ}$, which is absurd as both summands lie in the interval $\\left(0^{\\circ}, 180^{\\circ}\\right)$.\n\nThis proves $Q C=Q A$; so the triangles $Q A_{1} C$ and $Q C_{1} A$ are congruent their sides being equal, which in turn yields\n\n$$\n\\angle A_{1} Q C=\\angle C_{1} Q A .\n\\tag{2}\n$$\n\nFinally our Lemma ( $i i$ i) is applicable to the triangles $Q A_{1} B$ and $Q B_{1} A$. Indeed we have $Q A_{1}=Q B_{1}$ and $A_{1} B=B_{1} A$ as usual, and the angle condition $\\angle A_{1} B Q+\\angle Q A B_{1}=180^{\\circ}$ holds as $A$ and $B$ lie on different sides of the chord $Q C$ in $\\Omega$. Consequently we have\n\n$$\n\\angle B Q A_{1}=\\angle B_{1} Q A .\n\\tag{3}\n$$\n\nFrom (1) and (3) we get\n\n$$\n\\left(\\angle B_{1} Q C+\\angle B_{1} Q A\\right)+\\left(\\angle C_{1} Q B-\\angle B Q A_{1}\\right)=180^{\\circ},\n$$\n\ni.e. $\\angle C Q A+\\angle A_{1} Q C_{1}=180^{\\circ}$. In light of (2) this may be rewritten as $2 \\angle C Q A=180^{\\circ}$ and as $Q$ lies on $\\Omega$ this implies that the triangle $A B C$ has a right angle at $B$.\n\n\n\nFigure 3']",,True,,, 2075,Number Theory,,"Let $\mathbb{Z}_{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $$ m^{2}+f(n) \mid m f(m)+n $$ for all positive integers $m$ and $n$.","['Setting $m=n=2$ tells us that $4+f(2) \\mid 2 f(2)+2$. Since $2 f(2)+2<2(4+f(2))$, we must have $2 f(2)+2=4+f(2)$, so $f(2)=2$. Plugging in $m=2$ then tells us that $4+f(n) \\mid 4+n$, which implies that $f(n) \\leqslant n$ for all $n$.\n\nSetting $m=n$ gives $n^{2}+f(n) \\mid n f(n)+n$, so $n f(n)+n \\geqslant n^{2}+f(n)$ which we rewrite as $(n-1)(f(n)-n) \\geqslant 0$. Therefore $f(n) \\geqslant n$ for all $n \\geqslant 2$. This is trivially true for $n=1$ also.\n\nIt follows that $f(n)=n$ for all $n$. This function obviously satisfies the desired property.', 'Setting $m=f(n)$ we get $f(n)(f(n)+1) \\mid f(n) f(f(n))+n$. This implies that $f(n) \\mid n$ for all $n$.\n\nNow let $m$ be any positive integer, and let $p>2 m^{2}$ be a prime number. Note that $p>m f(m)$ also. Plugging in $n=p-m f(m)$ we learn that $m^{2}+f(n)$ divides $p$. Since $m^{2}+f(n)$ cannot equal 1, it must equal $p$. Therefore $p-m^{2}=f(n) \\mid n=p-m f(m)$. But $p-m f(m)|f(n)-n|$. It follows that $f$ is the identity function.']",['$f(n)=n$'],False,,Expression, 2076,Number Theory,,"Prove that for any pair of positive integers $k$ and $n$ there exist $k$ positive integers $m_{1}, m_{2}, \ldots, m_{k}$ such that $$ 1+\frac{2^{k}-1}{n}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{k}}\right) $$","['We proceed by induction on $k$. For $k=1$ the statement is trivial. Assuming we have proved it for $k=j-1$, we now prove it for $k=j$.\n\nCase 1. $n=2 t-1$ for some positive integer $t$.\n\nObserve that\n\n$$\n1+\\frac{2^{j}-1}{2 t-1}=\\frac{2\\left(t+2^{j-1}-1\\right)}{2 t} \\cdot \\frac{2 t}{2 t-1}=\\left(1+\\frac{2^{j-1}-1}{t}\\right)\\left(1+\\frac{1}{2 t-1}\\right) .\n$$\n\nBy the induction hypothesis we can find $m_{1}, \\ldots, m_{j-1}$ such that\n\n$$\n1+\\frac{2^{j-1}-1}{t}=\\left(1+\\frac{1}{m_{1}}\\right)\\left(1+\\frac{1}{m_{2}}\\right) \\cdots\\left(1+\\frac{1}{m_{j-1}}\\right)\n$$\n\nso setting $m_{j}=2 t-1$ gives the desired expression.\n\nCase 2. $n=2 t$ for some positive integer $t$.\n\nNow we have\n\n$$\n1+\\frac{2^{j}-1}{2 t}=\\frac{2 t+2^{j}-1}{2 t+2^{j}-2} \\cdot \\frac{2 t+2^{j}-2}{2 t}=\\left(1+\\frac{1}{2 t+2^{j}-2}\\right)\\left(1+\\frac{2^{j-1}-1}{t}\\right),\n$$\n\nnoting that $2 t+2^{j}-2>0$. Again, we use that\n\n$$\n1+\\frac{2^{j-1}-1}{t}=\\left(1+\\frac{1}{m_{1}}\\right)\\left(1+\\frac{1}{m_{2}}\\right) \\cdots\\left(1+\\frac{1}{m_{j-1}}\\right)\n$$\n\nSetting $m_{j}=2 t+2^{j}-2$ then gives the desired expression.', 'Consider the base 2 expansions of the residues of $n-1$ and $-n$ modulo $2^{k}$ :\n\n$$\n\\begin{aligned}\n& n-1 \\equiv 2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{r}}\\left(\\bmod 2^{k}\\right) \\quad \\text { where } 0 \\leqslant a_{1}q_{n-1} \\text { and } q_{n}>q_{n+1}\\right\\}\n$$\n\nis infinite, since for each $n \\in S$ one has\n\n$$\np_{n}=\\max \\left\\{q_{n}, q_{n-1}\\right\\}=q_{n}=\\max \\left\\{q_{n}, q_{n+1}\\right\\}=p_{n+1}\n$$\n\nSuppose on the contrary that $S$ is finite. Since $q_{2}=7<13=q_{3}$ and $q_{3}=13>7=q_{4}$, the set $S$ is non-empty. Since it is finite, we can consider its largest element, say $m$.\n\nNote that it is impossible that $q_{m}>q_{m+1}>q_{m+2}>\\ldots$ because all these numbers are positive integers, so there exists a $k \\geqslant m$ such that $q_{k}q_{\\ell+1}$. By the minimality of $\\ell$ we have $q_{\\ell-1}k \\geqslant m$, this contradicts the maximality of $m$, and hence $S$ is indeed infinite.']",,True,,, 2078,Number Theory,,"Prove that there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \ldots$ and a positive integer $N$ such that for every integer $k>N$, the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square.","['Assume that $a_{1}, a_{2}, a_{3}, \\ldots$ is such a sequence. For each positive integer $k$, let $y_{k}=$ $\\overline{a_{k} a_{k-1} \\ldots a_{1}}$. By the assumption, for each $k>N$ there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$.\n\nI. For every $n$, let $5^{\\gamma_{n}}$ be the greatest power of 5 dividing $x_{n}$. Let us show first that $2 \\gamma_{n} \\geqslant n$ for every positive integer $n>N$.\n\nAssume, to the contrary, that there exists a positive integer $n>N$ such that $2 \\gamma_{n}N$.\n\nII. Consider now any integer $k>\\max \\{N / 2,2\\}$. Since $2 \\gamma_{2 k+1} \\geqslant 2 k+1$ and $2 \\gamma_{2 k+2} \\geqslant 2 k+2$, we have $\\gamma_{2 k+1} \\geqslant k+1$ and $\\gamma_{2 k+2} \\geqslant k+1$. So, from $y_{2 k+2}=a_{2 k+2} \\cdot 10^{2 k+1}+y_{2 k+1}$ we obtain $5^{2 k+2} \\mid y_{2 k+2}-y_{2 k+1}=a_{2 k+2} \\cdot 10^{2 k+1}$ and thus $5 \\mid a_{2 k+2}$, which implies $a_{2 k+2}=5$. Therefore,\n\n$$\n\\left(x_{2 k+2}-x_{2 k+1}\\right)\\left(x_{2 k+2}+x_{2 k+1}\\right)=x_{2 k+2}^{2}-x_{2 k+1}^{2}=y_{2 k+2}-y_{2 k+1}=5 \\cdot 10^{2 k+1}=2^{2 k+1} \\cdot 5^{2 k+2} .\n$$\n\nSetting $A_{k}=x_{2 k+2} / 5^{k+1}$ and $B_{k}=x_{2 k+1} / 5^{k+1}$, which are integers, we obtain\n\n$$\n\\left(A_{k}-B_{k}\\right)\\left(A_{k}+B_{k}\\right)=2^{2 k+1} .\n\\tag{1}\n$$\n\nBoth $A_{k}$ and $B_{k}$ are odd, since otherwise $y_{2 k+2}$ or $y_{2 k+1}$ would be a multiple of 10 which is false by $a_{1} \\neq 0$; so one of the numbers $A_{k}-B_{k}$ and $A_{k}+B_{k}$ is not divisible by 4 . Therefore (1) yields $A_{k}-B_{k}=2$ and $A_{k}+B_{k}=2^{2 k}$, hence $A_{k}=2^{2 k-1}+1$ and thus\n\n$$\nx_{2 k+2}=5^{k+1} A_{k}=10^{k+1} \\cdot 2^{k-2}+5^{k+1}>10^{k+1},\n$$\n\nsince $k \\geqslant 2$. This implies that $y_{2 k+2}>10^{2 k+2}$ which contradicts the fact that $y_{2 k+2}$ contains $2 k+2$ digits. The desired result follows.', ""Again, we assume that a sequence $a_{1}, a_{2}, a_{3}, \\ldots$ satisfies the problem conditions, introduce the numbers $x_{k}$ and $y_{k}$ as in the previous solution, and notice that\n\n$$\ny_{k+1}-y_{k}=\\left(x_{k+1}-x_{k}\\right)\\left(x_{k+1}+x_{k}\\right)=10^{k} a_{k+1}\n\\tag{2}\n$$\n\nfor all $k>N$. Consider any such $k$. Since $a_{1} \\neq 0$, the numbers $x_{k}$ and $x_{k+1}$ are not multiples of 10 , and therefore the numbers $p_{k}=x_{k+1}-x_{k}$ and $q_{k}=x_{k+1}+x_{k}$ cannot be simultaneously multiples of 20 , and hence one of them is not divisible either by 4 or by 5 . In view of (2), this means that the other one is divisible by either $5^{k}$ or by $2^{k-1}$. Notice also that $p_{k}$ and $q_{k}$ have the same parity, so both are even.\n\nOn the other hand, we have $x_{k+1}^{2}=x_{k}^{2}+10^{k} a_{k+1} \\geqslant x_{k}^{2}+10^{k}>2 x_{k}^{2}$, so $x_{k+1} / x_{k}>\\sqrt{2}$, which implies that\n\n$$\n1<\\frac{q_{k}}{p_{k}}=1+\\frac{2}{x_{k+1} / x_{k}-1}<1+\\frac{2}{\\sqrt{2}-1}<6\n\\tag{3}\n$$\n\nThus, if one of the numbers $p_{k}$ and $q_{k}$ is divisible by $5^{k}$, then we have\n\n$$\n10^{k+1}>10^{k} a_{k+1}=p_{k} q_{k} \\geqslant \\frac{\\left(5^{k}\\right)^{2}}{6}\n$$\n\nand hence $(5 / 2)^{k}<60$ which is false for sufficiently large $k$. So, assuming that $k$ is large, we get that $2^{k-1}$ divides one of the numbers $p_{k}$ and $q_{k}$. Hence\n\n$$\n\\left\\{p_{k}, q_{k}\\right\\}=\\left\\{2^{k-1} \\cdot 5^{r_{k}} b_{k}, 2 \\cdot 5^{k-r_{k}} c_{k}\\right\\} \\quad \\text { with nonnegative integers } b_{k}, c_{k}, r_{k} \\text { such that } b_{k} c_{k}=a_{k+1} \\text {. }\n$$\n\nMoreover, from (3) we get\n\n$$\n6>\\frac{2^{k-1} \\cdot 5^{r_{k}} b_{k}}{2 \\cdot 5^{k-r_{k}} c_{k}} \\geqslant \\frac{1}{36} \\cdot\\left(\\frac{2}{5}\\right)^{k} \\cdot 5^{2 r_{k}} \\quad \\text { and } \\quad 6>\\frac{2 \\cdot 5^{k-r_{k}} c_{k}}{2^{k-1} \\cdot 5^{r_{k}} b_{k}} \\geqslant \\frac{4}{9} \\cdot\\left(\\frac{5}{2}\\right)^{k} \\cdot 5^{-2 r_{k}}\n$$\n\nSo\n\n$$\n\\alpha k+c_{1}c_{1} .\n\\tag{4}\n$$\n\nConsequently, for $C=c_{2}-c_{1}+1-\\alpha>0$ we have\n\n$$\n(k+1)-r_{k+1} \\leqslant k-r_{k}+C\n\\tag{5}\n$$\n\nNext, we will use the following easy lemma.\n\nLemma. Let $s$ be a positive integer. Then $5^{s+2^{s}} \\equiv 5^{s}\\left(\\bmod 10^{s}\\right)$.\n\nProof. Euler's theorem gives $5^{2^{s}} \\equiv 1\\left(\\bmod 2^{s}\\right)$, so $5^{s+2^{s}}-5^{s}=5^{s}\\left(5^{2^{s}}-1\\right)$ is divisible by $2^{s}$ and $5^{s}$.\n\nNow, for every large $k$ we have\n\n$$\nx_{k+1}=\\frac{p_{k}+q_{k}}{2}=5^{r_{k}} \\cdot 2^{k-2} b_{k}+5^{k-r_{k}} c_{k} \\equiv 5^{k-r_{k}} c_{k} \\quad\\left(\\bmod 10^{r_{k}}\\right)\n\\tag{6}\n$$\n\nsince $r_{k} \\leqslant k-2$ by $(4)$; hence $y_{k+1} \\equiv 5^{2\\left(k-r_{k}\\right)} c_{k}^{2}\\left(\\bmod 10^{r_{k}}\\right)$. Let us consider some large integer $s$, and choose the minimal $k$ such that $2\\left(k-r_{k}\\right) \\geqslant s+2^{s}$; it exists by (4). Set $d=2\\left(k-r_{k}\\right)-\\left(s+2^{s}\\right)$. By (4) we have $2^{s}<2\\left(k-r_{k}\\right)<\\left(\\frac{2}{\\alpha}-2\\right) r_{k}-\\frac{2 c_{1}}{\\alpha}$; if $s$ is large this implies $r_{k}>s$, so (6) also holds modulo $10^{s}$. Then (6) and the lemma give\n\n$$\ny_{k+1} \\equiv 5^{2\\left(k-r_{k}\\right)} c_{k}^{2}=5^{s+2^{s}} \\cdot 5^{d} c_{k}^{2} \\equiv 5^{s} \\cdot 5^{d} c_{k}^{2} \\quad\\left(\\bmod 10^{s}\\right)\n\\tag{7}\n$$\n\nBy (5) and the minimality of $k$ we have $d \\leqslant 2 C$, so $5^{d} c_{k}^{2} \\leqslant 5^{2 C} \\cdot 81=D$. Using $5^{4}<10^{3}$ we obtain\n\n$$\n5^{s} \\cdot 5^{d} c_{k}^{2}<10^{3 s / 4} D<10^{s-1}\n$$\n\nfor sufficiently large $s$. This, together with (7), shows that the sth digit from the right in $y_{k+1}$, which is $a_{s}$, is zero. This contradicts the problem condition.""]",,True,,, 2079,Number Theory,,"Fix an integer $k \geqslant 2$. Two players, called Ana and Banana, play the following game of numbers: Initially, some integer $n \geqslant k$ gets written on the blackboard. Then they take moves in turn, with Ana beginning. A player making a move erases the number $m$ just written on the blackboard and replaces it by some number $m^{\prime}$ with $k \leqslant m^{\prime}m \\geqslant k$ and $\\operatorname{gcd}(m, n)=1$, then $n$ itself is bad, for Ana has the following winning strategy in the game with initial number $n$ : She proceeds by first playing $m$ and then using Banana's strategy for the game with starting number $m$.\n\nOtherwise, if some integer $n \\geqslant k$ has the property that every integer $m$ with $n>m \\geqslant k$ and $\\operatorname{gcd}(m, n)=1$ is bad, then $n$ is good. Indeed, if Ana can make a first move at all in the game with initial number $n$, then she leaves it in a position where the first player has a winning strategy, so that Banana can defeat her.\n\nIn particular, this implies that any two good numbers have a non-trivial common divisor. Also, $k$ itself is good.\n\nFor brevity, we say that $n \\longrightarrow x$ is a move if $n$ and $x$ are two coprime integers with $n>x \\geqslant k$.\n\nClaim 1. If $n$ is good and $n^{\\prime}$ is a multiple of $n$, then $n^{\\prime}$ is also good.\n\nProof. If $n^{\\prime}$ were bad, there would have to be some move $n^{\\prime} \\longrightarrow x$, where $x$ is good. As $n^{\\prime}$ is a multiple of $n$ this implies that the two good numbers $n$ and $x$ are coprime, which is absurd.\n\nClaim 2. If $r$ and $s$ denote two positive integers for which $r s \\geqslant k$ is bad, then $r^{2} s$ is also bad. Proof. Since $r s$ is bad, there is a move $r s \\longrightarrow x$ for some good $x$. Evidently $x$ is coprime to $r^{2} s$ as well, and hence the move $r^{2} s \\longrightarrow x$ shows that $r^{2} s$ is indeed bad.\n\nClaim 3. If $p>k$ is prime and $n \\geqslant k$ is bad, then $n p$ is also bad.\n\nProof. Otherwise we choose a counterexample with $n$ being as small as possible. In particular, $n p$ is good. Since $n$ is bad, there is a move $n \\longrightarrow x$ for some good $x$. Now $n p \\longrightarrow x$ cannot be a valid move, which tells us that $x$ has to be divisible by $p$. So we can write $x=p^{r} y$, where $r$ and $y$ denote some positive integers, the latter of which is not divisible by $p$.\n\nNote that $y=1$ is impossible, for then we would have $x=p^{r}$ and the move $x \\longrightarrow k$ would establish that $x$ is bad. In view of this, there is a least power $y^{\\alpha}$ of $y$ that is at least as large as $k$. Since the numbers $n p$ and $y^{\\alpha}$ are coprime and the former is good, the latter has to be bad. Moreover, the minimality of $\\alpha$ implies $y^{\\alpha}k$, but now we get the same contradiction using Claim 3 instead of Claim 2. Thereby the problem is solved."", 'We use the same analysis of the game of numbers as in the first five paragraphs of the first solution. Let us call a prime number $p$ small in case $p \\leqslant k$ and big otherwise. We again call two integers similar if their sets of small prime factors coincide.\n\nClaim 4. For each integer $b \\geqslant k$ having some small prime factor, there exists an integer $x$ similar to it with $b \\geqslant x \\geqslant k$ and having no big prime factors.\n\nProof. Unless $b$ has a big prime factor we may simply choose $x=b$. Now let $p$ and $q$ denote a small and a big prime factor of $b$, respectively. Let $a$ be the product of all small prime factors of $b$. Further define $n$ to be the least non-negative integer for which the number $x=p^{n} a$ is at least as large as $k$. It suffices to show that $b>x$. This is clear in case $n=0$, so let us assume $n>0$ from now on. Then we have $x

0}$.","['I. We start by verifying that these functions do indeed satisfy (1). This is clear for all constant functions. Now consider any triple $(x, a, b) \\in \\mathbb{Q} \\times \\mathbb{Z} \\times \\mathbb{Z}_{>0}$ and set\n\n$$\nq=\\left\\lfloor\\frac{x+a}{b}\\right\\rfloor .\n$$\n\nThis means that $q$ is an integer and $b q \\leqslant x+a0}$. According to the behaviour of the restriction of $f$ to the integers we distinguish two cases.\n\nCase 1: There is some $m \\in \\mathbb{Z}$ such that $f(m) \\neq m$.\n\nWrite $f(m)=C$ and let $\\eta \\in\\{-1,+1\\}$ and $b$ denote the sign and absolute value of $f(m)-m$, respectively. Given any integer $r$, we may plug the triple $(m, r b-C, b)$ into (1), thus getting $f(r)=f(r-\\eta)$. Starting with $m$ and using induction in both directions, we deduce from this that the equation $f(r)=C$ holds for all integers $r$. Now any rational number $y$ can be written in the form $y=\\frac{p}{q}$ with $(p, q) \\in \\mathbb{Z} \\times \\mathbb{Z}_{>0}$, and substituting $(C-p, p-C, q)$ into (1) we get $f(y)=f(0)=C$. Thus $f$ is the constant function whose value is always $C$.\n\nCase 2: One has $f(m)=m$ for all integers $m$.\n\nNote that now the special case $b=1$ of (1) takes a particularly simple form, namely\n\n$$\nf(x)+a=f(x+a) \\quad \\text { for all }(x, a) \\in \\mathbb{Q} \\times \\mathbb{Z}\n\\tag{2}\n$$\n\nDefining $f\\left(\\frac{1}{2}\\right)=\\omega$ we proceed in three steps.\n\nStep A. We show that $\\omega \\in\\{0,1\\}$.\n\nIf $\\omega \\leqslant 0$, we may plug $\\left(\\frac{1}{2},-\\omega, 1-2 \\omega\\right)$ into (1), obtaining $0=f(0)=f\\left(\\frac{1}{2}\\right)=\\omega$. In the contrary case $\\omega \\geqslant 1$ we argue similarly using the triple $\\left(\\frac{1}{2}, \\omega-1,2 \\omega-1\\right)$.\n\nStep B. We show that $f(x)=\\omega$ for all rational numbers $x$ with $0b \\geqslant k+\\omega$, which is absurd. Similarly, $m \\geqslant r$ leads to $r a-m b0} \\text { and } \\frac{p}{q} \\in S \\text {, then } \\frac{p}{q+1} \\in S \\text { holds as well. }\n\\tag{7}\n$$\n\nTo see this we just plug $\\left(\\frac{p}{q}, p, q+1\\right)$ into (1), thus getting $f\\left(\\frac{p}{q+1}\\right)=f\\left(\\frac{p}{q}\\right)=0$.\n\nFrom this we get that\n\n$$\n\\text { if } x, y \\in \\mathbb{Q}, x>y>0 \\text {, and } x \\in S \\text {, then } y \\in S \\text {. }\n\\tag{8}\n$$\n\nIndeed, if we write $x=\\frac{p}{q}$ and $y=\\frac{r}{s}$ with $p, q, r, s \\in \\mathbb{Z}_{>0}$, then $p s>q r$ and (7) tells us\n\n$$\n0=f\\left(\\frac{p}{q}\\right)=f\\left(\\frac{p r}{q r}\\right)=f\\left(\\frac{p r}{q r+1}\\right)=\\ldots=f\\left(\\frac{p r}{p s}\\right)=f\\left(\\frac{r}{s}\\right)\n$$\n\nEssentially the same argument also establishes that\n\n$$\n\\text { if } x, y \\in \\mathbb{Q}, x2 m$. Such a pair is necessarily a child of $(c, d-c)$, and thus a descendant of some pair $\\left(c, d^{\\prime}\\right)$ with $m\\{(b+k a) \\nu\\}=\\left\\{\\left(b+k_{0} a\\right) \\nu\\right\\}+\\left(k-k_{0}\\right)\\{a \\nu\\}$ for all $k>k_{0}$, which is absurd.\n\nSimilarly, one can prove that $S_{m}$ contains finitely many pairs $(c, d)$ with $c>2 m$, thus finitely many elements at all.\n\nWe are now prepared for proving the following crucial lemma.\n\nLemma. Consider any pair $(a, b)$ with $f(a, b) \\neq m$. Then the number $g(a, b)$ of its $m$-excellent descendants is equal to the number $h(a, b)$ of ways to represent the number $t=m-f(a, b)$ as $t=k a+\\ell b$ with $k$ and $\\ell$ being some nonnegative integers.\n\nProof. We proceed by induction on the number $N$ of descendants of $(a, b)$ in $S_{m}$. If $N=0$ then clearly $g(a, b)=0$. Assume that $h(a, b)>0$; without loss of generality, we have $a \\leqslant b$. Then, clearly, $m-f(a, b) \\geqslant a$, so $f(a, b+a) \\leqslant f(a, b)+a \\leqslant m$ and $a \\leqslant m$, hence $(a, b+a) \\in S_{m}$ which is impossible. Thus in the base case we have $g(a, b)=h(a, b)=0$, as desired.\n\nNow let $N>0$. Assume that $f(a+b, b)=f(a, b)+b$ and $f(a, b+a)=f(a, b)$ (the other case is similar). If $f(a, b)+b \\neq m$, then by the induction hypothesis we have\n\n$$\ng(a, b)=g(a+b, b)+g(a, b+a)=h(a+b, b)+h(a, b+a)\n$$\n\nNotice that both pairs $(a+b, b)$ and $(a, b+a)$ are descendants of $(a, b)$ and thus each of them has strictly less descendants in $S_{m}$ than $(a, b)$ does.\n\nNext, each one of the $h(a+b, b)$ representations of $m-f(a+b, b)=m-b-f(a, b)$ as the sum $k^{\\prime}(a+b)+\\ell^{\\prime} b$ provides the representation $m-f(a, b)=k a+\\ell b$ with $k=k^{\\prime}1} d=\\sum_{d \\mid m} d\n$$\n\nas required.']",,True,,, 2082,Algebra,,"Find the largest possible integer $k$, such that the following statement is true: Let 2009 arbitrary non-degenerated triangles be given. In every triangle the three sides are colored, such that one is blue, one is red and one is white. Now, for every color separately, let us sort the lengths of the sides. We obtain $$ \begin{aligned} b_{1} \leq b_{2} \leq \ldots \leq b_{2009} & \text { the lengths of the blue sides } \\ r_{1} \leq r_{2} \leq \ldots \leq r_{2009} & \text { the lengths of the red sides, } \\ \text { and } \quad & w_{1} \leq w_{2} \leq \ldots \leq w_{2009} \quad \text { the lengths of the white sides. } \end{aligned} $$ Then there exist $k$ indices $j$ such that we can form a non-degenerated triangle with side lengths $b_{j}, r_{j}, w_{j}$.","['We will prove that the largest possible number $k$ of indices satisfying the given condition is one.\n\nFirstly we prove that $b_{2009}, r_{2009}, w_{2009}$ are always lengths of the sides of a triangle. Without loss of generality we may assume that $w_{2009} \\geq r_{2009} \\geq b_{2009}$. We show that the inequality $b_{2009}+r_{2009}>w_{2009}$ holds. Evidently, there exists a triangle with side lengths $w, b, r$ for the white, blue and red side, respectively, such that $w_{2009}=w$. By the conditions of the problem we have $b+r>w, b_{2009} \\geq b$ and $r_{2009} \\geq r$. From these inequalities it follows\n\n$$\nb_{2009}+r_{2009} \\geq b+r>w=w_{2009}\n$$\n\nSecondly we will describe a sequence of triangles for which $w_{j}, b_{j}, r_{j}$ with $j<2009$ are not the lengths of the sides of a triangle. Let us define the sequence $\\Delta_{j}, j=1,2, \\ldots, 2009$, of triangles, where $\\Delta_{j}$ has\n\na blue side of length $2 j$,\n\na red side of length $j$ for all $j \\leq 2008$ and 4018 for $j=2009$,\n\nand a white side of length $j+1$ for all $j \\leq 2007,4018$ for $j=2008$ and 1 for $j=2009$. Since\n\n$$\n\\begin{aligned}\n(j+1)+j>2 j & \\geq j+1>j, & & \\text { if } \\quad j \\leq 2007 \\\\\n2 j+j>4018 & >2 j \\quad>j, & & \\text { if } j=2008, \\\\\n4018+1>2 j & =4018>1, & & \\text { if } j=2009\n\\end{aligned}\n$$\n\nsuch a sequence of triangles exists. Moreover, $w_{j}=j, r_{j}=j$ and $b_{j}=2 j$ for $1 \\leq j \\leq 2008$. Then\n\n$$\nw_{j}+r_{j}=j+j=2 j=b_{j},\n$$\n\ni.e., $b_{j}, r_{j}$ and $w_{j}$ are not the lengths of the sides of a triangle for $1 \\leq j \\leq 2008$.']",['1'],False,,Numerical, 2083,Algebra,,"Let $a, b, c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that $$ \frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leq \frac{3}{16} . $$","['For positive real numbers $x, y, z$, from the arithmetic-geometric-mean inequality,\n\n$$\n2 x+y+z=(x+y)+(x+z) \\geq 2 \\sqrt{(x+y)(x+z)}\n$$\n\nwe obtain\n\n$$\n\\frac{1}{(2 x+y+z)^{2}} \\leq \\frac{1}{4(x+y)(x+z)}\n$$\n\nApplying this to the left-hand side terms of the inequality to prove, we get\n\n$$\n\\begin{aligned}\n\\frac{1}{(2 a+b+c)^{2}} & +\\frac{1}{(2 b+c+a)^{2}}+\\frac{1}{(2 c+a+b)^{2}} \\\\\n& \\leq \\frac{1}{4(a+b)(a+c)}+\\frac{1}{4(b+c)(b+a)}+\\frac{1}{4(c+a)(c+b)} \\\\\n& =\\frac{(b+c)+(c+a)+(a+b)}{4(a+b)(b+c)(c+a)}=\\frac{a+b+c}{2(a+b)(b+c)(c+a)} .\n\\end{aligned}\n\\tag{1}\n$$\n\nA second application of the inequality of the arithmetic-geometric mean yields\n\n$$\na^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b \\geq 6 a b c\n$$\n\nor, equivalently,\n\n$$\n9(a+b)(b+c)(c+a) \\geq 8(a+b+c)(a b+b c+c a)\n\\tag{2}\n$$\n\nThe supposition $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=a+b+c$ can be written as\n\n$$\na b+b c+c a=a b c(a+b+c) .\n\\tag{3}\n$$\n\nApplying the arithmetic-geometric-mean inequality $x^{2} y^{2}+x^{2} z^{2} \\geq 2 x^{2} y z$ thrice, we get\n\n$$\na^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \\geq a^{2} b c+a b^{2} c+a b c^{2}\n$$\n\nwhich is equivalent to\n\n$$\n(a b+b c+c a)^{2} \\geq 3 a b c(a+b+c) .\n\\tag{4}\n$$\n\n\n\nCombining (1), (2), (3), and (4), we will finish the proof:\n\n$$\n\\begin{aligned}\n\\frac{a+b+c}{2(a+b)(b+c)(c+a)} & =\\frac{(a+b+c)(a b+b c+c a)}{2(a+b)(b+c)(c+a)} \\cdot \\frac{a b+b c+c a}{a b c(a+b+c)} \\cdot \\frac{a b c(a+b+c)}{(a b+b c+c a)^{2}} \\\\\n& \\leq \\frac{9}{2 \\cdot 8} \\cdot 1 \\cdot \\frac{1}{3}=\\frac{3}{16}\n\\end{aligned}\n$$', ""Equivalently, we prove the homogenized inequality\n\n$$\n\\frac{(a+b+c)^{2}}{(2 a+b+c)^{2}}+\\frac{(a+b+c)^{2}}{(a+2 b+c)^{2}}+\\frac{(a+b+c)^{2}}{(a+b+2 c)^{2}} \\leq \\frac{3}{16}(a+b+c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\n$$\n\nfor all positive real numbers $a, b, c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a, b, c>0$, fulfilling this condition, the inequality\n\n$$\n\\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}}+\\frac{1}{(1+c)^{2}} \\leq \\frac{3}{16}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\n\\tag{5}\n$$\n\nApplying JENSEN's inequality to the function $f(x)=\\frac{x}{(1+x)^{2}}$, which is concave for $0 \\leq x \\leq 2$ and increasing for $0 \\leq x \\leq 1$, we obtain\n\n$$\n\\alpha \\frac{a}{(1+a)^{2}}+\\beta \\frac{b}{(1+b)^{2}}+\\gamma \\frac{c}{(1+c)^{2}} \\leq(\\alpha+\\beta+\\gamma) \\frac{A}{(1+A)^{2}}, \\quad \\text { where } \\quad A=\\frac{\\alpha a+\\beta b+\\gamma c}{\\alpha+\\beta+\\gamma}\n$$\n\nChoosing $\\alpha=\\frac{1}{a}, \\beta=\\frac{1}{b}$, and $\\gamma=\\frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality\n\n$$\nA=\\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}} \\leq \\frac{a+b+c}{3}=\\frac{1}{3}<1\n$$\n\nFinally we prove (5):\n\n$$\n\\begin{aligned}\n\\frac{1}{(1+a)^{2}}+\\frac{1}{(1+b)^{2}}+\\frac{1}{(1+c)^{2}} & \\leq\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\frac{A}{(1+A)^{2}} \\\\\n& \\leq\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\frac{\\frac{1}{3}}{\\left(1+\\frac{1}{3}\\right)^{2}}=\\frac{3}{16}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) .\n\\end{aligned}\n$$""]",,True,,, 2084,Algebra,,"Determine all functions $f$ from the set of positive integers into the set of positive integers such that for all $x$ and $y$ there exists a non degenerated triangle with sides of lengths $$ x, \quad f(y) \text { and } f(y+f(x)-1) . $$","['The identity function $f(x)=x$ is the only solution of the problem.\n\nIf $f(x)=x$ for all positive integers $x$, the given three lengths are $x, y=f(y)$ and $z=$ $f(y+f(x)-1)=x+y-1$. Because of $x \\geq 1, y \\geq 1$ we have $z \\geq \\max \\{x, y\\}>|x-y|$ and $z1$ we would conclude $f(y)=f(y+m)$ for all $y$ considering the triangle with the side lengths $1, f(y)$ and $f(y+m)$. Thus, $f$ would be $m$-periodic and, consequently, bounded. Let $B$ be a bound, $f(x) \\leq B$. If we choose $x>2 B$ we obtain the contradiction $x>2 B \\geq f(y)+f(y+f(x)-1)$.\n\nStep 2. For all positive integers $z$, we have $f(f(z))=z$.\n\nSetting $x=z$ and $y=1$ this follows immediately from Step 1 .\n\nStep 3. For all integers $z \\geq 1$, we have $f(z) \\leq z$.\n\nLet us show, that the contrary leads to a contradiction. Assume $w+1=f(z)>z$ for some $z$. From Step 1 we know that $w \\geq z \\geq 2$. Let $M=\\max \\{f(1), f(2), \\ldots, f(w)\\}$ be the largest value of $f$ for the first $w$ integers. First we show, that no positive integer $t$ exists with\n\n$$\nf(t)>\\frac{z-1}{w} \\cdot t+M,\n\\tag{1}\n$$\n\notherwise we decompose the smallest value $t$ as $t=w r+s$ where $r$ is an integer and $1 \\leq s \\leq w$. Because of the definition of $M$, we have $t>w$. Setting $x=z$ and $y=t-w$ we get from the triangle inequality\n\n$$\nz+f(t-w)>f((t-w)+f(z)-1)=f(t-w+w)=f(t)\n$$\n\nHence,\n\n$$\nf(t-w) \\geq f(t)-(z-1)>\\frac{z-1}{w}(t-w)+M\n$$\n\na contradiction to the minimality of $t$.\n\nTherefore the inequality (1) fails for all $t \\geq 1$, we have proven\n\n$$\nf(t) \\leq \\frac{z-1}{w} \\cdot t+M\n\\tag{2}\n$$\n\ninstead.\n\n\n\nNow, using (2), we finish the proof of Step 3. Because of $z \\leq w$ we have $\\frac{z-1}{w}<1$ and we can choose an integer $t$ sufficiently large to fulfill the condition\n\n$$\n\\left(\\frac{z-1}{w}\\right)^{2} t+\\left(\\frac{z-1}{w}+1\\right) My f(x)+x . $$","['Assume that\n\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text { for all real } x, y\n\\tag{1}\n$$\n\nLet $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n\n$$\nf(y) \\leq y+a \\text { for all real } y\n\\tag{2}\n$$\n\nSetting $x=f(y)$ in (1) yields in view of (2)\n\n$$\na=f(0) \\leq y f(f(y))+f(y) \\leq y f(f(y))+y+a .\n$$\n\nThis implies $0 \\leq y(f(f(y))+1)$ and thus\n\n$$\nf(f(y)) \\geq-1 \\text { for all } y>0 \\text {. }\n\\tag{3}\n$$\n\nFrom (2) and (3) we obtain $-1 \\leq f(f(y)) \\leq f(y)+a$ for all $y>0$, so\n\n$$\nf(y) \\geq-a-1 \\text { for all } y>0\n\\tag{4}\n$$\n\nNow we show that\n\n$$\nf(x) \\leq 0 \\text { for all real } x \\text {. }\n\\tag{5}\n$$\n\nAssume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that\n\n$$\ny0\n$$\n\nand with (1) and (4) we obtain\n\n$$\ny f(x)+x \\geq f(x-f(y)) \\geq-a-1,\n$$\n\nwhence\n\n$$\ny \\geq \\frac{-a-x-1}{f(x)}\n$$\n\ncontrary to our choice of $y$. Thereby, we have established (5).\n\nSetting $x=0$ in (5) leads to $a=f(0) \\leq 0$ and (2) then yields\n\n$$\nf(x) \\leq x \\quad \\text { for all real } x\n\\tag{6}\n$$\n\nNow choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (1), (5) and\n\n\n\n(6) we obtain\n\n$$\nf(-1)=f(x-f(y)) \\leq y f(x)+x=y f(f(y)-1)+f(y)-1 \\leq y(f(y)-1)-1 \\leq-y-1,\n$$\n\ni.e. $y \\leq-f(-1)-1$, a contradiction to the choice of $y$.', 'Assume that\n\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text { for all real } x, y\n\\tag{7}\n$$\n\nLet $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n\n$$\nf(y) \\leq y+a \\text { for all real } y\n\\tag{8}\n$$\n\nNow we show that\n\n$$\nf(z) \\geq 0 \\text { for all } z \\geq 1 \\text {. }\n\\tag{9}\n$$\n\nLet $z \\geq 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives\n\n$$\nf(w)-z f(w+b) \\leq w+b \\quad \\text { for all real } w\n\\tag{10}\n$$\n\nApplying to $w, w+b, \\ldots, w+(n-1) b$, where $n=1,2, \\ldots$, leads to\n\n$$\n\\begin{gathered}\nf(w)-z^{n} f(w+n b)=(f(w)-z f(w+b))+z(f(w+b)-z f(w+2 b)) \\\\\n+\\cdots+z^{n-1}(f(w+(n-1) b)-z f(w+n b)) \\\\\n\\leq(w+b)+z(w+2 b)+\\cdots+z^{n-1}(w+n b) .\n\\end{gathered}\n$$\n\nFrom (8) we obtain\n\n$$\nf(w+n b) \\leq w+n b+a\n$$\n\nand, thus, we have for all positive integers $n$\n\n$$\nf(w) \\leq\\left(1+z+\\cdots+z^{n-1}+z^{n}\\right) w+\\left(1+2 z+\\cdots+n z^{n-1}+n z^{n}\\right) b+z^{n} a .\n\\tag{11}\n$$\n\nWith $w=0$ we get\n\n$$\na \\leq\\left(1+2 z+\\cdots+n z^{n-1}+n z^{n}\\right) b+a z^{n} .\n\\tag{12}\n$$\n\nIn view of the assumption $b<0$ we find some $n$ such that\n\n$$\na>(n b+a) z^{n}\n\\tag{13}\n$$\n\nbecause the right hand side tends to $-\\infty$ as $n \\rightarrow \\infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality\n\n$$\nf(1)>0 \\text {. }\n\\tag{14}\n$$\n\nIndeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to\n\n$$\nf(-1) \\leq-(n+1)+a\n$$\n\nwhich is false if $n$ is sufficiently large.\n\nTo complete the proof we set $t=\\min \\{-a,-2 / f(1)\\}$. Setting $x=1$ and $y=t$ in (7) gives\n\n$$\nf(1-f(t)) \\leq t f(1)+1 \\leq-2+1=-1 .\n\\tag{15}\n$$\n\n\n\nOn the other hand, by (8) and the choice of $t$ we have $f(t) \\leq t+a \\leq 0$ and hence $1-f(t) \\geq 1$. The inequality (9) yields\n\n$$\nf(1-f(t)) \\geq 0\n$$\n\nwhich contradicts (15).']",,True,,, 2087,Algebra,,"Suppose that $s_{1}, s_{2}, s_{3}, \ldots$ is a strictly increasing sequence of positive integers such that the subsequences $$ s_{s_{1}}, s_{s_{2}}, s_{s_{3}}, \ldots \quad \text { and } \quad s_{s_{1}+1}, s_{s_{2}+1}, s_{s_{3}+1}, \ldots $$ are both arithmetic progressions. Prove that $s_{1}, s_{2}, s_{3}, \ldots$ is itself an arithmetic progression.","['Let $D$ be the common difference of the progression $s_{s_{1}}, s_{s_{2}}, \\ldots$. Let for $n=$ $1,2, \\ldots$\n\n$$\nd_{n}=s_{n+1}-s_{n} \\text {. }\n$$\n\nWe have to prove that $d_{n}$ is constant. First we show that the numbers $d_{n}$ are bounded. Indeed, by supposition $d_{n} \\geq 1$ for all $n$. Thus, we have for all $n$\n\n$$\nd_{n}=s_{n+1}-s_{n} \\leq d_{s_{n}}+d_{s_{n}+1}+\\cdots+d_{s_{n+1}-1}=s_{s_{n+1}}-s_{s_{n}}=D .\n$$\n\nThe boundedness implies that there exist\n\n$$\nm=\\min \\left\\{d_{n}: n=1,2, \\ldots\\right\\} \\quad \\text { and } \\quad M=\\max \\left\\{d_{n}: n=1,2, \\ldots\\right\\} \\text {. }\n$$\n\nIt suffices to show that $m=M$. Assume that $mn$ if $d_{n}=n$, because in the case $s_{n}=n$ we would have $m=d_{n}=d_{s_{n}}=M$ in contradiction to the assumption $mn$. Consequently, there is a strictly increasing sequence $n_{1}, n_{2}, \\ldots$ such that\n\n$$\nd_{s_{n_{1}}}=M, \\quad d_{s_{n_{2}}}=m, \\quad d_{s_{n_{3}}}=M, \\quad d_{s_{n_{4}}}=m, \\quad \\ldots\n$$\n\nThe sequence $d_{s_{1}}, d_{s_{2}}, \\ldots$ is the sequence of pairwise differences of $s_{s_{1}+1}, s_{s_{2}+1}, \\ldots$ and $s_{s_{1}}, s_{s_{2}}, \\ldots$, hence also an arithmetic progression. Thus $m=M$', '$\\quad$ Let the integers $D$ and $E$ be the common differences of the progressions $s_{s_{1}}, s_{s_{2}}, \\ldots$ and $s_{s_{1}+1}, s_{s_{2}+1}, \\ldots$, respectively. Let briefly $A=s_{s_{1}}-D$ and $B=s_{s_{1}+1}-E$. Then, for all positive integers $n$,\n\n$$\ns_{s_{n}}=A+n D, \\quad s_{s_{n}+1}=B+n E .\n$$\n\nSince the sequence $s_{1}, s_{2}, \\ldots$ is strictly increasing, we have for all positive integers $n$\n\n$$\ns_{s_{n}}s_{n}+m$. Then $\\left.m(m+1) \\leq m\\left(s_{n+1}-s_{n}\\right) \\leq s_{s_{n+1}}-s_{s_{n}}=(A+(n+1) D)-(A+n D)\\right)=D=m(B-A)=m^{2}$, a contradiction. Hence $s_{1}, s_{2}, \\ldots$ is an arithmetic progression with common difference $m$.']",,True,,, 2088,Algebra,,"Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity $$ f(x f(x+y))=f(y f(x))+x^{2} . $$","['It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions.\n\nLet $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0) \\neq 0$. For any real $t$, substituting $(x, y)=\\left(0, \\frac{t}{f(0)}\\right)$ into the given functional equation, we obtain\n\n$$\nf(0)=f(t)\n\\tag{1}\n$$\n\ncontradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x, y)=(t, 0)$ and $(x, y)=(t,-t)$ into the given equation, we get\n\n$$\nf(t f(t))=f(0)+t^{2}=t^{2},\n$$\n\nand\n\n$$\nf(t f(0))=f(-t f(t))+t^{2}\n$$\n\nrespectively. Therefore, we conclude that\n\n$$\nf(t f(t))=t^{2}, \\quad f(-t f(t))=-t^{2}, \\quad \\text { for every real } t\n\\tag{2}\n$$\n\nConsequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(t f(t))=t^{2}$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$.\n\nWe next show that for any real number $s$,\n\n$$\nf(-s)=-f(s) \\text {. }\n\\tag{3}\n$$\n\nThis is clear if $f(s)=0$. Suppose now $f(s)<0$, then we can find a number $t$ for which $f(s)=-t^{2}$. As $t \\neq 0$ implies $f(t) \\neq 0$, we can also find number $a$ such that $a f(t)=s$. Substituting $(x, y)=(t, a)$ into the given equation, we get\n\n$$\nf(t f(t+a))=f(a f(t))+t^{2}=f(s)+t^{2}=0\n$$\n\nand therefore, $t f(t+a)=0$, which implies $t+a=0$, and hence $s=-t f(t)$. Consequently, $f(-s)=f(t f(t))=t^{2}=-\\left(-t^{2}\\right)=-f(s)$ holds in this case.\n\nFinally, suppose $f(s)>0$ holds. Then there exists a real number $t \\neq 0$ for which $f(s)=t^{2}$. Choose a number $a$ such that $t f(a)=s$. Substituting $(x, y)=(t, a-t)$ into the given equation, we get $f(s)=f(t f(a))=f((a-t) f(t))+t^{2}=f((a-t) f(t))+f(s)$. So we have $f((a-t) f(t))=0$, from which we conclude that $(a-t) f(t)=0$. Since $f(t) \\neq 0$, we get $a=t$ so that $s=t f(t)$ and thus we see $f(-s)=f(-t f(t))=-t^{2}=-f(s)$ holds in this case also. This observation finishes the proof of (3).\n\nBy substituting $(x, y)=(s, t),(x, y)=(t,-s-t)$ and $(x, y)=(-s-t, s)$ into the given equation,\n\n\n\nwe obtain\n\n$$\n\\begin{array}{r}\nf(s f(s+t)))=f(t f(s))+s^{2} \\\\\nf(t f(-s))=f((-s-t) f(t))+t^{2}\n\\end{array}\n$$\n\nand\n\n$$\nf((-s-t) f(-t))=f(s f(-s-t))+(s+t)^{2},\n$$\n\nrespectively. Using the fact that $f(-x)=-f(x)$ holds for all $x$ to rewrite the second and the third equation, and rearranging the terms, we obtain\n\n$$\n\\begin{aligned}\nf(t f(s))-f(s f(s+t)) & =-s^{2} \\\\\nf(t f(s))-f((s+t) f(t)) & =-t^{2} \\\\\nf((s+t) f(t))+f(s f(s+t)) & =(s+t)^{2} .\n\\end{aligned}\n$$\n\nAdding up these three equations now yields $2 f(t f(s))=2 t s$, and therefore, we conclude that $f(t f(s))=t s$ holds for every pair of real numbers $s, t$. By fixing $s$ so that $f(s)=1$, we obtain $f(x)=s x$. In view of the given equation, we see that $s= \\pm 1$. It is easy to check that both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation, so these are the desired solutions.', 'It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions.\n\nLet $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0) \\neq 0$. For any real $t$, substituting $(x, y)=\\left(0, \\frac{t}{f(0)}\\right)$ into the given functional equation, we obtain\n\n$$\nf(0)=f(t)\n\\tag{1}\n$$\n\ncontradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x, y)=(t, 0)$ and $(x, y)=(t,-t)$ into the given equation, we get\n\n$$\nf(t f(t))=f(0)+t^{2}=t^{2},\n$$\n\nand\n\n$$\nf(t f(0))=f(-t f(t))+t^{2}\n$$\n\nrespectively. Therefore, we conclude that\n\n$$\nf(t f(t))=t^{2}, \\quad f(-t f(t))=-t^{2}, \\quad \\text { for every real } t\n\\tag{2}\n$$\n\nConsequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(t f(t))=t^{2}$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$.\n\nWe next show that for any real number $s$,\n\n$$\nf(-s)=-f(s) \\text {. }\n\\tag{3}\n$$\n\nThis is clear if $f(s)=0$.\n\nNow we prove that $f$ is injective. For this purpose, let us assume that $f(r)=f(s)$ for some $r \\neq s$. Then, by $(2)$\n\n$$\nr^{2}=f(r f(r))=f(r f(s))=f((s-r) f(r))+r^{2},\n$$\n\nwhere the last statement follows from the given functional equation with $x=r$ and $y=s-r$. Hence, $h=(s-r) f(r)$ satisfies $f(h)=0$ which implies $h^{2}=f(h f(h))=f(0)=0$, i.e., $h=0$. Then, by $s \\neq r$ we have $f(r)=0$ which implies $r=0$, and finally $f(s)=f(r)=f(0)=0$. Analogously, it follows that $s=0$ which gives the contradiction $r=s$.\n\nTo prove $|f(1)|=1$ we apply (2) with $t=1$ and also with $t=f(1)$ and obtain $f(f(1))=1$ and $(f(1))^{2}=f(f(1) \\cdot f(f(1)))=f(f(1))=1$.\n\nNow we choose $\\eta \\in\\{-1,1\\}$ with $f(1)=\\eta$. Using that $f$ is odd and the given equation with $x=1, y=z$ (second equality) and with $x=-1, y=z+2$ (fourth equality) we obtain\n\n$$\n\\begin{aligned}\n& f(z)+2 \\eta=\\eta(f(z \\eta)+2)=\\eta(f(f(z+1))+1)=\\eta(-f(-f(z+1))+1) \\\\\n& \\quad=-\\eta f((z+2) f(-1))=-\\eta f((z+2)(-\\eta))=\\eta f((z+2) \\eta)=f(z+2) .\n\\end{aligned}\n$$\n\nHence,\n\n$$\nf(z+2 \\eta)=\\eta f(\\eta z+2)=\\eta(f(\\eta z)+2 \\eta)=f(z)+2 .\n$$\n\nUsing this argument twice we obtain\n\n$$\nf(z+4 \\eta)=f(z+2 \\eta)+2=f(z)+4 .\n$$\n\nSubstituting $z=2 f(x)$ we have\n\n$$\nf(2 f(x))+4=f(2 f(x)+4 \\eta)=f(2 f(x+2)),\n$$\n\n\n\nwhere the last equality follows from (4). Applying the given functional equation we proceed to\n\n$$\nf(2 f(x+2))=f(x f(2))+4=f(2 \\eta x)+4\n$$\n\nwhere the last equality follows again from (4) with $z=0$, i.e., $f(2)=2 \\eta$. Finally, $f(2 f(x))=$ $f(2 \\eta x)$ and by injectivity of $f$ we get $2 f(x)=2 \\eta x$ and hence the two solutions.']",,True,,, 2089,Combinatorics,,"Consider 2009 cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. Does the game necessarily end?","['We interpret a card showing black as the digit 0 and a card showing gold as the digit 1. Thus each position of the 2009 cards, read from left to right, corresponds bijectively to a nonnegative integer written in binary notation of 2009 digits, where leading zeros are allowed. Each move decreases this integer, so the game must end.']",,True,,, 2090,Combinatorics,,"Consider 2009 cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. Does there exist a winning strategy for the starting player?","['We show that there is no winning strategy for the starting player. We label the cards from right to left by $1, \\ldots, 2009$ and consider the set $S$ of cards with labels $50 i, i=1,2, \\ldots, 40$. Let $g_{n}$ be the number of cards from $S$ showing gold after $n$ moves. Obviously, $g_{0}=40$. Moreover, $\\left|g_{n}-g_{n+1}\\right|=1$ as long as the play goes on. Thus, after an odd number of moves, the nonstarting player finds a card from $S$ showing gold and hence can make a move. Consequently, this player always wins.']",,True,,, 2091,Combinatorics,,"For any integer $n \geq 2$, let $N(n)$ be the maximal number of triples $\left(a_{i}, b_{i}, c_{i}\right), i=1, \ldots, N(n)$, consisting of nonnegative integers $a_{i}, b_{i}$ and $c_{i}$ such that the following two conditions are satisfied: (1) $a_{i}+b_{i}+c_{i}=n$ for all $i=1, \ldots, N(n)$, (2) If $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$. Determine $N(n)$ for all $n \geq 2$.","['Let $n \\geq 2$ be an integer and let $\\left\\{T_{1}, \\ldots, T_{N}\\right\\}$ be any set of triples of nonnegative integers satisfying the conditions (1) and (2). Since the $a$-coordinates are pairwise distinct we have\n\n$$\n\\sum_{i=1}^{N} a_{i} \\geq \\sum_{i=1}^{N}(i-1)=\\frac{N(N-1)}{2}\n$$\n\nAnalogously,\n\n$$\n\\sum_{i=1}^{N} b_{i} \\geq \\frac{N(N-1)}{2} \\text { and } \\quad \\sum_{i=1}^{N} c_{i} \\geq \\frac{N(N-1)}{2}\n$$\n\nSumming these three inequalities and applying (1) yields\n\n$$\n3 \\frac{N(N-1)}{2} \\leq \\sum_{i=1}^{N} a_{i}+\\sum_{i=1}^{N} b_{i}+\\sum_{i=1}^{N} c_{i}=\\sum_{i=1}^{N}\\left(a_{i}+b_{i}+c_{i}\\right)=n N\n$$\n\nhence $3 \\frac{N-1}{2} \\leq n$ and, consequently,\n\n$$\nN \\leq\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1\n$$\n\nBy constructing examples, we show that this upper bound can be attained, so $N(n)=\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1$.\n\n\nWe distinguish the cases $n=3 k-1, n=3 k$ and $n=3 k+1$ for $k \\geq 1$ and present the extremal examples in form of a table.\n\n| $n=3 k-1$ | | |\n| :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1=2 k$ | | |\n| $a_{i}$ | $b_{i}$ | $c_{i}$ |\n| 0 | $k+1$ | $2 k-2$ |\n| 1 | $k+2$ | $2 k-4$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $k-1$ | $2 k$ | 0 |\n| $k$ | 0 | $2 k-1$ |\n| $k+1$ | 1 | $2 k-3$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $2 k-1$ | $k-1$ | 1 |\n\n\n| $n=3 k$ | | |\n| :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1=2 k+1$ | | |\n| $a_{i}$ | $b_{i}$ | $c_{i}$ |\n| 0 | $k$ | $2 k$ |\n| 1 | $k+1$ | $2 k-2$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $k$ | $2 k$ | 0 |\n| $k+1$ | 0 | $2 k-1$ |\n| $k+2$ | 1 | $2 k-3$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $2 k$ | $k-1$ | 1 |\n\n\n| | $=3 k$ | |\n| :---: | :---: | :---: |\n| $\\frac{2 n}{3}$ | $+1=$ | $k+1$ |\n| $a_{i}$ | $b_{i}$ | $c_{i}$ |\n| 0 | $k$ | $2 k+1$ |\n| 1 | $k+1$ | $2 k-1$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $k$ | $2 k$ | 1 |\n| $k+1$ | 0 | $2 k$ |\n| $k+2$ | 1 | $2 k-2$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $2 k$ | $k-1$ | 2 |\n\nIt can be easily seen that the conditions (1) and (2) are satisfied and that we indeed have $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1$ triples in each case.']",['$N(n)=\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1$'],False,,Expression, 2092,Combinatorics,,"Let $n$ be a positive integer. Given a sequence $\varepsilon_{1}, \ldots, \varepsilon_{n-1}$ with $\varepsilon_{i}=0$ or $\varepsilon_{i}=1$ for each $i=1, \ldots, n-1$, the sequences $a_{0}, \ldots, a_{n}$ and $b_{0}, \ldots, b_{n}$ are constructed by the following rules: $$ \begin{gathered} a_{0}=b_{0}=1, \quad a_{1}=b_{1}=7, \\ a_{i+1}=\left\{\begin{array}{ll} 2 a_{i-1}+3 a_{i}, & \text { if } \varepsilon_{i}=0, \\ 3 a_{i-1}+a_{i}, & \text { if } \varepsilon_{i}=1, \end{array} \text { for each } i=1, \ldots, n-1,\right. \\ b_{i+1}=\left\{\begin{array}{ll} 2 b_{i-1}+3 b_{i}, & \text { if } \varepsilon_{n-i}=0, \\ 3 b_{i-1}+b_{i}, & \text { if } \varepsilon_{n-i}=1, \end{array} \text { for each } i=1, \ldots, n-1 .\right. \end{gathered} $$ Prove that $a_{n}=b_{n}$.","['For a binary word $w=\\sigma_{1} \\ldots \\sigma_{n}$ of length $n$ and a letter $\\sigma \\in\\{0,1\\}$ let $w \\sigma=$ $\\sigma_{1} \\ldots \\sigma_{n} \\sigma$ and $\\sigma w=\\sigma \\sigma_{1} \\ldots \\sigma_{n}$. Moreover let $\\bar{w}=\\sigma_{n} \\ldots \\sigma_{1}$ and let $\\emptyset$ be the empty word (of length 0 and with $\\bar{\\emptyset}=\\emptyset)$. Let $(u, v)$ be a pair of two real numbers. For binary words $w$ we define recursively the numbers $(u, v)^{w}$ as follows:\n\n$$\n\\begin{array}{r}\n(u, v)^{\\emptyset}=v, \\quad(u, v)^{0}=2 u+3 v, \\quad(u, v)^{1}=3 u+v \\\\\n(u, v)^{w \\sigma \\varepsilon}= \\begin{cases}2(u, v)^{w}+3(u, v)^{w \\sigma}, & \\text { if } \\varepsilon=0 \\\\\n3(u, v)^{w}+(u, v)^{w \\sigma}, & \\text { if } \\varepsilon=1\\end{cases}\n\\end{array}\n$$\n\nIt easily follows by induction on the length of $w$ that for all real numbers $u_{1}, v_{1}, u_{2}, v_{2}, \\lambda_{1}$ and $\\lambda_{2}$\n\n$$\n\\left(\\lambda_{1} u_{1}+\\lambda_{2} u_{2}, \\lambda_{1} v_{1}+\\lambda_{2} v_{2}\\right)^{w}=\\lambda_{1}\\left(u_{1}, v_{1}\\right)^{w}+\\lambda_{2}\\left(u_{2}, v_{2}\\right)^{w}\n\\tag{1}\n$$\n\nand that for $\\varepsilon \\in\\{0,1\\}$\n\n$$\n(u, v)^{\\varepsilon w}=\\left(v,(u, v)^{\\varepsilon}\\right)^{w} .\n\\tag{2}\n$$\n\nObviously, for $n \\geq 1$ and $w=\\varepsilon_{1} \\ldots \\varepsilon_{n-1}$, we have $a_{n}=(1,7)^{w}$ and $b_{n}=(1,7)^{\\bar{w}}$. Thus it is sufficient to prove that\n\n$$\n(1,7)^{w}=(1,7)^{\\bar{w}}\n\\tag{3}\n$$\n\nfor each binary word $w$. We proceed by induction on the length of $w$. The assertion is obvious if $w$ has length 0 or 1 . Now let $w \\sigma \\varepsilon$ be a binary word of length $n \\geq 2$ and suppose that the assertion is true for all binary words of length at most $n-1$.\n\nNote that $(2,1)^{\\sigma}=7=(1,7)^{\\emptyset}$ for $\\sigma \\in\\{0,1\\},(1,7)^{0}=23$, and $(1,7)^{1}=10$.\n\nFirst let $\\varepsilon=0$. Then in view of the induction hypothesis and the equalities (1) and (2), we obtain\n\n$$\n\\begin{aligned}\n(1,7)^{w \\sigma 0}=2(1,7)^{w}+3(1,7)^{w \\sigma}=2(1,7)^{\\bar{w}}+3(1,7)^{\\sigma \\bar{w}}=2(2,1)^{\\sigma \\bar{w}}+ & 3(1,7)^{\\sigma \\bar{w}} \\\\\n& =(7,23)^{\\sigma \\bar{w}}=(1,7)^{0 \\sigma \\bar{w}}\n\\end{aligned}\n$$\n\n\n\nNow let $\\varepsilon=1$. Analogously, we obtain\n\n$$\n\\begin{aligned}\n&(1,7)^{w \\sigma 1}=3(1,7)^{w}+(1,7)^{w \\sigma}=3(1,7)^{\\bar{w}}+(1,7)^{\\sigma \\bar{w}}=3(2,1)^{\\sigma \\bar{w}}+(1,7)^{\\sigma \\bar{w}} \\\\\n&=(7,10)^{\\sigma \\bar{w}}=(1,7)^{1 \\sigma \\bar{w}}\n\\end{aligned}\n$$\n\nThus the induction step is complete, (3) and hence also $a_{n}=b_{n}$ are proved.']",,True,,, 2093,Combinatorics,,"For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1 . Determine the smallest possible sum of rectangle perimeters in such a partition.","[""For a $k \\times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{i j}$ in row $i$ and column $j$ has vertices $(i-1, j-1),(i-$ $1, j),(i, j-1),(i, j)$, where $i, j \\in\\{1, \\ldots, k\\}$. Without loss of generality assume that the cells $C_{i i}$, $i=1, \\ldots, k$, form a separate rectangle. Then we may consider the boards $B_{k}=\\bigcup_{10$, JENSEN's inequality immediately shows that the minimum of the right hand sight of (1) is attained for $i=k / 2$. Hence the total perimeter of the optimal partition of $B_{k}$ is at least $2 k+2 k / 2 \\log _{2} k / 2+2(k / 2) \\log _{2}(k / 2)=D_{k}$."", ""We start as in Solution 1 and present another proof that $m 2^{m+1}$ is a lower bound for the total perimeter of a partition of $B_{2^{m}}$ into $n$ rectangles. Let briefly $M=2^{m}$. For $1 \\leq i \\leq M$, let $r_{i}$ denote the number of rectangles in the partition that cover some cell from row $i$ and let $c_{j}$ be the number of rectangles that cover some cell from column $j$. Note that the total perimeter $p$ of all rectangles in the partition is\n\n$$\np=2\\left(\\sum_{i=1}^{M} r_{i}+\\sum_{i=1}^{M} c_{i}\\right)\n$$\n\nNo rectangle can simultaneously cover cells from row $i$ and from column $i$ since otherwise it would also cover the cell $C_{i i}$. We classify subsets $S$ of rectangles of the partition as follows. We say that $S$ is of type $i, 1 \\leq i \\leq M$, if $S$ contains all $r_{i}$ rectangles that cover some cell from row $i$, but none of the $c_{i}$ rectangles that cover some cell from column $i$. Altogether there are $2^{n-r_{i}-c_{i}}$ subsets of type $i$. Now we show that no subset $S$ can be simultaneously of type $i$ and of type $j$ if $i \\neq j$. Assume the contrary and let without loss of generality $i1$. By condition $\\left(2^{\\prime}\\right)$, after the Stepmother has distributed her water we have $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \\leq \\frac{5}{2}$. Therefore,\n\n$$\n\\left(y_{0}+y_{2}\\right)+\\left(y_{1}+y_{3}\\right)+\\left(y_{2}+y_{4}\\right)+\\left(y_{3}+y_{0}\\right)+\\left(y_{4}+y_{1}\\right)=2\\left(y_{0}+y_{1}+y_{2}+y_{3}+y_{4}\\right) \\leq 5\n$$\n\nand hence there is a pair of non-neighboring buckets which is not critical, say $\\left(B_{0}, B_{2}\\right)$. Now, if both of the pairs $\\left(B_{3}, B_{0}\\right)$ and $\\left(B_{2}, B_{4}\\right)$ are critical, we must have $y_{1}<\\frac{1}{2}$ and Cinderella can empty the buckets $B_{3}$ and $B_{4}$. This clearly leaves no critical pair of buckets and the total contents of all the buckets is then $y_{1}+\\left(y_{0}+y_{2}\\right) \\leq \\frac{3}{2}$. Therefore, conditions $\\left(1^{\\prime}\\right)$ and $\\left(2^{\\prime}\\right)$ are fulfilled.\n\nNow suppose that without loss of generality the pair $\\left(B_{3}, B_{0}\\right)$ is not critical. If in this case $y_{0} \\leq \\frac{1}{2}$, then one of the inequalities $y_{0}+y_{1}+y_{2} \\leq \\frac{3}{2}$ and $y_{0}+y_{3}+y_{4} \\leq \\frac{3}{2}$ must hold. But then Cinderella can empty $B_{3}$ and $B_{4}$ or $B_{1}$ and $B_{2}$, respectively and clearly fulfill the conditions.\n\nFinally consider the case $y_{0}>\\frac{1}{2}$. By $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \\leq \\frac{5}{2}$, at least one of the pairs $\\left(B_{1}, B_{3}\\right)$ and $\\left(B_{2}, B_{4}\\right)$ is not critical. Without loss of generality let this be the pair $\\left(B_{1}, B_{3}\\right)$. Since the pair $\\left(B_{3}, B_{0}\\right)$ is not critical and $y_{0}>\\frac{1}{2}$, we must have $y_{3} \\leq \\frac{1}{2}$. But then, as before, Cinderella can maintain the two conditions at the beginning of the next round by either emptying $B_{1}$ and $B_{2}$ or $B_{4}$ and $B_{0}$.']",,True,,, 2095,Combinatorics,,"On a $999 \times 999$ board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A nonintersecting route of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called cyclic, if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit?","['First we show that this number is an upper bound for the number of cells a limp rook can visit. To do this we color the cells with four colors $A, B, C$ and $D$ in the following way: for $(i, j) \\equiv(0,0) \\bmod 2$ use $A$, for $(i, j) \\equiv(0,1) \\bmod 2$ use $B$, for $(i, j) \\equiv(1,0) \\bmod 2$ use $C$ and for $(i, j) \\equiv(1,1) \\bmod 2$ use $D$. From an $A$-cell the rook has to move to a $B$-cell or a $C$-cell. In the first case, the order of the colors of the cells visited is given by $A, B, D, C, A, B, D, C, A, \\ldots$, in the second case it is $A, C, D, B, A, C, D, B, A, \\ldots$ Since the route is closed it must contain the same number of cells of each color. There are only $499^{2} A$-cells. In the following we will show that the rook cannot visit all the $A$-cells on its route and hence the maximum possible number of cells in a route is $4 \\cdot\\left(499^{2}-1\\right)$.\n\nAssume that the route passes through every single $A$-cell. Color the $A$-cells in black and white in a chessboard manner, i.e. color any two $A$-cells at distance 2 in different color. Since the number of $A$-cells is odd the rook cannot always alternate between visiting black and white $A$-cells along its route. Hence there are two $A$-cells of the same color which are four rook-steps apart that are visited directly one after the other. Let these two $A$-cells have row and column numbers $(a, b)$ and $(a+2, b+2)$ respectively.\n\n\n\nThere is up to reflection only one way the rook can take from $(a, b)$ to $(a+2, b+2)$. Let this way be $(a, b) \\rightarrow(a, b+1) \\rightarrow(a+1, b+1) \\rightarrow(a+1, b+2) \\rightarrow(a+2, b+2)$. Also let without loss of generality the color of the cell $(a, b+1)$ be $B$ (otherwise change the roles of columns and rows).\n\nNow consider the $A$-cell $(a, b+2)$. The only way the rook can pass through it is via $(a-1, b+2) \\rightarrow$ $(a, b+2) \\rightarrow(a, b+3)$ in this order, since according to our assumption after every $A$-cell the rook passes through a $B$-cell. Hence, to connect these two parts of the path, there must be\n\n\n\na path connecting the cell $(a, b+3)$ and $(a, b)$ and also a path connecting $(a+2, b+2)$ and $(a-1, b+2)$.\n\nBut these four cells are opposite vertices of a convex quadrilateral and the paths are outside of that quadrilateral and hence they must intersect. This is due to the following fact:\n\nThe path from $(a, b)$ to $(a, b+3)$ together with the line segment joining these two cells form a closed loop that has one of the cells $(a-1, b+2)$ and $(a+2, b+2)$ in its inside and the other one on the outside. Thus the path between these two points must cross the previous path.\n\nBut an intersection is only possible if a cell is visited twice. This is a contradiction.\n\nHence the number of cells visited is at most $4 \\cdot\\left(499^{2}-1\\right)$.\n\nThe following picture indicates a recursive construction for all $n \\times n$-chessboards with $n \\equiv 3$ mod 4 which clearly yields a path that misses exactly one $A$-cell (marked with a dot, the center cell of the $15 \\times 15$-chessboard) and hence, in the case of $n=999$ crosses exactly $4 \\cdot\\left(499^{2}-1\\right)$ cells.\n\n']",['996000'],False,,Numerical, 2096,Combinatorics,," A grasshopper jumps along the real axis. He starts at point 0 and makes 2009 jumps to the right with lengths $1,2, \ldots, 2009$ in an arbitrary order. Let $M$ be a set of 2008 positive integers less than $1005 \cdot 2009$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$.","['We construct the set of landing points of the grasshopper.\n\nCase 1. $M$ does not contain numbers divisible by 2009.\n\nWe fix the numbers $2009 k$ as landing points, $k=1,2, \\ldots, 1005$. Consider the open intervals $I_{k}=(2009(k-1), 2009 k), k=1,2, \\ldots, 1005$. We show that we can choose exactly one point outside of $M$ as a landing point in 1004 of these intervals such that all lengths from 1 to 2009 are realized. Since there remains one interval without a chosen point, the length 2009 indeed will appear. Each interval has length 2009, hence a new landing point in an interval yields with a length $d$ also the length $2009-d$. Thus it is enough to implement only the lengths from $D=\\{1,2, \\ldots, 1004\\}$. We will do this in a greedy way. Let $n_{k}, k=1,2, \\ldots, 1005$, be the number of elements of $M$ that belong to the interval $I_{k}$. We order these numbers in a decreasing way, so let $p_{1}, p_{2}, \\ldots, p_{1005}$ be a permutation of $\\{1,2, \\ldots, 1005\\}$ such that $n_{p_{1}} \\geq n_{p_{2}} \\geq \\cdots \\geq n_{p_{1005}}$. In $I_{p_{1}}$ we do not choose a landing point. Assume that landing points have already been chosen in the intervals $I_{p_{2}}, \\ldots, I_{p_{m}}$ and the lengths $d_{2}, \\ldots, d_{m}$ from $D$ are realized, $m=1, \\ldots, 1004$. We show that there is some $d \\in D \\backslash\\left\\{d_{2}, \\ldots, d_{m}\\right\\}$ that can be implemented with a new landing point in $I_{p_{m+1}}$. Assume the contrary. Then the $1004-(m-1)$ other lengths are obstructed by the $n_{p_{m+1}}$ points of $M$ in $I_{p_{m+1}}$. Each length $d$ can be realized by two landing points, namely $2009\\left(p_{m+1}-1\\right)+d$ and $2009 p_{m+1}-d$, hence\n\n$$\nn_{p_{m+1}} \\geq 2(1005-m)\n\\tag{1}\n$$\n\nMoreover, since $|M|=2008=n_{1}+\\cdots+n_{1005}$,\n\n$$\n2008 \\geq n_{p_{1}}+n_{p_{2}}+\\cdots+n_{p_{m+1}} \\geq(m+1) n_{p_{m+1}}\n\\tag{2}\n$$\n\nConsequently, by (1) and (2),\n\n$$\n2008 \\geq 2(m+1)(1005-m)\n$$\n\nThe right hand side of the last inequality obviously attains its minimum for $m=1004$ and this minimum value is greater than 2008 , a contradiction.\n\nCase 2. $M$ does contain a number $\\mu$ divisible by 2009.\n\nBy the pigeonhole principle there exists some $r \\in\\{1, \\ldots, 2008\\}$ such that $M$ does not contain numbers with remainder $r$ modulo 2009. We fix the numbers $2009(k-1)+r$ as landing points, $k=1,2, \\ldots, 1005$. Moreover, $1005 \\cdot 2009$ is a landing point. Consider the open intervals\n\n\n\n$I_{k}=(2009(k-1)+r, 2009 k+r), k=1,2, \\ldots, 1004$. Analogously to Case 1 , it is enough to show that we can choose in 1003 of these intervals exactly one landing point outside of $M \\backslash\\{\\mu\\}$ such that each of the lengths of $D=\\{1,2, \\ldots, 1004\\} \\backslash\\{r\\}$ are implemented. Note that $r$ and $2009-r$ are realized by the first and last jump and that choosing $\\mu$ would realize these two differences again. Let $n_{k}, k=1,2, \\ldots, 1004$, be the number of elements of $M \\backslash\\{\\mu\\}$ that belong to the interval $I_{k}$ and $p_{1}, p_{2}, \\ldots, p_{1004}$ be a permutation of $\\{1,2, \\ldots, 1004\\}$ such that $n_{p_{1}} \\geq n_{p_{2}} \\geq \\cdots \\geq n_{p_{1004}}$. With the same reasoning as in Case 1 we can verify that a greedy choice of the landing points in $I_{p_{2}}, I_{p_{3}}, \\ldots, I_{p_{1004}}$ is possible. We only have to replace (1) by\n\n$$\nn_{p_{m+1}} \\geq 2(1004-m)\n$$\n\n( $D$ has one element less) and (2) by\n\n$$\n2007 \\geq n_{p_{1}}+n_{p_{2}}+\\cdots+n_{p_{m+1}} \\geq(m+1) n_{p_{m+1}}\n$$']",,True,,, 2097,Combinatorics,," Let $n$ be a nonnegative integer. A grasshopper jumps along the real axis. He starts at point 0 and makes $n+1$ jumps to the right with pairwise different positive integral lengths $a_{1}, a_{2}, \ldots, a_{n+1}$ in an arbitrary order. Let $M$ be a set of $n$ positive integers in the interval $(0, s)$, where $s=a_{1}+a_{2}+\cdots+a_{n+1}$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$.","['First of all we remark that the statement in the problem implies a strengthening of itself: Instead of $|M|=n$ it is sufficient to suppose that $|M \\cap(0, s-\\bar{a}]| \\leq n$, where $\\bar{a}=\\min \\left\\{a_{1}, a_{2}, \\ldots, a_{n+1}\\right\\}$. This fact will be used in the proof.\n\nWe prove the statement by induction on $n$. The case $n=0$ is obvious. Let $n>0$ and let the assertion be true for all nonnegative integers less than $n$. Moreover let $a_{1}, a_{2}, \\ldots, a_{n+1}, s$ and $M$ be given as in the problem. Without loss of generality we may assume that $a_{n+1}1$. If $T_{\\bar{k}-1} \\in M$, then (4) follows immediately by the minimality of $\\bar{k}$. If $T_{\\bar{k}-1} \\notin M$, by the smoothness of $\\bar{k}-1$, we obtain a situation as in Claim 1 with $m=\\bar{k}-1$ provided that $\\left|M \\cap\\left(0, T_{\\bar{k}-1}\\right]\\right| \\geq \\bar{k}-1$. Hence, we may even restrict ourselves to $\\left|M \\cap\\left(0, T_{\\bar{k}-1}\\right]\\right| \\leq \\bar{k}-2$ in this case and Claim 3 is proved.\n\nChoose an integer $v \\geq 0$ with $\\left|M \\cap\\left(0, T_{\\bar{k}}\\right)\\right|=\\bar{k}+v$. Let $r_{1}>r_{2}>\\cdots>r_{l}$ be exactly those indices $r$ from $\\{\\bar{k}+1, \\bar{k}+2, \\ldots, n+1\\}$ for which $T_{\\bar{k}}+a_{r} \\notin M$. Then\n\n$$\nn=|M|=\\left|M \\cap\\left(0, T_{\\bar{k}}\\right)\\right|+1+\\left|M \\cap\\left(T_{\\bar{k}}, s\\right)\\right| \\geq \\bar{k}+v+1+(n+1-\\bar{k}-l)\n$$\n\nand consequently $l \\geq v+2$. Note that\n\n$T_{\\bar{k}}+a_{r_{1}}-a_{1}f_{i}(\\varepsilon)$ for all $i=0, \\ldots, k$.\n\nWe will show the essential fact:\n\nFact 3. $f_{0}(n)>f_{0}(h(n))$.\n\nThen the empty string will necessarily be reached after a finite number of applications of $h$. But starting from a string without leading zeros, $\\varepsilon$ can only be reached via the strings $1 \\rightarrow 00 \\rightarrow 0 \\rightarrow \\varepsilon$. Hence also the number 1 will appear after a finite number of applications of $h$.\n\nProof of Fact 3. If the last digit $r$ of $n$ is 0 , then we write $n=x_{0} 0 \\ldots 0 x_{m-1} 0 \\varepsilon$ where the $x_{i}$ do not contain the digit 0 . Then $h(n)=x_{0} 0 \\ldots 0 x_{m-1}$ and $f_{0}(n)-f_{0}(h(n))=f_{0}(\\varepsilon)>0$.\n\nSo let the last digit $r$ of $n$ be at least 1. Let $L=y k$ and $R=z r$ be the corresponding left and right parts where $y$ is some string, $k \\leq r-1$ and the string $z$ consists only of digits not less\n\n\n\nthan $r$. Then $n=y k z r$ and $h(n)=y k z(r-1) z(r-1)$. Let $d(y)$ be the smallest digit of $y$. We consider two cases which do not exclude each other.\n\nCase 1. $d(y) \\geq k$.\n\nThen\n\n$$\nf_{k}(n)-f_{k}(h(n))=f_{k}(z r)-f_{k}(z(r-1) z(r-1)) .\n$$\n\nIn view of Fact 1 this difference is positive if and only if\n\n$$\nf_{r-1}(z r)-f_{r-1}(z(r-1) z(r-1))>0 .\n$$\n\nWe have, using Fact 2,\n\n$$\nf_{r-1}(z r)=4^{f_{r}(z r)}=4^{f_{r}(z)+4^{f_{r+1}(\\varepsilon)}} \\geq 4 \\cdot 4^{f_{r}(z)}>4^{f_{r}(z)}+4^{f_{r}(z)}+4^{f_{r}(\\varepsilon)}=f_{r-1}(z(r-1) z(r-1)) .\n$$\n\nHere we use the additional definition $f_{10}(\\varepsilon)=0$ if $r=9$. Consequently, $f_{k}(n)-f_{k}(h(n))>0$ and according to Fact $1, f_{0}(n)-f_{0}(h(n))>0$.\n\nCase 2. $d(y) \\leq k$.\n\nWe prove by induction on $d(y)=k, k-1, \\ldots, 0$ that $f_{i}(n)-f_{i}(h(n))>0$ for all $i=0, \\ldots, d(y)$. By Fact 1, it suffices to do so for $i=d(y)$. The initialization $d(y)=k$ was already treated in Case 1. Let $t=d(y)0 .\n$$\n\nThus the inequality $f_{d(y)}(n)-f_{d(y)}(h(n))>0$ is established and from Fact 1 it follows that $f_{0}(n)-f_{0}(h(n))>0$.', 'We identify integers $n \\geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from 0 to 9. Moreover, let us define that the empty string, $\\varepsilon$, is being mapped to the empty string. In the following all functions map the set of strings into the set of strings. For two functions $f$ and $g$ let $g \\circ f$ be defined by $(g \\circ f)(x)=g(f(x))$ for all strings $x$ and let, for non-negative integers $n, f^{n}$ denote the $n$-fold application of $f$. For any string $x$ let $s(x)$ be the smallest digit of $x$, and for the empty string let $s(\\varepsilon)=\\infty$. We define nine functions $g_{1}, \\ldots, g_{9}$ as follows: Let $k \\in\\{1, \\ldots, 9\\}$ and let $x$ be a string. If $x=\\varepsilon$ then $g_{k}(x)=\\varepsilon$. Otherwise, write $x$ in the form $x=y z r$ where $y$ is either the empty string or ends with a digit smaller than $k, s(z) \\geq k$ and $r$ is the rightmost digit of $x$. Then $g_{k}(x)=z r$.\n\nLemma 1. We have $g_{k} \\circ h=g_{k} \\circ h \\circ g_{k}$ for all $k=1, \\ldots, 9$.\n\nProof of Lemma 1. Let $x=y z r$ be as in the definition of $g_{k}$. If $y=\\varepsilon$, then $g_{k}(x)=x$, whence\n\n$$\ng_{k}(h(x))=g_{k}\\left(h\\left(g_{k}(x)\\right) .\\right.\n\\tag{1}\n$$\n\nSo let $y \\neq \\varepsilon$.\n\nCase 1. $z$ contains a digit smaller than $r$.\n\nLet $z=u a v$ where $a0\\end{cases}\n$$\n\n\n\nand\n\n$$\nh\\left(g_{k}(x)\\right)=h(z r)=h(u a v r)= \\begin{cases}u a v & \\text { if } r=0, \\\\ \\operatorname{uav}(r-1) v(r-1) & \\text { if } r>0 .\\end{cases}\n$$\n\nSince $y$ ends with a digit smaller than $k$, (1) is obviously true.\n\nCase 2. $z$ does not contain a digit smaller than $r$.\n\nLet $y=u v$ where $u$ is either the empty string or ends with a digit smaller than $r$ and $s(v) \\geq r$. We have\n\n$$\nh(x)= \\begin{cases}u v z & \\text { if } r=0 \\\\ u v z(r-1) v z(r-1) & \\text { if } r>0\\end{cases}\n$$\n\nand\n\n$$\nh\\left(g_{k}(x)\\right)=h(z r)= \\begin{cases}z & \\text { if } r=0 \\\\ z(r-1) z(r-1) & \\text { if } r>0\\end{cases}\n$$\n\nRecall that $y$ and hence $v$ ends with a digit smaller than $k$, but all digits of $v$ are at least $r$. Now if $r>k$, then $v=\\varepsilon$, whence the terminal digit of $u$ is smaller than $k$, which entails\n\n$$\ng_{k}(h(x))=z(r-1) z(r-1)=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right) .\n$$\n\nIf $r \\leq k$, then\n\n$$\ng_{k}(h(x))=z(r-1)=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right),\n$$\n\nso that in both cases (1) is true. Thus Lemma 1 is proved.\n\nLemma 2. Let $k \\in\\{1, \\ldots, 9\\}$, let $x$ be a non-empty string and let $n$ be a positive integer. If $h^{n}(x)=\\varepsilon$ then $\\left(g_{k} \\circ h\\right)^{n}(x)=\\varepsilon$.\n\nProof of Lemma 2. We proceed by induction on $n$. If $n=1$ we have\n\n$$\n\\varepsilon=h(x)=g_{k}(h(x))=\\left(g_{k} \\circ h\\right)(x) .\n$$\n\nNow consider the step from $n-1$ to $n$ where $n \\geq 2$. Let $h^{n}(x)=\\varepsilon$ and let $y=h(x)$. Then $h^{n-1}(y)=\\varepsilon$ and by the induction hypothesis $\\left(g_{k} \\circ h\\right)^{n-1}(y)=\\varepsilon$. In view of Lemma 1,\n\n$$\n\\begin{aligned}\n\\varepsilon=\\left(g_{k} \\circ h\\right)^{n-2}( & \\left.\\left(g_{k} \\circ h\\right)(y)\\right)=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}(h(y))\\right. \\\\\n& =\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}\\left(h\\left(g_{k}(y)\\right)\\right)=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}\\left(h\\left(g_{k}(h(x))\\right)\\right)=\\left(g_{k} \\circ h\\right)^{n}(x) .\\right.\\right.\n\\end{aligned}\n$$\n\nThus the induction step is complete and Lemma 2 is proved.\n\nWe say that the non-empty string $x$ terminates if $h^{n}(x)=\\varepsilon$ for some non-negative integer $n$.\n\nLemma 3. Let $x=y z r$ where $s(y) \\geq k, s(z) \\geq k, y$ ends with the digit $k$ and $z$ is possibly empty. If $y$ and $z r$ terminate then also $x$ terminates.\n\nProof of Lemma 3. Suppose that $y$ and $z r$ terminate. We proceed by induction on $k$. Let $k=0$. Obviously, $h(y w)=y h(w)$ for any non-empty string $w$. Let $h^{n}(z r)=\\epsilon$. It follows easily by induction on $m$ that $h^{m}(y z r)=y h^{m}(z r)$ for $m=1, \\ldots, n$. Consequently, $h^{n}(y z r)=y$. Since $y$ terminates, also $x=y z r$ terminates.\n\nNow let the assertion be true for all nonnegative integers less than $k$ and let us prove it for $k$ where $k \\geq 1$. It turns out that it is sufficient to prove that $y g_{k}(h(z r))$ terminates. Indeed:\n\nCase 1. $r=0$.\n\nThen $h(y z r)=y z=y g_{k}(h(z r))$.\n\nCase 2. $0k$.\n\nThen $h(y z r)=y h(z r)=y g_{k}(h(z r))$.\n\nNote that $y g_{k}(h(z r))$ has the form $y z^{\\prime} r^{\\prime}$ where $s\\left(z^{\\prime}\\right) \\geq k$. By the same arguments it is sufficient to prove that $y g_{k}\\left(h\\left(z^{\\prime} r^{\\prime}\\right)\\right)=y\\left(g_{k} \\circ h\\right)^{2}(z r)$ terminates and, by induction, that $y\\left(g_{k} \\circ h\\right)^{m}(z r)$ terminates for some positive integer $m$. In view of Lemma 2 there is some $m$ such that $\\left(g_{k} \\circ\\right.$ $h)^{m}(z r)=\\epsilon$, so $x=y z r$ terminates if $y$ terminates. Thus Lemma 3 is proved.\n\nNow assume that there is some string $x$ that does not terminate. We choose $x$ minimal. If $x \\geq 10$, we can write $x$ in the form $x=y z r$ of Lemma 3 and by this lemma $x$ terminates since $y$ and $z r$ are smaller than $x$. If $x \\leq 9$, then $h(x)=(x-1)(x-1)$ and $h(x)$ terminates again by Lemma 3 and the minimal choice of $x$.', 'We commence by introducing some terminology. Instead of integers, we will consider the set $S$ of all strings consisting of the digits $0,1, \\ldots, 9$, including the empty string $\\epsilon$. If $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ is a nonempty string, we let $\\rho(a)=a_{n}$ denote the terminal digit of $a$ and $\\lambda(a)$ be the string with the last digit removed. We also define $\\lambda(\\epsilon)=\\epsilon$ and denote the set of non-negative integers by $\\mathbb{N}_{0}$.\n\nNow let $k \\in\\{0,1,2, \\ldots, 9\\}$ denote any digit. We define a function $f_{k}: S \\longrightarrow S$ on the set of strings: First, if the terminal digit of $n$ belongs to $\\{0,1, \\ldots, k\\}$, then $f_{k}(n)$ is obtained from $n$ by deleting this terminal digit, i.e $f_{k}(n)=\\lambda(n)$. Secondly, if the terminal digit of $n$ belongs to $\\{k+1, \\ldots, 9\\}$, then $f_{k}(n)$ is obtained from $n$ by the process described in the problem. We also define $f_{k}(\\epsilon)=\\epsilon$. Note that up to the definition for integers $n \\leq 1$, the function $f_{0}$ coincides with the function $h$ in the problem, through interpreting integers as digit strings. The argument will be roughly as follows. We begin by introducing a straightforward generalization of our claim about $f_{0}$. Then it will be easy to see that $f_{9}$ has all these stronger properties, which means that is suffices to show for $k \\in\\{0,1, \\ldots, 8\\}$ that $f_{k}$ possesses these properties provided that $f_{k+1}$ does.\n\nWe continue to use $k$ to denote any digit. The operation $f_{k}$ is said to be separating, if the followings holds: Whenever $a$ is an initial segment of $b$, there is some $N \\in \\mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. The following two notions only apply to the case where $f_{k}$ is indeed separating, otherwise they remain undefined. For every $a \\in S$ we denote the least $N \\in \\mathbb{N}_{0}$ for which $f_{k}^{N}(a)=\\epsilon$ occurs by $g_{k}(a)$ (because $\\epsilon$ is an initial segment of $a$, such an $N$ exists if $f_{k}$ is separating). If for every two strings $a$ and $b$ such that $a$ is a terminal segment of $b$ one has $g_{k}(a) \\leq g_{k}(b)$, we say that $f_{k}$ is coherent. In case that $f_{k}$ is separating and coherent we call the digit $k$ seductive.\n\nAs $f_{9}(a)=\\lambda(a)$ for all $a$, it is obvious that 9 is seductive. Hence in order to show that 0 is seductive, which clearly implies the statement of the problem, it suffices to take any $k \\in\\{0,1, \\ldots, 8\\}$ such that $k+1$ is seductive and to prove that $k$ has to be seductive as well. Note that in doing so, we have the function $g_{k+1}$ at our disposal. We have to establish two things and we begin with\n\nStep 1. $f_{k}$ is separating.\n\nBefore embarking on the proof of this, we record a useful observation which is easily proved by induction on $M$.\n\n\n\nClaim 1. For any strings $A, B$ and any positive integer $M$ such that $f_{k}^{M-1}(B) \\neq \\epsilon$, we have\n\n$$\nf_{k}^{M}(A k B)=A k f_{k}^{M}(B)\n$$\n\nNow we call a pair $(a, b)$ of strings wicked provided that $a$ is an initial segment of $b$, but there is no $N \\in \\mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. We need to show that there are none, so assume that there were such pairs. Choose a wicked pair $(a, b)$ for which $g_{k+1}(b)$ attains its minimal possible value. Obviously $b \\neq \\epsilon$ for any wicked pair $(a, b)$. Let $z$ denote the terminal digit of $b$. Observe that $a \\neq b$, which means that $a$ is also an initial segment of $\\lambda(b)$. To facilitate the construction of the eventual contradiction, we prove\n\nClaim 2. There cannot be an $N \\in \\mathbb{N}_{0}$ such that\n\n$$\nf_{k}^{N}(b)=\\lambda(b)\n$$\n\nProof of Claim 2. For suppose that such an $N$ existed. Because $g_{k+1}(\\lambda(b))k+1$ is impossible: Set $B=f_{k}(b)$. Then also $f_{k+1}(b)=B$, but $g_{k+1}(B)g_{k}(b)$. Observe that if $f_{k}$ was incoherent, which we shall assume from now on, then such pairs existed. Now among all aggressive pairs we choose one, say $(a, b)$, for which $g_{k}(b)$ attains its least possible value. Obviously $f_{k}(a)$ cannot be injectible into $f_{k}(b)$, for otherwise the pair $\\left(f_{k}(a), f_{k}(b)\\right)$ was aggressive and contradicted our choice of $(a, b)$. Let $\\left(A_{1}, A_{2}, \\ldots, A_{m}\\right)$ and $\\left(B_{1}, B_{2}, \\ldots, B_{n}\\right)$ be the decompositions of $a$ and $b$ and take a function $H:\\{1,2, \\ldots, m\\} \\longrightarrow\\{1,2, \\ldots, n\\}$ exemplifying that $a$ is indeed injectible into $b$. If we had $H(m)\n\nFigure 1\n\n\n\nFigure 2\n\nLet $P, Q$ and $R$ be the points where the incircle of triangle $A D C$ touches the sides $A D, D C$ and $C A$, respectively. Since $K$ and $I$ lie on the angle bisector of $\\angle A C D$, the segments $I D$ and $I F$ are symmetric with respect to the line $I C$. Hence there is a point $S$ on $I F$ where the incircle of triangle $A D C$ touches the segment $I F$. Then segments $K P, K Q, K R$ and $K S$ all have the same length and are perpendicular to $A D, D C, C A$ and $I F$, respectively. So - regardless of the value of $\\angle B E K$ - the quadrilateral $K R F S$ is a square and $\\angle S F K=\\angle K F C=45^{\\circ}$.\n\nConsider the case $\\angle B A C=60^{\\circ}$ (see Figure 1). Then triangle $A B C$ is equilateral. Furthermore we have $F=E$, hence $\\angle B E K=\\angle I F K=\\angle S E K=45^{\\circ}$. So $60^{\\circ}$ is a possible value for $\\angle B A C$.\n\nNow consider the case $\\angle B A C=90^{\\circ}$ (see Figure 2). Then $\\angle C B A=\\angle A C B=45^{\\circ}$. Furthermore, $\\angle K I E=\\frac{1}{2} \\angle C B A+\\frac{1}{2} \\angle A C B=45^{\\circ}, \\angle A E B=180^{\\circ}-90^{\\circ}-22.5^{\\circ}=67.5^{\\circ}$ and $\\angle E I A=\\angle B I D=180^{\\circ}-90^{\\circ}-22.5^{\\circ}=67.5^{\\circ}$. Hence triangle $I E A$ is isosceles and a reflection of the bisector of $\\angle I A E$ takes $I$ to $E$ and $K$ to itself. So triangle $I K E$ is symmetric with respect to this axis, i.e. $\\angle K I E=\\angle I E K=\\angle B E K=45^{\\circ}$. So $90^{\\circ}$ is a possible value for $\\angle B A C$, too.\n\nIf, on the other hand, $\\angle B E K=45^{\\circ}$ then $\\angle B E K=\\angle I E K=\\angle I F K=45^{\\circ}$. Then\n\n- either $F=E$, which makes the angle bisector $B I$ be an altitude, i.e., which makes triangle $A B C$ isosceles with base $A C$ and hence equilateral and so $\\angle B A C=60^{\\circ}$,\n- or $E$ lies between $F$ and $C$, which makes the points $K, E, F$ and $I$ concyclic, so $45^{\\circ}=$ $\\angle K F C=\\angle K F E=\\angle K I E=\\angle C B I+\\angle I C B=2 \\cdot \\angle I C B=90^{\\circ}-\\frac{1}{2} \\angle B A C$, and so $\\angle B A C=90^{\\circ}$,\n\n\n\n- or $F$ lies between $E$ and $C$, then again, $K, E, F$ and $I$ are concyclic, so $45^{\\circ}=\\angle K F C=$ $180^{\\circ}-\\angle K F E=\\angle K I E$, which yields the same result $\\angle B A C=90^{\\circ}$. (However, for $\\angle B A C=90^{\\circ} E$ lies, in fact, between $F$ and $C$, see Figure 2. So this case does not occur.)\n\nThis proves $90^{\\circ}$ and $60^{\\circ}$ to be the only possible values for $\\angle B A C$.', 'Denote angles at $A, B$ and $C$ as usual by $\\alpha, \\beta$ and $\\gamma$. Since triangle $A B C$ is isosceles, we have $\\beta=\\gamma=90^{\\circ}-\\frac{\\alpha}{2}<90^{\\circ}$, so $\\angle E C K=45^{\\circ}-\\frac{\\alpha}{4}=\\angle K C D$. Since $K$ is the incenter of triangle $A D C$, we have $\\angle C D K=\\angle K D A=45^{\\circ}$; furthermore $\\angle D I C=45^{\\circ}+\\frac{\\alpha}{4}$. Now, if $\\angle B E K=45^{\\circ}$, easy calculations within triangles $B C E$ and $K C E$ yield\n\n$$\n\\begin{aligned}\n& \\angle K E C=180^{\\circ}-\\frac{\\beta}{2}-45^{\\circ}-\\beta=135^{\\circ}-\\frac{3}{2} \\beta=\\frac{3}{2}\\left(90^{\\circ}-\\beta\\right)=\\frac{3}{4} \\alpha, \\\\\n& \\angle I K E=\\frac{3}{4} \\alpha+45^{\\circ}-\\frac{\\alpha}{4}=45^{\\circ}+\\frac{\\alpha}{2} .\n\\end{aligned}\n$$\n\nSo in triangles $I C E, I K E, I D K$ and $I D C$ we have (see Figure 3)\n\n$$\n\\begin{array}{ll}\n\\frac{I C}{I E}=\\frac{\\sin \\angle I E C}{\\sin \\angle E C I}=\\frac{\\sin \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right)}{\\sin \\left(45^{\\circ}-\\frac{\\alpha}{4}\\right)}, & \\frac{I E}{I K}=\\frac{\\sin \\angle E K I}{\\sin \\angle I E K}=\\frac{\\sin \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)}{\\sin 45^{\\circ}} \\\\\n\\frac{I K}{I D}=\\frac{\\sin \\angle K D I}{\\sin \\angle I K D}=\\frac{\\sin 45^{\\circ}}{\\sin \\left(90^{\\circ}-\\frac{\\alpha}{4}\\right)}, & \\frac{I D}{I C}=\\frac{\\sin \\angle I C D}{\\sin \\angle C D I}=\\frac{\\sin \\left(45^{\\circ}-\\frac{\\alpha}{4}\\right)}{\\sin 90^{\\circ}} .\n\\end{array}\n$$\n\n\n\nFigure 3\n\nMultiplication of these four equations yields\n\n$$\n1=\\frac{\\sin \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\sin \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)}{\\sin \\left(90^{\\circ}-\\frac{\\alpha}{4}\\right)} .\n$$\n\nBut, since\n\n$$\n\\begin{aligned}\n\\sin \\left(90^{\\circ}-\\frac{\\alpha}{4}\\right) & =\\cos \\frac{\\alpha}{4}=\\cos \\left(\\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right)-\\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)\\right) \\\\\n& =\\cos \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\cos \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)+\\sin \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\sin \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)\n\\end{aligned}\n$$\n\nthis is equivalent to\n\n$$\n\\sin \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\sin \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)=\\cos \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\cos \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)+\\sin \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\sin \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)\n$$\n\nand finally\n\n$$\n\\cos \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right) \\cos \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)=0 .\n$$\n\n\n\nBut this means $\\cos \\left(45^{\\circ}+\\frac{3}{4} \\alpha\\right)=0$, hence $45^{\\circ}+\\frac{3}{4} \\alpha=90^{\\circ}$, i.e. $\\alpha=60^{\\circ}$ or $\\cos \\left(45^{\\circ}+\\frac{\\alpha}{2}\\right)=0$, hence $45^{\\circ}+\\frac{\\alpha}{2}=90^{\\circ}$, i.e. $\\alpha=90^{\\circ}$. So these values are the only two possible values for $\\alpha$.\n\nOn the other hand, both $\\alpha=90^{\\circ}$ and $\\alpha=60^{\\circ}$ yield $\\angle B E K=45^{\\circ}$.']","['$90^{\\circ}$,$60^{\\circ}$']",True,,Numerical, 2100,Geometry,,"Let $A B C$ be a triangle with circumcenter $O$. The points $P$ and $Q$ are interior points of the sides $C A$ and $A B$, respectively. The circle $k$ passes through the midpoints of the segments $B P$, $C Q$, and $P Q$. Prove that if the line $P Q$ is tangent to circle $k$ then $O P=O Q$.","['Let $K, L, M, B^{\\prime}, C^{\\prime}$ be the midpoints of $B P, C Q, P Q, C A$, and $A B$, respectively (see Figure 1). Since $C A \\| L M$, we have $\\angle L M P=\\angle Q P A$. Since $k$ touches the segment $P Q$ at $M$, we find $\\angle L M P=\\angle L K M$. Thus $\\angle Q P A=\\angle L K M$. Similarly it follows from $A B \\| M K$ that $\\angle P Q A=\\angle K L M$. Therefore, triangles $A P Q$ and $M K L$ are similar, hence\n\n$$\n\\frac{A P}{A Q}=\\frac{M K}{M L}=\\frac{\\frac{Q B}{2}}{\\frac{P C}{2}}=\\frac{Q B}{P C}\n\\tag{1}\n$$\n\nNow (1) is equivalent to $A P \\cdot P C=A Q \\cdot Q B$ which means that the power of points $P$ and $Q$ with respect to the circumcircle of $\\triangle A B C$ are equal, hence $O P=O Q$.\n\n\n\nFigure 1', 'Again, denote by $K, L, M$ the midpoints of segments $B P, C Q$, and $P Q$, respectively. Let $O, S, T$ be the circumcenters of triangles $A B C, K L M$, and $A P Q$, respectively (see Figure 2). Note that $M K$ and $L M$ are the midlines in triangles $B P Q$ and $C P Q$, respectively, so $\\overrightarrow{M K}=\\frac{1}{2} \\overrightarrow{Q B}$ and $\\overrightarrow{M L}=\\frac{1}{2} \\overrightarrow{P C}$. Denote by $\\operatorname{pr}_{l}(\\vec{v})$ the projection of vector $\\vec{v}$ onto line $l$. Then $\\operatorname{pr}_{A B}(\\overrightarrow{O T})=\\operatorname{pr}_{A B}(\\overrightarrow{O A}-\\overrightarrow{T A})=\\frac{1}{2} \\overrightarrow{B A}-\\frac{1}{2} \\overrightarrow{Q A}=\\frac{1}{2} \\overrightarrow{B Q}=\\overrightarrow{K M}$ and $\\operatorname{pr}_{A B}(\\overrightarrow{S M})=\\operatorname{pr}_{M K}(\\overrightarrow{S M})=$ $\\frac{1}{2} \\overrightarrow{K M}=\\frac{1}{2} \\operatorname{pr}_{A B}(\\overrightarrow{O T})$. Analogously we get $\\operatorname{pr}_{C A}(\\overrightarrow{S M})=\\frac{1}{2} \\operatorname{pr}_{C A}(\\overrightarrow{O T})$. Since $A B$ and $C A$ are not parallel, this implies that $\\overrightarrow{S M}=\\frac{1}{2} \\overrightarrow{O T}$.\n\n\n\nFigure 2\n\nNow, since the circle $k$ touches $P Q$ at $M$, we get $S M \\perp P Q$, hence $O T \\perp P Q$. Since $T$ is equidistant from $P$ and $Q$, the line $O T$ is a perpendicular bisector of segment $P Q$, and hence $O$ is equidistant from $P$ and $Q$ which finishes the proof.']",,True,,, 2101,Geometry,,"Let $A B C$ be a triangle. The incircle of $A B C$ touches the sides $A B$ and $A C$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $B Y$ and $C Z$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $B C Y R$ and $B C S Z$ are parallelograms. Prove that $G R=G S$.","['Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $A B C$. Let $k$ and $k_{a}$ touch the side $B C$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $A B$ and $A C$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parallelogram are of equal length, that points of contact of the excircle and incircle to a side of a triangle lie symmetric with respect to the midpoint of this side and that segments on two tangents to a circle defined by the points of contact and their point of intersection have the same length. So we conclude\n\n$$\n\\begin{gathered}\nZ P=Z B+B P=X B+B T=B X+C X=Z S \\text { and } \\\\\nC Q=C T=B X=B Z=C S .\n\\end{gathered}\n$$\n\n\n\nSo for each of the points $Z, C$, their distances to $S$ equal the length of a tangent segment from this point to $k_{a}$. It is well-known, that all points with this property lie on the line $Z C$, which is the radical axis of $S$ and $k_{a}$. Similar arguments yield that $B Y$ is the radical axis of $R$ and $k_{a}$. So the point of intersection of $Z C$ and $B Y$, which is $G$ by definition, is the radical center of $R, S$ and $k_{a}$, from which the claim $G R=G S$ follows immediately.', ""Denote $x=A Z=A Y, y=B Z=B X, z=C X=C Y, p=Z G, q=G C$. Several lengthy calculations (MENELAOs' theorem in triangle $A Z C$, law of Cosines in triangles $A B C$ and $A Z C$ and STEWART's theorem in triangle $Z C S$ ) give four equations for $p, q, \\cos \\alpha$\n\n\n\nand $G S$ in terms of $x, y$, and $z$ that can be resolved for $G S$. The result is symmetric in $y$ and $z$, so $G R=G S$. More in detail this means:\n\nThe line $B Y$ intersects the sides of triangle $A Z C$, so MEnELaOS' theorem yields $\\frac{p}{q} \\cdot \\frac{z}{x} \\cdot \\frac{x+y}{y}=1$, hence\n\n$$\n\\frac{p}{q}=\\frac{x y}{y z+z x}\n\\tag{1}\n$$\n\nSince we only want to show that the term for $G S$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N=x y+y z+z x$. So (1) implies\n\n$$\n\\frac{p}{p+q}=\\frac{x y}{x y+y z+z x}=\\frac{x y}{N} \\quad \\text { and } \\quad \\frac{q}{p+q}=\\frac{y z+z x}{x y+y z+z x}=\\frac{y z+z x}{N}\n\\tag{2}\n$$\n\nNow the law of Cosines in triangle $A B C$ yields\n\n$$\n\\cos \\alpha=\\frac{(x+y)^{2}+(x+z)^{2}-(y+z)^{2}}{2(x+y)(x+z)}=\\frac{2 x^{2}+2 x y+2 x z-2 y z}{2(x+y)(x+z)}=1-\\frac{2 y z}{(x+y)(x+z)}\n$$\n\nWe use this result to apply the law of Cosines in triangle $A Z C$ :\n\n$$\n\\begin{aligned}\n(p+q)^{2} & =x^{2}+(x+z)^{2}-2 x(x+z) \\cos \\alpha \\\\\n& =x^{2}+(x+z)^{2}-2 x(x+z) \\cdot\\left(1-\\frac{2 y z}{(x+y)(x+z)}\\right) \\\\\n& =z^{2}+\\frac{4 x y z}{x+y} .\n\\end{aligned}\n\\tag{3}\n$$\n\nNow in triangle $Z C S$ the segment $G S$ is a cevian, so with STEWART's theorem we have $p y^{2}+q(y+z)^{2}=(p+q)\\left(G S^{2}+p q\\right)$, hence\n\n$$\nG S^{2}=\\frac{p}{p+q} \\cdot y^{2}+\\frac{q}{p+q} \\cdot(y+z)^{2}-\\frac{p}{p+q} \\cdot \\frac{q}{p+q} \\cdot(p+q)^{2}\n$$\n\nReplacing the $p$ 's and $q$ 's herein by (2) and (3) yields\n\n$$\n\\begin{aligned}\nG S^{2} & =\\frac{x y}{N} y^{2}+\\frac{y z+z x}{N}(y+z)^{2}-\\frac{x y}{N} \\cdot \\frac{y z+z x}{N} \\cdot\\left(z^{2}+\\frac{4 x y z}{x+y}\\right) \\\\\n& =\\frac{x y^{3}}{N}+\\underbrace{\\frac{y z(y+z)^{2}}{N}}_{M_{1}}+\\frac{z x(y+z)^{2}}{N}-\\frac{x y z^{3}(x+y)}{N^{2}}-\\underbrace{\\frac{4 x^{2} y^{2} z^{2}}{N^{2}}}_{M_{2}} \\\\\n& =\\frac{x y^{3}+z x(y+z)^{2}}{N}-\\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\\\\n& =\\underbrace{\\frac{x\\left(y^{3}+y^{2} z+y z^{2}+z^{3}\\right)}{N}}_{M_{3}}+\\frac{x y z^{2} N}{N^{2}}-\\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\\\\n& =\\frac{x^{2} y^{2} z^{2}+x y^{2} z^{3}+x^{2} y z^{3}-x^{2} y z^{3}-x y^{2} z^{3}}{N^{2}}+M_{1}-M_{2}+M_{3} \\\\\n& =\\frac{x^{2} y^{2} z^{2}}{N^{2}}+M_{1}-M_{2}+M_{3},\n\\end{aligned}\n$$\n\na term that is symmetric in $y$ and $z$, indeed.""]",,True,,, 2102,Geometry,,"Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$.","['It suffices to show that $\\angle H E F=\\angle H G E$ (see Figure 1), since in circle $E G H$ the angle over the chord $E H$ at $G$ equals the angle between the tangent at $E$ and $E H$.\n\nFirst, $\\angle B A D=180^{\\circ}-\\angle D C B=\\angle F C D$. Since triangles $F A B$ and $F C D$ have also a common interior angle at $F$, they are similar.\n\n\n\nFigure 1\n\nDenote by $\\mathcal{T}$ the transformation consisting of a reflection at the bisector of $\\angle D F C$ followed by a dilation with center $F$ and factor of $\\frac{F A}{F C}$. Then $\\mathcal{T}$ maps $F$ to $F, C$ to $A, D$ to $B$, and $H$ to $G$. To see this, note that $\\triangle F C A \\sim \\triangle F D B$, so $\\frac{F A}{F C}=\\frac{F B}{F D}$. Moreover, as $\\angle A D B=\\angle A C B$, the image of the line $D E$ under $\\mathcal{T}$ is parallel to $A C$ (and passes through $B$ ) and similarly the image of $C E$ is parallel to $D B$ and passes through $A$. Hence $E$ is mapped to the point $X$ which is the fourth vertex of the parallelogram $B E A X$. Thus, in particular $\\angle H E F=\\angle F X G$.\n\nAs $G$ is the midpoint of the diagonal $A B$ of the parallelogram $B E A X$, it is also the midpoint of $E X$. In particular, $E, G, X$ are collinear, and $E X=2 \\cdot E G$.\n\nDenote by $Y$ the fourth vertex of the parallelogram $D E C Y$. By an analogous reasoning as before, it follows that $\\mathcal{T}$ maps $Y$ to $E$, thus $E, H, Y$ are collinear with $E Y=2 \\cdot E H$. Therefore, by the intercept theorem, $H G \\| X Y$.\n\nFrom the construction of $\\mathcal{T}$ it is clear that the lines $F X$ and $F E$ are symmetric with respect to the bisector of $\\angle D F C$, as are $F Y$ and $F E$. Thus, $F, X, Y$ are collinear, which together with $H G \\| X Y$ implies $\\angle F X E=\\angle H G E$. This completes the proof.', 'We use the following\n\nLemma (Gauß). Let $A B C D$ be a quadrilateral. Let $A B$ and $C D$ intersect at $P$, and $B C$ and $D A$ intersect at $Q$. Then the midpoints $K, L, M$ of $A C, B D$, and $P Q$, respectively, are collinear.\n\nProof: Let us consider the points $Z$ that fulfill the equation\n\n$$\n(A B Z)+(C D Z)=(B C Z)+(D A Z),\n\\tag{1}\n$$\n\nwhere $(R S T)$ denotes the oriented area of the triangle $R S T$ (see Figure 2).\n\n\n\nFigure 2\n\nAs (1) is linear in $Z$, it can either characterize a line, or be contradictory, or be trivially fulfilled for all $Z$ in the plane. If (1) was fulfilled for all $Z$, then it would hold for $Z=A, Z=B$, which gives $(C D A)=(B C A),(C D B)=(D A B)$, respectively, i.e. the diagonals of $A B C D$ would bisect each other, thus $A B C D$ would be a parallelogram. This contradicts the hypothesis that $A D$ and $B C$ intersect. Since $E, F, G$ fulfill (1), it is the equation of a line which completes the proof of the lemma.\n\nNow consider the parallelograms $E A X B$ and $E C Y D$ (see Figure 1). Then $G, H$ are the midpoints of $E X, E Y$, respectively. Let $M$ be the midpoint of $E F$. By applying the Lemma to the (re-entrant) quadrilateral $A D B C$, it is evident that $G, H$, and $M$ are collinear. A dilation by a factor of 2 with center $E$ shows that $X, Y, F$ are collinear. Since $A X \\| D E$ and $B X \\| C E$, we have pairwise equal interior angles in the quadrilaterals $F D E C$ and $F B X A$. Since we have also $\\angle E B A=\\angle D C A=\\angle C D Y$, the quadrilaterals are similar. Thus, $\\angle F X A=\\angle C E F$.\n\nClearly the parallelograms $E C Y D$ and $E B X A$ are similar, too, thus $\\angle E X A=\\angle C E Y$. Consequently, $\\angle F X E=\\angle F X A-\\angle E X A=\\angle C E F-\\angle C E Y=\\angle Y E F$. By the converse of the tangent-chord angle theorem $E F$ is tangent to the circle $X E Y$. A dilation by a factor of $\\frac{1}{2}$ completes the proof.', 'As in Solution 2, G, H, M are proven to be collinear. It suffices to show that $M E^{2}=M G \\cdot M H$. If $\\boldsymbol{p}=\\overrightarrow{O P}$ denotes the vector from circumcenter $O$ to point $P$, the claim becomes\n\n$$\n\\left(\\frac{\\boldsymbol{e}-\\boldsymbol{f}}{2}\\right)^{2}=\\left(\\frac{\\boldsymbol{e}+\\boldsymbol{f}}{2}-\\frac{\\boldsymbol{a}+\\boldsymbol{b}}{2}\\right)\\left(\\frac{\\boldsymbol{e}+\\boldsymbol{f}}{2}-\\frac{\\boldsymbol{c}+\\boldsymbol{d}}{2}\\right)\n$$\n\nor equivalently\n\n$$\n4 \\boldsymbol{e} \\boldsymbol{f}-(\\boldsymbol{e}+\\boldsymbol{f})(\\boldsymbol{a}+\\boldsymbol{b}+\\boldsymbol{c}+\\boldsymbol{d})+(\\boldsymbol{a}+\\boldsymbol{b})(\\boldsymbol{c}+\\boldsymbol{d})=0\n\\tag{2}\n$$\n\nWith $R$ as the circumradius of $A B C D$, we obtain for the powers $\\mathcal{P}(E)$ and $\\mathcal{P}(F)$ of $E$ and $F$, respectively, with respect to the circumcircle\n\n$$\n\\begin{aligned}\n& \\mathcal{P}(E)=(\\boldsymbol{e}-\\boldsymbol{a})(\\boldsymbol{e}-\\boldsymbol{c})=(\\boldsymbol{e}-\\boldsymbol{b})(\\boldsymbol{e}-\\boldsymbol{d})=\\boldsymbol{e}^{2}-R^{2} \\\\\n& \\mathcal{P}(F)=(\\boldsymbol{f}-\\boldsymbol{a})(\\boldsymbol{f}-\\boldsymbol{d})=(\\boldsymbol{f}-\\boldsymbol{b})(\\boldsymbol{f}-\\boldsymbol{c})=\\boldsymbol{f}^{2}-R^{2}\n\\end{aligned}\n$$\n\nhence\n\n$$\n(\\boldsymbol{e}-\\boldsymbol{a})(\\boldsymbol{e}-\\boldsymbol{c})=\\boldsymbol{e}^{2}-R^{2}\n\\tag{3}\n$$\n$$\n(\\boldsymbol{e}-\\boldsymbol{b})(\\boldsymbol{e}-\\boldsymbol{d})=\\boldsymbol{e}^{2}-R^{2}\n\\tag{4}\n$$\n$$\n(\\boldsymbol{f}-\\boldsymbol{a})(\\boldsymbol{f}-\\boldsymbol{d})=\\boldsymbol{f}^{2}-R^{2}\n\\tag{5}\n$$\n$$\n(\\boldsymbol{f}-\\boldsymbol{b})(\\boldsymbol{f}-\\boldsymbol{c})=\\boldsymbol{f}^{2}-R^{2}\n\\tag{6}\n$$\n\nSince $F$ lies on the polar to $E$ with respect to the circumcircle, we have\n\n$$\n4 \\boldsymbol{e} \\boldsymbol{f}=4 R^{2}\n\\tag{7}\n$$\n\nAdding up (3) to (7) yields (2), as desired.']",,True,,, 2103,Geometry,,"Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have $$ \frac{|R|}{|P|} \leq \sqrt{2} $$ where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively.","['We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $\\left|R_{1}\\right| \\leq \\sqrt{2}|P|$ and $\\left|R_{3}\\right| \\leq \\sqrt{2}|P|$ holds (see Figure 1).\n\nFirst we will construct a parallelogram $R_{1} \\supseteq P$ with the property that the midpoints of the sides of $R_{1}$ are points of the boundary of $P$.\n\nChoose two points $A$ and $B$ of $P$ such that the triangle $O A B$ has maximal area. Let $a$ be the line through $A$ parallel to $O B$ and $b$ the line through $B$ parallel to $O A$. Let $A^{\\prime}, B^{\\prime}, a^{\\prime}$ and $b^{\\prime}$ be the points or lines, that are symmetric to $A, B, a$ and $b$, respectively, with respect to $O$. Now let $R_{1}$ be the parallelogram defined by $a, b, a^{\\prime}$ and $b^{\\prime}$.\n\n\n\nFigure 1\n\nObviously, $A$ and $B$ are located on the boundary of the polygon $P$, and $A, B, A^{\\prime}$ and $B^{\\prime}$ are midpoints of the sides of $R_{1}$. We note that $P \\subseteq R_{1}$. Otherwise, there would be a point $Z \\in P$ but $Z \\notin R_{1}$, i.e., one of the lines $a, b, a^{\\prime}$ or $b^{\\prime}$ were between $O$ and $Z$. If it is $a$, we have $|O Z B|>|O A B|$, which is contradictory to the choice of $A$ and $B$. If it is one of the lines $b, a^{\\prime}$ or $b^{\\prime}$ almost identical arguments lead to a similar contradiction.\n\nLet $R_{2}$ be the parallelogram $A B A^{\\prime} B^{\\prime}$. Since $A$ and $B$ are points of $P$, segment $A B \\subset P$ and so $R_{2} \\subset R_{1}$. Since $A, B, A^{\\prime}$ and $B^{\\prime}$ are midpoints of the sides of $R_{1}$, an easy argument yields\n\n$$\n\\left|R_{1}\\right|=2 \\cdot\\left|R_{2}\\right|\n\\tag{1}\n$$\n\nLet $R_{3}$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $A B$ and $B A^{\\prime}$. Obviously $R_{2} \\subset R_{3}$ and every side of $R_{3}$ contains at least one point of the boundary of $P$. Denote by $C$ the intersection point of $a$ and $b$, by $X$ the intersection point of $A B$ and $O C$, and by $X^{\\prime}$ the intersection point of $X C$ and the boundary of $R_{3}$. In a similar way denote by $D$\n\n\n\nthe intersection point of $b$ and $a^{\\prime}$, by $Y$ the intersection point of $A^{\\prime} B$ and $O D$, and by $Y^{\\prime}$ the intersection point of $Y D$ and the boundary of $R_{3}$.\n\nNote that $O C=2 \\cdot O X$ and $O D=2 \\cdot O Y$, so there exist real numbers $x$ and $y$ with $1 \\leq x, y \\leq 2$ and $O X^{\\prime}=x \\cdot O X$ and $O Y^{\\prime}=y \\cdot O Y$. Corresponding sides of $R_{3}$ and $R_{2}$ are parallel which yields\n\n$$\n\\left|R_{3}\\right|=x y \\cdot\\left|R_{2}\\right| .\n\\tag{2}\n$$\n\nThe side of $R_{3}$ containing $X^{\\prime}$ contains at least one point $X^{*}$ of $P$; due to the convexity of $P$ we have $A X^{*} B \\subset P$. Since this side of the parallelogram $R_{3}$ is parallel to $A B$ we have $\\left|A X^{*} B\\right|=\\left|A X^{\\prime} B\\right|$, so $\\left|O A X^{\\prime} B\\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A$. In a similar way we conclude that $\\left|O B^{\\prime} Y^{\\prime} A^{\\prime}\\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A^{\\prime}$. Putting things together we have $\\left|O A X^{\\prime} B\\right|=x \\cdot|O A B|,\\left|O B D A^{\\prime}\\right|=y \\cdot\\left|O B A^{\\prime}\\right|$. Since $|O A B|=\\left|O B A^{\\prime}\\right|$, we conclude that $|P| \\geq 2 \\cdot\\left|A X^{\\prime} B Y^{\\prime} A^{\\prime}\\right|=2 \\cdot\\left(x \\cdot|O A B|+y \\cdot\\left|O B A^{\\prime}\\right|\\right)=4 \\cdot \\frac{x+y}{2} \\cdot|O A B|=\\frac{x+y}{2} \\cdot R_{2}$; this is in short\n\n$$\n\\frac{x+y}{2} \\cdot\\left|R_{2}\\right| \\leq|P| \\text {. }\n\\tag{3}\n$$\n\nSince all numbers concerned are positive, we can combine (1)-(3). Using the arithmeticgeometric-mean inequality we obtain\n\n$$\n\\left|R_{1}\\right| \\cdot\\left|R_{3}\\right|=2 \\cdot\\left|R_{2}\\right| \\cdot x y \\cdot\\left|R_{2}\\right| \\leq 2 \\cdot\\left|R_{2}\\right|^{2}\\left(\\frac{x+y}{2}\\right)^{2} \\leq 2 \\cdot|P|^{2}\n$$\n\nThis implies immediately the desired result $\\left|R_{1}\\right| \\leq \\sqrt{2} \\cdot|P|$ or $\\left|R_{3}\\right| \\leq \\sqrt{2} \\cdot|P|$.', 'We construct the parallelograms $R_{1}, R_{2}$ and $R_{3}$ in the same way as in Solution 1 and will show that $\\frac{\\left|R_{1}\\right|}{|P|} \\leq \\sqrt{2}$ or $\\frac{\\left|R_{3}\\right|}{|P|} \\leq \\sqrt{2}$.\n\n\n\nFigure 2\n\nRecall that affine one-to-one maps of the plane preserve the ratio of areas of subsets of the plane. On the other hand, every parallelogram can be transformed with an affine map onto a square. It follows that without loss of generality we may assume that $R_{1}$ is a square (see Figure 2).\n\nThen $R_{2}$, whose vertices are the midpoints of the sides of $R_{1}$, is a square too, and $R_{3}$, whose sides are parallel to the diagonals of $R_{1}$, is a rectangle.\n\nLet $a>0, b \\geq 0$ and $c \\geq 0$ be the distances introduced in Figure 2. Then $\\left|R_{1}\\right|=2 a^{2}$ and\n\n\n\n$\\left|R_{3}\\right|=(a+2 b)(a+2 c)$.\n\nPoints $A, A^{\\prime}, B$ and $B^{\\prime}$ are in the convex polygon $P$. Hence the square $A B A^{\\prime} B^{\\prime}$ is a subset of $P$. Moreover, each of the sides of the rectangle $R_{3}$ contains a point of $P$, otherwise $R_{3}$ would not be minimal. It follows that\n\n$$\n|P| \\geq a^{2}+2 \\cdot \\frac{a b}{2}+2 \\cdot \\frac{a c}{2}=a(a+b+c)\n$$\n\nNow assume that both $\\frac{\\left|R_{1}\\right|}{|P|}>\\sqrt{2}$ and $\\frac{\\left|R_{3}\\right|}{|P|}>\\sqrt{2}$, then\n\n$$\n2 a^{2}=\\left|R_{1}\\right|>\\sqrt{2} \\cdot|P| \\geq \\sqrt{2} \\cdot a(a+b+c)\n$$\n\nand\n\n$$\n(a+2 b)(a+2 c)=\\left|R_{3}\\right|>\\sqrt{2} \\cdot|P| \\geq \\sqrt{2} \\cdot a(a+b+c)\n$$\n\nAll numbers concerned are positive, so after multiplying these inequalities we get\n\n$$\n2 a^{2}(a+2 b)(a+2 c)>2 a^{2}(a+b+c)^{2} .\n$$\n\nBut the arithmetic-geometric-mean inequality implies the contradictory result\n\n$$\n2 a^{2}(a+2 b)(a+2 c) \\leq 2 a^{2}\\left(\\frac{(a+2 b)+(a+2 c)}{2}\\right)^{2}=2 a^{2}(a+b+c)^{2}\n$$\n\nHence $\\frac{\\left|R_{1}\\right|}{|P|} \\leq \\sqrt{2}$ or $\\frac{\\left|R_{3}\\right|}{|P|} \\leq \\sqrt{2}$, as desired.']",,True,,, 2104,Geometry,,"Let the sides $A D$ and $B C$ of the quadrilateral $A B C D$ (such that $A B$ is not parallel to $C D$ ) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $A B P$ and $D C P$, respectively. Denote the midpoints of segments $O_{1} H_{1}$ and $O_{2} H_{2}$ by $E_{1}$ and $E_{2}$, respectively. Prove that the perpendicular from $E_{1}$ on $C D$, the perpendicular from $E_{2}$ on $A B$ and the line $H_{1} H_{2}$ are concurrent.","[""We keep triangle $A B P$ fixed and move the line $C D$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $\\mathrm{O}_{2}, \\mathrm{H}_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$ on $A B$ moves uniformly. Obviously, the points $O_{1}, H_{1}, E_{1}$ and the perpendicular from $E_{1}$ on $C D$ do not move at all. Hence, the intersection point $S$ of these two perpendiculars moves uniformly. Since $H_{1}$ does not move, while $H_{2}$ and $S$ move uniformly along parallel lines (both are perpendicular to $C D$ ), it is sufficient to prove their collinearity for two different positions of $C D$.\n\n\n\nFigure 1\n\nLet $C D$ pass through either point $A$ or point $B$. Note that by hypothesis these two cases are different. We will consider the case $A \\in C D$, i.e. $A=D$. So we have to show that the perpendiculars from $E_{1}$ on $A C$ and from $E_{2}$ on $A B$ intersect on the altitude $A H$ of triangle $A B C$ (see Figure 2).\n\n\n\n\n\nFigure 2\n\nTo this end, we consider the midpoints $A_{1}, B_{1}, C_{1}$ of $B C, C A, A B$, respectively. As $E_{1}$ is the center of FEUERBACH's circle (nine-point circle) of $\\triangle A B P$, we have $E_{1} C_{1}=E_{1} H$. Similarly, $E_{2} B_{1}=E_{2} H$. Note further that a point $X$ lies on the perpendicular from $E_{1}$ on $A_{1} C_{1}$ if and only if\n\n$$\nX C_{1}^{2}-X A_{1}^{2}=E_{1} C_{1}^{2}-E_{1} A_{1}^{2}\n$$\n\nSimilarly, the perpendicular from $E_{2}$ on $A_{1} B_{1}$ is characterized by\n\n$$\nX A_{1}^{2}-X B_{1}^{2}=E_{2} A_{1}^{2}-E_{2} B_{1}^{2}\n$$\n\nThe line $H_{1} H_{2}$, which is perpendicular to $B_{1} C_{1}$ and contains $A$, is given by\n\n$$\nX B_{1}^{2}-X C_{1}^{2}=A B_{1}^{2}-A C_{1}^{2}\n$$\n\nThe three lines are concurrent if and only if\n\n$$\n\\begin{aligned}\n0 & =X C_{1}^{2}-X A_{1}^{2}+X A_{1}^{2}-X B_{1}^{2}+X B_{1}^{2}-X C_{1}^{2} \\\\\n& =E_{1} C_{1}^{2}-E_{1} A_{1}^{2}+E_{2} A_{1}^{2}-E_{2} B_{1}^{2}+A B_{1}^{2}-A C_{1}^{2} \\\\\n& =-E_{1} A_{1}^{2}+E_{2} A_{1}^{2}+E_{1} H^{2}-E_{2} H^{2}+A B_{1}^{2}-A C_{1}^{2}\n\\end{aligned}\n$$\n\ni.e. it suffices to show that\n\n$$\nE_{1} A_{1}^{2}-E_{2} A_{1}^{2}-E_{1} H^{2}+E_{2} H^{2}=\\frac{A C^{2}-A B^{2}}{4}\n$$\n\nWe have\n\n$$\n\\frac{A C^{2}-A B^{2}}{4}=\\frac{H C^{2}-H B^{2}}{4}=\\frac{(H C+H B)(H C-H B)}{4}=\\frac{H A_{1} \\cdot B C}{2}\n$$\n\nLet $F_{1}, F_{2}$ be the projections of $E_{1}, E_{2}$ on $B C$. Obviously, these are the midpoints of $H P_{1}$,\n\n\n\n$H P_{2}$, where $P_{1}, P_{2}$ are the midpoints of $P B$ and $P C$ respectively. Then\n\n$$\n\\begin{aligned}\n& E_{1} A_{1}^{2}-E_{2} A_{1}^{2}-E_{1} H^{2}+E_{2} H^{2} \\\\\n& =F_{1} A_{1}^{2}-F_{1} H^{2}-F_{2} A_{1}^{2}+F_{2} H^{2} \\\\\n& =\\left(F_{1} A_{1}-F_{1} H\\right)\\left(F_{1} A_{1}+F_{1} H\\right)-\\left(F_{2} A_{1}-F_{2} H\\right)\\left(F_{2} A_{1}+F_{2} H\\right) \\\\\n& =A_{1} H \\cdot\\left(A_{1} P_{1}-A_{1} P_{2}\\right) \\\\\n& =\\frac{A_{1} H \\cdot B C}{2} \\\\\n& =\\frac{A C^{2}-A B^{2}}{4}\n\\end{aligned}\n$$\n\nwhich proves the claim."", 'Let the perpendicular from $E_{1}$ on $C D$ meet $P H_{1}$ at $X$, and the perpendicular from $E_{2}$ on $A B$ meet $P H_{2}$ at $Y$ (see Figure 3). Let $\\varphi$ be the intersection angle of $A B$ and $C D$. Denote by $M, N$ the midpoints of $P H_{1}, P H_{2}$ respectively.\n\n\n\nFigure 3\n\nWe will prove now that triangles $E_{1} X M$ and $E_{2} Y N$ have equal angles at $E_{1}, E_{2}$, and supplementary angles at $X, Y$.\n\nIn the following, angles are understood as oriented, and equalities of angles modulo $180^{\\circ}$.\n\nLet $\\alpha=\\angle H_{2} P D, \\psi=\\angle D P C, \\beta=\\angle C P H_{1}$. Then $\\alpha+\\psi+\\beta=\\varphi, \\angle E_{1} X H_{1}=\\angle H_{2} Y E_{2}=\\varphi$, thus $\\angle M X E_{1}+\\angle N Y E_{2}=180^{\\circ}$.\n\nBy considering the FEUERBACH circle of $\\triangle A B P$ whose center is $E_{1}$ and which goes through $M$, we have $\\angle E_{1} M H_{1}=\\psi+2 \\beta$. Analogous considerations with the FEUERBACH circle of $\\triangle D C P$ yield $\\angle H_{2} N E_{2}=\\psi+2 \\alpha$. Hence indeed $\\angle X E_{1} M=\\varphi-(\\psi+2 \\beta)=(\\psi+2 \\alpha)-\\varphi=\\angle Y E_{2} N$.\n\nIt follows now that\n\n$$\n\\frac{X M}{M E_{1}}=\\frac{Y N}{N E_{2}}\n$$\n\nFurthermore, $M E_{1}$ is half the circumradius of $\\triangle A B P$, while $P H_{1}$ is the distance of $P$ to the orthocenter of that triangle, which is twice the circumradius times the cosine of $\\psi$. Together\n\n\n\nwith analogous reasoning for $\\triangle D C P$ we have\n\n$$\n\\frac{M E_{1}}{P H_{1}}=\\frac{1}{4 \\cos \\psi}=\\frac{N E_{2}}{P H_{2}}\n$$\n\nBy multiplication,\n\n$$\n\\frac{X M}{P H_{1}}=\\frac{Y N}{P H_{2}}\n$$\n\nand therefore\n\n$$\n\\frac{P X}{X H_{1}}=\\frac{H_{2} Y}{Y P}\n$$\n\nLet $E_{1} X, E_{2} Y$ meet $H_{1} H_{2}$ in $R, S$ respectively.\n\nApplying the intercept theorem to the parallels $E_{1} X, P H_{2}$ and center $H_{1}$ gives\n\n$$\n\\frac{H_{2} R}{R H_{1}}=\\frac{P X}{X H_{1}}\n$$\n\nwhile with parallels $E_{2} Y, P H_{1}$ and center $H_{2}$ we obtain\n\n$$\n\\frac{H_{2} S}{S H_{1}}=\\frac{H_{2} Y}{Y P}\n$$\n\nCombination of the last three equalities yields that $R$ and $S$ coincide.']",,True,,, 2105,Geometry,,"Let $A B C$ be a triangle with incenter $I$ and let $X, Y$ and $Z$ be the incenters of the triangles $B I C, C I A$ and $A I B$, respectively. Let the triangle $X Y Z$ be equilateral. Prove that $A B C$ is equilateral too.","['$A Z, A I$ and $A Y$ divide $\\angle B A C$ into four equal angles; denote them by $\\alpha$. In the same way we have four equal angles $\\beta$ at $B$ and four equal angles $\\gamma$ at $C$. Obviously $\\alpha+\\beta+\\gamma=\\frac{180^{\\circ}}{4}=45^{\\circ}$; and $0^{\\circ}<\\alpha, \\beta, \\gamma<45^{\\circ}$.\n\n\n\nEasy calculations in various triangles yield $\\angle B I C=180^{\\circ}-2 \\beta-2 \\gamma=180^{\\circ}-\\left(90^{\\circ}-2 \\alpha\\right)=$ $90^{\\circ}+2 \\alpha$, hence (for $X$ is the incenter of triangle $B C I$, so $I X$ bisects $\\angle B I C$ ) we have $\\angle X I C=$ $\\angle B I X=\\frac{1}{2} \\angle B I C=45^{\\circ}+\\alpha$ and with similar aguments $\\angle C I Y=\\angle Y I A=45^{\\circ}+\\beta$ and $\\angle A I Z=\\angle Z I B=45^{\\circ}+\\gamma$. Furthermore, we have $\\angle X I Y=\\angle X I C+\\angle C I Y=\\left(45^{\\circ}+\\alpha\\right)+$ $\\left(45^{\\circ}+\\beta\\right)=135^{\\circ}-\\gamma, \\angle Y I Z=135^{\\circ}-\\alpha$, and $\\angle Z I X=135^{\\circ}-\\beta$.\n\nNow we calculate the lengths of $I X, I Y$ and $I Z$ in terms of $\\alpha, \\beta$ and $\\gamma$. The perpendicular from $I$ on $C X$ has length $I X \\cdot \\sin \\angle C X I=I X \\cdot \\sin \\left(90^{\\circ}+\\beta\\right)=I X \\cdot \\cos \\beta$. But $C I$ bisects $\\angle Y C X$, so the perpendicular from $I$ on $C Y$ has the same length, and we conclude\n\n$$\nI X \\cdot \\cos \\beta=I Y \\cdot \\cos \\alpha\n$$\n\nTo make calculations easier we choose a length unit that makes $I X=\\cos \\alpha$. Then $I Y=\\cos \\beta$ and with similar arguments $I Z=\\cos \\gamma$.\n\nSince $X Y Z$ is equilateral we have $Z X=Z Y$. The law of Cosines in triangles $X Y I, Y Z I$ yields\n\n$$\n\\begin{aligned}\n& Z X^{2}=Z Y^{2} \\\\\n\\Longrightarrow & I Z^{2}+I X^{2}-2 \\cdot I Z \\cdot I X \\cdot \\cos \\angle Z I X=I Z^{2}+I Y^{2}-2 \\cdot I Z \\cdot I Y \\cdot \\cos \\angle Y I Z \\\\\n\\Longrightarrow & I X^{2}-I Y^{2}=2 \\cdot I Z \\cdot(I X \\cdot \\cos \\angle Z I X-I Y \\cdot \\cos \\angle Y I Z) \\\\\n\\Longrightarrow & \\underbrace{\\cos ^{2} \\alpha-\\cos ^{2} \\beta}_{\\text {L.H.S. }}=\\underbrace{2 \\cdot \\cos \\gamma \\cdot\\left(\\cos \\alpha \\cdot \\cos \\left(135^{\\circ}-\\beta\\right)-\\cos \\beta \\cdot \\cos \\left(135^{\\circ}-\\alpha\\right)\\right)}_{\\text {R.H.S. }} .\n\\end{aligned}\n$$\n\nA transformation of the left-hand side (L.H.S.) yields\n\n$$\n\\begin{aligned}\n\\text { L.H.S. } & =\\cos ^{2} \\alpha \\cdot\\left(\\sin ^{2} \\beta+\\cos ^{2} \\beta\\right)-\\cos ^{2} \\beta \\cdot\\left(\\sin ^{2} \\alpha+\\cos ^{2} \\alpha\\right) \\\\\n& =\\cos ^{2} \\alpha \\cdot \\sin ^{2} \\beta-\\cos ^{2} \\beta \\cdot \\sin ^{2} \\alpha\n\\end{aligned}\n$$\n\n\n\n$$\n\\begin{aligned}\n& =(\\cos \\alpha \\cdot \\sin \\beta+\\cos \\beta \\cdot \\sin \\alpha) \\cdot(\\cos \\alpha \\cdot \\sin \\beta-\\cos \\beta \\cdot \\sin \\alpha) \\\\\n& =\\sin (\\beta+\\alpha) \\cdot \\sin (\\beta-\\alpha)=\\sin \\left(45^{\\circ}-\\gamma\\right) \\cdot \\sin (\\beta-\\alpha)\n\\end{aligned}\n$$\n\nwhereas a transformation of the right-hand side (R.H.S.) leads to\n\n$$\n\\begin{aligned}\n\\text { R.H.S. } & =2 \\cdot \\cos \\gamma \\cdot\\left(\\cos \\alpha \\cdot\\left(-\\cos \\left(45^{\\circ}+\\beta\\right)\\right)-\\cos \\beta \\cdot\\left(-\\cos \\left(45^{\\circ}+\\alpha\\right)\\right)\\right) \\\\\n& =2 \\cdot \\frac{\\sqrt{2}}{2} \\cdot \\cos \\gamma \\cdot(\\cos \\alpha \\cdot(\\sin \\beta-\\cos \\beta)+\\cos \\beta \\cdot(\\cos \\alpha-\\sin \\alpha)) \\\\\n& =\\sqrt{2} \\cdot \\cos \\gamma \\cdot(\\cos \\alpha \\cdot \\sin \\beta-\\cos \\beta \\cdot \\sin \\alpha) \\\\\n& =\\sqrt{2} \\cdot \\cos \\gamma \\cdot \\sin (\\beta-\\alpha)\n\\end{aligned}\n$$\n\nEquating L.H.S. and R.H.S. we obtain\n\n$$\n\\begin{aligned}\n& \\sin \\left(45^{\\circ}-\\gamma\\right) \\cdot \\sin (\\beta-\\alpha)=\\sqrt{2} \\cdot \\cos \\gamma \\cdot \\sin (\\beta-\\alpha) \\\\\n\\Longrightarrow & \\sin (\\beta-\\alpha) \\cdot\\left(\\sqrt{2} \\cdot \\cos \\gamma-\\sin \\left(45^{\\circ}-\\gamma\\right)\\right)=0 \\\\\n\\Longrightarrow & \\alpha=\\beta \\quad \\text { or } \\quad \\sqrt{2} \\cdot \\cos \\gamma=\\sin \\left(45^{\\circ}-\\gamma\\right) .\n\\end{aligned}\n$$\n\nBut $\\gamma<45^{\\circ}$; so $\\sqrt{2} \\cdot \\cos \\gamma>\\cos \\gamma>\\cos 45^{\\circ}=\\sin 45^{\\circ}>\\sin \\left(45^{\\circ}-\\gamma\\right)$. This leaves $\\alpha=\\beta$. With similar reasoning we have $\\alpha=\\gamma$, which means triangle $A B C$ must be equilateral.']",,True,,, 2106,Geometry,,"Let $A B C D$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $B C$ in $M$ and the line $C D$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\triangle A B M$, $\triangle M N C$, and $\triangle N D A$, respectively. Show that the orthocenter of $\triangle I_{1} I_{2} I_{3}$ lies on $g$.","['Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $A B M, M N C$, and $N D A$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $C B$ is also tangent to $k_{3}$.\n\n\n\nFigure 1\n\nTo this end, let $X$ denote the point of intersection of $g$ and $h$. Then $A B C X$ and $A B C D$ are circumscribed quadrilaterals, whence\n\n$$\nC D-C X=(A B+C D)-(A B+C X)=(B C+A D)-(B C+A X)=A D-A X\n$$\n\ni.e.\n\n$$\nA X+C D=C X+A D\n$$\n\nwhich in turn reveals that the quadrilateral $A X C D$ is also circumscribed. Thus $h$ touches indeed the circle $k_{3}$.\n\nMoreover, we find that $\\angle I_{3} C I_{1}=\\angle I_{3} C X+\\angle X C I_{1}=\\frac{1}{2}(\\angle D C X+\\angle X C B)=\\frac{1}{2} \\angle D C B=$ $\\frac{1}{2}\\left(180^{\\circ}-\\angle M C N\\right)=180^{\\circ}-\\angle M I_{2} N=\\angle I_{3} I_{2} I_{1}$, from which we conclude that $C, I_{1}, I_{2}, I_{3}$ are concyclic.\n\nLet now $L_{1}$ and $L_{3}$ be the reflection points of $C$ with respect to the lines $I_{2} I_{3}$ and $I_{1} I_{2}$ respectively. Since $I_{1} I_{2}$ is the angle bisector of $\\angle N M C$, it follows that $L_{3}$ lies on $g$. By analogous reasoning, $L_{1}$ lies on $g$.\n\nLet $H$ be the orthocenter of $\\triangle I_{1} I_{2} I_{3}$. We have $\\angle I_{2} L_{3} I_{1}=\\angle I_{1} C I_{2}=\\angle I_{1} I_{3} I_{2}=180^{\\circ}-\\angle I_{1} H I_{2}$, which entails that the quadrilateral $I_{2} H I_{1} L_{3}$ is cyclic. Analogously, $I_{3} H L_{1} I_{2}$ is cyclic.\n\n\n\nThen, working with oriented angles modulo $180^{\\circ}$, we have\n\n$$\n\\angle L_{3} H I_{2}=\\angle L_{3} I_{1} I_{2}=\\angle I_{2} I_{1} C=\\angle I_{2} I_{3} C=\\angle L_{1} I_{3} I_{2}=\\angle L_{1} H I_{2},\n$$\n\nwhence $L_{1}, L_{3}$, and $H$ are collinear. By $L_{1} \\neq L_{3}$, the claim follows.', 'We start by proving that $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic.\n\n\n\nFigure 2\n\nTo this end, notice first that $I_{2}, M, I_{1}$ are collinear, as are $N, I_{2}, I_{3}$ (see Figure 2). Denote by $\\alpha, \\beta, \\gamma, \\delta$ the internal angles of $A B C D$. By considerations in triangle $C M N$, it follows that $\\angle I_{3} I_{2} I_{1}=\\frac{\\gamma}{2}$. We will show that $\\angle I_{3} C I_{1}=\\frac{\\gamma}{2}$, too. Denote by $I$ the incenter of $A B C D$. Clearly, $I_{1} \\in B I, I_{3} \\in D I, \\angle I_{1} A I_{3}=\\frac{\\alpha}{2}$.\n\nUsing the abbreviation $[X, Y Z]$ for the distance from point $X$ to the line $Y Z$, we have because of $\\angle B A I_{1}=\\angle I A I_{3}$ and $\\angle I_{1} A I=\\angle I_{3} A D$ that\n\n$$\n\\frac{\\left[I_{1}, A B\\right]}{\\left[I_{1}, A I\\right]}=\\frac{\\left[I_{3}, A I\\right]}{\\left[I_{3}, A D\\right]}\n$$\n\nFurthermore, consideration of the angle sums in $A I B, B I C, C I D$ and $D I A$ implies $\\angle A I B+$ $\\angle C I D=\\angle B I C+\\angle D I A=180^{\\circ}$, from which we see\n\n$$\n\\frac{\\left[I_{1}, A I\\right]}{\\left[I_{3}, C I\\right]}=\\frac{I_{1} I}{I_{3} I}=\\frac{\\left[I_{1}, C I\\right]}{\\left[I_{3}, A I\\right]}\n$$\n\nBecause of $\\left[I_{1}, A B\\right]=\\left[I_{1}, B C\\right],\\left[I_{3}, A D\\right]=\\left[I_{3}, C D\\right]$, multiplication yields\n\n$$\n\\frac{\\left[I_{1}, B C\\right]}{\\left[I_{3}, C I\\right]}=\\frac{\\left[I_{1}, C I\\right]}{\\left[I_{3}, C D\\right]}\n$$\n\nBy $\\angle D C I=\\angle I C B=\\gamma / 2$ it follows that $\\angle I_{1} C B=\\angle I_{3} C I$ which concludes the proof of the\n\n\n\nabove statement.\n\nLet the perpendicular from $I_{1}$ on $I_{2} I_{3}$ intersect $g$ at $Z$. Then $\\angle M I_{1} Z=90^{\\circ}-\\angle I_{3} I_{2} I_{1}=$ $90^{\\circ}-\\gamma / 2=\\angle M C I_{2}$. Since we have also $\\angle Z M I_{1}=\\angle I_{2} M C$, triangles $M Z I_{1}$ and $M I_{2} C$ are similar. From this one easily proves that also $M I_{2} Z$ and $M C I_{1}$ are similar. Because $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic, $\\angle M Z I_{2}=\\angle M I_{1} C=\\angle N I_{3} C$, thus $N I_{2} Z$ and $N C I_{3}$ are similar, hence $N C I_{2}$ and $N I_{3} Z$ are similar. We conclude $\\angle Z I_{3} I_{2}=\\angle I_{2} C N=90^{\\circ}-\\gamma / 2$, hence $I_{1} I_{2} \\perp Z I_{3}$. This completes the proof.']",,True,,, 2107,Geometry,,"A social club has $n$ members. They have the membership numbers $1,2, \ldots, n$, respectively. From time to time members send presents to other members, including items they have already received as presents from other members. In order to avoid the embarrassing situation that a member might receive a present that he or she has sent to other members, the club adds the following rule to its statutes at one of its annual general meetings: ""A member with membership number $a$ is permitted to send a present to a member with membership number $b$ if and only if $a(b-1)$ is a multiple of $n . ""$ Prove that, if each member follows this rule, none will receive a present from another member that he or she has already sent to other members. Alternative formulation: Let $G$ be a directed graph with $n$ vertices $v_{1}, v_{2}, \ldots, v_{n}$, such that there is an edge going from $v_{a}$ to $v_{b}$ if and only if $a$ and $b$ are distinct and $a(b-1)$ is a multiple of $n$. Prove that this graph does not contain a directed cycle.","['Suppose there is an edge from $v_{i}$ to $v_{j}$. Then $i(j-1)=i j-i=k n$ for some integer $k$, which implies $i=i j-k n$. If $\\operatorname{gcd}(i, n)=d$ and $\\operatorname{gcd}(j, n)=e$, then $e$ divides $i j-k n=i$ and thus $e$ also divides $d$. Hence, if there is an edge from $v_{i}$ to $v_{j}$, then $\\operatorname{gcd}(j, n) \\mid \\operatorname{gcd}(i, n)$.\n\nIf there is a cycle in $G$, say $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$, then we have\n\n$$\n\\operatorname{gcd}\\left(i_{1}, n\\right)\\left|\\operatorname{gcd}\\left(i_{r}, n\\right)\\right| \\operatorname{gcd}\\left(i_{r-1}, n\\right)|\\ldots| \\operatorname{gcd}\\left(i_{2}, n\\right) \\mid \\operatorname{gcd}\\left(i_{1}, n\\right)\n$$\n\nwhich implies that all these greatest common divisors must be equal, say be equal to $t$.\n\nNow we pick any of the $i_{k}$, without loss of generality let it be $i_{1}$. Then $i_{r}\\left(i_{1}-1\\right)$ is a multiple of $n$ and hence also (by dividing by $t$ ), $i_{1}-1$ is a multiple of $\\frac{n}{t}$. Since $i_{1}$ and $i_{1}-1$ are relatively prime, also $t$ and $\\frac{n}{t}$ are relatively prime. So, by the Chinese remainder theorem, the value of $i_{1}$ is uniquely determined modulo $n=t \\cdot \\frac{n}{t}$ by the value of $t$. But, as $i_{1}$ was chosen arbitrarily among the $i_{k}$, this implies that all the $i_{k}$ have to be equal, a contradiction.', 'If $a, b, c$ are integers such that $a b-a$ and $b c-b$ are multiples of $n$, then also $a c-a=a(b c-b)+(a b-a)-(a b-a) c$ is a multiple of $n$. This implies that if there is an edge from $v_{a}$ to $v_{b}$ and an edge from $v_{b}$ to $v_{c}$, then there also must be an edge from $v_{a}$ to $v_{c}$. Therefore, if there are any cycles at all, the smallest cycle must have length 2 . But suppose the vertices $v_{a}$ and $v_{b}$ form such a cycle, i. e., $a b-a$ and $a b-b$ are both multiples of $n$. Then $a-b$ is also a multiple of $n$, which can only happen if $a=b$, which is impossible.', 'Suppose there was a cycle $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$. Then $i_{1}\\left(i_{2}-1\\right)$ is a multiple of $n$, i. e., $i_{1} \\equiv i_{1} i_{2} \\bmod n$. Continuing in this manner, we get $i_{1} \\equiv i_{1} i_{2} \\equiv$ $i_{1} i_{2} i_{3} \\equiv i_{1} i_{2} i_{3} \\ldots i_{r} \\bmod n$. But the same holds for all $i_{k}$, i. e., $i_{k} \\equiv i_{1} i_{2} i_{3} \\ldots i_{r} \\bmod n$. Hence $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\bmod n$, which means $i_{1}=i_{2}=\\cdots=i_{r}$, a contradiction.', 'Let $n=k$ be the smallest value of $n$ for which the corresponding graph has a cycle. We show that $k$ is a prime power.\n\nIf $k$ is not a prime power, it can be written as a product $k=d e$ of relatively prime integers greater than 1. Reducing all the numbers modulo $d$ yields a single vertex or a cycle in the corresponding graph on $d$ vertices, because if $a(b-1) \\equiv 0 \\bmod k$ then this equation also holds modulo $d$. But since the graph on $d$ vertices has no cycles, by the minimality of $k$, we must have that all the indices of the cycle are congruent modulo $d$. The same holds modulo $e$ and hence also modulo $k=d e$. But then all the indices are equal, which is a contradiction.\n\nThus $k$ must be a prime power $k=p^{m}$. There are no edges ending at $v_{k}$, so $v_{k}$ is not contained in any cycle. All edges not starting at $v_{k}$ end at a vertex belonging to a non-multiple of $p$, and all edges starting at a non-multiple of $p$ must end at $v_{1}$. But there is no edge starting at $v_{1}$. Hence there is no cycle.', 'Suppose there was a cycle $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$. Let $q=p^{m}$ be a prime power dividing $n$. We claim that either $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\equiv 0 \\bmod q$ or $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\equiv$ $1 \\bmod q$.\n\nSuppose that there is an $i_{s}$ not divisible by $q$. Then, as $i_{s}\\left(i_{s+1}-1\\right)$ is a multiple of $q, i_{s+1} \\equiv$ $1 \\bmod p$. Similarly, we conclude $i_{s+2} \\equiv 1 \\bmod p$ and so on. So none of the labels is divisible by $p$, but since $i_{s}\\left(i_{s+1}-1\\right)$ is a multiple of $q=p^{m}$ for all $s$, all $i_{s+1}$ are congruent to 1 modulo $q$. This proves the claim.\n\nNow, as all the labels are congruent modulo all the prime powers dividing $n$, they must all be equal by the Chinese remainder theorem. This is a contradiction.']",,True,,, 2108,Number Theory,,"A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$. Prove that there exist distinct positive integers $a$ and $b$ such that all the numbers $P(1), P(2)$, $\ldots, P(50)$ are balanced.","['Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \\equiv f(n)+f(m) \\bmod 2$ for all positive integers $n, m$.\n\nNow for each positive integer $n$ consider the binary sequence $(f(n+1), f(n+2), \\ldots, f(n+$ $50)$ ). As there are only $2^{50}$ different such sequences there are two different positive integers $a$ and $b$ such that\n\n$$\n(f(a+1), f(a+2), \\ldots, f(a+50))=(f(b+1), f(b+2), \\ldots, f(b+50)) \\text {. }\n$$\n\nBut this implies that for the polynomial $P(x)=(x+a)(x+b)$ all the numbers $P(1), P(2)$, $\\ldots, P(50)$ are balanced, since for all $1 \\leq k \\leq 50$ we have $f(P(k)) \\equiv f(a+k)+f(b+k) \\equiv$ $2 f(a+k) \\equiv 0 \\bmod 2$.']",,True,,, 2109,Number Theory,,"A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$. Prove that if $P(n)$ is balanced for all positive integers $n$, then $a=b$.","['Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \\equiv f(n)+f(m) \\bmod 2$ for all positive integers $n, m$.\n\nNow suppose $P(n)$ is balanced for all positive integers $n$ and $av_{p_{i}}(f(1))$ for all $i=1,2, \\ldots, m$, e.g. $a=\\left(p_{1} p_{2} \\ldots p_{m}\\right)^{\\alpha}$ with $\\alpha$ sufficiently large. Pick any such $a$. The condition of the problem then yields $a \\mid(f(a+1)-f(1))$. Assume $f(a+1) \\neq f(1)$. Then we must have $v_{p_{i}}(f(a+1)) \\neq$ $v_{p_{i}}(f(1))$ for at least one $i$. This yields $v_{p_{i}}(f(a+1)-f(1))=\\min \\left\\{v_{p_{i}}(f(a+1)), v_{p_{i}}(f(1))\\right\\} \\leq$ $v_{p_{1}}(f(1))p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1} \\cdot(f(r)+r)-r \\\\\n& \\geq p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1}+(f(r)+r)-r \\\\\n& >p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{m}^{\\alpha_{m}}+f(r) \\\\\n& \\geq|f(M)-f(r)| .\n\\end{aligned}\n$$\n\nBut since $M-r$ divides $f(M)-f(r)$ this can only be true if $f(r)=f(M)=f(1)$, which contradicts the choice of $r$.']",,True,,, 2111,Number Theory,,"Find all positive integers $n$ such that there exists a sequence of positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $$ a_{k+1}=\frac{a_{k}^{2}+1}{a_{k-1}+1}-1 $$ for every $k$ with $2 \leq k \leq n-1$.","[""Such a sequence exists for $n=1,2,3,4$ and no other $n$. Since the existence of such a sequence for some $n$ implies the existence of such a sequence for all smaller $n$, it suffices to prove that $n=5$ is not possible and $n=4$ is possible.\n\nAssume first that for $n=5$ there exists a sequence of positive integers $a_{1}, a_{2}, \\ldots, a_{5}$ satisfying the conditions\n\n$$\n\\begin{aligned}\n& a_{2}^{2}+1=\\left(a_{1}+1\\right)\\left(a_{3}+1\\right), \\\\\n& a_{3}^{2}+1=\\left(a_{2}+1\\right)\\left(a_{4}+1\\right), \\\\\n& a_{4}^{2}+1=\\left(a_{3}+1\\right)\\left(a_{5}+1\\right) .\n\\end{aligned}\n$$\n\nAssume $a_{1}$ is odd, then $a_{2}$ has to be odd as well and as then $a_{2}^{2}+1 \\equiv 2 \\bmod 4, a_{3}$ has to be even. But this is a contradiction, since then the even number $a_{2}+1$ cannot divide the odd number $a_{3}^{2}+1$.\n\nHence $a_{1}$ is even.\n\nIf $a_{2}$ is odd, $a_{3}^{2}+1$ is even (as a multiple of $a_{2}+1$ ) and hence $a_{3}$ is odd, too. Similarly we must have $a_{4}$ odd as well. But then $a_{3}^{2}+1$ is a product of two even numbers $\\left(a_{2}+1\\right)\\left(a_{4}+1\\right)$ and thus is divisible by 4 , which is a contradiction as for odd $a_{3}$ we have $a_{3}^{2}+1 \\equiv 2 \\bmod 4$.\n\nHence $a_{2}$ is even. Furthermore $a_{3}+1$ divides the odd number $a_{2}^{2}+1$ and so $a_{3}$ is even. Similarly, $a_{4}$ and $a_{5}$ are even as well.\n\nNow set $x=a_{2}$ and $y=a_{3}$. From the given condition we get $(x+1) \\mid\\left(y^{2}+1\\right)$ and $(y+1) \\mid\\left(x^{2}+1\\right)$. We will prove that there is no pair of positive even numbers $(x, y)$ satisfying these two conditions, thus yielding a contradiction to the assumption.\n\nAssume there exists a pair $\\left(x_{0}, y_{0}\\right)$ of positive even numbers satisfying the two conditions $\\left(x_{0}+1\\right) \\mid\\left(y_{0}^{2}+1\\right)$ and $\\left(y_{0}+1\\right) \\mid\\left(x_{0}^{2}+1\\right)$.\n\nThen one has $\\left(x_{0}+1\\right) \\mid\\left(y_{0}^{2}+1+x_{0}^{2}-1\\right)$, i.e., $\\left(x_{0}+1\\right) \\mid\\left(x_{0}^{2}+y_{0}^{2}\\right)$, and similarly $\\left(y_{0}+1\\right) \\mid\\left(x_{0}^{2}+y_{0}^{2}\\right)$. Any common divisor $d$ of $x_{0}+1$ and $y_{0}+1$ must hence also divide the number $\\left(x_{0}^{2}+1\\right)+\\left(y_{0}^{2}+1\\right)-\\left(x_{0}^{2}+y_{0}^{2}\\right)=2$. But as $x_{0}+1$ and $y_{0}+1$ are both odd, we must have $d=1$. Thus $x_{0}+1$ and $y_{0}+1$ are relatively prime and therefore there exists a positive integer $k$ such that\n\n$$\nk(x+1)(y+1)=x^{2}+y^{2}\n$$\n\nhas the solution $\\left(x_{0}, y_{0}\\right)$. We will show that the latter equation has no solution $(x, y)$ in positive even numbers.\n\nAssume there is a solution. Pick the solution $\\left(x_{1}, y_{1}\\right)$ with the smallest sum $x_{1}+y_{1}$ and assume $x_{1} \\geq y_{1}$. Then $x_{1}$ is a solution to the quadratic equation\n\n$$\nx^{2}-k\\left(y_{1}+1\\right) x+y_{1}^{2}-k\\left(y_{1}+1\\right)=0 \\text {. }\n$$\n\n\n\nLet $x_{2}$ be the second solution, which by VIETA's theorem fulfills $x_{1}+x_{2}=k\\left(y_{1}+1\\right)$ and $x_{1} x_{2}=y_{1}^{2}-k\\left(y_{1}+1\\right)$. If $x_{2}=0$, the second equation implies $y_{1}^{2}=k\\left(y_{1}+1\\right)$, which is impossible, as $y_{1}+1>1$ cannot divide the relatively prime number $y_{1}^{2}$. Therefore $x_{2} \\neq 0$.\n\nAlso we get $\\left(x_{1}+1\\right)\\left(x_{2}+1\\right)=x_{1} x_{2}+x_{1}+x_{2}+1=y_{1}^{2}+1$ which is odd, and hence $x_{2}$ must be even and positive. Also we have $x_{2}+1=\\frac{y_{1}^{2}+1}{x_{1}+1} \\leq \\frac{y_{1}^{2}+1}{y_{1}+1} \\leq y_{1} \\leq x_{1}$. But this means that the pair $\\left(x^{\\prime}, y^{\\prime}\\right)$ with $x^{\\prime}=y_{1}$ and $y^{\\prime}=x_{2}$ is another solution of $k(x+1)(y+1)=x^{2}+y^{2}$ in even positive numbers with $x^{\\prime}+y^{\\prime}y\n$$\n\nand similarly in the other case.\n\nNow, if $a_{3}$ was odd, then $\\left(a_{2}+1\\right)\\left(a_{4}+1\\right)=a_{3}^{2}+1 \\equiv 2 \\bmod 4$ would imply that one of $a_{2}$ or $a_{4}$ is even, this contradicts (1). Thus $a_{3}$ and hence also $a_{1}, a_{2}, a_{4}$ and $a_{5}$ are even. According to (1), one has $a_{1}a_{2}>a_{3}>a_{4}>a_{5}$ but due to the minimality of $a_{1}$ the first series of inequalities must hold.\n\nConsider the identity\n\n$\\left(a_{3}+1\\right)\\left(a_{1}+a_{3}\\right)=a_{3}^{2}-1+\\left(a_{1}+1\\right)\\left(a_{3}+1\\right)=a_{2}^{2}+a_{3}^{2}=a_{2}^{2}-1+\\left(a_{2}+1\\right)\\left(a_{4}+1\\right)=\\left(a_{2}+1\\right)\\left(a_{2}+a_{4}\\right)$.\n\nAny common divisor of the two odd numbers $a_{2}+1$ and $a_{3}+1$ must also divide $\\left(a_{2}+1\\right)\\left(a_{4}+\\right.$ $1)-\\left(a_{3}+1\\right)\\left(a_{3}-1\\right)=2$, so these numbers are relatively prime. Hence the last identity shows that $a_{1}+a_{3}$ must be a multiple of $a_{2}+1$, i.e. there is an integer $k$ such that\n\n$$\na_{1}+a_{3}=k\\left(a_{2}+1\\right) .\n\\tag{2}\n$$\n\nNow set $a_{0}=k\\left(a_{1}+1\\right)-a_{2}$. This is an integer and we have\n\n$$\n\\begin{aligned}\n\\left(a_{0}+1\\right)\\left(a_{2}+1\\right) & =k\\left(a_{1}+1\\right)\\left(a_{2}+1\\right)-\\left(a_{2}-1\\right)\\left(a_{2}+1\\right) \\\\\n& =\\left(a_{1}+1\\right)\\left(a_{1}+a_{3}\\right)-\\left(a_{1}+1\\right)\\left(a_{3}+1\\right)+2 \\\\\n& =\\left(a_{1}+1\\right)\\left(a_{1}-1\\right)+2=a_{1}^{2}+1\n\\end{aligned}\n$$\n\n\n\nThus $a_{0} \\geq 0$. If $a_{0}>0$, then by (1) we would have $a_{0}1$ is relatively prime to $a_{1}^{2}$ and thus cannot be a divisior of this number.\n\nHence $n \\geq 5$ is not possible.']","['1,2,3,4']",True,,Numerical, 2112,Number Theory,,"Let $P(x)$ be a non-constant polynomial with integer coefficients. Prove that there is no function $T$ from the set of integers into the set of integers such that the number of integers $x$ with $T^{n}(x)=x$ is equal to $P(n)$ for every $n \geq 1$, where $T^{n}$ denotes the $n$-fold application of $T$.","['Assume there is a polynomial $P$ of degree at least 1 with the desired property for a given function $T$. Let $A(n)$ denote the set of all $x \\in \\mathbb{Z}$ such that $T^{n}(x)=x$ and let $B(n)$ denote the set of all $x \\in \\mathbb{Z}$ for which $T^{n}(x)=x$ and $T^{k}(x) \\neq x$ for all $1 \\leq k\\sqrt{a b}, b^{\\ell_{0}}>\\sqrt{a b}$, we define the polynomial\n\n$$\nP(x)=\\prod_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1}\\left(a^{k} b^{\\ell} x-\\sqrt{a b}\\right)=: \\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x^{i}\n$$\n\nwith integer coefficients $d_{i}$. By our assumption, the zeros\n\n$$\n\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}, \\quad k=0, \\ldots, k_{0}-1, \\quad \\ell=0, \\ldots, \\ell_{0}-1\n$$\n\nof $P$ are pairwise distinct.\n\nFurthermore, we consider the integer sequence\n\n$$\ny_{n}=\\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x_{n+i}, \\quad n=1,2, \\ldots\n\\tag{3}\n$$\n\nBy the theory of linear recursions, we obtain\n\n$$\ny_{n}=\\sum_{\\substack{k, \\ell \\geq 0 \\\\ k \\geq k_{0} \\text { or } \\ell \\geq \\ell_{0}}} e_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}, \\quad n=1,2, \\ldots,\n\\tag{4}\n$$\n\nwith real numbers $e_{k, \\ell}$. We have\n\n$$\n\\left|y_{n}\\right| \\leq \\sum_{\\substack{k, \\ell \\geq 0 \\\\ k \\geq k_{0} \\text { or } \\ell \\geq \\ell_{0}}}\\left|e_{k, \\ell}\\right|\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}=: M_{n}\n$$\n\n\n\nBecause the series in (4) is obtained by a finite linear combination of the absolutely convergent series (1), we conclude that in particular $M_{1}<\\infty$. Since\n\n$$\n\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}} \\leq \\lambda:=\\max \\left\\{\\frac{\\sqrt{a b}}{a^{k_{0}}}, \\frac{\\sqrt{a b}}{b^{\\ell_{0}}}\\right\\} \\quad \\text { for all } k, \\ell \\geq 0 \\text { such that } k \\geq k_{0} \\text { or } \\ell \\geq \\ell_{0}\n$$\n\nwe get the estimates $M_{n+1} \\leq \\lambda M_{n}, n=1,2, \\ldots$ Our choice of $k_{0}$ and $\\ell_{0}$ ensures $\\lambda<1$, which implies $M_{n} \\rightarrow 0$ and consequently $y_{n} \\rightarrow 0$ as $n \\rightarrow \\infty$. It follows that $y_{n}=0$ for all sufficiently large $n$.\n\nSo, equation (3) reduces to $\\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x_{n+i}=0$.\n\nUsing the theory of linear recursions again, for sufficiently large $n$ we have\n\n$$\nx_{n}=\\sum_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1} f_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}\n$$\n\nfor certain real numbers $f_{k, \\ell}$.\n\nComparing with (2), we see that $f_{k, \\ell}=c_{k, \\ell}$ for all $k, \\ell \\geq 0$ with $kj_{0}$. But this means that\n\n$$\n\\left(1-x^{\\mu}\\right)^{\\frac{1}{2}}\\left(1-x^{\\nu}\\right)^{\\frac{1}{2}}=\\sum_{j=0}^{j_{0}} g_{j} x^{j}\n$$\n\nfor all real numbers $x \\in(0,1)$. Squaring, we see that\n\n$$\n\\left(1-x^{\\mu}\\right)\\left(1-x^{\\nu}\\right)\n$$\n\nis the square of a polynomial in $x$. In particular, all its zeros are of order at least 2 , which implies $\\mu=\\nu$ by looking at roots of unity. So we obtain $\\mu=\\nu=1$, i. e., $a=b$, a contradiction.', 'We set $a^{2}=A, b^{2}=B$, and $z_{n}=\\sqrt{\\left(A^{n}-1\\right)\\left(B^{n}-1\\right)}$. Let us assume that $z_{n}$ is an integer for $n=1,2, \\ldots$ Without loss of generality, we may suppose that $b0$ as $z_{n}>0$. As before, one obtains\n\n$$\n\\begin{aligned}\n& A^{n} B^{n}-A^{n}-B^{n}+1=z_{n}^{2} \\\\\n& =\\left\\{\\delta_{0}(a b)^{n}-\\delta_{1}\\left(\\frac{a}{b}\\right)^{n}-\\delta_{2}\\left(\\frac{a}{b^{3}}\\right)^{n}-\\cdots-\\delta_{k}\\left(\\frac{a}{b^{2 k-1}}\\right)^{n}\\right\\}^{2} \\\\\n& =\\delta_{0}^{2} A^{n} B^{n}-2 \\delta_{0} \\delta_{1} A^{n}-\\sum_{i=2}^{i=k}\\left(2 \\delta_{0} \\delta_{i}-\\sum_{j=1}^{j=i-1} \\delta_{j} \\delta_{i-j}\\right)\\left(\\frac{A}{B^{i-1}}\\right)^{n}+O\\left(\\frac{A}{B^{k}}\\right)^{n} .\n\\end{aligned}\n$$\n\nEasy asymptotic calculations yield $\\delta_{0}=1, \\delta_{1}=\\frac{1}{2}, \\delta_{i}=\\frac{1}{2} \\sum_{j=1}^{j=i-1} \\delta_{j} \\delta_{i-j}$ for $i=2,3, \\ldots, k-2$, and then $a=b^{k-1}$. It follows that $k>2$ and there is some $P \\in \\mathbb{Q}[X]$ for which $(X-1)\\left(X^{k-1}-1\\right)=$ $P(X)^{2}$. But this cannot occur, for instance as $X^{k-1}-1$ has no double zeros. Thus our\n\n\n\nassumption that $z_{n}$ was an integer for $n=1,2, \\ldots$ turned out to be wrong, which solves the problem.']",,True,,, 2115,Algebra,,"Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that $$ \frac{f(p)^{2}+f(q)^{2}}{f\left(r^{2}\right)+f\left(s^{2}\right)}=\frac{p^{2}+q^{2}}{r^{2}+s^{2}} $$ for all $p, q, r, s>0$ with $p q=r s$.","['Let $f$ satisfy the given condition. Setting $p=q=r=s=1$ yields $f(1)^{2}=f(1)$ and hence $f(1)=1$. Now take any $x>0$ and set $p=x, q=1, r=s=\\sqrt{x}$ to obtain\n\n$$\n\\frac{f(x)^{2}+1}{2 f(x)}=\\frac{x^{2}+1}{2 x}\n$$\n\nThis recasts into\n\n$$\n\\begin{gathered}\nx f(x)^{2}+x=x^{2} f(x)+f(x), \\\\\n(x f(x)-1)(f(x)-x)=0 .\n\\end{gathered}\n$$\n\nAnd thus,\n\n$$\n\\text { for every } x>0, \\text { either } f(x)=x \\text { or } f(x)=\\frac{1}{x} \\text {. }\n\\tag{1}\n$$\n\nObviously, if\n\n$$\nf(x)=x \\quad \\text { for all } x>0 \\quad \\text { or } \\quad f(x)=\\frac{1}{x} \\quad \\text { for all } x>0\\tag{2}\n$$\n\nthen the condition of the problem is satisfied. We show that actually these two functions are the only solutions.\n\nSo let us assume that there exists a function $f$ satisfying the requirement, other than those in (2). Then $f(a) \\neq a$ and $f(b) \\neq 1 / b$ for some $a, b>0$. By (1), these values must be $f(a)=1 / a, f(b)=b$. Applying now the equation with $p=a, q=b, r=s=\\sqrt{a b}$ we obtain $\\left(a^{-2}+b^{2}\\right) / 2 f(a b)=\\left(a^{2}+b^{2}\\right) / 2 a b ;$ equivalently,\n\n$$\nf(a b)=\\frac{a b\\left(a^{-2}+b^{2}\\right)}{a^{2}+b^{2}}\\tag{3}\n$$\n\nWe know however (see (1)) that $f(a b)$ must be either $a b$ or $1 / a b$. If $f(a b)=a b$ then by (3) $a^{-2}+b^{2}=a^{2}+b^{2}$, so that $a=1$. But, as $f(1)=1$, this contradicts the relation $f(a) \\neq a$. Likewise, if $f(a b)=1 / a b$ then (3) gives $a^{2} b^{2}\\left(a^{-2}+b^{2}\\right)=a^{2}+b^{2}$, whence $b=1$, in contradiction to $f(b) \\neq 1 / b$. Thus indeed the functions listed in (2) are the only two solutions.']",['$$\nf(x)=x \\quad \\text { for all } x>0 \\quad \\text { or } \\quad f(x)=\\frac{1}{x} \\quad \\text { for all } x>0\n$$'],False,,Need_human_evaluate, 2116,Algebra,," Prove the inequality $$ \frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geq 1 $$ for real numbers $x, y, z \neq 1$ satisfying the condition $x y z=1$.","['We start with the substitution\n\n$$\n\\frac{x}{x-1}=a, \\quad \\frac{y}{y-1}=b, \\quad \\frac{z}{z-1}=c, \\quad \\text { i.e., } \\quad x=\\frac{a}{a-1}, \\quad y=\\frac{b}{b-1}, \\quad z=\\frac{c}{c-1}\n$$\n\nThe inequality to be proved reads $a^{2}+b^{2}+c^{2} \\geq 1$. The new variables are subject to the constraints $a, b, c \\neq 1$ and the following one coming from the condition $x y z=1$,\n\n$$\n(a-1)(b-1)(c-1)=a b c .\n$$\n\nThis is successively equivalent to\n\n$$\n\\begin{aligned}\na+b+c-1 & =a b+b c+c a, \\\\\n2(a+b+c-1) & =(a+b+c)^{2}-\\left(a^{2}+b^{2}+c^{2}\\right), \\\\\na^{2}+b^{2}+c^{2}-2 & =(a+b+c)^{2}-2(a+b+c), \\\\\na^{2}+b^{2}+c^{2}-1 & =(a+b+c-1)^{2} .\n\\end{aligned}\n$$\n\nThus indeed $a^{2}+b^{2}+c^{2} \\geq 1$, as desired.', 'Without changing variables, just setting $z=1 / x y$ and clearing fractions, the proposed inequality takes the form\n\n$$\n(x y-1)^{2}\\left(x^{2}(y-1)^{2}+y^{2}(x-1)^{2}\\right)+(x-1)^{2}(y-1)^{2} \\geq(x-1)^{2}(y-1)^{2}(x y-1)^{2} .\n$$\n\nWith the notation $p=x+y, q=x y$ this becomes, after lengthy routine manipulation and a lot of cancellation\n\n$$\nq^{4}-6 q^{3}+2 p q^{2}+9 q^{2}-6 p q+p^{2} \\geq 0\n$$\n\nIt is not hard to notice that the expression on the left is just $\\left(q^{2}-3 q+p\\right)^{2}$, hence nonnegative.\n\n(Without introducing $p$ and $q$, one is of course led with some more work to the same expression, just written in terms of $x$ and $y$; but then it is not that easy to see that it is a square.)']",,True,,, 2117,Algebra,," Show that there are infinitely many triples of rational numbers $x, y, z$ for which this inequality turns into equality.","['From the equation $a^{2}+b^{2}+c^{2}-1=(a+b+c-1)^{2}$ we see that the proposed inequality becomes an equality if and only if both sums $a^{2}+b^{2}+c^{2}$ and $a+b+c$ have value 1 . The first of them is equal to $(a+b+c)^{2}-2(a b+b c+c a)$. So the instances of equality are described by the system of two equations\n\n$$\na+b+c=1, \\quad a b+b c+c a=0\n$$\n\nplus the constraint $a, b, c \\neq 1$. Elimination of $c$ leads to $a^{2}+a b+b^{2}=a+b$, which we regard as a quadratic equation in $b$,\n\n$$\nb^{2}+(a-1) b+a(a-1)=0\n$$\n\nwith discriminant\n\n$$\n\\Delta=(a-1)^{2}-4 a(a-1)=(1-a)(1+3 a)\n$$\n\nWe are looking for rational triples $(a, b, c)$; it will suffice to have $a$ rational such that $1-a$ and $1+3 a$ are both squares of rational numbers (then $\\Delta$ will be so too). Set $a=k / m$. We want $m-k$ and $m+3 k$ to be squares of integers. This is achieved for instance by taking $m=k^{2}-k+1$ (clearly nonzero); then $m-k=(k-1)^{2}, m+3 k=(k+1)^{2}$. Note that distinct integers $k$ yield distinct values of $a=k / m$.\n\nAnd thus, if $k$ is any integer and $m=k^{2}-k+1, a=k / m$ then $\\Delta=\\left(k^{2}-1\\right)^{2} / m^{2}$ and the quadratic equation has rational roots $b=\\left(m-k \\pm k^{2} \\mp 1\\right) /(2 m)$. Choose e.g. the larger root,\n\n$$\nb=\\frac{m-k+k^{2}-1}{2 m}=\\frac{m+(m-2)}{2 m}=\\frac{m-1}{m} .\n$$\n\n\n\nComputing $c$ from $a+b+c=1$ then gives $c=(1-k) / m$. The condition $a, b, c \\neq 1$ eliminates only $k=0$ and $k=1$. Thus, as $k$ varies over integers greater than 1 , we obtain an infinite family of rational triples $(a, b, c)$-and coming back to the original variables $(x=a /(a-1)$ etc. $)$-an infinite family of rational triples $(x, y, z)$ with the needed property. (A short calculation shows that the resulting triples are $x=-k /(k-1)^{2}, y=k-k^{2}, z=(k-1) / k^{2}$; but the proof was complete without listing them.)', 'To have equality, one needs $q^{2}-3 q+p=0$. Note that $x$ and $y$ are the roots of the quadratic trinomial (in a formal variable $t$ ): $t^{2}-p t+q$. When $q^{2}-3 q+p=0$, the discriminant equals\n\n$$\n\\delta=p^{2}-4 q=\\left(3 q-q^{2}\\right)^{2}-4 q=q(q-1)^{2}(q-4)\n$$\n\nNow it suffices to have both $q$ and $q-4$ squares of rational numbers (then $p=3 q-q^{2}$ and $\\sqrt{\\delta}$ are also rational, and so are the roots of the trinomial). On setting $q=(n / m)^{2}=4+(l / m)^{2}$ the requirement becomes $4 m^{2}+l^{2}=n^{2}$ (with $l, m, n$ being integers). This is just the Pythagorean equation, known to have infinitely many integer solutions.']",,True,,, 2120,Algebra,,"For an integer $m$, denote by $t(m)$ the unique number in $\{1,2,3\}$ such that $m+t(m)$ is a multiple of 3 . A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies $f(-1)=0, f(0)=1, f(1)=-1$ and $$ f\left(2^{n}+m\right)=f\left(2^{n}-t(m)\right)-f(m) \text { for all integers } m, n \geq 0 \text { with } 2^{n}>m \text {. } $$ Prove that $f(3 p) \geq 0$ holds for all integers $p \geq 0$.","['The given conditions determine $f$ uniquely on the positive integers. The signs of $f(1), f(2), \\ldots$ seem to change quite erratically. However values of the form $f\\left(2^{n}-t(m)\\right)$ are sufficient to compute directly any functional value. Indeed, let $n>0$ have base 2 representation $n=2^{a_{0}}+2^{a_{1}}+\\cdots+2^{a_{k}}, a_{0}>a_{1}>\\cdots>a_{k} \\geq 0$, and let $n_{j}=2^{a_{j}}+2^{a_{j-1}}+\\cdots+2^{a_{k}}, j=0, \\ldots, k$. Repeated applications of the recurrence show that $f(n)$ is an alternating sum of the quantities $f\\left(2^{a_{j}}-t\\left(n_{j+1}\\right)\\right)$ plus $(-1)^{k+1}$. (The exact formula is not needed for our proof.)\n\nSo we focus attention on the values $f\\left(2^{n}-1\\right), f\\left(2^{n}-2\\right)$ and $f\\left(2^{n}-3\\right)$. Six cases arise; more specifically,\n\n$t\\left(2^{2 k}-3\\right)=2, t\\left(2^{2 k}-2\\right)=1, t\\left(2^{2 k}-1\\right)=3, t\\left(2^{2 k+1}-3\\right)=1, t\\left(2^{2 k+1}-2\\right)=3, t\\left(2^{2 k+1}-1\\right)=2$.\n\nClaim. For all integers $k \\geq 0$ the following equalities hold:\n\n$$\n\\begin{array}{lll}\nf\\left(2^{2 k+1}-3\\right)=0, & f\\left(2^{2 k+1}-2\\right)=3^{k}, & f\\left(2^{2 k+1}-1\\right)=-3^{k}, \\\\\nf\\left(2^{2 k+2}-3\\right)=-3^{k}, & f\\left(2^{2 k+2}-2\\right)=-3^{k}, & f\\left(2^{2 k+2}-1\\right)=2 \\cdot 3^{k} .\n\\end{array}\n$$\n\nProof. By induction on $k$. The base $k=0$ comes down to checking that $f(2)=-1$ and $f(3)=2$; the given values $f(-1)=0, f(0)=1, f(1)=-1$ are also needed. Suppose the claim holds for $k-1$. For $f\\left(2^{2 k+1}-t(m)\\right)$, the recurrence formula and the induction hypothesis yield\n\n$$\n\\begin{aligned}\n& f\\left(2^{2 k+1}-3\\right)=f\\left(2^{2 k}+\\left(2^{2 k}-3\\right)\\right)=f\\left(2^{2 k}-2\\right)-f\\left(2^{2 k}-3\\right)=-3^{k-1}+3^{k-1}=0, \\\\\n& f\\left(2^{2 k+1}-2\\right)=f\\left(2^{2 k}+\\left(2^{2 k}-2\\right)\\right)=f\\left(2^{2 k}-1\\right)-f\\left(2^{2 k}-2\\right)=2 \\cdot 3^{k-1}+3^{k-1}=3^{k}, \\\\\n& f\\left(2^{2 k+1}-1\\right)=f\\left(2^{2 k}+\\left(2^{2 k}-1\\right)\\right)=f\\left(2^{2 k}-3\\right)-f\\left(2^{2 k}-1\\right)=-3^{k-1}-2 \\cdot 3^{k-1}=-3^{k} .\n\\end{aligned}\n$$\n\nFor $f\\left(2^{2 k+2}-t(m)\\right)$ we use the three equalities just established:\n\n$$\n\\begin{aligned}\n& f\\left(2^{2 k+2}-3\\right)=f\\left(2^{2 k+1}+\\left(2^{2 k+1}-3\\right)\\right)=f\\left(2^{2 k+1}-1\\right)-f\\left(2^{2 k+1}-3\\right)=-3^{k}-0=-3^{k}, \\\\\n& f\\left(2^{2 k+2}-2\\right)=f\\left(2^{2 k+1}+\\left(2^{2 k+1}-2\\right)\\right)=f\\left(2^{2 k+1}-3\\right)-f\\left(2^{2 k}-2\\right)=0-3^{k}=-3^{k}, \\\\\n& f\\left(2^{2 k+2}-1\\right)=f\\left(2^{2 k+1}+\\left(2^{2 k+1}-1\\right)\\right)=f\\left(2^{2 k+1}-2\\right)-f\\left(2^{2 k+1}-1\\right)=3^{k}+3^{k}=2 \\cdot 3^{k} .\n\\end{aligned}\n$$\n\nThe claim follows.\n\nA closer look at the six cases shows that $f\\left(2^{n}-t(m)\\right) \\geq 3^{(n-1) / 2}$ if $2^{n}-t(m)$ is divisible by 3 , and $f\\left(2^{n}-t(m)\\right) \\leq 0$ otherwise. On the other hand, note that $2^{n}-t(m)$ is divisible by 3 if and only if $2^{n}+m$ is. Therefore, for all nonnegative integers $m$ and $n$,\n\n(i) $f\\left(2^{n}-t(m)\\right) \\geq 3^{(n-1) / 2}$ if $2^{n}+m$ is divisible by 3 ;\n\n(ii) $f\\left(2^{n}-t(m)\\right) \\leq 0$ if $2^{n}+m$ is not divisible by 3 .\n\nOne more (direct) consequence of the claim is that $\\left|f\\left(2^{n}-t(m)\\right)\\right| \\leq \\frac{2}{3} \\cdot 3^{n / 2}$ for all $m, n \\geq 0$.\n\nThe last inequality enables us to find an upper bound for $|f(m)|$ for $m$ less than a given power of 2 . We prove by induction on $n$ that $|f(m)| \\leq 3^{n / 2}$ holds true for all integers $m, n \\geq 0$ with $2^{n}>m$.\n\n\n\nThe base $n=0$ is clear as $f(0)=1$. For the inductive step from $n$ to $n+1$, let $m$ and $n$ satisfy $2^{n+1}>m$. If $m<2^{n}$, we are done by the inductive hypothesis. If $m \\geq 2^{n}$ then $m=2^{n}+k$ where $2^{n}>k \\geq 0$. Now, by $\\left|f\\left(2^{n}-t(k)\\right)\\right| \\leq \\frac{2}{3} \\cdot 3^{n / 2}$ and the inductive assumption,\n\n$$\n|f(m)|=\\left|f\\left(2^{n}-t(k)\\right)-f(k)\\right| \\leq\\left|f\\left(2^{n}-t(k)\\right)\\right|+|f(k)| \\leq \\frac{2}{3} \\cdot 3^{n / 2}+3^{n / 2}<3^{(n+1) / 2} .\n$$\n\nThe induction is complete.\n\nWe proceed to prove that $f(3 p) \\geq 0$ for all integers $p \\geq 0$. Since $3 p$ is not a power of 2 , its binary expansion contains at least two summands. Hence one can write $3 p=2^{a}+2^{b}+c$ where $a>b$ and $2^{b}>c \\geq 0$. Applying the recurrence formula twice yields\n\n$$\nf(3 p)=f\\left(2^{a}+2^{b}+c\\right)=f\\left(2^{a}-t\\left(2^{b}+c\\right)\\right)-f\\left(2^{b}-t(c)\\right)+f(c) .\n$$\n\nSince $2^{a}+2^{b}+c$ is divisible by 3 , we have $f\\left(2^{a}-t\\left(2^{b}+c\\right)\\right) \\geq 3^{(a-1) / 2}$ by (i). Since $2^{b}+c$ is not divisible by 3 , we have $f\\left(2^{b}-t(c)\\right) \\leq 0$ by (ii). Finally $|f(c)| \\leq 3^{b / 2}$ as $2^{b}>c \\geq 0$, so that $f(c) \\geq-3^{b / 2}$. Therefore $f(3 p) \\geq 3^{(a-1) / 2}-3^{b / 2}$ which is nonnegative because $a>b$.']",,True,,, 2121,Algebra,,"Let $a, b, c, d$ be positive real numbers such that $$ a b c d=1 \quad \text { and } \quad a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} $$ Prove that $$ a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} $$","['We show that if $a b c d=1$, the sum $a+b+c+d$ cannot exceed a certain weighted mean of the expressions $\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}$ and $\\frac{b}{a}+\\frac{c}{b}+\\frac{d}{c}+\\frac{a}{d}$.\n\nBy applying the AM-GM inequality to the numbers $\\frac{a}{b}, \\frac{a}{b}, \\frac{b}{c}$ and $\\frac{a}{d}$, we obtain\n\n$$\na=\\sqrt[4]{\\frac{a^{4}}{a b c d}}=\\sqrt[4]{\\frac{a}{b} \\cdot \\frac{a}{b} \\cdot \\frac{b}{c} \\cdot \\frac{a}{d}} \\leq \\frac{1}{4}\\left(\\frac{a}{b}+\\frac{a}{b}+\\frac{b}{c}+\\frac{a}{d}\\right)\n$$\n\nAnalogously,\n\n$$\nb \\leq \\frac{1}{4}\\left(\\frac{b}{c}+\\frac{b}{c}+\\frac{c}{d}+\\frac{b}{a}\\right), \\quad c \\leq \\frac{1}{4}\\left(\\frac{c}{d}+\\frac{c}{d}+\\frac{d}{a}+\\frac{c}{b}\\right) \\quad \\text { and } \\quad d \\leq \\frac{1}{4}\\left(\\frac{d}{a}+\\frac{d}{a}+\\frac{a}{b}+\\frac{d}{c}\\right) .\n$$\n\nSumming up these estimates yields\n\n$$\na+b+c+d \\leq \\frac{3}{4}\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}\\right)+\\frac{1}{4}\\left(\\frac{b}{a}+\\frac{c}{b}+\\frac{d}{c}+\\frac{a}{d}\\right)\n$$\n\nIn particular, if $a+b+c+d>\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}$ then $a+b+c+d<\\frac{b}{a}+\\frac{c}{b}+\\frac{d}{c}+\\frac{a}{d}$.']",,True,,, 2122,Algebra,,"Let $f: \mathbb{R} \rightarrow \mathbb{N}$ be a function which satisfies $$ f\left(x+\frac{1}{f(y)}\right)=f\left(y+\frac{1}{f(x)}\right) \quad \text { for all } x, y \in \mathbb{R}\tag{1} $$ Prove that there is a positive integer which is not a value of $f$.","['Suppose that the statement is false and $f(\\mathbb{R})=\\mathbb{N}$. We prove several properties of the function $f$ in order to reach a contradiction.\n\nTo start with, observe that one can assume $f(0)=1$. Indeed, let $a \\in \\mathbb{R}$ be such that $f(a)=1$, and consider the function $g(x)=f(x+a)$. By substituting $x+a$ and $y+a$ for $x$ and $y$ in (1), we have\n\n$$\ng\\left(x+\\frac{1}{g(y)}\\right)=f\\left(x+a+\\frac{1}{f(y+a)}\\right)=f\\left(y+a+\\frac{1}{f(x+a)}\\right)=g\\left(y+\\frac{1}{g(x)}\\right) .\n$$\n\nSo $g$ satisfies the functional equation (1), with the additional property $g(0)=1$. Also, $g$ and $f$ have the same set of values: $g(\\mathbb{R})=f(\\mathbb{R})=\\mathbb{N}$. Henceforth we assume $f(0)=1$.\n\nClaim 1. For an arbitrary fixed $c \\in \\mathbb{R}$ we have $\\left\\{f\\left(c+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\mathbb{N}$.\n\nProof. Equation (1) and $f(\\mathbb{R})=\\mathbb{N}$ imply\n\n$f(\\mathbb{R})=\\left\\{f\\left(x+\\frac{1}{f(c)}\\right): x \\in \\mathbb{R}\\right\\}=\\left\\{f\\left(c+\\frac{1}{f(x)}\\right): x \\in \\mathbb{R}\\right\\} \\subset\\left\\{f\\left(c+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\} \\subset f(\\mathbb{R})$.\n\nThe claim follows.\n\nWe will use Claim 1 in the special cases $c=0$ and $c=1 / 3$ :\n\n$$\n\\left\\{f\\left(\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\left\\{f\\left(\\frac{1}{3}+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\mathbb{N}\\tag{2}\n$$\n\nClaim 2. If $f(u)=f(v)$ for some $u, v \\in \\mathbb{R}$ then $f(u+q)=f(v+q)$ for all nonnegative rational $q$. Furthermore, if $f(q)=1$ for some nonnegative rational $q$ then $f(k q)=1$ for all $k \\in \\mathbb{N}$.\n\nProof. For all $x \\in \\mathbb{R}$ we have by (1)\n\n$$\nf\\left(u+\\frac{1}{f(x)}\\right)=f\\left(x+\\frac{1}{f(u)}\\right)=f\\left(x+\\frac{1}{f(v)}\\right)=f\\left(v+\\frac{1}{f(x)}\\right) .\n$$\n\nSince $f(x)$ attains all positive integer values, this yields $f(u+1 / n)=f(v+1 / n)$ for all $n \\in \\mathbb{N}$. Let $q=k / n$ be a positive rational number. Then $k$ repetitions of the last step yield\n\n$$\nf(u+q)=f\\left(u+\\frac{k}{n}\\right)=f\\left(v+\\frac{k}{n}\\right)=f(v+q)\n$$\n\nNow let $f(q)=1$ for some nonnegative rational $q$, and let $k \\in \\mathbb{N}$. As $f(0)=1$, the previous conclusion yields successively $f(q)=f(2 q), f(2 q)=f(3 q), \\ldots, f((k-1) q)=f(k q)$, as needed.\n\nClaim 3. The equality $f(q)=f(q+1)$ holds for all nonnegative rational $q$.\n\nProof. Let $m$ be a positive integer such that $f(1 / m)=1$. Such an $m$ exists by (2). Applying the second statement of Claim 2 with $q=1 / m$ and $k=m$ yields $f(1)=1$.\n\nGiven that $f(0)=f(1)=1$, the first statement of Claim 2 implies $f(q)=f(q+1)$ for all nonnegative rational $q$.\n\n\n\nClaim 4. The equality $f\\left(\\frac{1}{n}\\right)=n$ holds for every $n \\in \\mathbb{N}$.\n\nProof. For a nonnegative rational $q$ we set $x=q, y=0$ in (1) and use Claim 3 to obtain\n\n$$\nf\\left(\\frac{1}{f(q)}\\right)=f\\left(q+\\frac{1}{f(0)}\\right)=f(q+1)=f(q)\n$$\n\nBy (2), for each $n \\in \\mathbb{N}$ there exists a $k \\in \\mathbb{N}$ such that $f(1 / k)=n$. Applying the last equation with $q=1 / k$, we have\n\n$$\nn=f\\left(\\frac{1}{k}\\right)=f\\left(\\frac{1}{f(1 / k)}\\right)=f\\left(\\frac{1}{n}\\right)\n$$\n\nNow we are ready to obtain a contradiction. Let $n \\in \\mathbb{N}$ be such that $f(1 / 3+1 / n)=1$. Such an $n$ exists by (2). Let $1 / 3+1 / n=s / t$, where $s, t \\in \\mathbb{N}$ are coprime. Observe that $t>1$ as $1 / 3+1 / n$ is not an integer. Choose $k, l \\in \\mathbb{N}$ so that that $k s-l t=1$.\n\nBecause $f(0)=f(s / t)=1$, Claim 2 implies $f(k s / t)=1$. Now $f(k s / t)=f(1 / t+l)$; on the other hand $f(1 / t+l)=f(1 / t)$ by $l$ successive applications of Claim 3 . Finally, $f(1 / t)=t$ by Claim 4, leading to the impossible $t=1$. The solution is complete.']",,True,,, 2123,Algebra,,"Prove that for any four positive real numbers $a, b, c, d$ the inequality $$ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geq 0 $$ holds. Determine all cases of equality.","['Denote the four terms by\n\n$$\nA=\\frac{(a-b)(a-c)}{a+b+c}, \\quad B=\\frac{(b-c)(b-d)}{b+c+d}, \\quad C=\\frac{(c-d)(c-a)}{c+d+a}, \\quad D=\\frac{(d-a)(d-b)}{d+a+b}\n$$\n\nThe expression $2 A$ splits into two summands as follows,\n\n$$\n2 A=A^{\\prime}+A^{\\prime \\prime} \\quad \\text { where } \\quad A^{\\prime}=\\frac{(a-c)^{2}}{a+b+c}, \\quad A^{\\prime \\prime}=\\frac{(a-c)(a-2 b+c)}{a+b+c}\n$$\n\nthis is easily verified. We analogously represent $2 B=B^{\\prime}+B^{\\prime \\prime}, 2 C=C^{\\prime}+C^{\\prime \\prime}, 2 B=D^{\\prime}+D^{\\prime \\prime}$ and examine each of the sums $A^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime}$ and $A^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}$ separately.\n\nWrite $s=a+b+c+d$; the denominators become $s-d, s-a, s-b, s-c$. By the CauchySchwarz inequality,\n\n$$\n\\begin{aligned}\n& \\left(\\frac{|a-c|}{\\sqrt{s-d}} \\cdot \\sqrt{s-d}+\\frac{|b-d|}{\\sqrt{s-a}} \\cdot \\sqrt{s-a}+\\frac{|c-a|}{\\sqrt{s-b}} \\cdot \\sqrt{s-b}+\\frac{|d-b|}{\\sqrt{s-c}} \\cdot \\sqrt{s-c}\\right)^{2} \\\\\n& \\quad \\leq\\left(\\frac{(a-c)^{2}}{s-d}+\\frac{(b-d)^{2}}{s-a}+\\frac{(c-a)^{2}}{s-b}+\\frac{(d-b)^{2}}{s-c}\\right)(4 s-s)=3 s\\left(A^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime}\\right) .\n\\end{aligned}\n$$\n\nHence\n\n$$\nA^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime} \\geq \\frac{(2|a-c|+2|b-d|)^{2}}{3 s} \\geq \\frac{16 \\cdot|a-c| \\cdot|b-d|}{3 s}\\tag{1}\n$$\n\nNext we estimate the absolute value of the other sum. We couple $A^{\\prime \\prime}$ with $C^{\\prime \\prime}$ to obtain\n\n$$\n\\begin{aligned}\nA^{\\prime \\prime}+C^{\\prime \\prime} & =\\frac{(a-c)(a+c-2 b)}{s-d}+\\frac{(c-a)(c+a-2 d)}{s-b} \\\\\n& =\\frac{(a-c)(a+c-2 b)(s-b)+(c-a)(c+a-2 d)(s-d)}{(s-d)(s-b)} \\\\\n& =\\frac{(a-c)(-2 b(s-b)-b(a+c)+2 d(s-d)+d(a+c))}{s(a+c)+b d} \\\\\n& =\\frac{3(a-c)(d-b)(a+c)}{M}, \\quad \\text { with } \\quad M=s(a+c)+b d .\n\\end{aligned}\n$$\n\nHence by cyclic shift\n\n$$\nB^{\\prime \\prime}+D^{\\prime \\prime}=\\frac{3(b-d)(a-c)(b+d)}{N}, \\quad \\text { with } \\quad N=s(b+d)+c a\n$$\n\nThus\n\n$$\nA^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}=3(a-c)(b-d)\\left(\\frac{b+d}{N}-\\frac{a+c}{M}\\right)=\\frac{3(a-c)(b-d) W}{M N}\\tag{2}\n$$\n\nwhere\n\n$$\nW=(b+d) M-(a+c) N=b d(b+d)-a c(a+c)\\tag{3}\n$$\n\n\n\nNote that\n\n$$\nM N>(a c(a+c)+b d(b+d)) s \\geq|W| \\cdot s\\tag{4}\n$$\n\nNow (2) and (4) yield\n\n$$\n\\left|A^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}\\right| \\leq \\frac{3 \\cdot|a-c| \\cdot|b-d|}{s}\\tag{5}\n$$\n\nCombined with (1) this results in\n\n$$\n\\begin{aligned}\n& 2(A+B+C+D)=\\left(A^{\\prime}+B^{\\prime}+C^{\\prime}+D^{\\prime}\\right)+\\left(A^{\\prime \\prime}+B^{\\prime \\prime}+C^{\\prime \\prime}+D^{\\prime \\prime}\\right) \\\\\n& \\quad \\geq \\frac{16 \\cdot|a-c| \\cdot|b-d|}{3 s}-\\frac{3 \\cdot|a-c| \\cdot|b-d|}{s}=\\frac{7 \\cdot|a-c| \\cdot|b-d|}{3(a+b+c+d)} \\geq 0\n\\end{aligned}\n$$\n\nThis is the required inequality. From the last line we see that equality can be achieved only if either $a=c$ or $b=d$. Since we also need equality in (1), this implies that actually $a=c$ and $b=d$ must hold simultaneously, which is obviously also a sufficient condition.', 'We keep the notations $A, B, C, D, s$, and also $M, N, W$ from the preceding solution; the definitions of $M, N, W$ and relations (3), (4) in that solution did not depend on the foregoing considerations. Starting from\n\n$$\n2 A=\\frac{(a-c)^{2}+3(a+c)(a-c)}{a+b+c}-2 a+2 c\n$$\n\nwe get\n\n$$\n\\begin{aligned}\n2(A & +C)=(a-c)^{2}\\left(\\frac{1}{s-d}+\\frac{1}{s-b}\\right)+3(a+c)(a-c)\\left(\\frac{1}{s-d}-\\frac{1}{s-b}\\right) \\\\\n& =(a-c)^{2} \\frac{2 s-b-d}{M}+3(a+c)(a-c) \\cdot \\frac{d-b}{M}=\\frac{p(a-c)^{2}-3(a+c)(a-c)(b-d)}{M}\n\\end{aligned}\n$$\n\nwhere $p=2 s-b-d=s+a+c$. Similarly, writing $q=s+b+d$ we have\n\n$$\n2(B+D)=\\frac{q(b-d)^{2}-3(b+d)(b-d)(c-a)}{N}\n$$\n\nspecific grouping of terms in the numerators has its aim. Note that $p q>2 s^{2}$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$,\n\n$$\n2(A+B+C+D)=\\frac{p(a-c)^{2}}{M}+\\frac{3(a-c)(b-d) W}{M N}+\\frac{q(b-d)^{2}}{N}\n$$\n\nwith $W$ defined by (3).\n\nSubstitution $x=(a-c) / M, y=(b-d) / N$ brings the required inequality to the form\n\n$$\n2(A+B+C+D)=M p x^{2}+3 W x y+N q y^{2} \\geq 0\\tag{6}\n$$\n\nIt will be enough to verify that the discriminant $\\Delta=9 W^{2}-4 M N p q$ of the quadratic trinomial $M p t^{2}+3 W t+N q$ is negative; on setting $t=x / y$ one then gets (6). The first inequality in (4) together with $p q>2 s^{2}$ imply $4 M N p q>8 s^{3}(a c(a+c)+b d(b+d))$. Since\n\n$$\n(a+c) s^{3}>(a+c)^{4} \\geq 4 a c(a+c)^{2} \\quad \\text { and likewise } \\quad(b+d) s^{3}>4 b d(b+d)^{2}\n$$\n\nthe estimate continues as follows,\n\n$$\n4 M N p q>8\\left(4(a c)^{2}(a+c)^{2}+4(b d)^{2}(b+d)^{2}\\right)>32(b d(b+d)-a c(a+c))^{2}=32 W^{2} \\geq 9 W^{2}\n$$\n\nThus indeed $\\Delta<0$. The desired inequality (6) hence results. It becomes an equality if and only if $x=y=0$; equivalently, if and only if $a=c$ and simultaneously $b=d$.', '$$\n\\begin{gathered}\n(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\\\\n=((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)= \\\\\n=\\left(a^{2}+a d-b^{2}-b d\\right)\\left(a^{2}+a d-c^{2}-c d\\right)(b+c+d)= \\\\\n=\\left(a^{4}+2 a^{3} d-a^{2} b^{2}-a^{2} b d-a^{2} c^{2}-a^{2} c d+a^{2} d^{2}-a b^{2} d-a b d^{2}-a c^{2} d-a c d^{2}+b^{2} c^{2}+b^{2} c d+b c^{2} d+b c d^{2}\\right)(b+c+d)= \\\\\n=a^{4} b+a^{4} c+a^{4} d+\\left(b^{3} c^{2}+a^{2} d^{3}\\right)-a^{2} c^{3}+\\left(2 a^{3} d^{2}-b^{3} a^{2}+c^{3} b^{2}\\right)+ \\\\\n+\\left(b^{3} c d-c^{3} d a-d^{3} a b\\right)+\\left(2 a^{3} b d+c^{3} d b-d^{3} a c\\right)+\\left(2 a^{3} c d-b^{3} d a+d^{3} b c\\right) \\\\\n+\\left(-a^{2} b^{2} c+3 b^{2} c^{2} d-2 a c^{2} d^{2}\\right)+\\left(-2 a^{2} b^{2} d+2 b c^{2} d^{2}\\right)+\\left(-a^{2} b c^{2}-2 a^{2} c^{2} d-2 a b^{2} d^{2}+2 b^{2} c d^{2}\\right)+ \\\\\n+\\left(-2 a^{2} b c d-a b^{2} c d-a b c^{2} d-2 a b c d^{2}\\right)\n\\end{gathered}\n$$\n\nIntroducing the notation $S_{x y z w}=\\sum_{c y c} a^{x} b^{y} c^{z} d^{w}$, one can write\n\n$$\n\\begin{gathered}\n\\sum_{c y c}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\\\\n=S_{4100}+S_{4010}+S_{4001}+2 S_{3200}-S_{3020}+2 S_{3002}-S_{3110}+2 S_{3101}+2 S_{3011}-3 S_{2120}-6 S_{2111}= \\\\\n+\\left(S_{4100}+S_{4001}+\\frac{1}{2} S_{3110}+\\frac{1}{2} S_{3011}-3 S_{2120}\\right)+ \\\\\n+\\left(S_{4010}-S_{3020}-\\frac{3}{2} S_{3110}+\\frac{3}{2} S_{3011}+\\frac{9}{16} S_{2210}+\\frac{9}{16} S_{2201}-\\frac{9}{8} S_{2111}\\right)+ \\\\\n+\\frac{9}{16}\\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\\right)+\\frac{23}{16}\\left(S_{3200}-2 S_{3101}+S_{3002}\\right)+\\frac{39}{8}\\left(S_{3101}-S_{2111}\\right),\n\\end{gathered}\n$$\n\nwhere the expressions\n\n$$\n\\begin{gathered}\nS_{4100}+S_{4001}+\\frac{1}{2} S_{3110}+\\frac{1}{2} S_{3011}-3 S_{2120}=\\sum_{c y c}\\left(a^{4} b+b c^{4}+\\frac{1}{2} a^{3} b c+\\frac{1}{2} a b c^{3}-3 a^{2} b c^{2}\\right), \\\\\nS_{4010}-S_{3020}-\\frac{3}{2} S_{3110}+\\frac{3}{2} S_{3011}+\\frac{9}{16} S_{2210}+\\frac{9}{16} S_{2201}-\\frac{9}{8} S_{2111}=\\sum_{c y c} a^{2} c\\left(a-c-\\frac{3}{4} b+\\frac{3}{4} d\\right)^{2}, \\\\\nS_{3200}-S_{2210}-S_{2201}+S_{3002}=\\sum_{c y c} b^{2}\\left(a^{3}-a^{2} c-a c^{2}+c^{3}\\right)=\\sum_{c y c} b^{2}(a+c)(a-c)^{2}, \\\\\nS_{3200}-2 S_{3101}+S_{3002}=\\sum_{c y c} a^{3}(b-d)^{2} \\quad \\text { and } \\quad S_{3101}-S_{2111}=\\frac{1}{3} \\sum_{c y c} b d\\left(2 a^{3}+c^{3}-3 a^{2} c\\right)\n\\end{gathered}\n$$\n\nare all nonnegative.']",,True,,, 2124,Combinatorics,,"In the plane we consider rectangles whose sides are parallel to the coordinate axes and have positive length. Such a rectangle will be called a box. Two boxes intersect if they have a common point in their interior or on their boundary. Find the largest $n$ for which there exist $n$ boxes $B_{1}, \ldots, B_{n}$ such that $B_{i}$ and $B_{j}$ intersect if and only if $i \not \equiv j \pm 1(\bmod n)$.","['The maximum number of such boxes is 6 . One example is shown in the figure.\n\n\n\nNow we show that 6 is the maximum. Suppose that boxes $B_{1}, \\ldots, B_{n}$ satisfy the condition. Let the closed intervals $I_{k}$ and $J_{k}$ be the projections of $B_{k}$ onto the $x$ - and $y$-axis, for $1 \\leq k \\leq n$.\n\nIf $B_{i}$ and $B_{j}$ intersect, with a common point $(x, y)$, then $x \\in I_{i} \\cap I_{j}$ and $y \\in J_{i} \\cap J_{j}$. So the intersections $I_{i} \\cap I_{j}$ and $J_{i} \\cap J_{j}$ are nonempty. Conversely, if $x \\in I_{i} \\cap I_{j}$ and $y \\in J_{i} \\cap J_{j}$ for some real numbers $x, y$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Putting it around, $B_{i}$ and $B_{j}$ are disjoint if and only if their projections on at least one coordinate axis are disjoint.\n\nFor brevity we call two boxes or intervals adjacent if their indices differ by 1 modulo $n$, and nonadjacent otherwise.\n\nThe adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect for each $k=1, \\ldots, n$. Hence $\\left(I_{k}, I_{k+1}\\right)$ or $\\left(J_{k}, J_{k+1}\\right)$ is a pair of disjoint intervals, $1 \\leq k \\leq n$. So there are at least $n$ pairs of disjoint intervals among $\\left(I_{1}, I_{2}\\right), \\ldots,\\left(I_{n-1}, I_{n}\\right),\\left(I_{n}, I_{1}\\right) ;\\left(J_{1}, J_{2}\\right), \\ldots,\\left(J_{n-1}, J_{n}\\right),\\left(J_{n}, J_{1}\\right)$.\n\nNext, every two nonadjacent boxes intersect, hence their projections on both axes intersect, too. Then the claim below shows that at most 3 pairs among $\\left(I_{1}, I_{2}\\right), \\ldots,\\left(I_{n-1}, I_{n}\\right),\\left(I_{n}, I_{1}\\right)$ are disjoint, and the same holds for $\\left(J_{1}, J_{2}\\right), \\ldots,\\left(J_{n-1}, J_{n}\\right),\\left(J_{n}, J_{1}\\right)$. Consequently $n \\leq 3+3=6$, as stated. Thus we are left with the claim and its justification.\n\nClaim. Let $\\Delta_{1}, \\Delta_{2}, \\ldots, \\Delta_{n}$ be intervals on a straight line such that every two nonadjacent intervals intersect. Then $\\Delta_{k}$ and $\\Delta_{k+1}$ are disjoint for at most three values of $k=1, \\ldots, n$.\n\nProof. Denote $\\Delta_{k}=\\left[a_{k}, b_{k}\\right], 1 \\leq k \\leq n$. Let $\\alpha=\\max \\left(a_{1}, \\ldots, a_{n}\\right)$ be the rightmost among the left endpoints of $\\Delta_{1}, \\ldots, \\Delta_{n}$, and let $\\beta=\\min \\left(b_{1}, \\ldots, b_{n}\\right)$ be the leftmost among their right endpoints. Assume that $\\alpha=a_{2}$ without loss of generality.\n\nIf $\\alpha \\leq \\beta$ then $a_{i} \\leq \\alpha \\leq \\beta \\leq b_{i}$ for all $i$. Every $\\Delta_{i}$ contains $\\alpha$, and thus no disjoint pair $\\left(\\Delta_{i}, \\Delta_{i+1}\\right)$ exists.\n\n\n\nIf $\\beta<\\alpha$ then $\\beta=b_{i}$ for some $i$ such that $a_{i}3$ and consider any nice permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of $\\{1,2, \\ldots, n\\}$. Then $n-1$ must be a divisor of the number\n\n$$\n\\begin{aligned}\n2\\left(a_{1}+a_{2}+\\cdots\\right. & \\left.+a_{n-1}\\right)=2\\left((1+2+\\cdots+n)-a_{n}\\right) \\\\\n& =n(n+1)-2 a_{n}=(n+2)(n-1)+\\left(2-2 a_{n}\\right) .\n\\end{aligned}\n$$\n\nSo $2 a_{n}-2$ must be divisible by $n-1$, hence equal to 0 or $n-1$ or $2 n-2$. This means that\n\n$$\na_{n}=1 \\quad \\text { or } \\quad a_{n}=\\frac{n+1}{2} \\quad \\text { or } \\quad a_{n}=n \\text {. }\n$$\n\nSuppose that $a_{n}=(n+1) / 2$. Since the permutation is nice, taking $k=n-2$ we get that $n-2$ has to be a divisor of\n\n$$\n\\begin{aligned}\n2\\left(a_{1}+a_{2}+\\cdots+a_{n-2}\\right) & =2\\left((1+2+\\cdots+n)-a_{n}-a_{n-1}\\right) \\\\\n& =n(n+1)-(n+1)-2 a_{n-1}=(n+2)(n-2)+\\left(3-2 a_{n-1}\\right) .\n\\end{aligned}\n$$\n\nSo $2 a_{n-1}-3$ should be divisible by $n-2$, hence equal to 0 or $n-2$ or $2 n-4$. Obviously 0 and $2 n-4$ are excluded because $2 a_{n-1}-3$ is odd. The remaining possibility $\\left(2 a_{n-1}-3=n-2\\right)$ leads to $a_{n-1}=(n+1) / 2=a_{n}$, which also cannot hold. This eliminates $(n+1) / 2$ as a possible value of $a_{n}$. Consequently $a_{n}=1$ or $a_{n}=n$.\n\nIf $a_{n}=n$ then $\\left(a_{1}, a_{2}, \\ldots, a_{n-1}\\right)$ is a nice permutation of $\\{1,2, \\ldots, n-1\\}$. There are $F_{n-1}$ such permutations. Attaching $n$ to any one of them at the end creates a nice permutation of $\\{1,2, \\ldots, n\\}$.\n\nIf $a_{n}=1$ then $\\left(a_{1}-1, a_{2}-1, \\ldots, a_{n-1}-1\\right)$ is a permutation of $\\{1,2, \\ldots, n-1\\}$. It is also nice because the number\n\n$$\n2\\left(\\left(a_{1}-1\\right)+\\cdots+\\left(a_{k}-1\\right)\\right)=2\\left(a_{1}+\\cdots+a_{k}\\right)-2 k\n$$\n\nis divisible by $k$, for any $k \\leq n-1$. And again, any one of the $F_{n-1}$ nice permutations $\\left(b_{1}, b_{2}, \\ldots, b_{n-1}\\right)$ of $\\{1,2, \\ldots, n-1\\}$ gives rise to a nice permutation of $\\{1,2, \\ldots, n\\}$ whose last term is 1 , namely $\\left(b_{1}+1, b_{2}+1, \\ldots, b_{n-1}+1,1\\right)$.\n\nThe bijective correspondences established in both cases show that there are $F_{n-1}$ nice permutations of $\\{1,2, \\ldots, n\\}$ with the last term 1 and also $F_{n-1}$ nice permutations of $\\{1,2, \\ldots, n\\}$ with the last term $n$. Hence follows the recurrence $F_{n}=2 F_{n-1}$. With the base value $F_{3}=6$ this gives the outcome formula $F_{n}=3 \\cdot 2^{n-2}$ for $n \\geq 3$.']",,True,,, 2126,Combinatorics,,"In the coordinate plane consider the set $S$ of all points with integer coordinates. For a positive integer $k$, two distinct points $A, B \in S$ will be called $k$-friends if there is a point $C \in S$ such that the area of the triangle $A B C$ is equal to $k$. A set $T \subset S$ will be called a $k$-clique if every two points in $T$ are $k$-friends. Find the least positive integer $k$ for which there exists a $k$-clique with more than 200 elements.","['To begin, let us describe those points $B \\in S$ which are $k$-friends of the point $(0,0)$. By definition, $B=(u, v)$ satisfies this condition if and only if there is a point $C=(x, y) \\in S$ such that $\\frac{1}{2}|u y-v x|=k$. (This is a well-known formula expressing the area of triangle $A B C$ when $A$ is the origin.)\n\nTo say that there exist integers $x, y$ for which $|u y-v x|=2 k$, is equivalent to saying that the greatest common divisor of $u$ and $v$ is also a divisor of $2 k$. Summing up, a point $B=(u, v) \\in S$ is a $k$-friend of $(0,0)$ if and only if $\\operatorname{gcd}(u, v)$ divides $2 k$.\n\nTranslation by a vector with integer coordinates does not affect $k$-friendship; if two points are $k$-friends, so are their translates. It follows that two points $A, B \\in S, A=(s, t), B=(u, v)$, are $k$-friends if and only if the point $(u-s, v-t)$ is a $k$-friend of $(0,0)$; i.e., if $\\operatorname{gcd}(u-s, v-t) \\mid 2 k$.\n\nLet $n$ be a positive integer which does not divide $2 k$. We claim that a $k$-clique cannot have more than $n^{2}$ elements.\n\nIndeed, all points $(x, y) \\in S$ can be divided into $n^{2}$ classes determined by the remainders that $x$ and $y$ leave in division by $n$. If a set $T$ has more than $n^{2}$ elements, some two points $A, B \\in T, A=(s, t), B=(u, v)$, necessarily fall into the same class. This means that $n \\mid u-s$ and $n \\mid v-t$. Hence $n \\mid d$ where $d=\\operatorname{gcd}(u-s, v-t)$. And since $n$ does not divide $2 k$, also $d$ does not divide $2 k$. Thus $A$ and $B$ are not $k$-friends and the set $T$ is not a $k$-clique.\n\nNow let $M(k)$ be the least positive integer which does not divide $2 k$. Write $M(k)=m$ for the moment and consider the set $T$ of all points $(x, y)$ with $0 \\leq x, y200$.\n\nBy the definition of $M(k), 2 k$ is divisible by the numbers $1,2, \\ldots, M(k)-1$, but not by $M(k)$ itself. If $M(k)^{2}>200$ then $M(k) \\geq 15$. Trying to hit $M(k)=15$ we get a contradiction immediately ( $2 k$ would have to be divisible by 3 and 5 , but not by 15 ).\n\nSo let us try $M(k)=16$. Then $2 k$ is divisible by the numbers $1,2, \\ldots, 15$, hence also by their least common multiple $L$, but not by 16 . And since $L$ is not a multiple of 16 , we infer that $k=L / 2$ is the least $k$ with $M(k)=16$.\n\nFinally, observe that if $M(k) \\geq 17$ then $2 k$ must be divisible by the least common multiple of $1,2, \\ldots, 16$, which is equal to $2 L$. Then $2 k \\geq 2 L$, yielding $k>L / 2$.\n\nIn conclusion, the least $k$ with the required property is equal to $L / 2=180180$.']",['180180'],False,,Numerical, 2127,Combinatorics,,"Let $n$ and $k$ be fixed positive integers of the same parity, $k \geq n$. We are given $2 n$ lamps numbered 1 through $2 n$; each of them can be on or off. At the beginning all lamps are off. We consider sequences of $k$ steps. At each step one of the lamps is switched (from off to on or from on to off). Let $N$ be the number of $k$-step sequences ending in the state: lamps $1, \ldots, n$ on, lamps $n+1, \ldots, 2 n$ off. Let $M$ be the number of $k$-step sequences leading to the same state and not touching lamps $n+1, \ldots, 2 n$ at all. Find the ratio $N / M$.","['A sequence of $k$ switches ending in the state as described in the problem statement (lamps $1, \\ldots, n$ on, lamps $n+1, \\ldots, 2 n$ off) will be called an admissible process. If, moreover, the process does not touch the lamps $n+1, \\ldots, 2 n$, it will be called restricted. So there are $N$ admissible processes, among which $M$ are restricted.\n\nIn every admissible process, restricted or not, each one of the lamps $1, \\ldots, n$ goes from off to on, so it is switched an odd number of times; and each one of the lamps $n+1, \\ldots, 2 n$ goes from off to off, so it is switched an even number of times.\n\nNotice that $M>0$; i.e., restricted admissible processes do exist (it suffices to switch each one of the lamps $1, \\ldots, n$ just once and then choose one of them and switch it $k-n$ times, which by hypothesis is an even number).\n\nConsider any restricted admissible process $\\mathbf{p}$. Take any lamp $\\ell, 1 \\leq \\ell \\leq n$, and suppose that it was switched $k_{\\ell}$ times. As noticed, $k_{\\ell}$ must be odd. Select arbitrarily an even number of these $k_{\\ell}$ switches and replace each of them by the switch of lamp $n+\\ell$. This can be done in $2^{k_{\\ell}-1}$ ways (because a $k_{\\ell}$-element set has $2^{k_{\\ell}-1}$ subsets of even cardinality). Notice that $k_{1}+\\cdots+k_{n}=k$.\n\nThese actions are independent, in the sense that the action involving lamp $\\ell$ does not affect the action involving any other lamp. So there are $2^{k_{1}-1} \\cdot 2^{k_{2}-1} \\cdots 2^{k_{n}-1}=2^{k-n}$ ways of combining these actions. In any of these combinations, each one of the lamps $n+1, \\ldots, 2 n$ gets switched an even number of times and each one of the lamps $1, \\ldots, n$ remains switched an odd number of times, so the final state is the same as that resulting from the original process $\\mathbf{p}$.\n\nThis shows that every restricted admissible process $\\mathbf{p}$ can be modified in $2^{k-n}$ ways, giving rise to $2^{k-n}$ distinct admissible processes (with all lamps allowed).\n\nNow we show that every admissible process $\\mathbf{q}$ can be achieved in that way. Indeed, it is enough to replace every switch of a lamp with a label $\\ell>n$ that occurs in $\\mathbf{q}$ by the switch of the corresponding lamp $\\ell-n$; in the resulting process $\\mathbf{p}$ the lamps $n+1, \\ldots, 2 n$ are not involved.\n\nSwitches of each lamp with a label $\\ell>n$ had occurred in $\\mathbf{q}$ an even number of times. So the performed replacements have affected each lamp with a label $\\ell \\leq n$ also an even number of times; hence in the overall effect the final state of each lamp has remained the same. This means that the resulting process $\\mathbf{p}$ is admissible - and clearly restricted, as the lamps $n+1, \\ldots, 2 n$ are not involved in it any more.\n\nIf we now take process $\\mathbf{p}$ and reverse all these replacements, then we obtain process $\\mathbf{q}$. These reversed replacements are nothing else than the modifications described in the foregoing paragraphs.\n\nThus there is a one-to- $\\left(2^{k-n}\\right)$ correspondence between the $M$ restricted admissible processes and the total of $N$ admissible processes. Therefore $N / M=2^{k-n}$.']",['$2^{k-n}$'],False,,Expression, 2128,Combinatorics,,"Let $S=\left\{x_{1}, x_{2}, \ldots, x_{k+\ell}\right\}$ be a $(k+\ell)$-element set of real numbers contained in the interval $[0,1] ; k$ and $\ell$ are positive integers. A $k$-element subset $A \subset S$ is called nice if $$ \left|\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{\ell} \sum_{x_{j} \in S \backslash A} x_{j}\right| \leq \frac{k+\ell}{2 k \ell} $$ Prove that the number of nice subsets is at least $\frac{2}{k+\ell}\left(\begin{array}{c}k+\ell \\ k\end{array}\right)$.","['For a $k$-element subset $A \\subset S$, let $f(A)=\\frac{1}{k} \\sum_{x_{i} \\in A} x_{i}-\\frac{1}{\\ell} \\sum_{x_{j} \\in S \\backslash A} x_{j}$. Denote $\\frac{k+\\ell}{2 k \\ell}=d$. By definition a subset $A$ is nice if $|f(A)| \\leq d$.\n\nTo each permutation $\\left(y_{1}, y_{2}, \\ldots, y_{k+\\ell}\\right)$ of the set $S=\\left\\{x_{1}, x_{2}, \\ldots, x_{k+\\ell}\\right\\}$ we assign $k+\\ell$ subsets of $S$ with $k$ elements each, namely $A_{i}=\\left\\{y_{i}, y_{i+1}, \\ldots, y_{i+k-1}\\right\\}, i=1,2, \\ldots, k+\\ell$. Indices are taken modulo $k+\\ell$ here and henceforth. In other words, if $y_{1}, y_{2}, \\ldots, y_{k+\\ell}$ are arranged around a circle in this order, the sets in question are all possible blocks of $k$ consecutive elements.\n\nClaim. At least two nice sets are assigned to every permutation of $S$.\n\nProof. Adjacent sets $A_{i}$ and $A_{i+1}$ differ only by the elements $y_{i}$ and $y_{i+k}, i=1, \\ldots, k+\\ell$. By the definition of $f$, and because $y_{i}, y_{i+k} \\in[0,1]$,\n\n$$\n\\left|f\\left(A_{i+1}\\right)-f\\left(A_{i}\\right)\\right|=\\left|\\left(\\frac{1}{k}+\\frac{1}{\\ell}\\right)\\left(y_{i+k}-y_{i}\\right)\\right| \\leq \\frac{1}{k}+\\frac{1}{\\ell}=2 d\n$$\n\nEach element $y_{i} \\in S$ belongs to exactly $k$ of the sets $A_{1}, \\ldots, A_{k+\\ell}$. Hence in $k$ of the expressions $f\\left(A_{1}\\right), \\ldots, f\\left(A_{k+\\ell}\\right)$ the coefficient of $y_{i}$ is $1 / k$; in the remaining $\\ell$ expressions, its coefficient is $-1 / \\ell$. So the contribution of $y_{i}$ to the sum of all $f\\left(A_{i}\\right)$ equals $k \\cdot 1 / k-\\ell \\cdot 1 / \\ell=0$. Since this holds for all $i$, it follows that $f\\left(A_{1}\\right)+\\cdots+f\\left(A_{k+\\ell}\\right)=0$.\n\nIf $f\\left(A_{p}\\right)=\\min f\\left(A_{i}\\right), f\\left(A_{q}\\right)=\\max f\\left(A_{i}\\right)$, we obtain in particular $f\\left(A_{p}\\right) \\leq 0, f\\left(A_{q}\\right) \\geq 0$. Let $pq$ is analogous; and the claim is true for $p=q$ as $f\\left(A_{i}\\right)=0$ for all $i$ ).\n\nWe are ready to prove that at least two of the sets $A_{1}, \\ldots, A_{k+\\ell}$ are nice. The interval $[-d, d]$ has length $2 d$, and we saw that adjacent numbers in the circular arrangement $f\\left(A_{1}\\right), \\ldots, f\\left(A_{k+\\ell}\\right)$ differ by at most $2 d$. Suppose that $f\\left(A_{p}\\right)<-d$ and $f\\left(A_{q}\\right)>d$. Then one of the numbers $f\\left(A_{p+1}\\right), \\ldots, f\\left(A_{q-1}\\right)$ lies in $[-d, d]$, and also one of the numbers $f\\left(A_{q+1}\\right), \\ldots, f\\left(A_{p-1}\\right)$ lies there. Consequently, one of the sets $A_{p+1}, \\ldots, A_{q-1}$ is nice, as well as one of the sets $A_{q+1}, \\ldots, A_{p-1}$. If $-d \\leq f\\left(A_{p}\\right)$ and $f\\left(A_{q}\\right) \\leq d$ then $A_{p}$ and $A_{q}$ are nice.\n\nLet now $f\\left(A_{p}\\right)<-d$ and $f\\left(A_{q}\\right) \\leq d$. Then $f\\left(A_{p}\\right)+f\\left(A_{q}\\right)<0$, and since $\\sum f\\left(A_{i}\\right)=0$, there is an $r \\neq q$ such that $f\\left(A_{r}\\right)>0$. We have $02\\left(u_{m-1}-u_{1}\\right)$ for all $m \\geq 3$.\n\nIndeed, assume that $u_{m}-u_{1} \\leq 2\\left(u_{m-1}-u_{1}\\right)$ holds for some $m \\geq 3$. This inequality can be written as $2\\left(u_{m}-u_{m-1}\\right) \\leq u_{m}-u_{1}$. Take the unique $k$ such that $2^{k} \\leq u_{m}-u_{1}<2^{k+1}$. Then $2\\left(u_{m}-u_{m-1}\\right) \\leq u_{m}-u_{1}<2^{k+1}$ yields $u_{m}-u_{m-1}<2^{k}$. However the elements $z=u_{m}, x=u_{1}$, $y=u_{m-1}$ of $S_{a}$ then satisfy $z-y<2^{k}$ and $z-x \\geq 2^{k}$, so that $z=u_{m}$ is $k$-good to $S_{a}$.\n\nThus each term of the sequence $u_{2}-u_{1}, u_{3}-u_{1}, \\ldots, u_{p}-u_{1}$ is more than twice the previous one. Hence $u_{p}-u_{1}>2^{p-1}\\left(u_{2}-u_{1}\\right) \\geq 2^{p-1}$. But $u_{p} \\in\\left\\{1,2,3, \\ldots, 2^{n+1}\\right\\}$, so that $u_{p} \\leq 2^{n+1}$. This yields $p-1 \\leq n$, i. e. $p \\leq n+1$.\n\nIn other words, each set $S_{a}$ contains at most $n+1$ elements that are not good to it.\n\nTo summarize the conclusions, mark with red all elements in the sets $S_{a}$ that are good to the respective set, and with blue the ones that are not good. Then the total number of red elements, counting multiplicities, is at most $n \\cdot 2^{n+1}$ (each $z \\in A$ can be marked red in at most $n$ sets). The total number of blue elements is at most $(n+1) 2^{n}$ (each set $S_{a}$ contains at most $n+1$ blue elements). Therefore the sum of cardinalities of $S_{1}, S_{2}, \\ldots, S_{2^{n}}$ does not exceed $(3 n+1) 2^{n}$. By averaging, the smallest set has at most $3 n+1$ elements.', 'We show that one of the sets $S_{a}$ has at most $2 n+1$ elements. In the sequel $|\\cdot|$ denotes the cardinality of a (finite) set.\n\nClaim. For $n \\geq 2$, suppose that $k$ subsets $S_{1}, \\ldots, S_{k}$ of $\\left\\{1,2, \\ldots, 2^{n}\\right\\}$ (not necessarily different) satisfy the condition of the problem. Then\n\n$$\n\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}\n$$\n\nProof. Observe that if the sets $S_{i}(1 \\leq i \\leq k)$ satisfy the condition then so do their arbitrary subsets $T_{i}(1 \\leq i \\leq k)$. The condition also holds for the sets $t+S_{i}=\\left\\{t+x \\mid x \\in S_{i}\\right\\}$ where $t$ is arbitrary.\n\nNote also that a set may occur more than once among $S_{1}, \\ldots, S_{k}$ only if its cardinality is less than 3, in which case its contribution to the sum $\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-n\\right.$ ) is nonpositive (as $n \\geq 2$ ).\n\nThe proof is by induction on $n$. In the base case $n=2$ we have subsets $S_{i}$ of $\\{1,2,3,4\\}$. Only the ones of cardinality 3 and 4 need to be considered by the remark above; each one of\n\n\n\nthem occurs at most once among $S_{1}, \\ldots, S_{k}$. If $S_{i}=\\{1,2,3,4\\}$ for some $i$ then no $S_{j}$ is a 3 -element subset in view of the condition, hence $\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-2\\right) \\leq 2$. By the condition again, it is impossible that $S_{i}=\\{1,3,4\\}$ and $S_{j}=\\{2,3,4\\}$ for some $i, j$. So if $\\left|S_{i}\\right| \\leq 3$ for all $i$ then at most 3 summands $\\left|S_{i}\\right|-2$ are positive, corresponding to 3 -element subsets. This implies $\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-2\\right) \\leq 3$, therefore the conclusion is true for $n=2$.\n\nSuppose that the claim holds for some $n \\geq 2$, and let the sets $S_{1}, \\ldots, S_{k} \\subseteq\\left\\{1,2, \\ldots, 2^{n+1}\\right\\}$ satisfy the given property. Denote $U_{i}=S_{i} \\cap\\left\\{1,2, \\ldots, 2^{n}\\right\\}, V_{i}=S_{i} \\cap\\left\\{2^{n}+1, \\ldots, 2^{n+1}\\right\\}$. Let\n\n$$\nI=\\left\\{i|1 \\leq i \\leq k,| U_{i} \\mid \\neq 0\\right\\}, \\quad J=\\{1, \\ldots, k\\} \\backslash I\n$$\n\nThe sets $S_{j}$ with $j \\in J$ are all contained in $\\left\\{2^{n}+1, \\ldots, 2^{n+1}\\right\\}$, so the induction hypothesis applies to their translates $-2^{n}+S_{j}$ which have the same cardinalities. Consequently, this gives $\\sum_{j \\in J}\\left(\\left|S_{j}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}$, so that\n\n$$\n\\sum_{j \\in J}\\left(\\left|S_{j}\\right|-(n+1)\\right) \\leq \\sum_{j \\in J}\\left(\\left|S_{j}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}\\tag{1}\n$$\n\nFor $i \\in I$, denote by $v_{i}$ the least element of $V_{i}$. Observe that if $V_{a}$ and $V_{b}$ intersect, with $a2^{n}$, which implies $z=v_{a}$.\n\nIt follows that if the element $v_{i}$ is removed from each $V_{i}$, a family of pairwise disjoint sets $W_{i}=V_{i} \\backslash\\left\\{v_{i}\\right\\}$ is obtained, $i \\in I$ (we assume $W_{i}=\\emptyset$ if $V_{i}=\\emptyset$ ). As $W_{i} \\subseteq\\left\\{2^{n}+1, \\ldots, 2^{n+1}\\right\\}$ for all $i$, we infer that $\\sum_{i \\in I}\\left|W_{i}\\right| \\leq 2^{n}$. Therefore $\\sum_{i \\in I}\\left(\\left|V_{i}\\right|-1\\right) \\leq \\sum_{i \\in I}\\left|W_{i}\\right| \\leq 2^{n}$.\n\nOn the other hand, the induction hypothesis applies directly to the sets $U_{i}, i \\in I$, so that $\\sum_{i \\in \\mathcal{I}}\\left(\\left|U_{i}\\right|-n\\right) \\leq(2 n-1) 2^{n-2}$. In summary,\n\n$$\n\\sum_{i \\in I}\\left(\\left|S_{i}\\right|-(n+1)\\right)=\\sum_{i \\in I}\\left(\\left|U_{i}\\right|-n\\right)+\\sum_{i \\in I}\\left(\\left|V_{i}\\right|-1\\right) \\leq(2 n-1) 2^{n-2}+2^{n}\\tag{2}\n$$\n\nThe estimates (1) and (2) are sufficient to complete the inductive step:\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k}\\left(\\left|S_{i}\\right|-(n+1)\\right) & =\\sum_{i \\in I}\\left(\\left|S_{i}\\right|-(n+1)\\right)+\\sum_{j \\in J}\\left(\\left|S_{j}\\right|-(n+1)\\right) \\\\\n& \\leq(2 n-1) 2^{n-2}+2^{n}+(2 n-1) 2^{n-2}=(2 n+1) 2^{n-1}\n\\end{aligned}\n$$\n\nReturning to the problem, consider $k=2^{n}$ subsets $S_{1}, S_{2}, \\ldots, S_{2^{n}}$ of $\\left\\{1,2,3, \\ldots, 2^{n+1}\\right\\}$. If they satisfy the given condition, the claim implies $\\sum_{i=1}^{2^{n}}\\left(\\left|S_{i}\\right|-(n+1)\\right) \\leq(2 n+1) 2^{n-1}$. By averaging again, we see that the smallest set has at most $2 n+1$ elements.']",,True,,, 2130,Geometry,,"In an acute-angled triangle $A B C$, point $H$ is the orthocentre and $A_{0}, B_{0}, C_{0}$ are the midpoints of the sides $B C, C A, A B$, respectively. Consider three circles passing through $H: \omega_{a}$ around $A_{0}, \omega_{b}$ around $B_{0}$ and $\omega_{c}$ around $C_{0}$. The circle $\omega_{a}$ intersects the line $B C$ at $A_{1}$ and $A_{2} ; \omega_{b}$ intersects $C A$ at $B_{1}$ and $B_{2} ; \omega_{c}$ intersects $A B$ at $C_{1}$ and $C_{2}$. Show that the points $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle.","['The perpendicular bisectors of the segments $A_{1} A_{2}, B_{1} B_{2}, C_{1} C_{2}$ are also the perpendicular bisectors of $B C, C A, A B$. So they meet at $O$, the circumcentre of $A B C$. Thus $O$ is the only point that can possibly be the centre of the desired circle.\n\nFrom the right triangle $O A_{0} A_{1}$ we get\n\n$$\nO A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2} .\\tag{1}\n$$\n\nLet $K$ be the midpoint of $A H$ and let $L$ be the midpoint of $C H$. Since $A_{0}$ and $B_{0}$ are the midpoints of $B C$ and $C A$, we see that $A_{0} L \\| B H$ and $B_{0} L \\| A H$. Thus the segments $A_{0} L$ and $B_{0} L$ are perpendicular to $A C$ and $B C$, hence parallel to $O B_{0}$ and $O A_{0}$, respectively. Consequently $O A_{0} L B_{0}$ is a parallelogram, so that $O A_{0}$ and $B_{0} L$ are equal and parallel. Also, the midline $B_{0} L$ of triangle $A H C$ is equal and parallel to $A K$ and $K H$.\n\nIt follows that $A K A_{0} O$ and $H A_{0} O K$ are parallelograms. The first one gives $A_{0} K=O A=R$, where $R$ is the circumradius of $A B C$. From the second one we obtain\n\n$$\n2\\left(O A_{0}^{2}+A_{0} H^{2}\\right)=O H^{2}+A_{0} K^{2}=O H^{2}+R^{2} .\\tag{2}\n$$\n\n(In a parallelogram, the sum of squares of the diagonals equals the sum of squares of the sides).\n\nFrom (1) and (2) we get $O A_{1}^{2}=\\left(O H^{2}+R^{2}\\right) / 2$. By symmetry, the same holds for the distances $O A_{2}, O B_{1}, O B_{2}, O C_{1}$ and $O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ all lie on a circle with centre at $O$ and radius $\\left(O H^{2}+R^{2}\\right) / 2$.\n\n', 'We are going to show again that the circumcentre $O$ is equidistant from the six points in question.\n\nLet $A^{\\prime}$ be the second intersection point of $\\omega_{b}$ and $\\omega_{c}$. The line $B_{0} C_{0}$, which is the line of centers of circles $\\omega_{b}$ and $\\omega_{c}$, is a midline in triangle $A B C$, parallel to $B C$ and perpendicular to the altitude $A H$. The points $A^{\\prime}$ and $H$ are symmetric with respect to the line of centers. Therefore $A^{\\prime}$ lies on the line $A H$.\n\nFrom the two circles $\\omega_{b}$ and $\\omega_{c}$ we obtain $A C_{1} \\cdot A C_{2}=A A^{\\prime} \\cdot A H=A B_{1} \\cdot A B_{2}$. So the quadrilateral $B_{1} B_{2} C_{1} C_{2}$ is cyclic. The perpendicular bisectors of the sides $B_{1} B_{2}$ and $C_{1} C_{2}$ meet at $O$. Hence $O$ is the circumcentre of $B_{1} B_{2} C_{1} C_{2}$ and so $O B_{1}=O B_{2}=O C_{1}=O C_{2}$.\n\nAnalogous arguments yield $O A_{1}=O A_{2}=O B_{1}=O B_{2}$ and $O A_{1}=O A_{2}=O C_{1}=O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle centred at $O$.\n\n', 'Let again $O$ and $R$ be the circumcentre and circumradius. Consider the vectors\n\n$$\n\\overrightarrow{O A}=\\mathbf{a}, \\quad \\overrightarrow{O B}=\\mathbf{b}, \\quad \\overrightarrow{O C}=\\mathbf{c}, \\quad \\text { where } \\quad \\mathbf{a}^{2}=\\mathbf{b}^{2}=\\mathbf{c}^{2}=R^{2}\n$$\n\nIt is well known that $\\overrightarrow{O H}=\\mathbf{a}+\\mathbf{b}+\\mathbf{c}$. Accordingly,\n\n$$\n\\overrightarrow{A_{0} H}=\\overrightarrow{O H}-\\overrightarrow{O A_{0}}=(\\mathbf{a}+\\mathbf{b}+\\mathbf{c})-\\frac{\\mathbf{b}+\\mathbf{c}}{2}=\\frac{2 \\mathbf{a}+\\mathbf{b}+\\mathbf{c}}{2}\n$$\n\nand\n\n$$\n\\begin{gathered}\nO A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2}=\\left(\\frac{\\mathbf{b}+\\mathbf{c}}{2}\\right)^{2}+\\left(\\frac{2 \\mathbf{a}+\\mathbf{b}+\\mathbf{c}}{2}\\right)^{2} \\\\\n=\\frac{1}{4}\\left(\\mathbf{b}^{2}+2 \\mathbf{b} \\mathbf{c}+\\mathbf{c}^{2}\\right)+\\frac{1}{4}\\left(4 \\mathbf{a}^{2}+4 \\mathbf{a} \\mathbf{b}+4 \\mathbf{a} \\mathbf{c}+\\mathbf{b}^{2}+2 \\mathbf{b} \\mathbf{c}+\\mathbf{c}^{2}\\right)=2 R^{2}+(\\mathbf{a b}+\\mathbf{a c}+\\mathbf{b c}) ;\n\\end{gathered}\n$$\n\nhere $\\mathbf{a b}, \\mathbf{b c}$, etc. denote dot products of vectors. We get the same for the distances $O A_{2}, O B_{1}$, $O B_{2}, O C_{1}$ and $O C_{2}$.']",,True,,, 2131,Geometry,,"Given trapezoid $A B C D$ with parallel sides $A B$ and $C D$, assume that there exist points $E$ on line $B C$ outside segment $B C$, and $F$ inside segment $A D$, such that $\angle D A E=\angle C B F$. Denote by $I$ the point of intersection of $C D$ and $E F$, and by $J$ the point of intersection of $A B$ and $E F$. Let $K$ be the midpoint of segment $E F$; assume it does not lie on line $A B$. Prove that $I$ belongs to the circumcircle of $A B K$ if and only if $K$ belongs to the circumcircle of $C D J$.","['Assume that the disposition of points is as in the diagram.\n\nSince $\\angle E B F=180^{\\circ}-\\angle C B F=180^{\\circ}-\\angle E A F$ by hypothesis, the quadrilateral $A E B F$ is cyclic. Hence $A J \\cdot J B=F J \\cdot J E$. In view of this equality, $I$ belongs to the circumcircle of $A B K$ if and only if $I J \\cdot J K=F J \\cdot J E$. Expressing $I J=I F+F J, J E=F E-F J$, and $J K=\\frac{1}{2} F E-F J$, we find that $I$ belongs to the circumcircle of $A B K$ if and only if\n\n$$\nF J=\\frac{I F \\cdot F E}{2 I F+F E}\n$$\n\nSince $A E B F$ is cyclic and $A B, C D$ are parallel, $\\angle F E C=\\angle F A B=180^{\\circ}-\\angle C D F$. Then $C D F E$ is also cyclic, yielding $I D \\cdot I C=I F \\cdot I E$. It follows that $K$ belongs to the circumcircle of $C D J$ if and only if $I J \\cdot I K=I F \\cdot I E$. Expressing $I J=I F+F J, I K=I F+\\frac{1}{2} F E$, and $I E=I F+F E$, we find that $K$ is on the circumcircle of $C D J$ if and only if\n\n$$\nF J=\\frac{I F \\cdot F E}{2 I F+F E}\n$$\n\nThe conclusion follows.\n\n']",,True,,, 2132,Geometry,,Let $A B C D$ be a convex quadrilateral and let $P$ and $Q$ be points in $A B C D$ such that $P Q D A$ and $Q P B C$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $P Q$ such that $\angle P A E=\angle Q D E$ and $\angle P B E=\angle Q C E$. Show that the quadrilateral $A B C D$ is cyclic.,"['Let $F$ be the point on the line $A D$ such that $E F \\| P A$. By hypothesis, the quadrilateral $P Q D A$ is cyclic. So if $F$ lies between $A$ and $D$ then $\\angle E F D=\\angle P A D=180^{\\circ}-\\angle E Q D$; the points $F$ and $Q$ are on distinct sides of the line $D E$ and we infer that $E F D Q$ is a cyclic quadrilateral. And if $D$ lies between $A$ and $F$ then a similar argument shows that $\\angle E F D=\\angle E Q D$; but now the points $F$ and $Q$ lie on the same side of $D E$, so that $E D F Q$ is a cyclic quadrilateral.\n\nIn either case we obtain the equality $\\angle E F Q=\\angle E D Q=\\angle P A E$ which implies that $F Q \\| A E$. So the triangles $E F Q$ and $P A E$ are either homothetic or parallel-congruent. More specifically, triangle $E F Q$ is the image of $P A E$ under the mapping $f$ which carries the points $P, E$ respectively to $E, Q$ and is either a homothety or translation by a vector. Note that $f$ is uniquely determined by these conditions and the position of the points $P, E, Q$ alone.\n\nLet now $G$ be the point on the line $B C$ such that $E G \\| P B$. The same reasoning as above applies to points $B, C$ in place of $A, D$, implying that the triangle $E G Q$ is the image of $P B E$ under the same mapping $f$. So $f$ sends the four points $A, P, B, E$ respectively to $F, E, G, Q$.\n\nIf $P E \\neq Q E$, so that $f$ is a homothety with a centre $X$, then the lines $A F, P E, B G$-i.e. the lines $A D, P Q, B C$-are concurrent at $X$. And since $P Q D A$ and $Q P B C$ are cyclic quadrilaterals, the equalities $X A \\cdot X D=X P \\cdot X Q=X B \\cdot X C$ hold, showing that the quadrilateral $A B C D$ is cyclic.\n\nFinally, if $P E=Q E$, so that $f$ is a translation, then $A D\\|P Q\\| B C$. Thus $P Q D A$ and $Q P B C$ are isosceles trapezoids. Then also $A B C D$ is an isosceles trapezoid, hence a cyclic quadrilateral.\n\n', 'Here is another way to reach the conclusion that the lines $A D, B C$ and $P Q$ are either concurrent or parallel. From the cyclic quadrilateral $P Q D A$ we get\n\n$$\n\\angle P A D=180^{\\circ}-\\angle P Q D=\\angle Q D E+\\angle Q E D=\\angle P A E+\\angle Q E D .\n$$\n\n\n\nHence $\\angle Q E D=\\angle P A D-\\angle P A E=\\angle E A D$. This in view of the tangent-chord theorem means that the circumcircle of triangle $E A D$ is tangent to the line $P Q$ at $E$. Analogously, the circumcircle of triangle $E B C$ is tangent to $P Q$ at $E$.\n\nSuppose that the line $A D$ intersects $P Q$ at $X$. Since $X E$ is tangent to the circle $(E A D)$, $X E^{2}=X A \\cdot X D$. Also, $X A \\cdot X D=X P \\cdot X Q$ because $P, Q, D, A$ lie on a circle. Therefore $X E^{2}=X P \\cdot X Q$.\n\nIt is not hard to see that this equation determines the position of the point $X$ on the line $P Q$ uniquely. Thus, if $B C$ also cuts $P Q$, say at $Y$, then the analogous equation for $Y$ yields $X=Y$, meaning that the three lines indeed concur. In this case, as well as in the case where $A D\\|P Q\\| B C$, the concluding argument is the same as in the first solution.\n\nIt remains to eliminate the possibility that e.g. $A D$ meets $P Q$ at $X$ while $B C \\| P Q$. Indeed, $Q P B C$ would then be an isosceles trapezoid and the angle equality $\\angle P B E=\\angle Q C E$ would force that $E$ is the midpoint of $P Q$. So the length of $X E$, which is the geometric mean of the lengths of $X P$ and $X Q$, should also be their arithmetic mean -impossible, as $X P \\neq X Q$. The proof is now complete.']",,True,,, 2133,Geometry,,In an acute triangle $A B C$ segments $B E$ and $C F$ are altitudes. Two circles passing through the points $A$ and $F$ are tangent to the line $B C$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $P E$ and $Q F$ intersect on the circumcircle of triangle $A E F$.,"['To approach the desired result we need some information about the slopes of the lines $P E$ and $Q F$; this information is provided by formulas (1) and (2) which we derive below.\n\nThe tangents $B P$ and $B Q$ to the two circles passing through $A$ and $F$ are equal, as $B P^{2}=B A \\cdot B F=B Q^{2}$. Consider the altitude $A D$ of triangle $A B C$ and its orthocentre $H$. From the cyclic quadrilaterals $C D F A$ and $C D H E$ we get $B A \\cdot B F=B C \\cdot B D=B E \\cdot B H$. Thus $B P^{2}=B E \\cdot B H$, or $B P / B H=B E / B P$, implying that the triangles $B P H$ and $B E P$ are similar. Hence\n\n$$\n\\angle B P E=\\angle B H P \\text {. }\\tag{1}\n$$\n\nThe point $P$ lies between $D$ and $C$; this follows from the equality $B P^{2}=B C \\cdot B D$. In view of this equality, and because $B P=B Q$,\n\n$$\nD P \\cdot D Q=(B P-B D) \\cdot(B P+B D)=B P^{2}-B D^{2}=B D \\cdot(B C-B D)=B D \\cdot D C\n$$\n\nAlso $A D \\cdot D H=B D \\cdot D C$, as is seen from the similar triangles $B D H$ and $A D C$. Combining these equalities we obtain $A D \\cdot D H=D P \\cdot D Q$. Therefore $D H / D P=D Q / D A$, showing that the triangles $H D P$ and $Q D A$ are similar. Hence $\\angle H P D=\\angle Q A D$, which can be rewritten as $\\angle B P H=\\angle B A D+\\angle B A Q$. And since $B Q$ is tangent to the circumcircle of triangle $F A Q$,\n\n$$\n\\angle B Q F=\\angle B A Q=\\angle B P H-\\angle B A D .\\tag{2}\n$$\n\nFrom (1) and (2) we deduce\n\n$$\n\\begin{aligned}\n& \\angle B P E+\\angle B Q F=(\\angle B H P+\\angle B P H)-\\angle B A D=\\left(180^{\\circ}-\\angle P B H\\right)-\\angle B A D \\\\\n& =\\left(90^{\\circ}+\\angle B C A\\right)-\\left(90^{\\circ}-\\angle A B C\\right)=\\angle B C A+\\angle A B C=180^{\\circ}-\\angle C A B .\n\\end{aligned}\n$$\n\nThus $\\angle B P E+\\angle B Q F<180^{\\circ}$, which means that the rays $P E$ and $Q F$ meet. Let $S$ be the point of intersection. Then $\\angle P S Q=180^{\\circ}-(\\angle B P E+\\angle B Q F)=\\angle C A B=\\angle E A F$.\n\nIf $S$ lies between $P$ and $E$ then $\\angle P S Q=180^{\\circ}-\\angle E S F$; and if $E$ lies between $P$ and $S$ then $\\angle P S Q=\\angle E S F$. In either case the equality $\\angle P S Q=\\angle E A F$ which we have obtained means that $S$ lies on the circumcircle of triangle $A E F$.\n\n', ""Let $H$ be the orthocentre of triangle $A B C$ and let $\\omega$ be the circle with diameter $A H$, passing through $E$ and $F$. Introduce the points of intersection of $\\omega$ with the following lines emanating from $P: P A \\cap \\omega=\\{A, U\\}, P H \\cap \\omega=\\{H, V\\}, P E \\cap \\omega=\\{E, S\\}$. The altitudes of triangle $A H P$ are contained in the lines $A V, H U, B C$, meeting at its orthocentre $Q^{\\prime}$.\n\nBy Pascal's theorem applied to the (tied) hexagon $A E S F H V$, the points $A E \\cap F H=C$, $E S \\cap H V=P$ and $S F \\cap V A$ are collinear, so $F S$ passes through $Q^{\\prime}$.\n\nDenote by $\\omega_{1}$ and $\\omega_{2}$ the circles with diameters $B C$ and $P Q^{\\prime}$, respectively. Let $D$ be the foot of the altitude from $A$ in triangle $A B C$. Suppose that $A D$ meets the circles $\\omega_{1}$ and $\\omega_{2}$ at the respective points $K$ and $L$.\n\nSince $H$ is the orthocentre of $A B C$, the triangles $B D H$ and $A D C$ are similar, and so $D A \\cdot D H=D B \\cdot D C=D K^{2}$; the last equality holds because $B K C$ is a right triangle. Since $H$ is the orthocentre also in triangle $A Q^{\\prime} P$, we analogously have $D L^{2}=D A \\cdot D H$. Therefore $D K=D L$ and $K=L$.\n\nAlso, $B D \\cdot B C=B A \\cdot B F$, from the similar triangles $A B D, C B F$. In the right triangle $B K C$ we have $B K^{2}=B D \\cdot B C$. Hence, and because $B A \\cdot B F=B P^{2}=B Q^{2}$ (by the definition of $P$ and $Q$ in the problem statement), we obtain $B K=B P=B Q$. It follows that $B$ is the centre of $\\omega_{2}$ and hence $Q^{\\prime}=Q$. So the lines $P E$ and $Q F$ meet at the point $S$ lying on the circumcircle of triangle $A E F$.\n\n""]",,True,,, 2134,Geometry,,"Let $k$ and $n$ be integers with $0 \leq k \leq n-2$. Consider a set $L$ of $n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $I$ the set of intersection points of lines in $L$. Let $O$ be a point in the plane not lying on any line of $L$. A point $X \in I$ is colored red if the open line segment $O X$ intersects at most $k$ lines in $L$. Prove that $I$ contains at least $\frac{1}{2}(k+1)(k+2)$ red points.","['There are at least $\\frac{1}{2}(k+1)(k+2)$ points in the intersection set $I$ in view of the condition $n \\geq k+2$.\n\nFor each point $P \\in I$, define its order as the number of lines that intersect the open line segment $O P$. By definition, $P$ is red if its order is at most $k$. Note that there is always at least one point $X \\in I$ of order 0 . Indeed, the lines in $L$ divide the plane into regions, bounded or not, and $O$ belongs to one of them. Clearly any corner of this region is a point of $I$ with order 0.\n\nClaim. Suppose that two points $P, Q \\in I$ lie on the same line of $L$, and no other line of $L$ intersects the open line segment $P Q$. Then the orders of $P$ and $Q$ differ by at most 1 .\n\nProof. Let $P$ and $Q$ have orders $p$ and $q$, respectively, with $p \\geq q$. Consider triangle $O P Q$. Now $p$ equals the number of lines in $L$ that intersect the interior of side $O P$. None of these lines intersects the interior of side $P Q$, and at most one can pass through $Q$. All remaining lines must intersect the interior of side $O Q$, implying that $q \\geq p-1$. The conclusion follows.\n\nWe prove the main result by induction on $k$. The base $k=0$ is clear since there is a point of order 0 which is red. Assuming the statement true for $k-1$, we pass on to the inductive step. Select a point $P \\in I$ of order 0 , and consider one of the lines $\\ell \\in L$ that pass through $P$. There are $n-1$ intersection points on $\\ell$, one of which is $P$. Out of the remaining $n-2$ points, the $k$ closest to $P$ have orders not exceeding $k$ by the Claim. It follows that there are at least $k+1$ red points on $\\ell$.\n\nLet us now consider the situation with $\\ell$ removed (together with all intersection points it contains). By hypothesis of induction, there are at least $\\frac{1}{2} k(k+1)$ points of order not exceeding $k-1$ in the resulting configuration. Restoring $\\ell$ back produces at most one new intersection point on each line segment joining any of these points to $O$, so their order is at most $k$ in the original configuration. The total number of points with order not exceeding $k$ is therefore at least $(k+1)+\\frac{1}{2} k(k+1)=\\frac{1}{2}(k+1)(k+2)$. This completes the proof.']",,True,,, 2135,Geometry,,"There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that $$ \angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ} \tag{1} $$ if and only if the diagonals $A C$ and $B D$ are perpendicular.","['For a point $P$ in $A B C D$ which satisfies (1), let $K, L, M, N$ be the feet of perpendiculars from $P$ to lines $A B, B C, C D, D A$, respectively. Note that $K, L, M, N$ are interior to the sides as all angles in (1) are acute. The cyclic quadrilaterals $A K P N$ and $D N P M$ give\n\n$$\n\\angle P A B+\\angle P D C=\\angle P N K+\\angle P N M=\\angle K N M .\n$$\n\nAnalogously, $\\angle P B C+\\angle P A D=\\angle L K N$ and $\\angle P C D+\\angle P B A=\\angle M L K$. Hence the equalities (1) imply $\\angle K N M=\\angle L K N=\\angle M L K=90^{\\circ}$, so that $K L M N$ is a rectangle. The converse also holds true, provided that $K, L, M, N$ are interior to sides $A B, B C, C D, D A$.\n\n(i) Suppose that there exists a point $P$ in $A B C D$ such that $K L M N$ is a rectangle. We show that $A C$ and $B D$ are parallel to the respective sides of $K L M N$.\n\nLet $O_{A}$ and $O_{C}$ be the circumcentres of the cyclic quadrilaterals $A K P N$ and $C M P L$. Line $O_{A} O_{C}$ is the common perpendicular bisector of $L M$ and $K N$, therefore $O_{A} O_{C}$ is parallel to $K L$ and $M N$. On the other hand, $O_{A} O_{C}$ is the midline in the triangle $A C P$ that is parallel to $A C$. Therefore the diagonal $A C$ is parallel to the sides $K L$ and $M N$ of the rectangle. Likewise, $B D$ is parallel to $K N$ and $L M$. Hence $A C$ and $B D$ are perpendicular.\n\n\n\n(ii) Suppose that $A C$ and $B D$ are perpendicular and meet at $R$. If $A B C D$ is a rhombus, $P$ can be chosen to be its centre. So assume that $A B C D$ is not a rhombus, and let $B R', 'For a point $P$ distinct from $A, B, C, D$, let circles $(A P D)$ and $(B P C)$ intersect again at $Q(Q=P$ if the circles are tangent). Next, let circles $(A Q B)$ and $(C Q D)$ intersect again at $R$. We show that if $P$ lies in $A B C D$ and satisfies (1) then $A C$ and $B D$ intersect at $R$ and are perpendicular; the converse is also true. It is convenient to use directed angles. Let $\\measuredangle(U V, X Y)$ denote the angle of counterclockwise rotation that makes line $U V$ parallel to line $X Y$. Recall that four noncollinear points $U, V, X, Y$ are concyclic if and only if $\\measuredangle(U X, V X)=\\measuredangle(U Y, V Y)$.\n\nThe definitions of points $P, Q$ and $R$ imply\n\n$$\n\\begin{aligned}\n\\measuredangle(A R, B R) & =\\measuredangle(A Q, B Q)=\\measuredangle(A Q, P Q)+\\measuredangle(P Q, B Q)=\\measuredangle(A D, P D)+\\measuredangle(P C, B C), \\\\\n\\measuredangle(C R, D R) & \\measuredangle(C Q, D Q)=\\measuredangle(C Q, P Q)+\\measuredangle(P Q, D Q)=\\measuredangle(C B, P B)+\\measuredangle(P A, D A), \\\\\n\\measuredangle(B R, C R) & =\\measuredangle(B R, R Q)+\\measuredangle(R Q, C R)=\\measuredangle(B A, A Q)+\\measuredangle(D Q, C D) \\\\\n& =\\measuredangle(B A, A P)+\\measuredangle(A P, A Q)+\\measuredangle(D Q, D P)+\\measuredangle(D P, C D) \\\\\n& =\\measuredangle(B A, A P)+\\measuredangle(D P, C D) .\n\\end{aligned}\n$$\n\nObserve that the whole construction is reversible. One may start with point $R$, define $Q$ as the second intersection of circles $(A R B)$ and $(C R D)$, and then define $P$ as the second intersection of circles $(A Q D)$ and $(B Q C)$. The equalities above will still hold true.\n\n\n\nAssume in addition that $P$ is interior to $A B C D$. Then\n\n$$\n\\begin{gathered}\n\\measuredangle(A D, P D)=\\angle P D A, \\measuredangle(P C, B C)=\\angle P C B, \\measuredangle(C B, P B)=\\angle P B C, \\measuredangle(P A, D A)=\\angle P A D \\\\\n\\measuredangle(B A, A P)=\\angle P A B, \\measuredangle(D P, C D)=\\angle P D C .\n\\end{gathered}\n$$\n\n(i) Suppose that $P$ lies in $A B C D$ and satisfies (1). Then $\\measuredangle(A R, B R)=\\angle P D A+\\angle P C B=90^{\\circ}$ and similarly $\\measuredangle(B R, C R)=\\measuredangle(C R, D R)=90^{\\circ}$. It follows that $R$ is the common point of lines $A C$ and $B D$, and that these lines are perpendicular.\n\n(ii) Suppose that $A C$ and $B D$ are perpendicular and intersect at $R$. We show that the point $P$ defined by the reverse construction (starting with $R$ and ending with $P$ ) lies in $A B C D$. This is enough to finish the solution, because then the angle equalities above will imply (1).\n\nOne can assume that $Q$, the second common point of circles $(A B R)$ and $(C D R)$, lies in $\\angle A R D$. Then in fact $Q$ lies in triangle $A D R$ as angles $A Q R$ and $D Q R$ are obtuse. Hence $\\angle A Q D$ is obtuse, too, so that $B$ and $C$ are outside circle $(A D Q)(\\angle A B D$ and $\\angle A C D$ are acute).\n\nNow $\\angle C A B+\\angle C D B=\\angle B Q R+\\angle C Q R=\\angle C Q B$ implies $\\angle C A B<\\angle C Q B$ and $\\angle C D B<$ $\\angle C Q B$. Hence $A$ and $D$ are outside circle $(B C Q)$. In conclusion, the second common point $P$ of circles $(A D Q)$ and $(B C Q)$ lies on their arcs $A D Q$ and $B C Q$.\n\nWe can assume that $P$ lies in $\\angle C Q D$. Since\n\n$$\n\\begin{gathered}\n\\angle Q P C+\\angle Q P D=\\left(180^{\\circ}-\\angle Q B C\\right)+\\left(180^{\\circ}-\\angle Q A D\\right)= \\\\\n=360^{\\circ}-(\\angle R B C+\\angle Q B R)-(\\angle R A D-\\angle Q A R)=360^{\\circ}-\\angle R B C-\\angle R A D>180^{\\circ},\n\\end{gathered}\n$$\n\npoint $P$ lies in triangle $C D Q$, and hence in $A B C D$. The proof is complete.\n\n']",,True,,, 2136,Geometry,,"Let $A B C D$ be a convex quadrilateral with $A B \neq B C$. Denote by $\omega_{1}$ and $\omega_{2}$ the incircles of triangles $A B C$ and $A D C$. Suppose that there exists a circle $\omega$ inscribed in angle $A B C$, tangent to the extensions of line segments $A D$ and $C D$. Prove that the common external tangents of $\omega_{1}$ and $\omega_{2}$ intersect on $\omega$.","['The proof below is based on two known facts.\n\nLemma 1. Given a convex quadrilateral $A B C D$, suppose that there exists a circle which is inscribed in angle $A B C$ and tangent to the extensions of line segments $A D$ and $C D$. Then $A B+A D=C B+C D$.\n\nProof. The circle in question is tangent to each of the lines $A B, B C, C D, D A$, and the respective points of tangency $K, L, M, N$ are located as with circle $\\omega$ in the figure. Then\n\n$$\nA B+A D=(B K-A K)+(A N-D N), \\quad C B+C D=(B L-C L)+(C M-D M) .\n$$\n\nAlso $B K=B L, D N=D M, A K=A N, C L=C M$ by equalities of tangents. It follows that $A B+A D=C B+C D$.\n\n\n\nFor brevity, in the sequel we write ""excircle $A C$ "" for the excircle of a triangle with side $A C$ which is tangent to line segment $A C$ and the extensions of the other two sides.\n\nLemma 2. The incircle of triangle $A B C$ is tangent to its side $A C$ at $P$. Let $P P^{\\prime}$ be the diameter of the incircle through $P$, and let line $B P^{\\prime}$ intersect $A C$ at $Q$. Then $Q$ is the point of tangency of side $A C$ and excircle $A C$.\n\nProof. Let the tangent at $P^{\\prime}$ to the incircle $\\omega_{1}$ meet $B A$ and $B C$ at $A^{\\prime}$ and $C^{\\prime}$. Now $\\omega_{1}$ is the excircle $A^{\\prime} C^{\\prime}$ of triangle $A^{\\prime} B C^{\\prime}$, and it touches side $A^{\\prime} C^{\\prime}$ at $P^{\\prime}$. Since $A^{\\prime} C^{\\prime} \\| A C$, the homothety with centre $B$ and ratio $B Q / B P^{\\prime}$ takes $\\omega_{1}$ to the excircle $A C$ of triangle $A B C$. Because this homothety takes $P^{\\prime}$ to $Q$, the lemma follows.\n\n\n\nRecall also that if the incircle of a triangle touches its side $A C$ at $P$, then the tangency point $Q$ of the same side and excircle $A C$ is the unique point on line segment $A C$ such that $A P=C Q$.\n\nWe pass on to the main proof. Let $\\omega_{1}$ and $\\omega_{2}$ touch $A C$ at $P$ and $Q$, respectively; then $A P=(A C+A B-B C) / 2, C Q=(C A+C D-A D) / 2$. Since $A B-B C=C D-A D$ by Lemma 1, we obtain $A P=C Q$. It follows that in triangle $A B C$ side $A C$ and excircle $A C$ are tangent at $Q$. Likewise, in triangle $A D C$ side $A C$ and excircle $A C$ are tangent at $P$. Note that $P \\neq Q$ as $A B \\neq B C$.\n\nLet $P P^{\\prime}$ and $Q Q^{\\prime}$ be the diameters perpendicular to $A C$ of $\\omega_{1}$ and $\\omega_{2}$, respectively. Then Lemma 2 shows that points $B, P^{\\prime}$ and $Q$ are collinear, and so are points $D, Q^{\\prime}$ and $P$.\n\nConsider the diameter of $\\omega$ perpendicular to $A C$ and denote by $T$ its endpoint that is closer to $A C$. The homothety with centre $B$ and ratio $B T / B P^{\\prime}$ takes $\\omega_{1}$ to $\\omega$. Hence $B, P^{\\prime}$ and $T$ are collinear. Similarly, $D, Q^{\\prime}$ and $T$ are collinear since the homothety with centre $D$ and ratio $-D T / D Q^{\\prime}$ takes $\\omega_{2}$ to $\\omega$.\n\nWe infer that points $T, P^{\\prime}$ and $Q$ are collinear, as well as $T, Q^{\\prime}$ and $P$. Since $P P^{\\prime} \\| Q Q^{\\prime}$, line segments $P P^{\\prime}$ and $Q Q^{\\prime}$ are then homothetic with centre $T$. The same holds true for circles $\\omega_{1}$ and $\\omega_{2}$ because they have $P P^{\\prime}$ and $Q Q^{\\prime}$ as diameters. Moreover, it is immediate that $T$ lies on the same side of line $P P^{\\prime}$ as $Q$ and $Q^{\\prime}$, hence the ratio of homothety is positive. In particular $\\omega_{1}$ and $\\omega_{2}$ are not congruent.\n\nIn summary, $T$ is the centre of a homothety with positive ratio that takes circle $\\omega_{1}$ to circle $\\omega_{2}$. This completes the solution, since the only point with the mentioned property is the intersection of the the common external tangents of $\\omega_{1}$ and $\\omega_{2}$.']",,True,,, 2137,Number Theory,,"Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a, b, c$ are integers (not necessarily positive) satisfying the equations $$ a^{n}+p b=b^{n}+p c=c^{n}+p a, $$ then $a=b=c$.","['If two of $a, b, c$ are equal, it is immediate that all the three are equal. So we may assume that $a \\neq b \\neq c \\neq a$. Subtracting the equations we get $a^{n}-b^{n}=-p(b-c)$ and two cyclic copies of this equation, which upon multiplication yield\n\n$$\n\\frac{a^{n}-b^{n}}{a-b} \\cdot \\frac{b^{n}-c^{n}}{b-c} \\cdot \\frac{c^{n}-a^{n}}{c-a}=-p^{3}\n\\tag{1}\n$$\n\nIf $n$ is odd then the differences $a^{n}-b^{n}$ and $a-b$ have the same sign and the product on the left is positive, while $-p^{3}$ is negative. So $n$ must be even.\n\nLet $d$ be the greatest common divisor of the three differences $a-b, b-c, c-a$, so that $a-b=d u, \\quad b-c=d v, \\quad c-a=d w ; \\quad \\operatorname{gcd}(u, v, w)=1, u+v+w=0$.\n\nFrom $a^{n}-b^{n}=-p(b-c)$ we see that $(a-b) \\mid p(b-c)$, i.e., $u \\mid p v$; and cyclically $v|p w, w| p u$. As $\\operatorname{gcd}(u, v, w)=1$ and $u+v+w=0$, at most one of $u, v, w$ can be divisible by $p$. Supposing that the prime $p$ does not divide any one of them, we get $u|v, v| w, w \\mid u$, whence $|u|=|v|=|w|=1$; but this quarrels with $u+v+w=0$.\n\nThus $p$ must divide exactly one of these numbers. Let e.g. $p \\mid u$ and write $u=p u_{1}$. Now we obtain, similarly as before, $u_{1}|v, v| w, w \\mid u_{1}$ so that $\\left|u_{1}\\right|=|v|=|w|=1$. The equation $p u_{1}+v+w=0$ forces that the prime $p$ must be even; i.e. $p=2$. Hence $v+w=-2 u_{1}= \\pm 2$, implying $v=w(= \\pm 1)$ and $u=-2 v$. Consequently $a-b=-2(b-c)$.\n\nKnowing that $n$ is even, say $n=2 k$, we rewrite the equation $a^{n}-b^{n}=-p(b-c)$ with $p=2$ in the form\n\n$$\n\\left(a^{k}+b^{k}\\right)\\left(a^{k}-b^{k}\\right)=-2(b-c)=a-b .\n$$\n\nThe second factor on the left is divisible by $a-b$, so the first factor $\\left(a^{k}+b^{k}\\right)$ must be \\pm 1 . Then exactly one of $a$ and $b$ must be odd; yet $a-b=-2(b-c)$ is even. Contradiction ends the proof.', 'The beginning is as in the first solution. Assuming that $a, b, c$ are not all equal, hence are all distinct, we derive equation (1) with the conclusion that $n$ is even. Write $n=2 k$.\n\nSuppose that $p$ is odd. Then the integer\n\n$$\n\\frac{a^{n}-b^{n}}{a-b}=a^{n-1}+a^{n-2} b+\\cdots+b^{n-1},\n$$\n\n\n\nwhich is a factor in (1), must be odd as well. This sum of $n=2 k$ summands is odd only if $a$ and $b$ have different parities. The same conclusion holding for $b, c$ and for $c$, $a$, we get that $a, b, c, a$ alternate in their parities, which is clearly impossible.\n\nThus $p=2$. The original system shows that $a, b, c$ must be of the same parity. So we may divide (1) by $p^{3}$, i.e. $2^{3}$, to obtain the following product of six integer factors:\n\n$$\n\\frac{a^{k}+b^{k}}{2} \\cdot \\frac{a^{k}-b^{k}}{a-b} \\cdot \\frac{b^{k}+c^{k}}{2} \\cdot \\frac{b^{k}-c^{k}}{b-c} \\cdot \\frac{c^{k}+a^{k}}{2} \\cdot \\frac{c^{k}-a^{k}}{c-a}=-1\n\\tag{2}\n$$\n\nEach one of the factors must be equal to \\pm 1 . In particular, $a^{k}+b^{k}= \\pm 2$. If $k$ is even, this becomes $a^{k}+b^{k}=2$ and yields $|a|=|b|=1$, whence $a^{k}-b^{k}=0$, contradicting (2).\n\nLet now $k$ be odd. Then the sum $a^{k}+b^{k}$, with value \\pm 2 , has $a+b$ as a factor. Since $a$ and $b$ are of the same parity, this means that $a+b= \\pm 2$; and cyclically, $b+c= \\pm 2, c+a= \\pm 2$. In some two of these equations the signs must coincide, hence some two of $a, b, c$ are equal. This is the desired contradiction.']",,True,,, 2138,Number Theory,,"Let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct positive integers, $n \geq 3$. Prove that there exist distinct indices $i$ and $j$ such that $a_{i}+a_{j}$ does not divide any of the numbers $3 a_{1}, 3 a_{2}, \ldots, 3 a_{n}$.","[""Without loss of generality, let $0a_{n-1}\\right)$. Thus $3 a_{n-1}=a_{n}+a_{n-1}$, i. e. $a_{n}=2 a_{n-1}$.\n\nSimilarly, if $j=n$ then $3 a_{n}=k\\left(a_{n}+a_{n-1}\\right)$ for some integer $k$, and only $k=2$ is possible. Hence $a_{n}=2 a_{n-1}$ holds true in both cases remaining, $j=n-1$ and $j=n$.\n\nNow $a_{n}=2 a_{n-1}$ implies that the sum $a_{n-1}+a_{1}$ is strictly between $a_{n} / 2$ and $a_{n}$. But $a_{n-1}$ and $a_{1}$ are distinct as $n \\geq 3$, so it follows from the above that $a_{n-1}+a_{1}$ divides $a_{n}$. This provides the desired contradiction.""]",,True,,, 2139,Number Theory,,"Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $\operatorname{gcd}\left(a_{i}, a_{i+1}\right)>a_{i-1}$. Prove that $a_{n} \geq 2^{n}$ for all $n \geq 0$.","['Since $a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, the sequence is strictly increasing. In particular $a_{0} \\geq 1, a_{1} \\geq 2$. For each $i \\geq 1$ we also have $a_{i+1}-a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, and consequently $a_{i+1} \\geq a_{i}+a_{i-1}+1$. Hence $a_{2} \\geq 4$ and $a_{3} \\geq 7$. The equality $a_{3}=7$ would force equalities in the previous estimates, leading to $\\operatorname{gcd}\\left(a_{2}, a_{3}\\right)=\\operatorname{gcd}(4,7)>a_{1}=2$, which is false. Thus $a_{3} \\geq 8$; the result is valid for $n=0,1,2,3$. These are the base cases for a proof by induction.\n\nTake an $n \\geq 3$ and assume that $a_{i} \\geq 2^{i}$ for $i=0,1, \\ldots, n$. We must show that $a_{n+1} \\geq 2^{n+1}$. Let $\\operatorname{gcd}\\left(a_{n}, a_{n+1}\\right)=d$. We know that $d>a_{n-1}$. The induction claim is reached immediately in the following cases:\n\n$$\n\\begin{array}{ll}\n\\text { if } \\quad a_{n+1} \\geq 4 d & \\text { then } a_{n+1}>4 a_{n-1} \\geq 4 \\cdot 2^{n-1}=2^{n+1} \\\\\n\\text { if } \\quad a_{n} \\geq 3 d & \\text { then } a_{n+1} \\geq a_{n}+d \\geq 4 d>4 a_{n-1} \\geq 4 \\cdot 2^{n-1}=2^{n+1} \\text {; } \\\\\n\\text { if } a_{n}=d \\quad \\text { then } a_{n+1} \\geq a_{n}+d=2 a_{n} \\geq 2 \\cdot 2^{n}=2^{n+1}\n\\end{array}\n$$\n\nThe only remaining possibility is that $a_{n}=2 d$ and $a_{n+1}=3 d$, which we assume for the sequel. So $a_{n+1}=\\frac{3}{2} a_{n}$.\n\nLet now $\\operatorname{gcd}\\left(a_{n-1}, a_{n}\\right)=d^{\\prime}$; then $d^{\\prime}>a_{n-2}$. Write $a_{n}=m d^{\\prime} \\quad(m$ an integer $)$. Keeping in mind that $d^{\\prime} \\leq a_{n-1}9 a_{n-2} \\geq 9 \\cdot 2^{n-2}>2^{n+1} ;$\n\nif $3 \\leq m \\leq 4$ then $a_{n-1}<\\frac{1}{2} \\cdot 4 d^{\\prime}$, and hence $a_{n-1}=d^{\\prime}$,\n\n$$\na_{n+1}=\\frac{3}{2} m a_{n-1} \\geq \\frac{3}{2} \\cdot 3 a_{n-1} \\geq \\frac{9}{2} \\cdot 2^{n-1}>2^{n+1} \\text {. }\n$$\n\nSo we are left with the case $m=5$, which means that $a_{n}=5 d^{\\prime}, a_{n+1}=\\frac{15}{2} d^{\\prime}, a_{n-1}a_{n-3}$. Because $d^{\\prime \\prime}$ is a divisor of $a_{n-1}$, hence also of $2 d^{\\prime}$, we may write $2 d^{\\prime}=m^{\\prime} d^{\\prime \\prime}$ ( $m^{\\prime}$ an integer). Since $d^{\\prime \\prime} \\leq a_{n-2}\\frac{75}{4} a_{n-3} \\geq \\frac{75}{4} \\cdot 2^{n-3}>2^{n+1}\n$$\n\n$$\n\\text { if } 3 \\leq m^{\\prime} \\leq 4 \\text { then } a_{n-2}<\\frac{1}{2} \\cdot 4 d^{\\prime \\prime} \\text {, and hence } a_{n-2}=d^{\\prime \\prime} \\text {, }\n$$\n\n$$\na_{n+1}=\\frac{15}{4} m^{\\prime} a_{n-2} \\geq \\frac{15}{4} \\cdot 3 a_{n-2} \\geq \\frac{45}{4} \\cdot 2^{n-2}>2^{n+1}\n$$\n\nBoth of them have produced the induction claim. But now there are no cases left. Induction is complete; the inequality $a_{n} \\geq 2^{n}$ holds for all $n$.']",,True,,, 2140,Number Theory,,"Let $n$ be a positive integer. Show that the numbers $$ \left(\begin{array}{c} 2^{n}-1 \\ 0 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 1 \end{array}\right), \quad\left(\begin{array}{c} 2^{n}-1 \\ 2 \end{array}\right), \quad \cdots, \quad\left(\begin{array}{c} 2^{n}-1 \\ 2^{n-1}-1 \end{array}\right) $$ are congruent modulo $2^{n}$ to $1,3,5, \ldots, 2^{n}-1$ in some order.","['It is well-known that all these numbers are odd. So the assertion that their remainders $\\left(\\bmod 2^{n}\\right)$ make up a permutation of $\\left\\{1,3, \\ldots, 2^{n}-1\\right\\}$ is equivalent just to saying that these remainders are all distinct. We begin by showing that\n\n$$\n\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k\n\\end{array}\\right)+\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k+1\n\\end{array}\\right) \\equiv 0 \\quad\\left(\\bmod 2^{n}\\right) \\quad \\text { and } \\quad\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k\n\\end{array}\\right) \\equiv(-1)^{k}\\left(\\begin{array}{c}\n2^{n-1}-1 \\\\\nk\n\\end{array}\\right) \\quad\\left(\\bmod 2^{n}\\right)\n\\tag{1}\n$$\n\nThe first relation is immediate, as the sum on the left is equal to $\\left(\\begin{array}{c}2^{n} \\\\ 2 k+1\\end{array}\\right)=\\frac{2^{n}}{2 k+1}\\left(\\begin{array}{c}2^{n}-1 \\\\ 2 k\\end{array}\\right)$, hence is divisible by $2^{n}$. The second relation:\n\n$$\n\\left(\\begin{array}{c}\n2^{n}-1 \\\\\n2 k\n\\end{array}\\right)=\\prod_{j=1}^{2 k} \\frac{2^{n}-j}{j}=\\prod_{i=1}^{k} \\frac{2^{n}-(2 i-1)}{2 i-1} \\cdot \\prod_{i=1}^{k} \\frac{2^{n-1}-i}{i} \\equiv(-1)^{k}\\left(\\begin{array}{c}\n2^{n-1}-1 \\\\\nk\n\\end{array}\\right) \\quad\\left(\\bmod 2^{n}\\right) .\n$$\n\nThis prepares ground for a proof of the required result by induction on $n$. The base case $n=1$ is obvious. Assume the assertion is true for $n-1$ and pass to $n$, denoting $a_{k}=\\left(\\begin{array}{c}2^{n-1}-1 \\\\ k\\end{array}\\right)$, $b_{m}=\\left(\\begin{array}{c}2^{n}-1 \\\\ m\\end{array}\\right)$. The induction hypothesis is that all the numbers $a_{k}\\left(0 \\leq k<2^{n-2}\\right)$ are distinct $\\left(\\bmod 2^{n-1}\\right)$; the claim is that all the numbers $b_{m}\\left(0 \\leq m<2^{n-1}\\right)$ are distinct $\\left(\\bmod 2^{n}\\right)$.\n\nThe congruence relations (1) are restated as\n\n$$\nb_{2 k} \\equiv(-1)^{k} a_{k} \\equiv-b_{2 k+1} \\quad\\left(\\bmod 2^{n}\\right)\n\\tag{2}\n$$\n\nShifting the exponent in the first relation of (1) from $n$ to $n-1$ we also have the congruence $a_{2 i+1} \\equiv-a_{2 i}\\left(\\bmod 2^{n-1}\\right)$. We hence conclude:\n$$\n\\text{If, for some j, }k<2^{n-2}, a_{k} \\equiv-a_{j}\\left(\\bmod 2^{n-1}\\right)\\text{, then }\\{j, k\\}=\\{2 i, 2 i+1\\}\\text{ for some i.}\\tag{3}\n$$\n\nThis is so because in the sequence $\\left(a_{k}: k<2^{n-2}\\right)$ each term $a_{j}$ is complemented to $0\\left(\\bmod 2^{n-1}\\right)$ by only one other term $a_{k}$, according to the induction hypothesis.\n\nFrom $(2)$ we see that $b_{4 i} \\equiv a_{2 i}$ and $b_{4 i+3} \\equiv a_{2 i+1}\\left(\\bmod 2^{n}\\right)$. Let\n\n$$\nM=\\left\\{m: 0 \\leq m<2^{n-1}, m \\equiv 0 \\text { or } 3(\\bmod 4)\\right\\}, \\quad L=\\left\\{l: 0 \\leq l<2^{n-1}, l \\equiv 1 \\text { or } 2(\\bmod 4)\\right\\}\n$$\n\nThe last two congruences take on the unified form\n\n$$\nb_{m} \\equiv a_{\\lfloor m / 2\\rfloor} \\quad\\left(\\bmod 2^{n}\\right) \\quad \\text { for all } \\quad m \\in M\\tag{4}\n$$\n\nThus all the numbers $b_{m}$ for $m \\in M$ are distinct $\\left(\\bmod 2^{n}\\right)$ because so are the numbers $a_{k}$ (they are distinct $\\left(\\bmod 2^{n-1}\\right)$, hence also $\\left(\\bmod 2^{n}\\right)$ ).\n\nEvery $l \\in L$ is paired with a unique $m \\in M$ into a pair of the form $\\{2 k, 2 k+1\\}$. So (2) implies that also all the $b_{l}$ for $l \\in L$ are distinct $\\left(\\bmod 2^{n}\\right)$. It remains to eliminate the possibility that $b_{m} \\equiv b_{l}\\left(\\bmod 2^{n}\\right)$ for some $m \\in M, l \\in L$.\n\nSuppose that such a situation occurs. Let $m^{\\prime} \\in M$ be such that $\\left\\{m^{\\prime}, l\\right\\}$ is a pair of the form $\\{2 k, 2 k+1\\}$, so that $(\\operatorname{see}(2)) b_{m^{\\prime}} \\equiv-b_{l}\\left(\\bmod 2^{n}\\right)$. Hence $b_{m^{\\prime}} \\equiv-b_{m}\\left(\\bmod 2^{n}\\right)$. Since both $m^{\\prime}$ and $m$ are in $M$, we have by (4) $b_{m^{\\prime}} \\equiv a_{j}, b_{m} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$ for $j=\\left\\lfloor m^{\\prime} / 2\\right\\rfloor, k=\\lfloor m / 2\\rfloor$.\n\nThen $a_{j} \\equiv-a_{k}\\left(\\bmod 2^{n}\\right)$. Thus, according to (3), $j=2 i, k=2 i+1$ for some $i$ (or vice versa). The equality $a_{2 i+1} \\equiv-a_{2 i}\\left(\\bmod 2^{n}\\right)$ now means that $\\left(\\begin{array}{c}2^{n-1}-1 \\\\ 2 i\\end{array}\\right)+\\left(\\begin{array}{c}2^{n-1}-1 \\\\ 2 i+1\\end{array}\\right) \\equiv 0\\left(\\bmod 2^{n}\\right)$. However, the sum on the left is equal to $\\left(\\begin{array}{l}2^{n-1} \\\\ 2 i+1\\end{array}\\right)$. A number of this form cannot be divisible by $2^{n}$. This is a contradiction which concludes the induction step and proves the result.', 'We again proceed by induction, writing for brevity $N=2^{n-1}$ and keeping notation $a_{k}=\\left(\\begin{array}{c}N-1 \\\\ k\\end{array}\\right), b_{m}=\\left(\\begin{array}{c}2 N-1 \\\\ m\\end{array}\\right)$. Assume that the result holds for the sequence $\\left(a_{0}, a_{1}, a_{2}, \\ldots, a_{N / 2-1}\\right)$. In view of the symmetry $a_{N-1-k}=a_{k}$ this sequence is a permutation of $\\left(a_{0}, a_{2}, a_{4}, \\ldots, a_{N-2}\\right)$. So the induction hypothesis says that this latter sequence, taken $(\\bmod N)$, is a permutation of $(1,3,5, \\ldots, N-1)$. Similarly, the induction claim is that $\\left(b_{0}, b_{2}, b_{4}, \\ldots, b_{2 N-2}\\right)$, taken $(\\bmod 2 N)$, is a permutation of $(1,3,5, \\ldots, 2 N-1)$.\n\nIn place of the congruence relations (2) we now use the following ones,\n\n$$\nb_{4 i} \\equiv a_{2 i} \\quad(\\bmod N) \\quad \\text { and } \\quad b_{4 i+2} \\equiv b_{4 i}+N \\quad(\\bmod 2 N)\\tag{5}\n$$\n\nGiven this, the conclusion is immediate: the first formula of (5) together with the induction hypothesis tells us that $\\left(b_{0}, b_{4}, b_{8}, \\ldots, b_{2 N-4}\\right)(\\bmod N)$ is a permutation of $(1,3,5, \\ldots, N-1)$. Then the second formula of $(5)$ shows that $\\left(b_{2}, b_{6}, b_{10}, \\ldots, b_{2 N-2}\\right)(\\bmod N)$ is exactly the same permutation; moreover, this formula distinguishes $(\\bmod 2 N)$ each $b_{4 i}$ from $b_{4 i+2}$.\n\nConsequently, these two sequences combined represent $(\\bmod 2 N)$ a permutation of the sequence $(1,3,5, \\ldots, N-1, N+1, N+3, N+5, \\ldots, N+N-1)$, and this is precisely the induction claim.\n\nNow we prove formulas (5); we begin with the second one. Since $b_{m+1}=b_{m} \\cdot \\frac{2 N-m-1}{m+1}$,\n\n$$\nb_{4 i+2}=b_{4 i} \\cdot \\frac{2 N-4 i-1}{4 i+1} \\cdot \\frac{2 N-4 i-2}{4 i+2}=b_{4 i} \\cdot \\frac{2 N-4 i-1}{4 i+1} \\cdot \\frac{N-2 i-1}{2 i+1} .\n$$\n\nThe desired congruence $b_{4 i+2} \\equiv b_{4 i}+N$ may be multiplied by the odd number $(4 i+1)(2 i+1)$, giving rise to a chain of successively equivalent congruences:\n\n$$\n\\begin{aligned}\nb_{4 i}(2 N-4 i-1)(N-2 i-1) & \\equiv\\left(b_{4 i}+N\\right)(4 i+1)(2 i+1) & & (\\bmod 2 N) \\\\\nb_{4 i}(2 i+1-N) & \\equiv\\left(b_{4 i}+N\\right)(2 i+1) & & (\\bmod 2 N) \\\\\n\\left(b_{4 i}+2 i+1\\right) N & \\equiv 0 & & (\\bmod 2 N)\n\\end{aligned}\n$$\n\nand the last one is satisfied, as $b_{4 i}$ is odd. This settles the second relation in (5).\n\nThe first one is proved by induction on $i$. It holds for $i=0$. Assume $b_{4 i} \\equiv a_{2 i}(\\bmod 2 N)$ and consider $i+1$ :\n\n$$\nb_{4 i+4}=b_{4 i+2} \\cdot \\frac{2 N-4 i-3}{4 i+3} \\cdot \\frac{2 N-4 i-4}{4 i+4} ; \\quad a_{2 i+2}=a_{2 i} \\cdot \\frac{N-2 i-1}{2 i+1} \\cdot \\frac{N-2 i-2}{2 i+2}\n$$\n\nBoth expressions have the fraction $\\frac{N-2 i-2}{2 i+2}$ as the last factor. Since $2 i+21 \\tag{1}\n$$\n\nDirect verification shows that this function meets the requirements.\n\nConversely, let $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfy (i) and (ii). Applying (i) for $x=1$ gives $d(f(1))=1$, so $f(1)=1$. In the sequel we prove that (1) holds for all $n>1$. Notice that $f(m)=f(n)$ implies $m=n$ in view of (i). The formula $d\\left(p_{1}^{b_{1}} \\cdots p_{k}^{b_{k}}\\right)=\\left(b_{1}+1\\right) \\cdots\\left(b_{k}+1\\right)$ will be used throughout.\n\nLet $p$ be a prime. Since $d(f(p))=p$, the formula just mentioned yields $f(p)=q^{p-1}$ for some prime $q$; in particular $f(2)=q^{2-1}=q$ is a prime. We prove that $f(p)=p^{p-1}$ for all primes $p$.\n\nSuppose that $p$ is odd and $f(p)=q^{p-1}$ for a prime $q$. Applying (ii) first with $x=2$, $y=p$ and then with $x=p, y=2$ shows that $f(2 p)$ divides both $(2-1) p^{2 p-1} f(2)=p^{2 p-1} f(2)$ and $(p-1) 2^{2 p-1} f(p)=(p-1) 2^{2 p-1} q^{p-1}$. If $q \\neq p$ then the odd prime $p$ does not divide $(p-1) 2^{2 p-1} q^{p-1}$, hence the greatest common divisor of $p^{2 p-1} f(2)$ and $(p-1) 2^{2 p-1} q^{p-1}$ is a divisor of $f(2)$. Thus $f(2 p)$ divides $f(2)$ which is a prime. As $f(2 p)>1$, we obtain $f(2 p)=f(2)$ which is impossible. So $q=p$, i. e. $f(p)=p^{p-1}$.\n\nFor $p=2$ the same argument with $x=2, y=3$ and $x=3, y=2$ shows that $f(6)$ divides both $3^{5} f(2)$ and $2^{6} f(3)=2^{6} 3^{2}$. If the prime $f(2)$ is odd then $f(6)$ divides $3^{2}=9$, so $f(6) \\in\\{1,3,9\\}$. However then $6=d(f(6)) \\in\\{d(1), d(3), d(9)\\}=\\{1,2,3\\}$ which is false. In conclusion $f(2)=2$.\n\nNext, for each $n>1$ the prime divisors of $f(n)$ are among the ones of $n$. Indeed, let $p$ be the least prime divisor of $n$. Apply (ii) with $x=p$ and $y=n / p$ to obtain that $f(n)$ divides $(p-1) y^{n-1} f(p)=(p-1) y^{n-1} p^{p-1}$. Write $f(n)=\\ell P$ where $\\ell$ is coprime to $n$ and $P$ is a product of primes dividing $n$. Since $\\ell$ divides $(p-1) y^{n-1} p^{p-1}$ and is coprime to $y^{n-1} p^{p-1}$, it divides $p-1$; hence $d(\\ell) \\leq \\ell1$ ). So $f\\left(p^{a}\\right)=p^{b}$ for some $b \\geq 1$, and (i) yields $p^{a}=d\\left(f\\left(p^{a}\\right)\\right)=d\\left(p^{b}\\right)=b+1$. Hence $f\\left(p^{a}\\right)=p^{p^{a}-1}$, as needed.\n\nLet us finally show that (1) is true for a general $n>1$ with prime factorization $n=p_{1}^{a_{1}} \\cdots p_{k}^{a_{k}}$. We saw that the prime factorization of $f(n)$ has the form $f(n)=p_{1}^{b_{1}} \\cdots p_{k}^{b_{k}}$. For $i=1, \\ldots, k$, set $x=p_{i}^{a_{i}}$ and $y=n / x$ in (ii) to infer that $f(n)$ divides $\\left(p_{i}^{a_{i}}-1\\right) y^{n-1} f\\left(p_{i}^{a_{i}}\\right)$. Hence $p_{i}^{b_{i}}$ divides $\\left(p_{i}^{a_{i}}-1\\right) y^{n-1} f\\left(p_{i}^{a_{i}}\\right)$, and because $p_{i}^{b_{i}}$ is coprime to $\\left(p_{i}^{a_{i}}-1\\right) y^{n-1}$, it follows that $p_{i}^{b_{i}}$ divides $f\\left(p_{i}^{a_{i}}\\right)=p_{i}^{p_{i}^{a_{i}}-1}$. So $b_{i} \\leq p_{i}^{a_{i}}-1$ for all $i=1, \\ldots, k$. Combined with (i), these conclusions imply\n\n$$\np_{1}^{a_{1}} \\cdots p_{k}^{a_{k}}=n=d(f(n))=d\\left(p_{1}^{b_{1}} \\cdots p_{k}^{b_{k}}\\right)=\\left(b_{1}+1\\right) \\cdots\\left(b_{k}+1\\right) \\leq p_{1}^{a_{1}} \\cdots p_{k}^{a_{k}} .\n$$\n\nHence all inequalities $b_{i} \\leq p_{i}^{a_{i}}-1$ must be equalities, $i=1, \\ldots, k$, implying that (1) holds true. The proof is complete.']",['the function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ defined by $f(1)=1$ and \n$$\nf(n)=p_{1}^{p_{1}^{a_{1}}-1} p_{2}^{p_{2}^{a_{2}}-1} \\cdots p_{k}^{p_{k}^{a_{k}}-1} \\text{where }n=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{k}^{a_{k}} \\text{ is the prime factorization of } n>1\n$$'],False,,Need_human_evaluate, 2142,Number Theory,,Prove that there exist infinitely many positive integers $n$ such that $n^{2}+1$ has a prime divisor greater than $2 n+\sqrt{2 n}$.,"['Let $p \\equiv 1(\\bmod 8)$ be a prime. The congruence $x^{2} \\equiv-1(\\bmod p)$ has two solutions in $[1, p-1]$ whose sum is $p$. If $n$ is the smaller one of them then $p$ divides $n^{2}+1$ and $n \\leq(p-1) / 2$. We show that $p>2 n+\\sqrt{10 n}$.\n\nLet $n=(p-1) / 2-\\ell$ where $\\ell \\geq 0$. Then $n^{2} \\equiv-1(\\bmod p)$ gives\n\n$$\n\\left(\\frac{p-1}{2}-\\ell\\right)^{2} \\equiv-1 \\quad(\\bmod p) \\quad \\text { or } \\quad(2 \\ell+1)^{2}+4 \\equiv 0 \\quad(\\bmod p)\n$$\n\nThus $(2 \\ell+1)^{2}+4=r p$ for some $r \\geq 0$. As $(2 \\ell+1)^{2} \\equiv 1 \\equiv p(\\bmod 8)$, we have $r \\equiv 5(\\bmod 8)$, so that $r \\geq 5$. Hence $(2 \\ell+1)^{2}+4 \\geq 5 p$, implying $\\ell \\geq(\\sqrt{5 p-4}-1) / 2$. Set $\\sqrt{5 p-4}=u$ for clarity; then $\\ell \\geq(u-1) / 2$. Therefore\n\n$$\nn=\\frac{p-1}{2}-\\ell \\leq \\frac{1}{2}(p-u)\n$$\n\nCombined with $p=\\left(u^{2}+4\\right) / 5$, this leads to $u^{2}-5 u-10 n+4 \\geq 0$. Solving this quadratic inequality with respect to $u \\geq 0$ gives $u \\geq(5+\\sqrt{40 n+9}) / 2$. So the estimate $n \\leq(p-u) / 2$ leads to\n\n$$\np \\geq 2 n+u \\geq 2 n+\\frac{1}{2}(5+\\sqrt{40 n+9})>2 n+\\sqrt{10 n}\n$$\n\nSince there are infinitely many primes of the form $8 k+1$, it follows easily that there are also infinitely many $n$ with the stated property.']",,True,,, 2143,Algebra,,"Find all the functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $$ f(a)^{2}+f(b)^{2}+f(c)^{2}=2 f(a) f(b)+2 f(b) f(c)+2 f(c) f(a) $$ for all integers $a, b, c$ satisfying $a+b+c=0$.","['The substitution $a=b=c=0$ gives $3 f(0)^{2}=6 f(0)^{2}$, hence\n\n$$\nf(0)=0 \\text {. }\\tag{1}\n$$\n\nThe substitution $b=-a$ and $c=0$ gives $\\left((f(a)-f(-a))^{2}=0\\right.$. Hence $f$ is an even function:\n\n$$\nf(a)=f(-a) \\quad \\text { for all } a \\in \\mathbb{Z}\\tag{2}\n$$\n\nNow set $b=a$ and $c=-2 a$ to obtain $2 f(a)^{2}+f(2 a)^{2}=2 f(a)^{2}+4 f(a) f(2 a)$. Hence\n\n$$\nf(2 a)=0 \\text { or } f(2 a)=4 f(a) \\text { for all } a \\in \\mathbb{Z}\\tag{3}\n$$\n\nIf $f(r)=0$ for some $r \\geq 1$ then the substitution $b=r$ and $c=-a-r$ gives $(f(a+r)-f(a))^{2}=0$. So $f$ is periodic with period $r$, i. e.\n\n$$\nf(a+r)=f(a) \\text { for all } a \\in \\mathbb{Z}\n$$\n\nIn particular, if $f(1)=0$ then $f$ is constant, thus $f(a)=0$ for all $a \\in \\mathbb{Z}$. This function clearly satisfies the functional equation. For the rest of the analysis, we assume $f(1)=k \\neq 0$.\n\nBy (3) we have $f(2)=0$ or $f(2)=4 k$. If $f(2)=0$ then $f$ is periodic of period 2 , thus $f($ even $)=0$ and $f($ odd $)=k$. This function is a solution for every $k$. We postpone the verification; for the sequel assume $f(2)=4 k \\neq 0$.\n\nBy (3) again, we have $f(4)=0$ or $f(4)=16 k$. In the first case $f$ is periodic of period 4 , and $f(3)=f(-1)=f(1)=k$, so we have $f(4 n)=0, f(4 n+1)=f(4 n+3)=k$, and $f(4 n+2)=4 k$ for all $n \\in \\mathbb{Z}$. This function is a solution too, which we justify later. For the rest of the analysis, we assume $f(4)=16 k \\neq 0$.\n\nWe show now that $f(3)=9 k$. In order to do so, we need two substitutions:\n\n$$\n\\begin{gathered}\na=1, b=2, c=-3 \\Longrightarrow f(3)^{2}-10 k f(3)+9 k^{2}=0 \\Longrightarrow f(3) \\in\\{k, 9 k\\} \\\\\na=1, b=3, c=-4 \\Longrightarrow f(3)^{2}-34 k f(3)+225 k^{2}=0 \\Longrightarrow f(3) \\in\\{9 k, 25 k\\}\n\\end{gathered}\n$$\n\nTherefore $f(3)=9 k$, as claimed. Now we prove inductively that the only remaining function is $f(x)=k x^{2}, x \\in \\mathbb{Z}$. We proved this for $x=0,1,2,3,4$. Assume that $n \\geq 4$ and that $f(x)=k x^{2}$ holds for all integers $x \\in[0, n]$. Then the substitutions $a=n, b=1, c=-n-1$ and $a=n-1$, $b=2, c=-n-1$ lead respectively to\n\n$$\nf(n+1) \\in\\left\\{k(n+1)^{2}, k(n-1)^{2}\\right\\} \\quad \\text { and } \\quad f(n+1) \\in\\left\\{k(n+1)^{2}, k(n-3)^{2}\\right\\}\n$$\n\nSince $k(n-1)^{2} \\neq k(n-3)^{2}$ for $n \\neq 2$, the only possibility is $f(n+1)=k(n+1)^{2}$. This completes the induction, so $f(x)=k x^{2}$ for all $x \\geq 0$. The same expression is valid for negative values of $x$ since $f$ is even. To verify that $f(x)=k x^{2}$ is actually a solution, we need to check the identity $a^{4}+b^{4}+(a+b)^{4}=2 a^{2} b^{2}+2 a^{2}(a+b)^{2}+2 b^{2}(a+b)^{2}$, which follows directly by expanding both sides.\n\n\n\nTherefore the only possible solutions of the functional equation are the constant function $f_{1}(x)=0$ and the following functions:\n\n$$\nf_{2}(x)=k x^{2} \\quad f_{3}(x)=\\left\\{\\begin{array}{ccc}\n0 & x \\text { even } \\\\\nk & x \\text { odd }\n\\end{array} \\quad f_{4}(x)=\\left\\{\\begin{array}{ccc}\n0 & x \\equiv 0 & (\\bmod 4) \\\\\nk & x \\equiv 1 & (\\bmod 2) \\\\\n4 k & x \\equiv 2 & (\\bmod 4)\n\\end{array}\\right.\\right.\n$$\n\nfor any non-zero integer $k$. The verification that they are indeed solutions was done for the first two. For $f_{3}$ note that if $a+b+c=0$ then either $a, b, c$ are all even, in which case $f(a)=f(b)=f(c)=0$, or one of them is even and the other two are odd, so both sides of the equation equal $2 k^{2}$. For $f_{4}$ we use similar parity considerations and the symmetry of the equation, which reduces the verification to the triples $(0, k, k),(4 k, k, k),(0,0,0),(0,4 k, 4 k)$. They all satisfy the equation.']",['$f_{1}(x)=0$ and the following functions:\n\n$$\nf_{2}(x)=k x^{2} \\quad f_{3}(x)=\\left\\{\\begin{array}{ccc}\n0 & x \\text { even } \\\\\nk & x \\text { odd }\n\\end{array} \\quad f_{4}(x)=\\left\\{\\begin{array}{ccc}\n0 & x \\equiv 0 & (\\bmod 4) \\\\\nk & x \\equiv 1 & (\\bmod 2) \\\\\n4 k & x \\equiv 2 & (\\bmod 4)\n\\end{array}\\right.\\right.\n$$'],False,,Need_human_evaluate, 2144,Algebra,,"Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively. Here $X+Y$ denotes the set $\{x+y \mid x \in X, y \in Y\}$, for $X, Y \subseteq \mathbb{Z}$ and $X, Y \subseteq \mathbb{Q}$. a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint? b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A, B, C$ such that the sets $A+B, B+C, C+A$ are disjoint?","['a) The residue classes modulo 3 yield such a partition:\n\n$$\nA=\\{3 k \\mid k \\in \\mathbb{Z}\\}, \\quad B=\\{3 k+1 \\mid k \\in \\mathbb{Z}\\}, \\quad C=\\{3 k+2 \\mid k \\in \\mathbb{Z}\\}\n$$\n\nb) The answer is no. Suppose that $\\mathbb{Q}$ can be partitioned into non-empty subsets $A, B, C$ as stated. Note that for all $a \\in A, b \\in B, c \\in C$ one has\n\n$$\na+b-c \\in C, \\quad b+c-a \\in A, \\quad c+a-b \\in B\\tag{1}\n$$\n\nIndeed $a+b-c \\notin A$ as $(A+B) \\cap(A+C)=\\emptyset$, and similarly $a+b-c \\notin B$, hence $a+b-c \\in C$. The other two relations follow by symmetry. Hence $A+B \\subset C+C, B+C \\subset A+A, C+A \\subset B+B$.\n\nThe opposite inclusions also hold. Let $a, a^{\\prime} \\in A$ and $b \\in B, c \\in C$ be arbitrary. By (1) $a^{\\prime}+c-b \\in B$, and since $a \\in A, c \\in C$, we use (1) again to obtain\n\n$$\na+a^{\\prime}-b=a+\\left(a^{\\prime}+c-b\\right)-c \\in C .\n$$\n\nSo $A+A \\subset B+C$ and likewise $B+B \\subset C+A, C+C \\subset A+B$. In summary\n\n$$\nB+C=A+A, \\quad C+A=B+B, \\quad A+B=C+C .\n$$\n\nFurthermore suppose that $0 \\in A$ without loss of generality. Then $B=\\{0\\}+B \\subset A+B$ and $C=\\{0\\}+C \\subset A+C$. So, since $B+C$ is disjoint with $A+B$ and $A+C$, it is also disjoint with $B$ and $C$. Hence $B+C$ is contained in $\\mathbb{Z} \\backslash(B \\cup C)=A$. Because $B+C=A+A$, we obtain $A+A \\subset A$. On the other hand $A=\\{0\\}+A \\subset A+A$, implying $A=A+A=B+C$.\n\nTherefore $A+B+C=A+A+A=A$, and now $B+B=C+A$ and $C+C=A+B$ yield $B+B+B=A+B+C=A, C+C+C=A+B+C=A$. In particular if $r \\in \\mathbb{Q}=A \\cup B \\cup C$ is arbitrary then $3 r \\in A$.\n\nHowever such a conclusion is impossible. Take any $b \\in B(B \\neq \\emptyset)$ and let $r=b / 3 \\in \\mathbb{Q}$. Then $b=3 r \\in A$ which is a contradiction.', 'We prove that the example for $\\mathbb{Z}$ from the first solution is unique, and then use this fact to solve part $b)$.\n\nLet $\\mathbb{Z}=A \\cup B \\cup C$ be a partition of $\\mathbb{Z}$ with $A, B, C \\neq \\emptyset$ and $A+B, B+C, C+A$ disjoint. We need the relations (1) which clearly hold for $\\mathbb{Z}$. Fix two consecutive integers from different sets, say $b \\in B$ and $c=b+1 \\in C$. For every $a \\in A$ we have, in view of (1), $a-1=a+b-c \\in C$ and $a+1=a+c-b \\in B$. So every $a \\in A$ is preceded by a number from $C$ and followed by a number from $B$.\n\nIn particular there are pairs of the form $c, c+1$ with $c \\in C, c+1 \\in A$. For such a pair and any $b \\in B$ analogous reasoning shows that each $b \\in B$ is preceded by a number from $A$ and followed by a number from $C$. There are also pairs $b, b-1$ with $b \\in B, b-1 \\in A$. We use them in a similar way to prove that each $c \\in C$ is preceded by a number from $B$ and followed by a number from $A$.\n\nBy putting the observations together we infer that $A, B, C$ are the three congruence classes modulo 3. Observe that all multiples of 3 are in the set of the partition that contains 0 .\n\n\n\nNow we turn to part b). Suppose that there is a partition of $\\mathbb{Q}$ with the given properties. Choose three rationals $r_{i}=p_{i} / q_{i}$ from the three sets $A, B, C, i=1,2,3$, and set $N=3 q_{1} q_{2} q_{3}$.\n\nLet $S \\subset \\mathbb{Q}$ be the set of fractions with denominators $N$ (irreducible or not). It is obtained through multiplication of every integer by the constant $1 / N$, hence closed under sums and differences. Moreover, if we identify each $k \\in \\mathbb{Z}$ with $k / N \\in S$ then $S$ is essentially the set $\\mathbb{Z}$ with respect to addition. The numbers $r_{i}$ belong to $S$ because\n\n$$\nr_{1}=\\frac{3 p_{1} q_{2} q_{3}}{N}, \\quad r_{2}=\\frac{3 p_{2} q_{3} q_{1}}{N}, \\quad r_{3}=\\frac{3 p_{3} q_{1} q_{2}}{N}\n$$\n\nThe partition $\\mathbb{Q}=A \\cup B \\cup C$ of $\\mathbb{Q}$ induces a partition $S=A^{\\prime} \\cup B^{\\prime} \\cup C^{\\prime}$ of $S$, with $A^{\\prime}=A \\cap S$, $B^{\\prime}=B \\cap S, C^{\\prime}=C \\cap S$. Clearly $A^{\\prime}+B^{\\prime}, B^{\\prime}+C^{\\prime}, C^{\\prime}+A^{\\prime}$ are disjoint, so this partition has the properties we consider.\n\nBy the uniqueness of the example for $\\mathbb{Z}$ the sets $A^{\\prime}, B^{\\prime}, C^{\\prime}$ are the congruence classes modulo 3 , multiplied by $1 / N$. Also all multiples of $3 / N$ are in the same set, $A^{\\prime}, B^{\\prime}$ or $C^{\\prime}$. This holds for $r_{1}, r_{2}, r_{3}$ in particular as they are all multiples of $3 / N$. However $r_{1}, r_{2}, r_{3}$ are in different sets $A^{\\prime}, B^{\\prime}, C^{\\prime}$ since they were chosen from different sets $A, B, C$. The contradiction ends the proof.']",,True,,, 2145,Algebra,,"Let $a_{2}, \ldots, a_{n}$ be $n-1$ positive real numbers, where $n \geq 3$, such that $a_{2} a_{3} \cdots a_{n}=1$. Prove that $$ \left(1+a_{2}\right)^{2}\left(1+a_{3}\right)^{3} \cdots\left(1+a_{n}\right)^{n}>n^{n} . $$","['The substitution $a_{2}=\\frac{x_{2}}{x_{1}}, a_{3}=\\frac{x_{3}}{x_{2}}, \\ldots, a_{n}=\\frac{x_{1}}{x_{n-1}}$ transforms the original problem into the inequality\n\n$$\n\\left(x_{1}+x_{2}\\right)^{2}\\left(x_{2}+x_{3}\\right)^{3} \\cdots\\left(x_{n-1}+x_{1}\\right)^{n}>n^{n} x_{1}^{2} x_{2}^{3} \\cdots x_{n-1}^{n}\\tag{*}\n$$\n\nfor all $x_{1}, \\ldots, x_{n-1}>0$. To prove this, we use the AM-GM inequality for each factor of the left-hand side as follows:\n\n$$\n\\begin{array}{rlcll}\n\\left(x_{1}+x_{2}\\right)^{2} & & & & \\geq 2^{2} x_{1} x_{2} \\\\\n\\left(x_{2}+x_{3}\\right)^{3} & = & \\left(2\\left(\\frac{x_{2}}{2}\\right)+x_{3}\\right)^{3} & & \\geq 3^{3}\\left(\\frac{x_{2}}{2}\\right)^{2} x_{3} \\\\\n\\left(x_{3}+x_{4}\\right)^{4} & = & \\left(3\\left(\\frac{x_{3}}{3}\\right)+x_{4}\\right)^{4} & & \\geq 4^{4}\\left(\\frac{x_{3}}{3}\\right)^{3} x_{4} \\\\\n& \\vdots & \\vdots && \\vdots \\\\\n\\left(x_{n-1}+x_{1}\\right)^{n} & = & \\left((n-1)\\left(\\frac{x_{n-1}}{n-1}\\right)+x_{1}\\right)^{n} && \\geq n^{n}\\left(\\frac{x_{n-1}}{n-1}\\right)^{n-1} x_{1}\n\\end{array}\n$$\n\nMultiplying these inequalities together gives $(*)$, with inequality sign $\\geq$ instead of $>$. However for the equality to occur it is necessary that $x_{1}=x_{2}, x_{2}=2 x_{3}, \\ldots, x_{n-1}=(n-1) x_{1}$, implying $x_{1}=(n-1) ! x_{1}$. This is impossible since $x_{1}>0$ and $n \\geq 3$. Therefore the inequality is strict.']",,True,,, 2146,Algebra,,Let $f$ and $g$ be two nonzero polynomials with integer coefficients and $\operatorname{deg} f>\operatorname{deg} g$. Suppose that for infinitely many primes $p$ the polynomial $p f+g$ has a rational root. Prove that $f$ has a rational root.,"['Since $\\operatorname{deg} f>\\operatorname{deg} g$, we have $|g(x) / f(x)|<1$ for sufficiently large $x$; more precisely, there is a real number $R$ such that $|g(x) / f(x)|<1$ for all $x$ with $|x|>R$. Then for all such $x$ and all primes $p$ we have\n\n$$\n|p f(x)+g(x)| \\geq|f(x)|\\left(p-\\frac{|g(x)|}{|f(x)|}\\right)>0\n$$\n\nHence all real roots of the polynomials $p f+g$ lie in the interval $[-R, R]$.\n\nLet $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{0}$ and $g(x)=b_{m} x^{m}+b_{m-1} x^{m-1}+\\cdots+b_{0}$ where $n>m, a_{n} \\neq 0$ and $b_{m} \\neq 0$. Upon replacing $f(x)$ and $g(x)$ by $a_{n}^{n-1} f\\left(x / a_{n}\\right)$ and $a_{n}^{n-1} g\\left(x / a_{n}\\right)$ respectively, we reduce the problem to the case $a_{n}=1$. In other words one can assume that $f$ is monic. Then the leading coefficient of $p f+g$ is $p$, and if $r=u / v$ is a rational root of $p f+g$ with $(u, v)=1$ and $v>0$, then either $v=1$ or $v=p$.\n\nFirst consider the case when $v=1$ infinitely many times. If $v=1$ then $|u| \\leq R$, so there are only finitely many possibilities for the integer $u$. Therefore there exist distinct primes $p$ and $q$ for which we have the same value of $u$. Then the polynomials $p f+g$ and $q f+g$ share this root, implying $f(u)=g(u)=0$. So in this case $f$ and $g$ have an integer root in common.\n\nNow suppose that $v=p$ infinitely many times. By comparing the exponent of $p$ in the denominators of $p f(u / p)$ and $g(u / p)$ we get $m=n-1$ and $p f(u / p)+g(u / p)=0$ reduces to an equation of the form\n\n$$\n\\left(u^{n}+a_{n-1} p u^{n-1}+\\ldots+a_{0} p^{n}\\right)+\\left(b_{n-1} u^{n-1}+b_{n-2} p u^{n-2}+\\ldots+b_{0} p^{n-1}\\right)=0 .\n$$\n\nThe equation above implies that $u^{n}+b_{n-1} u^{n-1}$ is divisible by $p$ and hence, since $(u, p)=1$, we have $u+b_{n-1}=p k$ with some integer $k$. On the other hand all roots of $p f+g$ lie in the interval $[-R, R]$, so that\n\n$$\n\\begin{gathered}\n\\frac{\\left|p k-b_{n-1}\\right|}{p}=\\frac{|u|}{p}5 / 4$ then $-x<5 / 4$ and so $g(2-x)-g(-x)=2$ by the above. On the other hand (3) gives $g(x)=2-g(2-x), g(x+2)=2-g(-x)$, so that $g(x+2)-g(x)=g(2-x)-g(-x)=2$. Thus (4) is true for all $x \\in \\mathbb{R}$.\n\nNow replace $x$ by $-x$ in (3) to obtain $g(-x)+g(2+x)=2$. In view of (4) this leads to $g(x)+g(-x)=0$, i. e. $g(-x)=-g(x)$ for all $x$. Taking this into account, we apply (1) to the pairs $(-x, y)$ and $(x,-y)$ :\n\n$g(1-x y)-g(-x+y)=(g(x)+1)(1-g(y)), \\quad g(1-x y)-g(x-y)=(1-g(x))(g(y)+1)$.\n\nAdding up yields $g(1-x y)=1-g(x) g(y)$. Then $g(1+x y)=1+g(x) g(y)$ by (3). Now the original equation (1) takes the form $g(x+y)=g(x)+g(y)$. Hence $g$ is additive.\n\nBy additvity $g(1+x y)=g(1)+g(x y)=1+g(x y)$; since $g(1+x y)=1+g(x) g(y)$ was shown above, we also have $g(x y)=g(x) g(y)$ ( $g$ is multiplicative). In particular $y=x$ gives $g\\left(x^{2}\\right)=g(x)^{2} \\geq 0$ for all $x$, meaning that $g(x) \\geq 0$ for $x \\geq 0$. Since $g$ is additive and bounded from below on $[0,+\\infty)$, it is linear; more exactly $g(x)=g(1) x=x$ for all $x \\in \\mathbb{R}$.\n\nIn summary $f(x)=x-1, x \\in \\mathbb{R}$. It is straightforward that this function satisfies the requirements.']",['$f(x)=x-1$'],False,,Expression, 2148,Algebra,,"Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function, and let $f^{m}$ be $f$ applied $m$ times. Suppose that for every $n \in \mathbb{N}$ there exists a $k \in \mathbb{N}$ such that $f^{2 k}(n)=n+k$, and let $k_{n}$ be the smallest such $k$. Prove that the sequence $k_{1}, k_{2}, \ldots$ is unbounded.","['We restrict attention to the set\n\n$$\nS=\\left\\{1, f(1), f^{2}(1), \\ldots\\right\\}\n$$\n\nObserve that $S$ is unbounded because for every number $n$ in $S$ there exists a $k>0$ such that $f^{2 k}(n)=n+k$ is in $S$. Clearly $f$ maps $S$ into itself; moreover $f$ is injective on $S$. Indeed if $f^{i}(1)=f^{j}(1)$ with $i \\neq j$ then the values $f^{m}(1)$ start repeating periodically from some point on, and $S$ would be finite.\n\nDefine $g: S \\rightarrow S$ by $g(n)=f^{2 k_{n}}(n)=n+k_{n}$. We prove that $g$ is injective too. Suppose that $g(a)=g(b)$ with $ak_{b}$. So, since $f$ is injective on $S$, we obtain\n\n$$\nf^{2\\left(k_{a}-k_{b}\\right)}(a)=b=a+\\left(k_{a}-k_{b}\\right)\n$$\n\nHowever this contradicts the minimality of $k_{a}$ as $0n$ for $n \\in S$, so $T$ is non-empty. For each $t \\in T$ denote $C_{t}=\\left\\{t, g(t), g^{2}(t), \\ldots\\right\\}$; call $C_{t}$ the chain starting at $t$. Observe that distinct chains are disjoint because $g$ is injective. Each $n \\in S \\backslash T$ has the form $n=g\\left(n^{\\prime}\\right)$ with $n^{\\prime}k$, i. e. $k_{n}>k$. In conclusion $k_{1}, k_{2}, \\ldots$ is unbounded.']",,True,,, 2149,Algebra,,"We say that a function $f: \mathbb{R}^{k} \rightarrow \mathbb{R}$ is a metapolynomial if, for some positive integers $m$ and $n$, it can be represented in the form $$ f\left(x_{1}, \ldots, x_{k}\right)=\max _{i=1, \ldots, m} \min _{j=1, \ldots, n} P_{i, j}\left(x_{1}, \ldots, x_{k}\right) $$ where $P_{i, j}$ are multivariate polynomials. Prove that the product of two metapolynomials is also a metapolynomial.","['We use the notation $f(x)=f\\left(x_{1}, \\ldots, x_{k}\\right)$ for $x=\\left(x_{1}, \\ldots, x_{k}\\right)$ and $[m]=\\{1,2, \\ldots, m\\}$. Observe that if a metapolynomial $f(x)$ admits a representation like the one in the statement for certain positive integers $m$ and $n$, then they can be replaced by any $m^{\\prime} \\geq m$ and $n^{\\prime} \\geq n$. For instance, if we want to replace $m$ by $m+1$ then it is enough to define $P_{m+1, j}(x)=P_{m, j}(x)$ and note that repeating elements of a set do not change its maximum nor its minimum. So one can assume that any two metapolynomials are defined with the same $m$ and $n$. We reserve letters $P$ and $Q$ for polynomials, so every function called $P, P_{i, j}, Q, Q_{i, j}, \\ldots$ is a polynomial function.\n\nWe start with a lemma that is useful to change expressions of the form $\\min \\max f_{i, j}$ to ones of the form $\\max \\min g_{i, j}$.\n\nLemma. Let $\\left\\{a_{i, j}\\right\\}$ be real numbers, for all $i \\in[m]$ and $j \\in[n]$. Then\n\n$$\n\\min _{i \\in[m]} \\max _{j \\in[n]} a_{i, j}=\\max _{j_{1}, \\ldots, j_{m} \\in[n]} \\min _{i \\in[m]} a_{i, j_{i}}\n$$\n\nwhere the max in the right-hand side is over all vectors $\\left(j_{1}, \\ldots, j_{m}\\right)$ with $j_{1}, \\ldots, j_{m} \\in[n]$.\n\nProof. We can assume for all $i$ that $a_{i, n}=\\max \\left\\{a_{i, 1}, \\ldots, a_{i, n}\\right\\}$ and $a_{m, n}=\\min \\left\\{a_{1, n}, \\ldots, a_{m, n}\\right\\}$. The left-hand side is $=a_{m, n}$ and hence we need to prove the same for the right-hand side. If $\\left(j_{1}, j_{2}, \\ldots, j_{m}\\right)=(n, n, \\ldots, n)$ then $\\min \\left\\{a_{1, j_{1}}, \\ldots, a_{m, j_{m}}\\right\\}=\\min \\left\\{a_{1, n}, \\ldots, a_{m, n}\\right\\}=a_{m, n}$ which implies that the right-hand side is $\\geq a_{m, n}$. It remains to prove the opposite inequality and this is equivalent to $\\min \\left\\{a_{1, j_{1}}, \\ldots, a_{m, j_{m}}\\right\\} \\leq a_{m, n}$ for all possible $\\left(j_{1}, j_{2}, \\ldots, j_{m}\\right)$. This is true because $\\min \\left\\{a_{1, j_{1}}, \\ldots, a_{m, j_{m}}\\right\\} \\leq a_{m, j_{m}} \\leq a_{m, n}$.\n\nWe need to show that the family $\\mathcal{M}$ of metapolynomials is closed under multiplication, but it turns out easier to prove more: that it is also closed under addition, maxima and minima.\n\nFirst we prove the assertions about the maxima and the minima. If $f_{1}, \\ldots, f_{r}$ are metapolynomials, assume them defined with the same $m$ and $n$. Then\n\n$$\nf=\\max \\left\\{f_{1}, \\ldots, f_{r}\\right\\}=\\max \\left\\{\\max _{i \\in[m]} \\min _{j \\in[n]} P_{i, j}^{1}, \\ldots, \\max _{i \\in[m]} \\min _{j \\in[n]} P_{i, j}^{r}\\right\\}=\\max _{s \\in[r], i \\in[m]} \\min _{j \\in[n]} P_{i, j}^{s}\n$$\n\nIt follows that $f=\\max \\left\\{f_{1}, \\ldots, f_{r}\\right\\}$ is a metapolynomial. The same argument works for the minima, but first we have to replace $\\min \\max$ by $\\max \\min$, and this is done via the lemma.\n\nAnother property we need is that if $f=\\max \\min P_{i, j}$ is a metapolynomial then so is $-f$. Indeed, $-f=\\min \\left(-\\min P_{i, j}\\right)=\\min \\max P_{i, j}$.\n\nTo prove $\\mathcal{M}$ is closed under addition let $f=\\max \\min P_{i, j}$ and $g=\\max \\min Q_{i, j}$. Then\n\n$$\n\\begin{gathered}\nf(x)+g(x)=\\max _{i \\in[m]} \\min _{j \\in[n]} P_{i, j}(x)+\\max _{i \\in[m]} \\min _{j \\in[n]} Q_{i, j}(x) \\\\\n=\\max _{i_{1}, i_{2} \\in[m]}\\left(\\min _{j \\in[n]} P_{i_{1}, j}(x)+\\min _{j \\in[n]} Q_{i_{2}, j}(x)\\right)=\\max _{i_{1}, i_{2} \\in[m]} \\min _{j_{1}, j_{2} \\in[n]}\\left(P_{i_{1}, j_{1}}(x)+Q_{i_{2}, j_{2}}(x)\\right),\n\\end{gathered}\n$$\n\nand hence $f(x)+g(x)$ is a metapolynomial.\n\nWe proved that $\\mathcal{M}$ is closed under sums, maxima and minima, in particular any function that can be expressed by sums, max, min, polynomials or even metapolynomials is in $\\mathcal{M}$.\n\nWe would like to proceed with multiplication along the same lines like with addition, but there is an essential difference. In general the product of the maxima of two sets is not equal\n\n\n\nto the maximum of the product of the sets. We need to deal with the fact that $ay$ and $x$ is to the left of $y$, and replaces the pair $(x, y)$ by either $(y+1, x)$ or $(x-1, x)$. Prove that she can perform only finitely many such iterations.","['Note first that the allowed operation does not change the maximum $M$ of the initial sequence. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be the numbers obtained at some point of the process. Consider the sum\n\n$$\nS=a_{1}+2 a_{2}+\\cdots+n a_{n} .\n$$\n\nWe claim that $S$ increases by a positive integer amount with every operation. Let the operation replace the pair $\\left(a_{i}, a_{i+1}\\right)$ by a pair $\\left(c, a_{i}\\right)$, where $a_{i}>a_{i+1}$ and $c=a_{i+1}+1$ or $c=a_{i}-1$. Then the new and the old value of $S$ differ by $d=\\left(i c+(i+1) a_{i}\\right)-\\left(i a_{i}+(i+1) a_{i+1}\\right)=a_{i}-a_{i+1}+i\\left(c-a_{i+1}\\right)$. The integer $d$ is positive since $a_{i}-a_{i+1} \\geq 1$ and $c-a_{i+1} \\geq 0$.\n\nOn the other hand $S \\leq(1+2+\\cdots+n) M$ as $a_{i} \\leq M$ for all $i=1, \\ldots, n$. Since $S$ increases by at least 1 at each step and never exceeds the constant $(1+2+\\cdots+n) M$, the process stops after a finite number of iterations.', 'Like in the first solution note that the operations do not change the maximum $M$ of the initial sequence. Now consider the reverse lexicographical order for $n$-tuples of integers. We say that $\\left(x_{1}, \\ldots, x_{n}\\right)<\\left(y_{1}, \\ldots, y_{n}\\right)$ if $x_{n}y$ and $y \\leq a \\leq x$, we see that $s_{i}$ decreases by at least 1 . This concludes the proof.""]",,True,,, 2151,Combinatorics,,"Let $n \geq 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{1,2, \ldots, n\}$ such that the sums of the different pairs are different integers not exceeding $n$ ?","['Consider $x$ such pairs in $\\{1,2, \\ldots, n\\}$. The sum $S$ of the $2 x$ numbers in them is at least $1+2+\\cdots+2 x$ since the pairs are disjoint. On the other hand $S \\leq n+(n-1)+\\cdots+(n-x+1)$ because the sums of the pairs are different and do not exceed $n$. This gives the inequality\n\n$$\n\\frac{2 x(2 x+1)}{2} \\leq n x-\\frac{x(x-1)}{2}\n$$\n\nwhich leads to $x \\leq \\frac{2 n-1}{5}$. Hence there are at most $\\left\\lfloor\\frac{2 n-1}{5}\\right\\rfloor$ pairs with the given properties.\n\nWe show a construction with exactly $\\left\\lfloor\\frac{2 n-1}{5}\\right\\rfloor$ pairs. First consider the case $n=5 k+3$ with $k \\geq 0$, where $\\left\\lfloor\\frac{2 n-1}{5}\\right\\rfloor=2 k+1$. The pairs are displayed in the following table.\n\n| Pairs | $3 k+1$ | $3 k$ | $\\cdots$ | $2 k+2$ | $4 k+2$ | $4 k+1$ | $\\cdots$ | $3 k+3$ | $3 k+2$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| | 2 | 4 | $\\cdots$ | $2 k$ | 1 | 3 | $\\cdots$ | $2 k-1$ | $2 k+1$ |\n| Sums | $3 k+3$ | $3 k+4$ | $\\cdots$ | $4 k+2$ | $4 k+3$ | $4 k+4$ | $\\cdots$ | $5 k+2$ | $5 k+3$ |\n\nThe $2 k+1$ pairs involve all numbers from 1 to $4 k+2$; their sums are all numbers from $3 k+3$ to $5 k+3$. The same construction works for $n=5 k+4$ and $n=5 k+5$ with $k \\geq 0$. In these cases the required number $\\left\\lfloor\\frac{2 n-1}{5}\\right\\rfloor$ of pairs equals $2 k+1$ again, and the numbers in the table do not exceed $5 k+3$. In the case $n=5 k+2$ with $k \\geq 0$ one needs only $2 k$ pairs. They can be obtained by ignoring the last column of the table (thus removing $5 k+3$ ). Finally, $2 k$ pairs are also needed for the case $n=5 k+1$ with $k \\geq 0$. Now it suffices to ignore the last column of the table and then subtract 1 from each number in the first row.']",['$\\lfloor\\frac{2 n-1}{5}\\rfloor$'],False,,Expression, 2152,Combinatorics,,"In a $999 \times 999$ square table some cells are white and the remaining ones are red. Let $T$ be the number of triples $\left(C_{1}, C_{2}, C_{3}\right)$ of cells, the first two in the same row and the last two in the same column, with $C_{1}$ and $C_{3}$ white and $C_{2}$ red. Find the maximum value $T$ can attain.","['We prove that in an $n \\times n$ square table there are at most $\\frac{4 n^{4}}{27}$ such triples.\n\nLet row $i$ and column $j$ contain $a_{i}$ and $b_{j}$ white cells respectively, and let $R$ be the set of red cells. For every red cell $(i, j)$ there are $a_{i} b_{j}$ admissible triples $\\left(C_{1}, C_{2}, C_{3}\\right)$ with $C_{2}=(i, j)$, therefore\n\n$$\nT=\\sum_{(i, j) \\in R} a_{i} b_{j}\n$$\n\nWe use the inequality $2 a b \\leq a^{2}+b^{2}$ to obtain\n\n$$\nT \\leq \\frac{1}{2} \\sum_{(i, j) \\in R}\\left(a_{i}^{2}+b_{j}^{2}\\right)=\\frac{1}{2} \\sum_{i=1}^{n}\\left(n-a_{i}\\right) a_{i}^{2}+\\frac{1}{2} \\sum_{j=1}^{n}\\left(n-b_{j}\\right) b_{j}^{2}\n$$\n\nThis is because there are $n-a_{i}$ red cells in row $i$ and $n-b_{j}$ red cells in column $j$. Now we maximize the right-hand side.\n\nBy the AM-GM inequality we have\n\n$$\n(n-x) x^{2}=\\frac{1}{2}(2 n-2 x) \\cdot x \\cdot x \\leq \\frac{1}{2}\\left(\\frac{2 n}{3}\\right)^{3}=\\frac{4 n^{3}}{27}\n$$\n\nwith equality if and only if $x=\\frac{2 n}{3}$. By putting everything together, we get\n\n$$\nT \\leq \\frac{n}{2} \\frac{4 n^{3}}{27}+\\frac{n}{2} \\frac{4 n^{3}}{27}=\\frac{4 n^{4}}{27}\n$$\n\nIf $n=999$ then any coloring of the square table with $x=\\frac{2 n}{3}=666$ white cells in each row and column attains the maximum as all inequalities in the previous argument become equalities. For example color a cell $(i, j)$ white if $i-j \\equiv 1,2, \\ldots, 666(\\bmod 999)$, and red otherwise.\n\nTherefore the maximum value $T$ can attain is $T=\\frac{4 \\cdot 999^{4}}{27}$.']",['$\\frac{4 \\cdot 999^{4}}{27}$'],False,,Expression, 2153,Combinatorics,,"Players $A$ and $B$ play a game with $N \geq 2012$ coins and 2012 boxes arranged around a circle. Initially $A$ distributes the coins among the boxes so that there is at least 1 coin in each box. Then the two of them make moves in the order $B, A, B, A, \ldots$ by the following rules: - On every move of his $B$ passes 1 coin from every box to an adjacent box. - On every move of hers $A$ chooses several coins that were not involved in $B$ 's previous move and are in different boxes. She passes every chosen coin to an adjacent box. Player $A$ 's goal is to ensure at least 1 coin in each box after every move of hers, regardless of how $B$ plays and how many moves are made. Find the least $N$ that enables her to succeed.","[""We argue for a general $n \\geq 7$ instead of 2012 and prove that the required minimum $N$ is $2 n-2$. For $n=2012$ this gives $N_{\\min }=4022$.\n\na) If $N=2 n-2$ player $A$ can achieve her goal. Let her start the game with a regular distribution: $n-2$ boxes with 2 coins and 2 boxes with 1 coin. Call the boxes of the two kinds red and white respectively. We claim that on her first move $A$ can achieve a regular distribution again, regardless of $B$ 's first move $M$. She acts according as the following situation $S$ occurs after $M$ or not: The initial distribution contains a red box $R$ with 2 white neighbors, and $R$ receives no coins from them on move $M$.\n\nSuppose that $S$ does not occur. Exactly one of the coins $c_{1}$ and $c_{2}$ in a given red box $X$ is involved in $M$, say $c_{1}$. If $M$ passes $c_{1}$ to the right neighbor of $X$, let $A$ pass $c_{2}$ to its left neighbor, and vice versa. By doing so with all red boxes $A$ performs a legal move $M^{\\prime}$. Thus $M$ and $M^{\\prime}$ combined move the 2 coins of every red box in opposite directions. Hence after $M$ and $M^{\\prime}$ are complete each neighbor of a red box $X$ contains exactly 1 coin that was initially in $X$. So each box with a red neighbor is non-empty after $M^{\\prime}$. If initially there is a box $X$ with 2 white neighbors ( $X$ is red and unique) then $X$ receives a coin from at least one of them on move $M$ since $S$ does not occur. Such a coin is not involved in $M^{\\prime}$, so $X$ is also non-empty after $M^{\\prime}$. Furthermore each box $Y$ has given away its initial content after $M$ and $M^{\\prime}$. A red neighbor of $Y$ adds 1 coin to it; a white neighbor adds at most 1 coin because it is not involved in $M^{\\prime}$. Hence each box contains 1 or 2 coins after $M^{\\prime}$. Because $N=2 n-2$, such a distribution is regular.\n\nNow let $S$ occur after move $M$. Then $A$ leaves untouched the exceptional red box $R$. With all remaining red boxes she proceeds like in the previous case, thus making a legal move $M^{\\prime \\prime}$. Box $R$ receives no coins from its neighbors on either move, so there is 1 coin in it after $M^{\\prime \\prime}$. Like above $M$ and $M^{\\prime \\prime}$ combined pass exactly 1 coin from every red box different from $R$ to each of its neighbors. Every box except $R$ has a red neighbor different from $R$, hence all boxes are non-empty after $M^{\\prime \\prime}$. Next, each box $Y$ except $R$ loses its initial content after $M$ and $M^{\\prime \\prime}$. A red neighbor of $Y$ adds at most 1 coin to it; a white neighbor also adds at most 1 coin as it does not participate in $M^{\\prime \\prime}$. Thus each box has 1 or 2 coins after $M^{\\prime \\prime}$, and the obtained distribution is regular.\n\nPlayer $A$ can apply the described strategy indefinitely, so $N=2 n-2$ enables her to succeed.\n\nb) For $N \\leq 2 n-3$ player $B$ can achieve an empty box after some move of $A$. Let $\\alpha$ be a set of $\\ell$ consecutive boxes containing a total of $N(\\alpha)$ coins. We call $\\alpha$ an $\\operatorname{arc}$ if $\\ell \\leq n-2$ and $N(\\alpha) \\leq 2 \\ell-3$. Note that $\\ell \\geq 2$ by the last condition. Moreover if both extremes of $\\alpha$ are non-empty boxes then $N(\\alpha) \\geq 2$, so that $N(\\alpha) \\leq 2 \\ell-3$ implies $\\ell \\geq 3$. Observe also that if an extreme $X$ of $\\alpha$ has more than 1 coin then ignoring $X$ yields a shorter arc. It follows that every arc contains an arc whose extremes have at most 1 coin each.\n\nGiven a clockwise labeling $1,2, \\ldots, n$ of the boxes, suppose that boxes $1,2, \\ldots, \\ell$ form an arc $\\alpha$, with $\\ell \\leq n-2$ and $N(\\alpha) \\leq 2 \\ell-3$. Suppose also that all $n \\geq 7$ boxes are non-empty. Then $B$ can move so that an arc $\\alpha^{\\prime}$ with $N\\left(\\alpha^{\\prime}\\right)2^{k}$ we show how Ben can find a number $y \\in T$ that is different from $x$. By performing this step repeatedly he can reduce $T$ to be of size $2^{k} \\leq n$ and thus win.\n\nSince only the size $m>2^{k}$ of $T$ is relevant, assume that $T=\\left\\{0,1, \\ldots, 2^{k}, \\ldots, m-1\\right\\}$. Ben begins by asking repeatedly whether $x$ is $2^{k}$. If Amy answers no $k+1$ times in a row, one of these answers is truthful, and so $x \\neq 2^{k}$. Otherwise Ben stops asking about $2^{k}$ at the first answer yes. He then asks, for each $i=1, \\ldots, k$, if the binary representation of $x$ has a 0 in the $i$ th digit. Regardless of what the $k$ answers are, they are all inconsistent with a certain number $y \\in\\left\\{0,1, \\ldots, 2^{k}-1\\right\\}$. The preceding answer yes about $2^{k}$ is also inconsistent with $y$. Hence $y \\neq x$. Otherwise the last $k+1$ answers are not truthful, which is impossible.\n\nEither way, Ben finds a number in $T$ that is different from $x$, and the claim is proven.']",,True,,, 2156,Combinatorics,,"Let $k$ and $n$ be fixed positive integers. In the liar's guessing game, Amy chooses integers $x$ and $N$ with $1 \leq x \leq N$. She tells Ben what $N$ is, but not what $x$ is. Ben may then repeatedly ask Amy whether $x \in S$ for arbitrary sets $S$ of integers. Amy will always answer with yes or no, but she might lie. The only restriction is that she can lie at most $k$ times in a row. After he has asked as many questions as he wants, Ben must specify a set of at most $n$ positive integers. If $x$ is in this set he wins; otherwise, he loses. Prove that: For sufficiently large $k$ there exist $n \geq 1.99^{k}$ such that Ben cannot guarantee a win.","['Consider an answer $A \\in\\{y e s, n o\\}$ to a question of the kind ""Is $x$ in the set $S$ ?"" We say that $A$ is inconsistent with a number $i$ if $A=y e s$ and $i \\notin S$, or if $A=n o$ and $i \\in S$. Observe that an answer inconsistent with the target number $x$ is a lie.\n\nWe prove that if $1<\\lambda<2$ and $n=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1$ then Ben cannot guarantee a win. To complete the proof, then it suffices to take $\\lambda$ such that $1.99<\\lambda<2$ and $k$ large enough so that\n\n$$\nn=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1 \\geq 1.99^{k}\n$$\n\nConsider the following strategy for Amy. First she chooses $N=n+1$ and $x \\in\\{1,2, \\ldots, n+1\\}$ arbitrarily. After every answer of hers Amy determines, for each $i=1,2, \\ldots, n+1$, the number $m_{i}$ of consecutive answers she has given by that point that are inconsistent with $i$. To decide on her next answer, she then uses the quantity\n\n$$\n\\phi=\\sum_{i=1}^{n+1} \\lambda^{m_{i}}\n$$\n\nNo matter what Ben\'s next question is, Amy chooses the answer which minimizes $\\phi$.\n\nWe claim that with this strategy $\\phi$ will always stay less than $\\lambda^{k+1}$. Consequently no exponent $m_{i}$ in $\\phi$ will ever exceed $k$, hence Amy will never give more than $k$ consecutive answers inconsistent with some $i$. In particular this applies to the target number $x$, so she will never lie more than $k$ times in a row. Thus, given the claim, Amy\'s strategy is legal. Since the strategy does not depend on $x$ in any way, Ben can make no deductions about $x$, and therefore he cannot guarantee a win.\n\nIt remains to show that $\\phi<\\lambda^{k+1}$ at all times. Initially each $m_{i}$ is 0 , so this condition holds in the beginning due to $1<\\lambda<2$ and $n=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1$. Suppose that $\\phi<\\lambda^{k+1}$ at some point, and Ben has just asked if $x \\in S$ for some set $S$. According as Amy answers yes or no, the new value of $\\phi$ becomes\n\n$$\n\\phi_{1}=\\sum_{i \\in S} 1+\\sum_{i \\notin S} \\lambda^{m_{i}+1} \\quad \\text { or } \\quad \\phi_{2}=\\sum_{i \\in S} \\lambda^{m_{i}+1}+\\sum_{i \\notin S} 1\n$$\n\n\n\nSince Amy chooses the option minimizing $\\phi$, the new $\\phi$ will equal $\\min \\left(\\phi_{1}, \\phi_{2}\\right)$. Now we have\n\n$$\n\\min \\left(\\phi_{1}, \\phi_{2}\\right) \\leq \\frac{1}{2}\\left(\\phi_{1}+\\phi_{2}\\right)=\\frac{1}{2}\\left(\\sum_{i \\in S}\\left(1+\\lambda^{m_{i}+1}\\right)+\\sum_{i \\notin S}\\left(\\lambda^{m_{i}+1}+1\\right)\\right)=\\frac{1}{2}(\\lambda \\phi+n+1)\n$$\n\nBecause $\\phi<\\lambda^{k+1}$, the assumptions $\\lambda<2$ and $n=\\left\\lfloor(2-\\lambda) \\lambda^{k+1}\\right\\rfloor-1$ lead to\n\n$$\n\\min \\left(\\phi_{1}, \\phi_{2}\\right)<\\frac{1}{2}\\left(\\lambda^{k+2}+(2-\\lambda) \\lambda^{k+1}\\right)=\\lambda^{k+1}\n$$\n\nThe claim follows, which completes the solution.']",,True,,, 2157,Combinatorics,,"There are given $2^{500}$ points on a circle labeled $1,2, \ldots, 2^{500}$ in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal.","['The proof is based on the following general fact.\n\nLemma. In a graph $G$ each vertex $v$ has degree $d_{v}$. Then $G$ contains an independent set $S$ of vertices such that $|S| \\geq f(G)$ where\n\n$$\nf(G)=\\sum_{v \\in G} \\frac{1}{d_{v}+1}\n$$\n\nProof. Induction on $n=|G|$. The base $n=1$ is clear. For the inductive step choose a vertex $v_{0}$ in $G$ of minimum degree $d$. Delete $v_{0}$ and all of its neighbors $v_{1}, \\ldots, v_{d}$ and also all edges with endpoints $v_{0}, v_{1}, \\ldots, v_{d}$. This gives a new graph $G^{\\prime}$. By the inductive assumption $G^{\\prime}$ contains an independent set $S^{\\prime}$ of vertices such that $\\left|S^{\\prime}\\right| \\geq f\\left(G^{\\prime}\\right)$. Since no vertex in $S^{\\prime}$ is a neighbor of $v_{0}$ in $G$, the set $S=S^{\\prime} \\cup\\left\\{v_{0}\\right\\}$ is independent in $G$.\n\nLet $d_{v}^{\\prime}$ be the degree of a vertex $v$ in $G^{\\prime}$. Clearly $d_{v}^{\\prime} \\leq d_{v}$ for every such vertex $v$, and also $d_{v_{i}} \\geq d$ for all $i=0,1, \\ldots, d$ by the minimal choice of $v_{0}$. Therefore\n\n$$\nf\\left(G^{\\prime}\\right)=\\sum_{v \\in G^{\\prime}} \\frac{1}{d_{v}^{\\prime}+1} \\geq \\sum_{v \\in G^{\\prime}} \\frac{1}{d_{v}+1}=f(G)-\\sum_{i=0}^{d} \\frac{1}{d_{v_{i}}+1} \\geq f(G)-\\frac{d+1}{d+1}=f(G)-1 .\n$$\n\nHence $|S|=\\left|S^{\\prime}\\right|+1 \\geq f\\left(G^{\\prime}\\right)+1 \\geq f(G)$, and the induction is complete.\n\nWe pass on to our problem. For clarity denote $n=2^{499}$ and draw all chords determined by the given $2 n$ points. Color each chord with one of the colors $3,4, \\ldots, 4 n-1$ according to the sum of the numbers at its endpoints. Chords with a common endpoint have different colors. For each color $c$ consider the following graph $G_{c}$. Its vertices are the chords of color $c$, and two chords are neighbors in $G_{c}$ if they intersect. Let $f\\left(G_{c}\\right)$ have the same meaning as in the lemma for all graphs $G_{c}$.\n\nEvery chord $\\ell$ divides the circle into two arcs, and one of them contains $m(\\ell) \\leq n-1$ given points. (In particular $m(\\ell)=0$ if $\\ell$ joins two consecutive points.) For each $i=0,1, \\ldots, n-2$ there are $2 n$ chords $\\ell$ with $m(\\ell)=i$. Such a chord has degree at most $i$ in the respective graph. Indeed let $A_{1}, \\ldots, A_{i}$ be all points on either arc determined by a chord $\\ell$ with $m(\\ell)=i$ and color $c$. Every $A_{j}$ is an endpoint of at most 1 chord colored $c, j=1, \\ldots, i$. Hence at most $i$ chords of color $c$ intersect $\\ell$.\n\nIt follows that for each $i=0,1, \\ldots, n-2$ the $2 n$ chords $\\ell$ with $m(\\ell)=i$ contribute at least $\\frac{2 n}{i+1}$ to the sum $\\sum_{c} f\\left(G_{c}\\right)$. Summation over $i=0,1, \\ldots, n-2$ gives\n\n$$\n\\sum_{c} f\\left(G_{c}\\right) \\geq 2 n \\sum_{i=1}^{n-1} \\frac{1}{i}\n$$\n\nBecause there are $4 n-3$ colors in all, averaging yields a color $c$ such that\n\n$$\nf\\left(G_{c}\\right) \\geq \\frac{2 n}{4 n-3} \\sum_{i=1}^{n-1} \\frac{1}{i}>\\frac{1}{2} \\sum_{i=1}^{n-1} \\frac{1}{i}\n$$\n\nBy the lemma there are at least $\\frac{1}{2} \\sum_{i=1}^{n-1} \\frac{1}{i}$ pairwise disjoint chords of color $c$, i. e. with the same sum $c$ of the pairs of numbers at their endpoints. It remains to show that $\\frac{1}{2} \\sum_{i=1}^{n-1} \\frac{1}{i} \\geq 100$ for $n=2^{499}$. Indeed we have\n\n$$\n\\sum_{i=1}^{n-1} \\frac{1}{i}>\\sum_{i=1}^{2^{400}} \\frac{1}{i}=1+\\sum_{k=1}^{400} \\sum_{i=2^{k-1+1}}^{2^{k}} \\frac{1}{i}>1+\\sum_{k=1}^{400} \\frac{2^{k-1}}{2^{k}}=201>200\n$$\n\nThis completes the solution.']",,True,,, 2158,Geometry,,"In the triangle $A B C$ the point $J$ is the center of the excircle opposite to $A$. This excircle is tangent to the side $B C$ at $M$, and to the lines $A B$ and $A C$ at $K$ and $L$ respectively. The lines $L M$ and $B J$ meet at $F$, and the lines $K M$ and $C J$ meet at $G$. Let $S$ be the point of intersection of the lines $A F$ and $B C$, and let $T$ be the point of intersection of the lines $A G$ and $B C$. Prove that $M$ is the midpoint of $S T$.","['Let $\\alpha=\\angle C A B, \\beta=\\angle A B C$ and $\\gamma=\\angle B C A$. The line $A J$ is the bisector of $\\angle C A B$, so $\\angle J A K=\\angle J A L=\\frac{\\alpha}{2}$. By $\\angle A K J=\\angle A L J=90^{\\circ}$ the points $K$ and $L$ lie on the circle $\\omega$ with diameter $A J$.\n\nThe triangle $K B M$ is isosceles as $B K$ and $B M$ are tangents to the excircle. Since $B J$ is the bisector of $\\angle K B M$, we have $\\angle M B J=90^{\\circ}-\\frac{\\beta}{2}$ and $\\angle B M K=\\frac{\\beta}{2}$. Likewise $\\angle M C J=90^{\\circ}-\\frac{\\gamma}{2}$ and $\\angle C M L=\\frac{\\gamma}{2}$. Also $\\angle B M F=\\angle C M L$, therefore\n\n$$\n\\angle L F J=\\angle M B J-\\angle B M F=\\left(90^{\\circ}-\\frac{\\beta}{2}\\right)-\\frac{\\gamma}{2}=\\frac{\\alpha}{2}=\\angle L A J\n$$\n\nHence $F$ lies on the circle $\\omega$. (By the angle computation, $F$ and $A$ are on the same side of $B C$.) Analogously, $G$ also lies on $\\omega$. Since $A J$ is a diameter of $\\omega$, we obtain $\\angle A F J=\\angle A G J=90^{\\circ}$.\n\n\n\nThe lines $A B$ and $B C$ are symmetric with respect to the external bisector $B F$. Because $A F \\perp B F$ and $K M \\perp B F$, the segments $S M$ and $A K$ are symmetric with respect to $B F$, hence $S M=A K$. By symmetry $T M=A L$. Since $A K$ and $A L$ are equal as tangents to the excircle, it follows that $S M=T M$, and the proof is complete.']",,True,,, 2159,Geometry,,"Let $A B C D$ be a cyclic quadrilateral whose diagonals $A C$ and $B D$ meet at $E$. The extensions of the sides $A D$ and $B C$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $E C G D$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $A D$. Prove that $D, H, F, G$ are concyclic.","['We show first that the triangles $F D G$ and $F B E$ are similar. Since $A B C D$ is cyclic, the triangles $E A B$ and $E D C$ are similar, as well as $F A B$ and $F C D$. The parallelogram $E C G D$ yields $G D=E C$ and $\\angle C D G=\\angle D C E$; also $\\angle D C E=\\angle D C A=\\angle D B A$ by inscribed angles. Therefore\n\n$$\n\\begin{gathered}\n\\angle F D G=\\angle F D C+\\angle C D G=\\angle F B A+\\angle A B D=\\angle F B E, \\\\\n\\frac{G D}{E B}=\\frac{C E}{E B}=\\frac{C D}{A B}=\\frac{F D}{F B}\n\\end{gathered}\n$$\n\nIt follows that $F D G$ and $F B E$ are similar, and so $\\angle F G D=\\angle F E B$.\n\n\n\nSince $H$ is the reflection of $E$ with respect to $F D$, we conclude that\n\n$$\n\\angle F H D=\\angle F E D=180^{\\circ}-\\angle F E B=180^{\\circ}-\\angle F G D .\n$$\n\nThis proves that $D, H, F, G$ are concyclic.']",,True,,, 2160,Geometry,,"In an acute triangle $A B C$ the points $D, E$ and $F$ are the feet of the altitudes through $A$, $B$ and $C$ respectively. The incenters of the triangles $A E F$ and $B D F$ are $I_{1}$ and $I_{2}$ respectively; the circumcenters of the triangles $A C I_{1}$ and $B C I_{2}$ are $O_{1}$ and $O_{2}$ respectively. Prove that $I_{1} I_{2}$ and $O_{1} O_{2}$ are parallel.","['Let $\\angle C A B=\\alpha, \\angle A B C=\\beta, \\angle B C A=\\gamma$. We start by showing that $A, B, I_{1}$ and $I_{2}$ are concyclic. Since $A I_{1}$ and $B I_{2}$ bisect $\\angle C A B$ and $\\angle A B C$, their extensions beyond $I_{1}$ and $I_{2}$ meet at the incenter $I$ of the triangle. The points $E$ and $F$ are on the circle with diameter $B C$, so $\\angle A E F=\\angle A B C$ and $\\angle A F E=\\angle A C B$. Hence the triangles $A E F$ and $A B C$ are similar with ratio of similitude $\\frac{A E}{A B}=\\cos \\alpha$. Because $I_{1}$ and $I$ are their incenters, we obtain $I_{1} A=I A \\cos \\alpha$ and $I I_{1}=I A-I_{1} A=2 I A \\sin ^{2} \\frac{\\alpha}{2}$. By symmetry $I I_{2}=2 I B \\sin ^{2} \\frac{\\beta}{2}$. The law of sines in the triangle $A B I$ gives $I A \\sin \\frac{\\alpha}{2}=I B \\sin \\frac{\\beta}{2}$. Hence\n\n$$\nI I_{1} \\cdot I A=2\\left(I A \\sin \\frac{\\alpha}{2}\\right)^{2}=2\\left(I B \\sin \\frac{\\beta}{2}\\right)^{2}=I I_{2} \\cdot I B .\n$$\n\nTherefore $A, B, I_{1}$ and $I_{2}$ are concyclic, as claimed.\n\n\n\nIn addition $I I_{1} \\cdot I A=I I_{2} \\cdot I B$ implies that $I$ has the same power with respect to the circles $\\left(A C I_{1}\\right),\\left(B C I_{2}\\right)$ and $\\left(A B I_{1} I_{2}\\right)$. Then $C I$ is the radical axis of $\\left(A C I_{1}\\right)$ and $\\left(B C I_{2}\\right)$; in particular $C I$ is perpendicular to the line of centers $O_{1} O_{2}$.\n\nNow it suffices to prove that $C I \\perp I_{1} I_{2}$. Let $C I$ meet $I_{1} I_{2}$ at $Q$, then it is enough to check that $\\angle I I_{1} Q+\\angle I_{1} I Q=90^{\\circ}$. Since $\\angle I_{1} I Q$ is external for the triangle $A C I$, we have\n\n$$\n\\angle I I_{1} Q+\\angle I_{1} I Q=\\angle I I_{1} Q+(\\angle A C I+\\angle C A I)=\\angle I I_{1} I_{2}+\\angle A C I+\\angle C A I .\n$$\n\nIt remains to note that $\\angle I I_{1} I_{2}=\\frac{\\beta}{2}$ from the cyclic quadrilateral $A B I_{1} I_{2}$, and $\\angle A C I=\\frac{\\gamma}{2}$, $\\angle C A I=\\frac{\\alpha}{2}$. Therefore $\\angle I I_{1} Q+\\angle I_{1} I Q=\\frac{\\alpha}{2}+\\frac{\\beta}{2}+\\frac{\\gamma}{2}=90^{\\circ}$, completing the proof.']",,True,,, 2161,Geometry,,Let $A B C$ be a triangle with $A B \neq A C$ and circumcenter $O$. The bisector of $\angle B A C$ intersects $B C$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $B C$. The lines through $D$ and $E$ perpendicular to $B C$ intersect the lines $A O$ and $A D$ at $X$ and $Y$ respectively. Prove that the quadrilateral $B X C Y$ is cyclic.,"['The bisector of $\\angle B A C$ and the perpendicular bisector of $B C$ meet at $P$, the midpoint of the minor arc $\\widehat{B C}$ (they are different lines as $A B \\neq A C$ ). In particular $O P$ is perpendicular to $B C$ and intersects it at $M$, the midpoint of $B C$.\n\nDenote by $Y^{\\prime}$ the reflexion of $Y$ with respect to $O P$. Since $\\angle B Y C=\\angle B Y^{\\prime} C$, it suffices to prove that $B X C Y^{\\prime}$ is cyclic.\n\n\n\nWe have\n\n$$\n\\angle X A P=\\angle O P A=\\angle E Y P .\n$$\n\nThe first equality holds because $O A=O P$, and the second one because $E Y$ and $O P$ are both perpendicular to $B C$ and hence parallel. But $\\left\\{Y, Y^{\\prime}\\right\\}$ and $\\{E, D\\}$ are pairs of symmetric points with respect to $O P$, it follows that $\\angle E Y P=\\angle D Y^{\\prime} P$ and hence\n\n$$\n\\angle X A P=\\angle D Y^{\\prime} P=\\angle X Y^{\\prime} P .\n$$\n\nThe last equation implies that $X A Y^{\\prime} P$ is cyclic. By the powers of $D$ with respect to the circles $\\left(X A Y^{\\prime} P\\right)$ and $(A B P C)$ we obtain\n\n$$\nX D \\cdot D Y^{\\prime}=A D \\cdot D P=B D \\cdot D C\n$$\n\nIt follows that $B X C Y^{\\prime}$ is cyclic, as desired.']",,True,,, 2162,Geometry,,"Let $A B C$ be a triangle with $\angle B C A=90^{\circ}$, and let $C_{0}$ be the foot of the altitude from $C$. Choose a point $X$ in the interior of the segment $C C_{0}$, and let $K, L$ be the points on the segments $A X, B X$ for which $B K=B C$ and $A L=A C$ respectively. Denote by $M$ the intersection of $A L$ and $B K$. Show that $M K=M L$.","['Let $C^{\\prime}$ be the reflection of $C$ in the line $A B$, and let $\\omega_{1}$ and $\\omega_{2}$ be the circles with centers $A$ and $B$, passing through $L$ and $K$ respectively. Since $A C^{\\prime}=A C=A L$ and $B C^{\\prime}=B C=B K$, both $\\omega_{1}$ and $\\omega_{2}$ pass through $C$ and $C^{\\prime}$. By $\\angle B C A=90^{\\circ}, A C$ is tangent to $\\omega_{2}$ at $C$, and $B C$ is tangent to $\\omega_{1}$ at $C$. Let $K_{1} \\neq K$ be the second intersection of $A X$ and $\\omega_{2}$, and let $L_{1} \\neq L$ be the second intersection of $B X$ and $\\omega_{1}$.\n\n\n\nBy the powers of $X$ with respect to $\\omega_{2}$ and $\\omega_{1}$,\n\n$$\nX K \\cdot X K_{1}=X C \\cdot X C^{\\prime}=X L \\cdot X L_{1}\n$$\n\nso the points $K_{1}, L, K, L_{1}$ lie on a circle $\\omega_{3}$.\n\nThe power of $A$ with respect to $\\omega_{2}$ gives\n\n$$\nA L^{2}=A C^{2}=A K \\cdot A K_{1},\n$$\n\nindicating that $A L$ is tangent to $\\omega_{3}$ at $L$. Analogously, $B K$ is tangent to $\\omega_{3}$ at $K$. Hence $M K$ and $M L$ are the two tangents from $M$ to $\\omega_{3}$ and therefore $M K=M L$.']",,True,,, 2163,Geometry,,"Let $A B C$ be a triangle with circumcenter $O$ and incenter $I$. The points $D, E$ and $F$ on the sides $B C, C A$ and $A B$ respectively are such that $B D+B F=C A$ and $C D+C E=A B$. The circumcircles of the triangles $B F D$ and $C D E$ intersect at $P \neq D$. Prove that $O P=O I$.","[""By MiqueL's theorem the circles $(A E F)=\\omega_{A},(B F D)=\\omega_{B}$ and $(C D E)=\\omega_{C}$ have a common point, for arbitrary points $D, E$ and $F$ on $B C, C A$ and $A B$. So $\\omega_{A}$ passes through the common point $P \\neq D$ of $\\omega_{B}$ and $\\omega_{C}$.\n\nLet $\\omega_{A}, \\omega_{B}$ and $\\omega_{C}$ meet the bisectors $A I, B I$ and $C I$ at $A \\neq A^{\\prime}, B \\neq B^{\\prime}$ and $C \\neq C^{\\prime}$ respectively. The key observation is that $A^{\\prime}, B^{\\prime}$ and $C^{\\prime}$ do not depend on the particular choice of $D, E$ and $F$, provided that $B D+B F=C A, C D+C E=A B$ and $A E+A F=B C$ hold true (the last equality follows from the other two). For a proof we need the following fact.\n\nLemma. Given is an angle with vertex $A$ and measure $\\alpha$. A circle $\\omega$ through $A$ intersects the angle bisector at $L$ and sides of the angle at $X$ and $Y$. Then $A X+A Y=2 A L \\cos \\frac{\\alpha}{2}$.\n\nProof. Note that $L$ is the midpoint of arc $\\widehat{X L Y}$ in $\\omega$ and set $X L=Y L=u, X Y=v$. By PtolemY's theorem $A X \\cdot Y L+A Y \\cdot X L=A L \\cdot X Y$, which rewrites as $(A X+A Y) u=A L \\cdot v$. Since $\\angle L X Y=\\frac{\\alpha}{2}$ and $\\angle X L Y=180^{\\circ}-\\alpha$, we have $v=2 \\cos \\frac{\\alpha}{2} u$ by the law of sines, and the claim follows.\n\n\n\nApply the lemma to $\\angle B A C=\\alpha$ and the circle $\\omega=\\omega_{A}$, which intersects $A I$ at $A^{\\prime}$. This gives $2 A A^{\\prime} \\cos \\frac{\\alpha}{2}=A E+A F=B C$; by symmetry analogous relations hold for $B B^{\\prime}$ and $C C^{\\prime}$. It follows that $A^{\\prime}, B^{\\prime}$ and $C^{\\prime}$ are independent of the choice of $D, E$ and $F$, as stated.\n\nWe use the lemma two more times with $\\angle B A C=\\alpha$. Let $\\omega$ be the circle with diameter $A I$. Then $X$ and $Y$ are the tangency points of the incircle of $A B C$ with $A B$ and $A C$, and hence $A X=A Y=\\frac{1}{2}(A B+A C-B C)$. So the lemma yields $2 A I \\cos \\frac{\\alpha}{2}=A B+A C-B C$. Next, if $\\omega$ is the circumcircle of $A B C$ and $A I$ intersects $\\omega$ at $M \\neq A$ then $\\{X, Y\\}=\\{B, C\\}$, and so $2 A M \\cos \\frac{\\alpha}{2}=A B+A C$ by the lemma. To summarize,\n\n$$\n2 A A^{\\prime} \\cos \\frac{\\alpha}{2}=B C, \\quad 2 A I \\cos \\frac{\\alpha}{2}=A B+A C-B C, \\quad 2 A M \\cos \\frac{\\alpha}{2}=A B+A C\n\\tag{*}\n$$\n\nThese equalities imply $A A^{\\prime}+A I=A M$, hence the segments $A M$ and $I A^{\\prime}$ have a common midpoint. It follows that $I$ and $A^{\\prime}$ are equidistant from the circumcenter $O$. By symmetry $O I=O A^{\\prime}=O B^{\\prime}=O C^{\\prime}$, so $I, A^{\\prime}, B^{\\prime}, C^{\\prime}$ are on a circle centered at $O$.\n\nTo prove $O P=O I$, now it suffices to show that $I, A^{\\prime}, B^{\\prime}, C^{\\prime}$ and $P$ are concyclic. Clearly one can assume $P \\neq I, A^{\\prime}, B^{\\prime}, C^{\\prime}$.\n\nWe use oriented angles to avoid heavy case distinction. The oriented angle between the lines $l$ and $m$ is denoted by $\\angle(l, m)$. We have $\\angle(l, m)=-\\angle(m, l)$ and $\\angle(l, m)+\\angle(m, n)=\\angle(l, n)$ for arbitrary lines $l, m$ and $n$. Four distinct non-collinear points $U, V, X, Y$ are concyclic if and only if $\\angle(U X, V X)=\\angle(U Y, V Y)$.\n\n\n\n\n\nSuppose for the moment that $A^{\\prime}, B^{\\prime}, P, I$ are distinct and noncollinear; then it is enough to check the equality $\\angle\\left(A^{\\prime} P, B^{\\prime} P\\right)=\\angle\\left(A^{\\prime} I, B^{\\prime} I\\right)$. Because $A, F, P, A^{\\prime}$ are on the circle $\\omega_{A}$, we have $\\angle\\left(A^{\\prime} P, F P\\right)=\\angle\\left(A^{\\prime} A, F A\\right)=\\angle\\left(A^{\\prime} I, A B\\right)$. Likewise $\\angle\\left(B^{\\prime} P, F P\\right)=\\angle\\left(B^{\\prime} I, A B\\right)$. Therefore\n\n$$\n\\angle\\left(A^{\\prime} P, B^{\\prime} P\\right)=\\angle\\left(A^{\\prime} P, F P\\right)+\\angle\\left(F P, B^{\\prime} P\\right)=\\angle\\left(A^{\\prime} I, A B\\right)-\\angle\\left(B^{\\prime} I, A B\\right)=\\angle\\left(A^{\\prime} I, B^{\\prime} I\\right) .\n$$\n\nHere we assumed that $P \\neq F$. If $P=F$ then $P \\neq D, E$ and the conclusion follows similarly (use $\\angle\\left(A^{\\prime} F, B^{\\prime} F\\right)=\\angle\\left(A^{\\prime} F, E F\\right)+\\angle(E F, D F)+\\angle\\left(D F, B^{\\prime} F\\right)$ and inscribed angles in $\\left.\\omega_{A}, \\omega_{B}, \\omega_{C}\\right)$.\n\nThere is no loss of generality in assuming $A^{\\prime}, B^{\\prime}, P, I$ distinct and noncollinear. If $A B C$ is an equilateral triangle then the equalities (*) imply that $A^{\\prime}, B^{\\prime}, C^{\\prime}, I, O$ and $P$ coincide, so $O P=O I$. Otherwise at most one of $A^{\\prime}, B^{\\prime}, C^{\\prime}$ coincides with $I$. If say $C^{\\prime}=I$ then $O I \\perp C I$ by the previous reasoning. It follows that $A^{\\prime}, B^{\\prime} \\neq I$ and hence $A^{\\prime} \\neq B^{\\prime}$. Finally $A^{\\prime}, B^{\\prime}$ and $I$ are noncollinear because $I, A^{\\prime}, B^{\\prime}, C^{\\prime}$ are concyclic.""]",,True,,, 2164,Geometry,,"Let $A B C$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $B C, C A, A B$ intersect $\ell$ at the points $X, Y, Z$ different from $P$. Prove that the circumcircles of the triangles $A X P, B Y P$ and $C Z P$ have a common point different from $P$ or are mutually tangent at $P$.","[""Let $\\omega_{A}, \\omega_{B}, \\omega_{C}$ and $\\omega$ be the circumcircles of triangles $A X P, B Y P, C Z P$ and $A B C$ respectively. The strategy of the proof is to construct a point $Q$ with the same power with respect to the four circles. Then each of $P$ and $Q$ has the same power with respect to $\\omega_{A}, \\omega_{B}, \\omega_{C}$ and hence the three circles are coaxial. In other words they have another common point $P^{\\prime}$ or the three of them are tangent at $P$.\n\nWe first give a description of the point $Q$. Let $A^{\\prime} \\neq A$ be the second intersection of $\\omega$ and $\\omega_{A}$; define $B^{\\prime}$ and $C^{\\prime}$ analogously. We claim that $A A^{\\prime}, B B^{\\prime}$ and $C C^{\\prime}$ have a common point. Once this claim is established, the point just constructed will be on the radical axes of the three pairs of circles $\\left\\{\\omega, \\omega_{A}\\right\\},\\left\\{\\omega, \\omega_{B}\\right\\},\\left\\{\\omega, \\omega_{C}\\right\\}$. Hence it will have the same power with respect to $\\omega, \\omega_{A}, \\omega_{B}, \\omega_{C}$.\n\n\n\nWe proceed to prove that $A A^{\\prime}, B B^{\\prime}$ and $C C^{\\prime}$ intersect at one point. Let $r$ be the circumradius of triangle $A B C$. Define the points $X^{\\prime}, Y^{\\prime}, Z^{\\prime}$ as the intersections of $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ with $\\ell$. Observe that $X^{\\prime}, Y^{\\prime}, Z^{\\prime}$ do exist. If $A A^{\\prime}$ is parallel to $\\ell$ then $\\omega_{A}$ is tangent to $\\ell$; hence $X=P$ which is a contradiction. Similarly, $B B^{\\prime}$ and $C C^{\\prime}$ are not parallel to $\\ell$.\n\nFrom the powers of the point $X^{\\prime}$ with respect to the circles $\\omega_{A}$ and $\\omega$ we get\n\n$$\nX^{\\prime} P \\cdot\\left(X^{\\prime} P+P X\\right)=X^{\\prime} P \\cdot X^{\\prime} X=X^{\\prime} A^{\\prime} \\cdot X^{\\prime} A=X^{\\prime} O^{2}-r^{2}\n$$\n\nhence\n\n$$\nX^{\\prime} P \\cdot P X=X^{\\prime} O^{2}-r^{2}-X^{\\prime} P^{2}=O P^{2}-r^{2}\n$$\n\nWe argue analogously for the points $Y^{\\prime}$ and $Z^{\\prime}$, obtaining\n\n$$\nX^{\\prime} P \\cdot P X=Y^{\\prime} P \\cdot P Y=Z^{\\prime} P \\cdot P Z=O P^{2}-r^{2}=k^{2} .\n\\tag{1}\n$$\n\nIn these computations all segments are regarded as directed segments. We keep the same convention for the sequel.\n\nWe prove that the lines $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ intersect at one point by CEvA's theorem. To avoid distracting remarks we interpret everything projectively, i. e. whenever two lines are parallel they meet at a point on the line at infinity.\n\n\n\nLet $U, V, W$ be the intersections of $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ with $B C, C A, A B$ respectively. The idea is that although it is difficult to calculate the ratio $\\frac{B U}{C U}$, it is easier to deal with the cross-ratio $\\frac{B U}{C U} / \\frac{B X}{C X}$ because we can send it to the line $\\ell$. With this in mind we apply MENELAUs' theorem to the triangle $A B C$ and obtain $\\frac{B X}{C X} \\cdot \\frac{C Y}{A Y} \\cdot \\frac{A Z}{B Z}=1$. Hence Ceva's ratio can be expressed as\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=\\frac{B U}{C U} / \\frac{B X}{C X} \\cdot \\frac{C V}{A V} / \\frac{C Y}{A Y} \\cdot \\frac{A W}{B W} / \\frac{A Z}{B Z}\n$$\n\n\n\nProject the line $B C$ to $\\ell$ from $A$. The cross-ratio between $B C$ and $U X$ equals the cross-ratio between $Z Y$ and $X^{\\prime} X$. Repeating the same argument with the lines $C A$ and $A B$ gives\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=\\frac{Z X^{\\prime}}{Y X^{\\prime}} / \\frac{Z X}{Y X} \\cdot \\frac{X Y^{\\prime}}{Z Y^{\\prime}} / \\frac{X Y}{Z Y} \\cdot \\frac{Y Z^{\\prime}}{X Z^{\\prime}} / \\frac{Y Z}{X Z}\n$$\n\nand hence\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=(-1) \\cdot \\frac{Z X^{\\prime}}{Y X^{\\prime}} \\cdot \\frac{X Y^{\\prime}}{Z Y^{\\prime}} \\cdot \\frac{Y Z^{\\prime}}{X Z^{\\prime}}\n$$\n\nThe equations (1) reduce the problem to a straightforward computation on the line $\\ell$. For instance, the transformation $t \\mapsto-k^{2} / t$ preserves cross-ratio and interchanges the points $X, Y, Z$ with the points $X^{\\prime}, Y^{\\prime}, Z^{\\prime}$. Then\n\n$$\n\\frac{B U}{C U} \\cdot \\frac{C V}{A V} \\cdot \\frac{A W}{B W}=(-1) \\cdot \\frac{Z X^{\\prime}}{Y X^{\\prime}} / \\frac{Z Z^{\\prime}}{Y Z^{\\prime}} \\cdot \\frac{X Y^{\\prime}}{Z Y^{\\prime}} / \\frac{X Z^{\\prime}}{Z Z^{\\prime}}=-1\n$$\n\nWe proved that CEva's ratio equals -1 , so $A A^{\\prime}, B B^{\\prime}, C C^{\\prime}$ intersect at one point $Q$."", 'First we prove that there is an inversion in space that takes $\\ell$ and $\\omega$ to parallel circles on a sphere. Let $Q R$ be the diameter of $\\omega$ whose extension beyond $Q$ passes through $P$. Let $\\Pi$ be the plane carrying our objects. In space, choose a point $O$ such that the line $Q O$ is perpendicular to $\\Pi$ and $\\angle P O R=90^{\\circ}$, and apply an inversion with pole $O$ (the radius of the inversion does not matter). For any object $\\mathcal{T}$ denote by $\\mathcal{T}^{\\prime}$ the image of $\\mathcal{T}$ under this inversion.\n\nThe inversion takes the plane $\\Pi$ to a sphere $\\Pi^{\\prime}$. The lines in $\\Pi$ are taken to circles through $O$, and the circles in $\\Pi$ also are taken to circles on $\\Pi^{\\prime}$.\n\n\n\nSince the line $\\ell$ and the circle $\\omega$ are perpendicular to the plane $O P Q$, the circles $\\ell^{\\prime}$ and $\\omega^{\\prime}$ also are perpendicular to this plane. Hence, the planes of the circles $\\ell^{\\prime}$ and $\\omega^{\\prime}$ are parallel.\n\nNow consider the circles $A^{\\prime} X^{\\prime} P^{\\prime}, B^{\\prime} Y^{\\prime} P^{\\prime}$ and $C^{\\prime} Z^{\\prime} P^{\\prime}$. We want to prove that either they have a common point (on $\\Pi^{\\prime}$ ), different from $P^{\\prime}$, or they are tangent to each other.\n\n\n\nThe point $X^{\\prime}$ is the second intersection of the circles $B^{\\prime} C^{\\prime} O$ and $\\ell^{\\prime}$, other than $O$. Hence, the lines $O X^{\\prime}$ and $B^{\\prime} C^{\\prime}$ are coplanar. Moreover, they lie in the parallel planes of $\\ell^{\\prime}$ and $\\omega^{\\prime}$. Therefore, $O X^{\\prime}$ and $B^{\\prime} C^{\\prime}$ are parallel. Analogously, $O Y^{\\prime}$ and $O Z^{\\prime}$ are parallel to $A^{\\prime} C^{\\prime}$ and $A^{\\prime} B^{\\prime}$.\n\nLet $A_{1}$ be the second intersection of the circles $A^{\\prime} X^{\\prime} P^{\\prime}$ and $\\omega^{\\prime}$, other than $A^{\\prime}$. The segments $A^{\\prime} A_{1}$ and $P^{\\prime} X^{\\prime}$ are coplanar, and therefore parallel. Now we know that $B^{\\prime} C^{\\prime}$ and $A^{\\prime} A_{1}$ are parallel to $O X^{\\prime}$ and $X^{\\prime} P^{\\prime}$ respectively, but these two segments are perpendicular because $O P^{\\prime}$ is a diameter in $\\ell^{\\prime}$. We found that $A^{\\prime} A_{1}$ and $B^{\\prime} C^{\\prime}$ are perpendicular, hence $A^{\\prime} A_{1}$ is the altitude in the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$, starting from $A$.\n\nAnalogously, let $B_{1}$ and $C_{1}$ be the second intersections of $\\omega^{\\prime}$ with the circles $B^{\\prime} P^{\\prime} Y^{\\prime}$ and $C^{\\prime} P^{\\prime} Z^{\\prime}$, other than $B^{\\prime}$ and $C^{\\prime}$ respectively. Then $B^{\\prime} B_{1}$ and $C^{\\prime} C_{1}$ are the other two altitudes in the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$.\n\n\n\nLet $H$ be the orthocenter of the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$. Let $W$ be the second intersection of the line $P^{\\prime} H$ with the sphere $\\Pi^{\\prime}$, other than $P^{\\prime}$. The point $W$ lies on the sphere $\\Pi^{\\prime}$, in the plane of the circle $A^{\\prime} P^{\\prime} X^{\\prime}$, so $W$ lies on the circle $A^{\\prime} P^{\\prime} X^{\\prime}$. Similarly, $W$ lies on the circles $B^{\\prime} P^{\\prime} Y^{\\prime}$ and $C^{\\prime} P^{\\prime} Z^{\\prime}$ as well; indeed $W$ is the second common point of the three circles.\n\nIf the line $P^{\\prime} H$ is tangent to the sphere then $W$ coincides with $P^{\\prime}$, and $P^{\\prime} H$ is the common tangent of the three circles.']",,True,,, 2165,Number Theory,,"Find all triples $(x, y, z)$ of positive integers such that $x \leq y \leq z$ and $$ x^{3}\left(y^{3}+z^{3}\right)=2012(x y z+2) \text {. } $$","[""First note that $x$ divides $2012 \\cdot 2=2^{3} \\cdot 503$. If $503 \\mid x$ then the right-hand side of the equation is divisible by $503^{3}$, and it follows that $503^{2} \\mid x y z+2$. This is false as $503 \\mid x$. Hence $x=2^{m}$ with $m \\in\\{0,1,2,3\\}$. If $m \\geq 2$ then $2^{6} \\mid 2012(x y z+2)$. However the highest powers of 2 dividing 2012 and $x y z+2=2^{m} y z+2$ are $2^{2}$ and $2^{1}$ respectively. So $x=1$ or $x=2$, yielding the two equations\n\n$$\ny^{3}+z^{3}=2012(y z+2), \\quad \\text { and } \\quad y^{3}+z^{3}=503(y z+1)\n$$\n\nIn both cases the prime $503=3 \\cdot 167+2$ divides $y^{3}+z^{3}$. We claim that $503 \\mid y+z$. This is clear if $503 \\mid y$, so let $503 \\nmid y$ and $503 \\nmid z$. Then $y^{502} \\equiv z^{502}(\\bmod 503)$ by FERMAT's little theorem. On the other hand $y^{3} \\equiv-z^{3}(\\bmod 503)$ implies $y^{3 \\cdot 167} \\equiv-z^{3 \\cdot 167}(\\bmod 503)$, i. e. $y^{501} \\equiv-z^{501}(\\bmod 503)$. It follows that $y \\equiv-z(\\bmod 503)$ as claimed.\n\nTherefore $y+z=503 k$ with $k \\geq 1$. In view of $y^{3}+z^{3}=(y+z)\\left((y-z)^{2}+y z\\right)$ the two equations take the form\n\n$$\nk(y-z)^{2}+(k-4) y z=8 \\tag{1}\n$$\n$$\nk(y-z)^{2}+(k-1) y z=1 \\tag{2}\n$$\n\nIn (1) we have $(k-4) y z \\leq 8$, which implies $k \\leq 4$. Indeed if $k>4$ then $1 \\leq(k-4) y z \\leq 8$, so that $y \\leq 8$ and $z \\leq 8$. This is impossible as $y+z=503 k \\geq 503$. Note next that $y^{3}+z^{3}$ is even in the first equation. Hence $y+z=503 k$ is even too, meaning that $k$ is even. Thus $k=2$ or $k=4$. Clearly (1) has no integer solutions for $k=4$. If $k=2$ then (1) takes the form $(y+z)^{2}-5 y z=4$. Since $y+z=503 k=503 \\cdot 2$, this leads to $5 y z=503^{2} \\cdot 2^{2}-4$. However $503^{2} \\cdot 2^{2}-4$ is not a multiple of 5 . Therefore (1) has no integer solutions.\n\nEquation (2) implies $0 \\leq(k-1) y z \\leq 1$, so that $k=1$ or $k=2$. Also $0 \\leq k(y-z)^{2} \\leq 1$, hence $k=2$ only if $y=z$. However then $y=z=1$, which is false in view of $y+z \\geq 503$. Therefore $k=1$ and (2) takes the form $(y-z)^{2}=1$, yielding $z-y=|y-z|=1$. Combined with $k=1$ and $y+z=503 k$, this leads to $y=251, z=252$.\n\nIn summary the triple $(2,251,252)$ is the only solution.""]","['$(2,251,252)$']",False,,Tuple, 2166,Number Theory,,"An integer $a$ is called friendly if the equation $\left(m^{2}+n\right)\left(n^{2}+m\right)=a(m-n)^{3}$ has a solution over the positive integers. Prove that there are at least 500 friendly integers in the set $\{1,2, \ldots, 2012\}$.","['Every $a$ of the form $a=4 k-3$ with $k \\geq 2$ is friendly. Indeed the numbers $m=2 k-1>0$ and $n=k-1>0$ satisfy the given equation with $a=4 k-3$ :\n\n$$\n\\left(m^{2}+n\\right)\\left(n^{2}+m\\right)=\\left((2 k-1)^{2}+(k-1)\\right)\\left((k-1)^{2}+(2 k-1)\\right)=(4 k-3) k^{3}=a(m-n)^{3} .\n$$\n\nHence $5,9, \\ldots, 2009$ are friendly and so $\\{1,2, \\ldots, 2012\\}$ contains at least 502 friendly numbers.']",,True,,, 2167,Number Theory,,"An integer $a$ is called friendly if the equation $\left(m^{2}+n\right)\left(n^{2}+m\right)=a(m-n)^{3}$ has a solution over the positive integers. Decide whether $a=2$ is friendly.","['We show that $a=2$ is not friendly. Consider the equation with $a=2$ and rewrite its left-hand side as a difference of squares:\n\n$$\n\\frac{1}{4}\\left(\\left(m^{2}+n+n^{2}+m\\right)^{2}-\\left(m^{2}+n-n^{2}-m\\right)^{2}\\right)=2(m-n)^{3}\n$$\n\nSince $m^{2}+n-n^{2}-m=(m-n)(m+n-1)$, we can further reformulate the equation as\n\n$$\n\\left(m^{2}+n+n^{2}+m\\right)^{2}=(m-n)^{2}\\left(8(m-n)+(m+n-1)^{2}\\right) .\n$$\n\nIt follows that $8(m-n)+(m+n-1)^{2}$ is a perfect square. Clearly $m>n$, hence there is an integer $s \\geq 1$ such that\n\n$$\n(m+n-1+2 s)^{2}=8(m-n)+(m+n-1)^{2} .\n$$\n\nSubtracting the squares gives $s(m+n-1+s)=2(m-n)$. Since $m+n-1+s>m-n$, we conclude that $s<2$. Therefore the only possibility is $s=1$ and $m=3 n$. However then the left-hand side of the given equation (with $a=2$ ) is greater than $m^{3}=27 n^{3}$, whereas its right-hand side equals $16 n^{3}$. The contradiction proves that $a=2$ is not friendly.']",,True,,, 2168,Number Theory,,"Let $x$ and $y$ be positive integers. If $x^{2^{n}}-1$ is divisible by $2^{n} y+1$ for every positive integer $n$, prove that $x=1$.","[""First we prove the following fact: For every positive integer $y$ there exist infinitely many primes $p \\equiv 3(\\bmod 4)$ such that $p$ divides some number of the form $2^{n} y+1$.\n\nClearly it is enough to consider the case $y$ odd. Let\n\n$$\n2 y+1=p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}}\n$$\n\nbe the prime factorization of $2 y+1$. Suppose on the contrary that there are finitely many primes $p_{r+1}, \\ldots, p_{r+s} \\equiv 3(\\bmod 4)$ that divide some number of the form $2^{n} y+1$ but do not divide $2 y+1$.\n\nWe want to find an $n$ such that $p_{i}^{e_{i}} \\| 2^{n} y+1$ for $1 \\leq i \\leq r$ and $p_{i} \\nmid 2^{n} y+1$ for $r+1 \\leq i \\leq r+s$. For this it suffices to take\n\n$$\nn=1+\\varphi\\left(p_{1}^{e_{1}+1} \\cdots p_{r}^{e_{r}+1} p_{r+1}^{1} \\cdots p_{r+s}^{1}\\right)\n$$\n\nbecause then\n\n$$\n2^{n} y+1 \\equiv 2 y+1 \\quad\\left(\\bmod p_{1}^{e_{1}+1} \\cdots p_{r}^{e_{r}+1} p_{r+1}^{1} \\cdots p_{r+s}^{1}\\right)\n$$\n\nThe last congruence means that $p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}$ divide exactly $2^{n} y+1$ and no prime $p_{r+1}, \\ldots, p_{r+s}$ divides $2^{n} y+1$. It follows that the prime factorization of $2^{n} y+1$ consists of the prime powers $p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}$ and powers of primes $\\equiv 1(\\bmod 4)$. Because $y$ is odd, we obtain\n\n$$\n2^{n} y+1 \\equiv p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}} \\equiv 2 y+1 \\equiv 3 \\quad(\\bmod 4)\n$$\n\nThis is a contradiction since $n>1$, and so $2^{n} y+1 \\equiv 1(\\bmod 4)$.\n\nNow we proceed to the problem. If $p$ is a prime divisor of $2^{n} y+1$ the problem statement implies that $x^{d} \\equiv 1(\\bmod p)$ for $d=2^{n}$. By FERMAT's little theorem the same congruence holds for $d=p-1$, so it must also hold for $d=\\left(2^{n}, p-1\\right)$. For $p \\equiv 3(\\bmod 4)$ we have $\\left(2^{n}, p-1\\right)=2$, therefore in this case $x^{2} \\equiv 1(\\bmod p)$.\n\nIn summary, we proved that every prime $p \\equiv 3(\\bmod 4)$ that divides some number of the form $2^{n} y+1$ also divides $x^{2}-1$. This is possible only if $x=1$, otherwise by the above $x^{2}-1$ would be a positive integer with infinitely many prime factors.""]",,True,,, 2169,Number Theory,,Prove that for every prime $p>100$ and every integer $r$ there exist two integers $a$ and $b$ such that $p$ divides $a^{2}+b^{5}-r$.,"['Throughout the solution, all congruence relations are meant modulo $p$.\n\nFix $p$, and let $\\mathcal{P}=\\{0,1, \\ldots, p-1\\}$ be the set of residue classes modulo $p$. For every $r \\in \\mathcal{P}$, let $S_{r}=\\left\\{(a, b) \\in \\mathcal{P} \\times \\mathcal{P}: a^{2}+b^{5} \\equiv r\\right\\}$, and let $s_{r}=\\left|S_{r}\\right|$. Our aim is to prove $s_{r}>0$ for all $r \\in \\mathcal{P}$.\n\nWe will use the well-known fact that for every residue class $r \\in \\mathcal{P}$ and every positive integer $k$, there are at most $k$ values $x \\in \\mathcal{P}$ such that $x^{k} \\equiv r$.\n\nLemma. Let $N$ be the number of quadruples $(a, b, c, d) \\in \\mathcal{P}^{4}$ for which $a^{2}+b^{5} \\equiv c^{2}+d^{5}$. Then\n\n$$\nN=\\sum_{r \\in \\mathcal{P}} s_{r}^{2}\\tag{a}\n$$\n\nand\n\n$$\nN \\leq p\\left(p^{2}+4 p-4\\right)\\tag{b}\n$$\n\nProof. (a) For each residue class $r$ there exist exactly $s_{r}$ pairs $(a, b)$ with $a^{2}+b^{5} \\equiv r$ and $s_{r}$ pairs $(c, d)$ with $c^{2}+d^{5} \\equiv r$. So there are $s_{r}^{2}$ quadruples with $a^{2}+b^{5} \\equiv c^{2}+d^{5} \\equiv r$. Taking the sum over all $r \\in \\mathcal{P}$, the statement follows.\n\n(b) Choose an arbitrary pair $(b, d) \\in \\mathcal{P}$ and look for the possible values of $a, c$.\n\n1. Suppose that $b^{5} \\equiv d^{5}$, and let $k$ be the number of such pairs $(b, d)$. The value $b$ can be chosen in $p$ different ways. For $b \\equiv 0$ only $d=0$ has this property; for the nonzero values of $b$ there are at most 5 possible values for $d$. So we have $k \\leq 1+5(p-1)=5 p-4$.\n\nThe values $a$ and $c$ must satisfy $a^{2} \\equiv c^{2}$, so $a \\equiv \\pm c$, and there are exactly $2 p-1$ such pairs $(a, c)$.\n\n2. Now suppose $b^{5} \\not \\equiv d^{5}$. In this case $a$ and $c$ must be distinct. By $(a-c)(a+c)=d^{5}-b^{5}$, the value of $a-c$ uniquely determines $a+c$ and thus $a$ and $c$ as well. Hence, there are $p-1$ suitable pairs $(a, c)$.\n\nThus, for each of the $k$ pairs $(b, d)$ with $b^{5} \\equiv d^{5}$ there are $2 p-1$ pairs $(a, c)$, and for each of the other $p^{2}-k$ pairs $(b, d)$ there are $p-1$ pairs $(a, c)$. Hence,\n\n$$\nN=k(2 p-1)+\\left(p^{2}-k\\right)(p-1)=p^{2}(p-1)+k p \\leq p^{2}(p-1)+(5 p-4) p=p\\left(p^{2}+4 p-4\\right)\n$$\n\nTo prove the statement of the problem, suppose that $S_{r}=\\emptyset$ for some $r \\in \\mathcal{P}$; obviously $r \\not \\equiv 0$. Let $T=\\left\\{x^{10}: x \\in \\mathcal{P} \\backslash\\{0\\}\\right\\}$ be the set of nonzero 10th powers modulo $p$. Since each residue class is the 10 th power of at most 10 elements in $\\mathcal{P}$, we have $|T| \\geq \\frac{p-1}{10} \\geq 4$ by $p>100$.\n\nFor every $t \\in T$, we have $S_{t r}=\\emptyset$. Indeed, if $(x, y) \\in S_{t r}$ and $t \\equiv z^{10}$ then\n\n$$\n\\left(z^{-5} x\\right)^{2}+\\left(z^{-2} y\\right)^{5} \\equiv t^{-1}\\left(x^{2}+y^{5}\\right) \\equiv r\n$$\n\nso $\\left(z^{-5} x, z^{-2} y\\right) \\in S_{r}$. So, there are at least $\\frac{p-1}{10} \\geq 4$ empty sets among $S_{1}, \\ldots, S_{p-1}$, and there are at most $p-4$ nonzero values among $s_{0}, s_{2}, \\ldots, s_{p-1}$. Then by the AM-QM inequality we obtain\n\n$$\nN=\\sum_{r \\in \\mathcal{P} \\backslash r T} s_{r}^{2} \\geq \\frac{1}{p-4}\\left(\\sum_{r \\in \\mathcal{P} \\backslash r T} s_{r}\\right)^{2}=\\frac{|\\mathcal{P} \\times \\mathcal{P}|^{2}}{p-4}=\\frac{p^{4}}{p-4}>p\\left(p^{2}+4 p-4\\right),\n$$\n\nwhich is impossible by the lemma.', 'If $5 \\nmid p-1$, then all modulo $p$ residue classes are complete fifth powers and the statement is trivial. So assume that $p=10 k+1$ where $k \\geq 10$. Let $g$ be a primitive root modulo $p$.\n\nWe will use the following facts:\n\n(F1) If some residue class $x$ is not quadratic then $x^{(p-1) / 2} \\equiv-1(\\bmod p)$.\n\n(F2) For every integer $d$, as a simple corollary of the summation formula for geometric progressions,\n\n$$\n\\sum_{i=0}^{2 k-1} g^{5 d i} \\equiv\\left\\{\\begin{array}{ll}\n2 k & \\text { if } 2 k \\mid d \\\\\n0 & \\text { if } 2 k \\nmid x d\n\\end{array} \\quad(\\bmod p)\\right.\n$$\n\nSuppose that, contrary to the statement, some modulo $p$ residue class $r$ cannot be expressed as $a^{2}+b^{5}$. Of course $r \\not \\equiv 0(\\bmod p)$. By $(\\mathrm{F} 1)$ we have $\\left(r-b^{5}\\right)^{(p-1) / 2}=\\left(r-b^{5}\\right)^{5 k} \\equiv-1(\\bmod p)$ for all residue classes $b$.\n\nFor $t=1,2 \\ldots, k-1$ consider the sums\n\n$$\nS(t)=\\sum_{i=0}^{2 k-1}\\left(r-g^{5 i}\\right)^{5 k} g^{5 t i}\n$$\n\nBy the indirect assumption and (F2),\n\n$$\nS(t)=\\sum_{i=0}^{2 k-1}\\left(r-\\left(g^{i}\\right)^{5}\\right)^{5 k} g^{5 t i} \\equiv \\sum_{i=0}^{2 k-1}(-1) g^{5 t i} \\equiv-\\sum_{i=0}^{2 k-1} g^{5 t i} \\equiv 0 \\quad(\\bmod p)\n$$\n\nbecause $2 k$ cannot divide $t$.\n\nOn the other hand, by the binomial theorem,\n\n$$\n\\begin{aligned}\nS(t) & =\\sum_{i=0}^{2 k-1}\\left(\\sum_{j=0}^{5 k}\\left(\\begin{array}{c}\n5 k \\\\\nj\n\\end{array}\\right) r^{5 k-j}\\left(-g^{5 i}\\right)^{j}\\right) g^{5 t i}=\\sum_{j=0}^{5 k}(-1)^{j}\\left(\\begin{array}{c}\n5 k \\\\\nj\n\\end{array}\\right) r^{5 k-j}\\left(\\sum_{i=0}^{2 k-1} g^{5(j+t) i}\\right) \\equiv \\\\\n& \\equiv \\sum_{j=0}^{5 k}(-1)^{j}\\left(\\begin{array}{c}\n5 k \\\\\nj\n\\end{array}\\right) r^{5 k-j}\\left\\{\\begin{array}{ll}\n2 k & \\text { if } 2 k \\mid j+t \\\\\n0 & \\text { if } 2 k \\nmid j+t\n\\end{array} \\quad(\\bmod p) .\\right.\n\\end{aligned}\n$$\n\nSince $1 \\leq j+t<6 k$, the number $2 k$ divides $j+t$ only for $j=2 k-t$ and $j=4 k-t$. Hence,\n\n$$\n\\begin{gathered}\n0 \\equiv S(t) \\equiv(-1)^{t}\\left(\\left(\\begin{array}{c}\n5 k \\\\\n2 k-t\n\\end{array}\\right) r^{3 k+t}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-t\n\\end{array}\\right) r^{k+t}\\right) \\cdot 2 k \\quad(\\bmod p) \\\\\n\\left(\\begin{array}{c}\n5 k \\\\\n2 k-t\n\\end{array}\\right) r^{2 k}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-t\n\\end{array}\\right) \\equiv 0 \\quad(\\bmod p) .\n\\end{gathered}\n$$\n\nTaking this for $t=1,2$ and eliminating $r$, we get\n\n$$\n\\begin{aligned}\n0 & \\equiv\\left(\\begin{array}{c}\n5 k \\\\\n2 k-2\n\\end{array}\\right)\\left(\\left(\\begin{array}{c}\n5 k \\\\\n2 k-1\n\\end{array}\\right) r^{2 k}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-1\n\\end{array}\\right)\\right)-\\left(\\begin{array}{c}\n5 k \\\\\n2 k-1\n\\end{array}\\right)\\left(\\left(\\begin{array}{c}\n5 k \\\\\n2 k-2\n\\end{array}\\right) r^{2 k}+\\left(\\begin{array}{c}\n5 k \\\\\n4 k-2\n\\end{array}\\right)\\right) \\\\\n& =\\left(\\begin{array}{c}\n5 k \\\\\n2 k-2\n\\end{array}\\right)\\left(\\begin{array}{c}\n5 k \\\\\n4 k-1\n\\end{array}\\right)-\\left(\\begin{array}{c}\n5 k \\\\\n2 k-1\n\\end{array}\\right)\\left(\\begin{array}{c}\n5 k \\\\\n4 k-2\n\\end{array}\\right) \\\\\n& =\\frac{(5 k) !^{2}}{(2 k-1) !(3 k+2) !(4 k-1) !(k+2) !}((2 k-1)(k+2)-(3 k+2)(4 k-1)) \\\\\n& =\\frac{-(5 k) !^{2} \\cdot 2 k(5 k+1)}{(2 k-1) !(3 k+2) !(4 k-1) !(k+2) !}(\\bmod p) .\n\\end{aligned}\n$$\n\nBut in the last expression none of the numbers is divisible by $p=10 k+1$, a contradiction.']",,True,,, 2170,Number Theory,,"According to Anna, the number 2021 is fantabulous. She states that if any element of the set $\{m, 2 m+1,3 m\}$ is fantabulous for a positive integer $m$, then they are all fantabulous. Is the number $2021^{2021}$ fantabulous?","['Consider the sequence of positive integers $m, 3 m, 6 m+1,12 m+3,4 m+1,2 m$. Since each number in the sequence is fantabulous if and only if the next one is, we deduce that $m$ is fantabulous if and only if $2 m$ is fantabulous.\n\nCombined with the fact that $m$ is fantabulous if and only if $2 m+1$ is fantabulous, this implies that\n\n$m>1$ is fantabulous if and only if $f(m)=\\left[\\frac{m}{2}\\right]$ is fantabulous. We can apply $f$ sufficiently many times to any positive integer $n$ to conclude that $n$ is fantabulous if and only if 1 is fantabulous. Therefore, the fact that 2021 is fantabulous implies that 1 is fantabulous, which in turn implies that $2021^{2021}$ is fantabulous.', 'Let $m>1$ be a fantabulous number. Note that at least one of the following four cases must hold.\n\n- Case 1. The number $m$ is odd;\n\nWe have $m=2 a+1$ for some positive integer $a$, so $a\n\nProof 1:\nLet $H$ be the orthocenter of the triangle $A B P$. Then\n\n$$\n\\angle B H P=180^{\\circ}-\\angle B A C=180^{\\circ}-\\angle B C P .\n$$\n\nSo $B H P C$ is cyclic. Then we get\n\n$$\nT K \\cdot T A=T C \\cdot T B=T P \\cdot T H .\n$$\n\nSo, $A H P K$ is also cyclic. But then\n\n$$\n\\angle A K P=180^{\\circ}-\\angle A H P=\\angle A B P .\n$$\n\n\n\n\nProof 2:\nConsider the symmetric points $B^{\\prime}$ and $C^{\\prime}$ of $B$ and $C$, respectively, with respect to the line $P T$. It is clear that\n\n$$\n\\mathrm{TC}^{\\prime} \\cdot T B^{\\prime}=T C \\cdot T B=T K \\cdot T A\n$$\n\nSo $B^{\\prime} C^{\\prime} K A$ is cyclic. Also, because of the symmetry we have\n\n$$\n\\angle P C^{\\prime} B^{\\prime}=\\angle P C B=\\angle P A B .\n$$\n\nSo $B^{\\prime} C^{\\prime} P A$ is also cyclic. Therefore, the points $B^{\\prime}, C^{\\prime}, K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have\n\n$$\n\\angle P K A=\\angle P B^{\\prime} A=\\angle P B A .\n$$\n\nSo, lemma is proved and now return to the problem.\n\n\n\nLet $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M^{\\prime}$ and $N^{\\prime}$ the intersection points of the circumcircle of the triangle $H E F$ with the segments $E C$ and $F B$, respectively. We are going to show that $M=M^{\\prime}$ and $N=N^{\\prime}$ and it will prove the points $E, F, N, M$ lie on a common circle.\n\nOf course, $A$ is an orthocenter of the triangle $B C H$. Therefore $\\angle B H A=\\angle B C A, \\angle C H A=\\angle C B A$ and $\\angle H B A=\\angle H C A$. Thus\n\n$$\n\\angle H E F=\\angle H B A+\\angle E A B=\\angle H C A+\\angle F A C=\\angle H F E .\n$$\n\nSo, the triangle $H E F$ is isosceles, $H E=H F$.\n\nBy using lemma, we get\n\n$$\n\\angle A M^{\\prime} E=\\angle A H E=\\angle A C B,\n$$\n\nand\n\n$$\n\\angle A N^{\\prime} F=\\angle A H F=\\angle A B C \\text {. }\n$$\n\n\n\nTherefore $M=M^{\\prime}$ and $N=N^{\\prime}$ and we are done.', 'Let $X, Y$ be projections of $B$ on $A C$, and $C$ on $A B$, respectively. Let $\\omega$ be circumcircle of $B X Y C$. Let $Z$ be intersection of $E C$ and $\\omega$ and $D$ be projection of $E$ on $B A$.\n\n$$\n\\angle M A C=\\angle A M E-\\angle M C A=\\angle X C B-\\angle X C E=\\angle Z C B=\\angle Z X B\n$$\n\nSince $B X Y C$ is cyclic $\\angle A C Y=\\angle X B A$, and since DEXA is cyclic\n\n$$\n\\angle E X D=\\angle E A D=\\angle F A C \\text {. }\n$$\n\nTherefore, we get that the quadrangles $B Z X D$ and $C M A F$ are similar. Hence $\\angle F M C=\\angle D Z B$. Since $Z E D B$ is cyclic,\n\n$$\n\\angle D Z B=\\angle D E B=\\angle X A B .\n$$\n\nThus $\\angle F M C=\\angle X A B$. Similarly, $\\angle E N B=\\angle Y A C$. We get that $\\angle F M C=\\angle E N B$ and it implies that the points $E, F, N, M$ lie on a circle.\n\n']",,True,,, 2173,Geometry,,Let $A B C$ be a triangle with incentre Iand let $D$ be an arbitrary point on the side $B C$. Let the line through $D$ perpendicular to $B I$ intersect $C I$ at $E$. Let the line through $D$ perpendicular to $C I$ intersect $B I$ at $F$. Prove that the reflection of $A$ in the line $E F$ lies on the line $B C$.,"['Let us consider the case when I lies inside of triangle EFD. For the other cases the proof is almost the same only with the slight difference.\n\nWe are going to prove that the intersection point of the circumcircles of $A E C$ and $A F B$ (denote it by $T$ ) lies on the line $B C$ and this point is the symmetric point of $A$ with respect to $E F$. First of all we prove that $A E I F$ is cyclic, which implies that $T$ lies on the line $B C$, because\n\n$$\n\\angle A T B+\\angle A T C=\\angle A F B+\\angle A E C=\\angle A F I+\\angle A E I=180^{\\circ} .\n$$\n\nDenote by $N$ an intersection point of the lines $D F$ and $A C$. Of course $N$ is the symmetric point with respect to $C I$. Thus, $\\angle I N A=\\angle I D B$. Also,\n\n$$\n\\angle I F D=\\angle N D C-\\angle I B C=90^{\\circ}-\\angle I C B-\\angle I B C=\\angle I A N .\n$$\n\nSo, we get that $A, I, N$ and $F$ lie on a common circle. Therefore, we have $\\angle A F I=\\angle I N A=\\angle I D B$. Analogously, $\\angle A E I=\\angle I D C$ and we have\n\n$$\n\\angle A F I+\\angle A E I=\\angle I D B+\\angle I D C\n$$\n\nSo $\\angle A F I+\\angle A E I=180^{\\circ}$, thus $A E I F$ is cyclic and $T$ lies on the line $B C$.\n\n\n\nBecause $E C$ bisects the angle $A C B$ and $A E T C$ is cyclic we get $E A=E T$. Because of the similar reasons we have $F A=F T$. Therefore $T$ is the symmetric point of $A$ with respect to the line $E F$ and it lies on the line $B C$.\n\n\n\n# Solution 2. \n\nLike to the first solution, consider the case when I lies inside of triangle EFD. we need to prove that $A E I F$ is cyclic. The finish of the proof is the same.\n\n\n\nfirst note that $\\triangle F D B \\sim \\triangle A I B$, because $\\angle F B D=\\angle A B I$, and\n\n$$\n\\angle B F D=\\angle F D C-\\angle I B C=90^{\\circ}-\\angle I C D-\\angle I B C=\\angle I A B .\n$$\n\nBecause of the similarity we have $\\frac{A B}{A F}=\\frac{B I}{B D}$. This equality of the length ratios with $\\angle I B D=\\angle A B F$ implies that $\\triangle A B F \\sim \\triangle I B D$. Therefore, we have $\\angle I D B=\\angle A F B$. Analogously, we can get $\\angle I D C=\\angle A E C$, thus\n\n$$\n\\angle A F I+\\angle A E I=\\angle I D B+\\angle I D C=180^{\\circ} .\n$$\n\nSo, $A E I F$ is cyclic and we are done.']",,True,,, 2174,Geometry,,"A plane has a special point $O$ called the origin. Let $P$ be a set of 2021 points in the plane, such that (i) no three points in $P$ lie on a line and (ii) no two points in $P$ lie on a line through the origin. A triangle with vertices in $P$ is $f a t$, if $O$ is strictly inside the triangle. Find the maximum number of fat triangles.","['We will count minimal number of triangles that are not fat. Let $F$ set of fat triangles, and $\\mathrm{S}$ set of triangles that are not fat. If triangle $X Y Z \\in S$, we call $X$ and $Z$ good vertices if $O Y$ is located between $O X$ and $O Z$. For $A \\in P$ let $S_{A} \\subseteq S$ be set of triangles in $S$ for which $A$ is one of the good vertex.\n\nIt is easy to see that\n\n$$\n2|S|=\\sum_{A \\in P}\\left|S_{A}\\right| \\tag{1}\n$$\n\n\n\nFor $A \\in P$, let $R_{A} \\subset P$ and $L_{A} \\subset P$ be parts of $P \\backslash\\{A\\}$ divided by $A O$. Suppose for $A X Y \\in S$ vertex $A$ is good, then clearly $X, Y \\in R_{A}$ or $X, Y \\in L_{A}$. On the other hand, if $X, Y \\in R_{A}$ or $X, Y \\in L_{A}$ then clearly $A X Y \\in S$ and $A$ is its good vertex. Therefore,\n\n$$\n\\left|S_{A}\\right|=\\left(\\begin{array}{c}\n\\left|R_{A}\\right| \\\\\n2\n\\end{array}\\right)+\\left(\\begin{array}{c}\n\\left|L_{A}\\right| \\\\\n2\n\\end{array}\\right) \\tag{2}\n$$\n\nIt is easy to show following identity:\n\n$$\n\\frac{x(x-1)}{2}+\\frac{y(y-1)}{2}-2 \\cdot \\frac{\\frac{x+y}{2}\\left(\\frac{x+y}{2}-1\\right)}{2}=\\frac{(x-y)^{2}}{4} \\tag{3}\n$$\n\nBy using (2) and (3) we get\n\n$$\n\\left|S_{A}\\right| \\geq 2 \\cdot\\left(\\begin{array}{c}\n\\frac{\\left|R_{A}\\right|+\\left|L_{A}\\right|}{2} \\\\\n2\n\\end{array}\\right)=2 \\cdot\\left(\\begin{array}{c}\n1010 \\\\\n2\n\\end{array}\\right)=1010 \\cdot 1009 \\tag{4}\n$$\n\nand the equality holds when $\\left|R_{A}\\right|=\\left|L_{A}\\right|=1010$. Hence\n\n$$\n|S|=\\frac{\\sum_{A \\in P}\\left|S_{A}\\right|}{2} \\geq \\frac{2021 \\cdot 1010 \\cdot 1009}{2}=2021 \\cdot 505 \\cdot 1009 \\tag{5}\n$$\n\nTherefore,\n\n$$\n|F|=\\left(\\begin{array}{c}\n2021 \\\\\n3\n\\end{array}\\right)-|S| \\leq 2021 \\cdot 1010 \\cdot 673-2021 \\cdot 505 \\cdot 1009=2021 \\cdot 505 \\cdot 337 \\tag{6}\n$$\n\nFor configuration of points on regular 2021-gon which is centered at $O$, inequalities in (4), (5), (6) become equalities. Hence $2021 \\cdot 505 \\cdot 337$ is indeed the answer.']",['$2021 \\cdot 505 \\cdot 337$'],False,,Numerical, 2175,Number Theory,,"Does there exist a nonnegative integer $a$ for which the equation $$ \left\lfloor\frac{m}{1}\right\rfloor+\left\lfloor\frac{m}{2}\right\rfloor+\left\lfloor\frac{m}{3}\right\rfloor+\cdots+\left\lfloor\frac{m}{m}\right\rfloor=n^{2}+a $$ has more than one million different solutions $(m, n)$ where $m$ and $n$ are positive integers? (The expression $\lfloor x\rfloor$ denotes the integer part(or floor) of the real number $x$. Thus $\lfloor\sqrt{2}\rfloor=1,\lfloor\pi\rfloor=\left\lfloor\frac{22}{7}\right\rfloor=$ $3,\lfloor 42\rfloor=42$ and $\lfloor 0\rfloor=0)$","['Denote the equation from the statement by (1). The left hand side of (1) depends only on $m$, and will throughout be denoted by $L(m)$. Fix an integer $q>10^{7}$ and note that for $m=q^{3}$\n\n\n\n$$\nL\\left(q^{3}\\right)=\\sum_{k=1}^{q^{3}}\\left[\\frac{q^{3}}{k}\\right] \\leq \\sum_{k=1}^{q^{3}} \\frac{q^{3}}{k} \\leq q^{3} \\cdot \\sum_{k=1}^{q^{3}} \\frac{1}{k} \\leq q^{3} \\cdot q=q^{4} \\tag{2}\n$$\n\nIndeed, the first inequality results from $[x] \\leq x$. The second inequality can be seen (for instance) as follows. We divide the terms in the sum $\\sum_{k=1}^{q^{3}} \\frac{1}{k}$ into several groups: For $j \\geq 0$, the $j$-th group contains the $2^{j}$ consecutive terms $\\frac{1}{2^{j}}, \\ldots, \\frac{1}{2^{j+1}-1}$. Since every term in the $j$-th group is bounded by $\\frac{1}{2^{j}}$, the overall contribution of the $j$-th group to the sum is at most 1 . Since the first $q$ groups together would contain $2^{q}-1>q^{3}$ terms, the number of groups does not exceed $q$, and hence the value of the sum under consideration is indeed bounded by $q$.\n\nCall an integer $m$ special, if it satisfies $1 \\leq L(m) \\leq q^{4}$. Denote by $g(m) \\geq 1$ the largest integer whose square is bounded by $L(m)$; in other words $g^{2}(m) \\leq L(m)<(g(m)+1)^{2}$. Note that $g(m) \\leq$ $q^{2}$ for all special $m$, which implies\n\n$$\n0 \\leq L(m)-g^{2}(m)<(g(m)+1)^{2}-g^{2}(m)=2 g(m)+1 \\leq 2 q^{2}+1 . \\tag{3}\n$$\n\nFinally, we do some counting. Inequality (2) and the monotonicity of $L(m)$ imply that there exist at least $q^{3}$ special integers. Because of (3), every special integer $m$ has $0 \\leq L(m)-g^{2}(m) \\leq 2 q^{2}+1$. By averaging, at least $\\frac{q^{3}}{2 q^{2}+2}>10^{6}$ special integers must yield the same value $L(m)-g^{2}(m)$. This frequently occurring value is our choice for $\\alpha$, which yields more than $10^{6}$ solutions $(m, g(m))$ to equation (1).']",,True,,, 2176,Geometry,,"Let $A B C D$ be a convex quadrilateral with $\angle D A B=\angle B C D=90^{\circ}$ and $\angle A B C>\angle C D A$. Let $Q$ and $R$ be points on the segments $B C$ and $C D$, respectively, such that the line $Q R$ intersects lines $A B$ and $A D$ at points $P$ and $S$, respectively. It is given that $P Q=R S$. Let the midpoint of $B D$ be $M$ and the midpoint of $Q R$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.","['Note that $N$ is also the midpoint of $P S$. From right-angled triangles $P A S$ and $C Q R$ we obtain $\\angle A N P=$ $2 \\angle A S P, \\angle C N Q=2 \\angle C R Q$, hence $\\angle A N C=\\angle A N P+\\angle C N Q=2(\\angle A S P+\\angle C R Q)=2(\\angle R S D+\\angle D R S)=$ $2 \\angle A D C$.\n\nSimilarly, using right-angled triangles $B A D$ and $B C D$, we obtain $\\angle A M C=2 \\angle A D C$.\n\nThus $\\angle A M C=\\angle A N C$, and the required statement follows.\n\n', 'In this proof we show that we have $\\angle N C M=\\angle N A M$ instead. From right-angled triangles $B C D$ and $Q C R$ we get $\\angle D R S=\\angle C R Q=\\angle R C N$ and $\\angle B D C=\\angle D C M$. Hence $\\angle N C M=\\angle D C M-\\angle R C N$. From rightangled triangle $A P S$ we get $\\angle P S A=\\angle S A N$. From right-angled triangle $B A D$ we have $\\angle M A D=\\angle B D A$. Moreover, $\\angle B D A=\\angle D R S+\\angle R S D-\\angle R D B$.\n\nTherefore $\\angle N A M=\\angle N A S-\\angle M A D=\\angle C D B-\\angle D R S=\\angle N C M$, and the required statement follows.', 'As $N$ is also the midpoint of $P S$, we can shrink triangle $A P S$ to a triangle $A_{0} Q R$ (where $P$ is sent to $Q$ and $S$ is sent to $R$ ). Then $A_{0}, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking the line $A_{0} R$ is parallel to the line $A D$. Therefore $\\angle C N A=\\angle C N A_{0}=2 \\angle C R A_{0}=2 \\angle C D A=\\angle C M A$. The required statement follows.']",,True,,, 2177,Combinatorics,,"Find the smallest positive integer $k$ for which there exist a colouring of the positive integers $\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ with the following two properties: (i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$. (ii) There are positive integers $m, n$ such that $f(m+n) \neq f(m)+f(n)$. In a colouring of $\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.","['The answer is $k=3$.\n\nFirst we show that there is such a function and coloring for $k=3$. Consider $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ given by $f(n)=n$ for all $n \\equiv 1$ or 2 modulo 3 , and $f(n)=2 n$ for $n \\equiv 0$ modulo 3 . Moreover, give a positive integer $n$ the $i$-th color if $n \\equiv i(3)$.\n\nBy construction we have $f(1+2)=6 \\neq 3=f(1)+f(2)$ and hence $f$ has property (ii).\n\nNow let $n, m$ be positive integers with the same color $i$. If $i=0$, then $n+m$ has color 0 , so $f(n+m)=$ $2(n+m)=2 n+2 m=f(n)+f(m)$. If $i=1$, then $n+m$ has color 2 , so $f(n+m)=n+m=f(n)+f(m)$. Finally, if $i=2$, then $n+m$ has color 1 , so $f(n+m)=n+m=f(n)+f(m)$. Therefore $f$ also satisfies condition (i).\n\nNext we show that there is no such function and coloring for $k=2$.\n\nConsider any coloring of $\\mathbb{Z}_{>0}$ with 2 colors and any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Then there exist positive integers $m$ and $n$ such that $f(m+n) \\neq f(m)+f(n)$. Choose $m$ and $n$ such that their sum is minimal among all such $m, n$ and define $a=m+n$. Then in particular for every $b1$, we have $\\frac{a+1}{2}0}$ with 2 colors and any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). We first notice with $m=n$ that $f(2 n)=2 f(n)$.\n\nLemma 3. For every $n \\in \\mathbb{Z}_{>0}, f(3 n)=3 f(n)$ holds.\n\nProof. Define $c=f(n), d=f(3 n)$. Then we have the relations\n\n$$\nf(2 n)=2 c, \\quad f(4 n)=4 c, \\quad f(6 n)=2 d\n$$\n\n- If $n$ and $2 n$ have the same color, then $f(3 n)=f(n)+f(2 n)=3 c=3 f(n)$.\n- If $n$ and $3 n$ have the same color, then $4 c=f(4 n)=f(n)+f(3 n)=c+f(3 n)$, so $f(3 n)=3 f(n)$.\n- If $2 n$ and $4 n$ have the same color, then $2 d=f(6 n)=f(2 n)+f(4 n)=2 c+4 c=6 c$, so $f(3 n)=d=3 c$.\n- Otherwise $n$ and $4 n$ have the same color, and $2 n$ and $3 n$ both have the opposite color to $n$. Therefore we compute $5 c=f(n)+f(4 n)=f(5 n)=f(2 n)+f(3 n)=2 c+f(3 n)$ so $f(3 n)=3 f(n)$.\n\nConsequently, for $k=2$ we necessarily have $f(3 n)=3 f(n)$.\n\nNow let $a$ be the smallest integer such that $f(a) \\neq a f(1)$. In particular $a$ is odd and $a>3$. Consider the three integers $a, \\frac{a-3}{2}, \\frac{a+3}{2}$. By pigeonhole principle two of them have the same color.\n\n- If $\\frac{a-3}{2}$ and $\\frac{a+3}{2}$ have the same color, then $f(a)=\\frac{a-3}{2} f(1)+\\frac{a+3}{2} f(1)=a f(1)$.\n- If $a$ and $\\frac{a-3}{2}$ have the same color, then $3 \\frac{a-1}{2} f(1)=3 f\\left(\\frac{a-1}{2}\\right)=f\\left(\\frac{3 a-3}{2}\\right)=f(a)+f\\left(\\frac{a-3}{2}\\right)=f(a)+$ $\\frac{a-3}{2} f(1)$, so $f(a)=a f(1)$.\n- If $a$ and $\\frac{a+3}{2}$ have the same color, then $3 \\frac{a+1}{2} f(1)=3 f\\left(\\frac{a+1}{2}\\right)=f\\left(\\frac{3 a+3}{2}\\right)=f(a)+f\\left(\\frac{a+3}{2}\\right)=f(a)+$ $\\frac{a+3}{2} f(1)$, so $f(a)=a f(1)$.\n\nIn the three cases we find a contradiction with $f(a) \\neq a f(1)$, so it finishes the proof.', 'As before we prove that $k \\leq 3$ and for any such function and colouring we have $f(2 n)=2 f(n)$.\n\nNow we show that there is no such function and coloring for $k=2$.\n\nConsider any coloring of $\\mathbb{Z}_{>0}$ with 2 colors and any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Say the two colors are white (W) and black (B). Pick $m, n$ any two integers such that $f(m+n)=f(m)+f(n)$. Without loss of generality we may assume that $m+n, m$ are black and $n$ is white.\n\nLemma 4. For all $l \\in \\mathbb{Z}_{>0}$ and every $x$ whose color is black, we have $x+l m$ is black and $f(x+\\operatorname{lm})=$ $f(x)+l f(m)$.\n\nProof. We proceed by induction. It is clearly true for $l=0$. If $x+l m$ is black and satisfies $f(x+l m)=$ $f(x)+l f(m)$, then $f(x+(l+1) m)=f(x+l m)+f(m)=f(x)+(l+1) f(m)$ and $f(x+(l+1) m+n)=$ $f(x+l m)+f(m+n)=f(x)+l f(m)+f(m+n) \\neq f(x)+(l+1) f(m)+f(n)=f(x+(l+1) m)+f(n)$, so $x+(l+1) m$ is not the same color of $n$, therefore $x+(l+1) m$ is black. Thjs completes the induction.\n\nIn particular we then must have that $2^{l} n$ is white for every $l$, because otherwise since $2^{l} m$ is black we would have $2^{l} f(m+n)=f\\left(2^{l} m+2^{l} n\\right)=f\\left(2^{l} m\\right)+f\\left(2^{l} n\\right)=2^{l}(f(m)+f(n))$, and consequently $f(m+n)=$ $f(m)+f(n)$.\n\nLemma 5. For every $l \\geq 1,2^{l} m+2^{l-1} n$ is black.\n\n\n\nProof. On the one hand we have $2^{l} f(m+n)=f\\left(2^{l} m+2^{l} n\\right)=f\\left(2^{l-1}(2 m+n)+2^{l-1} n\\right)$. On the other hand we have\n\n$\\left.2^{l} f(m+n)=2^{l-1} \\cdot 2 f(m+n) \\neq 2^{l-1}(f(m+n)+f(m)+f(n))=2^{l-1}(f(2 m+n)+f(n))=f\\left(2^{l} m+2^{l-1} n\\right)\\right)+f\\left(2^{l-1} n\\right)$.\n\nTherefore $2^{l} m+2^{l-1} n$ and $2^{l-1} n$ have different color, which means $2^{l} m+2^{l-1} n$ is black.\n\nCombining the two lemmas give $j m+2^{l-1} n$ is black for all $j \\geq 2^{l}$ and every $l \\geq 1$.\n\nNow write $m=2^{l-1} m^{\\prime}$ with $m^{\\prime}$ odd. Let $t$ be a number such that $\\frac{2^{t}-1}{m^{\\prime}}$ is an integer and $j=\\frac{2^{t}-1}{m^{\\prime}} n \\geq 2^{l}$, i.e. $t$ is some multiple of $\\phi\\left(m^{\\prime}\\right)$. Then we must have that $j m+2^{l-1} n$ is black, but by definition $j m+2^{l-1} n=$ $\\left(2^{t}-1\\right) 2^{l-1} n+2^{l-1} n=2^{t+l-1} n$ is white. This is a contradiction, so $k=2$ is impossible.']",['$k=3$'],False,,Numerical, 2178,Combinatorics,,"There are 2017 lines in a plane such that no 3 of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of 2 lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?","['We show that this is not possible.\n\nThe lines divide the plane into disjoint regions. We claim that there exists an alternating 2-coloring of these regions, that is each region can be colored in black or white, such that if two regions share a line segment, they have a different color. We show this inductively.\n\nIf there are no lines, this is obvious. Consider now an arrangement of $n$ lines in the plane and an alternating 2 -coloring of the regions. If we add a line $g$, we can simply switch the colors of all regions in one of the half planes divided by $g$ from white to black and vice versa. Any line segment not in $g$ will still be between two regions of different color. Any line segment in $g$ cuts a region determined by the $n$ lines in two, and since we switched colors on one side of $g$ this segment will also lie between two regions of different color.\n\nNow without loss of generality we may assume, that Turbo starts on a line segment with a white region on her left and a black one on her right. At any intersection, if she turns right, she will keep the black tile to her right. If she turns left, she will keep the white tile to her left. Thus wherever she goes, she will always have a white tile on her left and a black tile on her right. As a consequence, she can cross every line segment only in the direction where she has a white tile to her left, and never the opposite direction where she would have a black tile to the left.', ""Suppose the assumption is true.\n\nLet's label each segment in the snail's path with $\\mathbf{L}$ or $\\mathbf{R}$ depending on the direction that Turbo chose at the start point of this segment (one segment can have several labels if it has been visited several times).\n\nConsider the first segment that has been visited twice in different directions, name this segment $s_{1}$. Assume without loss of generality that it is labeled with $\\mathbf{L}$. Then next segment must be labeled with $\\mathbf{R}$, name this one $s_{2}$.\n\nLet's look at the label which $s_{1}$ can get on the second visit. If it gets $\\mathbf{L}$ then the previous segment in the path must be $s_{2}$. But in this case $s_{1}$ is not the first segment that has been visited twice in different directions because $s_{2}$ has been visited earlier. So the second label of $s_{1}$ must be $\\mathbf{R}$, and Turbo must have come from the opposite side of $s_{2}$.\n\n\n\n\n\nSince Turbo alters directions at each point, labels in her path also alter. And because two labels of $s_{1}$ are different, the number of visited segments between these two visits must be even.\n\nNow let's make the following observation: each segment in the path corresponds to exactly one line, and its previous and next segments are on opposite sides of this line.\n\n\n\nAgain consider the path between two visits of $s_{1}$.\n\nEach line intersecting this path must be crossed an even number of times because Turbo has to return to the initial side of each line. Therefore, an even number of segments of Turbo's path are contained on each of these lines. But the line containing $s_{1}$ must be crossed an odd number times. Since each crossing corresponds to exactly one segment in the path, the number of segments must be odd.\n\nHere we get the contradiction. Therefore, the assumption is false."", 'Suppose that the snail always slides slightly to the right of the line segments on her path. When turning to the right, she does not cross any line, whereas when turning to the left, she crosses exactly two lines. This means that at any time of her journey, she has crossed an even number of lines.\n\nAssuming that at some point she slides along a segment for the second time, but in the opposite direction, we argue that she needs to cross an odd number of lines. Let $\\ell$ be the line on which the revisit happens. In order to get to the other side of $\\ell$, the snail has to cross $\\ell$ an odd number of times. To visit the same segment of $\\ell$, she must cross every other line an even number of times.', 'Let us color in red all intersection points of the given lines and let us choose one of two possible directions on each segment (draw an arrow on each segment). Consider a red point $R$ where two given lines $a$ and $b$ meet, and the four segments $a_{1}, a_{2}, b_{1}, b_{2}$ with endpoint $R$ (so that $a_{i} \\subset a, b_{j} \\subset b$ ). $R$ is called a saddle if on $a_{1}, a_{2}$ the arrows go out of $R$ while on $b_{1}, b_{2}$ the arrows enter $R$, or visa versa, on $b_{1}, b_{2}$ the arrows go out of $R$ while on $a_{1}, a_{2}$ the arrows enter $R$. The set of arrows (chosen on all segments) is said to be good if all red points are saddles. It is sufficient to prove that there exists a good set of arrows. Indeed, if initially Turbo is moving along (or opposite) the arrow, then this condition holds after she turns at a red point.\n\nThe given lines cut the plane into regions. Further we need the following property of the good set of arrows (this property directly follows from the definition): the boundary of any bounded region is a directed cycle of arrows; the boundary of any unbounded region is a directed chain of arrows.\n\nWe construct a good set of arrows by induction on $n$ with trivial base $n=1$. Now erase one of $n$ given lines and assume we have a good set of arrows for remaining $n-1$ lines. Now restore the $n$-th line $\\ell$, assume that $\\ell$ is horizontal. Denote by $A_{1}, \\ldots, A_{n-1}$ all new red points on $\\ell$ from the left to the right. Each of $A_{i}$ belongs to some old segment $m_{i}$ of the line $\\ell_{i}$. Let us call $A_{i}$ ascending if the arrow on $m_{i}$ goes up, and descending if the arrow on $m_{i}$ goes down. Consider the region containing the segment $A_{i} A_{i+1}$. By the property, $A_{i}$ and $A_{i+1}$ can not be both ascending or both descending. Thus we can choose arrows on all pieces of $\\ell$ so that each arrow goes from a descending to an ascending vertex.\n\n\n\nEach of points $A_{i}$ cuts $m_{i}$ into two new pieces; the direction of new pieces supposed to be the same as on $m_{i}$. Now simultaneously change the direction of arrows on all pieces below the line $\\ell$. It is easy to see that $A_{1}, \\ldots, A_{n-1}$ become saddles, while the other red points remain saddles. This completes the induction step.']",,True,,, 2179,Combinatorics,,"Let $n \geq 1$ be an integer and let $t_{1}2$ elements $t_{1}<\\ldots1$ we distinguish the two cases $t_{1}>1$ and $t_{1}=1$. If $t_{1}>1$ there exists, by the induction hypothesis, a group $A$ of size $t_{n}$ that satisfies the conditions of the problem for $t_{1}^{\\prime}=t_{1}-1, \\ldots, t_{n}^{\\prime}=t_{n}-1$. Now add a new person to $A$ and let him/her play against everyone from $A$. The new group will be of size $t_{n}+1$ and there exists a person which has played $t$ games if and only if there exists a person that has played $t-1$ games within $A$, i.e. if and only if $t \\in\\left\\{t_{1}, \\ldots, t_{n}\\right\\}$. Hence the conditions of the problem are satisfied.\n\nIf $t_{1}=1$ there exists, by the induction hypothesis, a group $B$ of size $t_{n-1}$ that satisfies the conditions of the problem for $t_{1}^{\\prime}=t_{2}-1, \\ldots, t_{n-2}^{\\prime}=t_{n-1}-1$. Now add a new person $P$ and let him/her play with everyone from group $B$ and a group $C$ of size $t_{n}-t_{n-1}>0$ and let them play with $P$. The new group will be of size $t_{n-1}+1+\\left(t_{n}-t_{n-1}\\right)+1=t_{n}+1$. Since person $P$ has played against everyone he will have played $t_{n}$ games. The people in $C$ will have played $1=t_{1}$ games. There exists a person in $B$ that has played $t$ games if and only if there exist a person in $B$ that has played $t-1$ games within $B$, i.e. if and only if $t \\in\\left\\{t_{2}, \\ldots, t_{n-1}\\right\\}$. Hence the conditions of the problem are satisfied.']",,True,,, 2180,Algebra,,"Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that $$ \left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} . $$ Find all positive integers $n \geq 2$ for which there exists an expensive $n$-tuple.","['Notice that for odd integers $n>2$, the tuple $(1,1, \\ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.\n\nLemma 0.1. If an expensive $n$-tuple exists for some $n \\geq 4$, then also an expensive $n-2$-tuple.\n\nProof. In what follows all indices are considered modulo $n$. Let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities\n\n$$\n\\begin{gathered}\na_{t-1}+a_{t} \\leq 2 a_{t}<2\\left(a_{t}+a_{t+1}\\right)\n\\end{gathered} \\tag{1}\n$$\n$$\n\\begin{gathered}\na_{t}+a_{t+1} \\leq 2 a_{t}<2\\left(a_{t-1}+a_{t}\\right) .\n\\end{gathered} \\tag{2}\n$$\n\nSince both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and $(2)$\n\n$$\na_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}\n$$\n\nfor some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.\n\nConsider now the $n-2$-tuple $\\left(b_{1}, \\ldots, b_{n-2}\\right)$ obtained by removing $a_{t}$ and $a_{t+1}$ from $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$. By what we just said we have\n\n$$\n\\prod_{i=1}^{n-2}\\left(b_{i}+b_{i+1}\\right)=\\frac{\\prod_{i=1}^{n}\\left(a_{i}+a_{i+1}\\right)}{\\left(a_{t-1}+a_{t}\\right)\\left(a_{t}+a_{t+1}\\right)}=2^{2(k-r)-1}\n$$\n\nand hence $\\left(b_{1}, \\ldots, b_{n-2}\\right)$ is again expensive.\n\nFrom the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2 -tuple i.e.\n\n$$\n\\left(a_{1}+a_{2}\\right)^{2}=2^{2 k-1}\n$$\n\nfor some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.', 'For odd $n$ the tuple $(1,1, \\ldots, 1)$ is a solution.\n\nNow consider $n$ even. Since the product $\\prod\\left(a_{i}+a_{i+1}\\right)$ is a power of two, every factor needs to be a power of two. We are going to prove that for all tuples $\\left(a_{1}, \\ldots, a_{n}\\right)$ such that $a_{i}+a_{i+1}$ is always a power of two , it is the case that $\\prod\\left(a_{i}+a_{i+1}\\right)$ is equal to an even power of two. We are going to prove this with strong induction on $\\sum a_{i}$. When all $a_{i}$ are equal to one this is certainly the case. Since $a_{i}+a_{i+1}>1$ it is even and we conclude that the $a_{i}$ are either all odd or all even. In the case they are all even, then consider the tuple $\\left(b_{1}, \\ldots, b_{n}\\right)$ with $b_{i}=a_{i} / 2$. This tuple clearly satisfies the hypothesis as well and we have $\\sum b_{i}<\\sum a_{i}$. Furthermore we have $\\prod\\left(a_{i}+a_{i+1}\\right)=2^{n} \\prod\\left(b_{i}+b_{i+1}\\right)$ and since $n$ is even we are done in this case.\n\nNow all $a_{i}$ are odd. Suppose none of are one, then consider the tuple $\\left(b_{1}, \\ldots, b_{n}\\right)$ with $b_{i}=\\left(a_{i}+(-1)^{i}\\right) / 2$. Since all $a_{i}$ are odd and strictly larger than one, the $b_{i}$ are positive integers and satisfy $b_{i}+b_{i+1}=\\left(a_{i}+a_{i+1}\\right) / 2$, a power of two. Since $\\sum b_{i}<\\sum a_{i}$ and $\\prod\\left(a_{i}+a_{i+1}\\right)=2^{n} \\prod\\left(b_{i}+b_{i+1}\\right)$ we are done in this case again. Now there is at least one $a_{i}$ being one. We may assume $i=1$, because the condition is cyclic. Moreover we may also assume that $a_{2}>1$ since not all of the $a_{i}$ are equal to one. Let now $k$ be the smallest index larger than one such that $a_{k}$ is equal to one. We are not excluding the case $k=n+1$, yet. Now for $i=1, \\ldots, k-1$ we have $a_{i}+a_{i+1}>2$ and thus divisible by four. By induction it easily follows that $a_{i} \\equiv(-1)^{i+1} \\bmod (4)$ for $i=1, \\ldots, k-1$. In particular, since $a_{k}=1$ we find that $k$ is odd and at least three. Now consider the tuple $\\left(b_{1}, \\ldots, b_{n}\\right)$ with $b_{i}=\\left(a_{i}-(-1)^{i}\\right) / 2$ for $i=1, \\ldots, k$ and $b_{i}=a_{i}$ otherwise. This is again a tuple that satisfies the hypothesis, since $b_{1}=a_{1}=1=b_{k}=a_{k}$. Moreover $b_{2}2$'],False,,Need_human_evaluate, 2181,Algebra,,"Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that $$ \left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} . $$ Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple. There are exactly $n$ factors in the product on the left hand side.","['Notice that for odd integers $n>2$, the tuple $(1,1, \\ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.\n\nLemma 0.1. If an expensive $n$-tuple exists for some $n \\geq 4$, then also an expensive $n-2$-tuple.\n\nProof. In what follows all indices are considered modulo $n$. Let $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities\n\n$$\n\\begin{gathered}\na_{t-1}+a_{t} \\leq 2 a_{t}<2\\left(a_{t}+a_{t+1}\\right) \\\\\na_{t}+a_{t+1} \\leq 2 a_{t}<2\\left(a_{t-1}+a_{t}\\right) .\n\\end{gathered}\n$$\n\nSince both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and $(2)$\n\n$$\na_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}\n$$\n\nfor some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.\n\nConsider now the $n-2$-tuple $\\left(b_{1}, \\ldots, b_{n-2}\\right)$ obtained by removing $a_{t}$ and $a_{t+1}$ from $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$. By what we just said we have\n\n$$\n\\prod_{i=1}^{n-2}\\left(b_{i}+b_{i+1}\\right)=\\frac{\\prod_{i=1}^{n}\\left(a_{i}+a_{i+1}\\right)}{\\left(a_{t-1}+a_{t}\\right)\\left(a_{t}+a_{t+1}\\right)}=2^{2(k-r)-1}\n$$\n\nand hence $\\left(b_{1}, \\ldots, b_{n-2}\\right)$ is again expensive.\n\nFrom the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2 -tuple i.e.\n\n$$\n\\left(a_{1}+a_{2}\\right)^{2}=2^{2 k-1}\n$$\n\nfor some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.\n\nWe prove this by induction. From above we saw that 1 belongs to an expensive $n$-tuple. Assume now that all odd positive integers less that $2^{k}$ belong to an expensive $n$-tuple, for some $k \\geq 1$. Hence for any odd $r<2^{k}$ there is an integer $n$ and an expensive $n$-tuple $\\left(a_{1}, \\ldots, r, \\ldots, a_{n}\\right)$. We notice that then also $\\left(a_{1}, \\ldots, r, 2^{k+1}-\\right.$ $\\left.r, r, \\ldots, a_{n}\\right)$ is expensive. Since $2^{k+1}-r$ can take all odd values between $2^{k}$ and $2^{k+1}$ the induction step is complete.', 'Consider the operators $T_{ \\pm}(n)=2 n \\pm 1$. We claim that for every odd integer $m$ there is an integer $r$ and signs $\\epsilon_{i} \\in\\{+,-\\}$ for $i=1, \\ldots r$ such that $T_{\\epsilon_{r}} \\circ \\cdots \\circ T_{\\epsilon_{1}}(1)=m$. This is certainly true for $m=1$ and for $m>1$ we find that $m=T_{-}((m+1) / 2)$ if $m \\equiv 1 \\bmod (4)$ and $m=T_{+}((m-1) / 2)$ if $m \\equiv 3 \\bmod (4)$. Note that both $(m+1) / 2$ and $(m-1) / 2$ are odd integers in their respective cases and $(m-1) / 2 \\leq(m+1) / 21$. Therefore iterating the procedure will eventually terminate in one.\n\nFor the construction it is most convenient to set $n=2 l+1$ and label the tuple $\\left(a_{-l}, a_{-l+1}, \\ldots, a_{l}\\right)$. For $m=1$ we have the expensive tuple $(1,1,1)$. For $m>1$ we will define operators $T_{ \\pm}$on expensive tuples with the condition $a_{-l}=a_{l}=1$ that give rise to a new expensive tuple $\\left(b_{-l^{\\prime}}, \\ldots, b_{l^{\\prime}}\\right)$ with $b_{-l^{\\prime}}=b_{l^{\\prime}}=1$ and $b_{0}=T_{ \\pm} a_{0}$. It is then clear that $T_{\\epsilon_{r}} \\circ \\cdots \\circ T_{\\epsilon_{1}}((1,1,1))$ is an expensive tuple containing $m$. We define $T_{ \\pm}$as follows: set $l^{\\prime}=l+1$ and $b_{-l^{\\prime}}=b_{l^{\\prime}}=1$ and $b_{i}=T_{ \\pm(-1)^{i}}\\left(a_{i}\\right)$ for $i=-l, \\ldots, l$. Here we identify + with +1 and - with -1 . We are left to prove that the new tuple is indeed expensive. If $\\pm(-1)^{l}=-1$, then $\\prod\\left(b_{i}+b_{i+1}\\right)=4 \\cdot 2^{2 l} \\prod\\left(a_{i}+a_{i+1}\\right)$, and if $\\pm(-1)^{l}=+1$, then $\\prod\\left(b_{i}+b_{i+1}\\right)=4 \\cdot 2^{2 l+2} \\prod\\left(a_{i}+a_{i+1}\\right)$. In both cases we end up with an expensive tuple again.']",,True,,, 2182,Geometry,,"Let $A B C$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $A B C$ in its sides $B C, C A, A B$ are denoted by $G_{1}, G_{2}, G_{3}$, and $O_{1}, O_{2}, O_{3}$, respectively. Show that the circumcircles of the triangles $G_{1} G_{2} C, G_{1} G_{3} B, G_{2} G_{3} A, O_{1} O_{2} C, O_{1} O_{3} B, O_{2} O_{3} A$ and $A B C$ have a common point. The centroid of a triangle is the intersection point of the three medians. A median is a line connecting a vertex of the triangle to the midpoint of the opposite side.","['Let $H$ denote the orthocenter of $A B C$, and let $e$ denote its Euler line. Let $e_{1}, e_{2}, e_{3}$ denote the respective reflections of $e$ in $B C, C A, A B$. The proof naturally divides into two parts: we first show that pairwise intersections of the circles in question correspond to pairwise intersections of $e_{1}, e_{2}, e_{3}$, and then prove that $e_{1}, e_{2}, e_{3}$ intersect in a single point on the circumcircle of $A B C$.\n\n\n\n\n\nNow consider for example the circumcircles of $O_{1} O_{2} C$ and $G_{1} G_{2} C$. By construction, it is clear that $\\angle O_{2} C O_{1}=\\angle G_{2} C G_{1}=2 \\angle A C B$. Let $G_{1} O_{1}$ and $G_{2} O_{2}$ meet at $X$, and let $e$ meet $e_{1}, e_{2}$ at $E_{1}, E_{2}$, respectively, as shown in the diagram below. Chasing angles,\n\n$$\n\\begin{aligned}\n\\angle G_{2} X G_{1}=\\angle O_{2} X O_{1} & =\\angle E_{2} X E_{1}=180^{\\circ}-\\angle E_{1} E_{2} X-\\angle X E_{1} E_{2} \\\\\n& =180^{\\circ}-2 \\angle E_{1} E_{2} C-\\left(\\angle C E_{1} E_{2}-\\angle C E_{1} X\\right) .\n\\end{aligned}\n$$\n\nBut $\\angle C E_{1} X=\\angle B E_{1} O_{1}=\\angle B E_{1} O=180^{\\circ}-\\angle C E_{1} E_{2}$, and thus\n\n$$\n\\angle G_{2} X G_{1}=\\angle O_{2} X O_{1}=2\\left(180^{\\circ}-\\angle E_{1} E_{2} C-\\angle C E_{1} E_{2}\\right)=2 \\angle A C B .\n$$\n\n\n\nIt follows from this that $X$ lies on the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$. In other words, the second point of intersection of the circumcircles of $G_{1} G_{2} C$ and $O_{1} O_{2} C$ is the intersection of $e_{1}$ and $e_{2}$. Similarly, the circumcircles of $G_{1} G_{3} B$ and $O_{1} O_{3} B$ meet again at the intersection of $e_{1}$ and $e_{3}$, and those of $G_{2} G_{3} A$ and $O_{2} O_{3} A$ meet again at the intersection of $e_{2}$ and $e_{3}$.\n\n\n\n\n\nIt thus remains to show that $e_{1}, e_{2}, e_{3}$ are concurrent, and intersect on the circumcircle of $A B C$. Let $e$ meet the circumcircles of the triangles $B C H, A C H, A B H$ at $X_{1}, X_{2}, X_{3}$, respectively. It is well known that the reflections of $H$ in the sides of $A B C$ lie on the circumcircle of $A B C$. For this reason, the circumcircles of $B C H, A C H, A B H$ have the same radius as the circumcircle of $A B C$, and hence the reflections $X_{1}^{\\prime}, X_{2}^{\\prime}, X_{3}^{\\prime}$ of $X_{1}, X_{2}, X_{3}$ in the sides $[B C],[C A],[A B]$ lie on the circumcircle of $A B C$. By definition, $X_{1}^{\\prime}, X_{2}^{\\prime}, X_{3}^{\\prime}$ lie on $e_{1}, e_{2}, e_{3}$, respectively. It thus remains to show that they coincide.\n\nTo show that, for example, $X_{1}^{\\prime}=X_{2}^{\\prime}$, it will be sufficient to show that $\\angle X_{2} A C=\\angle X_{1}^{\\prime} A C$, since we have already shown that $X_{1}^{\\prime}$ and $X_{2}^{\\prime}$ lie on the circumcircle of $A B C$. But, chasing angles in the diagram above,\n\n$$\n\\begin{aligned}\n\\angle X_{1}^{\\prime} A C & =\\angle X_{1}^{\\prime} A B-\\angle B A C=\\left(180^{\\circ}-\\angle X_{1}^{\\prime} C B\\right)-\\angle B A C \\\\\n& =180^{\\circ}-\\angle X_{1} C B-\\angle B A C=\\left(180^{\\circ}-\\angle B A C\\right)-\\angle X_{1} H B \\\\\n& =\\angle B H C-\\angle X_{1} H B=\\angle X_{2} H C=\\angle X_{2} A C,\n\\end{aligned}\n$$\n\nwhere we have used the fact that $B H C X_{1}$ and $A H X_{2} C$ are cyclic by construction, and the fact that $\\angle B H C=180^{\\circ}-\\angle B A C$. This shows that $X_{1}^{\\prime}=X_{2}^{\\prime}$. Similarly, $X_{1}^{\\prime}=X_{3}^{\\prime}$, which completes the proof.', 'The proof consists of two parts. First, we show that if $P$ is any point inside the triangle $A B C$ and $P_{1}, P_{2}, P_{3}$ are its reflections in the sides $B C, C A, A B$, then the circumcircles of the triangles $P_{1} P_{2} C, P_{1} P_{3} B, P_{2} P_{3} A$ intersect in a point $T_{P}$ on the circumcircle of the triangle $A B C$. In the second part, we show that $T_{G}$ coincides with $T_{O}$.\n\nNow let $P$ be any point inside the triangle $A B C$ and let $P_{1}, P_{2}, P_{3}$ be the reflections in the sides as above. Let $T_{P}$ be the second intersection of the circumcircles of the triangles $P_{1} P_{2} C$ and $A B C$. We want to show that $T_{P}$ lies on the circumcircles of the triangles $P_{1} P_{3} B$ and $P_{2} P_{3} A$.\n\n\n\nBy construction, we have $P_{1} C=P_{2} C$, hence\n\n$$\n\\angle C P_{1} P_{2}=90^{\\circ}-\\frac{1}{2} \\angle P_{2} C P_{1}=90^{\\circ}-\\angle A C B .\n$$\n\nSimilarly, $\\angle P_{2} P_{3} A=90^{\\circ}-\\angle B A C$. This gives us\n\n$$\n\\begin{aligned}\n\\angle P_{2} T_{P} A & =\\angle C T_{P} A-\\angle C T_{P} P_{2}=\\angle C B A-\\angle C P_{1} P_{2} \\\\\n& =\\angle C B A-90^{\\circ}+\\angle A C B=90^{\\circ}-\\angle B A C=\\angle P_{2} P_{3} A,\n\\end{aligned}\n$$\n\nso $T_{P}$ lies on the circumcircle of the triangle $P_{2} P_{3} A$. Similarly, $T_{P}$ lies on the circumcircle of the triangle $P_{1} P_{3} B$ which completes the first part.\n\nNote that if $P_{2}$ is given, then $T_{P}$ is the unique point on the circumcircle of the triangle $A B C$ with $\\angle C T_{P} P_{2}=90^{\\circ}-\\angle A C B$. In the second part, we will use this as follows: If we can find a point $T$ on the circumcircle of the triangle $A B C$ with $\\angle C T G_{2}=\\angle C T O_{2}=90^{\\circ}-\\angle A C B$, then $T=T_{G}=T_{O}$ and we are done.\n\nLet $H$ be the orthocenter of the triangle $A B C$ and let $H_{2}$ be the reflection in the side $A C$. It is known that $\\mathrm{H}_{2}$ lies on the circumcircle of the triangle $A B C . G, O, H$ lie on the Euler line, so $G_{2}, \\mathrm{O}_{2}, \\mathrm{H}_{2}$ are collinear as well. Let $T$ be the second intersection of $G_{2} H_{2}$ and the circumcircle of the triangle $A B C$. We can now complete the proof by seeing that\n\n$$\n\\angle C T G_{2}=\\angle C T O_{2}=\\angle C T H_{2}=\\angle C B H_{2}=90^{\\circ}-\\angle A C B .\n$$\n\n', 'For every point $P$, let $p$ denote the corresponding complex number. Set $O$ to be the origin, so $o=0$, and without loss of generality we can assume that $a, b$ and $c$ lie on the unit circle. Then the centroid can be expressed as $g=\\frac{a+b+c}{3}$.\n\nThe segments $o o_{1}$ and $b c$ have a common midpoint, so $o_{1}+o=b+c$, and then $o_{1}=b+c$. Similarly $o_{2}=a+c$ and $o_{3}=a+b$. In order to compute $g_{1}$, define $y$ to be the projection of $g$ onto $b c$. Since $b$ and $c$ are on the unit circle, it is well known that $y$ can be expressed as\n\n$$\ny=\\frac{1}{2}(b+c+g-b c \\bar{g})\n$$\n\nBy using $\\bar{a}=\\frac{1}{a}, \\bar{b}=\\frac{1}{b}$ and $\\bar{c}=\\frac{1}{c}$ (points on the unit circle), we obtain\n\n$$\ng_{1}=b+c-\\frac{a b+b c+c a}{3 a}\n$$\n\nSimilarly, we get $g_{2}=a+c-\\frac{a b+b c+c a}{3 b}$ and $g_{3}=a+b-\\frac{a b+b c+c a}{3 c}$.\n\n1) Proof that circumcircles of triangles $a b c, o_{1} o_{2} c, o_{1} o_{3} b$ and $o_{2} o_{3} a$ have common point. Let $x$ be the point of intersection of circumcircles of triangles $o_{1} o_{2} c$ and $a b c(x \\neq c)$. We know that $x, o_{1}, o_{2}$ and $c$ are concyclic if and only if $\\frac{x-c}{o_{1}-c}: \\frac{x-o_{2}}{o_{1}-o_{2}}$ is real number, which is equivalent to\n\n$$\n\\frac{x-c}{\\bar{x}-\\bar{c}} \\cdot \\frac{o_{1}-o_{2}}{\\overline{o_{1}}-\\overline{o_{2}}}=\\frac{o_{1}-c}{\\overline{o_{1}}-\\bar{c}} \\cdot \\frac{x-o_{2}}{\\bar{x}-\\overline{o_{2}}}\n\\tag{1}\n$$\n\nSince $x$ and $c$ are on the unit circle $\\frac{x-c}{\\bar{x} - \\bar{c}} = -xc$. Also, $\\frac{o_1-o_2}{\\bar{o_1} - \\bar{o_2}} = \\frac{b-a}{\\bar{b} - \\bar{a}}=-ab$ and $\\frac{o_{1}-c}{\\overline{o_{1}}-\\bar{c}}=\\frac{b}{\\bar{b}}=b^{2}$. Since $\\bar{x}=\\frac{1}{x}$, from (1) and previous relations, we have:\n\n$$\nx=\\frac{a b+b c+c a}{a+b+c} \\text {. }\n$$\n\nThis formula is symmetric, so we conclude that $x$ also belongs to circumcircles of $o_{1} o_{3} b$ and $\\mathrm{o}_{2} \\mathrm{o}_{3} a$.\n\n2) Proof that $x$ belongs to circumcircles of $g_{1} g_{2} c, g_{1} g_{3} b$ and $g_{2} g_{3} a$.\n\nBecause of symmetry, it is enough to prove that $x$ belongs to circumcircle of $g_{1} g_{2} c$, i.e. to prove the following:\n\n$$\n\\frac{x-c}{\\bar{x}-\\bar{c}} \\cdot \\frac{g_{1}-g_{2}}{\\overline{g_{1}}-\\overline{g_{2}}}=\\frac{g_{1}-c}{\\overline{g_{1}}-\\bar{c}} \\cdot \\frac{x-g_{2}}{\\bar{x}-\\overline{g_{2}}} \\tag{2}\n$$\n\nEasy computations give that\n\n$$\ng_{1}-g_{2}=(b-a) \\frac{2 a b-b c-a c}{3 a b}, \\quad \\overline{g_{1}}-\\overline{g_{2}}=(\\bar{b}-\\bar{a}) \\frac{2-\\frac{a}{c}-\\frac{b}{c}}{3}\n$$\n\nand then\n\n$$\n\\frac{g_{1}-g_{2}}{\\overline{g_{1}}-\\overline{g_{2}}}=\\frac{c(b c+a c-2 a b)}{2 c-a-b}\n$$\n\n\n\nOn the other hand we have\n\n$$\ng_{1}-c=\\frac{2 a b-b c-a c}{3 a}, \\quad \\bar{g}_{1}-\\bar{c}=\\frac{2 c-a-b}{3 b c}\n$$\n\nThis implies\n\n$$\n\\frac{g_{1}-c}{{\\bar{g_{1}}-\\bar{c}}}=\\frac{2 a b-b c-a c}{2 c-a-b} \\cdot \\frac{b c}{a} .\n$$\n\nThen (2) is equivalent to\n\n$$\n\\begin{aligned}\n-x c \\cdot \\frac{c(b c+a c-2 a b)}{2 c-a-b}=\\frac{2 a b-b c-a c}{2 c-a-b} \\cdot \\frac{b c}{a} \\cdot \\frac{x-g_{2}}{\\bar{x}-\\overline{g_{2}}} \\\\\n\\Longleftrightarrow x c a\\left(\\bar{x}-\\overline{g_{2}}\\right)=b\\left(x-g_{2}\\right),\n\\end{aligned}\n$$\n\nwhich is also equivalent to\n\n$\\frac{a b+b c+c a}{a+b+c} \\cdot c a\\left(\\frac{a+b+c}{a b+b c+c a}-\\frac{1}{a}-\\frac{1}{c}+\\frac{a+b+c}{3 a c}\\right)=b \\cdot\\left(\\frac{a b+b c+c a}{a+b+c}-a-c+\\frac{a b+b c+c a}{3 b}\\right)$.\n\nThe last equality can easily be verified, which implies that $x$ belongs to circumcircle of triangle $g_{1} g_{2} c$. This concludes our proof.']",,True,,, 2183,Algebra,,"Let $n$ be an odd positive integer, and let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers. Show that $$ \min _{i=1, \ldots, n}\left(x_{i}^{2}+x_{i+1}^{2}\right) \leq \max _{j=1, \ldots, n}\left(2 x_{j} x_{j+1}\right) $$ where $x_{n+1}=x_{1}$.","['In what follows, indices are reduced modulo $n$. Consider the $n$ differences $x_{k+1}-x_{k}, k=1, \\ldots, n$. Since $n$ is odd, there exists an index $j$ such that $\\left(x_{j+1}-x_{j}\\right)\\left(x_{j+2}-x_{j+1}\\right) \\geq 0$. Without loss of generality, we may and will assume both factors non-negative, so $x_{j} \\leq x_{j+1} \\leq x_{j+2}$. Consequently,\n\n$$\n\\min _{k=1, \\ldots, n}\\left(x_{k}^{2}+x_{k+1}^{2}\\right) \\leq x_{j}^{2}+x_{j+1}^{2} \\leq 2 x_{j+1}^{2} \\leq 2 x_{j+1} x_{j+2} \\leq \\max _{k=1, \\ldots, n}\\left(2 x_{k} x_{k+1}\\right)\n$$']",,True,,, 2184,Geometry,,"Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$. ","['We are to prove that $\\angle E X Y=\\angle E F X$; alternatively, but equivalently, $\\angle A Y X+\\angle X A Y=\\angle B Y F+\\angle X B Y$.\n\nSince the quadrangle $A B C D$ is cyclic, the triangles $X A D$ and $X B C$ are similar, and since $A D_{1}$ and $B C_{1}$ are corresponding medians in these triangles, it follows that $\\angle X A Y=\\angle X A D_{1}=\\angle X B C_{1}=\\angle X B Y$.\n\nFinally, $\\angle A Y X=\\angle B Y F$, since $X$ and $M$ are corresponding points in the similar triangles $A B Y$ and $C_{1} D_{1} Y$ : indeed, $\\angle X A B=\\angle X D C=\\angle M C_{1} D_{1}$, and $\\angle X B A=\\angle X C D=\\angle M D_{1} C_{1}$.']",,True,,, 2184,Geometry,,"Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$. ![](https://cdn.mathpix.com/cropped/2023_12_21_2fe10796898d797e3ea2g-1.jpg?height=557&width=556&top_left_y=1943&top_left_x=750)","['We are to prove that $\\angle E X Y=\\angle E F X$; alternatively, but equivalently, $\\angle A Y X+\\angle X A Y=\\angle B Y F+\\angle X B Y$.\n\nSince the quadrangle $A B C D$ is cyclic, the triangles $X A D$ and $X B C$ are similar, and since $A D_{1}$ and $B C_{1}$ are corresponding medians in these triangles, it follows that $\\angle X A Y=\\angle X A D_{1}=\\angle X B C_{1}=\\angle X B Y$.\n\nFinally, $\\angle A Y X=\\angle B Y F$, since $X$ and $M$ are corresponding points in the similar triangles $A B Y$ and $C_{1} D_{1} Y$ : indeed, $\\angle X A B=\\angle X D C=\\angle M C_{1} D_{1}$, and $\\angle X B A=\\angle X C D=\\angle M D_{1} C_{1}$.']",['证明题，略'],True,,Need_human_evaluate, 2185,Combinatorics,,"Let $m$ be a positive integer. Consider a $4 m \times 4 m$ array of square unit cells. Two different cells are related to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are coloured blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells.","['The required minimum is $6 m$ and is achieved by a diagonal string of $m$ $4 \\times 4$ blocks of the form below (bullets mark centres of blue cells):\n\n\n\nIn particular, this configuration shows that the required minimum does not exceed $6 m$.\n\nWe now show that any configuration of blue cells satisfying the condition in the statement has cardinality at least $6 \\mathrm{~m}$.\n\nFix such a configuration and let $m_{1}^{r}$ be the number of blue cells in rows containing exactly one such, let $m_{2}^{r}$ be the number of blue cells in rows containing exactly two such, and let $m_{3}^{r}$ be the number of blue cells in rows containing at least three such; the numbers $m_{1}^{c}, m_{2}^{c}$ and $m_{3}^{c}$ are defined similarly.\n\nBegin by noticing that $m_{3}^{c} \\geq m_{1}^{r}$ and, similarly, $m_{3}^{r} \\geq m_{1}^{c}$. Indeed, if a blue cell is alone in its row, respectively column, then there are at least two other blue cells in its column, respectively row, and the claim follows.\n\nSuppose now, if possible, the total number of blue cells is less than $6 \\mathrm{~m}$. We will show that $m_{1}^{r}>m_{3}^{r}$ and $m_{1}^{c}>m_{3}^{c}$, and reach a contradiction by the preceding: $m_{1}^{r}>m_{3}^{r} \\geq m_{1}^{c}>m_{3}^{c} \\geq m_{1}^{r}$.\n\nWe prove the first inequality; the other one is dealt with similarly. To this end, notice that there are no empty rows - otherwise, each column would contain at least two blue cells, whence a total of at least $8 m>6 m$ blue cells, which is a contradiction. Next, count rows to get $m_{1}^{r}+m_{2}^{r} / 2+m_{3}^{r} / 3 \\geq 4 m$, and count blue cells to get $m_{1}^{r}+m_{2}^{r}+m_{3}^{r}<6 m$. Subtraction of the latter from the former multiplied by $3 / 2$ yields $m_{1}^{r}-m_{3}^{r}>m_{2}^{r} / 2 \\geq 0$, and the conclusion follows.', 'To prove that a minimal configuration of blue cells satisfying the condition in the statement has cardinality at least $6 m$, consider a bipartite graph whose vertex parts are the rows and the columns of the array, respectively, a row and a column being joined by an edge if and only if the two cross at a blue cell. Clearly, the number of blue cells is equal to the number of edges of this graph, and the relationship condition in the statement reads: for every row $r$ and every column $c, \\operatorname{deg} r+\\operatorname{deg} c-\\epsilon(r, c) \\geq 2$, where $\\epsilon(r, c)=2$ if $r$ and $c$ are joined by an edge, and $\\epsilon(r, c)=0$ otherwise.\n\nNotice that there are no empty rows/columns, so the graph has no isolated vertices. By the preceding, the cardinality of every connected component of the graph is at least 4 , so there are at most $2 \\cdot 4 m / 4=2 m$ such and, consequently, the graph has at least $8 m-2 m=6 m$ edges. This completes the proof.']",['$6m$'],False,,Expression, 2186,Geometry,,"Two circles, $\omega_{1}$ and $\omega_{2}$, of equal radius intersect at different points $X_{1}$ and $X_{2}$. Consider a circle $\omega$ externally tangent to $\omega_{1}$ at a point $T_{1}$, and internally tangent to $\omega_{2}$ at a point $T_{2}$. Prove that lines $X_{1} T_{1}$ and $X_{2} T_{2}$ intersect at a point lying on $\omega$.","['Let the line $X_{k} T_{k}$ and $\\omega$ meet again at $X_{k}^{\\prime}, k=1,2$, and notice that the tangent $t_{k}$ to $\\omega_{k}$ at $X_{k}$ and the tangent $t_{k}^{\\prime}$ to $\\omega$ at $X_{k}^{\\prime}$ are parallel. Since the $\\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so the $t_{k}^{\\prime}$ are parallel, and consequently the points $X_{1}^{\\prime}$ and $X_{2}^{\\prime}$ coincide (they are not antipodal, since they both lie on the same side of the line $T_{1} T_{2}$ ). The conclusion follows.', 'The circle $\\omega$ is the image of $\\omega_{k}$ under a homothety $h_{k}$ centred at $T_{k}, k=1,2$. The tangent to $\\omega$ at $X_{k}^{\\prime}=h_{k}\\left(X_{k}\\right)$ is therefore parallel to the tangent $t_{k}$ to $\\omega_{k}$ at $X_{k}$. Since the $\\omega_{k}$ have equal radii, the $t_{k}$ are parallel, so $X_{1}^{\\prime}=X_{2}^{\\prime}$; and since the points $X_{k}, T_{k}$ and $X_{k}^{\\prime}$ are collinear, the conclusion follows.\n', 'Invert from $X_{1}$ and use an asterisk to denote images under this inversion. Notice that $\\omega_{k}^{*}$ is the tangent from $X_{2}^{*}$ to $\\omega^{*}$ at $T_{k}^{*}$, and the pole $X_{1}$ lies on the bisectrix of the angle formed by the $\\omega_{k}^{*}$, not containing $\\omega^{*}$. Letting $X_{1} T_{1}^{*}$ and $\\omega^{*}$ meet again at $Y$, standard angle chase shows that $Y$ lies on the circle $X_{1} X_{2}^{*} T_{2}^{*}$, and the conclusion follows.']",,True,,, 2187,Combinatorics,,"Let $k$ and $n$ be integers such that $k \geq 2$ and $k \leq n \leq 2 k-1$. Place rectangular tiles, each of size $1 \times k$ or $k \times 1$, on an $n \times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.","['The required minimum is $n$ if $n=k$, and it is $\\min (n, 2 n-2 k+2)$ if $k', 'We prove that the hexagon $A B C D E F$ is the intersection of the equilateral triangles $P Q R$ and $X Y Z$.\n\nLet $d(S, A B)$ denote the signed distance from the point $S$ to the line $A B$, where the negative sign is taken if $A B$ separates $S$ and the hexagon. We define similarly the other distances $(d(S, B C)$, etc). Since $M \\in a$, we have $d(M, Z X)=d(M, Q R)$. In the same way, we have $d(M, X Y)=d(M, R P)$ and $d(M, Y Z)=d(M, P Q)$. Therefore $d(M, Z X)+d(M, X Y)+$ $d(M, Y Z)=d(M, Q R)+d(M, R P)+d(M, P Q)$.\n\nWe now use of the following well-known lemma (which can be easily proved using areas) to deduce that triangles $P Q R$ and $X Y Z$ are congruent.\n\nLemma. The sum of the signed distances from any point to the sidelines of an equilateral triangle (where the signs are taken such that all distances are positive inside the triangle) is constant and equals the length of the altitude.\n\nFor $N=b \\cap d$ we now find $d(N, Z X)=d(N, R P)$ and $d(N, X Y)=d(N, P Q)$. Using again the lemma for the point $N$, we get $d(N, Z X)+d(N, X Y)+d(N, Y Z)=d(N, Q R)+d(N, R P)+$ $d(N, P Q)$. Therefore $d(N, Y Z)=d(N, Q R)$, thus $N \\in f$.\n\nRemark. Instead of using the lemma, it is possible to use some equivalent observation in terms of signed areas.', 'We use the same notations as in Solution A. We will show that $a, c$ and $e$ are concurrent if and only if\n\n$$\nA B+C D+E F=B C+D E+F A,\n$$\n\nwhich clearly implies the problem statement by symmetry.\n\nLet $\\vec{a}$ be the vector of unit length parallel to $a$ directed from $A$ towards the interior of the hexagon. We define analogously $\\vec{b}$, etc. The angle conditions imply that opposite bisectors of the hexagon are parallel, so we have $\\vec{a}\\|\\vec{d}, \\vec{b}\\| \\vec{e}$ and $\\vec{c} \\| \\vec{f}$. Moreover, as in the previous solutions, we know that $\\vec{a}, \\vec{c}$ and $\\vec{e}$ make angles of $120^{\\circ}$ with each other. Let $M_{A}=c \\cap e, M_{C}=e \\cap a$ and $M_{E}=a \\cap c$. Then $M_{A}, M_{C}, M_{E}$ form an equilateral triangle with side length denoted by $s$. Note that the case $s=0$ is equivalent to $a, c$ and $e$ being concurrent.\n\nProjecting $\\overrightarrow{M_{E}} A+\\overrightarrow{A B}=\\overrightarrow{M_{E}} B=\\overrightarrow{M_{E} C}+\\overrightarrow{C B}$ onto $\\vec{e}=-\\vec{b}$, we obtain\n\n$$\n\\overrightarrow{A B} \\cdot \\vec{b}-\\overrightarrow{C B} \\cdot \\vec{b}=\\overrightarrow{M_{E} C} \\cdot \\vec{b}-\\overrightarrow{M_{E}} A \\cdot \\vec{b}=\\overrightarrow{M_{E} A} \\cdot \\vec{e}-\\overrightarrow{M_{E} C} \\cdot \\vec{e}\n$$\n\n\n\nWriting $\\varphi=\\frac{1}{2} \\angle B=\\frac{1}{2} \\angle D=\\frac{1}{2} \\angle F$, we know that $\\overrightarrow{A B} \\cdot \\vec{b}=-A B \\cdot \\cos (\\varphi)$, and similarly $\\overrightarrow{C B} \\cdot \\vec{b}=$ $-C B \\cdot \\cos (\\varphi)$. Because $M_{E} A$ and $M_{E} C$ intersect $e$ at $120^{\\circ}$ angles, we have $\\overrightarrow{M_{E}} A \\cdot \\vec{e}=\\frac{1}{2} M_{E} A$ and $\\overrightarrow{M_{E} C} \\cdot \\vec{e}=\\frac{1}{2} M_{E} C$. We conclude that\n\n$$\n2 \\cos (\\varphi)(A B-C B)=M_{E} C-M_{E} A\n$$\n\nAdding the analogous equalities $2 \\cos (\\varphi)(C D-E D)=M_{A} E-M_{A} C$ and $2 \\cos (\\varphi)(E F-A F)=$ $M_{C} A-M_{C} E$, we obtain\n\n$2 \\cos (\\varphi)(A B+C D+E F-C B-E D-A F)=M_{E} C-M_{E} A+M_{A} E-M_{A} C+M_{C} A-M_{C} E$.\n\nBecause $M_{A}, M_{C}$ and $M_{E}$ form an equilateral triangle with side length $s$, we have $M_{E} C-$ $M_{A} C= \\pm s, M_{C} A-M_{E} A= \\pm s$, and $M_{A} E-M_{C} E= \\pm s$. Therefore, the right hand side $M_{E} C-M_{E} A+M_{A} E-M_{A} C+M_{C} A-M_{C} E$ equals $\\pm s \\pm s \\pm s$, which (irrespective of the choices of the \\pm -signs) is 0 if and only if $s=0$. $\\operatorname{Because} \\cos (\\varphi) \\neq 0$, we conclude that\n\n$$\nA B+C D+E F=C B+E D+A F \\Longleftrightarrow s=0 \\Longleftrightarrow a, c, e \\text { concurrent, }\n$$\n\nas desired.\n\nRemark. Equalities used in the solution could appear in different forms, in particular, in terms of signed lengths.\n\nRemark. Similar solutions could be obtained by projecting onto the line perpendicular to $b$ instead of $b$.', ""We use the the same notations as in previous solutions and the fact that $a \\| d$, $b \\| e$ and $c \\| f$ make angles of $120^{\\circ}$. Also, we may assume that $E$ and $C$ are not symmetric in $a$ (if they are, the entire figure is symmetric and the conclusion is immediate).\n\nWe consider two mappings: the first one $s: a \\rightarrow B C \\rightarrow d$ sending $A^{\\prime} \\mapsto B^{\\prime} \\mapsto S$ is defined such that $A^{\\prime} B^{\\prime} \\| A B$ and $B^{\\prime} S \\| b$, and the second one $t: a \\rightarrow E F \\rightarrow d$ sending $A^{\\prime} \\mapsto F^{\\prime} \\mapsto T$ is defined such that $A^{\\prime} F^{\\prime} \\| A F$ and $F^{\\prime} T \\| f$. Both maps are affine linear since they are compositions of affine tranformations. We will prove that they coincide by finding two distinct points $A^{\\prime}, A^{\\prime \\prime} \\in a$ for which $s\\left(A^{\\prime}\\right)=t\\left(A^{\\prime}\\right)$ and $s\\left(A^{\\prime \\prime}\\right)=t\\left(A^{\\prime \\prime}\\right)$. Then we will obtain that $s(A)=t(A)$, which by construction implies that the bisectors of $\\angle B, \\angle D$ and $\\angle F$ are concurrent.\n\nWe will choose $A^{\\prime}$ to be the reflection of $C$ in $e$ and $A^{\\prime \\prime}$ to be the reflection of $E$ in $c$. They are distinct since otherwise $C$ and $E$ would be symmetric in $a$. Applying the above maps $a \\rightarrow B C$ and $a \\rightarrow E F$ to $A^{\\prime}$, we get points $B^{\\prime}$ and $F^{\\prime}$ such that $A^{\\prime} B^{\\prime} C D E F^{\\prime}$ satisfies the problem statement. However, this hexagon is symmetric in $e$, hence the bisectors of $\\angle B^{\\prime}, \\angle D, \\angle F^{\\prime}$ are concurrent and $s\\left(A^{\\prime}\\right)=t\\left(A^{\\prime}\\right)$. The same reasoning yields $s\\left(A^{\\prime \\prime}\\right)=t\\left(A^{\\prime \\prime}\\right)$, which finishes the solution.\n\nRemark. This solution is based on the fact that that two specific affine linear maps coincide. Here it was proved by exhibiting two points where they coincide. One could prove it in another way, exhibiting one such point and proving that the 'slopes' are equal.\n\nRemark. There are similar solutions where claims and proofs could be presented in more 'elementary' terms. For example, an elementary reformulation of the 'slopes' being equal is: if $b^{\\prime}$ passes through $B^{\\prime}$ parallel to $b$, and $f^{\\prime}$ passes through $F^{\\prime}$ parallel to $f$, then the line through $b \\cap f$ and $b^{\\prime} \\cap f^{\\prime}$ is parallel to $a$ (which is parallel to $d$ ).\n\nRemark. Circles $\\omega_{a}, \\omega_{c}$ and $\\omega_{e}$ could be helpful in some other solutions. In particular, the movement of $A$ along $a$ in current solution is equivalent to varying $r_{a}$."", 'We use the same notations as in previous solutions.\n\nSince the sum of the angles of a convex hexagon is $720^{\\circ}$, from the angle conditions we get $\\angle B+\\angle C=720^{\\circ} / 3=240^{\\circ}$. From $\\angle B+\\angle C=240^{\\circ}$ it follows that the angle between $c$ and $b$ equals $60^{\\circ}$. The same is analogously true for other pairs of bisectors of neighboring angles.\n\nConsider the points $O_{a} \\in a, O_{c} \\in c, O_{e} \\in e$, each at the same distance $d^{\\prime}$ from $M$, where $d^{\\prime}>\\max \\{M A, M C, M E\\}$, and such that the rays $A O_{a}, C O_{c}, E O_{e}$ point out of the hexagon. By construcion, $O_{a}$ and $O_{c}$ are symmetrical in $e$, hence $O_{a} O_{c} \\perp b$. Similarly, $O_{c} O_{e} \\perp d, O_{e} O_{a} \\perp f$. Thus it suffices to prove that perpendiculars from $B, D, F$ to the sidelines of $\\triangle O_{a} O_{c} O_{e}$ are concurrent. By a well-known criteria, this condition is equivalent to equality\n\n$$\nO_{a} B^{2}-O_{c} B^{2}+O_{c} D^{2}-O_{e} D^{2}+O_{e} F^{2}-O_{a} F^{2}=0 \\tag{*}\n$$\n\nTo prove $(*)$ consider a circle $\\omega_{a}$ centered at $O_{a}$ and tangent to $A B$ and $A F$ and define circles $\\omega_{c}$ and $\\omega_{e}$ in the same way. Rewrite $O_{a} B^{2}$ as $r_{a}^{2}+B_{a} B^{2}$, where $r_{a}$ is the radius of $\\omega_{a}$, and $B_{a}$ is the touch point of $\\omega_{a}$ with $A B$. Using similar notation for the other tangent points, transform $(*)$ into\n\n$$\nB_{a} B^{2}-B_{c} B^{2}+D_{c} D^{2}-D_{e} D^{2}+F_{e} F^{2}-F_{a} F^{2}=0 . \\tag{**}\n$$\n\nFurthermore, $\\angle O_{c} O_{a} B_{a}=\\angle M O_{a} B_{a}+\\angle O_{c} O_{a} M=\\left(90^{\\circ}-\\varphi\\right)+30^{\\circ}=120^{\\circ}-\\varphi$, where $\\varphi=$ $\\frac{1}{2} \\angle A$. (Note that $\\varphi>30^{\\circ}$, since $A B C D E F$ is convex.) By analogous arguments, $\\angle O_{a} O_{c} B_{c}=$ $\\angle O_{e} O_{c} D_{c}=\\angle O_{c} O_{e} D_{e}=\\angle O_{a} O_{e} F_{e}=\\angle O_{e} O_{a} F_{a}=120^{\\circ}-\\varphi$. It follows that rays $O_{a} B_{a}$ and $O_{c} B_{c}$ (being symmetrical in $e$ ) intersect at $U_{e} \\in e$ forming an isosceles triangle $\\triangle O_{a} U_{e} O_{c}$. Similarly define $\\triangle O_{c} U_{a} O_{e}$ and $\\triangle O_{e} U_{c} O_{a}$. These triangles are congruent (equal bases and corresponding angles). Therefore we have $O_{a} U_{c}=U_{c} O_{e}=O_{e} U_{a}=U_{a} O_{c}=O_{c} U_{e}=U_{e} O_{a}$. Moreover, we also have $B_{a} U_{e}=O_{a} U_{e}-r_{a}=O_{a} U_{c}-r_{a}=F_{a} U_{c}=x$, and thus similarly $D_{c} U_{a}=B_{c} U_{e}=y$, $F_{e} U_{c}=D_{e} U_{a}=z$.\n\nNow from quadrilateral $B B_{a} U_{e} B_{c}$ with two opposite right angles $B_{a} B^{2}-B_{c} B^{2}=B_{c} U_{e}^{2}-B_{a} U_{e}^{2}=$ $y^{2}-x^{2}$. Similarly $D_{c} D^{2}-D_{e} D^{2}=D_{e} U_{a}^{2}-D_{c} U_{a}^{2}=z^{2}-y^{2}$ and $F_{e} F^{2}-F_{a} F^{2}=F_{a} U_{c}^{2}-F_{e} U_{c}^{2}=$ $x^{2}-z^{2}$. Finally, we substitute this into $(* *)$, and the claim is proved.']",,True,,, 2191,Combinatorics,,"A permutation of the integers $1,2, \ldots, m$ is called fresh if there exists no positive integer $k90^{\circ}$. The circumcircle $\Gamma$ of $A B C$ has radius $R$. There is a point $P$ in the interior of the line segment $A B$ such that $P B=P C$ and the length of $P A$ is $R$. The perpendicular bisector of $P B$ intersects $\Gamma$ at the points $D$ and $E$. Prove that $P$ is the incentre of triangle $C D E$.","['The angle bisector of $\\angle E C D$ intersects the circumcircle of $C D E$ (which is $\\Gamma$ ) at the midpoint $M$ of arc $D B E$. It is well-known that the incentre is the intersection of the angle bisector segment $C M$ and the circle with centre at $M$ and passing through $D, E$. We will verify this property for $P$.\n\n\n\nBy the conditions we have $A P=O A=O B=O C=O D=O E=R$. Both lines $O M$ and $A P B$ are perpendicular to $E D$, therefore $A P \\| O M$; in the quadrilateral $A O M P$ we have $A P=O A=A M=R$ and $A P \\| O M$, so $A O M P$ is a rhombus and its fourth side is $P M=R$. In the convex quadrilateral $O M B P$ we have $O M \\| P B$, so $O M B P$ is a symmetric trapezoid; the perpendicular bisector of its bases $A O$ and $P B$ coincide. From this symmetry we obtain $M D=O D=R$ and $M E=O E=R$. (Note that the triangles $O E M$ ad $O M D$ are equilateral.) We already have $M P=M D=M E=R$, so $P$ indeed lies on the circle with center $M$ and passing through $D, E$. (Notice that this circle is the reflection of $\\Gamma$ about $D E$.)\n\nFrom $P B=P C$ and $O B=O C$ we know that $B$ and $C$ are symmetrical about $O P$; from the rhombus $A O M P$ we find that $A$ and $M$ are also symmetrical about $O P$. By reflecting the collinear points $B, P, A$ (with $P$ lying in the middle) we get that $C, P, M$ are collinear (and $P$ is in the middle). Hence, $P$ lies on the line segments $C M$.', 'Let $X$ be the second intersection of $C P$ with $\\Gamma$. Using the power of the point $P$ in the circle $\\Gamma$ and the fact that $P B=P C$, we find that $P X=P A=R$. The quadrilateral $A O X P$ has four sides of equal length, so it is a rhombus and in particular $O X$ is parallel to $A P$. This proves that $O X B P$ is a trapezoid, and because the diagonals $P X$ and $O B$ have equal length, this is even an isosceles trapezoid. Because of that, $D E$ is not only the perpendicular bisector of $P B$, but also of $O X$.\n\n\n\nIn particular we have $X D=X P=X O=X E=R$, which proves that $X$ is the middle of the arc $D E$ and $P$ belongs to the circle with center $X$ going through $D$ and $E$. These properties, together with the fact that $C, P, X$ are collinear, determine uniquely the incenter of $C D E$.', 'Let $Y$ be the circumcenter of triangle $B P C$. Then from $Y B=Y P$ it follows that $Y$ lies on $D E$ (we assume $D$ lies in between $Y$ and $E$ ), and from $Y B=Y C$ it follows that $Y$ lies on $O P$, where $O$ is the center of $\\Gamma$.\n\nFrom $\\angle A O C=2 \\angle A B C=\\angle A P C$ (because $\\angle P B C=\\angle P C B)$ we deduce that $A O P C$ is a cyclic quadrilateral, and from $A P=R$ it follows that $A O P C$ is an isosceles trapezoid. We now find that $\\angle Y C P=\\angle Y P C=180^{\\circ}-\\angle O P C=180^{\\circ}-\\angle A C P$, so $Y$ lies on $A C$.\n\n\n\nPower of a point gives $Y O \\cdot Y P=Y C \\cdot Y A=Y D \\cdot Y E$, so $D, P, O$ and $E$ are concyclic. It follows that $2 \\angle D A E=\\angle D O E=\\angle D P E=\\angle D B E=180^{\\circ}-\\angle D A E$, so $\\angle D A E=60^{\\circ}$. We can now finish the proof by angle chasing.\n\nFrom $A B \\perp D E$ we have $\\angle A O D+\\angle B O E=180^{\\circ}$ and from $\\angle D O E=2 \\angle D A E=120^{\\circ}$ it follows that $\\angle B O D+\\angle B O E=120^{\\circ}$. It follows that $\\angle A O D-\\angle B O D=180^{\\circ}-120^{\\circ}=60^{\\circ}$. Let $\\angle O A B=\\angle O B A=2 \\beta$; then $\\angle A O D+\\angle B O D=\\angle A O B=180^{\\circ}-4 \\beta$. Together with $\\angle A O D-\\angle B O D=60^{\\circ}$, this yields $\\angle A O D=120^{\\circ}-2 \\beta$ and $\\angle B O D=60^{\\circ}-2 \\beta$. We now find $\\angle A E D=\\frac{1}{2} \\angle A O D=60^{\\circ}-\\beta$, which together with $\\angle D A E=60^{\\circ}$ yields $\\angle A D E=60^{\\circ}+\\beta$. From the isosceles trapezoid $A O P C$ we have $\\angle C D A=\\angle C B A=\\frac{1}{2} \\angle C P A=\\frac{1}{2} \\angle P A O=\\beta$, so $\\angle C D E=\\angle C D A+\\angle A D E=\\beta+60^{\\circ}+\\beta=60^{\\circ}+2 \\beta$.\n\nFrom $\\angle B O D=60^{\\circ}-2 \\beta$ we deduce that $\\angle B E D=30^{\\circ}-\\beta$; together with $\\angle D B E=120^{\\circ}$ this yields $\\angle E D B=30^{\\circ}+\\beta$. We now see that $\\angle P D E=\\angle B D E=30^{\\circ}+\\beta=\\frac{1}{2} \\angle C D E$, so $P$ is on the angle bisector of $\\angle C D E$. Similarly, $P$ lies on the angle bisector of $\\angle C E D$, so $P$ is the incenter of $C D E$.', 'We draw the lines $D P$ and $E P$ and let $D^{\\prime}$ resp. $E^{\\prime}$ be the second intersection point with $\\Gamma$.\n\n\n\nThe triangles $A P D^{\\prime}$ and $D P B$ are similar, and the triangles $A P E^{\\prime}$ and $E P B$ are also similar, hence they are all isosceles and it follows that $E^{\\prime}, O, P, D^{\\prime}$ lie on a circle with center $A$. In particular $A O D^{\\prime}$ and $A O E^{\\prime}$ are equilateral triangles. Angle chasing gives\n\n$$\n\\begin{gathered}\n\\angle C D P=\\angle C D D^{\\prime}=\\frac{1}{2} \\angle C O D^{\\prime}=\\frac{1}{2}\\left(60^{\\circ}+\\angle C O A\\right) \\\\\n\\angle E D P=\\angle E D D^{\\prime}=\\angle E E^{\\prime} D^{\\prime}=\\angle P E^{\\prime} D^{\\prime}=\\frac{1}{2} \\angle P A D^{\\prime}=\\frac{1}{2}\\left(60^{\\circ}+\\angle P A O\\right)\n\\end{gathered}\n$$\n\nSimilarly we prove $\\angle C E P=\\frac{1}{2}\\left(60^{\\circ}-\\angle C O A\\right)$ and $\\angle D E P=\\frac{1}{2}\\left(60^{\\circ}-\\angle P A O\\right)$ so if we can prove that $\\angle C O A=\\angle P A O$, we will have proven that $P$ belongs to the angle bisector of $\\angle C E D$ and to the angle bisector of $\\angle C D E$, which is enough to prove that $P$ is the incenter of the triangle $C D E$.\n\nLet $\\beta=\\angle A B C=\\angle P C B$. We have $\\angle A P C=2 \\beta$ and $\\angle A O C=2 \\beta$, so $A O P C$ is an inscribed quadrilateral. Moreover, since the diagonals $A P$ and $C O$ have equal length, this is actually an isosceles trapezoid, and hence $\\angle P A O=\\angle C P A=\\angle C O A$ which concludes the proof.', 'Without loss of generality we assume that $D$ and $C$ are in the same half-plane regarding line $A B$.\n\nSince $P C=B P$ and $A B C D$ is inscribed quadrilateral we have $\\angle P C B=\\angle C B P=\\angle C E A=\\alpha$. As in the other solutions, $A O P C$ is an isosceles trapezoid and $2 \\alpha=\\angle C P A=\\angle P C O$\n\n\n\nLet $K$ be intersection of $E O$ and $\\Gamma$. Then $\\angle K D E=90^{\\circ}, A B \\| D K$ and $K D B A$ is isosceles trapezoid. We obtain $D P=B D=A K$, which implies that $D P A K$ is a parallelogram and hence $D K=P A=R=O D=O K$. We see that $D O K$ is an equilateral triangle. Then $\\angle E C D=\\angle E K D=60^{\\circ}$.\n\nFurther we prove that $P C$ bisects $\\angle E C D$ using $\\angle D C B=\\angle D K B=\\angle K D A$ (from isosceles trapeziod $D K A B$ ) and that $\\angle K E C=\\angle O E C=\\angle O C E$ (from isosceles triangle $O C E)$ :\n\n$$\n\\angle D C P=\\angle D C B+\\angle B C P=\\angle K D A+\\alpha=\\angle K E C+\\angle C E A+\\alpha=\\angle O C E+2 \\alpha=\\angle P C E\n$$\n\nFurther by $\\angle D C P=\\angle P C E=\\frac{1}{2} \\angle E C D=30^{\\circ}$ distances between $P$ and sides $\\triangle C D E$ are $\\frac{1}{2} P C=\\frac{1}{2} P B$ (as ratio between cathetus and hypotenuse in right triangle with angles $60^{\\circ}$ and $30^{\\circ}$ ). So, we have found incentre.', 'Assume $A B$ is parallel to the horizontal axis, and that $\\Gamma$ is the unit circle. Write $f(\\theta)$ for the point $(\\cos (\\theta), \\sin (\\theta))$ on $\\Gamma$. As in Solution $\\mathrm{C}$, assume that $\\angle A O B=180^{\\circ}-4 \\beta$; then we can take $B=f(2 \\beta)$ and $A=f\\left(180^{\\circ}-2 \\beta\\right)$. We observe that $A O P C$ is an isosceles trapezoid, which we use to deduce that $\\angle A B C=\\frac{1}{2} \\angle A P C=\\frac{1}{2} \\angle O A B=\\beta$. We now know that $C=f\\left(180^{\\circ}-4 \\beta\\right)$.\n\nThe point $P$ lies on $A B$ with $A P=R=1$, so $P=\\left(\\cos \\left(180^{\\circ}-2 \\beta\\right)+1, \\sin (2 \\beta)\\right)=(1-$ $\\cos (2 \\beta), \\sin (2 \\beta))$. The midpoint of $B P$ therefore has coordinates $\\left(\\frac{1}{2}, \\sin (2 \\beta)\\right)$, so $D$ and $E$ have $x$-coordinate $\\frac{1}{2}$. Without loss of generality, we take $D=f\\left(60^{\\circ}\\right)$ and $E=f\\left(-60^{\\circ}\\right)$.\n\nWe have now obtained coordinates for all points in the problem, with one free parameter $(\\beta)$. To show that $P$ is the incenter of $C D E$, we will show that $P$ lies on the bisector of $\\angle C D E$; analogously, one can show that $P$ lies on the bisector of angle $C E D$. The bisector of angle $C D E$ passes through the midpoint $M$ of the arc $C E$ not containing $D E$; because $C=f\\left(180^{\\circ}-4 \\beta\\right)$ and $E=f\\left(-60^{\\circ}\\right)$, we have $M=f\\left(240^{\\circ}-2 \\beta\\right)$.\n\nIt remains to show that $P=(1-\\cos (2 \\beta), \\sin (2 \\beta))$ lies on the line connecting the points $D=$ $\\left(\\cos \\left(60^{\\circ}\\right), \\sin \\left(60^{\\circ}\\right)\\right)$ and $M=\\left(\\cos \\left(240^{\\circ}-2 \\beta\\right), \\sin \\left(240^{\\circ}-2 \\beta\\right)\\right)=\\left(-\\cos \\left(60^{\\circ}-2 \\beta\\right),-\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right)$. The equation for the line $D M$ is\n\n$\\left(Y+\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right)\\left(\\cos \\left(60^{\\circ}\\right)+\\cos \\left(60^{\\circ}-2 \\beta\\right)\\right)=\\left(\\sin \\left(60^{\\circ}\\right)+\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right)\\left(X+\\cos \\left(60^{\\circ}-2 \\beta\\right)\\right)$,\n\nwhich, using the fact that $\\cos \\left(60^{\\circ}\\right)+\\cos \\left(60^{\\circ}-2 \\beta\\right)=2 \\cos (\\beta) \\cos \\left(60^{\\circ}-\\beta\\right)$ and $\\sin \\left(60^{\\circ}\\right)+\\sin \\left(60^{\\circ}-\\right.$ $2 \\beta)=2 \\cos (\\beta) \\sin \\left(60^{\\circ}-\\beta\\right)$, simplifies to\n\n$$\n\\left(Y+\\sin \\left(60^{\\circ}-2 \\beta\\right)\\right) \\cos \\left(60^{\\circ}-\\beta\\right)=\\left(X+\\cos \\left(60^{\\circ}-2 \\beta\\right)\\right) \\sin \\left(60^{\\circ}-\\beta\\right) \\text {. }\n$$\n\nBecause $\\cos \\left(60^{\\circ}-2 \\beta\\right) \\sin \\left(60^{\\circ}-\\beta\\right)-\\sin \\left(60^{\\circ}-2 \\beta\\right) \\cos \\left(60^{\\circ}-\\beta\\right)=\\sin (\\beta)$, this equation further simplifies to\n\n$$\nY \\cos \\left(60^{\\circ}-\\beta\\right)-X \\sin \\left(60^{\\circ}-\\beta\\right)=\\sin (\\beta) .\n$$\n\n\n\nPlugging in the coordinates of $P$, i.e., $X=1-\\cos (2 \\beta)$ and $Y=\\sin (2 \\beta)$, shows that $P$ is on this line: for this choice of $X$ and $Y$, the left hand side equals $\\sin \\left(60^{\\circ}+\\beta\\right)-\\sin \\left(60^{\\circ}-\\beta\\right)=$ $2 \\cos \\left(60^{\\circ}\\right) \\sin (\\beta)$, which is indeed equal to $\\sin (\\beta)$. So $P$ lies on the bisector $D M$ of $\\angle C D E$, as desired.', 'Let $\\Gamma$ be the complex unit circle and let $A B$ be parallel with the real line and $0<\\varphi=\\arg b<\\frac{\\pi}{2}$. Then\n\n$$\n|b|=1, \\quad a=-\\bar{b}, \\quad p=a+1=1-\\bar{b}\n$$\n\nFrom $\\operatorname{Re} d=\\operatorname{Re} e=\\operatorname{Re} \\frac{p+b}{2}=\\frac{1}{2}$ we get that $d=\\frac{1}{2}+\\frac{\\sqrt{3}}{2} i$ and $e=\\frac{1}{2}-\\frac{\\sqrt{3}}{2} i$ are conjugate 6th roots of unity; $d^{3}=e^{3}=-1, d+e=1, d^{2}=-e, e^{2}=-d$ etc.\n\nPoint $C$ is the reflection of $B$ in line $O P$. From $\\arg p=\\arg (1-\\bar{b})=\\frac{1}{2}(\\pi-\\varphi)$, we can get $\\arg c=2 \\arg p-\\arg b=\\pi-2 \\varphi$, so $c=-\\bar{b}^{2}$.\n\nNow we can verify that $E P$ bisects $\\angle C E D$. This happens if and only if $(p-e)^{2}(\\bar{c}-\\bar{e})(\\bar{d}-\\bar{e})$ is real. Since $\\bar{d}-\\bar{e}=-\\sqrt{3} i$, this is equivalent with $\\operatorname{Re}\\left[(p-e)^{2}(\\bar{c}-\\bar{e})\\right]=0$. Here\n\n$$\n\\begin{aligned}\n(p-e)^{2}(\\bar{c}-\\bar{e}) & =(1-\\bar{b}-e)^{2}\\left(-b^{2}-d\\right)=(d-\\bar{b})^{2}\\left(-b^{2}-d\\right) \\\\\n& =-|b|^{4}+2 d|b|^{2} b-d^{2} b^{2}-d \\bar{b}^{2}+2 d^{2} \\bar{b}-d^{3} \\\\\n& =-1-2 d b+\\bar{d} b^{2}-d \\bar{b}^{2}-2 \\overline{d b}+1 \\\\\n& =-2(d b-\\overline{d b})+\\left(\\bar{d} b^{2}-d \\bar{b}^{2}\\right)\n\\end{aligned}\n$$\n\nwhose real part is zero. It can be proved similarly that $D P$ bisects $\\angle E D C$.']",,True,,, 2193,Number Theory,,"Let $m>1$ be an integer. A sequence $a_{1}, a_{2}, a_{3}, \ldots$ is defined by $a_{1}=a_{2}=1$, $a_{3}=4$, and for all $n \geq 4$, $$ a_{n}=m\left(a_{n-1}+a_{n-2}\right)-a_{n-3} . $$ Determine all integers $m$ such that every term of the sequence is a square.","['Consider an integer $m>1$ for which the sequence defined in the problem statement contains only perfect squares. We shall first show that $m-1$ is a power of 3 .\n\nSuppose that $m-1$ is even. Then $a_{4}=5 m-1$ should be divisible by 4 and hence $m \\equiv 1(\\bmod 4)$. But then $a_{5}=5 m^{2}+3 m-1 \\equiv 3(\\bmod 4)$ cannot be a square, a contradiction. Therefore $m-1$ is odd.\n\nSuppose that an odd prime $p \\neq 3$ divides $m-1$. Note that $a_{n}-a_{n-1} \\equiv a_{n-2}-a_{n-3}(\\bmod p)$. It follows that modulo $p$ the sequence takes the form $1,1,4,4,7,7,10,10, \\ldots$; indeed, a simple induction shows that $a_{2 k} \\equiv a_{2 k-1} \\equiv 3 k-2(\\bmod p)$ for $k \\geq 1$. Since $\\operatorname{gcd}(p, 3)=1$ we get that the sequence $a_{n}(\\bmod p)$ contains all the residues modulo $p$, a contradiction since only $(p+1) / 2$ residues modulo $p$ are squares. This shows that $m-1$ is a power of 3 .\n\nLet $h, k$ be integers such that $m=3^{k}+1$ and $a_{4}=h^{2}$. We then have $5 \\cdot 3^{k}=(h-2)(h+2)$. Since $\\operatorname{gcd}(h-2, h+2)=1$, it follows that $h-2$ equals either $1,3^{k}$ or 5 , and $h+2$ equals either $5 \\cdot 3^{k}, 5$ or $3^{k}$, respectively. In the first two cases we get $k=0$ and in the last case we get $k=2$. This implies that either $m=2$ or $m=10$.\n\nWe now show the converse. Suppose that $m=2$ or $m=10$. Let $t=1$ or $t=3$ so that $m=t^{2}+1$. Let $b_{1}, b_{2}, b_{3}, \\ldots$ be a sequence of integers defined by $b_{1}=1, b_{2}=1, b_{3}=2$, and\n\n$$\nb_{n}=t b_{n-1}+b_{n-2}, \\quad \\text { for all } n \\geq 4\n$$\n\nClearly, $a_{n}=b_{n}^{2}$ for $n=1,2,3$. Note that if $m=2$ then $a_{4}=9$ and $b_{4}=3$, and if $m=10$ then $a_{4}=49$ and $b_{4}=7$. In both the cases we have $a_{4}=b_{4}^{2}$.\n\nIf $n \\geq 5$ then we have\n\n$$\nb_{n}^{2}+b_{n-3}^{2}=\\left(t b_{n-1}+b_{n-2}\\right)^{2}+\\left(b_{n-1}-t b_{n-2}\\right)^{2}=\\left(t^{2}+1\\right)\\left(b_{n-1}^{2}+b_{n-2}^{2}\\right)=m\\left(b_{n-1}^{2}+b_{n-2}^{2}\\right) .\n$$\n\nTherefore, it follows by induction that $a_{n}=b_{n}^{2}$ for all $n \\geq 1$. This completes the solution.', 'We present an alternate proof that $m=2$ and $m=10$ are the only possible values of $m$ with the required property.\n\nNote that\n\n$$\n\\begin{aligned}\n& a_{4}=5 m-1, \\\\\n& a_{5}=5 m^{2}+3 m-1, \\\\\n& a_{6}=5 m^{3}+8 m^{2}-2 m-4 .\n\\end{aligned}\n$$\n\nSince $a_{4}$ and $a_{6}$ are squares, so is $a_{4} a_{6}$. We have\n\n$$\n4 a_{4} a_{6}=100 m^{4}+140 m^{3}-72 m^{2}-72 m+16 .\n$$\n\nNotice that\n\n$$\n\\begin{aligned}\n& \\left(10 m^{2}+7 m-7\\right)^{2}=100 m^{4}+140 m^{3}-91 m^{2}-98 m+49<4 a_{4} a_{6}, \\\\\n& \\left(10 m^{2}+7 m-5\\right)^{2}=100 m^{4}+140 m^{3}-51 m^{2}-70 m+25>4 a_{4} a_{6},\n\\end{aligned}\n$$\n\nso we must have\n\n$$\n4 a_{4} a_{6}=\\left(10 m^{2}+7 m-6\\right)^{2}=100 m^{4}+140 m^{3}-71 m^{2}-84 m+36 .\n$$\n\nThis implies that $m^{2}-12 m+20=0$, so $m=2$ or $m=10$.']","['1,2']",True,,Numerical, 2194,Geometry,,"Let $A B C$ be a triangle with $C A=C B$ and $\angle A C B=120^{\circ}$, and let $M$ be the midpoint of $A B$. Let $P$ be a variable point on the circumcircle of $A B C$, and let $Q$ be the point on the segment $C P$ such that $Q P=2 Q C$. It is given that the line through $P$ and perpendicular to $A B$ intersects the line $M Q$ at a unique point $N$. Prove that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$.","['Let $O$ be the circumcenter of $A B C$. From the assumption that $\\angle A C B=120^{\\circ}$ it follows that $M$ is the midpoint of $C O$.\n\nLet $\\omega$ denote the circle with center in $C$ and radius $C O$. This circle in the image of the circumcircle of $A B C$ through the translation that sends $O$ to $C$. We claim that $N$ lies on $\\omega$.\n\n\nLet us consider the triangles $Q N P$ and $Q M C$. The angles in $Q$ are equal. Since $N P$ is parallel to $M C$ (both lines are perpendicular to $A B$ ), it turns out that $\\angle Q N P=\\angle Q M C$, and hence the two triangles are similar. Since $Q P=2 Q C$, it follows that\n\n$$\nN P=2 M C=C O\n$$\n\nwhich proves that $N$ lies on $\\omega$.', 'Let $M^{\\prime}$ denote the symmetric of $M$ with respect to $O$.\n\nLet us consider the quadrilateral $M M^{\\prime} P N$. The lines $M M^{\\prime}$ and $N P$ are parallel by construction. Also the lines $P M^{\\prime}$ and $N M$ are parallel (homothety from $C$ with coefficient 3). It follows that $M M^{\\prime} P N$ is a parallelogram, and hence $P N=M M^{\\prime}=O C$.', 'There are many computation approaches to this problem. For example, we can set Cartesian coordinates so that\n\n$$\nA=\\left(-\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right), \\quad B=\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right), \\quad C=(0,1), \\quad M=\\left(0, \\frac{1}{2}\\right)\n$$\n\nSetting $P=(a, b)$, we obtain that $Q=(a / 3,(2+b) / 3)$. The equation of the line through $P$ and perpendicular to $A B$ is $x=a$. The equation of the line $M Q$ (if $a \\neq 0$ ) is\n\n$$\ny-\\frac{1}{2}=\\frac{x}{a}\\left(\\frac{1}{2}+b\\right)\n$$\n\nThe intersection of the two lines is therefore\n\n$$\nN=(a, 1+b)=P+(0,1)\n$$\n\nThis shows that the map $P \\rightarrow N$ in the translation by the vector $(0,1)$. This result is independent of the position of $P$ (provided that $a \\neq 0$, because otherwise $N$ is not well-defined).\n\nWhen $P$ lies on the circumcircle of $A B C$, with the exception of the two points with $a=0$, then necessarily $N$ lies on the translated circle (which is the circle with center in $C$ and radius 1).']",,True,,, 2195,Algebra,,"Consider the set $$ A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\} $$ Prove that every integer $x \geq 2$ can be written as the product of one or more elements of $A$, which are not necessarily different.","['Every integer $x \\geq 2$ can be written as the telescopic product of $x-1$ elements of $A$ as\n\n$$\nx=\\left(1+\\frac{1}{x-1}\\right) \\cdot\\left(1+\\frac{1}{x-2}\\right) \\cdot \\ldots \\cdot\\left(1+\\frac{1}{2}\\right) \\cdot\\left(1+\\frac{1}{1}\\right),\n$$\n\nwhich is enough to establish the solution.']",,True,,, 2196,Algebra,,"Consider the set $$ A=\left\{1+\frac{1}{k}: k=1,2,3, \ldots\right\} $$ For every integer $x \geq 2$, let $f(x)$ denote the minimum integer such that $x$ can be written as the product of $f(x)$ elements of $A$, which are not necessarily different. Prove that there exist infinitely many pairs $(x, y)$ of integers with $x \geq 2, y \geq 2$, and $$ f(x y)4 k$. Since $y>2^{4 k-1}$, from (Q2.1) we already know that $f(y) \\geq 4 k$, and hence we just need to exclude that $f(y)=4 k$. Let us assume that we can represent $y$ in the form $a_{1} \\cdot \\ldots \\cdot a_{4 k}$ with every $a_{i}$ in $A$. At least one of the $a_{i}$ is not 2 (otherwise the product would be a power of 2 , while $y$ is odd), and hence it is less than or equal to $3 / 2$. It follows that\n\n$$\na_{1} \\cdot \\ldots \\cdot a_{4 k} \\leq 2^{4 k-1} \\cdot \\frac{3}{2}=15 \\cdot \\frac{2^{4 k-2}}{5}<\\frac{2^{4 k+2}}{5}2 k$. Since $y>2^{2 k-1}$ (for $k \\geq 1$ ), from (Q2.1) we already know that $f(y) \\geq 2 k$, and hence we just need to exclude that $f(y)=2 k$. Let us assume that we can represent $y$ in the form $a_{1} \\cdot \\ldots \\cdot a_{2 k}$. At least one of the factors is not 2 , and hence it is less than or equal to $3 / 2$. Thus when $k$ is large enough it follows that\n\n$$\na_{1} \\cdot \\ldots \\cdot a_{2 k} \\leq 2^{2 k-1} \\cdot \\frac{3}{2}=\\frac{3}{4} \\cdot 2^{2 k}<2^{2 k}-2^{k}7$, and any other product of at most three elements of $A$ does not exceed $2^{2} \\cdot \\frac{3}{2}=6<7$. On the other hand, $f(49) \\leq 7$ since $49=2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot \\frac{3}{2} \\cdot \\frac{49}{48}$.\n\nSuppose by contradiction that there exist only finitely many pairs $(x, y)$ that satisfy $f(x y)<$ $f(x)+f(y)$. This implies that there exists $M$ large enough so that whenever $a>M$ or $b>M$ holds we have $f(a b)=f(a)+f(b)$ (indeed, it is clear that the reverse inequality $f(a b) \\leq f(a)+f(b)$ is always satisfied).\n\nNow take any pair $(x, y)$ that satisfies $f(x y)M$ be any integer. We obtain\n\n$$\nf(n)+f(x y)=f(n x y)=f(n x)+f(y)=f(n)+f(x)+f(y),\n$$\n\nwhich contradicts $f(x y)\n\nLet us show now that at most $2^{n}-n-1$ moves are possible. To this end, let us identify a line of contestants with a permutation $\\sigma$ of $\\{1, \\ldots, n\\}$. To each permutation we associate the set of reverse pairs\n\n$$\nR(\\sigma):=\\{(i, j): 1 \\leq i\\sigma(j)\\}\n$$\n\nand the nonnegative integer\n\n$$\nW(\\sigma):=\\sum_{(i, j) \\in R(\\sigma)} 2^{i}\n$$\n\nwhich we call the total weight of the permutation. We claim that the total weight decreases after any move of the contestants. Indeed, let us assume that $C_{i}$ moves forward in the queue, let $\\sigma$ be the permutation before the move, and let $\\sigma^{\\prime}$ denote the permutation after the move. Since $C_{i}$ jumps over exactly $i$ contestants, necessarily she jumps over at least one contestant $C_{j}$ with index\n\n\n\n$j>i$. This means that the pair $(i, j)$ is reverse with respect to $\\sigma$ but not with respect to $\\sigma^{\\prime}$, and this yields a reduction of $2^{i}$ in the total weight. On the other hand, the move by $C_{i}$ can create new reverse pairs of the form $(k, i)$ with $ki$.\n\nLet us show now that the process ends after a finite number of moves. Let us assume that this is not the case. Then at least one contestant moves infinitely many times. Let $i_{0}$ be the largest index such that $C_{i_{0}}$ moves infinitely many times. Then necessarily $C_{i_{0}}$ jumps infinitely many times over some fixed $C_{j_{0}}$ with $j_{0}>i_{0}$. On the other hand, we know that $C_{j_{0}}$ makes only a finite number of moves, and therefore she can precede $C_{i_{0}}$ in the line only a finite number of times, which is absurd.\n\nIn order to estimate from above the maximal number of moves, we show that the contestant $C_{i}$ can make at most $2^{n-i}-1$ moves. Indeed, let us argue by ""backward extended induction"". To begin with, we observe that the estimate is trivially true for $C_{n}$ because she has no legal move.\n\nLet us assume now that the estimate has been proved for $C_{i}, C_{i+1}, \\ldots, C_{n}$, and let us prove it for $C_{i-1}$. When $C_{i-1}$ moves, at least one contestant $C_{j}$ with $j>i-1$ must precede her in the line. The initial configuration can provide at most $n-i$ contestants with larger index in front of $C_{i-1}$, which means at most $n-i$ moves for $C_{i-1}$. All other moves are possible only if some contestant in the range $C_{i}, C_{i+1}, \\ldots, C_{n}$ jumps over $C_{i-1}$ during her moves. As a consequence, the total number of moves of $C_{i-1}$ is at most\n\n$$\nn-i+\\sum_{k=i}^{n}\\left(2^{n-k}-1\\right)=2^{n-i+1}-1\n$$\n\nSumming over all indices we obtain that\n\n$$\n\\sum_{i=1}^{n}\\left(2^{n-i}-1\\right)=2^{n}-n-1\n$$\n\nwhich gives an estimate for the total number of moves.\n\n\nComment In every move of the example, the moving contestant jumps over exactly one contestant with larger index (and as a consequence over all contestants with smaller index).']",,True,,, 2198,Combinatorics,,"The $n$ contestants of an EGMO are named $C_{1}, \ldots, C_{n}$. After the competition they queue in front of the restaurant according to the following rules. - The Jury chooses the initial order of the contestants in the queue. - Every minute, the Jury chooses an integer $i$ with $1 \leq i \leq n$. - If contestant $C_{i}$ has at least $i$ other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions. - If contestant $C_{i}$ has fewer than $i$ other contestants in front of her, the restaurant opens and the process ends. Determine for every $n$ the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves.","['The maximal number of euros is $2^{n}-n-1$.\n\nTo begin with, we show that it is possible for the Jury to collect this number of euros. We argue by induction. Let us assume that the Jury can collect $M_{n}$ euros in a configuration with $n$ contestants. Then we show that the Jury can collect at least $2 M_{n}+n$ moves in a configuration with $n+1$ contestants. Indeed, let us begin with all the contestants lined up in reverse order. In the first $M_{n}$ moves the Jury keeps $C_{n+1}$ in first position and reverses the order of the remaining contestants, then in the next $n$ moves all contestants $C_{1}, \\ldots, C_{n}$ (in this order) jump over $C_{n+1}$ and end up in the first $n$ positions of the line in reverse order, and finally in the last $M_{n}$ moves the Jury rearranges the first $n$ positions.\n\nSince $M_{1}=0$ and $M_{n+1} \\geq 2 M_{n}+n$, an easy induction shows that $M_{n} \\geq 2^{n}-n-1$.\n\n\n\nLet us show now that at most $2^{n}-n-1$ moves are possible. To this end, let us identify a line of contestants with a permutation $\\sigma$ of $\\{1, \\ldots, n\\}$. To each permutation we associate the set of reverse pairs\n\n$$\nR(\\sigma):=\\{(i, j): 1 \\leq i\\sigma(j)\\}\n$$\n\nand the nonnegative integer\n\n$$\nW(\\sigma):=\\sum_{(i, j) \\in R(\\sigma)} 2^{i}\n$$\n\nwhich we call the total weight of the permutation. We claim that the total weight decreases after any move of the contestants. Indeed, let us assume that $C_{i}$ moves forward in the queue, let $\\sigma$ be the permutation before the move, and let $\\sigma^{\\prime}$ denote the permutation after the move. Since $C_{i}$ jumps over exactly $i$ contestants, necessarily she jumps over at least one contestant $C_{j}$ with index\n\n\n\n$j>i$. This means that the pair $(i, j)$ is reverse with respect to $\\sigma$ but not with respect to $\\sigma^{\\prime}$, and this yields a reduction of $2^{i}$ in the total weight. On the other hand, the move by $C_{i}$ can create new reverse pairs of the form $(k, i)$ with $k\n\nIf $n$ is any multiple of 3 , we can obtain a balanced configuration with $k=1$ by using $n / 3$ of these $3 \\times 3$ blocks along the principal diagonal of the board.\n\nThe following diagrams show balanced configurations with $k=3$ and $n \\in\\{4,5,6,7\\}$.\n\n\nAny $n \\geq 8$ can be written in the form $4 A+r$ where $A$ is a positive integer and $r \\in\\{4,5,6,7\\}$. Therefore, we can obtain a balanced configuration with $n \\geq 8$ and $k=3$ by using one block with size $r \\times r$, and $A$ blocks with size $4 \\times 4$ along the principal diagonal of the board. In particular, this construction covers all the cases where $n$ is not a multiple of 3 .']",,True,,, 2200,Geometry,,"Let $\Gamma$ be the circumcircle of triangle $A B C$. A circle $\Omega$ is tangent to the line segment $A B$ and is tangent to $\Gamma$ at a point lying on the same side of the line $A B$ as $C$. The angle bisector of $\angle B C A$ intersects $\Omega$ at two different points $P$ and $Q$. Prove that $\angle A B P=\angle Q B C$.","['Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\\Omega$ and $\\Gamma$, and let $U$ be the intersection of $\\Omega$ and $A B$.\n\n\n\nThe proof can be divided in two steps:\n\n1. Proving that $M P \\cdot M Q=M B^{2}$.\n\nIt is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\\Omega$ to $\\Gamma$ sends $U$ to the point of $\\Gamma$ where the tangent to $\\Gamma$ is parallel to $A B$, and this point is $M$ ), and\n\n$$\nM V \\cdot M U=M A^{2}=M B^{2} .\n$$\n\nThis follows from the similitude between the triangles $\\triangle M A V$ and $\\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \\cdot M Y$ is independent of the choice of the line.\n\nComputing the power of $M$ with respect to $\\Omega$ we obtain that\n\n$$\nM P \\cdot M Q=M U \\cdot M V=M B^{2}\n$$\n\n2. Conclude the proof given that $M P \\cdot M Q=M B^{2}$.\n\nThe relation $M P \\cdot M Q=M B^{2}$ in turn implies that triangle $\\triangle M B P$ is similar to triangle $\\triangle M Q B$, and in particular $\\angle M B P=\\angle M Q B$. Keeping into account that $\\angle M C B=$ $\\angle M B A$, we finally conclude that\n\n$$\n\\angle Q B C=\\angle M Q B-\\angle M C B=\\angle M B P-\\angle M B A=\\angle P B A,\n$$\n\nas required.', 'Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\\Omega$ and $\\Gamma$, and let $U$ be the intersection of $\\Omega$ and $A B$.\n\n\n\nThe proof can be divided in two steps:\n\n1. Proving that $M P \\cdot M Q=M B^{2}$.\n\nLet us consider the inversion with respect to circle with center $M$ and radius $M A=M B$. This inversion switches $A B$ and $\\Gamma$, and fixes the line passing through $M, U, V$. As a consequence, it keeps $\\Omega$ fixed, and therefore it switches $P$ and $Q$. This is because they are the intersections between the fixed line $M C$ and $\\Omega$, and the only fixed point on the segment $M C$ is its intersection with the inversion circle (thus $P$ and $Q$ are switched). This implies that $M P \\cdot M Q=M B^{2}$.\n\n2. Conclude the proof given that $M P \\cdot M Q=M B^{2}$.\n\nThe relation $M P \\cdot M Q=M B^{2}$ in turn implies that triangle $\\triangle M B P$ is similar to triangle $\\triangle M Q B$, and in particular $\\angle M B P=\\angle M Q B$. Keeping into account that $\\angle M C B=$ $\\angle M B A$, we finally conclude that\n\n$$\n\\angle Q B C=\\angle M Q B-\\angle M C B=\\angle M B P-\\angle M B A=\\angle P B A,\n$$\n\nas required.', 'Let $M$ be the midpoint of the arc $A B$ that does not contain $C$, let $V$ be the intersection of $\\Omega$ and $\\Gamma$, and let $U$ be the intersection of $\\Omega$ and $A B$.\n\n\n\nThe proof can be divided in two steps:\n\n1. Proving that $M P \\cdot M Q=M B^{2}$.\n\nIt is well-known that $V, U$ and $M$ are collinear (indeed the homothety with center in $V$ that sends $\\Omega$ to $\\Gamma$ sends $U$ to the point of $\\Gamma$ where the tangent to $\\Gamma$ is parallel to $A B$, and this point is $M$ ), and\n\n$$\nM V \\cdot M U=M A^{2}=M B^{2} .\n$$\n\nThis follows from the similitude between the triangles $\\triangle M A V$ and $\\triangle M U A$. Alternatively, it is a consequence of the following well-known lemma: Given a circle $\\Gamma$ with a chord $A B$, let $M$ be the middle point of one of the two arcs $A B$. Take a line through $M$ which intersects $\\Gamma$ again at $X$ and $A B$ at $Y$. Then $M X \\cdot M Y$ is independent of the choice of the line.\n\nComputing the power of $M$ with respect to $\\Omega$ we obtain that\n\n$$\nM P \\cdot M Q=M U \\cdot M V=M B^{2}\n$$\n\n2. Conclude the proof given that $M P \\cdot M Q=M B^{2}$.\n\nLet $I$ and $J$ be the incenter and the $C$-excenter of $\\triangle A B C$ respectively. It is well-known that $M A=M I=M J$, therefore the relation $M P \\cdot M Q=M A^{2}$ implies that $(P, Q, I, J)=-1$.\n\nNow observe that $\\angle I B J=90^{\\circ}$, thus $B I$ is the angle bisector of $\\angle P B Q$ as it is well-known from the theory of harmonic pencils, and this leads easily to the conclusion.', 'Let $D$ denote the intersection of $A B$ and $C M$. Let us consider an inversion with respect to $B$, and let us use primes to denote corresponding points in the transformed diagram, with the gentlemen agreement that $B^{\\prime}=B$.\n\n\n\nSince inversion preserves angles, it turns out that\n\n$$\n\\angle A^{\\prime} B^{\\prime} M^{\\prime}=\\angle A^{\\prime} M^{\\prime} B^{\\prime}=\\angle A C B,\n$$\n\nand in particular triangle $A^{\\prime} B^{\\prime} M^{\\prime}$ is isosceles with basis $B^{\\prime} M^{\\prime}$.\n\nThe image of $C M$ is the circumcircle of $B^{\\prime} C^{\\prime} M^{\\prime}$, which we denote by $\\omega^{\\prime}$. It follows that the centers of both $\\omega^{\\prime}$ and the image $\\Omega^{\\prime}$ of $\\Omega$ lie on the perpendicular bisector of $B^{\\prime} M^{\\prime}$. Therefore, the whole transformed diagram is symmetric with respect to the perpendicular bisector of $B^{\\prime} M^{\\prime}$, and in particular the $\\operatorname{arcs} D^{\\prime} P^{\\prime}$ and $Q^{\\prime} C^{\\prime}$ of $\\omega^{\\prime}$ are equal.\n\nThis is enough to conclude that $\\angle D^{\\prime} B^{\\prime} P^{\\prime}=\\angle Q^{\\prime} B^{\\prime} C^{\\prime}$, which implies the conclusion.']",,True,,, 2201,Algebra,," Prove that for every real number $t$ such that $0(1+t) s_{i}$ for every $i \\in\\{1, \\ldots, n-1\\}$, then by induction we obtain that\n\n$$\ns_{n}>(1+t)^{n-1} s_{1}\n$$\n\nAs a consequence, from (Q6.1) it follows that\n\n$$\n\\left|s_{1}\\right|=s_{1}<\\frac{1}{(1+t)^{n-1}} \\cdot s_{n} \\leq t s_{n}\n$$\n\nand therefore the required inequality is satisfied with $x=s_{1}, y=s_{n}$, and $m=0$.""]",,True,,, 2202,Algebra,," Determine whether for every real number $t$ such that $0t y $$ for every pair of different elements $x$ and $y$ of $S$ and every positive integer $m$ (i.e. $m>0$ ).","['We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form\n\n$$\n\\left|\\frac{x}{y}-m\\right|>t\n$$\n\nThis is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.\n\nNow we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{n}}>t \\quad \\forall n \\geq 1 \\tag{Q6.2}\n$$\n\nand such that for every $j>i$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{j}}<\\frac{1}{2} \\quad \\text { and } \\quad t<\\left\\{\\frac{s_{j}}{s_{i}}\\right\\}<\\frac{1}{2} \\tag{Q6.3}\n$$\n\nwhere $\\{\\alpha\\}$ denotes the fractional part of $\\alpha$. This is enough to show that the set $S:=\\left\\{s_{n}: n \\geq 1\\right\\}$ has the required property.\n\nTo this end, we consider the sequence defined recursively by\n\n$$\ns_{n+1}=\\frac{\\left(s_{1} \\cdot \\ldots \\cdot s_{n}\\right)^{2}+1}{2}\n$$\n\nwith $s_{1}$ large enough. An easy induction shows that this is an increasing sequence of odd positive integers. For every $i \\in\\{1, \\ldots, n\\}$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{n+1}} \\leq \\frac{2}{s_{i}} \\leq \\frac{2}{s_{1}}<\\frac{1}{2}\n$$\n\nbecause $s_{1}$ is large enough, which proves the first relation in (Q6.3). Moreover, it turns out that\n\n$$\n\\frac{s_{n+1}}{s_{i}}=\\frac{\\left(s_{1} \\cdot \\ldots \\cdot s_{n}\\right)^{2}}{2 s_{i}}+\\frac{1}{2 s_{i}}\n$$\n\nThe first term is a positive integer plus $1 / 2$, from which it follows that the distance of $s_{n+1} / s_{i}$ from the positive integers is greater than or equal to\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{i}} \\geq \\frac{1}{2}-\\frac{1}{2 s_{1}}\n$$\n\nwhich is greater than $t$ if $s_{1}$ is large enough. This proves the second relation in (Q6.3).', 'We claim that an infinite set with the required property exists. To this end, we rewrite the required condition in the form\n\n$$\n\\left|\\frac{x}{y}-m\\right|>t\n$$\n\nThis is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.\n\nNow we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{n}}>t \\quad \\forall n \\geq 1 \\tag{Q6.2}\n$$\n\nand such that for every $j>i$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{j}}<\\frac{1}{2} \\quad \\text { and } \\quad t<\\left\\{\\frac{s_{j}}{s_{i}}\\right\\}<\\frac{1}{2} \\tag{Q6.3}\n$$\n\nwhere $\\{\\alpha\\}$ denotes the fractional part of $\\alpha$. This is enough to show that the set $S:=\\left\\{s_{n}: n \\geq 1\\right\\}$ has the required property.\n\nWe produce an increasing sequence $s_{n}$ of odd and coprime positive integers that satisfies (Q6.3) every $j>i$. As in the previous solution, this is enough to conclude.\n\nWe argue by induction. To begin with, we choose $s_{1}$ to be any odd integer satisfying the inequality in (Q6.2). Let us assume now that $s_{1}, \\ldots, s_{n}$ have already been chosen, and let us choose $s_{n+1}$ in such a way that\n\n$$\ns_{n+1} \\equiv \\frac{s_{i}-1}{2} \\quad\\left(\\bmod s_{i}\\right) \\quad \\forall i \\in\\{1, \\ldots, n\\}\n$$\n\n\n\nWe can solve this system because the previously chosen integers are odd and coprime. Moreover, any solution of this system is coprime with $s_{1}, \\ldots, s_{n}$. Indeed, for every $1 \\leq i \\leq n$ it turns out that\n\n$$\ns_{n+1}=\\frac{s_{i}-1}{2}+k_{i} s_{i}\n$$\n\nfor some positive integer $k_{i}$. Therefore, any prime $p$ that divides both $s_{n+1}$ and $s_{i}$ divides also $\\left(2 k_{i}+1\\right) s_{i}-2 s_{n+1}=1$, which is absurd. Finally, we observe that we can assume that $s_{n+1}$ is odd and large enough. In this way we can guarantee that\n\n$$\n\\frac{s_{i}}{s_{n+1}}<\\frac{1}{2} \\quad \\forall i \\in\\{1, \\ldots, n\\}\n$$\n\nwhich is the first requirement in (Q6.3), and\n\n$$\nk_{i}+tt\n$$\n\nThis is equivalent to saying that the distance between the ratio $x / y$ and the set of positive integers is greater than $t$.\n\nNow we construct an increasing sequence $s_{n}$ of odd coprime positive integers satisfying\n\n$$\n\\frac{1}{2}-\\frac{1}{2 s_{n}}>t \\quad \\forall n \\geq 1 \\tag{Q6.2}\n$$\n\nand such that for every $j>i$ it turns out that\n\n$$\n\\frac{s_{i}}{s_{j}}<\\frac{1}{2} \\quad \\text { and } \\quad t<\\left\\{\\frac{s_{j}}{s_{i}}\\right\\}<\\frac{1}{2} \\tag{Q6.3}\n$$\n\nwhere $\\{\\alpha\\}$ denotes the fractional part of $\\alpha$. This is enough to show that the set $S:=\\left\\{s_{n}: n \\geq 1\\right\\}$ has the required property.\n\nAgain we produce an increasing sequence $s_{n}$ of positive integers that satisfies (Q6.3) every $j>i$.\n\nTo this end, for every positive integer $x$, we define its security region\n\n$$\nS(x):=\\bigcup_{n \\geq 1}\\left((n+t) x,\\left(n+\\frac{1}{2}\\right) x\\right)\n$$\n\nThe security region $S(x)$ is a periodic countable union of intervals of length $\\left(\\frac{1}{2}-t\\right) x$, whose left-hand or right-hand endpoints form an arithmetic sequence. It has the property that\n\n$$\nt<\\left\\{\\frac{y}{x}\\right\\}<\\frac{1}{2} \\quad \\forall y \\in S(x)\n$$\n\nNow we prove by induction that we can choose a sequence $s_{n}$ of positive integers satisfying (Q6.3) and in addition the fact that every interval of the security region $S\\left(s_{n}\\right)$ contains at least one interval of $S\\left(s_{n-1}\\right)$.\n\nTo begin with, we choose $s_{1}$ large enough so that the length of the intervals of $S\\left(s_{1}\\right)$ is larger than 1. This guarantees that any interval of $S\\left(s_{1}\\right)$ contains at least a positive integer. Now let us choose a positive integer $s_{2} \\in S\\left(s_{1}\\right)$ that is large enough. This guarantees that $s_{1} / s_{2}$ is small enough, that the fractional part of $s_{2} / s_{1}$ is in $(t, 1 / 2)$, and that every interval of the security region $S\\left(s_{2}\\right)$ contains at least one interval of $S\\left(s_{1}\\right)$, and hence at least one positive integer.\n\nLet us now assume that $s_{1}, \\ldots, s_{n}$ have been already chosen with the required properties. We know that every interval of $S\\left(s_{n}\\right)$ contains at least one interval of $S\\left(s_{n-1}\\right)$, which in turn contains an interval in $S\\left(s_{n-2}\\right)$, and so on up to $S\\left(s_{1}\\right)$. As a consequence, we can choose a large enough positive integer $s_{n+1}$ that lies in $S\\left(s_{k}\\right)$ for every $k \\in\\{1, \\ldots, n\\}$. Since $s_{n+1}$ is large enough, we are sure that\n\n$$\n\\frac{s_{k}}{s_{n+1}}\n\nor\n\n\n\n\n\nThe first option is that $A, B$ and $D$ are covered (marked with + in top row). Then the cells inside the starting square next to $A, B$ and $D$ are covered by the dominoes, but the cell in between them has now two adjacent cells with dominoes, contradiction. The second option is that $A, B$ and $C$ are covered. Then the cells inside the given square next to $A, B$ and $C$ are covered by the dominoes. But then the cell next to $\\mathrm{B}$ has two adjacent cells with dominoes, contradiction.\n\nNow we can split the border cells along one side in groups of 4 (leaving one group of 2 if $n$ is odd). So when $n$ is even, at most $n$ of the $2 n$ border cells along one side can be covered, and when $n$ is odd, at most $n+1$ out of the $2 n$ border cells can be covered. For all four borders together, this gives a contribution of $4 n$ when $n$ is even and $4 n+4$ when $n$ is odd. Adding $4 n^{2}$ and dividing by 8 we get the desired result.', 'Consider the number of pairs of adjacent cells, such that one of them is covered by a domino. Since each cell is adjacent to one covered cell, the number of such pairs is exactly $4 n^{2}$. On the other hand, let $n_{2}$ be the number of covered corner cells, $n_{3}$ the number of covered edge cells (cells with 3 neighbours), and $n_{4}$ be the number of covered interior cells (cells with 4 neighbours). Thus the number of pairs is $2 n_{2}+3 n_{3}+4 n_{4}=4 n^{2}$, whereas the number of dominoes is $m=\\frac{n_{2}+n_{3}+n_{4}}{2}$.\n\nConsidering only the outer frame (of corner and edge cells), observe that every covered cell dominates two others, so at most half of the cells are ccovered. The frame has a total of $4(2 n-1)$ cells, i.e. $n_{2}+n_{3} \\leq 4 n-2$. Additionally $n_{2} \\leq 4$ since there are only 4 corners, thus\n\n$8 m=4 n_{2}+4 n_{3}+4 n_{4}=\\left(2 n_{2}+3 n_{3}+4 n_{4}\\right)+\\left(n_{2}+n_{3}\\right)+n_{2} \\leq 4 n^{2}+(4 n-2)+4=4 n(n+1)+2$\n\nThus $m \\leq \\frac{n(n+1)}{2}+\\frac{1}{4}$, so in fact $m \\leq \\frac{n(n+1)}{2}$.', 'We prove that this is the upper bound (and also the lower bound!) by proving that any two configurations, say $A$ and $B$, must contain exactly the same number of dominoes.\n\nColour the board in a black and white checkboard colouring. Let $W$ be the set of white cells covered by dominoes of tiling $A$. For each cell $w \\in W$ let $N_{w}$ be the set of its adjacent (necessarily black) cells. Since each black cell has exactly one neighbour (necessarily white) covered by a domino of tiling $A$, it follows that each black cell is contained in exactly one $N_{w}$, i.e. the $N_{w}$ form a partition of the black cells. Since each white cell has exactly one (necessarily black) neighbour covered by a tile of $B$, each $B_{w}$ contains exactly one black tile covered by a domino of $B$. But, since each domino covers exactly one white and one black cell, we have\n\n$$\n|A|=|W|=\\left|\\left\\{N_{w}: w \\in W\\right\\}\\right|=|B|\n$$\n\nas claimed.']",['$\\frac{n(n+1)}{2}$'],False,,Expression, 2205,Geometry,,"Let $A B C$ be a triangle such that $\angle C A B>\angle A B C$, and let $I$ be its incentre. Let $D$ be the point on segment $B C$ such that $\angle C A D=\angle A B C$. Let $\omega$ be the circle tangent to $A C$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $A B C$. Prove that the angle bisectors of $\angle D A B$ and $\angle C X B$ intersect at a point on line $B C$.","['Let $S$ be the intersection point of $B C$ and the angle bisector of $\\angle B A D$, and let $T$ be the intersection point of $B C$ and the angle bisector of $\\angle B X C$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S=T$.\n\nFirstly denote by $M$ the middle of arc $A B$ of the circumcenter of $A B C$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $M A=M I=M B$ ); let it intersect $B C$ at $B$ and $S^{\\prime}$. Since $\\angle B A C>\\angle C B A$ it is easy to check that $S^{\\prime}$ lies on side $B C$. Denoting the angles in $A B C$ by $\\alpha, \\beta, \\gamma$ we get\n\n$$\n\\angle B A D=\\angle B A C-\\angle D A C=\\alpha-\\beta .\n$$\n\nMoreover since $\\angle M B C=\\angle M B A+\\angle A B C=\\frac{\\gamma}{2}+\\beta$, then\n\n$$\n\\angle B M S^{\\prime}=180^{\\circ}-2 \\angle M B C=180^{\\circ}-\\gamma-2 \\beta=\\alpha-\\beta .\n$$\n\nIt follows that $\\angle B A S^{\\prime}=2 \\angle B M S^{\\prime}=2 \\angle B A D$ which gives us $S=S^{\\prime}$.\n\n\nSecondly let $N$ be the middle of arc $B C$ of the circumcenter of $A B C$ which does not contain $A$. From $\\angle B A C>\\angle C B A$ we conclude that $X$ lies on the arc $A B$ of circumcircle of $A B C$ not containing $C$. Obviously both $A I$ and $X T$ are passing through $N$. Since $\\angle N B T=\\frac{\\alpha}{2}=\\angle B X N$ we obtain $\\triangle N B T \\sim \\triangle N X B$, therefore\n\n$$\nN T \\cdot N X=N B^{2}=N I^{2} .\n$$\n\nIt follows that $\\triangle N T I \\sim \\triangle N I X$. Keeping in mind that $\\angle N B C=\\angle N A C=\\angle I X A$ we get\n\n$$\n\\angle T I N=\\angle I X N=\\angle N X A-\\angle I X A=\\angle N B A-\\angle N B C=\\angle T B A .\n$$\n\nIt means that $A, I, B, T$ are concyclic which ends the proof.', ""Let $\\angle B A C=\\alpha, \\angle A B C=\\beta, \\angle B C A=\\gamma, \\angle A C X=\\phi$. Denote by $W_{1}$ and $W_{2}$ the intersections of segment $B C$ with the angle bisectors of $\\angle B X C$ and $\\angle B A D$ respectively. Then $B W_{1} / W_{1} C=B X / X C$ and $B W_{2} / W_{2} D=B A / A D$. We shall show that $B W_{1}=B W_{2}$.\n\n\n\nSince $\\angle D A C=\\angle C B A$, triangles $A D C$ and $B A C$ are similar and therefore\n\n$$\n\\frac{D C}{A C}=\\frac{A C}{B C}\n$$\n\nBy the Law of sines\n\n$$\n\\frac{B W_{2}}{W_{2} D}=\\frac{B A}{A D}=\\frac{B C}{A C}=\\frac{\\sin \\alpha}{\\sin \\beta}\n$$\n\nConsequently\n\n$$\n\\begin{gathered}\n\\frac{B D}{B W_{2}}=\\frac{W_{2} D}{B W_{2}}+1=\\frac{\\sin \\beta}{\\sin \\alpha}+1 \\\\\n\\frac{B C}{B W_{2}}=\\frac{B C}{B D} \\cdot \\frac{B D}{B W_{2}}=\\frac{1}{1-D C / B C} \\cdot \\frac{B D}{B W_{2}}=\\frac{1}{1-A C^{2} / B C^{2}} \\cdot \\frac{B D}{B W_{2}}= \\\\\n\\frac{\\sin ^{2} \\alpha}{\\sin ^{2} \\alpha-\\sin ^{2} \\beta} \\cdot \\frac{\\sin \\beta+\\sin \\alpha}{\\sin \\alpha}=\\frac{\\sin \\alpha}{\\sin \\alpha-\\sin \\beta} .\n\\end{gathered}\n$$\n\nNote that $A X B C$ is cyclic and so $\\angle B X C=\\angle B A C=\\alpha$. Hence, $\\angle X B C=180^{\\circ}-$ $\\angle B X C-\\angle B C X=180^{\\circ}-\\alpha-\\phi$. By the Law of sines for the triangle $B X C$, we have\n\n$$\n\\begin{gathered}\n\\frac{B C}{W_{1} B}=\\frac{W_{1} C}{W_{1} B}+1=\\frac{C X}{B X}+1=\\frac{\\sin \\angle C B X}{\\sin \\phi}+1= \\\\\n\\frac{\\sin (\\alpha+\\phi)}{\\sin \\phi}+1=\\sin \\alpha \\cot \\phi+\\cos \\alpha+1 .\n\\end{gathered}\n$$\n\nSo, it's enough to prove that\n\n$$\n\\frac{\\sin \\alpha}{\\sin \\alpha-\\sin \\beta}=\\sin \\alpha \\cot \\phi+\\cos \\alpha\n$$\n\nSince $A C$ is tangent to the circle $A I X$, we have $\\angle A X I=\\angle I A C=\\alpha / 2$. Moreover $\\angle X A I=\\angle X A B+\\angle B A I=\\phi+\\alpha / 2$ and $\\angle X I A=180^{\\circ}-\\angle X A I-\\angle A X I=180^{\\circ}-\\alpha-\\phi$. Applying the Law of sines again $X A C, X A I, I A C$ we obtain\n\n$$\n\\begin{gathered}\n\\frac{A X}{\\sin (\\alpha+\\phi)}=\\frac{A I}{\\sin \\alpha / 2}, \\\\\n\\frac{A X}{\\sin (\\gamma-\\phi)}=\\frac{A C}{\\sin \\angle A X C}=\\frac{A C}{\\sin \\beta}, \\\\\n\\frac{A I}{\\sin \\gamma / 2}=\\frac{A C}{\\sin (\\alpha / 2+\\gamma / 2)} .\n\\end{gathered}\n$$\n\nCombining the last three equalities we end up with\n\n$$\n\\begin{aligned}\n& \\frac{\\sin (\\gamma-\\phi)}{\\sin (\\alpha+\\phi)}=\\frac{A I}{A C} \\cdot \\frac{\\sin \\beta}{\\sin \\alpha / 2}=\\frac{\\sin \\beta}{\\sin \\alpha / 2} \\cdot \\frac{\\sin \\gamma / 2}{\\sin (\\alpha / 2+\\gamma / 2)} \\\\\n& \\frac{\\sin (\\gamma-\\phi)}{\\sin (\\alpha+\\phi)}=\\frac{\\sin \\gamma \\cot \\phi-\\cos \\gamma}{\\sin \\alpha \\cot \\phi+\\cos \\alpha}=\\frac{2 \\sin \\beta / 2 \\sin \\gamma / 2}{\\sin \\alpha / 2}\n\\end{aligned}\n$$\n\n\n\n$$\n\\frac{\\sin \\alpha \\sin \\gamma \\cot \\phi-\\sin \\alpha \\cos \\gamma}{\\sin \\gamma \\sin \\alpha \\cot \\phi+\\sin \\gamma \\cos \\alpha}=\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2}{\\cos \\gamma / 2}\n$$\n\nSubtracting 1 from both sides yields\n\n$$\n\\begin{gathered}\n\\frac{-\\sin \\alpha \\cos \\gamma-\\sin \\gamma \\cos \\alpha}{\\sin \\gamma \\sin \\alpha \\cot \\phi+\\sin \\gamma \\cos \\alpha}=\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2}{\\cos \\gamma / 2}-1= \\\\\n\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2-\\sin (\\alpha / 2+\\beta / 2)}{\\cos \\gamma / 2}=\\frac{\\sin \\beta / 2 \\cos \\alpha / 2-\\sin \\alpha / 2 \\cos \\beta / 2}{\\cos \\gamma / 2}, \\\\\n\\frac{-\\sin (\\alpha+\\gamma)}{\\sin \\gamma \\sin \\alpha \\cot \\phi+\\sin \\gamma \\cos \\alpha}=\\frac{\\sin (\\beta / 2-\\alpha / 2)}{\\cos \\gamma / 2}, \\\\\n\\frac{-\\sin \\beta}{\\sin \\alpha \\cot \\phi+\\cos \\alpha}=2 \\sin \\gamma / 2 \\sin (\\beta / 2-\\alpha / 2)= \\\\\n2 \\cos (\\beta / 2+\\alpha / 2) \\sin (\\beta / 2-\\alpha / 2)=\\sin \\beta-\\sin \\alpha,\n\\end{gathered}\n$$\n\nand the result follows. We are left to note that none of the denominators can vanish."", 'We first note that\n\n$$\n\\angle B A D=\\angle B A C-\\angle D A C=\\angle A-\\angle B .\n$$\n\nLet $C X$ and $A D$ meet at $K$. Then $\\angle C X A=\\angle A B C=\\angle K A C$. Also, we have $\\angle I X A=$ $\\angle A / 2$, since $\\omega$ is tangent to $A C$ at $A$. Therefore,\n\n$$\n\\angle D A I=|\\angle B-\\angle A / 2|=|\\angle K X A-\\angle I X A|=\\angle K X I\n$$\n\n(the absolute value depends on whether $\\angle B \\geq \\angle A / 2$ or not) which means that $X K I A$ is cyclic, i.e. $K$ lies also on $\\omega$.\n\nLet $I K$ meet $B C$ at $E$. (If $\\angle B=\\angle A / 2$, then $I K$ degenerates to the tangent line to $\\omega$ at I.) Note that $B E I A$ is cyclic, because\n\n$$\n\\angle E I A=180^{\\circ}-\\angle K X A=180^{\\circ}-\\angle A B E .\n$$\n\nWe have $\\angle E K A=180^{\\circ}-\\angle A X I=180^{\\circ}-\\angle A / 2$ and $\\angle A E I=\\angle A B I=\\angle B / 2$. Hence\n\n$$\n\\begin{aligned}\n\\angle E A K & =180^{\\circ}-\\angle E K A-\\angle A E I \\\\\n& =180^{\\circ}-\\left(180^{\\circ}-\\angle A / 2\\right)-\\angle B / 2 \\\\\n& =(\\angle A-\\angle B) / 2 \\\\\n& =\\angle B A D / 2 .\n\\end{aligned}\n$$\n\nThis means that $A E$ is the angle bisector of $\\angle B A D$. Next, let $M$ be the point of intersection of $A E$ and $B I$. Then\n\n$$\n\\angle E M I=180^{\\circ}-\\angle B / 2-\\angle B A D / 2=180^{\\circ}-\\angle A / 2 \\text {, }\n$$\n\n\n\nand so, its supplement is\n\n$$\n\\angle A M I=\\angle A / 2=\\angle A X I,\n$$\n\nso $X, M, K, I, A$ all lie on $\\omega$. Next, we have\n\n$$\n\\begin{aligned}\n\\angle X M A & =\\angle X K A \\\\\n& =180^{\\circ}-\\angle A D C-\\angle X C B \\\\\n& =180^{\\circ}-\\angle A-\\angle X C B \\\\\n& =\\angle B+\\angle X C A \\\\\n& =\\angle B+\\angle X B A \\\\\n& =\\angle X B E,\n\\end{aligned}\n$$\n\nand so $X, B, E, M$ are concyclic. Hence\n\n$$\n\\begin{aligned}\n\\angle E X C & =\\angle E X M+\\angle M X C \\\\\n& =\\angle M B E+\\angle M A K \\\\\n& =\\angle B / 2+\\angle B A D / 2 \\\\\n& =\\angle A / 2 \\\\\n& =\\angle B X C / 2 .\n\\end{aligned}\n$$\n\nThis means that $X E$ is the angle bisector of $\\angle B X C$ and so we are done!\n\n\n\n', 'It is $\\angle A B D=\\angle D A C$, and so $\\overline{A C}$ is tangent to the circumcircle of $\\triangle B A D$ at $A$. Hence $C A^{2}=C D \\cdot C B$.\n\n\n\n\n\nTriangle $\\triangle A B C$ is similar to triangle $\\triangle C A D$, because $\\angle C$ is a common angle and $\\angle C A D=\\angle A B C$, and so $\\angle A D C=\\angle B A C=2 \\varphi$.\n\nLet $Q$ be the point of intersection of $\\overline{A D}$ and $\\overline{C X}$. Since $\\angle B X C=\\angle B A C=2 \\varphi$, it follows that $B D Q X$ is cyclic. Therefore, $C D \\cdot C B=C Q \\cdot C X=C A^{2}$ which implies that $Q$ lies on $\\omega$.\n\nNext let $P$ be the point of intersection of $\\overline{A D}$ with the circumcircle of triangle $\\triangle A B C$. Then $\\angle P B C=\\angle P A C=\\angle A B C=\\angle A P C$ yielding $C A=C P$. So, let $T$ be on the side $\\overline{B C}$ such that $C T=C A=C P$. Then\n\n$$\n\\angle T A D=\\angle T A C-\\angle D A C=\\left(90^{\\circ}-\\frac{\\angle C}{2}\\right)-\\angle B=\\frac{\\angle A-\\angle B}{2}=\\frac{\\angle B A D}{2}\n$$\n\nthat is, line $\\overline{A T}$ is the angle bisector of $\\angle B A D$. We want to show that $\\overline{X T}$ is the angle bisector of $\\angle B X C$. To this end, it suffices to show that $\\angle T X C=\\varphi$.\n\nIt is $C T^{2}=C A^{2}=C Q \\cdot C X$, and so $\\overline{C T}$ is tangent to the circumcircle of $\\triangle X T Q$ at $T$. Since $\\angle T X Q=\\angle Q T C$ and $\\angle Q D C=2 \\varphi$, it suffices to show that $\\angle T Q D=\\varphi$, or, in other words, that $I, Q$, and $T$ are collinear.\n\nLet $T^{\\prime}$ is the point of intersection of $\\overline{I Q}$ and $\\overline{B C}$. Then $\\triangle A I C$ is congruent to $\\triangle T^{\\prime} I C$, since they share $\\overline{C I}$ as a common side, $\\angle A C I=\\angle T^{\\prime} C I$, and\n\n$$\n\\angle I T^{\\prime} D=2 \\varphi-\\angle T^{\\prime} Q D=2 \\varphi-\\angle I Q A=2 \\varphi-\\angle I X A=\\varphi=\\angle I A C .\n$$\n\nTherefore, $C T^{\\prime}=C A=C T$, which means that $T$ coincides with $T^{\\prime}$ and completes the proof.', 'Let $G$ be the point of intersection of $\\overline{A D}$ and $\\overline{C X}$. Since the quadrilateral $A X B C$ is cyclic, it is $\\angle A X C=\\angle A B C$.\n\n\n\n\n\nLet the line $\\overline{A D}$ meet $\\omega$ at $K$. Then it is $\\angle A X K=\\angle C A D=\\angle A B C$, because the angle that is formed by a chord and a tangent to the circle at an endpoint of the chord equals the inscribed angle to that chord. Therefore, $\\angle A X K=\\angle A X C=\\angle A X G$. This means that the point $G$ coincides with the point $K$ and so $G$ belongs to the circle $\\omega$.\n\nLet $E$ be the point of intersection of the angle bisector of $\\angle D A B$ with $\\overline{B C}$. It suffices to show that\n\n$$\n\\frac{C E}{B E}=\\frac{X C}{X B}\n$$\n\nLet $F$ be the second point of intersection of $\\omega$ with $\\overline{A B}$. Then we have $\\angle I A F=\\frac{\\angle C A B}{2}=$ $\\angle I X F$, where $I$ is the incenter of $\\triangle A B C$, because $\\angle I A F$ and $\\angle I X F$ are inscribed in the same arc of $\\omega$. Thus $\\triangle A I F$ is isosceles with $A I=I F$. Since $I$ is the incenter of $\\triangle A B C$, we have $A F=2(s-a)$, where $s=(a+b+c) / 2$ is the semiperimeter of $\\triangle A B C$. Also, it is $C E=A C=b$ because in triangle $\\triangle A C E$, we have\n\n$$\n\\begin{aligned}\n\\angle A E C & =\\angle A B C+\\angle B A E \\\\\n& =\\angle A B C+\\frac{\\angle B A D}{2} \\\\\n& =\\angle A B C+\\frac{\\angle B A C-\\angle A B C}{2} \\\\\n& =90^{\\circ}-\\frac{\\angle A C E}{2},\n\\end{aligned}\n$$\n\nand so $\\angle C A E=180^{\\circ}-\\angle A E C-\\angle A C E=90^{\\circ}-\\frac{\\angle A C E}{2}=\\angle A E C$. Hence\n\n$$\nB F=B A-A F=c-2(s-a)=a-b=C B-C E=B E \\text {. }\n$$\n\nMoreover, triangle $\\triangle C A X$ is similar to triangle $\\triangle B F X$, because $\\angle A C X=\\angle F B X$ and\n\n$$\n\\angle X F B=\\angle X A F+\\angle A X F=\\angle X A F+\\angle C A F=\\angle C A X .\n$$\n\nTherefore\n\n$$\n\\frac{C E}{B E}=\\frac{A C}{B F}=\\frac{X C}{X B}\n$$\n\nas desired. The proof is complete.', ""Let $\\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $Y$ be the second point of intersection of the circle $\\omega$ with the line $A D$. Let $L$ be the intersection of $B C$ with the angle bisector of $\\angle B A D$. We will prove $\\angle L X C=$ $1 / 2 \\angle B A C=1 / 2 \\angle B X C$.\n\nWe will refer to the angles of $\\triangle A B C$ as $\\angle A, \\angle B, \\angle C$. Thus $\\angle B A D=\\angle A-\\angle B$.\n\nOn the circumcircle of $\\triangle A B C$, we have $\\angle A X C=\\angle A B C=\\angle C A D$, and since $A C$ is tangent to $\\omega$, we have $\\angle C A D=\\angle C A Y=\\angle A X Y$. Hence $C, X, Y$ are collinear.\n\nAlso note that $\\triangle C A L$ is isosceles with $\\angle C A L=\\angle C L A=\\frac{1}{2}(\\angle B A D)+\\angle A B C=\\frac{1}{2}(\\angle A+$ $\\angle B)$ hence $A C=C L$. Moreover, $C I$ is angle bisector to $\\angle A C L$ so it's the symmetry axis for the triangle, hence $\\angle I L C=\\angle I A C=1 / 2 \\angle A$ and $\\angle A L I=\\angle L I A=1 / 2 \\angle B$. Since $A C$ is tangent to $\\omega$, we have $\\angle A Y I=\\angle I A C=1 / 2 \\angle A=\\angle L A Y+\\angle A L I$. Hence $L, Y, I$ are collinear.\n\nSince $A C$ is tangent to $\\omega$, we have $\\triangle C A Y \\sim \\triangle C X A$ hence $C A^{2}=C X \\cdot C Y$. However we proved $C A=C L$ hence $C L^{2}=C X \\cdot C Y$. Hence $\\triangle C L Y \\sim \\triangle C X L$ and hence $\\angle C X L=\\angle C L Y=\\angle C A I=1 / 2 \\angle A$.\n\n"", ""Let $M$ be the midpoint of the arc $B C$. Let $\\omega$ denote the circle through $A$ and $I$ tangent to $A C$. Let $N$ be the second point of intersection of $\\omega$ with $A B$ and $L$ the intersection of $B C$ with the angle bisector of $\\angle B A D$. We know $\\frac{D L}{L B}=\\frac{A D}{A B}$ and want to prove $\\frac{X B}{X C}=\\frac{L B}{L C}$.\n\nFirst note that $\\triangle C A L$ is isosceles with $\\angle C A L=\\angle C L A=\\frac{1}{2}(\\angle B A D)+\\angle A B C$ hence $A C=C L$ and $\\frac{L B}{L C}=\\frac{L B}{A C}$.\n\nNow we calculate $\\frac{X B}{X C}$ :\n\nComparing angles on the circles $\\omega$ and the circumcircle of $\\triangle A B C$ we get $\\triangle X I N \\sim$ $\\triangle X M B$ and hence also $\\triangle X I M \\sim \\triangle X N B$ (having equal angles at $X$ and proportional adjoint sides). Hence $\\frac{X B}{X M}=\\frac{N B}{I M}$.\n\nAlso comparing angles on the circles $\\omega$ and the circumcircle of $\\triangle A B C$ and using the tangent $A C$ we get $\\triangle X A I \\sim \\triangle X C M$ and hence also $\\triangle X A C \\sim \\triangle X I M$. Hence $\\frac{X C}{X M}=$ $\\frac{A C}{I M}$.\n\nComparing the last two equations we get $\\frac{X B}{X C}=\\frac{N B}{A C}$. Comparing with $\\frac{L B}{L C}=\\frac{L B}{A C}$, it remains to prove $N B=L B$.\n\n\n\n\n\nWe prove $\\triangle I N B \\equiv \\triangle I L B$ as follows:\n\nFirst, we note that $I$ is the circumcentre of $\\triangle A L N$. Indeed, $C I$ is angle bisector in the isosceles triangle $A C L$ so it's perpendicular bisector for $A L$. As well, $\\triangle I A N$ is isosceles with $\\angle I N A=\\angle C A I=\\angle I A B$ hence $I$ is also on the perpendicular bisector of $A N$.\n\nHence $I N=I L$ and also $\\angle N I L=2 \\angle N A L=\\angle A-\\angle B=2 \\angle N I B$ (the last angle is calculated using that the exterior angle of $\\triangle N I B$ is $\\angle I N A=\\angle A / 2$. Hence $\\angle N I B=$ $\\angle L I B$ and $\\triangle I N B \\equiv \\triangle I L B$ by SAS."", 'Let $M, N$ be the midpoints of arcs $B C, B A$ of the circumcircle $A B C$, respectively. Let $Y$ be the second intersection of $A D$ and circle $A B C$. Let $E$ be the incenter of triangle $A B Y$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $Y N$ (immediate), $B C$ (since $\\angle C B Y=\\angle C A Y=\\angle C A D=\\angle A B C)$ and the angle bisector of $\\angle D A B$; so the question reduces to showing that $E$ is also on $X M$, which is the angle bisector of $\\angle C X B$.\n\nWe claim that the three lines $C X, A D Y, I E$ are concurrent at a point $D^{\\prime}$. We will complete the proof using this fact, and the proof will appear at the end (and see the solution by HEL5 for an alternative proof of this fact).\n\nTo show that $X E M$ are collinear, we construct a projective transformation which projects $M$ to $X$ through center $E$. We produce it as a composition of three other projections. Let $O$ be the intersection of lines $A D^{\\prime} D Y$ and $C I N$. Projecting the points $Y N C M$ on the circle $A B C$ through the (concyclic) point $A$ to the line $C N$ yields the points $O N C I$. Projecting these points through $E$ to the line $A Y$ yields $O Y D D^{\\prime}$ (here we use the facts that $D^{\\prime}$ lies on $I E$ and $A Y$ ). Projecting these points to the circle $A B C$ through $C$ yields $N Y B X$ (here we use the fact that $D^{\\prime}$ lies on $C X$ ). Composing, we observe that we found a projection of the circle $A B C$ to itself sending $Y N C M$ to $N Y B X$. Since the projection of the circle through $E$ also sends $Y N C$ to $N Y B$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.\n\n\n\n\n\nLet $B^{\\prime}, D^{\\prime}$ be the intersections of $A B, A D$ with the circle $A X I$, respectively. We wish to show that this $D^{\\prime}$ is the concurrency point defined above, i.e. that $C D^{\\prime} X$ and $I D^{\\prime} E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $A B^{\\prime} E$.\n\nConsider the inversion with center $C$ and radius $C A$. The circles $A X I$ and $A B D$ are tangent to $C A$ at $A$ (the former by definition, the latter since $\\angle C A D=\\angle A B C$ ), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $C A B$ to the line $A D$. Thus $X$, which is the second intersection of circles $A B C$ and $A X I$, is sent by the inversion to the second intersection of $A D$ and circle $A X I$, which is $D^{\\prime}$. In particular $C D^{\\prime} X$ are collinear.\n\nIn the circle $A I B^{\\prime}, A I$ is the angle bisector of $B^{\\prime} A$ and the tangent at $A$, so $I$ is the midpoint of the $\\operatorname{arc} A B^{\\prime}$, and in particular $A I=I B^{\\prime}$. By angle chasing, we find that $A C E$ is an isosceles triangle:\n\n$\\angle C A E=\\angle C A D+\\angle D A E=\\angle A B C+\\angle E A B=\\angle A B E+\\angle E A B=\\angle A E B=\\angle A E C$,\n\nthus the angle bisector $C I$ is the perpendicular bisector of $A E$ and $A I=I E$. Thus $I$ is the circumcenter of $A B^{\\prime} E$.\n\nWe can now show that $I D^{\\prime} E$ are collinear by angle chasing:\n\n$$\n\\angle E I B^{\\prime}=2 \\angle E A B^{\\prime}=2 \\angle E A B=\\angle D A B=\\angle D^{\\prime} A B^{\\prime}=\\angle D^{\\prime} I B^{\\prime} .\n$$', 'Let $W$ be the midpoint of $\\operatorname{arc} B C$, let $D^{\\prime}$ be the second intersection point of $A D$ and the circle $A B C$. Let $P$ be the intersection of the angle bisector $X W$ of $\\angle C X B$ with $B C$; we wish to prove that $A P$ is the angle bisector of $D A B$. Denote $\\alpha=\\frac{\\angle C A B}{2}, \\beta=\\angle A B C$.\n\nLet $M$ be the intersection of $A D$ and $X C$. Angle chasing finds:\n\n$$\n\\begin{aligned}\n\\angle M X I & =\\angle A X I-\\angle A X M=\\angle C A I-\\angle A X C=\\angle C A I-\\angle A B C=\\alpha-\\beta \\\\\n& =\\angle C A I-\\angle C A D=\\angle D A I=\\angle M A I\n\\end{aligned}\n$$\n\nAnd in particular $M$ is on $\\omega$. By angle chasing we find\n\n$$\n\\angle X I A=\\angle I X A+\\angle X A I=\\angle I C A+\\angle X A I=\\angle X A C=\\angle X B C=\\angle X B P\n$$\n\nand $\\angle P X B=\\alpha=\\angle C A I=\\angle A X I$, and it follows that $\\triangle X I A \\sim \\triangle X B P$. Let $S$ be the second intersection point of the cirumcircles of $X I A$ and $X B P$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $I S B$ and $A S P$ are collinear.\n\n\n\nLet $L$ be the second intersection of $\\omega$ and $A B$. We want to prove that $A S P$ is the angle bisector of $\\angle D A B=\\angle M A L$, i.e. that $S$ is the midpoint of the $\\operatorname{arc} M L$ of $\\omega$. And this follows easily from chasing angular arc lengths in $\\omega$ :\n\n$$\n\\begin{aligned}\n& \\overparen{A I}=\\angle C A I=\\alpha \\\\\n& \\overparen{I L}=\\angle I A L=\\alpha \\\\\n& \\overparen{M I}=\\angle M X I=\\alpha-\\beta \\\\\n& \\overparen{A I}-\\overparen{S L}=\\angle A B I=\\frac{\\beta}{2}\n\\end{aligned}\n$$\n\nAnd thus\n\n$$\n\\overparen{M L}=\\overparen{M I}+\\overparen{I L}=2 \\alpha-\\beta=2\\left(\\overparen{A I}-\\frac{\\beta}{2}\\right)=2 \\overparen{S L}\n$$\n\n', 'Let $E$ be the intersection of the bisector of $\\angle B A D$ and $B C$, and $N$ be the middle point of arc $B C$ of the circumcircle of $A B C$. Then it suffices to show that $E$ is on line $X N$.\n\nWe consider the inversion at $A$. Let $P^{*}$ be the image of a point denoted by $P$. Then $A, B^{*}, C^{*}, E^{*}$ are concyclic, $X^{*}, B^{*}, C^{*}$ are colinear, and $X^{*} I^{*}$ and $A C^{*}$ are parallel. Now it suffices to show that $A, X^{*}, E^{*}, N^{*}$ are concyclic. Let $Y$ be the intersection of $B^{*} C^{*}$ and $A E^{*}$. Then, by the power of a point, we get\n\n$$\n\\begin{aligned}\nA, X^{*}, E^{*}, N^{*} \\text { are concyclic } & \\Longleftrightarrow Y X^{*} \\cdot Y N^{*}=Y A \\cdot Y E^{*} \\\\\n& \\Longleftrightarrow Y X^{*} \\cdot Y N^{*}=Y B^{*} \\cdot Y C^{*} \\\\\n& \\left(A, B^{*}, C^{*}, E^{*} \\text { are concyclic }\\right)\n\\end{aligned}\n$$\n\nHere, by the property of inversion, we have\n\n$$\n\\angle A I^{*} B^{*}=\\angle A B I=\\frac{1}{2} \\angle A B C=\\frac{1}{2} \\angle C^{*} A D^{*} \\text {. }\n$$\n\n\n\n\n\nDefine $Q, R$ as described in the figure, and we get by simple angle chasing\n\n$$\n\\angle Q A I^{*}=\\angle Q I^{*} A, \\quad \\angle R A I^{*}=\\angle B^{*} I^{*} A\n$$\n\nEspecially, $B^{*} R$ and $A I^{*}$ are parallel, so that we have\n\n$$\n\\frac{Y B^{*}}{Y N^{*}}=\\frac{Y R}{Y A}=\\frac{Y X^{*}}{Y C^{*}}\n$$\n\nand the proof is completed.']",,True,,, 2206,Geometry,,Let $A B C$ be a triangle with incentre $I$. The circle through $B$ tangent to $A I$ at $I$ meets side $A B$ again at $P$. The circle through $C$ tangent to $A I$ at $I$ meets side $A C$ again at $Q$. Prove that $P Q$ is tangent to the incircle of $A B C$.,"['Let $Q X, P Y$ be tangent to the incircle of $A B C$, where $X, Y$ lie on the incircle and do not lie on $A C, A B$. Denote $\\angle B A C=\\alpha, \\angle C B A=\\beta, \\angle A C B=\\gamma$.\n\nSince $A I$ is tangent to the circumcircle of $C Q I$ we get $\\angle Q I A=\\angle Q C I=\\frac{\\gamma}{2}$. Thus\n\n$$\n\\angle I Q C=\\angle I A Q+\\angle Q I A=\\frac{\\alpha}{2}+\\frac{\\gamma}{2} .\n$$\n\nBy the definition of $X$ we have $\\angle I Q C=\\angle X Q I$, therefore\n\n$$\n\\angle A Q X=180^{\\circ}-\\angle X Q C=180^{\\circ}-\\alpha-\\gamma=\\beta .\n$$\n\nSimilarly one can prove that $\\angle A P Y=\\gamma$. This means that $Q, P, X, Y$ are collinear which leads us to the conclusion that $X=Y$ and $Q P$ is tangent to the incircle at $X$.\n', 'By the power of a point we have\n\n$$\nA D \\cdot A C=A I^{2}=A P \\cdot A B, \\quad \\text { which means that } \\frac{A Q}{A P}=\\frac{A B}{A C}\n$$\n\nand therefore triangles $A D P, A B C$ are similar. Let $J$ be the incenter of $A Q P$. We obtain\n\n$$\n\\angle J P Q=\\angle I C B=\\angle Q C I=\\angle Q I J,\n$$\n\nthus $J, P, I, Q$ are concyclic. Let $S$ be the intersection of $A I$ and $B C$. It follows that\n\n$$\n\\angle I Q P=\\angle I J P=\\angle S I C=\\angle I Q C .\n$$\n\nThis means that $I Q$ is the angle bisector of $\\angle C Q P$, so $Q P$ is indeed tangent to the incircle of $A B C$.', 'Like before, notice that $A Q \\cdot A C=A P \\cdot A B=A I^{2}$. Consider the positive inversion $\\Psi$ with center $A$ and power $A I^{2}$. This maps $P$ to $B$ (and vice-versa), $Q$ to $C$ (and vice-versa), and keeps the incenter $I$ fixed. The problem statement will follow from the fact that the image of the incircle of triangle $A B C$ under $\\Psi$ is the so-called mixtilinear incircle of $A B C$, which is defined to be the circle tangent to the lines $A B, A C$, and the circumcircle of $A B C$. Indeed, since the image of the line $Q P$ is the circumcircle of $A B C$, and inversion preserves tangencies, this implies that $Q P$ is tangent to the incircle of $A B C$.\n\nWe justify the claim as follows: let $\\gamma$ be the incircle of $A B C$ and let $\\Gamma_{A}$ be the $A$-mixtilinear incircle of $A B C$. Let $K$ and $L$ be the tangency points of $\\gamma$ with the sides $A B$ and $A C$, and let $U$ and $V$ be the tangency points of $\\Gamma_{A}$ with the sides $A B$ and $A C$, respectively. It is well-known that the incenter $I$ is the midpoint of segment $U V$. In particular, since also $A I \\perp U V$, this implies that $A U=A V=\\frac{A I}{\\cos \\frac{A}{2}}$. Note that $A K=A L=A I \\cdot \\cos \\frac{A}{2}$. Therefore, $A U \\cdot A K=A V \\cdot A L=A I^{2}$, which means that $U$ and $V$ are the images of $K$ and $L$ under $\\Psi$. Since $\\Gamma_{A}$ is the unique circle simultaneously tangent to $A B$ at $U$ and to $A C$ at $V$, it follows that the image of $\\gamma$ under $\\Psi$ must be precisely $\\Gamma_{A}$, as claimed.', 'From the power of a point theorem, we have\n\n$$\nA P \\cdot A B=A I^{2}=A Q \\cdot A C .\n$$\n\nHence $P B C Q$ is cyclic, and so, $\\angle A P Q=\\angle B C A$. Let $K$ be the circumcenter of $\\triangle B I P$ and let $L$ be the circumcenter of $\\triangle Q I C$. Then $\\overline{K L}$ is perpendicular to $\\overline{A I}$ at $I$.\n\nLet $N$ be the point of intersection of line $\\overline{K L}$ with $\\overline{A B}$. Then in the right triangle $\\triangle N I A$, we have $\\angle A N I=90^{\\circ}-\\frac{\\angle B A C}{2}$ and from the external angle theorem for triangle $\\triangle B N I$, we have $\\angle A N I=\\frac{\\angle A B C}{2}+\\angle N I B$. Hence\n\n$$\n\\angle N I B=\\angle A N I-\\frac{\\angle A B C}{2}=\\left(90^{\\circ}-\\frac{\\angle B A C}{2}\\right)-\\frac{\\angle A B C}{2}=\\frac{\\angle B C A}{2} .\n$$\n\nSince $M I$ is tangent to the circumcircle of $\\triangle B I P$ at $I$, we have\n\n$$\n\\angle B P I=\\angle B I M=\\angle N I M-\\angle N I B=90^{\\circ}-\\frac{\\angle B C A}{2} .\n$$\n\nAlso, since $\\angle A P Q=\\angle B C A$, we have\n\n$\\angle Q P I=180^{\\circ}-\\angle A P Q-\\angle B P I=180^{\\circ}-\\angle B C A-\\left(90^{\\circ}-\\frac{\\angle B C A}{2}\\right)=90^{\\circ}-\\frac{\\angle B C A}{2}$,\n\nas well. Hence $I$ lies on the angle bisector of $\\angle B P Q$, and so it is equidistant from its sides $\\overline{P Q}$ and $\\overline{P B}$. Therefore, the distance of $I$ from $\\overline{P Q}$ equals the inradius of $\\triangle A B C$, as desired.\n\n\n\n', 'Let $D$ be the point of intersection of $\\overline{A I}$ and $\\overline{B C}$ and let $R$ be the point of intersection of $\\overline{A I}$ and $\\overline{P Q}$. We have $\\angle R I P=\\angle P B I=\\frac{\\angle B}{2}$, $\\angle R I Q=\\angle I C Q=\\frac{\\angle C}{2}, \\angle I Q C=\\angle D I C=x$ and $\\angle B P I=\\angle B I D=\\varphi$, since $\\overline{A I}$ is tangent to both circles.\n\n\n\nFrom the angle bisector theorem, we have\n\n$$\n\\frac{R Q}{R P}=\\frac{A Q}{A P} \\quad \\text { and } \\quad \\frac{A C}{A B}=\\frac{D C}{B D}\n、\n$$\n\n\n\nSince $\\overline{A I}$ is tangent to both circles at $I$, we have $A I^{2}=A Q \\cdot A C$ and $A I^{2}=A P \\cdot A B$. Therefore,\n\n$$\n\\frac{R Q}{R P} \\cdot \\frac{D C}{B D}=\\frac{A Q \\cdot A C}{A B \\cdot A P}=1\n\\tag{1}\n$$\n\nFrom the sine law in triangles $\\triangle Q R I$ and $\\triangle P R I$, it follows that $\\frac{R Q}{\\sin \\frac{\\angle C}{2}}=\\frac{R I}{\\sin y}$ and $\\frac{R P}{\\sin \\frac{\\angle B}{2}}=\\frac{R I}{\\sin \\omega}$, respectively. Hence\n\n$$\n\\frac{R Q}{R P} \\cdot \\frac{\\sin \\frac{\\angle B}{2}}{\\sin \\frac{\\angle C}{2}}=\\frac{\\sin \\omega}{\\sin y}\n\\tag{2}\n$$\n\nSimilarly, from the sine law in triangles $\\triangle I D C$ and $\\triangle I D B$, it is $\\frac{D C}{\\sin x}=\\frac{I D}{\\sin \\frac{\\angle C}{2}}$ and $\\frac{B D}{\\sin \\varphi}=\\frac{I D}{\\sin \\frac{\\angle B}{2}}$, and so\n\n$$\n\\frac{D C}{B D} \\cdot \\frac{\\sin \\varphi}{\\sin x}=\\frac{\\sin \\frac{\\angle B}{2}}{\\sin \\frac{\\angle C}{2}}\n\\tag{3}\n$$\n\nBy multiplying equations (2) with (3), we obtain $\\frac{R Q}{R P} \\cdot \\frac{D C}{B D} \\cdot \\frac{\\sin \\varphi}{\\sin x}=\\frac{\\sin \\omega}{\\sin y}$, which combined with (1) and cross-multiplying yields\n\n$$\n\\sin \\varphi \\cdot \\sin y=\\sin \\omega \\cdot \\sin x\n\\tag{4}\n$$\n\nLet $\\theta=90^{\\circ}+\\frac{\\angle A}{2}$. Since $I$ is the incenter of $\\triangle A B C$, we have $x=90^{\\circ}+\\frac{\\angle A}{2}-\\varphi=\\theta-\\phi$. Also, in triangle $\\triangle P I Q$, we see that $\\omega+y+\\frac{\\angle B}{2}+\\frac{\\angle C}{2}=180^{\\circ}$, and so $y=\\theta-\\omega$.\n\nTherefore, equation (4) yields\n\n$$\n\\sin \\varphi \\cdot \\sin (\\theta-\\omega)=\\sin \\omega \\cdot \\sin (\\theta-\\varphi)\n$$\n\nor\n\n$$\n\\frac{1}{2}(\\cos (\\varphi-\\theta+\\omega)-\\cos (\\varphi+\\theta-\\omega))=\\frac{1}{2}(\\cos (\\omega-\\theta+\\varphi)-\\cos (\\omega+\\theta-\\varphi)),\n$$\n\nwhich is equivalent to\n\n$$\n\\cos (\\varphi+\\theta-\\omega)=\\cos (\\omega+\\theta-\\varphi) .\n$$\n\nSo\n\n$$\n\\varphi+\\theta-\\omega=2 k \\cdot 180^{\\circ} \\pm(\\omega+\\theta-\\varphi), \\quad(k \\in \\mathbb{Z} .)\n$$\n\nIf $\\varphi+\\theta-\\omega=2 k \\cdot 180^{\\circ}+(\\omega+\\theta-\\varphi)$, then $2(\\varphi-\\omega)=2 k \\cdot 180^{\\circ}$, with $|\\varphi-\\omega|<180^{\\circ}$ forcing $k=0$ and $\\varphi=\\omega$. If $\\varphi+\\theta-\\omega=2 k \\cdot 180^{\\circ}-(\\omega+\\theta-\\varphi)$, then $2 \\theta=2 k \\cdot 180^{\\circ}$, which contradicts the fact that $0^{\\circ}<\\theta<180^{\\circ}$. Hence $\\varphi=\\omega$, and so $P I$ is the angle bisector of $\\angle Q P B$.\n\nTherefore the distance of $I$ from $\\overline{P Q}$ is the same with the distance of $I$ from $A B$, which is equal to the inradius of $\\triangle A B C$. Consequently, $\\overline{P Q}$ is tangent to the incircle of $\\triangle A B C$.']",,True,,, 2207,Number Theory,,"Let $n \geq 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive integers. Show that there exist positive integers $b_{1}, b_{2}, \ldots, b_{n}$ satisfying the following three conditions: 1. $a_{i} \leq b_{i}$ for $i=1,2, \ldots, n$; 2. the remainders of $b_{1}, b_{2}, \ldots, b_{n}$ on division by $n$ are pairwise different; and 3. $b_{1}+\cdots+b_{n} \leq n\left(\frac{n-1}{2}+\left\lfloor\frac{a_{1}+\cdots+a_{n}}{n}\right\rfloor\right)$. (Here, $\lfloor x\rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)","['We define the $b_{i}$ recursively by letting $b_{i}$ be the smallest integer such that $b_{i} \\geq a_{i}$ and such that $b_{i}$ is not congruent to any of $b_{1}, \\ldots, b_{i-1}$ modulo $n$. Then $b_{i}-a_{i} \\leq i-1$, since of the $i$ consecutive integers $a_{i}, a_{i}+1, \\ldots, a_{i}+i-1$, at most $i-1$ are congruent to one of $b_{1}, \\ldots, b_{i-1}$ modulo $n$. Since all $b_{i}$ are distinct modulo $n$, we have $\\sum_{i=1}^{n} b_{i} \\equiv \\sum_{i=1}^{n}(i-1)=\\frac{1}{2} n(n-1)$ modulo $n$, so $n$ divides $\\sum_{i=1}^{n} b_{i}-\\frac{1}{2} n(n-1)$. Moreover, we have $\\sum_{i=1}^{n} b_{i}-\\sum_{i=1}^{n} a_{i} \\leq \\sum_{i=1}^{n}(i-1)=\\frac{1}{2} n(n-1)$, hence $\\sum_{i=1}^{n} b_{i}-\\frac{1}{2} n(n-1) \\leq \\sum_{i=1}^{n}$. As the left hand side is divisible by $n$, we have\n\n$$\n\\frac{1}{n}\\left(\\sum_{i=1}^{n} b_{i}-\\frac{1}{2} n(n-1)\\right) \\leq\\left[\\frac{1}{n} \\sum_{i=1}^{n} a_{i}\\right]\n$$\n\nwhich we can rewrite as\n\n$$\n\\sum_{i=1}^{n} b_{i} \\leq n\\left(\\frac{n-1}{2}+\\left[\\frac{1}{n} \\sum_{i=1}^{n} a_{i}\\right]\\right)\n$$\n\nas required.', ""Note that the problem is invariant under each of the following operations:\n\n- adding a multiple of $n$ to some $a_{i}$ (and the corresponding $b_{i}$ );\n- adding the same integer to all $a_{i}$ (and all $b_{i}$ );\n- permuting the index set $1,2, \\ldots, n$.\n\nWe may therefore remove the restriction that our $a_{i}$ and $b_{i}$ be positive.\n\nFor each congruence class $\\bar{k}$ modulo $n(\\bar{k}=\\overline{0}, \\ldots, \\overline{n-1})$, let $h(k)$ be the number of $i$ such that $a_{i}$ belongs to $\\bar{k}$. We will now show that the problem is solved if we can find a $t \\in \\mathbb{Z}$ such that\n\n$$\n\\begin{aligned}\nh(t) & \\geq 1 \\\\\nh(t)+h(t+1) & \\geq 2 \\\\\nh(t)+h(t+1)+h(t+2) & \\geq 3\n\\end{aligned}\n$$\n\nIndeed, these inequalities guarantee the existence of elements $a_{i_{1}} \\in \\bar{t}, a_{i_{2}} \\in \\bar{t} \\cup \\overline{t+1}$, $a_{i_{3}} \\in \\bar{t} \\cup \\overline{t+1} \\cup \\overline{t+2}$, et cetera, where all $i_{k}$ are different. Subtracting appropriate multiples of $n$ and reordering our elements, we may assume $a_{1}=t, a_{2} \\in\\{t, t+1\\}$, $a_{3} \\in\\{t, t+1, t+2\\}$, et cetera. Finally subtracting $t$ from the complete sequence, we may assume $a_{1}=0, a_{2} \\in\\{0,1\\}, a_{3} \\in\\{0,1,2\\}$ et cetera. Now simply setting $b_{i}=i-1$ for all $i$ suffices, since $a_{i} \\leq b_{i}$ for all $i$, the $b_{i}$ are all different modulo $n$, and\n\n$$\n\\sum_{i=1}^{n} b_{i}=\\frac{n(n-1)}{2} \\leq \\frac{n(n-1)}{2}+n\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right]\n$$\n\nPut $x_{i}=h(i)-1$ for all $i=0, \\ldots, n-1$. Note that $x_{i} \\geq-1$, because $h(i) \\geq 0$. If we have $x_{i} \\geq 0$ for all $i=0, \\ldots, n-1$, then taking $t=0$ completes the proof. Otherwise, we can pick some index $j$ such that $x_{j}=-1$. Let $y_{i}=x_{i}$ where $i=0, \\ldots, j-1, j+1, \\ldots, n-1$ and $y_{j}=0$. For sequence $\\left\\{y_{i}\\right\\}$ we have\n\n$$\n\\sum_{i=0}^{n-1} y_{i}=\\sum_{i=0}^{n-1} x_{i}+1=\\sum_{i=0}^{n-1} h(i)-n+1=1\n$$\n\nso from Raney's lemma there exists index $k$ such that $\\sum_{i=k}^{k+j} y_{i}>0$ for all $j=0, \\ldots, n-1$ where $y_{n+j}=y_{j}$ for $j=0, \\ldots, k-1$. Taking $t=k$ we will have\n\n$$\n\\sum_{t=k}^{k+i} h(t)-(i+1)=\\sum_{t=k}^{k+i} x(t) \\geq \\sum_{t=k}^{k+i} y(t)-1 \\geq 0\n$$\n\nfor all $i=0, \\ldots, n-1$ and we are done."", 'Choose a random permutation $c_{1}, \\ldots, c_{n}$ of the integers $1,2, \\ldots, n$. Let $b_{i}=a_{i}+f\\left(c_{i}-a_{i}\\right)$, where $f(x) \\in\\{0, \\ldots, n-1\\}$ denotes a remainder of $x$ modulo $n$. Observe, that for such defined sequence the first two conditions hold. The expected value of $B:=b_{1}+\\ldots+b_{n}$ is easily seen to be equal to $a_{1}+\\ldots+a_{n}+n(n-1) / 2$. Indeed, for each $i$ the random number $c_{i}-a_{i}$ has uniform distribution modulo $n$, thus the expected value of $f\\left(c_{i}-a_{i}\\right)$ is $(0+\\ldots+(n-1)) / n=(n-1) / 2$. Therefore we may find such $c$ that $B \\leq a_{1}+\\ldots+a_{n}+n(n-1) / 2$. But $B-n(n-1) / 2$ is divisible by $n$ and therefore $B \\leq n\\left[\\left(a_{1}+\\ldots+a_{n}\\right) / n\\right]+n(n-1) / 2$ as needed.', ""We will prove the required statement for all sequences of non-negative integers $a_{i}$ by induction on $n$.\n\nCase $n=1$ is obvious, just set $b_{1}=a_{1}$.\n\nNow suppose that the statement is true for some $n \\geq 1$; we shall prove it for $n+1$.\n\nFirst note that, by subtracting a multiple of $n+1$ to each $a_{i}$ and possibly rearranging indices we can reduce the problem to the case where $0 \\leq a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n} \\leq a_{n+1}<$ $n+1$.\n\nNow, by the induction hypothesis there exists a sequence $d_{1}, d_{2}, \\ldots, d_{n}$ which satisfies the properties required by the statement in relation to the numbers $a_{1}, \\ldots, a_{n}$. Set $I=\\{i \\mid 1 \\leq$ $i \\leq n$ and $\\left.d_{i} \\bmod n \\geq a_{i}\\right\\}$ and construct $b_{i}$, for $i=1, \\ldots, n+1$, as follows:\n\n$$\nb_{i}=\\left\\{\\begin{array}{l}\nd_{i} \\bmod n, \\text { when } i \\in I, \\\\\nn+1+\\left(d_{i} \\bmod n\\right), \\text { when } i \\in\\{1, \\ldots, n\\} \\backslash I, \\\\\nn, \\text { for } i=n+1\n\\end{array}\\right.\n$$\n\n\n\nNow, $a_{i} \\leq d_{i} \\bmod n \\leq b_{i}$ for $i \\in I$, while for $i \\notin I$ we have $a_{i} \\leq n \\leq b_{i}$. Thus the sequence $\\left(b_{i}\\right)_{i=1}^{n+1}$ satisfies the first condition from the problem statement.\n\nBy the induction hypothesis, the numbers $d_{i} \\bmod n$ are distinct for $i \\in\\{1, \\ldots, n\\}$, so the values $b_{i} \\bmod (n+1)$ are distinct elements of $\\{0, \\ldots, n-1\\}$ for $i \\in\\{1, \\ldots, n\\}$. Since $b_{n+1}=n$, the second condition is also satisfied.\n\nDenote $k=|I|$. We have\n\n$$\n\\begin{gathered}\n\\sum_{i=1}^{n+1} b_{i}=\\sum_{i=1}^{n} b_{i}+n=\\sum_{i=1}^{n} d_{i} \\bmod n+(n-k)(n+1)+n= \\\\\n\\frac{n(n+1)}{2}+(n-k)(n+1)\n\\end{gathered}\n$$\n\nhence we need to show that\n\n$$\n\\frac{n(n+1)}{2}+(n-k)(n+1) \\leq \\frac{n(n+1)}{2}+(n+1)\\left[\\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\\right]\n$$\n\nequivalently, that\n\n$$\nn-k \\leq\\left[\\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\\right]\n$$\n\nNext, from the induction hypothesis we have\n\n$$\n\\begin{gathered}\n\\frac{n(n-1)}{2}+n\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right] \\geq \\sum_{i=1}^{n} d_{i}=\\sum_{i \\in I} d_{i}+\\sum_{i \\notin I} d_{i} \\geq \\\\\n\\sum_{i \\in I} d_{i} \\bmod n+\\sum_{i \\notin I}\\left(n+d_{i} \\bmod n\\right)=\\frac{n(n-1)}{2}+(n-k) n\n\\end{gathered}\n$$\n\nor\n\n$$\nn-k \\leq\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right]\n$$\n\nThus, it's enough to show that\n\n$$\n\\frac{\\sum_{i=1}^{n} a_{i}}{n} \\leq \\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\n$$\n\nbecause then\n\n$$\nn-k \\leq\\left[\\frac{\\sum_{i=1}^{n} a_{i}}{n}\\right] \\leq\\left[\\frac{\\sum_{i=1}^{n+1} a_{i}}{n+1}\\right]\n$$\n\nBut the required inequality is equivalent to $\\sum_{i=1}^{n} a_{i} \\leq n a_{n+1}$, which is obvious."", ""We can assume that all $a_{i} \\in\\{0,1, \\ldots, n-1\\}$, as we can deduct $n$ from both $a_{i}$ and $b_{i}$ for arbitrary $i$ without violating any of the three conditions from the problem statement. We shall also assume that $a_{1} \\leq \\ldots \\leq a_{n}$.\n\nNow let us provide an algorithm for constructing $b_{1}, \\ldots, b_{n}$.\n\n\n\nWe start at step 1 by choosing $f(1)$ to be the maximum $i$ in $\\{1, \\ldots, n\\}$ such that $a_{i} \\leq n-1$, that is $f(1)=n$. We set $b_{f(1)}=n-1$.\n\nHaving performed steps 1 through $j$, at step $j+1$ we set $f(j+1)$ to be the maximum $i$ in $\\{1, \\ldots, n\\} \\backslash\\{f(1), \\ldots, f(j)\\}$ such that $a_{i} \\leq n-j-1$, if such an index exists. If it does, we set $b_{f(j+1)}=n-j-1$. If there is no such index, then we define $T=j$ and assign to the terms $b_{i}$, where $i \\notin f(\\{1, \\ldots, j\\})$, the values $n, n+1 \\ldots, 2 n-j-1$, in any order, thus concluding the run of our algorithm.\n\nNotice that the sequence $\\left(b_{i}\\right)_{i=1}^{n}$ satisfies the first and second required conditions by construction. We wish to show that it also satisfies the third.\n\nNotice that, since the values chosen for the $b_{i}$ 's are those from $n-T$ to $2 n-T-1$, we have\n\n$$\n\\sum_{i=1}^{n} b_{i}=\\frac{n(n-1)}{2}+(n-T) n\n$$\n\nIt therefore suffices to show that\n\n$$\n\\left[\\frac{a_{1}+\\ldots+a_{n}}{n}\\right] \\geq n-T\n$$\n\nor (since the RHS is obviously an integer) $a_{1}+\\ldots+a_{n} \\geq(n-T) n$.\n\nFirst, we show that there exists $1 \\leq i \\leq T$ such that $n-i=b_{f(i)}=a_{f(i)}$.\n\nIndeed, this is true if $a_{n}=n-1$, so we may suppose $a_{n}t$ and in fact one can show $T=t+f(t+1)$ by proceeding inductively and using the fact that $t$ is the last time for which $a_{f(t)}=b_{f(t)}$.\n\nNow we get that, since $a_{f(t+1)+1} \\geq n-t$, then $\\sum_{i} a_{i} \\geq(n-t)(n-f(t+1))=(n-T+f(t+$ 1)) $(n-f(t+1))=n(n-T)+n f(t+1)-f(t+1)(n-T+f(t+1))=n(n-T)+t f(t+1) \\geq$ $n(n-T)$.""]",,True,,, 2208,Combinatorics,,"On a circle, Alina draws 2019 chords, the endpoints of which are all different. A point is considered marked if it is either (i) one of the 4038 endpoints of a chord; or (ii) an intersection point of at least two chords. Alina labels each marked point. Of the 4038 points meeting criterion (i), Alina labels 2019 points with a 0 and the other 2019 points with a 1 . She labels each point meeting criterion (ii) with an arbitrary integer (not necessarily positive). Along each chord, Alina considers the segments connecting two consecutive marked points. (A chord with $k$ marked points has $k-1$ such segments.) She labels each such segment in yellow with the sum of the labels of its two endpoints and in blue with the absolute value of their difference. Alina finds that the $N+1$ yellow labels take each value $0,1, \ldots, N$ exactly once. Show that at least one blue label is a multiple of 3 . (A chord is a line segment joining two different points on a circle.)","['First we prove the following:\n\nLemma: if we color all of the points white or black, then the number of white-black edges, which we denote $E_{W B}$, is equal modulo 2 to the number of white (or black) points on the circumference, which we denote $C_{W}$, resp. $C_{B}$.\n\nObserve that changing the colour of any interior point does not change the parity of $E_{W B}$, as each interior point has even degree, so it suffices to show the statement holds when all interior points are black. But then $E_{W B}=C_{W}$ so certainly the parities are equal.\n\nNow returning to the original problem, assume that no two adjacent vertex labels differ by a multiple of three, and three-colour the vertices according to the residue class of the labels modulo 3. Let $E_{01}$ denote the number of edges between 0 -vertices and 1-vertices, and $C_{0}$ denote the number of 0 -vertices on the boundary, and so on.\n\nThen, consider the two-coloring obtained by combining the 1-vertices and 2-vertices. By applying the lemma, we see that $E_{01}+E_{02} \\equiv C_{0} \\bmod 2$.\n\n$$\n\\text { Similarly } E_{01}+E_{12} \\equiv C_{1}, \\quad \\text { and } E_{02}+E_{12} \\equiv C_{2}, \\quad \\bmod 2 \\text {. }\n$$\n\nUsing the fact that $C_{0}=C_{1}=2019$ and $C_{2}=0$, we deduce that either $E_{02}$ and $E_{12}$ are even and $E_{01}$ is odd; or $E_{02}$ and $E_{12}$ are odd and $E_{01}$ is even.\n\nBut if the edge labels are the first $N$ non-negative integers, then $E_{01}=E_{12}$ unless $N \\equiv 0$ modulo 3 , in which case $E_{01}=E_{02}$. So however Alina chooses the vertex labels, it is not possible that the multiset of edge labels is $\\{0, \\ldots, N\\}$.\n\nHence in fact two vertex labels must differ by a multiple of 3 .', 'As before, colour vertices based on their label modulo 3.\n\nSuppose this gives a valid 3-colouring of the graph with 2019 0s and 2019 1s on the circumference. Identify pairs of 0-labelled vertices and pairs of 1-labelled vertices on the circumference, with one 0 and one 1 left over. The resulting graph has even degrees except these two leaves. So the connected component $\\mathcal{C}$ containing these leaves has an Eulerian path, and any other component has an Eulerian cycle.\n\nLet $E_{01}^{*}$ denote the number of edges between 0 -vertices and 1-vertices in $\\mathcal{C}$, and let $E_{01}^{\\prime}$ denote the number of such edges in the other components, and so on. By studying whether a given vertex has label congruent to 0 modulo 3 or not as we go along the Eulerian path in $\\mathcal{C}$, we find $E_{01}^{*}+E_{02}^{*}$ is odd, and similarly $E_{01}^{*}+E_{12}^{*}$ is odd. Since neither start nor end vertex is a 2 -vertex, $E_{02}^{*}+E_{12}^{*}$ must be even.\n\nApplying the same argument for the Eulerian cycle in each other component and adding up, we find that $E_{01}^{\\prime}+E_{02}^{\\prime}, E_{01}^{\\prime}+E_{12}^{\\prime}, E_{02}^{\\prime}+E_{12}^{\\prime}$ are all even. So, again we find $E_{01}+E_{02}$, $E_{01}+E_{12}$ are odd, and $E_{02}+E_{12}$ is even, and we finish as in the original solution.']",,True,,, 2209,Geometry,,"Let $A B C$ be an acute-angled triangle in which $B C\n\nWe show that $T$ and $H$ are both on the angle bisector $\\ell$ of $\\angle B S C$.\n\nWe first prove that $H \\in \\ell$. The altitude $C H$ in triangle $A B C$ is also the altitude in isosceles triangle $P B C$ with $C P=C B$. Therefore, $C H$ is also the angle bisector of $\\angle P C B$ and hence also of $\\angle S C B$. Analogously, $B H$ is the angle bisector of $\\angle S B C$. We conclude that $H$, as the intersection of two angle bisectors of $\\triangle B S C$, is also on the third angle bisector, which is $\\ell$.\n\nWe now prove that $T \\in \\ell$.\n\nVariant 1. In the isosceles triangles $\\triangle B C P$ and $\\triangle C B Q$ we see that $\\angle B C P=180^{\\circ}-2 \\angle B$ and $\\angle C B Q=180^{\\circ}-2 \\angle C$. This yields $\\angle P S Q=\\angle B S C=180^{\\circ}-\\left(180^{\\circ}-2 \\angle B\\right)-\\left(180^{\\circ}-2 \\angle C\\right)=$ $180^{\\circ}-2 \\angle A$. Furthermore, $\\angle P T Q=2 \\angle P A Q=2 \\angle A$ ( $T$ being circumcentre of $\\left.\\triangle A P Q\\right)$. Now $\\angle P T Q+\\angle P S Q=180^{\\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\\angle P S T=\\angle P Q T=\\angle Q P T=\\angle Q S T$, so $T$ is on the angle bisector $\\angle P S Q$, which is also $\\ell$.\n\nWe conclude that $T, S$ and $H$ are collinear.\n\nVariant 2. Let $R$ be the second intersection of $B Q$ and $\\odot A P Q$.\n\n$A P R Q$ is cyclic quadrilateral, so $\\angle P R S=\\angle A, \\angle A P R=\\angle B Q C=\\angle C$. On the other hand $\\angle B P C=\\angle B$. Therefore $\\angle R P S=180^{\\circ}-\\angle B-\\angle C=\\angle A$. Hence, the triangle $P R S$ is isosceles with $S P=S R$; then $\\ell$ is the perpendicular bisector of the chord $P R$ in the circle that passes through $T$.\n\nNote that if $B Q$ is tangent to $\\odot A P Q$, then $C P$ is also tangent to $\\odot A P Q$ and triangle $A B C$ is isosceles, so $T, S$ and $H$ lie on the altitude from $A$.\n\n\n\n', '\n\nVariant 1. In the isosceles triangles $\\triangle B C P$ and $\\triangle C B Q$ we see that $\\angle B C P=180^{\\circ}-2 \\angle B$ and $\\angle C B Q=180^{\\circ}-2 \\angle C$. This yields $\\angle P S Q=\\angle B S C=180^{\\circ}-\\left(180^{\\circ}-2 \\angle B\\right)-\\left(180^{\\circ}-2 \\angle C\\right)=$ $180^{\\circ}-2 \\angle A$. Furthermore, $\\angle P T Q=2 \\angle P A Q=2 \\angle A$ ( $T$ being circumcentre of $\\left.\\triangle A P Q\\right)$. Now $\\angle P T Q+\\angle P S Q=180^{\\circ}$, so $P T Q S$ is a cyclic quadrilateral. From $P T=T Q$ we then obtain that $\\angle P S T=\\angle P Q T=\\angle Q P T=\\angle Q S T$, so $T$ is on the angle bisector $\\angle P S Q$, which is also $\\ell$.\nFrom above, we see that $\\angle P S Q=180^{\\circ}-2 \\angle A$, so $\\angle C S Q=2 \\angle A$. From the cyclic quadrilateral $A E H D$ (with $E$ and $D$ feet of the altitudes $C H$ and $B H$ ) we see that $\\angle D H C=\\angle D A E=\\angle A$. Since $B H$ is the perpendicular bisector of $C Q$, we have $\\angle D H Q=\\angle A$ as well, so $\\angle C H Q=2 \\angle A$. From $\\angle C H Q=2 \\angle A=\\angle C S Q$, we see $C H S Q$ is a cyclic quadrilateral. This means $\\angle Q H S=\\angle Q C S$.\n\nSince triangles $P T Q$ and $C H Q$ are both isosceles with apex $2 \\angle A$, we get $\\triangle P T Q \\sim \\triangle C H Q$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\\triangle Q T H \\sim \\triangle Q P C$. This means that $\\angle Q H T=\\angle Q C P$. Combining this with $\\angle Q H S=\\angle Q C S$, we see that $\\angle Q H T=\\angle Q C P=\\angle Q C S=\\angle Q H S$. So $\\angle Q H T=\\angle Q H S$, which means that $T, S$ and $H$ are collinear.', '\n\nLet us draw a parallel $f$ to $B C$ through $A$. Let $B^{\\prime}=B Q \\cap f, C^{\\prime}=C P \\cap f$. Then $A Q B^{\\prime} \\sim C Q B$ and $A P C^{\\prime} \\sim B P C$, therefore both $Q A B^{\\prime}$ and $A P C^{\\prime}$ will be isosceles.\n\nAlso, $B C S \\sim B^{\\prime} C^{\\prime} S$ with respect to the similarity with center $S$, therefore if we take the image of the line $B H$ (which is the perpendicular bisector of the segment $C Q$ ) through this transformation, it will go through the point $B^{\\prime}$, and be perpendicular to $A Q$ (as the image of $C Q$ is parallel to $C Q$ ). As $A Q B^{\\prime}$ is isosceles, this line is the perpendicular bisector of the segment $A Q$. This means it goes through $T$, the circumcenter of $A P Q$. Similarly on the other side the image of $C H$ also goes through $T$. This means that the image of $H$ with respect to the similarity though $S$ will be $T$, so $T, S, H$ are collinear.']",,True,,, 2210,Algebra,,"Let $\mathbb{N}=\{1,2,3, \ldots\}$ be the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold: (1) $f(a b)=f(a) f(b)$, and (2) at least two of the numbers $f(a), f(b)$ and $f(a+b)$ are equal.","['First, all such functions $f$ satisfy the conditions, as $v_{p}(a) \\neq v_{p}(b)$ implies $v_{p}(a+b)=$ $\\min \\left(v_{p}(a), v_{p}(b)\\right)$. Plugging $a, b=1$ into (1) gives $f(1)=1$. Also, a simple induction gives that $f\\left(\\prod_{i=1}^{k} p_{i}^{a_{i}}\\right)=\\prod_{i=1}^{k} f\\left(p_{i}\\right)^{a_{i}}$. Let $S$ be the set of all primes $p$ such that $f(p) \\neq 1$. If $|S|=1$, then $S=\\{p\\}$ and $f(p)=a$ for some $a \\neq 1$, and thus $f(n)=a^{v_{p}(n)}$ for all $n$. If $S=\\emptyset$ then $f(n)=1$ for all $n$, which can be also written in the above form for $a=1$.\n\nNow suppose that $S$ contains at least two primes. Let $pq$, and let $p^{t}=a q+b$ with $0 \\leq bp^{t-1}$, so $a1$ and $f(q)>1$, and these are the two smallest such primes.\n\n$11$ and $f(q)>1$ are equal, so $f(q)=f(p)=a$.\n\nIf $p^{2}1$ for some $m \\in \\mathbb{N}$, then there are less than $m$ different prime numbers $p_{i}$ with $f\\left(p_{i}\\right)>1$.\n\nProof. On the one hand, if $f(m)=c>1$, then for any $n \\in \\mathbb{N}$, there is some $k \\in\\{n+1, n+2 \\ldots, n+m\\}$ such that $f(k) \\in\\{1, c\\}$. To show this, it is enough to see that condition (2) implies that for any $n \\in \\mathbb{N}$\n\n- if $f(n)=c$, then $f(n+1) \\in\\{1, c\\}$;\n- if $f(n)=1$, then $f(n+m) \\in\\{1, c\\}$.\n\nOn the other hand, if we suppose that there are $m$ different prime numbers $p_{1}, p_{2}, \\ldots, p_{m}$, then we know (by the Chinese Remainder Theorem) that for any number $t \\in \\mathbb{N}$, there exists a number $n_{t} \\in \\mathbb{N}$ such that $p_{k}^{t} \\mid n_{t}+k$ for each $k \\in\\{1,2, \\ldots, m\\}$. Hence for each $k$, we have\n\n$$\nf\\left(n_{t}+k\\right) \\geq f\\left(p_{i}^{t}\\right)=f\\left(p_{i}\\right)^{t} \\geq 2^{t}\n$$\n\nIf $2^{t}>c$, then it means that $f(k)>c$ for all $k \\in\\left\\{n_{t}+1, n_{t}+2 \\ldots, n_{t}+m\\right\\}$.\n\nNow we can suppose that for some $m \\geq 2$, the set of primes with $f(p)>1$ is $S=\\left\\{p_{1}, p_{2}, \\ldots, p_{m}\\right\\}$. Now let $S_{1}$ and $S_{2}$ be two disjoint nonempty subsets such that $S_{1} \\cup S_{2}=S$. Let $a=\\prod_{p_{i} \\in S_{1}} p_{i}$ and $b=\\prod_{p_{j} \\in S_{2}} p_{j}$.\n\nNow for any $k, \\ell \\in \\mathbb{N}, a^{k}+b^{\\ell}$ is coprime to all primes in $S$, hence $f\\left(a^{k}+b^{\\ell}\\right)=1$. It is easy to see that we can choose $k$ and $\\ell$ such that $f\\left(a^{k}\\right) \\neq f\\left(b^{\\ell}\\right)$, which contradicts condition (2).', ""Let $p$ the minimal (prime) number with $f(p)=c>1$.\n\nClaim 4.1. For any $k \\in \\mathbb{N}, f\\left(1+p+p^{2}+\\ldots+p^{k}\\right) \\in\\left\\{1, c, c^{2}, \\ldots, c^{k}\\right\\}$.\n\nProof of the claim. Use condition (2) recursively.\n\nNow suppose that there exists some prime $q>p$ with $f(q)>1$, and consider the following three values of $f$ :\n\n- $f(1)=1$.\n- $f\\left(p^{q-1}-1\\right)=f(p-1) f\\left(p^{q-2}+p^{q-3}+\\ldots+p+1\\right) \\in\\left\\{1, c, c^{2}, \\ldots, c^{q-2}\\right\\}$ using our claim, (note that we know $f(p-1)=1$ by the minimality of $p$ ).\n\nBut we also know (by Fermat's little theorem) that $q \\mid p^{q-1}-1$, hence $f(q) \\mid f\\left(p^{q-1}-1\\right.$ ). Therefore we have that: $f\\left(p^{q-1}-1\\right) \\in\\left\\{c, c^{2}, \\ldots, c^{q-2}\\right\\}$.\n\n- $f\\left(p^{q-1}\\right)=c^{q-1}$.\n\nThese are three different values, hence we have a contradiction by condition (2)."", ""Let $p$ the minimal (prime) number with $f(p)=c>1$.\n\nLemma 4B.1. For any $k \\in \\mathbb{N}, f\\left(1+p+p^{2}+\\ldots+p^{k}\\right)=1$.\n\nProof of the Lemma, by induction. $k=0$ is trivial.\n\n$k=1$ is also easy, as $p+1$ cannot be a prime, except in the case of $p=2$, which is also easy. (If $f(2)=c>1$, then $f(3) \\in\\{1, c\\}$ by $3=1+2$ and $f(3) \\in\\left\\{1, c^{2}\\right\\}$ by $2^{2}=1+3$.)\n\nFrom now we suppose $k \\geq 2$ and $f\\left(1+p+\\ldots+p^{k-1}\\right)=1$. By condition (2), we have that:\n\n- As $f\\left(1+p+\\ldots+p^{k-1}\\right)=1$ and $f\\left(p^{k}\\right)=c^{k}$,\n\nwe have that $f\\left(1+p+\\ldots+p^{k-1}\\right) \\in\\left\\{1, c^{k}\\right\\}$\n\n- As $f(1)=1$ and $f\\left(p+\\ldots+p^{k}\\right)=f(p) f\\left(1+p+\\ldots+p^{k-1}\\right)=c$,\n\nwe also have that $f\\left(1+p+\\ldots+p^{k-1}\\right) \\in\\{1, c\\}$.\n\nIf $k>1, c \\neq c^{k}$, therefore $f\\left(1+p+\\ldots+p^{k}\\right)=1$.\n\nNow suppose that there exists some prime $q>p$ with $f(q)>1$. Then by condition (1),\n\n$$\nf\\left(p^{q-1}-1\\right)=f(p-1) f\\left(1+p+\\ldots+p^{q-2}\\right)\n$$\n\nhere $f(p-1)=1$ (by the minimality of $p$ ) and $f\\left(1+p+\\ldots+p^{q-2}\\right)=1$ by the lemma. But we also know (by Fermat's little theorem) that $q \\mid p^{q-1}-1$, hence $1m \\in \\mathbb{N}$ ).\n\nProof of the Lemma. By condition (2), $\\{f(n+m), f(n-m)\\} \\subseteq\\{f(n), f(m)\\}$. We also have that:\n\n$$\nf(n-m) f(n+m)=f\\left(n^{2}-m^{2}\\right) \\in\\left\\{f(n)^{2}, f(m)^{2}\\right\\}\n$$\n\nwhich is only possible if $f(n-m)=f(n+m)=f(n)$ or $f(n-m)=f(n+m)=f(m)$.\n\nNow let $p$ be the smallest (prime) number with $f(p)>1$.\n\nClaim 5.2. If $p$ does not divide $n$, then $f(n)=1$.\n\nProof of the Claim. Let $n$ be the minimal counterexample, and and let $n=d p+r$ with $0b_{n-1}=c_{n-1}-c_{n-2}\n$$\n\nFactoring,\n\n$$\n\\left(m_{n}-m_{n-1}\\right)\\left(m_{n}+m_{n-1}\\right)>\\left(m_{n-1}-m_{n-2}\\right)\\left(m_{n-1}+m_{n-2}\\right) .\n$$\n\nIn the case that $m_{n}-m_{n-1}=m_{n-1}-m_{n-2}$, this inequality clearly holds, as $m_{n}+m_{n-1}>m_{n-1}+m_{n-2}$. Thus, the minimal possible value of $m_{n}$ satisfies the claimed inequality.\n\nProof 2. Denote $c_{n-1}=x^{2}, c_{n-2}=(x-d)^{2}$. Then\n\n$$\n2 c_{n-1}-c_{n-2}=2 x^{2}-(x-d)^{2}=x^{2}+2 d x-d^{2}=(x+d)^{2}-2 d^{2}<(x+d)^{2} .\n$$\n\nIt follows that $c_{n} \\leq(x+d)^{2}$.\n\nAnd so the sequence of positive integers $\\sqrt{c_{n}}-\\sqrt{c_{n-1}}$ is decreasing. Any such sequence is eventually constant.\n\nAs a corollary, $c_{n}=(x+n d)^{2}$ for sufficiently large $n$, with fixed integers $x$ and $d$. Along the lines above, it becomes clear that $a_{n}=c_{n}-2 c_{n-1}+c_{n-2}=2 d^{2}$, so the sequence $\\left(a_{n}\\right)$ is constant.', 'We write:\n\n$$\ns_{n}^{2}=S_{n}=a_{1}+\\left(a_{1}+a_{2}\\right)+\\ldots+\\left(a_{1}+\\ldots+a_{n}\\right)\n$$\n\nSo, setting $b_{n}:=a_{1}+\\ldots+a_{n}$, we have $S_{n}=b_{1}+b_{2}+\\ldots+b_{n}$ and, in particular $S_{n+1}=S_{n}+b_{n+1}$. Now, we study the quantity $S_{n} b_{n}=+b_{1}+b_{2}+\\ldots+b_{n}+b_{n}$ in two different ways. Since $b_{n+1}$ is the smallest integer strictly greater than $b_{n}$ such that $b_{1}+\\ldots+b_{n}+b_{n+1}$ is a perfect square, we must have\n\n$$\nb_{1}+b_{2}+\\ldots+b_{n}+b_{n} \\geq\\left(s_{n+1}-1\\right)^{2}\n$$\n\nHowever, we also have\n\n$$\nb_{1}+b_{2}+\\ldots+b_{n}+b_{n}=S_{n}+b_{n}=2 S_{n}-S_{n-1} \\text {. }\n$$\n\nCombining, we obtain\n\n$$\ns_{n}^{2} \\geq \\frac{s_{n-1}^{2}+\\left(s_{n+1}-1\\right)^{2}}{2}>\\left(\\frac{s_{n-1}+s_{n+1}-1}{2}\\right)^{2}\n$$\n\nwhere the final inequality is strict since the sequence $\\left(s_{k}\\right)$ is strictly increasing. Taking a square root, and noting that all the $\\left(s_{n}\\right)$ are integers, one obtains $s_{n+1}-s_{n} \\leq s_{n}-s_{n-1}$.\n\nNow we focus on the sequence $d_{n}=s_{n+1}-s_{n}$. $\\left(s_{k}\\right)$ is strictly increasing thus $\\left(d_{k}\\right)$ is positive. However we proved that $d_{n+1} \\leq d_{n}$, so the sequence $\\left(d_{k}\\right)$ is eventually constant, so eventually $s_{n}=b n+c$ and $S_{n}=(b n+c)^{2}$ with some numbers $b, c$; then\n\n$$\na_{n+2}=S_{n+2}-2 S_{n+1}+S_{n}=(b(n+2)+c)^{2}-2(b(n+1)+c)^{2}+(b n+c)^{2}=2 b^{2} .\n$$']",,True,,, 2212,Algebra,,"Given a positive integer $n \geq 2$, determine the largest positive integer $N$ for which there exist $N+1$ real numbers $a_{0}, a_{1}, \ldots, a_{N}$ such that (1) $a_{0}+a_{1}=-\frac{1}{n}$, and (2) $\left(a_{k}+a_{k-1}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1}$ for $1 \leq k \leq N-1$.","[""$\\left(a_{k}+a_{k-1}\\right)\\left(a_{k}+a_{k+1}\\right)=a_{k-1}-a_{k+1}$ is equivalent to $\\left(a_{k}+a_{k-1}+1\\right)\\left(a_{k}+a_{k+1}-1\\right)=-1$. Let $b_{k}=a_{k}+a_{k+1}$. Thus we need $b_{0}, b_{1}, \\ldots$ the following way: $b_{0}=-\\frac{1}{n}$ and $\\left(b_{k-1}+1\\right)\\left(b_{k}-1\\right)=-1$. There is a proper sequence $b_{0}, b_{1}, \\ldots, b_{N-1}$ if and only if there is proper sequence $a_{0}, a_{1}, \\ldots, a_{N}$, because from a a proper $\\left(a_{k}\\right)$ sequence we can get a proper $\\left(b_{k}\\right)$ sequence with $b_{k}=a_{k}+a_{k+1}$ for $k=0,1, \\ldots, N-1$ and from a proper $\\left(b_{k}\\right)$ sequence we can get a proper $\\left(a_{k}\\right)$ sequence by arbitrarily setting $a_{0}$ and then inductively defining $a_{k}=b_{k-1}-a_{k-1}$ for $k=1,2, \\ldots, N$.\n\nWe prove by induction that $b_{k}=-\\frac{1}{n-k}$ for $k\n\n\n\nLet $A B$ and $C D$ meet at $Q$, and let $B C$ and $A D$ meet at $R$. Without loss of generality, suppose that $B$ lies between $A$ and $Q$, and $D$ lies between $A$ and $R$. Now we show that line $Q R$ is the radical axis between circles $\\Omega$ and $\\omega$.\n\nPoint $W$ is the intersection point of the bisectors of $\\angle A$ and $\\angle D$, so in triangle $A D Q$, point $W$ is the incentre. Similary, $B Y$ and $C Y$ are the external bisectors of $\\angle Q B C$ and $\\angle B C Q$, so in triangle $B C Q$, point $Y$ is the excentre opposite to $Q$. Hence, both $Y$ and $W$ lie on the bisector of $\\angle D Q A$.\n\nBy $\\angle D Q A=180^{\\circ}-\\angle D-\\angle A=\\angle B-\\angle A$ we get\n\n$$\n\\angle B Y Q=\\angle Y B A-\\angle Y Q B=\\frac{\\angle B}{2}-\\frac{\\angle D Q A}{2}=\\frac{\\angle B}{2}-\\frac{\\angle B-\\angle A}{2}=\\frac{\\angle A}{2}=\\angle B A W\n$$\n\nso the points $A, B, Y, W$ are concyclic, and therefore $Q A \\cdot Q B=Q Y \\cdot Q W$; hence, $Q$ has equal power with respect to $\\Omega$ and $\\omega$. It can be seen similarly that $R$ has equal power with respect to the circles $\\Omega$ and $\\omega$. Hence, the radical axis of $\\Omega$ and $\\omega$ is line $Q R$.\n\nLet lines $O P$ and $Q R$ meet at point $S$. It is well-known that with respect to circle $A B C D$, the diagonal triangle $P Q R$ is autopolar. As consequences we have $O P \\perp Q R$, and the points $P$ and $S$ are symmetric to circle $A B C D$, so $O S \\cdot O P=r^{2}$.\n\nNotice that $P$ and $O$ lie inside $\\Omega$, so the polar line $Q R$ lies entirely outside, so $S$ is different from $P$ and $O$. Moreover,\n\n$$\nS O \\cdot S P=O S \\cdot(O S-O P)=O S^{2}-O S \\cdot O P=S O^{2}-r^{2}\n$$\n\nso $S O \\cdot S P$ is equal to the power of $S$ with respect to $\\Omega$. Since $S$ lies on the radical axis $Q R$, it has equal power with respect to the two circles; therefore, $S O \\cdot S P$ is equal to the power of $S$ with respect to $\\omega$. From this, it follows that $O \\in \\omega$ if and only if $P \\in \\omega$.', 'We will prove that the points $X, Y, Z, W, O, P$ lie on a conic section $\\Gamma$. Five distinct points uniquely determine a conic section, therefore\n\n$X, Y, Z, W, O$ are concyclic $\\Leftrightarrow \\Gamma$ is a circle $\\Leftrightarrow X, Y, Z, W, P$ are concyclic.\n\n\n\n\n\nLet $\\alpha=\\angle A C B=\\angle A D B, \\beta=\\angle B D C=\\angle B A C, \\gamma=\\angle C A D=\\angle C B D$ and $\\delta=\\angle D B A=$ $\\angle D C A$, where of course $\\alpha+\\beta+\\gamma+\\delta=180^{\\circ}$. For every point $\\xi$ in the plane, let $a(\\xi), b(\\xi), c(\\xi)$ and $d(\\xi)$ be the signed distances between $\\xi$ and the lines $A B, B C, C D$ and $D A$, respectively, such that the quadrilateral lies on the positive sides of these lines.\n\nThe equations of the bisectors of $\\angle A X W, \\angle B X Y, \\angle C Y Z$ and $\\angle D Z W$ are $a(\\xi)-d(\\xi)=0$, $b(\\xi)-a(\\xi)=0, c(\\xi)-b(\\xi)=0$ and $d(\\xi)-c(\\xi)=0$, respectively. Notice that the points $X, Y, Z, W$ satisfy the quadratic equations $(a(\\xi)-d(\\xi))(b(\\xi)-c(\\xi))=0$ and $(a(\\xi)-b(\\xi))(c(\\xi)-d(\\xi))=0$, so\n\n$$\na(\\xi) b(\\xi)+c(\\xi) d(\\xi)=a(\\xi) c(\\xi)+b(\\xi) d(\\xi)=a(\\xi) d(\\xi)+b(\\xi) c(\\xi) \\quad \\text { for } \\xi=X, Y, Z, W \\tag{1}\n$$\n\nWithout loss of generality, suppose that $A B C D$ is inscribed in a unit circle. Then\n\n$$\na(O)=\\cos \\alpha, \\quad b(O)=\\cos \\beta, \\quad c(O)=\\cos \\gamma, \\quad d(O)=\\cos \\delta \\tag{2a}\n$$\n\nBy $\\frac{a(P)}{b(P)}=\\frac{B P \\cdot \\sin \\delta}{B P \\cdot \\sin \\gamma}=\\frac{1 / \\sin \\gamma}{1 / \\sin \\delta}$ and the analogous relations we can see that\n\n$$\na(P)=\\frac{k}{\\sin \\gamma}, \\quad b(P)=\\frac{k}{\\sin \\delta}, \\quad c(P)=\\frac{k}{\\sin \\alpha}, \\quad d(P)=\\frac{k}{\\sin \\beta} \\tag{2b}\n$$\n\nwith some positive number $k$.\n\nNow let $s=\\frac{\\sin \\alpha \\sin \\beta \\sin \\gamma \\sin \\delta}{k^{2}}$; then, by $(2 a)$ and $(2 b)$,\n\n$$\n\\begin{aligned}\na(O) b(O)-s \\cdot a(P) b(P) & =\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta=\\cos (\\alpha+\\beta) \\\\\nc(O) d(O)-s \\cdot c(P) d(P) & =\\cos (\\gamma+\\delta)=-\\cos (\\alpha+\\beta) \\\\\na(O) b(O)+c(O) d(O) & =s \\cdot(a(P) b(P)+c(P) d(P))\n\\end{aligned} \\tag{3a}\n$$\n\nanalogous calculation provides\n\n$$\n\\begin{aligned}\n& a(O) c(O)+b(O) d(O)=s \\cdot(a(P) c(P)+b(P) d(P))\n\\end{aligned} \\tag{3b}\n$$\n$$\n\\begin{aligned}\n& a(O) d(O)+b(O) c(O)=s \\cdot(a(P) d(P)+b(P) c(P))\n\\end{aligned} \\tag{3c}\n$$\n\nIn order to find the equation of the curve $\\Gamma$, choose real numbers $u, v, w$, not all zero, such that\n\n$$\n\\begin{gathered}\nu+v+w=0\n\\end{gathered}\n\\tag{4}\n$$\n$$\n\\begin{gathered}\nu \\cdot(a(P) b(P)+c(P) d(P))+v \\cdot(a(P) c(P)+b(P) d(P))+w \\cdot(a(P) d(P)+b(P) c(P))=0\n\\end{gathered} \\tag{5}\n$$\n\nThis is always possible, because this a system of two homogeneous linear equations with three variables. Then the equation of $\\Gamma$ will be\n\n$$\nf(\\xi)=u \\cdot(a(\\xi) b(\\xi)+c(\\xi) d(\\xi))+v \\cdot(a(\\xi) c(\\xi)+b(\\xi) d(\\xi))+w \\cdot(a(\\xi) d(\\xi)+b(\\xi) c(\\xi))=0 \\tag{*}\n$$\n\nAs can be seen, $f(X)=f(Y)=f(Z)=f(W)=0$ follows from (1) and (4), $f(P)=0$ follows from (5), then $f(O)=s \\cdot f(P)=0$ follows from $(3 a-3 c)$. So, the points $X, Y, Z, W, O, P$ all satisfy the equation $f(\\xi)=0$.\n\nNotice that $f(B)=u \\cdot c(B) d(B)$ and $f(A)=w \\cdot b(A) c(A)$; since at least one of $u$ and $w$ is nonzero, either $A$ or $B$ does not satisfy $(*)$. Therefore, the equation cannot degenerate to an identity.\n\nHence, the equation $f(\\xi)=0$ is at most quadratic, it is not an identity, but satisfied by $X, Y, Z, W, O, P$, so these six points lie on a conic section.', ""Let $A^{\\prime}, B^{\\prime}, C^{\\prime}$ and $D^{\\prime}$ be the second intersection of the angle bisectors of $\\angle D A C$, $\\angle A B C, \\angle B C D$ and $\\angle C D A$ with the circumcircle of $A B C D$, respectively. Notice that $A^{\\prime}$ and $C^{\\prime}$ are the midpoints of the two arcs of $B D$ and $B^{\\prime}$ and $D^{\\prime}$ are the midpoints of arcs corresponding to $A C$ so the lines $A^{\\prime} C^{\\prime}$ and $B^{\\prime} D^{\\prime}$ intersect in $O$.\n\n\n\nWe will prove that the points $X, Y, Z, W, P$ and $O$ lie on a conic section. This is a more general statement because five points uniquely determine a conic sections so if $X, Y, Z, W$ and $P$ lie on a circle then this circle is the conic section so $O$ also lies on this circle. In the same way if $X, Y, Z$, $W$ and $O$ are on a circle then it also contains $P$.\n\nLet $E$ be the intersection of the lines $A^{\\prime} B$ and $C D^{\\prime}$. Let $F$ be the intersection of the lines $A^{\\prime} D$ and $B^{\\prime} C$. Finally, let $G$ be the intesection of $X Y$ and $Z W$.\n\nNow we use Pascal's theorem five times:\n\n1. From $B D D^{\\prime} C A A^{\\prime}$ we see that the points $P, W$ and $E$ are collinear.\n2. From $B B^{\\prime} D^{\\prime} C C^{\\prime} A^{\\prime}$ we see that the points $Y, O$ and $E$ are collinear.\n3. From $B^{\\prime} B D A^{\\prime} A C$ we see that the points $X, P$ and $F$ are collinear.\n4. From $B^{\\prime} D^{\\prime} D A^{\\prime} C^{\\prime} C$ we see that the points $O, Z$ and $F$ are collinear.\n5. From $D^{\\prime} C B^{\\prime} B A^{\\prime} D$ we see that the points $E, F$ and $G$ are collinear.\n\nSo the opposite sides of the hexagon $P X Y O Z W$ intersect at points $F, G$ and $E$, which are collinear, so by the converse of Pascal's theorem the points $X, Y, Z, W, P$ and $O$ indeed lie on a conic section.""]",,True,,, 2215,Geometry,,"The side $B C$ of the triangle $A B C$ is extended beyond $C$ to $D$ so that $C D=B C$. The side $C A$ is extended beyond $A$ to $E$ so that $A E=2 C A$. Prove that if $A D=B E$, then the triangle $A B C$ is right-angled.","['Define $F$ so that $A B F D$ is a parallelogram. Then $E, A, C, F$ are collinear (as diagonals of a parallelogram bisect each other) and $B F=A D=B E$. Further, $A$ is the midpoint of $E F$, since $A F=2 A C$, and thus $A B$ is an altitude of the isosceles triangle $E B F$ with apex $B$. Therefore $A B \\perp A C$.\n\n', 'Notice that $A$ is the centroid of triangle $B D E$, since $C$ is the midpoint of $[B D]$ and $A E=2 C A$. Let $M$ be the midpoint of $[B E]$. Then $M, A, D$ lie on a line, and further, $A M=\\frac{1}{2} A D=\\frac{1}{2} B E$. This implies that $\\angle E A B=90^{\\circ}$.\n\n\n\n', 'Let $P$ be the midpoint $[A E]$. Since $C$ is the midpoint of $[B D]$, and, moreover, $A C=E P$, we have\n\n$$\n[A C D]=[A B C]=[E B P] .\n$$\n\nBut $A D=B E$, and, as mentioned previously, $A C=E P$, so this implies that\n\n$$\n\\angle B E P=\\angle C A D \\text { or } \\angle B E P=180^{\\circ}-\\angle C A D \\text {. }\n$$\n\nBut $\\angle B E P<\\angle B A C$ and $\\angle B A C+\\angle C A D=\\angle B A D<180^{\\circ}$, so we must be in the first case, i.e. $\\angle B E P=\\angle C A D$. It follows that triangles $B E P$ and $D A C$ are congruent, and thus $\\angle B P A=\\angle A C B$. But $A P=A C$, so $B A$ is a median of the isosceles triangle $B C P$. Thus $A B \\perp P C$, completing the proof.\n\n', 'Write $\\beta=\\angle E C B$, and let $x=A C, y=B C=C D, z=B E=A D$. Notice that $E C=3 x$. Then, using the cosine theorem,\n\n$$\n\\begin{array}{ll}\nz^{2}=x^{2}+y^{2}+2 x y \\cos \\beta & \\text { in triangle } A C D \\\\\nz^{2}=9 x^{2}+y^{2}-6 x y \\cos \\beta & \\text { in triangle } B C E\n\\end{array}\n$$\n\nHence $4 z^{2}=12 x^{2}+4 y^{2}$ or $z^{2}-y^{2}=3 x^{2}$. Let $H$ be the foot of the perpendicular through $B$ to $A C$, and write $h=B H$. Then\n\n$$\ny^{2}-h^{2}=C H^{2}, z^{2}-h^{2}=E H^{2} .\n$$\n\n\n\nHence $z^{2}-y^{2}=E H^{2}-C H^{2}$. Substituting from the above,\n\n$$\nE H^{2}-C H^{2}=3 x^{2}=E A^{2}-C A^{2}\n$$\n\nThus $H=A$, and hence the triangle $A B C$ is right-angled at $A$.', 'Writing $a=B C, b=C A, c=A B$, we have\n\n$$\n\\begin{array}{rlrl}\na^{2} & =b^{2}+c^{2}-2 b c \\cos \\angle A & & \\text { in triangle } A B C ;\\\\\nc^{2} & =a^{2}+b^{2}-2 a b \\cos \\angle C & & \\text { in triangle } A B C ;\\\\\nE B^{2} & =4 b^{2}+c^{2}+4 b c \\cos \\angle A & & \\text { in triangle } A E B ;\\\\\nA D^{2} & =a^{2}+b^{2}+2 a b \\cos \\angle C \\quad & & \\text { in triangle } A C D ;\n\\end{array}\n$$\n\nThus\n\n$$\n\\begin{aligned}\n6 b^{2}+3 c^{2}-2 a^{2} & =4 b^{2}+c^{2}+4 b c \\cos \\angle A=E B^{2}=A D^{2} \\\\\n& =a^{2}+b^{2}+2 a b \\cos \\angle C=2 a^{2}+2 b^{2}-c^{2}\n\\end{aligned}\n$$\n\nwhich gives $a^{2}=b^{2}+c^{2}$. Therefore $\\angle B A C$ is a right angle by the converse of the theorem of Pythagoras.', 'Let $\\overrightarrow{A C}=\\vec{x}$ and $\\overrightarrow{A B}=\\vec{y}$. Now $\\overrightarrow{A D}=2 \\vec{x}-\\vec{y}$ and $\\overrightarrow{E B}=2 \\vec{x}+\\vec{y}$. Then\n\n$$\n\\overrightarrow{B E} \\cdot \\overrightarrow{B E}=\\overrightarrow{A D} \\cdot \\overrightarrow{A D} \\quad \\Longleftrightarrow \\quad(2 \\vec{x}+\\vec{y})^{2}=(2 \\vec{x}-\\vec{y})^{2} \\Longleftrightarrow \\vec{x} \\cdot \\vec{y}=0\n$$\n\nand thus $A C \\perp A B$, whence triangle $A B C$ is right-angled at $A$.', 'Let $a, b, c, d$, $e$ denote the complex co-ordinates of the points $A, B, C, D$, $E$ and take the unit circle to be the circumcircle of $A B C$. We have\n\n$$\nd=b+2(c-b)=2 c-b \\quad \\text { and } \\quad e=c+3(a-c)=3 a-2 c \\text {. }\n$$\n\nThus $b-e=(d-a)+2(b-a)$, and hence\n\n$$\n\\begin{aligned}\nB E=A D & \\Longleftrightarrow(b-e)(\\overline{b-e})=(d-a)(\\overline{d-a}) \\\\\n& \\Longleftrightarrow 2(d-a)(\\overline{b-a})+2(\\overline{d-a})(b-a)+4(b-a)(\\overline{b-a})=0 \\\\\n& \\Longleftrightarrow 2(d-a)(a-b)+2(\\overline{d-a})(b-a) a b+4(b-a)(a-b)=0 \\\\\n& \\Longleftrightarrow(d-a)-(\\overline{d-a}) a b+2(b-a)=0 \\\\\n& \\Longleftrightarrow 2 c-b-a-2 \\bar{c} a b+a+b+2(b-a)=0 \\\\\n& \\Longleftrightarrow c^{2}-a b+b c-a c=0 \\\\\n& \\Longleftrightarrow(b+c)(c-a)=0,\n\\end{aligned}\n$$\n\nimplying $c=-b$ and that triangle $A B C$ is right-angled at $A$.', 'We use areal co-ordinates with reference to the triangle $A B C$. Recall that if $\\left(x_{1}, y_{1}, z_{1}\\right)$ and $\\left(x_{2}, y_{2}, z_{2}\\right)$ are points in the plane, then the square of the distance between these two points is $-a^{2} v w-b^{2} w u-c^{2} u v$, where $(u, v, w)=\\left(x_{1}-x_{2}, y_{1}-y_{2}, z_{1}-z_{2}\\right)$.\n\nIn our case $A=(1,0,0), B=(0,1,0), C=(0,0,1)$, so $E=(3,0,2)$ and, introducing point $F$ as in the first solution, $F=(-1,0,2)$. Then\n\n$$\nB E^{2}=A D^{2} \\Longleftrightarrow-2 a^{2}+6 b^{2}+3 c^{2}=2 a^{2}+2 b^{2}-c^{2},\n$$\n\nand thus $a^{2}=b^{2}+c^{2}$. Therefore $\\angle B A C$ is a right angle by the converse of the theorem of Pythagoras.']",,True,,, 2216,Combinatorics,,"Determine all integers $m$ for which the $m \times m$ square can be dissected into five rectangles, the side lengths of which are the integers $1,2,3, \ldots, 10$ in some order.","['The solution naturally divides into three different parts: we first obtain some bounds on $m$. We then describe the structure of possible dissections, and finally, we deal with the few remaining cases.\n\nIn the first part of the solution, we get rid of the cases with $m \\leqslant 10$ or $m \\geqslant 14$. Let $\\ell_{1}, \\ldots, \\ell_{5}$ and $w_{1}, \\ldots, w_{5}$ be the lengths and widths of the five rectangles. Then the rearrangement inequality yields the lower bound\n\n$$\n\\begin{aligned}\n\\ell_{1} w_{1} & +\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5} \\\\\n& =\\frac{1}{2}\\left(\\ell_{1} w_{1}+\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5}+w_{1} \\ell_{1}+w_{2} \\ell_{2}+w_{3} \\ell_{3}+w_{3} \\ell_{4}+w_{5} \\ell_{5}\\right) \\\\\n& \\geqslant \\frac{1}{2}(1 \\cdot 10+2 \\cdot 9+3 \\cdot 8+\\cdots+8 \\cdot 3+9 \\cdot 2+10 \\cdot 1)=110\n\\end{aligned}\n$$\n\nand the upper bound\n\n$$\n\\begin{aligned}\n\\ell_{1} w_{1} & +\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5} \\\\\n& =\\frac{1}{2}\\left(\\ell_{1} w_{1}+\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5}+w_{1} \\ell_{1}+w_{2} \\ell_{2}+w_{3} \\ell_{3}+w_{3} \\ell_{4}+w_{5} \\ell_{5}\\right) \\\\\n& \\leqslant \\frac{1}{2}(1 \\cdot 1+2 \\cdot 2+3 \\cdot 3+\\cdots+8 \\cdot 8+9 \\cdot 9+10 \\cdot 10)=192.5\n\\end{aligned}\n$$\n\nAs the area of the square is sandwiched between 110 and 192.5 , the only possible candidates for $m$ are 11, 12, and 13 .\n\nIn the second part of the solution, we show that a dissection of the square into five rectangles must consist of a single inner rectangle and four outer rectangles that each cover one of the four corners of the square. Indeed, if one of the sides the square had three rectangles adjacent to it, removing these three rectangles would leave a polygon with eight vertices, which is clearly not the union of two rectangles. Moreover, since $m>10$, each side of the square has at least two adjacent rectangles. Hence each side of the square has precisely two adjacent rectangles, and thus the only way of partitionning the square into five rectangles is to have a single inner rectangle and four outer rectangles each covering of the four corners of the square, as claimed.\n\nLet us now show that a square of size $12 \\times 12$ cannot be dissected in the desired way. Let $R_{1}, R_{2}, R_{3}$ and $R_{4}$ be the outer rectangles (in clockwise orientation along the\n\n\n\nboundary of the square). If an outer rectangle has a side of length $s$, then some adjacent outer rectangle must have a side of length $12-s$. Therefore, neither of $s=1$ or $s=6$ can be sidelengths of an outer rectangle, so the inner rectangle must have dimensions $1 \\times 6$. One of the outer rectangles (say $R_{1}$ ) must have dimensions $10 \\times x$, and an adjacent rectangle (say $R_{2}$ ) must thus have dimensions $2 \\times y$. Rectangle $R_{3}$ then has dimensions $(12-y) \\times z$, and rectangle $R_{4}$ has dimensions $(12-z) \\times(12-x)$. Note that exactly one of the three numbers $x, y, z$ is even (and equals 4 or 8 ), while the other two numbers are odd. Now, the total area of all five rectangles is\n\n$$\n144=6+10 x+2 y+(12-y) z+(12-z)(12-x)\n$$\n\nwhich simplifies to $(y-x)(z-2)=6$. As exactly one of the three numbers $x, y, z$ is even, the factors $y-x$ and $z-2$ are either both even or both odd, so their product cannot equal 6 , and thus there is no solution with $m=12$.\n\nFinally, we handle the cases $m=11$ and $m=13$, which indeed are solutions. The corresponding rectangle sets are $10 \\times 5,1 \\times 9,8 \\times 2,7 \\times 4$ and $3 \\times 6$ for $m=11$, and $10 \\times 5,9 \\times 8,4 \\times 6,3 \\times 7$ and $1 \\times 2$ for $m=13$. These sets can be found by trial and error. The corresponding partitions are shown in the figure below.\n']","['11,13']",True,,Numerical, 2217,Combinatorics,,"Let $n$ be a positive integer. Prove that there exists a set $S$ of $6 n$ pairwise different positive integers, such that the least common multiple of any two elements of $S$ is no larger than $32 n^{2}$.","['Let the set $A$ consist of the $4 n$ integers $1,2, \\ldots, 4 n$ and let the set $B$ consist of the $2 n$ even integers $4 n+2,4 n+4, \\ldots, 8 n$. We claim that the $6 n$-element set $S=A \\cup B$ has the desired property.\n\nIndeed, the least common multiple of two (even) elements of $B$ is no larger than $8 n \\cdot(8 n / 2)=32 n^{2}$, and the least common multiple of some element of $A$ and some element of $A \\cup B$ is at most their product, which is at most $4 n \\cdot 8 n=32 n^{2}$.']",,True,,, 2218,Combinatorics,,"Let $n$ be a positive integer. Prove that every set $T$ of $6 n$ pairwise different positive integers contains two elements the least common multiple of which is larger than $9 n^{2}$.","['We prove the following lemma: ""If a set $U$ contains $m+1$ integers, where $m \\geqslant 2$, that are all not less than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$.""\n\nLet the elements of $U$ be $u_{1}>u_{2}>\\cdots>u_{m+1} \\geqslant m$. Note that $1 / u_{1} \\leqslant 1 / u_{i} \\leqslant 1 / m$ for $1 \\leqslant i \\leqslant m+1$. We partition the interval $\\left[1 / u_{1} ; 1 / m\\right]$ into $m$ subintervals of equal length. By the pigeonhole principle, there exist indices $i, j$ with $1 \\leqslant im^{2}$, completing the proof of the lemma.\n\nApplying the lemma with $m=3 n$ to the $3 n+1$ largest elements of $T$, which are all not less than $3 n$, we arrive at the desired statement.\n\nA Variant. Alternatively, for part (b), we prove the following lemma: ""If a set $U$ contains $m \\geqslant 2$ integers that all are greater than $m$, then some two of its elements have least common multiple strictly larger than $\\mathrm{m}^{2} . ""$\n\nLet $u_{1}>u_{2}>\\cdots>u_{m}$ be the elements of $U$. Since $u_{m}>m=m^{2} / m$, there exists a smallest index $k$ such that $u_{k}>m^{2} / k$. If $k=1$, then $u_{1}>m^{2}$, and the least common multiple of $u_{1}$ and $u_{2}$ is strictly larger than $m^{2}$. So let us suppose $k>1$ from now on, so that we have $u_{k}>m^{2} / k$ and $u_{k-1} \\leqslant m^{2} /(k-1)$. The greatest common divisor $d$ of $u_{k-1}$ and $u_{k}$ satisfies\n\n$$\nd \\leqslant u_{k-1}-u_{k}<\\frac{m^{2}}{k-1}-\\frac{m^{2}}{k}=\\frac{m^{2}}{(k-1) k}\n$$\n\n\n\nThis implies $m^{2} /(d k)>k-1$ and $u_{k} / d>k-1$, and hence $u_{k} / d \\geqslant k$. But then the least common multiple of $u_{k-1}$ and $u_{k}$ equals\n\n$$\n\\frac{u_{k-1} u_{k}}{d} \\geqslant u_{k} \\cdot \\frac{u_{k}}{d}>\\frac{m^{2}}{k} \\cdot k=m^{2}\n$$\n\nand the proof of the lemma is complete.\n\nIf we remove the $3 n$ smallest elements from set $T$ and apply the lemma with $m=3 n$ to the remaining elements, we arrive at the desired statement.']",,True,,, 2219,Number Theory,,"Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial $$ P(n)=\frac{n^{5}+a}{b} $$ takes integer values.","['Denote the three consecutive integers by $x-1, x$, and $x+1$, so that\n\n$$\n(x-1)^{5}+a \\equiv 0 \\quad(\\bmod b), \\quad x^{5}+a \\equiv 0 \\quad(\\bmod b), \\quad(x+1)^{5}+a \\equiv 0 \\quad(\\bmod b) \n\\tag{1}\n$$\n\nBy computing the differences of the equations in (1) we get\n\n$$\n\\begin{aligned}\n& A:=(x+1)^{5}-(x-1)^{5}=10 x^{4}+20 x^{2}+2 \\equiv 0 \\quad(\\bmod b) \\\\\n\\end{aligned} \\tag{2}\n$$\n$$\n\\begin{aligned}\n& B:=(x+1)^{5}-x^{5}=5 x^{4}+10 x^{3}+10 x^{2}+5 x+1 \\equiv 0 \\quad(\\bmod b) .\n\\end{aligned} \\tag{3}\n$$\n\nAdding the first and third equation in (1) and subtracting twice the second equation yields\n\n$$\nC:=(x+1)^{5}+(x-1)^{5}-2 x^{5}=20 x^{3}+10 x \\equiv 0 \\quad(\\bmod b) . \\tag{4}\n$$\n\nNext, (2) and (4) together yield\n\n$$\nD:=4 x A-\\left(2 x^{2}+3\\right) C=-22 x \\equiv 0 \\quad(\\bmod b) \\tag{5}\n$$\n\nFinally we combine (3) and (5) to derive\n\n$$\n22 B+\\left(5 x^{3}+10 x^{2}+10 x+5\\right) D=22 \\equiv 0 \\quad(\\bmod b) .\n$$\n\nAs the positive integer $b$ divides 22 , we are left with the four cases $b=1, b=2, b=11$ and $b=22$.\n\nIf $b$ is even (i.e. $b=2$ or $b=22$ ), then we get a contradiction from (3), because the integer $B=2\\left(5 x^{3}+5 x^{2}\\right)+5\\left(x^{4}+x\\right)+1$ is odd, and hence not divisible by any even integer.\n\nFor $b=1$, it is trivial to see that a polynomial of the form $P(n)=n^{5}+a$, with $a$ any positive integer, has the desired property.\n\nFor $b=11$, we note that\n\n$$\n\\begin{aligned}\n& n \\equiv 0,1,2,3,4,5,6,7,8,9,10 \\quad(\\bmod 11) \\\\\n& \\Longrightarrow \\quad n^{5} \\equiv 0,1,-1,1,1,1,-1,-1,-1,1,-1 \\quad(\\bmod 11) .\n\\end{aligned}\n$$\n\nHence a polynomial of the form $P(n)=\\left(n^{5}+a\\right) / 11$ has the desired property if and only if $a \\equiv \\pm 1(\\bmod 11)$. This completes the proof.\n\n\n\nA Variant. We start by following the first solution up to equation (4). We note that $b=1$ is a trivial solution, and assume from now on that $b \\geqslant 2$. As $(x-1)^{5}+a$ and $x^{5}+a$ have different parity, $b$ must be odd. As $B$ in (3) is a multiple of $b$, we conclude that (i) $b$ is not divisible by 5 and that (ii) $b$ and $x$ are relatively prime. As $C=10 x\\left(2 x^{2}+1\\right)$ in (4) is divisible by $b$, we altogether derive\n\n$$\nE:=2 x^{2}+1 \\equiv 0 \\quad(\\bmod b)\n$$\n\nTogether with (2) this implies that\n\n$$\n5 E^{2}+10 E-2 A=11 \\equiv 0 \\quad(\\bmod b)\n$$\n\nHence $b=11$ is the only remaining candidate, and it is handled as in the first solution.', ""Let $p$ be a prime such that $p$ divides $b$. For some integer $x$, we have\n\n$$\n(x-1)^{5} \\equiv x^{5} \\equiv(x+1)^{5} \\quad(\\bmod p)\n$$\n\nNow, there is a primitive root $g$ modulo $p$, so there exist $u, v, w$ such that\n\n$$\nx-1 \\equiv g^{u} \\quad(\\bmod p), \\quad x \\equiv g^{v} \\quad(\\bmod p), \\quad x+1 \\equiv g^{w} \\quad(\\bmod p) \\tag{6}\n$$\n\nThe condition of the problem is thus\n\n$$\ng^{5 u} \\equiv g^{5 v} \\equiv g^{5 w} \\quad(\\bmod p) \\quad \\Longrightarrow \\quad 5 u \\equiv 5 v \\equiv 5 w \\quad(\\bmod p-1)\n$$\n\nIf $p \\not \\equiv 1(\\bmod 5)$, then 5 is invertible modulo $p-1$ and thus $u \\equiv v \\equiv w(\\bmod p-1)$, i.e. $x-1 \\equiv x \\equiv x+1(\\bmod p)$. This is a contradiction. Hence $p \\equiv 1(\\bmod 5)$ and thus $u \\equiv v \\equiv w\\left(\\bmod \\frac{p-1}{5}\\right)$. Thus, from (6), there exist integers $k, \\ell$ such that\n\n$$\n\\left.\\begin{array}{rl}\nx-1 \\equiv g^{v+k \\frac{p-1}{5}} \\equiv x t^{k} & (\\bmod p) \\\\\nx+1 \\equiv g^{v+\\ell \\frac{p-1}{5}} \\equiv x t^{\\ell} & (\\bmod p)\n\\end{array}\\right\\} \\quad \\text { where } t=g^{\\frac{p-1}{5}}\n$$\n\nLet $r=t^{k}$ and $s=t^{\\ell}$. In particular, the above yields $r, s \\not \\equiv 1(\\bmod p)$, and thus\n\n$$\nx \\equiv-(r-1)^{-1} \\equiv(s-1)^{-1} \\quad(\\bmod p)\n$$\n\nIt follows that\n\n$$\n(r-1)^{-1}+(s-1)^{-1} \\equiv 0 \\quad(\\bmod p) \\quad \\Longrightarrow \\quad r+s \\equiv 2 \\quad(\\bmod p)\n$$\n\nNow $t^{5} \\equiv 1(\\bmod p)$, so $r$ and $s$ must be congruent, modulo $p$, to some of the non-trivial fifth roots of unity $t, t^{2}, t^{3}, t^{4}$. Observe that, for any pair of these non-trivial roots of unity,\n\n\n\neither one is the other's inverse, or one is the other's square. In the first case, we have $r+r^{-1} \\equiv 2(\\bmod p)$, implying $r \\equiv 1(\\bmod p)$, a contradiction. Hence\n\n$$\nr+r^{2} \\equiv 2 \\quad(\\bmod p) \\quad \\Longrightarrow \\quad(r-1)(r+2) \\equiv 0 \\quad(\\bmod p)\n$$\n\nor\n\n$$\ns^{2}+s \\equiv 2 \\quad(\\bmod p) \\quad \\Longrightarrow \\quad(s-1)(s+2) \\equiv 0 \\quad(\\bmod p)\n$$\n\nThus, since $r, s \\not \\equiv 1(\\bmod p)$, we have $r \\equiv-2(\\bmod p)$ or $s \\equiv-2(\\bmod p)$, and thus $1 \\equiv r^{5} \\equiv-32(\\bmod p)$ or an analogous equation obtained from $s$. Hence $p \\mid 33$. Since $p \\equiv 1(\\bmod 5)$, it follows that $p=11$, i.e. $b$ is a power of 11 .\n\nExamining the fifth powers modulo 11 , we see that $b=11$ is indeed a solution with $a \\equiv \\pm 1(\\bmod 11)$ and, correspondingly, $x \\equiv \\pm 4(\\bmod 11)$. Now suppose, for the sake of contradiction, that $11^{2}$ divides $b$. Then, for some integer $m$, we must have\n\n$$\n(x-1, x, x+1) \\equiv \\pm(3+11 m, 4+11 m, 5+11 m) \\quad(\\bmod 121)\n$$\n\nand thus, substituting into the condition of the problem,\n\n$$\n\\begin{aligned}\n3^{5}+55 \\cdot 3^{4} m & \\equiv 4^{5}+55 \\cdot 4^{4} m \\equiv 5^{5}+55 \\cdot 5^{4} m \\quad(\\bmod 121) \\\\\n& \\Longrightarrow \\quad 1-22 m \\equiv 56+44 m \\equiv-21+11 m \\quad(\\bmod 121)\n\\end{aligned}\n$$\n\nHence $33 m \\equiv 22(\\bmod 121)$ and $33 m \\equiv 44(\\bmod 121)$, so $22 \\equiv 0(\\bmod 121)$, a contradiction. It follows that $b \\mid 11$.\n\nFinally, we conclude that the positive integers satisfying the original condition are $b=11$, with $a \\equiv \\pm 1(\\bmod 11)$, and $b=1$, for any positive integer $a$.""]","['$b=11$, with $a \\equiv \\pm 1(\\bmod 11)$, and $b=1$, for any positive integer $a$']",True,,Need_human_evaluate, 2220,Geometry,,"Let $\Omega$ be the circumcircle of the triangle $A B C$. The circle $\omega$ is tangent to the sides $A C$ and $B C$, and it is internally tangent to $\Omega$ at the point $P$. A line parallel to $A B$ and intersecting the interior of triangle $A B C$ is tangent to $\omega$ at $Q$. Prove that $\angle A C P=\angle Q C B$.","['Assume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively and let $P E, P F, P Q$ meet $\\Omega$ at $K, L, M$, respectively. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$, and consider the homethety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Now $K$ is the image of $E$ under $\\mathscr{H}$, and $E I \\perp A C$. Hence $O K \\perp A C$, and thus $K$ is the midpoint of the arc $C A$. Similarly, $L$ is the midpoint of the $\\operatorname{arc} B C$ and $M$ is the midpoint of the arc $B A$. It follows that arcs $L M$ and $C K$ are equal, because\n\n$$\n\\begin{aligned}\n\\overparen{B M}=\\overparen{M A} & \\Longrightarrow \\overparen{B L}+\\overparen{L M}=\\overparen{M K}+\\overparen{K A} \\Longrightarrow \\overparen{L C}+\\overparen{L M}=\\overparen{M K}+\\overparen{C K} \\\\\n& \\Longrightarrow 2 \\overparen{L M}+\\overparen{M C}=\\overparen{M C}+2 \\overparen{C K} \\Longrightarrow \\overparen{L M}=\\overparen{C K}\n\\end{aligned}\n$$\n\nThus arcs $F Q$ and $D E$ of $\\omega$ are equal, too, where $D$ is the intersection of $C P$ with $\\omega$. Since $C E$ and $C F$ are tangents to $\\omega$, this implies that $\\angle D E C=\\angle C F Q$. Further, $C E=C F$, and thus triangles $C E D$ and $C F Q$ are congruent. In particular, $\\angle E C D=\\angle Q C F$, as required.\n\n', 'Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. Observe that $C I$ is the angle bisector of angle $\\angle C$, because $\\omega$ is tangent to $A C$ and $B C$. Consider the homethety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Let $M$ be the image of $Q$ under $\\mathscr{H}$. By construction, $I Q \\perp A B$, so $O M \\perp A B$. Thus the diameter $O M$ of $\\Omega$ passes through the midpoint of the $\\operatorname{arc} A B$ of $\\Omega$, which also lies on the angle bisector $C I$. This implies that $\\angle I C M=90^{\\circ}$. We next show that $P, I, Q, C$ lie on a circle. Notice that\n\n$$\n\\begin{aligned}\n\\angle P Q I & =90^{\\circ}-\\frac{1}{2} \\angle Q I P=90^{\\circ}-\\frac{1}{2} \\angle M O P=90^{\\circ}-\\left(180^{\\circ}-\\angle P C M\\right) \\\\\n& =(\\angle P C I+\\angle I C M)-90^{\\circ}=\\angle P C I .\n\\end{aligned}\n$$\n\nHence $P, I, Q, C$ lie on a circle. But $P I=I Q$, so $C I$ is the angle bisector of $\\angle P C Q$. Since $C I$ is also the angle bisector of angle $\\angle C$, it follows that $\\angle A C P=\\angle Q C B$, as required.\n\n', 'Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. Let $D$ be the second point of intersection of $C P$ with $\\omega$, and let $\\ell$ denote the tangent to $\\omega$ at $D$, which meets $A C$ at $S$. Hence $I D \\perp \\ell$. By construction, $P, I, O$ lie one a line, and hence the isosceles triangles $P I D$ and $P O C$ are similar. In particular, it follows that $O C \\perp \\ell$, so $C$ is the midpoint of the arc of $\\Omega$ defined by the points of intersection of $\\ell$ with $\\Omega$. It is easy to see that this implies that\n\n$$\n\\angle D S C=\\angle A B C .\n$$\n\nUnder reflection in the angle bisector $C I$ of $\\angle C, \\ell$ is thus mapped to a tangent to $\\omega$ parallel to $A B$ and intersecting the interior of $A B C$, since $\\omega$ is mapped to itself under this reflection. In particular, $D$ is mapped to $Q$, and thus $\\angle Q C B=\\angle A C D$, as required.\n\n', 'Let the tangent to $\\omega$ at $Q$ meet $A C$ and $B C$ at $X$ and $Y$, respectively. Then $A C / X C=B C / Y C$, and thus there is a radius $r$ such that $r^{2}=A C \\cdot Y C=B C \\cdot X C$. Let $\\Gamma$ denote the circle with centre $C$ and radius $r$, and consider the inversion $\\mathscr{I}$ in the circle $\\Gamma$. Under $\\mathscr{I}$,\n\n$A \\longmapsto A^{\\prime}$, the point on the ray $C A$ satisfying $C A^{\\prime}=C Y$;\n\n$B \\longmapsto B^{\\prime}$, the point on the ray $C B$ satisfying $C B^{\\prime}=C X$;\n\n$\\Omega \\longmapsto$ the line $A^{\\prime} B^{\\prime} ;$\n\n$\\omega \\longmapsto \\omega^{\\prime}$, the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$;\n\n$P \\longmapsto P^{\\prime}$, the point where $\\omega^{\\prime}$ touches $A^{\\prime} B^{\\prime}$;\n\nIn particular, $\\omega^{\\prime}$, the image of $\\omega$, is a circle tangent to $A C, B C$ and $A^{\\prime} B^{\\prime}$, so it is either the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$, or the incircle of $C A^{\\prime} B^{\\prime}$. Let $\\omega$ be tangent to $B C$ at $F$, and let $F^{\\prime}$ be the image of $F$ under $\\mathscr{I}$. Then $C F \\cdot C F^{\\prime}=B C \\cdot X C$. Now $C FC X=C B^{\\prime}$. Hence $\\omega^{\\prime}$ cannot be the incircle, so $\\omega^{\\prime}$ is indeed the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$.\n\nNow note that $\\omega$ is the excircle of $C X Y$ opposite $C$. The reflection about the angle bisector of $\\angle C$ maps $X$ to $B^{\\prime}, Y$ to $A^{\\prime}$. It thus maps the triangle $C X Y$ to $C B^{\\prime} A^{\\prime}, \\omega$ to $\\omega^{\\prime}$ and, finally, $Q$ to $P^{\\prime}$. It follows that $\\angle A C P=\\angle A C P^{\\prime}=\\angle Q C B$, as required.\n\n', 'Let $r$ be the radius such that $r^{2}=A C \\cdot B C$. Let $\\mathscr{J}$ denote the composition of the inversion $\\mathscr{I}$ in the circle of centre $C$ and radius $r$, followed by the reflection in the angle bisector of $\\angle C$. Under $\\mathscr{J}$,\n\n$A \\longmapsto B, B \\mapsto A ;$\n\n$\\Omega \\longmapsto$ the line $A B$;\n\n$\\omega \\longmapsto \\omega^{\\prime}$, the excircle of $A B C$ opposite the vertex $C$;\n\n$P \\longmapsto Q^{\\prime}$, the point where $\\omega^{\\prime}$ touches $A B$;\n\nIn particular, note that the image $\\omega^{\\prime}$ of $\\omega$ under $\\mathscr{J}$ is a circle tangent to $A C, B C$ and $A B$, so it is either the incircle of $A B C$, or the excircle opposite vertex $C$. Observe that $r \\geqslant \\min \\{A C, B C\\}$, so the image of the points of tangency of $\\omega$ must lie outside $A B C$, and thus $\\omega^{\\prime}$ cannot be the incircle. Thus $\\omega^{\\prime}$ is the excircle opposite vertex $C$ as claimed. Further, the point of tangency $P$ is mapped to $Q^{\\prime}$.\n\nNow, since the line $C P$ is mapped to itself under the inversion $\\mathscr{I}$, and mapped onto $C Q^{\\prime}$ under $\\mathscr{J}, C P$ and $C Q^{\\prime}$ are images of each other under reflection in the angle bisector of $\\angle C$. But $C, Q, Q^{\\prime}$ lie on a line for there is a homothety with centre $C$ that maps $\\omega$ onto the excircle $\\omega^{\\prime}$. This completes the proof.\n\n', '$\\quad$ Assume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively. Assume that $C P$ meets $\\omega$ at $D$. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. To set up a solution in the complex plane, we take the circle $\\omega$ as the unit circle centered at the origin of the complex plane, and let $P O$ be the real axis with $o>0$, where we use the convention that lowercase letters denote complex coordinates of corresponding points in the plane denoted by uppercase letters.\n\n\n\nNow, a point $Z$ on the circle $\\Omega$ satisfies\n\n$$\n|z-o|^{2}=(o+1)^{2} \\Longleftrightarrow \\quad z z^{*}-o\\left(z+z^{*}\\right)-2 o-1=0\n$$\n\nThe triangle $A B C$ is defined by the points $E$ and $F$ on $\\omega$, the intersection $C$ of the corresponding tangents lying on $\\Omega$. Thus $c=2 e f /(e+f)$, and further\n\n$$\n|c-o|^{2}=(o+1)^{2} \\Longleftrightarrow \\quad c c^{*}-o\\left(c+c^{*}\\right)-2 o-1=0\n\\tag{1}\n$$\n\nand this is the equality defining $o$. The points $A$ and $B$ are the second intersection points of $\\Omega$ with the tangents to $\\omega$ at $E$ and $F$ respectively. A point $Z$ on the tangent through $E$ is given by $z=2 e-e^{2} z^{*}$, and thus $A$ and $C$ satisfy\n\n$$\n\\begin{aligned}\n\\left(2 e-e^{2} z^{*}\\right) z^{*} & -o\\left(2 e-e^{2} z^{*}+z^{*}\\right)-2 o-1=0 \\\\\n& \\Longleftrightarrow \\quad-e^{2} z^{* 2}+\\left(2 e+o e^{2}-o\\right) z^{*}-(2 e o+2 o+1)=0 \\\\\n& \\Longleftrightarrow \\quad z^{* 2}-\\left(2 e^{*}+o-o e^{* 2}\\right) z^{*}+\\left(2 e^{*} o+2 o e^{* 2}+e^{* 2}\\right)=0\n\\end{aligned}\n$$\n\nsince $|e|=1$. Thus\n\n$$\na^{*}+c^{*}=2 e^{*}+o-o e^{* 2} \\quad \\Longrightarrow \\quad a^{*}=\\frac{2 e^{*} f}{e+f}+o\\left(1-e^{* 2}\\right)\n$$\n\nand similarly\n\n$$\nb^{*}=\\frac{2 f^{*} e}{f+e}+o\\left(1-f^{* 2}\\right)\n$$\n\nThen\n\n$$\n\\begin{aligned}\nb^{*}-a^{*} & =\\frac{2\\left(e f^{*}-e^{*} f\\right)}{e+f}+o\\left(e^{* 2}-f^{* 2}\\right) \\\\\n& =\\frac{2 e f\\left(f^{* 2}-e^{* 2}\\right)}{e+f}+o\\left(e^{* 2}-f^{* 2}\\right) \\\\\n& =\\left(f^{* 2}-e^{* 2}\\right)\\left(\\frac{2 e f}{e+f}-o\\right)=\\left(f^{* 2}-e^{* 2}\\right)(c-o) .\n\\end{aligned}\n$$\n\nNow let $Z$ be a point on the tangent to $\\omega$ parallel to $A B$ passing through $Q$. Then\n\n$$\nz=2 q-q^{2} z^{*} \\quad \\Longleftrightarrow \\quad z-q=q-q^{2} z^{*}=-q^{2}\\left(z^{*}-q^{*}\\right)\n$$\n\nfor $|q|=1$, and thus\n\n$$\n\\frac{b-a}{b^{*}-a^{*}}=\\frac{z-q}{z^{*}-q^{*}}=\\frac{-q^{2}\\left(z^{*}-q^{*}\\right)}{z^{*}-q^{*}}=-q^{2}\n$$\n\n\n\nIt follows that\n\n$$\n\\begin{aligned}\nq^{2} & =-\\frac{b-a}{b^{*}-a^{*}}=-\\frac{\\left(f^{2}-e^{2}\\right)\\left(c^{*}-o\\right)}{\\left(f^{* 2}-e^{* 2}\\right)(c-o)}=e^{2} f^{2} \\frac{c^{*}-o}{c-o} \\\\\n& =e^{2} f^{2} \\frac{\\left(c^{*}-o\\right)^{2}}{|c-o|^{2}}=e^{2} f^{2} \\frac{\\left(c^{*}-o\\right)^{2}}{(1+o)^{2}}\n\\end{aligned}\n$$\n\nwhere we have used (1). In particular,\n\n$$\nq=e f \\frac{c^{*}-o}{1+o}\n$$\n\nwhere the choice of sign is to be justified a posteriori. Further, the point $D$ satisfies\n\n$$\n-d p=\\frac{d-p}{d^{*}-p^{*}}=\\frac{c-p}{c^{*}-p^{*}} \\quad \\Longrightarrow \\quad d=-\\frac{c-p}{c^{*} p-1}=\\frac{c+1}{c^{*}+1}\n$$\n\nusing $p=-1$ to obtain the final equality.\n\nNow, it suffices to show that (i) $D Q \\| E F \\perp C I$ and (ii) the midpoint of $[D Q]$ is on $C I$. The desired equality then follows by symmetry with respect to the angle bisector of the angle $\\angle A C B$. Notice that (i) is equivalent with\n\n$$\n\\frac{d-q}{d^{*}-q^{*}}=\\frac{e-f}{e^{*}-f^{*}} \\Longleftrightarrow d q=e f\n$$\n\nfor $[D Q]$ and $[E F]$ are chords of $\\omega$. But\n\n$$\n\\begin{aligned}\nd q=e f & \\Longleftrightarrow \\frac{c+1}{c^{*}+1} e f \\frac{c^{*}-o}{1+o}=e f \\quad \\Longleftrightarrow \\quad(c+1)\\left(c^{*}-o\\right)=\\left(c^{*}+1\\right)(1+o) \\\\\n& \\Longleftrightarrow \\quad c c^{*}-o\\left(c+c^{*}\\right)-2 o-1=0 .\n\\end{aligned}\n$$\n\nThe last equality is precisely the defining relation for $o$, (1). This proves (i). Further, the midpoint of $[D Q]$ is $\\frac{1}{2}(d+q)$, so it remains to check that\n\n$$\nd q=\\frac{d+q}{d^{*}+q^{*}}=\\frac{c}{c^{*}}=e f\n$$\n\nwhere the first equality expresses that $[D Q]$ is a chord of $\\omega$ (obviously) containing its midpoint, the second equality expresses the alignment of the midpoint of $[D Q], C$ and $I$, and the third equality follows from the expression for $c$. But we have just shown that $d q=e f$. This proves (ii), justifies the choice of sign for $q$ a posteriori, and thus completes the solution of the problem.']",,True,,, 2221,Combinatorics,,"Snow White and the Seven Dwarves are living in their house in the forest. On each of 16 consecutive days, some of the dwarves worked in the diamond mine while the remaining dwarves collected berries in the forest. No dwarf performed both types of work on the same day. On any two different (not necessarily consecutive) days, at least three dwarves each performed both types of work. Further, on the first day, all seven dwarves worked in the diamond mine. Prove that, on one of these 16 days, all seven dwarves were collecting berries.","['We define $V$ as the set of all 128 vectors of length 7 with entries in $\\{0,1\\}$. Every such vector encodes the work schedule of a single day: if the $i$-th entry is 0 then the $i$-th dwarf works in the mine, and if this entry is 1 then the $i$-th dwarf collects berries. The 16 working days correspond to 16 vectors $d_{1}, \\ldots, d_{16}$ in $V$, which we will call dayvectors. The condition imposed on any pair of distinct days means that any two distinct day-vectors $d_{i}$ and $d_{j}$ differ in at least three positions.\n\nWe say that a vector $x \\in V$ covers some vector $y \\in V$, if $x$ and $y$ differ in at most one position; note that every vector in $V$ covers exactly eight vectors. For each of the 16 day-vectors $d_{i}$ we define $B_{i} \\subset V$ as the set of the eight vectors that are covered by $d_{i}$. As, for $i \\neq j$, the day-vectors $d_{i}$ and $d_{j}$ differ in at least three positions, their corresponding sets $B_{i}$ and $B_{j}$ are disjoint. As the sets $B_{1}, \\ldots, B_{16}$ together contain $16 \\cdot 8=128=|V|$ distinct elements, they form a partition of $V$; in other words, every vector in $V$ is covered by precisely one day-vector.\n\nThe weight of a vector $v \\in V$ is defined as the number of 1-entries in $v$. For $k=0,1, \\ldots, 7$, the set $V$ contains $\\left(\\begin{array}{l}7 \\\\ k\\end{array}\\right)$ vectors of weight $k$. Let us analyse the 16 dayvectors $d_{1}, \\ldots, d_{16}$ by their weights, and let us discuss how the vectors in $V$ are covered by them.\n\n1. As all seven dwarves work in the diamond mine on the first day, the first day-vector is $d_{1}=(0000000)$. This day-vector covers all vectors in $V$ with weight 0 or 1.\n2. No day-vector can have weight 2 , as otherwise it would differ from $d_{1}$ in at most two positions. Hence each of the $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)=21$ vectors of weight 2 must be covered by some day-vector of weight 3 . As every vector of weight 3 covers three vectors of weight 2 , exactly $21 / 3=7$ day-vectors have weight 3 .\n3. How are the $\\left(\\begin{array}{l}7 \\\\ 3\\end{array}\\right)=35$ vectors of weight 3 covered by the day-vectors? Seven of them are day-vectors, and the remaining 28 ones must be covered by day-vectors of weight 4 . As every vector of weight 4 covers four vectors of weight 3 , exactly $28 / 4=7$ day-vectors have weight 4.\n\n\nTo summarize, one day-vector has weight 0 , seven have weight 3 , and seven have weight 4 . None of these 15 day-vectors covers any vector of weight 6 or 7 , so that the eight heavyweight vectors in $V$ must be covered by the only remaining day-vector; and this remaining vector must be (1111111). On the day corresponding to (1111111) all seven dwarves are collecting berries, and that is what we wanted to show.', 'If a dwarf $X$ performs the same type of work on three days $D_{1}, D_{2}, D_{3}$, then we say that this triple of days is monotonous for $X$. We claim that the following configuration cannot occur: There are three dwarves $X_{1}, X_{2}, X_{3}$ and three days $D_{1}, D_{2}$, $D_{3}$, such that the triple $\\left(D_{1}, D_{2}, D_{3}\\right)$ is monotonous for each of the dwarves $X_{1}, X_{2}, X_{3}$.\n\n(Proof: Suppose that such a configuration occurs. Then among the remaining dwarves there exist three dwarves $Y_{1}, Y_{2}, Y_{3}$ that performed both types of work on day $D_{1}$ and on day $D_{2}$; without loss of generality these three dwarves worked in the mine on day $D_{1}$ and collected berries on day $D_{2}$. On day $D_{3}$, two of $Y_{1}, Y_{2}, Y_{3}$ performed the same type of work, and without loss of generality $Y_{1}$ and $Y_{2}$ worked in the mine. But then on days $D_{1}$ and $D_{3}$, each of the five dwarves $X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}$ performed only one type of work; this is in contradiction with the problem statement.)\n\nNext we consider some fixed triple $X_{1}, X_{2}, X_{3}$ of dwarves. There are eight possible working schedules for $X_{1}, X_{2}, X_{3}$ (like mine-mine-mine, mine-mine-berries, mine-berriesmine, etc). As the above forbidden configuration does not occur, each of these eight working schedules must occur on exactly two of the sixteen days. In particular this implies that every dwarf worked exactly eight times in the mine and exactly eight times in the forest.\n\nFor $0 \\leqslant k \\leqslant 7$ we denote by $d(k)$ the number of days on which exactly $k$ dwarves were collecting berries. Since on the first day all seven dwarves were in the mine, on each of the remaining days at least three dwarves collected berries. This yields $d(0)=1$ and $d(1)=d(2)=0$. We assume, for the sake of contradiction, that $d(7)=0$ and hence\n\n$$\nd(3)+d(4)+d(5)+d(6)=15 \\tag{1}\n$$\n\nAs every dwarf collected berries exactly eight times, we get that, further,\n\n$$\n3 d(3)+4 d(4)+5 d(5)+6 d(6)=7 \\cdot 8=56 .\n\\tag{2}\n$$\n\nNext, let us count the number $q$ of quadruples $\\left(X_{1}, X_{2}, X_{3}, D\\right)$ for which $X_{1}, X_{2}, X_{3}$ are three pairwise distinct dwarves that all collected berries on day $D$. As there are $7 \\cdot 6 \\cdot 5=210$ triples of pairwise distinct dwarves, and as every working schedule for three fixed dwarves occurs on exactly two days, we get $q=420$. As every day on which $k$ dwarves collect berries contributes $k(k-1)(k-2)$ such quadruples, we also have\n\n$$\n3 \\cdot 2 \\cdot 1 \\cdot d(3)+4 \\cdot 3 \\cdot 2 \\cdot d(4)+5 \\cdot 4 \\cdot 3 \\cdot d(5)+6 \\cdot 5 \\cdot 4 \\cdot d(6)=q=420\n$$\n\n\n\nwhich simplifies to\n\n$$\nd(3)+4 d(4)+10 d(5)+20 d(6)=70 \\tag{3}\n$$\n\nFinally, we count the number $r$ of quadruples $\\left(X_{1}, X_{2}, X_{3}, D\\right)$ for which $X_{1}, X_{2}, X_{3}$ are three pairwise distinct dwarves that all worked in the mine on day $D$. Similarly as above we see that $r=420$ and that\n\n$$\n7 \\cdot 6 \\cdot 5 \\cdot d(0)+4 \\cdot 3 \\cdot 2 \\cdot d(3)+3 \\cdot 2 \\cdot 1 \\cdot d(4)=r=420 \n$$\n\nwhich simplifies to\n\n$$\n4 d(3)+d(4)=35 \\tag{4}\n$$\n\nMultiplying (1) by -40 , multiplying (2) by 10 , multiplying (3) by -1 , multiplying (4) by 4 , and then adding up the four resulting equations yields $5 d(3)=30$ and hence $d(3)=6$. Then (4) yields $d(4)=11$. As $d(3)+d(4)=17$, the total number of days cannot be 16 . We have reached the desired contradiction.']",,True,,, 2222,Geometry,,"Let $\triangle A B C$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\angle A B C$ intersects $C D$ at $E$ and meets the circumcircle $\omega$ of triangle $\triangle A D E$ again at $F$. If $\angle A D F=45^{\circ}$, show that $C F$ is tangent to $\omega$.","['\nSince $\\angle C D F=90^{\\circ}-45^{\\circ}=45^{\\circ}$, the line $D F$ bisects $\\angle C D A$, and so $F$ lies on the perpendicular bisector of segment $A E$, which meets $A B$ at $G$. Let $\\angle A B C=2 \\beta$. Since $A D E F$ is cyclic, $\\angle A F E=90^{\\circ}$, and hence $\\angle F A E=45^{\\circ}$. Further, as $B F$ bisects $\\angle A B C$, we have $\\angle F A B=90^{\\circ}-\\beta$, and thus\n\n$$\n\\angle E A B=\\angle A E G=45^{\\circ}-\\beta, \\quad \\text { and } \\quad \\angle A E D=45^{\\circ}+\\beta \\text {, }\n$$\n\nso $\\angle G E D=2 \\beta$. This implies that right-angled triangles $\\triangle E D G$ and $\\triangle B D C$ are similar, and so we have $|G D| /|C D|=|D E| /|D B|$. Thus the right-angled triangles $\\triangle D E B$ and $\\triangle D G C$ are similar, whence $\\angle G C D=\\angle D B E=\\beta$. But $\\angle D F E=\\angle D A E=45^{\\circ}-$ $\\beta$, then $\\angle G F D=45^{\\circ}-\\angle D F E=\\beta$. Hence $G D C F$ is cyclic, so $\\angle G F C=90^{\\circ}$, whence $C F$ is perpendicular to the radius $F G$ of $\\omega$.\n\nIt follows that $C F$ is a tangent to $\\omega$, as required.', 'As $\\angle A D F=45^{\\circ}$ line $D F$ is an exterior bisector of $\\angle C D B$. Since $B F$ bisects $\\angle D B C$ line $C F$ is an exterior bisector of $\\angle B C D$. Let $\\angle A B C=2 \\beta$, so $\\angle E C F=(\\angle D B C+\\angle C D B) / 2=45^{\\circ}+\\beta$. Hence $\\angle C F E=180^{\\circ}-\\angle E C F-\\angle B C E-\\angle E B C=180^{\\circ}-\\left(45^{\\circ}+\\beta+90^{\\circ}-2 \\beta+\\beta\\right)=45^{\\circ}$. It follows that $\\angle F D C=\\angle C F E$, then $C F$ is tangent to $\\omega$.', 'Note that $A E$ is diameter of circumcircle of $\\triangle A B C$ since $\\angle C D F=90^{\\circ}$. From $\\angle A E F=\\angle A D F=45^{\\circ}$ it follows that triangle $\\triangle A F E$ is right-angled and isosceles. Without loss of generality, let points $A, E$ and $F$ have coordinates $(-1,0),(1,0)$ and $(0,1)$ respectively. Points $F, E$, $B$ are collinear, hence $B$ have coordinates $(b, 1-b)$ for some $b \\neq-1$. Let point $C^{\\prime}$ be intersection of line tangent to circumcircle of $\\triangle A F E$ at $F$ with line $E D$. Thus $C^{\\prime}$ have coordinates $(c, 1)$ and from $\\overline{C^{\\prime} E} \\perp \\overline{A B}$ we get $c=2 b /(b+1)$. Now vector $\\overline{B C^{\\prime}}=(2 b /(b+1)-b, b)=b /(b+1) \\cdot(1-b, b+1)$, vector $\\overline{B F}=(-b, b)=(-1,1) \\cdot b$ and vector $\\overline{B A}=(-(b+1),-(1-b))$. Its clear that $(1-b, b+1)$ and $(-(b+1),-(1-b))$ are symmetric with respect to $\\overline{F E}=(-1,1)$, hence $B F$ bisects $\\angle C^{\\prime} B A$ and $C^{\\prime}=C$ which completes the proof.', '\n\nAgain $F$ lies on the perpendicular bisector of segment $A E$, so $\\triangle A F E$ is right-angled and isosceles. Let $M$ be an intersection of $B C$ and $A F$. Note that $\\triangle A M B$ is isosceles since $B F$ is a bisector and altitude in this triangle. Thus $B F$ is a symmetry line of $\\triangle A M B$. Then $\\angle F D A=\\angle F E A=\\angle M E F=45^{\\circ}, A F=F E=F M$ and $\\angle D A E=\\angle E M C$. Let us show that $E C=C M$. Indeed,\n\n$$\n\\begin{aligned}\n\\angle C E M & =180^{\\circ}-(\\angle A E D+\\angle F E A+\\angle M E F)=90^{\\circ}-\\angle A E D= \\\\\n& =\\angle D A E=\\angle E M C .\n\\end{aligned}\n$$\n\nIt follows that $F M C E$ is a kite, since $E F=F M$ and $M C=C E$. Hence $\\angle E F C=\\angle C F M=\\angle E D F=45^{\\circ}$, so $F C$ is tangent to $\\omega$.', 'Let the tangent to $\\omega$ at $F$ intersect $C D$ at $C^{\\prime}$. Let $\\angle A B F=$ $\\angle F B C=\\beta$. It follows that $\\angle C^{\\prime} F E=45^{\\circ}$ since $C^{\\prime} F$ is tangent. We have\n\n$$\n\\frac{\\sin \\angle B D C}{\\sin \\angle C D F} \\cdot \\frac{\\sin \\angle D F C^{\\prime}}{\\sin \\angle C^{\\prime} F B} \\cdot \\frac{\\sin \\angle F B C}{\\sin \\angle C B D}=\\frac{\\sin 90^{\\circ}}{\\sin 45^{\\circ}} \\cdot \\frac{\\sin \\left(90^{\\circ}-\\beta\\right)}{\\sin 45^{\\circ}} \\cdot \\frac{\\sin \\beta}{\\sin 2 \\beta}=\\frac{2 \\sin \\beta \\cos \\beta}{\\sin 2 \\beta}=1 .\n$$\n\nSo by trig Ceva on triangle $\\triangle B D F$, lines $F C^{\\prime}, D C$ and $B C$ are concurrent (at $C$ ), so $C=C^{\\prime}$. Hence $C F$ is tangent to $\\omega$.']",,True,,, 2223,Combinatorics,,A domino is a $2 \times 1$ or $1 \times 2$ tile. Determine in how many ways exactly $n^{2}$ dominoes can be placed without overlapping on a $2 n \times 2 n$ chessboard so that every $2 \times 2$ square contains at least two uncovered unit squares which lie in the same row or column.,"['The answer is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)^{2}$.\n\nDivide the schessboard into $2 \\times 2$ squares. There are exactly $n^{2}$ such squares on the chessboard. Each of these squares can have at most two unit squares covered by the dominos. As the dominos cover exactly $2 n^{2}$ squares, each of them must have exactly two unit squares which are covered, and these squares must lie in the same row or column.\n\nWe claim that these two unit squares are covered by the same domino tile. Suppose that this is not the case for some $2 \\times 2$ square and one of the tiles covering one of its unit squares sticks out to the left. Then considering one of the leftmost $2 \\times 2$ squares in this division with this property gives a contradiction.\n\nNow consider this $n \\times n$ chessboard consisting of $2 \\times 2$ squares of the original board. Define $A, B$, $C, D$ as the following configurations on the original chessboard, where the gray squares indicate the domino tile, and consider the covering this $n \\times n$ chessboard with the letters $A, B, C, D$ in such a\n![](https://cdn.mathpix.com/cropped/2023_12_21_521ea44181413e4b4507g-1.jpg?height=118&width=1446&top_left_y=1030&top_left_x=317)\n\nway that the resulting configuration on the original chessboard satisfies the condition of the question.\n\nNote that then a square below or to the right of one containing an $A$ or $B$ must also contain an $A$ or $B$. Therefore the (possibly empty) region consisting of all squares containing an $A$ or $B$ abuts the lower right corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a $C$ or $D$ by a path which goes from the lower left corner to the upper right corner of this chessboard and which moves up or right at each step.\n\nA similar reasoning shows that the (possibly empty) region consisting of all squares containing an $A$ or $D$ abuts the lower left corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a $B$ or $C$ by a path which goes from the upper left corner to the lower right corner of this chessboard and which moves down or right at each step.\n\n| $D$ | $D$ | $C$ | $C$ | $C$ | $C$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| $D$ | $D$ | $C$ | $C$ | $C$ | $B$ |\n| $D$ | $D$ | $D$ | $B$ | $B$ | $B$ |\n| $D$ | $D$ | $D$ | $A$ | $A$ | $B$ |\n| $D$ | $D$ | $D$ | $A$ | $A$ | $B$ |\n| $D$ | $A$ | $A$ | $A$ | $A$ | $B$ |\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_521ea44181413e4b4507g-1.jpg?height=426&width=436&top_left_y=1690&top_left_x=1278)\n\nTherefore the $n \\times n$ chessboard is divided by these two paths into four (possibly empty) regions that consist respectively of all squares containing $A$ or $B$ or $C$ or $D$. Conversely, choosing two such paths and filling the four regions separated by them with $A \\mathrm{~s}, B \\mathrm{~s}, C \\mathrm{~s}$ and $D \\mathrm{~s}$ counterclockwise starting at the bottom results in a placement of the dominos on the original board satisfying the condition of the question.\n\nAs each of these paths can be chosen in $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ ways, there are $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)^{2}$ ways the dominos can be placed.']",['$\\binom{2n}{n}^2$'],False,,Need_human_evaluate, 2224,Algebra,,"Let $n, m$ be integers greater than 1 , and let $a_{1}, a_{2}, \ldots, a_{m}$ be positive integers not greater than $n^{m}$. Prove that there exist positive integers $b_{1}, b_{2}, \ldots, b_{m}$ not greater than $n$, such that $$ \operatorname{gcd}\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{m}+b_{m}\right)1$, let $a_{1}, \\ldots, a_{m}$ be positive integers with at least one $a_{i} \\leq n^{2^{m-1}}$. Then there are integers $b_{1}, \\ldots, b_{m}$, each equal to 1 or 2 , such that $\\operatorname{gcd}\\left(a_{1}+b_{1}, \\ldots, a_{m}+b_{m}\\right)1$ are all pairwise coprime, since for any two of them, there is some $i>1$ with $a_{i}+1$ appearing in one and $a_{i}+2$ in the other. Since each of these $2^{m-1}$ integers divides $a_{1}+1$, and each is $\\geq n$ with at most one equal to $n$, it follows that $a_{1}+1 \\geq n(n+1)^{2^{m-1}-1}$ so $a_{1} \\geq n^{2^{m-1}}$. The same is true for each $a_{i}, i=1, \\ldots, n$, a contradiction.']",,True,,, 2225,Algebra,,"Determine whether there exists an infinite sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers which satisfies the equality $$ a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_{n}} $$ for every positive integer $n$.","['The answer is no.\n\nSuppose that there exists a sequence $\\left(a_{n}\\right)$ of positive integers satisfying the given condition. We will show that this will lead to a contradiction.\n\nFor each $n \\geq 2$ define $b_{n}=a_{n+1}-a_{n}$. Then, by assumption, for $n \\geq 2$ we get $b_{n}=\\sqrt{a_{n}+a_{n-1}}$ so that we have\n\n$$\nb_{n+1}^{2}-b_{n}^{2}=\\left(a_{n+1}+a_{n}\\right)-\\left(a_{n}+a_{n-1}\\right)=\\left(a_{n+1}-a_{n}\\right)+\\left(a_{n}-a_{n-1}\\right)=b_{n}+b_{n-1} .\n$$\n\nSince each $a_{n}$ is a positive integer we see that $b_{n}$ is positive integer for $n \\geq 2$ and the sequence $\\left(b_{n}\\right)$ is strictly increasing for $n \\geq 3$. Thus $b_{n}+b_{n-1}=\\left(b_{n+1}-b_{n}\\right)\\left(b_{n+1}+b_{n}\\right) \\geq b_{n+1}+b_{n}$, whence $b_{n-1} \\geq b_{n+1}$ - a contradiction to increasing of the sequence $\\left(b_{i}\\right)$.\n\nThus we conclude that there exists no sequence $\\left(a_{n}\\right)$ of positive integers satisfying the given condition of the problem.', 'Suppose that such a sequence exists. We will calculate its members one by one and get a contradiction.\n\nFrom the equality $a_{3}=a_{2}+\\sqrt{a_{2}+a_{1}}$ it follows that $a_{3}>a_{2}$. Denote positive integers $\\sqrt{a_{3}+a_{2}}$ by $b$ and $a_{3}$ by $a$, then we have $\\sqrt{2 a}>b$. Since $a_{4}=a+b$ and $a_{5}=a+b+\\sqrt{2 a+b}$ are positive integers, then $\\sqrt{2 a+b}$ is positive integer.\n\nConsider $a_{6}=a+b+\\sqrt{2 a+b}+\\sqrt{2 a+2 b+\\sqrt{2 a+b}}$. Number $c=\\sqrt{2 a+2 b+\\sqrt{2 a+b}}$ must be positive integer, obviously it is greater than $\\sqrt{2 a+b}$. But\n\n$$\n(\\sqrt{2 a+b}+1)^{2}=2 a+b+2 \\sqrt{2 a+b}+1=2 a+2 b+\\sqrt{2 a+b}+(\\sqrt{2 a+b}-b)+1>c^{2}\n$$\n\nSo $\\sqrt{2 a+b}5$ members and satisfies\n\n$$\na_{n+2}=a_{n+1}+\\sqrt{a_{n+1}+a_{n}} \\tag{1}\n$$\n\nfor all $n=1, \\ldots, N-2$. Moreover, we will describe all such sequences with five members.\n\nSince every $a_{i}$ is a positive integer it follows from (1) that there exists such positive integer $k$ (obviously $k$ depends on $n$ ) that\n\n$$\na_{n+1}+a_{n}=k^{2} \\tag{2}\n$$\n\nFrom (1) we have $\\left(a_{n+2}-a_{n+1}\\right)^{2}=a_{n+1}+a_{n}$, consider this equality as a quadratic equation with respect to $a_{n+1}$ :\n\n$$\na_{n+1}^{2}-\\left(2 a_{n+2}+1\\right) a_{n+1}+a_{n+2}^{2}-a_{n}=0\n$$\n\nObviously its solutions are $\\left(a_{n+1}\\right)_{1,2}=\\frac{2 a_{n+2}+1 \\pm \\sqrt{D}}{2}$, where\n\n$$\nD=4\\left(a_{n}+a_{n+2}\\right)+1 \\tag{3}\n$$\n\nSince $a_{n+2}>a_{n+1}$ we have\n\n$$\na_{n+1}=\\frac{2 a_{n+2}+1-\\sqrt{D}}{2}\n$$\n\n\n\nFrom the last equality, using that $a_{n+1}$ and $a_{n+2}$ are positive integers, we conclude that $D$ is a square of some odd number i.e. $D=(2 m+1)^{2}$ for some positive integer $m \\in \\mathbb{N}$, substitute this into (3):\n\n$$\na_{n}+a_{n+2}=m(m+1) \\text {. } \\tag{4}\n$$\n\nNow adding $a_{n}$ to both sides of (1) and using (2) and (4) we get $m(m+1)=k^{2}+k$ whence $m=k$. So\n\n$$\n\\left\\{\\begin{array}{l}\na_{n}+a_{n+1}=k^{2} \\\\\na_{n}+a_{n+2}=k^{2}+k\n\\end{array}\\right. \\tag{5}\n$$\n\nfor some positive integer $k$ (recall that $k$ depends on $n$ ).\n\nWrite equations (5) for $n=2$ and $n=3$, then for some positive integers $k$ and $\\ell$ we get\n\n$$\n\\left\\{\\begin{array}{l}\na_{2}+a_{3}=k^{2}, \\\\\na_{2}+a_{4}=k^{2}+k, \\\\\na_{3}+a_{4}=\\ell^{2}, \\\\\na_{3}+a_{5}=\\ell^{2}+\\ell .\n\\end{array}\\right. \\tag{6}\n$$\n\nSolution of this linear system is\n\n$$\na_{2}=\\frac{2 k^{2}-\\ell^{2}+k}{2}, \\quad a_{3}=\\frac{\\ell^{2}-k}{2}, \\quad a_{4}=\\frac{\\ell^{2}+k}{2}, \\quad a_{5}=\\frac{\\ell^{2}+2 \\ell+k}{2} . \\tag{7}\n$$\n\nFrom $a_{2}\\ell^{2}$ and $2\\left(\\ell^{2}-k^{2}-k\\right)^{2}>2 k^{2}-\\ell^{2}+k$.', ""It is easy to see that $\\left(a_{n}\\right)$ is increasing for large enough $n$. Hence\n\n$$\na_{n+1}1$. Anastasia partitions the integers $1,2, \ldots, 2 m$ into $m$ pairs. Boris then chooses one integer from each pair and finds the sum of these chosen integers. Prove that Anastasia can select the pairs so that Boris cannot make his sum equal to $n$.","['Define the following ordered partitions:\n\n$$\n\\begin{aligned}\n& P_{1}=(\\{1,2\\},\\{3,4\\}, \\ldots,\\{2 m-1,2 m\\}), \\\\\n& P_{2}=(\\{1, m+1\\},\\{2, m+2\\}, \\ldots,\\{m, 2 m\\}), \\\\\n& P_{3}=(\\{1,2 m\\},\\{2, m+1\\},\\{3, m+2\\}, \\ldots,\\{m, 2 m-1\\}) .\n\\end{aligned}\n$$\n\nFor each $P_{j}$ we will compute the possible values for the expression $s=a_{1}+\\ldots+a_{m}$, where $a_{i} \\in P_{j, i}$ are the chosen integers. Here, $P_{j, i}$ denotes the $i$-th coordinate of the ordered partition $P_{j}$.\n\nWe will denote by $\\sigma$ the number $\\sum_{i=1}^{m} i=\\left(m^{2}+m\\right) / 2$.\n\n- Consider the partition $P_{1}$ and a certain choice with corresponding sum $s$. We find that\n\n$$\nm^{2}=\\sum_{i=1}^{m}(2 i-1) \\leq s \\leq \\sum_{i=1}^{m} 2 i=m^{2}+m\n$$\n\nHence, if $nm^{2}+m$, this partition gives a positive answer.\n\n- Consider the partition $P_{2}$ and a certain choice with corresponding $s$. We find that\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\quad(\\bmod m)\n$$\n\nHence, if $m^{2} \\leq n \\leq m^{2}+m$ and $n \\not \\equiv \\sigma(\\bmod m)$, this partition solves the problem.\n\n- Consider the partition $P_{3}$ and a certain choice with corresponding $s$. We set\n\n$$\nd_{i}= \\begin{cases}0 & \\text { if } a_{i}=i \\\\ 1, & \\text { if } a_{i} \\neq i\\end{cases}\n$$\n\nWe also put $d=\\sum_{i=1}^{m} d_{i}$, and note that $0 \\leq d \\leq m$. Note also that if $a_{i} \\neq i$, then $a_{i} \\equiv i-1$ $(\\bmod m)$. Hence, for all $a_{i} \\in P_{3, i}$ it holds that\n\n$$\na_{i} \\equiv i-d_{i} \\quad(\\bmod m)\n$$\n\nHence,\n\n$$\ns \\equiv \\sum_{i=1}^{m} a_{i} \\equiv \\sum_{i=1}^{m}\\left(i-d_{i}\\right) \\equiv \\sigma-d \\quad(\\bmod m)\n$$\n\nwhich can only be congruent to $\\sigma$ modulo $m$ if all $d_{i}$ are equal, which forces $s=\\left(m^{2}+m\\right) / 2$ or $s=\\left(3 m^{2}+m\\right) / 2$. Since $m>1$, it holds that\n\n$$\n\\frac{m^{2}+m}{2}m^{2}+m$, this partition gives a positive answer.\n\n- Consider the partition $P_{2}$ and a certain choice with corresponding $s$. We find that\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\quad(\\bmod m)\n$$\n\nHence, if $m^{2} \\leq n \\leq m^{2}+m$ and $n \\not \\equiv \\sigma(\\bmod m)$, this partition solves the problem.\n\nGiven the analysis of $P_{1}$ and $P_{2}$ as above, we may conclude (noting that $\\sigma \\equiv m(m+1) / 2(\\bmod m))$ that if $m$ is odd then $m^{2}$ and $m^{2}+m$ are the only candidates for counterexamples $n$, while if $m$ is even then $m^{2}+\\frac{m}{2}$ is the only candidate.\n\nThere are now various ways to proceed as alternatives to the partition $P_{3}$.\n\nConsider the partition $(\\{1, m+2\\},\\{2, m+3\\}, \\ldots,\\{m-1,2 m\\},\\{m, m+1\\})$. We consider possible sums $\\bmod m+1$. For the first $m-1$ pairs, the elements of each pair are congruent $\\bmod m+1$, so the sum of one element of each pair is $(\\bmod m+1)$ congruent to $\\frac{1}{2} m(m+1)-m$, which is congruent to 1 if $m+1$ is odd and $1+\\frac{m+1}{2}$ if $m+1$ is even. Now the elements of the last pair are congruent to -1 and 0 , so any achievable value of $n$ is congruent to 0 or 1 if $m+1$ is odd, and to 0 or 1 plus $\\frac{m+1}{2}$ if $m+1$ is even. If $m$ is even then $m^{2}+\\frac{m}{2} \\equiv 1+\\frac{m}{2}$, which is not congruent to 0 or 1 . If $m$ is odd then $m^{2} \\equiv 1$ and $m^{2}+m \\equiv 0$, neither of which can equal 0 or 1 plus $\\frac{m+1}{2}$.', 'Define the following ordered partitions:\n\n$$\n\\begin{aligned}\n& P_{1}=(\\{1,2\\},\\{3,4\\}, \\ldots,\\{2 m-1,2 m\\}), \\\\\n& P_{2}=(\\{1, m+1\\},\\{2, m+2\\}, \\ldots,\\{m, 2 m\\}), \\\\\n& P_{3}=(\\{1,2 m\\},\\{2, m+1\\},\\{3, m+2\\}, \\ldots,\\{m, 2 m-1\\}) .\n\\end{aligned}\n$$\n\nFor each $P_{j}$ we will compute the possible values for the expression $s=a_{1}+\\ldots+a_{m}$, where $a_{i} \\in P_{j, i}$ are the chosen integers. Here, $P_{j, i}$ denotes the $i$-th coordinate of the ordered partition $P_{j}$.\n\nWe will denote by $\\sigma$ the number $\\sum_{i=1}^{m} i=\\left(m^{2}+m\\right) / 2$.\n\n- Consider the partition $P_{1}$ and a certain choice with corresponding sum $s$. We find that\n\n$$\nm^{2}=\\sum_{i=1}^{m}(2 i-1) \\leq s \\leq \\sum_{i=1}^{m} 2 i=m^{2}+m\n$$\n\nHence, if $nm^{2}+m$, this partition gives a positive answer.\n\n- Consider the partition $P_{2}$ and a certain choice with corresponding $s$. We find that\n\n$$\ns \\equiv \\sum_{i=1}^{m} i \\equiv \\sigma \\quad(\\bmod m)\n$$\n\nHence, if $m^{2} \\leq n \\leq m^{2}+m$ and $n \\not \\equiv \\sigma(\\bmod m)$, this partition solves the problem.\n\nGiven the analysis of $P_{1}$ and $P_{2}$ as above, we may conclude (noting that $\\sigma \\equiv m(m+1) / 2(\\bmod m))$ that if $m$ is odd then $m^{2}$ and $m^{2}+m$ are the only candidates for counterexamples $n$, while if $m$ is even then $m^{2}+\\frac{m}{2}$ is the only candidate.\n\nThere are now various ways to proceed as alternatives to the partition $P_{3}$.\nConsider the partition $(\\{1, m\\},\\{2, m+1\\}, \\ldots,\\{m-1,2 m-2\\},\\{2 m-1,2 m\\})$, this time considering sums of elements of pairs $\\bmod m-1$. If $m-1$ is odd, the sum is congruent to 1 or 2 ; if $m-1$ is even, to 1 or 2 plus $\\frac{m-1}{2}$. If $m$ is even then $m^{2}+\\frac{m}{2} \\equiv 1+\\frac{m}{2}$, and this can only be congruent to 1 or 2 when $m=2$. If $m$ is odd, $m^{2}$ and $m^{2}+m$ are congruent to 1 and 2 , and these can only be congruent to 1 or 2 plus $\\frac{m-1}{2}$ when $m=3$. Now the cases of $m=2$ and $m=3$ need considering separately (by finding explicit partitions excluding each $n$ ).', 'Define the following ordered partitions:\n\n$$\n\\begin{aligned}\n& P_{1}=(\\{1,2\\},\\{3,4\\}, \\ldots,\\{2 m-1,2 m\\})\n\\end{aligned}\n$$\n\nFor each $P_{j}$ we will compute the possible values for the expression $s=a_{1}+\\ldots+a_{m}$, where $a_{i} \\in P_{j, i}$ are the chosen integers. Here, $P_{j, i}$ denotes the $i$-th coordinate of the ordered partition $P_{j}$.\n\nWe will denote by $\\sigma$ the number $\\sum_{i=1}^{m} i=\\left(m^{2}+m\\right) / 2$.\n\n- Consider the partition $P_{1}$ and a certain choice with corresponding sum $s$. We find that\n\n$$\nm^{2}=\\sum_{i=1}^{m}(2 i-1) \\leq s \\leq \\sum_{i=1}^{m} 2 i=m^{2}+m\n$$\n\nHence, if $nm^{2}+m$, this partition gives a positive answer.\n\nThis solution does not use modulo arguments. Use only $P_{1}$ from above to conclude that $m^{2} \\leq n \\leq m^{2}+m$. Now consider the partition $(\\{1,2 m\\},\\{2,3\\},\\{4,5\\}, \\ldots,\\{2 m-$ $2,2 m-1\\})$. If 1 is chosen from the first pair, the sum is at most $m^{2}$; if $2 m$ is chosen, the sum is at least $m^{2}+m$. So either $n=m^{2}$ or $n=m^{2}+m$. Now consider the partition $(\\{1,2 m-$ $1\\},\\{2,2 m\\},\\{3,4\\},\\{5,6\\}, \\ldots,\\{2 m-3,2 m-2\\})$. Sums of one element from each of the last $m-2$ pairs are in the range from $(m-2) m=m^{2}-2 m$ to $(m-2)(m+1)=m^{2}-m-2$ inclusive. Sums of one element from each of the first two pairs are $3,2 m+1$ and $4 m-1$. In the first case we have $n \\leq m^{2}-m+1m^{2}+m$. So these three partitions together have eliminated all $n$.']",,True,,, 2227,Geometry,,"Let $H$ be the orthocenter and $G$ be the centroid of acute-angled triangle $\triangle A B C$ with $A B \neq A C$. The line $A G$ intersects the circumcircle of $\triangle A B C$ at $A$ and $P$. Let $P^{\prime}$ be the reflection of $P$ in the line $B C$. Prove that $\angle C A B=60^{\circ}$ if and only if $H G=G P^{\prime}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_01ffde02433573694518g-1.jpg?height=691&width=1008&top_left_y=411&top_left_x=580)","[""Let $\\omega$ be the circumcircle of $\\triangle A B C$. Reflecting $\\omega$ in line $B C$, we obtain circle $\\omega^{\\prime}$ which, obviously, contains points $H$ and $P^{\\prime}$. Let $M$ be the midpoint of $B C$. As triangle $\\triangle A B C$ is acute-angled, then $H$ and $O$ lie inside this triangle.\n\nLet us assume that $\\angle C A B=60^{\\circ}$. Since\n\n$$\n\\angle C O B=2 \\angle C A B=120^{\\circ}=180^{\\circ}-60^{\\circ}=180^{\\circ}-\\angle C A B=\\angle C H B,\n$$\n\nhence $O$ lies on $\\omega^{\\prime}$. Reflecting $O$ in line $B C$, we obtain point $O^{\\prime}$ which lies on $\\omega$ and this point is the center of $\\omega^{\\prime}$. Then $O O^{\\prime}=2 O M=2 R \\cos \\angle C A B=A H$, so $A H=O O^{\\prime}=H O^{\\prime}=A O=R$, where $R$ is the radius of $\\omega$ and, naturally, of $\\omega^{\\prime}$. Then quadrilateral $A H O^{\\prime} O$ is a rhombus, so $A$ and $O^{\\prime}$ are symmetric to each other with respect to $H O$. As $H, G$ and $O$ are collinear (Euler line), then $\\angle G A H=\\angle H O^{\\prime} G$. Diagonals of quadrilateral $G O P O^{\\prime}$ intersects at $M$. Since $\\angle B O M=60^{\\circ}$, so\n\n$$\nO M=M O^{\\prime}=\\operatorname{ctg} 60^{\\circ} \\cdot M B=\\frac{M B}{\\sqrt{3}}\n$$\n\nAs $3 M O \\cdot M O^{\\prime}=M B^{2}=M B \\cdot M C=M P \\cdot M A=3 M G \\cdot M P$, then $G O P O^{\\prime}$ is a cyclic. Since $B C$ is a perpendicular bisector of $O O^{\\prime}$, so the circumcircle of quadrilateral $G O P O^{\\prime}$ is symmetrical with respect to $B C$. Thus $P^{\\prime}$ also belongs to the circumcircle of $G O P O^{\\prime}$, hence $\\angle G O^{\\prime} P^{\\prime}=\\angle G P P^{\\prime}$. Note that $\\angle G P P^{\\prime}=\\angle G A H$ since $A H \\| P P^{\\prime}$. And as it was proved $\\angle G A H=\\angle H O^{\\prime} G$, then $\\angle H O^{\\prime} G=\\angle G O^{\\prime} P^{\\prime}$. Thus triangles $\\triangle H O^{\\prime} G$ and $\\triangle G O^{\\prime} P^{\\prime}$ are equal and hence $H G=G P^{\\prime}$.\n\nNow we will prove that if $H G=G P^{\\prime}$ then $\\angle C A B=60^{\\circ}$. Reflecting $A$ with respect to $M$, we get $A^{\\prime}$. Then, as it was said in the first part of solution, points $B, C, H$ and $P^{\\prime}$ belong to $\\omega^{\\prime}$. Also it is clear that $A^{\\prime}$ belongs to $\\omega^{\\prime}$. Note that $H C \\perp C A^{\\prime}$ since $A B \\| C A^{\\prime}$ and hence $H A^{\\prime}$ is a diameter of $\\omega^{\\prime}$. Obviously, the center $O^{\\prime}$ of circle $\\omega^{\\prime}$ is midpoint of $H A^{\\prime}$. From $H G=G P^{\\prime}$ it follows that $\\triangle H G O^{\\prime}$ is equal to $\\triangle P^{\\prime} G O^{\\prime}$. Therefore $H$ and $P^{\\prime}$ are symmetric with respect to $G O^{\\prime}$. Hence $G O^{\\prime} \\perp H P^{\\prime}$ and $G O^{\\prime} \\| A^{\\prime} P^{\\prime}$. Let $H G$ intersect $A^{\\prime} P^{\\prime}$ at $K$ and $K \\not \\equiv O$ since $A B \\neq A C$. We conclude that $H G=G K$, because line $G O^{\\prime}$ is midline of the triangle $\\triangle H K A^{\\prime}$. Note that $2 G O=H G$. since $H O$ is Euler line of triangle $A B C$. So $O$ is midpoint of segment $G K$. Because of $\\angle C M P=\\angle C M P^{\\prime}$, then $\\angle G M O=\\angle O M P^{\\prime}$. Line $O M$, that passes through $O^{\\prime}$, is an external angle bisector of $\\angle P^{\\prime} M A^{\\prime}$. Also we know that $P^{\\prime} O^{\\prime}=O^{\\prime} A^{\\prime}$, then $O^{\\prime}$ is the midpoint of arc $P^{\\prime} M A^{\\prime}$ of the circumcircle of triangle $\\triangle P^{\\prime} M A^{\\prime}$. It\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9abe1efc82013409d084g-1.jpg?height=743&width=1068&top_left_y=103&top_left_x=556)\n\nfollows that quadrilateral $P^{\\prime} M O^{\\prime} A^{\\prime}$ is cyclic, then $\\angle O^{\\prime} M A^{\\prime}=\\angle O^{\\prime} P^{\\prime} A^{\\prime}=\\angle O^{\\prime} A^{\\prime} P^{\\prime}$. Let $O M$ and $P^{\\prime} A^{\\prime}$ intersect at $T$. Triangles $\\triangle T O^{\\prime} A^{\\prime}$ and $\\triangle A^{\\prime} O^{\\prime} M$ are similar, hence $O^{\\prime} A^{\\prime} / O^{\\prime} M=O^{\\prime} T / O^{\\prime} A^{\\prime}$. In the other words, $O^{\\prime} M \\cdot O^{\\prime} T=O^{\\prime} A^{\\prime 2}$. Using Menelaus' theorem for triangle $\\triangle H K A^{\\prime}$ and line $T O^{\\prime}$, we obtain that\n\n$$\n\\frac{A^{\\prime} O^{\\prime}}{O^{\\prime} H} \\cdot \\frac{H O}{O K} \\cdot \\frac{K T}{T A^{\\prime}}=3 \\cdot \\frac{K T}{T A^{\\prime}}=1\n$$\n\nIt follows that $K T / T A^{\\prime}=1 / 3$ and $K A^{\\prime}=2 K T$. Using Menelaus' theorem for triangle $T O^{\\prime} A^{\\prime}$ and line $H K$ we get that\n\n$$\n1=\\frac{O^{\\prime} H}{H A^{\\prime}} \\cdot \\frac{A^{\\prime} K}{K T} \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{1}{2} \\cdot 2 \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{T O}{O O^{\\prime}}\n$$\n\nIt means that $T O=O O^{\\prime}$, so $O^{\\prime} A^{\\prime 2}=O^{\\prime} M \\cdot O^{\\prime} T=O O^{\\prime 2}$. Hence $O^{\\prime} A^{\\prime}=O O^{\\prime}$ and, consequently, $O \\in \\omega^{\\prime}$. Finally we conclude that $2 \\angle C A B=\\angle B O C=180^{\\circ}-\\angle C A B$, so $\\angle C A B=60^{\\circ}$.\n![](https://cdn.mathpix.com/cropped/2023_12_21_9abe1efc82013409d084g-1.jpg?height=820&width=1636&top_left_y=1560&top_left_x=302)"", 'Let $O^{\\prime}$ and $G^{\\prime}$ denote the reflection of $O$ and $G$, respectively, with respect to the line $B C$. We then need to show $\\angle C A B=60^{\\circ}$ iff $G^{\\prime} H^{\\prime}=G^{\\prime} P$. Note that $\\triangle H^{\\prime} O P$ is isosceles and hence\n\n\n\n$G^{\\prime} H^{\\prime}=G^{\\prime} P$ is equivalent to $G^{\\prime}$ lying on the bisector $\\angle H^{\\prime} O P$. Let $\\angle H^{\\prime} A P=\\varepsilon$. By the assumption $A B \\neq A C$, we have $\\varepsilon \\neq 0$. Then $\\angle H^{\\prime} O P=2 \\angle H^{\\prime} A P=2 \\varepsilon$, hence $G^{\\prime} H^{\\prime}=G^{\\prime} P$ iff $\\angle G^{\\prime} O H^{\\prime}=\\varepsilon$. But $\\angle G O^{\\prime} H=\\angle G^{\\prime} O H^{\\prime}$. Let $D$ be the midpoint of $O O^{\\prime}$. It is known that $\\angle G D O=\\angle G A H=\\varepsilon$. Let $F$ be the midpoint of $H G$. Then $H G=F O$ (Euler line). Let $\\angle G O^{\\prime} H=\\delta$. We then have to show $\\delta=\\varepsilon$ iff $\\angle C A B=60^{\\circ}$. But by similarity $\\left(\\triangle G D O \\sim \\triangle F O^{\\prime} O\\right.$ ) we have $\\angle F O^{\\prime} O=\\varepsilon$. Consider the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$. By the sine law and since the segments $H G$ and $F O$ are of equal length we deduce that the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$ are symmetric with respect to the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$. Obviously, $O^{\\prime}$ is the common point of these two circles. Hence $O^{\\prime}$ must be fixed after the symmetry about the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$ so we have $\\varepsilon=\\delta$ iff $\\triangle H O O^{\\prime}$ is isosceles. But $H O^{\\prime}=H^{\\prime} O=R$, and so\n\n$$\n\\varepsilon=\\delta \\Longleftrightarrow O O^{\\prime}=R \\Longleftrightarrow O D=\\frac{R}{2} \\Longleftrightarrow \\cos \\angle C A B=\\frac{1}{2} \\Longleftrightarrow \\angle C A B=60^{\\circ} \\text {. }\n$$']",['证明题，略'],True,,Need_human_evaluate, 2227,Geometry,,"Let $H$ be the orthocenter and $G$ be the centroid of acute-angled triangle $\triangle A B C$ with $A B \neq A C$. The line $A G$ intersects the circumcircle of $\triangle A B C$ at $A$ and $P$. Let $P^{\prime}$ be the reflection of $P$ in the line $B C$. Prove that $\angle C A B=60^{\circ}$ if and only if $H G=G P^{\prime}$. ","[""Let $\\omega$ be the circumcircle of $\\triangle A B C$. Reflecting $\\omega$ in line $B C$, we obtain circle $\\omega^{\\prime}$ which, obviously, contains points $H$ and $P^{\\prime}$. Let $M$ be the midpoint of $B C$. As triangle $\\triangle A B C$ is acute-angled, then $H$ and $O$ lie inside this triangle.\n\nLet us assume that $\\angle C A B=60^{\\circ}$. Since\n\n$$\n\\angle C O B=2 \\angle C A B=120^{\\circ}=180^{\\circ}-60^{\\circ}=180^{\\circ}-\\angle C A B=\\angle C H B,\n$$\n\nhence $O$ lies on $\\omega^{\\prime}$. Reflecting $O$ in line $B C$, we obtain point $O^{\\prime}$ which lies on $\\omega$ and this point is the center of $\\omega^{\\prime}$. Then $O O^{\\prime}=2 O M=2 R \\cos \\angle C A B=A H$, so $A H=O O^{\\prime}=H O^{\\prime}=A O=R$, where $R$ is the radius of $\\omega$ and, naturally, of $\\omega^{\\prime}$. Then quadrilateral $A H O^{\\prime} O$ is a rhombus, so $A$ and $O^{\\prime}$ are symmetric to each other with respect to $H O$. As $H, G$ and $O$ are collinear (Euler line), then $\\angle G A H=\\angle H O^{\\prime} G$. Diagonals of quadrilateral $G O P O^{\\prime}$ intersects at $M$. Since $\\angle B O M=60^{\\circ}$, so\n\n$$\nO M=M O^{\\prime}=\\operatorname{ctg} 60^{\\circ} \\cdot M B=\\frac{M B}{\\sqrt{3}}\n$$\n\nAs $3 M O \\cdot M O^{\\prime}=M B^{2}=M B \\cdot M C=M P \\cdot M A=3 M G \\cdot M P$, then $G O P O^{\\prime}$ is a cyclic. Since $B C$ is a perpendicular bisector of $O O^{\\prime}$, so the circumcircle of quadrilateral $G O P O^{\\prime}$ is symmetrical with respect to $B C$. Thus $P^{\\prime}$ also belongs to the circumcircle of $G O P O^{\\prime}$, hence $\\angle G O^{\\prime} P^{\\prime}=\\angle G P P^{\\prime}$. Note that $\\angle G P P^{\\prime}=\\angle G A H$ since $A H \\| P P^{\\prime}$. And as it was proved $\\angle G A H=\\angle H O^{\\prime} G$, then $\\angle H O^{\\prime} G=\\angle G O^{\\prime} P^{\\prime}$. Thus triangles $\\triangle H O^{\\prime} G$ and $\\triangle G O^{\\prime} P^{\\prime}$ are equal and hence $H G=G P^{\\prime}$.\n\nNow we will prove that if $H G=G P^{\\prime}$ then $\\angle C A B=60^{\\circ}$. Reflecting $A$ with respect to $M$, we get $A^{\\prime}$. Then, as it was said in the first part of solution, points $B, C, H$ and $P^{\\prime}$ belong to $\\omega^{\\prime}$. Also it is clear that $A^{\\prime}$ belongs to $\\omega^{\\prime}$. Note that $H C \\perp C A^{\\prime}$ since $A B \\| C A^{\\prime}$ and hence $H A^{\\prime}$ is a diameter of $\\omega^{\\prime}$. Obviously, the center $O^{\\prime}$ of circle $\\omega^{\\prime}$ is midpoint of $H A^{\\prime}$. From $H G=G P^{\\prime}$ it follows that $\\triangle H G O^{\\prime}$ is equal to $\\triangle P^{\\prime} G O^{\\prime}$. Therefore $H$ and $P^{\\prime}$ are symmetric with respect to $G O^{\\prime}$. Hence $G O^{\\prime} \\perp H P^{\\prime}$ and $G O^{\\prime} \\| A^{\\prime} P^{\\prime}$. Let $H G$ intersect $A^{\\prime} P^{\\prime}$ at $K$ and $K \\not \\equiv O$ since $A B \\neq A C$. We conclude that $H G=G K$, because line $G O^{\\prime}$ is midline of the triangle $\\triangle H K A^{\\prime}$. Note that $2 G O=H G$. since $H O$ is Euler line of triangle $A B C$. So $O$ is midpoint of segment $G K$. Because of $\\angle C M P=\\angle C M P^{\\prime}$, then $\\angle G M O=\\angle O M P^{\\prime}$. Line $O M$, that passes through $O^{\\prime}$, is an external angle bisector of $\\angle P^{\\prime} M A^{\\prime}$. Also we know that $P^{\\prime} O^{\\prime}=O^{\\prime} A^{\\prime}$, then $O^{\\prime}$ is the midpoint of arc $P^{\\prime} M A^{\\prime}$ of the circumcircle of triangle $\\triangle P^{\\prime} M A^{\\prime}$. It\n\n\n\n\n\nfollows that quadrilateral $P^{\\prime} M O^{\\prime} A^{\\prime}$ is cyclic, then $\\angle O^{\\prime} M A^{\\prime}=\\angle O^{\\prime} P^{\\prime} A^{\\prime}=\\angle O^{\\prime} A^{\\prime} P^{\\prime}$. Let $O M$ and $P^{\\prime} A^{\\prime}$ intersect at $T$. Triangles $\\triangle T O^{\\prime} A^{\\prime}$ and $\\triangle A^{\\prime} O^{\\prime} M$ are similar, hence $O^{\\prime} A^{\\prime} / O^{\\prime} M=O^{\\prime} T / O^{\\prime} A^{\\prime}$. In the other words, $O^{\\prime} M \\cdot O^{\\prime} T=O^{\\prime} A^{\\prime 2}$. Using Menelaus' theorem for triangle $\\triangle H K A^{\\prime}$ and line $T O^{\\prime}$, we obtain that\n\n$$\n\\frac{A^{\\prime} O^{\\prime}}{O^{\\prime} H} \\cdot \\frac{H O}{O K} \\cdot \\frac{K T}{T A^{\\prime}}=3 \\cdot \\frac{K T}{T A^{\\prime}}=1\n$$\n\nIt follows that $K T / T A^{\\prime}=1 / 3$ and $K A^{\\prime}=2 K T$. Using Menelaus' theorem for triangle $T O^{\\prime} A^{\\prime}$ and line $H K$ we get that\n\n$$\n1=\\frac{O^{\\prime} H}{H A^{\\prime}} \\cdot \\frac{A^{\\prime} K}{K T} \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{1}{2} \\cdot 2 \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{T O}{O O^{\\prime}}\n$$\n\nIt means that $T O=O O^{\\prime}$, so $O^{\\prime} A^{\\prime 2}=O^{\\prime} M \\cdot O^{\\prime} T=O O^{\\prime 2}$. Hence $O^{\\prime} A^{\\prime}=O O^{\\prime}$ and, consequently, $O \\in \\omega^{\\prime}$. Finally we conclude that $2 \\angle C A B=\\angle B O C=180^{\\circ}-\\angle C A B$, so $\\angle C A B=60^{\\circ}$.\n"", 'Let $O^{\\prime}$ and $G^{\\prime}$ denote the reflection of $O$ and $G$, respectively, with respect to the line $B C$. We then need to show $\\angle C A B=60^{\\circ}$ iff $G^{\\prime} H^{\\prime}=G^{\\prime} P$. Note that $\\triangle H^{\\prime} O P$ is isosceles and hence\n\n\n\n$G^{\\prime} H^{\\prime}=G^{\\prime} P$ is equivalent to $G^{\\prime}$ lying on the bisector $\\angle H^{\\prime} O P$. Let $\\angle H^{\\prime} A P=\\varepsilon$. By the assumption $A B \\neq A C$, we have $\\varepsilon \\neq 0$. Then $\\angle H^{\\prime} O P=2 \\angle H^{\\prime} A P=2 \\varepsilon$, hence $G^{\\prime} H^{\\prime}=G^{\\prime} P$ iff $\\angle G^{\\prime} O H^{\\prime}=\\varepsilon$. But $\\angle G O^{\\prime} H=\\angle G^{\\prime} O H^{\\prime}$. Let $D$ be the midpoint of $O O^{\\prime}$. It is known that $\\angle G D O=\\angle G A H=\\varepsilon$. Let $F$ be the midpoint of $H G$. Then $H G=F O$ (Euler line). Let $\\angle G O^{\\prime} H=\\delta$. We then have to show $\\delta=\\varepsilon$ iff $\\angle C A B=60^{\\circ}$. But by similarity $\\left(\\triangle G D O \\sim \\triangle F O^{\\prime} O\\right.$ ) we have $\\angle F O^{\\prime} O=\\varepsilon$. Consider the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$. By the sine law and since the segments $H G$ and $F O$ are of equal length we deduce that the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$ are symmetric with respect to the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$. Obviously, $O^{\\prime}$ is the common point of these two circles. Hence $O^{\\prime}$ must be fixed after the symmetry about the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$ so we have $\\varepsilon=\\delta$ iff $\\triangle H O O^{\\prime}$ is isosceles. But $H O^{\\prime}=H^{\\prime} O=R$, and so\n\n$$\n\\varepsilon=\\delta \\Longleftrightarrow O O^{\\prime}=R \\Longleftrightarrow O D=\\frac{R}{2} \\Longleftrightarrow \\cos \\angle C A B=\\frac{1}{2} \\Longleftrightarrow \\angle C A B=60^{\\circ} \\text {. }\n$$']",,True,,, 2228,Algebra,,"There are $n \geqslant 3$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$. For each $1 \leqslant i \leqslant n$ we let $b_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ (here we define $a_{0}$ to be $a_{n}$ and $a_{n+1}$ to be $a_{1}$ ). Assume that for all $i$ and $j$ in the range 1 to $n$, we have $a_{i} \leqslant a_{j}$ if and only if $b_{i} \leqslant b_{j}$. Prove that $a_{1}=a_{2}=\cdots=a_{n}$.","['Suppose that not all $a_{i}$ are equal. Consider an index $i$ such that $a_{i}$ is maximal and $a_{i+1}c_{\\ell-1} \\geqslant 1$\n\nFrom $c_{\\ell-1} c_{\\ell} c_{\\ell+1}c_{\\ell} \\geqslant 1$, which is a contradiction to our choice of $\\ell$.\n\nCase 2: $c_{\\ell-1} \\geqslant c_{\\ell} \\geqslant 1$\n\nOnce again looking at the inequality (5) we can find that\n\n$$\nc_{\\ell-2} c_{\\ell-1} c_{\\ell} \\geqslant c_{\\ell-1}^{2} c_{\\ell} \\Longrightarrow c_{\\ell-2} \\geqslant c_{\\ell-1} \\tag{7}\n$$\n\nNote that we only needed $c_{\\ell-1} \\geqslant c_{\\ell} \\geqslant 1$ to show $c_{\\ell-2} \\geqslant c_{\\ell-1} \\geqslant 1$. So using induction we can easily show $c_{\\ell-s-1} \\geqslant c_{\\ell-s}$ for all $s$.\n\nSo\n\n$$\nc_{1} \\leqslant c_{2} \\leqslant \\cdots \\leqslant c_{n} \\leqslant c_{1} \\tag{8}\n$$\n\na contradiction to our innitial assumption.\n\nSo our innitial assumtion must have been wrong, which implies that all the $a_{i}$ must have been equal from the start.']",,True,,, 2229,Geometry,,"We are given an acute triangle $A B C$. Let $D$ be the point on its circumcircle such that $A D$ is a diameter. Suppose that points $K$ and $L$ lie on segments $A B$ and $A C$, respectively, and that $D K$ and $D L$ are tangent to circle $A K L$. Show that line $K L$ passes through the orthocentre of $A B C$. The altitudes of a triangle meet at its orthocentre.","['\nLet $M$ be the midpoint of $K L$. We will prove that $M$ is the orthocentre of $A B C$. Since $D K$ and $D L$ are tangent to the same circle, $|D K|=|D L|$ and hence $D M \\perp K L$. The theorem of Thales in circle $A B C$ also gives $D B \\perp B A$ and $D C \\perp C A$. The right angles then give that quadrilaterals $B D M K$ and $D M L C$ are cyclic.\n\nIf $\\angle B A C=\\alpha$, then clearly $\\angle D K M=\\angle M L D=\\alpha$ by angle in the alternate segment of circle $A K L$, and so $\\angle M D K=\\angle L D M=\\frac{\\pi}{2}-\\alpha$, which thanks to cyclic quadrilaterals gives $\\angle M B K=\\angle L C M=\\frac{\\pi}{2}-\\alpha$. From this, we have $B M \\perp A C$ and $C M \\perp A B$, and so $M$ indeed is the orthocentre of $A B C$.', 'Let $A B C$ be a triangle with circumcircle $\\Gamma$. Let $X$ be a point in the plane. The Simson line (Wallace-Simson line) is defined via the following theorem. Drop perpendiculars from $X$ to each of the three side lines of $A B C$. The feet of these perpendiculars are collinear (on\n\n\n\nthe Simson line of $X$ ) if and only if $X$ lies of $\\Gamma$. The Simson line of $X$ in the circumcircle bisects the line segment $X H$ where $H$ is the orthocentre of triangle $A B C$. See Figure 2\n\n\n\nFigure 2: The Wallace-Simson configuration\n\nWhen $X$ is on $\\Gamma$, we can enlarge from $X$ with scale factor 2 (a homothety) to take the Simson line to the doubled Simson line which passes through the orthocentre $H$ and contains the reflections of $X$ in each of the three sides of $A B C$.\n\n\n\nFigure 3: Three circles do the work\n\nLet $\\Gamma$ be the circle $A B C, \\Sigma$ be the circle $A K L$ with centre $O$, and $\\Omega$ be the circle on diameter $O D$ so $K$ and $L$ are on this circle by converse of Thales. Let $\\Omega$ and $\\Gamma$ meet at $D$ and $F$. By Thales in both circles, $\\angle A F D$ and $\\angle O F D$ are both right angles so $A O F$ is a line. Let $A F$ meet $\\Sigma$ again at $T$ so $A T$ (containing $O$ ) is a diameter of this circle and by Thales, $T L \\perp A C$.\n\nLet $G$ (on $\\Sigma$ ) be the reflection of $K$ in $A F$. Now $A T$ is the internal angle bisector of $\\angle G A K$ so, by an upmarket use of angles in the same segment (of $\\Sigma$ ), $T L$ is the internal angle bisector of $\\angle G L K$. Thus the line $G L$ is the reflection of the line $K L$ in $T L$, and so also the reflection of $K L$ in the line $A C$ (internal and external angle bisectors).\n\nOur next project is to show that $L G F$ are collinear. Well $\\angle F L K=\\angle F O K$ (angles in the same segment of $\\Omega$ ) and $\\angle G L K=\\angle G A K$ (angles in the same segment of $\\Sigma$ ) $=2 \\angle O A K$ $(A K G$ is isosceles with apex $A)=\\angle T O K$ (since $O A K$ is isosceles with apex $O$, and this is an external angle at $O$ ). The point $T$ lies in the interior of the line segment $F O$ so $\\angle T O K=\\angle F O K$. Therefore $\\angle F L K=\\angle G L K$ so $L G F$ is a line.\n\nNow from the second paragraph, $F$ is on the reflection of $K L$ in $A C$. By symmetry, $F$ is also on the reflection of $K L$ in $A B$. Therefore the reflections of $F$ in $A B$ and $A C$ are both on $K L$ which must therefore be the doubled Wallace-Simson line of $F$. Therefore the orthocentre of $A B C$ lies on $K L$.', 'Let $H$ be the orthocentre of triangle $A B C$ and $\\Sigma$ the circumcircle of $A K L$ with centre $O$. Let $\\Omega$ be the circle with diameter $O D$, which contains $K$ and $L$ by Thales, and let $\\Gamma$ be the circumcircle of $A B C$ containing $D$. Denote the second intersection of $\\Omega$ and $\\Gamma$ by $F$. Since $O D$ and $A D$ are diameters of $\\Omega$ and $\\Gamma$ we have $\\angle O F D=\\frac{\\pi}{2}=\\angle A F D$, so the points $A, O, F$ are collinear. Let $M$ and $N$ be the second intersections of $C H$ and $B H$ with $\\Gamma$, respectively. It is well-known that $M$ and $N$ are the reflections of $H$ in $A B$ and $A C$, respectively (because $\\angle N C A=\\angle N B A=\\angle A C M=\\angle A B M$ ). By collinearity of $A, O, F$ and the angles in $\\Gamma$ we have\n\n$$\n\\angle N F O=\\angle N F A=\\angle N B A=\\frac{\\pi}{2}-\\angle B A C=\\frac{\\pi}{2}-\\angle K A L .\n$$\n\nSince $D L$ is tangent to $\\Sigma$ we obtain\n\n$$\n\\angle N F O=\\frac{\\pi}{2}-\\angle K L D=\\angle L D O,\n$$\n\nwhere the last equality follows from the fact that $O D$ is bisector of $\\angle L D K$ since $L D$\n\n\n\nFigure 4: Diagram to Solution 3\n\n\n\nand $K D$ are tangent to $\\Sigma$. Furthermore, $\\angle L D O=\\angle L F O$ since these are angles in $\\Omega$. Hence, $\\angle N F O=\\angle L F O$, which implies that points $N, L, F$ are collinear. Similarly points $M, K, F$ are collinear. Since $N$ and $M$ are reflections of $H$ in $A C$ and $A B$ we have\n\n$$\n\\angle L H N=\\angle H N L=\\angle B N F=\\angle B M F=\\angle B M K=\\angle K H B .\n$$\n\nHence,\n\n$$\n\\angle L H K=\\angle L H N+\\angle N H K=\\angle K H B+\\angle N H K=\\pi\n$$\n\nand the points $L, H, K$ are collinear.', 'Let $M$ and $N$ be the reflections of the orthocentre in $A B$ and $A C$. Let $\\angle B A C=\\alpha$. Then $\\angle N D M=\\pi-\\angle M A N=\\pi-2 \\alpha$.\n\nLet $M K$ and $N L$ intersect at $F$. See Figure 3.\n\nClaim. $\\angle N F M=\\pi-2 \\alpha$, so $F$ lies on the circumcircle.\n\nProof. Since $K D$ and $L D$ are tangents to circle $A K L$, we have $|D K|=|D L|$ and $\\angle D K L=\\angle K L D=\\alpha$, so $\\angle L D K=\\pi-2 \\alpha$.\n\nBy definition of $M, N$ and $D, \\angle M N D=\\angle A N D-\\angle A N M=\\frac{\\pi}{2}-\\left(\\frac{\\pi}{2}-\\alpha\\right)=\\alpha$ and analogously $\\angle D M N=\\alpha$. Hence $|D M|=|D N|$.\n\nFrom $\\angle N D M=\\angle L D K=\\pi-2 \\alpha$ if follows that $\\angle L D N=\\angle K D M$. Since $|D K|=|D L|$ and $|D M|=|D N|$, triangles $M D K$ and $N D L$ are related by a rotation about $D$ through angle $\\pi-2 \\alpha$, and hence the angle between $M K$ and $N L$ is $\\pi-2 \\alpha$, which proved the claim.\n\nWe now finish as below:\n\n$$\n\\begin{gathered}\n\\angle M H K=\\angle K M H=\\angle F M C=\\angle F A C, \\\\\n\\angle L H N=\\angle H N L=\\angle B N F=\\angle B A F .\n\\end{gathered}\n$$\n\nAs $\\angle B A F+\\angle F A C=\\alpha$, we have $\\angle L H K=\\alpha+\\angle N H M=\\alpha+\\pi-\\alpha=\\pi$, so $H$ lies on $K L$.', 'Since $A D$ is a diameter, it is well known that $D B H C$ is a parallelogram (indeed, both $B D$ and $C H$ are perpendicular to $A B$, hence parallel, and similarly for $D C \\| B H)$. Let $B^{\\prime}, C^{\\prime}$ be the reflections of $D$ in lines $A K B$ and $A L C$, respectively; since $A B D$ and $A C D$ are right angles, these are also the factor-2 homotheties of $B$ and $C$ with respect to $D$, hence $H$ is the midpoint of $B^{\\prime} C^{\\prime}$. We will prove that $B^{\\prime} K C^{\\prime} L$ is a parallelogram: it will then follow that the midpoint of $B^{\\prime} C^{\\prime}$, which is $H$, is also the midpoint of $K L$, and in particular is on the line, as we wanted to show.\n\nWe will prove $B^{\\prime} K C^{\\prime} L$ is a parallelogram by showing that $B^{\\prime} K$ and $C^{\\prime} L$ are the same length and direction. Indeed, for lengths we have $K B^{\\prime}=K D=L D=L C^{\\prime}$, where the first and last equalities arise from the reflections defining $B^{\\prime}$ and $C^{\\prime}$, and the middle one\n\n\n\nis equality of tangents. For directions, let $\\alpha, \\beta, \\gamma$ denote the angles of triangle $A K L$. Immediate angle chasing in the circle $A K L$, and the properties of the reflections, yield\n\n$$\n\\begin{gathered}\n\\angle C^{\\prime} L C=\\angle C L D=\\angle A K L=\\beta \\\\\n\\angle B K B^{\\prime}=\\angle D K B=\\angle K L A=\\gamma \\\\\n\\angle L D K=2 \\alpha-\\pi\n\\end{gathered}\n$$\n\nand therefore in directed angles $(\\bmod 2 \\pi)$ we have\n\n$\\angle\\left(C^{\\prime} L, B^{\\prime} K\\right)=\\angle C^{\\prime} L C+\\angle C L D+\\angle L D K+\\angle D K B+\\angle B K B^{\\prime}=2 \\alpha+2 \\beta+2 \\gamma-\\pi=\\pi$\n\nand hence $C^{\\prime} L$ and $B^{\\prime} K$ are parallel and in opposite directions, i.e. $C^{\\prime} L$ and $K B^{\\prime}$ are in the same direction, as claimed.', 'There are a number of ""phantom point"" arguments which define $K^{\\prime}$ and $L^{\\prime}$ in terms of angles and then deduce that these points are actually $K$ and $L$.\n\nNote: In these solutions it is necessary to show that $K$ and $L$ are uniquely determined by the conditions of the problem. One example of doing this is the following:\n\nTo prove uniqueness of $K$ and $L$, let us consider that there exist two other points $K^{\\prime}$ and $L^{\\prime}$ that satisfy the same properties ( $K^{\\prime}$ on $A B$ and $L^{\\prime}$ on $A C$ such that $D K^{\\prime}$ and $D L^{\\prime}$ are tangent to the circle $\\left.A K^{\\prime} L^{\\prime}\\right)$.\n\nThen, we have that $D K=D L$ and $D K^{\\prime}=D L^{\\prime}$. We also have that $\\angle K D L=\\angle K^{\\prime} D L^{\\prime}=$ $\\pi-2 \\angle A$. Hence, we deduce $\\angle K D K^{\\prime}=\\angle K D L-\\angle K^{\\prime} D L=\\angle K^{\\prime} D L^{\\prime}-\\angle K^{\\prime} D L=\\angle L D L^{\\prime}$ Thus we have that $\\triangle K D K^{\\prime} \\equiv \\triangle L D L^{\\prime}$, so we deduce $\\angle D K A=\\angle D K K^{\\prime}=\\angle D L L^{\\prime}=$ $\\pi-\\angle A L D$. This implies that $A K D L$ is concyclic, which is clearly a contradiction since $\\angle K A L+\\angle K D L=\\pi-\\angle B A C$.', 'We will use the usual complex number notation, where we will use a capital letter (like $Z$ ) to denote the point associated to a complex number (like $z$ ). Consider $\\triangle A K L$ on the unit circle. So, we have $a \\cdot \\bar{a}=k \\cdot \\bar{k}=l \\cdot \\bar{l}=1$ As point $D$ is the intersection of the tangents to the unit circle at $K$ and $L$, we have that\n\n$$\nd=\\frac{2 k l}{k+l} \\text { and } \\bar{d}=\\frac{2}{k+l}\n$$\n\nDefining $B$ as the foot of the perpendicular from $D$ on the line $A K$, and $C$ as the foot of the perpendicular from $D$ on the line $A L$, we have the formulas:\n\n$$\nb=\\frac{1}{2}\\left(d+\\frac{(a-k) \\bar{d}+\\bar{a} k-a \\bar{k}}{\\bar{a}-\\bar{k}}\\right)\n$$\n\n\n\n$$\nc=\\frac{1}{2}\\left(d+\\frac{(a-l) \\bar{d}+\\bar{a} l-a \\bar{l}}{\\bar{a}-\\bar{l}}\\right)\n$$\n\nSimplyfing these formulas, we get:\n\n$$\n\\begin{gathered}\nb=\\frac{1}{2}\\left(d+\\frac{(a-k) \\frac{2}{k+l}+\\frac{k}{a}-\\frac{a}{k}}{\\frac{1}{a}-\\frac{1}{k}}\\right)=\\frac{1}{2}\\left(d+\\frac{\\frac{2(a-k)}{k+l}+\\frac{k^{2}-a^{2}}{a k}}{\\frac{k-a}{a k}}\\right) \\\\\nb=\\frac{1}{2}\\left(\\frac{2 k l}{k+l}-\\frac{2 a k}{k+l}+(a+k)\\right)=\\frac{k(l-a)}{k+l}+\\frac{1}{2}(k+a) \\\\\nc=\\frac{1}{2}\\left(d+\\frac{(a-l) \\frac{2}{k+l}+\\frac{l}{a}-\\frac{a}{l}}{\\frac{1}{a}-\\frac{1}{l}}\\right)=\\frac{1}{2}\\left(d+\\frac{\\frac{2(a-l)}{k+l}+\\frac{l^{2}-a^{2}}{a l}}{\\frac{l-a}{a l}}\\right) \\\\\nc=\\frac{1}{2}\\left(\\frac{2 k l}{k+l}-\\frac{2 a l}{k+l}+(a+l)\\right)=\\frac{l(k-a)}{k+l}+\\frac{1}{2}(l+a)\n\\end{gathered}\n$$\n\nLet $O$ be the the circumcenter of triangle $\\triangle A B C$. As $A D$ is the diameter of this circle, we have that:\n\n$$\no=\\frac{a+d}{2}\n$$\n\nDefining $H$ as the orthocentre of the $\\triangle A B C$, we get that:\n\n$$\n\\begin{gathered}\nh=a+b+c-2 \\cdot o=a+\\left(\\frac{k(l-a)}{k+l}+\\frac{1}{2}(k+a)\\right)+\\left(\\frac{l(k-a)}{k+l}+\\frac{1}{2}(l+a)\\right)-(a+d) \\\\\nh=a+\\frac{2 k l}{k+l}-\\frac{a(k+l)}{k+l}+\\frac{1}{2} k++\\frac{1}{2} l++a-\\left(a+\\frac{2 k l}{k+l}\\right) \\\\\nh=\\frac{1}{2}(k+l)\n\\end{gathered}\n$$\n\nHence, we conclude that $H$ is the midpoint of $K L$, so $H, K, L$ are collinear.', 'Let us employ the barycentric coordinates. Set $A(1,0,0), K(0,1,0), L(0,0,1)$.\n\nThe tangent at $K$ of $(A K L)$ is $a^{2} z+c^{2} x=0$, and the tangent of of $L$ at $(A K L)$ is $a^{2} y+b^{2} x=0$. Their intersection is\n\n$$\nD\\left(-a^{2}: b^{2}: c^{2}\\right)\n$$\n\nSince $B \\in A K$, we can let $B(1-t, t, 0)$. Solving for $\\overrightarrow{A B} \\cdot \\overrightarrow{B D}=0$ gives\n\n$$\nt=\\frac{3 b^{2}+c^{2}-a^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)} \\Longrightarrow B=\\left(\\frac{-a^{2}-b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, \\frac{-a^{2}+3 b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, 0\\right)\n$$\n\nLikewise, $C$ has the coordinate\n\n$$\nC=\\left(\\frac{-a^{2}+b^{2}-c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, 0, \\frac{-a^{2}+b^{2}+3 c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right) .\n$$\n\n\n\nThe altitude from $B$ for triangle $A B C$ is\n\n$$\n-b^{2}\\left(x-z-\\frac{-a^{2}-b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)+\\left(c^{2}-a^{2}\\right)\\left(y-\\frac{-a^{2}+3 b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)=0 .\n$$\n\nAlso the altitude from $C$ for triangle $A B C$ is\n\n$$\n-c^{2}\\left(x-y-\\frac{-a^{2}+b^{2}-c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)+\\left(a^{2}-b^{2}\\right)\\left(z-\\frac{-a^{2}+b^{2}+3 c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)=0 .\n$$\n\nThe intersection of these two altitudes, which is the orthocenter of triangle $A B C$, has the barycentric coordinate\n\n$$\nH=(0,1 / 2,1 / 2)\n$$\n\nwhich is the midpoint of the segment $K L$.']",,True,,, 2230,Combinatorics,,"Let $k$ be a positive integer. Lexi has a dictionary $\mathcal{D}$ consisting of some $k$-letter strings containing only the letters $A$ and $B$. Lexi would like to write either the letter $A$ or the letter $B$ in each cell of a $k \times k$ grid so that each column contains a string from $\mathcal{D}$ when read from top-to-bottom and each row contains a string from $\mathcal{D}$ when read from left-to-right. What is the smallest integer $m$ such that if $\mathcal{D}$ contains at least $m$ different strings, then Lexi can fill her grid in this manner, no matter what strings are in $\mathcal{D}$ ?","['We claim the minimum value of $m$ is $2^{k-1}$.\n\nFirstly, we provide a set $\\mathcal{S}$ of size $2^{k-1}-1$ for which Lexi cannot fill her grid. Consider the set of all length- $k$ strings containing only $A \\mathrm{~s}$ and $B \\mathrm{~s}$ which end with a $B$, and remove the string consisting of $k$ $B \\mathrm{~s}$. Clearly there are 2 independent choices for each of the first $k-1$ letters and 1 for the last letter, and since exactly one string is excluded, there must be exactly $2^{k-1}-1$ strings in this set.\n\nSuppose Lexi tries to fill her grid. For each row to have a valid string, it must end in a $B$. But then the right column would necessarily contain $k B \\mathrm{~s}$, and not be in our set. Thus, Lexi cannot fill her grid with our set, and we must have $m \\geqslant 2^{k-1}$.\n\nNow, consider any set $\\mathcal{S}$ with at least $2^{k-1}$ strings. Clearly, if $\\mathcal{S}$ contained either the uniform string with $k A \\mathrm{~s}$ or the string with $k B \\mathrm{~s}$, then Lexi could fill her grid with all of the relevant letters and each row and column would contain that string.\n\nConsider the case where $\\mathcal{S}$ contains neither of those strings. Among all $2^{k}$ possible length$k$ strings with $A \\mathrm{~s}$ and $B \\mathrm{~s}$, each has a complement which corresponds to the string with $B$ s in every position where first string had $A$ s and vice-versa. Clearly, the string with all $A$ s is paired with the string with all $B$ s. We may assume that we do not take the two uniform strings and thus applying the pigeonhole principle to the remaining set of strings, we must have two strings which are complementary.\n\nLet this pair of strings be $\\ell, \\ell^{\\prime} \\in \\mathcal{S}$ in some order. Define the set of indices $\\mathcal{J}$ corresponding to the $A \\mathrm{~s}$ in $\\ell$ and thus the $B \\mathrm{~s}$ in $\\ell^{\\prime}$, and all other indices (not in $\\mathcal{J}$ ) correspond to $B \\mathrm{~s}$ in $\\ell$ (and thus $A$ s in $\\ell^{\\prime}$ ). Then, we claim that Lexi puts an $A$ in the cell in row $r$, column $c$ if $r, c \\in \\mathcal{J}$ or $r, c \\notin \\mathcal{J}$, and a $B$ otherwise, each row and column contains a string in $\\mathcal{S}$.\n\nWe illustrate this with a simple example: If $k=6$ and we have that $A A A B A B$ and $B B B A B A$ are both in the dictionary, then Lexi could fill the table as follows:\n\n| A | A | A | B | A | B |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| A | A | A | B | A | B |\n| A | A | A | B | A | B |\n| B | B | B | A | B | A |\n| A | A | A | B | A | B |\n| B | B | B | A | B | A |\n\nSuppose we are looking at row $i$ or column $i$ for $i \\in \\mathcal{J}$. Then by construction the string in this row/column contains $A \\mathrm{~s}$ at indices $k$ with $k \\in \\mathcal{J}$ and $B$ s elsewhere, and thus is precisely $\\ell$. Suppose instead we are looking at row $i$ or column $i$ for $i \\notin \\mathcal{J}$. Then again by construction the string in this row/column contains $A$ s at indices $k$ with $k \\notin \\mathcal{J}$ and $B$ s elsewhere, and thus is precisely $\\ell^{\\prime}$. So each row and column indeed contains a string in $\\mathcal{S}$.\n\nThus, for any $\\mathcal{S}$ with $|\\mathcal{S}| \\geqslant 2^{k-1}$, Lexi can definitely fill the grid appropriately. Since we know $m \\geqslant 2^{k-1}, 2^{k-1}$ is the minimum possible value of $m$ as claimed.']",['$2^{k-1}$'],False,,Expression, 2231,Geometry,,"Turbo the snail sits on a point on a circle with circumference 1. Given an infinite sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$. Turbo successively crawls distances $c_{1}, c_{2}, c_{3}, \ldots$ around the circle, each time choosing to crawl either clockwise or counterclockwise. For example, if the sequence $c_{1}, c_{2}, c_{3}, \ldots$ is $0.4,0.6,0.3, \ldots$, then Turbo may start crawling as follows: ![](https://cdn.mathpix.com/cropped/2023_12_21_b8bb930a7b439039cf6bg-1.jpg?height=306&width=1106&top_left_y=594&top_left_x=472) Determine the largest constant $C>0$ with the following property: for every sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$ with $c_{i}\\frac{1}{2}$, we write $C=\\frac{1}{2}+a$ with $a>0$, and we choose the sequence\n\n$$\n\\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\ldots\n$$\n\nIn other words, $c_{i}=\\frac{1}{2}$ if $i$ is odd and $c_{i}=\\frac{1+a}{2}1$ for all $i$. Therefore, we are left with the case that Turbo alternates crawling clockwise and crawling counterclockwise. If it, without loss of generality, starts by going clockwise, then it will always crawl a distance $\\frac{1}{2}$ clockwise followed by a distance $\\frac{1+a}{2}$ counterclockwise. The net effect is that it crawls a distance $\\frac{a}{2}$ counterclockwise. Because $\\frac{a}{2}$ is positive, there exists a positive integer $N$ such that $\\frac{a}{2} \\cdot N>1$. After $2 N$ crawls, Turbo will have crawled a distance $\\frac{a}{2}$ counterclockwise $N$ times, therefore having covered a total distance of $\\frac{a}{2} \\cdot N>1$, meaning that it must have crawled over all points on the circle.\n\nNote: Every sequence of the form $c_{i}=x$ if $i$ is odd, and $c_{i}=y$ if $i$ is even, where $00$ and let $[-1+\\varepsilon, 1-\\varepsilon]$ be a minimal interval, that Chet cannot be forced out of. Then we can force Chet arbitrarily close to $\\pm(1-\\varepsilon)$. In partiular, we can force Chet out of $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$ by minimality of $M$. This means that there exists a sequence $d_{1}, d_{2}, \\ldots$ for which Chet has to leave $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, which means he ends up either in the interval $\\left[-1+\\varepsilon,-1+\\frac{4}{3} \\varepsilon\\right)$ or in the interval $\\left(1-\\frac{4}{3} \\varepsilon, 1-\\varepsilon\\right]$.\n\nNow consider the sequence,\n\n$$\nd_{1}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{2}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{3}, \\ldots\n$$\n\nobtained by adding the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ in between every two steps. We claim that this sequence forces Chet to leave the larger interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed no two consecutive elements in the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ can have the same sign, because the sum of any two consecutive terms is larger than $2-2 \\varepsilon$ and Chet would leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. It follows that the $\\left(1-\\frac{7}{6} \\varepsilon\\right)$ 's and the $\\left(1-\\frac{2}{3} \\varepsilon\\right)$ 's cancel out, so the position after $d_{k}$ is the same as before $d_{k+1}$. Hence, the positions after each $d_{k}$ remain the same as in the original sequence. Thus, Chet is also forced to the boundary in the new sequence.\n\nIf Chet is outside the interval $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, then Chet has to move $1-\\frac{7}{6} \\varepsilon$ towards 0 , and ends in $\\left[-\\frac{1}{6} \\varepsilon, \\frac{1}{6} \\varepsilon\\right]$. Chet then has to move by $1-\\frac{2}{3} \\varepsilon$, which means that he has to leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed the absolute value of the final position is at least $1-\\frac{5}{6} \\varepsilon$. This contradicts the assumption, that we cannot force Chet out of $[-1+\\varepsilon, 1-\\varepsilon]$. Hence $M \\geqslant 2$ as needed.""]",['$\\frac{1}{2}$'],False,,Numerical, 2231,Geometry,,"Turbo the snail sits on a point on a circle with circumference 1. Given an infinite sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$. Turbo successively crawls distances $c_{1}, c_{2}, c_{3}, \ldots$ around the circle, each time choosing to crawl either clockwise or counterclockwise. For example, if the sequence $c_{1}, c_{2}, c_{3}, \ldots$ is $0.4,0.6,0.3, \ldots$, then Turbo may start crawling as follows: Determine the largest constant $C>0$ with the following property: for every sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$ with $c_{i}\\frac{1}{2}$, we write $C=\\frac{1}{2}+a$ with $a>0$, and we choose the sequence\n\n$$\n\\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\quad \\frac{1+a}{2}, \\quad \\frac{1}{2}, \\ldots\n$$\n\nIn other words, $c_{i}=\\frac{1}{2}$ if $i$ is odd and $c_{i}=\\frac{1+a}{2}1$ for all $i$. Therefore, we are left with the case that Turbo alternates crawling clockwise and crawling counterclockwise. If it, without loss of generality, starts by going clockwise, then it will always crawl a distance $\\frac{1}{2}$ clockwise followed by a distance $\\frac{1+a}{2}$ counterclockwise. The net effect is that it crawls a distance $\\frac{a}{2}$ counterclockwise. Because $\\frac{a}{2}$ is positive, there exists a positive integer $N$ such that $\\frac{a}{2} \\cdot N>1$. After $2 N$ crawls, Turbo will have crawled a distance $\\frac{a}{2}$ counterclockwise $N$ times, therefore having covered a total distance of $\\frac{a}{2} \\cdot N>1$, meaning that it must have crawled over all points on the circle.\n\nNote: Every sequence of the form $c_{i}=x$ if $i$ is odd, and $c_{i}=y$ if $i$ is even, where $0\n\nFigure 5: Chet and Turbo equivalence\n\nClaim: $M \\geqslant 2$.\n\nProof. Suppose not, so $M<2$. Say $M=2-2 \\varepsilon$ for some $\\varepsilon>0$ and let $[-1+\\varepsilon, 1-\\varepsilon]$ be a minimal interval, that Chet cannot be forced out of. Then we can force Chet arbitrarily close to $\\pm(1-\\varepsilon)$. In partiular, we can force Chet out of $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$ by minimality of $M$. This means that there exists a sequence $d_{1}, d_{2}, \\ldots$ for which Chet has to leave $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, which means he ends up either in the interval $\\left[-1+\\varepsilon,-1+\\frac{4}{3} \\varepsilon\\right)$ or in the interval $\\left(1-\\frac{4}{3} \\varepsilon, 1-\\varepsilon\\right]$.\n\nNow consider the sequence,\n\n$$\nd_{1}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{2}, 1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon, d_{3}, \\ldots\n$$\n\nobtained by adding the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ in between every two steps. We claim that this sequence forces Chet to leave the larger interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed no two consecutive elements in the sequence $1-\\frac{7}{6} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{2}{3} \\varepsilon, 1-\\frac{7}{6} \\varepsilon$ can have the same sign, because the sum of any two consecutive terms is larger than $2-2 \\varepsilon$ and Chet would leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. It follows that the $\\left(1-\\frac{7}{6} \\varepsilon\\right)$ 's and the $\\left(1-\\frac{2}{3} \\varepsilon\\right)$ 's cancel out, so the position after $d_{k}$ is the same as before $d_{k+1}$. Hence, the positions after each $d_{k}$ remain the same as in the original sequence. Thus, Chet is also forced to the boundary in the new sequence.\n\nIf Chet is outside the interval $\\left[-1+\\frac{4}{3} \\varepsilon, 1-\\frac{4}{3} \\varepsilon\\right]$, then Chet has to move $1-\\frac{7}{6} \\varepsilon$ towards 0 , and ends in $\\left[-\\frac{1}{6} \\varepsilon, \\frac{1}{6} \\varepsilon\\right]$. Chet then has to move by $1-\\frac{2}{3} \\varepsilon$, which means that he has to leave the interval $[-1+\\varepsilon, 1-\\varepsilon]$. Indeed the absolute value of the final position is at least $1-\\frac{5}{6} \\varepsilon$. This contradicts the assumption, that we cannot force Chet out of $[-1+\\varepsilon, 1-\\varepsilon]$. Hence $M \\geqslant 2$ as needed.""]",['$\\frac{1}{2}$'],False,,Numerical, 2232,Algebra,,"We are given a positive integer $s \geqslant 2$. For each positive integer $k$, we define its twist $k^{\prime}$ as follows: write $k$ as $a s+b$, where $a, b$ are non-negative integers and $bs^{2}$. Since the reminder when a power of $s$ is divided by $s^{2}-1$ is either 1 or $s$, there exists a positive integer $m$ such that $s^{m}-n$ is non-negative and divisible by $s^{2}-1$. By our assumption $m \\geqslant 3$. We also take the smallest such $m$, so that $n>s^{m-2}$. The quotient $\\frac{s^{m}-n}{s^{2}-1}$ is therefore smaller than $s^{m-2}$, so there exist $b_{1}, \\ldots, b_{m-2} \\in\\{0,1, \\ldots, s-1\\}$ such that $\\frac{s^{m}-n}{s^{2}-1}=\\sum_{i=1}^{m-2} b_{i} s^{i-1}$. It follows that\n\n$$\nn=s^{m}-\\sum_{i=1}^{m-2} b_{i}\\left(s^{i+1}-s^{i-1}\\right)\n$$\n\nWe now show that\n\n$$\nd_{j}=s^{m+1-j}-\\sum_{i=1}^{m-1-j} b_{i}\\left(s^{i+1}-s^{i-1}\\right) \\tag{10}\n$$\n\nfor $j=1,2, \\ldots, m-2$ by induction on $j$. For $j=1$ this follows from $d_{1}=n$. Assume now that (10) holds for some $j\n\nPart I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\nPart II: Consider the triangle $A S_{b} S_{c}$. Note that since $T$ is the intersection of the tangents at $S_{b}$ and $S_{c}$ to the circumcircle of $A S_{b} S_{c}$, it follows that $A T$ is the symmedian of $A$ in this triangle. Let $X$ denote the second intersection of the symmedian $A T$ with $\\Omega$. We wish to show that $X$ is also on $I N_{a}$.\n\nNote that $A N_{a}$ is the external angle bisector of angle $A$, and therefore it is parallel to $S_{b} S_{c}$. Let $M$ denote the midpoint of $S_{b} S_{c}$, and let $Y$ be the second intersection of $A M$ with $\\Omega$. Since in $A S_{b} S_{c}, A X T$ is the symmedian and $A M Y$ is the median, it follows that $X Y$ is also parallel to $S_{b} S_{c}$. Thus, reflecting in the perpendicular bisector of $S_{b} S_{c}$ sends the line $A M Y$ to line $N_{a} M X$.\n\nNext, consider the quadrilateral $A S_{b} I S_{c}$. From the trillium theorem we have $\\left|S_{b} A\\right|=\\left|S_{b} I\\right|$ and $\\left|S_{c} A\\right|=\\left|S_{c} I\\right|$, thus the quadrilateral is a kite, from which it follows that the reflection of the line $A M$ in $S_{b} S_{c}$ is the line $I M$. But previously we have seen that this is also the line $N_{a} M X$. Thus $M, I, N_{a}$ and $X$ are collinear, as we wanted to show.', '\n\nPart I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\nWe begin by showing that the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$ as above.\n\nPart II: We introduce the point $S_{a}$ with the obvious meaning. Observe that the incentre $I$ of $A B C$ is the orthocentre of $S_{a} S_{b} S_{c}$ either because this is well-known, or because of an angle argument that $A$ reflects in $S_{b} S_{c}$ to $I$ (and similar results by cyclic change of letters). Therefore $A S_{a}$ is perpendicular to $S_{b} S_{c}$.\n\n\n\nFigure 7: A reflections argument for Solution\n\nLet $M$ denote the midpoint of $S_{b} S_{c}$. Then $A$ is the reflection of $S_{a}$ in the diameter parallel to $S_{b} S_{c}$, so the reflection of $A$ in the diameter perpendicular to $S_{b} S_{c}$ is $N_{a}$, the antipode of $S_{a}$. Let the reflection of $X$ in $T M$ be $Y$, so $T Y$ passes through $N_{a}$ and is the reflection of $T X$ in $T M$.\n\nNow $S_{b} S_{c}$ is the polar line of $T$ with respect to $\\Omega$, so $A Y$ and $N_{a} X$ meet on this line, and by symmetry at its midpoint $M$. The line $N_{a} M X$ is therefore the reflection of the line $Y M A$ in $S_{b} S_{c}$, and so $N_{a} M X$ passes through $I$ (the reflection of $A$ in $S_{b} S_{c}$ ).\n\nThe triangle $A S_{c} S_{b}$ can be taken as generic, and from the argument above we can extract the fact that the symmedian point and the centroid are isogonal conjugates in that triangle.', 'Part I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\n\n\nFigure 8: Diagram to Solution\n\nPart II: By the trillelium theorem, $S_{c} S_{b}$ bisects $A I$, and since $N_{a} A \\| S_{b} S_{c}$, then $O T$ is a bisector of $A N_{a}$. This implies $\\left|M N_{a}\\right|=|M A|=|M I|$, since $M$ is the midpoint of $S_{c} S_{b}$ and lies also on $O T$. Hence, $M$ is the circumcentre of triangle $I A N_{a}$. But this triangle has a right angle at $A$ (since $A I$ and $A N_{a}$ are the inner and outer angle bisector at $A$ ), hence $M$ lies on $I N_{a}$.\n\nAgain, let $X$ be the second intersection of $T A$ and $\\Omega$. By the above, it suffices to prove that $X$ lies on the line $N_{a} M$. From the power of point $T$ with respect to $\\Omega$ we get $|T A| \\cdot|T X|=\\left|T S_{c}\\right|^{2}$. Since $M$ is the foot of the altitude of right triangle $T S_{c} O$, we obtain $\\left|T S_{c}\\right|^{2}=|T M| \\cdot|T O|$. Hence, $|T A| \\cdot|T X|=|T M| \\cdot|T O|$ so the points $O, M, A, X$ are concyclic. It follows that $\\angle M X A=\\angle M O A=\\frac{1}{2} \\angle N_{a} O A=\\angle N_{a} X A$. Hence, $X$ lies on the line $N_{a} M$.\n\nRemark. To show that $O M A X$ is cyclic, one can also invert the line $T A X$ in the circumcircle of the triangle $A B C$.\n\nSolution 4. \nPart I: First we show that A lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$.\n\nWe first note that the line through the intersections of two circles is the radical line of the two circles. Let the tangents to $\\Omega$ at $S_{b}$ and $S_{c}$ intersect at $T$. Clearly $T$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$ (and in fact is the radical centre of $\\omega_{b}, \\omega_{c}$ and $\\Omega$ ).\n\nWe next show that $A$ lies on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $P_{b}$ denote the point of tangency of $\\omega_{b}$ and $A B$, and let $P_{c}$ denote the point of tangency of $\\omega_{c}$ and $A C$. Furthermore, let $U$ be the intersection of the tangent to $\\Omega$ at $S_{b}$ with the line $A B$, and let $V$ be the intersection of the tangent to $\\Omega$ at $S_{c}$ with the line $A C$. Then $T V A U$ is parallelogram. Morover, due to equality of tangent segments we have $\\left|U S_{b}\\right|=\\left|U P_{b}\\right|,\\left|V P_{c}\\right|=\\left|V S_{c}\\right|$ and $\\left|T S_{b}\\right|=\\left|T S_{c}\\right|$. It follows that\n\n$$\n\\begin{aligned}\n\\left|A P_{b}\\right| & =\\left|U P_{b}\\right|-|U A|=\\left|U S_{b}\\right|-|T V|=\\left|T S_{b}\\right|-|T U|-|T V| \\\\\n& =\\left|T S_{s}\\right|-|T V|-|T U|=\\left|V S_{c}\\right|-|A V|=\\left|V P_{c}\\right|-|V A|=\\left|A P_{c}\\right| .\n\\end{aligned} \\tag{11}\n$$\n\n\n\nBut $\\left|A P_{b}\\right|,\\left|A P_{c}\\right|$ are exactly the square roots of powers of $A$ with respect to $\\omega_{b}$ and $\\omega_{c}$, hence $A$ is indeed on their radical axis.\n\nThus, the radical axis of $\\omega_{b}, \\omega_{c}$ is $A T$.\n\n\n\nFigure 9: Diagram to Solution\n\nPart II: We show that $A N_{a} \\| S_{b} S_{c}$. In particular, $\\angle N_{a} O T=\\angle T O A$. The conclusion of the problem trivially holds if $|A B|=|A C|$, therefore we assume without loss of generality that $|A C|>|A B|$. Let $S_{a}$ be the midpoint of the $\\operatorname{arc} B C$ which does not contain $A$. Then $N_{a} S_{a}$ is a diameter, so $\\angle S_{a} B N_{a}=\\frac{\\pi}{2}=\\angle O S_{c} T$. We also compute $\\angle B N_{a} S_{a}=\\angle B A S_{a}=\\frac{1}{2} \\angle B A C=\\frac{1}{2} \\angle S_{c} T S_{b}=\\angle S_{c} T O$. It follows that the triangles $T S_{c} O$ and $N_{a} B S_{a}$ are similar. In particular,\n\n$$\n\\frac{\\left|N_{a} B\\right|}{\\left|T S_{c}\\right|}=\\frac{\\left|S_{a} B\\right|}{\\left|O S_{c}\\right|} \\tag{12}\n$$\n\nNext we compute\n\n$$\n\\angle I S_{a} B=\\angle N_{a} S_{a} B-\\angle N_{a} S_{a} I=\\angle T O S_{c}-\\frac{1}{2} \\angle N_{a} O A=\\angle T O S_{c}-\\angle T O A=\\angle A O S_{c} \\tag{13}\n$$\n\n\n\nand\n\n$$\n\\angle I B N_{a}=\\angle C B N_{a}-\\angle C B I=\\frac{1}{2}\\left(\\pi-\\angle B N_{a} C\\right)-\\frac{1}{2} \\angle C B A=\\frac{1}{2} \\angle A C B=\\angle A C S_{c}=\\angle A S_{c} T \\tag{14}\n$$,\n\nhence\n\n$$\n\\angle S_{a} B I=\\frac{\\pi}{2}-\\angle I B N_{a}=\\frac{\\pi}{2}-\\angle A S_{c} T=\\angle O S_{c} A .\n$$\n\nTogether with (13) it follows that the triangles $I B S_{a}$ and $A S_{c} O$ are similar, so $\\frac{\\left|S_{a} B\\right|}{\\left|O S_{c}\\right|}=$ $\\frac{|I B|}{\\left|A S_{c}\\right|}$, and (12) implies $\\frac{\\left|N_{a} B\\right|}{\\left|T S_{c}\\right|}=\\frac{|I B|}{\\left|A S_{c}\\right|}$. Consequently, by (14) the triangles $T S_{c} A$ and $N_{a} B I$ are similar and therefore $\\angle S_{c} T A=\\angle B N_{a} I$. Now let $Q$ be the second intersection of $N_{a} I$ with $\\Omega$. Then $\\angle B N_{a} I=\\angle B N_{a} Q=\\angle B A Q$, so $\\angle S_{c} T A=\\angle B A Q$. Since $A B$ is parallel to $T S_{c}$, we get $A Q \\| T A$, i.e. $A, T, Q$ are collinear.\n\nRemark. After proving similarity of triangles $T S_{c} O$ and $N_{a} B S_{a}$ one can use spiral symmetry to show similarity of triangles $T S_{c} A$ and $N_{a} B I$.', 'Part I: First we show that $A$ lies on the radical axis between $\\omega_{b}$ and $\\omega_{c}$.\n\nLet $T$ be the radical center of the circumcircle, $\\omega_{b}$ and $\\omega_{c}$; then $T S_{b}$ and $T S_{c}$ are common tangents of the circles, as shown in Figure 5a. Moreover, let $P_{b}=A B \\cap S_{b} S_{c}$ and $P_{c}=$ $A C \\cap S_{b} S_{c}$. The triangle $T S_{c} S_{b}$ is isosceles: $A B \\| T S_{c}$ and $A C \\| T S_{b}$ so\n\n$$\n\\angle A P_{b} P_{c}=\\angle T S_{c} S_{b}=\\angle S_{c} S_{b} T=\\angle P_{b} P_{c} A .\n$$\n\nFrom these angles we can see that $\\omega_{b}$ passes through $P_{b}, \\omega_{c}$ passes through $P_{c}$, and finally $A P_{b}$ and $A P_{c}$ are equal tangents to $\\omega_{b}$ and $\\omega_{c}$, so $A$ lies on the radical axis.\n\n\n\nFigure 5a\n\n\n\nFigure $5 b$\n\n\n\nPart II. Let the radical axis $T A$ meet the circumcircle again at $X$, let $S_{a}$ be the midpoint of the $\\operatorname{arc} B C$ opposite to $A$, and let $X I$ meet the circumcirlce again at $N$. (See Figure 2.) For solving the problem, we have prove that $N_{a}=N$.\n\nThe triples of points $A, I, S_{a} ; B, I, S_{b}$ and $C, I, S_{c}$ are collinear because they lie on the angle bisectors of the triangle $A B C$.\n\nNotice that the quadrilateral $A S_{c} X S_{b}$ is harmonic, because the tangents at $S_{b}$ and $S_{c}$, and the line $A X$ are concurrent at $T$. This quadrilateral can be projected (or inverted) to the quadrilateral $S_{a} C N B$ through $I$. So, $S_{a} C N B$ also is a harmonic quadrilateral. Due to $S_{a} B=S_{a} C$, this implies $N B=N C$, so $N=N_{a}$. Done.', ""Part I: First let's show that this is equivalent to proving that $T A$ and $N_{a} I$ intersect in $\\Omega$.\n\nLemma: Let's recall that if we have two circles $\\omega_{1}$ and $\\omega_{2}$ which are internally tangent at point $X$ and if we have a line $A B$ tangent to $\\omega_{2}$ at $Y$. Let $M$ be the midpoint of the arc $A B$ not containing $Z$. We have that $Z, Y, M$ are collinear.\n\n\n\n\n\nLet $P_{b}=A B \\cap \\omega_{b}$ and $P_{c}=A C \\cap \\omega_{c}$. We can notice by the lemma that $S_{b}, P_{b}$ and $S_{c}$ are collinear, and similarly $S_{c}, P_{c}$ and $S_{b}$ are also collinear. Therefore $S_{c}, P_{b}, P_{c}$, and $S_{b}$ are collinear, and since $\\angle A P_{b} P_{c}=\\frac{\\angle A B C}{2}+\\frac{\\angle A C B}{2}=\\angle A P_{c} P_{b}$ then $A P_{b}=A P_{c}$ so $A$ is on the radical axis of $\\omega_{b}$ and $\\omega_{c}$. Let $T$ be the intersection of the tangent lines of $\\Omega$ through $S_{c}$ and $S_{b}$. Since $T S_{c}=T S_{b}$ then $A T$ is the radical axis between $\\omega_{b}$ and $\\omega_{c}$.\n\nPart II: $T A$ and $N_{a} I$ intersect in $\\Omega$.\n\n\n\nLet $\\omega_{a}$ the $A$-mixtilinear incircle (that is, the circle internally tangent to $\\Omega$, and tangent to both $A B$ and $A C$ ), and let $X=\\Omega \\cap \\omega_{a}$. It is known that $N_{a}, I, X$ are collinear.\n\nLet $M_{c}$ and $M_{b}$ be the tangent points of $\\omega_{A}$ to $A B$ and $A C$ respectively, then by the lemma $X, M_{c}, S_{c}$ are collinear and $X, M_{b}, S_{b}$ are collinear. We can see that $S_{c} T S_{b}$ and $M_{c} A M_{b}$ are homothetic with respect to $X$; therefore $T$ and $A$ are homothetic with respect to $X$, impying that $T, A, X$ are collinear.\n\n\n\nFigure 10: Diagram to Solution""]",,True,,, 2234,Algebra,,"In an increasing sequence of numbers with an odd number of terms, the difference between any two consecutive terms is a constant $d$, and the middle term is 302 . When the last 4 terms are removed from the sequence, the middle term of the resulting sequence is 296. What is the value of $d$ ?","['Let the number of terms in the sequence be $2 k+1$.\n\nWe label the terms $a_{1}, a_{2}, \\ldots, a_{2 k+1}$.\n\nThe middle term here is $a_{k+1}=302$.\n\nSince the difference between any two consecutive terms in this increasing sequence is $d$, $a_{m+1}-a_{m}=d$ for $m=1,2, \\ldots, 2 k$.\n\nWhen the last 4 terms are removed, the last term is now $a_{2 k-3}$ so the middle term is then $a_{k-1}=296$. (When four terms are removed from the end, the middle term shifts two terms to the left.)\n\nNow $6=a_{k+1}-a_{k-1}=\\left(a_{k+1}-a_{k}\\right)+\\left(a_{k}-a_{k-1}\\right)=d+d=2 d$.\n\nTherefore $d=3$.', 'If the last four terms are removed from the sequence this results in 302 shifting 2 terms to the left in the new sequence meaning that $302-296=2 d, d=3$.']",['3'],False,,Numerical, 2235,Algebra,,"There are two increasing sequences of five consecutive integers, each of which have the property that the sum of the squares of the first three integers in the sequence equals the sum of the squares of the last two. Determine these two sequences.","['Let $n$ be the smallest integer in one of these sequences.\n\nSo we want to solve the equation $n^{2}+(n+1)^{2}+(n+2)^{2}=(n+3)^{2}+(n+4)^{2}$ (translating the given problem into an equation).\n\nThus $n^{2}+n^{2}+2 n+1+n^{2}+4 n+4=n^{2}+6 n+9+n^{2}+8 n+16$\n\n\n\n$$\n\\begin{array}{r}\nn^{2}-8 n-20=0 \\\\\n(n-10)(n+2)=0\n\\end{array}\n$$\n\nSo $n=10$ or $n=-2$.\n\nTherefore, the sequences are 10, 11, 12, 13, 14 and $-2,-1,0,1,2$.\n\nVerification \n\n$(-2)^{2}+(-1)^{2}+0^{2}=1^{2}+2^{2}=5$ and $10^{2}+11^{2}+12^{2}=13^{2}+14^{2}=365$']","['10,11,12,13,14,-2,-1,0,1,2']",True,,Numerical, 2236,Algebra,,"If $f(t)=\sin \left(\pi t-\frac{\pi}{2}\right)$, what is the smallest positive value of $t$ at which $f(t)$ attains its minimum value?","['Since $t>0, \\pi t-\\frac{\\pi}{2}>-\\frac{\\pi}{2}$. So $\\sin \\left(\\pi t-\\frac{\\pi}{2}\\right)$ first attains its minimum value when\n\n$$\n\\begin{aligned}\n\\pi t-\\frac{\\pi}{2} & =\\frac{3 \\pi}{2} \\\\\nt & =2 .\n\\end{aligned}\n$$', 'Rewriting $f(t)$ as, $f(t)=\\sin \\left[\\pi\\left(t-\\frac{1}{2}\\right)\\right]$.\n\nThus $f(t)$ has a period $\\frac{2 \\pi}{\\pi}=2$ and appears in the diagram at the right.\n\nThus $f(t)$ attains its minimum at $t=2$. Note that $f(t)$ attains a minimum value at $t=0$ but since $t>0$, the required answer is $t=2$.\n\n']",['2'],False,,Numerical, 2237,Geometry,,"In the diagram, $\angle A B F=41^{\circ}, \angle C B F=59^{\circ}, D E$ is parallel to $B F$, and $E F=25$. If $A E=E C$, determine the length of $A E$, to 2 decimal places. ![](https://cdn.mathpix.com/cropped/2023_12_21_3d20c72c65b4fe186184g-1.jpg?height=626&width=331&top_left_y=305&top_left_x=1469)","['Let the length of $A E=E C$ be $x$.\n\nThen $A F=x-25$.\n\nIn, $\\triangle B C F, \\frac{x+25}{B F}=\\tan \\left(59^{\\circ}\\right)$.\n\nIn $\\triangle A B F, \\frac{x-25}{B F}=\\tan \\left(41^{\\circ}\\right)$.\n\nSolving for $B F$ in these two equations and equating,\n\n$$\nB F=\\frac{x+25}{\\tan 59^{\\circ}}=\\frac{x-25}{\\tan 41^{\\circ}}\n$$\n\nso $\\quad\\left(\\tan 41^{\\circ}\\right)(x+25)=\\left(\\tan 59^{\\circ}\\right)(x-25)$\n\n$$\n\\begin{aligned}\n25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right) & =x\\left(\\tan 59^{\\circ}-\\tan 41^{\\circ}\\right) \\\\\nx & =\\frac{25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right)}{\\tan 59^{\\circ}-\\tan 41^{\\circ}} \\\\\nx & \\doteq 79.67 .\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_204c58d31fdb1aa1cd53g-1.jpg?height=691&width=426&top_left_y=1064&top_left_x=1397)\n\nTherefore the length of $A E$ is 79.67 .']",['79.67'],False,,Numerical,1e-1 2237,Geometry,,"In the diagram, $\angle A B F=41^{\circ}, \angle C B F=59^{\circ}, D E$ is parallel to $B F$, and $E F=25$. If $A E=E C$, determine the length of $A E$, to 2 decimal places. ","['Let the length of $A E=E C$ be $x$.\n\nThen $A F=x-25$.\n\nIn, $\\triangle B C F, \\frac{x+25}{B F}=\\tan \\left(59^{\\circ}\\right)$.\n\nIn $\\triangle A B F, \\frac{x-25}{B F}=\\tan \\left(41^{\\circ}\\right)$.\n\nSolving for $B F$ in these two equations and equating,\n\n$$\nB F=\\frac{x+25}{\\tan 59^{\\circ}}=\\frac{x-25}{\\tan 41^{\\circ}}\n$$\n\nso $\\quad\\left(\\tan 41^{\\circ}\\right)(x+25)=\\left(\\tan 59^{\\circ}\\right)(x-25)$\n\n$$\n\\begin{aligned}\n25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right) & =x\\left(\\tan 59^{\\circ}-\\tan 41^{\\circ}\\right) \\\\\nx & =\\frac{25\\left(\\tan 59^{\\circ}+\\tan 41^{\\circ}\\right)}{\\tan 59^{\\circ}-\\tan 41^{\\circ}} \\\\\nx & \\doteq 79.67 .\n\\end{aligned}\n$$\n\n\n\nTherefore the length of $A E$ is 79.67 .']",['79.67'],False,,Numerical,1e-1 2238,Algebra,,Determine all integer values of $x$ such that $\left(x^{2}-3\right)\left(x^{2}+5\right)<0$.,"['Since $x^{2} \\geq 0$ for all $x, x^{2}+5>0$. Since $\\left(x^{2}-3\\right)\\left(x^{2}+5\\right)<0, x^{2}-3<0$, so $x^{2}<3$ or $-\\sqrt{3}","['Join $B Y$. Since $B C$ is a diameter, then $\\angle B Y C=90^{\\circ}$. Since $A B=B C, \\triangle A B C$ is isosceles and $B Y$ is an altitude in $\\triangle A B C$, then $A Y=Y C=15$.\n\nLet $\\angle B A C=\\theta$.\n\nSince $\\triangle A B C$ is isosceles, $\\angle B C A=\\theta$.\n\nSince $B C Y X$ is cyclic, $\\angle B X Y=180-\\theta$ and so $\\angle A X Y=\\theta$.\n\n\n\nThus $\\triangle A X Y$ is isosceles and so $X Y=A Y=15$.\n\nTherefore $X Y=15$.', 'Join $B Y . \\angle B Y C=90^{\\circ}$ since it is inscribed in a semicircle.\n\n$\\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.\n\nTherefore $B Y=\\sqrt{25^{2}-15^{2}}=20$.\n\nJoin $C X . \\angle C X B=90^{\\circ}$ since it is also inscribed in a semicircle.\n\n\n\nThe area of $\\triangle A B C$ is\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A C)(B Y) & =\\frac{1}{2}(A B)(C X) \\\\\n\\frac{1}{2}(30)(20) & =\\frac{1}{2}(25)(C X) \\\\\nC X & =\\frac{600}{25}=24 .\n\\end{aligned}\n$$\n\nFrom $\\triangle A B Y$ we conclude that $\\cos \\angle A B Y=\\frac{B Y}{A B}=\\frac{20}{25}=\\frac{4}{5}$.\n\nIn $\\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \\cos \\angle X B Y$.\n\nNow (by Pythagoras $\\triangle B X C$ ),\n\n\n\n$$\n\\begin{aligned}\nB X^{2} & =B C^{2}-C X^{2} \\\\\n& =25^{2}-24^{2} \\\\\n& =49 \\\\\nB X & =7 .\n\\end{aligned}\n$$\n\nTherefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \\frac{4}{5}$\n\n$$\n=49+400-224\n$$\n\n$$\n=225 \\text {. }\n$$\n\nTherefore $X Y=15$.']",['15'],False,,Numerical, 2241,Algebra,,What is the value of $x$ such that $\log _{2}\left(\log _{2}(2 x-2)\right)=2$ ?,['$$\n\\begin{aligned}\n\\log _{2}\\left(\\log _{2}(2 x-2)\\right) & =2 \\\\\n\\log _{2}(2 x-2) & =2^{2} \\\\\n2 x-2 & =2^{\\left(2^{2}\\right)} \\\\\n2 x-2 & =2^{4} \\\\\n2 x-2 & =16 \\\\\n2 x & =18 \\\\\nx & =9\n\\end{aligned}\n$$'],['9'],False,,Numerical, 2242,Algebra,,"Let $f(x)=2^{k x}+9$, where $k$ is a real number. If $f(3): f(6)=1: 3$, determine the value of $f(9)-f(3)$.","['From the given condition,\n\n$$\n\\begin{aligned}\n\\frac{f(3)}{f(6)}=\\frac{2^{3 k}+9}{2^{6 k}+9} & =\\frac{1}{3} \\\\\n3\\left(2^{3 k}+9\\right) & =2^{6 k}+9 \\\\\n0 & =2^{6 k}-3\\left(2^{3 k}\\right)-18 .\n\\end{aligned}\n$$\n\nWe treat this as a quadratic equation in the variable $x=2^{3 k}$, so\n\n$$\n\\begin{aligned}\n& 0=x^{2}-3 x-18 \\\\\n& 0=(x-6)(x+3)\n\\end{aligned}\n$$\n\nTherefore, $2^{3 k}=6$ or $2^{3 k}=-3$. Since $2^{a}>0$ for any $a$, then $2^{3 k} \\neq-3$.\n\nSo $2^{3 k}=6$. We could solve for $k$ here, but this is unnecessary.\n\n\n\nWe calculate $f(9)-f(3)=\\left(2^{9 k}+9\\right)-\\left(2^{3 k}+9\\right)$\n\n$$\n\\begin{aligned}\n& =2^{9 k}-2^{3 k} \\\\\n& =\\left(2^{3 k}\\right)^{3}-2^{3 k} \\\\\n& =6^{3}-6 \\\\\n& =210 .\n\\end{aligned}\n$$\n\nTherefore $f(9)-f(3)=210$.']",['210'],False,,Numerical, 2243,Algebra,,"Determine, with justification, all values of $k$ for which $y=x^{2}-4$ and $y=2|x|+k$ do not intersect.","['Since each of these two graphs is symmetric about the $y$-axis (i.e. both are even functions), then we only need to find $k$ so that there are no points of intersection with $x \\geq 0$.\n\nSo let $x \\geq 0$ and consider the intersection between $y=2 x+k$ and $y=x^{2}-4$.\n\nEquating, we have, $2 x+k=x^{2}-4$.\n\nRearranging, we want $x^{2}-2 x-(k+4)=0$ to have no solutions.\n\n\n\nFor no solutions, the discriminant is negative, i.e.\n\n$$\n\\begin{aligned}\n20+4 k & <0 \\\\\n4 k & <-20 \\\\\nk & <-5 .\n\\end{aligned}\n$$\n\nSo $y=x^{2}-4$ and $y=2|x|+k$ have no intersection points when $k<-5$.\n\n']","['(-\\infty,-5)']",False,,Interval, 2244,Geometry,,"Triangle $A B C$ is right-angled at $B$ and has side lengths which are integers. A second triangle, $P Q R$, is located inside $\triangle A B C$ as shown, such that its sides are parallel to the sides of $\triangle A B C$ and the distance between parallel lines is 2 . Determine the side lengths of all possible triangles $A B C$, such that the area of $\triangle A B C$ is 9 times that of $\triangle P Q R$. ![](https://cdn.mathpix.com/cropped/2023_12_21_eacf83d29c256cdd2ca3g-1.jpg?height=412&width=480&top_left_y=604&top_left_x=1362)","['Let the sides of $\\triangle A B C$ be $A B=c, B C=a, A C=b, a, b, c$ are all integers.\n\nSince the sides of $\\triangle P Q R$ are all parallel to the sides of $\\triangle A B C$, then $\\triangle A B C$ is similar to $\\triangle P Q R$.\n\nNow the ratio of areas of $\\triangle A B C$ to $\\triangle P Q R$ is $9=3^{2}$ to 1 , so the ratio of side lengths will be 3 to 1 .\n\nSo the sides of $\\triangle P Q R$ are $P Q=\\frac{c}{3}, Q R=\\frac{a}{3}, P R=\\frac{b}{3}$.\n\n\n\nSo we can label the diagram as indicated.\n\nWe join the corresponding vertices of the two triangles as\n\nArea of trapezoid $B Q R C$\n\nArea of trapezoid CRPA\n\nArea of trapezoid $A P Q B$\n\n$+\\quad$ Area of $\\triangle P Q R$\n\nArea of $\\triangle A B C$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_0e6430f266c4e3edca52g-1.jpg?height=450&width=534&top_left_y=282&top_left_x=1381)\n\nDoing so gives,\n\n$2\\left(\\frac{2}{3} a\\right)+2\\left(\\frac{2}{3} b\\right)+2\\left(\\frac{2}{3} c\\right)+\\frac{a c}{18}=\\frac{a c}{72}$\n\nOr upon simplifying $a c=3 a+3 b+3 c$ (Note that this relationship can be derived in a variety of ways.)\n\n$$\n\\begin{array}{rlrl}\na c & =3 c+3 b+3 a & & \\\\\na c-3 c-3 a & =3 b & \\\\\na c-3 c-3 a & =3 \\sqrt{a^{2}+c^{2}} & & \\left(\\text { since } b=\\sqrt{a^{2}+c^{2}}\\right) \\\\\na^{2} c^{2}+9 c^{2}+9 a^{2}-6 a c^{2}-6 a^{2} c+18 a c & =9\\left(a^{2}+c^{2}\\right) & & \\text { (squaring both sides) } \\\\\na c(a c-6 c-6 a+18) & =0 & \\\\\na c-6 c-6 a+18 & =0 & \\\\\nc(a-6) & =6 a-18 \\\\\nc & =\\frac{6 a-18}{a-6} \\\\\nc & =6+\\frac{18}{a-6} .\n\\end{array}\n$$\n\nSince $a$ is a side of a triangle, $a>0$. We are now looking for positive integer values such that $\\frac{18}{a-6}$ is also an integer.\n\nThe only possible values for $a$ are 3, 7, 8, 9, 12, 15 and 24 .\n\nTabulating the possibilities and calculating values for $b$ and $c$ gives,\n\n| $a$ | 3 | 7 | 8 | 9 | 12 | 15 | 24 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $c$ | 0 | 24 | 15 | 12 | 9 | 8 | 7 |\n| $b$ | - | 25 | 17 | 15 | 15 | 17 | 25 |\n\nThus the only possibilities for the triangle are $(7,24,25),(8,15,7)$ and $(9,12,15)$.', 'The two triangles are similar with areas in the ratio 1:9.\n\nTherefore the sides are in the ratio 1:3.\n\nLet $a=B C, b=C A, c=B A$.\n\n\n\nThen $\\frac{a}{3}=P Q, \\frac{b}{3}=Q R, \\frac{c}{3}=P R$.\n\nLocate points $K, L$ on $B C ; M, N$ on $C A$; and $T, S$ on $A B$ as shown.\n\n$$\n\\begin{gathered}\nB C=B K+K L+L C \\\\\na=B K+\\frac{a}{3}+2\n\\end{gathered}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_e2a5fb5a1f25f531825cg-1.jpg?height=504&width=594&top_left_y=274&top_left_x=1362)\n\nTherefore $B K=\\frac{2}{3} a-2$.\n\nIn a similar way, $A N=\\frac{2}{3} b-2$.\n\nNow $\\triangle B K P \\cong \\triangle B T P$ and $\\triangle A N R \\cong \\triangle A S R$, both by $H L$.\n\nTherefore $B T=B K=\\frac{2}{3} a-2$ and $A S=A N=\\frac{2}{3} b-2$.\n\nNow, $A B=A S+S T+B T$\n\n$$\n\\begin{aligned}\nc & =\\frac{2}{3} b-2+\\frac{c}{3}+\\frac{2}{3} a-2 \\\\\n\\frac{2}{3} c & =\\frac{2}{3} b+\\frac{2}{3} a-4 \\\\\nc & =b+a-6 \\\\\nb & =c+(6-a) .\n\\end{aligned}\n$$\n\nBy Pythagoras, $a^{2}+b^{2}=c^{2}$\n\n$$\n\\begin{aligned}\n& a^{2}+[c+(6-a)] 2=c^{2} \\\\\n& a^{2}+\\not \\not^{2}+2 c(6-a)+(6-a) 2=\\not \\not^{2} \\\\\n& a^{2}+\\quad(6-a)^{2}=-2 c(6-a) \\\\\n& 2 a^{2}-\\quad 12 a+36=2 c(a-6) \\\\\n& a^{2}-\\quad 6 a+18=c(a-6) \\\\\n& c=\\frac{a^{2}-6 a+18}{a-6} \\\\\n& c=\\frac{a(a-6)+18}{a-6} \\\\\n& c=a+\\frac{18}{a-6} \\text {. }\n\\end{aligned}\n$$\n\nSince $a$ and $c$ are integers, $a-6$ is a divisor of 18 .\n\nAlso since $b0$.\n\nThus $a-6$ can be $1,2,3,6,9,18$.\n\nThe values of $a$ are: 7, 8, 9, 12, 15, 24 .\n\nMatching values for $c: 25,17,15,15,17,25$\n\nMatching values for $b: 24,15,12,9,8,7$\n\n\n\nThe distinct triangles are $(7,24,25),(8,15,17)$ and $(9,12,15)$.']","['$(7,24,25),(8,15,7),(9,12,15)$']",True,,Need_human_evaluate, 2245,Geometry,,"Points $P$ and $Q$ are located inside the square $A B C D$ such that $D P$ is parallel to $Q B$ and $D P=Q B=P Q$. Determine the minimum possible value of $\angle A D P$. ![](https://cdn.mathpix.com/cropped/2023_12_21_eacf83d29c256cdd2ca3g-1.jpg?height=436&width=393&top_left_y=1105&top_left_x=1403)","['Placing the information on the coordinate axes, the diagram is indicated to the right.\n\nWe note that $P$ has coordinates $(a, b)$.\n\nBy symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.\n\nSince $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$\n\n$$\n\\begin{aligned}\n& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\\\\n& \\left(a-\\frac{4}{3}\\right)^{2}+\\left(b-\\frac{4}{3}\\right)^{2}=\\frac{8}{9}\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4d6291049b08a1afc7c4g-1.jpg?height=491&width=518&top_left_y=996&top_left_x=1324)\n\n$P$ is on a circle with centre $O\\left(\\frac{4}{3}, \\frac{4}{3}\\right)$ with $r=\\frac{2}{3} \\sqrt{2}$.\n\nThe minimum angle for $\\theta$ occurs when $D P$ is tangent to the circle.\n\nSo we have the diagram noted to the right.\n\nSince $O D$ makes an angle of $45^{\\circ}$ with the $x$-axis then $\\angle P D O=45-\\theta$ and $O D=\\frac{4}{3} \\sqrt{2}$.\n\nTherefore $\\sin (45-\\theta)=\\frac{\\frac{2}{3} \\sqrt{2}}{\\frac{4}{3} \\sqrt{2}}=\\frac{1}{2}$ which means $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.\n\nThus the minimum value for $\\theta$ is $15^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4d6291049b08a1afc7c4g-1.jpg?height=436&width=426&top_left_y=1717&top_left_x=1424)', 'Let $A B=B C=C D=D A=1$.\n\nJoin $D$ to $B$. Let $\\angle A D P=\\theta$. Therefore, $\\angle P D B=45-\\theta$.\n\nLet $P D=a$ and $P B=b$ and $P Q=\\frac{a}{2}$.\n\n\n\nWe now establish a relationship between $a$ and $b$.\n\nIn $\\triangle P D B, b^{2}=a^{2}+2-2(a)(\\sqrt{2}) \\cos (45-\\theta)$\n\n$$\n\\text { or, } \\quad \\cos (45-\\theta)=\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d35ff2cf67f11a9d36b7g-1.jpg?height=475&width=431&top_left_y=283&top_left_x=1324)\n\nIn $\\triangle P D R,\\left(\\frac{a}{2}\\right)^{2}=a^{2}+\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}-2 a \\frac{\\sqrt{2}}{2} \\cos (45-\\theta)$\n\nor, $\\cos (45-\\theta)=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$\n\nComparing (1) and (2) gives, $\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$.\n\nSimplifying this, $a^{2}+2 b^{2}=2$\n\n$$\n\\text { or, } \\quad b^{2}=\\frac{2-a^{2}}{2}\n$$\n\nNow $\\cos (45-\\theta)=\\frac{a^{2}+2-\\left(\\frac{2-a^{2}}{2}\\right)}{2 a \\sqrt{2}}=\\frac{1}{4 \\sqrt{2}}\\left(3 a+\\frac{2}{a}\\right)$.\n\nNow considering $3 a+\\frac{2}{a}$, we know $\\left(\\sqrt{3 a}-\\sqrt{\\frac{2}{a}}\\right)^{2} \\geq 0$\n\n$$\n\\text { or, } \\quad 3 a+\\frac{2}{a} \\geq 2 \\sqrt{6}\n$$\n\nThus, $\\cos (45-\\theta) \\geq \\frac{1}{4 \\sqrt{2}}(2 \\sqrt{6})=\\frac{\\sqrt{3}}{2}$\n\n$$\n\\cos (45-\\theta) \\geq \\frac{\\sqrt{3}}{2}\n$$\n\n$\\cos (45-\\theta)$ has a minimum value for $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.', 'Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\\angle A D P=\\theta$.\n\nBy interior alternate angles, $\\angle P=\\angle Q$ and $\\angle P D M=\\angle Q B M$.\n\nThus $\\triangle P D M \\cong \\triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.\n\nSo $M$ is the midpoint of $B D$ and the centre of the square.\n\n\n\nWithout loss of generality, let $P M=1$. Then $P D=2$.\n\nSince $\\theta+\\alpha=45^{\\circ}$ (see diagram), $\\theta$ will be minimized when $\\alpha$ is maximized.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_8b358656a56f9e807422g-1.jpg?height=485&width=440&top_left_y=278&top_left_x=1409)\n\nConsider $\\triangle P M D$.\n\nUsing the sine law, $\\frac{\\sin \\alpha}{1}=\\frac{\\sin (\\angle P M D)}{2}$.\n\nTo maximize $\\alpha$, we maximize $\\sin \\alpha$.\n\nBut $\\sin \\alpha=\\frac{\\sin (\\angle P M D)}{2}$, so it is maximized when $\\sin (\\angle P M D)=1$.\n\nIn this case, $\\sin \\alpha=\\frac{1}{2}$, so $\\alpha=30^{\\circ}$.\n\nTherefore, $\\theta=45^{\\circ}-30^{\\circ}=15^{\\circ}$, and so the minimum value of $\\theta$ is $15^{\\circ}$.', 'We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.\n\nLet $P D=P Q=Q B=a$, and $\\angle A D P=\\theta$.\n\nDrop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.\n\nThen $P X=a \\sin \\theta, D X=a \\cos \\theta$.\n\nTherefore the coordinates of $P$ are $(a \\sin \\theta, a \\cos \\theta)$.\n\nSince $P D \\| B Q$, then $\\angle Q B C=\\theta$.\n\nSo by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \\sin \\theta, 1+a \\cos \\theta)$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_8b358656a56f9e807422g-1.jpg?height=496&width=534&top_left_y=1324&top_left_x=1316)\n\nNow $(P Q)^{2}=a^{2}$, so $(1-2 a \\sin \\theta)^{2}+(1-2 a \\cos \\theta)^{2}=a^{2}$\n\n$$\n2+4 a^{2} \\sin ^{2} \\theta+4 a^{2} \\cos ^{2} \\theta-4 a(\\sin \\theta+\\cos \\theta)=a^{2}\n$$\n\n\n\n$$\n\\begin{aligned}\n2+4 a^{2}-a^{2} & =4 a(\\sin \\theta+\\cos \\theta) \\\\\n\\frac{2+3 a^{2}}{4 a} & =\\sin \\theta+\\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\frac{1}{\\sqrt{2}} \\sin \\theta+\\frac{1}{\\sqrt{2}} \\cos \\theta=\\cos \\left(45^{\\circ}\\right) \\sin \\theta+\\sin \\left(45^{\\circ}\\right) \\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\sin \\left(\\theta+45^{\\circ}\\right)\n\\end{aligned}\n$$\n\n$$\n\\text { Now } \\begin{aligned}\n\\left(a-\\sqrt{\\frac{2}{3}}\\right)^{2} & \\geq 0 \\\\\na^{2}-2 a \\sqrt{\\frac{2}{3}}+\\frac{2}{3} & \\geq 0 \\\\\n3 a^{2}-2 a \\sqrt{6}+2 & \\geq 0 \\\\\n3 a^{2}+2 & \\geq 2 a \\sqrt{6} \\\\\n\\frac{3 a^{2}+2}{4 \\sqrt{2} a} & \\geq \\frac{\\sqrt{3}}{2}\n\\end{aligned}\n$$\n\nand equality occurs when $a=\\sqrt{\\frac{2}{3}}$.\n\nSo $\\sin \\left(\\theta+45^{\\circ}\\right) \\geq \\frac{\\sqrt{3}}{2}$ and thus since $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, then $\\theta+45^{\\circ} \\geq 60^{\\circ}$ or $\\theta \\geq 15^{\\circ}$.\n\nTherefore the minimum possible value of $\\angle A D P$ is $15^{\\circ}$.']",['$15^{\\circ}$'],False,,Numerical, 2245,Geometry,,"Points $P$ and $Q$ are located inside the square $A B C D$ such that $D P$ is parallel to $Q B$ and $D P=Q B=P Q$. Determine the minimum possible value of $\angle A D P$. ","['Placing the information on the coordinate axes, the diagram is indicated to the right.\n\nWe note that $P$ has coordinates $(a, b)$.\n\nBy symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.\n\nSince $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$\n\n$$\n\\begin{aligned}\n& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\\\\n& \\left(a-\\frac{4}{3}\\right)^{2}+\\left(b-\\frac{4}{3}\\right)^{2}=\\frac{8}{9}\n\\end{aligned}\n$$\n\n\n\n$P$ is on a circle with centre $O\\left(\\frac{4}{3}, \\frac{4}{3}\\right)$ with $r=\\frac{2}{3} \\sqrt{2}$.\n\nThe minimum angle for $\\theta$ occurs when $D P$ is tangent to the circle.\n\nSo we have the diagram noted to the right.\n\nSince $O D$ makes an angle of $45^{\\circ}$ with the $x$-axis then $\\angle P D O=45-\\theta$ and $O D=\\frac{4}{3} \\sqrt{2}$.\n\nTherefore $\\sin (45-\\theta)=\\frac{\\frac{2}{3} \\sqrt{2}}{\\frac{4}{3} \\sqrt{2}}=\\frac{1}{2}$ which means $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.\n\nThus the minimum value for $\\theta$ is $15^{\\circ}$.\n\n', 'Let $A B=B C=C D=D A=1$.\n\nJoin $D$ to $B$. Let $\\angle A D P=\\theta$. Therefore, $\\angle P D B=45-\\theta$.\n\nLet $P D=a$ and $P B=b$ and $P Q=\\frac{a}{2}$.\n\n\n\nWe now establish a relationship between $a$ and $b$.\n\nIn $\\triangle P D B, b^{2}=a^{2}+2-2(a)(\\sqrt{2}) \\cos (45-\\theta)$\n\n$$\n\\text { or, } \\quad \\cos (45-\\theta)=\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}\n$$\n\n\n\nIn $\\triangle P D R,\\left(\\frac{a}{2}\\right)^{2}=a^{2}+\\left(\\frac{\\sqrt{2}}{2}\\right)^{2}-2 a \\frac{\\sqrt{2}}{2} \\cos (45-\\theta)$\n\nor, $\\cos (45-\\theta)=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$\n\nComparing (1) and (2) gives, $\\frac{a^{2}-b^{2}+2}{2 \\sqrt{2} a}=\\frac{\\frac{3}{4} a^{2}+\\frac{1}{2}}{a \\sqrt{2}}$.\n\nSimplifying this, $a^{2}+2 b^{2}=2$\n\n$$\n\\text { or, } \\quad b^{2}=\\frac{2-a^{2}}{2}\n$$\n\nNow $\\cos (45-\\theta)=\\frac{a^{2}+2-\\left(\\frac{2-a^{2}}{2}\\right)}{2 a \\sqrt{2}}=\\frac{1}{4 \\sqrt{2}}\\left(3 a+\\frac{2}{a}\\right)$.\n\nNow considering $3 a+\\frac{2}{a}$, we know $\\left(\\sqrt{3 a}-\\sqrt{\\frac{2}{a}}\\right)^{2} \\geq 0$\n\n$$\n\\text { or, } \\quad 3 a+\\frac{2}{a} \\geq 2 \\sqrt{6}\n$$\n\nThus, $\\cos (45-\\theta) \\geq \\frac{1}{4 \\sqrt{2}}(2 \\sqrt{6})=\\frac{\\sqrt{3}}{2}$\n\n$$\n\\cos (45-\\theta) \\geq \\frac{\\sqrt{3}}{2}\n$$\n\n$\\cos (45-\\theta)$ has a minimum value for $45^{\\circ}-\\theta=30^{\\circ}$ or $\\theta=15^{\\circ}$.', 'Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\\angle A D P=\\theta$.\n\nBy interior alternate angles, $\\angle P=\\angle Q$ and $\\angle P D M=\\angle Q B M$.\n\nThus $\\triangle P D M \\cong \\triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.\n\nSo $M$ is the midpoint of $B D$ and the centre of the square.\n\n\n\nWithout loss of generality, let $P M=1$. Then $P D=2$.\n\nSince $\\theta+\\alpha=45^{\\circ}$ (see diagram), $\\theta$ will be minimized when $\\alpha$ is maximized.\n\n\n\nConsider $\\triangle P M D$.\n\nUsing the sine law, $\\frac{\\sin \\alpha}{1}=\\frac{\\sin (\\angle P M D)}{2}$.\n\nTo maximize $\\alpha$, we maximize $\\sin \\alpha$.\n\nBut $\\sin \\alpha=\\frac{\\sin (\\angle P M D)}{2}$, so it is maximized when $\\sin (\\angle P M D)=1$.\n\nIn this case, $\\sin \\alpha=\\frac{1}{2}$, so $\\alpha=30^{\\circ}$.\n\nTherefore, $\\theta=45^{\\circ}-30^{\\circ}=15^{\\circ}$, and so the minimum value of $\\theta$ is $15^{\\circ}$.', 'We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.\n\nLet $P D=P Q=Q B=a$, and $\\angle A D P=\\theta$.\n\nDrop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.\n\nThen $P X=a \\sin \\theta, D X=a \\cos \\theta$.\n\nTherefore the coordinates of $P$ are $(a \\sin \\theta, a \\cos \\theta)$.\n\nSince $P D \\| B Q$, then $\\angle Q B C=\\theta$.\n\nSo by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \\sin \\theta, 1+a \\cos \\theta)$.\n\n\n\nNow $(P Q)^{2}=a^{2}$, so $(1-2 a \\sin \\theta)^{2}+(1-2 a \\cos \\theta)^{2}=a^{2}$\n\n$$\n2+4 a^{2} \\sin ^{2} \\theta+4 a^{2} \\cos ^{2} \\theta-4 a(\\sin \\theta+\\cos \\theta)=a^{2}\n$$\n\n\n\n$$\n\\begin{aligned}\n2+4 a^{2}-a^{2} & =4 a(\\sin \\theta+\\cos \\theta) \\\\\n\\frac{2+3 a^{2}}{4 a} & =\\sin \\theta+\\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\frac{1}{\\sqrt{2}} \\sin \\theta+\\frac{1}{\\sqrt{2}} \\cos \\theta=\\cos \\left(45^{\\circ}\\right) \\sin \\theta+\\sin \\left(45^{\\circ}\\right) \\cos \\theta \\\\\n\\frac{2+3 a^{2}}{4 \\sqrt{2} a} & =\\sin \\left(\\theta+45^{\\circ}\\right)\n\\end{aligned}\n$$\n\n$$\n\\text { Now } \\begin{aligned}\n\\left(a-\\sqrt{\\frac{2}{3}}\\right)^{2} & \\geq 0 \\\\\na^{2}-2 a \\sqrt{\\frac{2}{3}}+\\frac{2}{3} & \\geq 0 \\\\\n3 a^{2}-2 a \\sqrt{6}+2 & \\geq 0 \\\\\n3 a^{2}+2 & \\geq 2 a \\sqrt{6} \\\\\n\\frac{3 a^{2}+2}{4 \\sqrt{2} a} & \\geq \\frac{\\sqrt{3}}{2}\n\\end{aligned}\n$$\n\nand equality occurs when $a=\\sqrt{\\frac{2}{3}}$.\n\nSo $\\sin \\left(\\theta+45^{\\circ}\\right) \\geq \\frac{\\sqrt{3}}{2}$ and thus since $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, then $\\theta+45^{\\circ} \\geq 60^{\\circ}$ or $\\theta \\geq 15^{\\circ}$.\n\nTherefore the minimum possible value of $\\angle A D P$ is $15^{\\circ}$.']",['$15^{\\circ}$'],False,,Numerical, 2246,Geometry,,"In the diagram, $\angle E A D=90^{\circ}, \angle A C D=90^{\circ}$, and $\angle A B C=90^{\circ}$. Also, $E D=13, E A=12$, $D C=4$, and $C B=2$. Determine the length of $A B$. ![](https://cdn.mathpix.com/cropped/2023_12_21_be0b235a4c1314f85f7dg-1.jpg?height=271&width=556&top_left_y=2336&top_left_x=1251)","['By the Pythagorean Theorem in $\\triangle E A D$, we have $E A^{2}+A D^{2}=E D^{2}$ or $12^{2}+A D^{2}=13^{2}$, and so $A D=\\sqrt{169-144}=5$, since $A D>0$.\n\nBy the Pythagorean Theorem in $\\triangle A C D$, we have $A C^{2}+C D^{2}=A D^{2}$ or $A C^{2}+4^{2}=5^{2}$, and so $A C=\\sqrt{25-16}=3$, since $A C>0$.\n\n(We could also have determined the lengths of $A D$ and $A C$ by recognizing 3-4-5 and 5-12-13 right-angled triangles.)\n\nBy the Pythagorean Theorem in $\\triangle A B C$, we have $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+2^{2}=3^{2}$, and so $A B=\\sqrt{9-4}=\\sqrt{5}$, since $A B>0$.']",['$\\sqrt{5}$'],False,,Numerical, 2246,Geometry,,"In the diagram, $\angle E A D=90^{\circ}, \angle A C D=90^{\circ}$, and $\angle A B C=90^{\circ}$. Also, $E D=13, E A=12$, $D C=4$, and $C B=2$. Determine the length of $A B$. ","['By the Pythagorean Theorem in $\\triangle E A D$, we have $E A^{2}+A D^{2}=E D^{2}$ or $12^{2}+A D^{2}=13^{2}$, and so $A D=\\sqrt{169-144}=5$, since $A D>0$.\n\nBy the Pythagorean Theorem in $\\triangle A C D$, we have $A C^{2}+C D^{2}=A D^{2}$ or $A C^{2}+4^{2}=5^{2}$, and so $A C=\\sqrt{25-16}=3$, since $A C>0$.\n\n(We could also have determined the lengths of $A D$ and $A C$ by recognizing 3-4-5 and 5-12-13 right-angled triangles.)\n\nBy the Pythagorean Theorem in $\\triangle A B C$, we have $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+2^{2}=3^{2}$, and so $A B=\\sqrt{9-4}=\\sqrt{5}$, since $A B>0$.']",['$\\sqrt{5}$'],False,,Numerical, 2247,Algebra,,"If $2 \leq x \leq 5$ and $10 \leq y \leq 20$, what is the maximum value of $15-\frac{y}{x}$ ?","['Since we want to make $15-\\frac{y}{x}$ as large as possible, then we want to subtract as little as possible from 15.\n\nIn other words, we want to make $\\frac{y}{x}$ as small as possible.\n\nTo make a fraction with positive numerator and denominator as small as possible, we make the numerator as small as possible and the denominator as large as possible.\n\nSince $2 \\leq x \\leq 5$ and $10 \\leq y \\leq 20$, then we make $x=5$ and $y=10$.\n\nTherefore, the maximum value of $15-\\frac{y}{x}$ is $15-\\frac{10}{5}=13$.', 'Since $y$ is positive and $2 \\leq x \\leq 5$, then $15-\\frac{y}{x} \\leq 15-\\frac{y}{5}$ for any $x$ with $2 \\leq x \\leq 5$ and positive $y$.\n\nSince $10 \\leq y \\leq 20$, then $15-\\frac{y}{5} \\leq 15-\\frac{10}{5}$ for any $y$ with $10 \\leq y \\leq 20$.\n\nTherefore, for any $x$ and $y$ in these ranges, $15-\\frac{y}{x} \\leq 15-\\frac{10}{5}=13$, and so the maximum possible value is 13 (which occurs when $x=5$ and $y=10$ ).']",['13'],False,,Numerical, 2248,Algebra,,"The functions $f$ and $g$ satisfy $$ \begin{aligned} & f(x)+g(x)=3 x+5 \\ & f(x)-g(x)=5 x+7 \end{aligned} $$ for all values of $x$. Determine the value of $2 f(2) g(2)$.","['First, we add the two given equations to obtain\n\n$$\n(f(x)+g(x))+(f(x)-g(x))=(3 x+5)+(5 x+7)\n$$\n\nor $2 f(x)=8 x+12$ which gives $f(x)=4 x+6$.\n\nSince $f(x)+g(x)=3 x+5$, then $g(x)=3 x+5-f(x)=3 x+5-(4 x+6)=-x-1$.\n\n(We could also find $g(x)$ by subtracting the two given equations or by using the second of the given equations.)\n\nSince $f(x)=4 x+6$, then $f(2)=14$.\n\nSince $g(x)=-x-1$, then $g(2)=-3$.\n\nTherefore, $2 f(2) g(2)=2 \\times 14 \\times(-3)=-84$.', 'Since the two given equations are true for all values of $x$, then we can substitute $x=2$ to obtain\n\n$$\n\\begin{aligned}\n& f(2)+g(2)=11 \\\\\n& f(2)-g(2)=17\n\\end{aligned}\n$$\n\nNext, we add these two equations to obtain $2 f(2)=28$ or $f(2)=14$.\n\nSince $f(2)+g(2)=11$, then $g(2)=11-f(2)=11-14=-3$.\n\n(We could also find $g(2)$ by subtracting the two equations above or by using the second of these equations.)\n\nTherefore, $2 f(2) g(2)=2 \\times 14 \\times(-3)=-84$.']",['-84'],False,,Numerical, 2249,Combinatorics,,"Three different numbers are chosen at random from the set $\{1,2,3,4,5\}$. The numbers are arranged in increasing order. What is the probability that the resulting sequence is an arithmetic sequence? (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9 is an arithmetic sequence with four terms.)","['We consider choosing the three numbers all at once.\n\nWe list the possible sets of three numbers that can be chosen:\n\n$$\n\\{1,2,3\\}\\{1,2,4\\}\\{1,2,5\\} \\quad\\{1,3,4\\} \\quad\\{1,3,5\\} \\quad\\{1,4,5\\} \\quad\\{2,3,4\\} \\quad\\{2,3,5\\} \\quad\\{2,4,5\\} \\quad\\{3,4,5\\}\n$$\n\nWe have listed each in increasing order because once the numbers are chosen, we arrange them in increasing order.\n\nThere are 10 sets of three numbers that can be chosen.\n\nOf these 10, the 4 sequences 1,2,3 and 1,3,5 and 2,3,4 and 3,4,5 are arithmetic sequences. Therefore, the probability that the resulting sequence is an arithmetic sequence is $\\frac{4}{10}$ or $\\frac{2}{5}$.']",['$\\frac{2}{5}$'],False,,Numerical, 2250,Geometry,,"In the diagram, $A B C D$ is a quadrilateral with $A B=B C=C D=6, \angle A B C=90^{\circ}$, and $\angle B C D=60^{\circ}$. Determine the length of $A D$. ![](https://cdn.mathpix.com/cropped/2023_12_21_e4d45cb8a520c4aa5010g-1.jpg?height=455&width=477&top_left_y=1057&top_left_x=1258)","['Join $B$ to $D$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4e662a5ccdbfc2f8eac7g-1.jpg?height=453&width=477&top_left_y=283&top_left_x=932)\n\nConsider $\\triangle C B D$.\n\nSince $C B=C D$, then $\\angle C B D=\\angle C D B=\\frac{1}{2}\\left(180^{\\circ}-\\angle B C D\\right)=\\frac{1}{2}\\left(180^{\\circ}-60^{\\circ}\\right)=60^{\\circ}$.\n\nTherefore, $\\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.\n\nConsider $\\triangle D B A$.\n\nNote that $\\angle D B A=90^{\\circ}-\\angle C B D=90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\nSince $B D=B A=6$, then $\\angle B D A=\\angle B A D=\\frac{1}{2}\\left(180^{\\circ}-\\angle D B A\\right)=\\frac{1}{2}\\left(180^{\\circ}-30^{\\circ}\\right)=75^{\\circ}$.\n\nWe calculate the length of $A D$.\n\nMethod 1\n\nBy the Sine Law in $\\triangle D B A$, we have $\\frac{A D}{\\sin (\\angle D B A)}=\\frac{B A}{\\sin (\\angle B D A)}$.\n\nTherefore, $A D=\\frac{6 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(75^{\\circ}\\right)}=\\frac{6 \\times \\frac{1}{2}}{\\sin \\left(75^{\\circ}\\right)}=\\frac{3}{\\sin \\left(75^{\\circ}\\right)}$.\n\nMethod 2\n\nIf we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\\triangle B D A$ is isosceles. Thus, $A D=2 A P$.\n\nAlso, $B P$ bisects $\\angle D B A$, so $\\angle A B P=15^{\\circ}$.\n\nNow, $A P=B A \\sin (\\angle A B P)=6 \\sin \\left(15^{\\circ}\\right)$.\n\nTherefore, $A D=2 A P=12 \\sin \\left(15^{\\circ}\\right)$.\n\nMethod 3\n\nBy the Cosine Law in $\\triangle D B A$,\n\n$$\n\\begin{aligned}\nA D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \\cos (\\angle A B D) \\\\\n& =6^{2}+6^{2}-2(6)(6) \\cos \\left(30^{\\circ}\\right) \\\\\n& =72-72\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\\n& =72-36 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$ since $A D>0$.', 'Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_a94f238e941925a9b44ag-1.jpg?height=466&width=482&top_left_y=285&top_left_x=927)\n\nThen $C Q=C D \\cos (\\angle D C Q)=6 \\cos \\left(60^{\\circ}\\right)=6 \\times \\frac{1}{2}=3$.\n\nAlso, $D Q=C D \\sin (\\angle D C Q)=6 \\sin \\left(60^{\\circ}\\right)=6 \\times \\frac{\\sqrt{3}}{2}=3 \\sqrt{3}$.\n\nSince $B C=6$, then $B Q=B C-C Q=6-3=3$.\n\nNow quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.\n\nThus, $R D=B Q=3$ and $R B=D Q=3 \\sqrt{3}$.\n\nSince $A B=6$, then $A R=A B-R B=6-3 \\sqrt{3}$.\n\nSince $\\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain\n\n$$\nA D=\\sqrt{R D^{2}+A R^{2}}=\\sqrt{3^{2}+(6-3 \\sqrt{3})^{2}}=\\sqrt{9+36-36 \\sqrt{3}+27}=\\sqrt{72-36 \\sqrt{3}}\n$$\n\nwhich we can rewrite as $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$.']",['$6\\sqrt{2-\\sqrt{3}}$'],False,,Numerical, 2250,Geometry,,"In the diagram, $A B C D$ is a quadrilateral with $A B=B C=C D=6, \angle A B C=90^{\circ}$, and $\angle B C D=60^{\circ}$. Determine the length of $A D$. ","['Join $B$ to $D$.\n\n\n\nConsider $\\triangle C B D$.\n\nSince $C B=C D$, then $\\angle C B D=\\angle C D B=\\frac{1}{2}\\left(180^{\\circ}-\\angle B C D\\right)=\\frac{1}{2}\\left(180^{\\circ}-60^{\\circ}\\right)=60^{\\circ}$.\n\nTherefore, $\\triangle B C D$ is equilateral, and so $B D=B C=C D=6$.\n\nConsider $\\triangle D B A$.\n\nNote that $\\angle D B A=90^{\\circ}-\\angle C B D=90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\nSince $B D=B A=6$, then $\\angle B D A=\\angle B A D=\\frac{1}{2}\\left(180^{\\circ}-\\angle D B A\\right)=\\frac{1}{2}\\left(180^{\\circ}-30^{\\circ}\\right)=75^{\\circ}$.\n\nWe calculate the length of $A D$.\n\nMethod 1\n\nBy the Sine Law in $\\triangle D B A$, we have $\\frac{A D}{\\sin (\\angle D B A)}=\\frac{B A}{\\sin (\\angle B D A)}$.\n\nTherefore, $A D=\\frac{6 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(75^{\\circ}\\right)}=\\frac{6 \\times \\frac{1}{2}}{\\sin \\left(75^{\\circ}\\right)}=\\frac{3}{\\sin \\left(75^{\\circ}\\right)}$.\n\nMethod 2\n\nIf we drop a perpendicular from $B$ to $P$ on $A D$, then $P$ is the midpoint of $A D$ since $\\triangle B D A$ is isosceles. Thus, $A D=2 A P$.\n\nAlso, $B P$ bisects $\\angle D B A$, so $\\angle A B P=15^{\\circ}$.\n\nNow, $A P=B A \\sin (\\angle A B P)=6 \\sin \\left(15^{\\circ}\\right)$.\n\nTherefore, $A D=2 A P=12 \\sin \\left(15^{\\circ}\\right)$.\n\nMethod 3\n\nBy the Cosine Law in $\\triangle D B A$,\n\n$$\n\\begin{aligned}\nA D^{2} & =A B^{2}+B D^{2}-2(A B)(B D) \\cos (\\angle A B D) \\\\\n& =6^{2}+6^{2}-2(6)(6) \\cos \\left(30^{\\circ}\\right) \\\\\n& =72-72\\left(\\frac{\\sqrt{3}}{2}\\right) \\\\\n& =72-36 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$ since $A D>0$.', 'Drop perpendiculars from $D$ to $Q$ on $B C$ and from $D$ to $R$ on $B A$.\n\n\n\nThen $C Q=C D \\cos (\\angle D C Q)=6 \\cos \\left(60^{\\circ}\\right)=6 \\times \\frac{1}{2}=3$.\n\nAlso, $D Q=C D \\sin (\\angle D C Q)=6 \\sin \\left(60^{\\circ}\\right)=6 \\times \\frac{\\sqrt{3}}{2}=3 \\sqrt{3}$.\n\nSince $B C=6$, then $B Q=B C-C Q=6-3=3$.\n\nNow quadrilateral $B Q D R$ has three right angles, so it must have a fourth right angle and so must be a rectangle.\n\nThus, $R D=B Q=3$ and $R B=D Q=3 \\sqrt{3}$.\n\nSince $A B=6$, then $A R=A B-R B=6-3 \\sqrt{3}$.\n\nSince $\\triangle A R D$ is right-angled at $R$, then using the Pythagorean Theorem and the fact that $A D>0$, we obtain\n\n$$\nA D=\\sqrt{R D^{2}+A R^{2}}=\\sqrt{3^{2}+(6-3 \\sqrt{3})^{2}}=\\sqrt{9+36-36 \\sqrt{3}+27}=\\sqrt{72-36 \\sqrt{3}}\n$$\n\nwhich we can rewrite as $A D=\\sqrt{36(2-\\sqrt{3})}=6 \\sqrt{2-\\sqrt{3}}$.']",['$6\\sqrt{2-\\sqrt{3}}$'],False,,Numerical, 2251,Combinatorics,,What is the largest two-digit number that becomes $75 \%$ greater when its digits are reversed?,"['Let $n$ be the original number and $N$ be the number when the digits are reversed. Since we are looking for the largest value of $n$, we assume that $n>0$.\n\nSince we want $N$ to be $75 \\%$ larger than $n$, then $N$ should be $175 \\%$ of $n$, or $N=\\frac{7}{4} n$.\n\nSuppose that the tens digit of $n$ is $a$ and the units digit of $n$ is $b$. Then $n=10 a+b$.\n\nAlso, the tens digit of $N$ is $b$ and the units digit of $N$ is $a$, so $N=10 b+a$.\n\nWe want $10 b+a=\\frac{7}{4}(10 a+b)$ or $4(10 b+a)=7(10 a+b)$ or $40 b+4 a=70 a+7 b$ or $33 b=66 a$, and so $b=2 a$.\n\nThis tells us that that any two-digit number $n=10 a+b$ with $b=2 a$ has the required property.\n\nSince both $a$ and $b$ are digits then $b<10$ and so $a<5$, which means that the possible values of $n$ are 12, 24, 36, and 48 .\n\nThe largest of these numbers is 48.']",['48'],False,,Numerical, 2252,Geometry,,"A triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0","['We ""complete the rectangle"" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.\n\n\n\n\n\nSince $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.\n\nSince $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .\n\nSince $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .\n\nTherefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \\times 5=20$.\n\nNow rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .\n\nLet us examine each of these triangles:\n\n- $\\triangle A B C$ has area 8 (given information)\n- $\\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\\frac{1}{2} \\times 4 \\times 3=6$.\n- $\\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\\frac{1}{2} \\times k \\times 2=k$.\n- $\\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\\frac{1}{2} \\times(4-k) \\times 5=10-\\frac{5}{2} k$.\n\nSince the sum of the areas of these triangles is 20 , then $8+6+k+10-\\frac{5}{2} k=20$ or $4=\\frac{3}{2} k$ and so $k=\\frac{8}{3}$.']",['$\\frac{8}{3}$'],False,,Numerical, 2253,Geometry,,"Serge likes to paddle his raft down the Speed River from point $A$ to point $B$. The speed of the current in the river is always the same. When Serge paddles, he always paddles at the same constant speed. On days when he paddles with the current, it takes him 18 minutes to get from $A$ to $B$. When he does not paddle, the current carries him from $A$ to $B$ in 30 minutes. If there were no current, how long would it take him to paddle from $A$ to $B$ ?","['Suppose that the distance from point $A$ to point $B$ is $d \\mathrm{~km}$.\n\nSuppose also that $r_{c}$ is the speed at which Serge travels while not paddling (i.e. being carried by just the current), that $r_{p}$ is the speed at which Serge travels with no current (i.e. just from his paddling), and $r_{p+c}$ his speed when being moved by both his paddling and the current.\n\nIt takes Serge 18 minutes to travel from $A$ to $B$ while paddling with the current.\n\nThus, $r_{p+c}=\\frac{d}{18} \\mathrm{~km} / \\mathrm{min}$.\n\nIt takes Serge 30 minutes to travel from $A$ to $B$ with just the current.\n\nThus, $r_{c}=\\frac{d}{30} \\mathrm{~km} / \\mathrm{min}$.\n\nBut $r_{p}=r_{p+c}-r_{c}=\\frac{d}{18}-\\frac{d}{30}=\\frac{5 d}{90}-\\frac{3 d}{90}=\\frac{2 d}{90}=\\frac{d}{45} \\mathrm{~km} / \\mathrm{min}$.\n\nSince Serge can paddle the $d \\mathrm{~km}$ from $A$ to $B$ at a speed of $\\frac{d}{45} \\mathrm{~km} / \\mathrm{min}$, then it takes him 45 minutes to paddle from $A$ to $B$ with no current.', 'Suppose that the distance from point $A$ to point $B$ is $d \\mathrm{~km}$, the speed of the current of the river is $r \\mathrm{~km} / \\mathrm{h}$, and the speed that Serge can paddle is $s \\mathrm{~km} / \\mathrm{h}$.\n\nSince the current can carry Serge from $A$ to $B$ in 30 minutes (or $\\frac{1}{2} \\mathrm{~h}$ ), then $\\frac{d}{r}=\\frac{1}{2}$.\n\nWhen Serge paddles with the current, his speed equals his paddling speed plus the speed of the current, or $(s+r) \\mathrm{km} / \\mathrm{h}$.\n\nSince Serge can paddle with the current from $A$ to $B$ in 18 minutes (or $\\frac{3}{10} \\mathrm{~h}$ ), then $\\frac{d}{r+s}=\\frac{3}{10}$.\n\nThe time to paddle from $A$ to $B$ with no current would be $\\frac{d}{s} \\mathrm{~h}$.\n\n\n\nSince $\\frac{d}{r}=\\frac{1}{2}$, then $\\frac{r}{d}=2$.\n\nSince $\\frac{d}{r+s}=\\frac{3}{10}$, then $\\frac{r+s}{d}=\\frac{10}{3}$.\n\nTherefore, $\\frac{s}{d}=\\frac{r+s}{d}-\\frac{r}{d}=\\frac{10}{3}-2=\\frac{4}{3}$.\n\nThus, $\\frac{d}{s}=\\frac{3}{4}$, and so it would take Serge $\\frac{3}{4}$ of an hour, or 45 minutes, to paddle from $A$ to $B$ with no current.', 'Suppose that the distance from point $A$ to point $B$ is $d \\mathrm{~km}$, the speed of the current of the river is $r \\mathrm{~km} / \\mathrm{h}$, and the speed that Serge can paddle is $s \\mathrm{~km} / \\mathrm{h}$.\n\nSince the current can carry Serge from $A$ to $B$ in 30 minutes (or $\\frac{1}{2}$ h), then $\\frac{d}{r}=\\frac{1}{2}$ or $d=\\frac{1}{2} r$.\n\nWhen Serge paddles with the current, his speed equals his paddling speed plus the speed of the current, or $(s+r) \\mathrm{km} / \\mathrm{h}$.\n\nSince Serge can paddle with the current from $A$ to $B$ in 18 minutes (or $\\frac{3}{10} \\mathrm{~h}$ ), then $\\frac{d}{r+s}=\\frac{3}{10}$ or $d=\\frac{3}{10}(r+s)$.\n\nSince $d=\\frac{1}{2} r$ and $d=\\frac{3}{10}(r+s)$, then $\\frac{1}{2} r=\\frac{3}{10}(r+s)$ or $5 r=3 r+3 s$ and so $s=\\frac{2}{3} r$.\n\nTo travel from $A$ to $B$ with no current, the time in hours that it takes is $\\frac{d}{s}=\\frac{\\frac{1}{2} r}{\\frac{2}{3} r}=\\frac{3}{4}$, or 45 minutes.']",['45'],False,minute,Numerical, 2254,Geometry,,"Square $O P Q R$ has vertices $O(0,0), P(0,8), Q(8,8)$, and $R(8,0)$. The parabola with equation $y=a(x-2)(x-6)$ intersects the sides of the square $O P Q R$ at points $K, L, M$, and $N$. Determine all the values of $a$ for which the area of the trapezoid $K L M N$ is 36 .","['First, we note that $a \\neq 0$. (If $a=0$, then the ""parabola"" $y=a(x-2)(x-6)$ is actually the horizontal line $y=0$ which intersects the square all along $O R$.)\n\nSecond, we note that, regardless of the value of $a \\neq 0$, the parabola has $x$-intercepts 2 and 6 , and so intersects the $x$-axis at $(2,0)$ and $(6,0)$, which we call $K(2,0)$ and $L(6,0)$. This gives $K L=4$.\n\nThird, we note that since the $x$-intercepts of the parabola are 2 and 6 , then the axis of symmetry of the parabola has equation $x=\\frac{1}{2}(2+6)=4$.\n\nSince the axis of symmetry of the parabola is a vertical line of symmetry, then if the parabola intersects the two vertical sides of the square, it will intersect these at the same height, and if the parabola intersects the top side of the square, it will intersect it at two points that are symmetrical about the vertical line $x=4$.\n\nFourth, we recall that a trapezoid with parallel sides of lengths $a$ and $b$ and height $h$ has area $\\frac{1}{2} h(a+b)$.\n\nWe now examine three cases.\n\n\n\nCase 1: $a<0$\n\nHere, the parabola opens downwards.\n\nSince the parabola intersects the square at four points, it must intersect $P Q$ at points $M$ and $N$. (The parabola cannot intersect the vertical sides of the square since it gets ""narrower"" towards the vertex.)\n\n\n\nSince the parabola opens downwards, then $M N0 ; M$ and $N$ on $P Q$\n\nWe have the following configuration:\n\n\n\nHere, the height of the trapezoid is $8, K L=4$, and $M$ and $N$ are symmetric about $x=4$. Since the area of the trapezoid is 36 , then $\\frac{1}{2} h(K L+M N)=36$ or $\\frac{1}{2}(8)(4+M N)=36$ or $4+M N=9$ or $M N=5$.\n\nThus, $M$ and $N$ are each $\\frac{5}{2}$ units from $x=4$, and so $N$ has coordinates $\\left(\\frac{3}{2}, 8\\right)$.\n\nSince this point lies on the parabola with equation $y=a(x-2)(x-6)$, then $8=a\\left(\\frac{3}{2}-2\\right)\\left(\\frac{3}{2}-6\\right)$ or $8=a\\left(-\\frac{1}{2}\\right)\\left(-\\frac{9}{2}\\right)$ or $8=\\frac{9}{4} a$ or $a=\\frac{32}{9}$.\n\n\n\nCase 3: $a>0 ; M$ and $N$ on $Q R$ and $P O$\n\nWe have the following configuration:\n\n\n\nHere, $K L=4, M N=8$, and $M$ and $N$ have the same $y$-coordinate.\n\nSince the area of the trapezoid is 36 , then $\\frac{1}{2} h(K L+M N)=36$ or $\\frac{1}{2} h(4+8)=36$ or $6 h=36$ or $h=6$.\n\nThus, $N$ has coordinates $(0,6)$.\n\nSince this point lies on the parabola with equation $y=a(x-2)(x-6)$, then $6=a(0-2)(0-6)$ or $6=12 a$ or $a=\\frac{1}{2}$.\n\nTherefore, the possible values of $a$ are $\\frac{32}{9}$ and $\\frac{1}{2}$.']","['$\\frac{32}{9}$,$\\frac{1}{2}$']",True,,Numerical, 2255,Algebra,,"A 75 year old person has a $50 \%$ chance of living at least another 10 years. A 75 year old person has a $20 \%$ chance of living at least another 15 years. An 80 year old person has a $25 \%$ chance of living at least another 10 years. What is the probability that an 80 year old person will live at least another 5 years?","['Consider a population of 100 people, each of whom is 75 years old and who behave according to the probabilities given in the question.\n\nEach of the original 100 people has a $50 \\%$ chance of living at least another 10 years, so there will be $50 \\% \\times 100=50$ of these people alive at age 85 .\n\nEach of the original 100 people has a $20 \\%$ chance of living at least another 15 years, so there will be $20 \\% \\times 100=20$ of these people alive at age 90 .\n\nSince there is a $25 \\%$ ( or $\\frac{1}{4}$ ) chance that an 80 year old person will live at least another 10 years (that is, to age 90), then there should be 4 times as many of these people alive at age 80 than at age 90 .\n\nSince there are 20 people alive at age 90 , then there are $4 \\times 20=80$ of the original 100 people alive at age 80 .\n\nIn summary, of the initial 100 people of age 75, there are 80 alive at age 80,50 alive at age 85 , and 20 people alive at age 90 .\n\nBecause 50 of the 80 people alive at age 80 are still alive at age 85 , then the probability that an 80 year old person will live at least 5 more years (that is, to age 85 ) is $\\frac{50}{80}=\\frac{5}{8}$, or $62.5 \\%$.', 'Suppose that the probability that a 75 year old person lives to 80 is $p$, the probability that an 80 year old person lives to 85 is $q$, and the probability that an 85 year old person lives to 90 is $r$.\n\nWe want to the determine the value of $q$.\n\nFor a 75 year old person to live at least another 10 years, they must live another 5 years (to age 80) and then another 5 years (to age 85). The probability of this is equal to $p q$. We are told in the question that this is equal to $50 \\%$ or 0.5 .\n\nTherefore, $p q=0.5$.\n\n\n\nFor a 75 year old person to live at least another 15 years, they must live another 5 years (to age 80), then another 5 years (to age 85), and then another 5 years (to age 90). The probability of this is equal to $p q r$. We are told in the question that this is equal to $20 \\%$ or 0.2 .\n\nTherefore, $p q r=0.2$\n\nSimilarly, since the probability that an 80 year old person will live another 10 years is $25 \\%$, then $q r=0.25$.\n\nSince $p q r=0.2$ and $p q=0.5$, then $r=\\frac{p q r}{p q}=\\frac{0.2}{0.5}=0.4$.\n\nSince $q r=0.25$ and $r=0.4$, then $q=\\frac{q r}{r}=\\frac{0.25}{0.4}=0.625$.\n\nTherefore, the probability that an 80 year old man will live at least another 5 years is 0.625 , or $62.5 \\%$.']",['62.5%'],False,,Numerical, 2256,Algebra,,Determine all values of $x$ for which $2^{\log _{10}\left(x^{2}\right)}=3\left(2^{1+\log _{10} x}\right)+16$.,"['Using logarithm rules, the given equation is equivalent to $2^{2 \\log _{10} x}=3\\left(2 \\cdot 2^{\\log _{10} x}\\right)+16$ or $\\left(2^{\\log _{10} x}\\right)^{2}=6 \\cdot 2^{\\log _{10} x}+16$.\n\nSet $u=2^{\\log _{10} x}$. Then the equation becomes $u^{2}=6 u+16$ or $u^{2}-6 u-16=0$.\n\nFactoring, we obtain $(u-8)(u+2)=0$ and so $u=8$ or $u=-2$.\n\nSince $2^{a}>0$ for any real number $a$, then $u>0$ and so we can reject the possibility that $u=-2$.\n\nThus, $u=2^{\\log _{10} x}=8$ which means that $\\log _{10} x=3$.\n\nTherefore, $x=1000$.']",['1000'],False,,Numerical, 2257,Algebra,,"The Sieve of Sundaram uses the following infinite table of positive integers: | 4 | 7 | 10 | 13 | $\cdots$ | | :---: | :---: | :---: | :---: | :---: | | 7 | 12 | 17 | 22 | $\cdots$ | | 10 | 17 | 24 | 31 | $\cdots$ | | 13 | 22 | 31 | 40 | $\cdots$ | | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | | The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown. Determine the number in the 50th row and 40th column.","['First, we determine the first entry in the 50th row.\n\nSince the first column is an arithmetic sequence with common difference 3, then the 50th entry in the first column (the first entry in the 50th row) is $4+49(3)=4+147=151$.\n\nSecond, we determine the common difference in the 50th row by determining the second entry in the 50th row.\n\nSince the second column is an arithmetic sequence with common difference 5 , then the 50 th entry in the second column (that is, the second entry in the 50th row) is $7+49(5)$ or $7+245=252$.\n\nTherefore, the common difference in the 50th row must be $252-151=101$.\n\nThus, the 40th entry in the 50th row (that is, the number in the 50th row and the 40th column) is $151+39(101)=151+3939=4090$.']",['4090'],False,,Numerical, 2258,Algebra,,"The Sieve of Sundaram uses the following infinite table of positive integers: | 4 | 7 | 10 | 13 | $\cdots$ | | :---: | :---: | :---: | :---: | :---: | | 7 | 12 | 17 | 22 | $\cdots$ | | 10 | 17 | 24 | 31 | $\cdots$ | | 13 | 22 | 31 | 40 | $\cdots$ | | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | | The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown. Determine a formula for the number in the $R$ th row and $C$ th column.","['First, we determine the first entry in the $R$ th row.\n\nSince the first column is an arithmetic sequence with common difference 3 , then the $R$ th entry in the first column (that is, the first entry in the $R$ th row) is $4+(R-1)(3)$ or $4+3 R-3=3 R+1$.\n\nSecond, we determine the common difference in the $R$ th row by determining the second entry in the $R$ th row.\n\nSince the second column is an arithmetic sequence with common difference 5 , then the $R$ th entry in the second column (that is, the second entry in the $R$ th row) is $7+(R-1)(5)$ or $7+5 R-5=5 R+2$.\n\nTherefore, the common difference in the $R$ th row must be $(5 R+2)-(3 R+1)=2 R+1$. Thus, the $C$ th entry in the $R$ th row (that is, the number in the $R$ th row and the $C$ th column) is\n\n$$\n3 R+1+(C-1)(2 R+1)=3 R+1+2 R C+C-2 R-1=2 R C+R+C\n$$']",['$2RC+R+C$'],False,,Expression, 2259,Algebra,,"The Sieve of Sundaram uses the following infinite table of positive integers: | 4 | 7 | 10 | 13 | $\cdots$ | | :---: | :---: | :---: | :---: | :---: | | 7 | 12 | 17 | 22 | $\cdots$ | | 10 | 17 | 24 | 31 | $\cdots$ | | 13 | 22 | 31 | 40 | $\cdots$ | | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | | The numbers in each row in the table form an arithmetic sequence. The numbers in each column in the table form an arithmetic sequence. The first four entries in each of the first four rows and columns are shown. Prove that if $N$ is an entry in the table, then $2 N+1$ is composite.","['First, we prove that the number in the $R$ th row and $C$ th column equals $2RC+R+C$.\nProof. First, we determine the first entry in the $R$ th row.\n\nSince the first column is an arithmetic sequence with common difference 3 , then the $R$ th entry in the first column (that is, the first entry in the $R$ th row) is $4+(R-1)(3)$ or $4+3 R-3=3 R+1$.\n\nSecond, we determine the common difference in the $R$ th row by determining the second entry in the $R$ th row.\n\nSince the second column is an arithmetic sequence with common difference 5 , then the $R$ th entry in the second column (that is, the second entry in the $R$ th row) is $7+(R-1)(5)$ or $7+5 R-5=5 R+2$.\n\nTherefore, the common difference in the $R$ th row must be $(5 R+2)-(3 R+1)=2 R+1$. Thus, the $C$ th entry in the $R$ th row (that is, the number in the $R$ th row and the $C$ th column) is\n\n$$\n3 R+1+(C-1)(2 R+1)=3 R+1+2 R C+C-2 R-1=2 R C+R+C\n$$\n\nSuppose that $N$ is an entry in the table, say in the $R$ th row and $C$ th column.\n\nFrom the proof above, then $N=2 R C+R+C$ and so $2 N+1=4 R C+2 R+2 C+1$.\n\nNow $4 R C+2 R+2 C+1=2 R(2 C+1)+2 C+1=(2 R+1)(2 C+1)$.\n\nSince $R$ and $C$ are integers with $R \\geq 1$ and $C \\geq 1$, then $2 R+1$ and $2 C+1$ are each integers that are at least 3 .\n\nTherefore, $2 N+1=(2 R+1)(2 C+1)$ must be composite, since it is the product of two integers that are each greater than 1.']",,True,,, 2260,Number Theory,,"Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$. Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$. Determine the value of $g(2011)$.","['If $n=2011$, then $8 n-7=16081$ and so $\\sqrt{8 n-7} \\approx 126.81$.\n\nThus, $\\frac{1+\\sqrt{8 n-7}}{2} \\approx \\frac{1+126.81}{2} \\approx 63.9$.\n\nTherefore, $g(2011)=2(2011)+\\left\\lfloor\\frac{1+\\sqrt{8(2011)-7}}{2}\\right\\rfloor=4022+\\lfloor 63.9\\rfloor=4022+63=4085$.']",['4085'],False,,Numerical, 2261,Number Theory,,"Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$. Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$. Determine a value of $n$ for which $f(n)=100$.","['To determine a value of $n$ for which $f(n)=100$, we need to solve the equation\n\n$$\n2 n-\\left\\lfloor\\frac{1+\\sqrt{8 n-7}}{2}\\right\\rfloor=100\n$$\n\nWe first solve the equation\n\n$$\n2 x-\\frac{1+\\sqrt{8 x-7}}{2}=100 \\quad(* *)\n$$\n\nbecause the left sides of $(*)$ and $(* *)$ do not differ by much and so the solutions are likely close together. We will try integers $n$ in $(*)$ that are close to the solutions to $(* *)$.\n\nManipulating $(* *)$, we obtain\n\n$$\n\\begin{aligned}\n4 x-(1+\\sqrt{8 x-7}) & =200 \\\\\n4 x-201 & =\\sqrt{8 x-7} \\\\\n(4 x-201)^{2} & =8 x-7 \\\\\n16 x^{2}-1608 x+40401 & =8 x-7 \\\\\n16 x^{2}-1616 x+40408 & =0 \\\\\n2 x^{2}-202 x+5051 & =0\n\\end{aligned}\n$$\n\nBy the quadratic formula,\n\n$$\nx=\\frac{202 \\pm \\sqrt{202^{2}-4(2)(5051)}}{2(2)}=\\frac{202 \\pm \\sqrt{396}}{4}=\\frac{101 \\pm \\sqrt{99}}{2}\n$$\n\nand so $x \\approx 55.47$ or $x \\approx 45.53$.\n\nWe try $n=55$, which is close to 55.47 :\n\n$$\nf(55)=2(55)-\\left\\lfloor\\frac{1+\\sqrt{8(55)-7}}{2}\\right\\rfloor=110-\\left\\lfloor\\frac{1+\\sqrt{433}}{2}\\right\\rfloor\n$$\n\nSince $\\sqrt{433} \\approx 20.8$, then $\\frac{1+\\sqrt{433}}{2} \\approx 10.9$, which gives $\\left\\lfloor\\frac{1+\\sqrt{433}}{2}\\right\\rfloor=10$.\n\nThus, $f(55)=110-10=100$.\n\nTherefore, a value of $n$ for which $f(n)=100$ is $n=55$.']",['55'],False,,Numerical, 2262,Number Theory,,"Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. For example, $\lfloor 3.1\rfloor=3$ and $\lfloor-1.4\rfloor=-2$. Suppose that $f(n)=2 n-\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ and $g(n)=2 n+\left\lfloor\frac{1+\sqrt{8 n-7}}{2}\right\rfloor$ for each positive integer $n$. Suppose that $A=\{f(1), f(2), f(3), \ldots\}$ and $B=\{g(1), g(2), g(3), \ldots\}$; that is, $A$ is the range of $f$ and $B$ is the range of $g$. Prove that every positive integer $m$ is an element of exactly one of $A$ or $B$.","['We want to show that each positive integer $m$ is in the range of $f$ or the range of $g$, but not both.\n\nTo do this, we first try to better understand the ""complicated"" term of each of the functions - that is, the term involving the greatest integer function.\n\nIn particular, we start with a positive integer $k \\geq 1$ and try to determine the positive integers $n$ that give $\\left\\lfloor\\frac{1+\\sqrt{8 n-7}}{2}\\right\\rfloor=k$.\n\nBy definition of the greatest integer function, the equation $\\left\\lfloor\\frac{1+\\sqrt{8 n-7}}{2}\\right\\rfloor=k$ is equivalent to the inequality $k \\leq \\frac{1+\\sqrt{8 n-7}}{2}","['Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$.\n\n\n\nTherefore, $\\triangle H B L$ is isosceles and $B L=H B=400 \\mathrm{~m}$.\n\nIn $\\triangle B L P, B L=400 \\mathrm{~m}$ and $\\angle B L P=30^{\\circ}$, so $L P=B L \\cos \\left(30^{\\circ}\\right)=400\\left(\\frac{\\sqrt{3}}{2}\\right)=200 \\sqrt{3}$ m.\n\nTherefore, the distance between $L$ and $P$ is $200 \\sqrt{3} \\mathrm{~m}$.', 'Since $\\angle H L P=60^{\\circ}$ and $\\angle B L P=30^{\\circ}$, then $\\angle H L B=\\angle H L P-\\angle B L P=30^{\\circ}$.\n\nAlso, since $\\angle H L P=60^{\\circ}$ and $\\angle H P L=90^{\\circ}$, then $\\angle L H P=180^{\\circ}-90^{\\circ}-60^{\\circ}=30^{\\circ}$. Also, $\\angle L B P=60^{\\circ}$.\n\nLet $L P=x$.\n\n\n\nSince $\\triangle B L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $B P: L P=1: \\sqrt{3}$, so $B P=\\frac{1}{\\sqrt{3}} L P=\\frac{1}{\\sqrt{3}} x$.\n\n\n\nSince $\\triangle H L P$ is $30^{\\circ}-60^{\\circ}-90^{\\circ}$, then $H P: L P=\\sqrt{3}: 1$, so $H P=\\sqrt{3} L P=\\sqrt{3} x$.\n\nBut $H P=H B+B P$ so\n\n$$\n\\begin{aligned}\n\\sqrt{3} x & =400+\\frac{1}{\\sqrt{3}} x \\\\\n3 x & =400 \\sqrt{3}+x \\\\\n2 x & =400 \\sqrt{3} \\\\\nx & =200 \\sqrt{3}\n\\end{aligned}\n$$\n\nTherefore, the distance from $L$ to $P$ is $200 \\sqrt{3} \\mathrm{~m}$.']",['$200 \\sqrt{3}$ m'],False,,Numerical, 2265,Geometry,,"A goat starts at the origin $(0,0)$ and then makes several moves. On move 1 , it travels 1 unit up to $(0,1)$. On move 2 , it travels 2 units right to $(2,1)$. On move 3 , it travels 3 units down to $(2,-2)$. On move 4 , it travels 4 units to $(-2,-2)$. It continues in this fashion, so that on move $n$, it turns $90^{\circ}$ in a clockwise direction from its previous heading and travels $n$ units in this new direction. After $n$ moves, the goat has travelled a total of 55 units. Determine the coordinates of its position at this time.","['After 2 moves, the goat has travelled $1+2=3$ units.\n\nAfter 3 moves, the goat has travelled $1+2+3=6$ units.\n\nSimilarly, after $n$ moves, the goat has travelled a total of $1+2+3+\\cdots+n$ units.\n\nFor what value of $n$ is $1+2+3+\\cdots+n$ equal to 55 ?\n\nThe fastest way to determine the value of $n$ is by adding the first few integers until we obtain a sum of 55 . This will be $n=10$.\n\n(We could also do this by remembering that $1+2+3+\\cdots+n=\\frac{1}{2} n(n+1)$ and solving for $n$ this way.)\n\nSo we must determine the coordinates of the goat after 10 moves.\n\nWe consider first the $x$-coordinate.\n\nSince starting at $(0,0)$ the goat has moved 2 units in the positive $x$ direction, 4 units in the negative $x$ direction, 6 units in the positive $x$ direction, 8 units in the negative $x$ direction and 10 units in the positive $x$ direction, so its $x$ coordinate should be $2-4+6-8+10=6$. Similarly, its $y$-coordinate should be $1-3+5-7+9=5$.\n\nTherefore, after having travelled a distance of 55 units, the goat is at the point $(6,5)$.']","['(6,5)']",False,,Tuple, 2266,Algebra,,"Determine all possible values of $r$ such that the three term geometric sequence 4, $4 r, 4 r^{2}$ is also an arithmetic sequence. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9, 11 is an arithmetic sequence.)","['Since the sequence $4,4 r, 4 r^{2}$ is also arithmetic, then the difference between $4 r^{2}$ and $4 r$ equals the difference between $4 r$ and 4 , or\n\n$$\n\\begin{aligned}\n4 r^{2}-4 r & =4 r-4 \\\\\n4 r^{2}-8 r+4 & =0 \\\\\nr^{2}-2 r+1 & =0 \\\\\n(r-1)^{2} & =0\n\\end{aligned}\n$$\n\nTherefore, the only value of $r$ is $r=1$.', 'Since the sequence $4,4 r, 4 r^{2}$ is also arithmetic, then we can write $4 r=4+d$ and $4 r^{2}=4+2 d$ for some real number $d$. (Here, $d$ is the common difference in this arithmetic sequence.)\n\nThen $d=4 r-4$ and $2 d=4 r^{2}-4$ or $d=2 r^{2}-2$.\n\nTherefore, equating the two expressions for $d$, we obtain $2 r^{2}-2=4 r-4$ or $2 r^{2}-4 r+2=0$ or $r^{2}-2 r+1=0$ or $(r-1)^{2}=0$.\n\nTherefore, the only value of $r$ is $r=1$.']",['1'],False,,Numerical, 2267,Geometry,,"In the diagram, $A B C D$ is a quadrilateral in which $\angle A+\angle C=180^{\circ}$. What is the length of $C D$ ? ","['In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\\triangle B C D$.\n\nTo do this, we look at $\\angle A$ use the fact that $\\angle A+\\angle C=180^{\\circ}$.\n\n\n\n\n\nUsing the cosine law in $\\triangle A B D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =5^{2}+6^{2}-2(5)(6) \\cos (\\angle A) \\\\\n49 & =61-60 \\cos (\\angle A) \\\\\n\\cos (\\angle A) & =\\frac{1}{5}\n\\end{aligned}\n$$\n\nSince $\\cos (\\angle A)=\\frac{1}{5}$ and $\\angle A+\\angle C=180^{\\circ}$, then $\\cos (\\angle C)=-\\cos \\left(180^{\\circ}-\\angle A\\right)=-\\frac{1}{5}$.\n\n(We could have calculated the actual size of $\\angle A$ using $\\cos (\\angle A)=\\frac{1}{5}$ and then used this to calculate the size of $\\angle C$, but we would introduce the possibility of rounding error by doing this.)\n\nThen, using the cosine law in $\\triangle B C D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =4^{2}+C D^{2}-2(4)(C D) \\cos (\\angle C) \\\\\n49 & =16+C D^{2}-8(C D)\\left(-\\frac{1}{5}\\right) \\\\\n0 & =5 C D^{2}+8 C D-165 \\\\\n0 & =(5 C D+33)(C D-5)\n\\end{aligned}\n$$\n\nSo $C D=-\\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)\n\nSince $C D$ is a length, it must be positive, so $C D=5$.\n\n(We could have also proceeded by using the sine law in $\\triangle B C D$ to determine $\\angle B D C$ and then found the size of $\\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)']",['5'],False,,Numerical, 2267,Geometry,,"In the diagram, $A B C D$ is a quadrilateral in which $\angle A+\angle C=180^{\circ}$. What is the length of $C D$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_9babb378c9215bc0c2fdg-1.jpg?height=444&width=437&top_left_y=735&top_left_x=1256)","['In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\\triangle B C D$.\n\nTo do this, we look at $\\angle A$ use the fact that $\\angle A+\\angle C=180^{\\circ}$.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_72eb55817192542b5c6cg-1.jpg?height=442&width=439&top_left_y=153&top_left_x=946)\n\nUsing the cosine law in $\\triangle A B D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =5^{2}+6^{2}-2(5)(6) \\cos (\\angle A) \\\\\n49 & =61-60 \\cos (\\angle A) \\\\\n\\cos (\\angle A) & =\\frac{1}{5}\n\\end{aligned}\n$$\n\nSince $\\cos (\\angle A)=\\frac{1}{5}$ and $\\angle A+\\angle C=180^{\\circ}$, then $\\cos (\\angle C)=-\\cos \\left(180^{\\circ}-\\angle A\\right)=-\\frac{1}{5}$.\n\n(We could have calculated the actual size of $\\angle A$ using $\\cos (\\angle A)=\\frac{1}{5}$ and then used this to calculate the size of $\\angle C$, but we would introduce the possibility of rounding error by doing this.)\n\nThen, using the cosine law in $\\triangle B C D$, we obtain\n\n$$\n\\begin{aligned}\n7^{2} & =4^{2}+C D^{2}-2(4)(C D) \\cos (\\angle C) \\\\\n49 & =16+C D^{2}-8(C D)\\left(-\\frac{1}{5}\\right) \\\\\n0 & =5 C D^{2}+8 C D-165 \\\\\n0 & =(5 C D+33)(C D-5)\n\\end{aligned}\n$$\n\nSo $C D=-\\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)\n\nSince $C D$ is a length, it must be positive, so $C D=5$.\n\n(We could have also proceeded by using the sine law in $\\triangle B C D$ to determine $\\angle B D C$ and then found the size of $\\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)']",['5'],False,,Numerical, 2268,Algebra,,"If $f(x)=\sin ^{2} x-2 \sin x+2$, what are the minimum and maximum values of $f(x)$ ?","['We rewrite by completing the square as $f(x)=\\sin ^{2} x-2 \\sin x+2=(\\sin x-1)^{2}+1$.\n\nTherefore, since $(\\sin x-1)^{2} \\geq 0$, then $f(x) \\geq 1$, and in fact $f(x)=1$ when $\\sin x=1$ (which occurs for instance when $x=90^{\\circ}$ ).\n\nThus, the minimum value of $f(x)$ is 1 .\n\nTo maximize $f(x)$, we must maximize $(\\sin x-1)^{2}$.\n\nSince $-1 \\leq \\sin x \\leq 1$, then $(\\sin x-1)^{2}$ is maximized when $\\sin x=-1$ (for instance, when $\\left.x=270^{\\circ}\\right)$. In this case, $(\\sin x-1)^{2}=4$, so $f(x)=5$.\n\nThus, the maximum value of $f(x)$ is 5 .']","['5,1']",True,,Numerical, 2269,Geometry,,"In the diagram, the parabola $$ y=-\frac{1}{4}(x-r)(x-s) $$ intersects the axes at three points. The vertex of this parabola is the point $V$. Determine the value of $k$ and the coordinates of $V$. ![](https://cdn.mathpix.com/cropped/2023_12_21_9babb378c9215bc0c2fdg-1.jpg?height=550&width=566&top_left_y=1346&top_left_x=1251)","['From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_f83bb2c63ced8ec40925g-1.jpg?height=544&width=571&top_left_y=167&top_left_x=880)\n\nSince we are given that $y=-\\frac{1}{4}(x-r)(x-s)$, then the $x$-intercepts are $r$ and $s$, so $r$ and $s$ equal $-k$ and $3 k$ in some order.\n\nTherefore, we can rewrite the parabola as $y=-\\frac{1}{4}(x-(-k))(x-3 k)$.\n\nSince the point $(0,3 k)$ lies on the parabola, then $3 k=-\\frac{1}{4}(0+k)(0-3 k)$ or $12 k=3 k^{2}$ or $k^{2}-4 k=0$ or $k(k-4)=0$.\n\nThus, $k=0$ or $k=4$.\n\nSince the two roots are distinct, then we cannot have $k=0$ (otherwise both $x$-intercepts would be 0 ).\n\nThus, $k=4$.\n\nThis tells us that the equation of the parabola is $y=-\\frac{1}{4}(x+4)(x-12)$ or $y=-\\frac{1}{4} x^{2}+$ $2 x+12$.\n\nWe still have to determine the coordinates of the vertex, $V$.\n\nSince the $x$-intercepts of the parabola are -4 and 12 , then the $x$-coordinate of the vertex is the average of these intercepts, or 4.\n\n(We could have also used the fact that the $x$-coordinate is $-\\frac{b}{2 a}=-\\frac{2}{2\\left(-\\frac{1}{4}\\right)}$.)\n\nTherefore, the $y$-coordinate of the vertex is $y=-\\frac{1}{4}\\left(4^{2}\\right)+2(4)+12=16$.\n\nThus, the coordinates of the vertex are $(4,16)$.']","['$4,(4,16)$']",True,,"Numerical,Tuple", 2269,Geometry,,"In the diagram, the parabola $$ y=-\frac{1}{4}(x-r)(x-s) $$ intersects the axes at three points. The vertex of this parabola is the point $V$. Determine the value of $k$ and the coordinates of $V$. ","['From the diagram, the $x$-intercepts of the parabola are $x=-k$ and $x=3 k$.\n\n\n\n\n\nSince we are given that $y=-\\frac{1}{4}(x-r)(x-s)$, then the $x$-intercepts are $r$ and $s$, so $r$ and $s$ equal $-k$ and $3 k$ in some order.\n\nTherefore, we can rewrite the parabola as $y=-\\frac{1}{4}(x-(-k))(x-3 k)$.\n\nSince the point $(0,3 k)$ lies on the parabola, then $3 k=-\\frac{1}{4}(0+k)(0-3 k)$ or $12 k=3 k^{2}$ or $k^{2}-4 k=0$ or $k(k-4)=0$.\n\nThus, $k=0$ or $k=4$.\n\nSince the two roots are distinct, then we cannot have $k=0$ (otherwise both $x$-intercepts would be 0 ).\n\nThus, $k=4$.\n\nThis tells us that the equation of the parabola is $y=-\\frac{1}{4}(x+4)(x-12)$ or $y=-\\frac{1}{4} x^{2}+$ $2 x+12$.\n\nWe still have to determine the coordinates of the vertex, $V$.\n\nSince the $x$-intercepts of the parabola are -4 and 12 , then the $x$-coordinate of the vertex is the average of these intercepts, or 4.\n\n(We could have also used the fact that the $x$-coordinate is $-\\frac{b}{2 a}=-\\frac{2}{2\\left(-\\frac{1}{4}\\right)}$.)\n\nTherefore, the $y$-coordinate of the vertex is $y=-\\frac{1}{4}\\left(4^{2}\\right)+2(4)+12=16$.\n\nThus, the coordinates of the vertex are $(4,16)$.']","['$4,(4,16)$']",True,,"Numerical,Tuple", 2270,Geometry,,"In the diagram, two circles are tangent to each other at point $B$. A straight line is drawn through $B$ cutting the two circles at $A$ and $C$, as shown. Tangent lines are drawn to the circles at $A$ and $C$. Prove that these two tangent lines are parallel. ","['Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nDesignate two points $W$ and $X$ on either side of $A$ on one tangent line, and two points $Y$ and $Z$ on either side of $C$ on the other tangent line.\n\n\n\nLet $\\angle X A B=\\theta$.\n\nSince $W X$ is tangent to the circle with centre $O_{1}$ at $A$, then $O_{1} A$ is perpendicular to $W X$, so $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince $O_{1} A=O_{1} B$ because both are radii, then $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} B A=$ $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince the two circles are tangent at $B$, then the line segment joining $O_{1}$ and $O_{2}$ passes through $B$, ie. $O_{1} B O_{2}$ is a straight line segment.\n\nThus, $\\angle O_{2} B C=\\angle O_{1} B A=90^{\\circ}-\\theta$, by opposite angles.\n\nSince $O_{2} B=O_{2} C$, then similarly to above, $\\angle O_{2} C B=\\angle O_{2} B C=90^{\\circ}-\\theta$.\n\nSince $Y Z$ is tangent to the circle with centre $O_{2}$ at $C$, then $O_{2} C$ is perpendicular to $Y Z$. Thus, $\\angle Y C B=90^{\\circ}-\\angle O_{2} C B=\\theta$.\n\nSince $\\angle X A B=\\angle Y C B$, then $W X$ is parallel to $Y Z$, by alternate angles, as required.', 'Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nSince $A O_{1}$ and $B O_{1}$ are radii of the same circle, $A O_{1}=B O_{1}$ so $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} A B=\\angle O_{1} B A$.\n\n\n\nSince $\\mathrm{BO}_{2}$ and $\\mathrm{CO}_{2}$ are radii of the same circle, $B O_{2}=C_{2}$ so $\\triangle B O_{2} C$ is isosceles, so $\\angle O_{2} B C=\\angle O_{2} C B$.\n\nSince the two circles are tangent at $B$, then $O_{1} B O_{2}$ is a line segment (ie. the line segment joining $O_{1}$ and $O_{2}$ passes through the point of tangency of the two circles).\n\nSince $O_{1} B O_{2}$ is straight, then $\\angle O_{1} B A=\\angle O_{2} B C$, by opposite angles.\n\nThus, $\\angle O_{1} A B=\\angle O_{1} B A=\\angle O_{2} B C=\\angle O_{2} C B$.\n\nThis tells us that $\\triangle A O_{1} B$ is similar to $\\triangle B O_{2} C$, so $\\angle A O_{1} B=\\angle B O_{2} C$ or $\\angle A O_{1} O_{2}=$ $\\angle C O_{2} O_{1}$.\n\nTherefore, $A O_{1}$ is parallel to $C_{2}$, by alternate angles.\n\nBut $A$ and $C$ are points of tangency, $A O_{1}$ is perpendicular to the tangent line at $A$ and $\\mathrm{CO}_{2}$ is perpendicular to the tangent line at $C$.\n\nSince $A O_{1}$ and $C O_{2}$ are parallel, then the two tangent lines must be parallel.']",,True,,, 2270,Geometry,,"In the diagram, two circles are tangent to each other at point $B$. A straight line is drawn through $B$ cutting the two circles at $A$ and $C$, as shown. Tangent lines are drawn to the circles at $A$ and $C$. Prove that these two tangent lines are parallel. ![](https://cdn.mathpix.com/cropped/2023_12_21_7e10dbc5f8ae70c21b8eg-1.jpg?height=366&width=531&top_left_y=657&top_left_x=1252)","['Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nDesignate two points $W$ and $X$ on either side of $A$ on one tangent line, and two points $Y$ and $Z$ on either side of $C$ on the other tangent line.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_87da1d35d2fbe47e109eg-1.jpg?height=375&width=618&top_left_y=1474&top_left_x=859)\n\nLet $\\angle X A B=\\theta$.\n\nSince $W X$ is tangent to the circle with centre $O_{1}$ at $A$, then $O_{1} A$ is perpendicular to $W X$, so $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince $O_{1} A=O_{1} B$ because both are radii, then $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} B A=$ $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince the two circles are tangent at $B$, then the line segment joining $O_{1}$ and $O_{2}$ passes through $B$, ie. $O_{1} B O_{2}$ is a straight line segment.\n\nThus, $\\angle O_{2} B C=\\angle O_{1} B A=90^{\\circ}-\\theta$, by opposite angles.\n\nSince $O_{2} B=O_{2} C$, then similarly to above, $\\angle O_{2} C B=\\angle O_{2} B C=90^{\\circ}-\\theta$.\n\nSince $Y Z$ is tangent to the circle with centre $O_{2}$ at $C$, then $O_{2} C$ is perpendicular to $Y Z$. Thus, $\\angle Y C B=90^{\\circ}-\\angle O_{2} C B=\\theta$.\n\nSince $\\angle X A B=\\angle Y C B$, then $W X$ is parallel to $Y Z$, by alternate angles, as required.', 'Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nSince $A O_{1}$ and $B O_{1}$ are radii of the same circle, $A O_{1}=B O_{1}$ so $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} A B=\\angle O_{1} B A$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_60a3f40dbae5bbe7d26dg-1.jpg?height=371&width=548&top_left_y=430&top_left_x=889)\n\nSince $\\mathrm{BO}_{2}$ and $\\mathrm{CO}_{2}$ are radii of the same circle, $B O_{2}=C_{2}$ so $\\triangle B O_{2} C$ is isosceles, so $\\angle O_{2} B C=\\angle O_{2} C B$.\n\nSince the two circles are tangent at $B$, then $O_{1} B O_{2}$ is a line segment (ie. the line segment joining $O_{1}$ and $O_{2}$ passes through the point of tangency of the two circles).\n\nSince $O_{1} B O_{2}$ is straight, then $\\angle O_{1} B A=\\angle O_{2} B C$, by opposite angles.\n\nThus, $\\angle O_{1} A B=\\angle O_{1} B A=\\angle O_{2} B C=\\angle O_{2} C B$.\n\nThis tells us that $\\triangle A O_{1} B$ is similar to $\\triangle B O_{2} C$, so $\\angle A O_{1} B=\\angle B O_{2} C$ or $\\angle A O_{1} O_{2}=$ $\\angle C O_{2} O_{1}$.\n\nTherefore, $A O_{1}$ is parallel to $C_{2}$, by alternate angles.\n\nBut $A$ and $C$ are points of tangency, $A O_{1}$ is perpendicular to the tangent line at $A$ and $\\mathrm{CO}_{2}$ is perpendicular to the tangent line at $C$.\n\nSince $A O_{1}$ and $C O_{2}$ are parallel, then the two tangent lines must be parallel.']",['证明题，略'],True,,Need_human_evaluate, 2271,Geometry,,"The circle $(x-p)^{2}+y^{2}=r^{2}$ has centre $C$ and the circle $x^{2}+(y-p)^{2}=r^{2}$ has centre $D$. The circles intersect at two distinct points $A$ and $B$, with $x$-coordinates $a$ and $b$, respectively. Prove that $a+b=p$ and $a^{2}+b^{2}=r^{2}$.","['We have $(x-p)^{2}+y^{2}=r^{2}$ and $x^{2}+(y-p)^{2}=r^{2}$, so at the points of intersection,\n\n$$\n\\begin{aligned}\n(x-p)^{2}+y^{2} & =x^{2}+(y-p)^{2} \\\\\nx^{2}-2 p x+p^{2}+y^{2} & =x^{2}+y^{2}-2 p y+p^{2} \\\\\n-2 p x & =-2 p y\n\\end{aligned}\n$$\n\nand so $x=y$ (since we may assume that $p \\neq 0$ otherwise the two circles would coincide). Therefore, $a$ and $b$ are the two solutions of the equation $(x-p)^{2}+x^{2}=r^{2}$ or $2 x^{2}-2 p x+$ $\\left(p^{2}-r^{2}\\right)=0$ or $x^{2}-p x+\\frac{1}{2}\\left(p^{2}-r^{2}\\right)=0$.\n\nUsing the relationship between the sum and product of roots of a quadratic equation and its coefficients, we obtain that $a+b=p$ and $a b=\\frac{1}{2}\\left(p^{2}-r^{2}\\right)$.\n\n(We could have solved for $a$ and $b$ using the quadratic formula and calculated these directly.)\n\nSo we know that $a+b=p$.\n\nLastly, $a^{2}+b^{2}=(a+b)^{2}-2 a b=p^{2}-2\\left(\\frac{1}{2}\\left(p^{2}-r^{2}\\right)\\right)=r^{2}$, as required.', 'Since the circles are reflections of one another in the line $y=x$, then the two points of intersection must both lie on the line $y=x$, ie. $A$ has coordinates $(a, a)$ and $B$ has coordinates $(b, b)$.\n\nTherefore, $(a-p)^{2}+a^{2}=r^{2}$ and $(b-p)^{2}+b^{2}=r^{2}$, since these points lie on both circles.\n\n\n\nSubtracting the two equations, we get\n\n$$\n\\begin{aligned}\n(b-p)^{2}-(a-p)^{2}+b^{2}-a^{2} & =0 \\\\\n((b-p)-(a-p))((b-p)+(a-p))+(b-a)(b+a) & =0 \\\\\n(b-a)(a+b-2 p)+(b-a)(b+a) & =0 \\\\\n(b-a)(a+b-2 p+b+a) & =0 \\\\\n2(b-a)(a+b-p) & =0\n\\end{aligned}\n$$\n\nSince $a \\neq b$, then we must have $a+b=p$, as required.\n\nSince $a+b=p$, then $a-p=-b$, so substituting back into $(a-p)^{2}+a^{2}=r^{2}$ gives $(-b)^{2}+a^{2}=r^{2}$, or $a^{2}+b^{2}=r^{2}$, as required.']",,True,,, 2272,Geometry,,"The circle $(x-p)^{2}+y^{2}=r^{2}$ has centre $C$ and the circle $x^{2}+(y-p)^{2}=r^{2}$ has centre $D$. The circles intersect at two distinct points $A$ and $B$, with $x$-coordinates $a$ and $b$, respectively. If $r$ is fixed and $p$ is then found to maximize the area of quadrilateral $C A D B$, prove that either $A$ or $B$ is the origin.","['We first draw a diagram.\n\n\n\nWe know that $C$ has coordinates $(p, 0)$ and $D$ has coordinates $(0, p)$.\n\nThus, the slope of line segment $C D$ is -1 .\n\nSince the points $A$ and $B$ both lie on the line $y=x$, then the slope of line segment $A B$ is 1 .\n\nTherefore, $A B$ is perpendicular to $C D$, so $C A D B$ is a kite, and so its area is equal to $\\frac{1}{2}(A B)(C D)$.\n\n(We could derive this by breaking quadrilateral $C A D B$ into $\\triangle C A B$ and $\\triangle D A B$.)\n\nSince $C$ has coordinates $(p, 0)$ and $D$ has coordinates $(0, p)$, then $C D=\\sqrt{p^{2}+(-p)^{2}}=$ $\\sqrt{2 p^{2}}$.\n\n(We do not know if $p$ is positive, so this is not necessarily equal to $\\sqrt{2} p$.)\n\nWe know that $A$ has coordinates $(a, a)$ and $B$ has coordinates $(b, b)$, so\n\n$$\n\\begin{aligned}\nA B & =\\sqrt{(a-b)^{2}+(a-b)^{2}} \\\\\n& =\\sqrt{2 a^{2}-4 a b+2 b^{2}} \\\\\n& =\\sqrt{2\\left(a^{2}+b^{2}\\right)-4 a b} \\\\\n& =\\sqrt{2 r^{2}-4\\left(\\frac{1}{2}\\left(p^{2}-r^{2}\\right)\\right)} \\\\\n& =\\sqrt{4 r^{2}-2 p^{2}}\n\\end{aligned}\n$$\n\nTherefore, the area of quadrilateral $C A D B$ is\n\n$$\n\\frac{1}{2}(A B)(C D)=\\frac{1}{2} \\sqrt{4 r^{2}-2 p^{2}} \\sqrt{2 p^{2}}=\\sqrt{2 r^{2} p^{2}-p^{4}}\n$$\n\n\n\nTo maximize this area, we must maximize $2 r^{2} p^{2}-p^{4}=2 r^{2}\\left(p^{2}\\right)-\\left(p^{2}\\right)^{2}$.\n\nSince $r$ is fixed, we can consider this as a quadratic polynomial in $p^{2}$. Since the coefficient of $\\left(p^{2}\\right)^{2}$ is negative, then this is a parabola opening downwards, so we find its maximum value by finding its vertex.\n\nThe vertex of $2 r^{2}\\left(p^{2}\\right)-\\left(p^{2}\\right)^{2}$ is at $p^{2}=-\\frac{2 r^{2}}{2(-1)}=r^{2}$.\n\nSo the maximum area of the quadrilateral occurs when $p$ is chosen so that $p^{2}=r^{2}$.\n\nSince $p^{2}=r^{2}$, then $(a+b)^{2}=p^{2}=r^{2}$ so $a^{2}+2 a b+b^{2}=r^{2}$.\n\nSince $a^{2}+b^{2}=r^{2}$, then $2 a b=0$ so either $a=0$ or $b=0$, and so either $A$ has coordinates $(0,0)$ or $B$ has coordinates $(0,0)$, ie. either $A$ is the origin or $B$ is the origin.']",,True,,, 2273,Combinatorics,,"A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_7e10dbc5f8ae70c21b8eg-1.jpg?height=449&width=1283&top_left_y=1719&top_left_x=429) Determine $f(50)$.","['We proceed directly.\n\nOn the first pass from left to right, Josephine closes all of the even numbered lockers, leaving the odd ones open.\n\nThe second pass proceeds from right to left. Before the pass, the lockers which are open are $1,3, \\ldots, 47,49$.\n\nOn the second pass, she shuts lockers 47, 43, 39, .., 3 .\n\nThe third pass proceeds from left to right. Before the pass, the lockers which are open are $1,5, \\ldots, 45,49$.\n\nOn the third pass, she shuts lockers $5,13, \\ldots, 45$.\n\nThis leaves lockers 1, 9, 17, 25, 33, 41, 49 open.\n\nOn the fourth pass, from right to left, lockers 41, 25 and 9 are shut, leaving 1, 17, 33, 49.\n\nOn the fifth pass, from left to right, lockers 17 and 49 are shut, leaving 1 and 33 open.\n\nOn the sixth pass, from right to left, locker 1 is shut, leaving 33 open.\n\nThus, $f(50)=33$.']",['33'],False,,Numerical, 2273,Combinatorics,,"A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below: Determine $f(50)$.","['We proceed directly.\n\nOn the first pass from left to right, Josephine closes all of the even numbered lockers, leaving the odd ones open.\n\nThe second pass proceeds from right to left. Before the pass, the lockers which are open are $1,3, \\ldots, 47,49$.\n\nOn the second pass, she shuts lockers 47, 43, 39, .., 3 .\n\nThe third pass proceeds from left to right. Before the pass, the lockers which are open are $1,5, \\ldots, 45,49$.\n\nOn the third pass, she shuts lockers $5,13, \\ldots, 45$.\n\nThis leaves lockers 1, 9, 17, 25, 33, 41, 49 open.\n\nOn the fourth pass, from right to left, lockers 41, 25 and 9 are shut, leaving 1, 17, 33, 49.\n\nOn the fifth pass, from left to right, lockers 17 and 49 are shut, leaving 1 and 33 open.\n\nOn the sixth pass, from right to left, locker 1 is shut, leaving 33 open.\n\nThus, $f(50)=33$.']",['33'],False,,Numerical, 2274,Combinatorics,,"A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_7e10dbc5f8ae70c21b8eg-1.jpg?height=449&width=1283&top_left_y=1719&top_left_x=429) Prove that there is no positive integer $n$ such that $f(n)=2005$ and there are infinitely many positive integers $n$ such that $f(n)=f(2005)$.","['First, we calculate $f(n)$ for $n$ from 1 to 32 , to get a feeling for what happens. We obtain $1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11,1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11$. This will help us to establish some patterns.\n\nNext, we establish two recursive formulas for $f(n)$.\n\nFirst, from our pattern, it looks like $f(2 m)=f(2 m-1)$.\n\nWhy is this true in general?\n\nConsider a row of $2 m$ lockers.\n\nOn the first pass through, Josephine shuts all of the even numbered lockers, leaving open lockers $1,3, \\ldots, 2 m-1$.\n\nThese are exactly the same open lockers as if she had started with $2 m-1$ lockers in total. Thus, as she starts her second pass from right to left, the process will be the same now whether she started with $2 m$ lockers or $2 m-1$ lockers.\n\nTherefore, $f(2 m)=f(2 m-1)$.\n\nThis tells us that we need only focus on the values of $f(n)$ where $n$ is odd.\n\nSecondly, we show that $f(2 m-1)=2 m+1-2 f(m)$.\n\n(It is helpful to connect $n=2 m-1$ to a smaller case.)\n\nWhy is this formula true?\n\nStarting with $2 m-1$ lockers, the lockers left open after the first pass are $1,3, \\ldots, 2 m-1$, ie. $m$ lockers in total.\n\nSuppose $f(m)=p$. As Josephine begins her second pass, which is from right to left, we can think of this as being like the first pass through a row of $m$ lockers.\n\nThus, the last open locker will be the $p$ th locker, counting from the right hand end, from the list $1,3, \\ldots, 2 m-1$.\n\nThe first locker from the right is $2 m-1=2 m+1-2(1)$, the second is $2 m-3=2 m+1-2(2)$, and so on, so the $p$ th locker is $2 m+1-2 p$.\n\nTherefore, the final open locker is $2 m+1-2 p$, ie. $f(2 m-1)=2 m+1-2 p=2 m+1-2 f(m)$.\n\nUsing these two formulae repeatedly,\n\n$$\n\\begin{aligned}\nf(4 k+1) & =f(2(2 k+1)-1) \\\\\n& =2(2 k+1)+1-2 f(2 k+1) \\\\\n& =4 k+3-2 f(2(k+1)-1) \\\\\n& =4 k+3-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+3-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-3\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\nf(4 k+3) & =f(2(2 k+2)-1) \\\\\n& =2(2 k+2)+1-2 f(2 k+2) \\\\\n& =4 k+5-2 f(2 k+1) \\\\\n& =4 k+5-2 f(2(k+1)-1) \\\\\n& =4 k+5-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+5-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-1\n\\end{aligned}\n$$\n\nFrom our initial list of values of $f(n)$, it appears as if $f(n)$ cannot leave a remainder of 5 or 7 when divided by 8 . So we use these recursive relations once more to try to establish this:\n\n$$\n\\begin{aligned}\nf(8 l+1) & =4 f(2 l+1)-3 \\quad(\\text { since } 8 l+1=4(2 l)+1) \\\\\n& =4(2 l+3-2 f(l+1))-3 \\\\\n& =8 l+9-8 f(l+1) \\\\\n& =8(l-f(l+1))+9 \\\\\nf(8 l+3) & =4 f(2 l+1)-1 \\quad(\\text { since } 8 l+3=4(2 l)+3) \\\\\n& =4(2 l+3-2 f(l+1))-1 \\\\\n& =8 l+11-8 f(l+1) \\\\\n& =8(l-f(l+1))+11\n\\end{aligned}\n$$\n\nSimilarly, $f(8 l+5)=8 l+9-8 f(l+1)$ and $f(8 l+7)=8 l+11-8 f(l+1)$.\n\nTherefore, since any odd positive integer $n$ can be written as $8 l+1,8 l+3,8 l+5$ or $8 l+7$, then for any odd positive integer $n, f(n)$ is either 9 more or 11 more than a multiple of 8 . Therefore, for any odd positive integer $n, f(n)$ cannot be 2005, since 2005 is not 9 more or 11 more than a multiple of 8 .\n\nThus, for every positive integer $n, f(n) \\neq 2005$, since we only need to consider odd values of $n$.\n\nNext, we show that there are infinitely many positive integers $n$ such that $f(n)=f(2005)$. We do this by looking at the pattern we initially created and conjecturing that\n\n$$\nf(2005)=f\\left(2005+2^{2 a}\\right)\n$$\n\nif $2^{2 a}>2005$. (We might guess this by looking at the connection between $f(1)$ and $f(3)$ with $f(5)$ and $f(7)$ and then $f(1)$ through $f(15)$ with $f(17)$ through $f(31)$. In fact, it appears to be true that $f\\left(m+2^{2 a}\\right)=f(m)$ if $2^{2 a}>m$.)\n\n\n\nUsing our formulae from above,\n\n$$\n\\begin{aligned}\n& f\\left(2005+2^{2 a}\\right)=4 f\\left(502+2^{2 a-2}\\right)-3 \\quad\\left(2005+2^{2 a}=4\\left(501+2^{2 a-2}\\right)+1\\right) \\\\\n& =4 f\\left(501+2^{2 a-2}\\right)-3 \\\\\n& =4\\left(4 f\\left(126+2^{2 a-4}\\right)-3\\right)-3 \\quad\\left(501+2^{2 a-2}=4\\left(125+2^{2 a-4}\\right)+1\\right) \\\\\n& =16 f\\left(126+2^{2 a-4}\\right)-15 \\\\\n& =16 f\\left(125+2^{2 a-4}\\right)-15 \\\\\n& =16\\left(4 f\\left(32+2^{2 a-6}\\right)-3\\right)-15 \\\\\n& =64 f\\left(32+2^{2 a-6}\\right)-63 \\\\\n& =64 f\\left(31+2^{2 a-6}\\right)-63 \\\\\n& =64\\left(4 f\\left(8+2^{2 a-8}\\right)-1\\right)-63 \\\\\n& =256 f\\left(8+2^{2 a-8}\\right)-127 \\\\\n& =256 f\\left(7+2^{2 a-8}\\right)-127 \\\\\n& =256\\left(4 f\\left(2+2^{2 a-10}\\right)-1\\right)-127 \\quad\\left(7+2^{2 a-8}=4\\left(1+2^{2 a-10}\\right)+3\\right) \\\\\n& =1024 f\\left(2+2^{2 a-10}\\right)-383 \\\\\n& =1024 f\\left(1+2^{2 a-10}\\right)-383 \\\\\n& \\left(125+2^{2 a-4}=4\\left(31+2^{2 a-6}\\right)+1\\right) \\\\\n& \\left(31+2^{2 a-6}=4\\left(7+2^{2 a-8}\\right)+3\\right)\n\\end{aligned}\n$$\n\n(Notice that we could have removed the powers of 2 from inside the functions and used this same approach to show that $f(2005)=1024 f(1)-383=641$.)\n\nBut, $f\\left(2^{2 b}+1\\right)=1$ for every positive integer $b$.\n\nWhy is this true? We can prove this quickly by induction.\n\nFor $b=1$, we know $f(5)=1$.\n\nAssume that the result is true for $b=B-1$, for some positive integer $B \\geq 2$.\n\nThen $f\\left(2^{2 B}+1\\right)=f\\left(4\\left(2^{2 B-2}\\right)+1\\right)=4 f\\left(2^{2 B-2}+1\\right)-3=4(1)-3=1$ by our induction hypothesis.\n\nTherefore, if $a \\geq 6$, then $f\\left(1+2^{2 a-10}\\right)=f\\left(1+2^{2(a-5)}\\right)=1$ so\n\n$$\nf\\left(2005+2^{2 a}\\right)=1024(1)-383=641=f(2005)\n$$\n\nso there are infinitely many integers $n$ for which $f(n)=f(2005)$.']",['证明题，略'],True,,Need_human_evaluate, 2274,Combinatorics,,"A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below: Prove that there is no positive integer $n$ such that $f(n)=2005$ and there are infinitely many positive integers $n$ such that $f(n)=f(2005)$.","['First, we calculate $f(n)$ for $n$ from 1 to 32 , to get a feeling for what happens. We obtain $1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11,1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11$. This will help us to establish some patterns.\n\nNext, we establish two recursive formulas for $f(n)$.\n\nFirst, from our pattern, it looks like $f(2 m)=f(2 m-1)$.\n\nWhy is this true in general?\n\nConsider a row of $2 m$ lockers.\n\nOn the first pass through, Josephine shuts all of the even numbered lockers, leaving open lockers $1,3, \\ldots, 2 m-1$.\n\nThese are exactly the same open lockers as if she had started with $2 m-1$ lockers in total. Thus, as she starts her second pass from right to left, the process will be the same now whether she started with $2 m$ lockers or $2 m-1$ lockers.\n\nTherefore, $f(2 m)=f(2 m-1)$.\n\nThis tells us that we need only focus on the values of $f(n)$ where $n$ is odd.\n\nSecondly, we show that $f(2 m-1)=2 m+1-2 f(m)$.\n\n(It is helpful to connect $n=2 m-1$ to a smaller case.)\n\nWhy is this formula true?\n\nStarting with $2 m-1$ lockers, the lockers left open after the first pass are $1,3, \\ldots, 2 m-1$, ie. $m$ lockers in total.\n\nSuppose $f(m)=p$. As Josephine begins her second pass, which is from right to left, we can think of this as being like the first pass through a row of $m$ lockers.\n\nThus, the last open locker will be the $p$ th locker, counting from the right hand end, from the list $1,3, \\ldots, 2 m-1$.\n\nThe first locker from the right is $2 m-1=2 m+1-2(1)$, the second is $2 m-3=2 m+1-2(2)$, and so on, so the $p$ th locker is $2 m+1-2 p$.\n\nTherefore, the final open locker is $2 m+1-2 p$, ie. $f(2 m-1)=2 m+1-2 p=2 m+1-2 f(m)$.\n\nUsing these two formulae repeatedly,\n\n$$\n\\begin{aligned}\nf(4 k+1) & =f(2(2 k+1)-1) \\\\\n& =2(2 k+1)+1-2 f(2 k+1) \\\\\n& =4 k+3-2 f(2(k+1)-1) \\\\\n& =4 k+3-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+3-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-3\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\nf(4 k+3) & =f(2(2 k+2)-1) \\\\\n& =2(2 k+2)+1-2 f(2 k+2) \\\\\n& =4 k+5-2 f(2 k+1) \\\\\n& =4 k+5-2 f(2(k+1)-1) \\\\\n& =4 k+5-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+5-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-1\n\\end{aligned}\n$$\n\nFrom our initial list of values of $f(n)$, it appears as if $f(n)$ cannot leave a remainder of 5 or 7 when divided by 8 . So we use these recursive relations once more to try to establish this:\n\n$$\n\\begin{aligned}\nf(8 l+1) & =4 f(2 l+1)-3 \\quad(\\text { since } 8 l+1=4(2 l)+1) \\\\\n& =4(2 l+3-2 f(l+1))-3 \\\\\n& =8 l+9-8 f(l+1) \\\\\n& =8(l-f(l+1))+9 \\\\\nf(8 l+3) & =4 f(2 l+1)-1 \\quad(\\text { since } 8 l+3=4(2 l)+3) \\\\\n& =4(2 l+3-2 f(l+1))-1 \\\\\n& =8 l+11-8 f(l+1) \\\\\n& =8(l-f(l+1))+11\n\\end{aligned}\n$$\n\nSimilarly, $f(8 l+5)=8 l+9-8 f(l+1)$ and $f(8 l+7)=8 l+11-8 f(l+1)$.\n\nTherefore, since any odd positive integer $n$ can be written as $8 l+1,8 l+3,8 l+5$ or $8 l+7$, then for any odd positive integer $n, f(n)$ is either 9 more or 11 more than a multiple of 8 . Therefore, for any odd positive integer $n, f(n)$ cannot be 2005, since 2005 is not 9 more or 11 more than a multiple of 8 .\n\nThus, for every positive integer $n, f(n) \\neq 2005$, since we only need to consider odd values of $n$.\n\nNext, we show that there are infinitely many positive integers $n$ such that $f(n)=f(2005)$. We do this by looking at the pattern we initially created and conjecturing that\n\n$$\nf(2005)=f\\left(2005+2^{2 a}\\right)\n$$\n\nif $2^{2 a}>2005$. (We might guess this by looking at the connection between $f(1)$ and $f(3)$ with $f(5)$ and $f(7)$ and then $f(1)$ through $f(15)$ with $f(17)$ through $f(31)$. In fact, it appears to be true that $f\\left(m+2^{2 a}\\right)=f(m)$ if $2^{2 a}>m$.)\n\n\n\nUsing our formulae from above,\n\n$$\n\\begin{aligned}\n& f\\left(2005+2^{2 a}\\right)=4 f\\left(502+2^{2 a-2}\\right)-3 \\quad\\left(2005+2^{2 a}=4\\left(501+2^{2 a-2}\\right)+1\\right) \\\\\n& =4 f\\left(501+2^{2 a-2}\\right)-3 \\\\\n& =4\\left(4 f\\left(126+2^{2 a-4}\\right)-3\\right)-3 \\quad\\left(501+2^{2 a-2}=4\\left(125+2^{2 a-4}\\right)+1\\right) \\\\\n& =16 f\\left(126+2^{2 a-4}\\right)-15 \\\\\n& =16 f\\left(125+2^{2 a-4}\\right)-15 \\\\\n& =16\\left(4 f\\left(32+2^{2 a-6}\\right)-3\\right)-15 \\\\\n& =64 f\\left(32+2^{2 a-6}\\right)-63 \\\\\n& =64 f\\left(31+2^{2 a-6}\\right)-63 \\\\\n& =64\\left(4 f\\left(8+2^{2 a-8}\\right)-1\\right)-63 \\\\\n& =256 f\\left(8+2^{2 a-8}\\right)-127 \\\\\n& =256 f\\left(7+2^{2 a-8}\\right)-127 \\\\\n& =256\\left(4 f\\left(2+2^{2 a-10}\\right)-1\\right)-127 \\quad\\left(7+2^{2 a-8}=4\\left(1+2^{2 a-10}\\right)+3\\right) \\\\\n& =1024 f\\left(2+2^{2 a-10}\\right)-383 \\\\\n& =1024 f\\left(1+2^{2 a-10}\\right)-383 \\\\\n& \\left(125+2^{2 a-4}=4\\left(31+2^{2 a-6}\\right)+1\\right) \\\\\n& \\left(31+2^{2 a-6}=4\\left(7+2^{2 a-8}\\right)+3\\right)\n\\end{aligned}\n$$\n\n(Notice that we could have removed the powers of 2 from inside the functions and used this same approach to show that $f(2005)=1024 f(1)-383=641$.)\n\nBut, $f\\left(2^{2 b}+1\\right)=1$ for every positive integer $b$.\n\nWhy is this true? We can prove this quickly by induction.\n\nFor $b=1$, we know $f(5)=1$.\n\nAssume that the result is true for $b=B-1$, for some positive integer $B \\geq 2$.\n\nThen $f\\left(2^{2 B}+1\\right)=f\\left(4\\left(2^{2 B-2}\\right)+1\\right)=4 f\\left(2^{2 B-2}+1\\right)-3=4(1)-3=1$ by our induction hypothesis.\n\nTherefore, if $a \\geq 6$, then $f\\left(1+2^{2 a-10}\\right)=f\\left(1+2^{2(a-5)}\\right)=1$ so\n\n$$\nf\\left(2005+2^{2 a}\\right)=1024(1)-383=641=f(2005)\n$$\n\nso there are infinitely many integers $n$ for which $f(n)=f(2005)$.']",,True,,, 2275,Algebra,,What is the sum of the digits of the integer equal to $\left(10^{3}+1\right)^{2}$ ?,"['Using a calculator, we see that\n\n$$\n\\left(10^{3}+1\\right)^{2}=1001^{2}=1002001\n$$\n\nThe sum of the digits of this integer is $1+2+1$ which equals 4 .\n\nTo determine this integer without using a calculator, we can let $x=10^{3}$.\n\nThen\n\n$$\n\\begin{aligned}\n\\left(10^{3}+1\\right)^{2} & =(x+1)^{2} \\\\\n& =x^{2}+2 x+1 \\\\\n& =\\left(10^{3}\\right)^{2}+2\\left(10^{3}\\right)+1 \\\\\n& =1002001\n\\end{aligned}\n$$']",['1002001'],False,,Numerical, 2276,Algebra,,"A bakery sells small and large cookies. Before a price increase, the price of each small cookie is $\$ 1.50$ and the price of each large cookie is $\$ 2.00$. The price of each small cookie is increased by $10 \%$ and the price of each large cookie is increased by $5 \%$. What is the percentage increase in the total cost of a purchase of 2 small cookies and 1 large cookie?","['Before the price increase, the total cost of 2 small cookies and 1 large cookie is $2 \\cdot \\$ 1.50+\\$ 2.00=\\$ 5.00$.\n\n$10 \\%$ of $\\$ 1.50$ is $0.1 \\cdot \\$ 1.50=\\$ 0.15$. After the price increase, 1 small cookie costs $\\$ 1.50+\\$ 0.15=\\$ 1.65$.\n\n$5 \\%$ of $\\$ 2.00$ is $0.05 \\cdot \\$ 2.00=\\$ 0.10$. After the price increase, 1 large cookie costs $\\$ 2.00+\\$ 0.10=\\$ 2.10$.\n\nAfter the price increase, the total cost of 2 small cookies and 1 large cookie is $2 \\cdot \\$ 1.65+\\$ 2.10=\\$ 5.40$.\n\nThe percentage increase in the total cost is $\\frac{\\$ 5.40-\\$ 5.00}{\\$ 5.00} \\times 100 \\%=\\frac{40}{500} \\times 100 \\%=8 \\%$.']",['$8 \\%$'],False,,Numerical, 2277,Algebra,,"Qing is twice as old as Rayna. Qing is 4 years younger than Paolo. The average age of Paolo, Qing and Rayna is 13. Determine their ages.","[""Suppose that Rayna's age is $x$ years.\n\nSince Qing is twice as old as Rayna, Qing's age is $2 x$ years.\n\nSince Qing is 4 years younger than Paolo, Paolo's age is $2 x+4$ years.\n\nSince the average of their ages is 13 years, we obtain\n\n$$\n\\frac{x+(2 x)+(2 x+4)}{3}=13\n$$\n\nThis gives $5 x+4=39$ and so $5 x=35$ or $x=7$.\n\nTherefore, Rayna is 7 years old, Qing is 14 years old, and Paolo is 18 years old.\n\n(Checking, the average of 7,14 and 18 is $\\frac{7+14+18}{3}=\\frac{39}{3}=13$.)""]","['7,14,18']",True,,Numerical, 2278,Geometry,,"In the diagram, $P Q R S$ is a quadrilateral. What is its perimeter? ","['The length of $P Q$ is equal to $\\sqrt{(0-5)^{2}+(12-0)^{2}}=\\sqrt{(-5)^{2}+12^{2}}=13$.\n\nIn a similar way, we can see that $Q R=R S=S P=13$.\n\nTherefore, the perimeter of $P Q R S$ is $4 \\cdot 13=52$.\n\n(We can also see that if $O$ is the origin, then $\\triangle P O Q, \\triangle P O S, \\triangle R O Q$, and $\\triangle R O S$ are congruent because $O Q=O S$ and $O P=O R$, which means that $P Q=Q R=R S=S P$.)']",['52'],False,,Numerical, 2278,Geometry,,"In the diagram, $P Q R S$ is a quadrilateral. What is its perimeter? ![](https://cdn.mathpix.com/cropped/2023_12_21_620dad4369d1ef5c24f3g-1.jpg?height=442&width=498&top_left_y=191&top_left_x=1258)","['The length of $P Q$ is equal to $\\sqrt{(0-5)^{2}+(12-0)^{2}}=\\sqrt{(-5)^{2}+12^{2}}=13$.\n\nIn a similar way, we can see that $Q R=R S=S P=13$.\n\nTherefore, the perimeter of $P Q R S$ is $4 \\cdot 13=52$.\n\n(We can also see that if $O$ is the origin, then $\\triangle P O Q, \\triangle P O S, \\triangle R O Q$, and $\\triangle R O S$ are congruent because $O Q=O S$ and $O P=O R$, which means that $P Q=Q R=R S=S P$.)']",['52'],False,,Numerical, 2279,Geometry,,"In the diagram, $A$ has coordinates $(0,8)$. Also, the midpoint of $A B$ is $M(3,9)$ and the midpoint of $B C$ is $N(7,6)$. What is the slope of $A C$ ? ","['Suppose that $B$ has coordinates $(r, s)$ and $C$ has coordinates $(t, u)$.\n\nSince $M(3,9)$ is the midpoint of $A(0,8)$ and $B(r, s)$, then 3 is the average of 0 and $r$ (which gives $r=6)$ and 9 is the average of 8 and $s$ (which gives $s=10$ ).\n\nSince $N(7,6)$ is the midpoint of $B(6,10)$ and $C(t, u)$, then 7 is the average of 6 and $t$ (which gives $t=8$ ) and 6 is the average of 10 and $u$ (which gives $u=2$ ).\n\nThe slope of the line segment joining $A(0,8)$ and $C(8,2)$ is $\\frac{8-2}{0-8}$ which equals $-\\frac{3}{4}$.', 'Since $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$, then $M N$ is parallel to $A C$. Therefore, the slope of $A C$ equals the slope of the line segment joining $M(3,9)$ to $N(7,6)$, which is $\\frac{9-6}{3-7}$ or $-\\frac{3}{4}$.']",['$-\\frac{3}{4}$'],False,,Numerical, 2279,Geometry,,"In the diagram, $A$ has coordinates $(0,8)$. Also, the midpoint of $A B$ is $M(3,9)$ and the midpoint of $B C$ is $N(7,6)$. What is the slope of $A C$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_620dad4369d1ef5c24f3g-1.jpg?height=440&width=415&top_left_y=669&top_left_x=1256)","['Suppose that $B$ has coordinates $(r, s)$ and $C$ has coordinates $(t, u)$.\n\nSince $M(3,9)$ is the midpoint of $A(0,8)$ and $B(r, s)$, then 3 is the average of 0 and $r$ (which gives $r=6)$ and 9 is the average of 8 and $s$ (which gives $s=10$ ).\n\nSince $N(7,6)$ is the midpoint of $B(6,10)$ and $C(t, u)$, then 7 is the average of 6 and $t$ (which gives $t=8$ ) and 6 is the average of 10 and $u$ (which gives $u=2$ ).\n\nThe slope of the line segment joining $A(0,8)$ and $C(8,2)$ is $\\frac{8-2}{0-8}$ which equals $-\\frac{3}{4}$.', 'Since $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$, then $M N$ is parallel to $A C$. Therefore, the slope of $A C$ equals the slope of the line segment joining $M(3,9)$ to $N(7,6)$, which is $\\frac{9-6}{3-7}$ or $-\\frac{3}{4}$.']",['$-\\frac{3}{4}$'],False,,Numerical, 2280,Geometry,,"The parabola with equation $y=-2 x^{2}+4 x+c$ has vertex $V(1,18)$. The parabola intersects the $y$-axis at $D$ and the $x$-axis at $E$ and $F$. Determine the area of $\triangle D E F$.","['Since $V(1,18)$ is on the parabola, then $18=-2\\left(1^{2}\\right)+4(1)+c$ and so $c=18+2-4=16$.\n\nThus, the equation of the parabola is $y=-2 x^{2}+4 x+16$.\n\nThe $y$-intercept occurs when $x=0$, and so $y=16$. Thus, $D$ has coordinates $(0,16)$.\n\nThe $x$-intercepts occur when $y=0$. Here,\n\n$$\n\\begin{array}{r}\n-2 x^{2}+4 x+16=0 \\\\\n-2\\left(x^{2}-2 x-8\\right)=0 \\\\\n-2(x-4)(x+2)=0\n\\end{array}\n$$\n\n\n\nand so $x=4$ and $x=-2$.\n\nThis means that $E$ and $F$, in some order, have coordinates $(4,0)$ and $(-2,0)$.\n\nTherefore, $\\triangle D E F$ has base $E F$ of length $4-(-2)=6$ and height 16 (vertical distance from the $x$-axis to the point $D$ ).\n\nFinally, the area of $\\triangle D E F$ is $\\frac{1}{2} \\cdot 6 \\cdot 16=48$.']",['48'],False,,Numerical, 2281,Algebra,,"If $3\left(8^{x}\right)+5\left(8^{x}\right)=2^{61}$, what is the value of the real number $x$ ?","['We obtain successively\n\n$$\n\\begin{aligned}\n3\\left(8^{x}\\right)+5\\left(8^{x}\\right) & =2^{61} \\\\\n8\\left(8^{x}\\right) & =2^{61} \\\\\n8^{x+1} & =2^{61} \\\\\n\\left(2^{3}\\right)^{x+1} & =2^{61} \\\\\n2^{3(x+1)} & =2^{61}\n\\end{aligned}\n$$\n\nThus, $3(x+1)=61$ and so $3 x+3=61$ which gives $3 x=58$ or $x=\\frac{58}{3}$.']",['$\\frac{58}{3}$'],False,,Numerical, 2282,Number Theory,,"For some real numbers $m$ and $n$, the list $3 n^{2}, m^{2}, 2(n+1)^{2}$ consists of three consecutive integers written in increasing order. Determine all possible values of $m$.","['Since the list $3 n^{2}, m^{2}, 2(n+1)^{2}$ consists of three consecutive integers written in increasing order, then\n\n$$\n\\begin{aligned}\n2(n+1)^{2}-3 n^{2} & =2 \\\\\n2 n^{2}+4 n+2-3 n^{2} & =2 \\\\\n-n^{2}+4 n & =0 \\\\\n-n(n-4) & =0\n\\end{aligned}\n$$\n\nand so $n=0$ or $n=4$.\n\nIf $n=0$, the list becomes $0, m^{2}, 2$. This means that $m^{2}=1$ and so $m= \\pm 1$.\n\nIf $n=4$, we have $3 n^{2}=3 \\cdot 16=48$ and $2(n+1)^{2}=2 \\cdot 25=50$ giving the list $48, m^{2}, 50$. This means that $m^{2}=49$ and so $m= \\pm 7$.\n\nThus, the possible values for $m$ are $1,-1,7,-7$.']","['1,-1,7,-7']",True,,Numerical, 2283,Combinatorics,,"Chinara starts with the point $(3,5)$, and applies the following three-step process, which we call $\mathcal{P}$ : Step 1: Reflect the point in the $x$-axis. Step 2: Translate the resulting point 2 units upwards. Step 3: Reflect the resulting point in the $y$-axis. As she does this, the point $(3,5)$ moves to $(3,-5)$, then to $(3,-3)$, and then to $(-3,-3)$. Chinara then starts with a different point $S_{0}$. She applies the three-step process $\mathcal{P}$ to the point $S_{0}$ and obtains the point $S_{1}$. She then applies $\mathcal{P}$ to $S_{1}$ to obtain the point $S_{2}$. She applies $\mathcal{P}$ four more times, each time using the previous output of $\mathcal{P}$ to be the new input, and eventually obtains the point $S_{6}(-7,-1)$. What are the coordinates of the point $S_{0}$ ?","['Suppose that $S_{0}$ has coordinates $(a, b)$.\n\nStep 1 moves $(a, b)$ to $(a,-b)$.\n\nStep 2 moves $(a,-b)$ to $(a,-b+2)$.\n\nStep 3 moves $(a,-b+2)$ to $(-a,-b+2)$.\n\nThus, $S_{1}$ has coordinates $(-a,-b+2)$.\n\nStep 1 moves $(-a,-b+2)$ to $(-a, b-2)$.\n\nStep 2 moves $(-a, b-2)$ to $(-a, b)$.\n\nStep 3 moves $(-a, b)$ to $(a, b)$.\n\nThus, $S_{2}$ has coordinates $(a, b)$, which are the same coordinates as $S_{0}$.\n\nContinuing this process, $S_{4}$ will have the same coordinates as $S_{2}$ (and thus as $S_{0}$ ) and $S_{6}$ will have the same coordinates as $S_{4}, S_{2}$ and $S_{0}$.\n\nSince the coordinates of $S_{6}$ are $(-7,-1)$, the coordinates of $S_{0}$ are also $(-7,-1)$.', 'We work backwards from $S_{6}(-7,-1)$.\n\nTo do this, we undo the Steps of the process $\\mathcal{P}$ by applying them in reverse order.\n\n\n\nSince Step 3 reflects a point in the $y$-axis, its inverse does the same.\n\nSince Step 2 translates a point 2 units upwards, its inverse translates a point 2 units downwards.\n\nSince Step 1 reflects a point in the $x$-axis, its inverse does the same.\n\nApplying these inverse steps to $S_{6}(-7,-1)$, we obtain $(7,-1)$, then $(7,-3)$, then $(7,3)$.\n\nThus, $S_{5}$ has coordinates $(7,3)$.\n\nApplying the inverse steps to $S_{5}(7,3)$, we obtain $(-7,3)$, then $(-7,1)$, then $(-7,-1)$.\n\nThus, $S_{4}$ has coordinates $(-7,-1)$, which are the same coordinates as $S_{6}$.\n\nIf we apply these steps two more times, we will see that $S_{2}$ is the same point as $S_{4}$.\n\nTwo more applications tell us that $S_{0}$ is the same point as $S_{2}$.\n\nTherefore, the coordinates of $S_{0}$ are the same as the coordinates of $S_{6}$, which are $(-7,-1)$.']","['(-7,-1)']",False,,Tuple, 2284,Geometry,,"In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$. ![](https://cdn.mathpix.com/cropped/2023_12_21_3c4a0213f9fdb308210bg-1.jpg?height=244&width=428&top_left_y=274&top_left_x=1355) Determine the length of $A B$ in terms of $x$.","['We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4720ccb222a99329fc9eg-1.jpg?height=247&width=434&top_left_y=955&top_left_x=951)\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$']",['$2 \\sqrt{3} x$'],False,,Expression, 2284,Geometry,,"In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$. Determine the length of $A B$ in terms of $x$.","['We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$']",['$2 \\sqrt{3} x$'],False,,Expression, 2285,Geometry,,"In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$. Determine positive integers $r$ and $s$ for which $$ \frac{A C}{A D}=\sqrt{\frac{r}{s}} $$","['We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$\n\nNext, we determine $\\frac{A C}{A D}$.\n\nNow, $\\frac{A D}{B D}=\\frac{2}{1}$ and so $A D=2 B D=4 x$.\n\nSuppose that $M$ is the midpoint of $A E$ and $N$ is the midpoint of $B D$.\n\nSince $A E=B D=2 x$, then $A M=M E=B N=N D=x$.\n\nJoin $M$ to $N$ and $N$ to $C$ and $A$ to $C$.\n\n\n\nSince $A B D E$ is a rectangle, then $M N$ is parallel to $A B$ and so $M N$ is perpendicular to both $A E$ and $B D$.\n\nAlso, $M N=A B=2 \\sqrt{3} x$.\n\nSince $\\triangle B C D$ is equilateral, its median $C N$ is perpendicular to $B D$.\n\nSince $M N$ and $N C$ are perpendicular to $B D, M N C$ is actually a straight line segment and so $M C=M N+N C$.\n\nNow $\\triangle B N C$ is also a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $N C=\\sqrt{3} B N=\\sqrt{3} x$.\n\nThis means that $M C=2 \\sqrt{3} x+\\sqrt{3} x=3 \\sqrt{3} x$.\n\n\n\nFinally, $\\triangle A M C$ is right-angled at $M$ and so\n\n$$\nA C=\\sqrt{A M^{2}+M C^{2}}=\\sqrt{x^{2}+(3 \\sqrt{3} x)^{2}}=\\sqrt{x^{2}+27 x^{2}}=\\sqrt{28 x^{2}}=2 \\sqrt{7} x\n$$\n\nsince $x>0$.\n\nThis means that $\\frac{A C}{A D}=\\frac{2 \\sqrt{7} x}{4 x}=\\frac{\\sqrt{7}}{2}=\\sqrt{\\frac{7}{4}}$, which means that the integers $r=7$ and $s=4$ satisfy the conditions.']","['7,4']",True,,Numerical, 2285,Geometry,,"In the diagram, $A B D E$ is a rectangle, $\triangle B C D$ is equilateral, and $A D$ is parallel to $B C$. Also, $A E=2 x$ for some real number $x$. ![](https://cdn.mathpix.com/cropped/2023_12_21_3c4a0213f9fdb308210bg-1.jpg?height=244&width=428&top_left_y=274&top_left_x=1355) Determine positive integers $r$ and $s$ for which $$ \frac{A C}{A D}=\sqrt{\frac{r}{s}} $$","['We begin by determining the length of $A B$ in terms of $x$.\n\nSince $A B D E$ is a rectangle, $B D=A E=2 x$.\n\nSince $\\triangle B C D$ is equilateral, $\\angle D B C=60^{\\circ}$.\n\nJoin $A$ to $D$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4720ccb222a99329fc9eg-1.jpg?height=247&width=434&top_left_y=955&top_left_x=951)\n\nSince $A D$ and $B C$ are parallel, $\\angle A D B=\\angle D B C=60^{\\circ}$.\n\nConsider $\\triangle A D B$. This is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle since $\\angle A B D$ is a right angle.\n\nUsing ratios of side lengths, $\\frac{A B}{B D}=\\frac{\\sqrt{3}}{1}$ and so $A B=\\sqrt{3} B D=2 \\sqrt{3} x$\n\nNext, we determine $\\frac{A C}{A D}$.\n\nNow, $\\frac{A D}{B D}=\\frac{2}{1}$ and so $A D=2 B D=4 x$.\n\nSuppose that $M$ is the midpoint of $A E$ and $N$ is the midpoint of $B D$.\n\nSince $A E=B D=2 x$, then $A M=M E=B N=N D=x$.\n\nJoin $M$ to $N$ and $N$ to $C$ and $A$ to $C$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4720ccb222a99329fc9eg-1.jpg?height=250&width=447&top_left_y=1840&top_left_x=945)\n\nSince $A B D E$ is a rectangle, then $M N$ is parallel to $A B$ and so $M N$ is perpendicular to both $A E$ and $B D$.\n\nAlso, $M N=A B=2 \\sqrt{3} x$.\n\nSince $\\triangle B C D$ is equilateral, its median $C N$ is perpendicular to $B D$.\n\nSince $M N$ and $N C$ are perpendicular to $B D, M N C$ is actually a straight line segment and so $M C=M N+N C$.\n\nNow $\\triangle B N C$ is also a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $N C=\\sqrt{3} B N=\\sqrt{3} x$.\n\nThis means that $M C=2 \\sqrt{3} x+\\sqrt{3} x=3 \\sqrt{3} x$.\n\n\n\nFinally, $\\triangle A M C$ is right-angled at $M$ and so\n\n$$\nA C=\\sqrt{A M^{2}+M C^{2}}=\\sqrt{x^{2}+(3 \\sqrt{3} x)^{2}}=\\sqrt{x^{2}+27 x^{2}}=\\sqrt{28 x^{2}}=2 \\sqrt{7} x\n$$\n\nsince $x>0$.\n\nThis means that $\\frac{A C}{A D}=\\frac{2 \\sqrt{7} x}{4 x}=\\frac{\\sqrt{7}}{2}=\\sqrt{\\frac{7}{4}}$, which means that the integers $r=7$ and $s=4$ satisfy the conditions.']","['7,4']",True,,Numerical, 2286,Algebra,,"Suppose that $n>5$ and that the numbers $t_{1}, t_{2}, t_{3}, \ldots, t_{n-2}, t_{n-1}, t_{n}$ form an arithmetic sequence with $n$ terms. If $t_{3}=5, t_{n-2}=95$, and the sum of all $n$ terms is 1000 , what is the value of $n$ ? (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.)","['Since the sequence $t_{1}, t_{2}, t_{3}, \\ldots, t_{n-2}, t_{n-1}, t_{n}$ is arithmetic, then\n\n$$\nt_{1}+t_{n}=t_{2}+t_{n-1}=t_{3}+t_{n-2}\n$$\n\nThis is because, if $d$ is the common difference, we have $t_{2}=t_{1}+d$ and $t_{n-1}=t_{n}-d$, as well as having $t_{3}=t_{1}+2 d$ and $t_{n-2}=t_{n}-2 d$.\n\nSince the sum of all $n$ terms is 1000, using one formula for the sum of an arithmetic sequence gives\n\n$$\n\\begin{aligned}\n\\frac{n}{2}\\left(t_{1}+t_{n}\\right) & =1000 \\\\\nn\\left(t_{1}+t_{n}\\right) & =2000 \\\\\nn\\left(t_{3}+t_{n-2}\\right) & =2000 \\\\\nn(5+95) & =2000\n\\end{aligned}\n$$\n\nand so $n=20$.', 'Suppose that the arithmetic sequence with $n$ terms has first term $a$ and common difference $d$.\n\nThen $t_{3}=a+2 d=5$ and $t_{n-2}=a+(n-3) d=95$.\n\nSince the sum of the $n$ terms equals 1000, then\n\n$$\n\\frac{n}{2}(2 a+(n-1) d)=1000\n$$\n\nAdding the equations $a+2 d=5$ and $a+(n-3) d=95$, we obtain $2 a+(n-1) d=100$.\n\nSubstituting, we get $\\frac{n}{2}(100)=1000$ from which we obtain $n=20$.']",['20'],False,,Numerical, 2287,Algebra,,"Suppose that $a$ and $r$ are real numbers. A geometric sequence with first term $a$ and common ratio $r$ has 4 terms. The sum of this geometric sequence is $6+6 \sqrt{2}$. A second geometric sequence has the same first term $a$ and the same common ratio $r$, but has 8 terms. The sum of this second geometric sequence is $30+30 \sqrt{2}$. Determine all possible values for $a$. (A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3,-6,12,-24$ are the first four terms of a geometric sequence.)","['Since the sum of a geometric sequence with first term $a$, common ratio $r$ and 4 terms is $6+6 \\sqrt{2}$, then\n\n$$\na+a r+a r^{2}+a r^{3}=6+6 \\sqrt{2}\n$$\n\nSince the sum of a geometric sequence with first term $a$, common ratio $r$ and 8 terms is $30+30 \\sqrt{2}$, then\n\n$$\na+a r+a r^{2}+a r^{3}+a r^{4}+a r^{5}+a r^{6}+a r^{7}=30+30 \\sqrt{2}\n$$\n\nBut\n\n$$\n\\begin{aligned}\na+a r & +a r^{2}+a r^{3}+a r^{4}+a r^{5}+a r^{6}+a r^{7} \\\\\n& =\\left(a+a r+a r^{2}+a r^{3}\\right)+r^{4}\\left(a+a r+a r^{2}+a r^{3}\\right) \\\\\n& =\\left(1+r^{4}\\right)\\left(a+a r+a r^{2}+a r^{3}\\right)\n\\end{aligned}\n$$\n\n\n\nTherefore,\n\n$$\n\\begin{aligned}\n30+30 \\sqrt{2} & =\\left(1+r^{4}\\right)(6+6 \\sqrt{2}) \\\\\n\\frac{30+30 \\sqrt{2}}{6+6 \\sqrt{2}} & =1+r^{4} \\\\\n5 & =1+r^{4} \\\\\nr^{4} & =4 \\\\\nr^{2} & =2 \\quad\\left(\\text { since } r^{2}>0\\right) \\\\\nr & = \\pm \\sqrt{2}\n\\end{aligned}\n$$\n\nIf $r=\\sqrt{2}$,\n\n$a+a r+a r^{2}+a r^{3}=a+\\sqrt{2} a+a(\\sqrt{2})^{2}+a(\\sqrt{2})^{3}=a+\\sqrt{2} a+2 a+2 \\sqrt{2} a=a(3+3 \\sqrt{2})$\n\nSince $a+a r+a r^{2}+a r^{3}=6+6 \\sqrt{2}$, then $a(3+3 \\sqrt{2})=6+6 \\sqrt{2}$ and so $a=\\frac{6+6 \\sqrt{2}}{3+3 \\sqrt{2}}=2$.\n\nIf $r=-\\sqrt{2}$,\n\n$a+a r+a r^{2}+a r^{3}=a-\\sqrt{2} a+a(-\\sqrt{2})^{2}+a(-\\sqrt{2})^{3}=a-\\sqrt{2} a+2 a-2 \\sqrt{2} a=a(3-3 \\sqrt{2})$\n\nSince $a+a r+a r^{2}+a r^{3}=6+6 \\sqrt{2}$, then $a(3-3 \\sqrt{2})=6+6 \\sqrt{2}$ and so\n\n$$\na=\\frac{6+6 \\sqrt{2}}{3-3 \\sqrt{2}}=\\frac{2+2 \\sqrt{2}}{1-\\sqrt{2}}=\\frac{(2+2 \\sqrt{2})(1+\\sqrt{2})}{(1-\\sqrt{2})(1+\\sqrt{2})}=\\frac{2+2 \\sqrt{2}+2 \\sqrt{2}+4}{1-2}=-6-4 \\sqrt{2}\n$$\n\nTherefore, the possible values of $a$ are $a=2$ and $a=-6-4 \\sqrt{2}$.\n\nAn alternate way of arriving at the equation $1+r^{4}=5$ is to use the formula for the sum of a geometric sequence twice to obtain\n\n$$\n\\frac{a\\left(1-r^{4}\\right)}{1-r}=6+6 \\sqrt{2} \\quad \\frac{a\\left(1-r^{8}\\right)}{1-r}=30+30 \\sqrt{2}\n$$\n\nassuming that $r \\neq 1$. (Can you explain why $r \\neq 1$ and $r^{4} \\neq 1$ without knowing already that $r= \\pm \\sqrt{2}$ ?)\n\nDividing the second equation by the first, we obtain\n\n$$\n\\frac{a\\left(1-r^{8}\\right)}{1-r} \\cdot \\frac{1-r}{a\\left(1-r^{4}\\right)}=\\frac{30+30 \\sqrt{2}}{6+6 \\sqrt{2}}\n$$\n\nwhich gives\n\n$$\n\\frac{1-r^{8}}{1-r^{4}}=5\n$$\n\nSince $1-r^{8}=\\left(1+r^{4}\\right)\\left(1-r^{4}\\right)$, we obtain $1+r^{4}=5$. We then can proceed as above.']","['$a=2$, $a=-6-4 \\sqrt{2}$']",True,,Expression, 2288,Combinatorics,,"A bag contains 3 green balls, 4 red balls, and no other balls. Victor removes balls randomly from the bag, one at a time, and places them on a table. Each ball in the bag is equally likely to be chosen each time that he removes a ball. He stops removing balls when there are two balls of the same colour on the table. What is the probability that, when he stops, there is at least 1 red ball and at least 1 green ball on the table?","['Victor stops when there are either 2 green balls on the table or 2 red balls on the table. If the first 2 balls that Victor removes are the same colour, Victor will stop.\n\nIf the first 2 balls that Victor removes are different colours, Victor does not yet stop, but when he removes a third ball, its colour must match the colour of one of the first 2 balls and so Victor does stop.\n\nTherefore, the probability that he stops with at least 1 red ball and 1 green ball on the table is equal to the probability that the first 2 balls that he removes are different colours. Also, the probability that the first 2 balls that he removes are different colours is equal to 1 minus the probability that the first 2 balls that he removes are the same colour.\n\nThe probability that the first two balls that Victor draws are both green is $\\frac{3}{7} \\cdot \\frac{2}{6}$ because for the first ball there are 7 balls in the bag, 3 of which are green and for the second ball there are 6 balls in the bag, 2 of which are green.\n\nThe probability that the first two balls that Victor draws are both red is $\\frac{4}{7} \\cdot \\frac{3}{6}$ because for the first ball there are 7 balls in the bag, 4 of which are red and for the second ball there are 6 balls in the bag, 3 of which are red.\n\nThus, the probability that the first two balls that Victor removes are the same colour is\n\n$$\n\\frac{3}{7} \\cdot \\frac{2}{6}+\\frac{4}{7} \\cdot \\frac{3}{6}=\\frac{1}{7}+\\frac{2}{7}=\\frac{3}{7}\n$$\n\nThis means that the desired probability is $1-\\frac{3}{7}=\\frac{4}{7}$.']",['$\\frac{4}{7}$'],False,,Numerical, 2289,Algebra,,Suppose that $f(a)=2 a^{2}-3 a+1$ for all real numbers $a$ and $g(b)=\log _{\frac{1}{2}} b$ for all $b>0$. Determine all $\theta$ with $0 \leq \theta \leq 2 \pi$ for which $f(g(\sin \theta))=0$.,"['Using the definition of $f$, the following equations are equivalent:\n\n$$\n\\begin{aligned}\nf(a) & =0 \\\\\n2 a^{2}-3 a+1 & =0 \\\\\n(a-1)(2 a-1) & =0\n\\end{aligned}\n$$\n\nTherefore, $f(a)=0$ exactly when $a=1$ or $a=\\frac{1}{2}$.\n\nThus, $f(g(\\sin \\theta))=0$ exactly when $g(\\sin \\theta)=1$ or $g(\\sin \\theta)=\\frac{1}{2}$.\n\nUsing the definition of $g$,\n\n- $g(b)=1$ exactly when $\\log _{\\frac{1}{2}} b=1$, which gives $b=\\left(\\frac{1}{2}\\right)^{1}=\\frac{1}{2}$, and\n- $g(b)=1 / 2$ exactly when $\\log _{\\frac{1}{2}} b=1 / 2$, which gives $b=\\left(\\frac{1}{2}\\right)^{1 / 2}=\\frac{1}{\\sqrt{2}}$.\n\nTherefore, $f(g(\\sin \\theta))=0$ exactly when $\\sin \\theta=\\frac{1}{2}$ or $\\sin \\theta=\\frac{1}{\\sqrt{2}}$.\n\nSince $0 \\leq \\theta \\leq 2 \\pi$, the solutions are $\\theta=\\frac{1}{6} \\pi, \\frac{5}{6} \\pi, \\frac{1}{4} \\pi, \\frac{3}{4} \\pi$.']","['$\\frac{1}{6} \\pi, \\frac{5}{6} \\pi, \\frac{1}{4} \\pi, \\frac{3}{4} \\pi$']",True,,Numerical, 2290,Number Theory,,"Five distinct integers are to be chosen from the set $\{1,2,3,4,5,6,7,8\}$ and placed in some order in the top row of boxes in the diagram. Each box that is not in the top row then contains the product of the integers in the two boxes connected to it in the row directly above. Determine the number of ways in which the integers can be chosen and placed in the top row so that the integer in the bottom box is 9953280000 . ![](https://cdn.mathpix.com/cropped/2023_12_21_3c4a0213f9fdb308210bg-1.jpg?height=369&width=388&top_left_y=2060&top_left_x=1259)","['Suppose that the integers in the first row are, in order, $a, b, c, d, e$.\n\nUsing these, we calculate the integer in each of the boxes below the top row in terms of these variables, using the rule that each integer is the product of the integers in the two boxes above:\n\n$a$\n\n| $b$ | $c$ | $c$ | | $d$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $a b^{2} c$ | $b c$ | $c d$ | $c$ | |\n| | $a b^{3} c^{3} d$ | $b c^{2} d$ | $c d^{2} e$ | |$\\quad d e$\n\nTherefore, $a b^{4} c^{6} d^{4} e=9953280000$.\n\n\n\nNext, we determine the prime factorization of the integer 9953280000 :\n\n$$\n\\begin{aligned}\n9953280000 & =10^{4} \\cdot 995328 \\\\\n& =2^{4} \\cdot 5^{4} \\cdot 2^{3} \\cdot 124416 \\\\\n& =2^{7} \\cdot 5^{4} \\cdot 2^{3} \\cdot 15552 \\\\\n& =2^{10} \\cdot 5^{4} \\cdot 2^{3} \\cdot 1944 \\\\\n& =2^{13} \\cdot 5^{4} \\cdot 2^{3} \\cdot 243 \\\\\n& =2^{16} \\cdot 5^{4} \\cdot 3^{5} \\\\\n& =2^{16} \\cdot 3^{5} \\cdot 5^{4}\n\\end{aligned}\n$$\n\nThus, $a b^{4} c^{6} d^{4} e=2^{16} \\cdot 3^{5} \\cdot 5^{4}$.\n\nSince the right side is not divisible by 7 , none of $a, b, c, d$, $e$ can equal 7 .\n\nThus, $a, b, c, d, e$ are five distinct integers chosen from $\\{1,2,3,4,5,6,8\\}$.\n\nThe only one of these integers divisible by 5 is 5 itself.\n\nSince $2^{16} \\cdot 3^{5} \\cdot 5^{4}$ includes exactly 4 factors of 5 , then either $b=5$ or $d=5$. No other placement of the 5 can give exactly 4 factors of 5 .\n\nCase 1: $b=5$\n\nHere, $a c^{6} d^{4} e=2^{16} \\cdot 3^{5}$ and $a, c, d, e$ are four distinct integers chosen from $\\{1,2,3,4,6,8\\}$. Since $a c^{6} d^{4} e$ includes exactly 5 factors of 3 and the possible values of $a, c, d$, e that are divisible by 3 are 3 and 6 , then either $d=3$ and one of $a$ and $e$ is 6 , or $d=6$ and one of $a$ and $e$ is 3 . No other placements of the multiples of 3 can give exactly 5 factors of 3 .\n\nCase 1a: $b=5, d=3, a=6$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=6 \\cdot c^{6} \\cdot 3^{4} \\cdot e=2 \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{15}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=8$ is the only solution in this case. Here, $(a, b, c, d, e)=(6,5,4,3,8)$.\n\nCase 1b: $b=5, d=3, e=6$ We obtain $(a, b, c, d, e)=(8,5,4,3,6)$.\n\nCase 1c: $b=5, d=6, a=3$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=3 \\cdot c^{6} \\cdot 6^{4} \\cdot e=2^{4} \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{12}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=1$ is the only solution in this case. Here, $(a, b, c, d, e)=(3,5,4,6,1)$.\n\nCase 1d: $b=5, d=6, e=3$ We obtain $(a, b, c, d, e)=(1,5,4,6,3)$.\n\nCase 2: $d=5$ : A similar analysis leads to 4 further quintuples $(a, b, c, d, e)$.\n\nTherefore, there are 8 ways in which the integers can be chosen and placed in the top row to obtain the desired integer in the bottom box.']",['8'],False,,Numerical, 2290,Number Theory,,"Five distinct integers are to be chosen from the set $\{1,2,3,4,5,6,7,8\}$ and placed in some order in the top row of boxes in the diagram. Each box that is not in the top row then contains the product of the integers in the two boxes connected to it in the row directly above. Determine the number of ways in which the integers can be chosen and placed in the top row so that the integer in the bottom box is 9953280000 . ","['Suppose that the integers in the first row are, in order, $a, b, c, d, e$.\n\nUsing these, we calculate the integer in each of the boxes below the top row in terms of these variables, using the rule that each integer is the product of the integers in the two boxes above:\n\n$a$\n\n| $b$ | $c$ | $c$ | | $d$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $a b^{2} c$ | $b c$ | $c d$ | $c$ | |\n| | $a b^{3} c^{3} d$ | $b c^{2} d$ | $c d^{2} e$ | |$\\quad d e$\n\nTherefore, $a b^{4} c^{6} d^{4} e=9953280000$.\n\n\n\nNext, we determine the prime factorization of the integer 9953280000 :\n\n$$\n\\begin{aligned}\n9953280000 & =10^{4} \\cdot 995328 \\\\\n& =2^{4} \\cdot 5^{4} \\cdot 2^{3} \\cdot 124416 \\\\\n& =2^{7} \\cdot 5^{4} \\cdot 2^{3} \\cdot 15552 \\\\\n& =2^{10} \\cdot 5^{4} \\cdot 2^{3} \\cdot 1944 \\\\\n& =2^{13} \\cdot 5^{4} \\cdot 2^{3} \\cdot 243 \\\\\n& =2^{16} \\cdot 5^{4} \\cdot 3^{5} \\\\\n& =2^{16} \\cdot 3^{5} \\cdot 5^{4}\n\\end{aligned}\n$$\n\nThus, $a b^{4} c^{6} d^{4} e=2^{16} \\cdot 3^{5} \\cdot 5^{4}$.\n\nSince the right side is not divisible by 7 , none of $a, b, c, d$, $e$ can equal 7 .\n\nThus, $a, b, c, d, e$ are five distinct integers chosen from $\\{1,2,3,4,5,6,8\\}$.\n\nThe only one of these integers divisible by 5 is 5 itself.\n\nSince $2^{16} \\cdot 3^{5} \\cdot 5^{4}$ includes exactly 4 factors of 5 , then either $b=5$ or $d=5$. No other placement of the 5 can give exactly 4 factors of 5 .\n\nCase 1: $b=5$\n\nHere, $a c^{6} d^{4} e=2^{16} \\cdot 3^{5}$ and $a, c, d, e$ are four distinct integers chosen from $\\{1,2,3,4,6,8\\}$. Since $a c^{6} d^{4} e$ includes exactly 5 factors of 3 and the possible values of $a, c, d$, e that are divisible by 3 are 3 and 6 , then either $d=3$ and one of $a$ and $e$ is 6 , or $d=6$ and one of $a$ and $e$ is 3 . No other placements of the multiples of 3 can give exactly 5 factors of 3 .\n\nCase 1a: $b=5, d=3, a=6$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=6 \\cdot c^{6} \\cdot 3^{4} \\cdot e=2 \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{15}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=8$ is the only solution in this case. Here, $(a, b, c, d, e)=(6,5,4,3,8)$.\n\nCase 1b: $b=5, d=3, e=6$ We obtain $(a, b, c, d, e)=(8,5,4,3,6)$.\n\nCase 1c: $b=5, d=6, a=3$\n\nHere, $a \\cdot c^{6} \\cdot d^{4} \\cdot e=3 \\cdot c^{6} \\cdot 6^{4} \\cdot e=2^{4} \\cdot 3^{5} \\cdot c^{6} \\cdot e$.\n\nThis gives $c^{6} e=2^{12}$ and $c$ and $e$ are distinct integers from $\\{1,2,4,8\\}$.\n\nTrying the four possible values of $c$ shows that $c=4$ and $e=1$ is the only solution in this case. Here, $(a, b, c, d, e)=(3,5,4,6,1)$.\n\nCase 1d: $b=5, d=6, e=3$ We obtain $(a, b, c, d, e)=(1,5,4,6,3)$.\n\nCase 2: $d=5$ : A similar analysis leads to 4 further quintuples $(a, b, c, d, e)$.\n\nTherefore, there are 8 ways in which the integers can be chosen and placed in the top row to obtain the desired integer in the bottom box.']",['8'],False,,Numerical, 2291,Number Theory,,"Prove that the integer $\frac{(1 !)(2 !)(3 !) \cdots(398 !)(399 !)(400 !)}{200 !}$ is a perfect square. (In this fraction, the numerator is the product of the factorials of the integers from 1 to 400 , inclusive.)","['Let $N=\\frac{(1 !)(2 !)(3 !) \\cdots(398 !)(399 !)(400 !)}{200 !}$.\n\nFor each integer $k$ from 1 to 200 , inclusive, we rewrite $(2 k)$ ! as $2 k \\cdot(2 k-1)$ !.\n\nTherefore, $(2 k-1) !(2 k) !=(2 k-1) ! \\cdot 2 k \\cdot(2 k-1) !=2 k((2 k-1) !)^{2}$.\n\n(In particular, $(1 !)(2 !)=2(1 !)^{2},(3 !)(4 !)=4(3 !)^{2}$, and so on.)\n\nThus,\n\n$$\nN=\\frac{2(1 !)^{2} \\cdot 4(3 !)^{2} \\cdots \\cdot 398(397 !)^{2} \\cdot 400(399 !)^{2}}{200 !}\n$$\n\nRe-arranging the numerator of the expression, we obtain\n\n$$\nN=\\frac{(1 !)^{2}(3 !)^{2} \\cdots(397 !)^{2}(399 !)^{2} \\cdot(2 \\cdot 4 \\cdot \\cdots 398 \\cdot 400)}{200 !}\n$$\n\nWe can now re-write $2 \\cdot 4 \\cdot \\cdots 398 \\cdot 400$ as $(2 \\cdot 1) \\cdot(2 \\cdot 2) \\cdots(2 \\cdot 199) \\cdot(2 \\cdot 200)$.\n\nSince there are 200 sets of parentheses, we obtain\n\n$$\nN=\\frac{(1 !)^{2}(3 !)^{2} \\cdots(397 !)^{2}(399 !)^{2} \\cdot 2^{200} \\cdot(1 \\cdot 2 \\cdots \\cdots 199 \\cdot 200)}{200 !}\n$$\n\nSince $1 \\cdot 2 \\cdot \\cdots \\cdot 199 \\cdot 200=200$ !, we can conclude that\n\n$$\nN=2^{200}(1 !)^{2}(3 !)^{2} \\cdots(397 !)^{2}(399 !)^{2}\n$$\n\nTherefore,\n\n$$\n\\sqrt{N}=2^{100}(1 !)(3 !) \\cdots(397 !)(399 !)\n$$\n\nwhich is a product of integers and thus an integer itself.\n\nSince $\\sqrt{N}$ is an integer, $N$ is a perfect square, as required.']",,True,,, 2292,Number Theory,,"Suppose that $a=5$ and $b=4$. Determine all pairs of integers $(K, L)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.","['When $a=5$ and $b=4$, we obtain $a^{2}+b^{2}-a b=5^{2}+4^{2}-5 \\cdot 4=21$.\n\nTherefore, we want to find all pairs of integers $(K, L)$ with $K^{2}+3 L^{2}=21$.\n\nIf $L=0$, then $L^{2}=0$, which gives $K^{2}=21$ which has no integer solutions.\n\nIf $L= \\pm 1$, then $L^{2}=1$, which gives $K^{2}=18$ which has no integer solutions.\n\nIf $L= \\pm 2$, then $L^{2}=4$, which gives $K^{2}=9$ which gives $K= \\pm 3$.\n\nIf $L= \\pm 3$, then $L^{2}=9$. Since $3 L^{2}=27>21$, then there are no real solutions for $K$.\n\nSimilarly, if $L^{2}>9$, there are no real solutions for $K$.\n\nTherefore, the solutions are $(K, L)=(3,2),(-3,2),(3,-2),(-3,-2)$.']","['$(3,2),(-3,2),(3,-2),(-3,-2)$']",True,,Tuple, 2293,Number Theory,,"Prove that, for all integers $K$ and $L$, there is at least one pair of integers $(a, b)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.","['Suppose that $K$ and $L$ are integers.\n\nThen\n\n$$\n\\begin{aligned}\n& (K+L)^{2}+(K-L)^{2}-(K+L)(K-L) \\\\\n& \\quad=\\left(K^{2}+2 K L+L^{2}\\right)+\\left(K^{2}-2 K L+L^{2}\\right)-\\left(K^{2}-L^{2}\\right) \\\\\n& \\quad=K^{2}+3 L^{2}\n\\end{aligned}\n$$\n\nTherefore, the integers $a=K+L$ and $b=K-L$ satisfy the equation $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$, and so for all integers $K$ and $L$, there is at least one pair of integers $(a, b)$ that satisfy the equation.\n\nHow could we come up with this? One way to do this would be trying some small values\n\n\n\nof $K$ and $L$, calculating $K^{2}+3 L^{2}$ and using this to make a guess, which can then be proven algebraically as above. In particular, here are some values:\n\n| $K$ | $L$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $a$ | $b$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 1 | 1 | 4 | 2 | 0 |\n| 2 | 1 | 7 | 3 | 1 |\n| 3 | 1 | 12 | 4 | 2 |\n| 1 | 2 | 13 | 3 | -1 |\n| 2 | 2 | 16 | 4 | 0 |\n| 3 | 2 | 21 | 5 | 1 |\n\nThe columns for $a$ and $b$ might lead us to guess that $a=K+L$ and $b=K-L$, which we proved above does in fact work.']",,True,,, 2294,Number Theory,,"Prove that, for all integers $a$ and $b$, there is at least one pair of integers $(K, L)$ for which $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.","['Suppose that $a$ and $b$ are integers.\n\nIf $a$ is even, then $\\frac{a}{2}$ is an integer and so\n\n$$\n\\left(\\frac{a}{2}-b\\right)^{2}+3\\left(\\frac{a}{2}\\right)^{2}=\\frac{a^{2}}{4}-2 \\cdot \\frac{a}{2} \\cdot b+b^{2}+\\frac{3 a^{2}}{4}=a^{2}+b^{2}-a b\n$$\n\nThus, if $K=\\frac{a}{2}-b$ and $L=\\frac{a}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nIf $b$ is even, then $\\frac{b}{2}$ is an integer and so a similar algebraic argument shows that\n\n$$\n\\left(\\frac{b}{2}-a\\right)^{2}+3\\left(\\frac{b}{2}\\right)^{2}=a^{2}+b^{2}-a b\n$$\n\nand so if $K=\\frac{b}{2}-a$ and $L=\\frac{b}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nIf $a$ and $b$ are both odd, then $a+b$ and $a-b$ are both even, which means that $\\frac{a+b}{2}$ and $\\frac{a-b}{2}$ are both integers, and so\n\n$\\left(\\frac{a+b}{2}\\right)^{2}+3\\left(\\frac{a-b}{2}\\right)^{2}=\\frac{a^{2}+2 a b+b^{2}}{4}+\\frac{3 a^{2}-6 a b+3 b^{2}}{4}=\\frac{4 a^{2}+4 b^{2}-4 a b}{4}=a^{2}+b^{2}-a b$\n\nThus, if $K=\\frac{a+b}{2}$ and $L=\\frac{a-b}{2}$, we have $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nTherefore, in all cases, for all integers $a$ and $b$, there is at least one pair of integers $(K, L)$ with $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$.\n\nNow, trying some small cases might help us make a guess of possible expressions for $K$ and $L$ in terms of $a$ and $b$ :\n\n| $a$ | $b$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $K$ | $L$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 1 | 1 | 1 | 1 | 0 |\n| 2 | 1 | 3 | 0 | 1 |\n| 3 | 1 | 7 | 2 | 1 |\n| 4 | 1 | 13 | 1 | 2 |\n| 1 | 2 | 3 | 0 | 1 |\n| 2 | 2 | 4 | 1 | 1 |\n| 3 | 2 | 7 | 2 | 1 |\n| 4 | 2 | 12 | 3 | 1 |\n| 5 | 3 | 19 | 4 | 1 |\n\n\n\nWhile there might not initially seem to be useful patterns here, re-arranging the rows and adding some duplicates might help show a pattern:\n\n| $a$ | $b$ | $K^{2}+3 L^{2}=a^{2}+b^{2}-a b$ | $K$ | $L$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 2 | 1 | 3 | 0 | 1 |\n| 4 | 1 | 13 | 1 | 2 |\n| 2 | 2 | 4 | 1 | 1 |\n| 4 | 2 | 12 | 3 | 1 |\n| 1 | 2 | 3 | 0 | 1 |\n| 3 | 2 | 7 | 2 | 1 |\n| 4 | 2 | 12 | 3 | 1 |\n| 1 | 1 | 1 | 1 | 0 |\n| 3 | 1 | 7 | 2 | 1 |\n| 5 | 3 | 19 | 4 | 1 |']",,True,,, 2295,Geometry,,"In the diagram, eleven circles of four different radius 1, each circle labelled $X$ has radius 2, the circle labelled $Y$ has radius 4 , and the circle labelled $Z$ has radius $r$. Each of the circles labelled $W$ or $X$ is tangent to three other circles. The circle labelled $Y$ is tangent to all ten of the other circles. The circle labelled $Z$ is tangent to three other circles. Determine positive integers $s$ and $t$ for which $r=\frac{s}{t}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_dff3b889a8199883de5cg-1.jpg?height=346&width=536&top_left_y=857&top_left_x=1250)","['We label the centres of the outer circles, starting with the circle labelled $Z$ and proceeding clockwise, as $A, B, C, D, E, F, G, H, J$, and $K$, and the centre of the circle labelled $Y$ as $L$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_c26842f50f756ae98092g-1.jpg?height=613&width=954&top_left_y=1103&top_left_x=691)\n\nJoin $L$ to each of $A, B, C, D, E, F, G, H, J$, and $K$. Join $A$ to $B, B$ to $C, C$ to $D, D$ to $E, E$ to $F, F$ to $G, G$ to $H, H$ to $J, J$ to $K$, and $K$ to $A$.\n\nWhen two circles are tangent, the distance between their centres equals the sum of their radii.\n\nThus,\n\n$$\n\\begin{array}{r}\nB C=C D=D E=E F=F G=G H=H J=J K=2+1=3 \\\\\nB L=D L=F L=H L=K L=2+4=6 \\\\\nC L=E L=G L=J L=1+4=5 \\\\\nA B=A K=r+2 \\\\\nA L=r+4\n\\end{array}\n$$\n\nBy side-side-side congruence, the following triangles are congruent:\n\n$$\n\\triangle B L C, \\triangle D L C, \\triangle D L E, \\triangle F L E, \\triangle F L G, \\triangle H L G, \\triangle H L J, \\triangle K L J\n$$\n\nSimilarly, $\\triangle A L B$ and $\\triangle A L K$ are congruent by side-side-side.\n\nLet $\\angle A L B=\\theta$ and let $\\angle B L C=\\alpha$.\n\n\n\nBy congruent triangles, $\\angle A L K=\\theta$ and\n\n$$\n\\angle B L C=\\angle D L C=\\angle D L E=\\angle F L E=\\angle F L G=\\angle H L G=\\angle H L J=\\angle K L J=\\alpha\n$$\n\nThe angles around $L$ add to $360^{\\circ}$ and so $2 \\theta+8 \\alpha=360^{\\circ}$ which gives $\\theta+4 \\alpha=180^{\\circ}$ and so $\\theta=180^{\\circ}-4 \\alpha$.\n\nSince $\\theta=180^{\\circ}-4 \\alpha$, then $\\cos \\theta=\\cos \\left(180^{\\circ}-4 \\alpha\\right)=-\\cos 4 \\alpha$.\n\nConsider $\\triangle A L B$ and $\\triangle B L C$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b77e2de70eb587b4b2bfg-1.jpg?height=288&width=464&top_left_y=577&top_left_x=928)\n\nBy the cosine law in $\\triangle A L B$,\n\n$$\n\\begin{aligned}\nA B^{2} & =A L^{2}+B L^{2}-2 \\cdot A L \\cdot B L \\cdot \\cos \\theta \\\\\n(r+2)^{2} & =(r+4)^{2}+6^{2}-2(r+4)(6) \\cos \\theta \\\\\n12(r+4) \\cos \\theta & =r^{2}+8 r+16+36-r^{2}-4 r-4 \\\\\n\\cos \\theta & =\\frac{4 r+48}{12(r+4)} \\\\\n\\cos \\theta & =\\frac{r+12}{3 r+12}\n\\end{aligned}\n$$\n\nBy the cosine law in $\\triangle B L C$,\n\n$$\n\\begin{aligned}\nB C^{2} & =B L^{2}+C L^{2}-2 \\cdot B L \\cdot C L \\cdot \\cos \\alpha \\\\\n3^{2} & =6^{2}+5^{2}-2(6)(5) \\cos \\alpha \\\\\n60 \\cos \\alpha & =36+25-9 \\\\\n\\cos \\alpha & =\\frac{52}{60} \\\\\n\\cos \\alpha & =\\frac{13}{15}\n\\end{aligned}\n$$\n\nSince $\\cos \\alpha=\\frac{13}{15}$, then\n\n$$\n\\begin{aligned}\n\\cos 2 \\alpha & =2 \\cos ^{2} \\alpha-1 \\\\\n& =2 \\cdot \\frac{169}{225}-1 \\\\\n& =\\frac{338}{225}-\\frac{225}{225} \\\\\n& =\\frac{113}{225}\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\n\\cos 4 \\alpha & =2 \\cos ^{2} 2 \\alpha-1 \\\\\n& =2 \\cdot \\frac{113^{2}}{225^{2}}-1 \\\\\n& =\\frac{25538}{50625}-\\frac{50625}{50625} \\\\\n& =-\\frac{25087}{50625}\n\\end{aligned}\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n\\cos \\theta & =-\\cos 4 \\alpha \\\\\n\\frac{r+12}{3 r+12} & =\\frac{25087}{50625} \\\\\n\\frac{r+12}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{(r+4)+8}{r+4} & =\\frac{25087}{16875} \\\\\n1+\\frac{8}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{8}{r+4} & =\\frac{8212}{16875} \\\\\n\\frac{2}{r+4} & =\\frac{2053}{16875} \\\\\n\\frac{r+4}{2} & =\\frac{16875}{2053} \\\\\nr+4 & =\\frac{33750}{2053} \\\\\nr & =\\frac{25538}{2053}\n\\end{aligned}\n$$\n\nTherefore, the positive integers $s=25538$ and $t=2053$ satisfy the required conditions.']","['$25538$,$2053$']",True,,Numerical, 2295,Geometry,,"In the diagram, eleven circles of four different radius 1, each circle labelled $X$ has radius 2, the circle labelled $Y$ has radius 4 , and the circle labelled $Z$ has radius $r$. Each of the circles labelled $W$ or $X$ is tangent to three other circles. The circle labelled $Y$ is tangent to all ten of the other circles. The circle labelled $Z$ is tangent to three other circles. Determine positive integers $s$ and $t$ for which $r=\frac{s}{t}$. ","['We label the centres of the outer circles, starting with the circle labelled $Z$ and proceeding clockwise, as $A, B, C, D, E, F, G, H, J$, and $K$, and the centre of the circle labelled $Y$ as $L$.\n\n\n\nJoin $L$ to each of $A, B, C, D, E, F, G, H, J$, and $K$. Join $A$ to $B, B$ to $C, C$ to $D, D$ to $E, E$ to $F, F$ to $G, G$ to $H, H$ to $J, J$ to $K$, and $K$ to $A$.\n\nWhen two circles are tangent, the distance between their centres equals the sum of their radii.\n\nThus,\n\n$$\n\\begin{array}{r}\nB C=C D=D E=E F=F G=G H=H J=J K=2+1=3 \\\\\nB L=D L=F L=H L=K L=2+4=6 \\\\\nC L=E L=G L=J L=1+4=5 \\\\\nA B=A K=r+2 \\\\\nA L=r+4\n\\end{array}\n$$\n\nBy side-side-side congruence, the following triangles are congruent:\n\n$$\n\\triangle B L C, \\triangle D L C, \\triangle D L E, \\triangle F L E, \\triangle F L G, \\triangle H L G, \\triangle H L J, \\triangle K L J\n$$\n\nSimilarly, $\\triangle A L B$ and $\\triangle A L K$ are congruent by side-side-side.\n\nLet $\\angle A L B=\\theta$ and let $\\angle B L C=\\alpha$.\n\n\n\nBy congruent triangles, $\\angle A L K=\\theta$ and\n\n$$\n\\angle B L C=\\angle D L C=\\angle D L E=\\angle F L E=\\angle F L G=\\angle H L G=\\angle H L J=\\angle K L J=\\alpha\n$$\n\nThe angles around $L$ add to $360^{\\circ}$ and so $2 \\theta+8 \\alpha=360^{\\circ}$ which gives $\\theta+4 \\alpha=180^{\\circ}$ and so $\\theta=180^{\\circ}-4 \\alpha$.\n\nSince $\\theta=180^{\\circ}-4 \\alpha$, then $\\cos \\theta=\\cos \\left(180^{\\circ}-4 \\alpha\\right)=-\\cos 4 \\alpha$.\n\nConsider $\\triangle A L B$ and $\\triangle B L C$.\n\n\n\nBy the cosine law in $\\triangle A L B$,\n\n$$\n\\begin{aligned}\nA B^{2} & =A L^{2}+B L^{2}-2 \\cdot A L \\cdot B L \\cdot \\cos \\theta \\\\\n(r+2)^{2} & =(r+4)^{2}+6^{2}-2(r+4)(6) \\cos \\theta \\\\\n12(r+4) \\cos \\theta & =r^{2}+8 r+16+36-r^{2}-4 r-4 \\\\\n\\cos \\theta & =\\frac{4 r+48}{12(r+4)} \\\\\n\\cos \\theta & =\\frac{r+12}{3 r+12}\n\\end{aligned}\n$$\n\nBy the cosine law in $\\triangle B L C$,\n\n$$\n\\begin{aligned}\nB C^{2} & =B L^{2}+C L^{2}-2 \\cdot B L \\cdot C L \\cdot \\cos \\alpha \\\\\n3^{2} & =6^{2}+5^{2}-2(6)(5) \\cos \\alpha \\\\\n60 \\cos \\alpha & =36+25-9 \\\\\n\\cos \\alpha & =\\frac{52}{60} \\\\\n\\cos \\alpha & =\\frac{13}{15}\n\\end{aligned}\n$$\n\nSince $\\cos \\alpha=\\frac{13}{15}$, then\n\n$$\n\\begin{aligned}\n\\cos 2 \\alpha & =2 \\cos ^{2} \\alpha-1 \\\\\n& =2 \\cdot \\frac{169}{225}-1 \\\\\n& =\\frac{338}{225}-\\frac{225}{225} \\\\\n& =\\frac{113}{225}\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\n\\cos 4 \\alpha & =2 \\cos ^{2} 2 \\alpha-1 \\\\\n& =2 \\cdot \\frac{113^{2}}{225^{2}}-1 \\\\\n& =\\frac{25538}{50625}-\\frac{50625}{50625} \\\\\n& =-\\frac{25087}{50625}\n\\end{aligned}\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n\\cos \\theta & =-\\cos 4 \\alpha \\\\\n\\frac{r+12}{3 r+12} & =\\frac{25087}{50625} \\\\\n\\frac{r+12}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{(r+4)+8}{r+4} & =\\frac{25087}{16875} \\\\\n1+\\frac{8}{r+4} & =\\frac{25087}{16875} \\\\\n\\frac{8}{r+4} & =\\frac{8212}{16875} \\\\\n\\frac{2}{r+4} & =\\frac{2053}{16875} \\\\\n\\frac{r+4}{2} & =\\frac{16875}{2053} \\\\\nr+4 & =\\frac{33750}{2053} \\\\\nr & =\\frac{25538}{2053}\n\\end{aligned}\n$$\n\nTherefore, the positive integers $s=25538$ and $t=2053$ satisfy the required conditions.']","['$25538$,$2053$']",True,,Numerical, 2296,Number Theory,,"Suppose that $c$ is a positive integer. Define $f(c)$ to be the number of pairs $(a, b)$ of positive integers with $cc$ and $b>c$, the integers $a-c$ and $b-c$ are positive and form a positive divisor pair of the integer $2 c^{2}$.\n\nSince $a0, \\sqrt{2 c^{2}}=\\sqrt{2} c$ which is not an integer since $c$ is an integer, which means that $2 c^{2}$ is not a perfect square.\n\nTherefore, every pair $(a, b)$ corresponds to a positive divisor pair of $2 c^{2}$ (namely, $a-c$ and $b-c)$.\n\nSimilarly, every divisor pair $e$ and $g$ of $2 c^{2}$ with $e>g$ gives a pair of positive integers $(a, b)$ with $a","['We are told that when $a, b$ and $c$ are the numbers in consecutive sectors, then $b=a c$. This means that if $a$ and $b$ are the numbers in consecutive sectors, then the number in the next sector is $c=\\frac{b}{a}$. (That is, each number is equal to the previous number divided by the one before that.)\n\nStarting with the given 2 and 3 and proceeding clockwise, we obtain\n\n$$\n2,3, \\quad \\frac{3}{2}, \\frac{3 / 2}{3}=\\frac{1}{2}, \\frac{1 / 2}{3 / 2}=\\frac{1}{3}, \\frac{1 / 3}{1 / 2}=\\frac{2}{3}, \\frac{2 / 3}{1 / 3}=2, \\frac{2}{2 / 3}=3, \\quad \\frac{3}{2}, \\ldots\n$$\n\nAfter the first 6 terms, the first 2 terms ( 2 and 3) reappear, and so the first 6 terms will repeat again. (This is because each term comes from the previous two terms, so when two consecutive terms reappear, then the following terms are the same as when these two consecutive terms appeared earlier.)\n\nSince there are 36 terms in total, then the 6 terms repeat exactly $\\frac{36}{6}=6$ times.\n\nTherefore, the sum of the 36 numbers is $6\\left(2+3+\\frac{3}{2}+\\frac{1}{2}+\\frac{1}{3}+\\frac{2}{3}\\right)=6(2+3+2+1)=48$.']",['48'],False,,Numerical, 2298,Algebra,,Determine all values of $x$ for which $0<\frac{x^{2}-11}{x+1}<7$.,"['We consider two cases: $x>-1$ (that is, $x+1>0$ ) and $x<-1$ (that is, $x+1<0$ ). Note that $x \\neq-1$.\n\nCase 1: $x>-1$\n\nWe take the given inequality $0<\\frac{x^{2}-11}{x+1}<7$ and multiply through by $x+1$, which is positive, to obtain $00$ and $x^{2}-11<7 x+7$.\n\nFrom the first, we obtain $x^{2}>11$ and so $x>\\sqrt{11}$ or $x<-\\sqrt{11}$.\n\nSince $x>-1$, then $x>\\sqrt{11}$. (Note that $-\\sqrt{11}<-1$.)\n\nFrom the second, we obtain $x^{2}-7 x-18<0$ or $(x-9)(x+2)<0$. Thus, $-2-1$ and $-2\\sqrt{11}$ and $-1x^{2}-11>7 x+7$.\n\nThus, $x^{2}-11<0$ and $x^{2}-11>7 x+7$.\n\nFrom the first, we obtain $x^{2}<11$ and so $-\\sqrt{11}0$ or $(x-9)(x+2)>0$. Thus, $x<-2$ or $x>9$. (Since $y=x^{2}-7 x-18$ represents a parabola opening upwards, its $y$-values are positive outside its $x$-intercepts.)\n\nSince $x<-1$, we obtain $x<-2$.\n\nSince $-\\sqrt{11}","['Join $B E$.\n\n\n\nSince $\\triangle F B D$ is congruent to $\\triangle A E C$, then $F B=A E$.\n\nSince $\\triangle F A B$ and $\\triangle A F E$ are each right-angled, share a common side $A F$ and have equal hypotenuses $(F B=A E)$, then these triangles are congruent, and so $A B=F E$.\n\nNow $B A F E$ has two right angles at $A$ and $F$ (so $A B$ and $F E$ are parallel) and has equal sides $A B=F E$ so must be a rectangle.\n\nThis means that $B C D E$ is also a rectangle.\n\nNow the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal $A E$ of the rectangle splits the rectangle into two congruent triangles, which have equal area. The diagonals bisect each other, so the four smaller triangles all have equal area.) Since $\\frac{1}{4}$ of rectangle $A B E F$ is shaded and $\\frac{1}{4}$ of rectangle $B C D E$ is shaded, then $\\frac{1}{4}$ of the total area is shaded. (If the area of $A B E F$ is $x$ and the area of $B C D E$ is $y$, then the total shaded area is $\\frac{1}{4} x+\\frac{1}{4} y$, which is $\\frac{1}{4}$ of the total area $x+y$.)\n\nSince $A C=200$ and $C D=50$, then the area of rectangle $A C D F$ is $200(50)=10000$, so the total shaded area is $\\frac{1}{4}(10000)=2500$.']",['2500'],False,,Numerical, 2300,Algebra,,"The numbers $a_{1}, a_{2}, a_{3}, \ldots$ form an arithmetic sequence with $a_{1} \neq a_{2}$. The three numbers $a_{1}, a_{2}, a_{6}$ form a geometric sequence in that order. Determine all possible positive integers $k$ for which the three numbers $a_{1}, a_{4}, a_{k}$ also form a geometric sequence in that order. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9 are the first four terms of an arithmetic sequence. A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant. For example, $3,6,12$ is a geometric sequence with three terms.)","['Suppose that the arithmetic sequence $a_{1}, a_{2}, a_{3}, \\ldots$ has first term $a$ and common difference $d$.\n\nThen, for each positive integer $n, a_{n}=a+(n-1) d$.\n\nSince $a_{1}=a$ and $a_{2}=a+d$ and $a_{1} \\neq a_{2}$, then $d \\neq 0$.\n\nSince $a_{1}, a_{2}, a_{6}$ form a geometric sequence in that order, then $\\frac{a_{2}}{a_{1}}=\\frac{a_{6}}{a_{2}}$ or $\\left(a_{2}\\right)^{2}=a_{1} a_{6}$.\n\nSubstituting, we obtain\n\n$$\n\\begin{aligned}\n(a+d)^{2} & =a(a+5 d) \\\\\na^{2}+2 a d+d^{2} & =a^{2}+5 a d \\\\\nd^{2} & =3 a d \\\\\nd & =3 a \\quad(\\text { since } d \\neq 0)\n\\end{aligned}\n$$\n\nTherefore, $a_{n}=a+(n-1) d=a+(n-1)(3 a)=(3 n-2) a$ for each $n \\geq 1$.\n\nThus, $a_{4}=(3(4)-2) a=10 a$, and $a_{k}=(3 k-2) a$. (Note that $a_{1}=(3(1)-2) a=a$.)\n\nFor $a_{1}, a_{4}, a_{k}$ to also form a geometric sequence then, as above, $\\left(a_{4}\\right)^{2}=a_{1} a_{k}$, and so\n\n$$\n\\begin{aligned}\n(10 a)^{2} & =(a)((3 k-2) a) \\\\\n100 a^{2} & =(3 k-2) a^{2}\n\\end{aligned}\n$$\n\nSince $d \\neq 0$ and $d=3 a$, then $a \\neq 0$.\n\nSince $100 a^{2}=(3 k-2) a^{2}$ and $a \\neq 0$, then $100=3 k-2$ and so $3 k=102$ or $k=34$.\n\nChecking, we note that $a_{1}=a, a_{4}=10 a$ and $a_{34}=100 a$ which form a geometric sequence with common ratio 10 .\n\nTherefore, the only possible value of $k$ is $k=34$.']",['34'],False,,Numerical, 2301,Number Theory,,"For some positive integers $k$, the parabola with equation $y=\frac{x^{2}}{k}-5$ intersects the circle with equation $x^{2}+y^{2}=25$ at exactly three distinct points $A, B$ and $C$. Determine all such positive integers $k$ for which the area of $\triangle A B C$ is an integer.","['First, we note that since $k$ is a positive integer, then $k \\geq 1$.\n\nNext, we note that the given parabola passes through the point $(0,-5)$ as does the given circle. (This is because if $x=0$, then $y=\\frac{0^{2}}{k}-5=-5$ and if $(x, y)=(0,-5)$, then $x^{2}+y^{2}=0^{2}+(-5)^{2}=25$, so $(0,-5)$ satisfies each of the equations.)\n\nTherefore, for every positive integer $k$, the two graphs intersect in at least one point.\n\nIf $y=-5$, then $x^{2}+(-5)^{2}=25$ and so $x^{2}=0$ or $x=0$. In other words, there is one point on both parabola and circle with $y=-5$, namely $(0,-5)$.\n\nNow, the given circle with equation $x^{2}+y^{2}=25=5^{2}$ has centre $(0,0)$ and radius 5 .\n\nThis means that the $y$-coordinates of points on this circle satisfy $-5 \\leq y \\leq 5$.\n\nTo find the other points of intersection, we re-write $y=\\frac{x^{2}}{k}-5$ as $k y=x^{2}-5 k$ or $x^{2}=k y+5 k$ and substitute into $x^{2}+y^{2}=25$ to obtain\n\n$$\n\\begin{aligned}\n(k y+5 k)+y^{2} & =25 \\\\\ny^{2}+k y+(5 k-25) & =0 \\\\\n(y+5)(y+(k-5)) & =0\n\\end{aligned}\n$$\n\nand so $y=-5$ or $y=5-k$.\n\n(We note that since the two graphs intersect at $y=-5$, then $(y+5)$ was going to be a factor of the quadratic equation $y^{2}+k y+(5 k-25)=0$. If we had not seen this, we could have used the quadratic formula.)\n\nTherefore, for $y=5-k$ to give points on the circle, we need $-5 \\leq 5-k$ and $5-k \\leq 5$.\n\nThis gives $k \\leq 10$ and $k \\geq 0$.\n\nSince $k$ is a positive integer, the possible values of $k$ to this point are $k=1,2,3,4,5,6,7,8,9,10$.\n\nIf $k=1$, then $y=5-1=4$. In this case, $x^{2}+4^{2}=25$ or $x^{2}=9$ and so $x= \\pm 3$.\n\nThis gives the two points $(3,4)$ and $(-3,4)$ which lie on the parabola and circle.\n\nConsider the three points $A(3,4), B(-3,4)$ and $C(0,-5)$.\n\nNow $A B$ is horizontal with $A B=3-(-3)=6$. (This is the difference in $x$-coordinates.) The vertical distance from $A B$ to $C$ is $4-(-5)=9$. (This is the difference in $y$ coordinates.)\n\nTherefore, the area of $\\triangle A B C$ is $\\frac{1}{2}(6)(9)=27$, which is a positive integer.\n\nWe now repeat these calculations for each of the other values of $k$ by making a table:\n\n| $k$ | $y$ | $x= \\pm \\sqrt{25-y^{2}}$ | Base | Height | Area of triangle |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | 4 | $\\pm 3$ | $3-(-3)=6$ | $4-(-5)=9$ | 27 |\n| 2 | 3 | $\\pm 4$ | $4-(-4)=8$ | $3-(-5)=8$ | 32 |\n| 3 | 2 | $\\pm \\sqrt{21}$ | $2 \\sqrt{21}$ | 7 | $7 \\sqrt{21}$ |\n| 4 | 1 | $\\pm \\sqrt{24}$ | $2 \\sqrt{24}$ | 6 | $6 \\sqrt{24}$ |\n| 5 | 0 | $\\pm 5$ | 10 | 5 | 25 |\n| 6 | -1 | $\\pm \\sqrt{24}$ | $2 \\sqrt{24}$ | 4 | $4 \\sqrt{24}$ |\n| 7 | -2 | $\\pm \\sqrt{21}$ | $2 \\sqrt{21}$ | 3 | $3 \\sqrt{21}$ |\n| 8 | -3 | $\\pm 4$ | 8 | 2 | 8 |\n| 9 | -4 | $\\pm 3$ | 6 | 1 | 3 |\n| 10 | -5 | 0 | | | |\n\nWhen $k=10$, we have $y=5-k=-5$ and $x=0$ only, so there is only one point of intersection.\n\nFinally, the values of $k$ for which there are three points of intersection and for which the area of the resulting triangle is a positive integer are $k=1,2,5,8,9$.']","['1,2,5,8,9']",True,,Numerical, 2302,Geometry,,"In the diagram, $\triangle X Y Z$ is isosceles with $X Y=X Z=a$ and $Y Z=b$ where $b<2 a$. A larger circle of radius $R$ is inscribed in the triangle (that is, the circle is drawn so that it touches all three sides of the triangle). A smaller circle of radius $r$ is drawn so that it touches $X Y, X Z$ and the larger circle. Determine an expression for $\frac{R}{r}$ in terms of $a$ and $b$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ff40fe20a2d5d1ed9823g-1.jpg?height=474&width=333&top_left_y=1991&top_left_x=1254)","['Suppose that $M$ is the midpoint of $Y Z$.\n\nSuppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $X Y$ at $C$ and $X Z$ at $D$, and that the larger circle touches $X Y$ at $E$ and $X Z$ at $F$.\n\nJoin $O C, O D$ and $P E$.\n\nSince $O C$ and $P E$ are radii that join the centres of circles to points of tangency, then $O C$ and $P E$ are perpendicular to $X Y$.\n\nJoin $X M$. Since $\\triangle X Y Z$ is isosceles, then $X M$ (which is a median by construction) is an altitude (that is, $X M$ is perpendicular to $Y Z)$ and an angle bisector (that is, $\\angle M X Y=\\angle M X Z$ ).\n\nNow $X M$ passes through $O$ and $P$. (Since $X C$ and $X D$ are tangents from $X$ to the same circle, then $X C=X D$. This means that $\\triangle X C O$ is congruent to $\\triangle X D O$ by side-side-side. This means that $\\angle O X C=\\angle O X D$ and so $O$ lies on the angle bisector of $\\angle C X D$, and so $O$ lies on $X M$. Using a similar argument, $P$ lies on $X M$.)\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_04e57b63898bbe86db6ag-1.jpg?height=474&width=333&top_left_y=454&top_left_x=1579)\nDraw a perpendicular from $O$ to $T$ on $P E$. Note that $O T$ is parallel to $X Y$ (since each is perpendicular to $P E$ ) and that $O C E T$ is a rectangle (since it has three right angles).\n\nConsider $\\triangle X M Y$ and $\\triangle O T P$.\n\nEach triangle is right-angled (at $M$ and at $T$ ).\n\nAlso, $\\angle Y X M=\\angle P O T$. (This is because $O T$ is parallel to $X Y$, since both are perpendicular to $P E$.)\n\nTherefore, $\\triangle X M Y$ is similar to $\\triangle O T P$.\n\nThus, $\\frac{X Y}{Y M}=\\frac{O P}{P T}$.\n\nNow $X Y=a$ and $Y M=\\frac{1}{2} b$.\n\nAlso, $O P$ is the line segment joining the centres of two tangent circles, so $O P=r+R$.\n\nLastly, $P T=P E-E T=R-r$, since $P E=R, E T=O C=r$, and $O C E T$ is a rectangle. Therefore,\n\n$$\n\\begin{aligned}\n\\frac{a}{b / 2} & =\\frac{R+r}{R-r} \\\\\n\\frac{2 a}{b} & =\\frac{R+r}{R-r} \\\\\n2 a(R-r) & =b(R+r) \\\\\n2 a R-b R & =2 a r+b r \\\\\nR(2 a-b) & =r(2 a+b) \\\\\n\\frac{R}{r} & =\\frac{2 a+b}{2 a-b} \\quad(\\text { since } 2 a>b \\text { so } 2 a-b \\neq 0, \\text { and } r>0)\n\\end{aligned}\n$$\n\nTherefore, $\\frac{R}{r}=\\frac{2 a+b}{2 a-b}$.']",['$\\frac{2 a+b}{2 a-b}$'],False,,Expression, 2302,Geometry,,"In the diagram, $\triangle X Y Z$ is isosceles with $X Y=X Z=a$ and $Y Z=b$ where $b<2 a$. A larger circle of radius $R$ is inscribed in the triangle (that is, the circle is drawn so that it touches all three sides of the triangle). A smaller circle of radius $r$ is drawn so that it touches $X Y, X Z$ and the larger circle. Determine an expression for $\frac{R}{r}$ in terms of $a$ and $b$. ","['Suppose that $M$ is the midpoint of $Y Z$.\n\nSuppose that the centre of the smaller circle is $O$ and the centre of the larger circle is $P$. Suppose that the smaller circle touches $X Y$ at $C$ and $X Z$ at $D$, and that the larger circle touches $X Y$ at $E$ and $X Z$ at $F$.\n\nJoin $O C, O D$ and $P E$.\n\nSince $O C$ and $P E$ are radii that join the centres of circles to points of tangency, then $O C$ and $P E$ are perpendicular to $X Y$.\n\nJoin $X M$. Since $\\triangle X Y Z$ is isosceles, then $X M$ (which is a median by construction) is an altitude (that is, $X M$ is perpendicular to $Y Z)$ and an angle bisector (that is, $\\angle M X Y=\\angle M X Z$ ).\n\nNow $X M$ passes through $O$ and $P$. (Since $X C$ and $X D$ are tangents from $X$ to the same circle, then $X C=X D$. This means that $\\triangle X C O$ is congruent to $\\triangle X D O$ by side-side-side. This means that $\\angle O X C=\\angle O X D$ and so $O$ lies on the angle bisector of $\\angle C X D$, and so $O$ lies on $X M$. Using a similar argument, $P$ lies on $X M$.)\n\n\nDraw a perpendicular from $O$ to $T$ on $P E$. Note that $O T$ is parallel to $X Y$ (since each is perpendicular to $P E$ ) and that $O C E T$ is a rectangle (since it has three right angles).\n\nConsider $\\triangle X M Y$ and $\\triangle O T P$.\n\nEach triangle is right-angled (at $M$ and at $T$ ).\n\nAlso, $\\angle Y X M=\\angle P O T$. (This is because $O T$ is parallel to $X Y$, since both are perpendicular to $P E$.)\n\nTherefore, $\\triangle X M Y$ is similar to $\\triangle O T P$.\n\nThus, $\\frac{X Y}{Y M}=\\frac{O P}{P T}$.\n\nNow $X Y=a$ and $Y M=\\frac{1}{2} b$.\n\nAlso, $O P$ is the line segment joining the centres of two tangent circles, so $O P=r+R$.\n\nLastly, $P T=P E-E T=R-r$, since $P E=R, E T=O C=r$, and $O C E T$ is a rectangle. Therefore,\n\n$$\n\\begin{aligned}\n\\frac{a}{b / 2} & =\\frac{R+r}{R-r} \\\\\n\\frac{2 a}{b} & =\\frac{R+r}{R-r} \\\\\n2 a(R-r) & =b(R+r) \\\\\n2 a R-b R & =2 a r+b r \\\\\nR(2 a-b) & =r(2 a+b) \\\\\n\\frac{R}{r} & =\\frac{2 a+b}{2 a-b} \\quad(\\text { since } 2 a>b \\text { so } 2 a-b \\neq 0, \\text { and } r>0)\n\\end{aligned}\n$$\n\nTherefore, $\\frac{R}{r}=\\frac{2 a+b}{2 a-b}$.']",['$\\frac{2 a+b}{2 a-b}$'],False,,Expression, 2303,Algebra,,"Consider the following system of equations in which all logarithms have base 10: $$ \begin{aligned} (\log x)(\log y)-3 \log 5 y-\log 8 x & =a \\ (\log y)(\log z)-4 \log 5 y-\log 16 z & =b \\ (\log z)(\log x)-4 \log 8 x-3 \log 625 z & =c \end{aligned} $$ If $a=-4, b=4$, and $c=-18$, solve the system of equations.","['Using $\\log$ arithm rules $\\log (u v)=\\log u+\\log v$ and $\\log \\left(s^{t}\\right)=t \\log s$ for all $u, v, s>0$, the first equation becomes\n\n$$\n\\begin{aligned}\n(\\log x)(\\log y)-3 \\log 5-3 \\log y-\\log 8-\\log x & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-\\log 8-\\log 5^{3} & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-\\log (8 \\cdot 125) & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-\\log (1000) & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-3 & =a\n\\end{aligned}\n$$\n\nSimilarly, the second equation becomes\n\n$$\n\\begin{aligned}\n(\\log y)(\\log z)-4 \\log 5-4 \\log y-\\log 16-\\log z & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-4 \\log 5-\\log 16 & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-\\log \\left(5^{4} \\cdot 16\\right) & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-\\log (10000) & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-4 & =b\n\\end{aligned}\n$$\n\nAnd the third equation becomes\n\n$$\n\\begin{aligned}\n(\\log z)(\\log x)-4 \\log 8-4 \\log x-3 \\log 625-3 \\log z & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-4 \\log 8-3 \\log 625 & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-\\log \\left(8^{4} \\cdot 625^{3}\\right) & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-\\log \\left(2^{12} \\cdot 5^{12}\\right) & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-12 & =c\n\\end{aligned}\n$$\n\nSince each of the steps that we have made are reversible, the original system of equations is equivalent to the new system of equations\n\n$$\n\\begin{aligned}\n(\\log x)(\\log y)-\\log x-3 \\log y-3 & =a \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-4 & =b \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-12 & =c\n\\end{aligned}\n$$\n\nNext, we make the substitution $X=\\log x, Y=\\log y$ and $Z=\\log z$. (This is equivalent to saying $x=10^{X}, y=10^{Y}$ and $z=10^{Z}$.)\n\nThis transforms the system of equations to the equivalent system\n\n$$\n\\begin{aligned}\nX Y-X-3 Y-3 & =a \\\\\nY Z-4 Y-Z-4 & =b \\\\\nX Z-4 X-3 Z-12 & =c\n\\end{aligned}\n$$\n\nWe re-write the first of these three equations as $X(Y-1)-3 Y-3=a$ and then as $X(Y-1)-3(Y-1)-6=a$ and then as $(X-3)(Y-1)=a+6$.\n\nIn a similar way, we re-write the second and third of these equations to obtain the equivalent system\n\n$$\n\\begin{aligned}\n(X-3)(Y-1) & =a+6 \\\\\n(Y-1)(Z-4) & =b+8 \\\\\n(X-3)(Z-4) & =c+24\n\\end{aligned}\n$$\n\n\n\nNext, we make the substitution $p=X-3, q=Y-1$ and $r=Z-4$. (This is equivalent to saying $X=p+3, Y=q+1$ and $Z=r+4$, or $x=10^{p+3}, y=10^{q+1}$ and $z=10^{r+4}$.)\n\nThis transforms the original system of equations into the equivalent system\n\n$$\n\\begin{aligned}\np q & =a+6 \\\\\nq r & =b+8 \\\\\np r & =c+24\n\\end{aligned}\n$$\n\nWe again note that this system of equations is equivalent to the initial system of equations, and each solution of this system corresponds with a solution of the initial system.\n\nSuppose that $a=-4, b=4$ and $c=-18$.\n\nThen the last version of the system is\n\n$$\n\\begin{aligned}\np q & =2 \\\\\nq r & =12 \\\\\np r & =6\n\\end{aligned}\n$$\n\nMultiplying the three equations together gives $p^{2} q^{2} r^{2}=2 \\cdot 12 \\cdot 6=144$.\n\nSince $(p q r)^{2}=144$, then $p q r= \\pm 12$.\n\nTherefore, $r=\\frac{p q r}{p q}=\\frac{ \\pm 12}{2}= \\pm 6$ and $p=\\frac{p q r}{q r}=\\frac{ \\pm 12}{12}= \\pm 1$ and $q=\\frac{p q r}{p r}=\\frac{ \\pm 12}{6}= \\pm 2$.\n\nTherefore, the solutions to the last version of the system are $(p, q, r)=(1,2,6)$ and $(p, q, r)=(-1,-2,-6)$.\n\nConverting back to the original variables, we see that the solutions to the original system when $(a, b, c)=(-4,4,-18)$ are $(x, y, z)=\\left(10^{4}, 10^{3}, 10^{10}\\right)$ and $(x, y, z)=\\left(10^{2}, 10^{-1}, 10^{-2}\\right)$.\n\n']","['$(10^{4}, 10^{3}, 10^{10}),(10^{2}, 10^{-1}, 10^{-2})$']",True,,Tuple, 2304,Algebra,,"Consider the following system of equations in which all logarithms have base 10: $$ \begin{aligned} (\log x)(\log y)-3 \log 5 y-\log 8 x & =a \\ (\log y)(\log z)-4 \log 5 y-\log 16 z & =b \\ (\log z)(\log x)-4 \log 8 x-3 \log 625 z & =c \end{aligned} $$ Determine all triples $(a, b, c)$ of real numbers for which the system of equations has an infinite number of solutions $(x, y, z)$.","['Using $\\log$ arithm rules $\\log (u v)=\\log u+\\log v$ and $\\log \\left(s^{t}\\right)=t \\log s$ for all $u, v, s>0$, the first equation becomes\n\n$$\n\\begin{aligned}\n(\\log x)(\\log y)-3 \\log 5-3 \\log y-\\log 8-\\log x & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-\\log 8-\\log 5^{3} & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-\\log (8 \\cdot 125) & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-\\log (1000) & =a \\\\\n(\\log x)(\\log y)-\\log x-3 \\log y-3 & =a\n\\end{aligned}\n$$\n\nSimilarly, the second equation becomes\n\n$$\n\\begin{aligned}\n(\\log y)(\\log z)-4 \\log 5-4 \\log y-\\log 16-\\log z & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-4 \\log 5-\\log 16 & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-\\log \\left(5^{4} \\cdot 16\\right) & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-\\log (10000) & =b \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-4 & =b\n\\end{aligned}\n$$\n\nAnd the third equation becomes\n\n$$\n\\begin{aligned}\n(\\log z)(\\log x)-4 \\log 8-4 \\log x-3 \\log 625-3 \\log z & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-4 \\log 8-3 \\log 625 & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-\\log \\left(8^{4} \\cdot 625^{3}\\right) & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-\\log \\left(2^{12} \\cdot 5^{12}\\right) & =c \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-12 & =c\n\\end{aligned}\n$$\n\nSince each of the steps that we have made are reversible, the original system of equations is equivalent to the new system of equations\n\n$$\n\\begin{aligned}\n(\\log x)(\\log y)-\\log x-3 \\log y-3 & =a \\\\\n(\\log y)(\\log z)-4 \\log y-\\log z-4 & =b \\\\\n(\\log z)(\\log x)-4 \\log x-3 \\log z-12 & =c\n\\end{aligned}\n$$\n\nNext, we make the substitution $X=\\log x, Y=\\log y$ and $Z=\\log z$. (This is equivalent to saying $x=10^{X}, y=10^{Y}$ and $z=10^{Z}$.)\n\nThis transforms the system of equations to the equivalent system\n\n$$\n\\begin{aligned}\nX Y-X-3 Y-3 & =a \\\\\nY Z-4 Y-Z-4 & =b \\\\\nX Z-4 X-3 Z-12 & =c\n\\end{aligned}\n$$\n\nWe re-write the first of these three equations as $X(Y-1)-3 Y-3=a$ and then as $X(Y-1)-3(Y-1)-6=a$ and then as $(X-3)(Y-1)=a+6$.\n\nIn a similar way, we re-write the second and third of these equations to obtain the equivalent system\n\n$$\n\\begin{aligned}\n(X-3)(Y-1) & =a+6 \\\\\n(Y-1)(Z-4) & =b+8 \\\\\n(X-3)(Z-4) & =c+24\n\\end{aligned}\n$$\n\n\n\nNext, we make the substitution $p=X-3, q=Y-1$ and $r=Z-4$. (This is equivalent to saying $X=p+3, Y=q+1$ and $Z=r+4$, or $x=10^{p+3}, y=10^{q+1}$ and $z=10^{r+4}$.)\n\nThis transforms the original system of equations into the equivalent system\n\n$$\n\\begin{aligned}\np q & =a+6 \\\\\nq r & =b+8 \\\\\np r & =c+24\n\\end{aligned}\n$$\n\nWe again note that this system of equations is equivalent to the initial system of equations, and each solution of this system corresponds with a solution of the initial system.\n\nWe consider the various possibilities for the product, $(a+6)(b+8)(c+24)$, of the right sides of the equations in the final form of the system above: whether it is positive, negative or equal to 0 .\n\nCase 1: $(a+6)(b+8)(c+24)<0$\n\nAs in (a), we multiply the three equations together to obtain $(p q r)^{2}=(a+6)(b+8)(c+24)$. Since the left side is at least 0 and the right side is negative, then there are no solutions to the system of equations in this case.\n\nCase 2: $(a+6)(b+8)(c+24)>0$\n\nAs in (a), we multiply the three equations together to obtain $(p q r)^{2}=(a+6)(b+8)(c+24)$. Since $(p q r)^{2}=(a+6)(b+8)(c+24)$ and $(a+6)(b+8)(c+24)>0$, then $p q r= \\pm \\sqrt{(a+6)(b+8)(c+24)}$.\n\nSince $(a+6)(b+8)(c+24)>0$, then $\\sqrt{(a+6)(b+8)(c+24)}$ is well-defined.\n\nAlso, since $(a+6)(b+8)(c+24)>0$, then each of $a+6, b+8, c+24$ is non-zero, so we can divide by each of these quantities.\n\nAs we did in (a), we can solve to obtain\n\n$$\n\\begin{aligned}\n& p=\\frac{p q r}{q r}=\\frac{ \\pm \\sqrt{(a+6)(b+8)(c+24)}}{b+8} \\\\\n& q=\\frac{p q r}{p r}=\\frac{ \\pm \\sqrt{(a+6)(b+8)(c+24)}}{c+24} \\\\\n& r=\\frac{p q r}{p q}=\\frac{ \\pm \\sqrt{(a+6)(b+8)(c+24)}}{a+6}\n\\end{aligned}\n$$\n\n\n\nSince $(a+6)(b+8)(c+24)>0$, these are all valid fractions and there are exactly two triples $(p, q, r)$ that are solutions and so two triples $(x, y, z)$ that are solutions to the original system.\n\nCase 3: $(a+6)(b+8)(c+24)=0$\n\n$\\text { Suppose that exactly one of } a+6, b+8$ and $c+24$ equals 0 .\n\nWithout loss of generality, suppose that $a+6=0, b+8 \\neq 0$ and $c+24 \\neq 0$.\n\nSince $p q=a+6=0$, then $p=0$ or $q=0$.\n\nIn this case, either $q r=b+8$ or $p r=c+24$ will equal 0 , which contradicts our assumption that neither $b+8$ nor $c+24$ is 0 .\n\nTherefore, it cannot be the case that exactly one of $a+6, b+8$ and $c+24$ equals 0 .\n\nSuppose next that exactly two of $a+6, b+8$ and $c+24$ equal 0 .\n\nWithout loss of generality, suppose that $a+6=b+8=0$ and $c+24 \\neq 0$.\n\nSince $p r=c+24 \\neq 0$, then $p \\neq 0$ and $r \\neq 0$.\n\nSince $p q=a+6=0$ and $q r=b+8=0$ and $p \\neq 0$ and $r \\neq 0$, then $q=0$.\n\nIn this case, any triple $(p, q, r)$ with $q=0$ and $p r=c+24 \\neq 0$ is a solution to the system of equations.\n\nThus, when $a+6=b+8=0$ and $c+24 \\neq 0$ (that is, $(a, b, c)=(-6,-8, c)$ with $c \\neq 24)$, each triple $(p, q, r)=\\left(p, 0, \\frac{c+24}{p}\\right)$ with $p \\neq 0$ is a solution to the system of equations.\n\nEach of these solutions corresponds to a solution to the original system of equations in $(x, y, z)$, so if $(a, b, c)=(-6,-8, c)$ with $c \\neq 0$, then there are infinite number of solutions to the system of equations.\n\nSimilarly, if $(a, b, c)=(-6, b,-24)$ with $b \\neq-8$ (that is, if $p=a+6=0$ and $r=c+24=0$ but $q=b+8 \\neq 0)$ or $(a, b, c)=(a,-8,-24)$ with $a \\neq-6$, then there are infinitely many solutions $(x, y, z)$ to the original system of equations.\n\nFinally, we must consider the case of $a+6=b+8=c+24=0$.\n\nHere, we must solve the system of equations\n\n$$\n\\begin{aligned}\n& p q=0 \\\\\n& q r=0 \\\\\n& p r=0\n\\end{aligned}\n$$\n\nEach triple $(p, q, r)=(0,0, r)$ is a solution of this system and there are infinitely many such solutions. (This is not all of the solutions, but represents infinitely many solutions.) Therefore, when $(a, b, c)=(-6,-8,-24)$, there are also infinitely many solutions to the original system of equations.\n\nTherefore, the system of equations has an infinite number of solutions $(x, y, z)$ precisely when $(a, b, c)=(-6,-8, c)$ for some real number $c$ or $(a, b, c)=(-6, b,-24)$ for some real number $b$ or $(a, b, c)=(a,-8,-24)$ for some real number $a$ or $(a, b, c)=(-6,-8,-24)$. (This last triple is in fact included in each of the previous three families of triples.)']","['$(a, b, c)=(-6,-8, c)$ for some real number $c$ or $(a, b, c)=(-6, b,-24)$ for some real number $b$ or $(a, b, c)=(a,-8,-24)$ for some real number $a$']",False,,Need_human_evaluate, 2305,Combinatorics,,"For each positive integer $n \geq 1$, let $C_{n}$ be the set containing the $n$ smallest positive integers; that is, $C_{n}=\{1,2, \ldots, n-1, n\}$. For example, $C_{4}=\{1,2,3,4\}$. We call a set, $F$, of subsets of $C_{n}$ a Furoni family of $C_{n}$ if no element of $F$ is a subset of another element of $F$. Consider $A=\{\{1,2\},\{1,3\},\{1,4\}\}$. Note that $A$ is a Furoni family of $C_{4}$. Determine the two Furoni families of $C_{4}$ that contain all of the elements of $A$ and to which no other subsets of $C_{4}$ can be added to form a new (larger) Furoni family.","['The subsets of $C_{4}$ are:\n\n| | \\{\\} | $\\{1\\}$ | $\\{2\\}$ | $\\{3\\}$ | $\\{4\\}$ | |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\{1,2\\}$ | $\\{1,3\\}$ | $\\{1,4\\}$ | $\\{2,3\\}$ | $\\{2,4\\}$ | $\\{3,4\\}$ | |\n| $\\{1,2,3\\}$ | $\\{1,2,4\\}$ | $\\{1,3,4\\}$ | $\\{2,3,4\\}$ | $\\{1,2,3,4\\}$ | | |\n\n(There are 16 such subsets including the empty set \\{\\} and the complete set $C_{4}=\\{1,2,3,4\\}$.) Consider the Furoni family $A=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\}\\}$.\n\nEach of the following subsets of $C_{4}$ is already an element of $A:\\{1,2\\},\\{1,3\\},\\{1,4\\}$.\n\nEach of the following subsets of $C_{4}$ is a subset of one or more of the elements of $A$ : \\{\\}$,\\{1\\},\\{2\\},\\{3\\},\\{4\\}$.\n\nEach of the following subsets of $C_{4}$ has the property that one or more of the elements of $A$ is a subset of it: $\\{1,2,3\\},\\{1,2,4\\},\\{1,3,4\\},\\{1,2,3,4\\}$.\n\nSince a Furoni family of $C_{4}$ cannot contain two subsets of $C_{4}$ one of which is a subset of the other, none of the subsets in either of these two lists can be added to $A$ to form a larger Furoni family.\n\nThis leaves the following subsets of $C_{4}$ to consider as possible elements to add to $A$ : $\\{2,3\\},\\{2,4\\},\\{3,4\\},\\{2,3,4\\}$.\n\nIf $\\{2,3,4\\}$ is added to $A$ to form $A^{\\prime}=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3,4\\}\\}$, then $A^{\\prime}$ is still a Furoni family of $C_{4}$ and none of $\\{2,3\\},\\{2,4\\},\\{3,4\\}$ can be added, since each is a subset of $\\{2,3,4\\}$. Therefore, $A^{\\prime}$ is a Furoni family of $C_{4}$ to which no other subset can be added. If any of $\\{2,3\\},\\{2,4\\},\\{3,4\\}$ is added to $A$, then $\\{2,3,4\\}$ cannot be added (since each of these three two elements sets is a subset of $\\{2,3,4\\}$ ) but each of the remaining two element sets can be still added without violating the conditions for being a Furoni family.\n\nThus, $A^{\\prime \\prime}=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{3,4\\}\\}$ is a Furoni family of $C_{4}$ to which no other subset can be added.\n\nTherefore, the two Furoni families of $C_{4}$ that contain all of the elements of $A$ and to which no further subsets of $C_{4}$ can be added are\n\n$$\nA^{\\prime}=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3,4\\}\\} \\quad A^{\\prime \\prime}=\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{3,4\\}\\}\n$$']","['$\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3,4\\}\\},\\{\\{1,2\\},\\{1,3\\},\\{1,4\\},\\{2,3\\},\\{2,4\\},\\{3,4\\}\\}$']",True,,Need_human_evaluate, 2306,Combinatorics,,"For each positive integer $n \geq 1$, let $C_{n}$ be the set containing the $n$ smallest positive integers; that is, $C_{n}=\{1,2, \ldots, n-1, n\}$. For example, $C_{4}=\{1,2,3,4\}$. We call a set, $F$, of subsets of $C_{n}$ a Furoni family of $C_{n}$ if no element of $F$ is a subset of another element of $F$. Suppose that $n$ is a positive integer and that $F$ is a Furoni family of $C_{n}$. For each non-negative integer $k$, define $a_{k}$ to be the number of elements of $F$ that contain exactly $k$ integers. Prove that $$ \frac{a_{0}}{\left(\begin{array}{l} n \\ 0 \end{array}\right)}+\frac{a_{1}}{\left(\begin{array}{l} n \\ 1 \end{array}\right)}+\frac{a_{2}}{\left(\begin{array}{l} n \\ 2 \end{array}\right)}+\cdots+\frac{a_{n-1}}{\left(\begin{array}{c} n \\ n-1 \end{array}\right)}+\frac{a_{n}}{\left(\begin{array}{l} n \\ n \end{array}\right)} \leq 1 $$ (The sum on the left side includes $n+1$ terms.) (Note: If $n$ is a positive integer and $k$ is an integer with $0 \leq k \leq n$, then $\left(\begin{array}{l}n \\ k\end{array}\right)=\frac{n !}{k !(n-k) !}$ is the number of subsets of $C_{n}$ that contain exactly $k$ integers, where $0 !=1$ and, if $m$ is a positive integer, $m$ ! represents the product of the integers from 1 to $m$, inclusive.)","['Suppose that $n$ is a positive integer and $F$ is a Furoni family of $C_{n}$ that contains $a_{k}$ elements that contain exactly $k$ integers each, for each integer $k$ from 0 to $n$, inclusive.\n\nConsider each element $E$ of $F$.\n\nEach $E$ is a subset of $C_{n}$. Suppose that a particular choice for $E$ contains exactly $k$ elements.\n\nWe use $E$ to generate $k !(n-k)$ ! permutations $\\sigma$ of the integers in $C_{n}=\\{1,2,3, \\ldots, n\\}$ by starting with a permutation $\\alpha$ of the elements of $E$ and appending a permutation $\\beta$ of the elements in $C_{n}$ not in $E$.\n\nSince there are $k$ elements in $E$, there are $k$ ! possible permutations $\\alpha$.\n\nSince there are $n-k$ elements in $C_{n}$ that are not in $E$, there are $(n-k)$ ! possible permutations $\\beta$.\n\nEach possible $\\alpha$ can have each possible $\\beta$ appended to it, so there are $k !(n-k)$ ! possible permutations $\\sigma=\\alpha \\mid \\beta$. (The notation "" $\\alpha \\mid \\beta$ "" means the permutation of $C_{n}$ formed by writing out the permutation $\\alpha$ (of the elements of $E$ ) followed by writing out the permutation $\\beta$ (of the elements of $C_{n}$ not in $E$ ).)\n\nEach of these $k !(n-k)$ ! permutations generated by $E$ is indeed different, since if two permutations $\\sigma=\\alpha \\mid \\beta$ and $\\sigma^{\\prime}=\\alpha^{\\prime} \\mid \\beta^{\\prime}$ are equal, then since $\\alpha$ and $\\alpha^{\\prime}$ are both permutations of the elements of $E$, then they have the same length and so $\\alpha\\left|\\beta=\\alpha^{\\prime}\\right| \\beta^{\\prime}$ means $\\alpha=\\alpha^{\\prime}$.\n\n\n\nThis then means that $\\beta=\\beta^{\\prime}$ and so the permutations started out the same.\n\nWe repeat this process for each of the elements $E$ of $F$.\n\nSince, for each $k$, there are $a_{k}$ subsets of size $k$ in $F$, then the total number of permutations that this generates is\n\n$$\na_{0} 0 !(n-0) !+a_{1} 1 !(n-1) !+\\cdots+a_{n-1}(n-1) !(n-(n-1)) 1 !+a_{n} n !(n-n) !\n$$\n\nIf each of these permutations is different, then this total is at most $n$ !, since this is the total number of permutations of the elements of $C_{n}$.\n\nIs it possible that two elements $E$ and $G$ of $F$ generate identical permutations of the elements of $C_{n}$ in this way?\n\nSuppose that two permutations $\\sigma=\\alpha \\mid \\beta$ (generated by $E$ ) and $\\sigma^{\\prime}=\\alpha^{\\prime} \\mid \\beta^{\\prime}$ (generated by $G)$ are identical.\n\nSuppose that $E$ contains $k$ elements and $G$ contains $k^{\\prime}$ elements.\n\nEither $k \\leq k^{\\prime}$ or $k^{\\prime} \\leq k$ (or both, if they are equal).\n\nWithout loss of generality, suppose that $k \\leq k^{\\prime}$.\n\nThen the length of $\\alpha$ (which is $k$ ) is less than or equal to the length of $\\alpha^{\\prime}$ (which is $k^{\\prime}$ ). But $\\alpha\\left|\\beta=\\alpha^{\\prime}\\right| \\beta^{\\prime}$, so this means that the first $k$ entries in $\\alpha^{\\prime}$ are equal to the first $k$ entries in $\\alpha$.\n\nBut the entries in $\\alpha$ are the elements of $E$ and the entries of $\\alpha^{\\prime}$ are the elements of $G$, so this means that $E$ is a subset of $G$, which cannot be the case. This is a contradiction. Therefore, each of the permutations generated by each of the subsets of $C_{n}$ contained in $F$ is unique.\n\nTherefore,\n\n$$\na_{0} 0 !(n-0) !+a_{1} 1 !(n-1) !+\\cdots+a_{n-1}(n-1) !(n-(n-1)) 1 !+a_{n} n !(n-n) ! \\leq n !\n$$\n\nDividing both sides by $n$ !, we obtain successively\n\n$$\n\\begin{array}{r}\na_{0} 0 !(n-0) !+a_{1} 1 !(n-1) !+\\cdots+a_{n-1}(n-1) !(n-(n-1)) 1 !+a_{n} n !(n-n) ! \\leq n ! \\\\\na_{0} \\frac{0 !(n-0) !}{n !}+a_{1} \\frac{1 !(n-1) !}{n !}+\\cdots+a_{n-1} \\frac{(n-1) !(n-(n-1)) 1 !}{n !}+a_{n} \\frac{n !(n-n) !}{n !} \\leq 1 \\\\\na_{0} \\frac{1}{\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)}+a_{1} \\frac{1}{\\left(\\begin{array}{l}\nn \\\\\n1\n\\end{array}\\right)}+\\cdots+a_{n-1} \\frac{1}{\\left(\\begin{array}{c}\nn \\\\\nn-1\n\\end{array}\\right)}+a_{n} \\frac{1}{\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)} \\leq 1 \\\\\n\\frac{a_{0}}{\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)}+\\frac{a_{1}}{\\left(\\begin{array}{l}\nn \\\\\n1\n\\end{array}\\right)}+\\cdots+\\frac{a_{n-1}}{\\left(\\begin{array}{c}\nn \\\\\nn-1\n\\end{array}\\right)}+\\frac{a_{n}}{\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)} \\leq 1\n\\end{array}\n$$\n\nas required.', 'Suppose that $n$ is a positive integer and that $F$ is a randomly chosen Furoni family of $C_{n}$. Consider $L=\\{\\{\\},\\{1\\},\\{1,2\\},\\{1,2,3\\},\\{1,2,3, \\ldots, n\\}\\}$.\n\nThe probability that the intersection of $L$ and $F$ is non-empty is at most 1 .\n\nNote that since each element of $L$ is a subset of all of those to its right in the listing of $L$, then at most one of the elements of $L$ can be in $F$.\n\nIf $k$ is an integer with $k \\geq 0$, the probability that $\\{1,2,3, \\ldots, k\\}$ is an element of $F$ is $\\frac{a_{k}}{\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)}$, where $a_{k}$ is the number of elements in $F$ that contain exactly $k$ integers:\n\nThere are $\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)$ subsets of $C_{n}$ that contain exactly $k$ integer.\n\nThe probability that any particular one of these subsets is $\\{1,2,3, \\ldots, k\\}$ equals\n\n$\\frac{1}{\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)}$.\n\nSince $a_{k}$ of these subsets are in $F$, then the probability that one of these $a_{k}$ subsets is $\\{1,2,3, \\ldots, k\\}$ equals $\\frac{a_{k}}{\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)}$.\n\n(Note that we use the convention that if $k=0$, then $\\{1,2,3, \\ldots, k\\}=\\{\\}$.)\n\nThe probability that any of the elements of $L$ is in $F$ is the sum of the probability of each element being in $F$, since at most one of the elements in $L$ is in $F$.\n\nTherefore,\n\n$$\n\\frac{a_{0}}{\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)}+\\frac{a_{1}}{\\left(\\begin{array}{l}\nn \\\\\n1\n\\end{array}\\right)}+\\cdots+\\frac{a_{n-1}}{\\left(\\begin{array}{c}\nn \\\\\nn-1\n\\end{array}\\right)}+\\frac{a_{n}}{\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)} \\leq 1\n$$\n\nas required.']",,True,,, 2307,Geometry,,"In the diagram, what is the area of figure $A B C D E F$ ? ","['Because all of the angles in the figure are right angles, then $B C=D E=4$.\n\nThus, we can break up the figure into a 4 by 8 rectangle and a 4 by 4 square, by extending $B C$ to hit $F E$. Therefore, the area of the figure is $(8)(4)+(4)(4)=48$.']",['48'],False,,Numerical, 2307,Geometry,,"In the diagram, what is the area of figure $A B C D E F$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=364&width=338&top_left_y=680&top_left_x=1403)","['Because all of the angles in the figure are right angles, then $B C=D E=4$.\n\nThus, we can break up the figure into a 4 by 8 rectangle and a 4 by 4 square, by extending $B C$ to hit $F E$. Therefore, the area of the figure is $(8)(4)+(4)(4)=48$.']",['48'],False,,Numerical, 2308,Geometry,,"In the diagram, $A B C D$ is a rectangle with $A E=15, E B=20$ and $D F=24$. What is the length of $C F$ ? ","['By the Pythagorean Theorem in triangle $A B E$, $A B^{2}=15^{2}+20^{2}=625$, so $A B=25$.\n\nSince $A B C D$ is a rectangle, $C D=A B=25$, so by the Pythagorean Theorem in triangle $C F D$, we have $625=25^{2}=24^{2}+C F^{2}$, so $C F^{2}=625-576=49$, or $C F=7$.']",['7'],False,,Numerical, 2308,Geometry,,"In the diagram, $A B C D$ is a rectangle with $A E=15, E B=20$ and $D F=24$. What is the length of $C F$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=314&width=374&top_left_y=1079&top_left_x=1361)","['By the Pythagorean Theorem in triangle $A B E$, $A B^{2}=15^{2}+20^{2}=625$, so $A B=25$.\n\nSince $A B C D$ is a rectangle, $C D=A B=25$, so by the Pythagorean Theorem in triangle $C F D$, we have $625=25^{2}=24^{2}+C F^{2}$, so $C F^{2}=625-576=49$, or $C F=7$.']",['7'],False,,Numerical, 2309,Geometry,,"In the diagram, $A B C D$ is a square of side length 6. Points $E, F, G$, and $H$ are on $A B, B C, C D$, and $D A$, respectively, so that the ratios $A E: E B, B F: F C$, $C G: G D$, and $D H: H A$ are all equal to $1: 2$. What is the area of $E F G H$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=361&width=372&top_left_y=1432&top_left_x=1386)","['Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nThen the area of $E F G H$ is equal to the area of square $A B C D$ minus the combined area of the four triangles, or $6^{2}-4\\left[\\frac{1}{2}(2)(4)\\right]=36-16=20$ square units.', 'Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nBy the Pythagorean Theorem,\n\n$E F=F G=G H=H E=\\sqrt{2^{2}+4^{2}}=\\sqrt{20}$.\n\nSince the two triangles $H A E$ and $E B F$ are congruent (we know the lengths of all three sides of each), then $\\angle A H E=\\angle B E F$. But $\\angle A H E+\\angle A E H=90^{\\circ}$, so $\\angle B E F+\\angle A E H=90^{\\circ}$, so $\\angle H E F=90^{\\circ}$.\n\nIn a similar way, we can show that each of the four angles of $E F G H$ is a right-angle, and so $E F G H$ is a square of side length $\\sqrt{20}$.\n\nTherefore, the area of $E F G H$ is $(\\sqrt{20})^{2}=20$ square units.']",['20'],False,,Numerical, 2309,Geometry,,"In the diagram, $A B C D$ is a square of side length 6. Points $E, F, G$, and $H$ are on $A B, B C, C D$, and $D A$, respectively, so that the ratios $A E: E B, B F: F C$, $C G: G D$, and $D H: H A$ are all equal to $1: 2$. What is the area of $E F G H$ ? ","['Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nThen the area of $E F G H$ is equal to the area of square $A B C D$ minus the combined area of the four triangles, or $6^{2}-4\\left[\\frac{1}{2}(2)(4)\\right]=36-16=20$ square units.', 'Since $A B C D$ is a square of side length 6 and each of $A E: E B, B F: F C, C G: G D$, and $D H: H A$ is equal to $1: 2$, then $A E=B F=C G=D H=2$ and $E B=F C=G D=H A=4$.\n\nThus, each of the triangles $H A E, E B F, F C G$, and $G D H$ is right-angled, with one leg of length 2 and the other of length 4.\n\nBy the Pythagorean Theorem,\n\n$E F=F G=G H=H E=\\sqrt{2^{2}+4^{2}}=\\sqrt{20}$.\n\nSince the two triangles $H A E$ and $E B F$ are congruent (we know the lengths of all three sides of each), then $\\angle A H E=\\angle B E F$. But $\\angle A H E+\\angle A E H=90^{\\circ}$, so $\\angle B E F+\\angle A E H=90^{\\circ}$, so $\\angle H E F=90^{\\circ}$.\n\nIn a similar way, we can show that each of the four angles of $E F G H$ is a right-angle, and so $E F G H$ is a square of side length $\\sqrt{20}$.\n\nTherefore, the area of $E F G H$ is $(\\sqrt{20})^{2}=20$ square units.']",['20'],False,,Numerical, 2310,Geometry,,"In the diagram, line $A$ has equation $y=2 x$. Line $B$ is obtained by reflecting line $A$ in the $y$-axis. Line $C$ is perpendicular to line $B$. What is the slope of line $C$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_872e856c9056571e7b48g-1.jpg?height=626&width=613&top_left_y=1942&top_left_x=1206)","['When line $A$ with equation $y=2 x$ is reflected in the $y$-axis, the resulting line (line $B$ ) has equation $y=-2 x$. (Reflecting a line in the $y$-axis changes the sign of the slope.)\n\nSince the slope of line $B$ is -2 and line $C$ is perpendicular to line $B$, then the slope of line $C$ is $\\frac{1}{2}$ (the slopes of perpendicular lines are negative reciprocals).']",['$\\frac{1}{2}$'],False,,Numerical, 2310,Geometry,,"In the diagram, line $A$ has equation $y=2 x$. Line $B$ is obtained by reflecting line $A$ in the $y$-axis. Line $C$ is perpendicular to line $B$. What is the slope of line $C$ ? ","['When line $A$ with equation $y=2 x$ is reflected in the $y$-axis, the resulting line (line $B$ ) has equation $y=-2 x$. (Reflecting a line in the $y$-axis changes the sign of the slope.)\n\nSince the slope of line $B$ is -2 and line $C$ is perpendicular to line $B$, then the slope of line $C$ is $\\frac{1}{2}$ (the slopes of perpendicular lines are negative reciprocals).']",['$\\frac{1}{2}$'],False,,Numerical, 2311,Geometry,,"Three squares, each of side length 1 , are drawn side by side in the first quadrant, as shown. Lines are drawn from the origin to $P$ and $Q$. Determine, with explanation, the length of $A B$. ","['Consider the line through $O$ and $P$. To get from $O$ to $P$, we go right 2 and up 1. Since $B$ lies on this line and to get from $O$ to $B$ we go over 1, then we must go up $\\frac{1}{2}$, to keep the ratio constant.\n\nConsider the line through $O$ and $Q$. To get from $O$ to $Q$, we go right 3 and up 1. Since $A$ lies on this line and to get from $O$ to $A$ we go over 1, then we must go up $\\frac{1}{3}$, to keep the ratio constant.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.', 'Since the line through $P$ passes through the origin, then its equation is of the form $y=m x$. Since it passes through the point $(2,1)$, then $1=2 m$, so the line has equation $y=\\frac{1}{2} x$. Since $B$ has $x$-coordinate 1, then $y=\\frac{1}{2}(1)=\\frac{1}{2}$, so $B$ has coordinates $\\left(1, \\frac{1}{2}\\right)$. Similarly, we can determine that the equation of the line through $Q$ is $y=\\frac{1}{3} x$, and so $A$ has coordinates $\\left(1, \\frac{1}{3}\\right)$.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.']",['$\\frac{1}{6}$'],False,,Numerical, 2311,Geometry,,"Three squares, each of side length 1 , are drawn side by side in the first quadrant, as shown. Lines are drawn from the origin to $P$ and $Q$. Determine, with explanation, the length of $A B$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b2aa85be1fe1a84ee805g-1.jpg?height=423&width=615&top_left_y=260&top_left_x=1189)","['Consider the line through $O$ and $P$. To get from $O$ to $P$, we go right 2 and up 1. Since $B$ lies on this line and to get from $O$ to $B$ we go over 1, then we must go up $\\frac{1}{2}$, to keep the ratio constant.\n\nConsider the line through $O$ and $Q$. To get from $O$ to $Q$, we go right 3 and up 1. Since $A$ lies on this line and to get from $O$ to $A$ we go over 1, then we must go up $\\frac{1}{3}$, to keep the ratio constant.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.', 'Since the line through $P$ passes through the origin, then its equation is of the form $y=m x$. Since it passes through the point $(2,1)$, then $1=2 m$, so the line has equation $y=\\frac{1}{2} x$. Since $B$ has $x$-coordinate 1, then $y=\\frac{1}{2}(1)=\\frac{1}{2}$, so $B$ has coordinates $\\left(1, \\frac{1}{2}\\right)$. Similarly, we can determine that the equation of the line through $Q$ is $y=\\frac{1}{3} x$, and so $A$ has coordinates $\\left(1, \\frac{1}{3}\\right)$.\n\nTherefore, since $A$ and $B$ lie on the same vertical line, then $A B=\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$.']",['$\\frac{1}{6}$'],False,,Numerical, 2312,Combinatorics,,"Two fair dice, each having six faces numbered 1 to 6 , are thrown. What is the probability that the product of the two numbers on the top faces is divisible by 5 ?","['There are 36 possibilities for the pair of numbers on the faces when the dice are thrown. For the product of the two numbers, each of which is between 1 and 6 , to be divisible by 5 , one of the two numbers must be equal to 5 .\n\nTherefore, the possible pairs for the faces are\n\n$$\n(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(5,1),(5,2),(5,3),(5,4),(5,6)\n$$\n\nie. there are 11 possibilities.\n\nThus, the probability is $\\frac{11}{36}$.', 'For the product of the two numbers, each of which is between 1 and 6 , to be divisible by 5 , one of the two numbers must be equal to 5 .\n\nWhen the two dice are thrown, the probability that the first die has a 5 on the top face and any number appears on the second die has any number on the top face is $\\frac{1}{6} \\times 1=\\frac{1}{6}$.\n\nAlso, the probability that any number appears on the first die and a 5 appears on the second die is $1 \\times \\frac{1}{6}=\\frac{1}{6}$.\n\n\n\nIf we consider the sum of these probabilities, we have double-counted the possibility that a 5 occurs on both dice, which happens with probability $\\frac{1}{6} \\times \\frac{1}{6}=\\frac{1}{36}$.\n\nTherefore, the required probability is $\\frac{1}{6}+\\frac{1}{6}-\\frac{1}{36}=\\frac{11}{36}$.']",['$\\frac{11}{36}$'],False,,Numerical, 2313,Algebra,,"If $f(x)=x^{2}-x+2, g(x)=a x+b$, and $f(g(x))=9 x^{2}-3 x+2$, determine all possible ordered pairs $(a, b)$ which satisfy this relationship.","['First, we compute an expression for the composition of the two given functions:\n\n$$\n\\begin{aligned}\nf(g(x)) & =f(a x+b) \\\\\n& =(a x+b)^{2}-(a x+b)+2 \\\\\n& =a^{2} x^{2}+2 a b x+b^{2}-a x-b+2 \\\\\n& =a^{2} x^{2}+(2 a b-a) x+\\left(b^{2}-b+2\\right)\n\\end{aligned}\n$$\n\nBut we already know that $f(g(x))=9 x^{2}-3 x+2$, so comparing coefficients, we see that\n\n$$\n\\begin{aligned}\na^{2} & =9 \\\\\n2 a b-a & =-3 \\\\\nb^{2}-b+2 & =2\n\\end{aligned}\n$$\n\nFrom the first equation, $a=3$ or $a=-3$.\n\nFrom the third equation, $b^{2}-b=b(b-1)=0$ so $b=0$ or $b=1$.\n\nThere are thus 4 possible pairs $(a, b)$ which could solve the problem. We will check which pairs work by looking at the second equation.\n\nFrom the second equation, $a(2 b-1)=-3$, so if $a=3$ then $b=0$, and if $a=-3$ then $b=1$. Therefore, the possible ordered pairs $(a, b)$ are $(3,0)$ and $(-3,1)$.']","['$(3,0),(-3,1)$']",True,,Tuple, 2314,Geometry,,"In the diagram, the parabola with equation $y=x^{2}+t x-2$ intersects the $x$-axis at points $P$ and $Q$. Also, the line with equation $y=3 x+3$ intersects the parabola at points $P$ and $R$. Determine the value of $t$ and the area of triangle $P Q R$. ","['Point $P$ is the point where the line $y=3 x+3$ crosses the $x$ axis, and so has coordinates $(-1,0)$.\n\nTherefore, one of the roots of the parabola $y=x^{2}+t x-2$ is $x=-1$, so\n\n$$\n\\begin{aligned}\n0 & =(-1)^{2}+t(-1)-2 \\\\\n0 & =1-t-2 \\\\\nt & =-1\n\\end{aligned}\n$$\n\nThe parabola now has equation $y=x^{2}-x-2=(x+1)(x-2)$ (we already knew one of the roots so this helped with the factoring) and so its two $x$-intercepts are -1 and 2 , ie. $P$ has coordinates $(-1,0)$ and $Q$ has coordinates $(2,0)$.\n\nWe now have to find the coordinates of the point $R$. We know that $R$ is one of the two points of intersection of the line and the parabola, so we equate their equations:\n\n$$\n\\begin{aligned}\n3 x+3 & =x^{2}-x-2 \\\\\n0 & =x^{2}-4 x-5 \\\\\n0 & =(x+1)(x-5)\n\\end{aligned}\n$$\n\n(Again, we already knew one of the solutions to this equation $(x=-1)$ so this made factoring easier.) Since $R$ does not have $x$-coordinate -1 , then $R$ has $x$-coordinate $x=5$. Since $R$ lies on the line, then $y=3(5)+3=18$, so $R$ has coordinates $(5,18)$.\n\nWe can now calculate the area of triangle $P Q R$. This triangle has base of length 3 (from $P$ to $Q$ ) and height of length 18 (from the $x$-axis to $R$ ), and so has area $\\frac{1}{2}(3)(18)=27$.\n\nThus, $t=-1$ and the area of triangle $P Q R$ is 27 .']","['-1,27']",True,,Numerical, 2314,Geometry,,"In the diagram, the parabola with equation $y=x^{2}+t x-2$ intersects the $x$-axis at points $P$ and $Q$. Also, the line with equation $y=3 x+3$ intersects the parabola at points $P$ and $R$. Determine the value of $t$ and the area of triangle $P Q R$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b2aa85be1fe1a84ee805g-1.jpg?height=721&width=556&top_left_y=1488&top_left_x=1245)","['Point $P$ is the point where the line $y=3 x+3$ crosses the $x$ axis, and so has coordinates $(-1,0)$.\n\nTherefore, one of the roots of the parabola $y=x^{2}+t x-2$ is $x=-1$, so\n\n$$\n\\begin{aligned}\n0 & =(-1)^{2}+t(-1)-2 \\\\\n0 & =1-t-2 \\\\\nt & =-1\n\\end{aligned}\n$$\n\nThe parabola now has equation $y=x^{2}-x-2=(x+1)(x-2)$ (we already knew one of the roots so this helped with the factoring) and so its two $x$-intercepts are -1 and 2 , ie. $P$ has coordinates $(-1,0)$ and $Q$ has coordinates $(2,0)$.\n\nWe now have to find the coordinates of the point $R$. We know that $R$ is one of the two points of intersection of the line and the parabola, so we equate their equations:\n\n$$\n\\begin{aligned}\n3 x+3 & =x^{2}-x-2 \\\\\n0 & =x^{2}-4 x-5 \\\\\n0 & =(x+1)(x-5)\n\\end{aligned}\n$$\n\n(Again, we already knew one of the solutions to this equation $(x=-1)$ so this made factoring easier.) Since $R$ does not have $x$-coordinate -1 , then $R$ has $x$-coordinate $x=5$. Since $R$ lies on the line, then $y=3(5)+3=18$, so $R$ has coordinates $(5,18)$.\n\nWe can now calculate the area of triangle $P Q R$. This triangle has base of length 3 (from $P$ to $Q$ ) and height of length 18 (from the $x$-axis to $R$ ), and so has area $\\frac{1}{2}(3)(18)=27$.\n\nThus, $t=-1$ and the area of triangle $P Q R$ is 27 .']","['-1,27']",True,,Numerical, 2315,Algebra,,"Digital images consist of a very large number of equally spaced dots called pixels The resolution of an image is the number of pixels/cm in each of the horizontal and vertical directions. Thus, an image with dimensions $10 \mathrm{~cm}$ by $15 \mathrm{~cm}$ and a resolution of 75 pixels/cm has a total of $(10 \times 75) \times(15 \times 75)=843750$ pixels. If each of these dimensions was increased by $n \%$ and the resolution was decreased by $n \%$, the image would have 345600 pixels. Determine the value of $n$.","['When the dimensions were increased by $n \\%$ from 10 by 15 , the new dimensions were $10\\left(1+\\frac{n}{100}\\right)$ by $15\\left(1+\\frac{n}{100}\\right)$.\n\nWhen the resolution was decreased by $n$ percent, the new resolution was $75\\left(1-\\frac{n}{100}\\right)$.\n\n(Note that $n$ cannot be larger than 100, since the resolution cannot be decreased by more than $100 \\%$.)\n\nTherefore, the number of pixels in the new image is\n\n$$\n\\left[10\\left(1+\\frac{n}{100}\\right) \\times 75\\left(1-\\frac{n}{100}\\right)\\right] \\times\\left[15\\left(1+\\frac{n}{100}\\right) \\times 75\\left(1-\\frac{n}{100}\\right)\\right]\n$$\n\nSince we know that the number of pixels in the new image is 345600 , then\n\n$$\n\\begin{aligned}\n{\\left[10\\left(1+\\frac{n}{100}\\right) \\times 75\\left(1-\\frac{n}{100}\\right)\\right] \\times\\left[15\\left(1+\\frac{n}{100}\\right) \\times 75\\left(1-\\frac{n}{100}\\right)\\right] } & =345600 \\\\\n{[10 \\times 75] \\times[15 \\times 75] \\times\\left(1+\\frac{n}{100}\\right)^{2} \\times\\left(1-\\frac{n}{100}\\right)^{2} } & =345600 \\\\\n843750\\left(1+\\frac{n}{100}\\right)^{2}\\left(1-\\frac{n}{100}\\right)^{2} & =345600 \\\\\n\\left(1-\\frac{n^{2}}{100^{2}}\\right)^{2} & =0.4096 \\\\\n1-\\frac{n^{2}}{100^{2}} & = \\pm 0.64 \\\\\n1-\\frac{n^{2}}{100^{2}} & =0.64 \\\\\n\\frac{n^{2}}{100^{2}} & =0.36 \\\\\n\\frac{n}{100} & =0.6 \\\\\nn & =60\n\\end{aligned}\n$$\n\n$$\n\\begin{array}{rlrl}\n1-\\frac{n^{2}}{100^{2}} & =0.64 & & (n \\text { cannot be larger than } 100) \\\\\n\\frac{n^{2}}{100^{2}} & =0.36 & \\\\\n\\frac{n}{100} & =0.6 & & (\\text { since } n \\text { must be positive })\n\\end{array}\n$$\n\nThus, $n=60$.']",['60'],False,,Numerical, 2316,Geometry,,"In the diagram, $A C=B C, A D=7, D C=8$, and $\angle A D C=120^{\circ}$. What is the value of $x$ ? ","['We first calculate the length of $A C$ using the cosine law:\n\n$$\n\\begin{aligned}\nA C^{2} & =7^{2}+8^{2}-2(7)(8) \\cos \\left(120^{\\circ}\\right) \\\\\nA C^{2} & =49+64-112\\left(-\\frac{1}{2}\\right) \\\\\nA C^{2} & =169 \\\\\nA C & =13\n\\end{aligned}\n$$\n\nSince triangle $A B C$ is right-angled and isosceles, then $x=A B=\\sqrt{2}(A C)=13 \\sqrt{2}$.']",['$13 \\sqrt{2}$'],False,,Numerical, 2316,Geometry,,"In the diagram, $A C=B C, A D=7, D C=8$, and $\angle A D C=120^{\circ}$. What is the value of $x$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_bfecee96824679158031g-1.jpg?height=456&width=366&top_left_y=905&top_left_x=1389)","['We first calculate the length of $A C$ using the cosine law:\n\n$$\n\\begin{aligned}\nA C^{2} & =7^{2}+8^{2}-2(7)(8) \\cos \\left(120^{\\circ}\\right) \\\\\nA C^{2} & =49+64-112\\left(-\\frac{1}{2}\\right) \\\\\nA C^{2} & =169 \\\\\nA C & =13\n\\end{aligned}\n$$\n\nSince triangle $A B C$ is right-angled and isosceles, then $x=A B=\\sqrt{2}(A C)=13 \\sqrt{2}$.']",['$13 \\sqrt{2}$'],False,,Numerical, 2317,Algebra,,"If $T=x^{2}+\frac{1}{x^{2}}$, determine the values of $b$ and $c$ so that $x^{6}+\frac{1}{x^{6}}=T^{3}+b T+c$ for all non-zero real numbers $x$.","['Consider the right side of the given equation:\n\n$$\n\\begin{aligned}\nT^{3}+b T+c & =\\left(x^{2}+\\frac{1}{x^{2}}\\right)^{3}+b\\left(x^{2}+\\frac{1}{x^{2}}\\right)+c \\\\\n& =\\left(x^{4}+2+\\frac{1}{x^{4}}\\right)\\left(x^{2}+\\frac{1}{x^{2}}\\right)+b\\left(x^{2}+\\frac{1}{x^{2}}\\right)+c \\\\\n& =x^{6}+3 x^{2}+\\frac{3}{x^{2}}+\\frac{1}{x^{6}}+b\\left(x^{2}+\\frac{1}{x^{2}}\\right)+c \\\\\n& =x^{6}+\\frac{1}{x^{6}}+(b+3)\\left(x^{2}+\\frac{1}{x^{2}}\\right)+c\n\\end{aligned}\n$$\n\nFor this expression to be equal to $x^{6}+\\frac{1}{x^{6}}$ for all values of $x$, we want $b+3=0$ or $b=-3$ and $c=0$.']","['-3,0']",True,,Numerical, 2318,Combinatorics,,"A Skolem sequence of order $n$ is a sequence $\left(s_{1}, s_{2}, \ldots, s_{2 n}\right)$ of $2 n$ integers satisfying the conditions: i) for every $k$ in $\{1,2,3, \ldots, n\}$, there exist exactly two elements $s_{i}$ and $s_{j}$ with $s_{i}=s_{j}=k$, and ii) if $s_{i}=s_{j}=k$ with $i10$, which is not possible.\n\nTherefore, we must have $c=3$ and $b=2$, which gives $a=6$.']",['623'],False,,Numerical, 2322,Combinatorics,,"Eleanor has 100 marbles, each of which is black or gold. The ratio of the number of black marbles to the number of gold marbles is $1: 4$. How many gold marbles should she add to change this ratio to $1: 6$ ?","['Since Eleanor has 100 marbles which are black and gold in the ratio $1: 4$, then $\\frac{1}{5}$ of her marbles are black, which means that she has $\\frac{1}{5} \\cdot 100=20$ black marbles.\n\nWhen more gold marbles are added, the ratio of black to gold is $1: 6$, which means that she has $6 \\cdot 20=120$ gold marbles.\n\nEleanor now has $20+120=140$ marbles, which means that she added $140-100=40$ gold marbles.']",['40'],False,,Numerical, 2323,Number Theory,,Suppose that $n$ is a positive integer and that the value of $\frac{n^{2}+n+15}{n}$ is an integer. Determine all possible values of $n$.,"['First, we see that $\\frac{n^{2}+n+15}{n}=\\frac{n^{2}}{n}+\\frac{n}{n}+\\frac{15}{n}=n+1+\\frac{15}{n}$.\n\nThis means that $\\frac{n^{2}+n+15}{n}$ is an integer exactly when $n+1+\\frac{15}{n}$ is an integer.\n\nSince $n+1$ is an integer, then $\\frac{n^{2}+n+15}{n}$ is an integer exactly when $\\frac{15}{n}$ is an integer.\n\nThe expression $\\frac{15}{n}$ is an integer exactly when $n$ is a divisor of 15 .\n\nSince $n$ is a positive integer, then the possible values of $n$ are 1, 3, 5, and 15 .']","['1, 3, 5, 15']",True,,Numerical, 2324,Geometry,,"Donna has a laser at $C$. She points the laser beam at the point $E$. The beam reflects off of $D F$ at $E$ and then off of $F H$ at $G$, as shown, arriving at point $B$ on $A D$. If $D E=E F=1 \mathrm{~m}$, what is the length of $B D$, in metres? ","['First, we note that a triangle with one right angle and one angle with measure $45^{\\circ}$ is isosceles.\n\nThis is because the measure of the third angle equals $180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ which means that the triangle has two equal angles.\n\nIn particular, $\\triangle C D E$ is isosceles with $C D=D E$ and $\\triangle E F G$ is isosceles with $E F=F G$. Since $D E=E F=1 \\mathrm{~m}$, then $C D=F G=1 \\mathrm{~m}$.\n\nJoin $C$ to $G$.\n\n\n\nConsider quadrilateral $C D F G$. Since the angles at $D$ and $F$ are right angles and since $C D=G F$, it must be the case that $C D F G$ is a rectangle.\n\nThis means that $C G=D F=2 \\mathrm{~m}$ and that the angles at $C$ and $G$ are right angles.\n\nSince $\\angle C G F=90^{\\circ}$ and $\\angle D C G=90^{\\circ}$, then $\\angle B G C=180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ and $\\angle B C G=90^{\\circ}$.\n\nThis means that $\\triangle B C G$ is also isosceles with $B C=C G=2 \\mathrm{~m}$.\n\nFinally, $B D=B C+C D=2 \\mathrm{~m}+1 \\mathrm{~m}=3 \\mathrm{~m}$.']",['3'],False,,Numerical, 2324,Geometry,,"Donna has a laser at $C$. She points the laser beam at the point $E$. The beam reflects off of $D F$ at $E$ and then off of $F H$ at $G$, as shown, arriving at point $B$ on $A D$. If $D E=E F=1 \mathrm{~m}$, what is the length of $B D$, in metres? ![](https://cdn.mathpix.com/cropped/2023_12_21_55f6cf88d5d7c500e366g-1.jpg?height=456&width=317&top_left_y=276&top_left_x=1354)","['First, we note that a triangle with one right angle and one angle with measure $45^{\\circ}$ is isosceles.\n\nThis is because the measure of the third angle equals $180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ which means that the triangle has two equal angles.\n\nIn particular, $\\triangle C D E$ is isosceles with $C D=D E$ and $\\triangle E F G$ is isosceles with $E F=F G$. Since $D E=E F=1 \\mathrm{~m}$, then $C D=F G=1 \\mathrm{~m}$.\n\nJoin $C$ to $G$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_efd50db100807eeb16e0g-1.jpg?height=463&width=358&top_left_y=1384&top_left_x=989)\n\nConsider quadrilateral $C D F G$. Since the angles at $D$ and $F$ are right angles and since $C D=G F$, it must be the case that $C D F G$ is a rectangle.\n\nThis means that $C G=D F=2 \\mathrm{~m}$ and that the angles at $C$ and $G$ are right angles.\n\nSince $\\angle C G F=90^{\\circ}$ and $\\angle D C G=90^{\\circ}$, then $\\angle B G C=180^{\\circ}-90^{\\circ}-45^{\\circ}=45^{\\circ}$ and $\\angle B C G=90^{\\circ}$.\n\nThis means that $\\triangle B C G$ is also isosceles with $B C=C G=2 \\mathrm{~m}$.\n\nFinally, $B D=B C+C D=2 \\mathrm{~m}+1 \\mathrm{~m}=3 \\mathrm{~m}$.']",['3'],False,,Numerical, 2325,Combinatorics,,"Ada starts with $x=10$ and $y=2$, and applies the following process: Step 1: Add $x$ and $y$. Let $x$ equal the result. The value of $y$ does not change. Step 2: Multiply $x$ and $y$. Let $x$ equal the result. The value of $y$ does not change. Step 3: Add $y$ and 1. Let $y$ equal the result. The value of $x$ does not change. Ada keeps track of the values of $x$ and $y$ : | | $x$ | $y$ | | :---: | :---: | :---: | | Before Step 1 | 10 | 2 | | After Step 1 | 12 | 2 | | After Step 2 | 24 | 2 | | After Step 3 | 24 | 3 | Continuing now with $x=24$ and $y=3$, Ada applies the process two more times. What is the final value of $x$ ?","['We apply the process two more times:\n\n| | $x$ | $y$ |\n| :---: | :---: | :---: |\n| Before Step 1 | 24 | 3 |\n| After Step 1 | 27 | 3 |\n| After Step 2 | 81 | 3 |\n| After Step 3 | 81 | 4 |\n\n\n| | $x$ | $y$ |\n| :---: | :---: | :---: |\n| Before Step 1 | 81 | 4 |\n| After Step 1 | 85 | 4 |\n| After Step 2 | 340 | 4 |\n| After Step 3 | 340 | 5 |\n\nTherefore, the final value of $x$ is 340 .']",['340'],False,,Numerical, 2326,Number Theory,,"Determine all integers $k$, with $k \neq 0$, for which the parabola with equation $y=k x^{2}+6 x+k$ has two distinct $x$-intercepts.","['The parabola with equation $y=k x^{2}+6 x+k$ has two distinct $x$-intercepts exactly when the discriminant of the quadratic equation $k x^{2}+6 x+k=0$ is positive.\n\nHere, the disciminant equals $\\Delta=6^{2}-4 \\cdot k \\cdot k=36-4 k^{2}$.\n\nThe inequality $36-4 k^{2}>0$ is equivalent to $k^{2}<9$.\n\nSince $k$ is an integer and $k \\neq 0$, then $k$ can equal $-2,-1,1,2$.\n\n(If $k \\geq 3$ or $k \\leq-3$, we get $k^{2} \\geq 9$ so no values of $k$ in these ranges give the desired result.)']","['-2,-1,1,2']",True,,Numerical, 2327,Number Theory,,"The positive integers $a$ and $b$ have no common divisor larger than 1 . If the difference between $b$ and $a$ is 15 and $\frac{5}{9}<\frac{a}{b}<\frac{4}{7}$, what is the value of $\frac{a}{b}$ ?","['Since $\\frac{a}{b}<\\frac{4}{7}$ and $\\frac{4}{7}<1$, then $\\frac{a}{b}<1$.\n\nSince $a$ and $b$ are positive integers, then $a75$.\n\nFrom this, we see that $a>\\frac{75}{4}=18.75$.\n\nSince $a$ is an integer, then $a \\geq 19$.\n\nWe multiply both sides of the right inequality by $7(a+15)$ (which is positive) to obtain $7 a<4(a+15)$ from which we get $7 a<4 a+60$ and so $3 a<60$.\n\nFrom this, we see that $a<20$.\n\nSince $a$ is an integer, then $a \\leq 19$.\n\nSince $a \\geq 19$ and $a \\leq 19$, then $a=19$, which means that $\\frac{a}{b}=\\frac{19}{34}$.']",['$\\frac{19}{34}$'],False,,Numerical, 2328,Algebra,,"A geometric sequence has first term 10 and common ratio $\frac{1}{2}$. An arithmetic sequence has first term 10 and common difference $d$. The ratio of the 6th term in the geometric sequence to the 4th term in the geometric sequence equals the ratio of the 6th term in the arithmetic sequence to the 4 th term in the arithmetic sequence. Determine all possible values of $d$. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, 3, 5, 7, 9 are the first four terms of an arithmetic sequence. A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant, called the common ratio. For example, $3,6,12$ is a geometric sequence with three terms.)","['The first 6 terms of a geometric sequence with first term 10 and common ratio $\\frac{1}{2}$ are $10,5, \\frac{5}{2}, \\frac{5}{4}, \\frac{5}{8}, \\frac{5}{16}$.\n\nHere, the ratio of its 6 th term to its 4 th term is $\\frac{5 / 16}{5 / 4}$ which equals $\\frac{1}{4}$. (We could have determined this without writing out the sequence, since moving from the 4th term to the 6th involves multiplying by $\\frac{1}{2}$ twice.)\n\nThe first 6 terms of an arithmetic sequence with first term 10 and common difference $d$ are $10,10+d, 10+2 d, 10+3 d, 10+4 d, 10+5 d$.\n\nHere, the ratio of the 6 th term to the 4 th term is $\\frac{10+5 d}{10+3 d}$.\n\nSince these ratios are equal, then $\\frac{10+5 d}{10+3 d}=\\frac{1}{4}$, which gives $4(10+5 d)=10+3 d$ and so $40+20 d=10+3 d$ or $17 d=-30$ and so $d=-\\frac{30}{17}$.']",['$-\\frac{30}{17}$'],False,,Numerical, 2329,Algebra,,"For each positive real number $x$, define $f(x)$ to be the number of prime numbers $p$ that satisfy $x \leq p \leq x+10$. What is the value of $f(f(20))$ ?","['Let $a=f(20)$. Then $f(f(20))=f(a)$.\n\nTo calculate $f(f(20))$, we determine the value of $a$ and then the value of $f(a)$.\n\nBy definition, $a=f(20)$ is the number of prime numbers $p$ that satisfy $20 \\leq p \\leq 30$.\n\nThe prime numbers between 20 and 30, inclusive, are 23 and 29 , so $a=f(20)=2$.\n\nThus, $f(f(20))=f(a)=f(2)$.\n\nBy definition, $f(2)$ is the number of prime numbers $p$ that satisfy $2 \\leq p \\leq 12$.\n\nThe prime numbers between 2 and 12, inclusive, are $2,3,5,7,11$, of which there are 5 .\n\nTherefore, $f(f(20))=5$.']",['5'],False,,Numerical, 2330,Algebra,,"Determine all triples $(x, y, z)$ of real numbers that satisfy the following system of equations: $$ \begin{aligned} (x-1)(y-2) & =0 \\ (x-3)(z+2) & =0 \\ x+y z & =9 \end{aligned} $$","['Since $(x-1)(y-2)=0$, then $x=1$ or $y=2$.\n\nSuppose that $x=1$. In this case, the remaining equations become:\n\n$$\n\\begin{aligned}\n(1-3)(z+2) & =0 \\\\\n1+y z & =9\n\\end{aligned}\n$$\n\nor\n\n$$\n\\begin{array}{r}\n-2(z+2)=0 \\\\\ny z=8\n\\end{array}\n$$\n\nFrom the first of these equations, $z=-2$.\n\nFrom the second of these equations, $y(-2)=8$ and so $y=-4$.\n\nTherefore, if $x=1$, the only solution is $(x, y, z)=(1,-4,-2)$.\n\nSuppose that $y=2$. In this case, the remaining equations become:\n\n$$\n\\begin{aligned}\n(x-3)(z+2) & =0 \\\\\nx+2 z & =9\n\\end{aligned}\n$$\n\nFrom the first equation $x=3$ or $z=-2$.\n\nIf $x=3$, then $3+2 z=9$ and so $z=3$.\n\nIf $z=-2$, then $x+2(-2)=9$ and so $x=13$.\n\nTherefore, if $y=2$, the solutions are $(x, y, z)=(3,2,3)$ and $(x, y, z)=(13,2,-2)$.\n\nIn summary, the solutions to the system of equations are\n\n$$\n(x, y, z)=(1,-4,-2),(3,2,3),(13,2,-2)\n$$\n\nWe can check by substitution that each of these triples does indeed satisfy each of the equations.']","['(1,-4,-2),(3,2,3),(13,2,-2)']",True,,Tuple, 2331,Algebra,,Suppose that the function $g$ satisfies $g(x)=2 x-4$ for all real numbers $x$ and that $g^{-1}$ is the inverse function of $g$. Suppose that the function $f$ satisfies $g\left(f\left(g^{-1}(x)\right)\right)=2 x^{2}+16 x+26$ for all real numbers $x$. What is the value of $f(\pi)$ ?,"['Since the function $g$ is linear and has positive slope, then it is one-to-one and so invertible. This means that $g^{-1}(g(a))=a$ for every real number $a$ and $g\\left(g^{-1}(b)\\right)=b$ for every real number $b$.\n\nTherefore, $g\\left(f\\left(g^{-1}(g(a))\\right)\\right)=g(f(a))$ for every real number $a$.\n\nThis means that\n\n$$\n\\begin{aligned}\ng(f(a)) & =g\\left(f\\left(g^{-1}(g(a))\\right)\\right) \\\\\n& =2(g(a))^{2}+16 g(a)+26 \\\\\n& =2(2 a-4)^{2}+16(2 a-4)+26 \\\\\n& =2\\left(4 a^{2}-16 a+16\\right)+32 a-64+26 \\\\\n& =8 a^{2}-6\n\\end{aligned}\n$$\n\nFurthermore, if $b=f(a)$, then $g^{-1}(g(f(a)))=g^{-1}(g(b))=b=f(a)$.\n\nTherefore,\n\n$$\nf(a)=g^{-1}(g(f(a)))=g^{-1}\\left(8 a^{2}-6\\right)\n$$\n\nSince $g(x)=2 x-4$, then $y=2 g^{-1}(y)-4$ and so $g^{-1}(y)=\\frac{1}{2} y+2$.\n\nTherefore,\n\n$$\nf(a)=\\frac{1}{2}\\left(8 a^{2}-6\\right)+2=4 a^{2}-1\n$$\n\nand so $f(\\pi)=4 \\pi^{2}-1$.', 'Since the function $g$ is linear and has positive slope, then it is one-to-one and so invertible. To find a formula for $g^{-1}(y)$, we start with the equation $g(x)=2 x-4$, convert to $y=2 g^{-1}(y)-4$ and then solve for $g^{-1}(y)$ to obtain $2 g^{-1}(y)=y+4$ and so $g^{-1}(y)=\\frac{y+4}{2}$. We are given that $g\\left(f\\left(g^{-1}(x)\\right)\\right)=2 x^{2}+16 x+26$.\n\nWe can apply the function $g^{-1}$ to both sides to obtain successively:\n\n$$\n\\begin{aligned}\nf\\left(g^{-1}(x)\\right) & =g^{-1}\\left(2 x^{2}+16 x+26\\right) \\\\\nf\\left(g^{-1}(x)\\right) & \\left.=\\frac{\\left(2 x^{2}+16 x+26\\right)+4}{2} \\quad \\text { (knowing a formula for } g^{-1}\\right) \\\\\nf\\left(g^{-1}(x)\\right) & =x^{2}+8 x+15 \\\\\nf\\left(\\frac{x+4}{2}\\right) & \\left.=x^{2}+8 x+15 \\quad \\text { (knowing a formula for } g^{-1}\\right) \\\\\nf\\left(\\frac{x+4}{2}\\right) & =x^{2}+8 x+16-1 \\\\\nf\\left(\\frac{x+4}{2}\\right) & =(x+4)^{2}-1\n\\end{aligned}\n$$\n\nWe want to determine the value of $f(\\pi)$.\n\nThus, we can replace $\\frac{x+4}{2}$ with $\\pi$, which is equivalent to replacing $x+4$ with $2 \\pi$.\n\nThus, $f(\\pi)=(2 \\pi)^{2}-1=4 \\pi^{2}-1$.']",['$4 \\pi^{2}-1$'],False,,Numerical, 2332,Algebra,,"Determine all pairs of angles $(x, y)$ with $0^{\circ} \leq x<180^{\circ}$ and $0^{\circ} \leq y<180^{\circ}$ that satisfy the following system of equations: $$ \begin{aligned} \log _{2}(\sin x \cos y) & =-\frac{3}{2} \\ \log _{2}\left(\frac{\sin x}{\cos y}\right) & =\frac{1}{2} \end{aligned} $$","['Using logarithm laws, the given equations are equivalent to\n\n$$\n\\begin{aligned}\n& \\log _{2}(\\sin x)+\\log _{2}(\\cos y)=-\\frac{3}{2} \\\\\n& \\log _{2}(\\sin x)-\\log _{2}(\\cos y)=\\frac{1}{2}\n\\end{aligned}\n$$\n\nAdding these two equations, we obtain $2 \\log _{2}(\\sin x)=-1$ which gives $\\log _{2}(\\sin x)=-\\frac{1}{2}$ and so $\\sin x=2^{-1 / 2}=\\frac{1}{2^{1 / 2}}=\\frac{1}{\\sqrt{2}}$.\n\nSince $0^{\\circ} \\leq x<180^{\\circ}$, then $x=45^{\\circ}$ or $x=135^{\\circ}$.\n\nSince $\\log _{2}(\\sin x)+\\log _{2}(\\cos y)=-\\frac{3}{2}$ and $\\log _{2}(\\sin x)=-\\frac{1}{2}$, then $\\log _{2}(\\cos y)=-1$, which gives $\\cos y=2^{-1}=\\frac{1}{2}$.\n\nSince $0^{\\circ} \\leq y<180^{\\circ}$, then $y=60^{\\circ}$.\n\nTherefore, $(x, y)=\\left(45^{\\circ}, 60^{\\circ}\\right)$ or $(x, y)=\\left(135^{\\circ}, 60^{\\circ}\\right)$.', 'First, we note that $2^{1 / 2}=\\sqrt{2}$ and $2^{-3 / 2}=\\frac{1}{2^{3 / 2}}=\\frac{1}{2^{1} 2^{1 / 2}}=\\frac{1}{2 \\sqrt{2}}$.\n\nFrom the given equations, we obtain\n\n$$\n\\begin{aligned}\n\\sin x \\cos y & =2^{-3 / 2}=\\frac{1}{2 \\sqrt{2}} \\\\\n\\frac{\\sin x}{\\cos y} & =2^{1 / 2}=\\sqrt{2}\n\\end{aligned}\n$$\n\nMultiplying these two equations together, we obtain $(\\sin x)^{2}=\\frac{1}{2}$ which gives $\\sin x= \\pm \\frac{1}{\\sqrt{2}}$.\n\nSince $0^{\\circ} \\leq x<180^{\\circ}$, it must be the case that $\\sin x \\geq 0$ and so $\\sin x=\\frac{1}{\\sqrt{2}}$.\n\nSince $0^{\\circ} \\leq x<180^{\\circ}$, we obtain $x=45^{\\circ}$ or $x=135^{\\circ}$.\n\nSince $\\sin x \\cos y=\\frac{1}{2 \\sqrt{2}}$ and $\\sin x=\\frac{1}{\\sqrt{2}}$, we obtain $\\cos y=\\frac{1}{2}$.\n\nSince $0^{\\circ} \\leq y<180^{\\circ}$, then $y=60^{\\circ}$.\n\nTherefore, $(x, y)=\\left(45^{\\circ}, 60^{\\circ}\\right)$ or $(x, y)=\\left(135^{\\circ}, 60^{\\circ}\\right)$.']","['$(45^{\\circ}, 60^{\\circ}),(135^{\\circ}, 60^{\\circ})$']",True,,Tuple, 2333,Combinatorics,,"Four tennis players Alain, Bianca, Chen, and Dave take part in a tournament in which a total of three matches are played. First, two players are chosen randomly to play each other. The other two players also play each other. The winners of the two matches then play to decide the tournament champion. Alain, Bianca and Chen are equally matched (that is, when a match is played between any two of them, the probability that each player wins is $\frac{1}{2}$ ). When Dave plays each of Alain, Bianca and Chen, the probability that Dave wins is $p$, for some real number $p$. Determine the probability that Bianca wins the tournament, expressing your answer in the form $\frac{a p^{2}+b p+c}{d}$ where $a, b, c$, and $d$ are integers.","['Let $x$ be the probability that Bianca wins the tournament.\n\nBecause Alain, Bianca and Chen are equally matched and because their roles in the tournament are identical, then the probability that each of them wins will be the same.\n\nThus, the probability that Alain wins the tournament is $x$ and the probability that Chen wins the tournament is $x$.\n\nLet $y$ be the probability that Dave wins the tournament.\n\nSince exactly one of Alain, Bianca, Chen, and Dave wins the tournament, then $3 x+y=1$ and so $x=\\frac{1-y}{3}$. We can calculate $y$ in terms of $p$.\n\nIn order for Dave to win the tournament, he needs to win two matches.\n\nNo matter who Dave plays, his probability of winning each match is $p$.\n\nThus, the probability that he wins his two consecutive matches is $p^{2}$ and so the probability that he wins the tournament is $y=p^{2}$.\n\nThus, the probability that Bianca wins the tournament is $\\frac{1-p^{2}}{3}$.\n\n(We could rewrite this as $\\frac{-p^{2}+0 p+1}{3}$ to match the desired form.)', 'Let $x$ be the probability that Bianca wins the tournament.\n\nThere are three possible pairings for the first two matches:\n\n(i) Bianca versus Alain, and Chen versus Dave\n\n(ii) Bianca versus Chen, and Alain versus Dave\n\n(iii) Bianca versus Dave, and Alain versus Chen\n\nEach of these three pairings occurs with probability $\\frac{1}{3}$.\n\nIn (i), Bianca wins either if Bianca beats Alain, Chen beats Dave, and Bianca beats Chen, or if Bianca beats Alain, Dave beats Chen, and Bianca beats Dave.\n\nSince Bianca beats Alain with probability $\\frac{1}{2}$, Chen beats Dave with probability $1-p$, and Bianca beats Chen with probability $\\frac{1}{2}$, then the first possibility has probability $\\frac{1}{2} \\cdot(1-p) \\cdot \\frac{1}{2}$. Since Bianca beats Alain with probability $\\frac{1}{2}$, Dave beats Chen with probability $p$, and Bianca beats Dave with probability $1-p$, then the second possibility has probability $\\frac{1}{2} \\cdot p \\cdot(1-p)$.\n\nTherefore, the probability of Bianca winning, given that possibility (i) occurs, is $\\frac{1}{2} \\cdot(1-$ p) $\\cdot \\frac{1}{2}+\\frac{1}{2} \\cdot p \\cdot(1-p)$.\n\nIn (ii), Bianca wins either if Bianca beats Chen, Alain beats Dave, and Bianca beats Alain, or if Bianca beats Alain, Dave beats Alain, and Bianca beats Dave.\n\nThe combined probability of these is $\\frac{1}{2} \\cdot(1-p) \\cdot \\frac{1}{2}+\\frac{1}{2} \\cdot p \\cdot(1-p)$.\n\nIn (iii), Bianca wins either if Bianca beats Dave, Alain beats Chen, and Bianca beats Alain, or if Bianca beats Dave, Chen beats Alain, and Bianca beats Chen.\n\nThe combined probability of these is $(1-p) \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}+(1-p) \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nx & =\\frac{1}{3}\\left(\\frac{1}{4}(1-p)+\\frac{1}{2} p(1-p)+\\frac{1}{4}(1-p)+\\frac{1}{2} p(1-p)+\\frac{1}{4}(1-p)+\\frac{1}{4}(1-p)\\right) \\\\\n& =\\frac{1}{3}(p(1-p)+(1-p)) \\\\\n& =\\frac{1}{3}\\left(p-p^{2}+1-p\\right)\n\\end{aligned}\n$$\n\nThus, the probability that Bianca wins the tournament is $\\frac{1-p^{2}}{3}$.']",['$\\frac{1-p^{2}}{3}$'],False,,Expression, 2334,Geometry,,"Three microphones $A, B$ and $C$ are placed on a line such that $A$ is $1 \mathrm{~km}$ west of $B$ and $C$ is $2 \mathrm{~km}$ east of $B$. A large explosion occurs at a point $P$ not on this line. Each of the three microphones receives the sound. The sound travels at $\frac{1}{3} \mathrm{~km} / \mathrm{s}$. Microphone $B$ receives the sound first, microphone $A$ receives the sound $\frac{1}{2}$ s later, and microphone $C$ receives it $1 \mathrm{~s}$ after microphone $A$. Determine the distance from microphone $B$ to the explosion at $P$.","['Throughout this solution, we will mostly not include units, but will assume that all lengths are in kilometres, all times are in seconds, and all speeds are in kilometres per second.\n\nWe place the points in the coordinate plane with $B$ at $(0,0), A$ on the negative $x$-axis, and $C$ on the positive $x$-axis.\n\nWe put $A$ at $(-1,0)$ and $C$ at $(2,0)$.\n\nSuppose that $P$ has coordinates $(x, y)$ and that the distance from $P$ to $B$ is $d \\mathrm{~km}$.\n\n\n\nSince the sound arrives at $A \\frac{1}{2} \\mathrm{~s}$ after arriving at $B$ and sound travels at $\\frac{1}{3} \\mathrm{~km} / \\mathrm{s}$, then $A$ is $\\left(\\frac{1}{2} \\mathrm{~s}\\right) \\cdot\\left(\\frac{1}{3} \\mathrm{~km} / \\mathrm{s}\\right)=\\frac{1}{6} \\mathrm{~km}$ farther from $P$ than $B$ is.\n\nThus, the distance from $P$ to $A$ is $\\left(d+\\frac{1}{6}\\right) \\mathrm{km}$.\n\nSince the sound arrives at $C$ an additional 1 second later, then $C$ is an additional $\\frac{1}{3} \\mathrm{~km}$ farther, and so is $\\left(d+\\frac{1}{6}\\right) \\mathrm{km}+\\left(\\frac{1}{3} \\mathrm{~km}\\right)=\\left(d+\\frac{1}{2}\\right) \\mathrm{km}$ from $P$.\n\nSince the distance from $P$ to $B$ is $d \\mathrm{~km}$, then $(x-0)^{2}+(y-0)^{2}=d^{2}$.\n\nSince the distance from $P$ to $A$ is $\\left(d+\\frac{1}{6}\\right) \\mathrm{km}$, then $(x+1)^{2}+(y-0)^{2}=\\left(d+\\frac{1}{6}\\right)^{2}$.\n\nSince the distance from $P$ to $C$ is $\\left(d+\\frac{1}{2}\\right) \\mathrm{km}$, then $(x-2)^{2}+(y-0)^{2}=\\left(d+\\frac{1}{2}\\right)^{2}$.\n\nWhen these equations are expanded and simplified, we obtain\n\n$$\n\\begin{aligned}\nx^{2}+y^{2} & =d^{2} \\\\\nx^{2}+2 x+1+y^{2} & =d^{2}+\\frac{1}{3} d+\\frac{1}{36} \\\\\nx^{2}-4 x+4+y^{2} & =d^{2}+d+\\frac{1}{4}\n\\end{aligned}\n$$\n\nSubtracting the first equation from the second, we obtain\n\n$$\n2 x+1=\\frac{1}{3} d+\\frac{1}{36}\n$$\n\nSubtracting the first equation from the third, we obtain\n\n$$\n-4 x+4=d+\\frac{1}{4}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n2(2 x+1)+(-4 x+4) & =2\\left(\\frac{1}{3} d+\\frac{1}{36}\\right)+\\left(d+\\frac{1}{4}\\right) \\\\\n6 & =\\frac{2}{3} d+\\frac{1}{18}+d+\\frac{1}{4} \\\\\n216 & =24 d+2+36 d+9 \\quad \\text { (multiplying by } 36) \\\\\n205 & =60 d \\\\\nd & =\\frac{41}{12}\n\\end{aligned}\n$$\n\nTherefore, the distance from $B$ to $P$ is $\\frac{41}{12} \\mathrm{~km}$.']",['$\\frac{41}{12}$'],False,km,Numerical, 2335,Number Theory,,"An L shape is made by adjoining three congruent squares. The L is subdivided into four smaller L shapes, as shown. Each of the resulting L's is subdivided in this same way. After the third round of subdivisions, how many L's of the smallest size are there? ","[""After each round, each L shape is divided into 4 smaller $\\mathrm{L}$ shapes.\n\nThis means that the number of $\\mathrm{L}$ shapes increases by a factor of 4 after each round.\n\nAfter 1 round, there are $4 \\mathrm{~L}$ shapes.\n\nAfter 2 rounds, there are $4^{2}=16$ L's of the smallest size.\n\nAfter 3 rounds, there are $4^{3}=64$ L's of the smallest size.""]",['64'],False,,Numerical, 2335,Number Theory,,"An L shape is made by adjoining three congruent squares. The L is subdivided into four smaller L shapes, as shown. Each of the resulting L's is subdivided in this same way. After the third round of subdivisions, how many L's of the smallest size are there? ![](https://cdn.mathpix.com/cropped/2023_12_21_68ed10fc1401a8e73334g-1.jpg?height=226&width=472&top_left_y=1372&top_left_x=890)","[""After each round, each L shape is divided into 4 smaller $\\mathrm{L}$ shapes.\n\nThis means that the number of $\\mathrm{L}$ shapes increases by a factor of 4 after each round.\n\nAfter 1 round, there are $4 \\mathrm{~L}$ shapes.\n\nAfter 2 rounds, there are $4^{2}=16$ L's of the smallest size.\n\nAfter 3 rounds, there are $4^{3}=64$ L's of the smallest size.""]",['64'],False,,Numerical, 2336,Number Theory,,"Kerry has a list of $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$. Kerry calculates the pairwise sums of all $m=\frac{1}{2} n(n-1)$ possible pairs of integers in her list and orders these pairwise sums as $s_{1} \leq s_{2} \leq \ldots \leq s_{m}$. For example, if Kerry's list consists of the three integers $1,2,4$, the three pairwise sums are $3,5,6$. Suppose that $n=4$ and that the 6 pairwise sums are $s_{1}=8, s_{2}=104, s_{3}=106$, $s_{4}=110, s_{5}=112$, and $s_{6}=208$. Determine two possible lists $a_{1}, a_{2}, a_{3}, a_{4}$ that Kerry could have.","[""Here, the pairwise sums of the numbers $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4}$ are $s_{1} \\leq s_{2} \\leq s_{3} \\leq s_{4} \\leq s_{5} \\leq s_{6}$. The six pairwise sums of the numbers in the list can be expressed as\n\n$$\na_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}, a_{3}+a_{4}\n$$\n\nSince $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4}$, then the smallest sum must be the sum of the two smallest numbers. Thus, $s_{1}=a_{1}+a_{2}$.\n\nSimilarly, the largest sum must be the sum of the two largest numbers, and so $s_{6}=a_{3}+a_{4}$. Since $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4}$, then the second smallest sum is $a_{1}+a_{3}$. This is because $a_{1}+a_{3}$ is no greater than each of the four sums $a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}$, and $a_{3}+a_{4}$ :\n\nSince $a_{3} \\leq a_{4}$, then $a_{1}+a_{3} \\leq a_{1}+a_{4}$.\n\nSince $a_{1} \\leq a_{2}$, then $a_{1}+a_{3} \\leq a_{2}+a_{3}$.\n\nSince $a_{1} \\leq a_{2}$ and $a_{3} \\leq a_{4}$, then $a_{1}+a_{3} \\leq a_{2}+a_{4}$.\n\nSince $a_{1} \\leq a_{4}$, then $a_{1}+a_{3} \\leq a_{3}+a_{4}$.\n\nThus, $s_{2}=a_{1}+a_{3}$.\n\nUsing a similar argument, $s_{5}=a_{2}+a_{4}$.\n\nSo far, we have $s_{1}=a_{1}+a_{2}$ and $s_{2}=a_{1}+a_{3}$ and $s_{5}=a_{2}+a_{4}$ and $s_{6}=a_{3}+a_{4}$.\n\nThis means that $s_{3}$ and $s_{4}$ equal $a_{1}+a_{4}$ and $a_{2}+a_{3}$ in some order.\n\nIt turns out that either order is possible.\n\nCase 1: $s_{3}=a_{1}+a_{4}$ and $s_{4}=a_{2}+a_{3}$\n\nHere, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=110$.\n\nAdding these three equations gives\n\n$$\n\\left(a_{1}+a_{2}\\right)+\\left(a_{1}+a_{3}\\right)+\\left(a_{2}+a_{3}\\right)=8+104+110\n$$\n\nand so $2 a_{1}+2 a_{2}+2 a_{3}=222$ or $a_{1}+a_{2}+a_{3}=111$.\n\nSince $a_{2}+a_{3}=110$, then $a_{1}=\\left(a_{1}+a_{2}+a_{3}\\right)-\\left(a_{2}+a_{3}\\right)=111-110=1$.\n\nSince $a_{1}=1$ and $a_{1}+a_{2}=8$, then $a_{2}=7$.\n\nSince $a_{1}=1$ and $a_{1}+a_{3}=104$, then $a_{3}=103$.\n\nSince $a_{3}=103$ and $a_{3}+a_{4}=208$, then $a_{4}=105$.\n\nThus, $\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)=(1,7,103,105)$.\n\nCase 2: $s_{3}=a_{2}+a_{3}$ and $s_{4}=a_{1}+a_{4}$\n\nHere, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=106$.\n\nUsing the same process, $a_{1}+a_{2}+a_{3}=109$.\n\nFrom this, we obtain $\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)=(3,5,101,107)$.\n\nTherefore, Kerry's two possible lists are 1,7,103, 105 and 3, 5, 101, 107.""]","['1,7,103, 105 and 3, 5, 101, 107']",True,,Need_human_evaluate, 2336,Number Theory,,"Kerry has a list of $n$ integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$. Kerry calculates the pairwise sums of all $m=\frac{1}{2} n(n-1)$ possible pairs of integers in her list and orders these pairwise sums as $s_{1} \leq s_{2} \leq \ldots \leq s_{m}$. For example, if Kerry's list consists of the three integers $1,2,4$, the three pairwise sums are $3,5,6$. Suppose that $n=4$ and that the 6 pairwise sums are $s_{1}=8, s_{2}=104, s_{3}=106$, $s_{4}=110, s_{5}=112$, and $s_{6}=208$. Determine two possible lists $(a_{1}, a_{2}, a_{3}, a_{4})$ that Kerry could have.","[""Here, the pairwise sums of the numbers $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4}$ are $s_{1} \\leq s_{2} \\leq s_{3} \\leq s_{4} \\leq s_{5} \\leq s_{6}$. The six pairwise sums of the numbers in the list can be expressed as\n\n$$\na_{1}+a_{2}, a_{1}+a_{3}, a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}, a_{3}+a_{4}\n$$\n\nSince $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4}$, then the smallest sum must be the sum of the two smallest numbers. Thus, $s_{1}=a_{1}+a_{2}$.\n\nSimilarly, the largest sum must be the sum of the two largest numbers, and so $s_{6}=a_{3}+a_{4}$. Since $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4}$, then the second smallest sum is $a_{1}+a_{3}$. This is because $a_{1}+a_{3}$ is no greater than each of the four sums $a_{1}+a_{4}, a_{2}+a_{3}, a_{2}+a_{4}$, and $a_{3}+a_{4}$ :\n\nSince $a_{3} \\leq a_{4}$, then $a_{1}+a_{3} \\leq a_{1}+a_{4}$.\n\nSince $a_{1} \\leq a_{2}$, then $a_{1}+a_{3} \\leq a_{2}+a_{3}$.\n\nSince $a_{1} \\leq a_{2}$ and $a_{3} \\leq a_{4}$, then $a_{1}+a_{3} \\leq a_{2}+a_{4}$.\n\nSince $a_{1} \\leq a_{4}$, then $a_{1}+a_{3} \\leq a_{3}+a_{4}$.\n\nThus, $s_{2}=a_{1}+a_{3}$.\n\nUsing a similar argument, $s_{5}=a_{2}+a_{4}$.\n\nSo far, we have $s_{1}=a_{1}+a_{2}$ and $s_{2}=a_{1}+a_{3}$ and $s_{5}=a_{2}+a_{4}$ and $s_{6}=a_{3}+a_{4}$.\n\nThis means that $s_{3}$ and $s_{4}$ equal $a_{1}+a_{4}$ and $a_{2}+a_{3}$ in some order.\n\nIt turns out that either order is possible.\n\nCase 1: $s_{3}=a_{1}+a_{4}$ and $s_{4}=a_{2}+a_{3}$\n\nHere, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=110$.\n\nAdding these three equations gives\n\n$$\n\\left(a_{1}+a_{2}\\right)+\\left(a_{1}+a_{3}\\right)+\\left(a_{2}+a_{3}\\right)=8+104+110\n$$\n\nand so $2 a_{1}+2 a_{2}+2 a_{3}=222$ or $a_{1}+a_{2}+a_{3}=111$.\n\nSince $a_{2}+a_{3}=110$, then $a_{1}=\\left(a_{1}+a_{2}+a_{3}\\right)-\\left(a_{2}+a_{3}\\right)=111-110=1$.\n\nSince $a_{1}=1$ and $a_{1}+a_{2}=8$, then $a_{2}=7$.\n\nSince $a_{1}=1$ and $a_{1}+a_{3}=104$, then $a_{3}=103$.\n\nSince $a_{3}=103$ and $a_{3}+a_{4}=208$, then $a_{4}=105$.\n\nThus, $\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)=(1,7,103,105)$.\n\nCase 2: $s_{3}=a_{2}+a_{3}$ and $s_{4}=a_{1}+a_{4}$\n\nHere, $a_{1}+a_{2}=8$ and $a_{1}+a_{3}=104$ and $a_{2}+a_{3}=106$.\n\nUsing the same process, $a_{1}+a_{2}+a_{3}=109$.\n\nFrom this, we obtain $\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)=(3,5,101,107)$.\n\nTherefore, Kerry's two possible lists are 1,7,103, 105 and 3, 5, 101, 107.\n\n""]","['(1,7,103, 105), (3, 5, 101, 107)']",True,,Tuple, 2337,Geometry,,"Jimmy is baking two large identical triangular cookies, $\triangle A B C$ and $\triangle D E F$. Each cookie is in the shape of an isosceles right-angled triangle. The length of the shorter sides of each of these triangles is $20 \mathrm{~cm}$. He puts the cookies on a rectangular baking tray so that $A, B, D$, and $E$ are at the vertices of the rectangle, as shown. If the distance between parallel sides $A C$ and $D F$ is $4 \mathrm{~cm}$, what is the width $B D$ of the tray? ","['We note that $B D=B C+C D$ and that $B C=20 \\mathrm{~cm}$, so we need to determine $C D$.\n\nWe draw a line from $C$ to $P$ on $F D$ so that $C P$ is perpendicular to $D F$.\n\nSince $A C$ and $D F$ are parallel, then $C P$ is also perpendicular to $A C$.\n\nThe distance between $A C$ and $D F$ is $4 \\mathrm{~cm}$, so $C P=4 \\mathrm{~cm}$.\n\nSince $\\triangle A B C$ is isosceles and right-angled, then $\\angle A C B=45^{\\circ}$.\n\n\n\nThus, $\\angle P C D=180^{\\circ}-\\angle A C B-\\angle P C A=180^{\\circ}-45^{\\circ}-90^{\\circ}=45^{\\circ}$.\n\nSince $\\triangle C P D$ is right-angled at $P$ and $\\angle P C D=45^{\\circ}$, then $\\triangle C P D$ is also an isosceles right-angled triangle.\n\nTherefore, $C D=\\sqrt{2} C P=4 \\sqrt{2} \\mathrm{~cm}$.\n\nFinally, $B D=B C+C D=(20+4 \\sqrt{2}) \\mathrm{cm}$.']",['$(20+4 \\sqrt{2})$'],False,cm,Numerical, 2337,Geometry,,"Jimmy is baking two large identical triangular cookies, $\triangle A B C$ and $\triangle D E F$. Each cookie is in the shape of an isosceles right-angled triangle. The length of the shorter sides of each of these triangles is $20 \mathrm{~cm}$. He puts the cookies on a rectangular baking tray so that $A, B, D$, and $E$ are at the vertices of the rectangle, as shown. If the distance between parallel sides $A C$ and $D F$ is $4 \mathrm{~cm}$, what is the width $B D$ of the tray? ![](https://cdn.mathpix.com/cropped/2023_12_21_ac18afada1a016e79e7bg-1.jpg?height=387&width=480&top_left_y=278&top_left_x=1343)","['We note that $B D=B C+C D$ and that $B C=20 \\mathrm{~cm}$, so we need to determine $C D$.\n\nWe draw a line from $C$ to $P$ on $F D$ so that $C P$ is perpendicular to $D F$.\n\nSince $A C$ and $D F$ are parallel, then $C P$ is also perpendicular to $A C$.\n\nThe distance between $A C$ and $D F$ is $4 \\mathrm{~cm}$, so $C P=4 \\mathrm{~cm}$.\n\nSince $\\triangle A B C$ is isosceles and right-angled, then $\\angle A C B=45^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3c1abeb8d8eb989cf095g-1.jpg?height=388&width=469&top_left_y=1733&top_left_x=1530)\n\nThus, $\\angle P C D=180^{\\circ}-\\angle A C B-\\angle P C A=180^{\\circ}-45^{\\circ}-90^{\\circ}=45^{\\circ}$.\n\nSince $\\triangle C P D$ is right-angled at $P$ and $\\angle P C D=45^{\\circ}$, then $\\triangle C P D$ is also an isosceles right-angled triangle.\n\nTherefore, $C D=\\sqrt{2} C P=4 \\sqrt{2} \\mathrm{~cm}$.\n\nFinally, $B D=B C+C D=(20+4 \\sqrt{2}) \\mathrm{cm}$.']",['$(20+4 \\sqrt{2})$'],False,cm,Numerical, 2338,Algebra,,Determine all values of $x$ for which $\frac{x^{2}+x+4}{2 x+1}=\frac{4}{x}$.,"['Manipulating the given equation and noting that $x \\neq 0$ and $x \\neq-\\frac{1}{2}$ since neither denominator can equal 0 , we obtain\n\n$$\n\\begin{aligned}\n\\frac{x^{2}+x+4}{2 x+1} & =\\frac{4}{x} \\\\\nx\\left(x^{2}+x+4\\right) & =4(2 x+1) \\\\\nx^{3}+x^{2}+4 x & =8 x+4 \\\\\nx^{3}+x^{2}-4 x-4 & =0 \\\\\nx^{2}(x+1)-4(x+1) & =0 \\\\\n(x+1)\\left(x^{2}-4\\right) & =0 \\\\\n(x+1)(x-2)(x+2) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-1$ or $x=2$ or $x=-2$. We can check by substitution that each satisfies the original equation.']","['$-1$,$2$,$-2$']",True,,Numerical, 2339,Number Theory,,"Determine the number of positive divisors of 900, including 1 and 900, that are perfect squares. (A positive divisor of 900 is a positive integer that divides exactly into 900.)","['Since $900=30^{2}$ and $30=2 \\times 3 \\times 5$, then $900=2^{2} 3^{2} 5^{2}$.\n\nThe positive divisors of 900 are those integers of the form $d=2^{a} 3^{b} 5^{c}$, where each of $a, b, c$ is 0,1 or 2 .\n\nFor $d$ to be a perfect square, the exponent on each prime factor in the prime factorization of $d$ must be even.\n\nThus, for $d$ to be a perfect square, each of $a, b, c$ must be 0 or 2 .\n\nThere are two possibilities for each of $a, b, c$ so $2 \\times 2 \\times 2=8$ possibilities for $d$.\n\nThese are $2^{0} 3^{0} 5^{0}=1,2^{2} 3^{0} 5^{0}=4,2^{0} 3^{2} 5^{0}=9,2^{0} 3^{0} 5^{2}=25,2^{2} 3^{2} 5^{0}=36,2^{2} 3^{0} 5^{2}=100$, $2^{0} 3^{2} 5^{2}=225$, and $2^{2} 3^{2} 5^{2}=900$.\n\nThus, 8 of the positive divisors of 900 are perfect squares.', 'The positive divisors of 900 are\n\n$1,2,3,4,5,6,9,10,12,15,18,20,25,30,36,45,50,60,75,90,100,150,180,225,300,450,900$\n\nOf these, $1,4,9,25,36,100,225$, and 900 are perfect squares $\\left(1^{2}, 2^{2}, 3^{2}, 5^{2}, 6^{2}, 10^{2}, 15^{2}, 30^{2}\\right.$, respectively).\n\nThus, 8 of the positive divisors of 900 are perfect squares.']",['8'],False,,Numerical, 2340,Geometry,,"Points $A(k, 3), B(3,1)$ and $C(6, k)$ form an isosceles triangle. If $\angle A B C=\angle A C B$, determine all possible values of $k$.","['In isosceles triangle $A B C, \\angle A B C=\\angle A C B$, so the sides opposite these angles $(A C$ and $A B$, respectively) are equal in length.\n\nSince the vertices of the triangle are $A(k, 3), B(3,1)$ and $C(6, k)$, then we obtain\n\n$$\n\\begin{aligned}\nA C & =A B \\\\\n\\sqrt{(k-6)^{2}+(3-k)^{2}} & =\\sqrt{(k-3)^{2}+(3-1)^{2}} \\\\\n(k-6)^{2}+(3-k)^{2} & =(k-3)^{2}+(3-1)^{2} \\\\\n(k-6)^{2}+(k-3)^{2} & =(k-3)^{2}+2^{2} \\\\\n(k-6)^{2} & =4\n\\end{aligned}\n$$\n\nThus, $k-6=2$ or $k-6=-2$, and so $k=8$ or $k=4$.\n\nWe can check by substitution that each satisfies the original equation.']","['$8$,$4$']",True,,Numerical, 2341,Combinatorics,,"A chemist has three bottles, each containing a mixture of acid and water: - bottle A contains $40 \mathrm{~g}$ of which $10 \%$ is acid, - bottle B contains $50 \mathrm{~g}$ of which $20 \%$ is acid, and - bottle C contains $50 \mathrm{~g}$ of which $30 \%$ is acid. She uses some of the mixture from each of the bottles to create a mixture with mass $60 \mathrm{~g}$ of which $25 \%$ is acid. Then she mixes the remaining contents of the bottles to create a new mixture. What percentage of the new mixture is acid?","['Bottle A contains $40 \\mathrm{~g}$ of which $10 \\%$ is acid.\n\nThus, it contains $0.1 \\times 40=4 \\mathrm{~g}$ of acid and $40-4=36 \\mathrm{~g}$ of water.\n\nBottle B contains $50 \\mathrm{~g}$ of which $20 \\%$ is acid.\n\nThus, it contains $0.2 \\times 50=10 \\mathrm{~g}$ of acid and $50-10=40 \\mathrm{~g}$ of water.\n\nBottle C contains $50 \\mathrm{~g}$ of which $30 \\%$ is acid.\n\nThus, it contains $0.3 \\times 50=15 \\mathrm{~g}$ of acid and $50-15=35 \\mathrm{~g}$ of water.\n\nIn total, the three bottles contain $40+50+50=140 \\mathrm{~g}$, of which $4+10+15=29 \\mathrm{~g}$ is acid and $140-29=111 \\mathrm{~g}$ is water.\n\nThe new mixture has mass $60 \\mathrm{~g}$ of which $25 \\%$ is acid.\n\nThus, it contains $0.25 \\times 60=15 \\mathrm{~g}$ of acid and $60-15=45 \\mathrm{~g}$ of water.\n\nSince the total mass in the three bottles is initially $140 \\mathrm{~g}$ and the new mixture has mass $60 \\mathrm{~g}$, then the remaining contents have mass $140-60=80 \\mathrm{~g}$.\n\nSince the total mass of acid in the three bottles is initially $29 \\mathrm{~g}$ and the acid in the new mixture has mass $15 \\mathrm{~g}$, then the acid in the remaining contents has mass $29-15=14 \\mathrm{~g}$. This remaining mixture is thus $\\frac{14 \\mathrm{~g}}{80 \\mathrm{~g}} \\times 100 \\%=17.5 \\%$ acid.']",['17.5%'],False,,Numerical, 2342,Algebra,,Suppose that $x$ and $y$ are real numbers with $3 x+4 y=10$. Determine the minimum possible value of $x^{2}+16 y^{2}$.,"['Since $3 x+4 y=10$, then $4 y=10-3 x$.\n\nTherefore, when $3 x+4 y=10$,\n\n$$\n\\begin{aligned}\nx^{2}+16 y^{2} & =x^{2}+(4 y)^{2} \\\\\n& =x^{2}+(10-3 x)^{2} \\\\\n& =x^{2}+\\left(9 x^{2}-60 x+100\\right) \\\\\n& =10 x^{2}-60 x+100 \\\\\n& =10\\left(x^{2}-6 x+10\\right) \\\\\n& =10\\left(x^{2}-6 x+9+1\\right) \\\\\n& =10\\left((x-3)^{2}+1\\right) \\\\\n& =10(x-3)^{2}+10\n\\end{aligned}\n$$\n\nSince $(x-3)^{2} \\geq 0$, then the minimum possible value of $10(x-3)^{2}+10$ is $10(0)+10=10$. This occurs when $(x-3)^{2}=0$ or $x=3$.\n\nTherefore, the minimum possible value of $x^{2}+16 y^{2}$ when $3 x+4 y=10$ is 10 .']",['10'],False,,Numerical, 2343,Combinatorics,,"A bag contains 40 balls, each of which is black or gold. Feridun reaches into the bag and randomly removes two balls. Each ball in the bag is equally likely to be removed. If the probability that two gold balls are removed is $\frac{5}{12}$, how many of the 40 balls are gold?","['Suppose that the bag contains $g$ gold balls.\n\nWe assume that Feridun reaches into the bag and removes the two balls one after the other.\n\nThere are 40 possible balls that he could remove first and then 39 balls that he could remove second. In total, there are 40(39) pairs of balls that he could choose in this way.\n\nIf he removes 2 gold balls, then there are $g$ possible balls that he could remove first and then $g-1$ balls that he could remove second. In total, there are $g(g-1)$ pairs of gold balls that he could remove.\n\nWe are told that the probability of removing 2 gold balls is $\\frac{5}{12}$.\n\nSince there are $40(39)$ total pairs of balls that can be chosen and $g(g-1)$ pairs of gold balls that can be chosen in this way, then $\\frac{g(g-1)}{40(39)}=\\frac{5}{12}$ which is equivalent to $g(g-1)=\\frac{5}{12}(40)(39)=650$.\n\n\n\nTherefore, $g^{2}-g-650=0$ or $(g-26)(g+25)=0$, and so $g=26$ or $g=-25$.\n\nSince $g>0$, then $g=26$, so there are 26 gold balls in the bag.', 'Suppose that the bag contains $g$ gold balls.\n\nWe assume that Feridun reaches into the bag and removes the two balls together.\n\nSince there are 40 balls in the bag, there are $\\left(\\begin{array}{c}40 \\\\ 2\\end{array}\\right)$ pairs of balls that he could choose in this way.\n\nSince there are $g$ gold balls in the bag, then there are $\\left(\\begin{array}{l}g \\\\ 2\\end{array}\\right)$ pairs of gold balls that he could choose in this way.\n\nWe are told that the probability of removing 2 gold balls is $\\frac{5}{12}$.\n\nSince there are $\\left(\\begin{array}{c}40 \\\\ 2\\end{array}\\right)$ pairs in total that can be chosen and $\\left(\\begin{array}{l}g \\\\ 2\\end{array}\\right)$ pairs of gold balls that can be chosen in this way, then $\\frac{\\left(\\begin{array}{l}g \\\\ 2\\end{array}\\right)}{\\left(\\begin{array}{c}40 \\\\ 2\\end{array}\\right)}=\\frac{5}{12}$ which is equivalent to $\\left(\\begin{array}{l}g \\\\ 2\\end{array}\\right)=\\frac{5}{12}\\left(\\begin{array}{c}40 \\\\ 2\\end{array}\\right)$.\n\nSince $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)=\\frac{n(n-1)}{2}$, then this equation is equivalent to $\\frac{g(g-1)}{2}=\\frac{5}{12} \\frac{40(39)}{2}=325$.\n\nTherefore, $g(g-1)=650$ or $g^{2}-g-650=0$ or $(g-26)(g+25)=0$, and so $g=26$ or $g=-25$.\n\nSince $g>0$, then $g=26$, so there are 26 gold balls in the bag.']",['26'],False,,Numerical, 2344,Algebra,,"The geometric sequence with $n$ terms $t_{1}, t_{2}, \ldots, t_{n-1}, t_{n}$ has $t_{1} t_{n}=3$. Also, the product of all $n$ terms equals 59049 (that is, $t_{1} t_{2} \cdots t_{n-1} t_{n}=59049$ ). Determine the value of $n$. (A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)","['Suppose that the first term in the geometric sequence is $t_{1}=a$ and the common ratio in the sequence is $r$.\n\nThen the sequence, which has $n$ terms, is $a, a r, a r^{2}, a r^{3}, \\ldots, a r^{n-1}$.\n\nIn general, the $k$ th term is $t_{k}=a r^{k-1}$; in particular, the $n$th term is $t_{n}=a r^{n-1}$.\n\nSince $t_{1} t_{n}=3$, then $a \\cdot a r^{n-1}=3$ or $a^{2} r^{n-1}=3$.\n\nSince $t_{1} t_{2} \\cdots t_{n-1} t_{n}=59049$, then\n\n$$\n\\begin{aligned}\n(a)(a r) \\cdots\\left(a r^{n-2}\\right)\\left(a r^{n-1}\\right) & =59049 \\\\\na^{n} r r^{2} \\cdots r^{n-2} r^{n-1} & =59049 \\\\\na^{n} r^{1+2+\\cdots+(n-2)+(n-1)} & =59049 \\\\\na^{n} r^{\\frac{1}{2}(n-1)(n)} & =59049\n\\end{aligned}\n$$\n\n$$\na^{n} r r^{2} \\cdots r^{n-2} r^{n-1}=59049 \\quad \\text { (since there are } n \\text { factors of } a \\text { on the left side) }\n$$\n\nsince $1+2+\\cdots+(n-2)+(n-1)=\\frac{1}{2}(n-1)(n)$.\n\nSince $a^{2} r^{n-1}=3$, then $\\left(a^{2} r^{n-1}\\right)^{n}=3^{n}$ or $a^{2 n} r^{(n-1)(n)}=3^{n}$.\n\nSince $a^{n} r^{\\frac{1}{2}(n-1)(n)}=59049$, then $\\left(a^{n} r^{\\frac{1}{2}(n-1)(n)}\\right)^{2}=59049^{2}$ or $a^{2 n} r^{(n-1)(n)}=59049^{2}$.\n\nSince the left sides of these equations are the same, then $3^{n}=59049^{2}$.\n\nNow\n\n$$\n59049=3(19683)=3^{2}(6561)=3^{3}(2187)=3^{4}(729)=3^{5}(243)=3^{6}(81)=3^{6} 3^{4}=3^{10}\n$$\n\nSince $59049=3^{10}$, then $59049^{2}=3^{20}$ and so $3^{n}=3^{20}$, which gives $n=20$.']",['20'],False,,Numerical, 2345,Algebra,,"If $\frac{(x-2013)(y-2014)}{(x-2013)^{2}+(y-2014)^{2}}=-\frac{1}{2}$, what is the value of $x+y$ ?","['Let $a=x-2013$ and let $b=y-2014$.\n\nThe given equation becomes $\\frac{a b}{a^{2}+b^{2}}=-\\frac{1}{2}$, which is equivalent to $2 a b=-a^{2}-b^{2}$ and $a^{2}+2 a b+b^{2}=0$.\n\nThis is equivalent to $(a+b)^{2}=0$ which is equivalent to $a+b=0$.\n\nSince $a=x-2013$ and $b=y-2014$, then $x-2013+y-2014=0$ or $x+y=4027$.']",['4027'],False,,Numerical, 2346,Algebra,,"Determine all real numbers $x$ for which $$ \left(\log _{10} x\right)^{\log _{10}\left(\log _{10} x\right)}=10000 $$","['Let $a=\\log _{10} x$.\n\nThen $\\left(\\log _{10} x\\right)^{\\log _{10}\\left(\\log _{10} x\\right)}=10000$ becomes $a^{\\log _{10} a}=10^{4}$.\n\nTaking the base 10 logarithm of both sides and using the fact that $\\log _{10}\\left(a^{b}\\right)=b \\log _{10} a$, we obtain $\\left(\\log _{10} a\\right)\\left(\\log _{10} a\\right)=4$ or $\\left(\\log _{10} a\\right)^{2}=4$.\n\nTherefore, $\\log _{10} a= \\pm 2$ and so $\\log _{10}\\left(\\log _{10} x\\right)= \\pm 2$.\n\nIf $\\log _{10}\\left(\\log _{10} x\\right)=2$, then $\\log _{10} x=10^{2}=100$ and so $x=10^{100}$.\n\nIf $\\log _{10}\\left(\\log _{10} x\\right)=-2$, then $\\log _{10} x=10^{-2}=\\frac{1}{100}$ and so $x=10^{1 / 100}$.\n\nTherefore, $x=10^{100}$ or $x=10^{1 / 100}$.\n\nWe check these answers in the original equation.\n\nIf $x=10^{100}$, then $\\log _{10} x=100$.\n\nThus, $\\left(\\log _{10} x\\right)^{\\log _{10}\\left(\\log _{10} x\\right)}=100^{\\log _{10} 100}=100^{2}=10000$.\n\nIf $x=10^{1 / 100}$, then $\\log _{10} x=1 / 100=10^{-2}$.\n\nThus, $\\left(\\log _{10} x\\right)^{\\log _{10}\\left(\\log _{10} x\\right)}=\\left(10^{-2}\\right)^{\\log _{10}\\left(10^{-2}\\right)}=\\left(10^{-2}\\right)^{-2}=10^{4}=10000$.']","['$10^{100}$,$10^{1 / 100}$']",True,,Numerical, 2347,Geometry,,"In the diagram, $\angle A C B=\angle A D E=90^{\circ}$. If $A B=75, B C=21, A D=20$, and $C E=47$, determine the exact length of $B D$. ","['We use the cosine law in $\\triangle A B D$ to determine the length of $B D$ :\n\n$$\nB D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D)\n$$\n\nWe are given that $A B=75$ and $A D=20$, so we need to determine $\\cos (\\angle B A D)$.\n\nNow\n\n$$\n\\begin{aligned}\n\\cos (\\angle B A D) & =\\cos (\\angle B A C+\\angle E A D) \\\\\n& =\\cos (\\angle B A C) \\cos (\\angle E A D)-\\sin (\\angle B A C) \\sin (\\angle E A D) \\\\\n& =\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}\n\\end{aligned}\n$$\n\nsince $\\triangle A B C$ and $\\triangle A D E$ are right-angled.\n\nSince $A B=75$ and $B C=21$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}-B C^{2}}=\\sqrt{75^{2}-21^{2}}=\\sqrt{5625-441}=\\sqrt{5184}=72\n$$\n\nsince $A C>0$.\n\nSince $A C=72$ and $C E=47$, then $A E=A C-C E=25$.\n\nSince $A E=25$ and $A D=20$, then by the Pythagorean Theorem,\n\n$$\nE D=\\sqrt{A E^{2}-A D^{2}}=\\sqrt{25^{2}-20^{2}}=\\sqrt{625-400}=\\sqrt{225}=15\n$$\n\nsince $E D>0$.\n\nTherefore,\n\n$$\n\\cos (\\angle B A D)=\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}=\\frac{72}{75} \\frac{20}{25}-\\frac{21}{75} \\frac{15}{25}=\\frac{1440-315}{75(25)}=\\frac{1125}{75(25)}=\\frac{45}{75}=\\frac{3}{5}\n$$\n\n\n\nFinally,\n\n$$\n\\begin{aligned}\nB D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D) \\\\\n& =75^{2}+20^{2}-2(75)(20)\\left(\\frac{3}{5}\\right) \\\\\n& =5625+400-1800 \\\\\n& =4225\n\\end{aligned}\n$$\n\nSince $B D>0$, then $B D=\\sqrt{4225}=65$, as required.']",['65'],False,,Numerical, 2347,Geometry,,"In the diagram, $\angle A C B=\angle A D E=90^{\circ}$. If $A B=75, B C=21, A D=20$, and $C E=47$, determine the exact length of $B D$. ![](https://cdn.mathpix.com/cropped/2023_12_21_09b264570ad909ff6decg-1.jpg?height=380&width=379&top_left_y=588&top_left_x=1250)","['We use the cosine law in $\\triangle A B D$ to determine the length of $B D$ :\n\n$$\nB D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D)\n$$\n\nWe are given that $A B=75$ and $A D=20$, so we need to determine $\\cos (\\angle B A D)$.\n\nNow\n\n$$\n\\begin{aligned}\n\\cos (\\angle B A D) & =\\cos (\\angle B A C+\\angle E A D) \\\\\n& =\\cos (\\angle B A C) \\cos (\\angle E A D)-\\sin (\\angle B A C) \\sin (\\angle E A D) \\\\\n& =\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}\n\\end{aligned}\n$$\n\nsince $\\triangle A B C$ and $\\triangle A D E$ are right-angled.\n\nSince $A B=75$ and $B C=21$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}-B C^{2}}=\\sqrt{75^{2}-21^{2}}=\\sqrt{5625-441}=\\sqrt{5184}=72\n$$\n\nsince $A C>0$.\n\nSince $A C=72$ and $C E=47$, then $A E=A C-C E=25$.\n\nSince $A E=25$ and $A D=20$, then by the Pythagorean Theorem,\n\n$$\nE D=\\sqrt{A E^{2}-A D^{2}}=\\sqrt{25^{2}-20^{2}}=\\sqrt{625-400}=\\sqrt{225}=15\n$$\n\nsince $E D>0$.\n\nTherefore,\n\n$$\n\\cos (\\angle B A D)=\\frac{A C}{A B} \\frac{A D}{A E}-\\frac{B C}{A B} \\frac{E D}{A E}=\\frac{72}{75} \\frac{20}{25}-\\frac{21}{75} \\frac{15}{25}=\\frac{1440-315}{75(25)}=\\frac{1125}{75(25)}=\\frac{45}{75}=\\frac{3}{5}\n$$\n\n\n\nFinally,\n\n$$\n\\begin{aligned}\nB D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \\cos (\\angle B A D) \\\\\n& =75^{2}+20^{2}-2(75)(20)\\left(\\frac{3}{5}\\right) \\\\\n& =5625+400-1800 \\\\\n& =4225\n\\end{aligned}\n$$\n\nSince $B D>0$, then $B D=\\sqrt{4225}=65$, as required.']",['65'],False,,Numerical, 2348,Geometry,,"In the diagram, $C$ lies on $B D$. Also, $\triangle A B C$ and $\triangle E C D$ are equilateral triangles. If $M$ is the midpoint of $B E$ and $N$ is the midpoint of $A D$, prove that $\triangle M N C$ is equilateral. ![](https://cdn.mathpix.com/cropped/2023_12_21_09b264570ad909ff6decg-1.jpg?height=323&width=510&top_left_y=1020&top_left_x=1203)","['Consider $\\triangle B C E$ and $\\triangle A C D$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_12ef85aa7fb125863bf7g-1.jpg?height=311&width=520&top_left_y=728&top_left_x=908)\n\nSince $\\triangle A B C$ is equilateral, then $B C=A C$.\n\nSince $\\triangle E C D$ is equilateral, then $C E=C D$.\n\nSince $B C D$ is a straight line and $\\angle E C D=60^{\\circ}$, then $\\angle B C E=180^{\\circ}-\\angle E C D=120^{\\circ}$.\n\nSince $B C D$ is a straight line and $\\angle B C A=60^{\\circ}$, then $\\angle A C D=180^{\\circ}-\\angle B C A=120^{\\circ}$.\n\nTherefore, $\\triangle B C E$ is congruent to $\\triangle A C D$ (""side-angle-side"").\n\nSince $\\triangle B C E$ and $\\triangle A C D$ are congruent and $C M$ and $C N$ are line segments drawn from the corresponding vertex ( $C$ in both triangles) to the midpoint of the opposite side, then $C M=C N$.\n\nSince $\\angle E C D=60^{\\circ}$, then $\\triangle A C D$ can be obtained by rotating $\\triangle B C E$ through an angle of $60^{\\circ}$ clockwise about $C$.\n\nThis means that after this $60^{\\circ}$ rotation, $C M$ coincides with $C N$.\n\nIn other words, $\\angle M C N=60^{\\circ}$.\n\nBut since $C M=C N$ and $\\angle M C N=60^{\\circ}$, then\n\n$$\n\\angle C M N=\\angle C N M=\\frac{1}{2}\\left(180^{\\circ}-\\angle M C N\\right)=60^{\\circ}\n$$\n\nTherefore, $\\triangle M N C$ is equilateral, as required.', 'We prove that $\\triangle M N C$ is equilateral by introducing a coordinate system.\n\nSuppose that $C$ is at the origin $(0,0)$ with $B C D$ along the $x$-axis, with $B$ having coordinates $(-4 b, 0)$ and $D$ having coordinates $(4 d, 0)$ for some real numbers $b, d>0$.\n\nDrop a perpendicular from $E$ to $P$ on $C D$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_70b24695c6426f20ba6bg-1.jpg?height=445&width=680&top_left_y=455&top_left_x=820)\n\nSince $\\triangle E C D$ is equilateral, then $P$ is the midpoint of $C D$.\n\nSince $C$ has coordinates $(0,0)$ and $D$ has coordinates $(4 d, 0)$, then the coordinates of $P$ are $(2 d, 0)$.\n\nSince $\\triangle E C D$ is equilateral, then $\\angle E C D=60^{\\circ}$ and so $\\triangle E P C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle and so $E P=\\sqrt{3} C P=2 \\sqrt{3} d$.\n\nTherefore, the coordinates of $E$ are $(2 d, 2 \\sqrt{3} d)$.\n\nIn a similar way, we can show that the coordinates of $A$ are $(-2 b, 2 \\sqrt{3} b)$.\n\nNow $M$ is the midpoint of $B(-4 b, 0)$ and $E(2 d, 2 \\sqrt{3} d)$, so the coordinates of $M$ are $\\left(\\frac{1}{2}(-4 b+2 d), \\frac{1}{2}(0+2 \\sqrt{3} d)\\right)$ or $(-2 b+d, \\sqrt{3} d)$.\n\nAlso, $N$ is the midpoint of $A(-2 b, 2 \\sqrt{3} b)$ and $D(4 d, 0)$, so the coordinates of $N$ are $\\left(\\frac{1}{2}(-2 b+4 d), \\frac{1}{2}(2 \\sqrt{3} b+0)\\right)$ or $(-b+2 d, \\sqrt{3} b)$.\n\nTo show that $\\triangle M N C$ is equilateral, we show that $C M=C N=M N$ or equivalently that $C M^{2}=C N^{2}=M N^{2}$ :\n\n$$\n\\begin{aligned}\nC M^{2} & =(-2 b+d-0)^{2}+(\\sqrt{3} d-0)^{2} \\\\\n& =(-2 b+d)^{2}+(\\sqrt{3} d)^{2} \\\\\n& =4 b^{2}-4 b d+d^{2}+3 d^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nC N^{2} & =(-b+2 d-0)^{2}+(\\sqrt{3} b-0)^{2} \\\\\n& =(-b+2 d)^{2}+(\\sqrt{3} b)^{2} \\\\\n& =b^{2}-4 b d+4 d^{2}+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nM N^{2} & =((-2 b+d)-(-b+2 d))^{2}+(\\sqrt{3} d-\\sqrt{3} b)^{2} \\\\\n& =(-b-d)^{2}+3(d-b)^{2} \\\\\n& =b^{2}+2 b d+d^{2}+3 d^{2}-6 b d+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2}\n\\end{aligned}\n$$\n\nTherefore, $C M^{2}=C N^{2}=M N^{2}$ and so $\\triangle M N C$ is equilateral, as required.\n\n\n\n#']",['证明题，略'],True,,Need_human_evaluate, 2348,Geometry,,"In the diagram, $C$ lies on $B D$. Also, $\triangle A B C$ and $\triangle E C D$ are equilateral triangles. If $M$ is the midpoint of $B E$ and $N$ is the midpoint of $A D$, prove that $\triangle M N C$ is equilateral. ","['Consider $\\triangle B C E$ and $\\triangle A C D$.\n\n\n\nSince $\\triangle A B C$ is equilateral, then $B C=A C$.\n\nSince $\\triangle E C D$ is equilateral, then $C E=C D$.\n\nSince $B C D$ is a straight line and $\\angle E C D=60^{\\circ}$, then $\\angle B C E=180^{\\circ}-\\angle E C D=120^{\\circ}$.\n\nSince $B C D$ is a straight line and $\\angle B C A=60^{\\circ}$, then $\\angle A C D=180^{\\circ}-\\angle B C A=120^{\\circ}$.\n\nTherefore, $\\triangle B C E$ is congruent to $\\triangle A C D$ (""side-angle-side"").\n\nSince $\\triangle B C E$ and $\\triangle A C D$ are congruent and $C M$ and $C N$ are line segments drawn from the corresponding vertex ( $C$ in both triangles) to the midpoint of the opposite side, then $C M=C N$.\n\nSince $\\angle E C D=60^{\\circ}$, then $\\triangle A C D$ can be obtained by rotating $\\triangle B C E$ through an angle of $60^{\\circ}$ clockwise about $C$.\n\nThis means that after this $60^{\\circ}$ rotation, $C M$ coincides with $C N$.\n\nIn other words, $\\angle M C N=60^{\\circ}$.\n\nBut since $C M=C N$ and $\\angle M C N=60^{\\circ}$, then\n\n$$\n\\angle C M N=\\angle C N M=\\frac{1}{2}\\left(180^{\\circ}-\\angle M C N\\right)=60^{\\circ}\n$$\n\nTherefore, $\\triangle M N C$ is equilateral, as required.', 'We prove that $\\triangle M N C$ is equilateral by introducing a coordinate system.\n\nSuppose that $C$ is at the origin $(0,0)$ with $B C D$ along the $x$-axis, with $B$ having coordinates $(-4 b, 0)$ and $D$ having coordinates $(4 d, 0)$ for some real numbers $b, d>0$.\n\nDrop a perpendicular from $E$ to $P$ on $C D$.\n\n\n\nSince $\\triangle E C D$ is equilateral, then $P$ is the midpoint of $C D$.\n\nSince $C$ has coordinates $(0,0)$ and $D$ has coordinates $(4 d, 0)$, then the coordinates of $P$ are $(2 d, 0)$.\n\nSince $\\triangle E C D$ is equilateral, then $\\angle E C D=60^{\\circ}$ and so $\\triangle E P C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle and so $E P=\\sqrt{3} C P=2 \\sqrt{3} d$.\n\nTherefore, the coordinates of $E$ are $(2 d, 2 \\sqrt{3} d)$.\n\nIn a similar way, we can show that the coordinates of $A$ are $(-2 b, 2 \\sqrt{3} b)$.\n\nNow $M$ is the midpoint of $B(-4 b, 0)$ and $E(2 d, 2 \\sqrt{3} d)$, so the coordinates of $M$ are $\\left(\\frac{1}{2}(-4 b+2 d), \\frac{1}{2}(0+2 \\sqrt{3} d)\\right)$ or $(-2 b+d, \\sqrt{3} d)$.\n\nAlso, $N$ is the midpoint of $A(-2 b, 2 \\sqrt{3} b)$ and $D(4 d, 0)$, so the coordinates of $N$ are $\\left(\\frac{1}{2}(-2 b+4 d), \\frac{1}{2}(2 \\sqrt{3} b+0)\\right)$ or $(-b+2 d, \\sqrt{3} b)$.\n\nTo show that $\\triangle M N C$ is equilateral, we show that $C M=C N=M N$ or equivalently that $C M^{2}=C N^{2}=M N^{2}$ :\n\n$$\n\\begin{aligned}\nC M^{2} & =(-2 b+d-0)^{2}+(\\sqrt{3} d-0)^{2} \\\\\n& =(-2 b+d)^{2}+(\\sqrt{3} d)^{2} \\\\\n& =4 b^{2}-4 b d+d^{2}+3 d^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nC N^{2} & =(-b+2 d-0)^{2}+(\\sqrt{3} b-0)^{2} \\\\\n& =(-b+2 d)^{2}+(\\sqrt{3} b)^{2} \\\\\n& =b^{2}-4 b d+4 d^{2}+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nM N^{2} & =((-2 b+d)-(-b+2 d))^{2}+(\\sqrt{3} d-\\sqrt{3} b)^{2} \\\\\n& =(-b-d)^{2}+3(d-b)^{2} \\\\\n& =b^{2}+2 b d+d^{2}+3 d^{2}-6 b d+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2}\n\\end{aligned}\n$$\n\nTherefore, $C M^{2}=C N^{2}=M N^{2}$ and so $\\triangle M N C$ is equilateral, as required.\n\n\n\n#']",,True,,, 2349,Algebra,,"Without using a calculator, determine positive integers $m$ and $n$ for which $$ \sin ^{6} 1^{\circ}+\sin ^{6} 2^{\circ}+\sin ^{6} 3^{\circ}+\cdots+\sin ^{6} 87^{\circ}+\sin ^{6} 88^{\circ}+\sin ^{6} 89^{\circ}=\frac{m}{n} $$ (The sum on the left side of the equation consists of 89 terms of the form $\sin ^{6} x^{\circ}$, where $x$ takes each positive integer value from 1 to 89.)","['Let $S=\\sin ^{6} 1^{\\circ}+\\sin ^{6} 2^{\\circ}+\\sin ^{6} 3^{\\circ}+\\cdots+\\sin ^{6} 87^{\\circ}+\\sin ^{6} 88^{\\circ}+\\sin ^{6} 89^{\\circ}$.\n\nSince $\\sin \\theta=\\cos \\left(90^{\\circ}-\\theta\\right)$, then $\\sin ^{6} \\theta=\\cos ^{6}\\left(90^{\\circ}-\\theta\\right)$, and so\n\n$$\n\\begin{aligned}\nS= & \\sin ^{6} 1^{\\circ}+\\sin ^{6} 2^{\\circ}+\\cdots+\\sin ^{6} 44^{\\circ}+\\sin ^{6} 45^{\\circ} \\\\\n& \\quad+\\cos ^{6}\\left(90^{\\circ}-46^{\\circ}\\right)+\\cos ^{6}\\left(90^{\\circ}-47^{\\circ}\\right)+\\cdots+\\cos ^{6}\\left(90^{\\circ}-89^{\\circ}\\right) \\\\\n= & \\sin ^{6} 1^{\\circ}+\\sin ^{6} 2^{\\circ}+\\cdots+\\sin ^{6} 44^{\\circ}+\\sin ^{6} 45^{\\circ}+\\cos ^{6} 44^{\\circ}+\\cos ^{6} 43^{\\circ}+\\cdots+\\cos ^{6} 1^{\\circ} \\\\\n= & \\left(\\sin ^{6} 1^{\\circ}+\\cos ^{6} 1^{\\circ}\\right)+\\left(\\sin ^{6} 2^{\\circ}+\\cos ^{6} 2^{\\circ}\\right)+\\cdots+\\left(\\sin ^{6} 44^{\\circ}+\\cos ^{6} 44^{\\circ}\\right)+\\sin ^{6} 45^{\\circ}\n\\end{aligned}\n$$\n\nSince $\\sin 45^{\\circ}=\\frac{1}{\\sqrt{2}}$, then $\\sin ^{6} 45^{\\circ}=\\frac{1}{2^{3}}=\\frac{1}{8}$.\n\nAlso, since\n\n$$\nx^{3}+y^{3}=(x+y)\\left(x^{2}-x y+y^{2}\\right)=(x+y)\\left((x+y)^{2}-3 x y\\right)\n$$\n\nthen substituting $x=\\sin ^{2} \\theta$ and $y=\\cos ^{2} \\theta$, we obtain\n\n$$\n\\begin{aligned}\nx^{3}+y^{3} & =(x+y)\\left((x+y)^{2}-3 x y\\right) \\\\\n\\sin ^{6} \\theta+\\cos ^{6} \\theta & =\\left(\\sin ^{2} \\theta+\\cos ^{2} \\theta\\right)\\left(\\left(\\sin ^{2} \\theta+\\cos ^{2} \\theta\\right)^{2}-3 \\sin ^{2} \\theta \\cos ^{2} \\theta\\right) \\\\\n\\sin ^{6} \\theta+\\cos ^{6} \\theta & =1\\left(1-3 \\sin ^{2} \\theta \\cos ^{2} \\theta\\right)\n\\end{aligned}\n$$\n\nsince $\\sin ^{2} \\theta+\\cos ^{2} \\theta=1$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nS & =\\left(\\sin ^{6} 1^{\\circ}+\\cos ^{6} 1^{\\circ}\\right)+\\left(\\sin ^{6} 2^{\\circ}+\\cos ^{6} 2^{\\circ}\\right)+\\cdots+\\left(\\sin ^{6} 44^{\\circ}+\\cos ^{6} 44^{\\circ}\\right)+\\sin ^{6} 45^{\\circ} \\\\\n& =\\left(1-3 \\sin ^{2} 1^{\\circ} \\cos ^{2} 1^{\\circ}\\right)+\\left(1-3 \\sin ^{2} 2^{\\circ} \\cos ^{2} 2^{\\circ}\\right)+\\cdots+\\left(1-3 \\sin ^{2} 44^{\\circ} \\cos ^{2} 44^{\\circ}\\right)+\\frac{1}{8} \\\\\n& =44-\\left(3 \\sin ^{2} 1^{\\circ} \\cos ^{2} 1^{\\circ}+3 \\sin ^{2} 2^{\\circ} \\cos ^{2} 2^{\\circ}+\\cdots+3 \\sin ^{2} 44^{\\circ} \\cos ^{2} 44^{\\circ}\\right)+\\frac{1}{8} \\\\\n& =\\frac{353}{8}-\\frac{3}{4}\\left(4 \\sin ^{2} 1^{\\circ} \\cos ^{2} 1^{\\circ}+4 \\sin ^{2} 2^{\\circ} \\cos ^{2} 2^{\\circ}+\\cdots+4 \\sin ^{2} 44^{\\circ} \\cos ^{2} 44^{\\circ}\\right)\n\\end{aligned}\n$$\n\nSince $\\sin 2 \\theta=2 \\sin \\theta \\cos \\theta$, then $4 \\sin ^{2} \\theta \\cos ^{2} \\theta=\\sin ^{2} 2 \\theta$, which gives\n\n$$\n\\begin{aligned}\nS & =\\frac{353}{8}-\\frac{3}{4}\\left(4 \\sin ^{2} 1^{\\circ} \\cos ^{2} 1^{\\circ}+4 \\sin ^{2} 2^{\\circ} \\cos ^{2} 2^{\\circ}+\\cdots+4 \\sin ^{2} 44^{\\circ} \\cos ^{2} 44^{\\circ}\\right) \\\\\n& =\\frac{353}{8}-\\frac{3}{4}\\left(\\sin ^{2} 2^{\\circ}+\\sin ^{2} 4^{\\circ}+\\cdots+\\sin ^{2} 88^{\\circ}\\right) \\\\\n& =\\frac{353}{8}-\\frac{3}{4}\\left(\\sin ^{2} 2^{\\circ}+\\sin ^{2} 4^{\\circ}+\\cdots+\\sin ^{2} 44^{\\circ}+\\sin ^{2} 46^{\\circ}+\\cdots+\\sin ^{2} 86^{\\circ}+\\sin ^{2} 88^{\\circ}\\right) \\\\\n& =\\frac{353}{8}-\\frac{3}{4}\\left(\\sin ^{2} 2^{\\circ}+\\sin ^{2} 4^{\\circ}+\\cdots+\\sin ^{2} 44^{\\circ}+\\right. \\\\\n& \\left.\\cos ^{2}\\left(90^{\\circ}-46^{\\circ}\\right)+\\cdots+\\cos ^{2}\\left(90^{\\circ}-86^{\\circ}\\right)+\\cos ^{2}\\left(90^{\\circ}-88^{\\circ}\\right)\\right) \\\\\n& =\\frac{353}{8}-\\frac{3}{4}\\left(\\sin ^{2} 2^{\\circ}+\\sin ^{2} 4^{\\circ}+\\cdots+\\sin ^{2} 44^{\\circ}+\\cos ^{2} 44^{\\circ}+\\cdots+\\cos ^{2} 4^{\\circ}+\\cos ^{2} 2^{\\circ}\\right) \\\\\n& =\\frac{353}{8}-\\frac{3}{4}\\left(\\left(\\sin ^{2} 2^{\\circ}+\\cos ^{2} 2^{\\circ}\\right)+\\left(\\sin ^{2} 4^{\\circ}+\\cos ^{2} 4^{\\circ}\\right)+\\cdots+\\left(\\sin ^{2} 44^{\\circ}+\\cos ^{2} 44^{\\circ}\\right)\\right) \\\\\n& =\\frac{353}{8}-\\frac{3}{4}(22) \\quad\\left(\\operatorname{since} \\sin ^{2} \\theta+\\cos ^{2} \\theta=1\\right) \\\\\n& =\\frac{353}{8}-\\frac{132}{8} \\\\\n& =\\frac{221}{8}\n\\end{aligned}\n$$\n\nTherefore, since $S=\\frac{m}{n}$, then $m=221$ and $n=8$ satisfy the required equation.']","['$221,$8$']",True,,Numerical, 2350,Number Theory,,"Let $f(n)$ be the number of positive integers that have exactly $n$ digits and whose digits have a sum of 5. Determine, with proof, how many of the 2014 integers $f(1), f(2), \ldots, f(2014)$ have a units digit of 1 .","['First, we prove that $f(n)=\\frac{n(n+1)(n+2)(n+3)}{24}$ in two different ways.\n\nMethod 1\n\nIf an $n$-digit integer has digits with a sum of 5 , then there are several possibilities for the combination of non-zero digits used:\n\n$$\n5 \\quad 4,1 \\quad 3,2 \\quad 3,1,1 \\quad 2,2,1 \\quad 2,1,1,1 \\quad 1,1,1,1,1\n$$\n\nWe count the number of possible integers in each case by determining the number of arrangements of the non-zero digits; we call the number of ways of doing this $a$. (For example, the digits 4 and 1 can be arranged as 41 or 14 .) We then place the leftmost digit in such an arrangement as the leftmost digit of the $n$-digit integer (which must be nonzero) and choose the positions for the remaining non-zero digits among the remaining $n-1$ positions; we call the number of ways of doing this $b$. (For example, for the arrangement 14 , the digit 1 is in the leftmost position and the digit 4 can be in any of the remaining $n-1$ positions.) We fill the rest of the positions with 0s. The number of possible integers in each case will be $a b$, since this method will create all such integers and for each of the $a$ arrangements of the non-zero digits, there will be $b$ ways of arranging the digits after the first one. We make a chart to summarize the cases, expanding each total and writing it as a fraction with denominator 24 :\n\n| Case | $a$ | $b$ | $a b$ (expanded) |\n| :---: | :---: | :---: | :--- |\n| 5 | 1 | 1 | $1=\\frac{24}{24}$ |\n| 4,1 | 2 | $(n-1)$ | $2(n-1)=\\frac{48 n-48}{24}$ |\n| 3,2 | 2 | $(n-1)$ | $2(n-1)=\\frac{48 n-48}{24}$ |\n| $3,1,1$ | 3 | $(<>$)$ | $3(<>$)=\\frac{36 n^{2}-108 n+72}{24}$ |\n| $2,2,1$ | 3 | $(<>$)$ | $3(<>$)=\\frac{36 n^{2}-108 n+72}{24}$ |\n| $2,1,1,1$ | 4 | $(<>$)$ | $4(<>$)=\\frac{16 n^{3}-96 n^{2}+176 n-96}{24}$ |\n| $1,1,1,1,1$ | 1 | $(<>$)$ | $(<>$)=\\frac{n^{4}-10 n^{3}+35 n^{2}-50 n+24}{24}$ |\n\n(Note that in the second and third cases we need $n \\geq 2$, in the fourth and fifth cases we need $n \\geq 3$, in the sixth case we need $n \\geq 4$, and the seventh case we need $n \\geq 5$. In each case, though, the given formula works for smaller positive values of $n$ since it is equal to 0 in each case. Note also that we say $b=1$ in the first case since there is exactly 1 way of placing $0 \\mathrm{~s}$ in all of the remaining $n-1$ positions.)\n\n$f(n)$ is then the sum of the expressions in the last column of this table, and so\n\n$$\nf(n)=\\frac{n^{4}+6 n^{3}+11 n^{2}+6 n}{24}=\\frac{n(n+1)(n+2)(n+3)}{24}\n$$\n\nas required.\n\nMethod 2\n\nFirst, we create a correspondence between each integer with $n$ digits and whose digits have\n\n\n\na sum of 5 and an arrangement of five 1 s and $(n-1)$ Xs that begins with a 1 .\n\nWe can then count these integers by counting the arrangements.\n\nStarting with such an integer, we write down an arrangement of the above type using the following rule:\n\nThe number of 1 s to the left of the first $\\mathrm{X}$ is the first digit of the number, the number of 1 s between the first $\\mathrm{X}$ and second $\\mathrm{X}$ is the second digit of the number, and so on, with the number of 1 s to the right of the $(n-1)$ st $\\mathrm{X}$ representing the $n$th digit of the number.\n\nFor example, the integer 1010020001 would correspond to 1XX1XXX11XXXX1.\n\nIn this way, each such integer gives an arrangement of the above type.\n\nSimilarly, each arrangement of this type can be associated back to a unique integer with the required properties by counting the number of 1 s before the first $\\mathrm{X}$ and writing this down as the leftmost digit, counting the number of 1 s between the first and second Xs and writing this down as the second digit, and so on. Since a total of five 1s are used, then each arrangement corresponds with an integer with $n$ digits whose digits have a sum of 5 . Therefore, there is a one-to-one correspondence between the integers and arrangements with the desired properties.\n\nThus, $f(n)$, which equals the number of such integers, also equals the number of such arrangements.\n\nTo count the number of such arrangements, we note that there are four 1 s and $n-1 \\mathrm{Xs}$ to arrange in the final $4+(n-1)=n+3$ positions, since the first position is occupied by a 1 .\n\nThere are $\\left(\\begin{array}{c}n+3 \\\\ 4\\end{array}\\right)$ ways to choose the positions of the remaining four 1s, and so $\\left(\\begin{array}{c}n+3 \\\\ 4\\end{array}\\right)$ arrangements.\n\nThus, $f(n)=\\left(\\begin{array}{c}n+3 \\\\ 4\\end{array}\\right)=\\frac{(n+3) !}{4 !(n-1) !}=\\frac{(n+3)(n+2)(n+1)(n)}{4 !}=\\frac{n(n+1)(n+2)(n+3)}{24}$.\n\nNext, we need to determine the positive integers $n$ between 1 and 2014, inclusive, for which the units digit of $f(n)$ is 1 .\n\nNow $f(n)=\\frac{n(n+1)(n+2)(n+3)}{24}$ is an integer for all positive integers $n$, since it is counting the number of things with a certain property.\n\nIf the units digit of $n$ is 0 or 5 , then $n$ is a multiple of 5 .\n\nIf the units digit of $n$ is 2 or 7 , then $n+3$ is a multiple of 5 .\n\nIf the units digit of $n$ is 3 or 8 , then $n+2$ is a multiple of 5 .\n\nIf the units digit of $n$ is 4 or 9 , then $n+1$ is a multiple of 5 .\n\nThus, if the units digit of $n$ is $0,2,3,4,5,7,8$, or 9 , then $n(n+1)(n+2)(n+3)$\n\nis a multiple of 5 and so $f(n)=\\frac{n(n+1)(n+2)(n+3)}{24}$ is a multiple of 5 , since the denominator contains no factors of 5 that can divide the factor from the numerator.\n\nTherefore, if the units digit of $n$ is $0,2,3,4,5,7,8$, or 9 , then $f(n)$ is divisible by 5 , and so cannot have a units digit of 1 .\n\nSo we consider the cases where $n$ has a units digit of 1 or of 6 ; these are the only possible values of $n$ for which $f(n)$ can have a units digit of 1 .\n\nWe note that $3 f(n)=\\frac{n(n+1)(n+2)(n+3)}{8}$, which is a positive integer for all positive integers $n$.\n\n\n\nAlso, we note that if $f(n)$ has units digit 1 , then $3 f(n)$ has units digit 3 , and if $3 f(n)$ has units digit 3 , then $f(n)$ must have units digit 1 .\n\nTherefore, determining the values of $n$ for which $f(n)$ has units digit 1 is equivalent to determining the values of $n$ for which $\\frac{n(n+1)(n+2)(n+3)}{8}$ has units digit 3 .\n\nWe consider the integers $n$ in groups of 40 . (Intuitively, we do this because the problem seems to involve multiples of 5 and multiples of 8 , and $5 \\times 8=40$.)\n\nIf $n$ has units digit 1 , then $n=40 k+1$ or $n=40 k+11$ or $n=40 k+21$ or $n=40 k+31$ for some integer $k \\geq 0$.\n\nIf $n$ has units digit 6 , then $n=40 k+6$ or $n=40 k+16$ or $n=40 k+26$ or $n=40 k+36$ for some integer $k \\geq 0$.\n\nIf $n=40 k+1$, then\n\n$$\n\\begin{aligned}\n3 f(n) & =\\frac{n(n+1)(n+2)(n+3)}{8} \\\\\n& =\\frac{(40 k+1)(40 k+2)(40 k+3)(40 k+4)}{8} \\\\\n& =(40 k+1)(20 k+1)(40 k+3)(10 k+1)\n\\end{aligned}\n$$\n\nThe units digit of $40 k+1$ is 1 , the units digit of $20 k+1$ is 1 , the units digit of $40 k+3$ is 3 , and the units digit of $10 k+1$ is 1 , so the units digit of the product is the units digit of $(1)(1)(3)(1)$ or 3.\n\nIn a similar way, we treat the remaining seven cases and summarize all eight cases in a chart:\n\n| $n$ | $3 f(n)$ simplified | Units digit of $3 f(n)$ |\n| :---: | :---: | :---: |\n| $40 k+1$ | $(40 k+1)(20 k+1)(40 k+3)(10 k+1)$ | 3 |\n| $40 k+11$ | $(40 k+11)(10 k+3)(40 k+13)(20 k+7)$ | 3 |\n| $40 k+21$ | $(40 k+21)(20 k+11)(40 k+23)(10 k+6)$ | 8 |\n| $40 k+31$ | $(40 k+31)(10 k+8)(40 k+33)(20 k+17)$ | 8 |\n| $40 k+6$ | $(20 k+3)(40 k+7)(10 k+2)(40 k+9)$ | 8 |\n| $40 k+16$ | $(10 k+4)(40 k+17)(20 k+9)(40 k+19)$ | 8 |\n| $40 k+26$ | $(20 k+13)(40 k+27)(10 k+7)(40 k+29)$ | 3 |\n| $40 k+36$ | $(10 k+9)(40 k+37)(20 k+19)(40 k+39)$ | 3 |\n\n(Note that, for example, when $n=40 k+16$, the simplified version of $3 f(n)$ is $(10 k+4)(40 k+17)(20 k+9)(40 k+19)$, so the units digit of $3 f(n)$ is the units digit of $(4)(7)(9)(9)$ which is the units digit of 2268 , or 8 .)\n\nTherefore, $f(n)$ has units digit 1 whenever $n=40 k+1$ or $n=40 k+11$ or $n=40 k+26$ or $n=40 k+36$ for some integer $k \\geq 0$.\n\nThere are 4 such integers $n$ between each pair of consecutive multiples of 40 .\n\nSince $2000=50 \\times 40$, then 2000 is the 50 th multiple of 40 , so there are $50 \\times 4=200$ integers $n$ less than 2000 for which the units digit of $f(n)$ is 1 .\n\nBetween 2000 and 2014, inclusive, there are two additional integers: $n=40(50)+1=2001$ and $n=40(50)+11=2011$.\n\nIn total, 202 of the integers $f(1), f(2), \\ldots, f(2014)$ have a units digit of 1 .']",['202'],False,,Numerical, 2351,Algebra,,"If $\log _{10} x=3+\log _{10} y$, what is the value of $\frac{x}{y}$ ?",['$$\n\\begin{gathered}\n\\log _{10} x-\\log _{10} y=3 \\\\\n\\Leftrightarrow \\log _{10}\\left(\\frac{x}{y}\\right)=3 \\\\\n\\Leftrightarrow \\frac{x}{y}=10^{3}=1000\n\\end{gathered}\n$$'],['1000'],False,,Numerical, 2352,Algebra,,"If $x+\frac{1}{x}=\frac{13}{6}$, determine all values of $x^{2}+\frac{1}{x^{2}}$.","['$\\left(x+\\frac{1}{x}\\right)^{2}=\\left(\\frac{13}{6}\\right)^{2}$; squaring\n\n$x^{2}+2+\\frac{1}{x^{2}}=\\frac{169}{36}$\n\n$x^{2}+\\frac{1}{x^{2}}=\\frac{169}{32}-2$\n\n$x^{2}+\\frac{1}{x^{2}}=\\frac{169}{36}-\\frac{72}{36}=\\frac{97}{36}$', '$6 x\\left(x+\\frac{1}{x}\\right)=6 x\\left(\\frac{13}{6}\\right)$\n\n$6 x^{2}+6=13 x$\n\n$6 x^{2}-13 x+6=0$\n\n$(3 x-2)(2 x-3)=0$\n\n\n\n$x=\\frac{2}{3}$ or $x=\\frac{3}{2}$\n\nFor $x=\\frac{2}{3}, x^{2}+\\frac{1}{x^{2}}$\n\n$=\\left(\\frac{2}{3}\\right)^{2}+\\frac{1}{\\left(\\frac{2}{3}\\right)^{2}}$\n\n$=\\frac{4}{9}+\\frac{9}{4}$\n\nFor $x=\\frac{3}{2},\\left(\\frac{3}{2}\\right)^{2}+\\frac{1}{\\left(\\frac{3}{2}\\right)^{2}}$\n\n$=\\frac{9}{4}+\\frac{4}{9}$\n\n$=\\frac{97}{36}$\n\n$=\\frac{81+16}{36}$\n\n$=\\frac{97}{36}$']",['$\\frac{97}{36}$'],False,,Numerical, 2353,Geometry,,"A circle, with diameter $A B$ as shown, intersects the positive $y$-axis at point $D(0, d)$. Determine $d$. ","['The centre of the circle is $(3,0)$ and the circle has a radius of 5.\n\nThus $\\sqrt{d^{2}+3^{2}}=5$\n\n$$\n\\begin{aligned}\n& d^{2}=5^{2}-3^{2} \\\\\n& d^{2}=16\n\\end{aligned}\n$$\n\nTherefore $d=4$, since $d>0$.', 'Since $A B$ is a diameter of the circle, $\\angle A D B=90^{\\circ}$ and $\\angle A O D=90^{\\circ}$.\n\n$\\triangle A D O \\sim \\triangle D B O$\n\nTherefore, $\\frac{O D}{A O}=\\frac{B O}{O D}$\n\nand $d^{2}=2(8)$\n\n$$\n\\begin{aligned}\nd^{2} & =16 \\\\\nd & =4, \\text { since } d>0 .\n\\end{aligned}\n$$', '$\\angle A D B=\\angle A O D=\\angle B O D=90^{\\circ}$\n\nIn $\\triangle A O D, A D^{2}=4+d^{2}$.\n\nIn $\\triangle B O D, D B^{2}=64+d^{2}$.\n\nIn $\\Delta A D B,\\left(4+d^{2}\\right)+\\left(64+d^{2}\\right)=100$\n\n$$\n\\begin{aligned}\n2 d^{2} & =32 \\\\\nd & =4, d>0\n\\end{aligned}\n$$']",['4'],False,,Numerical, 2353,Geometry,,"A circle, with diameter $A B$ as shown, intersects the positive $y$-axis at point $D(0, d)$. Determine $d$. ![](https://cdn.mathpix.com/cropped/2023_12_21_47e3f6341f2fa24a72f4g-1.jpg?height=474&width=618&top_left_y=210&top_left_x=1190)","['The centre of the circle is $(3,0)$ and the circle has a radius of 5.\n\nThus $\\sqrt{d^{2}+3^{2}}=5$\n\n$$\n\\begin{aligned}\n& d^{2}=5^{2}-3^{2} \\\\\n& d^{2}=16\n\\end{aligned}\n$$\n\nTherefore $d=4$, since $d>0$.', 'Since $A B$ is a diameter of the circle, $\\angle A D B=90^{\\circ}$ and $\\angle A O D=90^{\\circ}$.\n\n$\\triangle A D O \\sim \\triangle D B O$\n\nTherefore, $\\frac{O D}{A O}=\\frac{B O}{O D}$\n\nand $d^{2}=2(8)$\n\n$$\n\\begin{aligned}\nd^{2} & =16 \\\\\nd & =4, \\text { since } d>0 .\n\\end{aligned}\n$$', '$\\angle A D B=\\angle A O D=\\angle B O D=90^{\\circ}$\n\nIn $\\triangle A O D, A D^{2}=4+d^{2}$.\n\nIn $\\triangle B O D, D B^{2}=64+d^{2}$.\n\nIn $\\Delta A D B,\\left(4+d^{2}\\right)+\\left(64+d^{2}\\right)=100$\n\n$$\n\\begin{aligned}\n2 d^{2} & =32 \\\\\nd & =4, d>0\n\\end{aligned}\n$$']",['4'],False,,Numerical, 2354,Geometry,,"A square $P Q R S$ with side of length $x$ is subdivided into four triangular regions as shown so that area (A) + area $(B)=\text{area}(C)$. If $P T=3$ and $R U=5$, determine the value of $x$. ![](https://cdn.mathpix.com/cropped/2023_12_21_47e3f6341f2fa24a72f4g-1.jpg?height=376&width=374&top_left_y=701&top_left_x=1361)","['Since the side length of the square is $x, T S=x-3$ and $V S=x-5$\n\nArea of triangle $A=\\frac{1}{2}(3)(x)$.\n\nArea of triangle $B=\\frac{1}{2}(5)(x)$\n\nArea of triangle $C=\\frac{1}{2}(x-5)(x-3)$.\n\nFrom the given information, $\\frac{1}{2}(3 x)+\\frac{1}{2}(5 x)=\\frac{1}{2}(x-5)(x-3)$. Labelled diagram\n\n$3 x+5 x=x^{2}-8 x+15$\n\n$x^{2}-16 x+15=0$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_ccfc71aa0aef1ed6786cg-1.jpg?height=440&width=484&top_left_y=1755&top_left_x=1388)\n\nThus $x=15$ or $x=1$.\n\nTherefore $x=15$ since $x=1$ is inadmissible.']",['15'],False,,Numerical, 2354,Geometry,,"A square $P Q R S$ with side of length $x$ is subdivided into four triangular regions as shown so that area (A) + area $(B)=\text{area}(C)$. If $P T=3$ and $R U=5$, determine the value of $x$. ","['Since the side length of the square is $x, T S=x-3$ and $V S=x-5$\n\nArea of triangle $A=\\frac{1}{2}(3)(x)$.\n\nArea of triangle $B=\\frac{1}{2}(5)(x)$\n\nArea of triangle $C=\\frac{1}{2}(x-5)(x-3)$.\n\nFrom the given information, $\\frac{1}{2}(3 x)+\\frac{1}{2}(5 x)=\\frac{1}{2}(x-5)(x-3)$. Labelled diagram\n\n$3 x+5 x=x^{2}-8 x+15$\n\n$x^{2}-16 x+15=0$\n\n\n\nThus $x=15$ or $x=1$.\n\nTherefore $x=15$ since $x=1$ is inadmissible.']",['15'],False,,Numerical, 2355,Combinatorics,,"A die, with the numbers $1,2,3,4,6$, and 8 on its six faces, is rolled. After this roll, if an odd number appears on the top face, all odd numbers on the die are doubled. If an even number appears on the top face, all the even numbers are halved. If the given die changes in this way, what is the probability that a 2 will appear on the second roll of the die?","[""There are only two possibilities on the first roll - it can either be even or odd.\n\nPossibility 1 'The first roll is odd'\n\nThe probability of an odd outcome on the first roll is $\\frac{1}{3}$.\n\nAfter doubling all the numbers, the possible outcomes on the second roll would now be 2, 2, 6, $4,6,8$ with the probability of a 2 being $\\frac{1}{3}$.\n\nThus the probability of a 2 on the second roll would be $\\frac{1}{3} \\times \\frac{1}{3}=\\frac{1}{9}$.\n\nPossibility 2 'The first is even'\n\nThe probability of an even outcome on the first roll is $\\frac{2}{3}$.\n\nAfter halving all the numbers, the possible outcomes on the second roll would be 1, 1, 3, 2, 3, 8 .\n\nThe probability of a 2 on the second die would now be $\\frac{1}{6}$.\n\nThus the probability of a 2 on the second roll is $\\frac{2}{3} \\times \\frac{1}{6}=\\frac{1}{9}$.\n\nThe probability of a 2 appear on the top face is $\\frac{1}{9}+\\frac{1}{9}=\\frac{2}{9}$.""]",['$\\frac{2}{9}$'],False,,Numerical, 2356,Algebra,,"The table below gives the final standings for seven of the teams in the English Cricket League in 1998. At the end of the year, each team had played 17 matches and had obtained the total number of points shown in the last column. Each win $W$, each draw $D$, each bonus bowling point $A$, and each bonus batting point $B$ received $w, d, a$ and $b$ points respectively, where $w, d, a$ and $b$ are positive integers. No points are given for a loss. Determine the values of $w, d, a$ and $b$ if total points awarded are given by the formula: Points $=w \times W+d \times D+a \times A+b \times B$. Final Standings | | $W$ | Losses | $D$ | $A$ | $B$ | Points | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | | Sussex | 6 | 7 | 4 | 30 | 63 | 201 | | Warks | 6 | 8 | 3 | 35 | 60 | 200 | | Som | 6 | 7 | 4 | 30 | 54 | 192 | | Derbys | 6 | 7 | 4 | 28 | 55 | 191 | | Kent | 5 | 5 | 7 | 18 | 59 | 178 | | Worcs | 4 | 6 | 7 | 32 | 59 | 176 | | Glam | 4 | 6 | 7 | 36 | 55 | 176 |","['There are a variety of ways to find the unknowns.\n\nThe most efficient way is to choose equations that have like coefficients. Here is one way to solve the problem using this method.\n\nFor Sussex: $\\quad 6 w+4 d+30 a+63 b=201$\n\nFor Som: $\\quad 6 w+4 d+30 a+54 b=192$\n\nSubtracting, $\\quad 9 b=9 b=1$\n\nIf $b=1$\n\nFor Derbys: $\\quad 6 w+4 d+28 a+55=191$ \n\n$$\n6 w+4 d+28 a=136 \\tag{1}\n$$\n\nFor Sussex: $\\quad 6 w+4 d+30 a+63=201$\n\n$$\n6 w+4 d+30 a=138 \\tag{2}\n$$\n\nSubtracting, (2) - (1)\n\n$$\n2 a=2\n$$\n\n$$\na=1 \\text {. }\n$$\n\nWe can now calculate $d$ and $w$ by substituting $a=1, b=1$ into a pair of equations.\n\nAn efficient way of doing this is by substituting $a=1, b=1$ into Som and Worcs.\n\nFor Som: $\\quad 6 w+4 d+84=192$\n\n$$\n6 w+4 d=108 \\tag{3}\n$$\n\nFor Worcs: $\\quad 6 w+3 d+85=200$\n\n$$\n6 w+3 d=105 \\tag{4}\n$$\n\nSubtracting, (3) - (4) $\\quad d=3$.\n\nSubstituting $d=3$ in either (3) or (4), $6 w+4(3)=108$ (substituting in (3))\n\n$$\n\\begin{aligned}\n6 w & =96 \\\\\nw & =16 .\n\\end{aligned}\n$$\n\nTherefore $w=16, d=3, a=b=1$.']","['16,3,1,1']",True,,Numerical, 2357,Geometry,,"In the diagram, $A D=D C, \sin \angle D B C=0.6$ and $\angle A C B=90^{\circ}$. What is the value of $\tan \angle A B C$ ? ","['Let $D B=10$.\n\nTherefore, $D C=A D=6$.\n\nBy the theorem of Pythagoras, $B C^{2}=10^{2}-6^{2}=64$.\n\nTherefore, $B C=8$.\n\n\n\nThus, $\\tan \\angle A B C=\\frac{12}{8}=\\frac{3}{2}$.']",['$\\frac{3}{2}$'],False,,Numerical, 2357,Geometry,,"In the diagram, $A D=D C, \sin \angle D B C=0.6$ and $\angle A C B=90^{\circ}$. What is the value of $\tan \angle A B C$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=366&width=357&top_left_y=207&top_left_x=1342)","['Let $D B=10$.\n\nTherefore, $D C=A D=6$.\n\nBy the theorem of Pythagoras, $B C^{2}=10^{2}-6^{2}=64$.\n\nTherefore, $B C=8$.\n\n\n\nThus, $\\tan \\angle A B C=\\frac{12}{8}=\\frac{3}{2}$.']",['$\\frac{3}{2}$'],False,,Numerical, 2358,Geometry,,"On a cross-sectional diagram of the Earth, the $x$ and $y$-axes are placed so that $O(0,0)$ is the centre of the Earth and $C(6.40,0.00)$ is the location of Cape Canaveral. A space shuttle is forced to land on an island at $A(5.43,3.39)$, as shown. Each unit represents $1000 \mathrm{~km}$. Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest $10 \mathrm{~km}$. ",['$\\tan \\angle A O C=\\frac{3.39}{5.43}$\n\n$\\angle A O C=\\tan ^{-1}\\left(\\frac{3.39}{5.43}\\right)=31.97^{\\circ}$\n\nThe arc length $\\overparen{A C}=\\frac{31.97}{360^{\\circ}}[(2 \\pi)(6.40)]=3.57$ units\n\nThe distance is approximately $3570 \\mathrm{~km}$.'],['3570'],False,km,Numerical, 2358,Geometry,,"On a cross-sectional diagram of the Earth, the $x$ and $y$-axes are placed so that $O(0,0)$ is the centre of the Earth and $C(6.40,0.00)$ is the location of Cape Canaveral. A space shuttle is forced to land on an island at $A(5.43,3.39)$, as shown. Each unit represents $1000 \mathrm{~km}$. Determine the distance from Cape Canaveral to the island, measured on the surface of the earth, to the nearest $10 \mathrm{~km}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=315&width=493&top_left_y=634&top_left_x=1315)",['$\\tan \\angle A O C=\\frac{3.39}{5.43}$\n\n$\\angle A O C=\\tan ^{-1}\\left(\\frac{3.39}{5.43}\\right)=31.97^{\\circ}$\n\nThe arc length $\\overparen{A C}=\\frac{31.97}{360^{\\circ}}[(2 \\pi)(6.40)]=3.57$ units\n\nThe distance is approximately $3570 \\mathrm{~km}$.'],['3570'],False,km,Numerical, 2359,Algebra,,"Let $\lfloor x\rfloor$ represent the greatest integer which is less than or equal to $x$. For example, $\lfloor 3\rfloor=3,\lfloor 2.6\rfloor=2$. If $x$ is positive and $x\lfloor x\rfloor=17$, what is the value of $x$ ?","['We deduce that $4","['The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.\n\nThe equation of the line is $y=x-2$.\n\nIntersection points, $x-2=-x^{2}+4$\n\n$$\n\\begin{array}{r}\nx^{2}+x-6=0 \\\\\n(x+3)(x-2)=0 .\n\\end{array}\n$$\n\nTherefore, $x=-3$ or $x=2$.\n\nFor $x=-3, y=-3-2=-5$. Thus $B^{\\prime}$ has coordinates $(-3,-5)$.\n\nIf $(-3,-5) \\rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \\rightarrow(x+5, y+5)$.\n\nPossibility 1\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nIf $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.\n\nPossibility 2\n\nIf $B^{\\prime}$ has coordinates $(-3,-5)$ then $C^{\\prime}$ is the reflection of $B^{\\prime}$ in the $y$-axis. Thus $C^{\\prime}$ has coordinates $(3,-5)$.\n\nIf we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.\n\nThus $C$ has coordinates $(8,0)$.\n\nPossibility 3\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nThe equation of the image parabola is $y=-(x-5)^{2}+9$.\n\n\n\nTo find its intercepts, $-(x-5)^{2}+9=0$\n\n$$\n\\begin{aligned}\n(x-5)^{2} & =9 \\\\\nx-5 & = \\pm 3 .\n\\end{aligned}\n$$\n\nTherefore $x=8$ or $x=2$.\n\nThus $C$ has coordinates $(8,0)$.', 'The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .\n\nThe pre-image of $B^{\\prime}$ is $(2-t,-t)$.\n\nSince $B^{\\prime}$ is on the parabola with vertex $P$, we have\n\n$$\n\\begin{aligned}\n-t & =-(2-t)^{2}+4 \\\\\n-t & =-4+4 t-t^{2}+4 \\\\\nt^{2}-5 t & =0 \\\\\nt(t-5) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=0$ or $t=5$.\n\nThus $B^{\\prime}$ is $(-3,-5)$.\n\nLet $C$ have coordinates $(c, 0)$.\n\nThe pre-image of $C$ is $(c-5,-5)$.\n\nTherefore, $-5=-(c-5)^{2}+4$.\n\nOr, $(c-5)^{2}=9$.\n\nTherefore $c-5=3$ or $c-5=-3$.\n\n$$\nc=8 \\text { or } \\quad c=2\n$$\n\nThus $C$ has coordinates $(8,0)$.']","['$(8,0)$']",False,,Tuple, 2360,Geometry,,"The parabola $y=-x^{2}+4$ has vertex $P$ and intersects the $x$-axis at $A$ and $B$. The parabola is translated from its original position so that its vertex moves along the line $y=x+4$ to the point $Q$. In this position, the parabola intersects the $x$-axis at $B$ and $C$. Determine the coordinates of $C$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=540&width=596&top_left_y=1340&top_left_x=1212)","['The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.\n\nThe equation of the line is $y=x-2$.\n\nIntersection points, $x-2=-x^{2}+4$\n\n$$\n\\begin{array}{r}\nx^{2}+x-6=0 \\\\\n(x+3)(x-2)=0 .\n\\end{array}\n$$\n\nTherefore, $x=-3$ or $x=2$.\n\nFor $x=-3, y=-3-2=-5$. Thus $B^{\\prime}$ has coordinates $(-3,-5)$.\n\nIf $(-3,-5) \\rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \\rightarrow(x+5, y+5)$.\n\nPossibility 1\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nIf $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.\n\nPossibility 2\n\nIf $B^{\\prime}$ has coordinates $(-3,-5)$ then $C^{\\prime}$ is the reflection of $B^{\\prime}$ in the $y$-axis. Thus $C^{\\prime}$ has coordinates $(3,-5)$.\n\nIf we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.\n\nThus $C$ has coordinates $(8,0)$.\n\nPossibility 3\n\nUsing the general translation, we find the coordinates of $Q$ to be, $P(0,4) \\rightarrow Q(0+5,4+5)=Q(5,9)$.\n\nThe equation of the image parabola is $y=-(x-5)^{2}+9$.\n\n\n\nTo find its intercepts, $-(x-5)^{2}+9=0$\n\n$$\n\\begin{aligned}\n(x-5)^{2} & =9 \\\\\nx-5 & = \\pm 3 .\n\\end{aligned}\n$$\n\nTherefore $x=8$ or $x=2$.\n\nThus $C$ has coordinates $(8,0)$.', 'The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .\n\nThe pre-image of $B^{\\prime}$ is $(2-t,-t)$.\n\nSince $B^{\\prime}$ is on the parabola with vertex $P$, we have\n\n$$\n\\begin{aligned}\n-t & =-(2-t)^{2}+4 \\\\\n-t & =-4+4 t-t^{2}+4 \\\\\nt^{2}-5 t & =0 \\\\\nt(t-5) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=0$ or $t=5$.\n\nThus $B^{\\prime}$ is $(-3,-5)$.\n\nLet $C$ have coordinates $(c, 0)$.\n\nThe pre-image of $C$ is $(c-5,-5)$.\n\nTherefore, $-5=-(c-5)^{2}+4$.\n\nOr, $(c-5)^{2}=9$.\n\nTherefore $c-5=3$ or $c-5=-3$.\n\n$$\nc=8 \\text { or } \\quad c=2\n$$\n\nThus $C$ has coordinates $(8,0)$.']","['$(8,0)$']",False,,Tuple, 2361,Geometry,,"A cube has edges of length $n$, where $n$ is an integer. Three faces, meeting at a corner, are painted red. The cube is then cut into $n^{3}$ smaller cubes of unit length. If exactly 125 of these cubes have no faces painted red, determine the value of $n$.","['If we remove the cubes which have red paint, we are left with a smaller cube with measurements, $(n-1) \\times(n-1) \\times(n-1)$\n\nThus, $(n-1)^{3}=125$\n\n$$\nn=6 \\text {. }\n$$']",['6'],False,,Numerical, 2362,Geometry,,"In the isosceles trapezoid $A B C D$, $A B=C D=x$. The area of the trapezoid is 80 and the circle with centre $O$ and radius 4 is tangent to the four sides of the trapezoid. Determine the value of $x$. ","['Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.\n\nArea of trapezoid $A B C D=\\frac{1}{2}(8)(B C+A D)$\n\n$$\n\\begin{aligned}\n& =4(2 b+2 x-2 b) \\\\\n& =8 x .\n\\end{aligned}\n$$\n\n\n\nThus, $8 x=80$.\n\nTherefore, $x=10$.']",['10'],False,,Numerical, 2362,Geometry,,"In the isosceles trapezoid $A B C D$, $A B=C D=x$. The area of the trapezoid is 80 and the circle with centre $O$ and radius 4 is tangent to the four sides of the trapezoid. Determine the value of $x$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b1dc13dda57202a677cdg-1.jpg?height=266&width=333&top_left_y=2182&top_left_x=1362)","['Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.\n\nArea of trapezoid $A B C D=\\frac{1}{2}(8)(B C+A D)$\n\n$$\n\\begin{aligned}\n& =4(2 b+2 x-2 b) \\\\\n& =8 x .\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4eb15a30463877039083g-1.jpg?height=287&width=482&top_left_y=1250&top_left_x=1250)\n\nThus, $8 x=80$.\n\nTherefore, $x=10$.']",['10'],False,,Numerical, 2363,Geometry,,"In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$. ![](https://cdn.mathpix.com/cropped/2023_12_21_df527143fd6a00cd2aabg-1.jpg?height=268&width=599&top_left_y=224&top_left_x=1205) Prove that $P Q R S$ is a rectangle.","['In a parallelogram opposite angles are equal.\n\nSince $D F$ and $B E$ bisect the two angles, let $\\angle A D F=\\angle C D F=\\angle A B E=\\angle C B E$\n\n$$\n=x \\text { (in degrees) }\n$$\n\nAlso $\\angle C D F=\\angle A F D=x$ (alternate angles)\n\nLet $\\angle D A M=\\angle B A M=\\angle D C N=\\angle B C N=y$ (in degrees)\n\n\n\nFor any parallelogram, any two consecutive angles add to $180^{\\circ}, \\therefore 2 x+2 y=180$ or, $x+y=90$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_2940ec3d1fef53148a02g-1.jpg?height=345&width=707&top_left_y=234&top_left_x=1121)\n\nTherefore in $\\triangle P A F, \\angle A P F=90^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_2940ec3d1fef53148a02g-1.jpg?height=222&width=407&top_left_y=637&top_left_x=1119)\n\nUsing similar reasoning and properties of parallel lines we get right angles at $Q, R$ and $S$.\n\nThus $P Q R S$ is a rectangle.']",['证明题，略'],True,,Need_human_evaluate, 2363,Geometry,,"In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$. Prove that $P Q R S$ is a rectangle.","['In a parallelogram opposite angles are equal.\n\nSince $D F$ and $B E$ bisect the two angles, let $\\angle A D F=\\angle C D F=\\angle A B E=\\angle C B E$\n\n$$\n=x \\text { (in degrees) }\n$$\n\nAlso $\\angle C D F=\\angle A F D=x$ (alternate angles)\n\nLet $\\angle D A M=\\angle B A M=\\angle D C N=\\angle B C N=y$ (in degrees)\n\n\n\nFor any parallelogram, any two consecutive angles add to $180^{\\circ}, \\therefore 2 x+2 y=180$ or, $x+y=90$.\n\n\n\nTherefore in $\\triangle P A F, \\angle A P F=90^{\\circ}$.\n\n\n\nUsing similar reasoning and properties of parallel lines we get right angles at $Q, R$ and $S$.\n\nThus $P Q R S$ is a rectangle.']",,True,,, 2364,Geometry,,"In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$. ![](https://cdn.mathpix.com/cropped/2023_12_21_df527143fd6a00cd2aabg-1.jpg?height=268&width=599&top_left_y=224&top_left_x=1205) Prove that $P R=a-b$.","['Since $A M$ is a bisector of $\\angle D A B$, let $\\angle D A M=\\angle B A M=y$.\n\nAlso, $\\angle D M A=y$ (alternate angles)\n\nThis implies that $\\triangle A D M$ is isosceles.\n\nUsing the same reasoning in $\\triangle C B N$, we see that it is also isosceles and so the diagram may now be labelled as:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_2940ec3d1fef53148a02g-1.jpg?height=431&width=718&top_left_y=1481&top_left_x=622)\n\n$A N=a-b$\n\nThus $\\triangle A D M$ and $\\triangle C B N$ are identical isosceles triangles.\n\nAlso, $A M \\| N C$ (corresponding angles)\n\nor, $A P \\| N R$.\n\nBy using properties of isosceles triangles (or congruency), $A P=N R$ implying that $A P R N$ is a parallelogram.\n\nThus $A N=P R$ and since $A N=a-b, P R=a-b$ (as required)']",['证明题，略'],True,,Need_human_evaluate, 2364,Geometry,,"In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$. Prove that $P R=a-b$.","['Since $A M$ is a bisector of $\\angle D A B$, let $\\angle D A M=\\angle B A M=y$.\n\nAlso, $\\angle D M A=y$ (alternate angles)\n\nThis implies that $\\triangle A D M$ is isosceles.\n\nUsing the same reasoning in $\\triangle C B N$, we see that it is also isosceles and so the diagram may now be labelled as:\n\n\n\n$A N=a-b$\n\nThus $\\triangle A D M$ and $\\triangle C B N$ are identical isosceles triangles.\n\nAlso, $A M \\| N C$ (corresponding angles)\n\nor, $A P \\| N R$.\n\nBy using properties of isosceles triangles (or congruency), $A P=N R$ implying that $A P R N$ is a parallelogram.\n\nThus $A N=P R$ and since $A N=a-b, P R=a-b$ (as required)']",,True,,, 2365,Number Theory,,"A permutation of the integers $1,2, \ldots, n$ is a listing of these integers in some order. For example, $(3,1,2)$ and $(2,1,3)$ are two different permutations of the integers 1, 2, 3. A permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the integers $1,2, \ldots, n$ is said to be ""fantastic"" if $a_{1}+a_{2}+\ldots+a_{k}$ is divisible by $k$, for each $k$ from 1 to $n$. For example, $(3,1,2)$ is a fantastic permutation of $1,2,3$ because 3 is divisible by $1,3+1$ is divisible by 2 , and $3+1+2$ is divisible by 3 . However, $(2,1,3)$ is not fantastic because $2+1$ is not divisible by 2 . Show that no fantastic permutation exists for $n=2000$.","['In our consideration of whether there is a fantastic permutation for $n=2000$, we start by looking at the 2000th position.\n\nUsing our definition of fantastic permutation, it is necessary that $2000 \\mid(1+2+3+\\cdots+2000)$.\n\nSince $1+2+3+\\cdots+2000=\\frac{(2000)(2001)}{2}=(1000)(2001)$, it is required that $2000 \\mid 1000(2001)$.\n\nThis is not possible and so no fantastic permutation exists for $n=2000$.']",,True,,, 2366,Number Theory,,"A permutation of the integers $1,2, \ldots, n$ is a listing of these integers in some order. For example, $(3,1,2)$ and $(2,1,3)$ are two different permutations of the integers 1, 2, 3. A permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the integers $1,2, \ldots, n$ is said to be ""fantastic"" if $a_{1}+a_{2}+\ldots+a_{k}$ is divisible by $k$, for each $k$ from 1 to $n$. For example, $(3,1,2)$ is a fantastic permutation of $1,2,3$ because 3 is divisible by $1,3+1$ is divisible by 2 , and $3+1+2$ is divisible by 3 . However, $(2,1,3)$ is not fantastic because $2+1$ is not divisible by 2 . Does a fantastic permutation exist for $n=2001$ ? Explain.","['The sum of the integers from 1 to 2001 is $\\frac{(2001)(2002)}{2}=(2001)(1001)$ which is divisible by 2001. If $t_{1}, t_{2}, \\ldots, t_{2001}$ is a fantastic permutation, when we remove $t_{2001}$ from the above sum, and what remains must be divisible by 2000 .\n\nWe now consider $t_{1}+t_{2}+\\cdots+t_{2000}$ and determine what integer is not included in the permutation.\n\n$$\n\\begin{aligned}\nt_{1}+t_{2}+\\cdots+t_{2000} & =\\frac{(2001)(2002)}{2}-t_{2001} \\\\\n& =(1001)(2001)-t_{2001}\n\\end{aligned}\n$$\n\nSince $(1001)(2001)=2003001, t_{2001}$ must be a number of the form $k 001$ where $k$ is odd. The only integer less than or equal to 2001 with this property is 1001 . Therefore $t_{2001}=1001$.\n\nSo the sum up to $t_{2000}$ is $2003001-1001=2002000$.\n\nWhen we remove $t_{2000}$ we must get a multiple of 1999 .\n\nThe largest multiple of 1999 less than 2002000 is (1999)(1001) $=2000$ 999. This would make $t_{2000}=2002000-2000999=1001$ which is impossible since $t_{2000} \\neq t_{2001}$. If we choose lesser multiples of 1999 to subtract from 2002000 we will get values of $t_{2000}$ which are greater than 2001, which is also not possible.\n\nThus, a fantastic permutation is not possible for $n=2001$.']",,True,,, 2367,Geometry,,"An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$. ![](https://cdn.mathpix.com/cropped/2023_12_21_df527143fd6a00cd2aabg-1.jpg?height=372&width=387&top_left_y=1126&top_left_x=1346)","['Let $\\angle R Q C=\\theta$ and from $S$ draw a line perpendicular to the base at $P$.\n\nThen $\\angle T Q B=180-(90+\\theta)=90-\\theta$.\n\nLet $s$ be the length of the side of the square.\n\nFrom $R$ draw a line perpendicular to $B C$ at $D$ and then through $S$ draw a line parallel to $B C$. From $R$ draw a line perpendicular to this line at E.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_216628493d6885b0bc57g-1.jpg?height=482&width=490&top_left_y=1190&top_left_x=1273)\n\nFrom $\\triangle R Q D, R D=s \\sin \\theta$.\n\nSince $\\angle Q R D=90-\\theta$ then $\\angle S R E=\\theta$.\n\nFrom $\\triangle S E R, E R=s \\cos \\theta$.\n\nThe perpendicular distance from $S$ to $B C$ is $R D+E R=s \\sin \\theta+s \\cos \\theta$ which we must now show is a constant.\n\nWe can now take each of the lengths $D C, D Q, P F, F B$ and express them in terms of $s$.\n\nFrom $\\triangle R D C$ which is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, $\\frac{D C}{R D}=\\frac{1}{\\sqrt{3}}$.\n\nSince $R D=s \\sin \\theta$ (from above)\n\n$$\nD C=\\frac{1}{\\sqrt{3}}(s \\sin \\theta)=\\frac{\\sqrt{3}}{3} s \\sin \\theta\n$$\n\n\n\nFrom $\\Delta R D Q, \\frac{Q D}{R Q}=\\cos \\theta, Q D=s \\cos \\theta$.\n\nFrom $\\triangle T F Q, \\sin \\theta=\\frac{F Q}{s}$ and $\\cos \\theta=\\frac{T F}{s}$. or, $F Q=s \\sin \\theta$ and $T F=s \\sin \\theta$.\n\nFrom $\\triangle T F B, \\frac{B F}{T F}=\\frac{1}{\\sqrt{3}}, B F=\\frac{1}{\\sqrt{3}} T F=\\frac{1}{\\sqrt{3}} s \\cos \\theta=\\frac{\\sqrt{3}}{3} s \\cos \\theta$.\n\nSince $D C+Q D+F Q+B F=2, \\frac{\\sqrt{3}}{3} s \\sin \\theta+s \\cos \\theta+s \\sin \\theta+\\frac{\\sqrt{3}}{3} s \\cos \\theta=2$.\n\n$$\n\\begin{aligned}\n\\frac{\\sqrt{3}}{3}(s \\cos \\theta+s \\sin \\theta)+(s \\cos \\theta+s \\sin \\theta) & =2 \\\\\ns \\cos \\theta+s \\sin \\theta & =\\frac{2}{\\left(\\frac{\\sqrt{3}}{3}+1\\right)}\n\\end{aligned}\n$$\n\nThus $s \\cos \\theta+s \\sin \\theta$ is a constant and the path traced out by $S$ is a straight line parallel to $B C$.']",['证明题，略'],True,,Need_human_evaluate, 2367,Geometry,,"An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$. ","['Let $\\angle R Q C=\\theta$ and from $S$ draw a line perpendicular to the base at $P$.\n\nThen $\\angle T Q B=180-(90+\\theta)=90-\\theta$.\n\nLet $s$ be the length of the side of the square.\n\nFrom $R$ draw a line perpendicular to $B C$ at $D$ and then through $S$ draw a line parallel to $B C$. From $R$ draw a line perpendicular to this line at E.\n\n\n\nFrom $\\triangle R Q D, R D=s \\sin \\theta$.\n\nSince $\\angle Q R D=90-\\theta$ then $\\angle S R E=\\theta$.\n\nFrom $\\triangle S E R, E R=s \\cos \\theta$.\n\nThe perpendicular distance from $S$ to $B C$ is $R D+E R=s \\sin \\theta+s \\cos \\theta$ which we must now show is a constant.\n\nWe can now take each of the lengths $D C, D Q, P F, F B$ and express them in terms of $s$.\n\nFrom $\\triangle R D C$ which is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, $\\frac{D C}{R D}=\\frac{1}{\\sqrt{3}}$.\n\nSince $R D=s \\sin \\theta$ (from above)\n\n$$\nD C=\\frac{1}{\\sqrt{3}}(s \\sin \\theta)=\\frac{\\sqrt{3}}{3} s \\sin \\theta\n$$\n\n\n\nFrom $\\Delta R D Q, \\frac{Q D}{R Q}=\\cos \\theta, Q D=s \\cos \\theta$.\n\nFrom $\\triangle T F Q, \\sin \\theta=\\frac{F Q}{s}$ and $\\cos \\theta=\\frac{T F}{s}$. or, $F Q=s \\sin \\theta$ and $T F=s \\sin \\theta$.\n\nFrom $\\triangle T F B, \\frac{B F}{T F}=\\frac{1}{\\sqrt{3}}, B F=\\frac{1}{\\sqrt{3}} T F=\\frac{1}{\\sqrt{3}} s \\cos \\theta=\\frac{\\sqrt{3}}{3} s \\cos \\theta$.\n\nSince $D C+Q D+F Q+B F=2, \\frac{\\sqrt{3}}{3} s \\sin \\theta+s \\cos \\theta+s \\sin \\theta+\\frac{\\sqrt{3}}{3} s \\cos \\theta=2$.\n\n$$\n\\begin{aligned}\n\\frac{\\sqrt{3}}{3}(s \\cos \\theta+s \\sin \\theta)+(s \\cos \\theta+s \\sin \\theta) & =2 \\\\\ns \\cos \\theta+s \\sin \\theta & =\\frac{2}{\\left(\\frac{\\sqrt{3}}{3}+1\\right)}\n\\end{aligned}\n$$\n\nThus $s \\cos \\theta+s \\sin \\theta$ is a constant and the path traced out by $S$ is a straight line parallel to $B C$.']",,True,,, 2368,Geometry,,"In the diagram, points $P(p, 4), B(10,0)$, and $O(0,0)$ are shown. If $\triangle O P B$ is right-angled at $P$, determine all possible values of $p$. ","['Since $\\angle O P B=90^{\\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .\n\nThe slope of $O P$ is $\\frac{4-0}{p-0}=\\frac{4}{p}$ and the slope of $P B$ is $\\frac{4-0}{p-10}=\\frac{4}{p-10}$.\n\nTherefore, we need\n\n$$\n\\begin{aligned}\n\\frac{4}{p} \\cdot \\frac{4}{p-10} & =-1 \\\\\n16 & =-p(p-10) \\\\\np^{2}-10 p+16 & =0 \\\\\n(p-2)(p-8) & =0\n\\end{aligned}\n$$\n\nand so $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.', 'Since $\\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.\n\nAlso, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,\n\n$$\n\\begin{aligned}\n\\left(p^{2}+16\\right)+\\left(p^{2}-20 p+116\\right) & =10^{2} \\\\\n2 p^{2}-20 p+32 & =0 \\\\\np^{2}-10 p+16 & =0\n\\end{aligned}\n$$\n\n\n\nand so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.']","['2,8']",True,,Numerical, 2368,Geometry,,"In the diagram, points $P(p, 4), B(10,0)$, and $O(0,0)$ are shown. If $\triangle O P B$ is right-angled at $P$, determine all possible values of $p$. ![](https://cdn.mathpix.com/cropped/2023_12_21_6612ae75d8ef169f4be5g-1.jpg?height=377&width=477&top_left_y=722&top_left_x=1255)","['Since $\\angle O P B=90^{\\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .\n\nThe slope of $O P$ is $\\frac{4-0}{p-0}=\\frac{4}{p}$ and the slope of $P B$ is $\\frac{4-0}{p-10}=\\frac{4}{p-10}$.\n\nTherefore, we need\n\n$$\n\\begin{aligned}\n\\frac{4}{p} \\cdot \\frac{4}{p-10} & =-1 \\\\\n16 & =-p(p-10) \\\\\np^{2}-10 p+16 & =0 \\\\\n(p-2)(p-8) & =0\n\\end{aligned}\n$$\n\nand so $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.', 'Since $\\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.\n\nAlso, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,\n\n$$\n\\begin{aligned}\n\\left(p^{2}+16\\right)+\\left(p^{2}-20 p+116\\right) & =10^{2} \\\\\n2 p^{2}-20 p+32 & =0 \\\\\np^{2}-10 p+16 & =0\n\\end{aligned}\n$$\n\n\n\nand so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.']","['2,8']",True,,Numerical, 2369,Combinatorics,,Thurka bought some stuffed goats and some toy helicopters. She paid a total of $\$ 201$. She did not buy partial goats or partial helicopters. Each stuffed goat cost $\$ 19$ and each toy helicopter cost $\$ 17$. How many of each did she buy?,"['Suppose that Thurka bought $x$ goats and $y$ helicopters.\n\nThen $19 x+17 y=201$.\n\nSince $x$ and $y$ are non-negative integers, then $19 x \\leq 201$ so $x \\leq 10$.\n\nIf $x=10$, then $17 y=201-19 x=11$, which does not have an integer solution because 11 is not divisible by 17 .\n\nIf $x=9$, then $17 y=201-19 x=30$, which does not have an integer solution.\n\nIf $x=8$, then $17 y=201-19 x=49$, which does not have an integer solution.\n\nIf $x=7$, then $17 y=201-19 x=68$, so $y=4$.\n\nTherefore, $19(7)+17(4)=201$, and so Thurka buys 7 goats and 4 helicopters.\n\n(We can check that $x=0,1,2,3,4,5,6$ do not give values of $y$ that work.)']","['7,4']",True,,Numerical, 2370,Algebra,,Determine all real values of $x$ for which $(x+8)^{4}=(2 x+16)^{2}$.,"['Manipulating algebraically,\n\n$$\n\\begin{aligned}\n(x+8)^{4} & =(2 x+16)^{2} \\\\\n(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\\\\n(x+8)^{2}\\left((x+8)^{2}-2^{2}\\right) & =0 \\\\\n(x+8)^{2}((x+8)+2)((x+8)-2) & =0 \\\\\n(x+8)^{2}(x+10)(x+6) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-8$ or $x=-10$ or $x=-6$.', 'Manipulating algebraically,\n\n$$\n\\begin{aligned}\n(x+8)^{4} & =(2 x+16)^{2} \\\\\n(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\\\\n(x+8)^{2}\\left((x+8)^{2}-2^{2}\\right) & =0 \\\\\n(x+8)^{2}\\left(x^{2}+16 x+64-4\\right) & =0 \\\\\n(x+8)^{2}\\left(x^{2}+16 x+60\\right) & =0 \\\\\n(x+8)^{2}(x+10)(x+6) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-8$ or $x=-10$ or $x=-6$.', 'Since $(x+8)^{4}=(2 x+16)^{2}$, then $(x+8)^{2}=2 x+16$ or $(x+8)^{2}=-(2 x+16)$.\n\nFrom the first equation, $x^{2}+16 x+64=2 x+16$ or $x^{2}+14 x+48=0$ or $(x+6)(x+8)=0$. From the second equation, $x^{2}+16 x+64=-2 x-16$ or $x^{2}+18 x+80=0$ or $(x+10)(x+8)=0$.\n\nTherefore, $x=-8$ or $x=-10$ or $x=-6$.']","['-6,-8,-10']",True,,Numerical, 2371,Algebra,,"If $f(x)=2 x+1$ and $g(f(x))=4 x^{2}+1$, determine an expression for $g(x)$.","['We use the fact that $g(x)=g\\left(f\\left(f^{-1}(x)\\right)\\right)$.\n\nSince $f(x)=2 x+1$, then to determine $f^{-1}(x)$ we solve $x=2 y+1$ for $y$ to get $2 y=x-1$ or $y=\\frac{1}{2}(x-1)$. Thus, $f^{-1}(x)=\\frac{1}{2}(x-1)$.\n\nSince $g(f(x))=4 x^{2}+1$, then\n\n$$\n\\begin{aligned}\ng(x) & =g\\left(f\\left(f^{-1}(x)\\right)\\right) \\\\\n& =g\\left(f\\left(\\frac{1}{2}(x-1)\\right)\\right) \\\\\n& =4\\left(\\frac{1}{2}(x-1)\\right)^{2}+1 \\\\\n& =4 \\cdot \\frac{1}{4}(x-1)^{2}+1 \\\\\n& =(x-1)^{2}+1 \\\\\n& =x^{2}-2 x+2\n\\end{aligned}\n$$', 'We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.\n\nSince $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.\n\nSince $f(x)=2 x+1$, then $(f(x))^{2}=4 x^{2}+4 x+1$.\n\nSince $g(f(x))$ has no term involving $x$, then we subtract $2 f(x)$ (to remove the $4 x$ term) to get\n\n$$\n(f(x))^{2}-2 f(x)=\\left(4 x^{2}+4 x+1\\right)-2(2 x+1)=4 x^{2}-1\n$$\n\nTo get $g(f(x))$ from this, we add 2 to get $4 x^{2}+1$.\n\nTherefore, $g(f(x))=(f(x))^{2}-2 f(x)+2$, and so an expression for $g(x)$ is $x^{2}-2 x+2$.', 'We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.\n\nSince $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.\n\nSuppose that $g(x)=a x^{2}+b x+c$ for some real numbers $a, b, c$.\n\nThen\n\n$$\n\\begin{aligned}\ng(f(x)) & =g(2 x+1) \\\\\n& =a(2 x+1)^{2}+b(2 x+1)+c \\\\\n& =a\\left(4 x^{2}+4 x+1\\right)+b(2 x+1)+c \\\\\n& =4 a x^{2}+(4 a+2 b) x+(a+b+c)\n\\end{aligned}\n$$\n\nSince we are told that $g(f(x))=4 x^{2}+1$, then we can compare coefficients to deduce that $4 a=4$ and $4 a+2 b=0$ and $a+b+c=1$.\n\nFrom the first equation, $a=1$.\n\nFrom the second equation, $b=-2 a=-2$.\n\nFrom the third equation, $c=1-a-b=2$.\n\nTherefore, an expression for $g(x)$ is $x^{2}-2 x+2$.']",['$g(x)=x^2-2x+2$'],False,,Expression, 2372,Number Theory,,"A geometric sequence has 20 terms. The sum of its first two terms is 40 . The sum of its first three terms is 76 . The sum of its first four terms is 130 . Determine how many of the terms in the sequence are integers. (A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)","['Since the sum of the first two terms is 40 and the sum of the first three terms is 76, then the third term is $76-40=36$.\n\nSince the sum of the first three terms is 76 and the sum of the first four terms is 130, then the fourth term is $130-76=54$.\n\nSince the third term is 36 and the fourth term is 54 , then the common ratio in the geometric sequence is $\\frac{54}{36}=\\frac{3}{2}$.\n\nTherefore, the fifth term is $54 \\cdot \\frac{3}{2}=81$ and the sixth term is $81 \\cdot \\frac{3}{2}=\\frac{243}{2}$.\n\n\n\nAlso, the second term is $36 \\div \\frac{3}{2}=36 \\cdot \\frac{2}{3}=24$ and the first term is $24 \\div \\frac{3}{2}=24 \\cdot \\frac{2}{3}=16$.\n\nThus, the first six terms of the sequence are $16,24,36,54,81, \\frac{243}{2}$.\n\nSince the first term equals $2^{4}$ and the common ratio is $\\frac{3}{2}$, then the $n$th term in the sequence is $2^{4}\\left(\\frac{3}{2}\\right)^{n-1}=\\frac{3^{n-1}}{2^{n-5}}$.\n\nWhen $n \\geq 6$, this is a fraction whose numerator is odd and whose denominator is even, and so, when $n \\geq 6$, the $n$th term is not an integer. (An odd integer is never divisible by an even integer.)\n\nTherefore, there will be 5 integers in the sequence.', 'Suppose that $a$ is the first term and $r$ is the common ratio between consecutive terms (so that $a r$ is the second term, $a r^{2}$ is the third term, and so on).\n\nFrom the given information, $a+a r=40$ and $a+a r+a r^{2}=76$ and $a+a r+a r^{2}+a r^{3}=130$.\n\nSubtracting the first equation from the second, we obtain $a r^{2}=36$.\n\nSubtracting the second equation from the third, we obtain $a r^{3}=54$.\n\nSince $a r^{3}=54$ and $a r^{2}=36$, then $r=\\frac{a r^{3}}{a r^{2}}=\\frac{54}{36}=\\frac{3}{2}$.\n\nSince $a r^{2}=36$ and $r=\\frac{3}{2}$, then $a\\left(\\frac{3}{2}\\right)^{2}=36$ or $\\frac{9}{4} a=36$ or $a=\\frac{4}{9} \\cdot 36=16$.\n\nSince $a=16$ and $r=\\frac{3}{2}$, then the first six terms of the sequence are 16, 24, 36, 54, 81, $\\frac{243}{2}$. Since the first term equals $2^{4}$ and the common ratio is $\\frac{3}{2}$, then the $n$th term in the sequence is $2^{4}\\left(\\frac{3}{2}\\right)^{n-1}=\\frac{3^{n-1}}{2^{n-5}}$.\n\nWhen $n \\geq 6$, this is a fraction whose numerator is odd and whose denominator is even, and so, when $n \\geq 6$, the $n$th term is not an integer. (An odd integer is never divisible by an even integer.)\n\nTherefore, there will be 5 integers in the sequence.']",['5'],False,,Numerical, 2373,Geometry,,"A snail's shell is formed from six triangular sections, as shown. Each triangle has interior angles of $30^{\circ}, 60^{\circ}$ and $90^{\circ}$. If $A B$ has a length of $1 \mathrm{~cm}$, what is the length of $A H$, in $\mathrm{cm}$ ? ","['In a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, the ratio of the side opposite the $90^{\\circ}$ to the side opposite the $60^{\\circ}$ angle is $2: \\sqrt{3}$.\n\nNote that each of $\\triangle A B C, \\triangle A C D, \\triangle A D E, \\triangle A E F, \\triangle A F G$, and $\\triangle A G H$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $\\frac{A H}{A G}=\\frac{A G}{A F}=\\frac{A F}{A E}=\\frac{A E}{A D}=\\frac{A D}{A C}=\\frac{A C}{A B}=\\frac{2}{\\sqrt{3}}$.\n\nThus, $A H=\\frac{2}{\\sqrt{3}} A G=\\left(\\frac{2}{\\sqrt{3}}\\right)^{2} A F=\\left(\\frac{2}{\\sqrt{3}}\\right)^{3} A E=\\left(\\frac{2}{\\sqrt{3}}\\right)^{4} A D=\\left(\\frac{2}{\\sqrt{3}}\\right)^{5} A C=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6} A B$.\n\n(In other words, to get from $A B=1$ to the length of $A H$, we multiply by the ""scaling factor"" $\\frac{2}{\\sqrt{3}}$ six times.)\n\nTherefore, $A H=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6}=\\frac{64}{27}$.']",['$\\frac{64}{27}$'],False,,Numerical, 2373,Geometry,,"A snail's shell is formed from six triangular sections, as shown. Each triangle has interior angles of $30^{\circ}, 60^{\circ}$ and $90^{\circ}$. If $A B$ has a length of $1 \mathrm{~cm}$, what is the length of $A H$, in $\mathrm{cm}$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_6612ae75d8ef169f4be5g-1.jpg?height=288&width=523&top_left_y=1905&top_left_x=1256)","['In a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, the ratio of the side opposite the $90^{\\circ}$ to the side opposite the $60^{\\circ}$ angle is $2: \\sqrt{3}$.\n\nNote that each of $\\triangle A B C, \\triangle A C D, \\triangle A D E, \\triangle A E F, \\triangle A F G$, and $\\triangle A G H$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $\\frac{A H}{A G}=\\frac{A G}{A F}=\\frac{A F}{A E}=\\frac{A E}{A D}=\\frac{A D}{A C}=\\frac{A C}{A B}=\\frac{2}{\\sqrt{3}}$.\n\nThus, $A H=\\frac{2}{\\sqrt{3}} A G=\\left(\\frac{2}{\\sqrt{3}}\\right)^{2} A F=\\left(\\frac{2}{\\sqrt{3}}\\right)^{3} A E=\\left(\\frac{2}{\\sqrt{3}}\\right)^{4} A D=\\left(\\frac{2}{\\sqrt{3}}\\right)^{5} A C=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6} A B$.\n\n(In other words, to get from $A B=1$ to the length of $A H$, we multiply by the ""scaling factor"" $\\frac{2}{\\sqrt{3}}$ six times.)\n\nTherefore, $A H=\\left(\\frac{2}{\\sqrt{3}}\\right)^{6}=\\frac{64}{27}$.']",['$\\frac{64}{27}$'],False,,Numerical, 2374,Geometry,,"In rectangle $A B C D$, point $E$ is on side $D C$. Line segments $A E$ and $B D$ are perpendicular and intersect at $F$. If $A F=4$ and $D F=2$, determine the area of quadrilateral $B C E F$. ","['Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n\n\n\n\nTherefore, $\\triangle B F A, \\triangle A F D$, and $\\triangle D F E$ are all similar as each is right-angled and has either an angle of $\\beta$ or an angle of $90^{\\circ}-\\beta$ (and hence both of these angles).\n\nTherefore, $\\frac{A B}{A F}=\\frac{D A}{D F}$ and so $A B=\\frac{4(2 \\sqrt{5})}{2}=4 \\sqrt{5}$.\n\nAlso, $\\frac{F E}{F D}=\\frac{F D}{F A}$ and so $F E=\\frac{2(2)}{4}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $B C E F$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$', 'Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$. Since $\\triangle B A F$ is right-angled at $F$, then $\\angle A B F=\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n\n\nLooking at $\\triangle A F D$, we see that $\\sin \\beta=\\frac{F D}{A D}=\\frac{2}{2 \\sqrt{5}}=\\frac{1}{\\sqrt{5}}, \\cos \\beta=\\frac{A F}{A D}=\\frac{4}{2 \\sqrt{5}}=\\frac{2}{\\sqrt{5}}$, and $\\tan \\beta=\\frac{F D}{A F}=\\frac{2}{4}=\\frac{1}{2}$.\n\nSince $A F=4$ and $\\angle A B F=\\beta$, then $A B=\\frac{A F}{\\sin \\beta}=\\frac{4}{\\frac{1}{\\sqrt{5}}}=4 \\sqrt{5}$.\n\nSince $F D=2$ and $\\angle F D E=\\beta$, then $F E=F D \\tan \\beta=2 \\cdot \\frac{1}{2}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $E F B C$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$']",['19'],False,,Numerical, 2374,Geometry,,"In rectangle $A B C D$, point $E$ is on side $D C$. Line segments $A E$ and $B D$ are perpendicular and intersect at $F$. If $A F=4$ and $D F=2$, determine the area of quadrilateral $B C E F$. ![](https://cdn.mathpix.com/cropped/2023_12_21_6612ae75d8ef169f4be5g-1.jpg?height=339&width=483&top_left_y=2243&top_left_x=1252)","['Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_98e0ff93ea7be169163dg-1.jpg?height=331&width=548&top_left_y=165&top_left_x=889)\n\nTherefore, $\\triangle B F A, \\triangle A F D$, and $\\triangle D F E$ are all similar as each is right-angled and has either an angle of $\\beta$ or an angle of $90^{\\circ}-\\beta$ (and hence both of these angles).\n\nTherefore, $\\frac{A B}{A F}=\\frac{D A}{D F}$ and so $A B=\\frac{4(2 \\sqrt{5})}{2}=4 \\sqrt{5}$.\n\nAlso, $\\frac{F E}{F D}=\\frac{F D}{F A}$ and so $F E=\\frac{2(2)}{4}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $B C E F$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$', 'Since $\\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{A F^{2}+F D^{2}}=\\sqrt{4^{2}+2^{2}}=\\sqrt{20}=2 \\sqrt{5}\n$$\n\nsince $A D>0$.\n\nLet $\\angle F A D=\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B A F=90^{\\circ}-\\beta$. Since $\\triangle B A F$ is right-angled at $F$, then $\\angle A B F=\\beta$.\n\nSince $\\triangle A F D$ is right-angled at $F$, then $\\angle A D F=90^{\\circ}-\\beta$.\n\nSince $A B C D$ is a rectangle, then $\\angle B D C=90^{\\circ}-\\left(90^{\\circ}-\\beta\\right)=\\beta$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_98e0ff93ea7be169163dg-1.jpg?height=328&width=544&top_left_y=1663&top_left_x=888)\n\nLooking at $\\triangle A F D$, we see that $\\sin \\beta=\\frac{F D}{A D}=\\frac{2}{2 \\sqrt{5}}=\\frac{1}{\\sqrt{5}}, \\cos \\beta=\\frac{A F}{A D}=\\frac{4}{2 \\sqrt{5}}=\\frac{2}{\\sqrt{5}}$, and $\\tan \\beta=\\frac{F D}{A F}=\\frac{2}{4}=\\frac{1}{2}$.\n\nSince $A F=4$ and $\\angle A B F=\\beta$, then $A B=\\frac{A F}{\\sin \\beta}=\\frac{4}{\\frac{1}{\\sqrt{5}}}=4 \\sqrt{5}$.\n\nSince $F D=2$ and $\\angle F D E=\\beta$, then $F E=F D \\tan \\beta=2 \\cdot \\frac{1}{2}=1$.\n\nSince $A B C D$ is a rectangle, then $B C=A D=2 \\sqrt{5}$, and $D C=A B=4 \\sqrt{5}$.\n\nFinally, the area of quadrilateral $E F B C$ equals the area of $\\triangle D C B$ minus the area $\\triangle D F E$. Thus, the required area is\n\n$$\n\\frac{1}{2}(D C)(C B)-\\frac{1}{2}(D F)(F E)=\\frac{1}{2}(4 \\sqrt{5})(2 \\sqrt{5})-\\frac{1}{2}(2)(1)=20-1=19\n$$']",['19'],False,,Numerical, 2375,Algebra,,Determine all real values of $x$ for which $3^{(x-1)} 9^{\frac{3}{2 x^{2}}}=27$.,"['Using the facts that $9=3^{2}$ and $27=3^{3}$, and the laws for manipulating exponents, we have\n\n$$\n\\begin{aligned}\n3^{x-1} 9^{\\frac{3}{2 x^{2}}} & =27 \\\\\n3^{x-1}\\left(3^{2}\\right)^{\\frac{3}{2 x^{2}}} & =3^{3} \\\\\n3^{x-1} 3^{\\frac{3}{x^{2}}} & =3^{3} \\\\\n3^{x-1+\\frac{3}{x^{2}}} & =3^{3}\n\\end{aligned}\n$$\n\nWhen two powers of 3 are equal, their exponents must be equal so\n\n$$\n\\begin{aligned}\nx-1+\\frac{3}{x^{2}} & =3 \\\\\nx^{3}-x^{2}+3 & \\left.=3 x^{2} \\quad \\text { (multiplying by } x^{2}\\right) \\\\\nx^{3}-4 x^{2}+3 & =0\n\\end{aligned}\n$$\n\nSince $x=1$ satisfies the equation, then $x-1$ is a factor of the left side. Using long division or synthetic division, we can factor this out to get $(x-1)\\left(x^{2}-3 x-3\\right)=0$.\n\nUsing the quadratic formula, the quadratic equation $x^{2}-3 x-3=0$ has roots\n\n$$\nx=\\frac{3 \\pm \\sqrt{(-3)^{2}-4(1)(-3)}}{2}=\\frac{3 \\pm \\sqrt{21}}{2}\n$$\n\nTherefore, the solutions to the original equation are $x=1$ and $x=\\frac{3 \\pm \\sqrt{21}}{2}$.']","['$1$,$\\frac{3 + \\sqrt{21}}{2}$,$\\frac{3 - \\sqrt{21}}{2}$']",True,,Numerical, 2376,Algebra,,"Determine all points $(x, y)$ where the two curves $y=\log _{10}\left(x^{4}\right)$ and $y=\left(\log _{10} x\right)^{3}$ intersect.","['To determine the points of intersection, we equate $y$ values of the two curves and obtain $\\log _{10}\\left(x^{4}\\right)=\\left(\\log _{10} x\\right)^{3}$.\n\nSince $\\log _{10}\\left(a^{b}\\right)=b \\log _{10} a$, the equation becomes $4 \\log _{10} x=\\left(\\log _{10} x\\right)^{3}$.\n\nWe set $u=\\log _{10} x$ and so the equation becomes $4 u=u^{3}$, or $u^{3}-4 u=0$.\n\nWe can factor the left side as $u^{3}-4 u=u\\left(u^{2}-4\\right)=u(u+2)(u-2)$.\n\nTherefore, $u(u+2)(u-2)=0$, and so $u=0$ or $u=-2$ or $u=2$.\n\nTherefore, $\\log _{10} x=0$ or $\\log _{10} x=-2$ or $\\log _{10} x=2$.\n\nTherefore, $x=1$ or $x=\\frac{1}{100}$ or $x=100$.\n\nFinally, we must calculate the $y$-coordinates of the points of intersection. Since one of the original curves is $y=\\left(\\log _{10} x\\right)^{3}$, we can calculate the corresponding values of $y$ by using the fact that $y=u^{3}$.\n\nThe corresponding values of $y$ are $y=0^{3}=0$ and $y=(-2)^{3}=-8$ and $y=2^{3}=8$.\n\nTherefore, the points of intersection are $(1,0),\\left(\\frac{1}{100},-8\\right)$ and $(100,8)$.']","['$(1,0),(\\frac{1}{100},-8),(100,8)$']",True,,Tuple, 2377,Combinatorics,,"Oi-Lam tosses three fair coins and removes all of the coins that come up heads. George then tosses the coins that remain, if any. Determine the probability that George tosses exactly one head.","['If Oi-Lam tosses 3 heads, then George has no coins to toss, so cannot toss exactly 1 head. If Oi-Lam tosses 2, 1 or 0 heads, then George has at least one coin to toss, so can toss exactly 1 head.\n\nTherefore, the following possibilities exist:\n\n* Oi-Lam tosses 2 heads out of 3 coins and George tosses 1 head out of 1 coin\n* Oi-Lam tosses 1 head out of 3 coins and George tosses 1 head out of 2 coins\n* Oi-Lam tosses 0 heads out of 3 coins and George tosses 1 head out of 3 coins\n\nWe calculate the various probabilities.\n\nIf 3 coins are tossed, there are 8 equally likely possibilities: $\\mathrm{HHH}, \\mathrm{HHT}, \\mathrm{HTH}, \\mathrm{THH}, \\mathrm{TTH}$, THT, HTT, TTT. Each of these possibilities has probability $\\left(\\frac{1}{2}\\right)^{3}=\\frac{1}{8}$. Therefore,\n\n\n\n* the probability of tossing 0 heads out of 3 coins is $\\frac{1}{8}$\n* the probability of tossing 1 head out of 3 coins is $\\frac{3}{8}$\n* the probability of tossing 2 heads out of 3 coins is $\\frac{3}{8}$\n* the probability of tossing 3 heads out of 3 coins is $\\frac{1}{8}$\n\nIf 2 coins are tossed, there are 4 equally likely possibilities: HH, HT, TH, TT. Each of these possibilities has probability $\\left(\\frac{1}{2}\\right)^{2}=\\frac{1}{4}$. Therefore, the probability of tossing 1 head out of 2 coins is $\\frac{2}{4}=\\frac{1}{2}$.\n\nIf 1 coin is tossed, the probability of tossing 1 head is $\\frac{1}{2}$.\n\nTo summarize, the possibilities are\n\n* Oi-Lam tosses 2 heads out of 3 coins (with probability $\\frac{3}{8}$ ) and George tosses 1 head out of 1 coin (with probability $\\frac{1}{2}$ )\n* Oi-Lam tosses 1 head out of 3 coins (with probability $\\frac{3}{8}$ ) and George tosses 1 head out of 2 coins (with probability $\\frac{1}{2}$ )\n* Oi-Lam tosses 0 heads out of 3 coins (with probability $\\frac{1}{8}$ ) and George tosses 1 head out of 3 coins (with probability $\\frac{3}{8}$ )\n\nTherefore, the overall probability is $\\frac{3}{8} \\cdot \\frac{1}{2}+\\frac{3}{8} \\cdot \\frac{1}{2}+\\frac{1}{8} \\cdot \\frac{3}{8}=\\frac{27}{64}$.']",['$\\frac{27}{64}$'],False,,Numerical, 2378,Geometry,,"In the diagram, points $B, P, Q$, and $C$ lie on line segment $A D$. The semi-circle with diameter $A C$ has centre $P$ and the semi-circle with diameter $B D$ has centre $Q$. The two semi-circles intersect at $R$. If $\angle P R Q=40^{\circ}$, determine the measure of $\angle A R D$. ","['Suppose $\\angle P A R=x^{\\circ}$ and $\\angle Q D R=y^{\\circ}$.\n\n\n\nSince $P R$ and $P A$ are radii of the larger circle, then $\\triangle P A R$ is isosceles.\n\nThus, $\\angle P R A=\\angle P A R=x^{\\circ}$.\n\nSince $Q D$ and $Q R$ are radii of the smaller circle, then $\\triangle Q R D$ is isosceles.\n\nThus, $\\angle Q R D=\\angle Q D R=y^{\\circ}$.\n\nIn $\\triangle A R D$, the sum of the angles is $180^{\\circ}$, so $x^{\\circ}+\\left(x^{\\circ}+40^{\\circ}+y^{\\circ}\\right)+y^{\\circ}=180^{\\circ}$ or $2 x+2 y=140$ or $x+y=70$.\n\nTherefore, $\\angle C P D=x^{\\circ}+40^{\\circ}+y^{\\circ}=(x+y+40)^{\\circ}=110^{\\circ}$.']",['$110^{\\circ}$'],False,,Numerical, 2378,Geometry,,"In the diagram, points $B, P, Q$, and $C$ lie on line segment $A D$. The semi-circle with diameter $A C$ has centre $P$ and the semi-circle with diameter $B D$ has centre $Q$. The two semi-circles intersect at $R$. If $\angle P R Q=40^{\circ}$, determine the measure of $\angle A R D$. ![](https://cdn.mathpix.com/cropped/2023_12_21_8305addb4aaa1260f985g-1.jpg?height=271&width=545&top_left_y=683&top_left_x=1256)","['Suppose $\\angle P A R=x^{\\circ}$ and $\\angle Q D R=y^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_96c935933a9e5a2ec6ebg-1.jpg?height=267&width=555&top_left_y=1127&top_left_x=885)\n\nSince $P R$ and $P A$ are radii of the larger circle, then $\\triangle P A R$ is isosceles.\n\nThus, $\\angle P R A=\\angle P A R=x^{\\circ}$.\n\nSince $Q D$ and $Q R$ are radii of the smaller circle, then $\\triangle Q R D$ is isosceles.\n\nThus, $\\angle Q R D=\\angle Q D R=y^{\\circ}$.\n\nIn $\\triangle A R D$, the sum of the angles is $180^{\\circ}$, so $x^{\\circ}+\\left(x^{\\circ}+40^{\\circ}+y^{\\circ}\\right)+y^{\\circ}=180^{\\circ}$ or $2 x+2 y=140$ or $x+y=70$.\n\nTherefore, $\\angle C P D=x^{\\circ}+40^{\\circ}+y^{\\circ}=(x+y+40)^{\\circ}=110^{\\circ}$.']",['$110^{\\circ}$'],False,,Numerical, 2379,Algebra,,"If $\theta$ is an angle whose measure is not an integer multiple of $90^{\circ}$, prove that $$ \cot \theta-\cot 2 \theta=\frac{1}{\sin 2 \theta} $$","['$$\n\\begin{aligned}\n\\mathrm{LS} & =\\cot \\theta-\\cot 2 \\theta \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{\\sin 2 \\theta} \\\\\n& =\\frac{\\sin 2 \\theta \\cos \\theta-\\cos 2 \\theta \\sin \\theta}{\\sin \\theta \\sin 2 \\theta} \\\\\n& =\\frac{\\sin (2 \\theta-\\theta)}{\\sin \\theta \\sin 2 \\theta} \\\\\n& =\\frac{\\sin \\theta}{\\sin \\theta \\sin 2 \\theta} \\\\\n& =\\frac{1}{\\sin 2 \\theta} \\\\\n& =\\mathrm{RS}\n\\end{aligned}\n$$\n\nas required.', '$$\n\\begin{aligned}\n\\mathrm{LS} & =\\cot \\theta-\\cot 2 \\theta \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{\\sin 2 \\theta} \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{2 \\sin \\theta \\cos \\theta} \\\\\n& =\\frac{2 \\cos ^{2} \\theta-\\cos 2 \\theta}{2 \\sin \\theta \\cos \\theta} \\\\\n& =\\frac{2 \\cos ^{2} \\theta-\\left(2 \\cos ^{2} \\theta-1\\right)}{\\sin 2 \\theta} \\\\\n& =\\frac{1}{\\sin 2 \\theta} \\\\\n& =\\mathrm{RS}\n\\end{aligned}\n$$\n\nas required.']",,True,,, 2380,Algebra,,"Ross starts with an angle of measure $8^{\circ}$ and doubles it 10 times until he obtains $8192^{\circ}$. He then adds up the reciprocals of the sines of these 11 angles. That is, he calculates $$ S=\frac{1}{\sin 8^{\circ}}+\frac{1}{\sin 16^{\circ}}+\frac{1}{\sin 32^{\circ}}+\cdots+\frac{1}{\sin 4096^{\circ}}+\frac{1}{\sin 8192^{\circ}} $$ Determine, without using a calculator, the measure of the acute angle $\alpha$ so that $S=\frac{1}{\sin \alpha}$.","['We first prove Lemma(i): If $\\theta$ is an angle whose measure is not an integer multiple of $90^{\\circ}$, then\n$$\n\\cot \\theta-\\cot 2 \\theta=\\frac{1}{\\sin 2 \\theta}\n$$\n\nProof. \n$$\n\\begin{aligned}\n\\mathrm{LS} & =\\cot \\theta-\\cot 2 \\theta \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{\\sin 2 \\theta} \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{2 \\sin \\theta \\cos \\theta} \\\\\n& =\\frac{2 \\cos ^{2} \\theta-\\cos 2 \\theta}{2 \\sin \\theta \\cos \\theta} \\\\\n& =\\frac{2 \\cos ^{2} \\theta-\\left(2 \\cos ^{2} \\theta-1\\right)}{\\sin 2 \\theta} \\\\\n& =\\frac{1}{\\sin 2 \\theta} \\\\\n& =\\mathrm{RS}\n\\end{aligned}\n$$\n\nas required.\n\nWe use (i) to note that $\\frac{1}{\\sin 8^{\\circ}}=\\cot 4^{\\circ}-\\cot 8^{\\circ}$ and $\\frac{1}{\\sin 16^{\\circ}}=\\cot 8^{\\circ}-\\cot 16^{\\circ}$ and so on. Thus,\n\n$$\n\\begin{aligned}\nS= & \\frac{1}{\\sin 8^{\\circ}}+\\frac{1}{\\sin 16^{\\circ}}+\\frac{1}{\\sin 32^{\\circ}}+\\cdots+\\frac{1}{\\sin 4096^{\\circ}}+\\frac{1}{\\sin 8192^{\\circ}} \\\\\n= & \\left(\\cot 4^{\\circ}-\\cot 8^{\\circ}\\right)+\\left(\\cot 8^{\\circ}-\\cot 16^{\\circ}\\right)+\\left(\\cot 16^{\\circ}-\\cot 32^{\\circ}\\right)+ \\\\\n& \\cdots+\\left(\\cot 2048^{\\circ}-\\cot 4096^{\\circ}\\right)+\\left(\\cot 4096^{\\circ}-\\cot 8192^{\\circ}\\right) \\\\\n= & \\cot 4^{\\circ}-\\cot 8192^{\\circ}\n\\end{aligned}\n$$\n\nsince the sum ""telescopes"".\n\nSince the cotangent function has a period of $180^{\\circ}$, and $8100^{\\circ}$ is a multiple of $180^{\\circ}$, then $\\cot 8192^{\\circ}=\\cot 92^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nS & =\\cot 4^{\\circ}-\\cot 92^{\\circ} \\\\\n& =\\frac{\\cos 4^{\\circ}}{\\sin 4^{\\circ}}-\\frac{\\cos 92^{\\circ}}{\\sin 92^{\\circ}} \\\\\n& =\\frac{\\cos 4^{\\circ}}{\\sin 4^{\\circ}}-\\frac{-\\sin 2^{\\circ}}{\\cos 2^{\\circ}} \\\\\n& =\\frac{\\cos 4^{\\circ}}{2 \\sin 2^{\\circ} \\cos 2^{\\circ}}+\\frac{\\sin 2^{\\circ}}{\\cos 2^{\\circ}} \\\\\n& =\\frac{\\cos 4^{\\circ}+2 \\sin ^{2} 2^{\\circ}}{2 \\sin 2^{\\circ} \\cos 2^{\\circ}} \\\\\n& =\\frac{\\left(1-2 \\sin ^{2} 2^{\\circ}\\right)+2 \\sin ^{2} 2^{\\circ}}{\\sin 4^{\\circ}} \\\\\n& =\\frac{1}{\\sin 4^{\\circ}}\n\\end{aligned}\n$$\n\nTherefore, $\\alpha=4^{\\circ}$.']",['$4^{\\circ}$'],False,,Numerical, 2381,Geometry,,"In $\triangle A B C, B C=a, A C=b, A B=c$, and $a<\frac{1}{2}(b+c)$. Prove that $\angle B A C<\frac{1}{2}(\angle A B C+\angle A C B)$.","['We use the notation $A=\\angle B A C, B=\\angle A B C$ and $C=\\angle A C B$.\n\nWe need to show that $A<\\frac{1}{2}(B+C)$. Since the sum of the angles in $\\triangle A B C$ is $180^{\\circ}$, then $B+C=180^{\\circ}-A$, and so this inequality is equivalent to $A<\\frac{1}{2}\\left(180^{\\circ}-A\\right)$ which is equivalent to $\\frac{3}{2} A<90^{\\circ}$ or $A<60^{\\circ}$.\n\nSo we need to show that $A<60^{\\circ}$.\n\nWe know that $a<\\frac{1}{2}(b+c)$. Thus, $2 a\\frac{1}{2} \\quad(\\text { since } 8 b c>0)\n\\end{aligned}\n$$\n\nSince $2 a\\frac{1}{2}$ implies $A<60^{\\circ}$, as required.', 'We use the notation $A=\\angle B A C, B=\\angle A B C$ and $C=\\angle A C B$.\n\nWe need to show that $A<\\frac{1}{2}(B+C)$. Since the sum of the angles in $\\triangle A B C$ is $180^{\\circ}$, then $B+C=180^{\\circ}-A$, and so this inequality is equivalent to $A<\\frac{1}{2}\\left(180^{\\circ}-A\\right)$ which is equivalent to $\\frac{3}{2} A<90^{\\circ}$ or $A<60^{\\circ}$.\n\nSo we need to show that $A<60^{\\circ}$.\n\nWe know that $a<\\frac{1}{2}(b+c)$ which implies $2 a0 \\text { for } 0^{\\circ}0$. Next, we use the trigonometric formula $\\sin B+\\sin C=2 \\sin \\left(\\frac{B+C}{2}\\right) \\cos \\left(\\frac{B-C}{2}\\right)$.\n\nSince $\\cos \\theta \\leq 1$ for any $\\theta$, then $\\sin B+\\sin C \\leq 2 \\sin \\left(\\frac{B+C}{2}\\right) \\cdot 1=2 \\sin \\left(\\frac{B+C}{2}\\right)$.\n\n\n\nTherefore,\n\n$$\n\\begin{aligned}\n2 \\sin A & <\\sin B+\\sin C \\leq 2 \\sin \\left(\\frac{B+C}{2}\\right) \\\\\n2 \\sin A & <2 \\sin \\left(\\frac{B+C}{2}\\right) \\\\\n2 \\sin A & <2 \\sin \\left(\\frac{180^{\\circ}-A}{2}\\right) \\\\\n4 \\sin \\left(\\frac{1}{2} A\\right) \\cos \\left(\\frac{1}{2} A\\right) & <2 \\sin \\left(90^{\\circ}-\\frac{1}{2} A\\right) \\\\\n2 \\sin \\left(\\frac{1}{2} A\\right) \\cos \\left(\\frac{1}{2} A\\right) & <\\cos \\left(\\frac{1}{2} A\\right)\n\\end{aligned}\n$$\n\nSince $0^{\\circ}0$, so $\\sin \\left(\\frac{1}{2} A\\right)<\\frac{1}{2}$.\n\nSince $2 a2010$.","['Denote the side lengths of a triangle by $a, b$ and $c$, with $02010$.\n\nThen $N$ must be odd:\n\nIf $N$ was even, then by (F1), $T(N-3)=T(N)>2010$ and so $n=N-3$ would be an integer smaller than $N$ with $T(n)>2010$. This contradicts the fact that $n=N$ is the smallest such integer.\n\nTherefore, we want to find the smallest odd positive integer $N$ for which $T(N)>2010$. Next, we note that if we can find an odd positive integer $n$ such that $T(n)>2010 \\geq$ $T(n-2)$, then we will have found the desired value of $n$ :\n\nThis is because $n$ and $n-2$ are both odd, and by property (F2), any smaller odd positive integer $k$ will give $T(k) \\leq T(n-2) \\leq 2010$ and any larger odd positive integer $m$ will give $T(m) \\geq T(n)>2010$.\n\nWe show that $N=309$ is the desired value of $N$ by showing that $T(309)>2010$ and $T(307) \\leq 2010$.\n\nCalculation of $T(309)$\n\nWe know that $\\frac{309}{3} \\leq c<\\frac{309}{2}$, so $103 \\leq c \\leq 154$.\n\nFor each admissible value of $c$, we need to count the number of pairs of positive integers $(a, b)$ with $a \\leq b \\leq c$ and $a+b=309-c$.\n\nFor example, if $c=154$, then we need $a \\leq b \\leq 154$ and $a+b=155$.\n\nThis gives pairs $(1,154),(2,153), \\ldots,(76,79),(77,78)$, of which there are 77 .\n\nAlso, if $c=153$, then we need $a \\leq b \\leq 153$ and $a+b=156$.\n\nThis gives pairs $(3,153), \\ldots,(77,79),(78,78)$, of which there are 76 .\n\nIn general, if $c$ is even, then the minimum possible value of $a$ occurs when $b$ is as large as possible - that is, when $b=c$, so $a \\geq 309-2 c$.\n\nAlso, the largest possible value of $a$ occurs when $a$ and $b$ are as close to equal as possible. Since $c$ is even, then $309-c$ is odd, so $a$ and $b$ cannot be equal, but they can differ by 1 . In this case, $a=154-\\frac{1}{2} c$ and $b=155-\\frac{1}{2} c$.\n\nTherefore, if $c$ is even, there are $\\left(154-\\frac{1}{2} c\\right)-(309-2 c)+1=\\frac{3}{2} c-154$ possible pairs $(a, b)$ and so $\\frac{3}{2} c-154$ possible triples.\n\nIn general, if $c$ is odd, then the minimum possible value of $a$ occurs when $b$ is as large as possible - that is, when $b=c$, so $a \\geq 309-2 c$.\n\nAlso, the largest possible value of $a$ occurs when $a$ and $b$ are as close to equal as possible.\n\nSince $c$ is odd, then $309-c$ is even, so $a$ and $b$ can be equal. In this case, $a=\\frac{1}{2}(309-c)$. Therefore, if $c$ is odd, there are $\\frac{1}{2}(309-c)-(309-2 c)+1=\\frac{3}{2} c-\\frac{307}{2}$ possible pairs $(a, b)$ and so $\\frac{3}{2} c-\\frac{307}{2}$ possible triples.\n\nThe possible even values of $c$ are 104,106,...,152,154 (there are 26 such values) and the possible odd values of $c$ are 103,105,...,151,153 (there are 26 such values).\n\n\n\nTherefore,\n\n$$\n\\begin{aligned}\nT(309)= & \\left(\\frac{3}{2}(104)-154\\right)+\\left(\\frac{3}{2}(106)-154\\right)+\\cdots+\\left(\\frac{3}{2}(154)-154\\right)+ \\\\\n& \\quad\\left(\\frac{3}{2}(103)-\\frac{307}{2}\\right)+\\left(\\frac{3}{2}(105)-\\frac{307}{2}\\right)+\\cdots+\\left(\\frac{3}{2}(153)-\\frac{307}{2}\\right) \\\\\n= & \\frac{3}{2}(104+106+\\cdots+154)-26 \\cdot 154+\\frac{3}{2}(103+105+\\cdots+153)-26 \\cdot \\frac{307}{2} \\\\\n= & \\frac{3}{2}(103+104+105+106+\\cdots+153+154)-26 \\cdot 154-26 \\cdot \\frac{307}{2} \\\\\n= & \\frac{3}{2} \\cdot \\frac{1}{2}(103+154)(52)-26 \\cdot 154-26 \\cdot \\frac{307}{2} \\\\\n= & \\frac{3}{2}(26)(257)-26 \\cdot 154-26 \\cdot \\frac{307}{2} \\\\\n= & 2028\n\\end{aligned}\n$$\n\nTherefore, $T(309)>2010$, as required.\n\nCalculation of $T(307)$\n\nWe know that $\\frac{307}{3} \\leq c<\\frac{307}{2}$, so $103 \\leq c \\leq 153$.\n\nFor each admissible value of $c$, we need to count the number of pairs of positive integers $(a, b)$ with $a \\leq b \\leq c$ and $a+b=307-c$.\n\nThis can be done in a similar way to the calculation of $T(309)$ above.\n\nIf $n$ is even, there are $\\frac{3}{2} c-153$ possible triples.\n\nIf $n$ is odd, there are $\\frac{3}{2} c-\\frac{305}{2}$ possible triples.\n\nThe possible even values of $c$ are $104,106, \\ldots, 150,152$ (there are 25 such values) and the possible odd values of $c$ are $103,105, \\ldots, 151,153$ (there are 26 such values).\n\nTherefore,\n\n$$\n\\begin{aligned}\nT(307)= & \\left(\\frac{3}{2}(104)-153\\right)+\\left(\\frac{3}{2}(106)-153\\right)+\\cdots+\\left(\\frac{3}{2}(152)-153\\right)+ \\\\\n& \\quad\\left(\\frac{3}{2}(103)-\\frac{305}{2}\\right)+\\left(\\frac{3}{2}(105)-\\frac{305}{2}\\right)+\\cdots+\\left(\\frac{3}{2}(153)-\\frac{305}{2}\\right) \\\\\n= & \\frac{3}{2}(104+106+\\cdots+152)-25 \\cdot 153+\\frac{3}{2}(103+105+\\cdots+153)-26 \\cdot \\frac{305}{2} \\\\\n= & \\frac{3}{2}(103+104+105+106+\\cdots+152+153)-25 \\cdot 153-26 \\cdot \\frac{305}{2} \\\\\n= & \\frac{3}{2} \\cdot \\frac{1}{2}(103+153)(51)-25 \\cdot 153-26 \\cdot \\frac{305}{2} \\\\\n= & \\frac{3}{2}(51)(128)-25 \\cdot 153-26 \\cdot \\frac{305}{2} \\\\\n= & 2002\n\\end{aligned}\n$$\n\nTherefore, $T(307)<2010$, as required.\n\nTherefore, the smallest positive integer $n$ such that $T(n)>2010$ is $n=309$.\n\nAs a final note, we discuss briefly how one could guess that the answer was near $N=309$.\n\nConsider the values of $T(n)$ for small odd positive integers $n$.\n\nIn (a), by considering the possible values of $c$ from smallest (roughly $\\frac{1}{3} n$ ) to largest (roughly $\\frac{1}{2} n$ ), we saw that $T(11)=1+3=4$.\n\nIf we continue to calculate $T(n)$ for a few more small odd values of $n$ we will see that:\n\n$$\n\\begin{aligned}\n& T(13)=2+3=5 \\\\\n& T(15)=1+2+4=7 \\\\\n& T(17)=1+3+4=8 \\\\\n& T(19)=2+3+5=10 \\\\\n& T(21)=1+2+4+5=12 \\\\\n& T(23)=1+3+4+6=14\n\\end{aligned}\n$$\n\n\n\nThe pattern that seems to emerge is that for $n$ odd, $T(n)$ is roughly equal to the sum of the integers from 1 to $\\frac{1}{4} n$, with one out of every three integers removed. Thus, $T(n)$ is roughly equal to $\\frac{2}{3}$ of the sum of the integers from 1 to $\\frac{1}{4} n$. Therefore, $T(n) \\approx \\frac{2}{3} \\cdot \\frac{1}{2}\\left(\\frac{1}{4} n\\right)\\left(\\frac{1}{4} n+1\\right) \\approx \\frac{2}{3} \\cdot \\frac{1}{2}\\left(\\frac{1}{4} n\\right)^{2} \\approx \\frac{1}{48} n^{2}$.\n\nIt makes sense to look for an odd positive integer $n$ with $T(n) \\approx 2010$.\n\nThus, we are looking for a value of $n$ that roughly satisfies $\\frac{1}{48} n^{2} \\approx 2010$ or $n^{2} \\approx 96480$ or $n \\approx 310$.\n\nSince $n$ is odd, then it makes sense to consider $n=309$, as in the solution above.']",['309'],False,,Numerical, 2385,Geometry,,"In the diagram, a line is drawn through points $P, Q$ and $R$. If $P Q=Q R$, what are the coordinates of $R$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=363&width=564&top_left_y=840&top_left_x=1255)","['To get from $P$ to $Q$, we move 3 units right and 4 units up.\n\nSince $P Q=Q R$ and $R$ lies on the line through $Q$, then we must use the same motion to get from $Q$ to $R$.\n\nTherefore, to get from $Q(0,4)$ to $R$, we move 3 units right and 4 units up, so the coordinates of $R$ are $(3,8)$.', 'The line through $P(-3,0)$ and $Q(0,4)$ has slope $\\frac{4-0}{0-(-3)}=\\frac{4}{3}$ and $y$-intercept 4 , so has equation $y=\\frac{4}{3} x+4$.\n\nThus, $R$ has coordinates $\\left(a, \\frac{4}{3} a+4\\right)$ for some $a>0$.\n\nSince $P Q=Q R$, then $P Q^{2}=Q R^{2}$, so\n\n$$\n\\begin{aligned}\n(-3)^{2}+4^{2} & =a^{2}+\\left(\\frac{4}{3} a+4-4\\right)^{2} \\\\\n25 & =a^{2}+\\frac{16}{9} a^{2} \\\\\n\\frac{25}{9} a^{2} & =25 \\\\\na^{2} & =9\n\\end{aligned}\n$$\n\nso $a=3$ since $a>0$.\n\nThus, $R$ has coordinates $\\left(3, \\frac{4}{3}(3)+4\\right)=(3,8)$.']","['(3,8)']",False,,Tuple, 2385,Geometry,,"In the diagram, a line is drawn through points $P, Q$ and $R$. If $P Q=Q R$, what are the coordinates of $R$ ? ","['To get from $P$ to $Q$, we move 3 units right and 4 units up.\n\nSince $P Q=Q R$ and $R$ lies on the line through $Q$, then we must use the same motion to get from $Q$ to $R$.\n\nTherefore, to get from $Q(0,4)$ to $R$, we move 3 units right and 4 units up, so the coordinates of $R$ are $(3,8)$.', 'The line through $P(-3,0)$ and $Q(0,4)$ has slope $\\frac{4-0}{0-(-3)}=\\frac{4}{3}$ and $y$-intercept 4 , so has equation $y=\\frac{4}{3} x+4$.\n\nThus, $R$ has coordinates $\\left(a, \\frac{4}{3} a+4\\right)$ for some $a>0$.\n\nSince $P Q=Q R$, then $P Q^{2}=Q R^{2}$, so\n\n$$\n\\begin{aligned}\n(-3)^{2}+4^{2} & =a^{2}+\\left(\\frac{4}{3} a+4-4\\right)^{2} \\\\\n25 & =a^{2}+\\frac{16}{9} a^{2} \\\\\n\\frac{25}{9} a^{2} & =25 \\\\\na^{2} & =9\n\\end{aligned}\n$$\n\nso $a=3$ since $a>0$.\n\nThus, $R$ has coordinates $\\left(3, \\frac{4}{3}(3)+4\\right)=(3,8)$.']","['(3,8)']",False,,Tuple, 2386,Geometry,,"In the diagram, $O A=15, O P=9$ and $P B=4$. Determine the equation of the line through $A$ and $B$. Explain how you got your answer. ","['Since $O P=9$, then the coordinates of $P$ are $(9,0)$.\n\nSince $O P=9$ and $O A=15$, then by the Pythagorean Theorem,\n\n$$\nA P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144\n$$\n\nso $A P=12$.\n\nSince $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.\n\nSince $P B=4$, then $B$ has coordinates $(13,0)$.\n\nThe line through $A(9,12)$ and $B(13,0)$ has slope $\\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$.']",['$y=-3 x+39$'],False,,Expression, 2386,Geometry,,"In the diagram, $O A=15, O P=9$ and $P B=4$. Determine the equation of the line through $A$ and $B$. Explain how you got your answer. ![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=366&width=444&top_left_y=1259&top_left_x=1250)","['Since $O P=9$, then the coordinates of $P$ are $(9,0)$.\n\nSince $O P=9$ and $O A=15$, then by the Pythagorean Theorem,\n\n$$\nA P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144\n$$\n\nso $A P=12$.\n\nSince $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.\n\nSince $P B=4$, then $B$ has coordinates $(13,0)$.\n\nThe line through $A(9,12)$ and $B(13,0)$ has slope $\\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$.']",['$y=-3 x+39$'],False,,Expression, 2387,Geometry,,"In the diagram, $\triangle A B C$ is right-angled at $B$ and $A B=10$. If $\cos (\angle B A C)=\frac{5}{13}$, what is the value of $\tan (\angle A C B)$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=385&width=284&top_left_y=1678&top_left_x=1251)","['Since $\\cos (\\angle B A C)=\\frac{A B}{A C}$ and $\\cos (\\angle B A C)=\\frac{5}{13}$ and $A B=10$, then $A C=\\frac{13}{5} A B=26$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem, $B C^{2}=A C^{2}-A B^{2}=26^{2}-10^{2}=576$ so $B C=24$ since $B C>0$.\n\nTherefore, $\\tan (\\angle A C B)=\\frac{A B}{B C}=\\frac{10}{24}=\\frac{5}{12}$.']",['$\\frac{5}{12}$'],False,,Numerical, 2387,Geometry,,"In the diagram, $\triangle A B C$ is right-angled at $B$ and $A B=10$. If $\cos (\angle B A C)=\frac{5}{13}$, what is the value of $\tan (\angle A C B)$ ? ","['Since $\\cos (\\angle B A C)=\\frac{A B}{A C}$ and $\\cos (\\angle B A C)=\\frac{5}{13}$ and $A B=10$, then $A C=\\frac{13}{5} A B=26$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem, $B C^{2}=A C^{2}-A B^{2}=26^{2}-10^{2}=576$ so $B C=24$ since $B C>0$.\n\nTherefore, $\\tan (\\angle A C B)=\\frac{A B}{B C}=\\frac{10}{24}=\\frac{5}{12}$.']",['$\\frac{5}{12}$'],False,,Numerical, 2388,Geometry,,Suppose $0^{\circ}0$ because $0^{\\circ}","['Since $\\triangle A B C$ is isosceles and right-angled, then $\\angle B A C=45^{\\circ}$.\n\nAlso, $A C=\\sqrt{2} A B=\\sqrt{2}(2 \\sqrt{2})=4$.\n\nSince $\\angle E A B=75^{\\circ}$ and $\\angle B A C=45^{\\circ}$, then $\\angle C A E=\\angle E A B-\\angle B A C=30^{\\circ}$.\n\nSince $\\triangle A E C$ is right-angled and has a $30^{\\circ}$ angle, then $\\triangle A E C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nThus, $E C=\\frac{1}{2} A C=2$ (since $E C$ is opposite the $30^{\\circ}$ angle) and $A E=\\frac{\\sqrt{3}}{2} A C=2 \\sqrt{3}$ (since $A E$ is opposite the $60^{\\circ}$ angle).\n\nIn $\\triangle C D E, E D=D C$ and $\\angle E D C=60^{\\circ}$, so $\\triangle C D E$ is equilateral.\n\nTherefore, $E D=C D=E C=2$.\n\nOverall, the perimeter of $A B C D E$ is\n\n$$\nA B+B C+C D+D E+E A=2 \\sqrt{2}+2 \\sqrt{2}+2+2+2 \\sqrt{3}=4+4 \\sqrt{2}+2 \\sqrt{3}\n$$']",['$4+4 \\sqrt{2}+2 \\sqrt{3}$'],False,,Numerical, 2389,Geometry,,"In the diagram, $A B=B C=2 \sqrt{2}, C D=D E$, $\angle C D E=60^{\circ}$, and $\angle E A B=75^{\circ}$. Determine the perimeter of figure $A B C D E$. Explain how you got your answer. ![](https://cdn.mathpix.com/cropped/2023_12_21_e7d09dafb0c4acfee33dg-1.jpg?height=333&width=464&top_left_y=2208&top_left_x=1313)","['Since $\\triangle A B C$ is isosceles and right-angled, then $\\angle B A C=45^{\\circ}$.\n\nAlso, $A C=\\sqrt{2} A B=\\sqrt{2}(2 \\sqrt{2})=4$.\n\nSince $\\angle E A B=75^{\\circ}$ and $\\angle B A C=45^{\\circ}$, then $\\angle C A E=\\angle E A B-\\angle B A C=30^{\\circ}$.\n\nSince $\\triangle A E C$ is right-angled and has a $30^{\\circ}$ angle, then $\\triangle A E C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nThus, $E C=\\frac{1}{2} A C=2$ (since $E C$ is opposite the $30^{\\circ}$ angle) and $A E=\\frac{\\sqrt{3}}{2} A C=2 \\sqrt{3}$ (since $A E$ is opposite the $60^{\\circ}$ angle).\n\nIn $\\triangle C D E, E D=D C$ and $\\angle E D C=60^{\\circ}$, so $\\triangle C D E$ is equilateral.\n\nTherefore, $E D=C D=E C=2$.\n\nOverall, the perimeter of $A B C D E$ is\n\n$$\nA B+B C+C D+D E+E A=2 \\sqrt{2}+2 \\sqrt{2}+2+2+2 \\sqrt{3}=4+4 \\sqrt{2}+2 \\sqrt{3}\n$$']",['$4+4 \\sqrt{2}+2 \\sqrt{3}$'],False,,Numerical, 2390,Algebra,,"The first term of a sequence is 2007. Each term, starting with the second, is the sum of the cubes of the digits of the previous term. What is the 2007th term?","['From the given information, the first term in the sequence is 2007 and each term starting with the second can be determined from the previous term.\n\nThe second term is $2^{3}+0^{3}+0^{3}+7^{3}=8+0+0+343=351$.\n\nThe third term is $3^{3}+5^{3}+1^{3}=27+125+1=153$.\n\nThe fourth term is $1^{3}+5^{3}+3^{3}=27+125+1=153$.\n\nSince two consecutive terms are equal, then every term thereafter will be equal, because each term depends only on the previous term and a term of 153 always makes the next term 153.\n\nThus, the 2007th term will be 153 .']",['153'],False,,Numerical, 2391,Algebra,,"Sequence A has $n$th term $n^{2}-10 n+70$. (The first three terms of sequence $\mathrm{A}$ are $61,54,49$. ) Sequence B is an arithmetic sequence with first term 5 and common difference 10. (The first three terms of sequence $\mathrm{B}$ are $5,15,25$.) Determine all $n$ for which the $n$th term of sequence $\mathrm{A}$ is equal to the $n$th term of sequence B. Explain how you got your answer.","['The $n$th term of sequence $\\mathrm{A}$ is $n^{2}-10 n+70$.\n\nSince sequence B is arithmetic with first term 5 and common difference 10 , then the $n$th term of sequence $\\mathrm{B}$ is equal to $5+10(n-1)=10 n-5$. (Note that this formula agrees with the first few terms.)\n\nFor the $n$th term of sequence $\\mathrm{A}$ to be equal to the $n$th term of sequence $\\mathrm{B}$, we must have\n\n$$\n\\begin{aligned}\nn^{2}-10 n+70 & =10 n-5 \\\\\nn^{2}-20 n+75 & =0 \\\\\n(n-5)(n-15) & =0\n\\end{aligned}\n$$\n\nTherefore, $n=5$ or $n=15$. That is, 5 th and 15 th terms of sequence $\\mathrm{A}$ and sequence $\\mathrm{B}$ are equal to each other.']","['5,15']",True,,Numerical, 2392,Algebra,,Determine all values of $x$ for which $2+\sqrt{x-2}=x-2$.,"['Rearranging and then squaring both sides,\n\n$$\n\\begin{aligned}\n2+\\sqrt{x-2} & =x-2 \\\\\n\\sqrt{x-2} & =x-4 \\\\\nx-2 & =(x-4)^{2} \\\\\nx-2 & =x^{2}-8 x+16 \\\\\n0 & =x^{2}-9 x+18 \\\\\n0 & =(x-3)(x-6)\n\\end{aligned}\n$$\n\nso $x=3$ or $x=6$.\n\nWe should check both solutions, because we may have introduced extraneous solutions by squaring.\n\nIf $x=3$, the left side equals $2+\\sqrt{1}=3$ and the right side equals 1 , so $x=3$ must be rejected.\n\nIf $x=6$, the left side equals $2+\\sqrt{4}=4$ and the right side equals 4 , so $x=6$ is the only solution.', 'Suppose $u=\\sqrt{x-2}$.\n\nThe equation becomes $2+u=u^{2}$ or $u^{2}-u-2=0$ or $(u-2)(u+1)=0$.\n\nTherefore, $u=2$ or $u=-1$.\n\nBut we cannot have $\\sqrt{x-2}=-1$ (as square roots are always non-negative).\n\nTherefore, $\\sqrt{x-2}=2$ or $x-2=4$ or $x=6$.']",['6'],False,,Numerical, 2393,Geometry,,"In the diagram, the parabola intersects the $x$-axis at $A(-3,0)$ and $B(3,0)$ and has its vertex at $C$ below the $x$-axis. The area of $\triangle A B C$ is 54 . Determine the equation of the parabola. Explain how you got your answer. ","['From the diagram, the parabola has $x$-intercepts $x=3$ and $x=-3$.\n\nTherefore, the equation of the parabola is of the form $y=a(x-3)(x+3)$ for some real number $a$.\n\nTriangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nSince the parabola passes through $C(0,-18)$, then this point must satisfy the equation of the parabola.\n\nTherefore, $-18=a(0-3)(0+3)$ or $-18=-9 a$ or $a=2$.\n\nThus, the equation of the parabola is $y=2(x-3)(x+3)=2 x^{2}-18$.', 'Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nTherefore, the parabola has vertex $C(0,-18)$, so has equation $y=a(x-0)^{2}-18$.\n\n(The vertex of the parabola must lie on the $y$-axis since its roots are equally distant from the $y$-axis, so $C$ must be the vertex.)\n\nSince the parabola passes through $B(3,0)$, then these coordinates satisfy the equation, so $0=3^{2} a-18$ or $9 a=18$ or $a=2$.\n\nTherefore, the equation of the parabola is $y=2 x^{2}-18$.']",['$y=2 x^{2}-18$'],False,,Expression, 2393,Geometry,,"In the diagram, the parabola intersects the $x$-axis at $A(-3,0)$ and $B(3,0)$ and has its vertex at $C$ below the $x$-axis. The area of $\triangle A B C$ is 54 . Determine the equation of the parabola. Explain how you got your answer. ![](https://cdn.mathpix.com/cropped/2023_12_21_84558404cbbb20558c95g-1.jpg?height=406&width=553&top_left_y=797&top_left_x=1255)","['From the diagram, the parabola has $x$-intercepts $x=3$ and $x=-3$.\n\nTherefore, the equation of the parabola is of the form $y=a(x-3)(x+3)$ for some real number $a$.\n\nTriangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nSince the parabola passes through $C(0,-18)$, then this point must satisfy the equation of the parabola.\n\nTherefore, $-18=a(0-3)(0+3)$ or $-18=-9 a$ or $a=2$.\n\nThus, the equation of the parabola is $y=2(x-3)(x+3)=2 x^{2}-18$.', 'Triangle $A B C$ can be considered as having base $A B$ (of length $3-(-3)=6$ ) and height $O C$ (where $O$ is the origin).\n\nSuppose $C$ has coordinates $(0,-c)$. Then $O C=c$.\n\nThus, the area of $\\triangle A B C$ is $\\frac{1}{2}(A B)(O C)=3 c$. But we know that the area of $\\triangle A B C$ is 54 , so $3 c=54$ or $c=18$.\n\nTherefore, the parabola has vertex $C(0,-18)$, so has equation $y=a(x-0)^{2}-18$.\n\n(The vertex of the parabola must lie on the $y$-axis since its roots are equally distant from the $y$-axis, so $C$ must be the vertex.)\n\nSince the parabola passes through $B(3,0)$, then these coordinates satisfy the equation, so $0=3^{2} a-18$ or $9 a=18$ or $a=2$.\n\nTherefore, the equation of the parabola is $y=2 x^{2}-18$.']",['$y=2 x^{2}-18$'],False,,Expression, 2394,Geometry,,"In the diagram, $A(0, a)$ lies on the $y$-axis above $D$. If the triangles $A O B$ and $B C D$ have the same area, determine the value of $a$. Explain how you got your answer. ","['$\\triangle A O B$ is right-angled at $O$, so has area $\\frac{1}{2}(A O)(O B)=\\frac{1}{2} a(1)=\\frac{1}{2} a$.\n\nWe next need to calculate the area of $\\triangle B C D$.\n\nMethod 1: Completing the trapezoid\n\nDrop a perpendicular from $C$ to $P(3,0)$ on the $x$-axis.\n\n\n\nThen $D O P C$ is a trapezoid with parallel sides $D O$ of length 1 and $P C$ of length 2 and height $O P$ (which is indeed perpendicular to the parallel sides) of length 3.\n\nThe area of the trapezoid is thus $\\frac{1}{2}(D O+P C)(O P)=\\frac{1}{2}(1+2)(3)=\\frac{9}{2}$.\n\nBut the area of $\\triangle B C D$ equals the area of trapezoid $D O P C$ minus the areas of $\\triangle D O B$ and $\\triangle B P C$.\n\n$\\triangle D O B$ is right-angled at $O$, so has area $\\frac{1}{2}(D O)(O B)=\\frac{1}{2}(1)(1)=\\frac{1}{2}$.\n\n$\\triangle B P C$ is right-angled at $P$, so has area $\\frac{1}{2}(B P)(P C)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{9}{2}-\\frac{1}{2}-2=2$.\n\n\n\n(A similar method for calculating the area of $\\triangle D B C$ would be to drop a perpendicular to $Q$ on the $y$-axis, creating a rectangle $Q O P C$.)\n\nMethod 2: $\\triangle D B C$ is right-angled\n\nThe slope of line segment $D B$ is $\\frac{1-0}{0-1}=-1$.\n\nThe slope of line segment $B C$ is $\\frac{2-0}{3-1}=1$.\n\nSince the product of these slopes is -1 (that is, their slopes are negative reciprocals), then $D B$ and $B C$ are perpendicular.\n\nTherefore, the area of $\\triangle D B C$ is $\\frac{1}{2}(D B)(B C)$.\n\nNow $D B=\\sqrt{(1-0)^{2}+(0-1)^{2}}=\\sqrt{2}$ and $B C=\\sqrt{(3-1)^{2}+(2-0)^{2}}=\\sqrt{8}$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{1}{2} \\sqrt{2} \\sqrt{8}=2$.\n\nSince the area of $\\triangle A O B$ equals the area of $\\triangle D B C$, then $\\frac{1}{2} a=2$ or $a=4$.']",['4'],False,,Numerical, 2394,Geometry,,"In the diagram, $A(0, a)$ lies on the $y$-axis above $D$. If the triangles $A O B$ and $B C D$ have the same area, determine the value of $a$. Explain how you got your answer. ![](https://cdn.mathpix.com/cropped/2023_12_21_84558404cbbb20558c95g-1.jpg?height=434&width=485&top_left_y=1621&top_left_x=1259)","['$\\triangle A O B$ is right-angled at $O$, so has area $\\frac{1}{2}(A O)(O B)=\\frac{1}{2} a(1)=\\frac{1}{2} a$.\n\nWe next need to calculate the area of $\\triangle B C D$.\n\nMethod 1: Completing the trapezoid\n\nDrop a perpendicular from $C$ to $P(3,0)$ on the $x$-axis.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_99bcabad849ac4365f35g-1.jpg?height=423&width=502&top_left_y=1664&top_left_x=909)\n\nThen $D O P C$ is a trapezoid with parallel sides $D O$ of length 1 and $P C$ of length 2 and height $O P$ (which is indeed perpendicular to the parallel sides) of length 3.\n\nThe area of the trapezoid is thus $\\frac{1}{2}(D O+P C)(O P)=\\frac{1}{2}(1+2)(3)=\\frac{9}{2}$.\n\nBut the area of $\\triangle B C D$ equals the area of trapezoid $D O P C$ minus the areas of $\\triangle D O B$ and $\\triangle B P C$.\n\n$\\triangle D O B$ is right-angled at $O$, so has area $\\frac{1}{2}(D O)(O B)=\\frac{1}{2}(1)(1)=\\frac{1}{2}$.\n\n$\\triangle B P C$ is right-angled at $P$, so has area $\\frac{1}{2}(B P)(P C)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{9}{2}-\\frac{1}{2}-2=2$.\n\n\n\n(A similar method for calculating the area of $\\triangle D B C$ would be to drop a perpendicular to $Q$ on the $y$-axis, creating a rectangle $Q O P C$.)\n\nMethod 2: $\\triangle D B C$ is right-angled\n\nThe slope of line segment $D B$ is $\\frac{1-0}{0-1}=-1$.\n\nThe slope of line segment $B C$ is $\\frac{2-0}{3-1}=1$.\n\nSince the product of these slopes is -1 (that is, their slopes are negative reciprocals), then $D B$ and $B C$ are perpendicular.\n\nTherefore, the area of $\\triangle D B C$ is $\\frac{1}{2}(D B)(B C)$.\n\nNow $D B=\\sqrt{(1-0)^{2}+(0-1)^{2}}=\\sqrt{2}$ and $B C=\\sqrt{(3-1)^{2}+(2-0)^{2}}=\\sqrt{8}$.\n\nThus, the area of $\\triangle D B C$ is $\\frac{1}{2} \\sqrt{2} \\sqrt{8}=2$.\n\nSince the area of $\\triangle A O B$ equals the area of $\\triangle D B C$, then $\\frac{1}{2} a=2$ or $a=4$.']",['4'],False,,Numerical, 2395,Geometry,,"The Little Prince lives on a spherical planet which has a radius of $24 \mathrm{~km}$ and centre $O$. He hovers in a helicopter $(H)$ at a height of $2 \mathrm{~km}$ above the surface of the planet. From his position in the helicopter, what is the distance, in kilometres, to the furthest point on the surface of the planet that he can see? ![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=363&width=290&top_left_y=274&top_left_x=1248)","['Suppose that $O$ is the centre of the planet, $H$ is the place where His Highness hovers in the helicopter, and $P$ is the furthest point on the surface of the planet that he can see.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_86b73a5af3c2f1283f30g-1.jpg?height=366&width=307&top_left_y=1216&top_left_x=1012)\n\nThen $H P$ must be a tangent to the surface of the planet (otherwise he could see further), so $O P$ (a radius) is perpendicular to $H P$ (a tangent).\n\nWe are told that $O P=24 \\mathrm{~km}$.\n\nSince the helicopter hovers at a height of $2 \\mathrm{~km}$, then $O H=24+2=26 \\mathrm{~km}$.\n\nTherefore, $H P^{2}=O H^{2}-O P^{2}=26^{2}-24^{2}=100$, so $H P=10 \\mathrm{~km}$.\n\nTherefore, the distance to the furthest point that he can see is $10 \\mathrm{~km}$.']",['10'],False,km,Numerical, 2395,Geometry,,"The Little Prince lives on a spherical planet which has a radius of $24 \mathrm{~km}$ and centre $O$. He hovers in a helicopter $(H)$ at a height of $2 \mathrm{~km}$ above the surface of the planet. From his position in the helicopter, what is the distance, in kilometres, to the furthest point on the surface of the planet that he can see? ","['Suppose that $O$ is the centre of the planet, $H$ is the place where His Highness hovers in the helicopter, and $P$ is the furthest point on the surface of the planet that he can see.\n\n\n\nThen $H P$ must be a tangent to the surface of the planet (otherwise he could see further), so $O P$ (a radius) is perpendicular to $H P$ (a tangent).\n\nWe are told that $O P=24 \\mathrm{~km}$.\n\nSince the helicopter hovers at a height of $2 \\mathrm{~km}$, then $O H=24+2=26 \\mathrm{~km}$.\n\nTherefore, $H P^{2}=O H^{2}-O P^{2}=26^{2}-24^{2}=100$, so $H P=10 \\mathrm{~km}$.\n\nTherefore, the distance to the furthest point that he can see is $10 \\mathrm{~km}$.']",['10'],False,km,Numerical, 2396,Geometry,,"In the diagram, points $A$ and $B$ are located on islands in a river full of rabid aquatic goats. Determine the distance from $A$ to $B$, to the nearest metre. (Luckily, someone has measured the angles shown in the diagram as well as the distances $C D$ and $D E$.) ","['Since we know the measure of $\\angle A D B$, then to find the distance $A B$, it is enough to find the distances $A D$ and $B D$ and then apply the cosine law.\n\nIn $\\triangle D B E$, we have $\\angle D B E=180^{\\circ}-20^{\\circ}-70^{\\circ}=90^{\\circ}$, so $\\triangle D B E$ is right-angled, giving $B D=100 \\cos \\left(20^{\\circ}\\right) \\approx 93.969$.\n\nIn $\\triangle D A C$, we have $\\angle D A C=180^{\\circ}-50^{\\circ}-45^{\\circ}=85^{\\circ}$.\n\nUsing the sine law, $\\frac{A D}{\\sin \\left(50^{\\circ}\\right)}=\\frac{C D}{\\sin \\left(85^{\\circ}\\right)}$, so $A D=\\frac{150 \\sin \\left(50^{\\circ}\\right)}{\\sin \\left(85^{\\circ}\\right)} \\approx 115.346$.\n\n\n\nFinally, using the cosine law in $\\triangle A B D$, we get\n\n$$\n\\begin{aligned}\nA B^{2} & =A D^{2}+B D^{2}-2(A D)(B D) \\cos (\\angle A D B) \\\\\nA B^{2} & \\approx(115.346)^{2}+(93.969)^{2}-2(115.346)(93.969) \\cos \\left(35^{\\circ}\\right) \\\\\nA B^{2} & \\approx 4377.379 \\\\\nA B & \\approx 66.16\n\\end{aligned}\n$$\n\nTherefore, the distance from $A$ to $B$ is approximately $66 \\mathrm{~m}$.']",['66'],False,m,Numerical, 2396,Geometry,,"In the diagram, points $A$ and $B$ are located on islands in a river full of rabid aquatic goats. Determine the distance from $A$ to $B$, to the nearest metre. (Luckily, someone has measured the angles shown in the diagram as well as the distances $C D$ and $D E$.) ![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=490&width=754&top_left_y=682&top_left_x=1076)","['Since we know the measure of $\\angle A D B$, then to find the distance $A B$, it is enough to find the distances $A D$ and $B D$ and then apply the cosine law.\n\nIn $\\triangle D B E$, we have $\\angle D B E=180^{\\circ}-20^{\\circ}-70^{\\circ}=90^{\\circ}$, so $\\triangle D B E$ is right-angled, giving $B D=100 \\cos \\left(20^{\\circ}\\right) \\approx 93.969$.\n\nIn $\\triangle D A C$, we have $\\angle D A C=180^{\\circ}-50^{\\circ}-45^{\\circ}=85^{\\circ}$.\n\nUsing the sine law, $\\frac{A D}{\\sin \\left(50^{\\circ}\\right)}=\\frac{C D}{\\sin \\left(85^{\\circ}\\right)}$, so $A D=\\frac{150 \\sin \\left(50^{\\circ}\\right)}{\\sin \\left(85^{\\circ}\\right)} \\approx 115.346$.\n\n\n\nFinally, using the cosine law in $\\triangle A B D$, we get\n\n$$\n\\begin{aligned}\nA B^{2} & =A D^{2}+B D^{2}-2(A D)(B D) \\cos (\\angle A D B) \\\\\nA B^{2} & \\approx(115.346)^{2}+(93.969)^{2}-2(115.346)(93.969) \\cos \\left(35^{\\circ}\\right) \\\\\nA B^{2} & \\approx 4377.379 \\\\\nA B & \\approx 66.16\n\\end{aligned}\n$$\n\nTherefore, the distance from $A$ to $B$ is approximately $66 \\mathrm{~m}$.']",['66'],False,m,Numerical, 2397,Geometry,,Determine all values of $x$ for which $(\sqrt{x})^{\log _{10} x}=100$.,"['Using rules for manipulating logarithms,\n\n$$\n\\begin{aligned}\n(\\sqrt{x})^{\\log _{10} x} & =100 \\\\\n\\log _{10}\\left((\\sqrt{x})^{\\log _{10} x}\\right) & =\\log _{10} 100 \\\\\n\\left(\\log _{10} x\\right)\\left(\\log _{10} \\sqrt{x}\\right) & =2 \\\\\n\\left(\\log _{10} x\\right)\\left(\\log _{10} x^{\\frac{1}{2}}\\right) & =2 \\\\\n\\left(\\log _{10} x\\right)\\left(\\frac{1}{2} \\log _{10} x\\right) & =2 \\\\\n\\left(\\log _{10} x\\right)^{2} & =4 \\\\\n\\log _{10} x & = \\pm 2 \\\\\nx & =10^{ \\pm 2}\n\\end{aligned}\n$$\n\nTherefore, $x=100$ or $x=\\frac{1}{100}$.\n\n(We can check by substitution that each is indeed a solution.)']","['$100,\\frac{1}{100}$']",True,,Numerical, 2398,Geometry,,"In the diagram, line segment $F C G$ passes through vertex $C$ of square $A B C D$, with $F$ lying on $A B$ extended and $G$ lying on $A D$ extended. Prove that $\frac{1}{A B}=\frac{1}{A F}+\frac{1}{A G}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=344&width=545&top_left_y=1319&top_left_x=1256)","['Without loss of generality, suppose that square $A B C D$ has side length 1 .\n\nSuppose next that $B F=a$ and $\\angle C F B=\\theta$.\n\nSince $\\triangle C B F$ is right-angled at $B$, then $\\angle B C F=90^{\\circ}-\\theta$.\n\nSince $G C F$ is a straight line, then $\\angle G C D=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nTherefore, $\\triangle G D C$ is similar to $\\triangle C B F$, since $\\triangle G D C$ is right-angled at $D$.\n\nThus, $\\frac{G D}{D C}=\\frac{B C}{B F}$ or $\\frac{G D}{1}=\\frac{1}{a}$ or $G D=\\frac{1}{a}$.\n\nSo $A F=A B+B F=1+a$ and $A G=A D+D G=1+\\frac{1}{a}=\\frac{a+1}{a}$.\n\nThus, $\\frac{1}{A F}+\\frac{1}{A G}=\\frac{1}{1+a}+\\frac{a}{a+1}=\\frac{a+1}{a+1}=1=\\frac{1}{A B}$, as required.', 'We attach a set of coordinate axes to the diagram, with $A$ at the origin, $A G$ lying along the positive $y$-axis and $A F$ lying along the positive $x$-axis.\n\nWithout loss of generality, suppose that square $A B C D$ has side length 1 , so that $C$ has coordinates $(1,1)$. (We can make this assumption without loss of generality, because if the square had a different side length, then each of the lengths in the problem would be scaled by the same factor.)\n\n\n\nSuppose that the line through $G$ and $F$ has slope $m$.\n\nSince this line passes through $(1,1)$, its equation is $y-1=m(x-1)$ or $y=m x+(1-m)$. The $y$-intercept of this line is $1-m$, so $G$ has coordinates $(0,1-m)$.\n\nThe $x$-intercept of this line is $\\frac{m-1}{m}$, so $F$ has coordinates $\\left(\\frac{m-1}{m}, 0\\right)$. (Note that $m \\neq 0$ as the line cannot be horizontal.)\n\nTherefore,\n\n$$\n\\frac{1}{A F}+\\frac{1}{A G}=\\frac{m}{m-1}+\\frac{1}{1-m}=\\frac{m}{m-1}+\\frac{-1}{m-1}=\\frac{m-1}{m-1}=1=\\frac{1}{A B}\n$$\n\nas required.', 'Join $A$ to $C$.\n\nWe know that the sum of the areas of $\\triangle G C A$ and $\\triangle F C A$ equals the area of $\\triangle G A F$.\n\nThe area of $\\triangle G C A$ (thinking of $A G$ as the base) is $\\frac{1}{2}(A G)(D C)$, since $D C$ is perpendicular to $A G$.\n\nSimilarly, the area of $\\triangle F C A$ is $\\frac{1}{2}(A F)(C B)$.\n\nAlso, the area of $\\triangle G A F$ is $\\frac{1}{2}(A G)(A F)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A G)(D C)+\\frac{1}{2}(A F)(C B) & =\\frac{1}{2}(A G)(A F) \\\\\n\\frac{(A G)(D C)}{(A G)(A F)(A B)}+\\frac{(A F)(C B)}{(A G)(A F)(A B)} & =\\frac{(A G)(A F)}{(A G)(A F)(A B)} \\\\\n\\frac{1}{A F}+\\frac{1}{A G} & =\\frac{1}{A B}\n\\end{aligned}\n$$\n\nas required, since $A B=D C=C B$.']",['证明题，略'],True,,Need_human_evaluate, 2398,Geometry,,"In the diagram, line segment $F C G$ passes through vertex $C$ of square $A B C D$, with $F$ lying on $A B$ extended and $G$ lying on $A D$ extended. Prove that $\frac{1}{A B}=\frac{1}{A F}+\frac{1}{A G}$. ","['Without loss of generality, suppose that square $A B C D$ has side length 1 .\n\nSuppose next that $B F=a$ and $\\angle C F B=\\theta$.\n\nSince $\\triangle C B F$ is right-angled at $B$, then $\\angle B C F=90^{\\circ}-\\theta$.\n\nSince $G C F$ is a straight line, then $\\angle G C D=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nTherefore, $\\triangle G D C$ is similar to $\\triangle C B F$, since $\\triangle G D C$ is right-angled at $D$.\n\nThus, $\\frac{G D}{D C}=\\frac{B C}{B F}$ or $\\frac{G D}{1}=\\frac{1}{a}$ or $G D=\\frac{1}{a}$.\n\nSo $A F=A B+B F=1+a$ and $A G=A D+D G=1+\\frac{1}{a}=\\frac{a+1}{a}$.\n\nThus, $\\frac{1}{A F}+\\frac{1}{A G}=\\frac{1}{1+a}+\\frac{a}{a+1}=\\frac{a+1}{a+1}=1=\\frac{1}{A B}$, as required.', 'We attach a set of coordinate axes to the diagram, with $A$ at the origin, $A G$ lying along the positive $y$-axis and $A F$ lying along the positive $x$-axis.\n\nWithout loss of generality, suppose that square $A B C D$ has side length 1 , so that $C$ has coordinates $(1,1)$. (We can make this assumption without loss of generality, because if the square had a different side length, then each of the lengths in the problem would be scaled by the same factor.)\n\n\n\nSuppose that the line through $G$ and $F$ has slope $m$.\n\nSince this line passes through $(1,1)$, its equation is $y-1=m(x-1)$ or $y=m x+(1-m)$. The $y$-intercept of this line is $1-m$, so $G$ has coordinates $(0,1-m)$.\n\nThe $x$-intercept of this line is $\\frac{m-1}{m}$, so $F$ has coordinates $\\left(\\frac{m-1}{m}, 0\\right)$. (Note that $m \\neq 0$ as the line cannot be horizontal.)\n\nTherefore,\n\n$$\n\\frac{1}{A F}+\\frac{1}{A G}=\\frac{m}{m-1}+\\frac{1}{1-m}=\\frac{m}{m-1}+\\frac{-1}{m-1}=\\frac{m-1}{m-1}=1=\\frac{1}{A B}\n$$\n\nas required.', 'Join $A$ to $C$.\n\nWe know that the sum of the areas of $\\triangle G C A$ and $\\triangle F C A$ equals the area of $\\triangle G A F$.\n\nThe area of $\\triangle G C A$ (thinking of $A G$ as the base) is $\\frac{1}{2}(A G)(D C)$, since $D C$ is perpendicular to $A G$.\n\nSimilarly, the area of $\\triangle F C A$ is $\\frac{1}{2}(A F)(C B)$.\n\nAlso, the area of $\\triangle G A F$ is $\\frac{1}{2}(A G)(A F)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A G)(D C)+\\frac{1}{2}(A F)(C B) & =\\frac{1}{2}(A G)(A F) \\\\\n\\frac{(A G)(D C)}{(A G)(A F)(A B)}+\\frac{(A F)(C B)}{(A G)(A F)(A B)} & =\\frac{(A G)(A F)}{(A G)(A F)(A B)} \\\\\n\\frac{1}{A F}+\\frac{1}{A G} & =\\frac{1}{A B}\n\\end{aligned}\n$$\n\nas required, since $A B=D C=C B$.']",,True,,, 2399,Geometry,,"In the $4 \times 4$ grid shown, three coins are randomly placed in different squares. Determine the probability that no two coins lie in the same row or column. ![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=274&width=279&top_left_y=1728&top_left_x=1259)","['We consider placing the three coins individually.\n\nPlace one coin randomly on the grid.\n\nWhen the second coin is placed (in any one of 15 squares), 6 of the 15 squares will leave two coins in the same row or column and 9 of the 15 squares will leave the two coins in different rows and different columns.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_75254b466303852df435g-1.jpg?height=277&width=284&top_left_y=2130&top_left_x=1018)\n\nTherefore, the probability that the two coins are in different rows and different columns is $\\frac{9}{15}=\\frac{3}{5}$.\n\nThere are 14 possible squares in which the third coin can be placed.\n\n\n\nOf these 14 squares, 6 lie in the same row or column as the first coin and an additional 4 lie the same row or column as the second coin. Therefore, the probability that the third coin is placed in a different row and a different column than each of the first two coins is $\\frac{4}{14}=\\frac{2}{7}$.\n\nTherefore, the probability that all three coins are placed in different rows and different columns is $\\frac{3}{5} \\times \\frac{2}{7}=\\frac{6}{35}$.']",['$\\frac{6}{35}$'],False,,Numerical, 2399,Geometry,,"In the $4 \times 4$ grid shown, three coins are randomly placed in different squares. Determine the probability that no two coins lie in the same row or column. ","['We consider placing the three coins individually.\n\nPlace one coin randomly on the grid.\n\nWhen the second coin is placed (in any one of 15 squares), 6 of the 15 squares will leave two coins in the same row or column and 9 of the 15 squares will leave the two coins in different rows and different columns.\n\n\n\nTherefore, the probability that the two coins are in different rows and different columns is $\\frac{9}{15}=\\frac{3}{5}$.\n\nThere are 14 possible squares in which the third coin can be placed.\n\n\n\nOf these 14 squares, 6 lie in the same row or column as the first coin and an additional 4 lie the same row or column as the second coin. Therefore, the probability that the third coin is placed in a different row and a different column than each of the first two coins is $\\frac{4}{14}=\\frac{2}{7}$.\n\nTherefore, the probability that all three coins are placed in different rows and different columns is $\\frac{3}{5} \\times \\frac{2}{7}=\\frac{6}{35}$.']",['$\\frac{6}{35}$'],False,,Numerical, 2400,Geometry,,"In the diagram, the area of $\triangle A B C$ is 1 . Trapezoid $D E F G$ is constructed so that $G$ is to the left of $F, D E$ is parallel to $B C$, $E F$ is parallel to $A B$ and $D G$ is parallel to $A C$. Determine the maximum possible area of trapezoid $D E F G$. ![](https://cdn.mathpix.com/cropped/2023_12_21_1ec73aaee00c97b3df16g-1.jpg?height=360&width=499&top_left_y=2048&top_left_x=1255)","['Suppose that $A B=c, A C=b$ and $B C=a$.\n\nSince $D G$ is parallel to $A C, \\angle B D G=\\angle B A C$ and $\\angle D G B=\\angle A C B$, so $\\triangle D G B$ is similar to $\\triangle A C B$.\n\n(Similarly, $\\triangle A E D$ and $\\triangle E C F$ are also both similar to $\\triangle A B C$.)\n\nSuppose next that $D B=k c$, with $0","['Suppose that $A B=c, A C=b$ and $B C=a$.\n\nSince $D G$ is parallel to $A C, \\angle B D G=\\angle B A C$ and $\\angle D G B=\\angle A C B$, so $\\triangle D G B$ is similar to $\\triangle A C B$.\n\n(Similarly, $\\triangle A E D$ and $\\triangle E C F$ are also both similar to $\\triangle A B C$.)\n\nSuppose next that $D B=k c$, with $089$.\n\nThis final inequality is equivalent to $n>\\frac{89}{4}=22 \\frac{1}{4}$.\n\nTherefore, the smallest positive integer that satisfies this inequality, and hence for which $f(x)$ has no real roots, is $n=23$.']",['23'],False,,Numerical, 2404,Geometry,,"In the diagram, $\triangle P Q R$ has $P Q=a, Q R=b, P R=21$, and $\angle P Q R=60^{\circ}$. Also, $\triangle S T U$ has $S T=a, T U=b, \angle T S U=30^{\circ}$, and $\sin (\angle T U S)=\frac{4}{5}$. Determine the values of $a$ and $b$. ![](https://cdn.mathpix.com/cropped/2023_12_21_4165463fb3d29f83c565g-1.jpg?height=228&width=792&top_left_y=1225&top_left_x=730)","['Using the cosine law in $\\triangle P Q R$,\n\n$$\n\\begin{aligned}\nP R^{2} & =P Q^{2}+Q R^{2}-2 \\cdot P Q \\cdot Q R \\cdot \\cos (\\angle P Q R) \\\\\n21^{2} & =a^{2}+b^{2}-2 a b \\cos \\left(60^{\\circ}\\right) \\\\\n441 & =a^{2}+b^{2}-2 a b \\cdot \\frac{1}{2} \\\\\n441 & =a^{2}+b^{2}-a b\n\\end{aligned}\n$$\n\nUsing the sine law in $\\triangle S T U$, we obtain $\\frac{S T}{\\sin (\\angle T U S)}=\\frac{T U}{\\sin (\\angle T S U)}$ and so $\\frac{a}{4 / 5}=\\frac{b}{\\sin \\left(30^{\\circ}\\right)}$. Therefore, $\\frac{a}{4 / 5}=\\frac{b}{1 / 2}$ and so $a=\\frac{4}{5} \\cdot 2 b=\\frac{8}{5} b$.\n\nSubstituting into the previous equation,\n\n$$\n\\begin{aligned}\n& 441=\\left(\\frac{8}{5} b\\right)^{2}+b^{2}-\\left(\\frac{8}{5} b\\right) b \\\\\n& 441=\\frac{64}{25} b^{2}+b^{2}-\\frac{8}{5} b^{2} \\\\\n& 441=\\frac{64}{25} b^{2}+\\frac{25}{25} b^{2}-\\frac{40}{25} b^{2} \\\\\n& 441=\\frac{49}{25} b^{2} \\\\\n& 225=b^{2}\n\\end{aligned}\n$$\n\nSince $b>0$, then $b=15$ and so $a=\\frac{8}{5} b=\\frac{8}{5} \\cdot 15=24$.']","['$24,15$']",True,,Numerical, 2404,Geometry,,"In the diagram, $\triangle P Q R$ has $P Q=a, Q R=b, P R=21$, and $\angle P Q R=60^{\circ}$. Also, $\triangle S T U$ has $S T=a, T U=b, \angle T S U=30^{\circ}$, and $\sin (\angle T U S)=\frac{4}{5}$. Determine the values of $a$ and $b$. ","['Using the cosine law in $\\triangle P Q R$,\n\n$$\n\\begin{aligned}\nP R^{2} & =P Q^{2}+Q R^{2}-2 \\cdot P Q \\cdot Q R \\cdot \\cos (\\angle P Q R) \\\\\n21^{2} & =a^{2}+b^{2}-2 a b \\cos \\left(60^{\\circ}\\right) \\\\\n441 & =a^{2}+b^{2}-2 a b \\cdot \\frac{1}{2} \\\\\n441 & =a^{2}+b^{2}-a b\n\\end{aligned}\n$$\n\nUsing the sine law in $\\triangle S T U$, we obtain $\\frac{S T}{\\sin (\\angle T U S)}=\\frac{T U}{\\sin (\\angle T S U)}$ and so $\\frac{a}{4 / 5}=\\frac{b}{\\sin \\left(30^{\\circ}\\right)}$. Therefore, $\\frac{a}{4 / 5}=\\frac{b}{1 / 2}$ and so $a=\\frac{4}{5} \\cdot 2 b=\\frac{8}{5} b$.\n\nSubstituting into the previous equation,\n\n$$\n\\begin{aligned}\n& 441=\\left(\\frac{8}{5} b\\right)^{2}+b^{2}-\\left(\\frac{8}{5} b\\right) b \\\\\n& 441=\\frac{64}{25} b^{2}+b^{2}-\\frac{8}{5} b^{2} \\\\\n& 441=\\frac{64}{25} b^{2}+\\frac{25}{25} b^{2}-\\frac{40}{25} b^{2} \\\\\n& 441=\\frac{49}{25} b^{2} \\\\\n& 225=b^{2}\n\\end{aligned}\n$$\n\nSince $b>0$, then $b=15$ and so $a=\\frac{8}{5} b=\\frac{8}{5} \\cdot 15=24$.']","['$24,15$']",True,,Numerical, 2405,Geometry,,"A triangle of area $770 \mathrm{~cm}^{2}$ is divided into 11 regions of equal height by 10 lines that are all parallel to the base of the triangle. Starting from the top of the triangle, every other region is shaded, as shown. What is the total area of the shaded regions? ","['We make two copies of the given triangle, labelling them $\\triangle A B C$ and $\\triangle D E F$, as shown:\n\n\nThe combined area of these two triangles is $2 \\cdot 770 \\mathrm{~cm}^{2}=1540 \\mathrm{~cm}^{2}$, and the shaded area in each triangle is the same.\n\nNext, we rotate $\\triangle D E F$ by $180^{\\circ}$ :\n\n\nand join the two triangles together:\n\n\n\nWe note that $B C$ and $A E$ (which was $F E$ ) are equal in length (since they were copies of each other) and parallel (since they are $180^{\\circ}$ rotations of each other). The same is true for $A B$ and $E C$.\n\nTherefore, $A B C E$ is a parallelogram.\n\nFurther, $A B C E$ is divided into 11 identical parallelograms (6 shaded and 5 unshaded) by the horizontal lines. (Since the sections of the two triangles are equal in height, the horizontal lines on both sides of $A C$ align.)\n\nThe total area of parallelogram $A B C E$ is $1540 \\mathrm{~cm}^{2}$.\n\nThus, the shaded area of $A B C E$ is $\\frac{6}{11} \\cdot 1540 \\mathrm{~cm}^{2}=840 \\mathrm{~cm}^{2}$.\n\nSince this shaded area is equally divided between the two halves of the parallelogram, then the combined area of the shaded regions of $\\triangle A B C$ is $\\frac{1}{2} \\cdot 840 \\mathrm{~cm}^{2}=420 \\mathrm{~cm}^{2}$.', 'We label the points where the horizontal lines touch $A B$ and $A C$ as shown:\n\n\n\nWe use the notation $|\\triangle A B C|$ to represent the area of $\\triangle A B C$ and use similar notation for the area of other triangles and quadrilaterals.\n\nLet $\\mathcal{A}$ be equal to the total area of the shaded regions.\n\nThus,\n\n$$\n\\mathcal{A}=\\left|\\triangle A B_{1} C_{1}\\right|+\\left|B_{2} B_{3} C_{3} C_{2}\\right|+\\left|B_{4} B_{5} C_{5} C_{4}\\right|+\\left|B_{6} B_{7} C_{7} C_{6}\\right|+\\left|B_{8} B_{9} C_{9} C_{8}\\right|+\\left|B_{10} B C C_{10}\\right|\n$$\n\nThe area of each of these quadrilaterals is equal to the difference of the area of two triangles. For example,\n\n$$\n\\left|B_{2} B_{3} C_{3} C_{2}\\right|=\\left|\\triangle A B_{3} C_{3}\\right|-\\left|\\triangle A B_{2} C_{2}\\right|=-\\left|\\triangle A B_{2} C_{2}\\right|+\\left|\\triangle A B_{3} C_{3}\\right|\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A}=\\mid & \\triangle A B_{1} C_{1}|-| \\triangle A B_{2} C_{2}|+| \\triangle A B_{3} C_{3}|-| \\triangle A B_{4} C_{4}|+| \\triangle A B_{5} C_{5} \\mid \\\\\n& \\quad-\\left|\\triangle A B_{6} C_{6}\\right|+\\left|\\triangle A B_{7} C_{7}\\right|-\\left|\\triangle A B_{8} C_{8}\\right|+\\left|\\triangle A B_{9} C_{9}\\right|-\\left|\\triangle A B_{10} C_{10}\\right|+|\\triangle A B C|\n\\end{aligned}\n$$\n\nEach of $\\triangle A B_{1} C_{1}, \\triangle A B_{2} C_{2}, \\ldots, \\triangle A B_{10} C_{10}$ is similar to $\\triangle A B C$ because their two base angles are equal due.\n\nSuppose that the height of $\\triangle A B C$ from $A$ to $B C$ is $h$.\n\nSince the height of each of the 11 regions is equal in height, then the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, and so on.\n\nWhen two triangles are similar, their heights are in the same ratio as their side lengths:\n\nTo see this, suppose that $\\triangle P Q R$ is similar to $\\triangle S T U$ and that altitudes are drawn from $P$ and $S$ to $V$ and $W$.\n\n\nSince $\\angle P Q R=\\angle S T U$, then $\\triangle P Q V$ is similar to $\\triangle S T W$ (equal angle; right angle), which means that $\\frac{P Q}{S T}=\\frac{P V}{S W}$. In other words, the ratio of sides is equal to the ratio of heights.\n\nSince the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, then $B_{1} C_{1}=\\frac{1}{11} B C$.\n\nTherefore, $\\left|\\triangle A B_{1} C_{1}\\right|=\\frac{1}{2} \\cdot B_{1} C_{1} \\cdot \\frac{1}{11} h=\\frac{1}{2} \\cdot \\frac{1}{11} B C \\cdot \\frac{1}{11} h=\\frac{1^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{1^{2}}{11^{2}}|\\triangle A B C|$.\n\nSimilarly, since the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, then $B_{2} C_{2}=\\frac{2}{11} B C$.\n\n\n\nTherefore, $\\left|\\triangle A B_{2} C_{2}\\right|=\\frac{1}{2} \\cdot B_{2} C_{2} \\cdot \\frac{2}{11} h=\\frac{1}{2} \\cdot \\frac{2}{11} B C \\cdot \\frac{2}{11} h=\\frac{2^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{2^{2}}{11^{2}}|\\triangle A B C|$.\n\nThis result continues for each of the triangles.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A} & =\\frac{1^{2}}{11^{2}}|\\triangle A B C|-\\frac{2^{2}}{11^{2}}|\\triangle A B C|+\\frac{3^{2}}{11^{2}}|\\triangle A B C|-\\frac{4^{2}}{11^{2}}|\\triangle A B C|+\\frac{5^{2}}{11^{2}}|\\triangle A B C| \\\\\n& \\quad-\\frac{6^{2}}{11^{2}}|\\triangle A B C|+\\frac{7^{2}}{11^{2}}|\\triangle A B C|-\\frac{8^{2}}{11^{2}}|\\triangle A B C|+\\frac{9^{2}}{11^{2}}|\\triangle A B C|-\\frac{10^{2}}{11^{2}}|\\triangle A B C|+\\frac{11^{2}}{11^{2}}|\\triangle A B C| \\\\\n& =\\frac{1}{11^{2}}|\\triangle A B C|\\left(11^{2}-10^{2}+9^{2}-8^{2}+7^{2}-6^{2}+5^{2}-4^{2}+3^{2}-2^{2}+1\\right) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)((11+10)(11-10)+(9+8)(9-8)+\\cdots+(3+2)(3-2)+1) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)(11+10+9+8+7+6+5+4+3+2+1) \\\\\n& =\\frac{1}{11}\\left(70 \\mathrm{~cm}^{2}\\right) \\cdot 66 \\\\\n& =420 \\mathrm{~cm}^{2}\n\\end{aligned}\n$$\n\nTherefore, the combined area of the shaded regions of $\\triangle A B C$ is $420 \\mathrm{~cm}^{2}$.']",['$420$'],False,,Numerical, 2405,Geometry,,"A triangle of area $770 \mathrm{~cm}^{2}$ is divided into 11 regions of equal height by 10 lines that are all parallel to the base of the triangle. Starting from the top of the triangle, every other region is shaded, as shown. What is the total area of the shaded regions? ![](https://cdn.mathpix.com/cropped/2023_12_21_4165463fb3d29f83c565g-1.jpg?height=322&width=588&top_left_y=1541&top_left_x=1213)","['We make two copies of the given triangle, labelling them $\\triangle A B C$ and $\\triangle D E F$, as shown:\n![](https://cdn.mathpix.com/cropped/2023_12_21_f320a9c49ee40ec25c67g-1.jpg?height=388&width=1320&top_left_y=316&top_left_x=503)\n\nThe combined area of these two triangles is $2 \\cdot 770 \\mathrm{~cm}^{2}=1540 \\mathrm{~cm}^{2}$, and the shaded area in each triangle is the same.\n\nNext, we rotate $\\triangle D E F$ by $180^{\\circ}$ :\n![](https://cdn.mathpix.com/cropped/2023_12_21_f320a9c49ee40ec25c67g-1.jpg?height=404&width=1310&top_left_y=958&top_left_x=514)\n\nand join the two triangles together:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_f320a9c49ee40ec25c67g-1.jpg?height=385&width=783&top_left_y=1504&top_left_x=758)\n\nWe note that $B C$ and $A E$ (which was $F E$ ) are equal in length (since they were copies of each other) and parallel (since they are $180^{\\circ}$ rotations of each other). The same is true for $A B$ and $E C$.\n\nTherefore, $A B C E$ is a parallelogram.\n\nFurther, $A B C E$ is divided into 11 identical parallelograms (6 shaded and 5 unshaded) by the horizontal lines. (Since the sections of the two triangles are equal in height, the horizontal lines on both sides of $A C$ align.)\n\nThe total area of parallelogram $A B C E$ is $1540 \\mathrm{~cm}^{2}$.\n\nThus, the shaded area of $A B C E$ is $\\frac{6}{11} \\cdot 1540 \\mathrm{~cm}^{2}=840 \\mathrm{~cm}^{2}$.\n\nSince this shaded area is equally divided between the two halves of the parallelogram, then the combined area of the shaded regions of $\\triangle A B C$ is $\\frac{1}{2} \\cdot 840 \\mathrm{~cm}^{2}=420 \\mathrm{~cm}^{2}$.', 'We label the points where the horizontal lines touch $A B$ and $A C$ as shown:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_61afe1613d150a896975g-1.jpg?height=382&width=639&top_left_y=297&top_left_x=843)\n\nWe use the notation $|\\triangle A B C|$ to represent the area of $\\triangle A B C$ and use similar notation for the area of other triangles and quadrilaterals.\n\nLet $\\mathcal{A}$ be equal to the total area of the shaded regions.\n\nThus,\n\n$$\n\\mathcal{A}=\\left|\\triangle A B_{1} C_{1}\\right|+\\left|B_{2} B_{3} C_{3} C_{2}\\right|+\\left|B_{4} B_{5} C_{5} C_{4}\\right|+\\left|B_{6} B_{7} C_{7} C_{6}\\right|+\\left|B_{8} B_{9} C_{9} C_{8}\\right|+\\left|B_{10} B C C_{10}\\right|\n$$\n\nThe area of each of these quadrilaterals is equal to the difference of the area of two triangles. For example,\n\n$$\n\\left|B_{2} B_{3} C_{3} C_{2}\\right|=\\left|\\triangle A B_{3} C_{3}\\right|-\\left|\\triangle A B_{2} C_{2}\\right|=-\\left|\\triangle A B_{2} C_{2}\\right|+\\left|\\triangle A B_{3} C_{3}\\right|\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A}=\\mid & \\triangle A B_{1} C_{1}|-| \\triangle A B_{2} C_{2}|+| \\triangle A B_{3} C_{3}|-| \\triangle A B_{4} C_{4}|+| \\triangle A B_{5} C_{5} \\mid \\\\\n& \\quad-\\left|\\triangle A B_{6} C_{6}\\right|+\\left|\\triangle A B_{7} C_{7}\\right|-\\left|\\triangle A B_{8} C_{8}\\right|+\\left|\\triangle A B_{9} C_{9}\\right|-\\left|\\triangle A B_{10} C_{10}\\right|+|\\triangle A B C|\n\\end{aligned}\n$$\n\nEach of $\\triangle A B_{1} C_{1}, \\triangle A B_{2} C_{2}, \\ldots, \\triangle A B_{10} C_{10}$ is similar to $\\triangle A B C$ because their two base angles are equal due.\n\nSuppose that the height of $\\triangle A B C$ from $A$ to $B C$ is $h$.\n\nSince the height of each of the 11 regions is equal in height, then the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, and so on.\n\nWhen two triangles are similar, their heights are in the same ratio as their side lengths:\n\nTo see this, suppose that $\\triangle P Q R$ is similar to $\\triangle S T U$ and that altitudes are drawn from $P$ and $S$ to $V$ and $W$.\n![](https://cdn.mathpix.com/cropped/2023_12_21_61afe1613d150a896975g-1.jpg?height=306&width=846&top_left_y=1923&top_left_x=738)\n\nSince $\\angle P Q R=\\angle S T U$, then $\\triangle P Q V$ is similar to $\\triangle S T W$ (equal angle; right angle), which means that $\\frac{P Q}{S T}=\\frac{P V}{S W}$. In other words, the ratio of sides is equal to the ratio of heights.\n\nSince the height of $\\triangle A B_{1} C_{1}$ is $\\frac{1}{11} h$, then $B_{1} C_{1}=\\frac{1}{11} B C$.\n\nTherefore, $\\left|\\triangle A B_{1} C_{1}\\right|=\\frac{1}{2} \\cdot B_{1} C_{1} \\cdot \\frac{1}{11} h=\\frac{1}{2} \\cdot \\frac{1}{11} B C \\cdot \\frac{1}{11} h=\\frac{1^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{1^{2}}{11^{2}}|\\triangle A B C|$.\n\nSimilarly, since the height of $\\triangle A B_{2} C_{2}$ is $\\frac{2}{11} h$, then $B_{2} C_{2}=\\frac{2}{11} B C$.\n\n\n\nTherefore, $\\left|\\triangle A B_{2} C_{2}\\right|=\\frac{1}{2} \\cdot B_{2} C_{2} \\cdot \\frac{2}{11} h=\\frac{1}{2} \\cdot \\frac{2}{11} B C \\cdot \\frac{2}{11} h=\\frac{2^{2}}{11^{2}} \\cdot \\frac{1}{2} \\cdot B C \\cdot h=\\frac{2^{2}}{11^{2}}|\\triangle A B C|$.\n\nThis result continues for each of the triangles.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\mathcal{A} & =\\frac{1^{2}}{11^{2}}|\\triangle A B C|-\\frac{2^{2}}{11^{2}}|\\triangle A B C|+\\frac{3^{2}}{11^{2}}|\\triangle A B C|-\\frac{4^{2}}{11^{2}}|\\triangle A B C|+\\frac{5^{2}}{11^{2}}|\\triangle A B C| \\\\\n& \\quad-\\frac{6^{2}}{11^{2}}|\\triangle A B C|+\\frac{7^{2}}{11^{2}}|\\triangle A B C|-\\frac{8^{2}}{11^{2}}|\\triangle A B C|+\\frac{9^{2}}{11^{2}}|\\triangle A B C|-\\frac{10^{2}}{11^{2}}|\\triangle A B C|+\\frac{11^{2}}{11^{2}}|\\triangle A B C| \\\\\n& =\\frac{1}{11^{2}}|\\triangle A B C|\\left(11^{2}-10^{2}+9^{2}-8^{2}+7^{2}-6^{2}+5^{2}-4^{2}+3^{2}-2^{2}+1\\right) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)((11+10)(11-10)+(9+8)(9-8)+\\cdots+(3+2)(3-2)+1) \\\\\n& =\\frac{1}{11^{2}}\\left(770 \\mathrm{~cm}^{2}\\right)(11+10+9+8+7+6+5+4+3+2+1) \\\\\n& =\\frac{1}{11}\\left(70 \\mathrm{~cm}^{2}\\right) \\cdot 66 \\\\\n& =420 \\mathrm{~cm}^{2}\n\\end{aligned}\n$$\n\nTherefore, the combined area of the shaded regions of $\\triangle A B C$ is $420 \\mathrm{~cm}^{2}$.']",['$420$'],False,,Numerical, 2406,Geometry,,"A square lattice of 16 points is constructed such that the horizontal and vertical distances between adjacent points are all exactly 1 unit. Each of four pairs of points are connected by a line segment, as shown. The intersections of these line segments are the vertices of square $A B C D$. Determine the area of square $A B C D$. ","['We label five additional points in the diagram:\n\n\n\nSince $P Q=Q R=R S=1$, then $P S=3$ and $P R=2$.\n\nSince $\\angle P S T=90^{\\circ}$, then $P T=\\sqrt{P S^{2}+S T^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$ by the Pythagorean Theorem.\n\nWe are told that $A B C D$ is a square.\n\nThus, $P T$ is perpendicular to $Q C$ and to $R B$.\n\nThus, $\\triangle P D Q$ is right-angled at $D$ and $\\triangle P A R$ is right-angled at $A$.\n\nSince $\\triangle P D Q, \\triangle P A R$ and $\\triangle P S T$ are all right-angled and all share an angle at $P$, then these three triangles are similar.\n\nThis tells us that $\\frac{P A}{P S}=\\frac{P R}{P T}$ and so $P A=\\frac{3 \\cdot 2}{\\sqrt{10}}$. Also, $\\frac{P D}{P S}=\\frac{P Q}{P T}$ and so $P D=\\frac{1 \\cdot 3}{\\sqrt{10}}$.\n\nTherefore,\n\n$$\nD A=P A-P D=\\frac{6}{\\sqrt{10}}-\\frac{3}{\\sqrt{10}}=\\frac{3}{\\sqrt{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\frac{3}{\\sqrt{10}}\\right)^{2}=\\frac{9}{10}$.', 'We add coordinates to the diagram as shown:\n\n\n\nWe determine the side length of square $A B C D$ by determining the coordinates of $D$ and $A$ and then calculating the distance between these points.\n\nThe slope of the line through $(0,3)$ and $(3,2)$ is $\\frac{3-2}{0-3}=-\\frac{1}{3}$.\n\nThis equation of this line can be written as $y=-\\frac{1}{3} x+3$.\n\nThe slope of the line through $(0,0)$ and $(1,3)$ is 3.\n\nThe equation of this line can be written as $y=3 x$.\n\nThe slope of the line through $(1,0)$ and $(2,3)$ is also 3.\n\nThe equation of this line can be written as $y=3(x-1)=3 x-3$.\n\nPoint $D$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x$ and so $\\frac{10}{3} x=3$ which gives $x=\\frac{9}{10}$.\n\nSince $y=3 x$, we get $y=\\frac{27}{10}$ and so the coordinates of $D$ are $\\left(\\frac{9}{10}, \\frac{27}{10}\\right)$.\n\nPoint $A$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x-3$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x-3$ and so $\\frac{10}{3} x=6$ which gives $x=\\frac{18}{10}$.\n\nSince $y=3 x-3$, we get $y=\\frac{24}{10}$ and so the coordinates of $A$ are $\\left(\\frac{18}{10}, \\frac{24}{10}\\right)$. (It is easier to not reduce these fractions.)\n\nTherefore,\n\n$$\nD A=\\sqrt{\\left(\\frac{9}{10}-\\frac{18}{10}\\right)^{2}+\\left(\\frac{27}{10}-\\frac{24}{10}\\right)^{2}}=\\sqrt{\\left(-\\frac{9}{10}\\right)^{2}+\\left(\\frac{3}{10}\\right)^{2}}=\\sqrt{\\frac{90}{100}}=\\sqrt{\\frac{9}{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\sqrt{\\frac{9}{10}}\\right)^{2}=\\frac{9}{10}$.']",['$\\frac{9}{10}$'],False,,Numerical, 2406,Geometry,,"A square lattice of 16 points is constructed such that the horizontal and vertical distances between adjacent points are all exactly 1 unit. Each of four pairs of points are connected by a line segment, as shown. The intersections of these line segments are the vertices of square $A B C D$. Determine the area of square $A B C D$. ![](https://cdn.mathpix.com/cropped/2023_12_21_4165463fb3d29f83c565g-1.jpg?height=436&width=466&top_left_y=1896&top_left_x=1233)","['We label five additional points in the diagram:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_49c4431a3b32fd4003c0g-1.jpg?height=469&width=501&top_left_y=1034&top_left_x=904)\n\nSince $P Q=Q R=R S=1$, then $P S=3$ and $P R=2$.\n\nSince $\\angle P S T=90^{\\circ}$, then $P T=\\sqrt{P S^{2}+S T^{2}}=\\sqrt{3^{2}+1^{2}}=\\sqrt{10}$ by the Pythagorean Theorem.\n\nWe are told that $A B C D$ is a square.\n\nThus, $P T$ is perpendicular to $Q C$ and to $R B$.\n\nThus, $\\triangle P D Q$ is right-angled at $D$ and $\\triangle P A R$ is right-angled at $A$.\n\nSince $\\triangle P D Q, \\triangle P A R$ and $\\triangle P S T$ are all right-angled and all share an angle at $P$, then these three triangles are similar.\n\nThis tells us that $\\frac{P A}{P S}=\\frac{P R}{P T}$ and so $P A=\\frac{3 \\cdot 2}{\\sqrt{10}}$. Also, $\\frac{P D}{P S}=\\frac{P Q}{P T}$ and so $P D=\\frac{1 \\cdot 3}{\\sqrt{10}}$.\n\nTherefore,\n\n$$\nD A=P A-P D=\\frac{6}{\\sqrt{10}}-\\frac{3}{\\sqrt{10}}=\\frac{3}{\\sqrt{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\frac{3}{\\sqrt{10}}\\right)^{2}=\\frac{9}{10}$.', 'We add coordinates to the diagram as shown:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_61920460287a32b417ceg-1.jpg?height=529&width=528&top_left_y=294&top_left_x=907)\n\nWe determine the side length of square $A B C D$ by determining the coordinates of $D$ and $A$ and then calculating the distance between these points.\n\nThe slope of the line through $(0,3)$ and $(3,2)$ is $\\frac{3-2}{0-3}=-\\frac{1}{3}$.\n\nThis equation of this line can be written as $y=-\\frac{1}{3} x+3$.\n\nThe slope of the line through $(0,0)$ and $(1,3)$ is 3.\n\nThe equation of this line can be written as $y=3 x$.\n\nThe slope of the line through $(1,0)$ and $(2,3)$ is also 3.\n\nThe equation of this line can be written as $y=3(x-1)=3 x-3$.\n\nPoint $D$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x$ and so $\\frac{10}{3} x=3$ which gives $x=\\frac{9}{10}$.\n\nSince $y=3 x$, we get $y=\\frac{27}{10}$ and so the coordinates of $D$ are $\\left(\\frac{9}{10}, \\frac{27}{10}\\right)$.\n\nPoint $A$ is the intersection point of the lines with equations $y=-\\frac{1}{3} x+3$ and $y=3 x-3$.\n\nEquating expressions for $y$, we obtain $-\\frac{1}{3} x+3=3 x-3$ and so $\\frac{10}{3} x=6$ which gives $x=\\frac{18}{10}$.\n\nSince $y=3 x-3$, we get $y=\\frac{24}{10}$ and so the coordinates of $A$ are $\\left(\\frac{18}{10}, \\frac{24}{10}\\right)$. (It is easier to not reduce these fractions.)\n\nTherefore,\n\n$$\nD A=\\sqrt{\\left(\\frac{9}{10}-\\frac{18}{10}\\right)^{2}+\\left(\\frac{27}{10}-\\frac{24}{10}\\right)^{2}}=\\sqrt{\\left(-\\frac{9}{10}\\right)^{2}+\\left(\\frac{3}{10}\\right)^{2}}=\\sqrt{\\frac{90}{100}}=\\sqrt{\\frac{9}{10}}\n$$\n\nThis means that the area of square $A B C D$ is equal to $D A^{2}=\\left(\\sqrt{\\frac{9}{10}}\\right)^{2}=\\frac{9}{10}$.']",['$\\frac{9}{10}$'],False,,Numerical, 2407,Combinatorics,,"A bag contains 3 red marbles and 6 blue marbles. Akshan removes one marble at a time until the bag is empty. Each marble that they remove is chosen randomly from the remaining marbles. Given that the first marble that Akshan removes is red and the third marble that they remove is blue, what is the probability that the last two marbles that Akshan removes are both blue?","[""Each possible order in which Akshan removes the marbles corresponds to a sequence of 9 colours, 3 of which are red and 6 of which are blue.\n\nWe write these as sequences of 3 R's and 6 B's.\n\nSince are told that the first marble is red and the third is blue, we would like to consider all sequences of the form\n\n$$\nR \\_B\\_\\_\\_\\_\\_\\_\n$$\n\nThe 7 blanks must be filled with the remaining 2 R's and 5 B's.\n\nThere are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)=\\frac{7 \\cdot 6}{2}=21$ ways of doing this, because 2 of the 7 blanks must be chosen in which to place the R's. (We could count these 21 ways directly by working systematically through the possible pairs of blanks.)\n\nOf these 21 ways, some have the last two marbles being blue.\n\nThese correspond to the sequences of the form\n\n$$\nR \\_B \\_\\_\\_\\_ B B\n$$\n\nIn these sequences, the 5 blanks must be filled with the remaining $2 \\mathrm{R}$ 's and 3 B's.\n\nThere are $\\left(\\begin{array}{l}5 \\\\ 2\\end{array}\\right)=\\frac{5 \\cdot 4}{2}=10$ ways of doing this, because 2 of the 5 blanks must be chosen in which to place the R's.\n\nTherefore, 10 of the 21 possible sequences end in two B's, and so the probability that the last two marbles removed are blue is $\\frac{10}{21}$.""]",['$\\frac{10}{21}$'],False,,Numerical, 2408,Number Theory,,"Determine the number of quadruples of positive integers $(a, b, c, d)$ with $a0$, then $\\sqrt{a^{2}}=a$.\n\nWhen $\\log _{2} x \\leq-1$, we know that $\\log _{2} x+1 \\leq 0$ and $\\log _{2} x-3<0$, and so\n\n$$\nf(x)=-\\left(\\log _{2} x+1\\right)-\\left(\\log _{2} x-3\\right)=2-2 \\log _{2} x\n$$\n\nWhen $-1<\\log _{2} x \\leq 3$, we know that $\\log _{2} x+1>0$ and $\\log _{2} x-3 \\leq 0$, and so\n\n$$\nf(x)=\\left(\\log _{2} x+1\\right)-\\left(\\log _{2} x-3\\right)=4\n$$\n\nWhen $\\log _{2} x>3$, we know that $\\log _{2} x+1 \\geq 0$ and $\\log _{2} x-3>0$, and so\n\n$$\nf(x)=\\left(\\log _{2} x+1\\right)+\\left(\\log _{2} x-3\\right)=2 \\log _{2} x-2\n$$\n\nWe want to find all values of $x$ for which $f(x)=4$.\n\nWhen $\\log _{2} x \\leq-1, f(x)=2-2 \\log _{2} x=4$ exactly when $\\log _{2} x=-1$.\n\nWhen $-1<\\log _{2} x \\leq 3, f(x)$ is always equal to 4 .\n\nWhen $\\log _{2} x>3, f(x)=2 \\log _{2} x-2=4$ exactly when $\\log _{2} x=3$.\n\nTherefore, $f(x)=4$ exactly when $-1 \\leq \\log _{2} x \\leq 3$, which is true exactly when $\\frac{1}{2} \\leq x \\leq 8$. (It seems surprising that the solution to this equation is actually an interval of values, rather than a finite number of specific values.)']","['$[\\frac{1}{2}, 8]$']",False,,Interval, 2411,Combinatorics,,"At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ed73f7f90f51d294b52fg-1.jpg?height=401&width=331&top_left_y=1304&top_left_x=1469) Determine all ways in which people can be seated around a table with 8 chairs so that the table is full, in each case giving the labels on the chairs in which people are sitting.","['If there are 5 or more people seated around a table with 8 chairs, then there are at most 3 empty chairs. But there must be an empty chair between each pair of people, and this is not possible with 5 people and 3 empty chairs.\n\nTherefore, there are at most 4 people seated.\n\nIf there were only 2 people seated, then there would be 6 empty chairs which would mean that at least one of the two ""gaps"" around the circular table had at least 3 empty chairs, and so another person could be seated, meaning that the table wasn\'t full.\n\nTherefore, there are at least 3 people seated.\n\nThis means that a full table with 8 chairs has either 3 or 4 people.\n\nIf there are 4 people, there are 4 empty chairs, and so there is exactly 1 empty chair between each pair of people.\n\nThus, people are seated in chairs $\\{1,3,5,7\\}$ or in chairs $\\{2,4,6,8\\}$.\n\nIf there are 3 people, there are 5 empty chairs.\n\nWith 3 people, there are 3 gaps totalling 5 chairs, and each gap has at most 2 chairs in it. Therefore, the gaps must be 1, 2, 2 in some order. This is the only list of three positive integers, each equal to 1 or 2 , that adds to 5 .\n\nThe gap of 1 can be between any pair of seats. In other words, the gap of 1 could be between $\\{1,3\\},\\{2,4\\}$, and so on. In each case, the position of the third person is completely determined because the remaining two gaps have 2 chairs each.\n\nThus, with 3 people, they are seated in chairs\n\n$$\n\\{1,3,6\\},\\{2,4,7\\},\\{3,5,8\\},\\{4,6,1\\},\\{5,7,2\\},\\{6,8,3\\},\\{7,1,4\\},\\{8,2,5\\}\n$$\n\nIn total, there are thus 10 ways to seat people at a table with 8 chairs:\n\n$\\{1,3,5,7\\},\\{2,4,6,8\\},\\{1,3,6\\},\\{2,4,7\\},\\{3,5,8\\},\\{4,6,1\\},\\{5,7,2\\},\\{6,8,3\\},\\{7,1,4\\},\\{8,2,5\\}$']","['If there are 5 or more people seated around a table with 8 chairs, then there are at most 3 empty chairs. But there must be an empty chair between each pair of people, and this is not possible with 5 people and 3 empty chairs.\n\nTherefore, there are at most 4 people seated.\n\nIf there were only 2 people seated, then there would be 6 empty chairs which would mean that at least one of the two ""gaps"" around the circular table had at least 3 empty chairs, and so another person could be seated, meaning that the table wasn\'t full.\n\nTherefore, there are at least 3 people seated.\n\nThis means that a full table with 8 chairs has either 3 or 4 people.\n\nIf there are 4 people, there are 4 empty chairs, and so there is exactly 1 empty chair between each pair of people.\n\nThus, people are seated in chairs $\\{1,3,5,7\\}$ or in chairs $\\{2,4,6,8\\}$.\n\nIf there are 3 people, there are 5 empty chairs.\n\nWith 3 people, there are 3 gaps totalling 5 chairs, and each gap has at most 2 chairs in it. Therefore, the gaps must be 1, 2, 2 in some order. This is the only list of three positive integers, each equal to 1 or 2 , that adds to 5 .\n\nThe gap of 1 can be between any pair of seats. In other words, the gap of 1 could be between $\\{1,3\\},\\{2,4\\}$, and so on. In each case, the position of the third person is completely determined because the remaining two gaps have 2 chairs each.\n\nThus, with 3 people, they are seated in chairs\n\n$$\n\\{1,3,6\\},\\{2,4,7\\},\\{3,5,8\\},\\{4,6,1\\},\\{5,7,2\\},\\{6,8,3\\},\\{7,1,4\\},\\{8,2,5\\}\n$$\n\nIn total, there are thus 10 ways to seat people at a table with 8 chairs:\n\n$\\{1,3,5,7\\},\\{2,4,6,8\\},\\{1,3,6\\},\\{2,4,7\\},\\{3,5,8\\},\\{4,6,1\\},\\{5,7,2\\},\\{6,8,3\\},\\{7,1,4\\},\\{8,2,5\\}$']",True,,Need_human_evaluate, 2412,Combinatorics,,"At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. A full table with $6 k+5$ chairs, for some positive integer $k$, has $t$ people seated in its chairs. Determine, in terms of $k$, the number of possible values of $t$.","['Suppose that $k$ is a positive integer.\n\nSuppose that $t$ people are seated at a table with $6 k+5$ chairs so that the table is full.\n\nWhen $t$ people are seated, there are $t$ gaps. Each gap consists of either 1 or 2 chairs. (A gap with 3 or more chairs can have an additional person seated in it, so the table is not full.)\n\nTherefore, there are between $t$ and $2 t$ empty chairs.\n\nThis means that the total number of chairs is between $t+t$ and $t+2 t$.\n\nIn other words, $2 t \\leq 6 k+5 \\leq 3 t$.\n\nSince $2 t \\leq 6 k+5$, then $t \\leq 3 k+\\frac{5}{2}$. Since $k$ and $t$ are integers, then $t \\leq 3 k+2$.\n\nWe note that it is possible to seat $3 k+2$ people around the table in seats\n\n$$\n\\{2,4,6, \\ldots, 6 k+2,6 k+4\\}\n$$\n\nThis table is full becase $3 k+1$ of the gaps consist of 1 chair and 1 gap consists of 2 chairs. Since $3 t \\geq 6 k+5$, then $t \\geq 2 k+\\frac{5}{3}$. Since $k$ and $t$ are integers, then $t \\geq 2 k+2$.\n\nWe note that it is possible to seat $2 k+2$ people around the table in seats\n\n$$\n\\{3,6,9, \\ldots, 6 k, 6 k+3,6 k+5\\}\n$$\n\nThis table is full becase $2 k+1$ of the gaps consist of 2 chairs and 1 gap consists of 1 chair.\n\nWe now know that, if there are $t$ people seated at a full table with $6 k+5$ chairs, then $2 k+2 \\leq t \\leq 3 k+2$.\n\nTo confirm that every such value of $t$ is possible, consider a table with $t$ people, $3 t-(6 k+5)$\n\n\n\ngaps of 1 chair, and $(6 k+5)-2 t$ gaps of 2 chairs.\n\nFrom the work above, we know that $3 t \\geq 6 k+5$ and so $3 t-(6 k+5) \\geq 0$, and that $2 t \\leq 6 k+5$ and so $(6 k+5)-2 t \\geq 0$.\n\nThe total number of gaps is $3 t-(6 k+5)+(6 k+5)-2 t=t$, since there are $t$ people seated.\n\nFinally, the total number of chairs is\n\n$$\nt+1 \\cdot(3 t-(6 k+5))+2 \\cdot((6 k+5)-2 t)=t+3 t-4 t-(6 k+5)+2(6 k+5)=6 k+5\n$$\n\nas expected.\n\nThis shows that every $t$ with $2 k+2 \\leq t \\leq 3 k+2$ can produce a full table.\n\nTherefore, the possible values of $t$ are those integers that satisfy $2 k+2 \\leq t \\leq 3 k+2$.\n\nThere are $(3 k+2)-(2 k+2)+1=k+1$ possible values of $t$.']",['$k+1$'],False,,Expression, 2412,Combinatorics,,"At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ed73f7f90f51d294b52fg-1.jpg?height=401&width=331&top_left_y=1304&top_left_x=1469) A full table with $6 k+5$ chairs, for some positive integer $k$, has $t$ people seated in its chairs. Determine, in terms of $k$, the number of possible values of $t$.","['Suppose that $k$ is a positive integer.\n\nSuppose that $t$ people are seated at a table with $6 k+5$ chairs so that the table is full.\n\nWhen $t$ people are seated, there are $t$ gaps. Each gap consists of either 1 or 2 chairs. (A gap with 3 or more chairs can have an additional person seated in it, so the table is not full.)\n\nTherefore, there are between $t$ and $2 t$ empty chairs.\n\nThis means that the total number of chairs is between $t+t$ and $t+2 t$.\n\nIn other words, $2 t \\leq 6 k+5 \\leq 3 t$.\n\nSince $2 t \\leq 6 k+5$, then $t \\leq 3 k+\\frac{5}{2}$. Since $k$ and $t$ are integers, then $t \\leq 3 k+2$.\n\nWe note that it is possible to seat $3 k+2$ people around the table in seats\n\n$$\n\\{2,4,6, \\ldots, 6 k+2,6 k+4\\}\n$$\n\nThis table is full becase $3 k+1$ of the gaps consist of 1 chair and 1 gap consists of 2 chairs. Since $3 t \\geq 6 k+5$, then $t \\geq 2 k+\\frac{5}{3}$. Since $k$ and $t$ are integers, then $t \\geq 2 k+2$.\n\nWe note that it is possible to seat $2 k+2$ people around the table in seats\n\n$$\n\\{3,6,9, \\ldots, 6 k, 6 k+3,6 k+5\\}\n$$\n\nThis table is full becase $2 k+1$ of the gaps consist of 2 chairs and 1 gap consists of 1 chair.\n\nWe now know that, if there are $t$ people seated at a full table with $6 k+5$ chairs, then $2 k+2 \\leq t \\leq 3 k+2$.\n\nTo confirm that every such value of $t$ is possible, consider a table with $t$ people, $3 t-(6 k+5)$\n\n\n\ngaps of 1 chair, and $(6 k+5)-2 t$ gaps of 2 chairs.\n\nFrom the work above, we know that $3 t \\geq 6 k+5$ and so $3 t-(6 k+5) \\geq 0$, and that $2 t \\leq 6 k+5$ and so $(6 k+5)-2 t \\geq 0$.\n\nThe total number of gaps is $3 t-(6 k+5)+(6 k+5)-2 t=t$, since there are $t$ people seated.\n\nFinally, the total number of chairs is\n\n$$\nt+1 \\cdot(3 t-(6 k+5))+2 \\cdot((6 k+5)-2 t)=t+3 t-4 t-(6 k+5)+2(6 k+5)=6 k+5\n$$\n\nas expected.\n\nThis shows that every $t$ with $2 k+2 \\leq t \\leq 3 k+2$ can produce a full table.\n\nTherefore, the possible values of $t$ are those integers that satisfy $2 k+2 \\leq t \\leq 3 k+2$.\n\nThere are $(3 k+2)-(2 k+2)+1=k+1$ possible values of $t$.']",['$k+1$'],False,,Expression, 2413,Combinatorics,,"At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. ![](https://cdn.mathpix.com/cropped/2023_12_21_ed73f7f90f51d294b52fg-1.jpg?height=401&width=331&top_left_y=1304&top_left_x=1469) Determine the number of different full tables when $n=19$.","['For each integer $n \\geq 3$, we define $f(n)$ to be the number of different full tables of size $n$. We can check that\n\n- $f(3)=3$ because the full tables when $n=3$ have people in chairs $\\{1\\},\\{2\\},\\{3\\}$,\n- $f(4)=2$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\}$, and\n- $f(5)=5$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\},\\{3,5\\}$, $\\{4,1\\},\\{5,2\\}$.\n\nIn the problem, we are told that $f(6)=5$ and in part (a), we determined that $f(8)=10$. This gives us the following table:\n\n| $n$ | $f(n)$ |\n| :---: | :---: |\n| 3 | 3 |\n| 4 | 2 |\n| 5 | 5 |\n| 6 | 5 |\n| 7 | $?$ |\n| 8 | 10 |\n\nBased on this information, we make the guess that for every integer $n \\geq 6$, we have $f(n)=f(n-2)+f(n-3)$.\n\nFor example, this would mean that $f(7)=f(5)+f(4)=5+2=7$ which we can verify is true.\n\nBased on this recurrence relation (which we have yet to prove), we deduce the values of $f(n)$ up to and including $n=19$ :\n\n| $n$ | $f(n)$ | $n$ | $f(n)$ |\n| :---: | :---: | :---: | :---: |\n| 3 | 3 | 11 | 22 |\n| 4 | 2 | 12 | 29 |\n| 5 | 5 | 13 | 39 |\n| 6 | 5 | 14 | 51 |\n| 7 | 7 | 15 | 68 |\n| 8 | 10 | 16 | 90 |\n| 9 | 12 | 17 | 119 |\n| 10 | 17 | 18 | 158 |\n| | | 19 | 209 |\n\nWe now need to prove that the equation $f(n)=f(n-2)+f(n-3)$ is true for all $n \\geq 6$.\n\n\n\nWe think about each full table as a string of 0 s and 1s, with 1 representing a chair that is occupied and 0 representing an empty chair.\n\nLet $a(n)$ be the number of full tables with someone in seat 1 (and thus nobody in seat 2). Let $b(n)$ be the number of full tables with someone in seat 2 (and thus nobody in seat 1). Let $c(n)$ be the number of full tables with nobody in seat 1 or in seat 2 .\n\nSince every full table must be in one of these categories, then $f(n)=a(n)+b(n)+c(n)$. A full table with $n$ seats $n \\geq 4$ must correspond to a string that starts with 10,01 or 00 . Since there cannot be more than two consecutive 0s, we can further specify this, namely to say that a full table with $n$ seats must correspond to a string that starts with 1010 or 1001 or 0100 or 0101 or 0010 . In each case, these are the first 4 characters of the string and correspond to full (1) and empty (0) chairs.\n\nConsider the full tables starting with 1010. Note that such strings end with 0 since the table is circular. Removing the 10 from positions 1 and 2 creates strings of length $n-2$ that begin 10. These strings will still correspond to a full table, and so there are $a(n-2)$ such strings. (We note that all possible strings starting 1010 of length $n$ will lead to all possible strings starting with 1010 of length $n-2$.)\n\nConsider the full tables starting with 1001. Note that such a string ends with 0 since the table is circular. Removing the 100 from positions 1, 2 and 3 creates strings of length $n-3$ that begin 10. (There must have been a 0 in position 5 after the 1 in position 4.) These strings will still correspond to full tables, and so there are $a(n-3)$ such strings. Consider the full tables starting with 0100 . Removing the 100 from positions 2,3 and 4 creates strings of length $n-3$ that begin 01 . (There must have been a 1 in position 5 after the 0 in position 4.) These strings will still correspond to full tables, and so there are $b(n-3)$ such strings.\n\nConsider the full tables starting with 0101. Removing the 01 from positions 3 and 4 creates strings of length $n-2$ that begin 01 . (The 1 in position 4 must have been followed by one or two 0s and so these strings maintains the desired properties.) These strings will still correspond to full tables, and so there are $b(n-2)$ such strings.\n\nConsider the full tables starting with 0010. These strings must begin with either 00100 or 00101.\n\nIf strings start 00100 , then they start 001001 and so we remove the 001 in positions 4,5 and 6 and obtain strings of length $n-3$ that start 001 (and thus start 00). There are $c(n-3)$ such strings.\n\nIf strings start 00101, we remove the 01 in positions 4 and 5 and obtain strings of length $n-2$ that start 001 (and thus start 00 ). There are $c(n-2)$ such strings.\n\nThese 6 cases and subcases count all strings counted by $f(n)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n) & =a(n-2)+a(n-3)+b(n-3)+b(n-2)+c(n-3)+c(n-2) \\\\\n& =a(n-2)+b(n-2)+c(n-2)+a(n-3)+b(n-3)+c(n-3) \\\\\n& =f(n-2)+f(n-3)\n\\end{aligned}\n$$\n\nas required, which means that the number of different full tables when $n=19$ is 209 .', 'Extending our approach from (b), the number of people seated at a full table with 19 chairs is at least $\\frac{19}{3}=6 \\frac{1}{3}$ and at most $\\frac{19}{2}=9 \\frac{1}{2}$.\n\nSince the number of people is an integer, there must be 7,8 or 9 people at the table, which means that the number of empty chairs is 12,11 or 10 , respectively.\n\n\n\nSuppose that there are 9 people and 9 gaps with a total of 10 empty chairs.\n\nIn this case, there is 1 gap with 2 empty chairs and 8 gaps with 1 empty chair.\n\nThere are 19 pairs of chairs in which we can put 2 people with a gap of 2 in between: $\\{1,4\\},\\{2,5\\}, \\ldots,\\{19,3\\}$.\n\nOnce we choose one of these pairs, the seat choice for the remaining 8 people is completely determined by placing people in every other chair.\n\nTherefore, there are 19 different full tables with 9 people.\n\nSuppose that there are 8 people and 8 gaps with a total of 11 empty chairs.\n\nIn this case, there are 3 gaps with 2 empty chairs and 5 gaps with 1 empty chair. There are 7 different circular orderings in which these 8 gaps can be arranged:\n\n$$\n\\begin{array}{lllllll}\n22211111 & 22121111 & 22112111 & 22111211 & 22111121 & 21212111 & 21211211\n\\end{array}\n$$\n\nWe note that ""22211111"" would be the same as, for example, ""11222111"" since these gaps are arranged around a circle.\n\nIf the three gaps of length 2 are consecutive, there is only one configuration (22211111). If there are exactly 2 consecutive gaps of length 2 , there are 4 relative places in which the third gap of length 2 can be placed.\n\nIf there are no consecutive gaps of length 2 , these gaps can either be separated by 1 gap each (21212111) with 3 gaps on the far side, or can be separated by 1 gap, 2 gaps, and 2 gaps (21211211). There is only one configuration for the gaps in this last situation.\n\nThere are 7 different circular orderings for these 8 gaps.\n\nEach of these 7 different orderings can be placed around the circle of 19 chairs in 19 different ways, because each can be started in 19 different places. Because 19 is prime, none of these orderings overlap.\n\nTherefore, there are $7 \\cdot 19=133$ different full tables with 8 people.\n\nSuppose that there are 7 people and 7 gaps with a total of 12 empty chairs.\n\nIn this case, there are 2 gaps with 1 empty chair and 5 gaps with 2 empty chairs.\n\nThe 2 gaps with 1 empty chair can be separated by 0 gaps with 2 empty chairs, 1 gap with 2 empty chairs, or 2 gaps with 2 empty chairs. Because the chairs are around a circle, if there were 3 , 4 or 5 gaps with 2 empty chairs between them, there would be 2,1 or 0 gaps going the other way around the circle.\n\nThis means that there are 3 different configurations for the gaps.\n\nEach of these configurations can be placed in 19 different ways around the circle of chairs. Therefore, there are $3 \\cdot 19=57$ full tables with 7 people.\n\nIn total, there are $19+133+57=209$ full tables with 19 chairs.', 'As in Solution 2, there must be 7, 8 or 9 people in chairs, and so there are 7,8 or 9 gaps. If there are 7 gaps, there are 2 gaps of 1 chair and 5 gaps of 2 chairs.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 3 gaps of 2 chairs.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 1 gap of 2 chairs.\n\nWe consider three mutually exclusive cases: (i) there is a person in chair 1 and not in chair 2 , (ii) there is a person in chair 2 and not in chair 1 , and (iii) there is nobody in chair 1 or in chair 2. Every full table fits into exactly one of these three cases.\n\nCase (i): there is a person in chair 1 and not in chair 2\n\nWe use the person in chair 1 to ""anchor"" the arrangement, by starting at chair 1 and\n\n\n\narranging the gaps (and thus the full chairs) clockwise around the table from chair 1.\n\nIf there are 7 gaps, we need to choose 2 of them to be of length 1 , and so there are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 8 gaps, we need to choose 3 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 9 gaps, we need to choose 1 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)+\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)=21+56+9=86$ full tables.\n\nCase (ii): there is a person in chair 2 and not in chair 1\n\nWe use the same reasoning starting with the person in chair 2 as the anchor.\n\nAgain, there are 86 full tables in this case.\n\nCase (iii): there is nobody in chair 1 or chair 2\n\nSince there is nobody in chair 1 or chair 2 , there must be a person in chair 3 and also in chair 19 , which fixes one gap of 2 chairs.\n\nHere, we use the person in chair 3 as the anchor.\n\nIf there are 7 gaps, there are 2 gaps of 1 chair and 4 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 2 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 0 gaps of 2 chairs left to place. There is 1 way to do this.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+1=15+21+1=37$ full tables.\n\nIn total, there are $86+86+37=209$ full tables with 19 chairs.']",['209'],False,,Numerical, 2413,Combinatorics,,"At the Canadian Eatery with Multiple Configurations, there are round tables, around which chairs are placed. When a table has $n$ chairs around it for some integer $n \geq 3$, the chairs are labelled $1,2,3, \ldots, n-1, n$ in order around the table. A table is considered full if no more people can be seated without having two people sit in neighbouring chairs. For example, when $n=6$, full tables occur when people are seated in chairs labelled $\{1,4\}$ or $\{2,5\}$ or $\{3,6\}$ or $\{1,3,5\}$ or $\{2,4,6\}$. Thus, there are 5 different full tables when $n=6$. Determine the number of different full tables when $n=19$.","['For each integer $n \\geq 3$, we define $f(n)$ to be the number of different full tables of size $n$. We can check that\n\n- $f(3)=3$ because the full tables when $n=3$ have people in chairs $\\{1\\},\\{2\\},\\{3\\}$,\n- $f(4)=2$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\}$, and\n- $f(5)=5$ because the full tables when $n=4$ have people in chairs $\\{1,3\\},\\{2,4\\},\\{3,5\\}$, $\\{4,1\\},\\{5,2\\}$.\n\nIn the problem, we are told that $f(6)=5$ and in part (a), we determined that $f(8)=10$. This gives us the following table:\n\n| $n$ | $f(n)$ |\n| :---: | :---: |\n| 3 | 3 |\n| 4 | 2 |\n| 5 | 5 |\n| 6 | 5 |\n| 7 | $?$ |\n| 8 | 10 |\n\nBased on this information, we make the guess that for every integer $n \\geq 6$, we have $f(n)=f(n-2)+f(n-3)$.\n\nFor example, this would mean that $f(7)=f(5)+f(4)=5+2=7$ which we can verify is true.\n\nBased on this recurrence relation (which we have yet to prove), we deduce the values of $f(n)$ up to and including $n=19$ :\n\n| $n$ | $f(n)$ | $n$ | $f(n)$ |\n| :---: | :---: | :---: | :---: |\n| 3 | 3 | 11 | 22 |\n| 4 | 2 | 12 | 29 |\n| 5 | 5 | 13 | 39 |\n| 6 | 5 | 14 | 51 |\n| 7 | 7 | 15 | 68 |\n| 8 | 10 | 16 | 90 |\n| 9 | 12 | 17 | 119 |\n| 10 | 17 | 18 | 158 |\n| | | 19 | 209 |\n\nWe now need to prove that the equation $f(n)=f(n-2)+f(n-3)$ is true for all $n \\geq 6$.\n\n\n\nWe think about each full table as a string of 0 s and 1s, with 1 representing a chair that is occupied and 0 representing an empty chair.\n\nLet $a(n)$ be the number of full tables with someone in seat 1 (and thus nobody in seat 2). Let $b(n)$ be the number of full tables with someone in seat 2 (and thus nobody in seat 1). Let $c(n)$ be the number of full tables with nobody in seat 1 or in seat 2 .\n\nSince every full table must be in one of these categories, then $f(n)=a(n)+b(n)+c(n)$. A full table with $n$ seats $n \\geq 4$ must correspond to a string that starts with 10,01 or 00 . Since there cannot be more than two consecutive 0s, we can further specify this, namely to say that a full table with $n$ seats must correspond to a string that starts with 1010 or 1001 or 0100 or 0101 or 0010 . In each case, these are the first 4 characters of the string and correspond to full (1) and empty (0) chairs.\n\nConsider the full tables starting with 1010. Note that such strings end with 0 since the table is circular. Removing the 10 from positions 1 and 2 creates strings of length $n-2$ that begin 10. These strings will still correspond to a full table, and so there are $a(n-2)$ such strings. (We note that all possible strings starting 1010 of length $n$ will lead to all possible strings starting with 1010 of length $n-2$.)\n\nConsider the full tables starting with 1001. Note that such a string ends with 0 since the table is circular. Removing the 100 from positions 1, 2 and 3 creates strings of length $n-3$ that begin 10. (There must have been a 0 in position 5 after the 1 in position 4.) These strings will still correspond to full tables, and so there are $a(n-3)$ such strings. Consider the full tables starting with 0100 . Removing the 100 from positions 2,3 and 4 creates strings of length $n-3$ that begin 01 . (There must have been a 1 in position 5 after the 0 in position 4.) These strings will still correspond to full tables, and so there are $b(n-3)$ such strings.\n\nConsider the full tables starting with 0101. Removing the 01 from positions 3 and 4 creates strings of length $n-2$ that begin 01 . (The 1 in position 4 must have been followed by one or two 0s and so these strings maintains the desired properties.) These strings will still correspond to full tables, and so there are $b(n-2)$ such strings.\n\nConsider the full tables starting with 0010. These strings must begin with either 00100 or 00101.\n\nIf strings start 00100 , then they start 001001 and so we remove the 001 in positions 4,5 and 6 and obtain strings of length $n-3$ that start 001 (and thus start 00). There are $c(n-3)$ such strings.\n\nIf strings start 00101, we remove the 01 in positions 4 and 5 and obtain strings of length $n-2$ that start 001 (and thus start 00 ). There are $c(n-2)$ such strings.\n\nThese 6 cases and subcases count all strings counted by $f(n)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n) & =a(n-2)+a(n-3)+b(n-3)+b(n-2)+c(n-3)+c(n-2) \\\\\n& =a(n-2)+b(n-2)+c(n-2)+a(n-3)+b(n-3)+c(n-3) \\\\\n& =f(n-2)+f(n-3)\n\\end{aligned}\n$$\n\nas required, which means that the number of different full tables when $n=19$ is 209 .', 'Extending our approach from (b), the number of people seated at a full table with 19 chairs is at least $\\frac{19}{3}=6 \\frac{1}{3}$ and at most $\\frac{19}{2}=9 \\frac{1}{2}$.\n\nSince the number of people is an integer, there must be 7,8 or 9 people at the table, which means that the number of empty chairs is 12,11 or 10 , respectively.\n\n\n\nSuppose that there are 9 people and 9 gaps with a total of 10 empty chairs.\n\nIn this case, there is 1 gap with 2 empty chairs and 8 gaps with 1 empty chair.\n\nThere are 19 pairs of chairs in which we can put 2 people with a gap of 2 in between: $\\{1,4\\},\\{2,5\\}, \\ldots,\\{19,3\\}$.\n\nOnce we choose one of these pairs, the seat choice for the remaining 8 people is completely determined by placing people in every other chair.\n\nTherefore, there are 19 different full tables with 9 people.\n\nSuppose that there are 8 people and 8 gaps with a total of 11 empty chairs.\n\nIn this case, there are 3 gaps with 2 empty chairs and 5 gaps with 1 empty chair. There are 7 different circular orderings in which these 8 gaps can be arranged:\n\n$$\n\\begin{array}{lllllll}\n22211111 & 22121111 & 22112111 & 22111211 & 22111121 & 21212111 & 21211211\n\\end{array}\n$$\n\nWe note that ""22211111"" would be the same as, for example, ""11222111"" since these gaps are arranged around a circle.\n\nIf the three gaps of length 2 are consecutive, there is only one configuration (22211111). If there are exactly 2 consecutive gaps of length 2 , there are 4 relative places in which the third gap of length 2 can be placed.\n\nIf there are no consecutive gaps of length 2 , these gaps can either be separated by 1 gap each (21212111) with 3 gaps on the far side, or can be separated by 1 gap, 2 gaps, and 2 gaps (21211211). There is only one configuration for the gaps in this last situation.\n\nThere are 7 different circular orderings for these 8 gaps.\n\nEach of these 7 different orderings can be placed around the circle of 19 chairs in 19 different ways, because each can be started in 19 different places. Because 19 is prime, none of these orderings overlap.\n\nTherefore, there are $7 \\cdot 19=133$ different full tables with 8 people.\n\nSuppose that there are 7 people and 7 gaps with a total of 12 empty chairs.\n\nIn this case, there are 2 gaps with 1 empty chair and 5 gaps with 2 empty chairs.\n\nThe 2 gaps with 1 empty chair can be separated by 0 gaps with 2 empty chairs, 1 gap with 2 empty chairs, or 2 gaps with 2 empty chairs. Because the chairs are around a circle, if there were 3 , 4 or 5 gaps with 2 empty chairs between them, there would be 2,1 or 0 gaps going the other way around the circle.\n\nThis means that there are 3 different configurations for the gaps.\n\nEach of these configurations can be placed in 19 different ways around the circle of chairs. Therefore, there are $3 \\cdot 19=57$ full tables with 7 people.\n\nIn total, there are $19+133+57=209$ full tables with 19 chairs.', 'As in Solution 2, there must be 7, 8 or 9 people in chairs, and so there are 7,8 or 9 gaps. If there are 7 gaps, there are 2 gaps of 1 chair and 5 gaps of 2 chairs.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 3 gaps of 2 chairs.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 1 gap of 2 chairs.\n\nWe consider three mutually exclusive cases: (i) there is a person in chair 1 and not in chair 2 , (ii) there is a person in chair 2 and not in chair 1 , and (iii) there is nobody in chair 1 or in chair 2. Every full table fits into exactly one of these three cases.\n\nCase (i): there is a person in chair 1 and not in chair 2\n\nWe use the person in chair 1 to ""anchor"" the arrangement, by starting at chair 1 and\n\n\n\narranging the gaps (and thus the full chairs) clockwise around the table from chair 1.\n\nIf there are 7 gaps, we need to choose 2 of them to be of length 1 , and so there are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 8 gaps, we need to choose 3 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIf there are 9 gaps, we need to choose 1 of them to be of length 2 , and so there are $\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)$ ways of arranging the gaps starting at chair 1.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)+\\left(\\begin{array}{l}9 \\\\ 1\\end{array}\\right)=21+56+9=86$ full tables.\n\nCase (ii): there is a person in chair 2 and not in chair 1\n\nWe use the same reasoning starting with the person in chair 2 as the anchor.\n\nAgain, there are 86 full tables in this case.\n\nCase (iii): there is nobody in chair 1 or chair 2\n\nSince there is nobody in chair 1 or chair 2 , there must be a person in chair 3 and also in chair 19 , which fixes one gap of 2 chairs.\n\nHere, we use the person in chair 3 as the anchor.\n\nIf there are 7 gaps, there are 2 gaps of 1 chair and 4 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 8 gaps, there are 5 gaps of 1 chair and 2 gaps of 2 chairs left to place. There are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)$ ways of doing this.\n\nIf there are 9 gaps, there are 8 gaps of 1 chair and 0 gaps of 2 chairs left to place. There is 1 way to do this.\n\nIn this case, there are a total of $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)+\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)+1=15+21+1=37$ full tables.\n\nIn total, there are $86+86+37=209$ full tables with 19 chairs.']",['209'],False,,Numerical, 2414,Number Theory,,"For every real number $x$, define $\lfloor x\rfloor$ to be equal to the greatest integer less than or equal to $x$. (We call this the ""floor"" of $x$.) For example, $\lfloor 4.2\rfloor=4,\lfloor 5.7\rfloor=5$, $\lfloor-3.4\rfloor=-4,\lfloor 0.4\rfloor=0$, and $\lfloor 2\rfloor=2$. Determine the integer equal to $\left\lfloor\frac{1}{3}\right\rfloor+\left\lfloor\frac{2}{3}\right\rfloor+\left\lfloor\frac{3}{3}\right\rfloor+\ldots+\left\lfloor\frac{59}{3}\right\rfloor+\left\lfloor\frac{60}{3}\right\rfloor$. (The sum has 60 terms.)","['Since $0<\\frac{1}{3}<\\frac{2}{3}<1$, then $\\left\\lfloor\\frac{1}{3}\\right\\rfloor=\\left\\lfloor\\frac{2}{3}\\right\\rfloor=0$.\n\nSince $1 \\leq \\frac{3}{3}<\\frac{4}{3}<\\frac{5}{3}<2$, then $\\left\\lfloor\\frac{3}{3}\\right\\rfloor=\\left\\lfloor\\frac{4}{3}\\right\\rfloor=\\left\\lfloor\\frac{5}{3}\\right\\rfloor=1$.\n\nThese fractions can continue to be grouped in groups of 3 with the last full group of 3 satisfying $19 \\leq \\frac{57}{3}<\\frac{58}{3}<\\frac{59}{3}<20$, which means that $\\left\\lfloor\\frac{57}{3}\\right\\rfloor=\\left\\lfloor\\frac{58}{3}\\right\\rfloor=\\left\\lfloor\\frac{59}{3}\\right\\rfloor=19$.\n\nThe last term is $\\left\\lfloor\\frac{60}{3}\\right\\rfloor=\\lfloor 20\\rfloor=20$.\n\n\n\nIf the given sum is $S$, we obtain\n\n$$\n\\begin{aligned}\nS & =2 \\cdot 0+3 \\cdot 1+3 \\cdot 2+\\cdots+3 \\cdot 19+1 \\cdot 20 \\\\\n& =0+3(1+2+\\cdot+19)+20 \\\\\n& =3 \\cdot \\frac{1}{2} \\cdot 19 \\cdot 20+20 \\\\\n& =570+20 \\\\\n& =590\n\\end{aligned}\n$$']",['590'],False,,Numerical, 2415,Number Theory,,"For every real number $x$, define $\lfloor x\rfloor$ to be equal to the greatest integer less than or equal to $x$. (We call this the ""floor"" of $x$.) For example, $\lfloor 4.2\rfloor=4,\lfloor 5.7\rfloor=5$, $\lfloor-3.4\rfloor=-4,\lfloor 0.4\rfloor=0$, and $\lfloor 2\rfloor=2$. Determine a polynomial $p(x)$ so that for every positive integer $m>4$, $$ \lfloor p(m)\rfloor=\left\lfloor\frac{1}{3}\right\rfloor+\left\lfloor\frac{2}{3}\right\rfloor+\left\lfloor\frac{3}{3}\right\rfloor+\ldots+\left\lfloor\frac{m-2}{3}\right\rfloor+\left\lfloor\frac{m-1}{3}\right\rfloor $$ (The sum has $m-1$ terms.) A polynomial $f(x)$ is an algebraic expression of the form $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ for some integer $n \geq 0$ and for some real numbers $a_{n}, a_{n-1}, \ldots, a_{1}, a_{0}$.","['For every positive integer $m>4$, let\n\n$$\nq(m)=\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{m-2}{3}\\right\\rfloor+\\left\\lfloor\\frac{m-1}{3}\\right\\rfloor\n$$\n\nExtending our work from (a), we know that $k-1 \\leq \\frac{3 k-3}{3}<\\frac{3 k-2}{3}<\\frac{3 k-1}{3}4$ can be written as $m=3 s$ or $m=3 s+1$ or $m=3 s+2$, for some positive integer $s$, depending on its remainder when divided by 3 .\n\nWe can thus write\n\n$$\n\\begin{aligned}\nq(3 s) & =\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{3 s-2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3 s-1}{3}\\right\\rfloor \\\\\n& =2 \\cdot 0+3(1+2+3+\\cdots+(s-1)) \\\\\n& =3 \\cdot \\frac{1}{2} \\cdot(s-1) s \\\\\n& =\\frac{3 s(s-1)}{2} \\\\\n& =\\frac{3 s(3 s-3)}{6} \\\\\nq(3 s+1) & =\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{3 s-2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3 s-1}{3}\\right\\rfloor+\\left\\lfloor\\frac{3 s}{3}\\right\\rfloor \\\\\n& =q(3 s)+s \\\\\n& =\\frac{3 s(3 s-3)}{6}+\\frac{3 s \\cdot 2}{6} \\\\\n& =\\frac{3 s(3 s-1)}{6} \\\\\nq(3 s+2) & =q(3 s+1)+\\left\\lfloor\\frac{3 s+1}{3}\\right\\rfloor \\\\\n& =\\frac{3 s(3 s-1)}{6}+s \\\\\n& =\\frac{3 s(3 s-1)}{6}+\\frac{3 s \\cdot 2}{6} \\\\\n& =\\frac{3 s(3 s+1)}{6}\n\\end{aligned}\n$$\n\nWe want to find a polynomial $p(x)$ for which $q(m)=\\lfloor p(m)\\rfloor$ for every positive integer $m>4$.\n\n\n\nIn other words, we want to find a polynomial $p(x)$ for which\n\n$$\n\\lfloor p(3 s)\\rfloor=\\frac{3 s(3 s-3)}{6} \\quad\\lfloor p(3 s+1)\\rfloor=\\frac{3 s(3 s-1)}{6} \\quad\\lfloor p(3 s+2)\\rfloor=\\frac{3 s(3 s+1)}{6}\n$$\n\nfor every positive integer $s$.\n\nWe will show that the polynomial $p(x)=\\frac{(x-1)(x-2)}{6}$ satisfies the desired conditions.\n\nIf $x=3 s+1$ for some positive integer $s$, then\n\n$$\n\\frac{(x-1)(x-2)}{6}=\\frac{(3 s+1-1)(3 s+1-2)}{6}=\\frac{3 s(3 s-1)}{6}\n$$\n\nWe note that $3 s$ is a multiple of 3 . Since $3 s$ and $3 s-1$ are consecutive integers, then one of these is a multiple of 2 . Thus, $3 s(3 s-1)$ is a multiple of 6 and so $\\frac{3 s(3 s-1)}{6}$ is an integer.\n\nThis means that $\\left\\lfloor\\frac{3 s(3 s-1)}{6}\\right\\rfloor=\\frac{3 s(3 s-1)}{6}$.\n\nTherefore, $q(3 s+1)=\\frac{3 s(3 s-1)}{6}=\\left\\lfloor\\frac{3 s(3 s-1)}{6}\\right\\rfloor=\\lfloor p(3 s+1)\\rfloor$.\n\nIf $x=3 s+2$ for some positive integer $s$, then\n\n$$\n\\frac{(x-1)(x-2)}{6}=\\frac{(3 s+2-1)(3 s+2-2)}{6}=\\frac{3 s(3 s+1)}{6}\n$$\n\nWe note that $3 s$ is a multiple of 3 . Since $3 s$ and $3 s+1$ are consecutive integers, then one of these is a multiple of 2 . Thus, $3 s(3 s+1)$ is a multiple of 6 and so $\\frac{3 s(3 s+1)}{6}$ is an integer.\n\nThis means that $\\left\\lfloor\\frac{3 s(3 s+1)}{6}\\right\\rfloor=\\frac{3 s(3 s+1)}{6}$.\n\nTherefore, $q(3 s+2)=\\frac{3 s(3 s+1)}{6}=\\left\\lfloor\\frac{3 s(3 s+1)}{6}\\right\\rfloor=\\lfloor p(3 s+2)\\rfloor$.\n\nIf $x=3 s$ for some positive integer $s$, then\n\n$$\n\\frac{(x-1)(x-2)}{6}=\\frac{(3 s-1)(3 s-2)}{6}=\\frac{9 s^{2}-9 s+2}{6}\n$$\n\nNow, $\\frac{9 s^{2}-9 s}{6}=\\frac{9 s(s-1)}{6}$ is an integer because $9 s$ is a multiple of 3 and one of $s$ and $s-1$ is even.\n\nSince $\\frac{9 s^{2}-9 s+2}{6}=\\frac{9 s^{2}-9 s}{6}+\\frac{1}{3}$, then $\\frac{9 s^{2}-9 s+2}{6}$ is $\\frac{1}{3}$ more than an integer which means that $\\left\\lfloor\\frac{9 s^{2}-9 s+2}{6}\\right\\rfloor=\\frac{9 s^{2}-9 s}{6}=\\frac{3 s(3 s-3)}{6}=q(3 s)$.\n\nTherefore, $q(3 s)=\\frac{3 s(3 s-3)}{6}=\\left\\lfloor\\frac{(3 s-1)(3 s-2)}{6}\\right\\rfloor=\\lfloor p(3 s)\\rfloor$.\n\nThis means that the polynomial $p(x)=\\frac{(x-1)(x-2)}{6}$ satisfies the required conditions.']",['$p(x)=\\frac{(x-1)(x-2)}{6}$'],False,,Expression, 2416,Number Theory,,"For every real number $x$, define $\lfloor x\rfloor$ to be equal to the greatest integer less than or equal to $x$. (We call this the ""floor"" of $x$.) For example, $\lfloor 4.2\rfloor=4,\lfloor 5.7\rfloor=5$, $\lfloor-3.4\rfloor=-4,\lfloor 0.4\rfloor=0$, and $\lfloor 2\rfloor=2$. For each integer $n \geq 1$, define $f(n)$ to be equal to an infinite sum: $$ f(n)=\left\lfloor\frac{n}{1^{2}+1}\right\rfloor+\left\lfloor\frac{2 n}{2^{2}+1}\right\rfloor+\left\lfloor\frac{3 n}{3^{2}+1}\right\rfloor+\left\lfloor\frac{4 n}{4^{2}+1}\right\rfloor+\left\lfloor\frac{5 n}{5^{2}+1}\right\rfloor+\cdots $$ (The sum contains the terms $\left\lfloor\frac{k n}{k^{2}+1}\right\rfloor$ for all positive integers $k$, and no other terms.) Suppose $f(t+1)-f(t)=2$ for some odd positive integer $t$. Prove that $t$ is a prime number.","['Before working on the specific question we have been asked, we simplify the given expression for $f(n)$ by noting that if $k \\geq n$, then $k n \\leq k^{2}1$. ( $t$ can be written in this form in at least one way, so we take one of these possibilities.)\n\nIn this case, consider $f(t+1)-f(t)$.\n\nWe can write this as\n\n$$\n\\begin{aligned}\nf(t+1)-f(t)= & \\left\\lfloor\\frac{t+1}{1^{2}+1}\\right\\rfloor+\\left\\lfloor\\frac{2(t+1)}{2^{2}+1}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{(t-1)(t+1)}{(t-1)^{2}+1}\\right\\rfloor+\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor \\\\\n& -\\left\\lfloor\\frac{t}{1^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{2 t}{2^{2}+1}\\right\\rfloor-\\cdots-\\left\\lfloor\\frac{(t-1) t}{(t-1)^{2}+1}\\right\\rfloor\n\\end{aligned}\n$$\n\nWe re-write this as\n\n$$\n\\begin{gathered}\nf(t+1)-f(t)=\\left(\\left\\lfloor\\frac{t+1}{1^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{t}{1^{2}+1}\\right\\rfloor\\right)+\\left(\\left\\lfloor\\frac{2(t+1)}{2^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{2 t}{2^{2}+1}\\right\\rfloor\\right)+\\cdots \\\\\n+\\left(\\left\\lfloor\\frac{(t-1)(t+1)}{(t-1)^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{(t-1) t}{(t-1)^{2}+1}\\right\\rfloor\\right)+\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor\n\\end{gathered}\n$$\n\nIn the $t-1$ sets of parentheses, we have terms of the form $\\left\\lfloor\\frac{k(t+1)}{k^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{k t}{k^{2}+1}\\right\\rfloor$ for each integer $k$ from 1 to $t-1$.\n\nWe know that $\\frac{k(t+1)}{k^{2}+1}>\\frac{k t}{k^{2}+1}$ because both $k$ and $t$ are positive, the denominators are\n\n\n\nequal and $k(t+1)>k t$.\n\nThus, $\\left.\\left\\lfloor\\frac{k(t+1)}{k^{2}+1}\\right\\rfloor \\geq \\frac{k t}{k^{2}+1}\\right\\rfloor$. (The greatest integer less than or equal to the first fraction must be at least as large as the greatest integer less than or equal to the second fraction.) This means that the $t-1$ differences in parentheses, each of which is an integer, is at least 0 .\n\nTo show that $f(t+1)-f(t) \\neq 2$, we show that there are at least 2 places where the difference is at least 1, and that the final term is at least 1 . This will tell us that $f(t+1)-f(t) \\geq 3$ and so $f(t+1)-f(t) \\neq 2$, which will tell us that $t$ cannot be composite, and so $t$ must be prime, as required.\n\nConsider $\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor$.\n\nSince $t(t+1)=t^{2}+t \\geq t^{2}+1$, then $\\frac{t(t+1)}{t^{2}+1} \\geq 1$, which means that $\\left\\lfloor\\frac{t(t+1)}{t^{2}+1}\\right\\rfloor \\geq 1$.\n\nConsider $\\left\\lfloor\\frac{t+1}{1^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{t}{1^{2}+1}\\right\\rfloor=\\left\\lfloor\\frac{t+1}{2}\\right\\rfloor-\\left\\lfloor\\frac{t}{2}\\right\\rfloor$.\n\nSince $t$ is odd, then we write $t=2 u+1$ for some positive integer $u$, which gives\n\n$$\n\\left\\lfloor\\frac{t+1}{2}\\right\\rfloor-\\left\\lfloor\\frac{t}{2}\\right\\rfloor=\\left\\lfloor\\frac{2 u+2}{2}\\right\\rfloor-\\left\\lfloor\\frac{2 u+1}{2}\\right\\rfloor=\\lfloor u+1\\rfloor-\\left\\lfloor u+\\frac{1}{2}\\right\\rfloor=(u+1)-u=1\n$$\n\nRecall that $t=r s$ with $r \\geq s>1$.\n\nConsider the term $\\left\\lfloor\\frac{r(t+1)}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r t}{r^{2}+1}\\right\\rfloor$.\n\nWe have\n\n$$\n\\left\\lfloor\\frac{r(t+1)}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r t}{r^{2}+1}\\right\\rfloor=\\left\\lfloor\\frac{r(r s+1)}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r \\cdot r s}{r^{2}+1}\\right\\rfloor=\\left\\lfloor\\frac{r^{2} s+r}{r^{2}+1}\\right\\rfloor-\\left\\lfloor\\frac{r^{2} s}{r^{2}+1}\\right\\rfloor\n$$\n\nWe note that $\\frac{r^{2} s+r}{r^{2}+1} \\geq \\frac{r^{2} s+s}{r^{2}+1}=s$ and $\\frac{r^{2} s}{r^{2}+1}<\\frac{r^{2} s+s}{r^{2}+1}=s$.\n\nThus, $\\left\\lfloor\\frac{r^{2} s+r}{r^{2}+1}\\right\\rfloor \\geq s$.\n\nAlso, $\\left\\lfloor\\frac{r^{2} s}{r^{2}+1}\\right\\rfloor\\frac{k t}{k^{2}+1}$.\n\n\n\nThis pair of inequalities is equivalent to the pair of inequalities $t+1 \\geq N \\cdot \\frac{k^{2}+1}{k}>t$ which is in turn equivalent to $t+1 \\geq N k+\\frac{N}{k}>t$.\n\nThe following three pairs $(N, k)$ of integers satisfy this equation:\n\n- $k=1$ and $N=\\frac{t+1}{2}$ (noting that $t$ is odd), which give $N k+\\frac{N}{k}=t+1$;\n- $k=r$ and $N=s$, which give $N k+\\frac{N}{k}=r s+\\frac{s}{r}$ (noting that $\\left.\\frac{s}{r}<1\\right)$\n- $k=t$ and $N=1$, which give $N k+\\frac{N}{k}=t+\\frac{1}{t}$.\n\nThis shows that $f(t+1)-f(t) \\geq 3$ when $t$ is odd and composite, as required.']",,True,,, 2417,Geometry,,"In the diagram, $\triangle A B C$ is right-angled at $B$ and $\triangle A C D$ is right-angled at $A$. Also, $A B=3, B C=4$, and $C D=13$. What is the area of quadrilateral $A B C D$ ? ","['The area of quadrilateral $A B C D$ is the sum of the areas of $\\triangle A B C$ and $\\triangle A C D$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area equals $\\frac{1}{2}(A B)(B C)=\\frac{1}{2}(3)(4)=6$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}+B C^{2}}=\\sqrt{3^{2}+4^{2}}=\\sqrt{25}=5\n$$\n\nbecause $A C>0$. (We could have also observed that $\\triangle A B C$ must be a ""3-4-5"" triangle.) Since $\\triangle A C D$ is right-angled at $A$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{C D^{2}-A C^{2}}=\\sqrt{13^{2}-5^{2}}=\\sqrt{144}=12\n$$\n\nbecause $A D>0$. (We could have also observed that $\\triangle A C D$ must be a "" $5-12-13$ "" triangle.) Thus, the area of $\\triangle A C D$ equals $\\frac{1}{2}(A C)(A D)=\\frac{1}{2}(5)(12)=30$.\n\nFinally, the area of quadrilateral $A B C D$ is thus $6+30=36$.']",['36'],False,,Numerical, 2417,Geometry,,"In the diagram, $\triangle A B C$ is right-angled at $B$ and $\triangle A C D$ is right-angled at $A$. Also, $A B=3, B C=4$, and $C D=13$. What is the area of quadrilateral $A B C D$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_c71fc327224b5fdf6510g-1.jpg?height=279&width=475&top_left_y=278&top_left_x=1256)","['The area of quadrilateral $A B C D$ is the sum of the areas of $\\triangle A B C$ and $\\triangle A C D$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area equals $\\frac{1}{2}(A B)(B C)=\\frac{1}{2}(3)(4)=6$.\n\nSince $\\triangle A B C$ is right-angled at $B$, then by the Pythagorean Theorem,\n\n$$\nA C=\\sqrt{A B^{2}+B C^{2}}=\\sqrt{3^{2}+4^{2}}=\\sqrt{25}=5\n$$\n\nbecause $A C>0$. (We could have also observed that $\\triangle A B C$ must be a ""3-4-5"" triangle.) Since $\\triangle A C D$ is right-angled at $A$, then by the Pythagorean Theorem,\n\n$$\nA D=\\sqrt{C D^{2}-A C^{2}}=\\sqrt{13^{2}-5^{2}}=\\sqrt{144}=12\n$$\n\nbecause $A D>0$. (We could have also observed that $\\triangle A C D$ must be a "" $5-12-13$ "" triangle.) Thus, the area of $\\triangle A C D$ equals $\\frac{1}{2}(A C)(A D)=\\frac{1}{2}(5)(12)=30$.\n\nFinally, the area of quadrilateral $A B C D$ is thus $6+30=36$.']",['36'],False,,Numerical, 2418,Geometry,,"Three identical rectangles $P Q R S$, WTUV and $X W V Y$ are arranged, as shown, so that $R S$ lies along $T X$. The perimeter of each of the three rectangles is $21 \mathrm{~cm}$. What is the perimeter of the whole shape? ![](https://cdn.mathpix.com/cropped/2023_12_21_c71fc327224b5fdf6510g-1.jpg?height=400&width=1510&top_left_y=610&top_left_x=294)","['Let the width of each of the identical rectangles be $a$.\n\nIn other words, $Q P=R S=T W=W X=U V=V Y=a$.\n\nLet the height of each of the identical rectangles be $b$.\n\nIn other words, $Q R=P S=T U=W V=X Y=b$.\n\nThe perimeter of the whole shape equals\n\n$$\nQ P+P S+S X+X Y+V Y+U V+T U+T R+Q R\n$$\n\nSubstituting for known lengths, we obtain\n\n$$\na+b+S X+b+a+a+b+T R+b\n$$\n\nor $3 a+4 b+(S X+T R)$.\n\nBut $S X+T R=(T R+R S+S X)-R S=(T W+W X)-R S=a+a-a=a$.\n\nTherefore, the perimeter of the whole shape equals $4 a+4 b$.\n\nThe perimeter of one rectangle is $2 a+2 b$, which we are told equals $21 \\mathrm{~cm}$.\n\nFinally, the perimeter of the whole shape is thus $2(2 a+2 b)$ which equals $42 \\mathrm{~cm}$.']",['42'],False,cm,Numerical, 2418,Geometry,,"Three identical rectangles $P Q R S$, WTUV and $X W V Y$ are arranged, as shown, so that $R S$ lies along $T X$. The perimeter of each of the three rectangles is $21 \mathrm{~cm}$. What is the perimeter of the whole shape? ","['Let the width of each of the identical rectangles be $a$.\n\nIn other words, $Q P=R S=T W=W X=U V=V Y=a$.\n\nLet the height of each of the identical rectangles be $b$.\n\nIn other words, $Q R=P S=T U=W V=X Y=b$.\n\nThe perimeter of the whole shape equals\n\n$$\nQ P+P S+S X+X Y+V Y+U V+T U+T R+Q R\n$$\n\nSubstituting for known lengths, we obtain\n\n$$\na+b+S X+b+a+a+b+T R+b\n$$\n\nor $3 a+4 b+(S X+T R)$.\n\nBut $S X+T R=(T R+R S+S X)-R S=(T W+W X)-R S=a+a-a=a$.\n\nTherefore, the perimeter of the whole shape equals $4 a+4 b$.\n\nThe perimeter of one rectangle is $2 a+2 b$, which we are told equals $21 \\mathrm{~cm}$.\n\nFinally, the perimeter of the whole shape is thus $2(2 a+2 b)$ which equals $42 \\mathrm{~cm}$.']",['42'],False,cm,Numerical, 2419,Geometry,,"One of the faces of a rectangular prism has area $27 \mathrm{~cm}^{2}$. Another face has area $32 \mathrm{~cm}^{2}$. If the volume of the prism is $144 \mathrm{~cm}^{3}$, determine the surface area of the prism in $\mathrm{cm}^{2}$.","['Suppose that the rectangular prism has dimensions $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ by $c \\mathrm{~cm}$.\n\nSuppose further that one of the faces that is $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ is the face with area $27 \\mathrm{~cm}^{2}$ and that one of the faces that is $a \\mathrm{~cm}$ by $c \\mathrm{~cm}$ is the face with area $32 \\mathrm{~cm}^{2}$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)\n\nTherefore, $a b=27$ and $a c=32$.\n\nFurther, we are told that the volume of the prism is $144 \\mathrm{~cm}^{3}$, and so $a b c=144$.\n\n\n\nThus, $b c=\\frac{a^{2} b^{2} c^{2}}{a^{2} b c}=\\frac{(a b c)^{2}}{(a b)(a c)}=\\frac{144^{2}}{(27)(32)}=24$.\n\n(We could also note that $a b c=144$ means $a^{2} b^{2} c^{2}=144^{2}$ or $(a b)(a c)(b c)=144^{2}$ and so $b c=\\frac{144^{2}}{(27)(32)}$.)\n\nIn other words, the third type of face of the prism has area $24 \\mathrm{~cm}^{2}$.\n\nThus, since the prism has two faces of each type, the surface area of the prism is equal to $2\\left(27 \\mathrm{~cm}^{2}+32 \\mathrm{~cm}^{2}+24 \\mathrm{~cm}^{2}\\right)$ or $166 \\mathrm{~cm}^{2}$.', 'Suppose that the rectangular prism has dimensions $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ by $c \\mathrm{~cm}$.\n\nSuppose further that one of the faces that is $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ is the face with area $27 \\mathrm{~cm}^{2}$ and that one of the faces that is $a \\mathrm{~cm}$ by $c \\mathrm{~cm}$ is the face with area $32 \\mathrm{~cm}^{2}$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)\n\nTherefore, $a b=27$ and $a c=32$.\n\nFurther, we are told that the volume of the prism is $144 \\mathrm{~cm}^{3}$, and so $a b c=144$.\n\nSince $a b c=144$ and $a b=27$, then $c=\\frac{144}{27}=\\frac{16}{3}$.\n\nSince $a b c=144$ and $a c=32$, then $b=\\frac{144}{32}=\\frac{9}{2}$.\n\nThis means that $b c=\\frac{16}{3} \\cdot \\frac{9}{2}=24$.\n\nIn $\\mathrm{cm}^{2}$, the surface area of the prism equals $2 a b+2 a c+2 b c=2(27)+2(32)+2(24)=166$. Thus, the surface area of the prism is $166 \\mathrm{~cm}^{2}$.']",['$166$'],False,$cm^2$,Numerical, 2420,Algebra,,"The equations $y=a(x-2)(x+4)$ and $y=2(x-h)^{2}+k$ represent the same parabola. What are the values of $a, h$ and $k$ ?","['We expand the right sides of the two equations, collecting like terms in each case:\n\n$$\n\\begin{aligned}\n& y=a(x-2)(x+4)=a\\left(x^{2}+2 x-8\\right)=a x^{2}+2 a x-8 a \\\\\n& y=2(x-h)^{2}+k=2\\left(x^{2}-2 h x+h^{2}\\right)+k=2 x^{2}-4 h x+\\left(2 h^{2}+k\\right)\n\\end{aligned}\n$$\n\nSince these two equations represent the same parabola, then the corresponding coefficients must be equal. That is, $a=2$ and $2 a=-4 h$ and $-8 a=2 h^{2}+k$.\n\nSince $a=2$ and $2 a=-4 h$, then $4=-4 h$ and so $h=-1$.\n\nSince $-8 a=2 h^{2}+k$ and $a=2$ and $h=-1$, then $-16=2+k$ and so $k=-18$.\n\nThus, $a=2, h=-1$, and $k=-18$.', 'From the equation $y=a(x-2)(x+4)$, we can find the axis of symmetry by calculating the midpoint of the $x$-intercepts.\n\nSince the $x$-intercepts are 2 and -4 , the axis of symmetry is at $x=\\frac{1}{2}(2+(-4))=-1$.\n\nSince the vertex of the parabola lies on the axis of symmetry, then the $x$-coordinate of the vertex is -1 .\n\nTo find the $y$-coordinate of the vertex, we substitute $x=-1$ back into the equation $y=a(x-2)(x+4)$ to obtain $y=a(-1-2)(-1+4)=-9 a$.\n\nThus, the vertex of the parabola is $(-1,-9 a)$.\n\nSince the second equation for the same parabola is in vertex form, $y=2(x-h)^{2}+k$, we can see that the vertex is at $(h, k)$ and $a=2$.\n\nSince $a=2$, the vertex has coordinates $(-1,-18)$, which means that $h=-1$ and $k=-18$. Thus, $a=2, h=-1$ and $k=-18$.']","['$2,-1,-18$']",True,,Numerical, 2421,Algebra,,"In an arithmetic sequence with 5 terms, the sum of the squares of the first 3 terms equals the sum of the squares of the last 2 terms. If the first term is 5 , determine all possible values of the fifth term. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3,5,7,9,11 is an arithmetic sequence with five terms.)","['Let the common difference in this arithmetic sequence be $d$.\n\nSince the first term in the sequence is 5 , then the 5 terms are $5,5+d, 5+2 d, 5+3 d, 5+4 d$.\n\nFrom the given information, $5^{2}+(5+d)^{2}+(5+2 d)^{2}=(5+3 d)^{2}+(5+4 d)^{2}$.\n\nManipulating, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n5^{2}+(5+d)^{2}+(5+2 d)^{2} & =(5+3 d)^{2}+(5+4 d)^{2} \\\\\n25+\\left(25+10 d+d^{2}\\right)+\\left(25+20 d+4 d^{2}\\right) & =\\left(25+30 d+9 d^{2}\\right)+\\left(25+40 d+16 d^{2}\\right) \\\\\n75+30 d+5 d^{2} & =50+70 d+25 d^{2} \\\\\n0 & =20 d^{2}+40 d-25 \\\\\n0 & =4 d^{2}+8 d-5 \\\\\n0 & =(2 d+5)(2 d-1)\n\\end{aligned}\n$$\n\nTherefore, $d=-\\frac{5}{2}$ or $d=\\frac{1}{2}$.\n\nThese give possible fifth terms of $5+4 d=5+4\\left(-\\frac{5}{2}\\right)=-5$ and $5+4 d=5+4\\left(\\frac{1}{2}\\right)=7$.\n\n(We note that, for these two values of $d$, the sequences are $5, \\frac{5}{2}, 0,-\\frac{5}{2},-5$ and $5, \\frac{11}{2}, 6, \\frac{13}{2}, 7$.)']","['-5,7']",True,,Numerical, 2422,Algebra,,Dan was born in a year between 1300 and 1400. Steve was born in a year between 1400 and 1500. Each was born on April 6 in a year that is a perfect square. Each lived for 110 years. In what year while they were both alive were their ages both perfect squares on April 7?,"['First, we determine the perfect squares between 1300 and 1400 and between 1400 and 1500.\n\nSince $\\sqrt{1300} \\approx 36.06$, then the first perfect square larger than 1300 is $37^{2}=1369$.\n\nThe next perfect squares are $38^{2}=1444$ and $39^{2}=1521$.\n\nSince Dan was born between 1300 and 1400 in a year that was a perfect square, then Dan was born in 1369.\n\nSince Steve was born between 1400 and 1500 in a year that was a perfect square, then Steve was born in 1444.\n\nSuppose that on April 7 in some year, Dan was $m^{2}$ years old and Steve was $n^{2}$ years old for some positive integers $m$ and $n$. Thus, Dan was $m^{2}$ years old in the year $1369+m^{2}$ and Steve was $n^{2}$ years old in the year $1444+n^{2}$.\n\nSince these represent the same years, then $1369+m^{2}=1444+n^{2}$, or $m^{2}-n^{2}=1444-$ $1369=75$.\n\nIn other words, we want to find two perfect squares less than 110 (since their ages are less than 110) whose difference is 75.\n\nThe perfect squares less than 110 are $1,4,9,16,25,36,49,64,81,100$.\n\nThe two that differ by 75 are 100 and 25 .\n\nThus, $m^{2}=100$ and $n^{2}=25$.\n\nThis means that the year in which the age of each of Dan and Steve was a perfect square was the year $1369+100=1469$.']",['1469'],False,,Numerical, 2423,Geometry,,"Determine all values of $k$ for which the points $A(1,2), B(11,2)$ and $C(k, 6)$ form a right-angled triangle.","['$\\triangle A B C$ is right-angled exactly when one of the following statements is true:\n\n- $A B$ is perpendicular to $B C$, or\n- $A B$ is perpendicular to $A C$, or\n- $A C$ is perpendicular to $B C$.\n\nSince $A(1,2)$ and $B(11,2)$ share a $y$-coordinate, then $A B$ is horizontal.\n\nFor $A B$ and $B C$ to be perpendicular, $B C$ must be vertical.\n\nThus, $B(11,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=11$.\n\nFor $A B$ and $A C$ to be perpendicular, $A C$ must be vertical.\n\nThus, $A(1,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=1$.\n\n\n\nFor $A C$ to be perpendicular to $B C$, their slopes must have a product of -1 .\n\nThe slope of $A C$ is $\\frac{6-2}{k-1}$, which equals $\\frac{4}{k-1}$.\n\nThe slope of $B C$ is $\\frac{6-2}{k-11}$, which equals $\\frac{4}{k-11}$.\n\nThus, $A C$ and $B C$ are perpendicular when $\\frac{4}{k-1} \\cdot \\frac{4}{k-11}=-1$.\n\nAssuming that $k \\neq 1$ and $k \\neq 11$, we manipulate to obtain $16=-(k-1)(k-11)$ or $16=-k^{2}+12 k-11$ or $k^{2}-12 k+27=0$.\n\nFactoring, we obtain $(k-3)(k-9)=0$ and so $A C$ and $B C$ are perpendicular when $k=3$ or $k=9$.\n\nIn summary, $\\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.', '$\\triangle A B C$ is right-angled exactly when its three side lengths satisfy the Pythagorean Theorem in some orientation. That is, $\\triangle A B C$ is right-angled exactly when $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+A C^{2}=B C^{2}$ or $A C^{2}+B C^{2}=A B^{2}$.\n\nUsing $A(1,2)$ and $B(11,2)$, we obtain $A B^{2}=(11-1)^{2}+(2-2)^{2}=100$.\n\nUsing $A(1,2)$ and $C(k, 6)$, we obtain $A C^{2}=(k-1)^{2}+(6-2)^{2}=(k-1)^{2}+16$.\n\nUsing $B(11,2)$ and $C(k, 6)$, we obtain $B C^{2}=(k-11)^{2}+(6-2)^{2}=(k-11)^{2}+16$.\n\nUsing the Pythagorean relationships above, $\\triangle A B C$ is right-angled when one of the following is true:\n\n(i)\n\n$$\n\\begin{aligned}\n100+\\left((k-11)^{2}+16\\right) & =(k-1)^{2}+16 \\\\\n100+k^{2}-22 k+121+16 & =k^{2}-2 k+1+16 \\\\\n220 & =20 k \\\\\nk & =11\n\\end{aligned}\n$$\n\n(ii)\n\n$$\n\\begin{aligned}\n100+\\left((k-1)^{2}+16\\right) & =(k-11)^{2}+16 \\\\\n100+k^{2}-2 k+1+16 & =k^{2}-22 k+121+16 \\\\\n20 k & =20 \\\\\nk & =1\n\\end{aligned}\n$$\n\n(iii)\n\n$$\n\\begin{aligned}\n\\left((k-1)^{2}+16\\right)+\\left((k-11)^{2}+16\\right) & =100 \\\\\nk^{2}-2 k+1+16+k^{2}-22 k+121+16 & =100 \\\\\n2 k^{2}-24 k+54 & =0 \\\\\nk^{2}-12 k+27 & =0 \\\\\n(k-3)(k-9) & =0\n\\end{aligned}\n$$\n\nand so $k=3$ or $k=9$.\n\nIn summary, $\\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.']","['1,3,9,11']",True,,Numerical, 2424,Geometry,,"The diagram shows two hills that meet at $O$. One hill makes a $30^{\circ}$ angle with the horizontal and the other hill makes a $45^{\circ}$ angle with the horizontal. Points $A$ and $B$ are on the hills so that $O A=O B=20 \mathrm{~m}$. Vertical poles $B D$ and $A C$ are connected by a straight cable $C D$. If $A C=6 \mathrm{~m}$, what is the length of $B D$ for which $C D$ is as short as possible? ![](https://cdn.mathpix.com/cropped/2023_12_21_c71fc327224b5fdf6510g-1.jpg?height=287&width=437&top_left_y=2095&top_left_x=1256)","['Extend $C A$ and $D B$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3361192d5aa7e57927a4g-1.jpg?height=287&width=423&top_left_y=239&top_left_x=951)\n\nSince $C A$ and $D B$ are vertical, then $\\angle C P O=\\angle D Q O=90^{\\circ}$.\n\nSince $O A=20 \\mathrm{~m}$, then $A P=O A \\sin 30^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{2}=10 \\mathrm{~m}$.\n\nSince $O B=20 \\mathrm{~m}$, then $B Q=O B \\sin 45^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{\\sqrt{2}}=10 \\sqrt{2} \\mathrm{~m}$.\n\nSince $A C=6 \\mathrm{~m}$, then $C P=A C+A P=16 \\mathrm{~m}$.\n\nFor $C D$ to be as short as possible and given that $C$ is fixed, then it must be the case that $C D$ is horizontal:\n\nIf $C D$ were not horizontal, then suppose that $X$ is on $D Q$, possibly extended, so that $C X$ is horizontal.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3361192d5aa7e57927a4g-1.jpg?height=282&width=421&top_left_y=981&top_left_x=955)\n\nThen $\\angle C X D=90^{\\circ}$ and so $\\triangle C X D$ is right-angled with hypotenuse $C D$.\n\nIn this case, $C D$ is longer than $C X$ or $X D$.\n\nIn particular, $C D>C X$, which means that if $D$ were at $X$, then $C D$ would be shorter.\n\nIn other words, a horizontal $C D$ makes $C D$ as short as possible.\n\nWhen $C D$ is horizontal, $C D Q P$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $D Q=C P=16 \\mathrm{~m}$.\n\nFinally, this means that $B D=D Q-B Q=(16-10 \\sqrt{2}) \\mathrm{m}$.']",['$(16-10 \\sqrt{2})$'],False,m,Numerical, 2424,Geometry,,"The diagram shows two hills that meet at $O$. One hill makes a $30^{\circ}$ angle with the horizontal and the other hill makes a $45^{\circ}$ angle with the horizontal. Points $A$ and $B$ are on the hills so that $O A=O B=20 \mathrm{~m}$. Vertical poles $B D$ and $A C$ are connected by a straight cable $C D$. If $A C=6 \mathrm{~m}$, what is the length of $B D$ for which $C D$ is as short as possible? ","['Extend $C A$ and $D B$ downwards until they meet the horizontal through $O$ at $P$ and $Q$, respectively.\n\n\n\nSince $C A$ and $D B$ are vertical, then $\\angle C P O=\\angle D Q O=90^{\\circ}$.\n\nSince $O A=20 \\mathrm{~m}$, then $A P=O A \\sin 30^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{2}=10 \\mathrm{~m}$.\n\nSince $O B=20 \\mathrm{~m}$, then $B Q=O B \\sin 45^{\\circ}=(20 \\mathrm{~m}) \\cdot \\frac{1}{\\sqrt{2}}=10 \\sqrt{2} \\mathrm{~m}$.\n\nSince $A C=6 \\mathrm{~m}$, then $C P=A C+A P=16 \\mathrm{~m}$.\n\nFor $C D$ to be as short as possible and given that $C$ is fixed, then it must be the case that $C D$ is horizontal:\n\nIf $C D$ were not horizontal, then suppose that $X$ is on $D Q$, possibly extended, so that $C X$ is horizontal.\n\n\n\nThen $\\angle C X D=90^{\\circ}$ and so $\\triangle C X D$ is right-angled with hypotenuse $C D$.\n\nIn this case, $C D$ is longer than $C X$ or $X D$.\n\nIn particular, $C D>C X$, which means that if $D$ were at $X$, then $C D$ would be shorter.\n\nIn other words, a horizontal $C D$ makes $C D$ as short as possible.\n\nWhen $C D$ is horizontal, $C D Q P$ is a rectangle, since it has two vertical and two horizontal sides. Thus, $D Q=C P=16 \\mathrm{~m}$.\n\nFinally, this means that $B D=D Q-B Q=(16-10 \\sqrt{2}) \\mathrm{m}$.']",['$(16-10 \\sqrt{2})$'],False,m,Numerical, 2425,Algebra,,"If $\cos \theta=\tan \theta$, determine all possible values of $\sin \theta$, giving your answer(s) as simplified exact numbers.","['Since $\\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}$, then we assume that $\\cos \\theta \\neq 0$.\n\nTherefore, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\cos \\theta & =\\tan \\theta \\\\\n\\cos \\theta & =\\frac{\\sin \\theta}{\\cos \\theta} \\\\\n\\cos ^{2} \\theta & =\\sin \\theta \\\\\n1-\\sin ^{2} \\theta & =\\sin \\theta \\\\\n0 & =\\sin ^{2} \\theta+\\sin \\theta-1\n\\end{aligned}\n$$\n\nLet $u=\\sin \\theta$. This quadratic equation becomes $u^{2}+u-1=0$\n\nBy the quadratic formula, $u=\\frac{-1 \\pm \\sqrt{1^{2}-4(1)(-1)}}{2(1)}=\\frac{-1 \\pm \\sqrt{5}}{2}$.\n\nTherefore, $\\sin \\theta=\\frac{-1+\\sqrt{5}}{2} \\approx 0.62$ or $\\sin \\theta=\\frac{-1-\\sqrt{5}}{2} \\approx-1.62$.\n\nSince $-1 \\leq \\sin \\theta \\leq 1$, then the second solution is inadmissible. Thus, $\\sin \\theta=\\frac{-1+\\sqrt{5}}{2}$.']",['$\\frac{-1+\\sqrt{5}}{2}$'],False,,Numerical, 2426,Algebra,,"Linh is driving at $60 \mathrm{~km} / \mathrm{h}$ on a long straight highway parallel to a train track. Every 10 minutes, she is passed by a train travelling in the same direction as she is. These trains depart from the station behind her every 3 minutes and all travel at the same constant speed. What is the constant speed of the trains, in $\mathrm{km} / \mathrm{h}$ ?","['Suppose that the trains are travelling at $v \\mathrm{~km} / \\mathrm{h}$.\n\nConsider two consecutive points in time at which the car is passed by a train.\n\nSince these points are 10 minutes apart, and 10 minutes equals $\\frac{1}{6}$ hour, and the car travels at $60 \\mathrm{~km} / \\mathrm{h}$, then the car travels $(60 \\mathrm{~km} / \\mathrm{h}) \\cdot\\left(\\frac{1}{6} \\mathrm{~h}\\right)=10 \\mathrm{~km}$.\n\nDuring these 10 minutes, each train travels $\\frac{1}{6} v \\mathrm{~km}$, since its speed is $v \\mathrm{~km} / \\mathrm{h}$.\n\nAt the first instance, Train A and the car are next to each other.\n\nAt this time, Train B is "" 3 minutes"" behind Train A.\n\n\n\nSince 3 minutes is $\\frac{1}{20}$ hour, then Train B is $\\frac{1}{20} v \\mathrm{~km}$ behind Train A and the car.\n\nTherefore, the distance from the location of Train B at the first instance to the location where it passes the car is $\\left(\\frac{1}{20} v+10\\right) \\mathrm{km}$.\n\nBut this distance also equals $\\frac{1}{6} v \\mathrm{~km}$, since Train B travels for 10 minutes.\n\nThus, $\\frac{1}{6} v=\\frac{1}{20} v+10$ or $\\frac{10}{60} v-\\frac{3}{60} v=10$ and so $\\frac{7}{60} v=10$ or $v=\\frac{600}{7}$.\n\nTherefore, the trains are travelling at $\\frac{600}{7} \\mathrm{~km} / \\mathrm{h}$.', 'Suppose that the trains are travelling at $v \\mathrm{~km} / \\mathrm{h}$.\n\nConsider the following three points in time: the instant when the car and Train A are next to each other, the instant when Train B is at the same location that the car and Train A were at in the previous instant, and the instant when the car and Train B are next to each other.\n\n\n\nFrom the first instant to the second, Train B ""catches up"" to where Train A was, so this must take a total of 3 minutes, because the trains leave the station 3 minutes apart.\n\nSince 3 minutes equals $\\frac{3}{60}$ hour and the car travels at $60 \\mathrm{~km} / \\mathrm{h}$, then the car travels $(60 \\mathrm{~km} / \\mathrm{h}) \\cdot\\left(\\frac{3}{60} \\mathrm{~h}\\right)=3 \\mathrm{~km}$ between these two instants.\n\nFrom the first instant to the third, 10 minutes passes, since these are consecutive points at which the car is passed by trains. In 10 minutes, the car travels $10 \\mathrm{~km}$.\n\nTherefore, between the second and third instants, $10-3=7$ minutes pass. During these 7 minutes, Train B travels $10 \\mathrm{~km}$.\n\nSince 7 minutes equals $\\frac{7}{60}$ hour, then $v \\mathrm{~km} / \\mathrm{h}=\\frac{10 \\mathrm{~km}}{7 / 60 \\mathrm{~h}}=\\frac{600}{7} \\mathrm{~km} / \\mathrm{h}$, and so the trains are travelling at $\\frac{600}{7} \\mathrm{~km} / \\mathrm{h}$.']",['$\\frac{600}{7}$'],False,km/h,Numerical, 2427,Algebra,,"Determine all pairs $(a, b)$ of real numbers that satisfy the following system of equations: $$ \begin{aligned} \sqrt{a}+\sqrt{b} & =8 \\ \log _{10} a+\log _{10} b & =2 \end{aligned} $$ Give your answer(s) as pairs of simplified exact numbers.","['From the first equation, we note that $a \\geq 0$ and $b \\geq 0$, since the argument of a square root must be non-negative.\n\nFrom the second equation, we note that $a>0$ and $b>0$, since the argument of a logarithm must be positive.\n\nCombining these restrictions, we see that $a>0$ and $b>0$.\n\nFrom the equation $\\log _{10} a+\\log _{10} b=2$, we obtain $\\log _{10}(a b)=2$ and so $a b=10^{2}=100$. From the first equation, obtain\n\n$$\n\\begin{aligned}\n(\\sqrt{a}+\\sqrt{b})^{2} & =8^{2} \\\\\na+2 \\sqrt{a b}+b & =64 \\\\\na+2 \\sqrt{100}+b & =64 \\\\\na+b & =64-2 \\sqrt{100}=44\n\\end{aligned}\n$$\n\nSince $a+b=44$, then $b=44-a$.\n\nSince $a b=100$, then $a(44-a)=100$ or $44 a-a^{2}=100$ and so $0=a^{2}-44 a+100$.\n\nBy the quadratic formula,\n\n$$\na=\\frac{44 \\pm \\sqrt{44^{2}-4(1)(100)}}{2 \\cdot 1}=\\frac{44 \\pm \\sqrt{1536}}{2}=\\frac{44 \\pm 16 \\sqrt{6}}{2}=22 \\pm 8 \\sqrt{6}\n$$\n\nSince $b=44-a$, then $b=44-(22 \\pm 8 \\sqrt{6})=22 \\mp 8 \\sqrt{6}$.\n\nTherefore, $(a, b)=(22+8 \\sqrt{6}, 22-8 \\sqrt{6})$ or $(a, b)=(22-8 \\sqrt{6}, 22+8 \\sqrt{6})$.\n\n(We note that $22+8 \\sqrt{6}>0$ and $22-8 \\sqrt{6}>0$, so the initial restrictions on $a$ and $b$ are satisfied.)']","['$(22+8 \\sqrt{6}, 22-8 \\sqrt{6})$,$(22-8 \\sqrt{6}, 22+8 \\sqrt{6})$']",True,,Tuple, 2428,Geometry,,"In the diagram, line segments $A C$ and $D F$ are tangent to the circle at $B$ and $E$, respectively. Also, $A F$ intersects the circle at $P$ and $R$, and intersects $B E$ at $Q$, as shown. If $\angle C A F=35^{\circ}, \angle D F A=30^{\circ}$, and $\angle F P E=25^{\circ}$, determine the measure of $\angle P E Q$. ","['Let $\\angle P E Q=\\theta$.\n\nJoin $P$ to $B$.\n\nWe use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.\n\nMore concretely, $\\angle D E P=\\angle P B E$ (using the chord $E P$ and the tangent through $E$ ) and $\\angle A B P=\\angle P E Q=\\theta$ (using the chord $B P$ and the tangent through $B$ ).\n\nNow $\\angle D E P$ is exterior to $\\triangle F E P$ and so $\\angle D E P=\\angle F P E+\\angle E F P=25^{\\circ}+30^{\\circ}$, and so $\\angle P B E=\\angle D E P=55^{\\circ}$.\n\nFurthermore, $\\angle A Q B$ is an exterior angle of $\\triangle P Q E$.\n\nThus, $\\angle A Q B=\\angle Q P E+\\angle P E Q=25^{\\circ}+\\theta$.\n\n\n\nIn $\\triangle A B Q$, we have $\\angle B A Q=35^{\\circ}, \\angle A B Q=\\theta+55^{\\circ}$, and $\\angle A Q B=25^{\\circ}+\\theta$.\n\nThus, $35^{\\circ}+\\left(\\theta+55^{\\circ}\\right)+\\left(25^{\\circ}+\\theta\\right)=180^{\\circ}$ or $115^{\\circ}+2 \\theta=180^{\\circ}$, and so $2 \\theta=65^{\\circ}$.\n\nTherefore $\\angle P E Q=\\theta=\\frac{1}{2}\\left(65^{\\circ}\\right)=32.5^{\\circ}$.\n\nAs an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.\n\nConsider a circle with centre $O$ and a chord $X Y$, with tangent $Z X$ meeting the circle at $X$. We prove that if $Z X$ is tangent to the circle, then $\\angle Z X Y$ equals $\\angle X W Y$ whenever $W$ is a point on the circle on the opposite side of $X Y$ as $X Z$ (that is, the angle subtended by $X Y$ on the opposite side of the circle).\n\nWe prove this in the case that $\\angle Z X Y$ is acute. The cases where $\\angle Z X Y$ is a right angle or an obtuse angle are similar.\n\nDraw diameter $X O V$ and join $V Y$.\n\n\n\nSince $\\angle Z X Y$ is acute, points $V$ and $W$ are on the same arc of chord $X Y$.\n\nThis means that $\\angle X V Y=\\angle X W Y$, since they are angles subtended by the same chord.\n\nSince $O X$ is a radius and $X Z$ is a tangent, then $\\angle O X Z=90^{\\circ}$.\n\nThus, $\\angle O X Y+\\angle Z X Y=90^{\\circ}$.\n\nSince $X V$ is a diameter, then $\\angle X Y V=90^{\\circ}$.\n\nFrom $\\triangle X Y V$, we see that $\\angle X V Y+\\angle V X Y=90^{\\circ}$.\n\nBut $\\angle O X Y+\\angle Z X Y=90^{\\circ}$ and $\\angle X V Y+\\angle V X Y=90^{\\circ}$ and $\\angle O X Y=\\angle V X Y$ tells us that $\\angle Z X Y=\\angle X V Y$.\n\nThis gives us that $\\angle Z X Y=\\angle X W Y$, as required.']",['$32.5^{\\circ}$'],False,,Numerical, 2428,Geometry,,"In the diagram, line segments $A C$ and $D F$ are tangent to the circle at $B$ and $E$, respectively. Also, $A F$ intersects the circle at $P$ and $R$, and intersects $B E$ at $Q$, as shown. If $\angle C A F=35^{\circ}, \angle D F A=30^{\circ}$, and $\angle F P E=25^{\circ}$, determine the measure of $\angle P E Q$. ![](https://cdn.mathpix.com/cropped/2023_12_21_e1c85d542a446b534fb3g-1.jpg?height=602&width=550&top_left_y=924&top_left_x=1232)","['Let $\\angle P E Q=\\theta$.\n\nJoin $P$ to $B$.\n\nWe use the fact that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord. We prove this fact below.\n\nMore concretely, $\\angle D E P=\\angle P B E$ (using the chord $E P$ and the tangent through $E$ ) and $\\angle A B P=\\angle P E Q=\\theta$ (using the chord $B P$ and the tangent through $B$ ).\n\nNow $\\angle D E P$ is exterior to $\\triangle F E P$ and so $\\angle D E P=\\angle F P E+\\angle E F P=25^{\\circ}+30^{\\circ}$, and so $\\angle P B E=\\angle D E P=55^{\\circ}$.\n\nFurthermore, $\\angle A Q B$ is an exterior angle of $\\triangle P Q E$.\n\nThus, $\\angle A Q B=\\angle Q P E+\\angle P E Q=25^{\\circ}+\\theta$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_47eae7df5f8f4c5855a9g-1.jpg?height=604&width=531&top_left_y=744&top_left_x=900)\n\nIn $\\triangle A B Q$, we have $\\angle B A Q=35^{\\circ}, \\angle A B Q=\\theta+55^{\\circ}$, and $\\angle A Q B=25^{\\circ}+\\theta$.\n\nThus, $35^{\\circ}+\\left(\\theta+55^{\\circ}\\right)+\\left(25^{\\circ}+\\theta\\right)=180^{\\circ}$ or $115^{\\circ}+2 \\theta=180^{\\circ}$, and so $2 \\theta=65^{\\circ}$.\n\nTherefore $\\angle P E Q=\\theta=\\frac{1}{2}\\left(65^{\\circ}\\right)=32.5^{\\circ}$.\n\nAs an addendum, we prove that the angle between a tangent to a circle and a chord in that circle that passes through the point of tangency equals the angle inscribed by that chord.\n\nConsider a circle with centre $O$ and a chord $X Y$, with tangent $Z X$ meeting the circle at $X$. We prove that if $Z X$ is tangent to the circle, then $\\angle Z X Y$ equals $\\angle X W Y$ whenever $W$ is a point on the circle on the opposite side of $X Y$ as $X Z$ (that is, the angle subtended by $X Y$ on the opposite side of the circle).\n\nWe prove this in the case that $\\angle Z X Y$ is acute. The cases where $\\angle Z X Y$ is a right angle or an obtuse angle are similar.\n\nDraw diameter $X O V$ and join $V Y$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_47eae7df5f8f4c5855a9g-1.jpg?height=409&width=374&top_left_y=1720&top_left_x=1583)\n\nSince $\\angle Z X Y$ is acute, points $V$ and $W$ are on the same arc of chord $X Y$.\n\nThis means that $\\angle X V Y=\\angle X W Y$, since they are angles subtended by the same chord.\n\nSince $O X$ is a radius and $X Z$ is a tangent, then $\\angle O X Z=90^{\\circ}$.\n\nThus, $\\angle O X Y+\\angle Z X Y=90^{\\circ}$.\n\nSince $X V$ is a diameter, then $\\angle X Y V=90^{\\circ}$.\n\nFrom $\\triangle X Y V$, we see that $\\angle X V Y+\\angle V X Y=90^{\\circ}$.\n\nBut $\\angle O X Y+\\angle Z X Y=90^{\\circ}$ and $\\angle X V Y+\\angle V X Y=90^{\\circ}$ and $\\angle O X Y=\\angle V X Y$ tells us that $\\angle Z X Y=\\angle X V Y$.\n\nThis gives us that $\\angle Z X Y=\\angle X W Y$, as required.']",['$32.5^{\\circ}$'],False,,Numerical, 2429,Geometry,,"In the diagram, $A B C D$ and $P N C D$ are squares of side length 2, and $P N C D$ is perpendicular to $A B C D$. Point $M$ is chosen on the same side of $P N C D$ as $A B$ so that $\triangle P M N$ is parallel to $A B C D$, so that $\angle P M N=90^{\circ}$, and so that $P M=M N$. Determine the volume of the convex solid $A B C D P M N$. ","['Draw a line segment through $M$ in the plane of $\\triangle P M N$ parallel to $P N$ and extend this line until it reaches the plane through $P, A$ and $D$ at $Q$ on one side and the plane through $N, B$ and $C$ at $R$ on the other side.\n\nJoin $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.\n\n\n\nSo the volume of solid $A B C D P M N$ equals the volume of solid $A B C D P Q R N$ minus the volumes of solids $P M Q A$ and $N M R B$.\n\nSolid $A B C D P Q R N$ is a trapezoidal prism. This is because $N R$ and $B C$ are parallel (since they lie in parallel planes), which makes $N R B C$ a trapezoid. Similarly, $P Q A D$ is a trapezoid. Also, $P N, Q R, D C$, and $A B$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.\n\nSolids $P M Q A$ and $N M R B$ are triangular-based pyramids. We can think of their bases as being $\\triangle P M Q$ and $\\triangle N M R$. Their heights are each equal to 2 , the height of the original solid. (The volume of a triangular-based pyramid equals $\\frac{1}{3}$ times the area of its base times its height.)\n\nThe volume of $A B C D P Q R N$ equals the area of trapezoid $N R B C$ times the width of the prism, which is 2.\n\nThat is, this volume equals $\\frac{1}{2}(N R+B C)(N C)(N P)=\\frac{1}{2}(N R+2)(2)(2)=2 \\cdot N R+4$.\n\nSo we need to find the length of $N R$.\n\nConsider quadrilateral $P N R Q$. This quadrilateral is a rectangle since $P N$ and $Q R$ are perpendicular to the two side planes of the original solid.\n\nThus, $N R$ equals the height of $\\triangle P M N$.\n\nJoin $M$ to the midpoint $T$ of $P N$.\n\nSince $\\triangle P M N$ is isosceles, then $M T$ is perpendicular to $P N$.\n\n\n\nSince $N T=\\frac{1}{2} P N=1$ and $\\angle P M N=90^{\\circ}$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\nTherefore, $N R=M T=1$ and so the volume of $A B C D P Q R N$ is $2 \\cdot 1+4=6$.\n\nThe volumes of solids $P M Q A$ and $N M R B$ are equal. Each has height 2 and their bases $\\triangle P M Q$ and $\\triangle N M R$ are congruent, because each is right-angled (at $Q$ and at $R$ ) with $P Q=N R=1$ and $Q M=M R=1$.\n\nThus, using the formula above, the volume of each is $\\frac{1}{3}\\left(\\frac{1}{2}(1)(1)\\right) 2=\\frac{1}{3}$.\n\nFinally, the volume of the original solid equals $6-2 \\cdot \\frac{1}{3}=\\frac{16}{3}$.', 'We determine the volume of $A B C D P M N$ by splitting it into two solids: $A B C D P N$ and $A B N P M$ by slicing along the plane of $A B N P$.\n\nSolid $A B C D P N$ is a triangular prism, since $\\triangle B C N$ and $\\triangle A D P$ are each right-angled (at $C$ and $D$ ), $B C=C N=A D=D P=2$, and segments $P N, D C$ and $A B$ are perpendicular to each of the triangular faces and equal in length.\n\nThus, the volume of $A B C D P N$ equals the area of $\\triangle B C N$ times the length of $D C$, or $\\frac{1}{2}(B C)(C N)(D C)=\\frac{1}{2}(2)(2)(2)=4$. (This solid can also be viewed as ""half"" of a cube.)\n\nSolid $A B N P M$ is a pyramid with rectangular base $A B N P$. (Note that $P N$ and $A B$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $P N=A B=2$ and $B N=A P=\\sqrt{2^{2}+2^{2}}=2 \\sqrt{2}$, by the Pythagorean Theorem.)\n\nTherefore, the volume of $A B N P M$ equals $\\frac{1}{3}(A B)(B N) h=\\frac{4 \\sqrt{2}}{3} h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $A B N P$ ).\n\nSo we need to calculate $h$.\n\nJoin $M$ to the midpoint, $T$, of $P N$ and to the midpoint, $S$, of $A B$. Join $S$ and $T$. By symmetry, $M$ lies directly above $S T$. Since $A B N P$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, then $S T=A P=2 \\sqrt{2}$.\n\nSince $\\triangle P M N$ is right-angled and isosceles, then $M T$ is perpendicular to $P N$. Since $N T=\\frac{1}{2} P N=1$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\n\n\nAlso, $M S$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $A B C D$ (a distance of 2) and joining $U$ to $S$. Since $M$ is 1 unit horizontally from $P N$, then $U S=1$.\n\nThus, $M S=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$ by the Pythagorean Theorem.\n\n\n\nWe can now consider $\\triangle S M T . h$ is the height of this triangle, from $M$ to base $S T$.\n\n\n\nNow $h=M T \\sin (\\angle M T S)=\\sin (\\angle M T S)$.\n\nBy the cosine law in $\\triangle S M T$, we have\n\n$$\nM S^{2}=S T^{2}+M T^{2}-2(S T)(M T) \\cos (\\angle M T S)\n$$\n\nTherefore, $5=8+1-4 \\sqrt{2} \\cos (\\angle M T S)$ or $4 \\sqrt{2} \\cos (\\angle M T S)=4$.\n\nThus, $\\cos (\\angle M T S)=\\frac{1}{\\sqrt{2}}$ and so $\\angle M T S=45^{\\circ}$ which gives $h=\\sin (\\angle M T S)=\\frac{1}{\\sqrt{2}}$.\n\n(Alternatively, we note that the plane of $A B C D$ is parallel to the plane of $P M N$, and so since the angle between plane $A B C D$ and plane $P N B A$ is $45^{\\circ}$, then the angle between plane $P N B A$ and plane $P M N$ is also $45^{\\circ}$, and so $\\angle M T S=45^{\\circ}$.)\n\nFinally, this means that the volume of $A B N P M$ is $\\frac{4 \\sqrt{2}}{3} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{4}{3}$, and so the volume of solid $A B C D P M N$ is $4+\\frac{4}{3}=\\frac{16}{3}$.']",['$\\frac{16}{3}$'],False,,Numerical, 2429,Geometry,,"In the diagram, $A B C D$ and $P N C D$ are squares of side length 2, and $P N C D$ is perpendicular to $A B C D$. Point $M$ is chosen on the same side of $P N C D$ as $A B$ so that $\triangle P M N$ is parallel to $A B C D$, so that $\angle P M N=90^{\circ}$, and so that $P M=M N$. Determine the volume of the convex solid $A B C D P M N$. ![](https://cdn.mathpix.com/cropped/2023_12_21_e1c85d542a446b534fb3g-1.jpg?height=417&width=461&top_left_y=1575&top_left_x=1252)","['Draw a line segment through $M$ in the plane of $\\triangle P M N$ parallel to $P N$ and extend this line until it reaches the plane through $P, A$ and $D$ at $Q$ on one side and the plane through $N, B$ and $C$ at $R$ on the other side.\n\nJoin $Q$ to $P$ and $A$. Join $R$ to $N$ and $B$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_32c5559f1e227b6da6b9g-1.jpg?height=415&width=441&top_left_y=432&top_left_x=945)\n\nSo the volume of solid $A B C D P M N$ equals the volume of solid $A B C D P Q R N$ minus the volumes of solids $P M Q A$ and $N M R B$.\n\nSolid $A B C D P Q R N$ is a trapezoidal prism. This is because $N R$ and $B C$ are parallel (since they lie in parallel planes), which makes $N R B C$ a trapezoid. Similarly, $P Q A D$ is a trapezoid. Also, $P N, Q R, D C$, and $A B$ are all perpendicular to the planes of these trapezoids and equal in length, since they equal the side lengths of the squares.\n\nSolids $P M Q A$ and $N M R B$ are triangular-based pyramids. We can think of their bases as being $\\triangle P M Q$ and $\\triangle N M R$. Their heights are each equal to 2 , the height of the original solid. (The volume of a triangular-based pyramid equals $\\frac{1}{3}$ times the area of its base times its height.)\n\nThe volume of $A B C D P Q R N$ equals the area of trapezoid $N R B C$ times the width of the prism, which is 2.\n\nThat is, this volume equals $\\frac{1}{2}(N R+B C)(N C)(N P)=\\frac{1}{2}(N R+2)(2)(2)=2 \\cdot N R+4$.\n\nSo we need to find the length of $N R$.\n\nConsider quadrilateral $P N R Q$. This quadrilateral is a rectangle since $P N$ and $Q R$ are perpendicular to the two side planes of the original solid.\n\nThus, $N R$ equals the height of $\\triangle P M N$.\n\nJoin $M$ to the midpoint $T$ of $P N$.\n\nSince $\\triangle P M N$ is isosceles, then $M T$ is perpendicular to $P N$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_32c5559f1e227b6da6b9g-1.jpg?height=247&width=382&top_left_y=1855&top_left_x=969)\n\nSince $N T=\\frac{1}{2} P N=1$ and $\\angle P M N=90^{\\circ}$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\nTherefore, $N R=M T=1$ and so the volume of $A B C D P Q R N$ is $2 \\cdot 1+4=6$.\n\nThe volumes of solids $P M Q A$ and $N M R B$ are equal. Each has height 2 and their bases $\\triangle P M Q$ and $\\triangle N M R$ are congruent, because each is right-angled (at $Q$ and at $R$ ) with $P Q=N R=1$ and $Q M=M R=1$.\n\nThus, using the formula above, the volume of each is $\\frac{1}{3}\\left(\\frac{1}{2}(1)(1)\\right) 2=\\frac{1}{3}$.\n\nFinally, the volume of the original solid equals $6-2 \\cdot \\frac{1}{3}=\\frac{16}{3}$.', 'We determine the volume of $A B C D P M N$ by splitting it into two solids: $A B C D P N$ and $A B N P M$ by slicing along the plane of $A B N P$.\n\nSolid $A B C D P N$ is a triangular prism, since $\\triangle B C N$ and $\\triangle A D P$ are each right-angled (at $C$ and $D$ ), $B C=C N=A D=D P=2$, and segments $P N, D C$ and $A B$ are perpendicular to each of the triangular faces and equal in length.\n\nThus, the volume of $A B C D P N$ equals the area of $\\triangle B C N$ times the length of $D C$, or $\\frac{1}{2}(B C)(C N)(D C)=\\frac{1}{2}(2)(2)(2)=4$. (This solid can also be viewed as ""half"" of a cube.)\n\nSolid $A B N P M$ is a pyramid with rectangular base $A B N P$. (Note that $P N$ and $A B$ are perpendicular to the planes of both of the side triangular faces of the original solid, that $P N=A B=2$ and $B N=A P=\\sqrt{2^{2}+2^{2}}=2 \\sqrt{2}$, by the Pythagorean Theorem.)\n\nTherefore, the volume of $A B N P M$ equals $\\frac{1}{3}(A B)(B N) h=\\frac{4 \\sqrt{2}}{3} h$, where $h$ is the height of the pyramid (that is, the distance that $M$ is above plane $A B N P$ ).\n\nSo we need to calculate $h$.\n\nJoin $M$ to the midpoint, $T$, of $P N$ and to the midpoint, $S$, of $A B$. Join $S$ and $T$. By symmetry, $M$ lies directly above $S T$. Since $A B N P$ is a rectangle and $S$ and $T$ are the midpoints of opposite sides, then $S T=A P=2 \\sqrt{2}$.\n\nSince $\\triangle P M N$ is right-angled and isosceles, then $M T$ is perpendicular to $P N$. Since $N T=\\frac{1}{2} P N=1$ and $\\angle T N M=45^{\\circ}$, then $\\triangle M T N$ is also right-angled and isosceles with $M T=T N=1$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_10d3fb8fb2f3aeb374f1g-1.jpg?height=431&width=445&top_left_y=874&top_left_x=1515)\n\nAlso, $M S$ is the hypotenuse of the triangle formed by dropping a perpendicular from $M$ to $U$ in the plane of $A B C D$ (a distance of 2) and joining $U$ to $S$. Since $M$ is 1 unit horizontally from $P N$, then $U S=1$.\n\nThus, $M S=\\sqrt{2^{2}+1^{2}}=\\sqrt{5}$ by the Pythagorean Theorem.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_10d3fb8fb2f3aeb374f1g-1.jpg?height=418&width=458&top_left_y=1298&top_left_x=1517)\n\nWe can now consider $\\triangle S M T . h$ is the height of this triangle, from $M$ to base $S T$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_10d3fb8fb2f3aeb374f1g-1.jpg?height=204&width=483&top_left_y=1779&top_left_x=924)\n\nNow $h=M T \\sin (\\angle M T S)=\\sin (\\angle M T S)$.\n\nBy the cosine law in $\\triangle S M T$, we have\n\n$$\nM S^{2}=S T^{2}+M T^{2}-2(S T)(M T) \\cos (\\angle M T S)\n$$\n\nTherefore, $5=8+1-4 \\sqrt{2} \\cos (\\angle M T S)$ or $4 \\sqrt{2} \\cos (\\angle M T S)=4$.\n\nThus, $\\cos (\\angle M T S)=\\frac{1}{\\sqrt{2}}$ and so $\\angle M T S=45^{\\circ}$ which gives $h=\\sin (\\angle M T S)=\\frac{1}{\\sqrt{2}}$.\n\n(Alternatively, we note that the plane of $A B C D$ is parallel to the plane of $P M N$, and so since the angle between plane $A B C D$ and plane $P N B A$ is $45^{\\circ}$, then the angle between plane $P N B A$ and plane $P M N$ is also $45^{\\circ}$, and so $\\angle M T S=45^{\\circ}$.)\n\nFinally, this means that the volume of $A B N P M$ is $\\frac{4 \\sqrt{2}}{3} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{4}{3}$, and so the volume of solid $A B C D P M N$ is $4+\\frac{4}{3}=\\frac{16}{3}$.']",['$\\frac{16}{3}$'],False,,Numerical, 2430,Combinatorics,,"A permutation of a list of numbers is an ordered arrangement of the numbers in that list. For example, $3,2,4,1,6,5$ is a permutation of $1,2,3,4,5,6$. We can write this permutation as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, where $a_{1}=3, a_{2}=2, a_{3}=4, a_{4}=1, a_{5}=6$, and $a_{6}=5$. Determine the average value of $$ \left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right| $$ over all permutations $a_{1}, a_{2}, a_{3}, a_{4}$ of $1,2,3,4$.","['There are 4 ! $=4 \\cdot 3 \\cdot 2 \\cdot 1=24$ permutations of $1,2,3,4$.\n\nThis is because there are 4 possible choices for $a_{1}$, and for each of these there are 3 possible choices for $a_{2}$, and for each of these there are 2 possible choices for $a_{3}$, and then 1 possible choice for $a_{4}$.\n\nConsider the permutation $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4$. (We write this as $1,2,3,4$.)\n\nHere, $\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|=|1-2|+|3-4|=1+1=2$.\n\nThis value is the same as the value for each of $2,1,3,4$ and $1,2,4,3$ and $2,1,4,3$ and $3,4,1,2$ and 4,3,1,2 and 3,4,2,1 and 4,3,2,1.\n\nConsider the permutation $1,3,2,4$.\n\nHere, $\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|=|1-3|+|2-4|=2+2=4$.\n\nThis value is the same as the value for each of $3,1,2,4$ and $1,3,4,2$ and $3,1,4,2$ and $2,4,1,3$ and 4,2,1,3 and 2,4,3,1 and 4,2,3,1.\n\nConsider the permutation $1,4,2,3$.\n\nHere, $\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|=|1-4|+|2-3|=3+1=4$.\n\nThis value is the same as the value for each of 4,1,2,3 and 1,4,3,2 and 4,1,3,2 and 2,3,1,4 and $3,2,1,4$ and $2,3,4,1$ and $3,2,4,1$.\n\nThis accounts for all 24 permutations.\n\nTherefore, the average value is $\\frac{2 \\cdot 8+4 \\cdot 8+4 \\cdot 8}{24}=\\frac{80}{24}=\\frac{10}{3}$.']",['$\\frac{10}{3}$'],False,,Numerical, 2431,Combinatorics,,"A permutation of a list of numbers is an ordered arrangement of the numbers in that list. For example, $3,2,4,1,6,5$ is a permutation of $1,2,3,4,5,6$. We can write this permutation as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, where $a_{1}=3, a_{2}=2, a_{3}=4, a_{4}=1, a_{5}=6$, and $a_{6}=5$. Determine the average value of $$ a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7} $$ over all permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ of $1,2,3,4,5,6,7$.","['There are $7 !=7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$ permutations of $1,2,3,4,5,6,7$, because there are 7 choices for $a_{1}$, then 6 choices for $a_{2}$, and so on.\n\nWe determine the average value of $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}$ over all of these permutations by determining the sum of all 7 ! values of this expression and dividing by $7 !$.\n\nTo determine the sum of all 7 ! values, we determine the sum of the values of $a_{1}$ in each of these expressions and call this total $s_{1}$, the sum of the values of $a_{2}$ in each of these expressions and call this total $s_{2}$, and so on.\n\nThe sum of the 7 ! values of the original expression must equal $s_{1}-s_{2}+s_{3}-s_{4}+s_{5}-s_{6}+s_{7}$. This uses the fact that, when adding, the order in which we add the same set of numbers does not matter.\n\nBy symmetry, the sums of the values of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ will all be equal. That is, $s_{1}=s_{2}=s_{3}=s_{4}=s_{5}=s_{6}=s_{7}$.\n\nThis means that the desired average value equals\n\n$$\n\\frac{s_{1}-s_{2}+s_{3}-s_{4}+s_{5}-s_{6}+s_{7}}{7 !}=\\frac{\\left(s_{1}+s_{3}+s_{5}+s_{7}\\right)-\\left(s_{2}+s_{4}+s_{6}\\right)}{7 !}=\\frac{4 s_{1}-3 s_{1}}{7 !}=\\frac{s_{1}}{7 !}\n$$\n\nSo we need to determine the value of $s_{1}$.\n\nNow $a_{1}$ can equal each of $1,2,3,4,5,6,7$.\n\nIf $a_{1}=1$, there are 6 ! combinations of values for $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$, since there are still 6 choices for $a_{2}, 5$ for $a_{3}$, and so on.\n\nSimilarly, there are 6 ! combinations with $a_{1}$ equal to each of $2,3,4,5,6,7$.\n\nThus, $s_{1}=1 \\cdot 6 !+2 \\cdot 6 !+3 \\cdot 6 !+4 \\cdot 6 !+5 \\cdot 6 !+6 \\cdot 6 !+7 \\cdot 6 !=6 !(1+2+3+4+5+6+7)=28(6 !)$.\n\nTherefore, the average value of the expression is $\\frac{28(6 !)}{7 !}=\\frac{28(6 !)}{7(6 !)}=\\frac{28}{7}=4$.']",['4'],False,,Numerical, 2432,Combinatorics,,"A permutation of a list of numbers is an ordered arrangement of the numbers in that list. For example, $3,2,4,1,6,5$ is a permutation of $1,2,3,4,5,6$. We can write this permutation as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, where $a_{1}=3, a_{2}=2, a_{3}=4, a_{4}=1, a_{5}=6$, and $a_{6}=5$. Determine the average value of $$ \left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|+\cdots+\left|a_{197}-a_{198}\right|+\left|a_{199}-a_{200}\right| $$ over all permutations $a_{1}, a_{2}, a_{3}, \ldots, a_{199}, a_{200}$ of $1,2,3,4, \ldots, 199,200$. (The sum labelled (*) contains 100 terms of the form $\left|a_{2 k-1}-a_{2 k}\right|$.)","['There are 200! permutations of $1,2,3, \\ldots, 198,199,200$.\n\nWe determine the average value of\n\n$$\n\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|+\\cdots+\\left|a_{197}-a_{198}\\right|+\\left|a_{199}-a_{200}\\right|\n$$\n\nover all of these permutations by determining the sum of all 200! values of this expression and dividing by $200 !$.\n\nThen, we let $s_{1}$ be the sum of the values of $\\left|a_{1}-a_{2}\\right|$ in each of these expressions, $s_{2}$ be the sum of the values of $\\left|a_{3}-a_{4}\\right|$, and so on.\n\nThe sum of the 200 ! values of $(*)$ equals $s_{1}+s_{2}+\\cdots+s_{99}+s_{100}$.\n\nBy symmetry, $s_{1}=s_{2}=\\cdots=s_{99}=s_{100}$.\n\nTherefore, the average value of $(*)$ equals $\\frac{100 s_{1}}{200 !}$. So we need to determine the value of $s_{1}$.\n\nSuppose that $a_{1}=i$ and $a_{2}=j$ for some integers $i$ and $j$ between 1 and 200, inclusive.\n\nThere are 198! permutations with $a_{1}=i$ and $a_{2}=j$ because there are still 198 choices for $a_{3}, 197$ choices for $a_{4}$, and so on.\n\nSimilarly, there are 198! permutations with $a_{1}=j$ and $a_{2}=i$.\n\nSince $|i-j|=|j-i|$, then there are 2(198!) permutations with $\\left|a_{1}-a_{2}\\right|=|i-j|$ that come from $a_{1}$ and $a_{2}$ equalling $i$ and $j$ in some order.\n\nTherefore, we may assume that $i>j$ and note that $s_{1}$ equals 2(198!) times the sum of $i-j$ over all possible pairs $i>j$.\n\n(Note that there are $\\left(\\begin{array}{c}200 \\\\ 2\\end{array}\\right)=\\frac{200(199)}{2}$ choices for the pair of integers $(i, j)$ with $i>j$. For each of these choices, there are 2(198!) choices for the remaining entries in the permutation, which gives $\\frac{200(199)}{2} \\cdot 2(198 !)=200(199)(198 !)=200$ ! permutations, as expected.)\n\nSo to determine $s_{1}$, we need to determine the sum of the values of $i-j$.\n\nWe calculate this sum, which we call $D$, by letting $j=1,2,3, \\ldots, 198,199$ and for each of these, we let $i$ be the possible integers with $j4$.\n\nAlso, since $(t+2)^{2}=t^{2}+4 t+4$ and so $(t+2)^{2}$ and $t^{2}+4$ differ by a multiple of $n=2 t$, then $\\operatorname{rem}\\left((t+2)^{2}, 2 t\\right)=\\operatorname{rem}\\left(t^{2}+4,2 t\\right)$.\n\nSimilarly, since $(t+1)^{2}=t^{2}+2 t+1$, then $\\operatorname{rem}\\left((t+1)^{2}, 2 t\\right)=\\operatorname{rem}\\left(t^{2}+1,2 t\\right)$.\n\nTherefore, we need to show that $\\operatorname{rem}\\left(t^{2}+4,2 t\\right)-\\operatorname{rem}\\left(t^{2}+1,2 t\\right)=4-1=3$.\n\nSince $t \\geq 5$, then $t^{2}+t>t^{2}+4$.\n\nThis means that $t^{2}N>1$.\n\n\n\nSet $a=\\frac{1}{2}(M+N), b=n-a, c=\\frac{1}{2}(M-N)$, and $d=n-c$.\n\nSince $M$ and $N$ are both odd, then $M+N$ and $M-N$ are even, and so $a, b, c, d$ are integers.\n\nSince $M>N>0$, then $a>c>0$.\n\nSince $N \\geq 3$, then $n=M N \\geq 3 M>2 M$ and so $M<\\frac{1}{2} n$.\n\nSince $M>N$, then $a=\\frac{1}{2}(M+N)<\\frac{1}{2}(M+M)=M<\\frac{1}{2} n$.\n\nTherefore, $0","['Since $\\sin C=\\frac{A B}{A C}$, then $A B=A C \\sin C=20\\left(\\frac{3}{5}\\right)=12$.\n\nBy Pythagoras, $B C^{2}=A C^{2}-A B^{2}=20^{2}-12^{2}=256$ or $B C=16$.', 'Using the standard trigonometric ratios, $B C=A C \\cos C$.\n\nSince $\\sin C=\\frac{3}{5}$, then $\\cos ^{2} C=1-\\sin ^{2} C=1-\\frac{9}{25}=\\frac{16}{25}$ or $\\cos C=\\frac{4}{5}$. (Notice that $\\cos C$ is positive since angle $C$ is acute in triangle $A B C$.)\n\nTherefore, $B C=20\\left(\\frac{4}{5}\\right)=16$.']",['16'],False,,Numerical, 2437,Geometry,,"A helicopter is flying due west over level ground at a constant altitude of $222 \mathrm{~m}$ and at a constant speed. A lazy, stationary goat, which is due west of the helicopter, takes two measurements of the angle between the ground and the helicopter. The first measurement the goat makes is $6^{\circ}$ and the second measurement, which he makes 1 minute later, is $75^{\circ}$. If the helicopter has not yet passed over the goat, as shown, how fast is the helicopter travelling to the nearest kilometre per hour? ","['Let $G$ be the point where the goat is standing, $H$ the position of the helicopter when the goat first measures the angle, $P$ the point directly below the helicopter at this time, $J$ the position of the helicopter one minute later, and $Q$ the point directly below the helicopter at this time.\n\n\n\nUsing the initial position of the helicopter, $\\tan \\left(6^{\\circ}\\right)=\\frac{H P}{P G}$ or $P G=\\frac{222}{\\tan \\left(6^{\\circ}\\right)} \\approx 2112.19 \\mathrm{~m}$.\n\nUsing the second position of the helicopter, $\\tan \\left(75^{\\circ}\\right)=\\frac{J Q}{Q G}$ or $Q G=\\frac{222}{\\tan \\left(75^{\\circ}\\right)} \\approx 59.48 \\mathrm{~m}$.\n\nSo in the one minute that has elapsed, the helicopter has travelled\n\n$2112.19 \\mathrm{~m}-59.48 \\mathrm{~m}=2052.71 \\mathrm{~m}$ or $2.0527 \\mathrm{~km}$.\n\nTherefore, in one hour, the helicopter will travel $60(2.0527)=123.162 \\mathrm{~km}$.\n\nThus, the helicopter is travelling $123 \\mathrm{~km} / \\mathrm{h}$.']",['123'],False,km/h,Numerical, 2437,Geometry,,"A helicopter is flying due west over level ground at a constant altitude of $222 \mathrm{~m}$ and at a constant speed. A lazy, stationary goat, which is due west of the helicopter, takes two measurements of the angle between the ground and the helicopter. The first measurement the goat makes is $6^{\circ}$ and the second measurement, which he makes 1 minute later, is $75^{\circ}$. If the helicopter has not yet passed over the goat, as shown, how fast is the helicopter travelling to the nearest kilometre per hour? ![](https://cdn.mathpix.com/cropped/2023_12_21_1f6e0e123b672977527ag-1.jpg?height=379&width=881&top_left_y=2182&top_left_x=641)","['Let $G$ be the point where the goat is standing, $H$ the position of the helicopter when the goat first measures the angle, $P$ the point directly below the helicopter at this time, $J$ the position of the helicopter one minute later, and $Q$ the point directly below the helicopter at this time.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_94ca7e940013ecf0c964g-1.jpg?height=498&width=808&top_left_y=1426&top_left_x=1103)\n\nUsing the initial position of the helicopter, $\\tan \\left(6^{\\circ}\\right)=\\frac{H P}{P G}$ or $P G=\\frac{222}{\\tan \\left(6^{\\circ}\\right)} \\approx 2112.19 \\mathrm{~m}$.\n\nUsing the second position of the helicopter, $\\tan \\left(75^{\\circ}\\right)=\\frac{J Q}{Q G}$ or $Q G=\\frac{222}{\\tan \\left(75^{\\circ}\\right)} \\approx 59.48 \\mathrm{~m}$.\n\nSo in the one minute that has elapsed, the helicopter has travelled\n\n$2112.19 \\mathrm{~m}-59.48 \\mathrm{~m}=2052.71 \\mathrm{~m}$ or $2.0527 \\mathrm{~km}$.\n\nTherefore, in one hour, the helicopter will travel $60(2.0527)=123.162 \\mathrm{~km}$.\n\nThus, the helicopter is travelling $123 \\mathrm{~km} / \\mathrm{h}$.']",['123'],False,km/h,Numerical, 2438,Algebra,,"The function $f(x)$ has the property that $f(2 x+3)=2 f(x)+3$ for all $x$. If $f(0)=6$, what is the value of $f(9)$ ?","['Since we are looking for the value of $f(9)$, then it makes sense to use the given equation and to set $x=3$ in order to obtain $f(9)=2 f(3)+3$.\n\nSo we need to determine the value of $f(3)$. We use the equation again and set $x=0$ since we will then get $f(3)$ on the left side and $f(0)$ (whose value we already know) on the right side, ie.\n\n$$\nf(3)=2 f(0)+3=2(6)+3=15\n$$\n\nThus, $f(9)=2(15)+3=33$.']",['33'],False,,Numerical, 2439,Algebra,,"Suppose that the functions $f(x)$ and $g(x)$ satisfy the system of equations $$ \begin{aligned} f(x)+3 g(x) & =x^{2}+x+6 \\ 2 f(x)+4 g(x) & =2 x^{2}+4 \end{aligned} $$ for all $x$. Determine the values of $x$ for which $f(x)=g(x)$.","['We solve the system of equations for $f(x)$ and $g(x)$.\n\nDividing out the common factor of 2 from the second equation, we get\n\n$f(x)+2 g(x)=x^{2}+2$.\n\nSubtracting from the first equation, we get $g(x)=x+4$.\n\nThus, $f(x)=x^{2}+2-2 g(x)=x^{2}+2-2(x+4)=x^{2}-2 x-6$.\n\nEquating $f(x)$ and $g(x)$, we obtain\n\n$$\n\\begin{aligned}\nx^{2}-2 x-6 & =x+4 \\\\\nx^{2}-3 x-10 & =0 \\\\\n(x-5)(x+2) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=5$ or $x=-2$.', 'Instead of considering the equation $f(x)=g(x)$, we consider the equation $f(x)-g(x)=0$, and we try to obtain an expression for $f(x)-g(x)$ by manipulating the two given equations.\n\nIn fact, after some experimentation, we can see that\n\n$$\n\\begin{aligned}\nf(x)-g(x) & =2(2 f(x)+4 g(x))-3(f(x)+3 g(x)) \\\\\n& =2\\left(2 x^{2}+4\\right)-3\\left(x^{2}+x+6\\right) \\\\\n& =x^{2}-3 x-10\n\\end{aligned}\n$$\n\nSo to solve $f(x)-g(x)=0$, we solve $x^{2}-3 x-10=0$ or $(x-5)(x+2)=0$. Therefore, $x=5$ or $x=-2$.']","['$5,-2$']",True,,Numerical, 2440,Combinatorics,,"In a short-track speed skating event, there are five finalists including two Canadians. The first three skaters to finish the race win a medal. If all finalists have the same chance of finishing in any position, what is the probability that neither Canadian wins a medal?","[""We label the 5 skaters A, B, C, D, and E, where D and E are the two Canadians.\n\nThere are then $5 !=5 \\times 4 \\times 3 \\times 2 \\times 1=120$ ways of arranging these skaters in their order of finish (for example, $\\mathrm{ADBCE}$ indicates that A finished first, $\\mathrm{D}$ second, etc.), because there are 5 choices for the winner, 4 choices for the second place finisher, 3 choices for the third place finisher, etc.\n\n\n\nIf the two Canadians finish without winning medals, then they must finish fourth and fifth. So the $\\mathrm{D}$ and $\\mathrm{E}$ are in the final two positions, and $\\mathrm{A}, \\mathrm{B}$ and $\\mathrm{C}$ in the first three. There are $3 !=6$ ways of arranging the $\\mathrm{A}, \\mathrm{B}$ and $\\mathrm{C}$, and $2 !=2$ ways to arrange the $\\mathrm{D}$ and E. Thus, there are $6 \\times 2=12$ ways or arranging the skaters so that neither Canadian wins a medal.\n\nTherefore, the probability that neither Canadian wins a medal is\n\n$$\n\\frac{\\# \\text { of ways where Canadians don't win medals }}{\\text { Total } \\# \\text { of arrangements }}=\\frac{12}{120}=\\frac{1}{10}\n$$"", 'We label the 5 skaters as A, B, C, D, and E, where D and E are the two Canadians. In any race, two of the skaters finish fourth and fifth. Also, any pair of skaters are equally as likely to finish fourth and fifth, since the probability of every skater is equally likely to finish in a given position.\n\nHow many pairs of 2 skaters can we form from the 5 skaters? There are ten such pairs:\n\n$$\n\\{A, B\\},\\{A, C\\},\\{A, D\\},\\{A, E\\},\\{B, C\\},\\{B, D\\},\\{B, E\\},\\{C, D\\},\\{C, E\\},\\{D, E\\}\n$$\n\nOnly one of these ten pairs is made up of the two Canadians. Therefore, the probability is $\\frac{1}{10}$, since one out of ten choices gives the desired result.']",['$\\frac{1}{10}$'],False,,Numerical, 2441,Number Theory,,"Determine the number of positive integers less than or equal to 300 that are multiples of 3 or 5 , but are not multiples of 10 or 15 .","['Since the least common multiple of $3,5,10$ and 15 is 30 , then we can count the number of positive integers less than or equal to 30 satisfying these conditions, and multiply the total by 10 to obtain the number less than 300. (This is because each group of 30 consecutive integers starting with 1 more than a multiple of 30 will have the same number of integers having these properties, because we can subtract 30 from each one and not change these properties.)\n\nSo from 1 to 30, we have:\n\n$$\n3,5,6,9,12,18,21,24,25,27\n$$\n\nThus there are 10 less than or equal to 30 , and so 100 such positive integers less than or equal to 300 .', 'We proceed by doing a (careful!) count.\n\nThe number of positive multiples of 3 less than or equal to 300 is 100.\n\nThe number of positive multiples of 5 less than or equal to 300 is 60 .\n\nThus, we have 160 candidates, but have included multiples of 15 twice (since 15 is a multiple of each of 3 and 5), and have also included multiples of 10.\n\nThe number of multiples of 15 less than or equal to 300 is 20 , so to remove the multiples of 15 , we must remove 40 from 160 to get 120 positive integers less than or equal to 300 which are multiples of 3 or 5 but not of 15 .\n\n\n\nThis total still included some multiples of 10 that are less or equal to 300 (but not all, since we have already removed 30 , for instance).\n\nIn fact, there are 30 multiples of 10 less than or equal 300,10 of which are multiples of 15 as well (that is, the multiples of 30). So we must remove 20 from the total of 120. We then obtain that there are 100 positive integers less than or equal to 300 which are multiples of 3 or 5 , but not of 10 or 15 .']",['100'],False,,Numerical, 2442,Algebra,,"In the series of odd numbers $1+3+5-7-9-11+13+15+17-19-21-23 \ldots$ the signs alternate every three terms, as shown. What is the sum of the first 300 terms of the series?","['Since the signs alternate every three terms, it makes sense to look at the terms in groups of 6 .\n\nThe sum of the first 6 terms is $1+3+5-7-9-11=-18$.\n\nThe sum of the next 6 terms is $13+15+17-19-21-23=-18$.\n\nIn fact, the sum of each group of 6 terms will be the same, since in each group, 12 has been added to the numerical value of each term when compared to the previous group of 6 , so overall 12 has been added three times and subtracted three times.\n\nSince we are looking for the sum of the first 300 terms, then we are looking at 50 groups of 6 terms, so the sum must be $50(-18)=-900$.']",['-900'],False,,Numerical, 2443,Algebra,,"A two-digit number has the property that the square of its tens digit plus ten times its units digit equals the square of its units digit plus ten times its tens digit. Determine all two-digit numbers which have this property, and are prime numbers.","['Let the two digit integer have tens digit $a$ and units digit $b$. Then the given information tells us\n\n$$\n\\begin{aligned}\na^{2}+10 b & =b^{2}+10 a \\\\\na^{2}-b^{2}-10 a+10 b & =0 \\\\\n(a+b)(a-b)-10(a-b) & =0 \\\\\n(a-b)(a+b-10) & =0\n\\end{aligned}\n$$\n\nand so $a=b$ or $a+b=10$.\n\nSo the possibilities for the integer are 11, 22, 33, 44, 55, 66, 77, 88, 99, 19, 28, 37, 46, 55, $64,73,82,91$. We now must determine which integers in this list are prime.\n\nWe can quickly reject all multiples of 11 bigger than 11 and all of the even integers, to reduce the list to $11,19,37,73,91$.\n\nAll of these are prime, except for $91=13 \\times 7$.\n\nTherefore, the required integers are 11, 19, 37, and 73 .']","['11,19,37,73']",True,,Numerical, 2444,Algebra,,"A lead box contains samples of two radioactive isotopes of iron. Isotope A decays so that after every 6 minutes, the number of atoms remaining is halved. Initially, there are twice as many atoms of isotope $\mathrm{A}$ as of isotope $\mathrm{B}$, and after 24 minutes there are the same number of atoms of each isotope. How long does it take the number of atoms of isotope B to halve?","['In 24 minutes, the number of atoms of isotope $\\mathrm{A}$ has halved 4 times, so the initial number of atoms is $2^{4}=16$ times the number of atoms of isotope $\\mathrm{A}$ at time 24 minutes.\n\nBut there were initially half as many atoms of isotope B as of isotope B, so there was 8 times the final number of atoms. Therefore, the number of atoms of isotope B halves 3 times in the 24 minutes, so it takes 8 minutes for the number of atoms of isotope B to halve.', 'Initially, there is twice as many atoms of isotope A as of isotope B, so let the original numbers of atoms of each be $2 x$ and $x$, respectively.\n\nConsidering isotope A, after 24 minutes, if it loses half of its atoms every 6 minutes, there will be $2 x\\left(\\frac{1}{2}\\right)^{\\frac{24}{6}}$ atoms remaining.\n\nSimilarly for isotope B, after 24 minutes, there will be $x\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}}$ atoms remaining, where $T$ is the length of time (in minutes) that it takes for the number of atoms to halve.\n\nFrom the given information,\n\n$$\n\\begin{aligned}\n2 x\\left(\\frac{1}{2}\\right)^{\\frac{24}{6}} & =x\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}} \\\\\n2\\left(\\frac{1}{2}\\right)^{4} & =\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}} \\\\\n\\left(\\frac{1}{2}\\right)^{3} & =\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}} \\\\\n\\frac{24}{T} & =3 \\\\\nT & =8\n\\end{aligned}\n$$\n\nTherefore, it takes 8 minutes for the number of atoms of isotope B to halve.']",['8'],False,min,Numerical, 2445,Algebra,,"Solve the system of equations: $$ \begin{aligned} & \log _{10}\left(x^{3}\right)+\log _{10}\left(y^{2}\right)=11 \\ & \log _{10}\left(x^{2}\right)-\log _{10}\left(y^{3}\right)=3 \end{aligned} $$","[""Using the facts that $\\log _{10} A+\\log _{10} B=\\log _{10} A B$ and that $\\log _{10} A-\\log _{10} B=\\log _{10} \\frac{A}{B}$, then we can convert the two equations to\n\n$$\n\\begin{aligned}\n\\log _{10}\\left(x^{3} y^{2}\\right) & =11 \\\\\n\\log _{10}\\left(\\frac{x^{2}}{y^{3}}\\right) & =3\n\\end{aligned}\n$$\n\nRaising both sides to the power of 10 , we obtain\n\n$$\n\\begin{aligned}\nx^{3} y^{2} & =10^{11} \\\\\n\\frac{x^{2}}{y^{3}} & =10^{3}\n\\end{aligned}\n$$\n\nTo eliminate the $y$ 's, we raise the first equation to the power 3 and the second to the power 2 to obtain\n\n$$\n\\begin{aligned}\nx^{9} y^{6} & =10^{33} \\\\\n\\frac{x^{4}}{y^{6}} & =10^{6}\n\\end{aligned}\n$$\n\nand multiply to obtain $x^{9} x^{4}=x^{13}=10^{39}=10^{33} 10^{6}$.\n\nTherefore, since $x^{13}=10^{39}$, then $x=10^{3}$.\n\n\n\nSubstituting back into $x^{3} y^{2}=10^{11}$, we get $y^{2}=10^{2}$, and so $y= \\pm 10$. However, substituting into $\\frac{x^{2}}{y^{3}}=10^{3}$ we see that $y$ must be positive, so $y=10$.\n\nTherefore, the solution to the system of equation is $x=10^{3}$ and $y=10$."", 'Since the domain of the logarithm is the positive real numbers, then the quantities $\\log _{10}\\left(x^{3}\\right)$ and $\\log _{10}\\left(y^{3}\\right)$ tell us that $x$ and $y$ are positive.\n\nUsing the fact that $\\log _{10}\\left(a^{b}\\right)=b \\log _{10}(a)$, we rewrite the equations as\n\n$$\n\\begin{aligned}\n& 3 \\log _{10} x+2 \\log _{10} y=11 \\\\\n& 2 \\log _{10} x-3 \\log _{10} y=3\n\\end{aligned}\n$$\n\nWe solve the system of equations for $\\log _{10} x$ and $\\log _{10} y$ by multiplying the first equation by 3 and adding two times the second equation in order to eliminate $\\log _{10} y$. Thus we obtain $13 \\log _{10} x=39$ or $\\log _{10} x=3$.\n\nSubstituting back into the first equation, we obtain $\\log _{10} y=1$.\n\nTherefore, $x=10^{3}$ and $y=10$.']","['$10^{3},10$']",True,,Numerical, 2446,Geometry,,"A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal. In the diagram, $A B C D E F$ is a regular hexagon with an area of 36. The region common to the equilateral triangles $A C E$ and $B D F$ is a hexagon, which is shaded as shown. What is the area of the shaded hexagon? ![](https://cdn.mathpix.com/cropped/2023_12_21_fe07897436972c8371a1g-1.jpg?height=331&width=352&top_left_y=1992&top_left_x=1385)","['We label the vertices of the shaded hexagon $U, V, W, X$, $Y$, and $Z$.\n\nBy symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles. In order to determine the area of the inner hexagon, we determine the ratio of the side lengths of the two hexagons.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_31cf9a291a17adc4339dg-1.jpg?height=439&width=485&top_left_y=1298&top_left_x=1362)\n\nLet the side length of the inner hexagon be $x$. Then $A U=U F=x$. Then triangle $A U F$ has a $120^{\\circ}$ between the two sides of length $x$. If we draw a perpendicular from $U$ to point $P$ on side $A F$, then $U P$ divides\n\n$\\triangle A U F$ into two $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles. Thus, $F P=P A=\\frac{\\sqrt{3}}{2} x$ and so $A F=\\sqrt{3} x$.\n\nSo the ratio of the side lengths of the hexagons is $\\sqrt{3}: 1$, and so the ratio of\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_31cf9a291a17adc4339dg-1.jpg?height=401&width=810&top_left_y=1865&top_left_x=1075)\ntheir areas is $(\\sqrt{3})^{2}: 1=3: 1$.\n\nSince the area of the larger hexagon is 36 , then the area of the inner hexagon is 12.', 'We label the vertices of the hexagon $U, V, W, X, Y$, and $Z$. By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles.\n\nWe also join the opposite vertices of the inner hexagon, ie. we join $U$ to $X, V$ to $Y$, and $W$ to $Z$. (These 3 line segments all meet at a single point, say $O$.) This divides the inner hexagon into 6 small equilateral triangles identical to the\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_78c42ff33d676905b437g-1.jpg?height=439&width=485&top_left_y=301&top_left_x=1400)\nsix earlier mentioned equilateral triangles.\n\nLet the area of one of these triangles be $a$. Then we can label the 12 small equilateral triangles as all having area $a$.\n\nBut triangle $A U F$ also has area $a$, because if we consider triangle $A F V$, then $A U$ is a median (since $F U=A U=U V$ by symmetry) and so divides triangle $A F V$ into two triangles of equal area. Since the area of\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_78c42ff33d676905b437g-1.jpg?height=230&width=369&top_left_y=915&top_left_x=1363)\ntriangle $A U V$ is $a$, then the area of triangle $A U F$ is also $a$.\n\nTherefore, hexagon $A B C D E F$ is divided into 18 equal areas. Thus, $a=2$ since the area of the large hexagon is 36.\n\nSince the area of $U V W X Y Z$ is $6 a$, then its area is 12 .']",['12'],False,,Numerical, 2446,Geometry,,"A regular hexagon is a six-sided figure which has all of its angles equal and all of its side lengths equal. In the diagram, $A B C D E F$ is a regular hexagon with an area of 36. The region common to the equilateral triangles $A C E$ and $B D F$ is a hexagon, which is shaded as shown. What is the area of the shaded hexagon? ","['We label the vertices of the shaded hexagon $U, V, W, X$, $Y$, and $Z$.\n\nBy symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles. In order to determine the area of the inner hexagon, we determine the ratio of the side lengths of the two hexagons.\n\n\n\nLet the side length of the inner hexagon be $x$. Then $A U=U F=x$. Then triangle $A U F$ has a $120^{\\circ}$ between the two sides of length $x$. If we draw a perpendicular from $U$ to point $P$ on side $A F$, then $U P$ divides\n\n$\\triangle A U F$ into two $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangles. Thus, $F P=P A=\\frac{\\sqrt{3}}{2} x$ and so $A F=\\sqrt{3} x$.\n\nSo the ratio of the side lengths of the hexagons is $\\sqrt{3}: 1$, and so the ratio of\n\n\ntheir areas is $(\\sqrt{3})^{2}: 1=3: 1$.\n\nSince the area of the larger hexagon is 36 , then the area of the inner hexagon is 12.', 'We label the vertices of the hexagon $U, V, W, X, Y$, and $Z$. By symmetry, all of the six triangles with two vertices on the inner hexagon and one on the outer hexagon (eg. triangle $U V A$ ) are congruent equilateral triangles.\n\nWe also join the opposite vertices of the inner hexagon, ie. we join $U$ to $X, V$ to $Y$, and $W$ to $Z$. (These 3 line segments all meet at a single point, say $O$.) This divides the inner hexagon into 6 small equilateral triangles identical to the\n\n\nsix earlier mentioned equilateral triangles.\n\nLet the area of one of these triangles be $a$. Then we can label the 12 small equilateral triangles as all having area $a$.\n\nBut triangle $A U F$ also has area $a$, because if we consider triangle $A F V$, then $A U$ is a median (since $F U=A U=U V$ by symmetry) and so divides triangle $A F V$ into two triangles of equal area. Since the area of\n\n\ntriangle $A U V$ is $a$, then the area of triangle $A U F$ is also $a$.\n\nTherefore, hexagon $A B C D E F$ is divided into 18 equal areas. Thus, $a=2$ since the area of the large hexagon is 36.\n\nSince the area of $U V W X Y Z$ is $6 a$, then its area is 12 .']",['12'],False,,Numerical, 2447,Geometry,,"At the Big Top Circus, Herc the Human Cannonball is fired out of the cannon at ground level. (For the safety of the spectators, the cannon is partially buried in the sand floor.) Herc's trajectory is a parabola until he catches the vertical safety net, on his way down, at point $B$. Point $B$ is $64 \mathrm{~m}$ directly above point $C$ on the floor of the tent. If Herc reaches a maximum height of $100 \mathrm{~m}$, directly above a point $30 \mathrm{~m}$ from the cannon, determine the horizontal distance from the cannon to the net. ","['We assign coordinates to the diagram, with the mouth of the cannon at the point $(0,0)$, with the positive $x$-axis in the horizontal direction towards the safety net from the cannon, and the positive $y$ axis upwards from $(0,0)$.\n\nSince Herc reaches his maximum height when his horizontal distance is $30 \\mathrm{~m}$, then the axis of symmetry of the parabola is the line $x=30$. Since the parabola has a root at $x=0$, then the other root must be at $x=60$.\n\nTherefore, the parabola has the form $y=\\operatorname{ax}(x-60)$.\n\nIn order to determine the value of $a$, we note that Herc passes through the point $(30,100)$, and so\n\n\n\n\n\n$$\n\\begin{aligned}\n100 & =30 a(-30) \\\\\na & =-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the equation of the parabola is $y=-\\frac{1}{9} x(x-60)$.\n\n(Alternatively, we could say that since the parabola has its maximum point at $(30,100)$, then it must be of the form $y=a(x-30)^{2}+100$.\n\nSince the parabola passes through $(0,0)$, then we have\n\n$$\n\\begin{aligned}\n& 0=a(0-30)^{2}+100 \\\\\n& 0=900 a+100 \\\\\n& a=-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the parabola has the equation $y=-\\frac{1}{9}(x-30)^{2}+100$.)\n\nWe would like to find the points on the parabola which have $y$-coordinate 64 , so we solve\n\n$$\n\\begin{aligned}\n64 & =-\\frac{1}{9} x(x-60) \\\\\n0 & =x^{2}-60 x+576 \\\\\n0 & =(x-12)(x-48)\n\\end{aligned}\n$$\n\nSince we want a point after Herc has passed his highest point, then $x=48$, ie. the horizontal distance from the cannon to the safety net is $48 \\mathrm{~m}$.']",['48'],False,m,Numerical, 2447,Geometry,,"At the Big Top Circus, Herc the Human Cannonball is fired out of the cannon at ground level. (For the safety of the spectators, the cannon is partially buried in the sand floor.) Herc's trajectory is a parabola until he catches the vertical safety net, on his way down, at point $B$. Point $B$ is $64 \mathrm{~m}$ directly above point $C$ on the floor of the tent. If Herc reaches a maximum height of $100 \mathrm{~m}$, directly above a point $30 \mathrm{~m}$ from the cannon, determine the horizontal distance from the cannon to the net. ![](https://cdn.mathpix.com/cropped/2023_12_21_8c2daaa43e7e62803ac0g-1.jpg?height=794&width=683&top_left_y=202&top_left_x=1122)","['We assign coordinates to the diagram, with the mouth of the cannon at the point $(0,0)$, with the positive $x$-axis in the horizontal direction towards the safety net from the cannon, and the positive $y$ axis upwards from $(0,0)$.\n\nSince Herc reaches his maximum height when his horizontal distance is $30 \\mathrm{~m}$, then the axis of symmetry of the parabola is the line $x=30$. Since the parabola has a root at $x=0$, then the other root must be at $x=60$.\n\nTherefore, the parabola has the form $y=\\operatorname{ax}(x-60)$.\n\nIn order to determine the value of $a$, we note that Herc passes through the point $(30,100)$, and so\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_78c42ff33d676905b437g-1.jpg?height=870&width=651&top_left_y=1468&top_left_x=1233)\n\n\n\n$$\n\\begin{aligned}\n100 & =30 a(-30) \\\\\na & =-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the equation of the parabola is $y=-\\frac{1}{9} x(x-60)$.\n\n(Alternatively, we could say that since the parabola has its maximum point at $(30,100)$, then it must be of the form $y=a(x-30)^{2}+100$.\n\nSince the parabola passes through $(0,0)$, then we have\n\n$$\n\\begin{aligned}\n& 0=a(0-30)^{2}+100 \\\\\n& 0=900 a+100 \\\\\n& a=-\\frac{1}{9}\n\\end{aligned}\n$$\n\nThus, the parabola has the equation $y=-\\frac{1}{9}(x-30)^{2}+100$.)\n\nWe would like to find the points on the parabola which have $y$-coordinate 64 , so we solve\n\n$$\n\\begin{aligned}\n64 & =-\\frac{1}{9} x(x-60) \\\\\n0 & =x^{2}-60 x+576 \\\\\n0 & =(x-12)(x-48)\n\\end{aligned}\n$$\n\nSince we want a point after Herc has passed his highest point, then $x=48$, ie. the horizontal distance from the cannon to the safety net is $48 \\mathrm{~m}$.']",['48'],False,m,Numerical, 2448,Geometry,,"A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$. ","['Since both the circle with its centre on the $y$-axis and the graph of $y=|x|$ are symmetric about the $y$-axis, then for each point of intersection between these two graphs, there should be a corresponding point of intersection symmetrically located across the $y$-axis. Thus, since there are exactly three points of intersection, then one of these points must be on the $y$-axis, ie. has $x$-coordinate 0 . Since this point is on the graph of $y=|x|$, then this point must be $(0,0)$.\n\nSince the circle has centre on the $y$-axis (say, has coordinates $(0, b)$ ), then its radius is equal to $b$ (and $b$ must be positive for there to be three points of intersection). So the circle has equation $x^{2}+(y-b)^{2}=b^{2}$. Where are the other two points of intersection? We consider the points with $x$ positive and use symmetry to get the other point of intersection.\n\n\n\nWhen $x \\geq 0$, then $y=|x|$ has equation $y=x$. Substituting into the equation of the circle,\n\n$$\n\\begin{aligned}\nx^{2}+(x-b)^{2} & =b^{2} \\\\\n2 x^{2}-2 b x & =0 \\\\\n2 x(x-b) & =0\n\\end{aligned}\n$$\n\nTherefore, the points of intersection are $(0,0)$ and $(b, b)$ on the positive side of the $y$-axis, and so at the point $(-b, b)$ on the negative side of the $y$ axis.\n\nThus the points $O, A$ and $B$ are the points $(0,0)$, $(b, b)$ and $(-b, b)$.\n\n\n\nSince the radius of the circle is $b$, then the area of the circle is $\\pi b^{2}$.\n\nTriangle $O A B$ has a base from $(-b, b)$ to $(b, b)$ of length $2 b$, and a height from the line $y=b$ to the point $(0,0)$ of length $b$, and so an area of $\\frac{1}{2} b(2 b)=b^{2}$.\n\nTherefore, the ratio of the area of the triangle to the area of the circle is $b^{2}: \\pi b^{2}=1: \\pi$.']",,True,,, 2448,Geometry,,"A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$. ![](https://cdn.mathpix.com/cropped/2023_12_21_8c2daaa43e7e62803ac0g-1.jpg?height=412&width=483&top_left_y=1062&top_left_x=1336)","['Since both the circle with its centre on the $y$-axis and the graph of $y=|x|$ are symmetric about the $y$-axis, then for each point of intersection between these two graphs, there should be a corresponding point of intersection symmetrically located across the $y$-axis. Thus, since there are exactly three points of intersection, then one of these points must be on the $y$-axis, ie. has $x$-coordinate 0 . Since this point is on the graph of $y=|x|$, then this point must be $(0,0)$.\n\nSince the circle has centre on the $y$-axis (say, has coordinates $(0, b)$ ), then its radius is equal to $b$ (and $b$ must be positive for there to be three points of intersection). So the circle has equation $x^{2}+(y-b)^{2}=b^{2}$. Where are the other two points of intersection? We consider the points with $x$ positive and use symmetry to get the other point of intersection.\n\n\n\nWhen $x \\geq 0$, then $y=|x|$ has equation $y=x$. Substituting into the equation of the circle,\n\n$$\n\\begin{aligned}\nx^{2}+(x-b)^{2} & =b^{2} \\\\\n2 x^{2}-2 b x & =0 \\\\\n2 x(x-b) & =0\n\\end{aligned}\n$$\n\nTherefore, the points of intersection are $(0,0)$ and $(b, b)$ on the positive side of the $y$-axis, and so at the point $(-b, b)$ on the negative side of the $y$ axis.\n\nThus the points $O, A$ and $B$ are the points $(0,0)$, $(b, b)$ and $(-b, b)$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d64fac5709fb52027973g-1.jpg?height=401&width=613&top_left_y=510&top_left_x=1233)\n\nSince the radius of the circle is $b$, then the area of the circle is $\\pi b^{2}$.\n\nTriangle $O A B$ has a base from $(-b, b)$ to $(b, b)$ of length $2 b$, and a height from the line $y=b$ to the point $(0,0)$ of length $b$, and so an area of $\\frac{1}{2} b(2 b)=b^{2}$.\n\nTherefore, the ratio of the area of the triangle to the area of the circle is $b^{2}: \\pi b^{2}=1: \\pi$.']",['证明题，略'],True,,Need_human_evaluate, 2449,Geometry,,"In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$. ","['Since $M$ is the midpoint of a diameter of the circle, $M$ is the centre of the circle.\n\nJoin $P$ to $M$. Since $Q P$ is tangent to the circle, $P M$ is perpendicular to $Q P$.\n\nSince $P M$ and $B M$ are both radii of the circle, then $P M=M B$.\n\n\n\nTherefore, $\\triangle Q P M$ and $\\triangle Q B M$ are congruent (Hypotenuse - Side).\n\nThus, let $\\angle M Q B=\\angle M Q P=\\theta$. So $\\angle Q M B=\\angle Q M P=90^{\\circ}-\\theta$\n\nThen $\\angle P M C=180^{\\circ}-\\angle P M Q-\\angle B M Q=180^{\\circ}-\\left(90^{\\circ}-\\theta\\right)-\\left(90^{\\circ}-\\theta\\right)=2 \\theta$.\n\nBut $\\triangle P M C$ is isosceles with $P M=M C$ since $P M$ and $M C$ are both radii.\n\nTherefore, $\\angle C P M=\\frac{1}{2}\\left(180^{\\circ}-\\angle P M C\\right)=90^{\\circ}-\\theta$.\n\nBut then $\\angle C P M=\\angle P M Q$, and since $P M$ is a transversal between $A C$ and $Q M$, then $Q M$ is parallel to $A C$ because of equal alternating angles.', 'Join $M$ to $P$ and $B$ to $P$.\n\nSince $Q P$ and $Q B$ are tangents to the circle coming from the same point, they have the same length. Since $Q M$ joins the point of intersection of the tangents to the centre of the circle, then by symmetry, $\\angle P Q M=\\angle B Q M$ and $\\angle P M Q=\\angle B M Q$. So let $\\angle P Q M=\\angle B Q M=x$ and $\\angle P M Q=\\angle B M Q=y$.\n\n\n\nLooking at $\\triangle Q M B$, we see that $x+y=90^{\\circ}$, since $\\triangle Q M B$ is right-angled.\n\nNow if we consider the chord $P B$, we see that its central angle is $2 y$, so any angle that it subtends on the circle (eg. $\\angle P C B$ ) is equal to $y$.\n\nThus, $\\angle A C B=\\angle Q M B$, so $Q M$ is parallel to $A C$.', 'Join $P B$.\n\nSince $Q P$ is tangent to the circle, then by the Tangent-Chord Theorem, $\\angle Q P B=\\angle P C B=x$ (ie. the inscribed angle of a chord is equal to the angle between the tangent and chord.\n\n\n\nSince $B C$ is a diameter of the circle, then $\\angle C P B=90^{\\circ}$ and so $\\angle A P B=90^{\\circ}$, whence\n\n$\\angle A P Q=90^{\\circ}-\\angle Q P B=90^{\\circ}-x$.\n\nLooking at $\\triangle A B C$, we see that $\\angle P A Q=90^{\\circ}-x$, so $\\angle P A Q=\\angle A P Q$, and so $A Q=Q P$.\n\nBut $Q P$ and $Q B$ are both tangents to the circle $(Q B$ is tangent since it is perpendicular to a radius), so $Q P=Q B$.\n\nBut then $A Q=Q B$ and $B M=M C$, so $Q$ is the midpoint of $A B$ and $M$ is the midpoint of $B C$. Thus we can conclude that $Q M$ is parallel to $A C$.\n\n(To justify this last statement, we can show very easily that $\\triangle Q B M$ is similar to $\\triangle A B C$, and so show that $\\angle C A B=\\angle M Q B$.)']",,True,,, 2449,Geometry,,"In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$. ![](https://cdn.mathpix.com/cropped/2023_12_21_8c2daaa43e7e62803ac0g-1.jpg?height=290&width=414&top_left_y=1514&top_left_x=1365)","['Since $M$ is the midpoint of a diameter of the circle, $M$ is the centre of the circle.\n\nJoin $P$ to $M$. Since $Q P$ is tangent to the circle, $P M$ is perpendicular to $Q P$.\n\nSince $P M$ and $B M$ are both radii of the circle, then $P M=M B$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d64fac5709fb52027973g-1.jpg?height=338&width=468&top_left_y=1300&top_left_x=1357)\n\nTherefore, $\\triangle Q P M$ and $\\triangle Q B M$ are congruent (Hypotenuse - Side).\n\nThus, let $\\angle M Q B=\\angle M Q P=\\theta$. So $\\angle Q M B=\\angle Q M P=90^{\\circ}-\\theta$\n\nThen $\\angle P M C=180^{\\circ}-\\angle P M Q-\\angle B M Q=180^{\\circ}-\\left(90^{\\circ}-\\theta\\right)-\\left(90^{\\circ}-\\theta\\right)=2 \\theta$.\n\nBut $\\triangle P M C$ is isosceles with $P M=M C$ since $P M$ and $M C$ are both radii.\n\nTherefore, $\\angle C P M=\\frac{1}{2}\\left(180^{\\circ}-\\angle P M C\\right)=90^{\\circ}-\\theta$.\n\nBut then $\\angle C P M=\\angle P M Q$, and since $P M$ is a transversal between $A C$ and $Q M$, then $Q M$ is parallel to $A C$ because of equal alternating angles.', 'Join $M$ to $P$ and $B$ to $P$.\n\nSince $Q P$ and $Q B$ are tangents to the circle coming from the same point, they have the same length. Since $Q M$ joins the point of intersection of the tangents to the centre of the circle, then by symmetry, $\\angle P Q M=\\angle B Q M$ and $\\angle P M Q=\\angle B M Q$. So let $\\angle P Q M=\\angle B Q M=x$ and $\\angle P M Q=\\angle B M Q=y$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_07eb8e1abd34a8bdc722g-1.jpg?height=341&width=477&top_left_y=301&top_left_x=1298)\n\nLooking at $\\triangle Q M B$, we see that $x+y=90^{\\circ}$, since $\\triangle Q M B$ is right-angled.\n\nNow if we consider the chord $P B$, we see that its central angle is $2 y$, so any angle that it subtends on the circle (eg. $\\angle P C B$ ) is equal to $y$.\n\nThus, $\\angle A C B=\\angle Q M B$, so $Q M$ is parallel to $A C$.', 'Join $P B$.\n\nSince $Q P$ is tangent to the circle, then by the Tangent-Chord Theorem, $\\angle Q P B=\\angle P C B=x$ (ie. the inscribed angle of a chord is equal to the angle between the tangent and chord.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_07eb8e1abd34a8bdc722g-1.jpg?height=338&width=482&top_left_y=1029&top_left_x=1296)\n\nSince $B C$ is a diameter of the circle, then $\\angle C P B=90^{\\circ}$ and so $\\angle A P B=90^{\\circ}$, whence\n\n$\\angle A P Q=90^{\\circ}-\\angle Q P B=90^{\\circ}-x$.\n\nLooking at $\\triangle A B C$, we see that $\\angle P A Q=90^{\\circ}-x$, so $\\angle P A Q=\\angle A P Q$, and so $A Q=Q P$.\n\nBut $Q P$ and $Q B$ are both tangents to the circle $(Q B$ is tangent since it is perpendicular to a radius), so $Q P=Q B$.\n\nBut then $A Q=Q B$ and $B M=M C$, so $Q$ is the midpoint of $A B$ and $M$ is the midpoint of $B C$. Thus we can conclude that $Q M$ is parallel to $A C$.\n\n(To justify this last statement, we can show very easily that $\\triangle Q B M$ is similar to $\\triangle A B C$, and so show that $\\angle C A B=\\angle M Q B$.)']",['证明题，略'],True,,Need_human_evaluate, 2450,Geometry,,"Cyclic quadrilateral $A B C D$ has $A B=A D=1, C D=\cos \angle A B C$, and $\cos \angle B A D=-\frac{1}{3}$. Prove that $B C$ is a diameter of the circumscribed circle.","['Consider $\\triangle B A D$. Since we know the lengths of sides $B A$ and $A D$ and the cosine of the angle between them, we can calculate the length of $B D$ using the cosine law:\n\n$$\n\\begin{aligned}\nB D & =\\sqrt{B A^{2}+A D^{2}-2(B A)(A D) \\cos \\angle B A D} \\\\\n& =\\sqrt{2-2\\left(-\\frac{1}{3}\\right)} \\\\\n& =\\sqrt{\\frac{8}{3}}\n\\end{aligned}\n$$\n\n\n\nNext, let $x=\\cos \\angle A B C$. Note that $D C=x$.\n\n\n\nSince $A B C D$ is a cyclic quadrilateral, then $\\angle A D C=180^{\\circ}-\\angle A B C$, and so $\\cos \\angle A D C=-\\cos \\angle A B C=-x$.\n\nSimilarly, $\\cos \\angle B C D=-\\cos \\angle B A D=\\frac{1}{3}$ (since $A B C D$ is a cyclic quadrilateral).\n\nSo we can now use the cosine law simultaneously in $\\triangle A D C$ and $\\triangle A B C$ (since side $A C$ is common) in order to try to solve for $B C$ :\n\n$$\n\\begin{aligned}\n1^{2}+x^{2}-2(1)(x) \\cos \\angle A D C & =1^{2}+B C^{2}-2(1)(B C) \\cos \\angle A B C \\\\\n1^{2}+x^{2}-2(1)(x)(-x) & =1^{2}+B C^{2}-2(1)(B C)(x) \\\\\n0 & =B C^{2}-2(B C) x-3 x^{2} \\\\\n0 & =(B C-3 x)(B C+x)\n\\end{aligned}\n$$\n\nSince $x$ is already a side length, then $x$ must be positive (ie. $\\angle A B C$ is acute), so $B C=3 x$.\n\nSince $\\cos \\angle B C D=\\frac{1}{3}$ and sides $D C$ and $B C$ are in the ratio $1: 3$, then $\\triangle B C D$ must indeed be right-angled at $D$. (We could prove this by using the cosine law to calculate $B D^{2}=8 x^{2}$ and then noticing that $D C^{2}+B D^{2}=B C^{2}$.)\n\nSince $\\triangle B C D$ is right-angled at $D$, then $B C$ is a diameter of the circle.', 'Let $x=\\cos \\angle A B C=C D$, and let $B C=y$.\n\nSince the opposite angles in a cyclic quadrilateral are supplementary, their cosines are negatives of each other. Thus, $\\cos \\angle A D C=-x$ and $\\cos \\angle B C D=\\frac{1}{3}$.\n\n\n\nNext, we use the cosine law four times: twice to calculate $A C^{2}$ in the two triangles $A B C$ and $A D C$, and then twice to calculate $B D^{2}$ in the triangles $A D B$ and $C D B$ to obtain:\n\n$$\n\\begin{aligned}\n1^{2}+x^{2}-2(1)(x) \\cos \\angle A D C & =1^{2}+y^{2}-2(1)(y) \\cos \\angle A B C \\\\\n1+x^{2}-2 x(-x) & =1+y^{2}-2 y(x) \\\\\n0 & =y^{2}-2 x y-3 x^{2} \\\\\n0 & =(y-3 x)(y+x)\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\n1^{2}+1^{2}-2(1)(1) \\cos \\angle B A D & =x^{2}+y^{2}-2 x y \\cos \\angle B C D \\\\\n2-2\\left(-\\frac{1}{3}\\right) & =x^{2}+y^{2}-2 x y\\left(\\frac{1}{3}\\right) \\\\\n\\frac{8}{3} & =x^{2}+y^{2}-\\frac{2}{3} x y\n\\end{aligned}\n$$\n\nFrom the first equation, since $x$ is already a side length and so is positive, we must have that $y=3 x$.\n\nSubstituting into the second equation, we obtain\n\n\n\n$$\n\\begin{aligned}\n& \\frac{8}{3}=x^{2}+(3 x)^{2}-\\frac{2}{3} x(3 x) \\\\\n& \\frac{8}{3}=8 x^{2} \\\\\n& x=\\frac{1}{\\sqrt{3}}\n\\end{aligned}\n$$\n\nsince $x$ must be positive. Thus, since $y=3 x$, then $y=\\sqrt{3}$.\n\nLooking then at $\\triangle B D C$, we have side lengths $B C=\\sqrt{3}, C D=\\frac{1}{\\sqrt{3}}$ and $B D=\\sqrt{\\frac{8}{3}}$. (The last is from the left side of the second cosine law equation.) Thus, $B C^{2}=C D^{2}+B D^{2}$, and so $\\triangle B D C$ is right-angled at $D$, whence $B C$ is a diameter of the circle.']",,True,,, 2451,Algebra,,"A positive integer $n$ is called ""savage"" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that i) the sum of the elements in each of $A, B$, and $C$ is the same, ii) $A$ contains only odd numbers, iii) $B$ contains only even numbers, and iv) C contains every multiple of 3 (and possibly other numbers). Show that 8 is a savage integer.","['To show that 8 is a savage integer, we must partition the set $\\{1,2,3,4,5,6,7,8\\}$ according to the given criteria.\n\nSince the sum of the integers from 1 to 8 is 36 , then the sum of the elements in each of the sets $A, B$, and $C$ must be 12 .\n\n$C$ must contain both 3 and 6 .\n\n$A$ can contain only the numbers $1,5,7$, and may not contain all of these.\n\n$B$ can contain only the numbers $2,4,8$, and may not contain all of these.\n\nSo if we let $C=\\{1,2,3,6\\}, A=\\{5,7\\}$ and $B=\\{4,8\\}$, then these sets have the desired properties.\n\nTherefore, 8 is a savage integer.']",,True,,, 2452,Algebra,,"A positive integer $n$ is called ""savage"" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that i) the sum of the elements in each of $A, B$, and $C$ is the same, ii) $A$ contains only odd numbers, iii) $B$ contains only even numbers, and iv) C contains every multiple of 3 (and possibly other numbers). Prove that if $n$ is an even savage integer, then $\frac{n+4}{12}$ is an integer.","[""We use the strategy of putting all of the multiples of 3 between 1 and $n$ in the set $C$, all of the remaining even numbers in the set $B$, and all of the remaining numbers in the set $A$. The sums of these sets will not likely all be equal, but we then try to adjust the sums to by moving elements out of $A$ and $B$ into $C$ to try to make these sums equal. (Notice that we can't move elements either into $A$ or $B$, or out of $C$.) We will use the notation $|C|$ to denote the sum of the elements of $C$.\n\nSince we are considering the case of $n$ even and we want to examine multiples of 3 less than or equal to $n$, it makes sense to consider $n$ as having one of the three forms $6 k$, $6 k+2$ or $6 k+4$. (These forms allow us to quickly tell what the greatest multiple of 3 less than $n$ is.)\n\nCase 1: $n=6 k$\n\nIn this case, $C$ contains at least the integers $3,6,9, \\ldots, 6 k$, and so the sum of $C$ is greater than one-third of the sum of the integers from 1 to $n$, since if we divide the integers from 1 to $n=6 k$ into groups of 3 consecutive integers starting with 1,2, 3 , then the set $C$ will always contain the largest of the 3 .\n\n\n\nCase 2: $n=6 k+4$\n\nHere, the sum of the integers from 1 to $n=6 k+4$ is $\\frac{1}{2}(6 k+4)(6 k+5)=18 k^{2}+27 k+10=3\\left(6 k^{2}+9 k+3\\right)+1$, which is never divisible by 3 . Therefore, $n$ cannot be savage in this case because the integers from 1 to $n$ cannot be partitioned into 3 sets with equal sums.\n\nCase 3: $n=6 k+2$\n\nHere, the sum of the integers from 1 to $n=6 k+2$ is\n\n$\\frac{1}{2}(6 k+2)(6 k+3)=18 k^{2}+15 k+3$, so the sum of the elements of each of the sets $A, B$ and $C$ should be $6 k^{2}+5 k+1$, so that the sums are equal.\n\nIn this case $C$, contains at least the integers $3,6,9, \\ldots, 6 k$, and so $|C| \\geq 3+6+9+\\cdots 6 k=3(1+2+3+\\cdots+2 k)=3\\left(\\frac{1}{2}(2 k)(2 k+1)\\right)=6 k^{2}+3 k$\n\nThe set $A$ contains at most the integers $1,3,5,7, \\ldots, 6 k+1$, but does not contain the odd multiples of 3 less than $n$, ie. the integers $3,9,15, \\ldots, 6 k-3$. Therefore, $|A| \\leq(1+3+5+\\cdots+6 k+1)-(3+9+\\cdots+6 k-3)$\n\n$=\\frac{1}{2}(3 k+1)[1+6 k+1]-\\frac{1}{2}(k)[3+6 k-3]$\n\n$=(3 k+1)(3 k+1)-k(3 k)$\n\n$=6 k^{2}+6 k+1$\n\n(To compute the sum of each of these arithmetic sequences, we use the fact that the sum of an arithmetic sequence is equal to half of the number of terms times the sum of the first and last terms.)\n\nThe set $B$ contains at most the integers $2,4,6,8, \\ldots, 6 k+2$, but does not contain the even multiples of 3 less than $n$, ie. the integers $6,12, \\ldots, 6 k$. Therefore, $|B| \\leq(2+4+6+\\cdots+6 k+2)-(6+12+\\cdots+6 k)$\n\n$=\\frac{1}{2}(3 k+1)[2+6 k+2]-\\frac{1}{2}(k)[6+6 k]$\n\n$=(3 k+1)(3 k+2)-k(3 k+3)$\n\n$=6 k^{2}+6 k+2$\n\nThus, the set $C$ is $2 k+1$ short of the desired sum, while the set $A$ has a sum that is $k$ too big and the set $B$ has a sum that is $k+1$ too big.\n\nSo in order to correct this, we would like to move elements from $A$ adding to $k$, and elements from $B$ which add to $k+1$ all to set $C$.\n\n\n\nSince we are assuming that $n$ is savage, then this is possible, which means that $k+1$ must be even since every element in $B$ is even, so the sum of any number of elements of $B$ is even.\n\nTherefore, $k$ is odd, and so $k=2 l+1$ for some integer $l$, and so\n\n$n=6(2 l+1)+2=12 l+8$, ie. $\\frac{n+4}{12}$ is an integer.\n\nHaving examined all cases, we see that if $n$ is an even savage integer, then $\\frac{n+4}{12}$ is an integer.""]",,True,,, 2453,Algebra,,"A positive integer $n$ is called ""savage"" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that i) the sum of the elements in each of $A, B$, and $C$ is the same, ii) $A$ contains only odd numbers, iii) $B$ contains only even numbers, and iv) C contains every multiple of 3 (and possibly other numbers). Determine all even savage integers less than 100.","[""First, we prove lemma (b): if $n$ is an even savage integer, then $\\frac{n+4}{12}$ is an integer.\n\nProof of lemma (b):\nWe use the strategy of putting all of the multiples of 3 between 1 and $n$ in the set $C$, all of the remaining even numbers in the set $B$, and all of the remaining numbers in the set $A$. The sums of these sets will not likely all be equal, but we then try to adjust the sums to by moving elements out of $A$ and $B$ into $C$ to try to make these sums equal. (Notice that we can't move elements either into $A$ or $B$, or out of $C$.) We will use the notation $|C|$ to denote the sum of the elements of $C$.\n\nSince we are considering the case of $n$ even and we want to examine multiples of 3 less than or equal to $n$, it makes sense to consider $n$ as having one of the three forms $6 k$, $6 k+2$ or $6 k+4$. (These forms allow us to quickly tell what the greatest multiple of 3 less than $n$ is.)\n\nCase 1: $n=6 k$\n\nIn this case, $C$ contains at least the integers $3,6,9, \\ldots, 6 k$, and so the sum of $C$ is greater than one-third of the sum of the integers from 1 to $n$, since if we divide the integers from 1 to $n=6 k$ into groups of 3 consecutive integers starting with 1,2, 3 , then the set $C$ will always contain the largest of the 3 .\n\n\n\nCase 2: $n=6 k+4$\n\nHere, the sum of the integers from 1 to $n=6 k+4$ is $\\frac{1}{2}(6 k+4)(6 k+5)=18 k^{2}+27 k+10=3\\left(6 k^{2}+9 k+3\\right)+1$, which is never divisible by 3 . Therefore, $n$ cannot be savage in this case because the integers from 1 to $n$ cannot be partitioned into 3 sets with equal sums.\n\nCase 3: $n=6 k+2$\n\nHere, the sum of the integers from 1 to $n=6 k+2$ is\n\n$\\frac{1}{2}(6 k+2)(6 k+3)=18 k^{2}+15 k+3$, so the sum of the elements of each of the sets $A, B$ and $C$ should be $6 k^{2}+5 k+1$, so that the sums are equal.\n\nIn this case $C$, contains at least the integers $3,6,9, \\ldots, 6 k$, and so $|C| \\geq 3+6+9+\\cdots 6 k=3(1+2+3+\\cdots+2 k)=3\\left(\\frac{1}{2}(2 k)(2 k+1)\\right)=6 k^{2}+3 k$\n\nThe set $A$ contains at most the integers $1,3,5,7, \\ldots, 6 k+1$, but does not contain the odd multiples of 3 less than $n$, ie. the integers $3,9,15, \\ldots, 6 k-3$. Therefore, $|A| \\leq(1+3+5+\\cdots+6 k+1)-(3+9+\\cdots+6 k-3)$\n\n$=\\frac{1}{2}(3 k+1)[1+6 k+1]-\\frac{1}{2}(k)[3+6 k-3]$\n\n$=(3 k+1)(3 k+1)-k(3 k)$\n\n$=6 k^{2}+6 k+1$\n\n(To compute the sum of each of these arithmetic sequences, we use the fact that the sum of an arithmetic sequence is equal to half of the number of terms times the sum of the first and last terms.)\n\nThe set $B$ contains at most the integers $2,4,6,8, \\ldots, 6 k+2$, but does not contain the even multiples of 3 less than $n$, ie. the integers $6,12, \\ldots, 6 k$. Therefore, $|B| \\leq(2+4+6+\\cdots+6 k+2)-(6+12+\\cdots+6 k)$\n\n$=\\frac{1}{2}(3 k+1)[2+6 k+2]-\\frac{1}{2}(k)[6+6 k]$\n\n$=(3 k+1)(3 k+2)-k(3 k+3)$\n\n$=6 k^{2}+6 k+2$\n\nThus, the set $C$ is $2 k+1$ short of the desired sum, while the set $A$ has a sum that is $k$ too big and the set $B$ has a sum that is $k+1$ too big.\n\nSo in order to correct this, we would like to move elements from $A$ adding to $k$, and elements from $B$ which add to $k+1$ all to set $C$.\n\n\n\nSince we are assuming that $n$ is savage, then this is possible, which means that $k+1$ must be even since every element in $B$ is even, so the sum of any number of elements of $B$ is even.\n\nTherefore, $k$ is odd, and so $k=2 l+1$ for some integer $l$, and so\n\n$n=6(2 l+1)+2=12 l+8$, ie. $\\frac{n+4}{12}$ is an integer.\n\nHaving examined all cases, we see that if $n$ is an even savage integer, then $\\frac{n+4}{12}$ is an integer.\n\n\nFrom the proof of (b) above, the only possible even savage integers less than 100 are those satisfying the condition that $\\frac{n+4}{12}$ is an integer, ie. $8,20,32,44,56,68,80,92$. We already know that 8 is savage, so we examine the remaining 7 possibilities.\n\nWe make a table of the possibilities, using the notation from the proof of (b):\n\n| $n$ | $k$ | Sum of elements
to remove from $A$ | Sum of elements
to remove from $B$ | Possible? |\n| :---: | :---: | :---: | :---: | :---: |\n| 20 | 3 | 3 | 4 | No - cannot remove a sum of 3 from
A. |\n| 32 | 5 | 5 | 6 | Yes - remove 5 from $A, 2$ and 4
from $B$ |\n| 44 | 7 | 7 | 8 | Yes - remove 7 from $A, 8$ from $B$ |\n| 56 | 9 | 9 | 10 | No - cannot remove a sum of 9 from
A. |\n| 68 | 11 | 11 | 12 | Yes - remove 11 from $A, 4$ and 8
from $B$ |\n| 80 | 13 | 13 | 14 | Yes - remove 13 from $A, 14$ from $B$ |\n| 92 | 15 | 15 | 16 | No - cannot remove a sum of 15
from $A$ (since could only use $1,5,7$,
11,13 ) |\n\nTherefore, the only even savage integers less than 100 are 8, 32, 44, 68 and 80.""]","['8,32,44,68,80']",True,,Numerical, 2454,Combinatorics,,"Tanner has two identical dice. Each die has six faces which are numbered 2, 3, 5, $7,11,13$. When Tanner rolls the two dice, what is the probability that the sum of the numbers on the top faces is a prime number?","['We make a table of the 36 possible combinations of rolls and the resulting sums:\n\n| | 2 | 3 | 5 | 7 | 11 | 13 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 4 | 5 | 7 | 9 | 13 | 15 |\n| 3 | 5 | 6 | 8 | 10 | 14 | 16 |\n| 5 | 7 | 8 | 10 | 12 | 16 | 18 |\n| 7 | 9 | 10 | 12 | 14 | 18 | 20 |\n| 11 | 13 | 14 | 16 | 18 | 22 | 24 |\n| 13 | 15 | 16 | 18 | 20 | 24 | 26 |\n\nOf the 36 entries in the table, 6 are prime numbers (two entries each of 5, 7 and 13).\n\nTherefore, the probability that the sum is a prime number is $\\frac{6}{36}$ or $\\frac{1}{6}$.\n\n(Note that each sum is at least 4 and so must be odd to be prime. Since odd plus odd equals even, then the only possibilities that really need to be checked are even plus odd and odd plus even (that is, the first row and first column of the table).)']",['$\\frac{1}{6}$'],False,,Numerical, 2455,Geometry,,"In the diagram, $V$ is the vertex of the parabola with equation $y=-x^{2}+4 x+1$. Also, $A$ and $B$ are the points of intersection of the parabola and the line with equation $y=-x+1$. Determine the value of $A V^{2}+B V^{2}-A B^{2}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_859e4fcfbb4e54a90d70g-1.jpg?height=431&width=521&top_left_y=847&top_left_x=1298)","['First, we find the coordinates of $V$.\n\nTo do this, we use the given equation for the parabola and complete the square:\n\n$y=-x^{2}+4 x+1=-\\left(x^{2}-4 x-1\\right)=-\\left(x^{2}-4 x+2^{2}-2^{2}-1\\right)=-\\left((x-2)^{2}-5\\right)=-(x-2)^{2}+5$\n\nTherefore, the coordinates of the vertex $V$ are $(2,5)$.\n\nNext, we find the coordinates of $A$ and $B$.\n\nNote that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^{2}+4 x+1$.\n\nWe equate $y$-values to obtain $-x+1=-x^{2}+4 x+1$ or $x^{2}-5 x=0$ or $x(x-5)=0$.\n\nTherefore, $x=0$ or $x=5$.\n\nIf $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.\n\nIf $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.\n\n\n\nWe now have the points $V(2,5), A(0,1), B(5,-4)$.\n\nThis gives\n\n$$\n\\begin{aligned}\nA V^{2} & =(0-2)^{2}+(1-5)^{2}=20 \\\\\nB V^{2} & =(5-2)^{2}+(-4-5)^{2}=90 \\\\\nA B^{2} & =(0-5)^{2}+(1-(-4))^{2}=50\n\\end{aligned}\n$$\n\nand so $A V^{2}+B V^{2}-A B^{2}=20+90-50=60$.']",['60'],False,,Numerical, 2455,Geometry,,"In the diagram, $V$ is the vertex of the parabola with equation $y=-x^{2}+4 x+1$. Also, $A$ and $B$ are the points of intersection of the parabola and the line with equation $y=-x+1$. Determine the value of $A V^{2}+B V^{2}-A B^{2}$. ","['First, we find the coordinates of $V$.\n\nTo do this, we use the given equation for the parabola and complete the square:\n\n$y=-x^{2}+4 x+1=-\\left(x^{2}-4 x-1\\right)=-\\left(x^{2}-4 x+2^{2}-2^{2}-1\\right)=-\\left((x-2)^{2}-5\\right)=-(x-2)^{2}+5$\n\nTherefore, the coordinates of the vertex $V$ are $(2,5)$.\n\nNext, we find the coordinates of $A$ and $B$.\n\nNote that $A$ and $B$ are the points of intersection of the line with equation $y=-x+1$ and the parabola with equation $y=-x^{2}+4 x+1$.\n\nWe equate $y$-values to obtain $-x+1=-x^{2}+4 x+1$ or $x^{2}-5 x=0$ or $x(x-5)=0$.\n\nTherefore, $x=0$ or $x=5$.\n\nIf $x=0$, then $y=-x+1=1$, and so $A$ (which is on the $y$-axis) has coordinates $(0,1)$.\n\nIf $x=5$, then $y=-x+1=-4$, and so $B$ has coordinates $(5,-4)$.\n\n\n\nWe now have the points $V(2,5), A(0,1), B(5,-4)$.\n\nThis gives\n\n$$\n\\begin{aligned}\nA V^{2} & =(0-2)^{2}+(1-5)^{2}=20 \\\\\nB V^{2} & =(5-2)^{2}+(-4-5)^{2}=90 \\\\\nA B^{2} & =(0-5)^{2}+(1-(-4))^{2}=50\n\\end{aligned}\n$$\n\nand so $A V^{2}+B V^{2}-A B^{2}=20+90-50=60$.']",['60'],False,,Numerical, 2456,Geometry,,"In the diagram, $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$. The piece of pizza is placed on a circular pan with $A, B$ and $C$ touching the circumference of the pan, as shown. What fraction of the pan is covered by the piece of pizza? ","['Since $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \\mathrm{~cm}$, then $A C=A B=20 \\mathrm{~cm}$.\n\nWe are also told that $\\angle C A B=90^{\\circ}$ (one-quarter of $360^{\\circ}$ ).\n\nSince $\\angle C A B=90^{\\circ}$ and $A, B$ and $C$ are all on the circumference of the circle, then $C B$ is a diameter of the pan. (This is a property of circles: if $X, Y$ and $Z$ are three points on a circle with $\\angle Z X Y=90^{\\circ}$, then $Y Z$ must be a diameter of the circle.)\n\nSince $\\triangle C A B$ is right-angled and isosceles, then $C B=\\sqrt{2} A C=20 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the radius of the circular plate is $\\frac{1}{2} C B$ or $10 \\sqrt{2} \\mathrm{~cm}$.\n\nThus, the area of the circular pan is $\\pi(10 \\sqrt{2} \\mathrm{~cm})^{2}=200 \\pi \\mathrm{cm}^{2}$.\n\nThe area of the slice of pizza is one-quarter of the area of a circle with radius $20 \\mathrm{~cm}$, or $\\frac{1}{4} \\pi(20 \\mathrm{~cm})^{2}=100 \\pi \\mathrm{cm}^{2}$.\n\nFinally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\\frac{100 \\pi \\mathrm{cm}^{2}}{200 \\pi \\mathrm{cm}^{2}}=\\frac{1}{2}$.']",['$\\frac{1}{2}$'],False,,Numerical, 2456,Geometry,,"In the diagram, $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \mathrm{~cm}$. The piece of pizza is placed on a circular pan with $A, B$ and $C$ touching the circumference of the pan, as shown. What fraction of the pan is covered by the piece of pizza? ![](https://cdn.mathpix.com/cropped/2023_12_21_859e4fcfbb4e54a90d70g-1.jpg?height=401&width=415&top_left_y=1336&top_left_x=1256)","['Since $A B C$ is a quarter of a circular pizza with centre $A$ and radius $20 \\mathrm{~cm}$, then $A C=A B=20 \\mathrm{~cm}$.\n\nWe are also told that $\\angle C A B=90^{\\circ}$ (one-quarter of $360^{\\circ}$ ).\n\nSince $\\angle C A B=90^{\\circ}$ and $A, B$ and $C$ are all on the circumference of the circle, then $C B$ is a diameter of the pan. (This is a property of circles: if $X, Y$ and $Z$ are three points on a circle with $\\angle Z X Y=90^{\\circ}$, then $Y Z$ must be a diameter of the circle.)\n\nSince $\\triangle C A B$ is right-angled and isosceles, then $C B=\\sqrt{2} A C=20 \\sqrt{2} \\mathrm{~cm}$.\n\nTherefore, the radius of the circular plate is $\\frac{1}{2} C B$ or $10 \\sqrt{2} \\mathrm{~cm}$.\n\nThus, the area of the circular pan is $\\pi(10 \\sqrt{2} \\mathrm{~cm})^{2}=200 \\pi \\mathrm{cm}^{2}$.\n\nThe area of the slice of pizza is one-quarter of the area of a circle with radius $20 \\mathrm{~cm}$, or $\\frac{1}{4} \\pi(20 \\mathrm{~cm})^{2}=100 \\pi \\mathrm{cm}^{2}$.\n\nFinally, the fraction of the pan that is covered is the area of the slice of pizza divided by the area of the pan, or $\\frac{100 \\pi \\mathrm{cm}^{2}}{200 \\pi \\mathrm{cm}^{2}}=\\frac{1}{2}$.']",['$\\frac{1}{2}$'],False,,Numerical, 2457,Geometry,,"The deck $A B$ of a sailboat is $8 \mathrm{~m}$ long. Rope extends at an angle of $60^{\circ}$ from $A$ to the top $(M)$ of the mast of the boat. More rope extends at an angle of $\theta$ from $B$ to a point $P$ that is $2 \mathrm{~m}$ below $M$, as shown. Determine the height $M F$ of the mast, in terms of $\theta$. ","['Suppose that the length of $A F$ is $x \\mathrm{~m}$.\n\nSince the length of $A B$ is $8 \\mathrm{~m}$, then the length of $F B$ is $(8-x) \\mathrm{m}$.\n\nSince $\\triangle M A F$ is right-angled and has an angle of $60^{\\circ}$, then it is $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $M F=\\sqrt{3} A F$, since $M F$ is opposite the $60^{\\circ}$ angle and $A F$ is opposite the $30^{\\circ}$ angle.\n\nThus, $M F=\\sqrt{3} x \\mathrm{~m}$.\n\nSince $M P=2 \\mathrm{~m}$, then $P F=M F-M P=(\\sqrt{3} x-2) \\mathrm{m}$.\n\nWe can now look at $\\triangle B F P$ which is right-angled at $F$.\n\nWe have\n\n$$\n\\tan \\theta=\\frac{P F}{F B}=\\frac{(\\sqrt{3} x-2) \\mathrm{m}}{(8-x) \\mathrm{m}}=\\frac{\\sqrt{3} x-2}{8-x}\n$$\n\nTherefore, $(8-x) \\tan \\theta=\\sqrt{3} x-2$ or $8 \\tan \\theta+2=\\sqrt{3} x+(\\tan \\theta) x$.\n\nThis gives $8 \\tan \\theta+2=x(\\sqrt{3}+\\tan \\theta)$ or $x=\\frac{8 \\tan \\theta+2}{\\tan \\theta+\\sqrt{3}}$.\n\nFinally, $M F=\\sqrt{3} x=\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}} \\mathrm{~m}$.\n\n']",['$\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}}$'],False,\mathrm{~m},Expression, 2457,Geometry,,"The deck $A B$ of a sailboat is $8 \mathrm{~m}$ long. Rope extends at an angle of $60^{\circ}$ from $A$ to the top $(M)$ of the mast of the boat. More rope extends at an angle of $\theta$ from $B$ to a point $P$ that is $2 \mathrm{~m}$ below $M$, as shown. Determine the height $M F$ of the mast, in terms of $\theta$. ![](https://cdn.mathpix.com/cropped/2023_12_21_859e4fcfbb4e54a90d70g-1.jpg?height=377&width=371&top_left_y=1774&top_left_x=1254)","['Suppose that the length of $A F$ is $x \\mathrm{~m}$.\n\nSince the length of $A B$ is $8 \\mathrm{~m}$, then the length of $F B$ is $(8-x) \\mathrm{m}$.\n\nSince $\\triangle M A F$ is right-angled and has an angle of $60^{\\circ}$, then it is $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nTherefore, $M F=\\sqrt{3} A F$, since $M F$ is opposite the $60^{\\circ}$ angle and $A F$ is opposite the $30^{\\circ}$ angle.\n\nThus, $M F=\\sqrt{3} x \\mathrm{~m}$.\n\nSince $M P=2 \\mathrm{~m}$, then $P F=M F-M P=(\\sqrt{3} x-2) \\mathrm{m}$.\n\nWe can now look at $\\triangle B F P$ which is right-angled at $F$.\n\nWe have\n\n$$\n\\tan \\theta=\\frac{P F}{F B}=\\frac{(\\sqrt{3} x-2) \\mathrm{m}}{(8-x) \\mathrm{m}}=\\frac{\\sqrt{3} x-2}{8-x}\n$$\n\nTherefore, $(8-x) \\tan \\theta=\\sqrt{3} x-2$ or $8 \\tan \\theta+2=\\sqrt{3} x+(\\tan \\theta) x$.\n\nThis gives $8 \\tan \\theta+2=x(\\sqrt{3}+\\tan \\theta)$ or $x=\\frac{8 \\tan \\theta+2}{\\tan \\theta+\\sqrt{3}}$.\n\nFinally, $M F=\\sqrt{3} x=\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}} \\mathrm{~m}$.']",['$\\frac{8 \\sqrt{3} \\tan \\theta+2 \\sqrt{3}}{\\tan \\theta+\\sqrt{3}} \\mathrm{~m}$'],False,,Need_human_evaluate, 2458,Algebra,,"If $\frac{1}{\cos x}-\tan x=3$, what is the numerical value of $\sin x$ ?","['Beginning with the given equation, we have\n\n$$\n\\begin{aligned}\n\\frac{1}{\\cos x}-\\tan x & =3 \\\\\n\\frac{1}{\\cos x}-\\frac{\\sin x}{\\cos x} & =3 \\\\\n1-\\sin x & =3 \\cos x \\quad(\\text { since } \\cos x \\neq 0) \\\\\n(1-\\sin x)^{2} & =9 \\cos ^{2} x \\quad \\text { (squaring both sides) } \\\\\n1-2 \\sin x+\\sin ^{2} x & =9\\left(1-\\sin ^{2} x\\right) \\\\\n10 \\sin ^{2} x-2 \\sin x-8 & =0 \\\\\n5 \\sin ^{2} x-\\sin x-4 & =0 \\\\\n(5 \\sin x+4)(\\sin x-1) & =0\n\\end{aligned}\n$$\n\nTherefore, $\\sin x=-\\frac{4}{5}$ or $\\sin x=1$.\n\nIf $\\sin x=1$, then $\\cos x=0$ and $\\tan x$ is undefined, which is inadmissible in the original equation.\n\nTherefore, $\\sin x=-\\frac{4}{5}$.\n\n(We can check that if $\\sin x=-\\frac{4}{5}$, then $\\cos x= \\pm \\frac{3}{5}$ and the possibility that $\\cos x=\\frac{3}{5}$ satisfies the original equation, since in this case $\\frac{1}{\\cos x}=\\frac{5}{3}$ and $\\tan x=-\\frac{4}{3}$ and the difference between these fractions is 3 .)']",['$-\\frac{4}{5}$'],False,,Numerical, 2459,Algebra,,"Determine all linear functions $f(x)=a x+b$ such that if $g(x)=f^{-1}(x)$ for all values of $x$, then $f(x)-g(x)=44$ for all values of $x$. (Note: $f^{-1}$ is the inverse function of $f$.)","['Since $f(x)=a x+b$, we can determine an expression for $g(x)=f^{-1}(x)$ by letting $y=f(x)$ to obtain $y=a x+b$. We then interchange $x$ and $y$ to obtain $x=a y+b$ which we solve for $y$ to obtain $a y=x-b$ or $y=\\frac{x}{a}-\\frac{b}{a}$.\n\nTherefore, $f^{-1}(x)=\\frac{x}{a}-\\frac{b}{a}$.\n\nNote that $a \\neq 0$. (This makes sense since the function $f(x)=b$ has a graph which is a horizontal line, and so cannot be invertible.)\n\nTherefore, the equation $f(x)-g(x)=44$ becomes $(a x+b)-\\left(\\frac{x}{a}-\\frac{b}{a}\\right)=44$ or $\\left(a-\\frac{1}{a}\\right) x+\\left(b+\\frac{b}{a}\\right)=44=0 x+44$, and this equation is true for all $x$.\n\nWe can proceed in two ways.\n\nMethod \\#1: Comparing coefficients\n\nSince the equation\n\n$$\n\\left(a-\\frac{1}{a}\\right) x+\\left(b+\\frac{b}{a}\\right)=0 x+44\n$$\n\nis true for all $x$, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.\n\nTherefore, $a-\\frac{1}{a}=0$ and $b+\\frac{b}{a}=44$.\n\nFrom the first of these equations, we obtain $a=\\frac{1}{a}$ or $a^{2}=1$, which gives $a=1$ or $a=-1$. If $a=1$, the equation $b+\\frac{b}{a}=44$ becomes $b+b=44$, which gives $b=22$.\n\n\n\nIf $a=-1$, the equation $b+\\frac{b}{a}=44$ becomes $b-b=44$, which is not possible.\n\nTherefore, we must have $a=1$ and $b=22$, and so $f(x)=x+22$.\n\nMethod \\#2: Trying specific values for $x$\n\nSince the equation\n\n$$\n\\left(a-\\frac{1}{a}\\right) x+\\left(b+\\frac{b}{a}\\right)=0 x+44\n$$\n\nis true for all values of $x$, then it must be true for any specific values of $x$ that we choose.\n\nChoosing $x=0$, we obtain $0+\\left(b+\\frac{b}{a}\\right)=44$ or $b+\\frac{b}{a}=44$.\n\nChoosing $x=b$, we obtain $\\left(a-\\frac{1}{a}\\right) b+\\left(b+\\frac{b}{a}\\right)=44$ or $a b+b=44$.\n\nWe can rearrange the first of these equations to get $\\frac{a b+b}{a}=44$.\n\nUsing the second equation, we obtain $\\frac{44}{a}=44$ or $a=1$.\n\nSince $a=1$, then $a b+b=44$ gives $2 b=44$ or $b=22$.\n\nThus, $f(x)=x+22$.\n\nIn summary, the only linear function $f$ for which the given equation is true for all $x$ is $f(x)=x+22$.']",['$f(x)=x+22$'],False,,Expression, 2460,Number Theory,,"Determine all pairs $(a, b)$ of positive integers for which $a^{3}+2 a b=2013$.","['First, we factor the left side of the given equation to obtain $a\\left(a^{2}+2 b\\right)=2013$.\n\nNext, we factor the integer 2013 as $2013=3 \\times 671=3 \\times 11 \\times 61$. Note that each of 3,11 and 61 is prime, so we can factor 2013 no further. (We can find the factors of 3 and 11 using tests for divisibility by 3 and 11, or by systematic trial and error.)\n\nSince $2013=3 \\times 11 \\times 61$, then the positive divisors of 2013 are\n\n$$\n1,3,11,33,61,183,671,2013\n$$\n\nSince $a$ and $b$ are positive integers, then $a$ and $a^{2}+2 b$ are both positive integers.\n\nSince $a$ and $b$ are positive integers, then $a^{2} \\geq a$ and $2 b>0$, so $a^{2}+2 b>a$.\n\nSince $a\\left(a^{2}+2 b\\right)=2013$, then $a$ and $a^{2}+2 b$ must be a divisor pair of 2013 (that is, a pair of positive integers whose product is 2013) with $a0$ and $b>0$, then $a=2 b$ which gives $2^{x}=2 \\cdot 3^{x}$.\n\nTaking $\\log$ of both sides, we obtain $x \\log 2=\\log 2+x \\log 3$ and so $x(\\log 2-\\log 3)=\\log 2$ or $x=\\frac{\\log 2}{\\log 2-\\log 3}$.', 'We successively manipulate the given equation to produce equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{2}\\left(2^{x-1}+3^{x+1}\\right) & =2 x-\\log _{2}\\left(3^{x}\\right) \\\\\n\\log _{2}\\left(2^{x-1}+3^{x+1}\\right)+\\log _{2}\\left(3^{x}\\right) & =2 x \\\\\n\\log _{2}\\left(\\left(2^{x-1}+3^{x+1}\\right) 3^{x}\\right) & =2 x \\quad\\left(\\text { using } \\log _{2} A+\\log _{2} B=\\log _{2} A B\\right) \\\\\n\\left(2^{x-1}+3^{x+1}\\right) 3^{x} & =2^{2 x} \\quad \\text { (exponentiating both sides) } \\\\\n2^{-1} 2^{x} 3^{x}+3^{1} 3^{x} 3^{x} & =2^{2 x} \\\\\n\\frac{1}{2} \\cdot 2^{x} 3^{x}+3 \\cdot 3^{2 x} & =2^{2 x} \\\\\n2^{x} 3^{x}+6 \\cdot 3^{2 x} & \\left.=2 \\cdot 2^{2 x} \\quad \\text { (multiplying by } 2\\right) \\\\\n2^{x} 3^{x} 2^{-2 x}+6 \\cdot 3^{2 x} 2^{-2 x} & \\left.=2 \\quad \\text { (dividing both sides by } 2^{2 x} \\neq 0\\right) \\\\\n2^{-x} 3^{x}+6 \\cdot 3^{2 x} 2^{-2 x} & =2 \\\\\n\\left(\\frac{3}{2}\\right)^{x}+6\\left(\\frac{3}{2}\\right)^{2 x} & =2\n\\end{aligned}\n$$\n\nNext, we make the substitution $t=\\left(\\frac{3}{2}\\right)^{x}$, noting that $\\left(\\frac{3}{2}\\right)^{2 x}=\\left(\\left(\\frac{3}{2}\\right)^{x}\\right)^{2}=t^{2}$.\n\nThus, we obtain the equivalent equations\n\n$$\n\\begin{aligned}\nt+6 t^{2} & =2 \\\\\n6 t^{2}+t-2 & =0 \\\\\n(3 t+2)(2 t-1) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=-\\frac{2}{3}$ or $t=\\frac{1}{2}$.\n\nSince $t=\\left(\\frac{3}{2}\\right)^{x}>0$, then we must have $t=\\left(\\frac{3}{2}\\right)^{x}=\\frac{1}{2}$.\n\nThus,\n\n$$\nx=\\log _{3 / 2}(1 / 2)=\\frac{\\log (1 / 2)}{\\log (3 / 2)}=\\frac{\\log 1-\\log 2}{\\log 3-\\log 2}=\\frac{-\\log 2}{\\log 3-\\log 2}=\\frac{\\log 2}{\\log 2-\\log 3}\n$$']",['$\\frac{\\log 2}{\\log 2-\\log 3}$'],False,,Numerical, 2462,Geometry,,"A large square $A B C D$ is drawn, with a second smaller square $P Q R S$ completely inside it so that the squares do not touch. Line segments $A P, B Q, C R$, and $D S$ are drawn, dividing the region between the squares into four nonoverlapping convex quadrilaterals, as shown. If the sides of $P Q R S$ are not parallel to the sides of $A B C D$, prove that the sum of the areas of quadrilaterals $A P S D$ and $B C R Q$ equals the sum of the areas of quadrilaterals $A B Q P$ and $C D S R$. (Note: A convex quadrilateral is a quadrilateral in which the measure of each of the four interior angles is less than $180^{\circ}$.) ","['We begin by ""boxing in"" square $P Q R S$ by drawing horizontal and vertical lines through its vertices to form rectangle $W X Y Z$, as shown. (Because the four quadrilaterals $A B Q P$, $B C R Q, C D S R$, and $D A P S$ are convex, there will not be any configurations that look substantially different from this the diagram below.) We also label the various areas.\n\n\n\nSince $W X$ is parallel to $A B$, then quadrilateral $A B X W$ is a trapezoid. Similarly, quadrilaterals $B C Y X, C D Z Y$, and $D A W Z$ are trapezoids.\n\nWe use the notation $|A B Q P|$ to denote the area of quadrilateral $A B Q P$, and similar notation for other areas.\n\nSuppose that the side length of square $A B C D$ is $x$ and the side length of square $P Q R S$ is $y$.\n\nAlso, we let $\\angle W P Q=\\theta$.\n\nSince each of $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ is right-angled and each of the four angles of square $P Q R S$ is $90^{\\circ}$, then $\\angle W P Q=\\angle X Q R=\\angle Y R S=\\angle Z S P=\\theta$. This is because, for example,\n\n$\\angle X Q R=180^{\\circ}-\\angle P Q R-\\angle W Q P=90^{\\circ}-\\left(180^{\\circ}-\\angle P W Q-\\angle W P Q\\right)=90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$\n\nThis fact, together with the fact that $P Q=Q R=R S=S P=y$, allows us to conclude that the four triangles $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ are congruent.\n\nIn particular, this tells us\n\n* the four areas labelled $e, f, g$ and $h$ are equal (that is, $e=f=g=h$ ),\n* $P Z=Q W=R X=S Y=y \\sin \\theta$, and\n\n$* W P=X Q=Y R=Z S=y \\cos \\theta$.\n\n\n\nCombining these last two facts tells us that $W Z=X W=Y X=Z Y$, since, for example, $W Z=W P+P Z=Z S+S Y=Z Y$. In other words, $W X Y Z$ is a square, with side length $z$, say.\n\nNext, we show that $(a+r)+(c+n)$ is equal to $(b+m)+(d+s)$.\n\nNote that the sum of these two quantities is the total area between square $A B C D$ and square $W X Y Z$, so equals $x^{2}-z^{2}$.\n\nThus, to show that the quantities are equal, it is enough to show that $(a+r)+(c+n)$ equals $\\frac{1}{2}\\left(x^{2}-z^{2}\\right)$.\n\nLet the height of trapezoid $A B X W$ be $k$ and the height of trapezoid $Z Y C D$ be $l$.\n\nThen $|A B X W|=a+r=\\frac{1}{2} k(A B+W X)=\\frac{1}{2} k(x+z)$.\n\nAlso, $|Z Y C D|=c+n=\\frac{1}{2} l(D C+Z Y)=\\frac{1}{2} l(x+z)$.\n\nSince $A B, W X, Z Y$, and $D C$ are parallel, then the sum of the heights of trapezoid $A B X W$, square $W X Y Z$, and trapezoid $Z Y C D$ equals the height of square $A B C D$, so $k+z+l=x$, or $k+l=x-z$.\n\nTherefore,\n\n$$\n(a+r)+(c+n)=\\frac{1}{2} k(x+z)+\\frac{1}{2} l(x+z)=\\frac{1}{2}(x+z)(k+l)=\\frac{1}{2}(x+z)(x-z)=\\frac{1}{2}\\left(x^{2}-z^{2}\\right)\n$$\n\nas required.\n\nTherefore, $(a+r)+(c+n)=(b+m)+(d+s)$. We label this equation $(*)$.\n\nNext, we show that $r+n=m+s$.\n\nNote that $r=|\\triangle Q X B|$. This triangle can be viewed as having base $Q X$ and height equal to the height of trapezoid $A B X W$, or $k$.\n\nThus, $r=\\frac{1}{2}(y \\cos \\theta) k$.\n\nNote that $n=|\\triangle S Z D|$. This triangle can be viewed as having base $S Z$ and height equal to the height of trapezoid $Z Y C D$, or $l$.\n\nThus, $n=\\frac{1}{2}(y \\cos \\theta) l$.\n\nCombining these facts, we obtain\n\n$$\nn+r=\\frac{1}{2}(y \\cos \\theta) k+\\frac{1}{2}(y \\cos \\theta) l=\\frac{1}{2} y \\cos \\theta(k+l)=\\frac{1}{2} y \\cos \\theta(x-z)\n$$\n\nWe note that this sum depends only on the side lengths of the squares and the angle of rotation of the inner square, so is independent of the position of the inner square within the outer square.\n\nThis means that we can repeat this analysis to obtain the same expression for $m+s$.\n\nTherefore, $n+r=m+s$. We label this equation $(* *)$.\n\nWe subtract $(*)-(* *)$ to obtain $a+c=b+d$.\n\nFinally, we can combine all of this information:\n\n$$\n\\begin{aligned}\n& (|A B Q P|+|C D S R|)-(|B C R Q|+|A P S D|) \\\\\n& \\quad=(a+e+s+c+g+m)-(b+f+r+d+h+n) \\\\\n& \\quad=((a+c)-(b+d))+((m+s)-(n+r))+((e+g)-(f+h)) \\\\\n& \\quad=0+0+0\n\\end{aligned}\n$$\n\nsince $a+c=b+d$ and $n+r=m+s$ and $e=f=g=h$.\n\nTherefore, $|A B Q P|+|C D S R|=|B C R Q|+|A P S D|$, as required.']",,True,,, 2462,Geometry,,"A large square $A B C D$ is drawn, with a second smaller square $P Q R S$ completely inside it so that the squares do not touch. Line segments $A P, B Q, C R$, and $D S$ are drawn, dividing the region between the squares into four nonoverlapping convex quadrilaterals, as shown. If the sides of $P Q R S$ are not parallel to the sides of $A B C D$, prove that the sum of the areas of quadrilaterals $A P S D$ and $B C R Q$ equals the sum of the areas of quadrilaterals $A B Q P$ and $C D S R$. (Note: A convex quadrilateral is a quadrilateral in which the measure of each of the four interior angles is less than $180^{\circ}$.) ![](https://cdn.mathpix.com/cropped/2023_12_21_67f99a025aa829c0a7f6g-1.jpg?height=466&width=461&top_left_y=968&top_left_x=1298)","['We begin by ""boxing in"" square $P Q R S$ by drawing horizontal and vertical lines through its vertices to form rectangle $W X Y Z$, as shown. (Because the four quadrilaterals $A B Q P$, $B C R Q, C D S R$, and $D A P S$ are convex, there will not be any configurations that look substantially different from this the diagram below.) We also label the various areas.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_dc7a2c8c0b7b39460d9fg-1.jpg?height=651&width=678&top_left_y=962&top_left_x=821)\n\nSince $W X$ is parallel to $A B$, then quadrilateral $A B X W$ is a trapezoid. Similarly, quadrilaterals $B C Y X, C D Z Y$, and $D A W Z$ are trapezoids.\n\nWe use the notation $|A B Q P|$ to denote the area of quadrilateral $A B Q P$, and similar notation for other areas.\n\nSuppose that the side length of square $A B C D$ is $x$ and the side length of square $P Q R S$ is $y$.\n\nAlso, we let $\\angle W P Q=\\theta$.\n\nSince each of $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ is right-angled and each of the four angles of square $P Q R S$ is $90^{\\circ}$, then $\\angle W P Q=\\angle X Q R=\\angle Y R S=\\angle Z S P=\\theta$. This is because, for example,\n\n$\\angle X Q R=180^{\\circ}-\\angle P Q R-\\angle W Q P=90^{\\circ}-\\left(180^{\\circ}-\\angle P W Q-\\angle W P Q\\right)=90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$\n\nThis fact, together with the fact that $P Q=Q R=R S=S P=y$, allows us to conclude that the four triangles $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ are congruent.\n\nIn particular, this tells us\n\n* the four areas labelled $e, f, g$ and $h$ are equal (that is, $e=f=g=h$ ),\n* $P Z=Q W=R X=S Y=y \\sin \\theta$, and\n\n$* W P=X Q=Y R=Z S=y \\cos \\theta$.\n\n\n\nCombining these last two facts tells us that $W Z=X W=Y X=Z Y$, since, for example, $W Z=W P+P Z=Z S+S Y=Z Y$. In other words, $W X Y Z$ is a square, with side length $z$, say.\n\nNext, we show that $(a+r)+(c+n)$ is equal to $(b+m)+(d+s)$.\n\nNote that the sum of these two quantities is the total area between square $A B C D$ and square $W X Y Z$, so equals $x^{2}-z^{2}$.\n\nThus, to show that the quantities are equal, it is enough to show that $(a+r)+(c+n)$ equals $\\frac{1}{2}\\left(x^{2}-z^{2}\\right)$.\n\nLet the height of trapezoid $A B X W$ be $k$ and the height of trapezoid $Z Y C D$ be $l$.\n\nThen $|A B X W|=a+r=\\frac{1}{2} k(A B+W X)=\\frac{1}{2} k(x+z)$.\n\nAlso, $|Z Y C D|=c+n=\\frac{1}{2} l(D C+Z Y)=\\frac{1}{2} l(x+z)$.\n\nSince $A B, W X, Z Y$, and $D C$ are parallel, then the sum of the heights of trapezoid $A B X W$, square $W X Y Z$, and trapezoid $Z Y C D$ equals the height of square $A B C D$, so $k+z+l=x$, or $k+l=x-z$.\n\nTherefore,\n\n$$\n(a+r)+(c+n)=\\frac{1}{2} k(x+z)+\\frac{1}{2} l(x+z)=\\frac{1}{2}(x+z)(k+l)=\\frac{1}{2}(x+z)(x-z)=\\frac{1}{2}\\left(x^{2}-z^{2}\\right)\n$$\n\nas required.\n\nTherefore, $(a+r)+(c+n)=(b+m)+(d+s)$. We label this equation $(*)$.\n\nNext, we show that $r+n=m+s$.\n\nNote that $r=|\\triangle Q X B|$. This triangle can be viewed as having base $Q X$ and height equal to the height of trapezoid $A B X W$, or $k$.\n\nThus, $r=\\frac{1}{2}(y \\cos \\theta) k$.\n\nNote that $n=|\\triangle S Z D|$. This triangle can be viewed as having base $S Z$ and height equal to the height of trapezoid $Z Y C D$, or $l$.\n\nThus, $n=\\frac{1}{2}(y \\cos \\theta) l$.\n\nCombining these facts, we obtain\n\n$$\nn+r=\\frac{1}{2}(y \\cos \\theta) k+\\frac{1}{2}(y \\cos \\theta) l=\\frac{1}{2} y \\cos \\theta(k+l)=\\frac{1}{2} y \\cos \\theta(x-z)\n$$\n\nWe note that this sum depends only on the side lengths of the squares and the angle of rotation of the inner square, so is independent of the position of the inner square within the outer square.\n\nThis means that we can repeat this analysis to obtain the same expression for $m+s$.\n\nTherefore, $n+r=m+s$. We label this equation $(* *)$.\n\nWe subtract $(*)-(* *)$ to obtain $a+c=b+d$.\n\nFinally, we can combine all of this information:\n\n$$\n\\begin{aligned}\n& (|A B Q P|+|C D S R|)-(|B C R Q|+|A P S D|) \\\\\n& \\quad=(a+e+s+c+g+m)-(b+f+r+d+h+n) \\\\\n& \\quad=((a+c)-(b+d))+((m+s)-(n+r))+((e+g)-(f+h)) \\\\\n& \\quad=0+0+0\n\\end{aligned}\n$$\n\nsince $a+c=b+d$ and $n+r=m+s$ and $e=f=g=h$.\n\nTherefore, $|A B Q P|+|C D S R|=|B C R Q|+|A P S D|$, as required.']",['证明题，略'],True,,Need_human_evaluate, 2463,Combinatorics,,"A multiplicative partition of a positive integer $n \geq 2$ is a way of writing $n$ as a product of one or more integers, each greater than 1. Note that we consider a positive integer to be a multiplicative partition of itself. Also, the order of the factors in a partition does not matter; for example, $2 \times 3 \times 5$ and $2 \times 5 \times 3$ are considered to be the same partition of 30 . For each positive integer $n \geq 2$, define $P(n)$ to be the number of multiplicative partitions of $n$. We also define $P(1)=1$. Note that $P(40)=7$, since the multiplicative partitions of 40 are $40,2 \times 20,4 \times 10$, $5 \times 8,2 \times 2 \times 10,2 \times 4 \times 5$, and $2 \times 2 \times 2 \times 5$. (In each part, we use ""partition"" to mean ""multiplicative partition"". We also call the numbers being multiplied together in a given partition the ""parts"" of the partition.) Determine the value of $P(64)$.","[""We determine the multiplicative partitions of 64 by considering the number of parts in the various partitions. Note that 64 is a power of 2 so any divisor of 64 is also a power of 2 . In each partition, since the order of parts is not important, we list the parts in increasing order to make it easier to systematically find all of these.\n\n* One part. There is one possibility: 64.\n* Two parts. There are three possibilities: $64=2 \\times 32=4 \\times 16=8 \\times 8$.\n* Three parts. We start with the smallest possible first and second parts. We keep the first part fixed while adjusting the second and third parts. We then increase the first part and repeat.\n\nWe get: $64=2 \\times 2 \\times 16=2 \\times 4 \\times 8=4 \\times 4 \\times 4$.\n\n* Four parts. A partition of 64 with four parts must include at least two $2 \\mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With two $2 \\mathrm{~s}$, the remaining two parts have a product of 16 .\n\nWe get: $64=2 \\times 2 \\times 2 \\times 8=2 \\times 2 \\times 4 \\times 4$.\n\n* Five parts. A partition of 64 with five parts must include at least three $2 \\mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With three $2 \\mathrm{~s}$, the remaining two parts have a product of 8 .\n\nWe get: $64=2 \\times 2 \\times 2 \\times 2 \\times 4$.\n\n$*$ Six parts. Since $64=2^{6}$, there is only one possibility: $64=2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2$.\n\nTherefore, $P(64)=1+3+3+2+1+1=11$.""]",['11'],False,,Numerical, 2464,Combinatorics,,"A multiplicative partition of a positive integer $n \geq 2$ is a way of writing $n$ as a product of one or more integers, each greater than 1. Note that we consider a positive integer to be a multiplicative partition of itself. Also, the order of the factors in a partition does not matter; for example, $2 \times 3 \times 5$ and $2 \times 5 \times 3$ are considered to be the same partition of 30 . For each positive integer $n \geq 2$, define $P(n)$ to be the number of multiplicative partitions of $n$. We also define $P(1)=1$. Note that $P(40)=7$, since the multiplicative partitions of 40 are $40,2 \times 20,4 \times 10$, $5 \times 8,2 \times 2 \times 10,2 \times 4 \times 5$, and $2 \times 2 \times 2 \times 5$. (In each part, we use ""partition"" to mean ""multiplicative partition"". We also call the numbers being multiplied together in a given partition the ""parts"" of the partition.) Determine the value of $P(1000)$.","['First, we note that $1000=10^{3}=(2 \\cdot 5)^{3}=2^{3} 5^{3}$.\n\nWe calculate the value of $P\\left(p^{3} q^{3}\\right)$ for two distinct prime numbers $p$ and $q$. It will turn out that this value does not depend on $p$ and $q$. This value will be the value of $P(1000)$, since 1000 has this form of prime factorization.\n\nLet $n=p^{3} q^{3}$ for distinct prime numbers $p$ and $q$.\n\nThe integer $n$ has three prime factors equal to $p$.\n\nIn a given partition, these can be all together in one part (as $p^{3}$ ), can be split between two different parts (as $p$ and $p^{2}$ ), or can be split between three different parts (as $p, p$ and $p)$. There are no other ways to divide up three divisors of $p$.\n\nSimilarly, $n$ has three prime factors equal to $q$ which can be divided in similar ways.\n\nWe determine $P\\left(p^{3} q^{3}\\right)$ by considering the possible combination of the number of parts divisible by $p$ and the number of parts divisible by $q$ and counting partitions in each case. In other words, we complete the following table:\n\n\n\nWe note that the table is symmetric, since the factors of $p$ and $q$ are interchangeable.\n\nWe proceed to consider cases, considering only those on the top left to bottom right diagonal and and those below this diagonal in the table.\n\n\n\nCase 1: One part divisible by $p$, one part divisible by $q$\n\nThe partition must be $p^{3} q^{3}$ ( $n$ itself) or $p^{3} \\times q^{3}$.\n\nThere are two partitions in this case.\n\nCase 2: One part divisible by $p$, two parts divisible by $q$\n\nThe three factors of $p$ occur together as $p^{3}$. The three factors of $q$ occur as $q$ and $q^{2}$.\n\nThe $p^{3}$ can occur in one of the parts divisible by $q$ or not.\n\nThis gives partitions $p^{3} \\times q \\times q^{2}$ and $p^{3} q \\times q^{2}$ and $q \\times p^{3} q^{2}$.\n\nThere are three partitions in this case. Similarly, there are three partitions with one part divisible by $q$ and two parts divisible by $p$.\n\nCase 3: One part divisible by $p$, three parts divisible by $q$\n\nThe three factors of $p$ occur together as $p^{3}$. The three factors of $q$ occur as $q, q$ and $q$.\n\nThe $p^{3}$ can occur in one of the parts divisible by $q$ or not.\n\nThis gives partitions $p^{3} \\times q \\times q \\times q$ and $p^{3} q \\times q \\times q$.\n\n(Note that the three divisors of $q$ are interchangeable so $p^{3}$ only needs to be placed with one of them.)\n\nThere are two partitions in this case. Similarly, there are two partitions with one part divisible by $q$ and three parts divisible by $p$.\n\nCase 4: Two parts divisible by $p$, two parts divisible by $q$\n\nThe three factors of $p$ occur as $p$ and $p^{2}$. The three factors of $q$ occur as $q$ and $q^{2}$.\n\nEach of $p$ and $p^{2}$ can occur in one of the parts divisible by $q$ or not.\n\nIf no part is a multiple of both $p$ and $q$, we have one partition: $p \\times p^{2} \\times q \\times q^{2}$.\n\nIf one part is a multiple of both $p$ and $q$, there are two choices for which power of $p$ to include in this part and two choices for which power of $q$ to include. (There is no choice for the remaining parts.) Thus, there are $2 \\times 2=4$ such partitions:\n\n$$\np^{2} q^{2} \\times p \\times q \\quad p q^{2} \\times p^{2} \\times q \\quad p^{2} q \\times p \\times q^{2} \\quad p q \\times p^{2} \\times q^{2}\n$$\n\nIf two parts are a multiple of both $p$ and $q$, there are two ways to choose the power of $p$ in the part containing just $q$, so there are two such partitions: $p q \\times p^{2} q^{2}$ and $p^{2} q \\times p q^{2}$. There are seven partitions in this case.\n\nCase 5: Two parts divisible by $p$, three parts divisible by $q$\n\nThe three factors of $p$ occur as $p$ and $p^{2}$. The three factors of $q$ occur as $q, q$ and $q$.\n\nEach of $p$ and $p^{2}$ can occur in one of the parts divisible by $q$ or not.\n\nIf no part is a multiple of both $p$ and $q$, we have one partition: $p \\times p^{2} \\times q \\times q \\times q$.\n\nIf one part is a multiple of both $p$ and $q$, there are two choices for which power of $p$ to include in this part (since all powers of $q$ are identical).\n\nThus, there are 2 such partitions: $p^{2} q \\times p \\times q \\times q$ and $p q \\times p^{2} \\times q \\times q$.\n\nIf two parts are a multiple of both $p$ and $q$, there is one partition, since all of the powers of $q$ are identical: $p q \\times p^{2} q \\times q$.\n\nThere are four partitions in this case. Similarly, there are four partitions with two parts divisible by $q$ and three parts divisible by $p$.\n\nCase 6: Three parts divisible by $p$, three parts divisible by $q$\n\nThe three factors of $p$ as $p, p$ and $p$. The three factors of $q$ appear as $q, q$ and $q$.\n\nHere, the number of parts in the partition that are multiples of both $p$ and $q$ can be 0 , 1,2 or 3 . Since all of the powers of $p$ and $q$ are identical, the partitions are completely determined by this and are\n\n$$\np \\times p \\times p \\times q \\times q \\times q \\quad p \\times p \\times p q \\times q \\times q \\quad p \\times p q \\times p q \\times q \\quad p q \\times p q \\times p q\n$$\n\nThere are four partitions in this case.\n\n\n\nFinally, we complete the table:\n\nNumber of parts divisible by $p$ (Column)\n\nNumber of parts divisible by $q$ (Row)\n\n| | 1 | 2 | 3 |\n| :--- | :--- | :--- | :--- |\n| 1 | 2 | 3 | 2 |\n| 2 | 3 | 7 | 4 |\n| 3 | 2 | 4 | 4 |\n\nAdding the entries in the table, we obtain $P\\left(p^{3} q^{3}\\right)=31$.\n\nThus, $P(1000)=31$.']",['31'],False,,Numerical, 2465,Geometry,,"In the diagram, triangle ABC is right-angled at B. MT is the perpendicular bisector of $B C$ with $M$ on $B C$ and $T$ on $A C$. If $A T=A B$, what is the size of $\angle A C B$ ? ","['Since $M T$ is the perpendicular bisector of $B C$, then\n\n$B M=M C$, and $T M$ is perpendicular to $B C$.\n\nTherefore, $\\triangle C M T$ is similar to $\\triangle C B A$, since they share a common angle and each have a right angle.\n\n\n\nBut $\\frac{C M}{C B}=\\frac{1}{2}$ so $\\frac{C T}{C A}=\\frac{C M}{C B}=\\frac{1}{2}$, and thus $C T=A T=A B$, ie. $\\frac{A B}{A C}=\\frac{1}{2}$ or $\\sin (\\angle A C B)=\\frac{1}{2}$.\n\nTherefore, $\\angle A C B=30^{\\circ}$.', 'Since $T M \\| A B$, and $C M=M B$, then $C T=T A=A B$.\n\nJoin $T$ to $B$.\n\nSince $\\angle A B C=90^{\\circ}$, then $A C$ is the diameter of a circle passing through $A, C$ and $B$, with $T$ as its centre.\n\n\n\nThus, $T A=A B=B T$ (all radii), and so $\\triangle A B T$ is equilateral. Therefore, $\\angle B A C=60^{\\circ}$, and so $\\angle A C B=30^{\\circ}$.']",['$30^{\\circ}$'],False,,Numerical, 2465,Geometry,,"In the diagram, triangle ABC is right-angled at B. MT is the perpendicular bisector of $B C$ with $M$ on $B C$ and $T$ on $A C$. If $A T=A B$, what is the size of $\angle A C B$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_d7fe03f50cac4fc44fe1g-1.jpg?height=227&width=415&top_left_y=209&top_left_x=1386)","['Since $M T$ is the perpendicular bisector of $B C$, then\n\n$B M=M C$, and $T M$ is perpendicular to $B C$.\n\nTherefore, $\\triangle C M T$ is similar to $\\triangle C B A$, since they share a common angle and each have a right angle.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_0c20f2050e7a1226112ag-1.jpg?height=253&width=488&top_left_y=833&top_left_x=1407)\n\nBut $\\frac{C M}{C B}=\\frac{1}{2}$ so $\\frac{C T}{C A}=\\frac{C M}{C B}=\\frac{1}{2}$, and thus $C T=A T=A B$, ie. $\\frac{A B}{A C}=\\frac{1}{2}$ or $\\sin (\\angle A C B)=\\frac{1}{2}$.\n\nTherefore, $\\angle A C B=30^{\\circ}$.', 'Since $T M \\| A B$, and $C M=M B$, then $C T=T A=A B$.\n\nJoin $T$ to $B$.\n\nSince $\\angle A B C=90^{\\circ}$, then $A C$ is the diameter of a circle passing through $A, C$ and $B$, with $T$ as its centre.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_0c20f2050e7a1226112ag-1.jpg?height=260&width=488&top_left_y=1431&top_left_x=1407)\n\nThus, $T A=A B=B T$ (all radii), and so $\\triangle A B T$ is equilateral. Therefore, $\\angle B A C=60^{\\circ}$, and so $\\angle A C B=30^{\\circ}$.']",['$30^{\\circ}$'],False,,Numerical, 2466,Algebra,,"What are all values of $x$ such that $$ \log _{5}(x+3)+\log _{5}(x-1)=1 ? $$","['Combining the logarithms,\n\n$$\n\\begin{aligned}\n\\log _{5}(x+3)+\\log _{5}(x-1) & =1 \\\\\n\\log _{5}((x+3)(x-1)) & =1 \\\\\n\\log _{5}\\left(x^{2}+2 x-3\\right) & =1 \\\\\nx^{2}+2 x-3 & =5 \\\\\nx^{2}+2 x-8 & =0 \\\\\n(x+4)(x-2) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-4$ or $x=2$. Substituting the two values for $x$ back into the original equation, we see that $x=2$ works, but that $x=-4$ does not, since we cannot take the logarithm of a negative number.']",['2'],False,,Numerical, 2467,Algebra,,"A chef aboard a luxury liner wants to cook a goose. The time $t$ in hours to cook a goose at $180^{\circ} \mathrm{C}$ depends on the mass of the goose $m$ in kilograms according to the formula $$ t=a m^{b} $$ where $a$ and $b$ are constants. The table below gives the times observed to cook a goose at $180^{\circ} \mathrm{C}$. | Mass, $m(\mathrm{~kg})$ | Time, $t(\mathrm{~h})$ | | :---: | :---: | | 3.00 | 2.75 | | 6.00 | 3.75 | Using the data in the table, determine both $a$ and $b$ to two decimal places.","['From the table we have two pieces of information, so we substitute both of these into the given formula.\n\n$$\n\\begin{aligned}\n& 2.75=a(3.00)^{b} \\\\\n& 3.75=a(6.00)^{b}\n\\end{aligned}\n$$\n\nWe can now proceed in either of two ways to solve for $b$.\n\nMethod 1 to find $b$\n\nDividing the second equation by the first, we obtain\n\n$$\n\\frac{3.75}{2.75}=\\frac{a(6.00)^{b}}{a(3.00)^{b}}=\\frac{(6.00)^{b}}{(3.00)^{b}}=\\left(\\frac{6.00}{3.00}\\right)^{b}=2^{b}\n$$\n\nor\n\n$$\n2^{b} \\approx 1.363636\n$$\n\nTaking logarithms of both sides,\n\n\n\n$$\n\\begin{aligned}\n\\log \\left(2^{b}\\right) & \\approx \\log (1.363636) \\\\\nb \\log (2) & \\approx \\log (1.363636) \\\\\nb & \\approx \\frac{\\log (1.363636)}{\\log (2)} \\\\\nb & \\approx 0.4475\n\\end{aligned}\n$$\n\nMethod 2 to find $b$ \n\nTaking logarithms of both sides of the above equations, we obtain\n\n$$\n\\begin{aligned}\n\\log (2.75) & =\\log \\left(a(3.00)^{b}\\right) \\\\\n& =\\log (a)+\\log \\left((3.00)^{b}\\right) \\\\\n& =\\log (a)+b \\log (3.00)\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\n\\log (3.75)=\\log (a)+b \\log (6.00)\n$$\n\nSubtracting the first equation from the second, we obtain\n\n$$\n\\begin{aligned}\n\\log (3.75)-\\log (2.75) & =b(\\log (6.00)-\\log (3.00)) \\\\\nb & =\\frac{\\log (3.75)-\\log (2.75)}{\\log (6.00)-\\log (3.00)} \\\\\nb & \\approx 0.4475\n\\end{aligned}\n$$\n\nWe now continue in the same way for both methods.\n\nSubstituting this value for $b$ back into the first equation above,\n\n$$\n\\begin{aligned}\n2.75 & \\approx a(3.00)^{0.4475} \\\\\na & \\approx \\frac{2.75}{(3.00)^{0.4475}} \\\\\na & \\approx 1.6820\n\\end{aligned}\n$$\n\nTherefore, to two decimal places, $a=1.68$ and $b=0.45$.']","['$1.68,0.45$']",True,,Numerical, 2468,Geometry,,"In the diagram, $A B C D E F$ is a regular hexagon with a side length of 10 . If $X, Y$ and $Z$ are the midpoints of $A B, C D$ and $E F$, respectively, what is the length of $X Z$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_1d3f148ee017cf69bf7eg-1.jpg?height=260&width=301&top_left_y=214&top_left_x=1432)","['Extend $X A$ and $Z F$ to meet at point $T$.\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$ and $\\angle T A F=\\angle T F A=60^{\\circ}$, and so $\\triangle T A F$ and $\\triangle T X Z$ are both equilateral triangles.\n\nSince $A F=10$, then $T A=10$, which means\n\n$T X=10+5=15$, and so $X Z=T X=15$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_a74c1425f4891f5bece8g-1.jpg?height=365&width=455&top_left_y=365&top_left_x=1407)', 'We look at the quadrilateral $A X Z F$.\n\nSince $A B C D E F$ is a regular hexagon, then $\\angle F A X=\\angle A F Z=120^{\\circ}$.\n\nNote that $A F=10$, and also $A X=F Z=5$ since $X$ and $Z$ are midpoints of their respective sides.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_a74c1425f4891f5bece8g-1.jpg?height=238&width=510&top_left_y=846&top_left_x=1320)\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$, and so $A X Z F$ is a trapezoid.\n\nDrop perpendiculars from $A$ and $F$ to $P$ and $Q$, respectively, on $X Z$.\n\nBy symmetry again, $P X=Q Z$. Now, $P X=A X \\cos 60^{\\circ}=5\\left(\\frac{1}{2}\\right)=\\frac{5}{2}$.\n\nSince $A P Q F$ is a rectangle, then $P Q=10$.\n\nTherefore, $X Z=X P+P Q+Q Z=\\frac{5}{2}+10+\\frac{5}{2}=15$.']",['15'],False,,Numerical, 2468,Geometry,,"In the diagram, $A B C D E F$ is a regular hexagon with a side length of 10 . If $X, Y$ and $Z$ are the midpoints of $A B, C D$ and $E F$, respectively, what is the length of $X Z$ ? ","['Extend $X A$ and $Z F$ to meet at point $T$.\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$ and $\\angle T A F=\\angle T F A=60^{\\circ}$, and so $\\triangle T A F$ and $\\triangle T X Z$ are both equilateral triangles.\n\nSince $A F=10$, then $T A=10$, which means\n\n$T X=10+5=15$, and so $X Z=T X=15$.\n\n', 'We look at the quadrilateral $A X Z F$.\n\nSince $A B C D E F$ is a regular hexagon, then $\\angle F A X=\\angle A F Z=120^{\\circ}$.\n\nNote that $A F=10$, and also $A X=F Z=5$ since $X$ and $Z$ are midpoints of their respective sides.\n\n\n\nBy symmetry, $\\angle A X Z=\\angle F Z X=60^{\\circ}$, and so $A X Z F$ is a trapezoid.\n\nDrop perpendiculars from $A$ and $F$ to $P$ and $Q$, respectively, on $X Z$.\n\nBy symmetry again, $P X=Q Z$. Now, $P X=A X \\cos 60^{\\circ}=5\\left(\\frac{1}{2}\\right)=\\frac{5}{2}$.\n\nSince $A P Q F$ is a rectangle, then $P Q=10$.\n\nTherefore, $X Z=X P+P Q+Q Z=\\frac{5}{2}+10+\\frac{5}{2}=15$.']",['15'],False,,Numerical, 2469,Geometry,,A circle passes through the origin and the points of intersection of the parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. Determine the coordinates of the centre of this circle.,"['We first determine the three points through which the circle passes.\n\nThe first point is the origin $(0,0)$.\n\nThe second and third points are found by determining the points of intersection of the two parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. We do this by setting the $y$ values equal.\n\n$$\nx^{2}-3=-x^{2}-2 x+9\n$$\n\n$2 x^{2}+2 x-12=0$\n\n$x^{2}+x-6=0$\n\n$(x+3)(x-2)=0$\n\nso $x=-3$ or $x=2$.\n\n\n\nWe determine the points of intersection by substituting into the first parabola.\n\nIf $x=2, y=2^{2}-3=1$, so the point of intersection is $(2,1)$.\n\nIf $x=-3, y=(-3)^{2}-3=6$, so the point of intersection is $(-3,6)$.\n\nTherefore, the circle passes through the three points $A(0,0), B(2,1)$ and $C(-3,6)$.\n\nLet the centre of the circle be the point $Q(a, b)$.\n\n\n\nFinding the centre of the circle can be done in a variety of ways.\n\nWe use the fact $Q$ is of equal distance from each of the points $A, B$ and $C$. In particular $Q A^{2}=Q B^{2}=Q C^{2}$ or $x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2}=(x+3)^{2}+(y-6)^{2}$\n\nFrom the first equality,\n\n$$\n\\begin{aligned}\n& x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2} \\\\\n& 4 x+2 y=5\n\\end{aligned}\n$$\n\n\n\n\n\nFrom the second equality,\n\n$$\n\\begin{aligned}\n(x-2)^{2}+(y-1)^{2} & =(x+3)^{2}+(y-6)^{2} \\\\\n-10 x+10 y & =40 \\\\\ny & =x+4\n\\end{aligned}\n$$\n\nSubstituting the equation above into into $4 x+2 y=5$, we obtain $4 x+2(x+4)=5$ or $6 x=-3$ or $x=-\\frac{1}{2}$. Thus, $y=-\\frac{1}{2}+4=\\frac{7}{2}$, and so the centre of the circle is $\\left(-\\frac{1}{2}, \\frac{7}{2}\\right)$.']","['$(-\\frac{1}{2}, \\frac{7}{2})$']",True,,Tuple, 2470,Geometry,,"In the diagram, $A C=2 x, B C=2 x+1$ and $\angle A C B=30^{\circ}$. If the area of $\triangle A B C$ is 18 , what is the value of $x$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_1d3f148ee017cf69bf7eg-1.jpg?height=239&width=314&top_left_y=691&top_left_x=1407)","['Using a known formula for the area of a triangle, $A=\\frac{1}{2} a b \\sin C$,\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(2 x) \\sin 30^{\\circ} \\\\\n36 & =(2 x+1)(2 x)\\left(\\frac{1}{2}\\right) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$. Since $x$ is positive, then $x=4$.', 'Draw a perpendicular from $A$ to $P$ on $B C$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_86d8b0958ad843aa94fbg-1.jpg?height=298&width=369&top_left_y=369&top_left_x=1363)\n\nUsing $\\triangle A P C, A P=A C \\sin 30^{\\circ}=2 x\\left(\\frac{1}{2}\\right)=x$.\n\nNow $A P$ is the height of $\\triangle A B C$, so Area $=\\frac{1}{2}(B C)(A P)$.\n\nThen\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(x) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$.\n\nSince $x$ is positive, then $x=4$.']",['4'],False,,Numerical, 2470,Geometry,,"In the diagram, $A C=2 x, B C=2 x+1$ and $\angle A C B=30^{\circ}$. If the area of $\triangle A B C$ is 18 , what is the value of $x$ ? ","['Using a known formula for the area of a triangle, $A=\\frac{1}{2} a b \\sin C$,\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(2 x) \\sin 30^{\\circ} \\\\\n36 & =(2 x+1)(2 x)\\left(\\frac{1}{2}\\right) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$. Since $x$ is positive, then $x=4$.', 'Draw a perpendicular from $A$ to $P$ on $B C$.\n\n\n\nUsing $\\triangle A P C, A P=A C \\sin 30^{\\circ}=2 x\\left(\\frac{1}{2}\\right)=x$.\n\nNow $A P$ is the height of $\\triangle A B C$, so Area $=\\frac{1}{2}(B C)(A P)$.\n\nThen\n\n$$\n\\begin{aligned}\n18 & =\\frac{1}{2}(2 x+1)(x) \\\\\n0 & =2 x^{2}+x-36 \\\\\n0 & =(2 x+9)(x-4)\n\\end{aligned}\n$$\n\nand so $x=4$ or $x=-\\frac{9}{2}$.\n\nSince $x$ is positive, then $x=4$.']",['4'],False,,Numerical, 2471,Geometry,,"A ladder, $A B$, is positioned so that its bottom sits on horizontal ground and its top rests against a vertical wall, as shown. In this initial position, the ladder makes an angle of $70^{\circ}$ with the horizontal. The bottom of the ladder is then pushed $0.5 \mathrm{~m}$ away from the wall, moving the ladder to position $A^{\prime} B^{\prime}$. In this new position, the ladder makes an angle of $55^{\circ}$ with the horizontal. Calculate, to the nearest centimetre, the distance that the ladder slides down the wall (that is, the length of $B B^{\prime}$ ). ![](https://cdn.mathpix.com/cropped/2023_12_21_1d3f148ee017cf69bf7eg-1.jpg?height=499&width=331&top_left_y=992&top_left_x=1401)","['Let the length of the ladder be $L$.\n\nThen $A C=L \\cos 70^{\\circ}$ and $B C=L \\sin 70^{\\circ}$. Also, $A^{\\prime} C=L \\cos 55^{\\circ}$ and $B^{\\prime} C=L \\sin 55^{\\circ}$.\n\nSince $A^{\\prime} A=0.5$, then\n\n$$\n0.5=L \\cos 55^{\\circ}-L \\cos 70^{\\circ}\n$$\n\n$$\nL=\\frac{0.5}{\\cos 55^{\\circ}-\\cos 70^{\\circ}}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nB B^{\\prime} & =B C-B^{\\prime} C \\\\\n& =L \\sin 70^{\\circ}-L \\sin 55^{\\circ} \\\\\n& =L\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right) \\\\\n& =\\frac{(0.5)\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right)}{\\left(\\cos 55^{\\circ}-\\cos 70^{\\circ}\\right)} \\quad(\\text { from }(*)) \\\\\n& \\approx 0.2603 \\mathrm{~m}\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_86d8b0958ad843aa94fbg-1.jpg?height=686&width=634&top_left_y=1039&top_left_x=1231)\n\nTherefore, to the nearest centimetre, the distance that the ladder slides down the wall is $26 \\mathrm{~cm}$.']",['26'],False,cm,Numerical, 2471,Geometry,,"A ladder, $A B$, is positioned so that its bottom sits on horizontal ground and its top rests against a vertical wall, as shown. In this initial position, the ladder makes an angle of $70^{\circ}$ with the horizontal. The bottom of the ladder is then pushed $0.5 \mathrm{~m}$ away from the wall, moving the ladder to position $A^{\prime} B^{\prime}$. In this new position, the ladder makes an angle of $55^{\circ}$ with the horizontal. Calculate, to the nearest centimetre, the distance that the ladder slides down the wall (that is, the length of $B B^{\prime}$ ). ","['Let the length of the ladder be $L$.\n\nThen $A C=L \\cos 70^{\\circ}$ and $B C=L \\sin 70^{\\circ}$. Also, $A^{\\prime} C=L \\cos 55^{\\circ}$ and $B^{\\prime} C=L \\sin 55^{\\circ}$.\n\nSince $A^{\\prime} A=0.5$, then\n\n$$\n0.5=L \\cos 55^{\\circ}-L \\cos 70^{\\circ}\n$$\n\n$$\nL=\\frac{0.5}{\\cos 55^{\\circ}-\\cos 70^{\\circ}}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nB B^{\\prime} & =B C-B^{\\prime} C \\\\\n& =L \\sin 70^{\\circ}-L \\sin 55^{\\circ} \\\\\n& =L\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right) \\\\\n& =\\frac{(0.5)\\left(\\sin 70^{\\circ}-\\sin 55^{\\circ}\\right)}{\\left(\\cos 55^{\\circ}-\\cos 70^{\\circ}\\right)} \\quad(\\text { from }(*)) \\\\\n& \\approx 0.2603 \\mathrm{~m}\n\\end{aligned}\n$$\n\n\n\nTherefore, to the nearest centimetre, the distance that the ladder slides down the wall is $26 \\mathrm{~cm}$.']",['26'],False,cm,Numerical, 2472,Algebra,,"In a soccer league with 5 teams, each team plays 20 games(that is, 5 games with each of the other 4 teams). For each team, every game ends in a win (W), a loss (L), or a tie (T). The numbers of wins, losses and ties for each team at the end of the season are shown in the table. Determine the values of $x, y$ and $z$. | Team | W | L | T | | :---: | ---: | ---: | ---: | | A | 2 | 15 | 3 | | B | 7 | 9 | 4 | | C | 6 | 12 | 2 | | D | 10 | 8 | 2 | | E | $x$ | $y$ | $z$ |","['In total, there are $\\frac{1}{2} \\times 5 \\times 20=50$ games played, since each of 5 teams plays 20 games (we divide by 2 since each game is double-counted).\n\nIn each game, there is either a loss or a tie.\n\nThe number of games with a loss is $44+y$ from the second column, and the number of games with a tie is $\\frac{1}{2}(11+z)$ (since any game ending in a tie has 2 ties).\n\n\n\nSo\n\n$$\n\\begin{aligned}\n50 & =44+y+\\frac{1}{2}(11+z) \\\\\n100 & =88+2 y+11+z \\\\\n1 & =2 y+z\n\\end{aligned}\n$$\n\nSince $y$ and $z$ are non-negative integers, $z=1$ and $y=0$. So $x=19$ since Team E plays 20 games.', 'In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.\n\n$$\n\\begin{aligned}\n25+x & =44+y \\\\\nx-y & =19\n\\end{aligned}\n$$\n\nAlso, since team E plays 20 games, then\n\n$$\nx+y+z=20\n$$\n\nSo from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.\n\nLastly, we know that the total of all of the teams numbers of ties must be even, ie. $11+z$ is even, ie. $z$ is odd.\n\nSince $x$ is at least 19, then $z$ can be at most 1 by (2).\n\nTherefore, $z=1$. Thus, $x=19$ and $y=0$.', 'In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.\n\n$$\n\\begin{aligned}\n25+x & =44+y \\\\\nx-y & =19\n\\end{aligned}\n\\tag{1}\n$$\n\nAlso, since team E plays 20 games, then\n\n$$\nx+y+z=20\n\\tag{2}\n$$\n\nSo from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.\n\nConsider the possibility that $x=20$. From (2), then $y=z=0$, which does not agree with (1).\n\nThus, the only possibility is $x=19$. From (1), $y=0$, and so $z=1$ from (2). (These three values agree with both equations (1) and (2).)']","['19,0,1']",True,,Numerical, 2473,Algebra,,"Prove that it is not possible to create a sequence of 4 numbers $a, b, c, d$, such that the sum of any two consecutive terms is positive, and the sum of any three consecutive terms is negative.","['Assume such a sequence $a, b, c, d$ exists. (We proceed by contradiction.)\n\nSince the sum of any two consecutive terms is positive, $a+b>0, b+c>0$, and\n\n$c+d>0$. Adding these three inequalities, $(a+b)+(b+c)+(c+d)>0$ or\n\n$a+2 b+2 c+d>0$.\n\nWe are going to show that this statement contradicts the facts that are known about the sequence. We are told that the sum of any three consecutive terms is negative, ie.\n\n$a+b+c<0$ and $b+c+d<0$. Adding these two inequalities, $(a+b+c)+(b+c+d)<0$ or $a+2 b+2 c+d<0$.\n\nThis is a contradiction, since the two conditions $a+2 b+2 c+d>0$ and $a+2 b+2 c+d<0$ cannot occur simultaneously.\n\nTherefore, our original assumption is false, and so no such sequence exists.', 'Assume such a sequence $a, b, c, d$ exists. (We proceed by contradiction.)\n\nWe consider two cases.\n\nCase 1: $a \\leq 0$\n\nIn this case, $b>0$ since $a+b>0$\n\nThen, since $a+b+c<0$, we must have that $c<0$.\n\nBut $c+d>0$, so $d>0$.\n\nThis means that we have $b>0$ and $c+d>0$, ie. $b+c+d>0$.\n\nBut from the conditions on the sequence, $b+c+d<0$, a contradiction.\n\nTherefore, no such sequence exists with $a \\leq 0$.\n\nCase 2: $a>0$\n\nIn this case, it is not immediately clear whether $b$ has to be positive or negative.\n\nHowever, we do know that $a+b>0$ and $a+b+c<0$, so it must be true that $c<0$.\n\nThen since $b+c>0$ and $c+d>0$, we must have both $b>0$ and $d>0$. But then $b+c+d=b+(c+d)>0$ since $c+d>0$ and $b>0$.\n\nThis is again a contradiction.\n\nTherefore, no such sequence exists with $a>0$.']",,True,,, 2474,Geometry,,"In triangle $A B C, \angle A B C=90^{\circ}$. Rectangle $D E F G$ is inscribed in $\triangle A B C$, as shown. Squares $J K G H$ and $M L F N$ are inscribed in $\triangle A G D$ and $\triangle C F E$, respectively. If the side length of $J H G K$ is $v$, the side length of $M L F N$ is $w$, and $D G=u$, prove that $u=v+w$. ","['Let $\\angle B A C=\\theta$. Then by parallel lines, $\\angle D J H=\\angle B D E=\\theta$.\n\nThus, $\\angle B E D=90^{\\circ}-\\theta$ and so\n\n$\\angle N E M=\\theta$ since\n\n$\\angle D E F=90^{\\circ}$.\n\nSince $D G=u$ and $H G=v$,\n\nthen $D H=u-v$.\n\n\n\nSimilarly, $E N=u-w$.\n\nLooking at $\\triangle D H J$ and $\\triangle M N E$, we see that $\\tan \\theta=\\frac{u-v}{v}$ and $\\tan \\theta=\\frac{w}{u-w}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{u-v}{v} & =\\frac{w}{u-w} \\\\\n(u-v)(u-w) & =v w \\\\\nu^{2}-u v-u w+v w & =v w \\\\\nu(u-v-w) & =0\n\\end{aligned}\n$$\n\nand since $u \\neq 0$, we must have $u-v-w=0$ or $u=v+w$.\n\n[Note: If $u=0$, then the height of rectangle $D E F G$ is 0 , ie. $D$ coincides with point $A$ and $E$ coincides with point $C$, which says that we must also have $v=w=0$, ie. the squares have no place to go!]']",,True,,, 2474,Geometry,,"In triangle $A B C, \angle A B C=90^{\circ}$. Rectangle $D E F G$ is inscribed in $\triangle A B C$, as shown. Squares $J K G H$ and $M L F N$ are inscribed in $\triangle A G D$ and $\triangle C F E$, respectively. If the side length of $J H G K$ is $v$, the side length of $M L F N$ is $w$, and $D G=u$, prove that $u=v+w$. ![](https://cdn.mathpix.com/cropped/2023_12_21_02b1e9456f98fdbe5dd9g-1.jpg?height=274&width=477&top_left_y=199&top_left_x=1298)","['Let $\\angle B A C=\\theta$. Then by parallel lines, $\\angle D J H=\\angle B D E=\\theta$.\n\nThus, $\\angle B E D=90^{\\circ}-\\theta$ and so\n\n$\\angle N E M=\\theta$ since\n\n$\\angle D E F=90^{\\circ}$.\n\nSince $D G=u$ and $H G=v$,\n\nthen $D H=u-v$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_98602fddd2dfab207e62g-1.jpg?height=450&width=856&top_left_y=306&top_left_x=930)\n\nSimilarly, $E N=u-w$.\n\nLooking at $\\triangle D H J$ and $\\triangle M N E$, we see that $\\tan \\theta=\\frac{u-v}{v}$ and $\\tan \\theta=\\frac{w}{u-w}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{u-v}{v} & =\\frac{w}{u-w} \\\\\n(u-v)(u-w) & =v w \\\\\nu^{2}-u v-u w+v w & =v w \\\\\nu(u-v-w) & =0\n\\end{aligned}\n$$\n\nand since $u \\neq 0$, we must have $u-v-w=0$ or $u=v+w$.\n\n[Note: If $u=0$, then the height of rectangle $D E F G$ is 0 , ie. $D$ coincides with point $A$ and $E$ coincides with point $C$, which says that we must also have $v=w=0$, ie. the squares have no place to go!]']",['证明题，略'],True,,Need_human_evaluate, 2475,Geometry,,"Three thin metal rods of lengths 9,12 and 15 are welded together to form a right-angled triangle, which is held in a horizontal position. A solid sphere of radius 5 rests in the triangle so that it is tangent to each of the three sides. Assuming that the thickness of the rods can be neglected, how high above the plane of the triangle is the top of the sphere?","['Consider the cross-section of the sphere in the plane defined by the triangle. This crosssection will be a circle, since any cross-section of a sphere is a circle. This circle will be tangent to the three sides of the triangle, ie. will be the inscribed circle (or incircle) of the triangle. Let the centre of this circle be $O$, and its radius be $r$. We calculate the value of $r$.\n\n\n\nJoin $O$ to the three points of tangency, $P, Q, R$, and to the three vertices $A, B, C$. Then $O P, O Q$ and $O R$ (radii) will form right angles with the three sides of the triangle. Consider the three triangles $\\triangle A O B$, $\\triangle B O C$ and $\\triangle C O A$. Each of these triangles has a height of $r$ and they have bases 15, 9 and 12, respectively. Since the area of $\\triangle A B C$ is equal to the sum of the areas of $\\triangle A O B, \\triangle B O C$, and $\\triangle C O A$, So comparing areas,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(9)(12) & =\\frac{1}{2}(9)(r)+\\frac{1}{2}(12)(r)+\\frac{1}{2}(15)(r) \\\\\n54 & =\\frac{1}{2} r(9+12+15) \\\\\nr & =3\n\\end{aligned}\n$$\n\n\n\nNow join the centre of the cross-sectional circle to the centre of the sphere and let this distance be $h$. Now, the line joining the centre of the circle to the centre of the sphere will be perpendicular to the plane of the triangle, so we can form a right-angled triangle by joining the centre of the sphere to any point on the circumference of the cross-sectional circle. By Pythagoras,\n\n$$\n\\begin{aligned}\nh^{2}+r^{2} & =25 \\\\\nh & =4\n\\end{aligned}\n$$\n\nThis tells us that the top of the sphere is 9 units above the plane of the triangle, since the top of the sphere is 5 units above the centre of the sphere.']",['5'],False,,Numerical, 2476,Geometry,,"A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer. Show that every Pythagorean triangle is Heronian.","['Consider a Pythagorean triangle with integer side lengths $a, b, c$ satisfying $a^{2}+b^{2}=c^{2}$. To show that this triangle is Heronian, we must show that it has an integer area. Now we know that the area is equal to $\\frac{1}{2} a b$, so we must show that either $a$ or $b$ is an even integer.\n\n\n\nSuppose that both $a$ and $b$ are odd. (We proceed by contradiction.)\n\nIn this case, let $a=2 k+1$ and $b=2 l+1$. Then both $a^{2}$ and $b^{2}$ are odd, and so $c^{2}$ is even since $a^{2}+b^{2}=c^{2}$. Therefore, $c$ itself must be even, so let $c=2 m$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n(2 k+1)^{2}+(2 l+1)^{2} & =(2 m)^{2} \\\\\n4 k^{2}+4 k+1+4 l^{2}+4 l+1 & =4 m^{2} \\\\\n4\\left(k^{2}+k+l^{2}+l\\right)+2 & =4\\left(m^{2}\\right)\n\\end{aligned}\n$$\n\nBut the right side is a multiple of 4 , and the left side is not a multiple of 4 . This is a contradiction.\n\nTherefore, one of $a$ or $b$ must be even, and so the area of the triangle is an integer. Thus, any Pythagorean triangle is Heronian.']",,True,,, 2477,Geometry,,"A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer. Show that every odd integer greater than 1 is a side length of some Pythagorean triangle.","['We examine the first few smallest Pythagorean triples:\n\n$$\n345 \\quad\\left(3^{2}=4+5\\right)\n$$\n\n\n\n$$\n\\begin{array}{llll}\n5 & 12 & 13 & \\left(5^{2}=12+13\\right) \\\\\n6 & 8 & 10 & \\text { (Does not fit pattern) } \\\\\n7 & 24 & 25 & \\left(7^{2}=24+25\\right)\n\\end{array}\n$$\n\nIt appears from the first few examples that perhaps we can form a Pythagorean triple by using any odd number greater than 1 as its shortest leg.\n\nNext, we notice from the pattern that the sum of the second leg and the hypotenuse is the square of the shortest leg, and that these two side lengths differ by 1 .\n\nWill this pattern always hold?\n\nLet $a=2 k+1$ with $k \\geq 1$. (This formula will generate all odd integers greater than or equal to 3.) Can we always find $b$ so that $c=b+1$ and $a^{2}+b^{2}=c^{2}$ ?\n\nConsider the equation\n\n$$\n\\begin{aligned}\n(2 k+1)^{2}+b^{2} & =(b+1)^{2} \\\\\n4 k^{2}+4 k+1+b^{2} & =b^{2}+2 b+1 \\\\\n4 k^{2}+4 k & =2 b \\\\\nb & =2 k^{2}+2 k\n\\end{aligned}\n$$\n\nSo we can always find a $b$ to make the equation true. Therefore, since $a$ can be any odd integer greater than or equal to 3 , then we can make any odd number the shortest leg of a Pythagorean triangle, namely the Pythagorean triangle $a=2 k+1, b=2 k^{2}+2 k$, $c=2 k^{2}+2 k+1$. (Check that $a^{2}+b^{2}=c^{2}$ does indeed hold here!)']",,True,,, 2478,Geometry,,"A triangle is called Heronian if each of its side lengths is an integer and its area is also an integer. A triangle is called Pythagorean if it is right-angled and each of its side lengths is an integer. Find a Heronian triangle which has all side lengths different, and no side length divisible by $3,5,7$ or 11 .","['We consider forming a triangle by joining two Pythagorean triangles along a common side. Since any Pythagorean triangle is Heronian, then the triangle that is formed by joining two Pythagorean triangles in the manner shown will have integer side lengths and will have integer area, thus making it Heronian.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d5600eed21b7775d92fdg-1.jpg?height=184&width=328&top_left_y=1529&top_left_x=1489)\n\nSo again, we make a list of Pythagorean triples\n\n| 3 | 4 | 5 |\n| :--- | :--- | :--- |\n| 5 | 12 | 13 |\n| 6 | 8 | 10 |\n| 7 | 24 | 25 |\n| 8 | 15 | 17 |\n| 9 | 40 | 41 |\n| 10 | 24 | 26 |\n| 11 | 60 | 61 |\n\n\n\nWe notice that we can scale any Pythagorean triangle by an integer factor and obtain another Pythagorean triangle. This will enable us to create two Pythagorean triangles with a common side length.\n\nAlso, we note that when joining two Pythagorean triangles, the hypotenuse of each triangle becomes a side length in the new triangle. Since we cannot have a side length divisible by $3,5,7$ or 11 , this eliminates the $3-4-5,6-8-10$, and $7-24-25$ triangles from the list above.\n\nSuppose we scale the 8-15-17 triangle by a factor of 4 to obtain $32-60-68$ and join to the 11-60-61 triangle in the manner shown.\n\nThus we obtain a 43-61-68 triangle, which has integer area because its height is an even integer.\n\nTherefore, a 43-61-68 triangle is Heronian.']",['a 43-61-68 triangle'],False,,Need_human_evaluate, 2479,Geometry,,"Triangle $A B C$ has vertices $A(0,5), B(3,0)$ and $C(8,3)$. Determine the measure of $\angle A C B$.","['First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nSince $A B=B C$ and $A C=\\sqrt{2} A B=\\sqrt{2} B C$, then $\\triangle A B C$ is an isosceles right-angled triangle, with the\n\n\nright angle at $B$.\n\nTherefore, $\\angle A C B=45^{\\circ}$.', 'First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nLine segment $A B$ has slope $\\frac{5-0}{0-3}=-\\frac{5}{3}$.\n\nLine segment $B C$ has slope $\\frac{0-3}{3-8}=\\frac{3}{5}$.\n\nSince the product of these two slopes is -1 , then $A B$ and $B C$ are perpendicular.\n\nTherefore, $\\triangle A B C$ is right-angled at $B$.\n\nSince $A B=B C$, then $\\triangle A B C$ is an isosceles right-angled triangle, so $\\angle A C B=45^{\\circ}$.', 'First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nUsing the cosine law,\n\n$$\n\\begin{aligned}\nA B^{2} & =A C^{2}+B C^{2}-2(A C)(B C) \\cos (\\angle A C B) \\\\\n34 & =68+34-2(\\sqrt{68})(\\sqrt{34}) \\cos (\\angle A C B) \\\\\n0 & =68-2(\\sqrt{2} \\sqrt{34})(\\sqrt{34}) \\cos (\\angle A C B) \\\\\n0 & =68-68 \\sqrt{2} \\cos (\\angle A C B) \\\\\n68 \\sqrt{2} \\cos (\\angle A C B) & =68 \\\\\n\\cos (\\angle A C B) & =\\frac{1}{\\sqrt{2}}\n\\end{aligned}\n$$\n\nSince $\\cos (\\angle A C B)=\\frac{1}{\\sqrt{2}}$ and $0^{\\circ}<\\angle A C B<180^{\\circ}$, then $\\angle A C B=45^{\\circ}$.']",['$45^{\\circ}$'],False,,Numerical, 2480,Geometry,,"In the diagram, $P Q R S$ is an isosceles trapezoid with $P Q=7, P S=Q R=8$, and $S R=15$. Determine the length of the diagonal $P R$. ![](https://cdn.mathpix.com/cropped/2023_12_21_040d04737e70e698092cg-1.jpg?height=287&width=607&top_left_y=770&top_left_x=1212)","['Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $S R$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_8bce42a3784597e137dag-1.jpg?height=298&width=609&top_left_y=241&top_left_x=861)\n\nSince $P Q$ is parallel to $S R$ (because $P Q R S$ is a trapezoid) and $P X$ and $Q Y$ are perpendicular to $S R$, then $P Q Y X$ is a rectangle.\n\nThus, $X Y=P Q=7$ and $P X=Q Y$.\n\nSince $\\triangle P X S$ and $\\triangle Q Y R$ are right-angled with $P S=Q R$ and $P X=Q Y$, then these triangles are congruent, and so $S X=Y R$.\n\nSince $X Y=7$ and $S R=15$, then $S X+7+Y R=15$ or $2 \\times S X=8$ and so $S X=4$.\n\nBy the Pythagorean Theorem in $\\triangle P X S$,\n\n$$\nP X^{2}=P S^{2}-S X^{2}=8^{2}-4^{2}=64-16=48\n$$\n\nNow $P R$ is the hypotenuse of right-angled $\\triangle P X R$.\n\nSince $P R>0$, then by the Pythagorean Theorem,\n\n$$\nP R=\\sqrt{P X^{2}+X R^{2}}=\\sqrt{48+(7+4)^{2}}=\\sqrt{48+11^{2}}=\\sqrt{48+121}=\\sqrt{169}=13\n$$\n\nTherefore, $P R=13$.']",['13'],False,,Numerical, 2480,Geometry,,"In the diagram, $P Q R S$ is an isosceles trapezoid with $P Q=7, P S=Q R=8$, and $S R=15$. Determine the length of the diagonal $P R$. ","['Draw perpendiculars from $P$ and $Q$ to $X$ and $Y$, respectively, on $S R$.\n\n\n\nSince $P Q$ is parallel to $S R$ (because $P Q R S$ is a trapezoid) and $P X$ and $Q Y$ are perpendicular to $S R$, then $P Q Y X$ is a rectangle.\n\nThus, $X Y=P Q=7$ and $P X=Q Y$.\n\nSince $\\triangle P X S$ and $\\triangle Q Y R$ are right-angled with $P S=Q R$ and $P X=Q Y$, then these triangles are congruent, and so $S X=Y R$.\n\nSince $X Y=7$ and $S R=15$, then $S X+7+Y R=15$ or $2 \\times S X=8$ and so $S X=4$.\n\nBy the Pythagorean Theorem in $\\triangle P X S$,\n\n$$\nP X^{2}=P S^{2}-S X^{2}=8^{2}-4^{2}=64-16=48\n$$\n\nNow $P R$ is the hypotenuse of right-angled $\\triangle P X R$.\n\nSince $P R>0$, then by the Pythagorean Theorem,\n\n$$\nP R=\\sqrt{P X^{2}+X R^{2}}=\\sqrt{48+(7+4)^{2}}=\\sqrt{48+11^{2}}=\\sqrt{48+121}=\\sqrt{169}=13\n$$\n\nTherefore, $P R=13$.']",['13'],False,,Numerical, 2481,Combinatorics,,"Blaise and Pierre will play 6 games of squash. Since they are equally skilled, each is equally likely to win any given game. (In squash, there are no ties.) The probability that each of them will win 3 of the 6 games is $\frac{5}{16}$. What is the probability that Blaise will win more games than Pierre?","['There are two possibilities: either each player wins three games or one player wins more games than the other.\n\nSince the probability that each player wins three games is $\\frac{5}{16}$, then the probability that any one player wins more games than the other is $1-\\frac{5}{16}=\\frac{11}{16}$.\n\nSince each of Blaise and Pierre is equally likely to win any given game, then each must be equally likely to win more games than the other.\n\nTherefore, the probability that Blaise wins more games than Pierre is $\\frac{1}{2} \\times \\frac{11}{16}=\\frac{11}{32}$.', 'We consider the results of the 6 games as a sequence of 6 Bs or Ps, with each letter a B if Blaise wins the corresponding game or $\\mathrm{P}$ if Pierre wins.\n\nSince the two players are equally skilled, then the probability that each wins a given game is $\\frac{1}{2}$. This means that the probability of each letter being a $B$ is $\\frac{1}{2}$ and the probability of each letter being a $\\mathrm{P}$ is also $\\frac{1}{2}$.\n\nSince each sequence consists of 6 letters, then the probability of a particular sequence occurring is $\\left(\\frac{1}{2}\\right)^{6}=\\frac{1}{64}$, because each of the letters is specified.\n\nSince they play 6 games in total, then the probability that Blaise wins more games than Pierre is the sum of the probabilities that Blaise wins 4 games, that Blaise wins 5 games, and that Blaise wins 6 games.\n\nIf Blaise wins 6 games, then the sequence consists of 6 Bs. The probability of this is $\\frac{1}{64}$, since there is only one way to arrange $6 \\mathrm{Bs}$.\n\nIf Blaise wins 5 games, then the sequence consists of $5 \\mathrm{Bs}$ and $1 \\mathrm{P}$. The probability of this is $6 \\times \\frac{1}{64}=\\frac{6}{64}$, since there are 6 possible positions in the list for the $1 \\mathrm{P}$ (eg. PBBBBB,BPBBBB, BBPBBB, BBBPBB, BBBBPB, BBBBBP).\n\nThe probability that Blaise wins 4 games is $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right) \\times \\frac{1}{64}=\\frac{15}{64}$, since there are $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)=15$ ways for 4 Bs and 2 Ps to be arranged.\n\nTherefore, the probability that Blaise wins more games than Pierre is $\\frac{1}{64}+\\frac{6}{64}+\\frac{15}{64}=\\frac{22}{64}=\\frac{11}{32}$.']",['$\\frac{11}{32}$'],False,,Numerical, 2482,Algebra,,"Determine all real values of $x$ for which $$ 3^{x+2}+2^{x+2}+2^{x}=2^{x+5}+3^{x} $$","['Using exponent rules and arithmetic, we manipulate the given equation:\n\n$$\n\\begin{aligned}\n3^{x+2}+2^{x+2}+2^{x} & =2^{x+5}+3^{x} \\\\\n3^{x} 3^{2}+2^{x} 2^{2}+2^{x} & =2^{x} 2^{5}+3^{x} \\\\\n9\\left(3^{x}\\right)+4\\left(2^{x}\\right)+2^{x} & =32\\left(2^{x}\\right)+3^{x} \\\\\n8\\left(3^{x}\\right) & =27\\left(2^{x}\\right) \\\\\n\\frac{3^{x}}{2^{x}} & =\\frac{27}{8} \\\\\n\\left(\\frac{3}{2}\\right)^{x} & =\\left(\\frac{3}{2}\\right)^{3}\n\\end{aligned}\n$$\n\nSince the two expressions are equal and the bases are equal, then the exponents must be equal, so $x=3$.']",['3'],False,,Numerical, 2483,Geometry,,"In the diagram, $\triangle A B C$ has $A B=A C$ and $\angle B A C<60^{\circ}$. Point $D$ is on $A C$ with $B C=B D$. Point $E$ is on $A B$ with $B E=E D$. If $\angle B A C=\theta$, determine $\angle B E D$ in terms of $\theta$. ","['Since $A B=A C$, then $\\triangle A B C$ is isosceles and $\\angle A B C=\\angle A C B$. Note that $\\angle B A C=\\theta$.\n\n\n\nThe angles in $\\triangle A B C$ add to $180^{\\circ}$, so $\\angle A B C+\\angle A C B+\\angle B A C=180^{\\circ}$.\n\nThus, $2 \\angle A C B+\\theta=180^{\\circ}$ or $\\angle A B C=\\angle A C B=\\frac{1}{2}\\left(180^{\\circ}-\\theta\\right)=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nNow $\\triangle B C D$ is isosceles as well with $B C=B D$ and so $\\angle C D B=\\angle D C B=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nSince the angles in $\\triangle B C D$ add to $180^{\\circ}$, then\n\n$$\n\\angle C B D=180^{\\circ}-\\angle D C B-\\angle C D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)=\\theta\n$$\n\nNow $\\angle E B D=\\angle A B C-\\angle D B C=\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\theta=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nSince $B E=E D$, then $\\angle E D B=\\angle E B D=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nTherefore, $\\angle B E D=180^{\\circ}-\\angle E B D-\\angle E D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)=3 \\theta$.']",['$3 \\theta$'],False,,Expression, 2483,Geometry,,"In the diagram, $\triangle A B C$ has $A B=A C$ and $\angle B A C<60^{\circ}$. Point $D$ is on $A C$ with $B C=B D$. Point $E$ is on $A B$ with $B E=E D$. If $\angle B A C=\theta$, determine $\angle B E D$ in terms of $\theta$. ![](https://cdn.mathpix.com/cropped/2023_12_21_040d04737e70e698092cg-1.jpg?height=496&width=455&top_left_y=1628&top_left_x=1320)","['Since $A B=A C$, then $\\triangle A B C$ is isosceles and $\\angle A B C=\\angle A C B$. Note that $\\angle B A C=\\theta$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_920fb565baecc5ceb3fcg-1.jpg?height=494&width=460&top_left_y=1260&top_left_x=930)\n\nThe angles in $\\triangle A B C$ add to $180^{\\circ}$, so $\\angle A B C+\\angle A C B+\\angle B A C=180^{\\circ}$.\n\nThus, $2 \\angle A C B+\\theta=180^{\\circ}$ or $\\angle A B C=\\angle A C B=\\frac{1}{2}\\left(180^{\\circ}-\\theta\\right)=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nNow $\\triangle B C D$ is isosceles as well with $B C=B D$ and so $\\angle C D B=\\angle D C B=90^{\\circ}-\\frac{1}{2} \\theta$.\n\nSince the angles in $\\triangle B C D$ add to $180^{\\circ}$, then\n\n$$\n\\angle C B D=180^{\\circ}-\\angle D C B-\\angle C D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)=\\theta\n$$\n\nNow $\\angle E B D=\\angle A B C-\\angle D B C=\\left(90^{\\circ}-\\frac{1}{2} \\theta\\right)-\\theta=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nSince $B E=E D$, then $\\angle E D B=\\angle E B D=90^{\\circ}-\\frac{3}{2} \\theta$.\n\nTherefore, $\\angle B E D=180^{\\circ}-\\angle E B D-\\angle E D B=180^{\\circ}-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)-\\left(90^{\\circ}-\\frac{3}{2} \\theta\\right)=3 \\theta$.']",['$3 \\theta$'],False,,Expression, 2484,Geometry,,"In the diagram, the ferris wheel has a diameter of $18 \mathrm{~m}$ and rotates at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he is $1 \mathrm{~m}$ above the ground. When Kolapo is at point $P$ that is $16 \mathrm{~m}$ above the ground and is rising, it takes him 4 seconds to reach the highest point, $T$. He continues to travel for another 8 seconds reaching point $Q$. Determine Kolapo's height above the ground when he reaches point $Q$. ","[""Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.\n\nSince the radius of the ferris wheel is $9 \\mathrm{~m}$ (half of the diameter of $18 \\mathrm{~m}$ ) and $B$ is $1 \\mathrm{~m}$ above the ground, then $O$ is $9+1=10 \\mathrm{~m}$ above the ground.\n\nLet $\\angle T O P=\\theta$.\n\n\n\nSince the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\\angle T O Q=2 \\theta$.\n\nDraw a perpendicular from $P$ to $R$ on $T B$ and from $Q$ to $G$ on $T B$.\n\nSince $P$ is $16 \\mathrm{~m}$ above the ground and $O$ is $10 \\mathrm{~m}$ above the ground, then $O R=6 \\mathrm{~m}$.\n\nSince $O P$ is a radius of the circle, then $O P=9 \\mathrm{~m}$.\n\nLooking at right-angled $\\triangle O R P$, we see that $\\cos \\theta=\\frac{O R}{O P}=\\frac{6}{9}=\\frac{2}{3}$.\n\nSince $\\cos \\theta=\\frac{2}{3}<\\frac{1}{\\sqrt{2}}=\\cos \\left(45^{\\circ}\\right)$, then $\\theta>45^{\\circ}$.\n\nThis means that $2 \\theta>90^{\\circ}$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.\n\nSince $\\angle T O Q=2 \\theta$, then $\\angle Q O G=180^{\\circ}-2 \\theta$.\n\nKolapo's height above the ground at $Q$ equals $1 \\mathrm{~m}$ plus the length of $B G$.\n\nNow $B G=O B-O G$. We know that $O B=9 \\mathrm{~m}$.\n\nAlso, considering right-angled $\\triangle Q O G$, we have\n\n$$\nO G=O Q \\cos (\\angle Q O G)=9 \\cos \\left(180^{\\circ}-2 \\theta\\right)=-9 \\cos (2 \\theta)=-9\\left(2 \\cos ^{2} \\theta-1\\right)\n$$\n\nSince $\\cos \\theta=\\frac{2}{3}$, then $O G=-9\\left(2\\left(\\frac{2}{3}\\right)^{2}-1\\right)=-9\\left(\\frac{8}{9}-1\\right)=1 \\mathrm{~m}$.\n\nTherefore, $B G=9-1=8 \\mathrm{~m}$ and so $Q$ is $1+8=9 \\mathrm{~m}$ above the ground.""]",['9'],False,m,Numerical, 2484,Geometry,,"In the diagram, the ferris wheel has a diameter of $18 \mathrm{~m}$ and rotates at a constant rate. When Kolapo rides the ferris wheel and is at its lowest point, he is $1 \mathrm{~m}$ above the ground. When Kolapo is at point $P$ that is $16 \mathrm{~m}$ above the ground and is rising, it takes him 4 seconds to reach the highest point, $T$. He continues to travel for another 8 seconds reaching point $Q$. Determine Kolapo's height above the ground when he reaches point $Q$. ![](https://cdn.mathpix.com/cropped/2023_12_21_040d04737e70e698092cg-1.jpg?height=439&width=591&top_left_y=2125&top_left_x=1258)","[""Let $O$ be the centre of the ferris wheel and $B$ the lowest point on the wheel.\n\nSince the radius of the ferris wheel is $9 \\mathrm{~m}$ (half of the diameter of $18 \\mathrm{~m}$ ) and $B$ is $1 \\mathrm{~m}$ above the ground, then $O$ is $9+1=10 \\mathrm{~m}$ above the ground.\n\nLet $\\angle T O P=\\theta$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_09e23e682ca3ac23cf59g-1.jpg?height=441&width=555&top_left_y=384&top_left_x=880)\n\nSince the ferris wheel rotates at a constant speed, then in 8 seconds, the angle through which the wheel rotates is twice the angle through which it rotates in 4 seconds. In other words, $\\angle T O Q=2 \\theta$.\n\nDraw a perpendicular from $P$ to $R$ on $T B$ and from $Q$ to $G$ on $T B$.\n\nSince $P$ is $16 \\mathrm{~m}$ above the ground and $O$ is $10 \\mathrm{~m}$ above the ground, then $O R=6 \\mathrm{~m}$.\n\nSince $O P$ is a radius of the circle, then $O P=9 \\mathrm{~m}$.\n\nLooking at right-angled $\\triangle O R P$, we see that $\\cos \\theta=\\frac{O R}{O P}=\\frac{6}{9}=\\frac{2}{3}$.\n\nSince $\\cos \\theta=\\frac{2}{3}<\\frac{1}{\\sqrt{2}}=\\cos \\left(45^{\\circ}\\right)$, then $\\theta>45^{\\circ}$.\n\nThis means that $2 \\theta>90^{\\circ}$, which means that $Q$ is below the horizontal diameter through $O$ and so $G$ is below $O$.\n\nSince $\\angle T O Q=2 \\theta$, then $\\angle Q O G=180^{\\circ}-2 \\theta$.\n\nKolapo's height above the ground at $Q$ equals $1 \\mathrm{~m}$ plus the length of $B G$.\n\nNow $B G=O B-O G$. We know that $O B=9 \\mathrm{~m}$.\n\nAlso, considering right-angled $\\triangle Q O G$, we have\n\n$$\nO G=O Q \\cos (\\angle Q O G)=9 \\cos \\left(180^{\\circ}-2 \\theta\\right)=-9 \\cos (2 \\theta)=-9\\left(2 \\cos ^{2} \\theta-1\\right)\n$$\n\nSince $\\cos \\theta=\\frac{2}{3}$, then $O G=-9\\left(2\\left(\\frac{2}{3}\\right)^{2}-1\\right)=-9\\left(\\frac{8}{9}-1\\right)=1 \\mathrm{~m}$.\n\nTherefore, $B G=9-1=8 \\mathrm{~m}$ and so $Q$ is $1+8=9 \\mathrm{~m}$ above the ground.""]",['9'],False,m,Numerical, 2485,Algebra,,"On Saturday, Jimmy started painting his toy helicopter between 9:00 a.m. and 10:00 a.m. When he finished between 10:00 a.m. and 11:00 a.m. on the same morning, the hour hand was exactly where the minute hand had been when he started, and the minute hand was exactly where the hour hand had been when he started. Jimmy spent $t$ hours painting. Determine the value of $t$. ","['The hour hand and minute hand both turn at constant rates. Since the hour hand moves $\\frac{1}{12}$ of the way around the clock in 1 hour and the minute hand moves all of the way around the clock in 1 hour, then the minute hand turns 12 times as quickly as the hour hand.\n\n\nSuppose also that the hour hand moves through an angle of $x^{\\circ}$ between Before and After. Therefore, the minute hand moves through an angle of $\\left(360^{\\circ}-x^{\\circ}\\right)$ between Before and After, since these two angles add to $360^{\\circ}$.\n\n\n\nSince the minute hand moves 12 times as quickly as the hour hand, then $\\frac{360^{\\circ}-x^{\\circ}}{x^{\\circ}}=12$ or $360-x=12 x$ and so $13 x=360$, or $x=\\frac{360}{13}$.\n\nIn one hour, the hour hand moves through $\\frac{1}{12} \\times 360^{\\circ}=30^{\\circ}$.\n\nSince the hour hand is moving for $t$ hours, then we have $30^{\\circ} t=\\left(\\frac{360}{13}\\right)^{\\circ}$ and so $t=\\frac{360}{30(13)}=\\frac{12}{13}$.', ""Suppose that Jimmy starts painting $x$ hours after 9:00 a.m. and finishes painting $y$ hours after 10:00 a.m., where $0","['We use the notation $|P M Q N|$ to represent the area of quadrilateral $|P M Q N|,|\\triangle A P D|$ to represent the area of $\\triangle A P D$, and so on.\n\nWe want to show that $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$.\n\nThis is equivalent to showing\n\n$$\n|P M Q N|+|\\triangle D P N|+|\\triangle C Q N|=|\\triangle A P D|+|\\triangle D P N|+|\\triangle B Q C|+|\\triangle C Q N|\n$$\n\nwhich is equivalent to showing\n\n$$\n|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|\n$$\n\nsince combining quadrilateral $P M Q N$ with $\\triangle D P N$ and $\\triangle C Q N$ gives $\\triangle D M C$, combining $\\triangle A P D$ with $\\triangle D P N$ gives $\\triangle D A N$, and combining $\\triangle B Q C$ with $\\triangle C Q N$ gives $\\triangle C B N$. Suppose that $D C$ has length $x$ and $D N$ has length $t x$ for some $t$ with $0\n\nFigure 1\n\nThen $|\\triangle D A N|=\\frac{1}{2}(t x)(a)$ and $|\\triangle C B N|=\\frac{1}{2}((1-t) x) b$ so\n\n$$\n|\\triangle D A N|+|\\triangle C B N|=\\frac{1}{2}(t x a+(1-t) x b)=\\frac{1}{2} x(t a+(1-t) b)\n$$\n\nAlso, $|\\triangle D M C|=\\frac{1}{2} x m$.\n\nIn order to prove that $|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|$, we need to show that $\\frac{1}{2} x m$ equals\n\n\n\n$\\frac{1}{2} x(t a+(1-t) b)$ which is equivalent to showing that $m$ is equal to $t a+(1-t) b$.\n\nIn Figure 2, we draw a horizontal line from $A$ to $B G$, meeting $M F$ at $R$ and $B G$ at $S$.\n\nSince $M F$ and $B G$ are vertical and $A R S$ is horizontal, then these line segments are perpendicular.\n\nSince $A E=a, M F=m$ and $B G=b$, then $M R=m-a$ and $B S=b-a$.\n\n\n\nFigure 2\n\nNow $\\triangle A R M$ is similar to $\\triangle A S B$, since each is right-angled and they share a common angle at $A$.\n\nTherefore, $\\frac{M R}{B S}=\\frac{A M}{A B}=\\frac{N C}{D C}$.\n\nSince $M R=m-a$ and $B S=b-a$, then $\\frac{M R}{B S}=\\frac{m-a}{b-a}$.\n\nSince $\\frac{A M}{A B}=\\frac{N C}{D C}$, then $\\frac{M R}{B S}=\\frac{(1-t) x}{x}=1-t$.\n\nComparing these two expressions, we obtain $\\frac{m-a}{b-a}=(1-t)$ or $m-a=(b-a)(1-t)$ or $m=a+b(1-t)+(t-1) a=t a+(1-t) b$, as required.\n\nThis concludes the proof, and so $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$, as required.', 'Let $A M=x$ and $M B=y$. Then $A B=x+y$ and so $\\frac{A M}{A B}=\\frac{x}{x+y}$.\n\nLet $N C=n x$ for some real number $n$.\n\nSince $\\frac{N C}{D C}=\\frac{A M}{A B}$, then $\\frac{n x}{D C}=\\frac{x}{x+y}$ and so $D C=n(x+y)$.\n\nThis tells us that $D N=D C-N C=n(x+y)-n x=n y$.\n\nJoin $M$ to $N$ and label the areas as shown in the diagram:\n\n\n\nWe repeatedly use the fact that triangles with a common height have areas in proportion to the lengths of their bases.\n\nFor example, $\\triangle M D N$ and $\\triangle M N C$ have a common height from line segment to $D C$ to $M$ and so the ratio of their areas equals the ratio of the lengths of their bases.\n\nIn other words, $\\frac{w+r}{u+v}=\\frac{n x}{n y}=\\frac{x}{y}$. Thus, $w+r=\\frac{x}{y}(u+v)$.\n\n\n\nAlso, the ratio of the area of $\\triangle N A M$ to the area of $\\triangle N M B$ equals the ratio of $A M$ to $M B$.\n\nThis gives $\\frac{k+v}{s+w}=\\frac{x}{y}$ or $k+v=\\frac{x}{y}(s+w)$.\n\nNext, we join $A$ to $C$ and relabel the areas divided by this new line segment as shown:\n\n\n\n(The unlabelled triangle adjacent to the one labelled $k_{1}$ has area $k_{2}$ and the unlabelled triangle adjacent to the one labelled $r_{2}$ has area $r_{1}$.)\n\nConsider $\\triangle A N C$ and $\\triangle A D N$.\n\nAs above, the ratio of their areas equals the ratio of their bases.\n\nThus, $\\frac{k_{2}+v_{2}+w_{2}+r_{2}}{z+u}=\\frac{n x}{n y}=\\frac{x}{y}$, and so $k_{2}+v_{2}+w_{2}+r_{2}=\\frac{x}{y}(z+u)$.\n\nConsider $\\triangle C A M$ and $\\triangle C M B$.\n\nAs above, the ratio of their areas equals the ratio of their bases.\n\nThus, $\\frac{k_{1}+v_{1}+w_{1}+r_{1}}{s+t}=\\frac{x}{y}$, and so $k_{1}+v_{1}+w_{1}+r_{1}=\\frac{x}{y}(s+t)$.\n\nAdding $k_{2}+v_{2}+w_{2}+r_{2}=\\frac{x}{y}(z+u)$ and $k_{1}+v_{1}+w_{1}+r_{1}=\\frac{x}{y}(s+t)$ gives\n\n$$\n\\left(k_{1}+k_{2}\\right)+\\left(v_{1}+v_{2}\\right)+\\left(w_{1}+w_{2}\\right)+\\left(r_{1}+r_{2}\\right)=\\frac{x}{y}(s+t+z+u)\n$$\n\nor\n\n$$\nk+v+w+r=\\frac{x}{y}(s+t+z+u)\n$$\n\nSince $k+v=\\frac{x}{y}(s+w)$ and $w+r=\\frac{x}{y}(u+v)$, then\n\n$$\n\\frac{x}{y}(s+w)+\\frac{x}{y}(u+v)=\\frac{x}{y}(s+t+z+u)\n$$\n\nwhich gives\n\n$$\ns+w+u+v=s+t+z+u\n$$\n\nor\n\n$$\nw+v=t+z\n$$\n\nBut $w+v$ is the area of quadrilateral $P M Q N, z$ is the area of $\\triangle A P D$ and $t$ is the area of $\\triangle B Q C$. In other words, the area of quadrilateral $P M Q N$ equals the sum of the areas of $\\triangle A P D$ and $\\triangle P Q C$, as required.']",,True,,, 2488,Geometry,,"In the diagram, quadrilateral $A B C D$ has points $M$ and $N$ on $A B$ and $D C$, respectively, with $\frac{A M}{A B}=\frac{N C}{D C}$. Line segments $A N$ and $D M$ intersect at $P$, while $B N$ and $C M$ intersect at $Q$. Prove that the area of quadrilateral $P M Q N$ equals the sum of the areas of $\triangle A P D$ and $\triangle B Q C$. ![](https://cdn.mathpix.com/cropped/2023_12_21_4e0d9d101cf875a61e98g-1.jpg?height=358&width=629&top_left_y=1035&top_left_x=1168)","['We use the notation $|P M Q N|$ to represent the area of quadrilateral $|P M Q N|,|\\triangle A P D|$ to represent the area of $\\triangle A P D$, and so on.\n\nWe want to show that $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$.\n\nThis is equivalent to showing\n\n$$\n|P M Q N|+|\\triangle D P N|+|\\triangle C Q N|=|\\triangle A P D|+|\\triangle D P N|+|\\triangle B Q C|+|\\triangle C Q N|\n$$\n\nwhich is equivalent to showing\n\n$$\n|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|\n$$\n\nsince combining quadrilateral $P M Q N$ with $\\triangle D P N$ and $\\triangle C Q N$ gives $\\triangle D M C$, combining $\\triangle A P D$ with $\\triangle D P N$ gives $\\triangle D A N$, and combining $\\triangle B Q C$ with $\\triangle C Q N$ gives $\\triangle C B N$. Suppose that $D C$ has length $x$ and $D N$ has length $t x$ for some $t$ with $0","['Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nTherefore, the area of the part of the circle inside the first quadrant is equal to the area of $\\triangle A O B$ plus the area of the semi-circle above $A B$.\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the semi-circle is $\\frac{1}{2} \\pi(\\sqrt{2})^{2}=\\pi$.\n\nThe area of $\\triangle A O B$ is $\\frac{1}{2}(O B)(A O)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of the part of the circle inside the first quadrant is $\\pi+2$.', 'Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nThus, $A O=B O$.\n\nWe ""complete the square"" by adding point $D(2,2)$, which is on the circle, by symmetry.\n\n\n\n\n\nThe area of the square is 4 .\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the circle is $\\pi(\\sqrt{2})^{2}=2 \\pi$.\n\nThe area of the portion of the circle outside the square is thus $2 \\pi-4$. This area is divided into four equal sections (each of area $\\frac{1}{4}(2 \\pi-4)=\\frac{1}{2} \\pi-1$ ), two of which are the only portions of the circle outside the first quadrant.\n\nTherefore, the area of the part of the circle inside the the first quadrant is $2 \\pi-2\\left(\\frac{1}{2} \\pi-1\\right)=$ $\\pi+2$.\n\nTwo additional ways to find the coordinates of $A$ :\n\n$*$ The length of $O C$ is $\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$.\n\nSince $C$ is the centre of the circle and $O$ lies on the circle, then the circle has radius $\\sqrt{2}$.\n\nSince the circle has centre $(1,1)$ and radius $\\sqrt{2}$, its equation is $(x-1)^{2}+(y-1)^{2}=2$. To find the coordinates of $A$, we substitute $x=0$ to obtain $(0-1)^{2}+(y-1)^{2}=2$ or $(y-1)^{2}=1$, and so $y=0$ or $y=2$.\n\nSince $y=0$ gives us the point $O$, then $y=2$ gives us $A$, ie. $A$ has coordinates $(0,2)$.\n\n* Since $O$ and $A$ are both on the circle and each has a horizontal distance of 1 from $C$, then their vertical distances from $C$ must be same, ie. must each be 1.\n\nThus, $A$ has coordinates $(0,2)$.']","['$(0,2),\\pi+2$']",True,,"Tuple,Numerical", 2491,Geometry,,"In the diagram, the circle with centre $C(1,1)$ passes through the point $O(0,0)$, intersects the $y$-axis at $A$, and intersects the $x$-axis at $B(2,0)$. Determine, with justification, the coordinates of $A$ and the area of the part of the circle that lies in the first quadrant. ![](https://cdn.mathpix.com/cropped/2023_12_21_09a69cbcf9a187d86706g-1.jpg?height=539&width=547&top_left_y=281&top_left_x=1258)","['Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9efb0d708494864661dbg-1.jpg?height=534&width=561&top_left_y=335&top_left_x=869)\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nTherefore, the area of the part of the circle inside the first quadrant is equal to the area of $\\triangle A O B$ plus the area of the semi-circle above $A B$.\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the semi-circle is $\\frac{1}{2} \\pi(\\sqrt{2})^{2}=\\pi$.\n\nThe area of $\\triangle A O B$ is $\\frac{1}{2}(O B)(A O)=\\frac{1}{2}(2)(2)=2$.\n\nThus, the area of the part of the circle inside the first quadrant is $\\pi+2$.', 'Since $\\angle A O B=90^{\\circ}, A B$ is a diameter of the circle.\n\nJoin $A B$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_9efb0d708494864661dbg-1.jpg?height=540&width=551&top_left_y=1535&top_left_x=863)\n\nSince $C$ is the centre of the circle and $A B$ is a diameter, then $C$ is the midpoint of $A B$, so $A$ has coordinates $(0,2)$.\n\nThus, $A O=B O$.\n\nWe ""complete the square"" by adding point $D(2,2)$, which is on the circle, by symmetry.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_dded398d413ac384716fg-1.jpg?height=553&width=572&top_left_y=157&top_left_x=858)\n\nThe area of the square is 4 .\n\nThe radius of the circle is equal to the distance from $C$ to $B$, or $\\sqrt{(1-2)^{2}+(1-0)^{2}}=\\sqrt{2}$, so the area of the circle is $\\pi(\\sqrt{2})^{2}=2 \\pi$.\n\nThe area of the portion of the circle outside the square is thus $2 \\pi-4$. This area is divided into four equal sections (each of area $\\frac{1}{4}(2 \\pi-4)=\\frac{1}{2} \\pi-1$ ), two of which are the only portions of the circle outside the first quadrant.\n\nTherefore, the area of the part of the circle inside the the first quadrant is $2 \\pi-2\\left(\\frac{1}{2} \\pi-1\\right)=$ $\\pi+2$.\n\nTwo additional ways to find the coordinates of $A$ :\n\n$*$ The length of $O C$ is $\\sqrt{1^{2}+1^{2}}=\\sqrt{2}$.\n\nSince $C$ is the centre of the circle and $O$ lies on the circle, then the circle has radius $\\sqrt{2}$.\n\nSince the circle has centre $(1,1)$ and radius $\\sqrt{2}$, its equation is $(x-1)^{2}+(y-1)^{2}=2$. To find the coordinates of $A$, we substitute $x=0$ to obtain $(0-1)^{2}+(y-1)^{2}=2$ or $(y-1)^{2}=1$, and so $y=0$ or $y=2$.\n\nSince $y=0$ gives us the point $O$, then $y=2$ gives us $A$, ie. $A$ has coordinates $(0,2)$.\n\n* Since $O$ and $A$ are both on the circle and each has a horizontal distance of 1 from $C$, then their vertical distances from $C$ must be same, ie. must each be 1.\n\nThus, $A$ has coordinates $(0,2)$.']","['$(0,2),\\pi+2$']",True,,"Tuple,Numerical", 2492,Combinatorics,,"If $a$ is chosen randomly from the set $\{1,2,3,4,5\}$ and $b$ is chosen randomly from the set $\{6,7,8\}$, what is the probability that $a^{b}$ is an even number?","['Since there are 5 choices for $a$ and 3 choices for $b$, there are fifteen possible ways of choosing $a$ and $b$.\n\nIf $a$ is even, $a^{b}$ is even; if $a$ is odd, $a^{b}$ is odd.\n\nSo the choices of $a$ and $b$ which give an even value for $a^{b}$ are those where $a$ is even, or 6 of the choices (since there are two even choices for $a$ and three ways of choosing $b$ for each of these). (Notice that in fact the value of $b$ does not affect whether $a^{b}$ is even or odd, so the probability depends only on the choice of $a$.)\n\nThus, the probability is $\\frac{6}{15}=\\frac{2}{5}$.']",['$\\frac{2}{5}$'],False,,Numerical, 2493,Combinatorics,,"A bag contains some blue and some green hats. On each turn, Julia removes one hat without looking, with each hat in the bag being equally likely to be chosen. If it is green, she adds a blue hat into the bag from her supply of extra hats, and if it is blue, she adds a green hat to the bag. The bag initially contains 4 blue hats and 2 green hats. What is the probability that the bag again contains 4 blue hats and 2 green hats after two turns?","['Starting with 4 blue hats and 2 green hats, the probability that Julia removes a blue hat is $\\frac{4}{6}=\\frac{2}{3}$. The result would be 3 blue hats and 3 green hats, since a blue hat is replaced with a green hat.\n\nIn order to return to 4 blue hats and 2 green hats from 3 blue and 3 green, Julia would need remove a green hat (which would be replaced by a blue hat). The probability of her\n\n\n\nremoving a green hat from 3 blue and 3 green is $\\frac{3}{6}=\\frac{1}{2}$.\n\nSummarizing, the probability of choosing a blue hat and then a green hat is $\\frac{2}{3} \\times \\frac{1}{2}=\\frac{1}{3}$.\n\nStarting with 4 blue hats and 2 green hats, the probability that Julia removes a green hat is $\\frac{2}{6}=\\frac{1}{3}$. The result would be 5 blue hats and 1 green hat, since a green hat is replaced with a blue hat.\n\nIn order to return to 4 blue hats and 2 green hats from 5 blue and 1 green, Julia would need remove a blue hat (which would be replaced by a green hat). The probability of her removing a green hat from 5 blue and 1 green is $\\frac{5}{6}$.\n\nSummarizing, the probability of choosing a green hat and then a blue hat is $\\frac{1}{3} \\times \\frac{5}{6}=\\frac{5}{18}$.\n\nThese are the only two ways to return to 4 blue hats and 2 green hats after two turns removing a blue hat then a green, or removing a green then a blue.\n\nTherefore, the total probability of returning to 4 blue hats and 2 green hats after two turns is $\\frac{1}{3}+\\frac{5}{18}=\\frac{11}{18}$.']",['$\\frac{11}{18}$'],False,,Numerical, 2494,Combinatorics,,"Suppose that, for some angles $x$ and $y$, $$ \begin{aligned} & \sin ^{2} x+\cos ^{2} y=\frac{3}{2} a \\ & \cos ^{2} x+\sin ^{2} y=\frac{1}{2} a^{2} \end{aligned} $$ Determine the possible value(s) of $a$.","['Adding the two equations, we obtain\n\n$$\n\\begin{aligned}\n\\sin ^{2} x+\\cos ^{2} x+\\sin ^{2} y+\\cos ^{2} y & =\\frac{3}{2} a+\\frac{1}{2} a^{2} \\\\\n2 & =\\frac{3}{2} a+\\frac{1}{2} a^{2} \\\\\n4 & =3 a+a^{2} \\\\\n0 & =a^{2}+3 a-4 \\\\\n0 & =(a+4)(a-1)\n\\end{aligned}\n$$\n\nand so $a=-4$ or $a=1$.\n\nHowever, $a=-4$ is impossible, since this would give $\\sin ^{2} x+\\cos ^{2} y=-6$, whose left side is non-negative and whose right side is negative.\n\nTherefore, the only possible value for $a$ is $a=1$.\n\n(We can check that angles $x=90^{\\circ}$ and $y=45^{\\circ}$ give $\\sin ^{2} x+\\cos ^{2} y=\\frac{3}{2}$ and $\\cos ^{2} x+\\sin ^{2} y=$ $\\frac{1}{2}$, so $a=1$ is indeed possible.)']",['1'],False,,Numerical, 2495,Geometry,,"Survivors on a desert island find a piece of plywood $(A B C)$ in the shape of an equilateral triangle with sides of length $2 \mathrm{~m}$. To shelter their goat from the sun, they place edge $B C$ on the ground, lift corner $A$, and put in a vertical post $P A$ which is $h \mathrm{~m}$ long above ground. When the sun is directly overhead, the shaded region $(\triangle P B C)$ on the ground directly underneath the plywood is an isosceles triangle with largest angle $(\angle B P C)$ equal to $120^{\circ}$. Determine the value of $h$, to the nearest centimetre. ![](https://cdn.mathpix.com/cropped/2023_12_21_09a69cbcf9a187d86706g-1.jpg?height=317&width=376&top_left_y=1603&top_left_x=1449)","['From the given information, $P C=P B$.\n\nIf we can calculate the length of $P C$, we can calculate the value of $h$, since we already know the length of $A C$.\n\nNow $\\triangle C P B$ is isosceles with $P C=P B, B C=2$ and $\\angle B P C=120^{\\circ}$.\n\nSince $\\triangle C P B$ is isosceles, $\\angle P C B=\\angle P B C=30^{\\circ}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6944921028a13ba96297g-1.jpg?height=206&width=463&top_left_y=2274&top_left_x=934)\n\nJoin $P$ to the midpoint, $M$, of $B C$.\n\nThen $P M$ is perpendicular to $B C$, since $\\triangle P C B$ is isosceles.\n\n\n\nTherefore, $\\triangle P M C$ is right-angled, has $\\angle P C M=30^{\\circ}$ and has $C M=1$.\n\nThus, $P C=\\frac{2}{\\sqrt{3}}$.\n\n(There are many other techniques that we can use to calculate the length of $P C$.)\n\nReturning to $\\triangle A P C$, we see $A P^{2}=A C^{2}-P C^{2}$ or $h^{2}=2^{2}-\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}=4-\\frac{4}{3}=\\frac{8}{3}$, and so $h=\\sqrt{\\frac{8}{3}}=2 \\sqrt{\\frac{2}{3}}=\\frac{2 \\sqrt{6}}{3} \\approx 1.630$.\n\nTherefore, the height is approximately $1.63 \\mathrm{~m}$ or $163 \\mathrm{~cm}$.']",['163'],False,cm,Numerical, 2495,Geometry,,"Survivors on a desert island find a piece of plywood $(A B C)$ in the shape of an equilateral triangle with sides of length $2 \mathrm{~m}$. To shelter their goat from the sun, they place edge $B C$ on the ground, lift corner $A$, and put in a vertical post $P A$ which is $h \mathrm{~m}$ long above ground. When the sun is directly overhead, the shaded region $(\triangle P B C)$ on the ground directly underneath the plywood is an isosceles triangle with largest angle $(\angle B P C)$ equal to $120^{\circ}$. Determine the value of $h$, to the nearest centimetre. ","['From the given information, $P C=P B$.\n\nIf we can calculate the length of $P C$, we can calculate the value of $h$, since we already know the length of $A C$.\n\nNow $\\triangle C P B$ is isosceles with $P C=P B, B C=2$ and $\\angle B P C=120^{\\circ}$.\n\nSince $\\triangle C P B$ is isosceles, $\\angle P C B=\\angle P B C=30^{\\circ}$.\n\n\n\nJoin $P$ to the midpoint, $M$, of $B C$.\n\nThen $P M$ is perpendicular to $B C$, since $\\triangle P C B$ is isosceles.\n\n\n\nTherefore, $\\triangle P M C$ is right-angled, has $\\angle P C M=30^{\\circ}$ and has $C M=1$.\n\nThus, $P C=\\frac{2}{\\sqrt{3}}$.\n\n(There are many other techniques that we can use to calculate the length of $P C$.)\n\nReturning to $\\triangle A P C$, we see $A P^{2}=A C^{2}-P C^{2}$ or $h^{2}=2^{2}-\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}=4-\\frac{4}{3}=\\frac{8}{3}$, and so $h=\\sqrt{\\frac{8}{3}}=2 \\sqrt{\\frac{2}{3}}=\\frac{2 \\sqrt{6}}{3} \\approx 1.630$.\n\nTherefore, the height is approximately $1.63 \\mathrm{~m}$ or $163 \\mathrm{~cm}$.']",['163'],False,cm,Numerical, 2496,Algebra,,"The sequence $2,5,10,50,500, \ldots$ is formed so that each term after the second is the product of the two previous terms. The 15 th term ends with exactly $k$ zeroes. What is the value of $k$ ?","['We calculate the first 15 terms, writing each as an integer times a power of 10:\n\n$$\n\\begin{gathered}\n2,5,10,5 \\times 10,5 \\times 10^{2}, 5^{2} \\times 10^{3}, 5^{3} \\times 10^{5}, 5^{5} \\times 10^{8}, 5^{8} \\times 10^{13}, 5^{13} \\times 10^{21}, 5^{21} \\times 10^{34} \\\\\n5^{34} \\times 10^{55}, 5^{55} \\times 10^{89}, 5^{89} \\times 10^{144}, 5^{144} \\times 10^{233}\n\\end{gathered}\n$$\n\nSince the 15 th term equals an odd integer times $10^{233}$, then the 15 th term ends with 233 zeroes.', 'To obtain the 6 th term, we calculate $50 \\times 500=25 \\times 1000$.\n\nEach of the 4th and 5th terms equals an odd integer followed by a number of zeroes, so the 6th term also equals an odd integer followed by a number of zeroes, where the number of zeroes is the sum of the numbers of zeroes at the ends of the 4th and 5th terms.\n\nThis pattern will continue. Thus, starting with the 6th term, the number of zeroes at the end of the term will be the sum of the number of zeroes at the ends of the two previous terms.\n\nThis tells us that, starting with the 4th term, the number of zeroes at the ends of the terms is\n\n$$\n1,2,3,5,8,13,21,34,55,89,144,233\n$$\n\nTherefore, the 15 th term ends with 233 zeroes.']",['233'],False,,Numerical, 2497,Algebra,,"Suppose that $a, b, c$ are three consecutive terms in an arithmetic sequence. Prove that $a^{2}-b c, b^{2}-a c$, and $c^{2}-a b$ are also three consecutive terms in an arithmetic sequence. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms.)","['Since $a, b$ and $c$ are consecutive terms in an arithmetic sequence, then $b=a+d$ and $c=a+2 d$ for some number $d$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n& a^{2}-b c=a^{2}-(a+d)(a+2 d)=a^{2}-a^{2}-3 a d-2 d^{2}=-3 a d-2 d^{2} \\\\\n& b^{2}-a c=(a+d)^{2}-a(a+2 d)=a^{2}+2 a d+d^{2}-a^{2}-2 a d=d^{2} \\\\\n& c^{2}-a b=(a+2 d)^{2}-a(a+d)=a^{2}+4 a d+4 d^{2}-a^{2}-a d=3 a d+4 d^{2}\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=d^{2}-\\left(-3 a d-2 d^{2}\\right)=3 d^{2}+3 a d\n$$\n\nand\n\n$$\n\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)=\\left(3 a d+4 d^{2}\\right)-d^{2}=3 d^{2}+3 a d\n$$\n\nTherefore, $\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)$, so the sequence $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ is arithmetic.', 'Since $a, b$ and $c$ are consecutive terms in an arithmetic sequence, then $a=b-d$ and $c=b+d$ for some number $d$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n& a^{2}-b c=(b-d)^{2}-b(b+d)=b^{2}-2 b d+d^{2}-b^{2}-b d=-3 b d+d^{2} \\\\\n& b^{2}-a c=b^{2}-(b-d)(b+d)=b^{2}-b^{2}+d^{2}=d^{2} \\\\\n& c^{2}-a b=(b+d)^{2}-(b-d) b=b^{2}+2 b d+d^{2}-b^{2}+b d=3 b d+d^{2}\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=d^{2}-\\left(-3 b d+d^{2}\\right)=3 b d\n$$\n\nand\n\n$$\n\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)=\\left(3 b d+d^{2}\\right)-d^{2}=3 b d\n$$\n\nTherefore, $\\left(b^{2}-a c\\right)-\\left(a^{2}-b c\\right)=\\left(c^{2}-a b\\right)-\\left(b^{2}-a c\\right)$, so the sequence $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ is arithmetic.', 'To show that $a^{2}-b c, b^{2}-a c$ and $c^{2}-a b$ form an arithmetic sequence, we can show that $\\left(c^{2}-a b\\right)+\\left(a^{2}-b c\\right)=2\\left(b^{2}-a c\\right)$.\n\nSince $a, b$ and $c$ form an arithmetic sequence, then $a+c=2 b$.\n\nNow\n\n$$\n\\begin{aligned}\n\\left(c^{2}-a b\\right)+\\left(a^{2}-b c\\right) & =c^{2}+a^{2}-b(a+c) \\\\\n& =c^{2}+a^{2}+2 a c-b(a+c)-2 a c \\\\\n& =(c+a)^{2}-b(a+c)-2 a c \\\\\n& =(c+a)(a+c-b)-2 a c \\\\\n& =2 b(2 b-b)-2 a c \\\\\n& =2 b^{2}-2 a c \\\\\n& =2\\left(b^{2}-a c\\right)\n\\end{aligned}\n$$\n\nas required.']",,True,,, 2498,Algebra,,"If $\log _{2} x-2 \log _{2} y=2$, determine $y$, as a function of $x$","['We use logarithm rules to rearrange the equation to solve for $y$ :\n\n$$\n\\begin{aligned}\n\\log _{2} x-2 \\log _{2} y & =2 \\\\\n\\log _{2} x-\\log _{2}\\left(y^{2}\\right) & =2 \\\\\n\\log _{2}\\left(\\frac{x}{y^{2}}\\right) & =2 \\\\\n\\frac{x}{y^{2}} & =2^{2} \\\\\n\\frac{1}{4} x & =y^{2} \\\\\ny & = \\pm \\frac{1}{2} \\sqrt{x}\n\\end{aligned}\n$$\n\nBut since the domain of the $\\log _{2}$ function is all positive real numbers, we must have $x>0$ and $y>0$, so we can reject the negative square root to obtain\n\n$$\ny=\\frac{1}{2} \\sqrt{x}, \\quad x>0\n$$']","['$\\frac{1}{2},\\sqrt{x}$']",True,,Expression, 2499,Geometry,,"In the diagram, $A B$ and $B C$ are chords of the circle with $A B","['Join $A$ to $E$ and $C$, and $B$ to $E$.\n\n\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nNow $\\angle E A C=\\angle E B C$ since both are subtended by $E C$.\n\nTherefore, $\\angle E A C+\\angle A B C=\\angle E B C+\\angle A B C=\\angle E B A$ which is indeed equal to $90^{\\circ}$ as required, since $A E$ is a diameter.', 'Join $A$ to $E$ and $C$.\n\n\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nThus, $\\angle E C A=90^{\\circ}$.\n\nNow $\\angle A B C=\\angle A E C$ since both are subtended by $A C$.\n\nNow $\\angle E A C+\\angle A B C=\\angle E A C+\\angle A E C=180^{\\circ}-\\angle E C A$ using the sum of the angles in $\\triangle A E C$.\n\nBut $\\angle E C A=90^{\\circ}$, so $\\angle E A C+\\angle A E C=90^{\\circ}$.', 'Join $A$ to $E$ and $C$, and $C$ to $D$.\n\n\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nNow $\\angle A B C=\\angle A D C$ since both are subtended by $A C$.\n\nAlso $\\angle E A C=\\angle E D C$ since both are subtended by $E C$.\n\nSo $\\angle E A C+\\angle A B C=\\angle E D C+\\angle A D C=\\angle A D E=90^{\\circ}$.']",,True,,, 2500,Algebra,,"Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$. Determine all real numbers $k$ for which $f(x)$ is constant for all values of $x$.","['Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\sin ^{2} x+\\sin ^{4} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n\\end{aligned}\n$$\n\nTherefore, if $3+2 k=0$ or $k=-\\frac{3}{2}$, then $f(x)=1+k=-\\frac{1}{2}$ for all $x$ and so is constant. (If $k \\neq-\\frac{3}{2}$, then we get\n\n$$\n\\begin{aligned}\nf(0) & =1+k \\\\\nf\\left(\\frac{1}{4} \\pi\\right) & =(1+k)-(3+2 k)\\left(\\frac{1}{2}\\right)+(3+2 k)\\left(\\frac{1}{4}\\right)=\\frac{1}{4}+\\frac{1}{2} k \\\\\nf\\left(\\frac{1}{6} \\pi\\right) & =(1+k)-(3+2 k)\\left(\\frac{1}{4}\\right)+(3+2 k)\\left(\\frac{1}{16}\\right)=\\frac{7}{16}+\\frac{5}{8} k\n\\end{aligned}\n$$\n\nwhich cannot be all equal for any single value of $k$, so $f(x)$ is not constant if $k \\neq-\\frac{3}{2}$.)', 'Since $\\sin ^{2} x+\\cos ^{2} x=1$, then\n\n$$\n\\begin{aligned}\nf(x) & =\\left(\\sin ^{2} x+\\cos ^{2} x\\right)\\left(\\sin ^{4} x-\\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x\\right)+k\\left(\\sin ^{4} x+\\cos ^{4} x\\right) \\\\\n& =\\left(\\sin ^{4}+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-3 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& \\quad k\\left(\\sin ^{4} x+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-3 \\sin ^{2} x \\cos ^{2} x\\right)+k\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =1-3 \\sin ^{2} x \\cos ^{2} x+k\\left(1-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n\\end{aligned}\n$$\n\nTherefore, if $3+2 k=0$ or $k=-\\frac{3}{2}$, then $f(x)=1+k=-\\frac{1}{2}$ for all $x$ and so is constant.', 'For $f(x)$ to be constant, we need $f^{\\prime}(x)=0$ for all values of $x$.\n\nCalculating using the Chain Rule,\n\n$$\n\\begin{aligned}\nf^{\\prime}(x) & =6 \\sin ^{5} x \\cos x-6 \\cos ^{5} x \\sin x+k\\left(4 \\sin ^{3} x \\cos x-4 \\cos ^{3} x \\sin x\\right) \\\\\n& =2 \\sin x \\cos x\\left(3\\left(\\sin ^{4} x-\\cos ^{4} x\\right)+2 k\\left(\\sin ^{2} x-\\cos ^{2} x\\right)\\right) \\\\\n& =2 \\sin x \\cos x\\left(\\sin ^{2} x-\\cos ^{2} x\\right)\\left(3\\left(\\sin ^{2} x+\\cos ^{2} x\\right)+2 k\\right) \\\\\n& =2 \\sin x \\cos x\\left(\\sin ^{2} x-\\cos ^{2} x\\right)(3+2 k)\n\\end{aligned}\n$$\n\nIf $3+2 k=0$ or $k=-\\frac{3}{2}$, then $f^{\\prime}(x)=0$ for all $x$, so $f(x)$ is constant.\n\n(If $3+2 k \\neq 0$, then choosing $x=\\frac{1}{6} \\pi$ for example gives $f^{\\prime}(x) \\neq 0$ so $f(x)$ is not constant.)']",['$-\\frac{3}{2}$'],False,,Numerical, 2501,Algebra,,"Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$. If $k=-0.7$, determine all solutions to the equation $f(x)=0$.","['Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\sin ^{2} x+\\sin ^{4} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n$$\n\nand so we want to solve\n\n$$\n\\begin{array}{r}\n0.3-(1.6) \\sin ^{2} x+(1.6) \\sin ^{4} x=0 \\\\\n16 \\sin ^{4} x-16 \\sin ^{2} x+3=0 \\\\\n\\left(4 \\sin ^{2} x-3\\right)\\left(4 \\sin ^{2} x-1\\right)=0\n\\end{array}\n$$\n\nTherefore, $\\sin ^{2} x=\\frac{1}{4}, \\frac{3}{4}$, and so $\\sin x= \\pm \\frac{1}{2}, \\pm \\frac{\\sqrt{3}}{2}$.\n\nTherefore,\n\n$$\nx=\\frac{1}{6} \\pi+2 \\pi k, \\frac{5}{6} \\pi+2 \\pi k, \\frac{7}{6} \\pi+2 \\pi k, \\frac{11}{6} \\pi+2 \\pi k, \\frac{1}{3} \\pi+2 \\pi k, \\frac{2}{3} \\pi+2 \\pi k, \\frac{4}{3} \\pi+2 \\pi k, \\frac{5}{3} \\pi+2 \\pi k\n$$\n\nfor $k \\in \\mathbb{Z}$.', 'Since $\\sin ^{2} x+\\cos ^{2} x=1$, then\n\n$$\n\\begin{aligned}\nf(x) & =\\left(\\sin ^{2} x+\\cos ^{2} x\\right)\\left(\\sin ^{4} x-\\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x\\right)+k\\left(\\sin ^{4} x+\\cos ^{4} x\\right) \\\\\n& =\\left(\\sin ^{4}+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-3 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& \\quad k\\left(\\sin ^{4} x+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-3 \\sin ^{2} x \\cos ^{2} x\\right)+k\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =1-3 \\sin ^{2} x \\cos ^{2} x+k\\left(1-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n$$\n\nUsing the fact that $\\sin 2 x=2 \\sin x \\cos x$, we can further simplify $f(x)$ to\n\n$$\nf(x)=(1+k)-\\frac{1}{4}(3+2 k) \\sin ^{2} 2 x\n$$\n\n\n\nand so we want to solve\n\n$$\n\\begin{aligned}\n0.3-\\frac{1}{4}(1.6) \\sin ^{2} 2 x & =0 \\\\\n4 \\sin ^{2} 2 x & =3 \\\\\n\\sin ^{2} 2 x & =\\frac{3}{4}\n\\end{aligned}\n$$\n\nand so $\\sin 2 x= \\pm \\frac{\\sqrt{3}}{2}$.\n\nTherefore,\n\n$$\n2 x=\\frac{1}{3} \\pi+2 \\pi k, \\frac{2}{3} \\pi+2 \\pi k, \\frac{4}{3} \\pi+2 \\pi k, \\frac{5}{3} \\pi+2 \\pi k\n$$\n\nfor $k \\in \\mathbb{Z}$, and so\n\n$$\nx=\\frac{1}{6} \\pi+\\pi k, \\frac{1}{3} \\pi+\\pi k, \\frac{2}{3} \\pi+\\pi k, \\frac{5}{6} \\pi+\\pi k\n$$\n\nfor $k \\in \\mathbb{Z}$.']","['$x=\\frac{1}{6} \\pi+\\pi k, \\frac{1}{3} \\pi+\\pi k, \\frac{2}{3} \\pi+\\pi k, \\frac{5}{6} \\pi+\\pi k$']",True,,Expression, 2502,Algebra,,"Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$. Determine all real numbers $k$ for which there exists a real number $c$ such that $f(c)=0$.","['Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\sin ^{2} x+\\sin ^{4} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n$$\n\nWe want to determine the values of $k$ for which there is an $a$ such that $f(a)=0$.\n\nFrom (a), if $k=-\\frac{3}{2}, f(x)$ is constant and equal to $-\\frac{1}{2}$, so has no roots.\n\nLet $u=\\sin ^{2} x$.\n\nThen $u$ takes all values between 0 and 1 as $\\sin x$ takes all values between -1 and 1 . Then we want to determine for which $k$ the equation\n\n$$\n(3+2 k) u^{2}-(3+2 k) u+(1+k)=0\n$$\n\nhas a solution for $u$ with $0 \\leq u \\leq 1$.\n\nFirst, we must ensure that the equation $(*)$ has real solutions, ie.\n\n$$\n\\begin{aligned}\n(3+2 k)^{2}-4(3+2 k)(1+k) & \\geq 0 \\\\\n(3+2 k)(3+2 k-4(1+k)) & \\geq 0 \\\\\n(3+2 k)(-1-2 k) & \\geq 0 \\\\\n(3+2 k)(1+2 k) & \\leq 0\n\\end{aligned}\n$$\n\nThis is true if and only if $-\\frac{3}{2}-\\frac{3}{2}$ then $3+2 k>0$.\n\nFor $u$ to be between 0 and 1, we need to have\n\n$$\n0 \\leq \\sqrt{-\\frac{1+2 k}{3+2 k}} \\leq 1\n$$\n\nThus\n\n$$\n0 \\leq-\\frac{1+2 k}{3+2 k} \\leq 1\n$$\n\nSince $-\\frac{3}{2}0$ and $1+2 k \\leq 0$, so the left inequality is true.\n\nTherefore, we need $-\\frac{1+2 k}{3+2 k} \\leq 1$ or $-(1+2 k) \\leq(3+2 k)$ (we can multiply by $(3+2 k)$ since it is positive), and so $-4 \\leq 4 k$ or $k \\geq-1$.\n\nCombining with $-\\frac{3}{2}0$, we can multiply the inequality by $3+2 k$ to obtain\n\n$$\n0 \\leq 4(1+k) \\leq 3+2 k\n$$\n\nand so we get $k \\geq-1$ from the left inequality and $k \\leq-\\frac{1}{2}$ from the right inequality.\n\nCombining these with $-\\frac{3}{2} In this case, one of the arcs of the circle cut off by one of the sides of the triangle would have to be a major arc, which cannot happen, because of the above. Therefore, the centre is contained inside the triangle. If $N=7$, what is the probability that the triangle is acute? (A triangle is acute if each of its three interior angles is less than $90^{\circ}$.)","[""Since there are $N=7$ points from which the triangle's vertices can be chosen, there are $\\left(\\begin{array}{l}7 \\\\ 3\\end{array}\\right)=35$ triangles in total.\n\nWe compute the number of acute triangles.\n\nFix one of the vertices of such a triangle at $A_{1}$.\n\nWe construct the triangle by choosing the other two vertices in ascending subscript order. We choose the vertices by considering the arc length from the previous vertex - each of\n\n\n\nthese arc lengths must be smaller than half the total circumference of the circle.\n\nSince there are 7 equally spaced points on the circle, we assume the circumference is 7 , so the arc length formed by each side must be at most 3 .\n\nSince the first arc length is at most 3 , the second point can be only $A_{2}, A_{3}$ or $A_{4}$.\n\nIf the second point is $A_{2}$, then since the second and third arc lengths are each at most 3 , then the third point must be $A_{5}$. (Since the second arc length is at most 3, then the third point cannot be any further along than $A_{5}$. However, the arc length from $A_{5}$ around to $A_{1}$ is 3 , so it cannot be any closer than $A_{5}$.)\n\n\n\nIf the second point is $A_{3}$, the third point must be $A_{5}$ or $A_{6}$. If the second point is $A_{4}$, the third point must be $A_{5}$ or $A_{6}$ or $A_{7}$. Therefore, there are 6 acute triangles which include $A_{1}$ as one of its vertices.\n\nHow many acute triangles are there in total?\n\nWe can repeat the above process for each of the 6 other points, giving $7 \\times 6=42$ acute triangles.\n\nBut each triangle is counted three times here, as it has been counted once for each of its vertices.\n\nThus, there are $\\frac{7 \\times 6}{3}=14$ acute triangles.\n\nTherefore, the probability that a randomly chosen triangle is acute if $\\frac{14}{35}=\\frac{2}{5}$.""]",['$\\frac{2}{5}$'],False,,Numerical, 2503,Geometry,,"Points $A_{1}, A_{2}, \ldots, A_{N}$ are equally spaced around the circumference of a circle and $N \geq 3$. Three of these points are selected at random and a triangle is formed using these points as its vertices. Through this solution, we will use the following facts: When an acute triangle is inscribed in a circle: - each of the three angles of the triangle is the angle inscribed in the major arc defined by the side of the triangle by which it is subtended, - each of the three arcs into which the circle is divided by the vertices of the triangles is less than half of the circumference of the circle, and - it contains the centre of the circle. Why are these facts true? - Consider a chord of a circle which is not a diameter. Then the angle subtended in the major arc of this circle is an acute angle and the angle subtended in the minor arc is an obtuse angle. Now consider an acute triangle inscribed in a circle. Since each angle of the triangle is acute, then each of the three angles is inscribed in the major arc defined by the side of the triangle by which it is subtended. - It follows that each arc of the circle that is outside the triangle must be a minor arc, thus less than the circumference of the circle. - Lastly, if the centre was outside the triangle, then we would be able to draw a diameter of the circle with the triangle entirely on one side of the diameter. ![](https://cdn.mathpix.com/cropped/2023_12_21_cdeab4e3df6759515a2bg-1.jpg?height=382&width=382&top_left_y=1614&top_left_x=969) In this case, one of the arcs of the circle cut off by one of the sides of the triangle would have to be a major arc, which cannot happen, because of the above. Therefore, the centre is contained inside the triangle. If $N=7$, what is the probability that the triangle is acute? (A triangle is acute if each of its three interior angles is less than $90^{\circ}$.)","[""Since there are $N=7$ points from which the triangle's vertices can be chosen, there are $\\left(\\begin{array}{l}7 \\\\ 3\\end{array}\\right)=35$ triangles in total.\n\nWe compute the number of acute triangles.\n\nFix one of the vertices of such a triangle at $A_{1}$.\n\nWe construct the triangle by choosing the other two vertices in ascending subscript order. We choose the vertices by considering the arc length from the previous vertex - each of\n\n\n\nthese arc lengths must be smaller than half the total circumference of the circle.\n\nSince there are 7 equally spaced points on the circle, we assume the circumference is 7 , so the arc length formed by each side must be at most 3 .\n\nSince the first arc length is at most 3 , the second point can be only $A_{2}, A_{3}$ or $A_{4}$.\n\nIf the second point is $A_{2}$, then since the second and third arc lengths are each at most 3 , then the third point must be $A_{5}$. (Since the second arc length is at most 3, then the third point cannot be any further along than $A_{5}$. However, the arc length from $A_{5}$ around to $A_{1}$ is 3 , so it cannot be any closer than $A_{5}$.)\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_1bb09f07aaff2d2f7098g-1.jpg?height=412&width=437&top_left_y=583&top_left_x=947)\n\nIf the second point is $A_{3}$, the third point must be $A_{5}$ or $A_{6}$. If the second point is $A_{4}$, the third point must be $A_{5}$ or $A_{6}$ or $A_{7}$. Therefore, there are 6 acute triangles which include $A_{1}$ as one of its vertices.\n\nHow many acute triangles are there in total?\n\nWe can repeat the above process for each of the 6 other points, giving $7 \\times 6=42$ acute triangles.\n\nBut each triangle is counted three times here, as it has been counted once for each of its vertices.\n\nThus, there are $\\frac{7 \\times 6}{3}=14$ acute triangles.\n\nTherefore, the probability that a randomly chosen triangle is acute if $\\frac{14}{35}=\\frac{2}{5}$.""]",['$\\frac{2}{5}$'],False,,Numerical, 2504,Geometry,,"Hexagon $A B C D E F$ has vertices $A(0,0), B(4,0), C(7,2), D(7,5), E(3,5)$, $F(0,3)$. What is the area of hexagon $A B C D E F$ ?","['Let $P$ be the point with coordinates $(7,0)$ and let $Q$ be the point with coordinates $(0,5)$.\n\n\n\nThen $A P D Q$ is a rectangle with width 7 and height 5 , and so it has area $7 \\cdot 5=35$.\n\nHexagon $A B C D E F$ is formed by removing two triangles from rectangle $A P D Q$, namely $\\triangle B P C$ and $\\triangle E Q F$.\n\nEach of $\\triangle B P C$ and $\\triangle E Q F$ is right-angled, because each shares an angle with rectangle $A P D Q$.\n\nEach of $\\triangle B P C$ and $\\triangle E Q F$ has a base of length 3 and a height of 2.\n\nThus, their combined area is $2 \\cdot \\frac{1}{2} \\cdot 3 \\cdot 2=6$.\n\nThis means that the area of hexagon $A B C D E F$ is $35-6=29$.']",['29'],False,,Numerical, 2505,Geometry,,"In the diagram, $\triangle P Q S$ is right-angled at $P$ and $\triangle Q R S$ is right-angled at $Q$. Also, $P Q=x, Q R=8, R S=x+8$, and $S P=x+3$ for some real number $x$. Determine all possible values of the perimeter of quadrilateral $P Q R S$. ","['Since $\\triangle P Q S$ is right-angled at $P$, then by the Pythagorean Theorem,\n\n$$\nS Q^{2}=S P^{2}+P Q^{2}=(x+3)^{2}+x^{2}\n$$\n\nSince $\\triangle Q R S$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain\n\n$$\n\\begin{aligned}\nR S^{2} & =S Q^{2}+Q R^{2} \\\\\n(x+8)^{2} & =\\left((x+3)^{2}+x^{2}\\right)+8^{2} \\\\\nx^{2}+16 x+64 & =x^{2}+6 x+9+x^{2}+64 \\\\\n0 & =x^{2}-10 x+9 \\\\\n0 & =(x-1)(x-9)\n\\end{aligned}\n$$\n\nand so $x=1$ or $x=9$.\n\n(We can check that if $x=1, \\triangle P Q S$ has sides of lengths 4,1 and $\\sqrt{17}$ and $\\triangle Q R S$ has sides of lengths $\\sqrt{17}, 8$ and 9 , both of which are right-angled, and if $x=9, \\triangle P Q S$ has sides of lengths 12,9 and 15 and $\\triangle Q R S$ has sides of lengths 15,8 and 17 , both of which are right-angled.)\n\nIn terms of $x$, the perimeter of $P Q R S$ is $x+8+(x+8)+(x+3)=3 x+19$.\n\nThus, the possible perimeters of $P Q R S$ are 22 (when $x=1$ ) and 46 (when $x=9$ ).']","['22,46']",True,,Numerical, 2505,Geometry,,"In the diagram, $\triangle P Q S$ is right-angled at $P$ and $\triangle Q R S$ is right-angled at $Q$. Also, $P Q=x, Q R=8, R S=x+8$, and $S P=x+3$ for some real number $x$. Determine all possible values of the perimeter of quadrilateral $P Q R S$. ![](https://cdn.mathpix.com/cropped/2023_12_21_1ca8be37cce6eab426b2g-1.jpg?height=336&width=458&top_left_y=1244&top_left_x=1256)","['Since $\\triangle P Q S$ is right-angled at $P$, then by the Pythagorean Theorem,\n\n$$\nS Q^{2}=S P^{2}+P Q^{2}=(x+3)^{2}+x^{2}\n$$\n\nSince $\\triangle Q R S$ is right-angled at $Q$, then by the Pythagorean Theorem, we obtain\n\n$$\n\\begin{aligned}\nR S^{2} & =S Q^{2}+Q R^{2} \\\\\n(x+8)^{2} & =\\left((x+3)^{2}+x^{2}\\right)+8^{2} \\\\\nx^{2}+16 x+64 & =x^{2}+6 x+9+x^{2}+64 \\\\\n0 & =x^{2}-10 x+9 \\\\\n0 & =(x-1)(x-9)\n\\end{aligned}\n$$\n\nand so $x=1$ or $x=9$.\n\n(We can check that if $x=1, \\triangle P Q S$ has sides of lengths 4,1 and $\\sqrt{17}$ and $\\triangle Q R S$ has sides of lengths $\\sqrt{17}, 8$ and 9 , both of which are right-angled, and if $x=9, \\triangle P Q S$ has sides of lengths 12,9 and 15 and $\\triangle Q R S$ has sides of lengths 15,8 and 17 , both of which are right-angled.)\n\nIn terms of $x$, the perimeter of $P Q R S$ is $x+8+(x+8)+(x+3)=3 x+19$.\n\nThus, the possible perimeters of $P Q R S$ are 22 (when $x=1$ ) and 46 (when $x=9$ ).']","['22,46']",True,,Numerical, 2506,Algebra,,"A list $a_{1}, a_{2}, a_{3}, a_{4}$ of rational numbers is defined so that if one term is equal to $r$, then the next term is equal to $1+\frac{1}{1+r}$. For example, if $a_{3}=\frac{41}{29}$, then $a_{4}=1+\frac{1}{1+(41 / 29)}=\frac{99}{70}$. If $a_{3}=\frac{41}{29}$, what is the value of $a_{1} ?$","['If $r$ is a term in the sequence and $s$ is the next term, then $s=1+\\frac{1}{1+r}$.\n\nThis means that $s-1=\\frac{1}{1+r}$ and so $\\frac{1}{s-1}=1+r$ which gives $r=\\frac{1}{s-1}-1$.\n\nTherefore, since $a_{3}=\\frac{41}{29}$, then\n\n$$\na_{2}=\\frac{1}{a_{3}-1}-1=\\frac{1}{(41 / 29)-1}-1=\\frac{1}{12 / 29}-1=\\frac{29}{12}-1=\\frac{17}{12}\n$$\n\nFurther, since $a_{2}=\\frac{17}{12}$, then\n\n$$\na_{1}=\\frac{1}{a_{2}-1}-1=\\frac{1}{(17 / 12)-1}-1=\\frac{1}{5 / 12}-1=\\frac{12}{5}-1=\\frac{7}{5}\n$$']",['$\\frac{7}{5}$'],False,,Numerical, 2507,Geometry,,A hollow cylindrical tube has a radius of $10 \mathrm{~mm}$ and a height of $100 \mathrm{~mm}$. The tube sits flat on one of its circular faces on a horizontal table. The tube is filled with water to a depth of $h \mathrm{~mm}$. A solid cylindrical rod has a radius of $2.5 \mathrm{~mm}$ and a height of $150 \mathrm{~mm}$. The rod is inserted into the tube so that one of its circular faces sits flat on the bottom of the tube. The height of the water in the tube is now $64 \mathrm{~mm}$. Determine the value of $h$.,"[""Initially, the water in the hollow tube forms a cylinder with radius $10 \\mathrm{~mm}$ and height $h \\mathrm{~mm}$. Thus, the volume of the water is $\\pi(10 \\mathrm{~mm})^{2}(h \\mathrm{~mm})=100 \\pi h \\mathrm{~mm}^{3}$.\n\nAfter the rod is inserted, the level of the water rises to $64 \\mathrm{~mm}$. Note that this does not overflow the tube, since the tube's height is $100 \\mathrm{~mm}$.\n\nUp to the height of the water, the tube is a cylinder with radius $10 \\mathrm{~mm}$ and height 64 mm.\n\nThus, the volume of the tube up to the height of the water is\n\n$$\n\\pi(10 \\mathrm{~mm})^{2}(64 \\mathrm{~mm})=6400 \\pi \\mathrm{mm}^{3}\n$$\n\nThis volume consists of the water that is in the tube (whose volume, which has not changed, is $100 \\pi h \\mathrm{~mm}^{3}$ ) and the rod up to a height of $64 \\mathrm{~mm}$.\n\n\nSince the radius of the rod is $2.5 \\mathrm{~mm}$, the volume of the rod up to a height of $64 \\mathrm{~mm}$ is $\\pi(2.5 \\mathrm{~mm})^{2}(64 \\mathrm{~mm})=400 \\pi \\mathrm{mm}^{3}$.\n\nComparing volumes, $6400 \\pi \\mathrm{mm}^{3}=100 \\pi h \\mathrm{~mm}^{3}+400 \\pi \\mathrm{mm}^{3}$ and so $100 h=6000$ which gives $h=60$.""]",['60'],False,,Numerical, 2508,Algebra,,A function $f$ has the property that $f\left(\frac{2 x+1}{x}\right)=x+6$ for all real values of $x \neq 0$. What is the value of $f(4) ?$,"['We note that $\\frac{2 x+1}{x}=\\frac{2 x}{x}+\\frac{1}{x}=2+\\frac{1}{x}$.\n\nTherefore, $\\frac{2 x+1}{x}=4$ exactly when $2+\\frac{1}{x}=4$ or $\\frac{1}{x}=2$ and so $x=\\frac{1}{2}$.\n\nAlternatively, we could solve $\\frac{2 x+1}{x}=4$ directly to obtain $2 x+1=4 x$, which gives $2 x=1$ and so $x=\\frac{1}{2}$.\n\nThus, to determine the value of $f(4)$, we substitute $x=\\frac{1}{2}$ into the given equation $f\\left(\\frac{2 x+1}{x}\\right)=x+6$ and obtain $f(4)=\\frac{1}{2}+6=\\frac{13}{2}$.']",['$\\frac{13}{2}$'],False,,Numerical, 2509,Algebra,,"Determine all real numbers $a, b$ and $c$ for which the graph of the function $y=\log _{a}(x+b)+c$ passes through the points $P(3,5), Q(5,4)$ and $R(11,3)$.","['Since the graph passes through $(3,5),(5,4)$ and $(11,3)$, we can substitute these three points and obtain the following three equations:\n\n$$\n\\begin{aligned}\n& 5=\\log _{a}(3+b)+c \\\\\n& 4=\\log _{a}(5+b)+c \\\\\n& 3=\\log _{a}(11+b)+c\n\\end{aligned}\n$$\n\nSubtracting the second equation from the first and the third equation from the second, we obtain:\n\n$$\n\\begin{aligned}\n& 1=\\log _{a}(3+b)-\\log _{a}(5+b) \\\\\n& 1=\\log _{a}(5+b)-\\log _{a}(11+b)\n\\end{aligned}\n$$\n\nEquating right sides and manipulating, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{a}(5+b)-\\log _{a}(11+b) & =\\log _{a}(3+b)-\\log _{a}(5+b) \\\\\n2 \\log _{a}(5+b) & =\\log _{a}(3+b)+\\log _{a}(11+b) \\\\\n\\log _{a}\\left((5+b)^{2}\\right) & =\\log _{a}((3+b)(11+b)) \\quad(\\text { using log laws }) \\\\\n(5+b)^{2} & =(3+b)(11+b) \\quad \\text { (raising both sides to the power of } a) \\\\\n25+10 b+b^{2} & =33+14 b+b^{2} \\quad \\\\\n-8 & =4 b \\\\\nb & =-2\n\\end{aligned}\n$$\n\nSince $b=-2$, the equation $1=\\log _{a}(3+b)-\\log _{a}(5+b)$ becomes $1=\\log _{a} 1-\\log _{a} 3$.\n\nSince $\\log _{a} 1=0$ for every admissible value of $a$, then $\\log _{a} 3=-1$ which gives $a=3^{-1}=\\frac{1}{3}$.\n\nFinally, the equation $5=\\log _{a}(3+b)+c$ becomes $5=\\log _{1 / 3}(1)+c$ and so $c=5$.\n\nTherefore, $a=\\frac{1}{3}, b=-2$, and $c=5$, which gives $y=\\log _{1 / 3}(x-2)+5$.\n\nChecking:\n\n- When $x=3$, we obtain $y=\\log _{1 / 3}(3-2)+5=\\log _{1 / 3} 1+5=0+5=5$.\n- When $x=5$, we obtain $y=\\log _{1 / 3}(5-2)+5=\\log _{1 / 3} 3+5=-1+5=4$.\n- When $x=11$, we obtain $y=\\log _{1 / 3}(11-2)+5=\\log _{1 / 3} 9+5=-2+5=3$.']","['$\\frac{1}{3},-2,5$']",True,,Numerical, 2510,Algebra,,"A computer is programmed to choose an integer between 1 and 99, inclusive, so that the probability that it selects the integer $x$ is equal to $\log _{100}\left(1+\frac{1}{x}\right)$. Suppose that the probability that $81 \leq x \leq 99$ is equal to 2 times the probability that $x=n$ for some integer $n$. What is the value of $n$ ?","['The probability that the integer $n$ is chosen is $\\log _{100}\\left(1+\\frac{1}{n}\\right)$.\n\nThe probability that an integer between 81 and 99 , inclusive, is chosen equals the sum of the probabilities that the integers $81,82, \\ldots, 98,99$ are selected, which equals\n\n$$\n\\log _{100}\\left(1+\\frac{1}{81}\\right)+\\log _{100}\\left(1+\\frac{1}{82}\\right)+\\cdots+\\log _{100}\\left(1+\\frac{1}{98}\\right)+\\log _{100}\\left(1+\\frac{1}{99}\\right)\n$$\n\nSince the second probability equals 2 times the first probability, the following equations are equivalent:\n\n$$\n\\begin{aligned}\n\\log _{100}\\left(1+\\frac{1}{81}\\right)+\\log _{100}\\left(1+\\frac{1}{82}\\right)+\\cdots+\\log _{100}\\left(1+\\frac{1}{98}\\right)+\\log _{100}\\left(1+\\frac{1}{99}\\right) & =2 \\log _{100}\\left(1+\\frac{1}{n}\\right) \\\\\n\\log _{100}\\left(\\frac{82}{81}\\right)+\\log _{100}\\left(\\frac{83}{82}\\right)+\\cdots+\\log _{100}\\left(\\frac{99}{98}\\right)+\\log _{100}\\left(\\frac{100}{99}\\right) & =2 \\log _{100}\\left(1+\\frac{1}{n}\\right)\n\\end{aligned}\n$$\n\nUsing logarithm laws, these equations are further equivalent to\n\n$$\n\\begin{aligned}\n\\log _{100}\\left(\\frac{82}{81} \\cdot \\frac{83}{82} \\cdots \\cdot \\frac{99}{98} \\cdot \\frac{100}{99}\\right) & =\\log _{100}\\left(1+\\frac{1}{n}\\right)^{2} \\\\\n\\log _{100}\\left(\\frac{100}{81}\\right) & =\\log _{100}\\left(1+\\frac{1}{n}\\right)^{2}\n\\end{aligned}\n$$\n\nSince logarithm functions are invertible, we obtain $\\frac{100}{81}=\\left(1+\\frac{1}{n}\\right)^{2}$.\n\nSince $n>0$, then $1+\\frac{1}{n}=\\sqrt{\\frac{100}{81}}=\\frac{10}{9}$, and so $\\frac{1}{n}=\\frac{1}{9}$, which gives $n=9$.']",['9'],False,,Numerical, 2511,Geometry,,"In the diagram, $\triangle A B D$ has $C$ on $B D$. Also, $B C=2, C D=1, \frac{A C}{A D}=\frac{3}{4}$, and $\cos (\angle A C D)=-\frac{3}{5}$. Determine the length of $A B$. ![](https://cdn.mathpix.com/cropped/2023_12_21_31c9558111b63654ac46g-1.jpg?height=271&width=417&top_left_y=859&top_left_x=1258)","['Since $\\frac{A C}{A D}=\\frac{3}{4}$, then we let $A C=3 t$ and $A D=4 t$ for some real number $t>0$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_8df3bd851dc8f5cdf24eg-1.jpg?height=274&width=418&top_left_y=1592&top_left_x=951)\n\nUsing the cosine law in $\\triangle A C D$, the following equations are equivalent:\n\n$$\n\\begin{aligned}\nA D^{2} & =A C^{2}+C D^{2}-2 \\cdot A C \\cdot C D \\cdot \\cos (\\angle A C D) \\\\\n(4 t)^{2} & =(3 t)^{2}+1^{2}-2(3 t)(1)\\left(-\\frac{3}{5}\\right) \\\\\n16 t^{2} & =9 t^{2}+1+\\frac{18}{5} t \\\\\n80 t^{2} & =45 t^{2}+5+18 t \\\\\n35 t^{2}-18 t-5 & =0 \\\\\n(7 t-5)(5 t+1) & =0\n\\end{aligned}\n$$\n\nSince $t>0$, then $t=\\frac{5}{7}$.\n\nThus, $A C=3 t=\\frac{15}{7}$.\n\nUsing the cosine law in $\\triangle A C B$ and noting that\n\n$$\n\\cos (\\angle A C B)=\\cos \\left(180^{\\circ}-\\angle A C D\\right)=-\\cos (\\angle A C D)=\\frac{3}{5}\n$$\n\n\n\nthe following equations are equivalent:\n\n$$\n\\begin{aligned}\nA B^{2} & =A C^{2}+B C^{2}-2 \\cdot A C \\cdot B C \\cdot \\cos (\\angle A C B) \\\\\n& =\\left(\\frac{15}{7}\\right)^{2}+2^{2}-2\\left(\\frac{15}{7}\\right)(2)\\left(\\frac{3}{5}\\right) \\\\\n& =\\frac{225}{49}+4-\\frac{36}{7} \\\\\n& =\\frac{225}{49}+\\frac{196}{49}-\\frac{252}{49} \\\\\n& =\\frac{169}{49}\n\\end{aligned}\n$$\n\nSince $A B>0$, then $A B=\\frac{13}{7}$.']",['$\\frac{13}{7}$'],False,,Numerical, 2511,Geometry,,"In the diagram, $\triangle A B D$ has $C$ on $B D$. Also, $B C=2, C D=1, \frac{A C}{A D}=\frac{3}{4}$, and $\cos (\angle A C D)=-\frac{3}{5}$. Determine the length of $A B$. ","['Since $\\frac{A C}{A D}=\\frac{3}{4}$, then we let $A C=3 t$ and $A D=4 t$ for some real number $t>0$.\n\n\n\nUsing the cosine law in $\\triangle A C D$, the following equations are equivalent:\n\n$$\n\\begin{aligned}\nA D^{2} & =A C^{2}+C D^{2}-2 \\cdot A C \\cdot C D \\cdot \\cos (\\angle A C D) \\\\\n(4 t)^{2} & =(3 t)^{2}+1^{2}-2(3 t)(1)\\left(-\\frac{3}{5}\\right) \\\\\n16 t^{2} & =9 t^{2}+1+\\frac{18}{5} t \\\\\n80 t^{2} & =45 t^{2}+5+18 t \\\\\n35 t^{2}-18 t-5 & =0 \\\\\n(7 t-5)(5 t+1) & =0\n\\end{aligned}\n$$\n\nSince $t>0$, then $t=\\frac{5}{7}$.\n\nThus, $A C=3 t=\\frac{15}{7}$.\n\nUsing the cosine law in $\\triangle A C B$ and noting that\n\n$$\n\\cos (\\angle A C B)=\\cos \\left(180^{\\circ}-\\angle A C D\\right)=-\\cos (\\angle A C D)=\\frac{3}{5}\n$$\n\n\n\nthe following equations are equivalent:\n\n$$\n\\begin{aligned}\nA B^{2} & =A C^{2}+B C^{2}-2 \\cdot A C \\cdot B C \\cdot \\cos (\\angle A C B) \\\\\n& =\\left(\\frac{15}{7}\\right)^{2}+2^{2}-2\\left(\\frac{15}{7}\\right)(2)\\left(\\frac{3}{5}\\right) \\\\\n& =\\frac{225}{49}+4-\\frac{36}{7} \\\\\n& =\\frac{225}{49}+\\frac{196}{49}-\\frac{252}{49} \\\\\n& =\\frac{169}{49}\n\\end{aligned}\n$$\n\nSince $A B>0$, then $A B=\\frac{13}{7}$.']",['$\\frac{13}{7}$'],False,,Numerical, 2512,Algebra,,"Suppose that $a>\frac{1}{2}$ and that the parabola with equation $y=a x^{2}+2$ has vertex $V$. The parabola intersects the line with equation $y=-x+4 a$ at points $B$ and $C$, as shown. If the area of $\triangle V B C$ is $\frac{72}{5}$, determine the value of $a$. ","['The parabola with equation $y=a x^{2}+2$ is symmetric about the $y$-axis.\n\nThus, its vertex occurs when $x=0$ (which gives $y=a \\cdot 0^{2}+2=2$ ) and so $V$ has coordinates $(0,2)$.\n\nTo find the coordinates of $B$ and $C$, we use the equations of the parabola and line to obtain\n\n$$\n\\begin{aligned}\na x^{2}+2 & =-x+4 a \\\\\na x^{2}+x+(2-4 a) & =0\n\\end{aligned}\n$$\n\nUsing the quadratic formula,\n\n$$\nx=\\frac{-1 \\pm \\sqrt{1^{2}-4 a(2-4 a)}}{2 a}=\\frac{-1 \\pm \\sqrt{1-8 a+16 a^{2}}}{2 a}\n$$\n\nSince $1-8 a+16 a^{2}=(4 a-1)^{2}$ and $4 a-1>0\\left(\\right.$ since $\\left.a>\\frac{1}{2}\\right)$, then $\\sqrt{1-8 a+16 a^{2}}=4 a-1$ and so\n\n$$\nx=\\frac{-1 \\pm(4 a-1)}{2 a}\n$$\n\nwhich means that $x=\\frac{4 a-2}{2 a}=\\frac{2 a-1}{a}=2-\\frac{1}{a}$ or $x=\\frac{-4 a}{2 a}=-2$.\n\nWe can use the equation of the line to obtain the $y$-coordinates of $B$ and $C$.\n\nWhen $x=-2$ (corresponding to point $B$ ), we obtain $y=-(-2)+4 a=4 a+2$.\n\nWhen $x=2-\\frac{1}{a}$ (corresponding to point $C$ ), we obtain $y=-\\left(2-\\frac{1}{a}\\right)+4 a=4 a-2+\\frac{1}{a}$.\n\nLet $P$ and $Q$ be the points on the horizontal line through $V$ so that $B P$ and $C Q$ are perpendicular to $P Q$.\n\n\n\n\n\nThen the area of $\\triangle V B C$ is equal to the area of trapezoid $P B C Q$ minus the areas of right-angled $\\triangle B P V$ and right-angled $\\triangle C Q V$.\n\nSince $B$ has coordinates $(-2,4 a+2), P$ has coordinates $(-2,2), V$ has coordiantes $(0,2)$, $Q$ has coordinates $\\left(2-\\frac{1}{a}, 2\\right)$, and $C$ has coordinates $\\left(2-\\frac{1}{a}, 4 a-2+\\frac{1}{a}\\right)$, then\n\n$$\n\\begin{aligned}\nB P & =(4 a+2)-2=4 a \\\\\nC Q & =\\left(4 a-2+\\frac{1}{a}\\right)-2=4 a-4+\\frac{1}{a} \\\\\nP V & =0-(-2)=2 \\\\\nQ V & =2-\\frac{1}{a}-0=2-\\frac{1}{a} \\\\\nP Q & =P V+Q V=2+2-\\frac{1}{a}=4-\\frac{1}{a}\n\\end{aligned}\n$$\n\nTherefore, the area of trapezoid $P B C Q$ is\n\n$$\n\\frac{1}{2}(B P+C Q)(P Q)=\\frac{1}{2}\\left(4 a+4 a-4+\\frac{1}{a}\\right)\\left(4-\\frac{1}{a}\\right)=\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)\n$$\n\nAlso, the area of $\\triangle B P V$ is $\\frac{1}{2} \\cdot B P \\cdot P V=\\frac{1}{2}(4 a)(2)=4 a$.\n\nFurthermore, the area of $\\triangle C Q V$ is\n\n$$\n\\frac{1}{2} \\cdot C Q \\cdot Q V=\\frac{1}{2}\\left(4 a-4+\\frac{1}{a}\\right)\\left(2-\\frac{1}{a}\\right)=\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)\n$$\n\nFrom the given information,\n\n$$\n\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)-4 a-\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)=\\frac{72}{5}\n$$\n\nMultiplying both sides by $2 a^{2}$, which we distribute through the factors on the left side as $2 a \\cdot a$, we obtain\n\n$$\n\\left(8 a^{2}-4 a+1\\right)(4 a-1)-8 a^{3}-\\left(4 a^{2}-4 a+1\\right)(2 a-1)=\\frac{144}{5} a^{2}\n$$\n\nMultiplying both sides by 5 , we obtain\n\n$$\n5\\left(8 a^{2}-4 a+1\\right)(4 a-1)-40 a^{3}-5\\left(4 a^{2}-4 a+1\\right)(2 a-1)=144 a^{2}\n$$\n\nExpanding and simplifying, we obtain\n\n$$\n\\begin{aligned}\n\\left(160 a^{3}-120 a^{2}+40 a-5\\right)-40 a^{3}-\\left(40 a^{3}-60 a^{2}+30 a-5\\right) & =144 a^{2} \\\\\n80 a^{3}-204 a^{2}+10 a & =0 \\\\\n2 a\\left(40 a^{2}-102 a+5\\right) & =0 \\\\\n2 a(20 a-1)(2 a-5) & =0\n\\end{aligned}\n$$\n\nand so $a=0$ or $a=\\frac{1}{20}$ or $a=\\frac{5}{2}$. Since $a>\\frac{1}{2}$, then $a=\\frac{5}{2}$.']",['$\\frac{5}{2}$'],False,,Numerical, 2512,Algebra,,"Suppose that $a>\frac{1}{2}$ and that the parabola with equation $y=a x^{2}+2$ has vertex $V$. The parabola intersects the line with equation $y=-x+4 a$ at points $B$ and $C$, as shown. If the area of $\triangle V B C$ is $\frac{72}{5}$, determine the value of $a$. ![](https://cdn.mathpix.com/cropped/2023_12_21_31c9558111b63654ac46g-1.jpg?height=442&width=539&top_left_y=1297&top_left_x=1259)","['The parabola with equation $y=a x^{2}+2$ is symmetric about the $y$-axis.\n\nThus, its vertex occurs when $x=0$ (which gives $y=a \\cdot 0^{2}+2=2$ ) and so $V$ has coordinates $(0,2)$.\n\nTo find the coordinates of $B$ and $C$, we use the equations of the parabola and line to obtain\n\n$$\n\\begin{aligned}\na x^{2}+2 & =-x+4 a \\\\\na x^{2}+x+(2-4 a) & =0\n\\end{aligned}\n$$\n\nUsing the quadratic formula,\n\n$$\nx=\\frac{-1 \\pm \\sqrt{1^{2}-4 a(2-4 a)}}{2 a}=\\frac{-1 \\pm \\sqrt{1-8 a+16 a^{2}}}{2 a}\n$$\n\nSince $1-8 a+16 a^{2}=(4 a-1)^{2}$ and $4 a-1>0\\left(\\right.$ since $\\left.a>\\frac{1}{2}\\right)$, then $\\sqrt{1-8 a+16 a^{2}}=4 a-1$ and so\n\n$$\nx=\\frac{-1 \\pm(4 a-1)}{2 a}\n$$\n\nwhich means that $x=\\frac{4 a-2}{2 a}=\\frac{2 a-1}{a}=2-\\frac{1}{a}$ or $x=\\frac{-4 a}{2 a}=-2$.\n\nWe can use the equation of the line to obtain the $y$-coordinates of $B$ and $C$.\n\nWhen $x=-2$ (corresponding to point $B$ ), we obtain $y=-(-2)+4 a=4 a+2$.\n\nWhen $x=2-\\frac{1}{a}$ (corresponding to point $C$ ), we obtain $y=-\\left(2-\\frac{1}{a}\\right)+4 a=4 a-2+\\frac{1}{a}$.\n\nLet $P$ and $Q$ be the points on the horizontal line through $V$ so that $B P$ and $C Q$ are perpendicular to $P Q$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_930dd11e235e243baa6cg-1.jpg?height=439&width=523&top_left_y=2103&top_left_x=904)\n\n\n\nThen the area of $\\triangle V B C$ is equal to the area of trapezoid $P B C Q$ minus the areas of right-angled $\\triangle B P V$ and right-angled $\\triangle C Q V$.\n\nSince $B$ has coordinates $(-2,4 a+2), P$ has coordinates $(-2,2), V$ has coordiantes $(0,2)$, $Q$ has coordinates $\\left(2-\\frac{1}{a}, 2\\right)$, and $C$ has coordinates $\\left(2-\\frac{1}{a}, 4 a-2+\\frac{1}{a}\\right)$, then\n\n$$\n\\begin{aligned}\nB P & =(4 a+2)-2=4 a \\\\\nC Q & =\\left(4 a-2+\\frac{1}{a}\\right)-2=4 a-4+\\frac{1}{a} \\\\\nP V & =0-(-2)=2 \\\\\nQ V & =2-\\frac{1}{a}-0=2-\\frac{1}{a} \\\\\nP Q & =P V+Q V=2+2-\\frac{1}{a}=4-\\frac{1}{a}\n\\end{aligned}\n$$\n\nTherefore, the area of trapezoid $P B C Q$ is\n\n$$\n\\frac{1}{2}(B P+C Q)(P Q)=\\frac{1}{2}\\left(4 a+4 a-4+\\frac{1}{a}\\right)\\left(4-\\frac{1}{a}\\right)=\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)\n$$\n\nAlso, the area of $\\triangle B P V$ is $\\frac{1}{2} \\cdot B P \\cdot P V=\\frac{1}{2}(4 a)(2)=4 a$.\n\nFurthermore, the area of $\\triangle C Q V$ is\n\n$$\n\\frac{1}{2} \\cdot C Q \\cdot Q V=\\frac{1}{2}\\left(4 a-4+\\frac{1}{a}\\right)\\left(2-\\frac{1}{a}\\right)=\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)\n$$\n\nFrom the given information,\n\n$$\n\\left(4 a-2+\\frac{1}{2 a}\\right)\\left(4-\\frac{1}{a}\\right)-4 a-\\left(2 a-2+\\frac{1}{2 a}\\right)\\left(2-\\frac{1}{a}\\right)=\\frac{72}{5}\n$$\n\nMultiplying both sides by $2 a^{2}$, which we distribute through the factors on the left side as $2 a \\cdot a$, we obtain\n\n$$\n\\left(8 a^{2}-4 a+1\\right)(4 a-1)-8 a^{3}-\\left(4 a^{2}-4 a+1\\right)(2 a-1)=\\frac{144}{5} a^{2}\n$$\n\nMultiplying both sides by 5 , we obtain\n\n$$\n5\\left(8 a^{2}-4 a+1\\right)(4 a-1)-40 a^{3}-5\\left(4 a^{2}-4 a+1\\right)(2 a-1)=144 a^{2}\n$$\n\nExpanding and simplifying, we obtain\n\n$$\n\\begin{aligned}\n\\left(160 a^{3}-120 a^{2}+40 a-5\\right)-40 a^{3}-\\left(40 a^{3}-60 a^{2}+30 a-5\\right) & =144 a^{2} \\\\\n80 a^{3}-204 a^{2}+10 a & =0 \\\\\n2 a\\left(40 a^{2}-102 a+5\\right) & =0 \\\\\n2 a(20 a-1)(2 a-5) & =0\n\\end{aligned}\n$$\n\nand so $a=0$ or $a=\\frac{1}{20}$ or $a=\\frac{5}{2}$. Since $a>\\frac{1}{2}$, then $a=\\frac{5}{2}$.']",['$\\frac{5}{2}$'],False,,Numerical, 2513,Geometry,,"Consider the following statement: There is a triangle that is not equilateral whose side lengths form a geometric sequence, and the measures of whose angles form an arithmetic sequence. Show that this statement is true by finding such a triangle or prove that it is false by demonstrating that there cannot be such a triangle.","['We prove that there cannot be such a triangle.\n\nWe prove this by contradiction. That is, we suppose that there is such a triangle and prove that there is then a logical contradiction.\n\nSuppose that $\\triangle A B C$ is not equilateral, has side lengths that form a geometric sequence, and angles whose measures form an arithmetic sequence.\n\nSuppose that $\\triangle A B C$ has side lengths $B C=a, A C=a r$, and $A B=a r^{2}$, for some real numbers $a>0$ and $r>1$. (These lengths form a geometric sequence, and we can assume that this sequence is increasing, and that the sides are labelled in this particular order.) Since $B C\n\nWe could proceed using the cosine law:\n\n$$\n\\begin{aligned}\nA C^{2} & =B C^{2}+A B^{2}-2 \\cdot B C \\cdot A B \\cdot \\cos (\\angle A B C) \\\\\n(a r)^{2} & =a^{2}+\\left(a r^{2}\\right)^{2}-2 a\\left(a r^{2}\\right) \\cos \\left(60^{\\circ}\\right) \\\\\na^{2} r^{2} & =a^{2}+a^{2} r^{4}-2 a^{2} r^{2} \\cdot \\frac{1}{2} \\\\\na^{2} r^{2} & =a^{2}+a^{2} r^{4}-a^{2} r^{2} \\\\\n0 & =a^{2} r^{4}-2 a^{2} r^{2}+a^{2} \\\\\n0 & =a^{2}\\left(r^{4}-2 r^{2}+1\\right) \\\\\n0 & =a^{2}\\left(r^{2}-1\\right)^{2}\n\\end{aligned}\n$$\n\nThis tells us that $a=0$ (which is impossible) or $r^{2}=1$ (and thus $r= \\pm 1$, which is impossible).\n\nTherefore, we have reached a logical contradiction and so such a triangle cannot exist.\n\nAlternatively, we could proceed using the sine law, noting that\n\n$$\n\\begin{aligned}\n& \\angle B A C=\\theta=(\\theta+\\delta)-\\delta=60^{\\circ}-\\delta \\\\\n& \\angle A C B=\\theta+2 \\delta=(\\theta+\\delta)+\\delta=60^{\\circ}+\\delta\n\\end{aligned}\n$$\n\nBy the sine law,\n\n$$\n\\frac{B C}{\\sin (\\angle B A C)}=\\frac{A C}{\\sin (\\angle A B C)}=\\frac{A B}{\\sin (\\angle A C B)}\n$$\n\nfrom which we obtain\n\n$$\n\\frac{a}{\\sin \\left(60^{\\circ}-\\delta\\right)}=\\frac{a r}{\\sin \\left(60^{\\circ}\\right)}=\\frac{a r^{2}}{\\sin \\left(60^{\\circ}+\\delta\\right)}\n$$\n\n\n\nSince $a \\neq 0$, from the first two parts,\n\n$$\nr=\\frac{a r}{a}=\\frac{\\sin 60^{\\circ}}{\\sin \\left(60^{\\circ}-\\delta\\right)}\n$$\n\nSince $a r \\neq 0$, from the second two parts,\n\n$$\nr=\\frac{a r^{2}}{a r}=\\frac{\\sin \\left(60^{\\circ}+\\delta\\right)}{\\sin 60^{\\circ}}\n$$\n\nEquating expressions for $r$, we obtain successively\n\n$$\n\\begin{aligned}\n\\frac{\\sin 60^{\\circ}}{\\sin \\left(60^{\\circ}-\\delta\\right)} & =\\frac{\\sin \\left(60^{\\circ}+\\delta\\right)}{\\sin 60^{\\circ}} \\\\\n\\sin ^{2} 60^{\\circ} & =\\sin \\left(60^{\\circ}-\\delta\\right) \\sin \\left(60^{\\circ}+\\delta\\right) \\\\\n\\left(\\frac{\\sqrt{3}}{2}\\right)^{2} & =\\left(\\sin 60^{\\circ} \\cos \\delta-\\cos 60^{\\circ} \\sin \\delta\\right)\\left(\\sin 60^{\\circ} \\cos \\delta+\\cos 60^{\\circ} \\sin \\delta\\right) \\\\\n\\frac{3}{4} & =\\left(\\frac{\\sqrt{3}}{2} \\cos \\delta-\\frac{1}{2} \\sin \\delta\\right)\\left(\\frac{\\sqrt{3}}{2} \\cos \\delta+\\frac{1}{2} \\sin \\delta\\right) \\\\\n\\frac{3}{4} & =\\frac{3}{4} \\cos ^{2} \\delta-\\frac{1}{4} \\sin ^{2} \\delta \\\\\n\\frac{3}{4} & =\\frac{3}{4} \\cos ^{2} \\delta+\\frac{3}{4} \\sin ^{2} \\delta-\\sin ^{2} \\delta \\\\\n\\frac{3}{4} & =\\frac{3}{4}\\left(\\cos ^{2} \\delta+\\sin ^{2} \\delta\\right)-\\sin ^{2} \\delta \\\\\n\\frac{3}{4} & =\\frac{3}{4}-\\sin ^{2} \\delta \\\\\n\\sin ^{2} \\delta & =0\n\\end{aligned}\n$$\n\nwhich means that $\\delta=0^{\\circ}$. (Any other angle $\\delta$ with $\\sin \\delta=0$ would not produce angles in a triangle.)\n\nTherefore, all three angles in the triangle are $60^{\\circ}$, which means that the triangle is equilateral, which it cannot be.\n\nTherefore, we have reached a logical contradiction and so such a triangle cannot exist.\n\n#']",,True,,, 2514,Number Theory,,"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is ![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=151&width=891&top_left_y=453&top_left_x=644) The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. Determine the sum of the terms in the $(4,2)$-sawtooth sequence.","['The $(4,2)$-sawtooth sequence consists of the terms\n\n$$\n1, \\quad 2,3,4,3,2,1, \\quad 2,3,4,3,2,1\n$$\n\nwhose sum is 31 .']",['31'],False,,Numerical, 2514,Number Theory,,"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. Determine the sum of the terms in the $(4,2)$-sawtooth sequence.","['The $(4,2)$-sawtooth sequence consists of the terms\n\n$$\n1, \\quad 2,3,4,3,2,1, \\quad 2,3,4,3,2,1\n$$\n\nwhose sum is 31 .']",['31'],False,,Numerical, 2515,Number Theory,,"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is ![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=151&width=891&top_left_y=453&top_left_x=644) The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. For each positive integer $m \geq 2$, determine a simplified expression for the sum of the terms in the $(m, 3)$-sawtooth sequence.","['Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nWhen we write the ascending portion directly above the descending portion, we obtain\n\n$$\n\\begin{aligned}\n& 2, \\quad 3, \\quad 4, \\quad \\ldots, \\quad m-1, \\quad m \\\\\n& m-1, \\quad m-2, \\quad m-3, \\quad \\ldots, \\quad 2, \\quad 1\n\\end{aligned}\n$$\n\nFrom this presentation, we can see $m-1$ pairs of terms, the sum of each of which is $m+1$. $($ Note that $2+(m-1)=3+(m-2)=4+(m-3)=\\cdots=(m-1)+2=m+1$ and as we move from left to right, the terms on the top increase by 1 at each step and the terms on the bottom decrease by 1 at each step, so their sum is indeed constant.) Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$. This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.', 'Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nThis tooth includes one 1 , two $2 \\mathrm{~s}$, two $3 \\mathrm{~s}$, and so on, until we reach two $(m-1) \\mathrm{s}$, and one $m$.\n\nThe sum of these numbers is\n\n$$\n1(1)+2(2)+2(3)+\\cdots+2(m-1)+m\n$$\n\nwhich can be rewritten as\n\n$$\n2(1+2+3+\\cdots+(m-1)+m)-1-m=2 \\cdot \\frac{1}{2} m(m+1)-m-1=m^{2}+m-m-1=m^{2}-1\n$$\n\nTherefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$.\n\nThis means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.']",['$3 m^{2}-2$'],False,,Expression, 2515,Number Theory,,"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. For each positive integer $m \geq 2$, determine a simplified expression for the sum of the terms in the $(m, 3)$-sawtooth sequence.","['Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nWhen we write the ascending portion directly above the descending portion, we obtain\n\n$$\n\\begin{aligned}\n& 2, \\quad 3, \\quad 4, \\quad \\ldots, \\quad m-1, \\quad m \\\\\n& m-1, \\quad m-2, \\quad m-3, \\quad \\ldots, \\quad 2, \\quad 1\n\\end{aligned}\n$$\n\nFrom this presentation, we can see $m-1$ pairs of terms, the sum of each of which is $m+1$. $($ Note that $2+(m-1)=3+(m-2)=4+(m-3)=\\cdots=(m-1)+2=m+1$ and as we move from left to right, the terms on the top increase by 1 at each step and the terms on the bottom decrease by 1 at each step, so their sum is indeed constant.) Therefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$. This means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.', 'Suppose that $m \\geq 2$.\n\nThe $(m, 3)$-sawtooth sequence consists of an initial 1 followed by 3 teeth, each of which goes from 2 to $m$ to 1 .\n\nConsider one of these teeth whose terms are\n\n$$\n2,3,4, \\ldots, m-1, m, m-1, m-2, m-3, \\ldots, 2,1\n$$\n\nThis tooth includes one 1 , two $2 \\mathrm{~s}$, two $3 \\mathrm{~s}$, and so on, until we reach two $(m-1) \\mathrm{s}$, and one $m$.\n\nThe sum of these numbers is\n\n$$\n1(1)+2(2)+2(3)+\\cdots+2(m-1)+m\n$$\n\nwhich can be rewritten as\n\n$$\n2(1+2+3+\\cdots+(m-1)+m)-1-m=2 \\cdot \\frac{1}{2} m(m+1)-m-1=m^{2}+m-m-1=m^{2}-1\n$$\n\nTherefore, the sum of the numbers in one of the teeth is $(m-1)(m+1)=m^{2}-1$.\n\nThis means that the sum of the terms in the $(m, 3)$-sawtooth sequence is $1+3\\left(m^{2}-1\\right)$, which equals $3 m^{2}-2$.']",['$3 m^{2}-2$'],False,,Expression, 2516,Number Theory,,"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer.","['In an $(m, n)$-sawtooth sequence, the sum of the terms is $n\\left(m^{2}-1\\right)+1$.\n\nIn each tooth, there are $(m-1)+(m-1)=2 m-2$ terms (from 2 to $m$, inclusive, and from $m-1$ to 1 , inclusive).\n\nThis means that there are $n(2 m-2)+1$ terms in the sequence.\n\nThus, the average of the terms in the sequence is $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}$.\n\nWe need to prove that this is not an integer for all pairs of positive integers $(m, n)$ with $m \\geq 2$.\n\nSuppose that $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$ for some integer $k$. We will show, by contradiction, that this is not possible.\n\nSince $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$, then\n\n$$\n\\begin{aligned}\n\\frac{m^{2} n-n+1}{2 m n-2 n+1} & =k \\\\\nm^{2} n-n+1 & =2 m n k-2 n k+k \\\\\nm^{2} n-2 m n k+(2 n k-n-k+1) & =0\n\\end{aligned}\n$$\n\nWe treat this as a quadratic equation in $m$.\n\nSince $m$ is an integer, then this equation has integer roots, and so its discriminant must be a perfect square.\n\nThe discriminant of this quadratic equation is\n\n$$\n\\begin{aligned}\n\\Delta & =(-2 n k)^{2}-4 n(2 n k-n-k+1) \\\\\n& =4 n^{2} k^{2}-8 n^{2} k+4 n^{2}+4 n k-4 n \\\\\n& =4 n^{2}\\left(k^{2}-2 k+1\\right)+4 n(k-1) \\\\\n& =4 n^{2}(k-1)^{2}+4 n(k-1) \\\\\n& =(2 n(k-1))^{2}+2(2 n(k-1))+1-1 \\\\\n& =(2 n(k-1)+1)^{2}-1\n\\end{aligned}\n$$\n\nWe note that $(2 n(k-1)+1)^{2}$ is a perfect square and $\\Delta$ is supposed to be a perfect square. But these perfect squares differ by 1 , and the only two perfect squares that differ by 1 are\n\n\n\n1 and 0.\n\n(To justify this last fact, we could look at the equation $a^{2}-b^{2}=1$ where $a$ and $b$ are non-negative integers, and factor this to obtain $(a+b)(a-b)=1$ which would give $a+b=a-b=1$ from which we get $a=1$ and $b=0$.)\n\nSince $(2 n(k-1)+1)^{2}=1$ and $2 n(k-1)+1$ is non-negative, then $2 n(k-1)+1=1$ and so $2 n(k-1)=0$.\n\nSince $n$ is positive, then $k-1=0$ or $k=1$.\n\nTherefore, the only possible way in which the average is an integer is if the average is 1.\n\nIn this case, we get\n\n$$\n\\begin{aligned}\nm^{2} n-2 m n+(2 n-n-1+1) & =0 \\\\\nm^{2} n-2 m n+n & =0 \\\\\nn\\left(m^{2}-2 m+1\\right) & =0 \\\\\nn(m-1)^{2} & =0\n\\end{aligned}\n$$\n\nSince $n$ and $m$ are positive integers with $m \\geq 2$, then $n(m-1)^{2} \\neq 0$, which is a contradiction.\n\nTherefore, the average of the terms in an $(m, n)$-sawtooth sequence cannot be an integer.']",,True,,, 2516,Number Theory,,"Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is ![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=151&width=891&top_left_y=453&top_left_x=644) The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$. Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer.","['In an $(m, n)$-sawtooth sequence, the sum of the terms is $n\\left(m^{2}-1\\right)+1$.\n\nIn each tooth, there are $(m-1)+(m-1)=2 m-2$ terms (from 2 to $m$, inclusive, and from $m-1$ to 1 , inclusive).\n\nThis means that there are $n(2 m-2)+1$ terms in the sequence.\n\nThus, the average of the terms in the sequence is $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}$.\n\nWe need to prove that this is not an integer for all pairs of positive integers $(m, n)$ with $m \\geq 2$.\n\nSuppose that $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$ for some integer $k$. We will show, by contradiction, that this is not possible.\n\nSince $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$, then\n\n$$\n\\begin{aligned}\n\\frac{m^{2} n-n+1}{2 m n-2 n+1} & =k \\\\\nm^{2} n-n+1 & =2 m n k-2 n k+k \\\\\nm^{2} n-2 m n k+(2 n k-n-k+1) & =0\n\\end{aligned}\n$$\n\nWe treat this as a quadratic equation in $m$.\n\nSince $m$ is an integer, then this equation has integer roots, and so its discriminant must be a perfect square.\n\nThe discriminant of this quadratic equation is\n\n$$\n\\begin{aligned}\n\\Delta & =(-2 n k)^{2}-4 n(2 n k-n-k+1) \\\\\n& =4 n^{2} k^{2}-8 n^{2} k+4 n^{2}+4 n k-4 n \\\\\n& =4 n^{2}\\left(k^{2}-2 k+1\\right)+4 n(k-1) \\\\\n& =4 n^{2}(k-1)^{2}+4 n(k-1) \\\\\n& =(2 n(k-1))^{2}+2(2 n(k-1))+1-1 \\\\\n& =(2 n(k-1)+1)^{2}-1\n\\end{aligned}\n$$\n\nWe note that $(2 n(k-1)+1)^{2}$ is a perfect square and $\\Delta$ is supposed to be a perfect square. But these perfect squares differ by 1 , and the only two perfect squares that differ by 1 are\n\n\n\n1 and 0.\n\n(To justify this last fact, we could look at the equation $a^{2}-b^{2}=1$ where $a$ and $b$ are non-negative integers, and factor this to obtain $(a+b)(a-b)=1$ which would give $a+b=a-b=1$ from which we get $a=1$ and $b=0$.)\n\nSince $(2 n(k-1)+1)^{2}=1$ and $2 n(k-1)+1$ is non-negative, then $2 n(k-1)+1=1$ and so $2 n(k-1)=0$.\n\nSince $n$ is positive, then $k-1=0$ or $k=1$.\n\nTherefore, the only possible way in which the average is an integer is if the average is 1.\n\nIn this case, we get\n\n$$\n\\begin{aligned}\nm^{2} n-2 m n+(2 n-n-1+1) & =0 \\\\\nm^{2} n-2 m n+n & =0 \\\\\nn\\left(m^{2}-2 m+1\\right) & =0 \\\\\nn(m-1)^{2} & =0\n\\end{aligned}\n$$\n\nSince $n$ and $m$ are positive integers with $m \\geq 2$, then $n(m-1)^{2} \\neq 0$, which is a contradiction.\n\nTherefore, the average of the terms in an $(m, n)$-sawtooth sequence cannot be an integer.']",['证明题，略'],True,,Need_human_evaluate, 2517,Combinatorics,,"At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. For a 2-topping pizza, determine the probability that at least $\frac{1}{4}$ of the pizza is covered by both toppings.","['Assume that the first topping is placed on the top half of the pizza. (We can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the horizontal diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 360^{\\circ}$.\n\nWhen $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, the angle of the sector covered by both toppings is at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nWhen $90^{\\circ}<\\theta \\leq 180^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $\\theta$ moves past $180^{\\circ}$, the left-hand portion of the upper half circle starts to be covered with both toppings again. When $180^{\\circ} \\leq \\theta<270^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $270^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the angle of the sector covered by both toppings at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nTherefore, if $\\theta$ is chosen randomly between $0^{\\circ}$ and $360^{\\circ}$, the combined length of the intervals in which at least $\\frac{1}{4}$ of the pizza is covered with both toppings is $180^{\\circ}$.\n\nTherefore, the probability is $\\frac{180^{\\circ}}{360^{\\circ}}$, or $\\frac{1}{2}$.']",['$\\frac{1}{2}$'],False,,Numerical, 2517,Combinatorics,,"At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected ![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=288&width=525&top_left_y=1157&top_left_x=1193) uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. For a 2-topping pizza, determine the probability that at least $\frac{1}{4}$ of the pizza is covered by both toppings.","['Assume that the first topping is placed on the top half of the pizza. (We can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the horizontal diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n![](https://cdn.mathpix.com/cropped/2023_12_21_2f637db46c3cc012c3e8g-1.jpg?height=300&width=616&top_left_y=435&top_left_x=856)\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 360^{\\circ}$.\n\nWhen $0^{\\circ} \\leq \\theta \\leq 90^{\\circ}$, the angle of the sector covered by both toppings is at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nWhen $90^{\\circ}<\\theta \\leq 180^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $\\theta$ moves past $180^{\\circ}$, the left-hand portion of the upper half circle starts to be covered with both toppings again. When $180^{\\circ} \\leq \\theta<270^{\\circ}$, the angle of the sector covered by both toppings is less than $90^{\\circ}$ (and so is less than a quarter of the circle).\n\nWhen $270^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the angle of the sector covered by both toppings at least $90^{\\circ}$ (and so is at least a quarter of the circle).\n\nTherefore, if $\\theta$ is chosen randomly between $0^{\\circ}$ and $360^{\\circ}$, the combined length of the intervals in which at least $\\frac{1}{4}$ of the pizza is covered with both toppings is $180^{\\circ}$.\n\nTherefore, the probability is $\\frac{180^{\\circ}}{360^{\\circ}}$, or $\\frac{1}{2}$.']",['$\\frac{1}{2}$'],False,,Numerical, 2518,Combinatorics,,"At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected ![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=288&width=525&top_left_y=1157&top_left_x=1193) uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. For a 3-topping pizza, determine the probability that some region of the pizza with non-zero area is covered by all 3 toppings. (The diagram above shows an example where no region is covered by all 3 toppings.)","['Suppose that the first topping is placed on the top half of the pizza. (Again, we can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$. If $180^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the resulting pizza can be seen as a reflection of the one shown.\n![](https://cdn.mathpix.com/cropped/2023_12_21_2f637db46c3cc012c3e8g-1.jpg?height=302&width=696&top_left_y=1822&top_left_x=818)\n\nConsider the third diameter added, shown dotted in the diagram above. Suppose that its angle with the horizontal is $\\alpha$. (In the diagram, $\\alpha<90^{\\circ}$.) We assume that the topping is added on the half pizza clockwise beginning at the angle of $\\alpha$, and that this topping stays in the same relative position as the diameter sweeps around the circle.\n\nFor what angles $\\alpha$ will there be a portion of the pizza covered with all three toppings?\n\nIf $0^{\\circ} \\leq \\alpha<180^{\\circ}$, there will be a portion covered with three toppings; this portion is above the right half of the horizontal diameter.\n\nIf $180^{\\circ} \\leq \\alpha<180^{\\circ}+\\theta$, the third diameter will pass through the two regions with angle $\\theta$ and the third topping will be below this diameter, so there will not be a region covered\n\n\n\nwith three toppings.\n\nIf $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$, the third topping starts to cover the leftmost part of the region currently covered with two toppings, and so a region is covered with three toppings.\n\nTherefore, for an angle $\\theta$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$, a region of the pizza is covered with three toppings when $0^{\\circ} \\leq \\alpha<180^{\\circ}$ and when $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$.\n\nTo determine the desired probability, we graph points $(\\theta, \\alpha)$. A particular choice of diameters corresponds to a choice of angles $\\theta$ and $\\alpha$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$ and $0^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$, which corresponds to a point on the graph below.\n\nThe probability that we are looking for then equals the area of the region of this graph where three toppings are in a portion of the pizza divided by the total allowable area of the graph.\n\nThe shaded region of the graph corresponds to instances where a portion of the pizza will be covered by three toppings.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_e5e5956d98741d27d95bg-1.jpg?height=453&width=488&top_left_y=852&top_left_x=924)\n\nThis shaded region consists of the entire portion of the graph where $0^{\\circ} \\leq \\alpha \\leq 180^{\\circ}$ (regardless of $\\theta$ ) as well as the region above the line with equation $\\alpha=\\theta+180^{\\circ}$ (that is, the region with $\\theta+180^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$ ).\n\nSince the slope of the line is 1 , it divides the upper half of the region, which is a square, into two pieces of equal area.\n\nTherefore, $\\frac{3}{4}$ of the graph is shaded, which means that the probability that a region of the pizza is covered by all three toppings is $\\frac{3}{4}$.']",['$\\frac{3}{4}$'],False,,Numerical, 2518,Combinatorics,,"At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. For a 3-topping pizza, determine the probability that some region of the pizza with non-zero area is covered by all 3 toppings. (The diagram above shows an example where no region is covered by all 3 toppings.)","['Suppose that the first topping is placed on the top half of the pizza. (Again, we can rotate the pizza so that this is the case.)\n\nAssume that the second topping is placed on the half of the pizza that is above the diameter that makes an angle of $\\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\\theta$ to $\\theta+180^{\\circ}$.\n\nWe may assume that $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$. If $180^{\\circ} \\leq \\theta \\leq 360^{\\circ}$, the resulting pizza can be seen as a reflection of the one shown.\n\n\nConsider the third diameter added, shown dotted in the diagram above. Suppose that its angle with the horizontal is $\\alpha$. (In the diagram, $\\alpha<90^{\\circ}$.) We assume that the topping is added on the half pizza clockwise beginning at the angle of $\\alpha$, and that this topping stays in the same relative position as the diameter sweeps around the circle.\n\nFor what angles $\\alpha$ will there be a portion of the pizza covered with all three toppings?\n\nIf $0^{\\circ} \\leq \\alpha<180^{\\circ}$, there will be a portion covered with three toppings; this portion is above the right half of the horizontal diameter.\n\nIf $180^{\\circ} \\leq \\alpha<180^{\\circ}+\\theta$, the third diameter will pass through the two regions with angle $\\theta$ and the third topping will be below this diameter, so there will not be a region covered\n\n\n\nwith three toppings.\n\nIf $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$, the third topping starts to cover the leftmost part of the region currently covered with two toppings, and so a region is covered with three toppings.\n\nTherefore, for an angle $\\theta$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$, a region of the pizza is covered with three toppings when $0^{\\circ} \\leq \\alpha<180^{\\circ}$ and when $180^{\\circ}+\\theta \\leq \\alpha \\leq 360^{\\circ}$.\n\nTo determine the desired probability, we graph points $(\\theta, \\alpha)$. A particular choice of diameters corresponds to a choice of angles $\\theta$ and $\\alpha$ with $0^{\\circ} \\leq \\theta \\leq 180^{\\circ}$ and $0^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$, which corresponds to a point on the graph below.\n\nThe probability that we are looking for then equals the area of the region of this graph where three toppings are in a portion of the pizza divided by the total allowable area of the graph.\n\nThe shaded region of the graph corresponds to instances where a portion of the pizza will be covered by three toppings.\n\n\n\nThis shaded region consists of the entire portion of the graph where $0^{\\circ} \\leq \\alpha \\leq 180^{\\circ}$ (regardless of $\\theta$ ) as well as the region above the line with equation $\\alpha=\\theta+180^{\\circ}$ (that is, the region with $\\theta+180^{\\circ} \\leq \\alpha \\leq 360^{\\circ}$ ).\n\nSince the slope of the line is 1 , it divides the upper half of the region, which is a square, into two pieces of equal area.\n\nTherefore, $\\frac{3}{4}$ of the graph is shaded, which means that the probability that a region of the pizza is covered by all three toppings is $\\frac{3}{4}$.']",['$\\frac{3}{4}$'],False,,Numerical, 2519,Combinatorics,,"At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected ![](https://cdn.mathpix.com/cropped/2023_12_21_381b11c03c23278b095cg-1.jpg?height=288&width=525&top_left_y=1157&top_left_x=1193) uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. Suppose that $N$ is a positive integer. For an $N$-topping pizza, determine the probability, in terms of $N$, that some region of the pizza with non-zero area is covered by all $N$ toppings.","['The main idea of this solution is that the toppings all overlap exactly when there is one topping with the property that all other toppings ""begin"" somewhere in that toppings semi-circle. In the rest of this solution, we determine the probability using this fact and then justify this fact.\n\nSuppose that, for $1 \\leq j \\leq N$, topping $j$ is put on the semi-circle that starts at an angle of $\\theta_{j}$ clockwise from the horizontal left-hand radius and continues to an angle of $\\theta_{j}+180^{\\circ}$, where $0^{\\circ} \\leq \\theta_{j}<360^{\\circ}$. By establishing these variables and this convention, we are fixing both the angle of the diameter and the semi-circle defined by this diameter on which the topping is placed.\n\nSuppose that there is some region of the pizza with non-zero area that is covered by all $N$ toppings.\n\nThis region will be a sector with two bounding radii, each of which must be half of a diameter that defines one of the toppings.\n\nSuppose that the radius at the clockwise ""end"" of the sector is the end of the semi-circle where topping $X$ is placed, and that the radius at the counter-clockwise ""beginning"" of the sector is the start of the semi-circle where topping $Y$ is placed.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_e3ee936c0c7c9be18322g-1.jpg?height=447&width=615&top_left_y=167&top_left_x=858)\n\nThis means that each of the other $N-2$ toppings begins between (in the clockwise sense) the points where topping $X$ begins and where topping $Y$ begins.\n\nConsider the beginning angle for topping $X, \\theta_{X}$.\n\nTo say that the other $N-1$ toppings begin at some point before topping $X$ ends is the same as saying that each $\\theta_{j}$ with $j \\neq X$ is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nHere, we can allow for the possibility that $\\theta_{X}+180^{\\circ}$ is greater than $360^{\\circ}$ by saying that an angle equivalent to $\\theta_{j}$ (which is either $\\theta_{j}$ or $\\theta_{j}+360^{\\circ}$ ) is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$. For each $j \\neq X$, the angle $\\theta_{j}$ is randomly, uniformly and independently chosen on the circle, so there is a probability of $\\frac{1}{2}$ that this angle (or its equivalent) will be in the semicircle between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nSince there are $N-1$ such angles, the probability that all are between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$ is $\\frac{1}{2^{N-1}}$.\n\nSince there are $N$ possible selections for the first topping that can end the common sector, then the desired probability will be $\\frac{N}{2^{N-1}}$ as long as we can show that no set of angles can give two different sectors that are both covered with all toppings.\n\nTo show this last fact, we suppose without loss of generality that\n\n$$\n0^{\\circ}=\\theta_{1}<\\theta_{2}<\\theta_{3}<\\cdots<\\theta_{N-1}<\\theta_{N}<180^{\\circ}\n$$\n\n(We can relabel the toppings if necessary to obtain this order and rotate the pizza so that topping 1 begins at $0^{\\circ}$.)\n\nWe need to show that it is not possible to have a $Z$ for which $\\theta_{Z}, \\theta_{Z+1}, \\ldots, \\theta_{N}, \\theta_{1}, \\theta_{2}, \\ldots, \\theta_{Z-1}$ all lie in a semi-circle starting with $\\theta_{Z}$.\n\nSince $\\theta_{Z}<180^{\\circ}$ and $\\theta_{1}$ can be thought of as $360^{\\circ}$, then this is not possible as $\\theta_{1}$ and the angles after it are all not within $180^{\\circ}$ of $\\theta_{Z}$.\n\nTherefore, it is not possible to have two such regions with the same set of angles, and so the desired probability is $\\frac{N}{2^{N-1}}$.']",['$\\frac{N}{2^{N-1}}$'],False,,Expression, 2519,Combinatorics,,"At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping. Suppose that $N$ is a positive integer. For an $N$-topping pizza, determine the probability, in terms of $N$, that some region of the pizza with non-zero area is covered by all $N$ toppings.","['The main idea of this solution is that the toppings all overlap exactly when there is one topping with the property that all other toppings ""begin"" somewhere in that toppings semi-circle. In the rest of this solution, we determine the probability using this fact and then justify this fact.\n\nSuppose that, for $1 \\leq j \\leq N$, topping $j$ is put on the semi-circle that starts at an angle of $\\theta_{j}$ clockwise from the horizontal left-hand radius and continues to an angle of $\\theta_{j}+180^{\\circ}$, where $0^{\\circ} \\leq \\theta_{j}<360^{\\circ}$. By establishing these variables and this convention, we are fixing both the angle of the diameter and the semi-circle defined by this diameter on which the topping is placed.\n\nSuppose that there is some region of the pizza with non-zero area that is covered by all $N$ toppings.\n\nThis region will be a sector with two bounding radii, each of which must be half of a diameter that defines one of the toppings.\n\nSuppose that the radius at the clockwise ""end"" of the sector is the end of the semi-circle where topping $X$ is placed, and that the radius at the counter-clockwise ""beginning"" of the sector is the start of the semi-circle where topping $Y$ is placed.\n\n\n\n\n\nThis means that each of the other $N-2$ toppings begins between (in the clockwise sense) the points where topping $X$ begins and where topping $Y$ begins.\n\nConsider the beginning angle for topping $X, \\theta_{X}$.\n\nTo say that the other $N-1$ toppings begin at some point before topping $X$ ends is the same as saying that each $\\theta_{j}$ with $j \\neq X$ is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nHere, we can allow for the possibility that $\\theta_{X}+180^{\\circ}$ is greater than $360^{\\circ}$ by saying that an angle equivalent to $\\theta_{j}$ (which is either $\\theta_{j}$ or $\\theta_{j}+360^{\\circ}$ ) is between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$. For each $j \\neq X$, the angle $\\theta_{j}$ is randomly, uniformly and independently chosen on the circle, so there is a probability of $\\frac{1}{2}$ that this angle (or its equivalent) will be in the semicircle between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$.\n\nSince there are $N-1$ such angles, the probability that all are between $\\theta_{X}$ and $\\theta_{X}+180^{\\circ}$ is $\\frac{1}{2^{N-1}}$.\n\nSince there are $N$ possible selections for the first topping that can end the common sector, then the desired probability will be $\\frac{N}{2^{N-1}}$ as long as we can show that no set of angles can give two different sectors that are both covered with all toppings.\n\nTo show this last fact, we suppose without loss of generality that\n\n$$\n0^{\\circ}=\\theta_{1}<\\theta_{2}<\\theta_{3}<\\cdots<\\theta_{N-1}<\\theta_{N}<180^{\\circ}\n$$\n\n(We can relabel the toppings if necessary to obtain this order and rotate the pizza so that topping 1 begins at $0^{\\circ}$.)\n\nWe need to show that it is not possible to have a $Z$ for which $\\theta_{Z}, \\theta_{Z+1}, \\ldots, \\theta_{N}, \\theta_{1}, \\theta_{2}, \\ldots, \\theta_{Z-1}$ all lie in a semi-circle starting with $\\theta_{Z}$.\n\nSince $\\theta_{Z}<180^{\\circ}$ and $\\theta_{1}$ can be thought of as $360^{\\circ}$, then this is not possible as $\\theta_{1}$ and the angles after it are all not within $180^{\\circ}$ of $\\theta_{Z}$.\n\nTherefore, it is not possible to have two such regions with the same set of angles, and so the desired probability is $\\frac{N}{2^{N-1}}$.']",['$\\frac{N}{2^{N-1}}$'],False,,Expression, 2520,Number Theory,,What is the smallest positive integer $x$ for which $\frac{1}{32}=\frac{x}{10^{y}}$ for some positive integer $y$ ?,"['Since $10^{y} \\neq 0$, the equation $\\frac{1}{32}=\\frac{x}{10^{y}}$ is equivalent to $10^{y}=32 x$.\n\nSo the given question is equivalent to asking for the smallest positive integer $x$ for which $32 x$ equals a positive integer power of 10 .\n\nNow $32=2^{5}$ and so $32 x=2^{5} x$.\n\nFor $32 x$ to equal a power of 10, each factor of 2 must be matched with a factor of 5 .\n\nTherefore, $x$ must be divisible by $5^{5}$ (that is, $x$ must include at least 5 powers of 5 ), and so $x \\geq 5^{5}=3125$.\n\nBut $32\\left(5^{5}\\right)=2^{5} 5^{5}=10^{5}$, and so if $x=5^{5}=3125$, then $32 x$ is indeed a power of 10 , namely $10^{5}$.\n\nThis tells us that the smallest positive integer $x$ for which $\\frac{1}{32}=\\frac{x}{10^{y}}$ for some positive integer $y$ is $x=5^{5}=3125$.']",['3125'],False,,Numerical, 2521,Number Theory,,"Determine all possible values for the area of a right-angled triangle with one side length equal to 60 and with the property that its side lengths form an arithmetic sequence. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.)","['Since the three side lengths of a right-angled triangle form an arithemetic sequence and must include 60 , then the three side lengths are $60,60+d, 60+2 d$ or $60-d, 60,60+d$ or $60-2 d, 60-d, 60$, for some $d \\geq 0$.\n\nFor a triangle with sides of length $60,60+d, 60+2 d$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:\n\n$$\n\\begin{aligned}\n60^{2}+(60+d)^{2} & =(60+2 d)^{2} \\\\\n3600+3600+120 d+d^{2} & =3600+240 d+4 d^{2} \\\\\n0 & =3 d^{2}+120 d-3600 \\\\\n0 & =d^{2}+40 d-1200 \\\\\n0 & =(d+60)(d-20)\n\\end{aligned}\n$$\n\n(Note that, since $d \\geq 0$, then $60+2 d$ must be the hypotenuse of the triangle.) Since $d \\geq 0$, then $d=20$, which gives the triangle with side lengths $60,80,100$.\n\nThe longest side length is the hypotenuse and the shorter two sides meet at right angles, giving an area of $\\frac{1}{2}(60)(80)=2400$.\n\n\n\nFor a triangle with sides of length $60-d, 60,60+d$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:\n\n$$\n\\begin{aligned}\n(60-d)^{2}+60^{2} & =(60+d)^{2} \\\\\n3600-120 d+d^{2}+3600 & =3600+120 d+d^{2} \\\\\n3600 & =240 d \\\\\nd & =15\n\\end{aligned}\n$$\n\nSince $d \\geq 0$, then $d=15$ is admissible, which gives the triangle with side lengths 45, 60,75. Using a similar analysis, the area of this triangle is $\\frac{1}{2}(45)(60)=1350$.\n\nFor a triangle with sides of length $60-2 d, 60-d, 60$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:\n\n$$\n\\begin{aligned}\n(60-2 d)^{2}+(60-d)^{2} & =60^{2} \\\\\n3600-240 d+4 d^{2}+3600-120 d+d^{2} & =3600 \\\\\n5 d^{2}-360 d+3600 & =0 \\\\\nd^{2}-72 d+720 & =0 \\\\\n(d-60)(d-12) & =0\n\\end{aligned}\n$$\n\nSince $d \\geq 0$, then $d=60$ or $d=12$, which give possible side lengths of $-60,0,60$ (which do not form a triangle) and 36,48,60 (which do form a triangle).\n\nUsing a similar analysis, the area of this triangle is $\\frac{1}{2}(36)(48)=864$.\n\nTherefore, the possible values for the area of such a triangle are 2400, 1350, and 864.', 'Suppose that a triangle has side lengths in arithemetic sequence.\n\nThen the side lengths can be written as $a-d, a, a+d$ for some $a>0$ and $d \\geq 0$.\n\nNote that $a-d \\leq a \\leq a+d$.\n\nFor such a triangle to be right-angled, by the Pythagorean Theorem, the following equivalent equations are true:\n\n$$\n\\begin{aligned}\n(a-d)^{2}+a^{2} & =(a+d)^{2} \\\\\na^{2}-2 a d+d^{2}+a^{2} & =a^{2}+2 a d+d^{2} \\\\\na^{2} & =4 a d\n\\end{aligned}\n$$\n\nSince $a>0$, then $a=4 d$, and so the side lengths of the triangle are $a-d=3 d, a=4 d$, and $a+d=5 d$ for some $d \\geq 0$.\n\n(Note that such triangles are all similar to the 3-4-5 triangle.)\n\nIf such a triangle has 60 as a side length, then there are three possibilities:\n\n(i) $3 d=60$ : This gives $d=20$ and side lengths $60,80,100$.\n\nSince the triangle is right-angled and its hypotenuse has length 100, then its area will equal $\\frac{1}{2}(60)(80)=2400$.\n\n(ii) $4 d=60$ : This gives $d=15$ and side lengths $45,60,75$.\n\nIn a similar way to case (i), its area will equal $\\frac{1}{2}(45)(60)=1350$.\n\n(iii) $5 d=60$ : This gives $d=12$ and side lengths $36,48,60$.\n\nIn a similar way to case (i), its area will equal $\\frac{1}{2}(36)(48)=864$.\n\nTherefore, the possible values for the area of such a triangle are 2400, 1350, and 864 .']","['2400, 1350, 864']",True,,Numerical, 2522,Algebra,,"Amrita and Zhang cross a lake in a straight line with the help of a one-seat kayak. Each can paddle the kayak at $7 \mathrm{~km} / \mathrm{h}$ and swim at $2 \mathrm{~km} / \mathrm{h}$. They start from the same point at the same time with Amrita paddling and Zhang swimming. After a while, Amrita stops the kayak and immediately starts swimming. Upon reaching the kayak (which has not moved since Amrita started swimming), Zhang gets in and immediately starts paddling. They arrive on the far side of the lake at the same time, 90 minutes after they began. Determine the amount of time during these 90 minutes that the kayak was not being paddled.","['Suppose that Amrita paddles the kayak for $p \\mathrm{~km}$ and swims for $s \\mathrm{~km}$.\n\nSince Amrita leaves the kayak in the lake and it does not move, then Zhang swims $p \\mathrm{~km}$ and paddles the kayak for $s \\mathrm{~km}$.\n\nNote that each paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and each swims at $2 \\mathrm{~km} / \\mathrm{h}$ and each takes exactly 90 minutes (or 1.5 hours) to complete the trip.\n\nIf $sp$, then Zhang would paddle farther and swim less distance than Amrita and so would reach the other side in less time than Amrita.\n\nSince they each take 90 minutes, then we must have $s=p$.\n\nAlternatively, since each paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and each swims at $2 \\mathrm{~km} / \\mathrm{h}$ and each takes exactly 90 minutes (or 1.5 hours) to complete the trip, then we obtain the two equations\n\n$$\n\\frac{p}{7}+\\frac{s}{2}=1.5 \\quad \\frac{p}{2}+\\frac{s}{7}=1.5\n$$\n\nUsing the fact that the right sides of these equations are equal, we obtain\n\n$$\n\\begin{aligned}\n\\frac{p}{7}+\\frac{s}{2} & =\\frac{s}{7}+\\frac{p}{2} \\\\\n\\frac{s}{2}-\\frac{s}{7} & =\\frac{p}{2}-\\frac{p}{7} \\\\\ns\\left(\\frac{1}{2}-\\frac{1}{7}\\right) & =p\\left(\\frac{1}{2}-\\frac{1}{7}\\right) \\\\\ns & =p\n\\end{aligned}\n$$\n\nTherefore, $\\frac{p}{7}+\\frac{p}{2}=1.5$ or $\\frac{9}{14} p=1.5=\\frac{3}{2}$ and so $p=\\frac{7}{3}$.\n\nFor Amrita to paddle these $\\frac{7}{3} \\mathrm{~km}$ at $7 \\mathrm{~km} / \\mathrm{h}$, it takes $\\frac{7 / 3}{7}=\\frac{1}{3}$ hour, or 20 minutes.\n\nFor Zhang to swim these $\\frac{7}{3} \\mathrm{~km}$ at $2 \\mathrm{~km} / \\mathrm{h}$, it takes $\\frac{7 / 3}{2}=\\frac{7}{6}$ hour, or 70 minutes.\n\nThe kayak is not being paddled for the period of time from when Amrita stops paddling to the time when Zhang stops swimming, which is a period of $70-20=50$ minutes.', 'Let $t_{1}$ hours be the length of time during which Amrita paddles and Zhang swims.\n\nLet $t_{2}$ hours be the length of time during which Amrita swims and Zhang swims; the kayak is not moving during this time.\n\nLet $t_{3}$ hours be the length of time during which Amrita swims and Zhang paddles.\n\nLet $d \\mathrm{~km}$ be the total distance across the lake.\n\nSince Amrita paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and swims at $2 \\mathrm{~km} / \\mathrm{h}$, then $7 t_{1}+2 t_{2}+2 t_{3}=d$.\n\nSince Zhang paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and swims at $2 \\mathrm{~km} / \\mathrm{h}$, then $2 t_{1}+2 t_{2}+7 t_{3}=d$.\n\nSince the kayak travels at $7 \\mathrm{~km} / \\mathrm{h}$ and does not move while both Amrita and Zhang are swimming, then $7 t_{1}+0 t_{2}+7 t_{3}=d$.\n\nSince Amrita and Zhang each take 90 minutes ( $\\frac{3}{2}$ hours) to cross the lake, then the total time gives $t_{1}+t_{2}+t_{3}=\\frac{3}{2}$.\n\nFrom $7 t_{1}+2 t_{2}+2 t_{3}=d$ and $2 t_{1}+2 t_{2}+7 t_{3}=d$, we obtain $7 t_{1}+2 t_{2}+2 t_{3}=2 t_{1}+2 t_{2}+7 t_{3}$ or $5 t_{1}=5 t_{3}$ and so $t_{1}=t_{3}$.\n\nSince $7 t_{1}+2 t_{2}+2 t_{3}=d$ and $7 t_{1}+0 t_{2}+7 t_{3}=d$ and $t_{1}=t_{3}$, then $7 t_{1}+2 t_{2}+2 t_{1}=7 t_{1}+7 t_{1}$ or $2 t_{2}=5 t_{1}$ or $t_{2}=\\frac{5}{2} t_{1}$.\n\nSince $t_{1}+t_{2}+t_{3}=\\frac{3}{2}$, then $t_{1}+\\frac{5}{2} t_{1}+t_{1}=\\frac{3}{2}$ or $\\frac{9}{2} t_{1}=\\frac{3}{2}$ and so $t_{1}=\\frac{1}{3}$.\n\nThus, $t_{2}=\\frac{5}{2} \\cdot \\frac{1}{3}=\\frac{5}{6}$ hours (or 50 minutes) is the period of time that the kayak is not moving.']",['50'],False,minutes,Numerical, 2523,Algebra,,"Determine all pairs $(x, y)$ of real numbers that satisfy the system of equations $$ \begin{aligned} x\left(\frac{1}{2}+y-2 x^{2}\right) & =0 \\ y\left(\frac{5}{2}+x-y\right) & =0 \end{aligned} $$","['From the first equation, $x\\left(\\frac{1}{2}+y-2 x^{2}\\right)=0$, we obtain $x=0$ or $\\frac{1}{2}+y-2 x^{2}=0$.\n\nFrom the second equation, $y\\left(\\frac{5}{2}+x-y\\right)=0$, we obtain $y=0$ or $\\frac{5}{2}+x-y=0$.\n\nIf $x=0$, the first equation is satisified.\n\nFor the second equation to be true in this case, we need $y=0$ (giving the solution $(0,0)$ ) or $\\frac{5}{2}+0-y=0$. The second equation gives $y=\\frac{5}{2}$ (giving the solution $\\left(0, \\frac{5}{2}\\right)$ ).\n\nIf $y=0$, the second equation is satisified.\n\nFor the first equation to be true in this case, we need $x=0$ (giving the solution $(0,0)$ ) or $\\frac{1}{2}+0-2 x^{2}=0$. The second equation gives $x^{2}=\\frac{1}{4}$ or $x= \\pm \\frac{1}{2}$ (giving the solutions $\\left(\\frac{1}{2}, 0\\right)$ and $\\left.\\left(-\\frac{1}{2}, 0\\right)\\right)$.\n\nSo far, we have accounted for all solutions with $x=0$ or $y=0$.\n\nIf $x \\neq 0$ and $y \\neq 0$, then for both equations to be true, we need $\\frac{1}{2}+y-2 x^{2}=0$ (or $1+2 y-4 x^{2}=0$ ) and $\\frac{5}{2}+x-y=0$ ( or $5+2 x-2 y=0$ ).\n\nAdding these two equations, we obtain $6+2 x-4 x^{2}=0$.\n\nThis is equivalent to $2 x^{2}-x-3=0$ or $(2 x-3)(x+1)=0$, whose solutions are $x=\\frac{3}{2}$ and $x=-1$.\n\nThe equation $\\frac{5}{2}+x-y=0$ tells us that $y=x+\\frac{5}{2}$.\n\nIf $x=\\frac{3}{2}$, then $y=4$; if $x=-1$, then $y=\\frac{3}{2}$.\n\nTherefore, the complete list of pairs that satisfy the given system of equations is\n\n$$\n(x, y)=(0,0),\\left(0, \\frac{5}{2}\\right),\\left(\\frac{1}{2}, 0\\right),\\left(-\\frac{1}{2}, 0\\right),\\left(\\frac{3}{2}, 4\\right),\\left(-1, \\frac{3}{2}\\right)\n$$']","['$(0,0),(0, \\frac{5}{2}),(\\frac{1}{2}, 0),(-\\frac{1}{2}, 0),(\\frac{3}{2}, 4),(-1, \\frac{3}{2})$']",True,,Tuple, 2524,Algebra,,"Determine all real numbers $x>0$ for which $$ \log _{4} x-\log _{x} 16=\frac{7}{6}-\log _{x} 8 $$","['Note that $x \\neq 1$ since 1 cannot be the base of a logarithm. This tells us that $\\log x \\neq 0$. Using the fact that $\\log _{a} b=\\frac{\\log b}{\\log a}$ and then using other logarithm laws, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{4} x-\\log _{x} 16 & =\\frac{7}{6}-\\log _{x} 8 \\\\\n\\frac{\\log x}{\\log 4}-\\frac{\\log 16}{\\log x} & =\\frac{7}{6}-\\frac{\\log 8}{\\log x} \\quad(\\text { note that } x \\neq 1, \\text { so } \\log x \\neq 0) \\\\\n\\frac{\\log x}{\\log 4} & =\\frac{7}{6}+\\frac{\\log 16-\\log 8}{\\log x} \\\\\n\\frac{\\log x}{\\log \\left(2^{2}\\right)} & =\\frac{7}{6}+\\frac{\\log \\left(\\frac{16}{8}\\right)}{\\log x} \\\\\n\\frac{\\log x}{2 \\log 2} & =\\frac{7}{6}+\\frac{\\log 2}{\\log x} \\\\\n\\frac{1}{2}\\left(\\frac{\\log x}{\\log 2}\\right) & =\\frac{7}{6}+\\frac{\\log 2}{\\log x}\n\\end{aligned}\n$$\n\nLetting $t=\\frac{\\log x}{\\log 2}=\\log _{2} x$ and noting that $t \\neq 0$ since $x \\neq 1$, we obtain the following equations equivalent to the previous ones:\n\n$$\n\\begin{aligned}\n\\frac{t}{2} & =\\frac{7}{6}+\\frac{1}{t} \\\\\n3 t^{2} & =7 t+6 \\quad(\\text { multiplying both sides by } 6 t) \\\\\n3 t^{2}-7 t-6 & =0 \\\\\n(3 t+2)(t-3) & =0\n\\end{aligned}\n$$\n\nTherefore, the original equation is equivalent to $t=-\\frac{2}{3}$ or $t=3$.\n\nConverting back to the variable $x$, we obtain $\\log _{2} x=-\\frac{2}{3}$ or $\\log _{2} x=3$, which gives $x=2^{-2 / 3}$ or $x=2^{3}=8$.']","['$2^{-2 / 3}$, $8$']",True,,Numerical, 2525,Combinatorics,,"The string $A A A B B B A A B B$ is a string of ten letters, each of which is $A$ or $B$, that does not include the consecutive letters $A B B A$. The string $A A A B B A A A B B$ is a string of ten letters, each of which is $A$ or $B$, that does include the consecutive letters $A B B A$. Determine, with justification, the total number of strings of ten letters, each of which is $A$ or $B$, that do not include the consecutive letters $A B B A$.","['There are $2^{10}=1024$ strings of ten letters, each of which is $A$ or $B$, because there are 2 choices for each of the 10 positions in the string.\n\nWe determine the number of these strings that do not include the ""substring"" $A B B A$ (that is, that do not include consecutive letters $A B B A$ ) by counting the number of strings that do include the substring $A B B A$ and subtracting this total from 1024.\n\nIf a string includes the substring $A B B A$, there are 7 possible positions in which this substring could start ( $A B B A x x x x x x, x A B B A x x x x x, \\ldots, \\operatorname{xxxxxxABBA).}$\n\nThere are 2 choices for each of the remaining 6 letters in such a string, so there are $7 \\cdot 2^{6}=448$ occurrences of the substring $A B B A$ among the 1024 strings.\n\nThis does not mean that there are 448 strings that contain the substring $A B B A$. Since $A B B A$ can appear multiple times in a single string, this total will count some strings more than once. (For example, the string $A B B A A A A B B A$ is included in both the first and seventh of these categories, so this string is counted twice.)\n\nSo we must ""correct"" this total of 448 by accounting for the strings in which $A B B A$ occurs more than once.\n\nWe note that, since two substrings of $A B B A$ can overlap in 0 letters (for example, $A B B A A B B A x x$ ) or in 1 letter (for example, $A B B A B B A x x x$ ), then the maximum number of times that the substring $A B B A$ can appear is 3 , and there is only one such string: $A B B A B B A B B A$.\n\nIf a string contains two copies of $A B B A$ that overlap, then it must be of one of the following forms:\n\n$A B B A B B A x x \\quad x A B B A B B A x x \\quad x x A B B A B B A x \\quad x x x A B B A B B A$\n\nSince there are 4 choices for the starting position of $A B B A B B A$ and 2 choices for each of the three unknown letters, then there are $4 \\cdot 2^{3}=32$ occurrences of $A B B A B B A$ among all of these strings.\n\nBut the string $A B B A B B A B B A$ is counted in each of the first and last categories, so we subtract 2 occurrences from this total to obtain 30 , the total number of strings of ten letters that included exactly two overlapping copies of $A B B A$. (We\'ll count the string $A B B A B B A B B A$ later.)\n\nIf a string contains exactly two substrings of $A B B A$ and these do not overlap, then it must be of one of the following forms:\n\n$$\n\\begin{array}{lll}\nA B B A A B B A x x & A B B A x A B B A x & A B B A x x A B B A \\\\\nx A B B A A B B A x & x A B B A x A B B A & x x A B B A A B B A\n\\end{array}\n$$\n\nSince there are 6 such forms and 2 choices for each of the 2 unknown letters in each case, then there appear to be $6 \\cdot 2^{2}=24$ such strings.\n\nThis total includes the string $A B B A B B A B B A$ in the third category, so we subtract 1 from this total to obtain 23 , the total number of strings of ten letters that include exactly two copies of $A B B A$ which do not overlap.\n\nSo there are 30 strings that contain exactly two overlapping substrings $A B B A, 23$ strings that contain exactly two non-overlapping substrings $A B B A$, and 1 string that contains exactly three substrings $A B B A$.\n\nTo get the total number of strings with one or more substrings $A B B A$ we take the total number of occurrences of $A B B A$ (448), subtract the number of strings with exactly two substrings $A B B A$ (since these were included twice in the original count), and subtract two times the number of strings with exactly three substrings $A B B A$ (since these were included three times in the original count).\n\nTherefore, there are $448-23-30-2 \\cdot 1=393$ strings that include at least one substring\n$A B B A$, and so there are $1024-393=631$ strings of ten letters that do not include the substring $A B B A$.']",['631'],False,,Numerical, 2526,Geometry,,"In the diagram, $A B C D$ is a square. Points $E$ and $F$ are chosen on $A C$ so that $\angle E D F=45^{\circ}$. If $A E=x, E F=y$, and $F C=z$, prove that $y^{2}=x^{2}+z^{2}$. ","['Rotate $\\triangle D F C$ through an angle of $90^{\\circ}$ counterclockwise about $D$, so that $D C$ now lies along $D A$ and $F^{\\prime}$ is outside the square, as shown.\n\nJoin $F^{\\prime}$ to $E$.\n\n\n\nSince $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nSince $\\angle E A D=45^{\\circ}$ and $\\angle F^{\\prime} A D=\\angle F C D=45^{\\circ}$, then $\\angle F^{\\prime} A E=45^{\\circ}+45^{\\circ}=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle F^{\\prime} A E$, we have $F^{\\prime} E^{2}=F^{\\prime} A^{2}+A E^{2}$.\n\nBut $F^{\\prime} A=F C=z$ and $A E=x$, so $F^{\\prime} E^{2}=z^{2}+x^{2}$.\n\nTo show that $y^{2}=x^{2}+z^{2}$, it is sufficient to show that $F^{\\prime} E=y$.\n\nConsider $\\triangle F^{\\prime} D E$ and $\\triangle F D E$.\n\nNote that $F^{\\prime} D=F D$ and $\\angle F^{\\prime} D A=\\angle F D C$ by construction and $E D=E D$.\n\nAlso, $\\angle F^{\\prime} D E=\\angle F^{\\prime} D A+\\angle E D A=\\angle F D C+\\angle E D A=90^{\\circ}-\\angle E D F=45^{\\circ}$, which tells us that $\\angle F^{\\prime} D E=\\angle F D E=45^{\\circ}$.\n\nTherefore, $\\triangle F^{\\prime} D E$ is congruent to $\\triangle F D E$ (side-angle-side), and so $F^{\\prime} E=F E=y$.\n\nThis gives us the desired conclusion that $y^{2}=x^{2}+z^{2}$.', 'Since $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nLet $\\angle A D E=\\theta$.\n\nSince the angles in a triangle have a sum of $180^{\\circ}$, then\n\n$$\n\\angle A E D=180^{\\circ}-\\angle E A D-\\angle A D E=180^{\\circ}-45^{\\circ}-\\theta=135^{\\circ}-\\theta\n$$\n\nSince $A E F$ is a straight angle, then $\\angle D E F=180^{\\circ}-\\angle A E D=180^{\\circ}-\\left(135^{\\circ}-\\theta\\right)=45^{\\circ}+\\theta$. Continuing in this way, we find that $\\angle E F D=90^{\\circ}-\\theta, \\angle D F C=90^{\\circ}+\\theta$, and $\\angle F D C=45^{\\circ}-\\theta$.\n\n\n\nUsing the sine law in $\\triangle A E D$, we see that $\\frac{A E}{\\sin \\angle A D E}=\\frac{E D}{\\sin \\angle E A D}$ or $\\frac{x}{\\sin \\theta}=\\frac{E D}{\\sin 45^{\\circ}}$.\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{E D}{\\sin \\angle E F D}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{E D}{\\sin \\left(90^{\\circ}-\\theta\\right)}$.\n\n\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{F D}{\\sin \\angle D E F}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{F D}{\\sin \\left(45^{\\circ}+\\theta\\right)}$.\n\nUsing the sine law in $\\triangle D F C$, we get $\\frac{F C}{\\sin \\angle F D C}=\\frac{F D}{\\sin \\angle D C F}$ or $\\frac{z}{\\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{F D}{\\sin 45^{\\circ}}$.\n\nDividing the first of these equations by the second, we obtain $\\frac{x \\sin 45^{\\circ}}{y \\sin \\theta}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{x}{y}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right) \\sin \\theta}{\\sin ^{2} 45^{\\circ}}$.\n\nDividing the fourth of these equations by the third, we obtain $\\frac{z \\sin 45^{\\circ}}{y \\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{z}{y}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)}{\\sin ^{2} 45^{\\circ}}$.\n\nSince $\\sin \\left(90^{\\circ}-\\alpha\\right)=\\cos \\alpha$ for every angle $\\alpha$, then $\\sin \\left(90^{\\circ}-\\theta\\right)=\\cos \\theta$.\n\nAlso, $\\sin \\left(45^{\\circ}+\\theta\\right)=\\sin \\left(90^{\\circ}-\\left(45^{\\circ}-\\theta\\right)\\right)=\\cos \\left(45^{\\circ}-\\theta\\right)$.\n\nUsing this and the fact that $\\frac{1}{\\sin ^{2} 45^{\\circ}}=\\frac{1}{(1 / \\sqrt{2})^{2}}=2$, the expressions for $\\frac{x}{y}$ and $\\frac{z}{y}$ become\n\n$$\n\\frac{x}{y}=2 \\cos \\theta \\sin \\theta=\\sin 2 \\theta\n$$\n\nand\n\n$$\n\\frac{z}{y}=2 \\cos \\left(45^{\\circ}-\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)=\\sin \\left(2\\left(45^{\\circ}-\\theta\\right)\\right)=\\sin \\left(90^{\\circ}-2 \\theta\\right)=\\cos 2 \\theta\n$$\n\nFinally, this tells us that\n\n$$\n\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=\\left(\\frac{x}{y}\\right)^{2}+\\left(\\frac{z}{y}\\right)^{2}=\\sin ^{2} 2 \\theta+\\cos ^{2} 2 \\theta=1\n$$\n\nSince $\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=1$, then $x^{2}+z^{2}=y^{2}$, as required.\n\n\n\n\n#']",,True,,, 2526,Geometry,,"In the diagram, $A B C D$ is a square. Points $E$ and $F$ are chosen on $A C$ so that $\angle E D F=45^{\circ}$. If $A E=x, E F=y$, and $F C=z$, prove that $y^{2}=x^{2}+z^{2}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c9c4cc88d8dad4fbf43ag-1.jpg?height=455&width=480&top_left_y=1862&top_left_x=1251)","['Rotate $\\triangle D F C$ through an angle of $90^{\\circ}$ counterclockwise about $D$, so that $D C$ now lies along $D A$ and $F^{\\prime}$ is outside the square, as shown.\n\nJoin $F^{\\prime}$ to $E$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b51b3abf2eae4fca87eag-1.jpg?height=458&width=569&top_left_y=384&top_left_x=884)\n\nSince $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nSince $\\angle E A D=45^{\\circ}$ and $\\angle F^{\\prime} A D=\\angle F C D=45^{\\circ}$, then $\\angle F^{\\prime} A E=45^{\\circ}+45^{\\circ}=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle F^{\\prime} A E$, we have $F^{\\prime} E^{2}=F^{\\prime} A^{2}+A E^{2}$.\n\nBut $F^{\\prime} A=F C=z$ and $A E=x$, so $F^{\\prime} E^{2}=z^{2}+x^{2}$.\n\nTo show that $y^{2}=x^{2}+z^{2}$, it is sufficient to show that $F^{\\prime} E=y$.\n\nConsider $\\triangle F^{\\prime} D E$ and $\\triangle F D E$.\n\nNote that $F^{\\prime} D=F D$ and $\\angle F^{\\prime} D A=\\angle F D C$ by construction and $E D=E D$.\n\nAlso, $\\angle F^{\\prime} D E=\\angle F^{\\prime} D A+\\angle E D A=\\angle F D C+\\angle E D A=90^{\\circ}-\\angle E D F=45^{\\circ}$, which tells us that $\\angle F^{\\prime} D E=\\angle F D E=45^{\\circ}$.\n\nTherefore, $\\triangle F^{\\prime} D E$ is congruent to $\\triangle F D E$ (side-angle-side), and so $F^{\\prime} E=F E=y$.\n\nThis gives us the desired conclusion that $y^{2}=x^{2}+z^{2}$.', 'Since $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nLet $\\angle A D E=\\theta$.\n\nSince the angles in a triangle have a sum of $180^{\\circ}$, then\n\n$$\n\\angle A E D=180^{\\circ}-\\angle E A D-\\angle A D E=180^{\\circ}-45^{\\circ}-\\theta=135^{\\circ}-\\theta\n$$\n\nSince $A E F$ is a straight angle, then $\\angle D E F=180^{\\circ}-\\angle A E D=180^{\\circ}-\\left(135^{\\circ}-\\theta\\right)=45^{\\circ}+\\theta$. Continuing in this way, we find that $\\angle E F D=90^{\\circ}-\\theta, \\angle D F C=90^{\\circ}+\\theta$, and $\\angle F D C=45^{\\circ}-\\theta$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b51b3abf2eae4fca87eag-1.jpg?height=450&width=479&top_left_y=1968&top_left_x=926)\n\nUsing the sine law in $\\triangle A E D$, we see that $\\frac{A E}{\\sin \\angle A D E}=\\frac{E D}{\\sin \\angle E A D}$ or $\\frac{x}{\\sin \\theta}=\\frac{E D}{\\sin 45^{\\circ}}$.\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{E D}{\\sin \\angle E F D}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{E D}{\\sin \\left(90^{\\circ}-\\theta\\right)}$.\n\n\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{F D}{\\sin \\angle D E F}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{F D}{\\sin \\left(45^{\\circ}+\\theta\\right)}$.\n\nUsing the sine law in $\\triangle D F C$, we get $\\frac{F C}{\\sin \\angle F D C}=\\frac{F D}{\\sin \\angle D C F}$ or $\\frac{z}{\\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{F D}{\\sin 45^{\\circ}}$.\n\nDividing the first of these equations by the second, we obtain $\\frac{x \\sin 45^{\\circ}}{y \\sin \\theta}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{x}{y}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right) \\sin \\theta}{\\sin ^{2} 45^{\\circ}}$.\n\nDividing the fourth of these equations by the third, we obtain $\\frac{z \\sin 45^{\\circ}}{y \\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{z}{y}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)}{\\sin ^{2} 45^{\\circ}}$.\n\nSince $\\sin \\left(90^{\\circ}-\\alpha\\right)=\\cos \\alpha$ for every angle $\\alpha$, then $\\sin \\left(90^{\\circ}-\\theta\\right)=\\cos \\theta$.\n\nAlso, $\\sin \\left(45^{\\circ}+\\theta\\right)=\\sin \\left(90^{\\circ}-\\left(45^{\\circ}-\\theta\\right)\\right)=\\cos \\left(45^{\\circ}-\\theta\\right)$.\n\nUsing this and the fact that $\\frac{1}{\\sin ^{2} 45^{\\circ}}=\\frac{1}{(1 / \\sqrt{2})^{2}}=2$, the expressions for $\\frac{x}{y}$ and $\\frac{z}{y}$ become\n\n$$\n\\frac{x}{y}=2 \\cos \\theta \\sin \\theta=\\sin 2 \\theta\n$$\n\nand\n\n$$\n\\frac{z}{y}=2 \\cos \\left(45^{\\circ}-\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)=\\sin \\left(2\\left(45^{\\circ}-\\theta\\right)\\right)=\\sin \\left(90^{\\circ}-2 \\theta\\right)=\\cos 2 \\theta\n$$\n\nFinally, this tells us that\n\n$$\n\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=\\left(\\frac{x}{y}\\right)^{2}+\\left(\\frac{z}{y}\\right)^{2}=\\sin ^{2} 2 \\theta+\\cos ^{2} 2 \\theta=1\n$$\n\nSince $\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=1$, then $x^{2}+z^{2}=y^{2}$, as required.\n\n\n\n\n#']",['证明题，略'],True,,Need_human_evaluate, 2527,Combinatorics,,"Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$. Calculate $P(10)$.","['Here, $k=10$ and so there are 10 balls in each bag.\n\nSince there are 10 balls in each bag, there are $10 \\cdot 10=100$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen. To determine $P(10)$, we count the number of pairs $(a, b)$ for which $a b$ is divisible by 10 .\n\nIf the number of pairs is $m$, then $P(10)=\\frac{m}{100}$.\n\nFor $a b$ to be divisible by 10, at least one of $a$ and $b$ must be a multiple of 5 and at least one of $a$ and $b$ must be even.\n\nIf $a=10$ or $b=10$, then the pair $(a, b)$ gives a product $a b$ divisible by 10 .\n\nIn this case, we obtain the 19 pairs\n\n$$\n(a, b)=(1,10),(2,10), \\ldots,(9,10),(10,10),(10,9), \\ldots,(10,2),(10,1)\n$$\n\nIf neither $a$ nor $b$ equals 10 , then either $a=5$ or $b=5$ in order for $a$ or $b$ to be divisible by 5 .\n\nIn this case, the other of $a$ and $b$ must be even and not equal to 10. (We have already counted the pairs where $a=10$ or $b=10$.)\n\nIn this case, we obtain the 8 pairs\n\n$$\n(a, b)=(5,2),(5,4),(5,6),(5,8),(2,5),(4,5),(6,5),(8,5)\n$$\n\nFrom our work above, there are no additional pairs for which $a b$ is divisible by 10 .\n\nThus, there are $19+8=27$ pairs $(a, b)$ for which $a b$ is divisible by 10 , which means that $P(10)=\\frac{27}{100}$.\n\n(We note that we could have made a 10 by 10 table that listed all possible combinations of $a$ and $b$ and their product, from which we could obtain $P(10)$.)']",['$\\frac{27}{100}$'],False,,Numerical, 2528,Combinatorics,,"Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$. Determine, with justification, a polynomial $f(n)$ for which - $P(n) \geq \frac{f(n)}{n^{2}}$ for all positive integers $n$ with $n \geq 2$, and - $P(n)=\frac{f(n)}{n^{2}}$ for infinitely many positive integers $n$ with $n \geq 2$. (A polynomial $f(x)$ is an algebraic expression of the form $f(x)=a_{m} x^{m}+a_{m-1} x^{m-1}+\cdots+a_{1} x+a_{0}$ for some integer $m \geq 0$ and for some real numbers $a_{m}, a_{m-1}, \ldots, a_{1}, a_{0}$.)","['Let $n$ be a positive integer with $n \\geq 2$.\n\nConsider $f(n)=2 n-1$. This is a polynomial in $n$.\n\nWe demonstrate that $P(n) \\geq \\frac{2 n-1}{n^{2}}$ for all positive integers $n$ with $n \\geq 2$ and that $P(n)=\\frac{2 n-1}{n^{2}}$ for infinitely many positive integers $n$ with $n \\geq 2$.\n\nSuppose that there are two bags, each containing $n$ balls labelled from 1 to $n$.\n\nSince there are $n$ balls in each bag, there are $n^{2}$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen.\n\nThe pairs\n\n$$\n(a, b)=(1, n),(2, n), \\ldots,(n-1, n),(n, n),(n, n-1), \\ldots,(n, 2),(n, 1)\n$$\n\neach have the property that $a b$ is divisible by $n$.\n\nThere are $(n-1)+1+(n-1)=2 n-1$ of these pairs.\n\nTherefore, at least $2 n-1$ of the pairs of balls that can be chosen have labels whose product is divisible by $n$.\n\nSince there are $n^{2}$ pairs that can be chosen and the number of pairs of balls with the desired property is at least $2 n-1$, then $P(n) \\geq \\frac{2 n-1}{n^{2}}$.\n\nThis proves the first part of what we needed to prove.\n\n\n\nNext, suppose that $n=p$, a prime number.\n\nFor $a b$ to be divisible by $p$, then either $a$ is divisible by $p$ or $b$ is divisible by $p$ (or both). (This property is not true when $p$ is not a prime number; for example, $2 \\cdot 2$ is divisible by 4 even though neither factor is divisible by 4.)\n\nSince $1 \\leq a \\leq p$ and $1 \\leq b \\leq p$, then if either $a$ is divisible by $p$ or $b$ is divisible by $p$ (or both), we must have either $a=p$ or $b=p$, or both.\n\nIn other words, $a b$ is divisible by $p$ exactly when $(a, b)$ is in the list\n\n$$\n(1, p),(2, p), \\ldots,(p-1, p),(p, p),(p, p-1), \\ldots,(p, 2),(p, 1)\n$$\n\nThere are $2 p-1$ pairs in this list and these are the only pairs for which $a b$ is divisible by $p$.\n\nTherefore, $P(n)=\\frac{2 n-1}{n^{2}}$ when $n$ is a prime number.\n\nSince there are infinitely many prime numbers, then $P(n)=\\frac{2 n-1}{n^{2}}$ for infinitely many positive integers $n$ with $n \\geq 2$.\n\nThus, $f(n)=2 n-1$ is a polynomial with the desired properties.']",,True,,, 2529,Combinatorics,,"Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$. Prove there exists a positive integer $m$ for which $P(m)>\frac{2016}{m}$.","['Let $N=2^{k}$, where $k$ is a positive integer with $k \\geq 2$.\n\nSuppose that there are two bags, each containing $N$ balls labelled from 1 to $N$.\n\nSince there are $N$ balls in each bag, there are $N^{2}$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen.\n\nLet $j$ be a positive integer with $1 \\leq j \\leq k-1$.\n\nConsider pairs of the form $(a, b)=\\left(2^{j} x, 2^{k-j} y\\right)$ where $x$ and $y$ are odd positive integers that keep $a$ and $b$ in the desired range.\n\nNote that, in each case, $a b=\\left(2^{j} x\\right)\\left(2^{k-j} y\\right)=2^{k} x y$ which is divisible by $N=2^{k}$.\n\nSince $1 \\leq a \\leq 2^{k}$, then $1 \\leq 2^{j} x \\leq 2^{k}$ and so $x \\leq 2^{k-j}$.\n\nSince half of the positive integers from 1 to $2^{k-j}$ are odd, then there are $\\frac{1}{2} 2^{k-j}=2^{k-j-1}$ choices for $x$.\n\nSimilarly, there are $2^{k-(k-j)-1}=2^{j-1}$ choices for $y$.\n\nNote that each choice of $x$ and $y$ gives a unique pair $(a, b)$.\n\nFor any fixed $j$, there are $2^{k-j-1}$ choices for $x$ and $2^{j-1}$ choices for $y$.\n\nThus, there are $2^{k-j-1} \\cdot 2^{j-1}=2^{k-2}$ choices of this form for the pair $(a, b)$.\n\nSo for a fixed $j$ with $1 \\leq j \\leq k-1$, this method gives $2^{k-2}$ pairs $(a, b)$ for which $a b$ is divisible by $N$.\n\nSince there are $k-1$ different values for $j$, then there are at least $(k-1) 2^{k-2}$ pairs $(a, b)$ for which $a b$ is divisible by $N$. (Note that two pairs that come from different values of $j$ will in fact be different since the number of factors of 2 in their values of $a$ will be different.) Since there are $N^{2}$ choices for $(a, b)$, then\n\n$$\nP(N) \\geq \\frac{(k-1) 2^{k-2}}{N^{2}}=\\frac{(k-1) 2^{k} 2^{-2}}{N^{2}}=\\frac{k-1}{4} \\cdot \\frac{1}{N}\n$$\n\nWhen $\\frac{k-1}{4}>2016$, we will have $P(N)>2016 \\cdot \\frac{1}{N}$.\n\nThe inequality $\\frac{k-1}{4}>2016$ is equivalent to $k-1>8064$ or $k>8065$.\n\nWe want to show that there exists a positive integer $m$ with $P(m)>\\frac{2016}{m}$.\n\nSet $m=2^{8066}$.\n\nBy the work above, $P(m) \\geq \\frac{8065}{4} \\cdot \\frac{1}{m}>\\frac{2016}{m}$, as required.']",,True,,, 2530,Geometry,,"In rectangle $A B C D, F$ is on diagonal $B D$ so that $A F$ is perpendicular to $B D$. Also, $B C=30, C D=40$ and $A F=x$. Determine the value of $x$. ","['Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nWe calculate the area of $\\triangle A D B$ is two different ways.\n\nFirst, using $A B$ as base and $A D$ as height, we obtain an area of $\\frac{1}{2}(40)(30)=600$.\n\nNext, using $D B$ as base and $A F$ as height, we obtain an area of $\\frac{1}{2}(50) x=25 x$.\n\nWe must have $25 x=600$ and so $x=\\frac{600}{25}=24$.', 'Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nSince $\\triangle D A B$ is right-angled at $A$, then $\\sin (\\angle A D B)=\\frac{A B}{B D}=\\frac{40}{50}=\\frac{4}{5}$.\n\nBut $\\triangle A D F$ is right-angled at $F$ and $\\angle A D F=\\angle A D B$.\n\nTherefore, $\\sin (\\angle A D F)=\\frac{A F}{A D}=\\frac{x}{30}$.\n\nThus, $\\frac{x}{30}=\\frac{4}{5}$ and so $x=\\frac{4}{5}(30)=24$.\n\nSolution 3\n\nSince $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nNote that $\\triangle B F A$ is similar to $\\triangle B A D$, since each is right-angled and they share a common angle at $B$.\n\nThus, $\\frac{A F}{A B}=\\frac{A D}{B D}$ and so $\\frac{x}{30}=\\frac{40}{50}$ which gives $x=\\frac{30(40)}{50}=24$.']",['24'],False,,Numerical, 2530,Geometry,,"In rectangle $A B C D, F$ is on diagonal $B D$ so that $A F$ is perpendicular to $B D$. Also, $B C=30, C D=40$ and $A F=x$. Determine the value of $x$. ![](https://cdn.mathpix.com/cropped/2023_12_21_a9a67760d94609d98542g-1.jpg?height=271&width=380&top_left_y=604&top_left_x=1401)","['Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nWe calculate the area of $\\triangle A D B$ is two different ways.\n\nFirst, using $A B$ as base and $A D$ as height, we obtain an area of $\\frac{1}{2}(40)(30)=600$.\n\nNext, using $D B$ as base and $A F$ as height, we obtain an area of $\\frac{1}{2}(50) x=25 x$.\n\nWe must have $25 x=600$ and so $x=\\frac{600}{25}=24$.', 'Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nSince $\\triangle D A B$ is right-angled at $A$, then $\\sin (\\angle A D B)=\\frac{A B}{B D}=\\frac{40}{50}=\\frac{4}{5}$.\n\nBut $\\triangle A D F$ is right-angled at $F$ and $\\angle A D F=\\angle A D B$.\n\nTherefore, $\\sin (\\angle A D F)=\\frac{A F}{A D}=\\frac{x}{30}$.\n\nThus, $\\frac{x}{30}=\\frac{4}{5}$ and so $x=\\frac{4}{5}(30)=24$.\n\nSolution 3\n\nSince $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.\n\nBy the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then\n\n$$\nB D=\\sqrt{30^{2}+40^{2}}=\\sqrt{900+1600}=\\sqrt{2500}=50\n$$\n\nNote that $\\triangle B F A$ is similar to $\\triangle B A D$, since each is right-angled and they share a common angle at $B$.\n\nThus, $\\frac{A F}{A B}=\\frac{A D}{B D}$ and so $\\frac{x}{30}=\\frac{40}{50}$ which gives $x=\\frac{30(40)}{50}=24$.']",['24'],False,,Numerical, 2531,Combinatorics,,"In an arithmetic sequence, the first term is 1 and the last term is 19 . The sum of all the terms in the sequence is 70 . How many terms does the sequence have? (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, 3, 5, 7, 9 is an arithmetic sequence with four terms.)","['The sum of the terms in an arithmetic sequence is equal to the average of the first and last terms times the number of terms.\n\nIf $n$ is the number of terms in the sequence, then $\\frac{1}{2}(1+19) n=70$ or $10 n=70$ and so $n=7$.', 'Let $n$ be the number of terms in the sequence and $d$ the common difference.\n\nSince the first term is 1 and the $n$th term equals 19 , then $1+(n-1) d=19$ and so $(n-1) d=18$.\n\nSince the sum of the terms in the sequence is 70 , then $\\frac{1}{2} n(1+1+(n-1) d)=70$.\n\nThus, $\\frac{1}{2} n(2+18)=70$ or $10 n=70$ and so $n=7$.']",['7'],False,,Numerical, 2532,Algebra,,Suppose that $a(x+b(x+3))=2(x+6)$ for all values of $x$. Determine $a$ and $b$.,"['Since the given equation is true for all values of $x$, then it is true for any particular value of $x$ that we try.\n\nIf $x=-3$, the equation becomes $a(-3+b(0))=2(3)$ or $-3 a=6$ and so $a=-2$.\n\nIf $x=0$, the equation becomes $-2(0+b(3))=2(6)$ or $-6 b=12$ and so $b=-2$.\n\nTherefore, $a=-2$ and $b=-2$.', 'We expand both sides of the equation:\n\n$$\n\\begin{aligned}\na(x+b(x+3)) & =2(x+6) \\\\\na(x+b x+3 b) & =2 x+12 \\\\\na x+a b x+3 a b & =2 x+12 \\\\\n(a+a b) x+3 a b & =2 x+12\n\\end{aligned}\n$$\n\nSince this equation is true for all values of $x$, then the coefficients on the left side and right side must be equal, so $a+a b=2$ and $3 a b=12$.\n\nFrom the second equation, $a b=4$ so the first equation becomes $a+4=2$ or $a=-2$.\n\nSince $a b=4$, then $-2 b=4$ and so $b=-2$.\n\nThus, $a=b=-2$.']","['-2,-2']",True,,Numerical, 2533,Geometry,,"In the diagram, $\triangle A B C$ is right-angled at $C$. Also, $2 \sin B=3 \tan A$. Determine the measure of angle $A$. ","['Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\cos A$.\n\nTherefore, $2 \\cos A=3 \\tan A=\\frac{3 \\sin A}{\\cos A}$ or $2 \\cos ^{2} A=3 \\sin A$.\n\nUsing the fact that $\\cos ^{2} A=1-\\sin ^{2} A$, this becomes $2-2 \\sin ^{2} A=3 \\sin A$\n\nor $2 \\sin ^{2} A+3 \\sin A-2=0$ or $(2 \\sin A-1)(\\sin A+2)=0$.\n\nSince $\\sin A$ is between -1 and 1 , then $\\sin A=\\frac{1}{2}$.\n\nSince $A$ is an acute angle, then $A=30^{\\circ}$.', 'Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\frac{b}{c}$ and $\\tan A=\\frac{a}{b}$.\n\nThus, the given equation is $\\frac{2 b}{c}=\\frac{3 a}{b}$ or $2 b^{2}=3 a c$.\n\nUsing the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.\n\nFactoring, we obtain $(c-2 a)(2 c+a)=0$.\n\nSince $a$ and $c$ must both be positive, then $c=2 a$.\n\nSince $\\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\\triangle A B C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, with $A=30^{\\circ}$.']",['$30^{\\circ}$'],False,,Numerical, 2533,Geometry,,"In the diagram, $\triangle A B C$ is right-angled at $C$. Also, $2 \sin B=3 \tan A$. Determine the measure of angle $A$. ![](https://cdn.mathpix.com/cropped/2023_12_21_a9a67760d94609d98542g-1.jpg?height=285&width=374&top_left_y=1535&top_left_x=1320)","['Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\cos A$.\n\nTherefore, $2 \\cos A=3 \\tan A=\\frac{3 \\sin A}{\\cos A}$ or $2 \\cos ^{2} A=3 \\sin A$.\n\nUsing the fact that $\\cos ^{2} A=1-\\sin ^{2} A$, this becomes $2-2 \\sin ^{2} A=3 \\sin A$\n\nor $2 \\sin ^{2} A+3 \\sin A-2=0$ or $(2 \\sin A-1)(\\sin A+2)=0$.\n\nSince $\\sin A$ is between -1 and 1 , then $\\sin A=\\frac{1}{2}$.\n\nSince $A$ is an acute angle, then $A=30^{\\circ}$.', 'Since $\\triangle A B C$ is right-angled at $C$, then $\\sin B=\\frac{b}{c}$ and $\\tan A=\\frac{a}{b}$.\n\nThus, the given equation is $\\frac{2 b}{c}=\\frac{3 a}{b}$ or $2 b^{2}=3 a c$.\n\nUsing the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.\n\nFactoring, we obtain $(c-2 a)(2 c+a)=0$.\n\nSince $a$ and $c$ must both be positive, then $c=2 a$.\n\nSince $\\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\\triangle A B C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, with $A=30^{\\circ}$.']",['$30^{\\circ}$'],False,,Numerical, 2534,Combinatorics,,"An integer $n$, with $100 \leq n \leq 999$, is chosen at random. What is the probability that the sum of the digits of $n$ is 24 ?","[""The number of integers between 100 and 999 inclusive is $999-100+1=900$.\n\nAn integer $n$ in this range has three digits, say $a, b$ and $c$, with the hundreds digit equal to $a$.\n\nNote that $0 \\leq b \\leq 9$ and $0 \\leq c \\leq 9$ and $1 \\leq a \\leq 9$.\n\nTo have $a+b+c=24$, then the possible triples for $a, b, c$ in some order are $9,9,6 ; 9,8,7$; $8,8,8$. (There cannot be three 9's. If there are two 9's, the the other digit equals 6 . If there is one 9 , the second and third digits add to 15 but are both less than 9 , so must equal 8 and 7 . If there are zero 9's, the maximum for each digit is 8 , and so each digt must be 8 in order for the sum of all three to equal 24.)\n\nIf the digits are 9, 9 and 6, there are 3 arrangements: 996, 969, 699.\n\n\n\nIf the digits are 9, 8 and 7, there are 6 arrangements: 987, 978, 897, 879, 798, 789.\n\nIf the digits are 8,8 and 8 , there is only 1 arrangement: 888 .\n\nTherefore, there are $3+6+1=10$ integers $n$ in the range 100 to 999 with the sum of the digits of $n$ equal to 24 .\n\nThe required probability equals the number of possible values of $n$ with the sum of digits equal to 24 divided by the total number of integers in the range, or $\\frac{10}{900}=\\frac{1}{90}$.""]",['$\\frac{1}{90}$'],False,,Numerical, 2535,Geometry,,"Alice drove from town $E$ to town $F$ at a constant speed of $60 \mathrm{~km} / \mathrm{h}$. Bob drove from $F$ to $E$ along the same road also at a constant speed. They started their journeys at the same time and passed each other at point $G$. ![](https://cdn.mathpix.com/cropped/2023_12_21_a9a67760d94609d98542g-1.jpg?height=157&width=482&top_left_y=2141&top_left_x=884) Alice drove from $G$ to $F$ in 45 minutes. Bob drove from $G$ to $E$ in 20 minutes. Determine Bob's constant speed.","[""Since Alice drives at $60 \\mathrm{~km} / \\mathrm{h}$, then she drives $1 \\mathrm{~km}$ every minute.\n\nSince Alice drove from $G$ to $F$ in 45 minutes, then the distance from $G$ to $F$ is $45 \\mathrm{~km}$.\n\nLet the distance from $E$ to $G$ be $d \\mathrm{~km}$ and let Bob's speed be $B \\mathrm{~km} / \\mathrm{h}$.\n\nSince Bob drove from $G$ to $E$ in 20 minutes (or $\\frac{1}{3}$ of an hour), then $\\frac{d}{B}=\\frac{1}{3}$. Thus, $d=\\frac{1}{3} B$.\n\nThe time that it took Bob to drive from $F$ to $G$ was $\\frac{45}{B}$ hours.\n\nThe time that it took Alice to drive from $E$ to $G$ was $\\frac{d}{60}$ hours.\n\nSince the time that it took each of Alice and Bob to reach $G$ was the same, then $\\frac{d}{60}=\\frac{45}{B}$\n\nand so $B d=45(60)=2700$.\n\nThus, $B\\left(\\frac{1}{3} B\\right)=2700$ so $B^{2}=8100$ or $B=90$ since $B>0$.\n\nTherefore, Bob's speed was $90 \\mathrm{~km} / \\mathrm{h}$.""]",['90'],False,km/h,Numerical, 2535,Geometry,,"Alice drove from town $E$ to town $F$ at a constant speed of $60 \mathrm{~km} / \mathrm{h}$. Bob drove from $F$ to $E$ along the same road also at a constant speed. They started their journeys at the same time and passed each other at point $G$. Alice drove from $G$ to $F$ in 45 minutes. Bob drove from $G$ to $E$ in 20 minutes. Determine Bob's constant speed.","[""Since Alice drives at $60 \\mathrm{~km} / \\mathrm{h}$, then she drives $1 \\mathrm{~km}$ every minute.\n\nSince Alice drove from $G$ to $F$ in 45 minutes, then the distance from $G$ to $F$ is $45 \\mathrm{~km}$.\n\nLet the distance from $E$ to $G$ be $d \\mathrm{~km}$ and let Bob's speed be $B \\mathrm{~km} / \\mathrm{h}$.\n\nSince Bob drove from $G$ to $E$ in 20 minutes (or $\\frac{1}{3}$ of an hour), then $\\frac{d}{B}=\\frac{1}{3}$. Thus, $d=\\frac{1}{3} B$.\n\nThe time that it took Bob to drive from $F$ to $G$ was $\\frac{45}{B}$ hours.\n\nThe time that it took Alice to drive from $E$ to $G$ was $\\frac{d}{60}$ hours.\n\nSince the time that it took each of Alice and Bob to reach $G$ was the same, then $\\frac{d}{60}=\\frac{45}{B}$\n\nand so $B d=45(60)=2700$.\n\nThus, $B\\left(\\frac{1}{3} B\\right)=2700$ so $B^{2}=8100$ or $B=90$ since $B>0$.\n\nTherefore, Bob's speed was $90 \\mathrm{~km} / \\mathrm{h}$.""]",['90'],False,km/h,Numerical, 2536,Geometry,,The parabola $y=x^{2}-2 x+4$ is translated $p$ units to the right and $q$ units down. The $x$-intercepts of the resulting parabola are 3 and 5 . What are the values of $p$ and $q$ ?,"['Completing the square on the original parabola, we obtain\n\n$$\ny=x^{2}-2 x+4=x^{2}-2 x+1-1+4=(x-1)^{2}+3\n$$\n\nTherefore, the vertex of the original parabola is $(1,3)$.\n\nSince the new parabola is a translation of the original parabola and has $x$-intercepts 3 and 5 , then its equation is $y=1(x-3)(x-5)=x^{2}-8 x+15$.\n\nCompleting the square here, we obtain\n\n$$\ny=x^{2}-8 x+15=x^{2}-8 x+16-16+15=(x-4)^{2}-1\n$$\n\nTherefore, the vertex of the new parabola is $(4,-1)$.\n\nThus, the point $(1,3)$ is translated $p$ units to the right and $q$ units down to reach $(4,-1)$, so $p=3$ and $q=4$.']","['3,4']",True,,Numerical, 2537,Geometry,,"In the diagram, $D$ is the vertex of a parabola. The parabola cuts the $x$-axis at $A$ and at $C(4,0)$. The parabola cuts the $y$-axis at $B(0,-4)$. The area of $\triangle A B C$ is 4. Determine the area of $\triangle D B C$. ","['First, we determine the coordinates of $A$.\n\nThe area of $\\triangle A B C$ is 4 . We can think of $A C$ as its base, and its height being the distance from $B$ to the $x$-axis.\n\nIf the coordinates of $A$ are $(a, 0)$, then the base has length $4-a$ and the height is 4 .\n\nThus, $\\frac{1}{2}(4-a)(4)=4$, so $4-a=2$ and so $a=2$.\n\nTherefore, the coordinates of $A$ are $(2,0)$.\n\nNext, we determine the equation of the parabola.\n\nThe parabola has $x$-intercepts 2 and 4 , so has equation $y=k(x-2)(x-4)$.\n\nSince the parabola passes through $(0,-4)$ as well, then $-4=k(-2)(-4)$ so $k=-\\frac{1}{2}$.\n\nTherefore, the parabola has equation $y=-\\frac{1}{2}(x-2)(x-4)$.\n\nNext, we determine the coordinates of $D$, the vertex of the parabola.\n\nSince the $x$-intercepts are 2 and 4 , then the $x$-coordinate of the vertex is the average of these, or 3.\n\n\n\nThe $y$-coordinate of $D$ can be obtained from the equation of the parabola; we obtain $y=-\\frac{1}{2}(3-2)(3-4)=-\\frac{1}{2}(1)(-1)=\\frac{1}{2}$.\n\nThus, the coordinates of $D$ are $\\left(3, \\frac{1}{2}\\right)$.\n\nLastly, we determine the area of $\\triangle B D C$, whose vertices have coordinates $B(0,-4)$, $D\\left(3, \\frac{1}{2}\\right)$, and $C(4,0)$.\n\nMethod 1 \n\nWe proceed be ""completing the rectangle"". That is, we draw the rectangle with horizontal sides along the lines $y=\\frac{1}{2}$ and $y=-4$ and vertical sides along the lines $x=0$ and $x=4$. We label this rectangle as $B P Q R$.\n\n\n\nThe area of $\\triangle B D C$ equals the area of the rectangle minus the areas of $\\triangle B P D, \\triangle D Q C$ and $\\triangle C R B$.\n\nRectangle $B P Q R$ has height $4+\\frac{1}{2}=\\frac{9}{2}$ and width 4 .\n\n$\\triangle B P D$ has height $\\frac{9}{2}$ and base 3 .\n\n$\\triangle D Q C$ has height $\\frac{1}{2}$ and base 1.\n\n$\\triangle C R B$ has height 4 and base 4.\n\nTherefore, the area of $\\triangle B D C$ is $4\\left(\\frac{9}{2}\\right)-\\frac{1}{2}\\left(\\frac{9}{2}\\right)(3)-\\frac{1}{2}\\left(\\frac{1}{2}\\right)(1)-\\frac{1}{2}(4)(4)=18-\\frac{27}{4}-\\frac{1}{4}-8=3$.\n\nMethod 2\n\nWe determine the coordinates of $E$, the point where $B D$ crosses the $x$-axis.\n\n\n\nOnce we have done this, then the area of $\\triangle B D C$ equals the sum of the areas of $\\triangle E C B$ and $\\triangle E C D$.\n\nSince $B$ has coordinates $(0,-4)$ and $D$ has coordinates $\\left(3, \\frac{1}{2}\\right)$, then the slope of $B D$ is $\\frac{\\frac{1}{2}-(-4)}{3-0}=\\frac{\\frac{9}{2}}{3}=\\frac{3}{2}$.\n\nSince $B$ is on the $y$-axis, then the equation of the line through $B$ and $D$ is $y=\\frac{3}{2} x-4$. To find the $x$-coordinate of $E$, we set $y=0$ to obtain $0=\\frac{3}{2} x-4$ or $\\frac{3}{2} x=4$ or $x=\\frac{8}{3}$.\n\nWe think of $E C$ as the base of each of the two smaller triangles. Note that $E C=4-\\frac{8}{3}=\\frac{4}{3}$. Thus, the area of $\\triangle E C D$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)\\left(\\frac{1}{2}\\right)=\\frac{1}{3}$.\n\nAlso, the area of $\\triangle E C B$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)(4)=\\frac{8}{3}$.\n\nTherefore, the area of $\\triangle B D C$ is $\\frac{1}{3}+\\frac{8}{3}=3$.']",['3'],False,,Numerical, 2537,Geometry,,"In the diagram, $D$ is the vertex of a parabola. The parabola cuts the $x$-axis at $A$ and at $C(4,0)$. The parabola cuts the $y$-axis at $B(0,-4)$. The area of $\triangle A B C$ is 4. Determine the area of $\triangle D B C$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b301ea0eb47ddd66a1f3g-1.jpg?height=431&width=567&top_left_y=424&top_left_x=1256)","['First, we determine the coordinates of $A$.\n\nThe area of $\\triangle A B C$ is 4 . We can think of $A C$ as its base, and its height being the distance from $B$ to the $x$-axis.\n\nIf the coordinates of $A$ are $(a, 0)$, then the base has length $4-a$ and the height is 4 .\n\nThus, $\\frac{1}{2}(4-a)(4)=4$, so $4-a=2$ and so $a=2$.\n\nTherefore, the coordinates of $A$ are $(2,0)$.\n\nNext, we determine the equation of the parabola.\n\nThe parabola has $x$-intercepts 2 and 4 , so has equation $y=k(x-2)(x-4)$.\n\nSince the parabola passes through $(0,-4)$ as well, then $-4=k(-2)(-4)$ so $k=-\\frac{1}{2}$.\n\nTherefore, the parabola has equation $y=-\\frac{1}{2}(x-2)(x-4)$.\n\nNext, we determine the coordinates of $D$, the vertex of the parabola.\n\nSince the $x$-intercepts are 2 and 4 , then the $x$-coordinate of the vertex is the average of these, or 3.\n\n\n\nThe $y$-coordinate of $D$ can be obtained from the equation of the parabola; we obtain $y=-\\frac{1}{2}(3-2)(3-4)=-\\frac{1}{2}(1)(-1)=\\frac{1}{2}$.\n\nThus, the coordinates of $D$ are $\\left(3, \\frac{1}{2}\\right)$.\n\nLastly, we determine the area of $\\triangle B D C$, whose vertices have coordinates $B(0,-4)$, $D\\left(3, \\frac{1}{2}\\right)$, and $C(4,0)$.\n\nMethod 1 \n\nWe proceed be ""completing the rectangle"". That is, we draw the rectangle with horizontal sides along the lines $y=\\frac{1}{2}$ and $y=-4$ and vertical sides along the lines $x=0$ and $x=4$. We label this rectangle as $B P Q R$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d9b0e20a7552d07ba134g-1.jpg?height=271&width=266&top_left_y=710&top_left_x=1035)\n\nThe area of $\\triangle B D C$ equals the area of the rectangle minus the areas of $\\triangle B P D, \\triangle D Q C$ and $\\triangle C R B$.\n\nRectangle $B P Q R$ has height $4+\\frac{1}{2}=\\frac{9}{2}$ and width 4 .\n\n$\\triangle B P D$ has height $\\frac{9}{2}$ and base 3 .\n\n$\\triangle D Q C$ has height $\\frac{1}{2}$ and base 1.\n\n$\\triangle C R B$ has height 4 and base 4.\n\nTherefore, the area of $\\triangle B D C$ is $4\\left(\\frac{9}{2}\\right)-\\frac{1}{2}\\left(\\frac{9}{2}\\right)(3)-\\frac{1}{2}\\left(\\frac{1}{2}\\right)(1)-\\frac{1}{2}(4)(4)=18-\\frac{27}{4}-\\frac{1}{4}-8=3$.\n\nMethod 2\n\nWe determine the coordinates of $E$, the point where $B D$ crosses the $x$-axis.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_d9b0e20a7552d07ba134g-1.jpg?height=277&width=309&top_left_y=1542&top_left_x=1011)\n\nOnce we have done this, then the area of $\\triangle B D C$ equals the sum of the areas of $\\triangle E C B$ and $\\triangle E C D$.\n\nSince $B$ has coordinates $(0,-4)$ and $D$ has coordinates $\\left(3, \\frac{1}{2}\\right)$, then the slope of $B D$ is $\\frac{\\frac{1}{2}-(-4)}{3-0}=\\frac{\\frac{9}{2}}{3}=\\frac{3}{2}$.\n\nSince $B$ is on the $y$-axis, then the equation of the line through $B$ and $D$ is $y=\\frac{3}{2} x-4$. To find the $x$-coordinate of $E$, we set $y=0$ to obtain $0=\\frac{3}{2} x-4$ or $\\frac{3}{2} x=4$ or $x=\\frac{8}{3}$.\n\nWe think of $E C$ as the base of each of the two smaller triangles. Note that $E C=4-\\frac{8}{3}=\\frac{4}{3}$. Thus, the area of $\\triangle E C D$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)\\left(\\frac{1}{2}\\right)=\\frac{1}{3}$.\n\nAlso, the area of $\\triangle E C B$ is $\\frac{1}{2}\\left(\\frac{4}{3}\\right)(4)=\\frac{8}{3}$.\n\nTherefore, the area of $\\triangle B D C$ is $\\frac{1}{3}+\\frac{8}{3}=3$.']",['3'],False,,Numerical, 2538,Geometry,,"In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$. ","['Join $O$ to $A, B$ and $C$.\n\n\n\nSince $A B$ is tangent to the circle at $A$, then $\\angle O A B=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle O A B$, we get $O A^{2}+A B^{2}=O B^{2}$ or $r^{2}+p^{2}=O B^{2}$.\n\nIn $\\triangle O D C$, we have $O D=D C=q$ and $O C=r$.\n\nBy the cosine law,\n\n$$\n\\begin{aligned}\nO C^{2} & =O D^{2}+D C^{2}-2(O D)(D C) \\cos (\\angle O D C) \\\\\nr^{2} & =q^{2}+q^{2}-2 q^{2} \\cos (\\angle O D C) \\\\\n\\cos (\\angle O D C) & =\\frac{2 q^{2}-r^{2}}{2 q^{2}}\n\\end{aligned}\n$$\n\nIn $\\triangle O D B$, we have $\\angle O D B=\\angle O D C$.\n\nThus, using the cosine law again,\n\n$$\n\\begin{aligned}\nO B^{2} & =O D^{2}+D B^{2}-2(O D)(D B) \\cos (\\angle O D B) \\\\\n& =q^{2}+(2 q)^{2}-2(q)(2 q)\\left(\\frac{2 q^{2}-r^{2}}{2 q^{2}}\\right) \\\\\n& =q^{2}+4 q^{2}-2\\left(2 q^{2}-r^{2}\\right) \\\\\n& =q^{2}+2 r^{2}\n\\end{aligned}\n$$\n\nSo $O B^{2}=r^{2}+p^{2}=q^{2}+2 r^{2}$, which gives $p^{2}=q^{2}+r^{2}$, as required.']",,True,,, 2538,Geometry,,"In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b301ea0eb47ddd66a1f3g-1.jpg?height=314&width=353&top_left_y=1274&top_left_x=1344)","['Join $O$ to $A, B$ and $C$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_dc1d6915b9aeb263a0c9g-1.jpg?height=306&width=338&top_left_y=1332&top_left_x=991)\n\nSince $A B$ is tangent to the circle at $A$, then $\\angle O A B=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle O A B$, we get $O A^{2}+A B^{2}=O B^{2}$ or $r^{2}+p^{2}=O B^{2}$.\n\nIn $\\triangle O D C$, we have $O D=D C=q$ and $O C=r$.\n\nBy the cosine law,\n\n$$\n\\begin{aligned}\nO C^{2} & =O D^{2}+D C^{2}-2(O D)(D C) \\cos (\\angle O D C) \\\\\nr^{2} & =q^{2}+q^{2}-2 q^{2} \\cos (\\angle O D C) \\\\\n\\cos (\\angle O D C) & =\\frac{2 q^{2}-r^{2}}{2 q^{2}}\n\\end{aligned}\n$$\n\nIn $\\triangle O D B$, we have $\\angle O D B=\\angle O D C$.\n\nThus, using the cosine law again,\n\n$$\n\\begin{aligned}\nO B^{2} & =O D^{2}+D B^{2}-2(O D)(D B) \\cos (\\angle O D B) \\\\\n& =q^{2}+(2 q)^{2}-2(q)(2 q)\\left(\\frac{2 q^{2}-r^{2}}{2 q^{2}}\\right) \\\\\n& =q^{2}+4 q^{2}-2\\left(2 q^{2}-r^{2}\\right) \\\\\n& =q^{2}+2 r^{2}\n\\end{aligned}\n$$\n\nSo $O B^{2}=r^{2}+p^{2}=q^{2}+2 r^{2}$, which gives $p^{2}=q^{2}+r^{2}$, as required.']",['证明题，略'],True,,Need_human_evaluate, 2539,Algebra,,"If $\log _{2} x,\left(1+\log _{4} x\right)$, and $\log _{8} 4 x$ are consecutive terms of a geometric sequence, determine the possible values of $x$. (A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)","['First, we convert each of the logarithms to a logarithm with base 2:\n\n$$\n\\begin{aligned}\n1+\\log _{4} x & =1+\\frac{\\log _{2} x}{\\log _{2} 4}=1+\\frac{\\log _{2} x}{2}=1+\\frac{1}{2} \\log _{2} x \\\\\n\\log _{8} 4 x & =\\frac{\\log _{2} 4 x}{\\log _{2} 8}=\\frac{\\log _{2} 4+\\log _{2} x}{3}=\\frac{2}{3}+\\frac{1}{3} \\log _{2} x\n\\end{aligned}\n$$\n\nLet $y=\\log _{2} x$. Then the three terms are $y, 1+\\frac{1}{2} y$, and $\\frac{2}{3}+\\frac{1}{3} y$. Since these three are in geometric sequence, then\n\n$$\n\\begin{aligned}\n\\frac{y}{1+\\frac{1}{2} y} & =\\frac{1+\\frac{1}{2} y}{\\frac{2}{3}+\\frac{1}{3} y} \\\\\ny\\left(\\frac{2}{3}+\\frac{1}{3} y\\right) & =\\left(1+\\frac{1}{2} y\\right)^{2} \\\\\n\\frac{2}{3} y+\\frac{1}{3} y^{2} & =1+y+\\frac{1}{4} y^{2} \\\\\n8 y+4 y^{2} & =12+12 y+3 y^{2} \\\\\ny^{2}-4 y-12 & =0 \\\\\n(y-6)(y+2) & =0\n\\end{aligned}\n$$\n\nTherefore, $y=\\log _{2} x=6$ or $y=\\log _{2} x=-2$, which gives $x=2^{6}=64$ or $x=2^{-2}=\\frac{1}{4}$.']","['$64,\\frac{1}{4}$']",True,,Numerical, 2540,Geometry,,"In the diagram, $P Q R S$ is a square with sides of length 4. Points $T$ and $U$ are on sides $Q R$ and $R S$ respectively such that $\angle U P T=45^{\circ}$. Determine the maximum possible perimeter of $\triangle R U T$. ","['Rotate a copy of $\\triangle P S U$ by $90^{\\circ}$ counterclockwise around $P$, forming a new triangle $P Q V$. Note that $V$ lies on the extension of $R Q$.\n\n\n\nThen $P V=P U$ by rotation.\n\nAlso, $\\angle V P T=\\angle V P Q+\\angle Q P T=\\angle U P S+\\angle Q P T=90^{\\circ}-\\angle U P T=90^{\\circ}-45^{\\circ}$.\n\nThis tells us that $\\triangle P T U$ is congruent to $\\triangle P T V$, by ""side-angle-side"".\n\nThus, the perimeter of $\\triangle R U T$ equals\n\n$$\n\\begin{aligned}\nU R+R T+U T & =U R+R T+T V \\\\\n& =U R+R T+T Q+Q V \\\\\n& =U R+R Q+S U \\\\\n& =S U+U R+R Q \\\\\n& =S R+R Q \\\\\n& =8\n\\end{aligned}\n$$\n\nThat is, the perimeter of $\\triangle R U T$ always equals 8 , so the maximum possible perimeter is 8 .', 'Let $\\angle S P U=\\theta$. Note that $0^{\\circ} \\leq \\theta \\leq 45^{\\circ}$.\n\nThen $\\tan \\theta=\\frac{S U}{P S}$, so $S U=4 \\tan \\theta$.\n\nSince $S R=4$, then $U R=S R-S U=4-4 \\tan \\theta$.\n\nSince $\\angle U P T=45^{\\circ}$, then $\\angle Q P T=90^{\\circ}-45^{\\circ}-\\theta=45^{\\circ}-\\theta$.\n\nThus, $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{Q T}{P Q}$ and so $Q T=4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nSince $Q R=4$, then $R T=4-4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nBut $\\tan (A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$, so $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{\\tan \\left(45^{\\circ}\\right)-\\tan \\theta}{1+\\tan \\left(45^{\\circ}\\right) \\tan \\theta}=\\frac{1-\\tan \\theta}{1+\\tan \\theta}$, since $\\tan \\left(45^{\\circ}\\right)=1$.\n\nThis gives $R T=4-4\\left(\\frac{1-\\tan \\theta}{1+\\tan \\theta}\\right)=\\frac{4+4 \\tan \\theta}{1+\\tan \\theta}-\\frac{4-4 \\tan \\theta}{1+\\tan \\theta}=\\frac{8 \\tan \\theta}{1+\\tan \\theta}$.\n\nBy the Pythagorean Theorem in $\\triangle U R T$, we obtain\n\n$$\n\\begin{aligned}\nU T & =\\sqrt{U R^{2}+R T^{2}} \\\\\n& =\\sqrt{(4-4 \\tan \\theta)^{2}+\\left(\\frac{8 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{(1-\\tan \\theta)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}\\right)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =\\sqrt[4]{\\frac{1-2 \\tan ^{2} \\theta+\\tan ^{4} \\theta+4 \\tan ^{2} \\theta}{\\left(1+\\tan ^{2}\\right)^{2}}} \\\\\n& =\\sqrt[4]{\\frac{1+2 \\tan ^{2} \\theta+\\tan ^{4} \\theta}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4 \\sqrt{\\frac{\\left(1+\\tan ^{2} \\theta\\right)^{2}}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan ^{2}}\\right)\n\\end{aligned}\n$$\n\nTherefore, the perimeter of $\\triangle U R T$ is\n\n$$\n\\begin{aligned}\nU R+R T+U T & =4-4 \\tan \\theta+\\frac{8 \\tan \\theta}{1+\\tan \\theta}+4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}+\\frac{2 \\tan \\theta}{1+\\tan \\theta}+\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{2+2 \\tan \\theta}{1+\\tan \\theta}\\right) \\\\\n& =8\n\\end{aligned}\n$$\n\nThus, the perimeter is always 8 , regardless of the value of $\\theta$, so the maximum possible perimeter is 8 .']",['8'],False,,Numerical, 2540,Geometry,,"In the diagram, $P Q R S$ is a square with sides of length 4. Points $T$ and $U$ are on sides $Q R$ and $R S$ respectively such that $\angle U P T=45^{\circ}$. Determine the maximum possible perimeter of $\triangle R U T$. ![](https://cdn.mathpix.com/cropped/2023_12_21_b301ea0eb47ddd66a1f3g-1.jpg?height=398&width=414&top_left_y=1926&top_left_x=1300)","['Rotate a copy of $\\triangle P S U$ by $90^{\\circ}$ counterclockwise around $P$, forming a new triangle $P Q V$. Note that $V$ lies on the extension of $R Q$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4d95b82665f44d964a68g-1.jpg?height=585&width=420&top_left_y=1285&top_left_x=950)\n\nThen $P V=P U$ by rotation.\n\nAlso, $\\angle V P T=\\angle V P Q+\\angle Q P T=\\angle U P S+\\angle Q P T=90^{\\circ}-\\angle U P T=90^{\\circ}-45^{\\circ}$.\n\nThis tells us that $\\triangle P T U$ is congruent to $\\triangle P T V$, by ""side-angle-side"".\n\nThus, the perimeter of $\\triangle R U T$ equals\n\n$$\n\\begin{aligned}\nU R+R T+U T & =U R+R T+T V \\\\\n& =U R+R T+T Q+Q V \\\\\n& =U R+R Q+S U \\\\\n& =S U+U R+R Q \\\\\n& =S R+R Q \\\\\n& =8\n\\end{aligned}\n$$\n\nThat is, the perimeter of $\\triangle R U T$ always equals 8 , so the maximum possible perimeter is 8 .', 'Let $\\angle S P U=\\theta$. Note that $0^{\\circ} \\leq \\theta \\leq 45^{\\circ}$.\n\nThen $\\tan \\theta=\\frac{S U}{P S}$, so $S U=4 \\tan \\theta$.\n\nSince $S R=4$, then $U R=S R-S U=4-4 \\tan \\theta$.\n\nSince $\\angle U P T=45^{\\circ}$, then $\\angle Q P T=90^{\\circ}-45^{\\circ}-\\theta=45^{\\circ}-\\theta$.\n\nThus, $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{Q T}{P Q}$ and so $Q T=4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nSince $Q R=4$, then $R T=4-4 \\tan \\left(45^{\\circ}-\\theta\\right)$.\n\nBut $\\tan (A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$, so $\\tan \\left(45^{\\circ}-\\theta\\right)=\\frac{\\tan \\left(45^{\\circ}\\right)-\\tan \\theta}{1+\\tan \\left(45^{\\circ}\\right) \\tan \\theta}=\\frac{1-\\tan \\theta}{1+\\tan \\theta}$, since $\\tan \\left(45^{\\circ}\\right)=1$.\n\nThis gives $R T=4-4\\left(\\frac{1-\\tan \\theta}{1+\\tan \\theta}\\right)=\\frac{4+4 \\tan \\theta}{1+\\tan \\theta}-\\frac{4-4 \\tan \\theta}{1+\\tan \\theta}=\\frac{8 \\tan \\theta}{1+\\tan \\theta}$.\n\nBy the Pythagorean Theorem in $\\triangle U R T$, we obtain\n\n$$\n\\begin{aligned}\nU T & =\\sqrt{U R^{2}+R T^{2}} \\\\\n& =\\sqrt{(4-4 \\tan \\theta)^{2}+\\left(\\frac{8 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{(1-\\tan \\theta)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =4 \\sqrt{\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}\\right)^{2}+\\left(\\frac{2 \\tan \\theta}{1+\\tan \\theta}\\right)^{2}} \\\\\n& =\\sqrt[4]{\\frac{1-2 \\tan ^{2} \\theta+\\tan ^{4} \\theta+4 \\tan ^{2} \\theta}{\\left(1+\\tan ^{2}\\right)^{2}}} \\\\\n& =\\sqrt[4]{\\frac{1+2 \\tan ^{2} \\theta+\\tan ^{4} \\theta}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4 \\sqrt{\\frac{\\left(1+\\tan ^{2} \\theta\\right)^{2}}{\\left(1+\\tan ^{2}\\right.}} \\\\\n& =4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan ^{2}}\\right)\n\\end{aligned}\n$$\n\nTherefore, the perimeter of $\\triangle U R T$ is\n\n$$\n\\begin{aligned}\nU R+R T+U T & =4-4 \\tan \\theta+\\frac{8 \\tan \\theta}{1+\\tan \\theta}+4\\left(\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{1-\\tan ^{2} \\theta}{1+\\tan \\theta}+\\frac{2 \\tan \\theta}{1+\\tan \\theta}+\\frac{1+\\tan ^{2} \\theta}{1+\\tan \\theta}\\right) \\\\\n& =4\\left(\\frac{2+2 \\tan \\theta}{1+\\tan \\theta}\\right) \\\\\n& =8\n\\end{aligned}\n$$\n\nThus, the perimeter is always 8 , regardless of the value of $\\theta$, so the maximum possible perimeter is 8 .']",['8'],False,,Numerical, 2541,Algebra,,"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Determine the value of $f(7,3)$.","['Suppose that $p_{1}=1$.\n\nThen $p_{2}=p_{7}=0$, so the string is of the form $10 p_{3} p_{4} p_{5} p_{6} 0$.\n\nSince $k=3$, then 2 of $p_{3}, p_{4}, p_{5}, p_{6}$ equal 1 , but in such a way that no two adjacent digits are both 1 .\n\nThe possible strings in this case are 1010100, 1010010 and 1001010.\n\nSuppose that $p_{1}=0$. Then $p_{2}$ can equal 1 or 0 .\n\nIf $p_{2}=1$, then $p_{3}=0$ as well. This means that the string is of the form $010 p_{4} p_{5} p_{6} p_{7}$, which is the same as the general string in the first case, but shifted by 1 position around the circle, so there are again 3 possibilities.\n\nIf $p_{2}=0$, then the string is of the form $00 p_{3} p_{4} p_{5} p_{6} p_{7}$ and 3 of the digits $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ equal 1 in such a way that no 2 adjacent digits equal 1.\n\nThere is only 1 way in which this can happen: 0010101.\n\nOverall, this gives 7 possible configurations, so $f(7,3)=7$.']",['7'],False,,Numerical, 2541,Algebra,,"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_1bbe653ae31ae7a70c3cg-1.jpg?height=344&width=674&top_left_y=454&top_left_x=754) Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Determine the value of $f(7,3)$.","['Suppose that $p_{1}=1$.\n\nThen $p_{2}=p_{7}=0$, so the string is of the form $10 p_{3} p_{4} p_{5} p_{6} 0$.\n\nSince $k=3$, then 2 of $p_{3}, p_{4}, p_{5}, p_{6}$ equal 1 , but in such a way that no two adjacent digits are both 1 .\n\nThe possible strings in this case are 1010100, 1010010 and 1001010.\n\nSuppose that $p_{1}=0$. Then $p_{2}$ can equal 1 or 0 .\n\nIf $p_{2}=1$, then $p_{3}=0$ as well. This means that the string is of the form $010 p_{4} p_{5} p_{6} p_{7}$, which is the same as the general string in the first case, but shifted by 1 position around the circle, so there are again 3 possibilities.\n\nIf $p_{2}=0$, then the string is of the form $00 p_{3} p_{4} p_{5} p_{6} p_{7}$ and 3 of the digits $p_{3}, p_{4}, p_{5}, p_{6}, p_{7}$ equal 1 in such a way that no 2 adjacent digits equal 1.\n\nThere is only 1 way in which this can happen: 0010101.\n\nOverall, this gives 7 possible configurations, so $f(7,3)=7$.']",['7'],False,,Numerical, 2542,Algebra,,"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Prove that $f(n, k)=f(n-1, k)+f(n-2, k-1)$ for all integers $n \geq 3$ and $k \geq 2$.","[""An allowable string $p_{1} p_{2} \\cdots p_{n-1} p_{n}$ has $\\left(p_{1}, p_{n}\\right)=(1,0),(0,1)$, or $(0,0)$.\n\nDefine $g(n, k, 1,0)$ to be the number of allowable strings of length $n$, containing $k 1$ 's, and with $\\left(p_{1}, p_{n}\\right)=(1,0)$.\n\nWe define $g(n, k, 0,1)$ and $g(n, k, 0,0)$ in a similar manner.\n\nNote that $f(n, k)=g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0)$.\n\nConsider the strings counted by $g(n, k, 0,1)$.\n\nSince $p_{n}=1$, then $p_{n-1}=0$. Since $p_{1}=0$, then $p_{2}$ can equal 0 or 1 .\n\nWe remove the first and last digits of these strings.\n\nWe obtain strings $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ that is strings of length $n-2$ containing $k-11$ 's.\n\nSince $p_{n-1}=0$, then the first and last digits of these strings are not both 1 . Also, since the original strings did not contain two consecutive 1's, then these new strings does not either.\n\nTherefore, $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ are allowable strings of length $n-2$ containing $k-1$ 1's, with $p_{n-1}=0$ and $p_{2}=1$ or $p_{2}=0$.\n\nThe number of such strings with $p_{2}=1$ and $p_{n-1}=0$ is $g(n-2, k-1,1,0)$ and the number of such strings with $p_{2}=0$ and $p_{n-1}=0$ is $g(n-2, k-1,0,0)$.\n\nThus, $g(n, k, 0,1)=g(n-2, k-1,1,0)+g(n-2, k-1,0,0)$.\n\nConsider the strings counted by $g(n, k, 0,0)$.\n\nSince $p_{1}=0$ and $p_{n}=0$, then we can remove $p_{n}$ to obtain strings $p_{1} p_{2} \\cdots p_{n-1}$ of length $n-1$ containing $k 1$ 's. These strings are allowable since $p_{1}=0$ and the original strings were allowable.\n\nNote that we have $p_{1}=0$ and $p_{n-1}$ is either 0 or 1 .\n\nSo the strings $p_{1} p_{2} \\cdots p_{n-1}$ are allowable strings of length $n-1$ containing $k$ 1's, starting with 0 , and ending with 0 or 1 .\n\nThe number of such strings with $p_{1}=0$ and $p_{n-1}=0$ is $g(n-1, k, 0,0)$ and the number of such strings with $p_{1}=0$ and $p_{n-1}=1$ is $g(n-1, k, 0,1)$.\n\nThus, $g(n, k, 0,0)=g(n-1, k, 0,0)+g(n-1, k, 0,1)$.\n\n\n\nConsider the strings counted by $g(n, k, 1,0)$.\n\nHere, $p_{1}=1$ and $p_{n}=0$. Thus, $p_{n-1}$ can equal 0 or 1 . We consider these two sets separately.\n\nIf $p_{n-1}=0$, then the string $p_{1} p_{2} \\cdots p_{n-1}$ is an allowable string of length $n-1$, containing $k 1$ 's, beginning with 1 and ending with 0 .\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=0$ is equal to $g(n-1, k, 1,0)$.\n\nIf $p_{n-1}=1$, then the string $p_{2} p_{3} \\cdots p_{n-1}$ is of length $n-2$, begins with 0 and ends with 1 . Also, it contains $k-1$ 1's (having removed the original leading 1) and is allowable since the original string was.\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=1$ is equal to $g(n-2, k-1,0,1)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n, k)= & g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0) \\\\\n= & (g(n-1, k, 1,0)+g(n-2, k-1,0,1)) \\\\\n& \\quad+(g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n& \\quad+(g(n-1, k, 0,0)+g(n-1, k, 0,1)) \\\\\n= & (g(n-1, k, 1,0)+g(n-1, k, 0,1)+g(n-1, k, 0,0)) \\\\\n& \\quad+(g(n-2, k-1,0,1)+g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n= & f(n-1, k)+f(n-2, k-1)\n\\end{aligned}\n$$\n\nas required."", ""We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nTo prove the desired fact, we will use the fact that $\\left(\\begin{array}{c}m \\\\ r\\end{array}\\right)=\\left(\\begin{array}{c}m-1 \\\\ r\\end{array}\\right)+\\left(\\begin{array}{c}m-1 \\\\ r-1\\end{array}\\right)$, which we prove at the end.\n\nNow\n\n$$\n\\begin{aligned}\nf & n-1, k)+f(n-2, k-1) \\\\\n& =\\left(\\begin{array}{c}\n(n-1)-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-1)-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1) \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1)-1 \\\\\n(k-1)-1\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\quad \\text { (using the identity above) } \\\\\n& =f(n, k)\n\\end{aligned}\n$$\n\nas required.\n\nTo prove the identity, we expand the terms on the right side:\n\n$$\n\\begin{aligned}\n\\left(\\begin{array}{c}\nm-1 \\\\\nr\n\\end{array}\\right)+\\left(\\begin{array}{c}\nm-1 \\\\\nr-1\n\\end{array}\\right) & =\\frac{(m-1) !}{r !(m-r-1) !}+\\frac{(m-1) !}{(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r-1) !(m-r)}+\\frac{r(m-1) !}{r(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r) !}+\\frac{r(m-1) !}{r !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r+r)}{r !(m-r) !} \\\\\n& =\\frac{(m-1) ! m}{r !(m-r) !} \\\\\n& =\\frac{m !}{r !(m-r) !} \\\\\n& =\\left(\\begin{array}{c}\nm \\\\\nr\n\\end{array}\\right)\n\\end{aligned}\n$$\n\nas required.""]",,True,,, 2542,Algebra,,"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_1bbe653ae31ae7a70c3cg-1.jpg?height=344&width=674&top_left_y=454&top_left_x=754) Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Prove that $f(n, k)=f(n-1, k)+f(n-2, k-1)$ for all integers $n \geq 3$ and $k \geq 2$.","[""An allowable string $p_{1} p_{2} \\cdots p_{n-1} p_{n}$ has $\\left(p_{1}, p_{n}\\right)=(1,0),(0,1)$, or $(0,0)$.\n\nDefine $g(n, k, 1,0)$ to be the number of allowable strings of length $n$, containing $k 1$ 's, and with $\\left(p_{1}, p_{n}\\right)=(1,0)$.\n\nWe define $g(n, k, 0,1)$ and $g(n, k, 0,0)$ in a similar manner.\n\nNote that $f(n, k)=g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0)$.\n\nConsider the strings counted by $g(n, k, 0,1)$.\n\nSince $p_{n}=1$, then $p_{n-1}=0$. Since $p_{1}=0$, then $p_{2}$ can equal 0 or 1 .\n\nWe remove the first and last digits of these strings.\n\nWe obtain strings $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ that is strings of length $n-2$ containing $k-11$ 's.\n\nSince $p_{n-1}=0$, then the first and last digits of these strings are not both 1 . Also, since the original strings did not contain two consecutive 1's, then these new strings does not either.\n\nTherefore, $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ are allowable strings of length $n-2$ containing $k-1$ 1's, with $p_{n-1}=0$ and $p_{2}=1$ or $p_{2}=0$.\n\nThe number of such strings with $p_{2}=1$ and $p_{n-1}=0$ is $g(n-2, k-1,1,0)$ and the number of such strings with $p_{2}=0$ and $p_{n-1}=0$ is $g(n-2, k-1,0,0)$.\n\nThus, $g(n, k, 0,1)=g(n-2, k-1,1,0)+g(n-2, k-1,0,0)$.\n\nConsider the strings counted by $g(n, k, 0,0)$.\n\nSince $p_{1}=0$ and $p_{n}=0$, then we can remove $p_{n}$ to obtain strings $p_{1} p_{2} \\cdots p_{n-1}$ of length $n-1$ containing $k 1$ 's. These strings are allowable since $p_{1}=0$ and the original strings were allowable.\n\nNote that we have $p_{1}=0$ and $p_{n-1}$ is either 0 or 1 .\n\nSo the strings $p_{1} p_{2} \\cdots p_{n-1}$ are allowable strings of length $n-1$ containing $k$ 1's, starting with 0 , and ending with 0 or 1 .\n\nThe number of such strings with $p_{1}=0$ and $p_{n-1}=0$ is $g(n-1, k, 0,0)$ and the number of such strings with $p_{1}=0$ and $p_{n-1}=1$ is $g(n-1, k, 0,1)$.\n\nThus, $g(n, k, 0,0)=g(n-1, k, 0,0)+g(n-1, k, 0,1)$.\n\n\n\nConsider the strings counted by $g(n, k, 1,0)$.\n\nHere, $p_{1}=1$ and $p_{n}=0$. Thus, $p_{n-1}$ can equal 0 or 1 . We consider these two sets separately.\n\nIf $p_{n-1}=0$, then the string $p_{1} p_{2} \\cdots p_{n-1}$ is an allowable string of length $n-1$, containing $k 1$ 's, beginning with 1 and ending with 0 .\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=0$ is equal to $g(n-1, k, 1,0)$.\n\nIf $p_{n-1}=1$, then the string $p_{2} p_{3} \\cdots p_{n-1}$ is of length $n-2$, begins with 0 and ends with 1 . Also, it contains $k-1$ 1's (having removed the original leading 1) and is allowable since the original string was.\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=1$ is equal to $g(n-2, k-1,0,1)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n, k)= & g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0) \\\\\n= & (g(n-1, k, 1,0)+g(n-2, k-1,0,1)) \\\\\n& \\quad+(g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n& \\quad+(g(n-1, k, 0,0)+g(n-1, k, 0,1)) \\\\\n= & (g(n-1, k, 1,0)+g(n-1, k, 0,1)+g(n-1, k, 0,0)) \\\\\n& \\quad+(g(n-2, k-1,0,1)+g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n= & f(n-1, k)+f(n-2, k-1)\n\\end{aligned}\n$$\n\nas required."", ""We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nTo prove the desired fact, we will use the fact that $\\left(\\begin{array}{c}m \\\\ r\\end{array}\\right)=\\left(\\begin{array}{c}m-1 \\\\ r\\end{array}\\right)+\\left(\\begin{array}{c}m-1 \\\\ r-1\\end{array}\\right)$, which we prove at the end.\n\nNow\n\n$$\n\\begin{aligned}\nf & n-1, k)+f(n-2, k-1) \\\\\n& =\\left(\\begin{array}{c}\n(n-1)-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-1)-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1) \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1)-1 \\\\\n(k-1)-1\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\quad \\text { (using the identity above) } \\\\\n& =f(n, k)\n\\end{aligned}\n$$\n\nas required.\n\nTo prove the identity, we expand the terms on the right side:\n\n$$\n\\begin{aligned}\n\\left(\\begin{array}{c}\nm-1 \\\\\nr\n\\end{array}\\right)+\\left(\\begin{array}{c}\nm-1 \\\\\nr-1\n\\end{array}\\right) & =\\frac{(m-1) !}{r !(m-r-1) !}+\\frac{(m-1) !}{(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r-1) !(m-r)}+\\frac{r(m-1) !}{r(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r) !}+\\frac{r(m-1) !}{r !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r+r)}{r !(m-r) !} \\\\\n& =\\frac{(m-1) ! m}{r !(m-r) !} \\\\\n& =\\frac{m !}{r !(m-r) !} \\\\\n& =\\left(\\begin{array}{c}\nm \\\\\nr\n\\end{array}\\right)\n\\end{aligned}\n$$\n\nas required.""]",['证明题，略'],True,,Need_human_evaluate, 2543,Algebra,,"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_1bbe653ae31ae7a70c3cg-1.jpg?height=344&width=674&top_left_y=454&top_left_x=754) Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Determine the smallest possible value of $n+k$ among all possible ordered pairs of integers $(n, k)$ for which $f(n, k)$ is a positive multiple of 2009 , where $n \geq 3$ and $k \geq 2$.","[""We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nIn order to look at divisibility, we need to first simplify the formula:\n\n$$\n\\begin{aligned}\nf(n, k) & =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\\\\n& =\\frac{(n-k) !}{k !(n-k-k) !}+\\frac{(n-k-1) !}{(k-1) !((n-k-1)-(k-1)) !} \\\\\n& =\\frac{(n-k) !}{k !(n-2 k) !}+\\frac{(n-k-1) !}{(k-1) !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k)}{k !(n-2 k) !}+\\frac{(n-k-1) ! k}{k !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k+k)}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1) !}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1)(n-k-2) \\cdots(n-2 k+2)(n-2 k+1)}{k !}\n\\end{aligned}\n$$\n\nNow that we have written $f(n, k)$ as a product, it is significantly easier to look at divisibility.\n\nNote that $2009=41 \\times 49=7^{2} \\times 41$, so we need $f(n, k)$ to be divisible by 41 and by 7 twice. For this to be the case, the numerator of $f(n, k)$ must have at least one more factor of 41 and at least two more factors of 7 than the denominator.\n\nAlso, we want to minimize $n+k$, so we work to keep $n$ and $k$ as small as possible.\n\nIf $n=49$ and $k=5$, then\n\n$$\nf(49,5)=\\frac{49(43)(42)(41)(40)}{5 !}=\\frac{49(43)(42)(41)(40)}{5(4)(3)(2)(1)}=49(43)(14)\n$$\n\nwhich is divisible by 2009 .\n\nWe show that this pair minimizes the value of $n+k$ with a value of 54 .\n\nWe consider the possible cases by looking separately at the factors of 41 and 7 that must occur. We focus on the factor of 41 first.\n\nFor the numerator to contain a factor of 41 , either $n$ is divisible by 41 or one of the terms in the product $(n-k-1)(n-k-2) \\cdots(n-2 k+1)$ is divisible by 41 .\n\nCase 1: $n$ is divisible by 41\n\nWe already know that $n=82$ is too large, so we consider $n=41$. From the original interpretation of $f(n, k)$, we see that $k \\leq 20$, as there can be no more than 20 gifts placed on 41 plates.\n\nHere, the numerator becomes 41 times the product of $k-1$ consecutive integers, the largest of which is $40-k$.\n\nNow the numerator must also contain at least two factors of 7 more than the denominator. But the denominator is the product of $k$ consecutive integers. Since the numerator contains the product of $k-1$ consecutive integers and the denominator contains the product of $k$ consecutive integers, then the denominator will always include at least as many multiples of 7 as the numerator (since there are more consecutive integers in the product in the denominator). Thus, it is impossible for the numerator to contain even one more\n\n\n\nadditional factor of 7 than the denominator.\n\nTherefore, if $n=41$, then $f(n, k)$ cannot be divisible by 2009 .\n\nCase 2: $n$ is not divisible by 41\n\nThis means that the factor of 41 in the numerator must occur in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nIn this case, the integer 41 must occur in this product, since an occurrence of 82 would make $n$ greater than 82 , which does not minimize $n+k$.\n\nSo we try to find values of $n$ and $k$ that include the integer 41 in this list.\n\nNote that $n-k-1$ is the largest factor in the product and $n-2 k+1$ is the smallest.\n\nSince 41 is contained somewhere in the product, then $n-2 k+1 \\leq 41$ (giving $n \\leq 40+2 k$ ) and $41 \\leq n-k-1$ (giving $n \\geq 42+k$ ).\n\nCombining these restrictions, we get $42+k \\leq n \\leq 40+2 k$.\n\nNow, we focus on the factors of 7 .\n\nEither $n$ is not divisible by 7 or $n$ is divisible by 7 .\n\n* If $n$ is not divisible by 7 , then at least two factors of 7 must be included in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nwhich means that either $k \\geq 8$ (to give two multiples of 7 in this list of $k-1$ consecutive integers) or one of the factors is divisible by 49 .\n\n- If $k \\geq 8$, then $n \\geq 42+k \\geq 50$ so $n+k \\geq 58$, which is not minimal.\n- If one of the factors is a multiple of 49 , then 49 must be included in the list so $n-2 k+1 \\leq 49$ (giving $n \\leq 48+2 k$ ) and $49 \\leq n-k-1$ (giving $n \\geq 50+k$ ). In this case, we already know that $42+k \\leq n \\leq 40+2 k$ and now we also have $50+k \\leq n \\leq 48+2 k$.\n\nFor these ranges to overlap, we need $50+k \\leq 40+2 k$ and so $k \\geq 10$, which means that $n \\geq 50+k \\geq 60$, and so $n+k \\geq 70$, which is not minimal.\n\n* Next, we consider the case where $n$ is a multiple of 7 .\n\nHere, $42+k \\leq n \\leq 40+2 k$ (to include 41 in the product) and $n$ is a multiple of 7 .\n\nSince $k$ is at least 2 by definition, then $n \\geq 42+k \\geq 44$, so $n$ is at least 49 .\n\nIf $n$ was 56 or more, we do not get a minimal value for $n+k$.\n\nThus, we need to have $n=49$. In this case, we do not need to look for another factor of 7 in the list.\n\nTo complete this case, we need to find the smallest value of $k$ for which 49 is in the range from $42+k$ to $40+2 k$ because we need to have $42+k \\leq n \\leq 40+2 k$.\n\nThis value of $k$ is $k=5$, which gives $n+k=49+5=54$.\n\nSince $f(49,5)$ is divisible by 2009 , as determined above, then this is the case that minimizes $n+k$, giving a value of 54 .""]",['54'],False,,Numerical, 2543,Algebra,,"Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below. Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also ""adjacent"" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$. Determine the smallest possible value of $n+k$ among all possible ordered pairs of integers $(n, k)$ for which $f(n, k)$ is a positive multiple of 2009 , where $n \geq 3$ and $k \geq 2$.","[""We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nIn order to look at divisibility, we need to first simplify the formula:\n\n$$\n\\begin{aligned}\nf(n, k) & =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\\\\n& =\\frac{(n-k) !}{k !(n-k-k) !}+\\frac{(n-k-1) !}{(k-1) !((n-k-1)-(k-1)) !} \\\\\n& =\\frac{(n-k) !}{k !(n-2 k) !}+\\frac{(n-k-1) !}{(k-1) !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k)}{k !(n-2 k) !}+\\frac{(n-k-1) ! k}{k !(n-2 k) !} \\\\\n& =\\frac{(n-k-1) !(n-k+k)}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1) !}{k !(n-2 k) !} \\\\\n& =\\frac{n(n-k-1)(n-k-2) \\cdots(n-2 k+2)(n-2 k+1)}{k !}\n\\end{aligned}\n$$\n\nNow that we have written $f(n, k)$ as a product, it is significantly easier to look at divisibility.\n\nNote that $2009=41 \\times 49=7^{2} \\times 41$, so we need $f(n, k)$ to be divisible by 41 and by 7 twice. For this to be the case, the numerator of $f(n, k)$ must have at least one more factor of 41 and at least two more factors of 7 than the denominator.\n\nAlso, we want to minimize $n+k$, so we work to keep $n$ and $k$ as small as possible.\n\nIf $n=49$ and $k=5$, then\n\n$$\nf(49,5)=\\frac{49(43)(42)(41)(40)}{5 !}=\\frac{49(43)(42)(41)(40)}{5(4)(3)(2)(1)}=49(43)(14)\n$$\n\nwhich is divisible by 2009 .\n\nWe show that this pair minimizes the value of $n+k$ with a value of 54 .\n\nWe consider the possible cases by looking separately at the factors of 41 and 7 that must occur. We focus on the factor of 41 first.\n\nFor the numerator to contain a factor of 41 , either $n$ is divisible by 41 or one of the terms in the product $(n-k-1)(n-k-2) \\cdots(n-2 k+1)$ is divisible by 41 .\n\nCase 1: $n$ is divisible by 41\n\nWe already know that $n=82$ is too large, so we consider $n=41$. From the original interpretation of $f(n, k)$, we see that $k \\leq 20$, as there can be no more than 20 gifts placed on 41 plates.\n\nHere, the numerator becomes 41 times the product of $k-1$ consecutive integers, the largest of which is $40-k$.\n\nNow the numerator must also contain at least two factors of 7 more than the denominator. But the denominator is the product of $k$ consecutive integers. Since the numerator contains the product of $k-1$ consecutive integers and the denominator contains the product of $k$ consecutive integers, then the denominator will always include at least as many multiples of 7 as the numerator (since there are more consecutive integers in the product in the denominator). Thus, it is impossible for the numerator to contain even one more\n\n\n\nadditional factor of 7 than the denominator.\n\nTherefore, if $n=41$, then $f(n, k)$ cannot be divisible by 2009 .\n\nCase 2: $n$ is not divisible by 41\n\nThis means that the factor of 41 in the numerator must occur in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nIn this case, the integer 41 must occur in this product, since an occurrence of 82 would make $n$ greater than 82 , which does not minimize $n+k$.\n\nSo we try to find values of $n$ and $k$ that include the integer 41 in this list.\n\nNote that $n-k-1$ is the largest factor in the product and $n-2 k+1$ is the smallest.\n\nSince 41 is contained somewhere in the product, then $n-2 k+1 \\leq 41$ (giving $n \\leq 40+2 k$ ) and $41 \\leq n-k-1$ (giving $n \\geq 42+k$ ).\n\nCombining these restrictions, we get $42+k \\leq n \\leq 40+2 k$.\n\nNow, we focus on the factors of 7 .\n\nEither $n$ is not divisible by 7 or $n$ is divisible by 7 .\n\n* If $n$ is not divisible by 7 , then at least two factors of 7 must be included in the product\n\n$$\n(n-k-1)(n-k-2) \\cdots(n-2 k+1)\n$$\n\nwhich means that either $k \\geq 8$ (to give two multiples of 7 in this list of $k-1$ consecutive integers) or one of the factors is divisible by 49 .\n\n- If $k \\geq 8$, then $n \\geq 42+k \\geq 50$ so $n+k \\geq 58$, which is not minimal.\n- If one of the factors is a multiple of 49 , then 49 must be included in the list so $n-2 k+1 \\leq 49$ (giving $n \\leq 48+2 k$ ) and $49 \\leq n-k-1$ (giving $n \\geq 50+k$ ). In this case, we already know that $42+k \\leq n \\leq 40+2 k$ and now we also have $50+k \\leq n \\leq 48+2 k$.\n\nFor these ranges to overlap, we need $50+k \\leq 40+2 k$ and so $k \\geq 10$, which means that $n \\geq 50+k \\geq 60$, and so $n+k \\geq 70$, which is not minimal.\n\n* Next, we consider the case where $n$ is a multiple of 7 .\n\nHere, $42+k \\leq n \\leq 40+2 k$ (to include 41 in the product) and $n$ is a multiple of 7 .\n\nSince $k$ is at least 2 by definition, then $n \\geq 42+k \\geq 44$, so $n$ is at least 49 .\n\nIf $n$ was 56 or more, we do not get a minimal value for $n+k$.\n\nThus, we need to have $n=49$. In this case, we do not need to look for another factor of 7 in the list.\n\nTo complete this case, we need to find the smallest value of $k$ for which 49 is in the range from $42+k$ to $40+2 k$ because we need to have $42+k \\leq n \\leq 40+2 k$.\n\nThis value of $k$ is $k=5$, which gives $n+k=49+5=54$.\n\nSince $f(49,5)$ is divisible by 2009 , as determined above, then this is the case that minimizes $n+k$, giving a value of 54 .""]",['54'],False,,Numerical, 2544,Number Theory,,"Determine the two pairs of positive integers $(a, b)$ with $a","['The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon and join the points from which the angles that measure $a^{\\circ}$ and $b^{\\circ}$ emanate to form a hexagon.\n\n\n\nThis polygon has 6 sides, and so the sum of its 6 angles is $4 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nThe remaining two angles have measures $a^{\\circ}+c^{\\circ}$ and $b^{\\circ}+d^{\\circ}$.\n\nWe are told that $a^{\\circ}+b^{\\circ}=88^{\\circ}$.\n\nAlso, the angles that measure $c^{\\circ}$ and $d^{\\circ}$ are two angles in a triangle whose third angle is $108^{\\circ}$.\n\nThus, $c^{\\circ}+d^{\\circ}=180^{\\circ}-108^{\\circ}=72^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+72^{\\circ} & =4 \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\left(4-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\frac{4 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{160^{\\circ}}{180^{\\circ}} & =\\frac{8}{n} \\\\\n\\frac{8}{9} & =\\frac{8}{n}\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .', ""The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon.\n\n\n\nThis polygon has 7 sides, and so the sum of its 7 angles is $5 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nTwo of its angles are the angles equal to $a^{\\circ}$ and $b^{\\circ}$, whose sum is $88^{\\circ}$.\n\nIts seventh angle is the reflex angle corresponding to the pentagon's angle of $108^{\\circ}$, which equals $360^{\\circ}-108^{\\circ}$ or $252^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+252^{\\circ} & =5 \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\left(5-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\frac{5 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{340^{\\circ}}{180^{\\circ}} & =\\frac{n+8}{n} \\\\\n\\frac{17}{9} & =\\frac{n+8}{n} \\\\\n17 n & =9(n+8) \\\\\n17 n & =9 n+72 \\\\\n8 n & =72\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .""]",['9'],False,,Numerical, 2546,Geometry,,"A regular pentagon covers part of another regular polygon, as shown. This regular polygon has $n$ sides, five of which are completely or partially visible. In the diagram, the sum of the measures of the angles marked $a^{\circ}$ and $b^{\circ}$ is $88^{\circ}$. Determine the value of $n$. (The side lengths of a regular polygon are all equal, as are the measures of its interior angles.) ![](https://cdn.mathpix.com/cropped/2023_12_21_17bd3319dc6e6567eee6g-1.jpg?height=620&width=1544&top_left_y=268&top_left_x=296)","['The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon and join the points from which the angles that measure $a^{\\circ}$ and $b^{\\circ}$ emanate to form a hexagon.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_0936dd99cfd990581a98g-1.jpg?height=399&width=309&top_left_y=1042&top_left_x=997)\n\nThis polygon has 6 sides, and so the sum of its 6 angles is $4 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nThe remaining two angles have measures $a^{\\circ}+c^{\\circ}$ and $b^{\\circ}+d^{\\circ}$.\n\nWe are told that $a^{\\circ}+b^{\\circ}=88^{\\circ}$.\n\nAlso, the angles that measure $c^{\\circ}$ and $d^{\\circ}$ are two angles in a triangle whose third angle is $108^{\\circ}$.\n\nThus, $c^{\\circ}+d^{\\circ}=180^{\\circ}-108^{\\circ}=72^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+72^{\\circ} & =4 \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\left(4-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n160^{\\circ} & =\\frac{4 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{160^{\\circ}}{180^{\\circ}} & =\\frac{8}{n} \\\\\n\\frac{8}{9} & =\\frac{8}{n}\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .', ""The angles in a polygon with $n$ sides have a sum of $(n-2) \\cdot 180^{\\circ}$.\n\nThis means that the angles in a pentagon have a sum of $3 \\cdot 180^{\\circ}$ or $540^{\\circ}$, which means that each interior angle in a regular pentagon equals $\\frac{1}{5} \\cdot 540^{\\circ}$ or $108^{\\circ}$.\n\nAlso, each interior angle in a regular polygon with $n$ sides equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$. (This is the general version of the statement in the previous sentence.)\n\nConsider the portion of the regular polygon with $n$ sides that lies outside the pentagon.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_c36108b31727ccbc67b7g-1.jpg?height=404&width=309&top_left_y=638&top_left_x=997)\n\nThis polygon has 7 sides, and so the sum of its 7 angles is $5 \\cdot 180^{\\circ}$.\n\nFour of its angles are the original angles from the $n$-sided polygon, so each equals $\\frac{n-2}{n} \\cdot 180^{\\circ}$.\n\nTwo of its angles are the angles equal to $a^{\\circ}$ and $b^{\\circ}$, whose sum is $88^{\\circ}$.\n\nIts seventh angle is the reflex angle corresponding to the pentagon's angle of $108^{\\circ}$, which equals $360^{\\circ}-108^{\\circ}$ or $252^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n4 \\cdot \\frac{n-2}{n} \\cdot 180^{\\circ}+88^{\\circ}+252^{\\circ} & =5 \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\left(5-\\frac{4(n-2)}{n}\\right) \\cdot 180^{\\circ} \\\\\n340^{\\circ} & =\\frac{5 n-(4 n-8)}{n} \\cdot 180^{\\circ} \\\\\n\\frac{340^{\\circ}}{180^{\\circ}} & =\\frac{n+8}{n} \\\\\n\\frac{17}{9} & =\\frac{n+8}{n} \\\\\n17 n & =9(n+8) \\\\\n17 n & =9 n+72 \\\\\n8 n & =72\n\\end{aligned}\n$$\n\nand so the value of $n$ is 9 .""]",['9'],False,,Numerical, 2547,Geometry,,"In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$. (A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant.) ","['Since the lengths of $A D, A B$ and $B C$ form a geometric sequence, we suppose that these lengths are $a$, ar and $a r^{2}$, respectively, for some real numbers $a>0$ and $r>0$.\n\nSince the angles at $A$ and $B$ are both right angles, we assign coordinates to the diagram, putting $B$ at the origin ( 0,0$), C$ on the positive $x$-axis at $\\left(a r^{2}, 0\\right), A$ on the positive $y$-axis at $(0, a r)$, and $D$ at $(a, a r)$.\n\n\n\nTherefore, the slope of the line segment joining $B(0,0)$ and $D(a, a r)$ is $\\frac{a r-0}{a-0}=r$.\n\nAlso, the slope of the line segment joining $A(0, a r)$ and $C\\left(a r^{2}, 0\\right)$ is $\\frac{a r-0}{0-a r^{2}}=-\\frac{1}{r}$.\n\nSince the product of the slopes of these two line segments is -1 , then the segments are perpendicular, as required.']",,True,,, 2547,Geometry,,"In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$. (A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant.) ![](https://cdn.mathpix.com/cropped/2023_12_21_17bd3319dc6e6567eee6g-1.jpg?height=271&width=401&top_left_y=819&top_left_x=1255)","['Since the lengths of $A D, A B$ and $B C$ form a geometric sequence, we suppose that these lengths are $a$, ar and $a r^{2}$, respectively, for some real numbers $a>0$ and $r>0$.\n\nSince the angles at $A$ and $B$ are both right angles, we assign coordinates to the diagram, putting $B$ at the origin ( 0,0$), C$ on the positive $x$-axis at $\\left(a r^{2}, 0\\right), A$ on the positive $y$-axis at $(0, a r)$, and $D$ at $(a, a r)$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_edb67d8383f857f7729ag-1.jpg?height=366&width=591&top_left_y=451&top_left_x=867)\n\nTherefore, the slope of the line segment joining $B(0,0)$ and $D(a, a r)$ is $\\frac{a r-0}{a-0}=r$.\n\nAlso, the slope of the line segment joining $A(0, a r)$ and $C\\left(a r^{2}, 0\\right)$ is $\\frac{a r-0}{0-a r^{2}}=-\\frac{1}{r}$.\n\nSince the product of the slopes of these two line segments is -1 , then the segments are perpendicular, as required.']",['证明题，略'],True,,Need_human_evaluate, 2548,Algebra,,Determine all real numbers $x$ for which $2 \log _{2}(x-1)=1-\log _{2}(x+2)$.,"['Using logarithm and exponent laws, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n2 \\log _{2}(x-1) & =1-\\log _{2}(x+2) \\\\\n2 \\log _{2}(x-1)+\\log _{2}(x+2) & =1 \\\\\n\\log _{2}\\left((x-1)^{2}\\right)+\\log _{2}(x+2) & =1 \\\\\n\\log _{2}\\left((x-1)^{2}(x+2)\\right) & =1 \\\\\n(x-1)^{2}(x+2) & =2^{1} \\\\\n\\left(x^{2}-2 x+1\\right)(x+2) & =2 \\\\\nx^{3}-3 x+2 & =2 \\\\\nx^{3}-3 x & =0 \\\\\nx\\left(x^{2}-3\\right) & =0\n\\end{aligned}\n$$\n\nand so $x=0$ or $x=\\sqrt{3}$ or $x=-\\sqrt{3}$.\n\nNote that if $x=0$, then $x-1=-1<0$ and so $\\log _{2}(x-1)$ is not defined. Thus, $x \\neq 0$. Note that if $x=-\\sqrt{3}$, then $x-1=-\\sqrt{3}-1<0$ and so $\\log _{2}(x-1)$ is not defined. Thus, $x \\neq-\\sqrt{3}$.\n\nIf $x=\\sqrt{3}$, we can verify that both logarithms in the original equation are defined and that the original equation is true. We could convince ourselves of this with a calculator or we could algebraically verify that raising 2 to the power of both sides gives the same number, so the expressions must actually be equal.\n\nTherefore, $x=\\sqrt{3}$ is the only solution.']",['$\\sqrt{3}$'],False,,Numerical, 2549,Algebra,,Consider the function $f(x)=x^{2}-2 x$. Determine all real numbers $x$ that satisfy the equation $f(f(f(x)))=3$.,"['Let $a=f(f(x))$.\n\nThus, the equation $f(f(f(x)))=3$ is equivalent to $f(a)=3$.\n\nSince $f(a)=a^{2}-2 a$, then we obtain the equation $a^{2}-2 a=3$ which gives $a^{2}-2 a-3=0$ and $(a-3)(a+1)=0$.\n\nThus, $a=3$ or $a=-1$ which means that $f(f(x))=3$ or $f(f(x))=-1$.\n\nLet $b=f(x)$.\n\nThus, the equations $f(f(x))=3$ and $f(f(x))=-1$ become $f(b)=3$ and $f(b)=-1$.\n\nIf $f(b)=3$, then $b=f(x)=3$ or $b=f(x)=-1$ using similar reasoning to above when $f(a)=3$.\n\nIf $f(b)=-1$, then $b^{2}-2 b=-1$ and so $b^{2}-2 b+1=0$ or $(b-1)^{2}=0$ which means that $b=f(x)=1$.\n\nThus, $f(x)=3$ or $f(x)=-1$ or $f(x)=1$.\n\nIf $f(x)=3$, then $x=3$ or $x=-1$ as above.\n\nIf $f(x)=-1$, then $x=1$ as above.\n\nIf $f(x)=1$, then $x^{2}-2 x=1$ and so $x^{2}-2 x-1=0$.\n\nBy the quadratic formula,\n\n$$\nx=\\frac{-(-2) \\pm \\sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\\frac{2 \\pm \\sqrt{8}}{2}=1 \\pm \\sqrt{2}\n$$\n\nTherefore, the solutions to the equation $f(f(f(x)))=3$ are $x=3,1,-1,1+\\sqrt{2}, 1-\\sqrt{2}$.']","['$3,1,-1,1+\\sqrt{2}, 1-\\sqrt{2}$']",True,,Numerical, 2550,Geometry,,"A circle has centre $O$ and radius 1. Quadrilateral $A B C D$ has all 4 sides tangent to the circle at points $P, Q, S$, and $T$, as shown. Also, $\angle A O B=\angle B O C=\angle C O D=\angle D O A$. If $A O=3$, determine the length of $D S$. ","['Since $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A$ and these angles form a complete circle around $O$, then $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A=\\frac{1}{4} \\cdot 360^{\\circ}=90^{\\circ}$.\n\nJoin point $O$ to $P, B, Q, C, S, D, T$, and $A$.\n\n\n\nSince $P, Q, S$, and $T$ are points of tangency, then the radii meet the sides of $A B C D$ at right angles at these points.\n\nSince $A O=3$ and $O T=1$ and $\\angle O T A=90^{\\circ}$, then by the Pythagorean Theorem, $A T=\\sqrt{A O^{2}-O T^{2}}=\\sqrt{8}=2 \\sqrt{2}$.\n\nSince $\\triangle O T A$ is right-angled at $T$, then $\\angle T A O+\\angle A O T=90^{\\circ}$.\n\nSince $\\angle D O A=90^{\\circ}$, then $\\angle A O T+\\angle D O T=90^{\\circ}$.\n\nThus, $\\angle T A O=\\angle D O T$.\n\nThis means that $\\triangle A T O$ is similar to $\\triangle O T D$.\n\nThus, $\\frac{D T}{O T}=\\frac{O T}{A T}$ and so $D T=\\frac{O T^{2}}{A T}=\\frac{1}{2 \\sqrt{2}}$.\n\nSince $D S$ and $D T$ are tangents to the circle from the same point, then $D S=D T=\\frac{1}{2 \\sqrt{2}}$.']",['$\\frac{1}{2 \\sqrt{2}}$'],False,,Numerical, 2550,Geometry,,"A circle has centre $O$ and radius 1. Quadrilateral $A B C D$ has all 4 sides tangent to the circle at points $P, Q, S$, and $T$, as shown. Also, $\angle A O B=\angle B O C=\angle C O D=\angle D O A$. If $A O=3$, determine the length of $D S$. ![](https://cdn.mathpix.com/cropped/2023_12_21_17bd3319dc6e6567eee6g-1.jpg?height=279&width=412&top_left_y=1660&top_left_x=1298)","['Since $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A$ and these angles form a complete circle around $O$, then $\\angle A O B=\\angle B O C=\\angle C O D=\\angle D O A=\\frac{1}{4} \\cdot 360^{\\circ}=90^{\\circ}$.\n\nJoin point $O$ to $P, B, Q, C, S, D, T$, and $A$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_e945ab0e1a14a95ca3e9g-1.jpg?height=274&width=464&top_left_y=1473&top_left_x=928)\n\nSince $P, Q, S$, and $T$ are points of tangency, then the radii meet the sides of $A B C D$ at right angles at these points.\n\nSince $A O=3$ and $O T=1$ and $\\angle O T A=90^{\\circ}$, then by the Pythagorean Theorem, $A T=\\sqrt{A O^{2}-O T^{2}}=\\sqrt{8}=2 \\sqrt{2}$.\n\nSince $\\triangle O T A$ is right-angled at $T$, then $\\angle T A O+\\angle A O T=90^{\\circ}$.\n\nSince $\\angle D O A=90^{\\circ}$, then $\\angle A O T+\\angle D O T=90^{\\circ}$.\n\nThus, $\\angle T A O=\\angle D O T$.\n\nThis means that $\\triangle A T O$ is similar to $\\triangle O T D$.\n\nThus, $\\frac{D T}{O T}=\\frac{O T}{A T}$ and so $D T=\\frac{O T^{2}}{A T}=\\frac{1}{2 \\sqrt{2}}$.\n\nSince $D S$ and $D T$ are tangents to the circle from the same point, then $D S=D T=\\frac{1}{2 \\sqrt{2}}$.']",['$\\frac{1}{2 \\sqrt{2}}$'],False,,Numerical, 2551,Algebra,,"Suppose that $x$ satisfies $00$, prove that $p_{1}=p_{6}$ and $q_{1}=q_{6}$.","['For the sum of the two digits printed to be 2, each digit must equal 1.\n\nThus, $S(2)=p_{1} q_{1}$.\n\nFor the sum of the two digits printed to be 12, each digit must equal 6 .\n\nThus, $S(12)=p_{6} q_{6}$.\n\nFor the sum of the two digits printed to be 7 , the digits must be 1 and 6 , or 2 and 5 , or 3 and 4 , or 4 and 3 , or 5 and 2 , or 6 and 1 .\n\nThus, $S(7)=p_{1} q_{6}+p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}+p_{6} q_{1}$.\n\nSince $S(2)=S(12)$, then $p_{1} q_{1}=p_{6} q_{6}$.\n\nSince $S(2)>0$ and $S(12)>0$, then $p_{1}, q_{1}, p_{6}, q_{6}>0$.\n\nIf $p_{1}=p_{6}$, then we can divide both sides of $p_{1} q_{1}=p_{6} q_{6}$ by $p_{1}=p_{6}$ to obtain $q_{1}=q_{6}$.\n\nIf $q_{1}=q_{6}$, then we can divide both sides of $p_{1} q_{1}=p_{6} q_{6}$ by $q_{1}=q_{6}$ to obtain $p_{1}=p_{6}$.\n\nTherefore, if we can show that either $p_{1}=p_{6}$ or $q_{1}=q_{6}$, our result will be true.\n\nSuppose that $p_{1} \\neq p_{6}$ and $q_{1} \\neq q_{6}$.\n\nSince $S(2)=\\frac{1}{2} S(7)$ and $S(12)=\\frac{1}{2} S(7)$, then\n\n$$\n\\begin{aligned}\nS(7)-\\frac{1}{2} S(7)-\\frac{1}{2} S(7) & =0 \\\\\nS(7)-S(2)-S(12) & =0 \\\\\np_{1} q_{6}+p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}+p_{6} q_{1}-p_{1} q_{1}-p_{6} q_{6} & =0 \\\\\np_{1} q_{6}+p_{6} q_{1}-p_{1} q_{1}-p_{6} q_{6}+\\left(p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}\\right) & =0 \\\\\n\\left(p_{1}-p_{6}\\right)\\left(q_{6}-q_{1}\\right)+\\left(p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2}\\right) & =0 \\\\\np_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2} & =-\\left(p_{1}-p_{6}\\right)\\left(q_{6}-q_{1}\\right) \\\\\np_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2} & =\\left(p_{1}-p_{6}\\right)\\left(q_{1}-q_{6}\\right)\n\\end{aligned}\n$$\n\nSince $p_{2}, p_{3}, p_{4}, p_{5}, q_{2}, q_{3}, q_{4}, q_{5} \\geq 0$, then $p_{2} q_{5}+p_{3} q_{4}+p_{4} q_{3}+p_{5} q_{2} \\geq 0$.\n\nFrom this, $\\left(p_{1}-p_{6}\\right)\\left(q_{1}-q_{6}\\right) \\geq 0$.\n\nSince $p_{1} \\neq p_{6}$, then either $p_{1}>p_{6}$ or $p_{1}p_{6}$, then $p_{1}-p_{6}>0$ and so $\\left(p_{1}-p_{6}\\right)\\left(q_{1}-q_{6}\\right) \\geq 0$ tells us that $q_{1}-q_{6}>0$ which means $q_{1}>q_{6}$.\n\nBut we know that $p_{1} q_{1}=p_{6} q_{6}$ and $p_{1}, q_{1}, p_{6}, q_{6}>0$ so we cannot have $p_{1}>p_{6}$ and $q_{1}>q_{6}$. If $p_{1}0$ so we cannot have $p_{1}p_{6}$ or $p_{1}0,-7 n+1103<0$\n\n$$\nn>157 \\frac{4}{7}\n$$\n\nTherefore the smallest value of $n$ is 158 .', 'For this series we want, $\\sum_{k=1}^{n} t_{k}<0$, or $\\sum_{k=1}^{n}(555-7 k)<0$.\n\nRewriting, $555 n-7 \\frac{(n)(n+1)}{2}<0$\n\n$$\n\\begin{aligned}\n1110 n-7 n^{2}-7 n & <0 \\\\\n7 n^{2}-1103 n & >0 \\\\\n\\text { or, } n & >\\frac{1103}{7} .\n\\end{aligned}\n$$\n\nThe smallest value of $n$ is 158 .', 'We generate the series as $548,541,534, \\ldots, 2,-5, \\ldots,-544,-551$.\n\nIf we pair the series from front to back the sum of each pair is -3 .\n\nIncluding all the pairs $548-551,541-544$ and so on there would be 79 pairs which give a sum of -237 .\n\nIf the last term, -551 , were omitted we would have a positive sum.\n\nTherefore we need all 79 pairs or 158 terms.']",['158'],False,,Numerical, 2560,Algebra,,"If $x$ and $y$ are real numbers, determine all solutions $(x, y)$ of the system of equations $$ \begin{aligned} & x^{2}-x y+8=0 \\ & x^{2}-8 x+y=0 \end{aligned} $$","['Subtracting,\n\n$$\n\\begin{array}{r}\nx^{2}-x y+8=0 \\\\\nx^{2}-8 x+y=0 \\\\\n\\hline-x y+8 x+8-y=0 \\\\\n8(1+x)-y(1+x)=0 \\\\\n(8-y)(1+x)=0 \\\\\ny=8 \\text { or } x=-1\n\\end{array}\n$$\n\n\n\nIf $y=8$, both equations become $x^{2}-8 x+8=0, x=4 \\pm 2 \\sqrt{2}$.\n\nIf $x=-1$ both equations become $y+9=0, y=-9$.\n\nThe solutions are $(-1,-9),(4+2 \\sqrt{2}, 8)$ and $(4-2 \\sqrt{2}, 8)$.', 'If $x^{2}-x y+8=0, y=\\frac{x^{2}+8}{x}$.\n\nAnd $x^{2}-8 x+y=0$ implies $y=8 x-x^{2}$.\n\nEquating, $\\frac{x^{2}+8}{x}=8 x-x^{2}$\n\n$$\n\\text { or, } x^{3}-7 x^{2}+8=0 \\text {. }\n$$\n\nBy inspection, $x=-1$ is a root.\n\nBy division, $x^{3}-7 x^{2}+8=(x+1)\\left(x^{2}-8 x+8\\right)$.\n\nAs before, the solutions are $(-1,-9),(4 \\pm 2 \\sqrt{2}, 8)$.']","['$(-1,-9),(4+2 \\sqrt{2}, 8),(4-2 \\sqrt{2}, 8)$']",True,,Tuple, 2561,Geometry,,"In the graph, the parabola $y=x^{2}$ has been translated to the position shown. Prove that $d e=f$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c810d95f38a712e9ddf7g-1.jpg?height=458&width=569&top_left_y=129&top_left_x=1228)","['Since the given graph is congruent to $y=x^{2}$ and has $x$-intercepts $-d$ and $e$, its general form is $y=(x+d)(x-e)$.\n\nTo find the $y$-intercept, let $x=0$. Therefore $y$-intercept $=-d e$.\n\nWe are given that the $y$-intercept is $-f$.\n\nTherefore $-f=-d e$ or $f=d e$.']",['证明题，略'],True,,Need_human_evaluate, 2561,Geometry,,"In the graph, the parabola $y=x^{2}$ has been translated to the position shown. Prove that $d e=f$. ","['Since the given graph is congruent to $y=x^{2}$ and has $x$-intercepts $-d$ and $e$, its general form is $y=(x+d)(x-e)$.\n\nTo find the $y$-intercept, let $x=0$. Therefore $y$-intercept $=-d e$.\n\nWe are given that the $y$-intercept is $-f$.\n\nTherefore $-f=-d e$ or $f=d e$.']",,True,,, 2562,Geometry,,"In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$. ","['Establish a coordinate system with $K(0,0), D(2 a, 0)$ on the $x$-axes. Let $W$ be $(2 b, 2 c)$ and $A$ be $(2 d, 2 e)$.\n\nThus $M$ is $(b, c)$ and $N$ is $(a+d, e)$.\n\n$K D$ has slope 0 and slope $W A=\\frac{e-c}{d-b}$.\n\nSince $M N=\\frac{1}{2}(A W+D K)$\n\n$$\n\\begin{aligned}\n& \\sqrt{(a+d-b)^{2}+(e-c)^{2}} \\\\\n= & \\frac{1}{2}\\left(2 a+\\sqrt{(2 d-2 b)^{2}+(2 e-2 c)^{2}}\\right) \\\\\n= & \\frac{1}{2}\\left(2 a+2 \\sqrt{(d-b)^{2}+(e-c)^{2}}\\right)\n\\end{aligned}\n$$\n\n\n\nSquaring both sides gives,\n\n$$\n\\begin{aligned}\n& (a+d-b)^{2}+(e-c)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}+(e-c)^{2} \\\\\n& a^{2}+2 a(d-b)+(d-b)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}\n\\end{aligned}\n$$\n\nSimplifying and dividing by $2 a$ we have, $d-b=\\sqrt{(d-b)^{2}+(e-c)^{2}}$.\n\nSquaring, $(d-b)^{2}=(d-b)^{2}+(e-c)^{2}$.\n\nTherefore $(e-c)^{2}=0$ or $e=c$.\n\nSince $e=c$ then slope of $W A$ is 0 and $K D \\| A W$.', 'Join $A$ to $K$ and call $P$ the mid-point of $A K$.\n\nJoin $M$ to $P, N$ to $P$ and $M$ to $N$.\n\nIn $\\triangle K A W, P$ and $M$ are the mid-points of $K A$ and $K W$.\n\nTherefore, $M P=\\frac{1}{2} W A$.\n\nSimilarly in $\\triangle K A D, P N=\\frac{1}{2} K D$.\n\nTherefore $M P+P N=M N$.\n\n\n\nAs a result $M, P$ and $N$ cannot form the vertices of a triangle but must form a straight line.\n\nSo if $M P N$ is a straight line with $M P \\| W A$ and $P N \\| K D$ then $W A \\| K D$ as required.', 'We are given that $\\overrightarrow{A N}=\\overrightarrow{N D}$ and $\\overrightarrow{W M}=\\overrightarrow{M K}$.\n\nUsing vectors,\n\n(1) $\\overrightarrow{M N}=\\overrightarrow{M W}+\\overrightarrow{W A}+\\overrightarrow{A N}$ (from quad. $M W A N$ )\n\n(2) $\\overrightarrow{M N}=\\overrightarrow{M K}+\\overrightarrow{K D}+\\overrightarrow{D N} \\quad$ (from quad. $K M N D$ )\n\nIt is also possible to write, $\\overrightarrow{M N}=-\\overrightarrow{M W}+\\overrightarrow{K D}-\\overrightarrow{A N}$,\n\n(3) (This comes from taking statement (2) and making appropriate substitutions.)\n\n\n\nIf we add (1) and (3) we find, $2 \\overrightarrow{M N}=\\overrightarrow{W A}+\\overrightarrow{K D}$.\n\nBut it is given that $2|\\overrightarrow{M N}|=|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nFrom these two previous statements, $\\overrightarrow{M N}$ must be parallel to $\\overrightarrow{W A}$ and $\\overrightarrow{K D}$ otherwise $2|\\overrightarrow{M N}|<|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nTherefore, $W A \\| K D$.']",,True,,, 2562,Geometry,,"In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c810d95f38a712e9ddf7g-1.jpg?height=404&width=488&top_left_y=644&top_left_x=1271)","['Establish a coordinate system with $K(0,0), D(2 a, 0)$ on the $x$-axes. Let $W$ be $(2 b, 2 c)$ and $A$ be $(2 d, 2 e)$.\n\nThus $M$ is $(b, c)$ and $N$ is $(a+d, e)$.\n\n$K D$ has slope 0 and slope $W A=\\frac{e-c}{d-b}$.\n\nSince $M N=\\frac{1}{2}(A W+D K)$\n\n$$\n\\begin{aligned}\n& \\sqrt{(a+d-b)^{2}+(e-c)^{2}} \\\\\n= & \\frac{1}{2}\\left(2 a+\\sqrt{(2 d-2 b)^{2}+(2 e-2 c)^{2}}\\right) \\\\\n= & \\frac{1}{2}\\left(2 a+2 \\sqrt{(d-b)^{2}+(e-c)^{2}}\\right)\n\\end{aligned}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_821212715b4b9a6167d5g-1.jpg?height=423&width=613&top_left_y=320&top_left_x=1336)\n\nSquaring both sides gives,\n\n$$\n\\begin{aligned}\n& (a+d-b)^{2}+(e-c)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}+(e-c)^{2} \\\\\n& a^{2}+2 a(d-b)+(d-b)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}\n\\end{aligned}\n$$\n\nSimplifying and dividing by $2 a$ we have, $d-b=\\sqrt{(d-b)^{2}+(e-c)^{2}}$.\n\nSquaring, $(d-b)^{2}=(d-b)^{2}+(e-c)^{2}$.\n\nTherefore $(e-c)^{2}=0$ or $e=c$.\n\nSince $e=c$ then slope of $W A$ is 0 and $K D \\| A W$.', 'Join $A$ to $K$ and call $P$ the mid-point of $A K$.\n\nJoin $M$ to $P, N$ to $P$ and $M$ to $N$.\n\nIn $\\triangle K A W, P$ and $M$ are the mid-points of $K A$ and $K W$.\n\nTherefore, $M P=\\frac{1}{2} W A$.\n\nSimilarly in $\\triangle K A D, P N=\\frac{1}{2} K D$.\n\nTherefore $M P+P N=M N$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_821212715b4b9a6167d5g-1.jpg?height=423&width=613&top_left_y=1439&top_left_x=1336)\n\nAs a result $M, P$ and $N$ cannot form the vertices of a triangle but must form a straight line.\n\nSo if $M P N$ is a straight line with $M P \\| W A$ and $P N \\| K D$ then $W A \\| K D$ as required.', 'We are given that $\\overrightarrow{A N}=\\overrightarrow{N D}$ and $\\overrightarrow{W M}=\\overrightarrow{M K}$.\n\nUsing vectors,\n\n(1) $\\overrightarrow{M N}=\\overrightarrow{M W}+\\overrightarrow{W A}+\\overrightarrow{A N}$ (from quad. $M W A N$ )\n\n(2) $\\overrightarrow{M N}=\\overrightarrow{M K}+\\overrightarrow{K D}+\\overrightarrow{D N} \\quad$ (from quad. $K M N D$ )\n\nIt is also possible to write, $\\overrightarrow{M N}=-\\overrightarrow{M W}+\\overrightarrow{K D}-\\overrightarrow{A N}$,\n\n(3) (This comes from taking statement (2) and making appropriate substitutions.)\n\n\n\nIf we add (1) and (3) we find, $2 \\overrightarrow{M N}=\\overrightarrow{W A}+\\overrightarrow{K D}$.\n\nBut it is given that $2|\\overrightarrow{M N}|=|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nFrom these two previous statements, $\\overrightarrow{M N}$ must be parallel to $\\overrightarrow{W A}$ and $\\overrightarrow{K D}$ otherwise $2|\\overrightarrow{M N}|<|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nTherefore, $W A \\| K D$.']",['证明题，略'],True,,Need_human_evaluate, 2563,Number Theory,,"Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal. ![](https://cdn.mathpix.com/cropped/2023_12_21_c810d95f38a712e9ddf7g-1.jpg?height=214&width=1366&top_left_y=1205&top_left_x=431)","[""The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, k(2 n-k+1), \\ldots, p(2 n-p+1), \\ldots, n(n+1)$.\n\nIn essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'\n\n$$\n\\begin{aligned}\nk(2 n-k+1) & =p(2 n-p+1) \\\\\n2 n k-k^{2}+k=2 n p-p^{2}+p & \\text { (supposing them to be equal) } \\\\\np^{2}-k^{2}+2 n k-2 n p+k-p & =0 \\\\\n(p-k)(p+k)+2 n(k-p)+(k-p) & =0 \\\\\n(p-k)[(p+k)-2 n-1] & =0 \\\\\n(p-k)(p+k-2 n-1) & =0\n\\end{aligned}\n$$\n\nSince $p$ and $k$ are both less than or equal to $n$, it follows $p+k-2 n-1 \\neq 0$. Therefore $p=k$ and they represent the same pair. Thus the required is proven."", 'The products are $1(2 n+1-1), 2(2 n+1-2), 3(2 n+1-3), \\ldots, n(2 n+1-n)$.\n\nConsider the function, $y=x(2 n+1-x)=-x^{2}+(2 n+1) x=f(x)$.\n\n\n\nThe graph of this function is a parabola, opening down, with its vertex at $x=n+\\frac{1}{2}$.\n\nThe products are the $y$-coordinates of the points on the parabola corresponding to $x=1,2,3, \\ldots, n$. Since all the points are to the left of the vertex, no two have the same $y$ coordinate.\n\nThus the products are distinct.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_f4caa869f0e75f41fb17g-1.jpg?height=523&width=515&top_left_y=226&top_left_x=1336)', 'The sum of these numbers is $\\frac{2 n(2 n+1)}{2}$ or $n(2 n+1)$.\n\nTheir average is $\\frac{n(2 n+1)}{2 n}=n+\\frac{1}{2}$.\n\nThe $2 n$ numbers can be rewritten as,\n\n$$\nn+\\frac{1}{2}-\\left(\\frac{2 n-1}{2}\\right), \\cdots, n+\\frac{1}{2}-\\frac{3}{2}, n+\\frac{1}{2}-\\frac{1}{2}, n+\\frac{1}{2}+\\frac{1}{2}, n+\\frac{1}{2}+\\frac{3}{2}, \\cdots, n+\\frac{1}{2}+\\left(\\frac{2 n-1}{2}\\right) .\n$$\n\nThe product pairs, starting from the middle and working outward are\n\n$$\n\\begin{gathered}\nP_{1}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{1}{4} \\\\\nP_{2}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{9}{4} \\\\\n\\vdots \\\\\nP_{n}=\\left(n+\\frac{1}{2}\\right)^{2}-\\left(\\frac{2 n-1}{2}\\right)^{2}\n\\end{gathered}\n$$\n\nEach of the numbers $\\left(\\frac{2 k-1}{2}\\right)^{2}$ is distinct for $k=1,2,3, \\ldots, n$ and hence no terms of $P_{k}$ are equal.', 'The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, n[2 n-(n-1)]$.\n\nThis sequence has exactly $n$ terms.\n\nWhen the $k$ th term is subtracted from the $(k+1)$ th term the difference is $(k+1)[2 n-k]-k[2 n-(k-1)]=2(n-k)$. Since $n>k$, this is a positive difference.\n\nTherefore each term is greater than the term before, so no two terms are equal.']",['证明题，略'],True,,Need_human_evaluate, 2563,Number Theory,,"Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal. ","[""The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, k(2 n-k+1), \\ldots, p(2 n-p+1), \\ldots, n(n+1)$.\n\nIn essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'\n\n$$\n\\begin{aligned}\nk(2 n-k+1) & =p(2 n-p+1) \\\\\n2 n k-k^{2}+k=2 n p-p^{2}+p & \\text { (supposing them to be equal) } \\\\\np^{2}-k^{2}+2 n k-2 n p+k-p & =0 \\\\\n(p-k)(p+k)+2 n(k-p)+(k-p) & =0 \\\\\n(p-k)[(p+k)-2 n-1] & =0 \\\\\n(p-k)(p+k-2 n-1) & =0\n\\end{aligned}\n$$\n\nSince $p$ and $k$ are both less than or equal to $n$, it follows $p+k-2 n-1 \\neq 0$. Therefore $p=k$ and they represent the same pair. Thus the required is proven."", 'The products are $1(2 n+1-1), 2(2 n+1-2), 3(2 n+1-3), \\ldots, n(2 n+1-n)$.\n\nConsider the function, $y=x(2 n+1-x)=-x^{2}+(2 n+1) x=f(x)$.\n\n\n\nThe graph of this function is a parabola, opening down, with its vertex at $x=n+\\frac{1}{2}$.\n\nThe products are the $y$-coordinates of the points on the parabola corresponding to $x=1,2,3, \\ldots, n$. Since all the points are to the left of the vertex, no two have the same $y$ coordinate.\n\nThus the products are distinct.\n\n', 'The sum of these numbers is $\\frac{2 n(2 n+1)}{2}$ or $n(2 n+1)$.\n\nTheir average is $\\frac{n(2 n+1)}{2 n}=n+\\frac{1}{2}$.\n\nThe $2 n$ numbers can be rewritten as,\n\n$$\nn+\\frac{1}{2}-\\left(\\frac{2 n-1}{2}\\right), \\cdots, n+\\frac{1}{2}-\\frac{3}{2}, n+\\frac{1}{2}-\\frac{1}{2}, n+\\frac{1}{2}+\\frac{1}{2}, n+\\frac{1}{2}+\\frac{3}{2}, \\cdots, n+\\frac{1}{2}+\\left(\\frac{2 n-1}{2}\\right) .\n$$\n\nThe product pairs, starting from the middle and working outward are\n\n$$\n\\begin{gathered}\nP_{1}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{1}{4} \\\\\nP_{2}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{9}{4} \\\\\n\\vdots \\\\\nP_{n}=\\left(n+\\frac{1}{2}\\right)^{2}-\\left(\\frac{2 n-1}{2}\\right)^{2}\n\\end{gathered}\n$$\n\nEach of the numbers $\\left(\\frac{2 k-1}{2}\\right)^{2}$ is distinct for $k=1,2,3, \\ldots, n$ and hence no terms of $P_{k}$ are equal.', 'The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, n[2 n-(n-1)]$.\n\nThis sequence has exactly $n$ terms.\n\nWhen the $k$ th term is subtracted from the $(k+1)$ th term the difference is $(k+1)[2 n-k]-k[2 n-(k-1)]=2(n-k)$. Since $n>k$, this is a positive difference.\n\nTherefore each term is greater than the term before, so no two terms are equal.']",,True,,, 2564,Algebra,,"The equations $x^{2}+5 x+6=0$ and $x^{2}+5 x-6=0$ each have integer solutions whereas only one of the equations in the pair $x^{2}+4 x+5=0$ and $x^{2}+4 x-5=0$ has integer solutions. Show that if $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions, then it is possible to find integers $a$ and $b$ such that $p^{2}=a^{2}+b^{2}$. (i.e. $(a, b, p)$ is a Pythagorean triple).","['We have that $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions.\n\nFor $x^{2}+p x+q=0$, its roots are $\\frac{-p \\pm \\sqrt{p^{2}-4 q}}{2}$.\n\nIn order that these roots be integers, $p^{2}-4 q$ must be a perfect square.\n\nTherefore, $p^{2}-4 q=m^{2}$ for some positive integer $m$.\n\nSimilarly for $x^{2}+p x-q=0$, it has roots $\\frac{-p \\pm \\sqrt{p^{2}+4 q}}{2}$ and in order that these roots be integers $p^{2}+4 q$ must be a perfect square.\n\nThus $p^{2}+4 q=n^{2}$ for some positive integer $n$.\n\nAdding gives $2 p^{2}=m^{2}+n^{2}$ (with $n \\geq m$ since $n^{2}=p^{2}+4 q$\n\n$$\n\\left.\\geq p^{2}-4 q=m^{2}\\right)\n$$\n\nAnd so $p^{2}=\\frac{1}{2} m^{2}+\\frac{1}{2} n^{2}=\\left(\\frac{n+m}{2}\\right)^{2}+\\left(\\frac{n-m}{2}\\right)^{2}$.\n\nWe note that $m$ and $n$ have the same parity since $m^{2}=p^{2}-4 q \\equiv p^{2}(\\bmod 2)$ and $n^{2} \\equiv p^{2}+4 q \\equiv p^{2}(\\bmod 2)$.\n\nSince $\\frac{n+m}{2}$ and $\\frac{n-m}{2}$ are positive integers then $p^{2}=a^{2}+b^{2}$ where $a=\\frac{n+m}{2}$ and $b=\\frac{n-m}{2}$.']",,True,,, 2565,Algebra,,"The equations $x^{2}+5 x+6=0$ and $x^{2}+5 x-6=0$ each have integer solutions whereas only one of the equations in the pair $x^{2}+4 x+5=0$ and $x^{2}+4 x-5=0$ has integer solutions. Determine $q$ in terms of $a$ and $b$.","['We have that $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions.\n\nFor $x^{2}+p x+q=0$, its roots are $\\frac{-p \\pm \\sqrt{p^{2}-4 q}}{2}$.\n\nIn order that these roots be integers, $p^{2}-4 q$ must be a perfect square.\n\nTherefore, $p^{2}-4 q=m^{2}$ for some positive integer $m$.\n\nSimilarly for $x^{2}+p x-q=0$, it has roots $\\frac{-p \\pm \\sqrt{p^{2}+4 q}}{2}$ and in order that these roots be integers $p^{2}+4 q$ must be a perfect square.\n\nThus $p^{2}+4 q=n^{2}$ for some positive integer $n$.\n\nAdding gives $2 p^{2}=m^{2}+n^{2}$ (with $n \\geq m$ since $n^{2}=p^{2}+4 q$\n\n$$\n\\left.\\geq p^{2}-4 q=m^{2}\\right)\n$$\n\nAnd so $p^{2}=\\frac{1}{2} m^{2}+\\frac{1}{2} n^{2}=\\left(\\frac{n+m}{2}\\right)^{2}+\\left(\\frac{n-m}{2}\\right)^{2}$.\n\nWe note that $m$ and $n$ have the same parity since $m^{2}=p^{2}-4 q \\equiv p^{2}(\\bmod 2)$ and $n^{2} \\equiv p^{2}+4 q \\equiv p^{2}(\\bmod 2)$.\n\nSince $\\frac{n+m}{2}$ and $\\frac{n-m}{2}$ are positive integers then $p^{2}=a^{2}+b^{2}$ where $a=\\frac{n+m}{2}$ and $b=\\frac{n-m}{2}$.\n\nFrom above, $a=\\frac{n+m}{2}$ and $b=\\frac{n-m}{2}$ or $n=a+b$ and $m=a-b$.\n\nFrom before, $p^{2}+4 q=n^{2}$\n\n$$\n\\begin{aligned}\n4 q^{2} & =n^{2}-p^{2} \\\\\n& =(a+b)^{2}-\\left(a^{2}+b^{2}\\right) \\\\\n4 q & =2 a b\n\\end{aligned}\n$$\n\nTherefore, $q=\\frac{a b}{2}$.']",['$\\frac{a b}{2}$'],False,,Expression, 2566,Geometry,,"In the diagram, $A B=21$ and $B C=16$. Also, $\angle A B C=60^{\circ}, \angle C A D=30^{\circ}$, and $\angle A C D=45^{\circ}$. Determine the length of $C D$, to the nearest tenth. ![](https://cdn.mathpix.com/cropped/2023_12_21_e753dfdb77a4e9b70a3dg-1.jpg?height=322&width=447&top_left_y=1655&top_left_x=1311)","['By the cosine law in $\\triangle C B A$,\n\n$$\n\\begin{aligned}\nC A^{2} & =C B^{2}+B A^{2}-2(C B)(B A) \\cos (\\angle C B A) \\\\\nC A^{2} & =16^{2}+21^{2}-2(16)(21) \\cos \\left(60^{\\circ}\\right) \\\\\nC A^{2} & =256+441-2(16)(21)\\left(\\frac{1}{2}\\right) \\\\\nC A^{2} & =256+441-(16)(21) \\\\\nC A^{2} & =361 \\\\\nC A & =\\sqrt{361}=19 \\quad(\\text { since } C A>0)\n\\end{aligned}\n$$\n\nIn $\\triangle C A D, \\angle C D A=180^{\\circ}-\\angle D C A-\\angle D A C=180^{\\circ}-45^{\\circ}-30^{\\circ}=105^{\\circ}$.\n\nBy the sine law in $\\triangle C D A$,\n\n$$\n\\begin{aligned}\n\\frac{C D}{\\sin (\\angle D A C)} & =\\frac{C A}{\\sin (\\angle C D A)} \\\\\nC D & =\\frac{19 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19\\left(\\frac{1}{2}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19}{2 \\sin \\left(105^{\\circ}\\right)} \\\\\nC D & \\approx 9.835\n\\end{aligned}\n$$\n\nso, to the nearest tenth, $C D$ equals 9.8 .\n\n(Note that we could have used\n\n$$\n\\begin{aligned}\n\\sin \\left(105^{\\circ}\\right) & =\\sin \\left(60^{\\circ}+45^{\\circ}\\right)=\\sin \\left(60^{\\circ}\\right) \\cos \\left(45^{\\circ}\\right)+\\cos \\left(60^{\\circ}\\right) \\sin \\left(45^{\\circ}\\right) \\\\\n& =\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}}+\\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\n\\end{aligned}\n$$\n\nto say that $C D=\\frac{19}{2\\left(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\right)}=\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$ exactly, and then evaluated this expression.)']",['$\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$'],False,,Numerical, 2566,Geometry,,"In the diagram, $A B=21$ and $B C=16$. Also, $\angle A B C=60^{\circ}, \angle C A D=30^{\circ}$, and $\angle A C D=45^{\circ}$. Determine the length of $C D$, to the nearest tenth. ","['By the cosine law in $\\triangle C B A$,\n\n$$\n\\begin{aligned}\nC A^{2} & =C B^{2}+B A^{2}-2(C B)(B A) \\cos (\\angle C B A) \\\\\nC A^{2} & =16^{2}+21^{2}-2(16)(21) \\cos \\left(60^{\\circ}\\right) \\\\\nC A^{2} & =256+441-2(16)(21)\\left(\\frac{1}{2}\\right) \\\\\nC A^{2} & =256+441-(16)(21) \\\\\nC A^{2} & =361 \\\\\nC A & =\\sqrt{361}=19 \\quad(\\text { since } C A>0)\n\\end{aligned}\n$$\n\nIn $\\triangle C A D, \\angle C D A=180^{\\circ}-\\angle D C A-\\angle D A C=180^{\\circ}-45^{\\circ}-30^{\\circ}=105^{\\circ}$.\n\nBy the sine law in $\\triangle C D A$,\n\n$$\n\\begin{aligned}\n\\frac{C D}{\\sin (\\angle D A C)} & =\\frac{C A}{\\sin (\\angle C D A)} \\\\\nC D & =\\frac{19 \\sin \\left(30^{\\circ}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19\\left(\\frac{1}{2}\\right)}{\\sin \\left(105^{\\circ}\\right)} \\\\\nC D & =\\frac{19}{2 \\sin \\left(105^{\\circ}\\right)} \\\\\nC D & \\approx 9.835\n\\end{aligned}\n$$\n\nso, to the nearest tenth, $C D$ equals 9.8 .\n\n(Note that we could have used\n\n$$\n\\begin{aligned}\n\\sin \\left(105^{\\circ}\\right) & =\\sin \\left(60^{\\circ}+45^{\\circ}\\right)=\\sin \\left(60^{\\circ}\\right) \\cos \\left(45^{\\circ}\\right)+\\cos \\left(60^{\\circ}\\right) \\sin \\left(45^{\\circ}\\right) \\\\\n& =\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}}+\\frac{1}{2} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\n\\end{aligned}\n$$\n\nto say that $C D=\\frac{19}{2\\left(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\right)}=\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$ exactly, and then evaluated this expression.)']",['$\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$'],False,,Numerical, 2567,Geometry,,"In the diagram, the large circle has radius 9 and centre $C(15,0)$. The small circles have radius 4 and centres $A$ and $B$ on the horizontal line $y=12$. Each of the two small circles is tangent to the large circle. It takes a bug 5 seconds to walk at a constant speed from $A$ to $B$ along the line $y=12$. How far does the bug walk in 1 second? ","['Consider $P$ on $A B$ with $C P$ perpendicular to $A B$. Note that $C P=12$.\n\nSince the small circle with centre $A$ is tangent to the large circle with centre $C$, then $A C$ equals the sum of the radii of these circles, or $A C=4+9=13$. Similarly, $B C=13$.\n\nThis tells us that $\\triangle A P C$ is congruent to $\\triangle B P C$ (they have equal hypotenuses and each is right-angled and has a common side), so $B P=A P$.\n\nBy the Pythagorean Theorem in $\\triangle A P C$,\n\n$$\nA P^{2}=A C^{2}-P C^{2}=13^{2}-12^{2}=169-144=25\n$$\n\nso $A P=5$ (since $A P>0)$.\n\nTherefore, $B P=A P=5$ and so $A B=10$.\n\nSince it takes the bug 5 seconds to walk this distance, then in 1 second, the bug walks a distance of $\\frac{10}{5}=2$.']",['2'],False,,Numerical, 2567,Geometry,,"In the diagram, the large circle has radius 9 and centre $C(15,0)$. The small circles have radius 4 and centres $A$ and $B$ on the horizontal line $y=12$. Each of the two small circles is tangent to the large circle. It takes a bug 5 seconds to walk at a constant speed from $A$ to $B$ along the line $y=12$. How far does the bug walk in 1 second? ![](https://cdn.mathpix.com/cropped/2023_12_21_e753dfdb77a4e9b70a3dg-1.jpg?height=445&width=542&top_left_y=2000&top_left_x=1255)","['Consider $P$ on $A B$ with $C P$ perpendicular to $A B$. Note that $C P=12$.\n\nSince the small circle with centre $A$ is tangent to the large circle with centre $C$, then $A C$ equals the sum of the radii of these circles, or $A C=4+9=13$. Similarly, $B C=13$.\n\nThis tells us that $\\triangle A P C$ is congruent to $\\triangle B P C$ (they have equal hypotenuses and each is right-angled and has a common side), so $B P=A P$.\n\nBy the Pythagorean Theorem in $\\triangle A P C$,\n\n$$\nA P^{2}=A C^{2}-P C^{2}=13^{2}-12^{2}=169-144=25\n$$\n\nso $A P=5$ (since $A P>0)$.\n\nTherefore, $B P=A P=5$ and so $A B=10$.\n\nSince it takes the bug 5 seconds to walk this distance, then in 1 second, the bug walks a distance of $\\frac{10}{5}=2$.']",['2'],False,,Numerical, 2568,Algebra,,"Determine all values of $k$, with $k \neq 0$, for which the parabola $$ y=k x^{2}+(5 k+3) x+(6 k+5) $$ has its vertex on the $x$-axis.","['For the parabola to have its vertex on the $x$-axis, the equation\n\n$$\ny=k x^{2}+(5 k+3) x+(6 k+5)=0\n$$\n\nmust have two equal real roots.\n\nThat is, its discriminant must equal 0 , and so\n\n$$\n\\begin{aligned}\n(5 k+3)^{2}-4 k(6 k+5) & =0 \\\\\n25 k^{2}+30 k+9-24 k^{2}-20 k & =0 \\\\\nk^{2}+10 k+9 & =0 \\\\\n(k+1)(k+9) & =0\n\\end{aligned}\n$$\n\nTherefore, $k=-1$ or $k=-9$.']","['$-1,-9$']",True,,Numerical, 2569,Algebra,,"The function $f(x)$ satisfies the equation $f(x)=f(x-1)+f(x+1)$ for all values of $x$. If $f(1)=1$ and $f(2)=3$, what is the value of $f(2008)$ ?","['Since $f(x)=f(x-1)+f(x+1)$, then $f(x+1)=f(x)-f(x-1)$, and so\n\n$$\n\\begin{aligned}\n& f(1)=1 \\\\\n& f(2)=3 \\\\\n& f(3)=f(2)-f(1)=3-1=2 \\\\\n& f(4)=f(3)-f(2)=2-3=-1 \\\\\n& f(5)=f(4)-f(3)=-1-2=-3 \\\\\n& f(6)=f(5)-f(4)=-3-(-1)=-2 \\\\\n& f(7)=f(6)-f(5)=-2-(-3)=1=f(1) \\\\\n& f(8)=f(7)-f(6)=1-(-2)=3=f(2)\n\\end{aligned}\n$$\n\nSince the value of $f$ at an integer depends only on the values of $f$ at the two previous integers, then the fact that the first several values form a cycle with $f(7)=f(1)$ and $f(8)=f(2)$ tells us that the values of $f$ will always repeat in sets of 6 .\n\nSince 2008 is 4 more than a multiple of 6 (as $2008=4+2004=4+6(334)$ ), then $f(2008)=f(2008-6(334))=f(4)=-1$.']",['-1'],False,,Numerical, 2570,Algebra,,"The numbers $a, b, c$, in that order, form a three term arithmetic sequence (see below) and $a+b+c=60$. The numbers $a-2, b, c+3$, in that order, form a three term geometric sequence. Determine all possible values of $a, b$ and $c$. (An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7$ is an arithmetic sequence with three terms. A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.) Present your answer in the form of coordinates (e.g. (1, 2, 3) for a=1, b=2, c=3).","['Since $a, b, c$ form an arithmetic sequence, then we can write $a=b-d$ and $c=b+d$ for some real number $d$.\n\nSince $a+b+c=60$, then $(b-d)+b+(b+d)=60$ or $3 b=60$ or $b=20$.\n\nTherefore, we can write $a, b, c$ as $20-d, 20,20+d$.\n\n(We could have written $a, b, c$ instead as $a, a+d, a+2 d$ and arrived at the same result.) Thus, $a-2=20-d-2=18-d$ and $c+3=20+d+3=23+d$, so we can write $a-2, b, c+3$ as $18-d, 20,23+d$.\n\n\n\nSince these three numbers form a geometric sequence, then\n\n$$\n\\begin{aligned}\n\\frac{20}{18-d} & =\\frac{23+d}{20} \\\\\n20^{2} & =(23+d)(18-d) \\\\\n400 & =-d^{2}-5 d+414 \\\\\nd^{2}+5 d-14 & =0 \\\\\n(d+7)(d-2) & =0\n\\end{aligned}\n$$\n\nTherefore, $d=-7$ or $d=2$.\n\nIf $d=-7$, then $a=27, b=20$ and $c=13$.\n\nIf $d=2$, then $a=18, b=20$ and $c=22$.\n\n(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)', 'Since $a, b, c$ form an arithmetic sequence, then $c-b=b-a$ or $a+c=2 b$.\n\nSince $a+b+c=60$, then $2 b+b=60$ or $3 b=60$ or $b=20$.\n\nThus, $a+c=40$, so $a=40-c$.\n\nTherefore, we can write $a, b, c$ as $40-c, 20, c$.\n\nAlso, $a-2=40-c-2=38-c$, so we can write $a-2, b, c+3$ as $38-c, 20, c+3$.\n\nSince these three numbers form a geometric sequence, then\n\n$$\n\\begin{aligned}\n\\frac{20}{38-c} & =\\frac{c+3}{20} \\\\\n20^{2} & =(38-c)(c+3) \\\\\n400 & =-c^{2}+35 c+114 \\\\\nc^{2}-35 d+286 & =0 \\\\\n(c-13)(c-22) & =0\n\\end{aligned}\n$$\n\nTherefore, $c=13$ or $c=22$.\n\nIf $c=13$, then $a=27$, so $a=27, b=20$ and $c=13$.\n\nIf $c=22$, then $a=18$, so $a=18, b=20$ and $c=22$.\n\n(We can check that, in each case, $a-2, b, c+3$ is a geometric sequence.)']","['$(27,20,13), (18,20,22)$']",True,,Tuple, 2571,Algebra,,"The average of three consecutive multiples of 3 is $a$. The average of four consecutive multiples of 4 is $a+27$. The average of the smallest and largest of these seven integers is 42 . Determine the value of $a$.","['Since the average of three consecutive multiples of 3 is $a$, then $a$ is the middle of these three integers, so the integers are $a-3, a, a+3$.\n\nSince the average of four consecutive multiples of 4 is $a+27$, then $a+27$ is halfway in between the second and third of these multiples (which differ by 4), so the second and third of the multiples are $(a+27)-2=a+25$ and $(a+27)+2=a+29$, so the four integers are $a+21, a+25, a+29, a+33$.\n\n(We have used in these two statements the fact that if a list contains an odd number of integers, then there is a middle integer in the list, and if the list contains an even number\n\n\n\nof integers, then the ""middle"" integer is between two integers from the list.)\n\nThe smallest of these seven integers is $a-3$ and the largest is $a+33$.\n\nThe average of these two integers is $\\frac{1}{2}(a-3+a+33)=\\frac{1}{2}(2 a+30)=a+15$.\n\nSince $a+15=42$, then $a=27$.']",['27'],False,,Numerical, 2572,Combinatorics,,Billy and Crystal each have a bag of 9 balls. The balls in each bag are numbered from 1 to 9. Billy and Crystal each remove one ball from their own bag. Let $b$ be the sum of the numbers on the balls remaining in Billy's bag. Let $c$ be the sum of the numbers on the balls remaining in Crystal's bag. Determine the probability that $b$ and $c$ differ by a multiple of 4 .,"['Suppose that Billy removes the ball numbered $x$ from his bag and that Crystal removes the ball numbered $y$ from her bag.\n\nThen $b=1+2+3+4+5+6+7+8+9-x=45-x$.\n\nAlso, $c=1+2+3+4+5+6+7+8+9-y=45-y$.\n\nHence, $b-c=(45-x)-(45-y)=y-x$.\n\nSince $1 \\leq x \\leq 9$ and $1 \\leq y \\leq 9$, then $-8 \\leq y-x \\leq 8$.\n\n(This is because $y-x$ is maximized when $y$ is largest (that is, $y=9$ ) and $x$ is smallest (that is, $x=1$ ), so $y-x \\leq 9-1=8$. Similarly, $y-x \\geq-8$.)\n\nSince $b-c=y-x$ is between -8 and 8 , then for it to be a multiple of $4, b-c=y-x$ can be $-8,-4,0,4$, or 8 .\n\nSince each of Billy and Crystal chooses 1 ball from 9 balls and each ball is equally likely to be chosen, then the probability of any specific ball being chosen from one of their bags is $\\frac{1}{9}$. Thus, the probability of any specific pair of balls being chosen (one from each bag) is $\\frac{1}{9} \\times \\frac{1}{9}=\\frac{1}{81}$.\n\nTherefore, to compute the desired probability, we must count the number of pairs $(x, y)$ where $y-x$ is $-8,-4,0,4,8$, and multiply this result by $\\frac{1}{81}$.\n\nMethod 1 \n\nIf $y-x=-8$, then $(x, y)$ must be $(9,1)$.\n\nIf $y-x=8$, then $(x, y)$ must be $(1,9)$.\n\nIf $y-x=-4$, then $(x, y)$ can be $(5,1),(6,2),(7,3),(8,4),(9,5)$.\n\nIf $y-x=4$, then $(x, y)$ can be $(1,5),(2,6),(3,7),(4,8),(5,9)$.\n\nIf $y-x=0$, then $(x, y)$ can be $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9)$.\n\nThere are thus 21 pairs $(x, y)$ that work, so the desired probability is $\\frac{21}{81}=\\frac{7}{27}$.\n\nMethod 2\n\nIf $x=9$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .\n\nIf $x=8$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .\n\nIf $x=7$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .\n\nIf $x=6$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .\n\nIf $x=5$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .\n\nIf $x=4$, then for $y-x$ to be a multiple of $4, y$ could be 8 or 4 .\n\nIf $x=3$, then for $y-x$ to be a multiple of $4, y$ could be 7 or 3 .\n\nIf $x=2$, then for $y-x$ to be a multiple of $4, y$ could be 6 or 2 .\n\nIf $x=1$, then for $y-x$ to be a multiple of $4, y$ could be 9,5 or 1 .\n\n\n\nThere are thus 21 pairs $(x, y)$ that work, so the desired probability is $\\frac{21}{81}=\\frac{7}{27}$.']",['$\\frac{7}{27}$'],False,,Numerical, 2573,Geometry,,"In the diagram, $A B C$ is a right-angled triangle with $P$ and $R$ on $A B$. Also, $Q$ is on $A C$, and $P Q$ is parallel to $B C$. If $R P=2$, $B R=3, B C=4$, and the area of $\triangle Q R C$ is 5 , determine the length of $A P$. ![](https://cdn.mathpix.com/cropped/2023_12_21_16c4f36a80a838603da9g-1.jpg?height=334&width=499&top_left_y=1931&top_left_x=1255)","['Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nConsider trapezoid $C B P Q$. We can think of this as having parallel bases of lengths 4 and $h$ and height 5 . Thus, its area is $\\frac{1}{2}(4+h)(5)$.\n\nHowever, we can also compute its area by adding the areas of $\\triangle C B R$ (which is $\\left.\\frac{1}{2}(4)(3)\\right)$, $\\triangle C R Q$ (which is given as 5), and $\\triangle R P Q$ (which is $\\left.\\frac{1}{2}(2)(h)\\right)$.\n\nThus,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(4+h)(5) & =\\frac{1}{2}(4)(3)+5+\\frac{1}{2}(2)(h) \\\\\n20+5 h & =12+10+2 h \\\\\n3 h & =2 \\\\\nh & =\\frac{2}{3}\n\\end{aligned}\n$$\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\n4 x & =\\frac{2}{3}(x+5) \\\\\n4 x & =\\frac{2}{3} x+\\frac{10}{3} \\\\\n\\frac{10}{3} x & =\\frac{10}{3} \\\\\nx & =1\n\\end{aligned}\n$$\n\n\n\nTherefore, $A P=x=1$.', 'Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area is $\\frac{1}{2}(4)(5+x)=10+2 x$.\n\nHowever, we can look at the area of the $\\triangle A B C$ in terms of its four triangular pieces:\n\n$\\triangle C B R$ (which has area $\\left.\\frac{1}{2}(4)(3)\\right), \\triangle C R Q$ (which has area 5), $\\triangle Q P R$ (which has area $\\left.\\frac{1}{2} h(2)\\right)$, and $\\triangle Q P A$ (which has area $\\frac{1}{2} x h$ ).\n\nTherefore, $10+2 x=6+5+h+\\frac{1}{2} x h$ so $x h-4 x+2 h+2=0$.\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\nx(4) & =h(x+5) \\\\\n4 x & =h x+5 h \\\\\n-5 h & =h x-4 x\n\\end{aligned}\n$$\n\nSubstituting this into the equation above, $x h+2 h-4 x+2=0$ becomes $-5 h+2 h+2=0$ or $3 h=2$ or $h=\\frac{2}{3}$.\n\nLastly, we solve for $x$ by subsituting our value for $h$ : $-5\\left(\\frac{2}{3}\\right)=\\frac{2}{3} x-4 x$ or $-\\frac{10}{3}=-\\frac{10}{3} x$ and so $x=1$.\n\nTherefore, $A P=x=1$.']",['1'],False,,Numerical, 2573,Geometry,,"In the diagram, $A B C$ is a right-angled triangle with $P$ and $R$ on $A B$. Also, $Q$ is on $A C$, and $P Q$ is parallel to $B C$. If $R P=2$, $B R=3, B C=4$, and the area of $\triangle Q R C$ is 5 , determine the length of $A P$. ","['Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nConsider trapezoid $C B P Q$. We can think of this as having parallel bases of lengths 4 and $h$ and height 5 . Thus, its area is $\\frac{1}{2}(4+h)(5)$.\n\nHowever, we can also compute its area by adding the areas of $\\triangle C B R$ (which is $\\left.\\frac{1}{2}(4)(3)\\right)$, $\\triangle C R Q$ (which is given as 5), and $\\triangle R P Q$ (which is $\\left.\\frac{1}{2}(2)(h)\\right)$.\n\nThus,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(4+h)(5) & =\\frac{1}{2}(4)(3)+5+\\frac{1}{2}(2)(h) \\\\\n20+5 h & =12+10+2 h \\\\\n3 h & =2 \\\\\nh & =\\frac{2}{3}\n\\end{aligned}\n$$\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\n4 x & =\\frac{2}{3}(x+5) \\\\\n4 x & =\\frac{2}{3} x+\\frac{10}{3} \\\\\n\\frac{10}{3} x & =\\frac{10}{3} \\\\\nx & =1\n\\end{aligned}\n$$\n\n\n\nTherefore, $A P=x=1$.', 'Let $A P=x$ and $Q P=h$.\n\nSince $Q P$ is parallel to $C B$, then $Q P$ is perpendicular to $B A$.\n\nSince $\\triangle A B C$ is right-angled at $B$, its area is $\\frac{1}{2}(4)(5+x)=10+2 x$.\n\nHowever, we can look at the area of the $\\triangle A B C$ in terms of its four triangular pieces:\n\n$\\triangle C B R$ (which has area $\\left.\\frac{1}{2}(4)(3)\\right), \\triangle C R Q$ (which has area 5), $\\triangle Q P R$ (which has area $\\left.\\frac{1}{2} h(2)\\right)$, and $\\triangle Q P A$ (which has area $\\frac{1}{2} x h$ ).\n\nTherefore, $10+2 x=6+5+h+\\frac{1}{2} x h$ so $x h-4 x+2 h+2=0$.\n\nNow, $\\triangle A P Q$ is similar to $\\triangle A B C$, as each has a right angle and they share a common angle at $A$. Thus,\n\n$$\n\\begin{aligned}\n\\frac{A P}{P Q} & =\\frac{A B}{B C} \\\\\n(A P)(B C) & =(P Q)(A B) \\\\\nx(4) & =h(x+5) \\\\\n4 x & =h x+5 h \\\\\n-5 h & =h x-4 x\n\\end{aligned}\n$$\n\nSubstituting this into the equation above, $x h+2 h-4 x+2=0$ becomes $-5 h+2 h+2=0$ or $3 h=2$ or $h=\\frac{2}{3}$.\n\nLastly, we solve for $x$ by subsituting our value for $h$ : $-5\\left(\\frac{2}{3}\\right)=\\frac{2}{3} x-4 x$ or $-\\frac{10}{3}=-\\frac{10}{3} x$ and so $x=1$.\n\nTherefore, $A P=x=1$.']",['1'],False,,Numerical, 2574,Algebra,,The equation $2^{x+2} 5^{6-x}=10^{x^{2}}$ has two real solutions. Determine these two solutions.,"['Rewriting the equation, we obtain\n\n$$\n\\begin{aligned}\n2^{x+2} 5^{6-x} & =2^{x^{2}} 5^{x^{2}} \\\\\n1 & =2^{x^{2}} 2^{-2-x} 5^{x^{2}} 5^{x-6} \\\\\n1 & =2^{x^{2}-x-2} 5^{x^{2}+x-6} \\\\\n0 & =\\left(x^{2}-x-2\\right) \\log _{10} 2+\\left(x^{2}+x-6\\right) \\log _{10} 5 \\\\\n0 & =(x-2)(x+1) \\log _{10} 2+(x-2)(x+3) \\log _{10} 5 \\\\\n0 & =(x-2)\\left[(x+1) \\log _{10} 2+(x+3) \\log _{10} 5\\right] \\\\\n0 & =(x-2)\\left[\\left(\\log _{10} 2+\\log _{10} 5\\right) x+\\left(\\log _{10} 2+3 \\log 105\\right)\\right] \\\\\n0 & =(x-2)\\left[\\left(\\log _{10} 10\\right) x+\\log _{10}\\left(2 \\cdot 5^{3}\\right)\\right] \\\\\n0 & =(x-2)\\left(x+\\log _{10} 250\\right)\n\\end{aligned}\n$$\n\nTherefore, $x=2$ or $x=-\\log _{10} 250$.', 'We take base 10 logarithms of both sides:\n\n$$\n\\begin{aligned}\n\\log _{10}\\left(2^{x+2} 5^{6-x}\\right) & =\\log _{10}\\left(10^{x^{2}}\\right) \\\\\n\\log _{10}\\left(2^{x+2}\\right)+\\log _{10}\\left(5^{6-x}\\right) & =x^{2} \\\\\n(x+2) \\log _{10} 2+(6-x) \\log _{10} 5 & =x^{2} \\\\\nx\\left(\\log _{10} 2-\\log _{10} 5\\right)+\\left(2 \\log _{10} 2+6 \\log _{10} 5\\right) & =x^{2} \\\\\nx^{2}-x\\left(\\log _{10} 2-\\log _{10} 5\\right)-\\left(2 \\log _{10} 2+6 \\log _{10} 5\\right) & =0\n\\end{aligned}\n$$\n\nNow, $\\log _{10} 2+\\log _{10} 5=\\log _{10} 10=1$ so $\\log _{10} 5=1-\\log _{10} 2$, so we can simplify the equation to\n\n$$\nx^{2}-x\\left(2 \\log _{10} 2-1\\right)-\\left(6-4 \\log _{10} 2\\right)=0\n$$\n\nThis is a quadratic equation in $x$, so should have at most 2 real solutions.\n\nBy the quadratic formula,\n\n$$\n\\begin{aligned}\nx & =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{\\left(2 \\log _{10} 2-1\\right)^{2}-4(1)\\left(-\\left(6-4 \\log _{10} 2\\right)\\right)}}{2(1)} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{4\\left(\\log _{10} 2\\right)^{2}-4\\left(\\log _{10} 2\\right)+1+24-16 \\log _{10} 2}}{2} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{4\\left(\\log _{10} 2\\right)^{2}-20\\left(\\log _{10} 2\\right)+25}}{2} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm \\sqrt{\\left(2 \\log _{10} 2-5\\right)^{2}}}{2} \\\\\n& =\\frac{\\left(2 \\log _{10} 2-1\\right) \\pm\\left(5-2 \\log _{10} 2\\right)}{2}\n\\end{aligned}\n$$\n\nsince $5-2 \\log _{10} 2>0$.\n\nTherefore,\n\n$$\nx=\\frac{\\left(2 \\log _{10} 2-1\\right)+\\left(5-2 \\log _{10} 2\\right)}{2}=\\frac{4}{2}=2\n$$\n\nor\n\n$$\nx=\\frac{\\left(2 \\log _{10} 2-1\\right)-\\left(5-2 \\log _{10} 2\\right)}{2}=\\frac{4 \\log _{10} 2-6}{2}=2 \\log _{10} 2-3\n$$\n\n(Note that at any point, we could have used a calculator to convert to decimal approximations and solve.)']","['$2,-\\log _{10} 250$']",True,,Numerical, 2575,Algebra,,"Determine all real solutions to the system of equations $$ \begin{aligned} & x+\log _{10} x=y-1 \\ & y+\log _{10}(y-1)=z-1 \\ & z+\log _{10}(z-2)=x+2 \end{aligned} $$ and prove that there are no more solutions.","['First, we rewrite the system as\n\n$$\n\\begin{aligned}\n& x+\\log _{10} x=y-1 \\\\\n& (y-1)+\\log _{10}(y-1)=z-2 \\\\\n& (z-2)+\\log _{10}(z-2)=x\n\\end{aligned}\n$$\n\nSecond, we make the substitution $a=x, b=y-1$ and $c=z-2$, allowing us to rewrite\n\n\n\nthe system as\n\n$$\n\\begin{aligned}\na+\\log _{10} a & =b \\\\\nb+\\log _{10} b & =c \\\\\nc+\\log _{10} c & =a\n\\end{aligned}\n$$\n\nThird, we observe that $(a, b, c)=(1,1,1)$ is a solution, since $1+\\log _{10} 1=1+0=1$.\n\nNext, if $a>1$, then $\\log _{10} a>0$, so from (1),\n\n$$\nb=a+\\log _{10} a>a+0=a>1\n$$\n\nso $\\log _{10} b>0$, so from $(2)$,\n\n$$\nc=b+\\log _{10} b>b+0=b>a>1\n$$\n\nso $\\log _{10} c>0$, so from (3),\n\n$$\na=c+\\log _{10} c>c+0=c>b>a>1\n$$\n\nBut this says that $a>c>b>a$, which is a contradiction.\n\nTherefore, $a$ cannot be larger than 1 .\n\nLastly, if $0c+0=c1$, then $f(n)=f(n-1)+1$. For example, $f(34)=f(17)$ and $f(17)=f(16)+1$. Determine the value of $f(50)$.","['We start with $f(50)$ and apply the given rules for the function until we reach $f(1)$ :\n\n$$\n\\begin{aligned}\nf(50) & =f(25) \\\\\n& =f(24)+1 \\\\\n& =f(12)+1 \\\\\n& =f(6)+1 \\\\\n& =f(3)+1 \\\\\n& =(f(2)+1)+1 \\\\\n& =f(1)+1+1 \\\\\n& =1+1+1 \\\\\n& =3\n\\end{aligned}\n$$\n\n(since 50 is even and $\\frac{1}{2}(50)=25$ )\n\n(since 25 is odd and $25-1=24$ )\n\n$$\n\\left(\\frac{1}{2}(24)=12\\right)\n$$\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{2}(12)\\right. & =6) \\\\\n\\left(\\frac{1}{2}(6)\\right. & =3) \\\\\n(3-1 & =2) \\\\\n\\left(\\frac{1}{2}(2)\\right. & =1) \\\\\n(f(1) & =1)\n\\end{aligned}\n$$\n\nTherefore, $f(50)=3$.']",['3'],False,,Numerical, 2578,Geometry,,The perimeter of equilateral $\triangle P Q R$ is 12. The perimeter of regular hexagon $S T U V W X$ is also 12. What is the ratio of the area of $\triangle P Q R$ to the area of $S T U V W X$ ?,"['Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2 .\n\nSince equilateral $\\triangle P Q R$ has perimeter 12 , then its side length is 4 .\n\nConsider equilateral triangles with side length 2.\n\nSix of these triangles can be combined to form a regular hexagon with side length 2 and four of these can be combined to form an equilateral triangle with side length 4 .\n\n\nNote that the six equilateral triangles around the centre of the hexagon give a total central angle of $6 \\cdot 60^{\\circ}=360^{\\circ}$ (a complete circle) and the three equilateral triangles along each side of the large equilateral triangle make a straight angle of $180^{\\circ}\\left(\\right.$ since $3 \\cdot 60^{\\circ}=180^{\\circ}$ ). Also, the length of each side of the hexagon is 2 and the measure of each internal angle is $120^{\\circ}$, which means that the hexagon is regular. Similarly, the triangle is equilateral.\n\nSince the triangle is made from four identical smaller triangles and the hexagon is made from six of these smaller triangles, the ratio of the area of the triangle to the hexagon is $4: 6$ which is equivalent to $2: 3$.']",['$\\frac{2}{3}$'],False,,Numerical, 2579,Geometry,,"In the diagram, sector $A O B$ is $\frac{1}{6}$ of an entire circle with radius $A O=B O=18$. The sector is cut into two regions with a single straight cut through $A$ and point $P$ on $O B$. The areas of the two regions are equal. Determine the length of $O P$. ","['Since sector $A O B$ is $\\frac{1}{6}$ of a circle with radius 18 , its area is $\\frac{1}{6}\\left(\\pi \\cdot 18^{2}\\right)$ or $54 \\pi$.\n\nFor the line $A P$ to divide this sector into two pieces of equal area, each piece has area $\\frac{1}{2}(54 \\pi)$ or $27 \\pi$.\n\nWe determine the length of $O P$ so that the area of $\\triangle P O A$ is $27 \\pi$.\n\nSince sector $A O B$ is $\\frac{1}{6}$ of a circle, then $\\angle A O B=\\frac{1}{6}\\left(360^{\\circ}\\right)=60^{\\circ}$.\n\nDrop a perpendicular from $A$ to $T$ on $O B$.\n\n\n\nThe area of $\\triangle P O A$ is $\\frac{1}{2}(O P)(A T)$.\n\n$\\triangle A O T$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nSince $A O=18$, then $A T=\\frac{\\sqrt{3}}{2}(A O)=9 \\sqrt{3}$.\n\nFor the area of $\\triangle P O A$ to equal $27 \\pi$, we have $\\frac{1}{2}(O P)(9 \\sqrt{3})=27 \\pi$ which gives $O P=\\frac{54 \\pi}{9 \\sqrt{3}}=\\frac{6 \\pi}{\\sqrt{3}}=2 \\sqrt{3} \\pi$.\n\n(Alternatively, we could have used the fact that the area of $\\triangle P O A$ is $\\frac{1}{2}(O A)(O P) \\sin (\\angle P O A)$.)']",['$2 \\sqrt{3} \\pi$'],False,,Numerical, 2579,Geometry,,"In the diagram, sector $A O B$ is $\frac{1}{6}$ of an entire circle with radius $A O=B O=18$. The sector is cut into two regions with a single straight cut through $A$ and point $P$ on $O B$. The areas of the two regions are equal. Determine the length of $O P$. ![](https://cdn.mathpix.com/cropped/2023_12_21_d6fc47bd20a6486a92ceg-1.jpg?height=273&width=288&top_left_y=1517&top_left_x=1271)","['Since sector $A O B$ is $\\frac{1}{6}$ of a circle with radius 18 , its area is $\\frac{1}{6}\\left(\\pi \\cdot 18^{2}\\right)$ or $54 \\pi$.\n\nFor the line $A P$ to divide this sector into two pieces of equal area, each piece has area $\\frac{1}{2}(54 \\pi)$ or $27 \\pi$.\n\nWe determine the length of $O P$ so that the area of $\\triangle P O A$ is $27 \\pi$.\n\nSince sector $A O B$ is $\\frac{1}{6}$ of a circle, then $\\angle A O B=\\frac{1}{6}\\left(360^{\\circ}\\right)=60^{\\circ}$.\n\nDrop a perpendicular from $A$ to $T$ on $O B$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4ac427a3237c2043180eg-1.jpg?height=268&width=276&top_left_y=1568&top_left_x=1033)\n\nThe area of $\\triangle P O A$ is $\\frac{1}{2}(O P)(A T)$.\n\n$\\triangle A O T$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle.\n\nSince $A O=18$, then $A T=\\frac{\\sqrt{3}}{2}(A O)=9 \\sqrt{3}$.\n\nFor the area of $\\triangle P O A$ to equal $27 \\pi$, we have $\\frac{1}{2}(O P)(9 \\sqrt{3})=27 \\pi$ which gives $O P=\\frac{54 \\pi}{9 \\sqrt{3}}=\\frac{6 \\pi}{\\sqrt{3}}=2 \\sqrt{3} \\pi$.\n\n(Alternatively, we could have used the fact that the area of $\\triangle P O A$ is $\\frac{1}{2}(O A)(O P) \\sin (\\angle P O A)$.)']",['$2 \\sqrt{3} \\pi$'],False,,Numerical, 2580,Number Theory,,For how many integers $k$ with $00$ and obtain the equivalent inequalities:\n\n$$\n\\begin{aligned}\n\\frac{5 \\sin \\theta-2}{\\sin ^{2} \\theta} & \\geq 2 \\\\\n5 \\sin \\theta-2 & \\geq 2 \\sin ^{2} \\theta \\\\\n0 & \\geq 2 \\sin ^{2} \\theta-5 \\sin \\theta+2 \\\\\n0 & \\geq(2 \\sin \\theta-1)(\\sin \\theta-2)\n\\end{aligned}\n$$\n\nSince $\\sin \\theta \\leq 1$, then $\\sin \\theta-2 \\leq-1<0$ for all $\\theta$.\n\nTherefore, $(2 \\sin \\theta-1)(\\sin \\theta-2) \\leq 0$ exactly when $2 \\sin \\theta-1 \\geq 0$.\n\nNote that $2 \\sin \\theta-1 \\geq 0$ exactly when $\\sin \\theta \\geq \\frac{1}{2}$.\n\nTherefore, the original inequality is true exactly when $\\frac{1}{2} \\leq \\sin \\theta \\leq 1$.\n\nNote that $\\sin 30^{\\circ}=\\sin 150^{\\circ}=\\frac{1}{2}$ and $0^{\\circ}<\\theta<180^{\\circ}$.\n\nWhen $\\theta=0^{\\circ}, \\sin \\theta=0$.\n\nFrom $\\theta=0^{\\circ}$ to $\\theta=30^{\\circ}, \\sin \\theta$ increases from 0 to $\\frac{1}{2}$.\n\nFrom $\\theta=30^{\\circ}$ to $\\theta=150^{\\circ}, \\sin \\theta$ increases from $\\frac{1}{2}$ to 1 and then decreases to $\\frac{1}{2}$.\n\nFrom $\\theta=150^{\\circ}$ to $\\theta=180^{\\circ}, \\sin \\theta$ decreases from $\\frac{1}{2}$ to 0 .\n\nTherefore, the original inequality is true exactly when $30^{\\circ} \\leq \\theta \\leq 150^{\\circ}$ which is equivalent to $30^{\\circ} \\leq 10 k^{\\circ} \\leq 150^{\\circ}$ and to $3 \\leq k \\leq 15$.\n\nThe integers $k$ in this range are $k=3,4,5,6, \\ldots, 12,13,14,15$, of which there are 13 .']",['13'],False,,Numerical, 2581,Algebra,,"In the diagram, a straight, flat road joins $A$ to $B$. ![](https://cdn.mathpix.com/cropped/2023_12_21_d6fc47bd20a6486a92ceg-1.jpg?height=78&width=575&top_left_y=2059&top_left_x=840) Karuna runs from $A$ to $B$, turns around instantly, and runs back to $A$. Karuna runs at $6 \mathrm{~m} / \mathrm{s}$. Starting at the same time as Karuna, Jorge runs from $B$ to $A$, turns around instantly, and runs back to $B$. Jorge runs from $B$ to $A$ at $5 \mathrm{~m} / \mathrm{s}$ and from $A$ to $B$ at $7.5 \mathrm{~m} / \mathrm{s}$. The distance from $A$ to $B$ is $297 \mathrm{~m}$ and each runner takes exactly $99 \mathrm{~s}$ to run their route. Determine the two values of $t$ for which Karuna and Jorge are at the same place on the road after running for $t$ seconds.","['Suppose that Karuna and Jorge meet for the first time after $t_{1}$ seconds and for the second time after $t_{2}$ seconds.\n\nWhen they meet for the first time, Karuna has run partway from $A$ to $B$ and Jorge has run partway from $B$ to $A$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_77d89371ac5d12b13590g-1.jpg?height=170&width=588&top_left_y=392&top_left_x=866)\n\nAt this instant, the sum of the distances that they have run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, she has run $6 t_{1} \\mathrm{~m}$.\n\nSince Jorge runs at $5 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, he has run $5 t_{1} \\mathrm{~m}$.\n\nTherefore, $6 t_{1}+5 t_{1}=297$ and so $11 t_{1}=297$ or $t_{1}=27$.\n\nWhen they meet for the second time, Karuna has run from $A$ to $B$ and is running back to $A$ and Jorge has run from $B$ to $A$ and is running back to $B$. This is because Jorge gets to $A$ halfway through his run before Karuna gets back to $A$ at the end of her run.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_77d89371ac5d12b13590g-1.jpg?height=236&width=588&top_left_y=1026&top_left_x=866)\n\nSince they each finish running after 99 seconds, then each has $99-t_{2}$ seconds left to run. At this instant, the sum of the distances that they have left to run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, she has to run $6\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nSince Jorge runs at $7.5 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, he has to run $7.5\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nTherefore, $6\\left(99-t_{2}\\right)+7.5\\left(99-t_{2}\\right)=297$ and so $13.5\\left(99-t_{2}\\right)=297$ or $99-t_{2}=22$ and so $t_{2}=77$.\n\nAlternatively, to calculate the value of $t_{2}$, we note that when Karuna and Jorge meet for the second time, they have each run the distance from $A$ to $B$ one full time and are on their return trips.\n\nThis means that they have each run the full distance from $A$ to $B$ once and the distances that they have run on their return trip add up to another full distance from $A$ to $B$, for a total distance of $3 \\cdot 297 \\mathrm{~m}=891 \\mathrm{~m}$.\n\nKaruna has run at $6 \\mathrm{~m} / \\mathrm{s}$ for $t_{2}$ seconds, for a total distance of $6 t_{2} \\mathrm{~m}$.\n\nJorge ran the first $297 \\mathrm{~m}$ at $5 \\mathrm{~m} / \\mathrm{s}$, which took $\\frac{297}{5} \\mathrm{~s}$ and ran the remaining $\\left(t_{2}-\\frac{297}{5}\\right)$ seconds at $7.5 \\mathrm{~m} / \\mathrm{s}$, for a total distance of $\\left(297+7.5\\left(t_{2}-\\frac{297}{5}\\right)\\right) \\mathrm{m}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n6 t_{2}+297+7.5\\left(t_{2}-\\frac{297}{5}\\right) & =891 \\\\\n13.5 t_{2} & =891-297+7.5 \\cdot \\frac{297}{5} \\\\\n13.5 t_{2} & =1039.5 \\\\\nt_{2} & =77\n\\end{aligned}\n$$\n\nTherefore, Karuna and Jorge meet after 27 seconds and after 77 seconds.']","['27, 77']",True,,Numerical, 2581,Algebra,,"In the diagram, a straight, flat road joins $A$ to $B$. Karuna runs from $A$ to $B$, turns around instantly, and runs back to $A$. Karuna runs at $6 \mathrm{~m} / \mathrm{s}$. Starting at the same time as Karuna, Jorge runs from $B$ to $A$, turns around instantly, and runs back to $B$. Jorge runs from $B$ to $A$ at $5 \mathrm{~m} / \mathrm{s}$ and from $A$ to $B$ at $7.5 \mathrm{~m} / \mathrm{s}$. The distance from $A$ to $B$ is $297 \mathrm{~m}$ and each runner takes exactly $99 \mathrm{~s}$ to run their route. Determine the two values of $t$ for which Karuna and Jorge are at the same place on the road after running for $t$ seconds.","['Suppose that Karuna and Jorge meet for the first time after $t_{1}$ seconds and for the second time after $t_{2}$ seconds.\n\nWhen they meet for the first time, Karuna has run partway from $A$ to $B$ and Jorge has run partway from $B$ to $A$.\n\n\n\nAt this instant, the sum of the distances that they have run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, she has run $6 t_{1} \\mathrm{~m}$.\n\nSince Jorge runs at $5 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, he has run $5 t_{1} \\mathrm{~m}$.\n\nTherefore, $6 t_{1}+5 t_{1}=297$ and so $11 t_{1}=297$ or $t_{1}=27$.\n\nWhen they meet for the second time, Karuna has run from $A$ to $B$ and is running back to $A$ and Jorge has run from $B$ to $A$ and is running back to $B$. This is because Jorge gets to $A$ halfway through his run before Karuna gets back to $A$ at the end of her run.\n\n\n\nSince they each finish running after 99 seconds, then each has $99-t_{2}$ seconds left to run. At this instant, the sum of the distances that they have left to run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, she has to run $6\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nSince Jorge runs at $7.5 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, he has to run $7.5\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nTherefore, $6\\left(99-t_{2}\\right)+7.5\\left(99-t_{2}\\right)=297$ and so $13.5\\left(99-t_{2}\\right)=297$ or $99-t_{2}=22$ and so $t_{2}=77$.\n\nAlternatively, to calculate the value of $t_{2}$, we note that when Karuna and Jorge meet for the second time, they have each run the distance from $A$ to $B$ one full time and are on their return trips.\n\nThis means that they have each run the full distance from $A$ to $B$ once and the distances that they have run on their return trip add up to another full distance from $A$ to $B$, for a total distance of $3 \\cdot 297 \\mathrm{~m}=891 \\mathrm{~m}$.\n\nKaruna has run at $6 \\mathrm{~m} / \\mathrm{s}$ for $t_{2}$ seconds, for a total distance of $6 t_{2} \\mathrm{~m}$.\n\nJorge ran the first $297 \\mathrm{~m}$ at $5 \\mathrm{~m} / \\mathrm{s}$, which took $\\frac{297}{5} \\mathrm{~s}$ and ran the remaining $\\left(t_{2}-\\frac{297}{5}\\right)$ seconds at $7.5 \\mathrm{~m} / \\mathrm{s}$, for a total distance of $\\left(297+7.5\\left(t_{2}-\\frac{297}{5}\\right)\\right) \\mathrm{m}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n6 t_{2}+297+7.5\\left(t_{2}-\\frac{297}{5}\\right) & =891 \\\\\n13.5 t_{2} & =891-297+7.5 \\cdot \\frac{297}{5} \\\\\n13.5 t_{2} & =1039.5 \\\\\nt_{2} & =77\n\\end{aligned}\n$$\n\nTherefore, Karuna and Jorge meet after 27 seconds and after 77 seconds.']","['27, 77']",True,,Numerical, 2582,Combinatorics,,"Eight people, including triplets Barry, Carrie and Mary, are going for a trip in four canoes. Each canoe seats two people. The eight people are to be randomly assigned to the four canoes in pairs. What is the probability that no two of Barry, Carrie and Mary will be in the same canoe?","['Among a group of $n$ people, there are $\\frac{n(n-1)}{2}$ ways of choosing a pair of these people:\n\nThere are $n$ people that can be chosen first.\n\nFor each of these $n$ people, there are $n-1$ people that can be chosen second.\n\nThis gives $n(n-1)$ orderings of two people.\n\nEach pair is counted twice (given two people A and B, we have counted both the\n\npair $\\mathrm{AB}$ and the pair $\\mathrm{BA})$, so the total number of pairs is $\\frac{n(n-1)}{2}$.\n\nWe label the four canoes W, X, Y, and Z.\n\nFirst, we determine the total number of ways to put the 8 people in the 4 canoes.\n\nWe choose 2 people to put in W. There are $\\frac{8 \\cdot 7}{2}$ pairs. This leaves 6 people for the remaining 3 canoes.\n\nNext, we choose 2 people to put in X. There are $\\frac{6 \\cdot 5}{2}$ pairs. This leaves 4 people for the remaining 2 canoes.\n\nNext, we choose 2 people to put in Y. There are $\\frac{4 \\cdot 3}{2}$ pairs. This leaves 2 people for the remaining canoe.\n\nThere is now 1 way to put the remaining people in $\\mathrm{Z}$.\n\nTherefore, there are\n\n$$\n\\frac{8 \\cdot 7}{2} \\cdot \\frac{6 \\cdot 5}{2} \\cdot \\frac{4 \\cdot 3}{2}=\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}{2^{3}}=7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3\n$$\n\nways to put the 8 people in the 4 canoes.\n\nNow, we determine the number of ways in which no two of Barry, Carrie and Mary will be in the same canoe.\n\nThere are 4 possible canoes in which Barry can go.\n\nThere are then 3 possible canoes in which Carrie can go, because she cannot go in the same canoe as Barry.\n\nThere are then 2 possible canoes in which Mary can go, because she cannot go in the same canoe as Barry or Carrie.\n\nThis leaves 5 people left to put in the canoes.\n\nThere are 5 choices of the person that can go with Barry, and then 4 choices of the person that can go with Carrie, and then 3 choices of the person that can go with Mary.\n\nThe remaining 2 people are put in the remaining empty canoe.\n\nThis means that there are $4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 \\cdot 3$ ways in which the 8 people can be put in 4 canoes so that no two of Barry, Carrie and Mary are in the same canoe.\n\nTherefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $\\frac{4 \\cdot 3 \\cdot 2 \\cdot 5 \\cdot 4 \\cdot 3}{7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3}=\\frac{4 \\cdot 3 \\cdot 2}{7 \\cdot 6}=\\frac{24}{42}=\\frac{4}{7}$.', 'Let $p$ be the probability that two of Barry, Carrie and Mary are in the same canoe.\n\nThe answer to the original problem will be $1-p$.\n\nLet $q$ be the probability that Barry and Carrie are in the same canoe.\n\nBy symmetry, the probability that Barry and Mary are in the same canoe also equals $q$ as does the probability that Carrie and Mary are in the same canoe.\n\nThis means that $p=3 q$.\n\nSo we calculate $q$.\n\nTo do this, we put Barry in a canoe. Since there are 7 possible people who can go in the canoe with him, then the probability that Carrie is in the canoe with him equals $\\frac{1}{7}$. The other 6 people can be put in the canoes in any way.\n\nThis means that the probability that Barry and Carrie are in the same canoe is $q=\\frac{1}{7}$.\n\nTherefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $1-3 \\cdot \\frac{1}{7}$ or $\\frac{4}{7}$.']",['$\\frac{4}{7}$'],False,,Numerical, 2583,Geometry,,Diagonal $W Y$ of square $W X Y Z$ has slope 2. Determine the sum of the slopes of $W X$ and $X Y$.,"['Suppose that $W Y$ makes an angle of $\\theta$ with the horizontal.\n\n\n\nSince the slope of $W Y$ is 2 , then $\\tan \\theta=2$, since the tangent of an angle equals the slope of a line that makes this angle with the horizontal.\n\nSince $\\tan \\theta=2>1=\\tan 45^{\\circ}$, then $\\theta>45^{\\circ}$.\n\nNow $W Y$ bisects $\\angle Z W X$, which is a right-angle.\n\nTherefore, $\\angle Z W Y=\\angle Y W X=45^{\\circ}$.\n\nTherefore, $W X$ makes an angle of $\\theta+45^{\\circ}$ with the horizontal and $W Z$ makes an angle of $\\theta-45^{\\circ}$ with the horizontal. Since $\\theta>45^{\\circ}$, then $\\theta-45^{\\circ}>0$ and $\\theta+45^{\\circ}>90^{\\circ}$.\n\nWe note that since $W Z$ and $X Y$ are parallel, then the slope of $X Y$ equals the slope of $W Z$.\n\nTo calculate the slopes of $W X$ and $W Z$, we can calculate $\\tan \\left(\\theta+45^{\\circ}\\right)$ and $\\tan \\left(\\theta-45^{\\circ}\\right)$.\n\nUsing the facts that $\\tan (A+B)=\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}$ and $\\tan (A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A \\tan B}$, we obtain:\n\n$$\n\\begin{aligned}\n& \\tan \\left(\\theta+45^{\\circ}\\right)=\\frac{\\tan \\theta+\\tan 45^{\\circ}}{1-\\tan \\theta \\tan 45^{\\circ}}=\\frac{2+1}{1-(2)(1)}=-3 \\\\\n& \\tan \\left(\\theta-45^{\\circ}\\right)=\\frac{\\tan \\theta-\\tan 45^{\\circ}}{1-\\tan \\theta \\tan 45^{\\circ}}=\\frac{2-1}{1+(2)(1)}=\\frac{1}{3}\n\\end{aligned}\n$$\n\nTherefore, the sum of the slopes of $W X$ and $X Y$ is $-3+\\frac{1}{3}=-\\frac{8}{3}$.', 'Consider a square $W X Y Z$ whose diagonal $W Y$ has slope 2 .\n\nTranslate this square so that $W$ is at the origin $(0,0)$. Translating a shape in the plane does not affect the slopes of any line segments.\n\nLet the coordinates of $Y$ be $(2 a, 2 b)$ for some non-zero numbers $a$ and $b$.\n\nSince the slope of $W Y$ is 2 , then $\\frac{2 b-0}{2 a-0}=2$ and so $2 b=4 a$ or $b=2 a$.\n\nThus, the coordinates of $Y$ can be written as $(2 a, 4 a)$.\n\nLet $C$ be the centre of square $W X Y Z$.\n\nThen $C$ is the midpoint of $W Y$, so $C$ has coordinates $(a, 2 a)$.\n\nWe find the slopes of $W X$ and $X Y$ by finding the coordinates of $X$.\n\nConsider the segment $X C$.\n\nSince the diagonals of a square are perpendicular, then $X C$ is perpendicular to $W C$.\n\nSince the slope of $W C$ is 2 , then the slopes of $X C$ and $Z C$ are $-\\frac{1}{2}$.\n\nSince the diagonals of a square are equal in length and $C$ is the midpoint of both diagonals, then $X C=W C$.\n\nSince $W C$ and $X C$ are perpendicular and equal in length, then the ""rise/run triangle"" above $X C$ will be a $90^{\\circ}$ rotation of the ""rise/run triangle"" below $W C$.\n\n\n\nThis is because these triangles are congruent (each is right-angled, their hypotenuses are of equal length, and their remaining angles are equal) and their hypotenuses are perpendicular.\n\nIn this diagram, we have assumed that $X$ is to the left of $W$ and $Z$ is to the right of $W$. Since the slopes of parallel sides are equal, it does not matter which vertex is labelled $X$ and which is labelled $Z$. We would obtain the same two slopes, but in a different order. To get from $W(0,0)$ to $C(a, 2 a)$, we go up $2 a$ and right $a$.\n\nThus, to get from $C(a, 2 a)$ to $X$, we go left $2 a$ and up $a$.\n\nTherefore, the coordinates of $X$ are $(a-2 a, 2 a+a)$ or $(-a, 3 a)$.\n\nThus, the slope of $W X$ is $\\frac{3 a-0}{-a-0}=-3$.\n\nSince $X Y$ is perpendicular to $W X$, then its slope is the negative reciprocal of -3 , which is $\\frac{1}{3}$.\n\nThe sum of the slopes of $W X$ and $X Y$ is $-3+\\frac{1}{3}=-\\frac{8}{3}$.']",['$-\\frac{8}{3}$'],False,,Numerical, 2584,Algebra,,Determine all values of $x$ such that $\log _{2 x}(48 \sqrt[3]{3})=\log _{3 x}(162 \sqrt[3]{2})$.,"['Since the base of a logarithm must be positive and cannot equal 1 , then $x>0$ and $x \\neq \\frac{1}{2}$ and $x \\neq \\frac{1}{3}$.\n\nThis tells us that $\\log 2 x$ and $\\log 3 x$ exist and do not equal 0 , which we will need shortly when we apply the change of base formula.\n\nWe note further that $48=2^{4} \\cdot 3$ and $162=3^{4} \\cdot 2$ and $\\sqrt[3]{3}=3^{1 / 3}$ and $\\sqrt[3]{2}=2^{1 / 3}$. Using logarithm rules, the following equations are equivalent:\n\n$$\n\\begin{aligned}\n\\log _{2 x}(48 \\sqrt[3]{3}) & =\\log _{3 x}(162 \\sqrt[3]{2}) \\\\\n\\frac{\\log \\left(2^{4} \\cdot 3 \\cdot 3^{1 / 3}\\right)}{\\log 2 x} & =\\frac{\\log \\left(3^{4} \\cdot 2 \\cdot 2^{1 / 3}\\right)}{\\log 3 x} \\quad \\text { (change of base formula) } \\\\\n\\frac{\\log \\left(2^{4} \\cdot 3^{4 / 3}\\right)}{\\log 2+\\log x} & =\\frac{\\log \\left(3^{4} \\cdot 2^{4 / 3}\\right)}{\\log 3+\\log x} \\quad(\\log a b=\\log a+\\log b) \\\\\n\\frac{\\log \\left(2^{4}\\right)+\\log \\left(3^{4 / 3}\\right)}{\\log 2+\\log x} & =\\frac{\\log \\left(3^{4}\\right)+\\log \\left(2^{4 / 3}\\right)}{\\log 3+\\log x} \\quad(\\log a b=\\log a+\\log b) \\\\\n\\frac{4 \\log 2+\\frac{4}{3} \\log 3}{\\log 2+\\log x} & =\\frac{4 \\log 3+\\frac{4}{3} \\log 2}{\\log 3+\\log x} \\quad\\left(\\log \\left(a^{c}\\right)=c \\log a\\right)\n\\end{aligned}\n$$\n\nCross-multiplying, we obtain\n\n$$\n\\left(4 \\log 2+\\frac{4}{3} \\log 3\\right)(\\log 3+\\log x)=\\left(4 \\log 3+\\frac{4}{3} \\log 2\\right)(\\log 2+\\log x)\n$$\n\nExpanding the left side, we obtain\n\n$$\n4 \\log 2 \\log 3+\\frac{4}{3}(\\log 3)^{2}+\\left(4 \\log 2+\\frac{4}{3} \\log 3\\right) \\log x\n$$\n\nExpanding the right side, we obtain\n\n$$\n4 \\log 3 \\log 2+\\frac{4}{3}(\\log 2)^{2}+\\left(4 \\log 3+\\frac{4}{3} \\log 2\\right) \\log x\n$$\n\nSimplifying and factoring, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\frac{4}{3}(\\log 3)^{2}-\\frac{4}{3}(\\log 2)^{2} & =\\log x\\left(4 \\log 3+\\frac{4}{3} \\log 2-4 \\log 2-\\frac{4}{3} \\log 3\\right) \\\\\n\\frac{4}{3}(\\log 3)^{2}-\\frac{4}{3}(\\log 2)^{2} & =\\log x\\left(\\frac{8}{3} \\log 3-\\frac{8}{3} \\log 2\\right) \\\\\n(\\log 3)^{2}-(\\log 2)^{2} & =2 \\log x(\\log 3-\\log 2) \\\\\n\\log x & =\\frac{(\\log 3)^{2}-(\\log 2)^{2}}{2(\\log 3-\\log 2)} \\\\\n\\log x & =\\frac{(\\log 3-\\log 2)(\\log 3+\\log 2)}{2(\\log 3-\\log 2)} \\\\\n\\log x & =\\frac{\\log 3+\\log 2}{2} \\\\\n\\log x & =\\frac{1}{2} \\log 6 \\\\\n\\log x & =\\log (\\sqrt{6})\n\\end{aligned}\n$$\n\nand so $x=\\sqrt{6}$.']",['$\\sqrt{6}$'],False,,Numerical, 2585,Geometry,,"In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$. ![](https://cdn.mathpix.com/cropped/2023_12_21_96cde5457e2677371bf4g-1.jpg?height=448&width=874&top_left_y=909&top_left_x=690) If $A B=718$ and $P Q=250$, determine the length of $B C$.","['Let $B C=x, P B=b$, and $B Q=a$.\n\nSince $B C=x$, then $A D=P S=Q R=x$.\n\nSince $B C=x$ and $B Q=a$, then $Q C=x-a$.\n\nSince $A B=718$ and $P B=b$, then $A P=718-b$.\n\nNote that $P Q=S R=250$.\n\nLet $\\angle B Q P=\\theta$.\n\nSince $\\triangle P B Q$ is right-angled at $B$, then $\\angle B P Q=90^{\\circ}-\\theta$.\n\nSince $B Q C$ is a straight angle and $\\angle P Q R=90^{\\circ}$, then $\\angle R Q C=180^{\\circ}-90^{\\circ}-\\theta=90^{\\circ}-\\theta$.\n\nSince $A P B$ is a straight angle and $\\angle S P Q=90^{\\circ}$, then $\\angle A P S=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nSince $\\triangle S A P$ and $\\triangle Q C R$ are each right-angled and have another angle in common with $\\triangle P B Q$, then these\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_4b861614b923867311b5g-1.jpg?height=634&width=659&top_left_y=206&top_left_x=1362)\nthree triangles are similar.\n\nContinuing in the same way, we can show that $\\triangle R D S$ is also similar to these three triangles.\n\nSince $R S=P Q$, then $\\triangle R D S$ is actually congruent to $\\triangle P B Q$ (angle-side-angle).\n\nSimilarly, $\\triangle S A P$ is congruent to $\\triangle Q C R$.\n\nIn particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.\n\nSince $\\triangle S A P$ and $\\triangle P B Q$ are similar, then $\\frac{S A}{P B}=\\frac{A P}{B Q}=\\frac{S P}{P Q}$.\n\nThus, $\\frac{x-a}{b}=\\frac{718-b}{a}=\\frac{x}{250}$.\n\nAlso, by the Pythagorean Theorem in $\\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.\n\nBy the Pythagorean Theorem in $\\triangle S A P$,\n\n$$\n\\begin{aligned}\nx^{2} & =(x-a)^{2}+(718-b)^{2} \\\\\nx^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\\\\n0 & =-2 a x+a^{2}+(718-b)^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.\n\nSince $\\frac{718-b}{a}=\\frac{x}{250}$, then $a x=250(718-b)$.\n\nTherefore, substituting into $(*)$, we obtain\n\n$$\n\\begin{aligned}\n0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\\\\nb^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\\\\nb^{2} & =((718-b)-250)^{2} \\quad\\left(\\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\\right) \\\\\nb^{2} & =(468-b)^{2} \\\\\nb & =468-b \\quad(\\text { since } b \\neq b-468) \\\\\n2 b & =468 \\\\\nb & =234\n\\end{aligned}\n$$\n\nTherefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \\cdot 16=22^{2} \\cdot 4^{2}=88^{2}$ and so $a=88$.\n\nFinally, $x=\\frac{250(718-b)}{a}=\\frac{250 \\cdot 484}{88}=1375$. Therefore, $B C=1375$.']",['1375'],False,,Numerical, 2585,Geometry,,"In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$. If $A B=718$ and $P Q=250$, determine the length of $B C$.","['Let $B C=x, P B=b$, and $B Q=a$.\n\nSince $B C=x$, then $A D=P S=Q R=x$.\n\nSince $B C=x$ and $B Q=a$, then $Q C=x-a$.\n\nSince $A B=718$ and $P B=b$, then $A P=718-b$.\n\nNote that $P Q=S R=250$.\n\nLet $\\angle B Q P=\\theta$.\n\nSince $\\triangle P B Q$ is right-angled at $B$, then $\\angle B P Q=90^{\\circ}-\\theta$.\n\nSince $B Q C$ is a straight angle and $\\angle P Q R=90^{\\circ}$, then $\\angle R Q C=180^{\\circ}-90^{\\circ}-\\theta=90^{\\circ}-\\theta$.\n\nSince $A P B$ is a straight angle and $\\angle S P Q=90^{\\circ}$, then $\\angle A P S=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nSince $\\triangle S A P$ and $\\triangle Q C R$ are each right-angled and have another angle in common with $\\triangle P B Q$, then these\n\n\nthree triangles are similar.\n\nContinuing in the same way, we can show that $\\triangle R D S$ is also similar to these three triangles.\n\nSince $R S=P Q$, then $\\triangle R D S$ is actually congruent to $\\triangle P B Q$ (angle-side-angle).\n\nSimilarly, $\\triangle S A P$ is congruent to $\\triangle Q C R$.\n\nIn particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.\n\nSince $\\triangle S A P$ and $\\triangle P B Q$ are similar, then $\\frac{S A}{P B}=\\frac{A P}{B Q}=\\frac{S P}{P Q}$.\n\nThus, $\\frac{x-a}{b}=\\frac{718-b}{a}=\\frac{x}{250}$.\n\nAlso, by the Pythagorean Theorem in $\\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.\n\nBy the Pythagorean Theorem in $\\triangle S A P$,\n\n$$\n\\begin{aligned}\nx^{2} & =(x-a)^{2}+(718-b)^{2} \\\\\nx^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\\\\n0 & =-2 a x+a^{2}+(718-b)^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.\n\nSince $\\frac{718-b}{a}=\\frac{x}{250}$, then $a x=250(718-b)$.\n\nTherefore, substituting into $(*)$, we obtain\n\n$$\n\\begin{aligned}\n0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\\\\nb^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\\\\nb^{2} & =((718-b)-250)^{2} \\quad\\left(\\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\\right) \\\\\nb^{2} & =(468-b)^{2} \\\\\nb & =468-b \\quad(\\text { since } b \\neq b-468) \\\\\n2 b & =468 \\\\\nb & =234\n\\end{aligned}\n$$\n\nTherefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \\cdot 16=22^{2} \\cdot 4^{2}=88^{2}$ and so $a=88$.\n\nFinally, $x=\\frac{250(718-b)}{a}=\\frac{250 \\cdot 484}{88}=1375$. Therefore, $B C=1375$.']",['1375'],False,,Numerical, 2586,Number Theory,,"An L-shaped triomino is composed of three unit squares, as shown: ![](https://cdn.mathpix.com/cropped/2023_12_21_96cde5457e2677371bf4g-1.jpg?height=130&width=146&top_left_y=1553&top_left_x=1518) Suppose that $H$ and $W$ are positive integers. An $H \times W$ rectangle can be tiled if the rectangle can be completely covered with non-overlapping copies of this triomino (each of which can be rotated and/or translated) and the sum of the areas of these non-overlapping triominos equals the area of the rectangle (that is, no triomino is partly outside the rectangle). If such a rectangle can be tiled, a tiling is a specific configuration of triominos that tile the rectangle. Determine, with justification, all integers $W$ for which a $6 \times W$ rectangle can be tiled.","['First, we note that it is possible to tile each of a $3 \\times 2$ and a $2 \\times 3$ rectangle:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3d1f6c0f030c1ddc8175g-1.jpg?height=208&width=349&top_left_y=590&top_left_x=991)\n\nNext, we note that it is not possible to tile a $6 \\times 1$ rectangle because each of the triominos needs a width of at least 2 to be placed.\n\nFinally, we show that it is possible to tile a $6 \\times W$ rectangle for every integer $W \\geq 2$.\n\nTo do this, we show that such a $6 \\times W$ rectangle can be made up from $3 \\times 2$ and $2 \\times 3$ rectangles. Since each of these types of rectangles can be tiled with triominos, then the larger rectangle can be tiled with triominos by combining these tilings.\n\nCase 1: $W$ is even\n\nSuppose that $W=2 k$ for some positive integer $k$.\n\nWe build a $6 \\times 2 k$ rectangle by placing $k 6 \\times 2$ rectangles side by side.\n\nEach $6 \\times 2$ rectangle is built by stacking two $3 \\times 2$ rectangles on top of each other.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3d1f6c0f030c1ddc8175g-1.jpg?height=409&width=525&top_left_y=1357&top_left_x=903)\n\nTherefore, each such rectangle can be tiled.\n\nCase 2: $W$ is odd, $W \\geq 3$\n\nSuppose that $W=2 k+1$ for some positive integer $k$.\n\nWe build a $6 \\times(2 k+1)$ rectangle by building a $6 \\times 3$ rectangle and then putting $k-1$ $6 \\times 2$ rectangles next to it. Note that $k-1 \\geq 0$ since $k \\geq 1$ and that $2 k+1=3+2(k-1)$. The $6 \\times 3$ rectangle is built by stacking three $2 \\times 3$ rectangles on top of each other. Each $6 \\times 2$ rectangle is built by stacking two $3 \\times 2$ rectangles on top of each other.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_3d1f6c0f030c1ddc8175g-1.jpg?height=415&width=697&top_left_y=2216&top_left_x=817)\n\n\n\nTherefore, each such rectangle can be tiled.\n\nThus, a $6 \\times W$ rectangle can be tiled with triominos exactly when $W \\geq 2$.']",['integer W when $W \\geq 2$'],False,,Need_human_evaluate, 2587,Number Theory,,"An L-shaped triomino is composed of three unit squares, as shown: ![](https://cdn.mathpix.com/cropped/2023_12_21_96cde5457e2677371bf4g-1.jpg?height=130&width=146&top_left_y=1553&top_left_x=1518) Suppose that $H$ and $W$ are positive integers. An $H \times W$ rectangle can be tiled if the rectangle can be completely covered with non-overlapping copies of this triomino (each of which can be rotated and/or translated) and the sum of the areas of these non-overlapping triominos equals the area of the rectangle (that is, no triomino is partly outside the rectangle). If such a rectangle can be tiled, a tiling is a specific configuration of triominos that tile the rectangle. Determine, with justification, all pairs $(H, W)$ of integers with $H \geq 4$ and $W \geq 4$ for which an $H \times W$ rectangle can be tiled.","['Suppose that $(H, W)$ is a pair of integers with $H \\geq 4$ and $W \\geq 4$.\n\nSince the area of each triomino is 3 , then the area of any rectangle that can be tiled must be a multiple of 3 since it is completely covered by triominos with area 3 .\n\nSince the area of an $H \\times W$ rectangle is $H W$, then we need $H W$ to be a multiple of 3 , which means that at least one of $H$ and $W$ is a multiple of 3.\n\nSince a rectangle that is $a \\times b$ can be tiled if and only if a rectangle that is $b \\times a$ can be tiled (we see this by rotating the tilings by $90^{\\circ}$ as we did with the $3 \\times 2$ and $2 \\times 3$ rectangles above), then we may assume without loss of generality that $H$ is divisible by 3 .\n\nWe show that if $H$ is divisible by 3 , then every $H \\times W$ rectangle with $H \\geq 4$ and $W \\geq 4$ can be tiled.\n\nCase 1: $H$ is divisible by $3, H$ is even\n\nHere, $H$ is a multiple of 6 , say $H=6 m$ for some positive integer $m$.\n\nSince $W \\geq 4$, we know that a $6 \\times W$ rectangle can be tiled.\n\nBy stacking $m 6 \\times W$ rectangles on top of each other, we obtain a $6 m \\times W$ rectangle.\n\nSince each $6 \\times W$ rectangle can be tiled, then the $6 m \\times W$ rectangle can be tiled.\n\nCase 2: $H$ is divisible by $3, H$ is odd, $W$ is even\n\nSuppose that $H=3 q$ for some odd positive integer $q$ and $W=2 r$ for some positive integer $r$.\n\nTo tile a $3 q \\times 2 r$ rectangle, we combine $q r 3 \\times 2$ rectangles in $q$ rows and $r$ columns:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_742b0b713146d818e937g-1.jpg?height=258&width=219&top_left_y=1324&top_left_x=1056)\n\nTherefore, every rectangle in this case can be tiled. (Note that in this case the fact that $q$ was odd was not important.)\n\nCase 3: $H$ is divisible by $3, H$ is odd, $W$ is odd\n\nSince $H \\geq 4$ and $W \\geq 4$, the rectangle with the smallest values of $H$ and $W$ is $9 \\times 5$ which can be tiled as shown:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_742b0b713146d818e937g-1.jpg?height=559&width=330&top_left_y=1886&top_left_x=995)\n\n(There are also other ways to tile this rectangle.)\n\nSince $H$ is an odd multiple of 3 and $H \\geq 4$, we can write $H=9+6 s$ for some integer $s \\geq 0$.\n\n\n\nSince $W$ is odd and $W \\geq 5$, we can write $W=5+2 t$ for some integer $t \\geq 0$.\n\nThus, the $H \\times W$ rectangle is $(9+6 s) \\times(5+2 t)$.\n\nWe break this rectangle into three rectangles - one that is $9 \\times 5$, one that is $9 \\times 2 t$, and one that is $6 s \\times W$ \n\n(If $s=0$ or $t=0$, there will be fewer than three rectangles.)\n\nThe $9 \\times 5$ rectangle can be tiled as we showed earlier.\n\nIf $t>0$, the $9 \\times 2 t$ rectangle can be tiled as seen in Case 2 .\n\nIf $s>0$, the $6 s \\times W$ rectangle can be tiled as seen in Case 1 .\n\nTherefore, the $H \\times W$ rectangle can be tiled.\n\nThrough these three cases, we have shown that any $H \\times W$ rectangle with $H \\geq 4$ and $W \\geq 4$ can be tiled when $H$ is a multiple of 3 .\n\nSince the roles of $H$ and $W$ can be interchanged and since at least one of $H$ and $W$ must be a multiple of 3 , then an $H \\times W$ rectangle with $H \\geq 4$ and $W \\geq 4$ can be tiled exactly when at least one of $H$ and $W$ is a multiple of 3 .']",['exactly when at least one of $H$ and $W$ is a multiple of 3'],False,,Need_human_evaluate, 2588,Algebra,,"In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown: | $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ | For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it. In one such array, $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$. Determine the value of $A_{4}$.","['We draw part of the array using the information that $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$ :\n\n$$\n\\begin{array}{l|l|l|l|l|l|l|lll|c|c|c|c|c|c}\n\\cdots & A_{0} & A_{1} & A_{2} & A_{3} & A_{4} & A_{5} & \\cdots & \\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \\cdots \\\\\n\\hline \\cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \\cdots & \\cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \\cdots\n\\end{array}\n$$\n\nSince $A_{1}$ is the average of $A_{0}, B_{1}$ and $A_{2}$, then $A_{1}=\\frac{A_{0}+B_{1}+A_{2}}{3}$ or $3 A_{1}=A_{0}+B_{1}+A_{2}$. Thus, $3(0)=0+B_{1}+0$ and so $B_{1}=0$.\n\nSince $A_{2}$ is the average of $A_{1}, B_{2}$ and $A_{3}$, then $3 A_{2}=A_{1}+B_{2}+A_{3}$ and so $3(0)=0+B_{2}+1$ which gives $B_{2}=-1$.\n\nSince $B_{2}$ is the average of $B_{1}, A_{2}$ and $B_{3}$, then $3 B_{2}=B_{1}+A_{2}+B_{3}$ and so $3(-1)=0+0+B_{3}$ which gives $B_{3}=-3$.\n\nSo far, this gives\n\n$$\n\\begin{array}{l|c|c|c|c|c|c|l}\n\\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \\cdots \\\\\n\\hline \\cdots & B_{0} & 0 & -1 & -3 & B_{4} & B_{5} & \\cdots\n\\end{array}\n$$\n\nSince $A_{3}$ is the average of $A_{2}, B_{3}$ and $A_{4}$, then $3 A_{3}=A_{2}+B_{3}+A_{4}$ and so $3(1)=$ $0+(-3)+A_{4}$ which gives $A_{4}=6$.']",['6'],False,,Numerical, 2589,Algebra,,"In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown: | $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ | For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it. In another such array, we define $S_{k}=A_{k}+B_{k}$ for each integer $k$. Prove that $S_{k+1}=2 S_{k}-S_{k-1}$ for each integer $k$.","['We draw part of the array:\n\n| $\\cdots$ | $A_{k-1}$ | $A_{k}$ | $A_{k+1}$ | $\\cdots$ |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\cdots$ | $B_{k-1}$ | $B_{k}$ | $B_{k+1}$ | $\\cdots$ |\n\nThen\n\n$$\n\\begin{aligned}\n3 S_{k} & =3 A_{k}+3 B_{k} \\\\\n& =3\\left(\\frac{A_{k-1}+B_{k}+A_{k+1}}{3}\\right)+3\\left(\\frac{B_{k-1}+A_{k}+B_{k+1}}{3}\\right) \\\\\n& =A_{k-1}+B_{k}+A_{k+1}+B_{k-1}+A_{k}+B_{k+1} \\\\\n& =\\left(A_{k-1}+B_{k-1}\\right)+\\left(A_{k}+B_{k}\\right)+\\left(A_{k+1}+B_{k+1}\\right) \\\\\n& =S_{k-1}+S_{k}+S_{k+1}\n\\end{aligned}\n$$\n\nSince $3 S_{k}=S_{k-1}+S_{k}+S_{k+1}$, then $S_{k+1}=2 S_{k}-S_{k-1}$.']",,True,,, 2590,Algebra,,"In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown: | $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ | For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it. Consider the following two statements about a third such array: (P) If each entry is a positive integer, then all of the entries in the array are equal. (Q) If each entry is a positive real number, then all of the entries in the array are equal. Prove statement (Q).","['First, we give a proof for lemma (b): we define $S_{k}=A_{k}+B_{k}$ for each integer $k$.Prove that $S_{k+1}=2 S_{k}-S_{k-1}$ for each integer $k$.\n\nProof for lemma (b):\nWe draw part of the array:\n\n| $\\cdots$ | $A_{k-1}$ | $A_{k}$ | $A_{k+1}$ | $\\cdots$ |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\cdots$ | $B_{k-1}$ | $B_{k}$ | $B_{k+1}$ | $\\cdots$ |\n\nThen\n\n$$\n\\begin{aligned}\n3 S_{k} & =3 A_{k}+3 B_{k} \\\\\n& =3\\left(\\frac{A_{k-1}+B_{k}+A_{k+1}}{3}\\right)+3\\left(\\frac{B_{k-1}+A_{k}+B_{k+1}}{3}\\right) \\\\\n& =A_{k-1}+B_{k}+A_{k+1}+B_{k-1}+A_{k}+B_{k+1} \\\\\n& =\\left(A_{k-1}+B_{k-1}\\right)+\\left(A_{k}+B_{k}\\right)+\\left(A_{k+1}+B_{k+1}\\right) \\\\\n& =S_{k-1}+S_{k}+S_{k+1}\n\\end{aligned}\n$$\n\nSince $3 S_{k}=S_{k-1}+S_{k}+S_{k+1}$, then $S_{k+1}=2 S_{k}-S_{k-1}$, which finish the proof of lemma (b).\n\n\nProof of statement (P)\n\nSuppose that all of the entries in the array are positive integers.\n\nAssume that not all of the entries in the array are equal.\n\nSince all of the entries are positive integers, there must be a minimum entry. Let $m$ be the minimum of all of the entries in the array.\n\nChoose an entry in the array equal to $m$, say $A_{r}=m$ for some integer $r$. The same argument can be applied with $B_{r}=m$ if there are no entries equal to $m$ in the top row.\n\nIf not all of the entries $A_{j}$ are equal to $m$, then by moving one direction or the other along the row we will get to some point where $A_{t}=m$ for some integer $t$ but one of its neighbours is not equal to $m$. (If this were not to happen, then all of the entries in both directions would be equal to $m$.)\n\nIf all of the entries $A_{j}$ are equal to $m$, then since not all of the entries in the array are equal to $m$, then there will be an entry $B_{t}$ which is not equal to $m$.\n\nIn other words, since not all of the entries in the array are equal, then there exists an integer $t$ for which $A_{t}=m$ and not all of $A_{t-1}, A_{t+1}, B_{t}$ are equal to $m$.\n\nBut $3 m=3 A_{t}$ and $3 A_{t}=A_{t-1}+B_{t}+A_{t+1}$ so $3 m=A_{t-1}+B_{t}+A_{t+1}$.\n\nSince not all of $A_{t-1}, B_{t}$ and $A_{t+1}$ are equal to $m$ and each is at least $m$, then one of these entries will be greater than $m$.\n\nThis means that $A_{t-1}+B_{t}+A_{t+1} \\geq m+m+(m+1)=3 m+1>3 m$, which is a contradiction.\n\nTherefore our assumption that not all of the entries are equal must be false, which means that all of the entries are equal, which proves statement $(\\mathrm{P})$.\n\nProof of statement (Q)\n\nSuppose that all of the entries are positive real numbers.\n\nAssume that not all of the entries in the array are equal.\n\nAs in (b), define $S_{k}=A_{k}+B_{k}$ for each integer $k$.\n\nAlso, define $D_{k}=A_{k}-B_{k}$ for each integer $k$.\n\nStep 1: Prove that the numbers $S_{k}$ form an arithmetic sequence\n\nFrom (b), $S_{k+1}=2 S_{k}-S_{k-1}$.\n\nRe-arranging, we see $S_{k+1}-S_{k}=S_{k}-S_{k-1}$ for each integer $k$, which means that the differences between consecutive pairs of terms are equal.\n\nSince this is true for all integers $k$, then the difference between each pair of consecutive\n\n\n\nterms through the whole sequence is constant, which means that the sequence is an arithmetic sequence.\n\nStep 2: Prove that $S_{k}$ is constant\n\nSuppose that $S_{0}=c$. Since $A_{0}>0$ and $B_{0}>0$, then $S_{0}=c>0$.\n\nSince the terms $S_{k}$ form an arithmetic sequence, then the sequence is either constant, increasing or decreasing.\n\nIf the sequence of terms $S_{k}$ is increasing, then the common difference $d=S_{1}-S_{0}$ is positive.\n\nNote that $S_{-1}=c-d, S_{-2}=c-2 d$, and so on.\n\nSince $c$ and $d$ are constant, then if we move far enough back along the sequence, eventually $S_{t}$ will be negative for some integer $t$. This is a contradiction since $A_{t}>0$ and $B_{t}>0$ and $S_{t}=A_{t}+B_{t}$.\n\nThus, the sequence cannot be increasing.\n\nIf the sequence of terms $S_{k}$ is decreasing, then the common difference $d=S_{1}-S_{0}$ is negative.\n\nNote that $S_{1}=c+d, S_{2}=c+2 d$, and so on.\n\nSince $c$ and $d$ are constant, then if we move far along the sequence, eventually $S_{t}$ will be negative for some integer $t$. This is also a contradiction since $A_{t}>0$ and $B_{t}>0$ and $S_{t}=A_{t}+B_{t}$.\n\nThus, the sequence cannot be decreasing.\n\nTherefore, since all of the entries are positive and the sequence $S_{k}$ is arithmetic, then $S_{k}$ is constant, say $S_{k}=c>0$ for all integers $k$.\n\nStep 3: Determine range of possible values for $D_{k}$\n\nWe note that $S_{k}=A_{k}+B_{k}=c$ for all integers $k$ and $A_{k}>0$ and $B_{k}>0$.\n\nSince $A_{k}>0$, then $B_{k}=S_{k}-A_{k}=c-A_{k}0-c=-c$.\n\nIn other words, $-c0$ or $D_{0}<0$.\n\nWe may assume that $D_{0}>0$. (If $D_{0}<0$, then we can switch the bottom and top rows of the array so that $D_{0}$ becomes positive.)\n\nSuppose that $D_{1} \\geq D_{0}>0$.\n\nThen $D_{2}=4 D_{1}-D_{0} \\geq 4 D_{1}-D_{1}=3 D_{1}$. Since $D_{1}>0$, this also means that $D_{2}>D_{1}>0$.\n\nSimilarly, $D_{3}=4 D_{2}-D_{1} \\geq 4 D_{2}-D_{2}=3 D_{2}>D_{2}>0$. Since $D_{2} \\geq 3 D_{1}$, then $D_{3} \\geq 9 D_{1}$. Continuing in this way, we see that $D_{4} \\geq 27 D_{1}$ and $D_{5} \\geq 81 D_{1}$ and so on, with $D_{k} \\geq 3^{k-1} D_{1}$ for each positive integer $k \\geq 2$. Since the value of $D_{1}$ is a fixed positive real number and $D_{k}4 D_{0}-D_{0}=3 D_{0}>D_{0}>0$.\n\nExtending this using a similar method, we see that $D_{-j}>3^{j} D_{0}$ for all positive integers $j$ which will lead to the same contradiction as above.\n\nTherefore, a contradiction is obtained in all cases and so it cannot be the case that $D_{k} \\neq 0$ for some integer $k$.\n\nSince $D_{k}=0$ and $S_{k}=c$ for all integers $k$, then $A_{k}=B_{k}=\\frac{1}{2} c$ for all integers $k$, which means that all entries in the array are equal.']",,True,,, 2591,Geometry,,"How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ? ","['If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.\n\n', 'This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area.']",['100'],False,,Numerical, 2591,Geometry,,"How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_2778c1b7df4d6803a6f0g-1.jpg?height=317&width=331&top_left_y=1988&top_left_x=1363)","['If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_7c9cbbc4456b37ad4aa9g-1.jpg?height=358&width=377&top_left_y=217&top_left_x=1297)', 'This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area.']",['100'],False,,Numerical, 2592,Algebra,,"The populations of Alphaville and Betaville were equal at the end of 1995. The population of Alphaville decreased by $2.9 \%$ during 1996, then increased by $8.9 \%$ during 1997 , and then increased by $6.9 \%$ during 1998 . The population of Betaville increased by $r \%$ in each of the three years. If the populations of the towns are equal at the end of 1998, determine the value of $r$ correct to one decimal place.","['If $P$ is the original population of Alphaville and Betaville,\n\n$$\n\\begin{aligned}\nP(.971)(1.089)(1.069) & =P\\left(1+\\frac{r}{100}\\right)^{3} \\\\\n1.1303 & =\\left(1+\\frac{r}{100}\\right)^{3}\n\\end{aligned}\n$$\n\nFrom here,\n\nPossibility 1\n\n$$\n\\begin{aligned}\n1+\\frac{r}{100} & =(1.1303)^{\\frac{1}{3}} \\\\\n1+\\frac{r}{100} & =1.0416 \\\\\nr & \\doteq 4.2 \\%\n\\end{aligned}\n$$\n\nOr, Possibility 2\n\n$$\n\\begin{aligned}\n3 \\log \\left(1+\\frac{r}{100}\\right) & =\\log 1.1303 \\\\\n\\log \\left(1+\\frac{r}{100}\\right) & =.01773 \\\\\n1+\\frac{r}{100} & =1.0416 \\\\\nr & \\doteq 4.2 \\%\n\\end{aligned}\n$$']",['4.2'],False,%,Numerical,1e-1 2593,Geometry,,"A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$ ![](https://cdn.mathpix.com/cropped/2023_12_21_609033623119f30122a6g-1.jpg?height=474&width=526&top_left_y=782&top_left_x=1293)","['If $P Q=\\frac{\\pi}{3}$, then by symmetry the coordinates of $R$\n\nare $\\left(\\frac{\\pi}{6}, k \\cos \\frac{\\pi}{6}\\right)$.\n\nArea of rectangle $P Q R S=\\frac{\\pi}{3}\\left(k \\cos \\frac{\\pi}{6}\\right)=\\frac{\\pi}{3}(k)\\left(\\frac{\\sqrt{3}}{2}\\right)$\n\nBut $\\frac{\\sqrt{3} k \\pi}{6}=\\frac{5 \\pi}{3} \\quad \\therefore k=\\frac{10}{\\sqrt{3}}$ or $\\frac{10}{3} \\sqrt{3}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_703c4ffbe2eb07053abcg-1.jpg?height=545&width=591&top_left_y=199&top_left_x=1298)']","['$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$']",True,,Numerical, 2593,Geometry,,"A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$ ","['If $P Q=\\frac{\\pi}{3}$, then by symmetry the coordinates of $R$\n\nare $\\left(\\frac{\\pi}{6}, k \\cos \\frac{\\pi}{6}\\right)$.\n\nArea of rectangle $P Q R S=\\frac{\\pi}{3}\\left(k \\cos \\frac{\\pi}{6}\\right)=\\frac{\\pi}{3}(k)\\left(\\frac{\\sqrt{3}}{2}\\right)$\n\nBut $\\frac{\\sqrt{3} k \\pi}{6}=\\frac{5 \\pi}{3} \\quad \\therefore k=\\frac{10}{\\sqrt{3}}$ or $\\frac{10}{3} \\sqrt{3}$.\n\n']","['$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$']",True,,Numerical, 2594,Geometry,,"In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre. ![](https://cdn.mathpix.com/cropped/2023_12_21_609033623119f30122a6g-1.jpg?height=385&width=548&top_left_y=1274&top_left_x=1271)","['In $\\triangle B A N, \\angle B N A=25^{\\circ}$\n\nUsing the Sine Law in $\\triangle B A N$,\n\n$\\frac{N B}{\\sin 108^{\\circ}}=\\frac{100}{\\sin 25^{\\circ}}$\n\nTherefore $N B=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\approx 225.04$,\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_703c4ffbe2eb07053abcg-1.jpg?height=431&width=635&top_left_y=1308&top_left_x=1054)\n\nNow in $\\triangle M N B, \\frac{M N}{N B}=\\tan 32^{\\circ}$\n\n$$\nM N=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\times \\tan 32^{\\circ} \\doteq 140.6\n$$\n\nThe tower is approximately $141 \\mathrm{~m}$ high.']",['141'],False,m,Numerical, 2594,Geometry,,"In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre. ","['In $\\triangle B A N, \\angle B N A=25^{\\circ}$\n\nUsing the Sine Law in $\\triangle B A N$,\n\n$\\frac{N B}{\\sin 108^{\\circ}}=\\frac{100}{\\sin 25^{\\circ}}$\n\nTherefore $N B=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\approx 225.04$,\n\n\n\nNow in $\\triangle M N B, \\frac{M N}{N B}=\\tan 32^{\\circ}$\n\n$$\nM N=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\times \\tan 32^{\\circ} \\doteq 140.6\n$$\n\nThe tower is approximately $141 \\mathrm{~m}$ high.']",['141'],False,m,Numerical, 2595,Geometry,,"The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ? ![](https://cdn.mathpix.com/cropped/2023_12_21_609033623119f30122a6g-1.jpg?height=642&width=677&top_left_y=1674&top_left_x=1144)","['$Q(4,0), Q(0,4)$\n\n$Q(2,0), Q(0,2)$\n\n$Q(-2,2), Q(2,-2)$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_648ed28d6b9965167750g-1.jpg?height=737&width=781&top_left_y=1212&top_left_x=366)']","['$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$']",True,,Tuple, 2595,Geometry,,"The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ? ","['$Q(4,0), Q(0,4)$\n\n$Q(2,0), Q(0,2)$\n\n$Q(-2,2), Q(2,-2)$\n\n']","['$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$']",True,,Tuple, 2596,Geometry,,Determine the coordinates of the points of intersection of the graphs of $y=\log _{10}(x-2)$ and $y=1-\log _{10}(x+1)$.,"['The intersection takes place where,\n\n$$\n\\begin{aligned}\n& \\log _{10}(x-2)=1-\\log _{10}(x+1) \\\\\n& \\log _{10}(x-2)+\\log _{10}(x+1)=1 \\\\\n& \\log _{10}\\left(x^{2}-x-2\\right)=1\n\\end{aligned}\n$$\n\n\n\n$$\n\\begin{aligned}\n& x^{2}-x-2=10 \\\\\n& x^{2}-x-12=0 \\\\\n& (x-4)(x+3)=0 \\\\\n& x=4 \\text { or }-3\n\\end{aligned}\n$$\n\nFor $x=-3, y$ is not defined.\n\nFor $x=4, y=\\log _{10} 2 \\doteq 0.3$.\n\nThe graphs therefore intersect at $\\left(4, \\log _{10} 2\\right)$.']","['$(4, \\log _{10} 2)$']",False,,Tuple, 2597,Geometry,,The equation $y=x^{2}+2 a x+a$ represents a parabola for all real values of $a$. Prove that each of these parabolas pass through a common point and determine the coordinates of this point.,"['Since $y=x^{2}+2 a x+a$ for all $a, a \\in R$, it must be true for $a=0$ and $a=1$.\n\nFor $a=0, y=x^{2}$; for $a=1, y=x^{2}+2 x+1$.\n\nBy comparison, (or substitution)\n\n$$\n\\begin{aligned}\n& x^{2}=x^{2}+2 x+1 \\\\\n& \\therefore x=\\frac{-1}{2} \\\\\n& \\Rightarrow y=\\frac{1}{4}\n\\end{aligned}\n$$\n\nWe must verify that $x=\\frac{-1}{2}, y=\\frac{1}{4}$ satisfies the original.\n\nVerification: $y=x^{2}+2 a x+a=\\left(\\frac{-1}{2}\\right)^{2}+2 a\\left(\\frac{-1}{2}\\right)+a=\\frac{1}{4}-a+a=\\frac{1}{4}$\n\n$$\n\\therefore\\left(\\frac{-1}{2}, \\frac{1}{4}\\right) \\text { is a point on } y=x^{2}+2 a x+a, a \\in R \\text {. }\n$$', 'If $y=x^{2}+2 a x+a$ represents a parabola for all real values of $a$ then it is true for all $a$ and $b$ where $a \\neq b$.\n\nSo, $y=x^{2}+2 a x+a$ and $y=x^{2}+2 b x+b$ (by substitution of $a$ and $b$ into $y=x^{2}+2 a x+a$ ) Since we are looking for common point, $x^{2}+2 a x+a=x^{2}+2 b x+b$\n\n$$\n\\begin{aligned}\n& 2 a x-2 b x+a-b=0 \\\\\n& a(2 x+1)-b(2 x+1)=0 \\\\\n& (a-b)(2 x+1)=0\n\\end{aligned}\n$$\n\nSince $a \\neq b, 2 x+1=0 \\Rightarrow x=\\frac{-1}{2}$ and $y=\\frac{1}{4}$.']",,True,,, 2598,Algebra,,"Charlie was born in the twentieth century. On his birthday in the present year (2014), he notices that his current age is twice the number formed by the rightmost two digits of the year in which he was born. Compute the four-digit year in which Charlie was born.","['Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. Then his current age is $100-N+14=114-N$. Setting this equal to $2 N$ and solving yields $N=38$, hence the answer is 1938 .', ""Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. The number of years from 1900 to 2014 can be thought of as the number of years before Charlie was born plus the number of years since he was born, or $N$ plus Charlie's age. Thus $N+2 N=114$, which leads to $N=38$, so the answer is 1938 .""]",['1938'],False,,Numerical, 2599,Combinatorics,,"Let $A, B$, and $C$ be randomly chosen (not necessarily distinct) integers between 0 and 4 inclusive. Pat and Chris compute the value of $A+B \cdot C$ by two different methods. Pat follows the proper order of operations, computing $A+(B \cdot C)$. Chris ignores order of operations, choosing instead to compute $(A+B) \cdot C$. Compute the probability that Pat and Chris get the same answer.","['If Pat and Chris get the same answer, then $A+(B \\cdot C)=(A+B) \\cdot C$, or $A+B C=A C+B C$, or $A=A C$. This equation is true if $A=0$ or $C=1$; the equation places no restrictions on $B$. There are 25 triples $(A, B, C)$ where $A=0,25$ triples where $C=1$, and 5 triples where $A=0$ and $C=1$. As all triples are equally likely, the answer is $\\frac{25+25-5}{5^{3}}=\\frac{45}{125}=\\frac{\\mathbf{9}}{\\mathbf{2 5}}$.']",['$\\frac{9}{25}$'],False,,Numerical, 2600,Combinatorics,,"Bobby, Peter, Greg, Cindy, Jan, and Marcia line up for ice cream. In an acceptable lineup, Greg is ahead of Peter, Peter is ahead of Bobby, Marcia is ahead of Jan, and Jan is ahead of Cindy. For example, the lineup with Greg in front, followed by Peter, Marcia, Jan, Cindy, and Bobby, in that order, is an acceptable lineup. Compute the number of acceptable lineups.","['There are 6 people, so there are $6 !=720$ permutations. However, for each arrangement of the boys, there are $3 !=6$ permutations of the girls, of which only one yields an acceptable lineup. The same logic holds for the boys. Thus the total number of permutations must be divided by $3 ! \\cdot 3 !=36$, yielding $6 ! /(3 ! \\cdot 3 !)=\\mathbf{2 0}$ acceptable lineups.', 'Once the positions of Greg, Peter, and Bobby are determined, the entire lineup is determined, because there is only one acceptable ordering of the three girls. Because the boys occupy three of the six positions, there are $\\left(\\begin{array}{l}6 \\\\ 3\\end{array}\\right)=\\mathbf{2 0}$ acceptable lineups.']",['20'],False,,Numerical, 2601,Geometry,,"In triangle $A B C, a=12, b=17$, and $c=13$. Compute $b \cos C-c \cos B$.","['Using the Law of Cosines, $a^{2}+b^{2}-2 a b \\cos C=c^{2}$ implies\n\n$$\nb \\cos C=\\frac{a^{2}+b^{2}-c^{2}}{2 a}\n$$\n\nSimilarly,\n\n$$\nc \\cos B=\\frac{a^{2}-b^{2}+c^{2}}{2 a}\n$$\n\nThus\n\n$$\n\\begin{aligned}\nb \\cos C-c \\cos B & =\\frac{a^{2}+b^{2}-c^{2}}{2 a}-\\frac{a^{2}-b^{2}+c^{2}}{2 a} \\\\\n& =\\frac{2 b^{2}-2 c^{2}}{2 a} \\\\\n& =\\frac{b^{2}-c^{2}}{a} .\n\\end{aligned}\n$$\n\n\n\nWith the given values, the result is $\\left(17^{2}-13^{2}\\right) / 12=120 / 12=\\mathbf{1 0}$.', 'Let $H$ be the foot of the altitude from $A$ to $\\overline{B C}$; let $B H=x$, $C H=y$, and $A H=h$. Then $b \\cos C=y, c \\cos B=x$, and the desired quantity is $Q=y-x$. However, $y+x=a$, so $y^{2}-x^{2}=a Q$. By the Pythagorean Theorem, $y^{2}=b^{2}-h^{2}$ and $x^{2}=c^{2}-h^{2}$, so $y^{2}-x^{2}=\\left(b^{2}-h^{2}\\right)-\\left(c^{2}-h^{2}\\right)=b^{2}-c^{2}$. Thus $a Q=b^{2}-c^{2}$, and $Q=\\frac{b^{2}-c^{2}}{a}$\n\nWith the given values, the result is $\\left(17^{2}-13^{2}\\right) / 12=120 / 12=\\mathbf{1 0}$.']",['10'],False,,Numerical, 2602,Algebra,,"The sequence of words $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=X, a_{2}=O$, and for $n \geq 3, a_{n}$ is $a_{n-1}$ followed by the reverse of $a_{n-2}$. For example, $a_{3}=O X, a_{4}=O X O, a_{5}=O X O X O$, and $a_{6}=O X O X O O X O$. Compute the number of palindromes in the first 1000 terms of this sequence.","[""Let $P$ denote a palindromic word, let $Q$ denote any word, and let $\\bar{R}$ denote the reverse of word $R$. Note that if two consecutive terms of the sequence are $a_{n}=P, a_{n+1}=Q$, then $a_{n+2}=Q \\bar{P}=Q P$ and $a_{n+3}=Q P \\bar{Q}$. Thus if $a_{n}$ is a palindrome, so is $a_{n+3}$. Because $a_{1}$ and $a_{2}$ are both palindromes, then so must be all terms in the subsequences $a_{4}, a_{7}, a_{10}, \\ldots$ and $a_{5}, a_{8}, a_{11}, \\ldots$\n\nTo show that the other terms are not palindromes, note that if $P^{\\prime}$ is not a palindrome, then $Q P^{\\prime} \\bar{Q}$ is also not a palindrome. Thus if $a_{n}$ is not a palindrome, then $a_{n+3}$ is not a palindrome either. Because $a_{3}=O X$ is not a palindrome, neither is any term of the subsequence $a_{6}, a_{9}, a_{12}, \\ldots$ (Alternatively, counting the number of $X$ 's in each word $a_{i}$ shows that the number of $X$ 's in $a_{3 k}$ is odd. So if $a_{3 k}$ were to be a palindrome, it would have to have an odd number of letters, with an $X$ in the middle. However, it can be shown that the length of $a_{3 k}$ is even. Thus $a_{3 k}$ cannot be a palindrome.)\n\nIn total there are $1000-333=\\mathbf{6 6 7}$ palindromes among the first 1000 terms.""]",['667'],False,,Numerical, 2603,Number Theory,,Compute the smallest positive integer $n$ such that $214 \cdot n$ and $2014 \cdot n$ have the same number of divisors.,"['Let $D(n)$ be the number of divisors of the integer $n$. Note that if $D(214 n)=D(2014 n)$ and if some $p$ divides $n$ and is relatively prime to both 214 and 2014 , then $D\\left(\\frac{214 n}{p}\\right)=D\\left(\\frac{2014 n}{p}\\right)$. Thus any prime divisor of the smallest possible positive $n$ will be a divisor of $214=2 \\cdot 107$ or $2014=2 \\cdot 19 \\cdot 53$. For the sake of convenience, write $n=2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}$, where $a, b, c, d \\geq 1$. Then $D(214 n)=(a+1) b c(d+1)$ and $D(2014 n)=(a+1)(b+1)(c+1) d$. Divide both sides by $a+1$ and expand to get $b c d+b c=b c d+b d+c d+d$, or $b c-b d-c d-d=0$.\n\nBecause the goal is to minimize $n$, try $d=1$ : $b c-b-c-1=0 \\Rightarrow(b-1)(c-1)=2$, which has solutions $(b, c)=(2,3)$ and $(3,2)$. The latter gives the smaller value for $n$, namely $19^{2} \\cdot 53=$ 19133. The only quadruples $(a, b, c, d)$ that satisfy $2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}<19133$ and $d>1$ are $(1,1,2,2),(1,2,1,2)$, and $(1,1,1,3)$. None of these quadruples satisfies $b c-b d-c d-d=0$, so the minimum value is $n=\\mathbf{1 9 1 3 3}$.']",['19133'],False,,Numerical, 2604,Number Theory,,"Let $N$ be the least integer greater than 20 that is a palindrome in both base 20 and base 14 . For example, the three-digit base-14 numeral (13)5(13) ${ }_{14}$ (representing $13 \cdot 14^{2}+5 \cdot 14^{1}+13 \cdot 14^{0}$ ) is a palindrome in base 14 , but not in base 20 , and the three-digit base-14 numeral (13)31 14 is not a palindrome in base 14 . Compute the base-10 representation of $N$.","['Because $N$ is greater than 20, the base-20 and base-14 representations of $N$ must be at least two digits long. The smallest possible case is that $N$ is a two-digit palindrome in both bases. Then $N=20 a+a=21 a$, where $1 \\leq a \\leq 19$. Similarly, in order to be a two-digit palindrome in base $14, N=14 b+b=15 b$, with $1 \\leq b \\leq 13$. So $N$ would have to be a multiple of both 21 and 15 . The least common multiple of 21 and 15 is 105 , which has the base 20 representation of $105=55_{20}$ and the base-14 representation of $105=77_{14}$, both of which are palindromes. Thus the answer is 105.']",['105'],False,,Numerical, 2605,Geometry,,"In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$. ","['Let $[A B C]=K$. Then $[B C D]=\\frac{2}{3} \\cdot K$. Let $\\overline{D E}$ be the bisector of $\\angle B D C$, as shown below.\n\n\n\nNotice that $\\mathrm{m} \\angle D B A=\\mathrm{m} \\angle B D C-\\mathrm{m} \\angle A=\\mathrm{m} \\angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\\overline{D E} \\| \\overline{A B}$, and by similar triangles, $[C D E]=\\frac{4}{9} \\cdot K$, which means $[B D E]=\\frac{2}{9} \\cdot K$. Because $[C D E]:[B D E]=2$ and $\\angle B D E \\cong \\angle C D E$, conclude that $\\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\\cos \\angle B D C=\\frac{1}{2} B D / C D=\\frac{1}{4}$. By the half-angle formula,\n\n$$\n\\sin A=\\sqrt{\\frac{1-\\cos \\angle B D C}{2}}=\\sqrt{\\frac{3}{8}}=\\frac{\\sqrt{6}}{\\mathbf{4}}\n$$']",['$\\frac{\\sqrt{6}}{4}$'],False,,Numerical, 2605,Geometry,,"In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$. ![](https://cdn.mathpix.com/cropped/2023_12_21_d5c02975ee9541bc9cf8g-1.jpg?height=287&width=591&top_left_y=431&top_left_x=816)","['Let $[A B C]=K$. Then $[B C D]=\\frac{2}{3} \\cdot K$. Let $\\overline{D E}$ be the bisector of $\\angle B D C$, as shown below.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_86abbbc5956c08debe19g-1.jpg?height=292&width=591&top_left_y=388&top_left_x=816)\n\nNotice that $\\mathrm{m} \\angle D B A=\\mathrm{m} \\angle B D C-\\mathrm{m} \\angle A=\\mathrm{m} \\angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\\overline{D E} \\| \\overline{A B}$, and by similar triangles, $[C D E]=\\frac{4}{9} \\cdot K$, which means $[B D E]=\\frac{2}{9} \\cdot K$. Because $[C D E]:[B D E]=2$ and $\\angle B D E \\cong \\angle C D E$, conclude that $\\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\\cos \\angle B D C=\\frac{1}{2} B D / C D=\\frac{1}{4}$. By the half-angle formula,\n\n$$\n\\sin A=\\sqrt{\\frac{1-\\cos \\angle B D C}{2}}=\\sqrt{\\frac{3}{8}}=\\frac{\\sqrt{6}}{\\mathbf{4}}\n$$']",['$\\frac{\\sqrt{6}}{4}$'],False,,Numerical, 2606,Number Theory,,$\quad$ Compute the greatest integer $k \leq 1000$ such that $\left(\begin{array}{c}1000 \\ k\end{array}\right)$ is a multiple of 7 .,"['The ratio of binomial coefficients $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right) /\\left(\\begin{array}{c}1000 \\\\ k+1\\end{array}\\right)=\\frac{k+1}{1000-k}$. Because 1000 is 1 less than a multiple of 7 , namely $1001=7 \\cdot 11 \\cdot 13$, either $1000-k$ and $k+1$ are both multiples of 7 or neither is. Hence whenever the numerator is divisible by 7, the denominator is also. Thus for the largest value of $k$ such that $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right)$ is a multiple of $7, \\frac{k+1}{1000-k}$ must equal $7 \\cdot \\frac{p}{q}$, where $p$ and $q$ are relatively prime integers and $7 \\nmid q$. The only way this can happen is when $k+1$ is a multiple of 49 , the greatest of which less than 1000 is 980 . Therefore the greatest value of $k$ satisfying the given conditions is $980-1=\\mathbf{9 7 9}$.', ""Rewrite 1000 in base 7: $1000=2626_{7}$. Let $k=\\underline{a} \\underline{b} \\underline{c}_{7}$. By Lucas's Theorem, $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right) \\equiv\\left(\\begin{array}{l}2 \\\\ a\\end{array}\\right)\\left(\\begin{array}{l}6 \\\\ b\\end{array}\\right)\\left(\\begin{array}{l}2 \\\\ c\\end{array}\\right)\\left(\\begin{array}{l}6 \\\\ d\\end{array}\\right) \\bmod 7$. The binomial coefficient $\\left(\\begin{array}{l}p \\\\ q\\end{array}\\right) \\stackrel{a}{=} 0$ only when $q>p$. Base 7 digits cannot exceed 6 , and $k \\leq 1000$, thus the greatest value of $k$ that works is $2566_{7}=\\mathbf{9 7 9}$. (Alternatively, the least value of $k$ that works is $30_{7}=21$; because $\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)=\\left(\\begin{array}{c}n \\\\ n-k\\end{array}\\right)$, the greatest such $k$ is $1000-21=979$.)""]",['979'],False,,Numerical, 2607,Number Theory,,An integer-valued function $f$ is called tenuous if $f(x)+f(y)>x^{2}$ for all positive integers $x$ and $y$. Let $g$ be a tenuous function such that $g(1)+g(2)+\cdots+g(20)$ is as small as possible. Compute the minimum possible value for $g(14)$.,"['For a tenuous function $g$, let $S_{g}=g(1)+g(2)+\\cdots+g(20)$. Then:\n\n$$\n\\begin{aligned}\nS_{g} & =(g(1)+g(20))+(g(2)+g(19))+\\cdots+(g(10)+g(11)) \\\\\n& \\geq\\left(20^{2}+1\\right)+\\left(19^{2}+1\\right)+\\cdots+\\left(11^{2}+1\\right) \\\\\n& =10+\\sum_{k=11}^{20} k^{2} \\\\\n& =2495 .\n\\end{aligned}\n$$\n\n\n\nThe following argument shows that if a tenuous function $g$ attains this sum, then $g(1)=$ $g(2)=\\cdots=g(10)$. First, if the sum equals 2495 , then $g(1)+g(20)=20^{2}+1, g(2)+g(19)=$ $19^{2}+1, \\ldots, g(10)+g(11)=11^{2}+1$. If $g(1)g(1)$, then $g(2)+g(20)<20^{2}+1$. Therefore $g(1)=g(2)$. Analogously, comparing $g(1)$ and $g(3), g(1)$ and $g(4)$, etc. shows that $g(1)=g(2)=g(3)=\\cdots=g(10)$.\n\nNow consider all functions $g$ for which $g(1)=g(2)=\\cdots=g(10)=a$ for some integer $a$. Then $g(n)=n^{2}+1-a$ for $n \\geq 11$. Because $g(11)+g(11)>11^{2}=121$, it is the case that $g(11) \\geq 61$. Thus $11^{2}+1-a \\geq 61 \\Rightarrow a \\leq 61$. Thus the smallest possible value for $g(14)$ is $14^{2}+1-61=\\mathbf{1 3 6}$.']",['136'],False,,Numerical, 2608,Geometry,,"Let $T=(0,0), N=(2,0), Y=(6,6), W=(2,6)$, and $R=(0,2)$. Compute the area of pentagon $T N Y W R$.","['Pentagon $T N Y W R$ fits inside square $T A Y B$, where $A=(6,0)$ and $B=(0,6)$. The region of $T A Y B$ not in $T N Y W R$ consists of triangles $\\triangle N A Y$ and $\\triangle W B R$, as shown below.\n\n\n\nThus\n\n$$\n\\begin{aligned}\n{[T N Y W R] } & =[T A Y B]-[N A Y]-[W B R] \\\\\n& =6^{2}-\\frac{1}{2} \\cdot 4 \\cdot 6-\\frac{1}{2} \\cdot 2 \\cdot 4 \\\\\n& =\\mathbf{2 0} .\n\\end{aligned}\n$$']",['20'],False,,Numerical, 2609,Geometry,,Let $T=20$. The lengths of the sides of a rectangle are the zeroes of the polynomial $x^{2}-3 T x+T^{2}$. Compute the length of the rectangle's diagonal.,"[""Let $r$ and $s$ denote the zeros of the polynomial $x^{2}-3 T x+T^{2}$. The rectangle's diagonal has length $\\sqrt{r^{2}+s^{2}}=\\sqrt{(r+s)^{2}-2 r s}$. Recall that for a quadratic polynomial $a x^{2}+b x+c$, the sum of its zeros is $-b / a$, and the product of its zeros is $c / a$. In this particular instance, $r+s=3 T$ and $r s=T^{2}$. Thus the length of the rectangle's diagonal is $\\sqrt{9 T^{2}-2 T^{2}}=T \\cdot \\sqrt{7}$. With $T=20$, the rectangle's diagonal is $\\mathbf{2 0} \\sqrt{\\mathbf{7}}$.""]",['$20 \\sqrt{7}$'],False,,Numerical, 2610,Algebra,,"Let $T=20 \sqrt{7}$. Let $w>0$ be a real number such that $T$ is the area of the region above the $x$-axis, below the graph of $y=\lceil x\rceil^{2}$, and between the lines $x=0$ and $x=w$. Compute $\lceil 2 w\rceil$.","['Write $w=k+\\alpha$, where $k$ is an integer, and $0 \\leq \\alpha<1$. Then\n\n$$\nT=1^{2}+2^{2}+\\cdots+k^{2}+(k+1)^{2} \\cdot \\alpha .\n$$\n\nComputing $\\lceil 2 w\\rceil$ requires computing $w$ to the nearest half-integer. First obtain the integer $k$. As $\\sqrt{7}>2$, with $T=20 \\sqrt{7}$, one obtains $T>40$. As $1^{2}+2^{2}+3^{2}+4^{2}=30$, it follows that $k \\geq 4$. To obtain an upper bound for $k$, note that $700<729$, so $10 \\sqrt{7}<27$, and $T=20 \\sqrt{7}<54$. As $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55$, it follows that $40.5$. To this end, one must determine whether $T>1^{2}+2^{2}+3^{2}+4^{2}+5^{2} / 2=42.5$. Indeed, note that $2.5^{2}=6.25<7$, so $T>(20)(2.5)=50$. It follows that $\\alpha>0.5$, so $4.5\\frac{21-12}{5}=1.8$. Because $2 w=2 k+2 \\alpha$, it follows that $\\lceil 2 w\\rceil=\\lceil 8+2 \\alpha\\rceil=\\mathbf{1 0}$, because $1.8<2 \\alpha<2$.']",['10'],False,,Numerical, 2611,Number Theory,,"Compute the least positive integer $n$ such that $\operatorname{gcd}\left(n^{3}, n !\right) \geq 100$.","['Note that if $p$ is prime, then $\\operatorname{gcd}\\left(p^{3}, p !\\right)=p$. A good strategy is to look for values of $n$ with several (not necessarily distinct) prime factors so that $n^{3}$ and $n$ ! will have many factors in common. For example, if $n=6, n^{3}=216=2^{3} \\cdot 3^{3}$ and $n !=720=2^{4} \\cdot 3^{2} \\cdot 5$, so $\\operatorname{gcd}(216,720)=2^{3} \\cdot 3^{2}=72$. Because 7 is prime, try $n=8$. Notice that $8^{3}=2^{9}$ while $8 !=2^{7} \\cdot 3^{2} \\cdot 5 \\cdot 7$. Thus $\\operatorname{gcd}(512,8 !)=2^{7}=128>100$, hence the smallest value of $n$ is $\\mathbf{8}$.']",['8'],False,,Numerical, 2612,Combinatorics,,"Let $T=8$. At a party, everyone shakes hands with everyone else exactly once, except Ed, who leaves early. A grand total of $20 T$ handshakes take place. Compute the number of people at the party who shook hands with Ed.","[""If there were $n$ people at the party, including Ed, and if Ed had not left early, there would have been $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$ handshakes. Because Ed left early, the number of handshakes is strictly less than that, but greater than $\\left(\\begin{array}{c}n-1 \\\\ 2\\end{array}\\right)$ (everyone besides Ed shook everyone else's hand). So find the least number $n$ such that $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right) \\geq 160$. The least such $n$ is 19 , because $\\left(\\begin{array}{c}18 \\\\ 2\\end{array}\\right)=153$ and $\\left(\\begin{array}{c}19 \\\\ 2\\end{array}\\right)=171$. Therefore there were 19 people at the party. However, $171-160=11$ handshakes never took place. Therefore the number of people who shook hands with Ed is $19-11-1=7$.""]",['7'],False,,Numerical, 2613,Algebra,,"Let $T=7$. Given the sequence $u_{n}$ such that $u_{3}=5, u_{6}=89$, and $u_{n+2}=3 u_{n+1}-u_{n}$ for integers $n \geq 1$, compute $u_{T}$.","['By the recursive definition, notice that $u_{6}=89=3 u_{5}-u_{4}$ and $u_{5}=3 u_{4}-u_{3}=3 u_{4}-5$. This is a linear system of equations. Write $3 u_{5}-u_{4}=89$ and $-3 u_{5}+9 u_{4}=15$ and add to obtain $u_{4}=13$. Now apply the recursive definition to obtain $u_{5}=34$ and $u_{7}=\\mathbf{2 3 3}$.', 'Notice that the given values are both Fibonacci numbers, and that in the Fibonacci sequence, $f_{1}=f_{2}=1, f_{5}=5$, and $f_{11}=89$. That is, 5 and 89 are six terms apart in the Fibonacci sequence, and only three terms apart in the given sequence. This relationship is not a coincidence: alternating terms in the Fibonacci sequence satisfy the given recurrence relation for the sequence $\\left\\{u_{n}\\right\\}$, that is, $f_{n+4}=3 f_{n+2}-f_{n}$. Proof: if $f_{n}=a$ and $f_{n+1}=b$, then $f_{n+2}=a+b, f_{n+3}=a+2 b$, and $f_{n+4}=2 a+3 b=3(a+b)-b=3 f_{n+2}-f_{n}$. To compute the final result, continue out the Fibonacci sequence to obtain $f_{12}=144$ and $u_{7}=f_{13}=233$.']",['233'],False,,Numerical, 2614,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. Compute $\operatorname{pop}\left(\mathcal{F}_{17}\right)$.","['There are $\\left(\\begin{array}{c}17 \\\\ 2\\end{array}\\right)=136$ possible pairs of dishes, so $\\mathcal{F}_{17}$ must have 136 people.']",['136'],False,,Numerical, 2615,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. Let $n=\operatorname{pop}\left(\mathcal{F}_{d}\right)$. In terms of $n$, compute $d$.","['With $d$ dishes there are $\\left(\\begin{array}{l}d \\\\ 2\\end{array}\\right)=\\frac{d^{2}-d}{2}$ possible pairs, so $n=\\frac{d^{2}-d}{2}$. Then $2 n=d^{2}-d$, or $d^{2}-d-2 n=0$. Using the quadratic formula yields $d=\\frac{1+\\sqrt{1+8 n}}{2}$ (ignoring the negative value).']",['$d=\\frac{1+\\sqrt{1+8 n}}{2}$'],False,,Numerical, 2616,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. Let $T$ be a full town and let $D \in \operatorname{dish}(T)$. Let $T^{\prime}$ be the town consisting of all residents of $T$ who do not know how to make $D$. Prove that $T^{\prime}$ is full.","['The town $T^{\\prime}$ consists of all residents of $T$ who do not know how to make $D$. Because $T$ is full, every pair of dishes $\\left\\{d_{i}, d_{j}\\right\\}$ in $\\operatorname{dish}(T)$ can be made by some resident $r_{i j}$ in $T$. If $d_{i} \\neq D$ and $d_{j} \\neq D$, then $r_{i j} \\in T^{\\prime}$. So every pair of dishes in $\\operatorname{dish}(T) \\backslash\\{D\\}$ can be made by some resident of $T^{\\prime}$. Hence $T^{\\prime}$ is full.']",,True,,, 2617,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. Show that $\operatorname{gr}($ ARMLton $)=2$.","['Paul and Arnold cannot be in the same group, because they both make pie, and Arnold and Kelly cannot be in the same group, because they both make salad. Hence there must be at least two groups. But Paul and Kelly make none of the same dishes, so they can be in the same group. Thus a valid group assignment is\n\n$$\n\\begin{aligned}\n\\text { Paul } & \\mapsto 1 \\\\\n\\text { Kelly } & \\mapsto 1 \\\\\n\\text { Arnold } & \\mapsto 2\n\\end{aligned}\n$$\n\nHence $\\operatorname{gr}($ ARMLton $)=2$.']",,True,,, 2618,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. Show that gr(ARMLville $)=3$.","['Sally and Ross both make calzones, Ross and David both make pancakes, and Sally and David both make steak. So no two of these people can be in the same group, and $\\operatorname{gr}($ ARMLville $)=3$.']",,True,,, 2619,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. Show that $\operatorname{gr}\left(\mathcal{F}_{4}\right)=3$.","['Let the dishes be $d_{1}, d_{2}, d_{3}, d_{4}$ and let resident $r_{i j}$ make dishes $d_{i}$ and $d_{j}$, where $i\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Compute the number of distinguishable resident cycles of length 6 in $\mathcal{F}_{8}$.","['Because the town is full, each pair of dishes is cooked by exactly one resident, so it is simplest to identify residents by the pairs of dishes they cook. Suppose the first resident cooks $\\left(d_{1}, d_{2}\\right)$, the second resident $\\left(d_{2}, d_{3}\\right)$, the third resident $\\left(d_{3}, d_{4}\\right)$, and so on, until the sixth resident, who cooks $\\left(d_{6}, d_{1}\\right)$. Then there are 8 choices for $d_{1}$ and 7 choices for $d_{2}$. There are only 6 choices for $d_{3}$, because $d_{3} \\neq d_{1}$ (otherwise two residents would cook the same pair of dishes). For $k>3$, the requirement that no two intermediate residents cook the same dishes implies that $d_{k+1}$ cannot equal any of $d_{1}, \\ldots, d_{k-1}$, and of course $d_{k}$ and $d_{k+1}$ must be distinct dishes. Hence there are $8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3=20,160$ six-person resident cycles, not accounting for different starting points in the cycle and the two different directions to go around the cycle. Taking these into account, there are $20,160 /(6 \\cdot 2)=1,680$ distinguishable resident cycles.']",['1680'],False,,Numerical, 2626,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. In terms of $k$ and $d$, find the number of distinguishable resident cycles of length $k$ in $\mathcal{F}_{d}$.","['First, we compute the number of distinguishable resident cycles of length 6 in $\\mathcal{F}_{8}$.\n\nBecause the town is full, each pair of dishes is cooked by exactly one resident, so it is simplest to identify residents by the pairs of dishes they cook. Suppose the first resident cooks $\\left(d_{1}, d_{2}\\right)$, the second resident $\\left(d_{2}, d_{3}\\right)$, the third resident $\\left(d_{3}, d_{4}\\right)$, and so on, until the sixth resident, who cooks $\\left(d_{6}, d_{1}\\right)$. Then there are 8 choices for $d_{1}$ and 7 choices for $d_{2}$. There are only 6 choices for $d_{3}$, because $d_{3} \\neq d_{1}$ (otherwise two residents would cook the same pair of dishes). For $k>3$, the requirement that no two intermediate residents cook the same dishes implies that $d_{k+1}$ cannot equal any of $d_{1}, \\ldots, d_{k-1}$, and of course $d_{k}$ and $d_{k+1}$ must be distinct dishes. Hence there are $8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3=20,160$ six-person resident cycles, not accounting for different starting points in the cycle and the two different directions to go around the cycle. Taking these into account, there are $20,160 /(6 \\cdot 2)=1,680$ distinguishable resident cycles.\n\nUsing the logic above, there are $d(d-1) \\cdots(d-k+1)$ choices for $d_{1}, d_{2}, \\ldots, d_{k}$. To account for indistinguishable cycles, divide by $k$ possible starting points and 2 possible directions, yielding $\\frac{d(d-1) \\cdots(d-k+1)}{2 k}$ or $\\frac{d !}{2 k(d-k) !}$ distinguishable resident cycles.']",['$\\frac{d !}{2 k(d-k) !}$'],False,,Numerical, 2627,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town with at least two residents that has a single resident cycle that contains every resident. Prove that $T$ is homogeneous if and only if $\operatorname{pop}(T)$ is even.","['Note that for every $D \\in \\operatorname{dish}(T), \\operatorname{chef}_{T}(D) \\leq 2$, because otherwise, $r_{1}, r_{2}, \\ldots, r_{n}$ could not be a resident cycle. Without loss of generality, assume the cycle is $r_{1}, r_{2}, \\ldots, r_{n}$. If $n$ is even, assign resident $r_{i}$ to group 1 if $i$ is odd, and to group 2 if $i$ is even. This is a valid group assignment, because the only pairs of residents who cook the same dish are $\\left(r_{i}, r_{i+1}\\right)$ for $i=1,2, \\ldots, n-1$ and $\\left(r_{n}, r_{1}\\right)$. In each case, the residents are assigned to different groups. This proves $\\operatorname{gr}(T)=2$, so $T$ is homogeneous.\n\nOn the other hand, if $n$ is odd, suppose for the sake of contradiction that there are only two groups. Then either $r_{1}$ and $r_{n}$ are in the same group, or for some $i, r_{i}$ and $r_{i+1}$ are in the\n\n\n\nsame group. In either case, two residents in the same group share a dish, contradicting the requirement that no members of a group have a common dish. Hence $\\operatorname{gr}(T) \\geq 3$ when $n$ is odd, making $T$ heterogeneous.']",,True,,, 2628,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=2$. Prove that there are finitely many resident cycles $C_{1}, C_{2}, \ldots, C_{j}$ in $T$ so that each resident belongs to exactly one of the $C_{i}$.","['First note that the condition $\\left|\\operatorname{chef}_{T}(D)\\right|=2$ for all $D$ implies that $\\operatorname{pop}(T)=|\\operatorname{dish}(T)|$, using the equation from problem 5 . So for the town in question, the population of the town equals the number of dishes in the town. Because no two chefs cook the same pair of dishes, it is impossible for such a town to have exactly two residents, and because each dish is cooked by exactly two chefs, it is impossible for such a town to have only one resident.\n\nThe claim is true for towns of three residents satisfying the conditions: such towns must have one resident who cooks dishes $d_{1}$ and $d_{2}$, one resident who cooks dishes $d_{2}$ and $d_{3}$, and one resident who cooks dishes $d_{3}$ and $d_{1}$, and those three residents form a cycle. So proceed by (modified) strong induction: assume that for some $n>3$ and for all positive integers $k$ such that $3 \\leq k0$, so by the inductive hypothesis, the residents of $T^{\\prime}$ can be divided into disjoint resident cycles.\n\nThus the statement is proved by strong induction.']",,True,,, 2629,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=2$. Prove that if $\operatorname{pop}(T)$ is odd, then $T$ is heterogeneous.","['In order for $T$ to be homogeneous, it must be possible to partition the residents into exactly two dining groups. First apply 10a to divide the town into finitely many resident cycles $C_{i}$, and assume towards a contradiction that such a group assignment $f: T \\rightarrow$ $\\{1,2\\}$ exists. If $\\operatorname{pop}(T)$ is odd, then at least one of the cycles $C_{i}$ must contain an odd number of residents; without loss of generality, suppose this cycle to be $C_{1}$, with residents $r_{1}, r_{2}, \\ldots, r_{2 k+1}$. (By the restrictions noted in part a, $k \\geq 1$.) Now because $r_{i}$ and $r_{i+1}$ cook a dish in common, $f\\left(r_{i}\\right) \\neq f\\left(r_{i+1}\\right)$ for all $i$. Thus if $f\\left(r_{1}\\right)=1$, it follows that $f\\left(r_{2}\\right)=2$, and that $f\\left(r_{3}\\right)=1$, etc. So $f\\left(r_{i}\\right)=f\\left(r_{1}\\right)$ if $i$ is odd and $f\\left(r_{i}\\right)=f\\left(r_{2}\\right)$ if $i$ is\n\n\n\neven; in particular, $f\\left(r_{2 k+1}\\right)=f(1)$. But that equation would imply that $r_{1}$ and $r_{2 k+1}$ cook no dishes in common, which is impossible if they are the first and last residents in a resident cycle. So no such group assignment can exist, and $\\operatorname{gr}(T) \\geq 3$. Hence $T$ is heterogeneous.']",,True,,, 2630,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=3$. Either find such a town $T$ for which $|\operatorname{dish}(T)|$ is odd, or show that no such town exists.","['First, we Prove that\n$$\n\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|=2 \\operatorname{pop}(T) .\n$$\n\nBecause each chef knows how to prepare exactly two dishes, and no two chefs know how to prepare the same two dishes, each chef is counted exactly twice in the sum $\\Sigma\\left|\\operatorname{chef}_{T}(D)\\right|$. More formally, consider the set of ""resident-dish pairs"":\n\n$$\nS=\\{(r, D) \\in T \\times \\operatorname{dish}(T) \\mid r \\text { makes } D\\}\n$$\n\nCount $|S|$ in two different ways. First, every dish $D$ is made by $\\left|\\operatorname{chef}_{T}(D)\\right|$ residents of $T$, so\n\n$$\n|S|=\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|\n$$\n\nSecond, each resident knows how to make exactly two dishes, so\n\n$$\n|S|=\\sum_{r \\in T} 2=2 \\operatorname{pop}(T)\n$$\n\n\nTherefore $\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|$ is even. But if $\\left|\\operatorname{chef}_{T}(D)\\right|=3$ for all $D \\in \\operatorname{dish}(T)$, then the sum is simply $3|\\operatorname{dish}(T)|$, so $|\\operatorname{dish}(T)|$ must be even.']",,True,,, 2631,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $T$ be a town such that, for each $D \in \operatorname{dish}(T),\left|\operatorname{chef}_{T}(D)\right|=3$. Prove that if $T$ contains a resident cycle such that for every dish $D \in \operatorname{dish}(T)$, there exists a chef in the cycle that can prepare $D$, then $\operatorname{gr}(T)=3$.","['First we prove that for any town $T$ and any $D \\in \\operatorname{dish}(T), \\operatorname{gr}(T) \\geq\\left|\\operatorname{chef}_{T}(D)\\right|$.\n\nLet $D \\in \\operatorname{dish}(T)$. Suppose that $f$ is a valid group assignment on $T$. Then for $r, s \\in \\operatorname{chef}_{T}(D)$, if $r \\neq s$, it follows that $f(r) \\neq f(s)$. Hence there must be at least $\\left|\\operatorname{chef}_{T}(D)\\right|$ distinct groups in the range of $f$, i.e., $\\operatorname{gr}(T) \\geq\\left|\\operatorname{chef}_{T}(D)\\right|$.\n\nBy the conclusion above, it must be the case that $\\operatorname{gr}(T) \\geq 3$. Let $C=\\left\\{r_{1}, r_{2}, \\ldots, r_{n}\\right\\}$ denote a resident cycle such that for every dish $D \\in \\operatorname{dish}(T)$, there exists a chef in $C$ that can prepare $D$. Each resident is a chef for two dishes, and every dish can be made by two residents in $C$ (although by three in $T$ ). Thus the number of residents in the resident cycle $C$ is equal to $|\\operatorname{dish}(T)|$, which was proved to be even in the previous part.\n\nDefine a group assignment by setting\n\n$$\nf(r)= \\begin{cases}1 & \\text { if } r \\notin C \\\\ 2 & \\text { if } r=r_{i} \\text { and } i \\text { is even } \\\\ 3 & \\text { if } r=r_{i} \\text { and } i \\text { is odd. }\\end{cases}\n$$\n\nFor any $D \\in \\operatorname{dish}(T)$, there are exactly three $D$-chefs, and exactly two of them belong to the resident cycle $C$. Hence exactly one of the $D$-chefs $r$ will have $f(r)=1$. The remaining two $D$-chefs will be $r_{i}$ and $r_{i+1}$ for some $i$, or $r_{1}$ and $r_{n}$. In either case, the group assignment $f$ will assign one of them to 2 and the other to 3 . Thus any two residents who make a common dish will be assigned different groups by $f$, so $f$ is a valid group assignment, proving that $\\operatorname{gr}(T)=3$.']",,True,,, 2632,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $k$ be a positive integer, and let $T$ be a town in which $\left|\operatorname{chef}_{T}(D)\right|=k$ for every dish $D \in \operatorname{dish}(T)$. Suppose further that $|\operatorname{dish}(T)|$ is odd. Show that $k$ is even.","['First, we Prove that\n$$\n\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|=2 \\operatorname{pop}(T) .\n$$\n\nBecause each chef knows how to prepare exactly two dishes, and no two chefs know how to prepare the same two dishes, each chef is counted exactly twice in the sum $\\Sigma\\left|\\operatorname{chef}_{T}(D)\\right|$. More formally, consider the set of ""resident-dish pairs"":\n\n$$\nS=\\{(r, D) \\in T \\times \\operatorname{dish}(T) \\mid r \\text { makes } D\\}\n$$\n\nCount $|S|$ in two different ways. First, every dish $D$ is made by $\\left|\\operatorname{chef}_{T}(D)\\right|$ residents of $T$, so\n\n$$\n|S|=\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|\n$$\n\nSecond, each resident knows how to make exactly two dishes, so\n\n$$\n|S|=\\sum_{r \\in T} 2=2 \\operatorname{pop}(T)\n$$\n\nFrom above prove,\n\n$$\n2 \\operatorname{pop}(T)=\\sum_{D \\in \\operatorname{dish}(T)}\\left|\\operatorname{chef}_{T}(D)\\right|\n$$\n\nBecause $\\left|\\operatorname{chef}_{T}(D)\\right|=k$ for all $D \\in \\operatorname{dish}(T)$, the sum is $k \\cdot \\operatorname{dish}(T)$. Thus $2 \\operatorname{pop} T=$ $k \\cdot \\operatorname{dish}(T)$, and so $k \\cdot \\operatorname{dish}(T)$ must be even. By assumption, $|\\operatorname{dish}(T)|$ is odd, so $k$ must be even.']",,True,,, 2633,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. Let $k$ be a positive integer, and let $T$ be a town in which $\left|\operatorname{chef}_{T}(D)\right|=k$ for every dish $D \in \operatorname{dish}(T)$. Suppose further that $|\operatorname{dish}(T)|$ is odd. Prove the following: for every group in $T$, there is some $\operatorname{dish} D \in \operatorname{dish}(T)$ such that no one in the group is a $D$-chef.","['Suppose for the sake of contradiction that there is some $n$ for which the group $R=\\{r \\in$ $T \\mid f(r)=n\\}$ has a $D$-chef for every dish $D$. Because $f$ is a group assignment and $f$ assigns every resident of $R$ to group $n$, no two residents of $R$ make the same dish. Thus for every $D \\in \\operatorname{dish}(T)$, exactly one resident of $R$ is a $D$-chef; and each $D$-chef cooks exactly one other dish, which itself is not cooked by anyone else in $R$. Thus the dishes come in pairs: for each dish $D$, there is another dish $D^{\\prime}$ cooked by the $D$-chef in $R$ and no one else in $R$. However, if the dishes can be paired off, there must be an even number of dishes, contradicting the assumption that $|\\operatorname{dish}(T)|$ is odd. Thus for every $n$, the set $\\{r \\in T \\mid f(r)=n\\}$ must be missing a $D$-chef for some dish $D$.']",,True,,, 2634,Combinatorics,,"In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group. It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make. | ARMLton | | | :--- | :--- | | Resident | Dishes | | Paul | pie, turkey | | Arnold | pie, salad | | Kelly | salad, broth | | ARMLville | | | :--- | :--- | | Resident | Dishes | | Sally | steak, calzones | | Ross | calzones, pancakes | | David | steak, pancakes | The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}. A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad. Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$. In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$. For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups. For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$. If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$. A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles. For each odd positive integer $d \geq 3$, prove that $\operatorname{gr}\left(\mathcal{F}_{d}\right)=d$.","['Fix $D \\in \\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)$. Then for every other dish $D^{\\prime} \\in \\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)$, there is exactly one chef who makes both $D$ and $D^{\\prime}$, hence $\\left|\\operatorname{chef}_{\\mathcal{F}_{d}}(D)\\right|=d-1$, which is even because $d$ is odd. Thus for each $D \\in \\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)$, $\\left|\\operatorname{chef}_{\\mathcal{F}_{d}}(D)\\right|$ is even. Because $\\left|\\operatorname{dish}\\left(\\mathcal{F}_{d}\\right)\\right|=d$ is odd and $\\left|\\operatorname{chef}_{\\mathcal{F}_{d}}(D)\\right|=d-1$ for every dish in $\\mathcal{F}_{d}$, problem 12c applies, hence $\\operatorname{gr}\\left(\\mathcal{F}_{d}\\right)>d-1$.\n\nLabel the dishes $D_{1}, D_{2}, \\ldots, D_{d}$, and label the residents $r_{i, j}$ for $1 \\leq i0}} \\frac{1}{n} & =\\sum_{\\substack{n \\mid 24 \\\\\nn>0}} \\frac{1}{24 / n} \\\\\n& =\\frac{1}{24} \\sum_{\\substack{n \\mid 24 \\\\\nn>0}} n\n\\end{aligned}\n$$\n\nBecause $24=2^{3} \\cdot 3$, the sum of the positive divisors of 24 is $\\left(1+2+2^{2}+2^{3}\\right)(1+3)=15 \\cdot 4=60$. Hence the sum is $60 / 24=\\mathbf{5} / \\mathbf{2}$.', 'Because $24=2^{3} \\cdot 3$, any positive divisor of 24 is of the form $2^{a} 3^{b}$ where $a=0,1,2$, or 3 , and $b=0$ or 1 . So the sum of the positive divisors of 24 can be represented as the product $(1+2+4+8)(1+3)$. Similarly, the sum of their reciprocals can be represented as the product $\\left(\\frac{1}{1}+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}\\right)\\left(\\frac{1}{1}+\\frac{1}{3}\\right)$. The first sum is $\\frac{15}{8}$ and the second is $\\frac{4}{3}$, so the product is $\\mathbf{5 / 2}$.']",['$\\frac{5}{2}$'],False,,Numerical, 2638,Number Theory,,"There exists a digit $Y$ such that, for any digit $X$, the seven-digit number $\underline{1} \underline{2} \underline{3} \underline{X} \underline{5} \underline{Y} \underline{7}$ is not a multiple of 11. Compute $Y$.","['Consider the ordered pairs of digits $(X, Y)$ for which $\\underline{1} \\underline{2} \\underline{3} \\underline{X} \\underline{5} \\underline{Y} \\underline{7}$ is a multiple of 11 . Recall that a number is a multiple of 11 if and only if the alternating sum of the digits is a multiple of 11 . Because $1+3+5+7=16$, the sum of the remaining digits, namely $2+X+Y$, must equal 5 or 16 . Thus $X+Y$ must be either 3 or 14 , making $X=3-Y$ (if $Y=0,1,2$, or 3 ) or $14-Y$ (if $Y=5,6,7,8$, or 9 ). Thus a solution $(X, Y)$ exists unless $Y=4$.']",['4'],False,,Numerical, 2639,Geometry,,"A point is selected at random from the interior of a right triangle with legs of length $2 \sqrt{3}$ and 4 . Let $p$ be the probability that the distance between the point and the nearest vertex is less than 2. Then $p$ can be written in the form $a+\sqrt{b} \pi$, where $a$ and $b$ are rational numbers. Compute $(a, b)$.","['Label the triangle as $\\triangle A B C$, with $A B=2 \\sqrt{3}$ and $B C=4$. Let $D$ and $E$ lie on $\\overline{A B}$ such that $D B=A E=2$. Let $F$ be the midpoint of $\\overline{B C}$, so that $B F=F C=2$. Let $G$ and $H$ lie on $\\overline{A C}$, with $A G=H C=2$. Now draw the arcs of radius 2 between $E$ and $G, D$ and $F$, and $F$ and $H$. Let the intersection of arc $D F$ and $\\operatorname{arc} E G$ be $J$. Finally, let $M$ be the midpoint of $\\overline{A B}$. The completed diagram is shown below.\n\n\n\nThe region $R$ consisting of all points within $\\triangle A B C$ that lie within 2 units of any vertex is the union of the three sectors $E A G, D B F$, and $F C H$. The angles of these sectors, being the angles $\\angle A, \\angle B$, and $\\angle C$, sum to $180^{\\circ}$, so the sum of their areas is $2 \\pi$. Computing the area of $R$ requires subtracting the areas of all intersections of the three sectors that make up $R$.\n\nThe only sectors that intersect are $E A G$ and $D B F$. Half this area of intersection, the part above $\\overline{M J}$, equals the difference between the areas of sector $D B J$ and of $\\triangle M B J$. Triangle $M B J$ is a $1: \\sqrt{3}: 2$ right triangle because $B M=\\sqrt{3}$ and $B J=2$, so the area of $\\triangle M B J$ is $\\frac{\\sqrt{3}}{2}$. Sector $D B J$ has area $\\frac{1}{12}(4 \\pi)=\\frac{\\pi}{3}$, because $\\mathrm{m} \\angle D B J=30^{\\circ}$. Therefore the area of intersection of the sectors is $2\\left(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{2}\\right)=\\frac{2 \\pi}{3}-\\sqrt{3}$. Hence the total area of $R$ is $2 \\pi-\\left(\\frac{2 \\pi}{3}-\\sqrt{3}\\right)=\\frac{4 \\pi}{3}+\\sqrt{3}$. The total area of $\\triangle A B C$ is $4 \\sqrt{3}$, therefore the desired probability is $\\frac{\\frac{4 \\pi}{3}+\\sqrt{3}}{4 \\sqrt{3}}=\\frac{\\pi}{3 \\sqrt{3}}+\\frac{1}{4}$. Then $a=\\frac{1}{4}$ and $b=\\left(\\frac{1}{3 \\sqrt{3}}\\right)^{2}=\\frac{1}{27}$, hence the answer is $\\left(\\frac{1}{4}, \\frac{1}{27}\\right)$.']","['$(\\frac{1}{4}, \\frac{1}{27})$']",True,,Numerical, 2640,Geometry,,"The square $A R M L$ is contained in the $x y$-plane with $A=(0,0)$ and $M=(1,1)$. Compute the length of the shortest path from the point $(2 / 7,3 / 7)$ to itself that touches three of the four sides of square $A R M L$.","['Consider repeatedly reflecting square $A R M L$ over its sides so that the entire plane is covered by copies of $A R M L$. A path starting at $(2 / 7,3 / 7)$ that touches one or more sides and returns to $(2 / 7,3 / 7)$ corresponds to a straight line starting at $(2 / 7,3 / 7)$ and ending at the image of $(2 / 7,3 / 7)$ in one of the copies of $A R M L$. To touch three sides, the path must cross three lines, at least one of which must be vertical and at least one of which must be horizontal.\n\n\n\nIf the path crosses two horizontal lines and the line $x=0$, it will have traveled a distance of 2 units vertically and $4 / 7$ units vertically for a total distance of $\\sqrt{2^{2}+(4 / 7)^{2}}$ units. Similarly, the total distance traveled when crossing two horizontal lines and $x=1$ is $\\sqrt{2^{2}+(10 / 7)^{2}}$, the total distance traveled when crossing two vertical lines and $y=0$ is $\\sqrt{2^{2}+(6 / 7)^{2}}$, and the total distance traveled when crossing two vertical lines and $y=1$ is $\\sqrt{2^{2}+(8 / 7)^{2}}$. The least of these is\n\n$$\n\\sqrt{2^{2}+(4 / 7)^{2}}=\\frac{2}{\\mathbf{7}} \\sqrt{\\mathbf{5 3}}\n$$']",['$\\frac{2}{7} \\sqrt{53}$'],False,,Numerical, 2641,Algebra,,"For each positive integer $k$, let $S_{k}$ denote the infinite arithmetic sequence of integers with first term $k$ and common difference $k^{2}$. For example, $S_{3}$ is the sequence $3,12,21, \ldots$ Compute the sum of all $k$ such that 306 is an element of $S_{k}$.","['If 306 is an element of $S_{k}$, then there exists an integer $m \\geq 0$ such that $306=k+m k^{2}$. Thus $k \\mid 306$ and $k^{2} \\mid 306-k$. The second relation can be rewritten as $k \\mid 306 / k-1$, which implies that $k \\leq \\sqrt{306}$ unless $k=306$. The prime factorization of 306 is $2 \\cdot 3^{2} \\cdot 17$, so the set of factors of 306 less than $\\sqrt{306}$ is $\\{1,2,3,6,9,17\\}$. Check each in turn:\n\n$$\n\\begin{aligned}\n306-1 & =305, & & 1^{2} \\mid 305 \\\\\n306-2 & =304, & & 2^{2} \\mid 304 \\\\\n306-3 & =303, & & 3^{2} \\nmid 303 \\\\\n306-6 & =300, & & 6^{2} \\nmid 300 \\\\\n306-9 & =297, & & 9^{2} \\nmid 297 \\\\\n306-17 & =289, & & 17^{2} \\mid 289 .\n\\end{aligned}\n$$\n\nThus the set of possible $k$ is $\\{1,2,17,306\\}$, and the sum is $1+2+17+306=\\mathbf{3 2 6}$.']",['326'],False,,Numerical, 2642,Algebra,,"Compute the sum of all values of $k$ for which there exist positive real numbers $x$ and $y$ satisfying the following system of equations. $$ \left\{\begin{aligned} \log _{x} y^{2}+\log _{y} x^{5} & =2 k-1 \\ \log _{x^{2}} y^{5}-\log _{y^{2}} x^{3} & =k-3 \end{aligned}\right. $$","['Let $\\log _{x} y=a$. Then the first equation is equivalent to $2 a+\\frac{5}{a}=2 k-1$, and the second equation is equivalent to $\\frac{5 a}{2}-\\frac{3}{2 a}=k-3$. Solving this system by eliminating $k$ yields the quadratic equation $3 a^{2}+5 a-8=0$, hence $a=1$ or $a=-\\frac{8}{3}$. Substituting each of these values\n\n\n\nof $a$ into either of the original equations and solving for $k$ yields $(a, k)=(1,4)$ or $\\left(-\\frac{8}{3},-\\frac{149}{48}\\right)$. Adding the values of $k$ yields the answer of $43 / 48$.', 'In terms of $a=\\log _{x} y$, the two equations become $2 a+\\frac{5}{a}=2 k-1$ and $\\frac{5 a}{2}-\\frac{3}{2 a}=k-3$. Eliminate $\\frac{1}{a}$ to obtain $31 a=16 k-33$; substitute this into either of the original equations and clear denominators to get $96 k^{2}-86 k-1192=0$. The sum of the two roots is $86 / 96=\\mathbf{4 3} / \\mathbf{4 8}$.']",['$\\frac{43}{48}$'],False,,Numerical, 2643,Geometry,,"Let $W=(0,0), A=(7,0), S=(7,1)$, and $H=(0,1)$. Compute the number of ways to tile rectangle $W A S H$ with triangles of area $1 / 2$ and vertices at lattice points on the boundary of WASH.","['Define a fault line to be a side of a tile other than its base. Any tiling of $W A S H$ can be represented as a sequence of tiles $t_{1}, t_{2}, \\ldots, t_{14}$, where $t_{1}$ has a fault line of $\\overline{W H}, t_{14}$ has a fault line of $\\overline{A S}$, and where $t_{k}$ and $t_{k+1}$ share a fault line for $1 \\leq k \\leq 13$. Also note that to determine the position of tile $t_{k+1}$, it is necessary and sufficient to know the fault line that $t_{k+1}$ shares with $t_{k}$, as well as whether the base of $t_{k+1}$ lies on $\\overline{W A}$ (abbreviated "" $\\mathrm{B}$ "" for ""bottom"") or on $\\overline{S H}$ (abbreviated ""T"" for ""top""). Because rectangle $W A S H$ has width 7 , precisely 7 of the 14 tiles must have their bases on $\\overline{W A}$. Thus any permutation of 7 B\'s and 7 T\'s determines a unique tiling $t_{1}, t_{2}, \\ldots, t_{14}$, and conversely, any tiling $t_{1}, t_{2}, \\ldots, t_{14}$ corresponds to a unique permutation of 7 B\'s and 7 T\'s. Thus the answer is $\\left(\\begin{array}{c}14 \\\\ 7\\end{array}\\right)=\\mathbf{3 4 3 2}$.', ""Let $T(a, b)$ denote the number of ways to triangulate the polygon with vertices at $(0,0),(b, 0),(a, 1),(0,1)$, where each triangle has area $1 / 2$ and vertices at lattice points. The problem is to compute $T(7,7)$. It is easy to see that $T(a, 0)=T(0, b)=1$ for all $a$ and $b$. If $a$ and $b$ are both positive, then either one of the triangles includes the edge from $(a-1,1)$ to $(b, 0)$ or one of the triangles includes the edge from $(a, 1)$ to $(b-1,0)$, but not both. (In fact, as soon as there is an edge from $(a, 1)$ to $(x, 0)$ with $x\\underline{A_{1}} \\underline{A_{2}} \\cdots \\underline{A_{n}}$. Suppose that $A_{2}=0$ and $A_{3}=A_{1}$, so that the number begins $\\underline{A_{1}} \\underline{0} \\underline{A_{1}} \\underline{A_{4}}$. If the number is to be fibbish, $A_{4} \\geq A_{1}>0$. Then if $A_{1} \\geq 2$ and $A_{4} \\geq 2$, because the number is fibbish, $A_{5} \\geq 4$, and $A_{6} \\geq 6$. In this case there can be no more digits, because $A_{5}+A_{6} \\geq 10$. So the largest possible fibbish number beginning with 20 is 202246. If $A_{1}=2$ and $A_{2}=1$, then $A_{3}$ must be at least 3 , and the largest possible number is 21459; changing $A_{3}$ to 3 does not increase the length. Now consider $A_{1}=1$. If $A_{2}=1$, then $A_{3} \\geq 2, A_{4} \\geq 3, A_{5} \\geq 5$, and $A_{6} \\geq 8$. There can be no seventh digit because that digit would have to be at least 13 . Increasing $A_{3}$ to 3 yields only two additional digits, because $A_{4} \\geq 4, A_{5} \\geq 7$. So $A_{3}=2$ yields a longer (and thus larger) number. Increasing $A_{4}$ to 4 yields only one additional digit, $A_{5} \\geq 6$, because $A_{4}+A_{5} \\geq 10$. But if $A_{4}=3$, increasing $A_{5}$ to 6 still allows $A_{6}=9$, yielding the largest possible number of digits (8) and the largest fibbish number with that many digits.']",['10112369'],False,,Numerical, 2664,Combinatorics,,"An ARMLbar is a $7 \times 7$ grid of unit squares with the center unit square removed. A portion of an ARMLbar is a square section of the bar, cut along the gridlines of the original bar. Compute the number of different ways there are to cut a single portion from an ARMLbar.","['Note that any portion of side length $m \\geq 4$ will overlap the center square, so consider only portions of side length 3 or less. If there were no hole in the candy bar, the number of portions could be counted by conditioning on the possible location of the upper-left corner of the portion. If the portion is of size $1 \\times 1$, then the corner can occupy any of the $7^{2}$ squares of the bar. If the portion is of size $2 \\times 2$, then the corner can occupy any of the top 6 rows and any of the left 6 columns, for $6^{2}=36$ possible $2 \\times 2$ portions. In general, the upper-left corner of an $m \\times m$ portion can occupy any of the top $8-m$ rows and any of the left $8-m$ columns. So the total number of portions from an intact bar would be $7^{2}+6^{2}+5^{2}$. Now when $m \\leq 3$, the number of $m \\times m$ portions that include the missing square is simply $m^{2}$, because the missing square could be any square of the portion. So the net number of portions is\n\n$$\n\\begin{aligned}\n7^{2}+6^{2}+5^{2}-3^{2}-2^{2}-1^{2} & =(49+36+25)-(9+4+1) \\\\\n& =110-14 \\\\\n& =\\mathbf{9 6}\n\\end{aligned}\n$$', 'First ignore the missing square. As in the previous solution, the number of $m \\times m$ portions that can fit in the bar is $(8-m)^{2}$. So the total number of portions of all sizes is simply\n\n$$\n7^{2}+6^{2}+\\cdots+1^{2}=\\frac{7(7+1)(2 \\cdot 7+1)}{6}=140\n$$\n\nTo exclude portions that overlap the missing center square, it is useful to consider the location of the missing square within the portion. If an $m \\times m$ portion includes the missing center\n\n\n\nsquare, and $m \\leq 4$, then the missing square could be any one of the $m^{2}$ squares in the portion. If $m=5$, then the missing square cannot be in the leftmost or rightmost columns of the portion, because then the entire bar would have to extend at least four squares past the hole, and it only extends three. By similar logic, the square cannot be in the top or bottom rows of the portion. So for $m=5$, there are $3 \\cdot 3=9$ possible positions. For $m=6$, the two left and two right columns are excluded, as are the two top and the two bottom rows, for $2 \\cdot 2=4$ possible positions for the portion. And in a $7 \\times 7$ square, the only possible location of the hole is in the center. So the total number of portions overlapping the missing square is\n\n$$\n1^{2}+2^{2}+3^{2}+4^{2}+3^{2}+2^{2}+1^{2}=44 .\n$$\n\nThe difference is thus $140-44=\\mathbf{9 6}$']",['96'],False,,Numerical, 2665,Geometry,,"Regular hexagon $A B C D E F$ and regular hexagon $G H I J K L$ both have side length 24 . The hexagons overlap, so that $G$ is on $\overline{A B}, B$ is on $\overline{G H}, K$ is on $\overline{D E}$, and $D$ is on $\overline{J K}$. If $[G B C D K L]=\frac{1}{2}[A B C D E F]$, compute $L F$.","['The diagram below shows the hexagons.\n\n\n\nThe area of hexagon $G B C D K L$ can be computed as $[G B C D K L]=[A B C D E F]-[A G L K E F]$, and $[A G L K E F]$ can be computed by dividing concave hexagon $A G L K E F$ into two parallelograms sharing $\\overline{F L}$. If $A B=s$, then the height $A E$ is $s \\sqrt{3}$, so the height of parallelogram $A G L F$ is $\\frac{s \\sqrt{3}}{2}$. Thus $[A G L F]=L F \\cdot \\frac{s \\sqrt{3}}{2}$ and $[A G L K E F]=L F \\cdot s \\sqrt{3}$. On the other hand, the area of a regular hexagon of side length $s$ is $\\frac{3 s^{2} \\sqrt{3}}{2}$. Because $[G B C D K L]=\\frac{1}{2}[A B C D E F]$, it follows that $[A G L K E F]=\\frac{1}{2}[A B C D E F]$, and\n\n$$\nL F \\cdot s \\sqrt{3}=\\frac{1}{2}\\left(\\frac{3 s^{2} \\sqrt{3}}{2}\\right)=\\frac{3 s^{2} \\sqrt{3}}{4}\n$$\n\nwhence $L F=\\frac{3}{4} s$. With $s=24$, the answer is $\\mathbf{1 8}$.', 'Compute $[B C D K L G]$ as twice the area of trapezoid $B C L G$. If $A B=s$, then $B G=s-L F$ and $C L=2 s-L F$, while the height of the trapezoid is $\\frac{s \\sqrt{3}}{2}$.[^0]\n\n\n[^0]: ${ }^{1}$ The answer 115 was also accepted for this problem because of an alternate (and unintended) reasonable interpretation of the problem statement. Some students also counted portions that contained the ""hole"", with the hole being strictly inside the portion, and not along its edges.\n\n\n\nThus the area of the trapezoid is:\n\n$$\n\\frac{1}{2}\\left(\\frac{s \\sqrt{3}}{2}\\right)((s-L F)+(2 s-L F))=\\frac{s \\sqrt{3}(3 s-2 L F)}{4}\n$$\n\nDouble that area to obtain\n\n$$\n[B C D K L G]=\\frac{s \\sqrt{3}(3 s-2 L F)}{2}\n$$\n\nOn the other hand, $[A B C D E F]=\\frac{3 s^{2} \\sqrt{3}}{2}$, so\n\n$$\n\\begin{aligned}\n\\frac{s \\sqrt{3}(3 s-2 L F)}{2} & =\\frac{3 s^{2} \\sqrt{3}}{4} \\\\\n3 s-2 L F & =\\frac{3 s}{2} \\\\\nL F & =\\frac{3}{4} s .\n\\end{aligned}\n$$\n\nSubstituting $s=24$ yields $L F=\\mathbf{1 8}$.']",['18'],False,,Numerical, 2666,Number Theory,,"Compute the largest base-10 integer $\underline{A} \underline{B} \underline{C} \underline{D}$, with $A>0$, such that $\underline{A} \underline{B} \underline{C} \underline{D}=B !+C !+D !$.","['Let $\\underline{A} \\underline{B} \\underline{C} \\underline{D}=N$. Because $7 !=5040$ and $8 !=40,320, N$ must be no greater than $7 !+6 !+6 !=6480$. This value of $N$ does not work, so work through the list of possible sums in decreasing order: $7 !+6 !+5 !, 7 !+6 !+4$ !, etc. The first value that works is $N=5762=7 !+6 !+2 !$.', 'Let $\\underline{A} \\underline{B} \\underline{C} \\underline{D}=N$. Because $7 !=5040$ and $8 !=40,320$, to find the maximal value, first consider values of $N$ that include 7 as a digit. Suppose then that $N=5040+X !+Y$ !. To force a 7 to appear in this sum with maximal $N$, let $X=6$, which yields $N=5040+720+Y !=5760+Y$ !. This value of $N$ has a 7 (and a 6 ), so search for values of $Y$ to find ones that satisfy the conditions of the problem. Only $Y=1$ and $Y=2$ will do, giving 5761 and 5762 . Hence $\\mathbf{5 7 6 2}$ is the maximum possible value of $N$.']",['5762'],False,,Numerical, 2667,Number Theory,,"Let $X$ be the number of digits in the decimal expansion of $100^{1000^{10,000}}$, and let $Y$ be the number of digits in the decimal expansion of $1000^{10,000^{100,000}}$. Compute $\left\lfloor\log _{X} Y\right\rfloor$.","['The number of digits of $n$ is $\\lfloor\\log n\\rfloor+1$. Because $100^{1000^{10,000}}=\\left(10^{2}\\right)^{1000^{10,000}}, X=2$. $1000^{10,000}+1$. Similarly, $Y=3 \\cdot 10,000^{100,000}+1$. Using the change-of-base formula,\n\n$$\n\\begin{aligned}\n\\log _{X} Y=\\frac{\\log Y}{\\log X} & \\approx \\frac{\\log 3+\\log 10,000^{100,000}}{\\log 2+\\log 1000^{10,000}} \\\\\n& =\\frac{\\log 3+100,000 \\log 10,000}{\\log 2+10,000 \\log 1000} \\\\\n& =\\frac{\\log 3+100,000 \\cdot 4}{\\log 2+10,000 \\cdot 3} \\\\\n& =\\frac{400,000+\\log 3}{30,000+\\log 2}\n\\end{aligned}\n$$\n\n\n\nBoth $\\log 3$ and $\\log 2$ are tiny compared to the integers to which they are being added. If the quotient 400,000/30,000 were an integer (or extremely close to an integer), the values of these logarithms might matter, but $400,000 / 30,000=40 / 3=13 . \\overline{3}$, so in this case, they are irrelevant. Hence\n\n$$\n\\left\\lfloor\\log _{X} Y\\right\\rfloor=\\left\\lfloor\\frac{400,000}{30,000}\\right\\rfloor=\\left\\lfloor\\frac{40}{3}\\right\\rfloor=13\n$$']",['13'],False,,Numerical, 2668,Geometry,,Compute the smallest possible value of $n$ such that two diagonals of a regular $n$-gon intersect at an angle of 159 degrees.,"['Let the vertices of the polygon be $A_{0}, A_{1}, \\ldots, A_{n-1}$. Considering the polygon as inscribed in a circle, the angle between diagonals $\\overline{A_{0} A_{i}}$ and $\\overline{A_{0} A_{j}}$ is $\\frac{1}{2} \\cdot\\left(\\frac{360^{\\circ}}{n}\\right) \\cdot|j-i|=\\left(\\frac{180|j-i|}{n}\\right)^{\\circ}$. The diagonal $\\overline{A_{k} A_{k+j}}$ can be considered as the rotation of $\\overline{A_{0} A_{j}}$ through $k / n$ of a circle, or $\\left(\\frac{360 k}{n}\\right)^{\\circ}$. So the diagonals $A_{0} A_{i}$ and $A_{k} A_{k+j}$ intersect at a combined angle of $\\left(\\frac{180|j-i|}{n}\\right)^{\\circ}+\\left(\\frac{360 k}{n}\\right)^{\\circ}$. Without loss of generality, assume $i\n\nTriangle $B Q R$ appears to be equilateral, and in fact, it is. Reflect the diagram in the tabletop $\\overline{A C}$ to obtain six mutually tangent congruent circles inside a larger circle:\n\n\n\nBecause the circles are congruent, their centers are equidistant from $B$, and the distances between adjacent centers are equal. So $Q$ can be obtained as the image of $R$ under a rotation of $360^{\\circ} / 6=60^{\\circ}$ counterclockwise around $B$. Then $P Q=r \\Rightarrow B Q=B R=2 r \\Rightarrow B D=$ $3 r$, hence $r=1 / 3$. Notice too that the height of the pyramid is simply the radius $r$ and the diagonal of the square base is twice the altitude of the equilateral triangle $B Q R$, that is, $2 \\cdot \\frac{r \\sqrt{3}}{2}=r \\sqrt{3}$. So the area of the base is $3 r^{2} / 2$. Thus the volume of the pyramid is $(1 / 3)\\left(3 r^{2} / 2\\right)(r)=r^{3} / 2$. Because $r=1 / 3$, the volume is $\\mathbf{1} / \\mathbf{5 4}$.']",['$\\frac{1}{54}$'],False,,Numerical, 2671,Number Theory,,Compute the smallest positive integer base $b$ for which $16_{b}$ is prime and $97_{b}$ is a perfect square.,"['Because 9 is used as a digit, $b \\geq 10$. The conditions require that $b+6$ be prime and $9 b+7$ be a perfect square. The numbers modulo 9 whose squares are congruent to 7 modulo 9 are 4 and 5. So $9 b+7=(9 k+4)^{2}$ or $(9 k+5)^{2}$ for some integer $k$. Finally, $b$ must be odd (otherwise $b+6$ is even), so $9 b+7$ must be even, which means that for any particular value of $k$, only one of $9 k+4$ and $9 k+5$ is possible. Taking these considerations together, $k=0$ is too small. Using $k=1$ makes $9 k+4$ odd, and while $(9 \\cdot 1+5)^{2}=196=9 \\cdot 21+7$ is even, because $21+6=27$ is composite, $b \\neq 21$. Using $k=2$ makes $9 k+4$ even, yielding $22^{2}=484=9 \\cdot 53+7$, and $53+6=59$ is prime. Thus $b=\\mathbf{5 3}$, and $53+6=59$ is prime while $9 \\cdot 53+7=484=22^{2}$.']",['53'],False,,Numerical, 2672,Algebra,,"For a positive integer $n$, let $C(n)$ equal the number of pairs of consecutive 1's in the binary representation of $n$. For example, $C(183)=C\left(10110111_{2}\right)=3$. Compute $C(1)+C(2)+$ $C(3)+\cdots+C(256)$.","[""Group values of $n$ according to the number of bits (digits) in their binary representations:\n\n| Bits | $C(n)$ values | Total |\n| :---: | :---: | :---: |\n| 1 | $C\\left(1_{2}\\right)=0$ | 0 |\n| 2 | $C\\left(10_{2}\\right)=0$
$C\\left(11_{2}\\right)=1$ | 1 |\n| 3 | $C\\left(100_{2}\\right)=0$ $C\\left(101_{2}\\right)=0$
$C\\left(110_{2}\\right)=1$ $C\\left(111_{2}\\right)=2$ | 3 |\n| 4 | $C\\left(1000_{2}\\right)=0$ $C\\left(1001_{2}\\right)=0$ $C\\left(1100_{2}\\right)=1$ $C\\left(1101_{2}\\right)=1$
$C\\left(1010_{2}\\right)=0$ $C\\left(1011_{2}\\right)=1$ $C\\left(1110_{2}\\right)=2$ $C\\left(1111_{2}\\right)=3$ | 8 |\n\nLet $B_{n}$ be the set of $n$-bit integers, and let $c_{n}=\\sum_{k \\in B_{n}} C(k)$ be the sum of the $C$-values for all $n$-bit integers. Observe that the integers in $B_{n+1}$ can be obtained by appending a 1 or a 0 to the integers in $B_{n}$. Appending a bit does not change the number of consecutive 1's in the previous (left) bits, but each number in $B_{n}$ generates two different numbers in $B_{n+1}$. Thus $c_{n+1}$ equals twice $2 c_{n}$ plus the number of new 11 pairs. Appending a 1 will create a new pair of consecutive 1's in (and only in) numbers that previously terminated in 1. The number of such numbers is half the number of elements in $B_{n}$. Because there are $2^{n-1}$ numbers in $B_{n}$, there are $2^{n-2}$ additional pairs of consecutive 1's among the elements in $B_{n+1}$. Thus for $n \\geq 2$, the sequence $\\left\\{c_{n}\\right\\}$ satisfies the recurrence relation\n\n$$\nc_{n+1}=2 c_{n}+2^{n-2}\n$$\n\n(Check: the table shows $c_{3}=3$ and $c_{4}=8$, and $8=2 \\cdot 3+2^{3-1}$.) Thus\n\n$$\n\\begin{aligned}\n& c_{5}=2 \\cdot c_{4}+2^{4-2}=2 \\cdot 8+4=20, \\\\\n& c_{6}=2 \\cdot c_{5}+2^{5-2}=2 \\cdot 20+8=48, \\\\\n& c_{7}=2 \\cdot c_{6}+2^{6-2}=2 \\cdot 48+16=112, \\text { and } \\\\\n& c_{8}=2 \\cdot c_{7}+2^{7-2}=2 \\cdot 112+32=256 .\n\\end{aligned}\n$$\n\nBecause $C(256)=0$, the desired sum is $c_{1}+c_{2}+c_{3}+c_{4}+c_{5}+c_{6}+c_{7}+c_{8}$, which equals 448 .""]",['448'],False,,Numerical, 2673,Combinatorics,,A set $S$ contains thirteen distinct positive integers whose sum is 120 . Compute the largest possible value for the median of $S$.,"[""Let $S_{L}$ be the set of the least six integers in $S$, let $m$ be the median of $S$, and let $S_{G}$ be the set of the greatest six integers in $S$. In order to maximize the median, the elements of $S_{L}$ should be as small as possible, so start with $S_{L}=\\{1,2,3,4,5,6\\}$. Then the sum of $S_{L}$ 's elements is 21, leaving 99 as the sum of $m$ and the six elements of $S_{G}$. If $m=11$ and $S_{G}=\\{12,13,14,15,16,17\\}$, then the sum of all thirteen elements of $S$ is 119 . It is impossible to increase $m$ any further, because then the smallest set of numbers for $S_{G}$ would be $\\{13,14,15,16,17,18\\}$, and the sum would be at least 126 . To get the sum to be exactly 120, simply increase either 6 to 7 or 17 to 18 . The answer is $\\mathbf{1 1 .}$""]",['11'],False,,Numerical, 2674,Number Theory,,"Let $T=11$. Compute the least positive integer $b$ such that, when expressed in base $b$, the number $T$ ! ends in exactly two zeroes.","['For any integers $n$ and $b$, define $d(n, b)$ to be the unique nonnegative integer $k$ such that $b^{k} \\mid n$ and $b^{k+1} \\nmid n$; for example, $d(9,3)=2, d(9,4)=0$, and $d(18,6)=1$. So the problem asks for the smallest value of $b$ such that $d(T !, b)=2$. If $p$ is a prime and $p \\mid b$, then $d(T !, b) \\leq d(T !, p)$, so the least value of $b$ such that $d(T !, b)=2$ must be prime. Also, if $b$ is prime, then $d(T !, b)=\\lfloor T / b\\rfloor+\\left\\lfloor T / b^{2}\\right\\rfloor+\\left\\lfloor T / b^{3}\\right\\rfloor+\\cdots$. The only way that $d(T, b)$ can equal 2 is if the first term $\\lfloor T / b\\rfloor$ equals 2 and all other terms equal zero. (If $T \\geq b^{2}$, then $b \\geq 2$ implies $T / b \\geq b \\geq 2$, which would mean the first two terms by themselves would have a sum of at least 3.) Thus $2 b \\leq T<3 b$, hence $b \\leq T / 2$ and $T / 3100 T$.","['Start by computing the first few terms of the sequence: $a_{1}=1, a_{2}=\\lceil\\sqrt{35}\\rceil=6, a_{3}=$ $\\lceil\\sqrt{70}\\rceil=9$, and $a_{4}=\\lceil\\sqrt{115}\\rceil=11$. Note that when $m \\geq 17,(m+1)^{2}=m^{2}+2 m+1>$ $m^{2}+34$, so if $a_{n} \\geq 17, a_{n+1}=\\left[\\sqrt{a_{n}^{2}+34}\\right\\rceil=a_{n}+1$. So it remains to continue the sequence until $a_{n} \\geq 17: a_{5}=13, a_{6}=15, a_{7}=17$. Then for $n>7, a_{n}=17+(n-7)=n+10$, and $a_{n}>100 T \\Rightarrow n>100 T-10$. With $T=5, n>490$, and the least value of $n$ is 491 .']",['491'],False,,Numerical, 2676,Geometry,,"Compute the smallest $n$ such that in the regular $n$-gon $A_{1} A_{2} A_{3} \cdots A_{n}, \mathrm{~m} \angle A_{1} A_{20} A_{13}<60^{\circ}$.","['If the polygon is inscribed in a circle, then the arc $\\overparen{A_{1} A_{13}}$ intercepted by $\\angle A_{1} A_{20} A_{13}$ has measure $12\\left(360^{\\circ} / n\\right)$, and thus $\\mathrm{m} \\angle A_{1} A_{20} A_{13}=6\\left(360^{\\circ} / n\\right)$. If $6(360 / n)<60$, then $n>6(360) / 60=$ 36. Thus the smallest value of $n$ is $\\mathbf{3 7}$.']",['37'],False,,Numerical, 2677,Geometry,,Let $T=37$. A cube has edges of length $T$. Square holes of side length 1 are drilled from the center of each face of the cube through the cube's center and across to the opposite face; the edges of each hole are parallel to the edges of the cube. Compute the surface area of the resulting solid.,"['After the holes have been drilled, each face of the cube has area $T^{2}-1$. The three holes meet in a $1 \\times 1 \\times 1$ cube in the center, forming six holes in the shape of rectangular prisms whose bases are $1 \\times 1$ squares and whose heights are $(T-1) / 2$. Each of these holes thus contributes $4(T-1) / 2=2(T-1)$ to the surface area, for a total of $12(T-1)$. Thus the total area is $6\\left(T^{2}-1\\right)+12(T-1)$, which can be factored as $6(T-1)(T+1+2)=6(T-1)(T+3)$. With $T=37$, the total surface area is $6(36)(40)=\\mathbf{8 6 4 0}$.']",['8640'],False,,Numerical, 2678,Algebra,,Let $T=8640$. Compute $\left\lfloor\log _{4}\left(1+2+4+\cdots+2^{T}\right)\right\rfloor$.,"['Let $S=\\log _{4}\\left(1+2+4+\\cdots+2^{T}\\right)$. Because $1+2+4+\\cdots+2^{T}=2^{T+1}-1$, the change-of-base formula yields\n\n$$\nS=\\frac{\\log _{2}\\left(2^{T+1}-1\\right)}{\\log _{2} 4}\n$$\n\n\n\nLet $k=\\log _{2}\\left(2^{T+1}-1\\right)$. Then $Td(b))$. Then $a-b=3^{\\alpha}\\left(a_{0}-3^{\\beta-\\alpha} b_{0}\\right)$. In this second factor, notice that the second term, $3^{\\beta-\\alpha} b_{0}$ is divisible by 3 but the first term $a_{0}$ is not, so their difference is not divisible by 3 . Thus $d(a, b)=3^{-\\alpha}=d(a)$. Therefore $d(a, b)=d(a)$ when $d(a)>d(b)$, and similarly $d(a, b)=d(b)$ when $d(b)>d(a)$. Hence $d(a, b) \\leq \\max \\{d(a), d(b)\\}$.']",,True,,, 2691,Geometry,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses. Prove that, for all $a, b$, and $c, d(a, c) \leq \max \{d(a, b), d(b, c)\}$.","['Note that $d(a, c)=d(a-c)=d((a-b)-(c-b))=d(a-b, c-b)$. Then $d(a, c) \\leq$ $\\max \\{d(a, b), d(c, b)\\}$ by part 5 a, and the fact that $d(c, b)=d(b, c)$.']",,True,,, 2692,Geometry,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses. Prove that, for all $a, b$, and $c, d(a, c) \leq d(a, b)+d(b, c)$.","['Because $d(x, y)>0$ for all $x \\neq y, \\max \\{d(a, b), d(b, c)\\}0$, so $d(a, c)1 / 9$ and $d(x)<1 / 9$. In the first case, $9 \\nmid x$ (either $x$ is not divisible by 3 , or $x$ is divisible by 3 and not 9 ). These houses do not move into $\\mathcal{N}(9): x \\not \\equiv 0 \\bmod 9 \\Rightarrow x+18 \\not \\equiv 0 \\bmod 9$. On the other hand, if $d(x)<1 / 9$, then $27 \\mid x$, i.e., $x=27 k$. Then $x+18=27 k+18$ which is divisible by 9 but not by 27 , so $d(x+18)=9$. So in fact every house $x$ of ARMLopolis with $d(x)<1 / 9$ will move into $\\mathcal{N}(9)$ !']",,True,,, 2698,Geometry,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. One day, Paul (house 23) and Sally (house 32), longtime residents of $\mathcal{N}(1)$, are discussing the "" 2 side"" of the neighborhood, that is, the set of houses $\{n \mid n=3 k+2, k \in \mathbb{Z}\}$. Paul says ""I feel like I'm at the center of the "" 2 side"": when I looked out my window, I realized that the ""2 side"" consists of exactly those houses whose distance from me is at most $1 / 3 . ""$ Prove that Paul is correct.","['If $n=3 k+2$, then $d(23, n)=d(3 k-21)$, and because $3 k-21=3(k-7), d(3 k-21) \\leq$ $1 / 3$. On the other hand, if $d(23, n) \\leq 1 / 3$, then $3 \\mid(n-23)$, so $n-23=3 k$, and $n=3 k+23=3(k+7)+2$.']",,True,,, 2699,Geometry,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. Sally (house 32). Prove that $\{n \mid n=3 k+2, k \in \mathbb{Z}\}$ consists exactly of those houses whose distance from Sally's house is at most $1 / 3$.","['If $n=3 k+2$, then $d(32, n)=d(3 k-30) \\leq 1 / 3$. Similarly, if $d(32, n) \\leq 1 / 3$, then $3 \\mid n-32 \\Rightarrow n=3 k+32=3(k+10)+2$.']",,True,,, 2700,Geometry,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. For any $x$, let $\mathcal{D}_{r}(x)=\{y \mid d(x, y) \leq r\}$, that is, $\mathcal{D}_{r}(x)$ is the disk of radius $r$. Prove that if $d(x, z)=r$, then $\mathcal{D}_{r}(x)=\mathcal{D}_{r}(z) . \quad[2 \mathrm{pts}]$","['Show that $\\mathcal{D}_{r}(x) \\subseteq \\mathcal{D}_{r}(z)$ and conversely. Suppose that $y \\in \\mathcal{D}_{r}(x)$. Then $d(x, y) \\leq r$. Because $d(x, z)=r$, by $5 \\mathrm{~b}, d(y, z) \\leq \\max \\{d(x, y), d(x, z)\\}=r$. So $y \\in \\mathcal{D}_{r}(z)$. Thus $\\mathcal{D}_{r}(x) \\subseteq \\mathcal{D}_{r}(z)$. Similarly, $\\mathcal{D}_{r}(z) \\subseteq \\mathcal{D}_{r}(x)$, so the two sets are equal.']",,True,,, 2701,Geometry,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35. Ross takes a walk starting at his house, which is number 34 . He first visits house $n_{1}$, such that $d\left(n_{1}, 34\right)=1 / 3$. He then goes to another house, $n_{2}$, such that $d\left(n_{1}, n_{2}\right)=1 / 3$. Continuing in that way, he visits houses $n_{3}, n_{4}, \ldots$, and each time, $d\left(n_{i}, n_{i+1}\right)=1 / 3$. At the end of the day, what is his maximum possible distance from his original house? Justify your answer.","['The maximum possible distance $d\\left(34, n_{k}\\right)$ is $1 / 3$. This can be proved by induction on $k: d\\left(n_{1}, 34\\right) \\leq 1 / 3$, and if both $d\\left(n_{k-1}, 34\\right) \\leq 1 / 3$ and $d\\left(n_{k-1}, n_{k}\\right) \\leq 1 / 3$, then $\\max \\left\\{d\\left(n_{k-1}, 34\\right), d\\left(n_{k-1}, n_{k}\\right)\\right\\} \\leq 1 / 3$ so by 5 b, $d\\left(34, n_{k}\\right) \\leq 1 / 3$.']",['$1/3$'],False,,Numerical, 2702,Algebra,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$. Compute $d(3 / 5), d(5 / 8)$, and $d(7 / 18)$.","['$\\frac{1}{3}, 1, 9$']","['$\\frac{1}{3}, 1, 9$']",True,,Numerical, 2703,Algebra,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$. Determine all pairs of relatively prime integers $p$ and $q$ such that $p / q \in \mathcal{N}(4 / 3)$.","['Because $d(4 / 3)=1 / 3, \\mathcal{N}(4 / 3)=\\{p / q$ in lowest terms such that $3 \\mid q$ and $9 \\nmid q\\}$. Thus the set of all possible $(p, q)$ consists precisely of those ordered pairs $\\left(p^{\\prime} r, q^{\\prime} r\\right)$, where $r$ is any non-zero integer, $p^{\\prime}$ and $q^{\\prime}$ are relatively prime integers, $3 \\mid q^{\\prime}$, and $9 \\nmid q^{\\prime}$.']","['$(p^{\\prime} r, q^{\\prime} r)$, where $r$ is any non-zero integer, $p^{\\prime}$ and $q^{\\prime}$ are relatively prime integers, $3 \\mid q^{\\prime}$, and $9 \\nmid q^{\\prime}$']",True,,Need_human_evaluate, 2704,Algebra,,"In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum ""location, location, location,"" this Power Question will refer to ""houses"" and ""house numbers"" interchangeably. Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$. The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$. Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$. A longtime resident of IMOpia moves to ARMLopolis and hopes to keep his same address. When asked, he says ""My old address was $e$, that is, the sum $\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\ldots$ I'd really like to keep that address, because the addresses of my friends here are all partial sums of this series: Alice's house is $\frac{1}{0 !}$, Bob's house is $\frac{1}{0 !}+\frac{1}{1 !}$, Carol's house is $\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}$, and so on. Just let me know what I have to do in order to be near my friends!"" After some head-scratching, the ARMLopolitan planning council announces that this request cannot be satisfied: there is no number, rational or otherwise, that corresponds to the (infinite) sum, or that is arbitrarily close to the houses in this sequence. Prove that the council is correct (and not just bureaucratic).","[""The houses in this sequence actually get arbitrarily far apart from each other, and so there's no single house which they approach. If $H_{n}=\\frac{1}{0 !}+\\frac{1}{1 !}+\\cdots+\\frac{1}{n !}$, then $d\\left(H_{n}, H_{n-1}\\right)=$ $d(1 / n !)=3^{k}$, where $k=\\lfloor n / 3\\rfloor+\\lfloor n / 9\\rfloor+\\lfloor n / 27\\rfloor \\ldots$. In particular, if $n !=3^{k} \\cdot n_{0}$, where $k>0$ and $3 \\nmid n_{0}$, then $d\\left(H_{n}, H_{n-1}\\right)=3^{k}$, which can be made arbitrarily large as $n$ increases.""]",,True,,, 2705,Geometry,,"Let $A R M L$ be a trapezoid with bases $\overline{A R}$ and $\overline{M L}$, such that $M R=R A=A L$ and $L R=$ $A M=M L$. Point $P$ lies inside the trapezoid such that $\angle R M P=12^{\circ}$ and $\angle R A P=6^{\circ}$. Diagonals $A M$ and $R L$ intersect at $D$. Compute the measure, in degrees, of angle $A P D$.","['First, determine the angles of $A R M L$. Let $\\mathrm{m} \\angle M=x$. Then $\\mathrm{m} \\angle L R M=x$ because $\\triangle L R M$ is isosceles, and $\\mathrm{m} \\angle R L M=180^{\\circ}-2 x$. Because $\\overline{A R} \\| \\overline{L M}, \\mathrm{~m} \\angle A R M=180^{\\circ}-x$ and $\\mathrm{m} \\angle A R L=180^{\\circ}-2 x$, as shown in the diagram below.\n\n\n\nHowever, $\\triangle A R L$ is also isosceles (because $A R=A L$ ), so $\\mathrm{m} \\angle A L R=180^{\\circ}-2 x$, yielding $\\mathrm{m} \\angle A L M=360^{\\circ}-4 x$. Because $\\mathrm{m} \\angle R M L=\\mathrm{m} \\angle A L M$, conclude that $360^{\\circ}-4 x=x$, so $x=72^{\\circ}$. Therefore the base angles $L$ and $M$ have measure $72^{\\circ}$ while the other base angles $A$ and $R$ have measure $108^{\\circ}$. Finally, the angle formed by diagonals $\\overline{A M}$ and $\\overline{L R}$ is as follows: $\\mathrm{m} \\angle R D M=180^{\\circ}-\\mathrm{m} \\angle L R M-\\mathrm{m} \\angle A M R=180^{\\circ}-72^{\\circ}-36^{\\circ}=72^{\\circ}$.\n\nNow construct equilateral $\\triangle R O M$ with $O$ on the exterior of the trapezoid, as shown below.\n\n\n\nBecause $A R=R M=R O$, triangle $O A R$ is isosceles with base $\\overline{A O}$. The measure of $\\angle A R O$ is $108^{\\circ}+60^{\\circ}=168^{\\circ}$, so $\\mathrm{m} \\angle R A O=(180-168)^{\\circ} / 2=6^{\\circ}$. Thus $P$ lies on $\\overline{A O}$. Additionally, $\\mathrm{m} \\angle P O M=\\mathrm{m} \\angle A O M=60^{\\circ}-6^{\\circ}=54^{\\circ}$, and $\\mathrm{m} \\angle P M O=60^{\\circ}+12^{\\circ}=72^{\\circ}$ by construction. Thus $\\mathrm{m} \\angle M P O=180^{\\circ}-72^{\\circ}-54^{\\circ}=54^{\\circ}$, hence $\\triangle P M O$ is isosceles with $P M=O M$. But because $O M=R M, \\triangle R M P$ is isosceles with $R M=M P$, and $R M=D M$ implies that $\\triangle P D M$ is also isosceles. But $\\mathrm{m} \\angle R M P=12^{\\circ}$ implies that $\\mathrm{m} \\angle P M D=36^{\\circ}-12^{\\circ}=24^{\\circ}$, so $\\mathrm{m} \\angle D P M=78^{\\circ}$. Thus $\\mathrm{m} \\angle A P D=180^{\\circ}-\\mathrm{m} \\angle O P M-\\mathrm{m} \\angle D P M=180^{\\circ}-54^{\\circ}-78^{\\circ}=48^{\\circ}$.']",['48'],False,,Numerical, 2706,Geometry,,A regular hexagon has side length 1. Compute the average of the areas of the 20 triangles whose vertices are vertices of the hexagon.,"['There are 6 triangles of side lengths $1,1, \\sqrt{3} ; 2$ equilateral triangles of side length $\\sqrt{3}$; and 12 triangles of side lengths $1, \\sqrt{3}, 2$. One triangle of each type is shown in the diagram below.\n\n\nEach triangle in the first set has area $\\sqrt{3} / 4$; each triangle in the second set has area $3 \\sqrt{3} / 4$; and each triangle in the third set has area $\\sqrt{3} / 2$. The average is\n\n$$\n\\frac{6\\left(\\frac{\\sqrt{3}}{4}\\right)+2\\left(\\frac{3 \\sqrt{3}}{4}\\right)+12\\left(\\frac{\\sqrt{3}}{2}\\right)}{20}=\\frac{\\frac{6 \\sqrt{3}}{4}+\\frac{6 \\sqrt{3}}{4}+\\frac{24 \\sqrt{3}}{4}}{20}=\\frac{\\mathbf{9} \\sqrt{\\mathbf{3}}}{\\mathbf{2 0}} .\n$$']",['$\\frac{9 \\sqrt{3}}{20}$'],False,,Numerical, 2707,Algebra,,"Paul was planning to buy 20 items from the ARML shop. He wanted some mugs, which cost $\$ 10$ each, and some shirts, which cost $\$ 6$ each. After checking his wallet he decided to put $40 \%$ of the mugs back. Compute the number of dollars he spent on the remaining items.","['The problem does not state the number of mugs Paul intended to buy, but the actual number is irrelevant. Suppose Paul plans to buy $M$ mugs and $20-M$ shirts. The total cost is $10 M+6(20-M)$ However, he puts back $40 \\%$ of the mugs, so he ends up spending $10(0.6 M)+$ $6(20-M)=6 M+120-6 M=\\mathbf{1 2 0}$ dollars.']",['120'],False,,Numerical, 2708,Number Theory,,"Let $x$ be the smallest positive integer such that $1584 \cdot x$ is a perfect cube, and let $y$ be the smallest positive integer such that $x y$ is a multiple of 1584 . Compute $y$.","['In order for $1584 \\cdot x$ to be a perfect cube, all of its prime factors must be raised to powers divisible by 3 . Because $1584=2^{4} \\cdot 3^{2} \\cdot 11$, $x$ must be of the form $2^{3 k+2} \\cdot 3^{3 m+1} \\cdot 11^{3 n+2} \\cdot r^{3}$, for nonnegative integers $k, m, n, r, r>0$. Thus the least positive value of $x$ is $2^{2} \\cdot 3 \\cdot 11^{2}=1452$. But in order for $x y$ to be a positive multiple of $1584, x y$ must be of the form $2^{a} \\cdot 3^{b} \\cdot 11^{c} \\cdot d$, where $a \\geq 4, b \\geq 2, c \\geq 1$, and $d \\geq 1$. Thus $y$ must equal $2^{2} \\cdot 3^{1}=\\mathbf{1 2}$.']",['12'],False,,Numerical, 2709,Algebra,,"Emma goes to the store to buy apples and peaches. She buys five of each, hands the shopkeeper one $\$ 5$ bill, but then has to give the shopkeeper another; she gets back some change. Jonah goes to the same store, buys 2 apples and 12 peaches, and tries to pay with a single $\$ 10$ bill. But that's not enough, so Jonah has to give the shopkeeper another $\$ 10$ bill, and also gets some change. Finally, Helen goes to the same store to buy 25 peaches. Assuming that the price in cents of each fruit is an integer, compute the least amount of money, in cents, that Helen can expect to pay.","['Let $a$ be the price of one apple and $p$ be the price of one peach, in cents. The first transaction shows that $500<5 a+5 p<1000$, hence $100\n\nThen $O P=10, P Q=O M=2$, and $O B=6$. Thus $M B=\\sqrt{6^{2}-2^{2}}=4 \\sqrt{2}$. Because $Q M=O P=10$, it follows that $Q B=10-4 \\sqrt{2}$ and $Q A=10+4 \\sqrt{2}$. So\n\n$$\n\\begin{aligned}\nP A^{2}+P B^{2} & =\\left(Q A^{2}+Q P^{2}\\right)+\\left(Q B^{2}+Q P^{2}\\right) \\\\\n& =(10+4 \\sqrt{2})^{2}+2^{2}+(10-4 \\sqrt{2})^{2}+2^{2} \\\\\n& =\\mathbf{2 7 2}\n\\end{aligned}\n$$']",['272'],False,,Numerical, 2711,Number Theory,,"A palindrome is a positive integer, not ending in 0 , that reads the same forwards and backwards. For example, 35253,171,44, and 2 are all palindromes, but 17 and 1210 are not. Compute the least positive integer greater than 2013 that cannot be written as the sum of two palindromes.","['If $a+b \\geq 2014$, then at least one of $a, b$ must be greater than 1006 . The palindromes greater than 1006 but less than 2014 are, in descending order, 2002, 1991, 1881, ..., 1111. Let a\n\n\n\nrepresent the larger of the two palindromes. Then for $n=2014, a=2002$ is impossible, because $2014-2002=12$. Any value of $a$ between 1111 and 2000 ends in 1 , so if $a+b=2014$, $b$ ends in 3 , and because $b<1000$, it follows that $303 \\leq b \\leq 393$. Subtracting 303 from 2014 yields 1711, and so $a \\leq 1711$. Thus $a=1661$ and $b=353$. A similar analysis shows the following results:\n\n$$\n\\begin{aligned}\n& 2015=1551+464 ; \\\\\n& 2016=1441+575 ; \\\\\n& 2017=1331+686 ; \\text { and } \\\\\n& 2018=1221+797\n\\end{aligned}\n$$\n\nBut 2019 cannot be expressed as the sum of two palindromes: $b$ would have to end in 8 , so $b=808+10 d$ for some digit $d$. Then $2019-898 \\leq a \\leq 2019-808$, hence $1121 \\leq a \\leq 1211$, and there is no palindrome in that interval.']",['2019'],False,,Numerical, 2712,Algebra,,"Positive integers $x, y, z$ satisfy $x y+z=160$. Compute the smallest possible value of $x+y z$.","['First consider the problem with $x, y, z$ positive real numbers. If $x y+z=160$ and $z$ is constant, then $y=\\frac{160-z}{x}$, yielding $x+y z=x+\\frac{z(160-z)}{x}$. For $a, x>0$, the quantity $x+\\frac{a}{x}$ is minimized when $x=\\sqrt{a}$ (proof: use the Arithmetic-Geometric Mean Inequality $\\frac{A+B}{2} \\geq \\sqrt{A B}$ with $A=x$ and $\\left.B=\\frac{a}{x}\\right)$; in this case, $x+\\frac{a}{x}=2 \\sqrt{a}$. Thus $x+y z \\geq 2 \\sqrt{z(160-z)}$. Considered as a function of $z$, this lower bound is increasing for $z<80$.\n\nThese results suggest the following strategy: begin with small values of $z$, and find a factorization of $160-z$ such that $x$ is close to $\\sqrt{z(160-z)}$. (Equivalently, such that $\\frac{x}{y}$ is close to $z$.) The chart below contains the triples $(x, y, z)$ with the smallest values of $x+y z$, conditional upon $z$.\n\n| $z$ | $(x, y, z)$ | $x+y z$ |\n| :---: | :---: | :---: |\n| 1 | $(53,3,1)$ | 56 |\n| 2 | $(79,2,2)$ | 83 |\n| 3 | $(157,1,3)$ | 160 |\n| 4 | $(26,6,4)$ | 50 |\n| 5 | $(31,5,5)$ | 56 |\n| 6 | $(22,7,6)$ | 64 |\n\nBecause $x+y z \\geq 2 \\sqrt{z(160-z)}$, it follows that $x+y z \\geq 64$ for $6 \\leq z \\leq 80$. And because $x+y z>80$ for $z \\geq 80$, the minimal value of $x+y z$ is $\\mathbf{5 0}$.']",['50'],False,,Numerical, 2713,Algebra,,Compute $\cos ^{3} \frac{2 \pi}{7}+\cos ^{3} \frac{4 \pi}{7}+\cos ^{3} \frac{8 \pi}{7}$.,"['The identity $\\cos 3 \\theta=4 \\cos ^{3} \\theta-3 \\cos \\theta$ can be rewritten into the power-reducing identity\n\n$$\n\\cos ^{3} \\theta=\\frac{1}{4} \\cos 3 \\theta+\\frac{3}{4} \\cos \\theta\n$$\n\n\n\nThus if $D$ is the desired sum,\n\n$$\n\\begin{aligned}\nD & =\\cos ^{3} \\frac{2 \\pi}{7}+\\cos ^{3} \\frac{4 \\pi}{7}+\\cos ^{3} \\frac{8 \\pi}{7} \\\\\n& =\\frac{1}{4}\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{12 \\pi}{7}+\\cos \\frac{24 \\pi}{7}\\right)+\\frac{3}{4}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right) .\n\\end{aligned}\n$$\n\nObserve that $\\cos \\frac{24 \\pi}{7}=\\cos \\frac{10 \\pi}{7}$, so\n\n$$\nD=\\frac{1}{4}\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{12 \\pi}{7}+\\cos \\frac{10 \\pi}{7}\\right)+\\frac{3}{4}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right) .\n$$\n\nNotice also that $\\cos \\theta=\\cos (2 \\pi-\\theta)$ implies $\\cos \\frac{12 \\pi}{7}=\\cos \\frac{2 \\pi}{7}, \\cos \\frac{10 \\pi}{7}=\\cos \\frac{4 \\pi}{7}$, and $\\cos \\frac{8 \\pi}{7}=$ $\\cos \\frac{6 \\pi}{7}$. Rewriting $D$ using the least positive equivalent angles yields\n\n$$\n\\begin{aligned}\nD & =\\frac{1}{4}\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}\\right)+\\frac{3}{4}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7}\\right) \\\\\n& =\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7} .\n\\end{aligned}\n$$\n\nTo evaluate this sum, use the identity $\\cos \\theta=\\cos (2 \\pi-\\theta)$ again to write\n\n$$\n2 D=\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7}+\\cos \\frac{8 \\pi}{7}+\\cos \\frac{10 \\pi}{7}+\\cos \\frac{12 \\pi}{7}\n$$\n\nIf $\\alpha=\\cos \\frac{2 \\pi}{7}+i \\sin \\frac{2 \\pi}{7}$, notice that the right side of the equation above is simply the real part of the sum $\\alpha+\\alpha^{2}+\\alpha^{3}+\\alpha^{4}+\\alpha^{5}+\\alpha^{6}$. Because $\\alpha^{n}$ is a solution to the equation $z^{7}=1$ for $n=0,1, \\ldots, 6$, the sum $1+\\alpha+\\alpha^{2}+\\cdots+\\alpha^{6}$ equals 0 . Hence $\\alpha+\\alpha^{2}+\\cdots+\\alpha^{6}=-1$ and $D=-1 / 2$.', 'Construct a cubic polynomial in $x$ for which $\\cos \\frac{2 \\pi}{7}, \\cos \\frac{4 \\pi}{7}$, and $\\cos \\frac{8 \\pi}{7}$ are zeros; then the sum of their cubes can be found using techniques from the theory of equations. In particular, suppose the three cosines are zeros of $x^{3}+b x^{2}+c x+d$. Then\n\n$$\n\\begin{aligned}\nb & =-\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right) \\\\\nc & =\\cos \\frac{2 \\pi}{7} \\cos \\frac{4 \\pi}{7}+\\cos \\frac{2 \\pi}{7} \\cos \\frac{8 \\pi}{7}+\\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7}, \\text { and } \\\\\nd & =-\\cos \\frac{2 \\pi}{7} \\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7}\n\\end{aligned}\n$$\n\nUse complex seventh roots of unity (as in the previous solution) to find $b=1 / 2$. To find $c$, use the product-to-sum formula $2 \\cos A \\cos B=\\cos (A+B)+\\cos (A-B)$ three times:\n\n$$\n\\begin{aligned}\n2 c & =\\left(\\cos \\frac{6 \\pi}{7}+\\cos \\frac{2 \\pi}{7}\\right)+\\left(\\cos \\frac{10 \\pi}{7}+\\cos \\frac{6 \\pi}{7}\\right)+\\left(\\cos \\frac{4 \\pi}{7}+\\cos \\frac{12 \\pi}{7}\\right) \\\\\n& \\left.=\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{6 \\pi}{7}+\\cos \\frac{8 \\pi}{7}+\\cos \\frac{10 \\pi}{7}+\\cos \\frac{12 \\pi}{7} \\text { [because } \\cos \\theta=\\cos (2 \\pi-\\theta)\\right] \\\\\n& =-1\n\\end{aligned}\n$$\n\n\n\nThus $c=-1 / 2$.\n\nTo compute $d$, multiply both sides by $\\sin \\frac{2 \\pi}{7}$ and use the identity $2 \\sin \\theta \\cos \\theta=\\sin 2 \\theta$ :\n\n$$\n\\begin{aligned}\nd \\sin \\frac{2 \\pi}{7} & =-\\sin \\frac{2 \\pi}{7} \\cos \\frac{2 \\pi}{7} \\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7} \\\\\n& =-\\frac{1}{2} \\sin \\frac{4 \\pi}{7} \\cos \\frac{4 \\pi}{7} \\cos \\frac{8 \\pi}{7} \\\\\n& =-\\frac{1}{4} \\sin \\frac{8 \\pi}{7} \\cos \\frac{8 \\pi}{7} \\\\\n& =-\\frac{1}{8} \\sin \\frac{16 \\pi}{7} .\n\\end{aligned}\n$$\n\nBecause $\\sin \\frac{16 \\pi}{7}=\\sin \\frac{2 \\pi}{7}$, the factors on both sides cancel, leaving\n\n$$\nd=-1 / 8\n$$\n\nThus $\\cos \\frac{2 \\pi}{7}, \\cos \\frac{4 \\pi}{7}$, and $\\cos \\frac{8 \\pi}{7}$ are roots of $x^{3}+\\frac{1}{2} x^{2}-\\frac{1}{2} x-\\frac{1}{8}$; so each value also satisfies the equation $x^{3}=-\\frac{1}{2} x^{2}+\\frac{1}{2} x+\\frac{1}{8}$. Hence the desired sum can be rewritten as\n\n$$\n\\begin{aligned}\n\\cos ^{3} \\frac{2 \\pi}{7}+\\cos ^{3} \\frac{4 \\pi}{7}+\\cos ^{3} \\frac{8 \\pi}{7} & =-\\frac{1}{2}\\left(\\cos ^{2} \\frac{2 \\pi}{7}+\\cos ^{2} \\frac{4 \\pi}{7}+\\cos ^{2} \\frac{8 \\pi}{7}\\right) \\\\\n& +\\frac{1}{2}\\left(\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}\\right)+\\frac{3}{8}\n\\end{aligned}\n$$\n\nPrevious work has already established that $\\cos \\frac{2 \\pi}{7}+\\cos \\frac{4 \\pi}{7}+\\cos \\frac{8 \\pi}{7}=-1 / 2$, so it remains to compute $\\cos ^{2} \\frac{2 \\pi}{7}+\\cos ^{2} \\frac{4 \\pi}{7}+\\cos ^{2} \\frac{8 \\pi}{7}$. The identity $A^{2}+B^{2}+C^{2}=(A+B+C)^{2}-2(A B+B C+A C)$ allows the use of previous results: $\\cos ^{2} \\frac{2 \\pi}{7}+\\cos ^{2} \\frac{4 \\pi}{7}+\\cos ^{2} \\frac{8 \\pi}{7}=(-1 / 2)^{2}-2(-1 / 2)=5 / 4$. Thus\n\n$$\n\\cos ^{3} \\frac{2 \\pi}{7}+\\cos ^{3} \\frac{4 \\pi}{7}+\\cos ^{3} \\frac{8 \\pi}{7}=-\\frac{1}{2}\\left(\\frac{5}{4}\\right)+\\frac{1}{2}\\left(-\\frac{1}{2}\\right)+\\frac{3}{8}=-\\frac{1}{2} .\n$$']",['$-\\frac{1}{2}$'],False,,Numerical, 2714,Geometry,,"In right triangle $A B C$ with right angle $C$, line $\ell$ is drawn through $C$ and is parallel to $\overline{A B}$. Points $P$ and $Q$ lie on $\overline{A B}$ with $P$ between $A$ and $Q$, and points $R$ and $S$ lie on $\ell$ with $C$ between $R$ and $S$ such that $P Q R S$ is a square. Let $\overline{P S}$ intersect $\overline{A C}$ in $X$, and let $\overline{Q R}$ intersect $\overline{B C}$ in $Y$. The inradius of triangle $A B C$ is 10 , and the area of square $P Q R S$ is 576 . Compute the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$.","['Note that in right triangle $A B C$ with right angle $C$, the inradius $r$ is equal to $\\frac{a+b-c}{2}$, where $a=B C, b=A C$, and $c=A B$, because the inradius equals the distance from the vertex of the right angle $C$ to (either) point of tangency along $\\overline{A C}$ or $\\overline{B C}$. Thus the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is equal to one-half the difference between the sum of the lengths of the legs of these triangles and the sum of the lengths of the hypotenuses of these triangles. Let $t$ be the side length of square $P Q R S$. Then the sum of the lengths of the legs of triangles $A X P, C X S, C Y R$, and $B Y Q$ is\n\n$$\n\\begin{aligned}\n& A P+P X+X S+S C+C R+R Y+Y Q+Q B \\\\\n= & A P+P S+S R+R Q+Q B \\\\\n= & A P+t+t+t+Q B \\\\\n= & A B-P Q+3 t \\\\\n= & c-t+3 t \\\\\n= & c+2 t .\n\\end{aligned}\n$$\n\n\n\nThe sum of the lengths of the hypotenuses of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $A X+X C+C Y+Y B=A C+C B=b+a$. Hence the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $\\frac{c+2 t-(a+b)}{2}=t-r$. Thus the desired sum equals $\\sqrt{576}-10=24-10=\\mathbf{1 4}$.']",['14'],False,,Numerical, 2715,Number Theory,,"Compute the sum of all real numbers $x$ such that $$ \left\lfloor\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{3}\right\rfloor=\frac{x}{7} $$","['Because the quantity on the left side is the difference of two integers, $x / 7$ must be an integer, hence $x$ is an integer (in fact a multiple of 7). Because the denominators on the left side are 2 and 3 , it is convenient to write $x=6 q+r$, where $0 \\leq r \\leq 5$, so that $\\lfloor x / 2\\rfloor=3 q+\\lfloor r / 2\\rfloor$ and $\\lfloor x / 3\\rfloor=2 q+\\lfloor r / 3\\rfloor$. Then for $r=0,1, \\ldots, 5$ these expressions can be simplified as shown in the table below.\n\n| $r$ | 0 | 1 | 2 | 3 | 4 | 5 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{x}{2}\\right\\rfloor$ | $3 q$ | $3 q$ | $3 q+1$ | $3 q+1$ | $3 q+2$ | $3 q+2$ |\n| $\\left\\lfloor\\frac{x}{3}\\right\\rfloor$ | $2 q$ | $2 q$ | $2 q$ | $2 q+1$ | $2 q+1$ | $2 q+1$ |\n| $\\left\\lfloor\\frac{x}{2}\\right\\rfloor-\\left\\lfloor\\frac{x}{3}\\right\\rfloor$ | $q$ | $q$ | $q+1$ | $q$ | $q+1$ | $q+1$ |\n\nNow proceed by cases:\n\n$r=0:$ Then $q=x / 6$. But from the statement of the problem, $q=x / 7$, so $x=0$.\n\n$r=1: \\quad$ Then $q=(x-1) / 6=x / 7 \\Rightarrow x=7$.\n\n$r=2: \\quad$ Then $q=(x-2) / 6$ and $q+1=x / 7$, so $(x+4) / 6=x / 7$, and $x=-28$.\n\n$r=3$ : Then $q=(x-3) / 6$ and $q=x / 7$, so $x=21$.\n\n$r=4: \\quad$ Then $q=(x-4) / 6$ and $q+1=x / 7$, so $(x+2) / 6=x / 7$, and $x=-14$.\n\n$r=5$ : Then $q=(x-5) / 6$ and $q+1=x / 7$, so $(x+1) / 6=x / 7$, and $x=-7$.\n\nThe sum of these values is $0+7+-28+21+-14+-7=\\mathbf{- 2 1}$.']",['-21'],False,,Numerical, 2716,Geometry,,"Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$. ![](https://cdn.mathpix.com/cropped/2023_12_21_71dab4a47f4ff807d6edg-1.jpg?height=393&width=330&top_left_y=497&top_left_x=949)","['First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\\frac{\\sqrt{3}}{2}$ and height $\\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}$, so its volume is $\\frac{1}{3} \\cdot 1^{2} \\cdot \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\\frac{\\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.\n\nFor convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6337ac12aa6d82067ee9g-1.jpg?height=407&width=395&top_left_y=249&top_left_x=903)\n\nLet $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\\frac{2}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\\frac{3}{36}+\\frac{1}{4}=\\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\\frac{2}{3}$, and $A P=\\frac{\\sqrt{6}}{3}$.\n\nNow let a vertical stretch take place along an axis parallel to $\\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\\frac{r \\sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\\overline{B P}$, the length $B P$ is unchanged, as can be seen below.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6337ac12aa6d82067ee9g-1.jpg?height=563&width=401&top_left_y=1239&top_left_x=903)\n\nThus $A B^{2}=\\frac{6 r^{2}}{9}+\\frac{1}{3}=\\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:\n\n$$\nr=\\frac{6}{\\frac{\\sqrt{2}}{3}}=\\frac{18}{\\sqrt{2}}=9 \\sqrt{2}\n$$\n\nThus $A B^{2}=\\frac{6(9 \\sqrt{2})^{2}+3}{9}=\\frac{975}{9}=\\frac{325}{3}$, and $A B=\\frac{5 \\sqrt{39}}{3}$.']",['$\\frac{5 \\sqrt{39}}{3}$'],False,,Numerical, 2716,Geometry,,"Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$. ","['First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\\frac{\\sqrt{3}}{2}$ and height $\\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}$, so its volume is $\\frac{1}{3} \\cdot 1^{2} \\cdot \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\\frac{\\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.\n\nFor convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.\n\n\n\n\n\nLet $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\\frac{2}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\\frac{3}{36}+\\frac{1}{4}=\\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\\frac{2}{3}$, and $A P=\\frac{\\sqrt{6}}{3}$.\n\nNow let a vertical stretch take place along an axis parallel to $\\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\\frac{r \\sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\\overline{B P}$, the length $B P$ is unchanged, as can be seen below.\n\n\n\nThus $A B^{2}=\\frac{6 r^{2}}{9}+\\frac{1}{3}=\\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:\n\n$$\nr=\\frac{6}{\\frac{\\sqrt{2}}{3}}=\\frac{18}{\\sqrt{2}}=9 \\sqrt{2}\n$$\n\nThus $A B^{2}=\\frac{6(9 \\sqrt{2})^{2}+3}{9}=\\frac{975}{9}=\\frac{325}{3}$, and $A B=\\frac{5 \\sqrt{39}}{3}$.']",['$\\frac{5 \\sqrt{39}}{3}$'],False,,Numerical, 2717,Algebra,,"Let $S=\{1,2, \ldots, 20\}$, and let $f$ be a function from $S$ to $S$; that is, for all $s \in S, f(s) \in S$. Define the sequence $s_{1}, s_{2}, s_{3}, \ldots$ by setting $s_{n}=\sum_{k=1}^{20} \underbrace{(f \circ \cdots \circ f)}_{n}(k)$. That is, $s_{1}=f(1)+$ $\cdots+f(20), s_{2}=f(f(1))+\cdots+f(f(20)), s_{3}=f(f(f(1)))+f(f(f(2)))+\cdots+f(f(f(20)))$, etc. Compute the smallest integer $p$ such that the following statement is true: The sequence $s_{1}, s_{2}, s_{3}, \ldots$ must be periodic after a certain point, and its period is at most $p$. (If the sequence is never periodic, then write $\infty$ as your answer.)","[""If $f$ is simply a permutation of $S$, then $\\left\\{s_{n}\\right\\}$ is periodic. To understand why, consider a smaller set $T=\\{1,2,3,4,5,6,7,8,9,10\\}$. If $f:[1,2,3,4,5,6,7,8,9,10] \\rightarrow[2,3,4,5,1,7,8,6,9,10]$, then $f$ has one cycle of period 5 and one cycle of period 3 , so the period of $f$ is 15 . However,\n\n$$\nf(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)=\n$$\n\n\n\n$$\n2+3+4+5+1+7+8+6+9+10=55,\n$$\n\nbecause $f$ just rearranges the order of the summands. So $s_{1}=s_{0}$, and for all $n, s_{n}=s_{n+1}$; in short, the period of $\\left\\{s_{n}\\right\\}$ is just 1 .\n\nIn order for $\\left\\{s_{n}\\right\\}$ to have a period greater than $1, f$ must be many-to-one, so that some values occur more than once (and some values do not occur at all) in the sum $f(1)+f(2)+\\cdots+f(10)$ (or, in the original problem, $f(1)+f(2)+\\cdots+f(20)$ ). For example, consider the function $f_{2}$ below:\n\n$$\nf_{2}:[1,2,3,4,5,6,7,8,9,10] \\rightarrow[2,3,4,5,1,10,9,10,7,3]\n$$\n\nNote that $s_{1}=2+3+4+5+1+10+9+10+7+3 \\neq 55$, so $\\left\\{s_{n}\\right\\}$ is not immediately periodic. But $\\left\\{s_{n}\\right\\}$ is eventually periodic, as the following argument shows. The function $f_{2}$ has two cycles: $1 \\rightarrow 2 \\rightarrow 3 \\rightarrow 4 \\rightarrow 5 \\rightarrow 1$, and $7 \\rightarrow 9 \\rightarrow 7$. There are also two paths that meet up with the first cycle: $6 \\rightarrow 10 \\rightarrow 3 \\rightarrow \\cdots$ and $8 \\rightarrow 10 \\rightarrow 3 \\rightarrow \\cdots$. Thus for all $k$ in $T, f_{2}\\left(f_{2}(k)\\right)$ is an element of one of these two extended cycles. Thus $\\left\\{s_{n}\\right\\}$ eventually becomes periodic.\n\nThe criterion that the function be many-to-one is necessary, but not sufficient, for $\\left\\{s_{n}\\right\\}$ to have period greater than 1 . To see why, consider the function $g:[1,2,3,4,5,6,7,8,9,10] \\rightarrow$ $[2,3,4,5,6,1,8,7,8,7]$. This function is many-to-one, and contains two cycles, $1 \\rightarrow 2 \\rightarrow$ $3 \\rightarrow 4 \\rightarrow 5 \\rightarrow 6 \\rightarrow 1$ and $7 \\rightarrow 8 \\rightarrow 7$. But because $g(9)=8$ and $g(10)=7$, the sum $s_{1}=2+3+4+5+6+1+8+7+8+7$, while $s_{2}=3+4+5+6+1+2+7+8+7+8$. In fact, for $n>1, s_{n+1}=s_{n}$, because applying $f$ only permutes the 6 -cycle and switches the two 7 's and two 8's. That is, in the list $\\underbrace{(g \\circ \\cdots \\circ g)}_{n}(1), \\ldots, \\underbrace{(g \\circ \\cdots \\circ g)}_{n}(10)$, the values 7 and 8 both show up exactly twice. This cycle is balanced: each of its elements shows up the same number of times for all $n$ in the list $\\underbrace{(g \\circ \\cdots \\circ g)}_{n}(1), \\ldots, \\underbrace{(g \\circ \\cdots \\circ g)}_{n}(10)$, for all $n$ after a certain point. The conclusion is that not all many-to-one functions produce unbalanced cycles.\n\nThere are two ways a function $g$ can produce balanced cycles. First, the cycles can be selfcontained, so no element outside of the cycle is ever absorbed into the cycle, as happens with the 6-cycle in the example above. Alternatively, the outside elements that are absorbed into a cycle can all arrive at different points of the cycle, so that each element of the cycle occurs equally often in each iteration of $g$. In the example above, the values $g(9)=7$ and $g(10)=8$ balance the $7 \\rightarrow 8 \\rightarrow 7$ cycle. On the other hand, in the function $f_{2}$ above, $f(f(6))=f(f(8))=f(f(1))=3$, making the large cycle unbalanced: in $s_{2}$, the value 3 appears three times in $s_{2}$, but the value 2 only appears once in $s_{2}$.\n\nThe foregoing shows that only unbalanced cycles can affect the periodicity of $\\left\\{s_{n}\\right\\}$. Because each element of a balanced cycle occurs equally often in each iteration, the period of that component of the sum $s_{n}$ attributed to the cycle is simply 1. (The case where $f$ is a permutation of $S$ is simply a special case of this result.) In the above example, the large cycle is\n\n\n\nunbalanced. Note the following results under $f_{2}$.\n\n| $n$ | $\\overbrace{\\left(f_{2} \\circ \\cdots \\circ f_{2}\\right)}^{n}(T)$ | $s_{n}$ |\n| :---: | :---: | :---: |\n| 1 | $[2,3,4,5,1,10,9,10,7,3]$ | 54 |\n| 2 | $[3,4,5,1,2,3,7,3,9,4]$ | 41 |\n| 3 | $[4,5,1,2,3,4,9,4,7,5]$ | 40 |\n| 4 | $[5,1,2,3,4,5,7,5,9,1]$ | 42 |\n| 5 | $[1,2,3,4,5,1,9,1,7,2]$ | 35 |\n| 6 | $[2,3,4,5,1,2,7,2,9,3]$ | 38 |\n| 7 | $[3,4,5,1,2,3,9,3,7,4]$ | 41 |\n| 8 | $[4,5,1,2,3,4,7,4,9,5]$ | 40 |\n| 9 | $[5,1,2,3,4,5,9,5,7,1]$ | 42 |\n\nThe period of $\\left\\{s_{n}\\right\\}$ for $f_{2}$ is 5 , the period of the unbalanced cycle.\n\nThe interested reader may inquire whether all unbalanced cycles affect the periodicity of $\\left\\{s_{n}\\right\\}$; we encourage those readers to explore the matter independently. For the purposes of solving this problem, it is sufficient to note that unbalanced cycles can affect $\\left\\{s_{n}\\right\\}$ 's periodicity.\n\nFinally, note that an unbalanced $k$-cycle actually requires at least $k+1$ elements: $k$ to form the cycle, plus at least 1 to be absorbed into the cycle and cause the imbalance. For the original set $S$, one way to create such an imbalance would be to have $f(20)=f(1)=$ $2, f(2)=3, f(3)=4, \\ldots, f(19)=1$. This arrangement creates an unbalanced cycle of length 19. But breaking up into smaller unbalanced cycles makes it possible to increase the period of $\\left\\{s_{n}\\right\\}$ even more, because then in most cases the period is the least common multiple of the periods of the unbalanced cycles. For example, $f:[1,2,3, \\ldots, 20]=$ $[2,3,4,5,6,7,8,9,1,1,12,13,14,15,16,17,18,11,11,11]$ has an unbalanced cycle of length 9 and an unbalanced cycle of length 8 , giving $\\left\\{s_{n}\\right\\}$ a period of 72 .\n\nSo the goal is to maximize $\\operatorname{lcm}\\left\\{k_{1}, k_{2}, \\ldots, k_{m}\\right\\}$ such that $k_{1}+k_{2}+\\cdots+k_{m}+m \\leq 20$. With $m=2$, the maximal period is 72 , achieved with $k_{1}=9$ and $k_{2}=8$. With $m=3$, $k_{1}+k_{2}+k_{3} \\leq 17$, but $\\operatorname{lcm}\\{7,6,4\\}=84<\\operatorname{lcm}\\{7,5,4\\}=140$. This last result can be obtained with unbalanced cycles of length 4,5 , and 7 , with the remaining four points entering the three cycles (or with one point forming a balanced cycle of length 1, i.e., a fixed point). Choosing larger values of $m$ decreases the values of $k$ so far that they no longer form long cycles: when $m=4, k_{1}+k_{2}+k_{3}+k_{4} \\leq 16$, and even if $k_{4}=2, k_{3}=3$, and $k_{2}=5$, for a period of 30 , the largest possible value of $k_{1}=6$, which does not alter the period. (Even $k_{1}=7, k_{2}=5$, and $k_{3}=k_{4}=2$ only yields a period of 70 .) Thus the maximum period of $s_{n}$ is $\\mathbf{1 4 0}$. One such function $f$ is given below.\n\n$$\n\\begin{array}{c|cccccccccccccccccccc}\nn & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\\\\n\\hline f(n) & 2 & 3 & 4 & 1 & 1 & 7 & 8 & 9 & 10 & 6 & 6 & 13 & 14 & 15 & 16 & 17 & 18 & 12 & 12 & 20\n\\end{array}\n$$""]",['140'],False,,Numerical, 2718,Number Theory,,Compute the smallest positive integer $n$ such that $n^{2}+n^{0}+n^{1}+n^{3}$ is a multiple of 13 .,"['Note that $n^{2}+n^{0}+n^{1}+n^{3}=n^{2}+1+n+n^{3}=\\left(n^{2}+1\\right)(1+n)$. Because 13 is prime, 13 must be a divisor of one of these factors. The smallest positive integer $n$ such that $13 \\mid 1+n$ is $n=12$, whereas the smallest positive integer $n$ such that $13 \\mid n^{2}+1$ is $n=\\mathbf{5}$.']",['5'],False,,Numerical, 2719,Geometry,,"Let $T=T N Y W R$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_7104b4f1e75a0066b281g-1.jpg?height=160&width=433&top_left_y=516&top_left_x=1472)","['For each point of tangency of consecutive circles, drop a perpendicular from that point to $\\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \\cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \\cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \\cdot 4=8$.']",['8'],False,,Numerical, 2719,Geometry,,"Let $T=T N Y W R$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$. ","['For each point of tangency of consecutive circles, drop a perpendicular from that point to $\\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \\cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \\cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \\cdot 4=8$.']",['8'],False,,Numerical, 2720,Algebra,,Let $T=T N Y W R$. Compute $2^{\log _{T} 8}-8^{\log _{T} 2}$.,['Let $\\log _{T} 8=x$. Then $T^{x}=8$. Thus the given expression equals $2^{x}-\\left(T^{x}\\right)^{\\log _{T} 2}=2^{x}-T^{x \\log _{T} 2}=$ $2^{x}-T^{\\log _{T} 2^{x}}=2^{x}-2^{x}=\\mathbf{0}$ (independent of $T$ ).'],['0'],False,,Numerical, 2721,Combinatorics,,"Let $T=T N Y W R$. At some point during a given week, a law enforcement officer had issued $T+2$ traffic warnings, 20 tickets, and had made $T+5$ arrests. How many more tickets must the officer issue in order for the combined number of tickets and arrests to be 20 times the number of warnings issued that week?","['The problem requests the value of $k$ such that $20+k+T+5=20(T+2)$, thus $k=19 T+15$. With $T=0$, it follows that $k=\\mathbf{1 5}$.']",['15'],False,,Numerical, 2722,Geometry,,"$\quad$ Let $T=T N Y W R$. In parallelogram $A R M L$, points $P$ and $Q$ trisect $\overline{A R}$ and points $W, X, Y, Z$ divide $\overline{M L}$ into fifths (where $W$ is closest to $M$, and points $X$ and $Y$ are both between $W$ and $Z$ ). If $[A R M L]=T$, compute $[P Q W Z]$.","['Let $h$ be the distance between $\\overline{A R}$ and $\\overline{M L}$, and for simplicity, let $A R=M L=15 n$. Then $[A R M L]=15 n h$, and $[P Q W Z]=(1 / 2)(P Q+W Z) h$. Note that $P Q=15 n / 3=5 n$ and $W Z=15 n-3 n-3 n=9 n$. Thus $[P Q W Z]=7 n h=(7 / 15) \\cdot[A R M L]=7 T / 15$. With $T=15$, the answer is 7 .']",['7'],False,,Numerical, 2723,Number Theory,,Let $T=T N Y W R$. Compute the number of positive perfect cubes that are divisors of $(T+10) !$.,"['Let $N=T+10$. In order for $k^{3}(k \\in \\mathbb{N})$ to be a divisor of $N$ !, the largest odd prime factor of $k$ (call it $p$ ) must be less than or equal to $N / 3$ so that there are at least three multiples of $p$ among the product of the first $N$ positive integers. If $p=3$, then the smallest possible value of $N$ is 9 , and the largest perfect cube factor of 9 ! is $2^{6} \\cdot 3^{3}$. Similarly, if $p=5$, then the smallest possible value of $N$ is 15 , and the largest perfect cube factor of 15 ! is $2^{9} \\cdot 3^{6} \\cdot 5^{3}$. With $T=7, N=17$, and the largest perfect cube factor of 17 ! is $2^{15} \\cdot 3^{6} \\cdot 5^{3}$. Thus $k^{3} \\mid 17$ ! if and only if $k \\mid 2^{5} \\cdot 3^{2} \\cdot 5^{1}$. Therefore $k=2^{x} 3^{y} 5^{z}$, where $x, y, z$ are nonnegative integers with $x \\leq 5, y \\leq 2, z \\leq 1$, yielding $6 \\cdot 3 \\cdot 2=\\mathbf{3 6}$ possible values of $k$.']",['36'],False,,Numerical, 2724,Geometry,,"Let $T=T N Y W R$. The graph of $y=x^{2}+2 x-T$ intersects the $x$-axis at points $A$ and $M$, which are diagonally opposite vertices of square $A R M L$. Compute $[A R M L]$.","['Note that the $x$-coordinates of $A$ and $M$ correspond to the two roots $r_{1}, r_{2}$ of $x^{2}+2 x-T$. If $s$ is the side length of square $A R M L$, then $A M=s \\sqrt{2}=\\left|r_{1}-r_{2}\\right|=\\sqrt{\\left(r_{1}-r_{2}\\right)^{2}}=$ $\\sqrt{\\left(r_{1}+r_{2}\\right)^{2}-4 r_{1} r_{2}}=\\sqrt{(-2)^{2}-4(-T)}=2 \\sqrt{1+T}$. Thus $[A R M L]=s^{2}=2(1+T)$. With $T=36,[A R M L]=\\mathbf{7 4}$.']",['74'],False,,Numerical, 2725,Number Theory,,"Let $S$ be the set of prime factors of the numbers you receive from positions 7 and 9 , and let $p$ and $q$ be the two least distinct elements of $S$, with $p2$, if $n$ is odd, then $a_{n}=a_{n-1}^{2}-a_{n-2}^{2}$, while if $n$ is even, then $a_{n}=2 a_{n-2} a_{n-3}$. Compute the sum of the squares of the first $T-3$ terms of the sequence.","['Using the identity $\\left(x^{2}-y^{2}\\right)^{2}+(2 x y)^{2}=\\left(x^{2}+y^{2}\\right)^{2}$, notice that $a_{2 n+1}^{2}+a_{2 n+2}^{2}=\\left(a_{2 n}^{2}-a_{2 n-1}^{2}\\right)^{2}+$ $\\left(2 a_{2 n} a_{2 n-1}\\right)^{2}=\\left(a_{2 n}^{2}+a_{2 n-1}^{2}\\right)^{2}$. So surprisingly, for all $n \\in \\mathbb{N}, a_{2 n+1}^{2}+a_{2 n+2}^{2}=1$. Thus if $n$ is even, the sum of the squares of the first $n$ terms is $n / 2$. With $T=19, T-3=16$, and the sum is 8 .']",['8'],False,,Numerical, 2729,Geometry,,Let $T=T N Y W R$. A regular $n$-gon has exactly $T$ more diagonals than a regular $(n-1)$-gon. Compute the value of $n$.,"['Using the formula $D(n)=\\frac{n(n-3)}{2}$ twice yields $D(n)-D(n-1)=\\frac{n^{2}-3 n}{2}-\\frac{n^{2}-5 n+4}{2}=\\frac{2 n-4}{2}=n-2$. So $T=n-2$, thus $n=T+2$, and with $T=17, n=19$.']",['19'],False,,Numerical, 2730,Algebra,,"Let $T=T N Y W R$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$, is arithmetic with $a_{16}=13$ and $a_{30}=20$. Compute the value of $k$ for which $a_{k}=T$.","['If $d$ is the common difference of the sequence, then the $n^{\\text {th }}$ term of the sequence is $a_{n}=$ $a_{16}+d(n-16)$. The values $a_{16}=13$ and $a_{30}=20$ yield $d=(20-13) /(30-16)=1 / 2$, hence $a_{n}=13+(1 / 2)(n-16)$. If $a_{n}=T$, then $n=2(T-13)+16=2 T-10$. With $T=27 / 2$, it follows that $n=\\mathbf{1 7}$.']",['17'],False,,Numerical, 2731,Geometry,,"Let $T=T N Y W R$. A rectangular prism has a length of 1 , a width of 3 , a height of $h$, and has a total surface area of $T$. Compute the value of $h$.","['The surface area is given by the expression $2 \\cdot 1 \\cdot 3+2 \\cdot 1 \\cdot h+2 \\cdot 3 \\cdot h=6+8 h$. Because $6+8 h=T, h=\\frac{T-6}{8}$. With $T=114, h=108 / 8=\\mathbf{2 7} / \\mathbf{2}$.']",['$\\frac{27}{2}$'],False,,Numerical, 2732,Algebra,,"The zeros of $x^{2}+b x+93$ are $r$ and $s$. If the zeros of $x^{2}-22 x+c$ are $r+1$ and $s+1$, compute $c$.","['Use sums and products of roots formulas: the desired quantity $c=(r+1)(s+1)=r s+r+s+1$. From the first equation, $r s=93$, while from the second equation, $(r+1)+(s+1)=r+s+2=$ 22. So $r s+r+s+1=93+22-1=\\mathbf{1 1 4}$.']",['114'],False,,Numerical, 2733,Number Theory,,"Let $N=888,888 \times 9,999,999$. Compute the sum of the digits of $N$.","['Write $N$ as\n\n$$\n\\begin{aligned}\n& (10,000,000-1) \\cdot 888,888 \\\\\n= & 8,888,880,000,000-888,888 \\\\\n= & 8,888,879,111,112 .\n\\end{aligned}\n$$\n\nThe sum of the digits of $N$ is 63 .']",['63'],False,,Numerical, 2734,Combinatorics,,Five equilateral triangles are drawn in the plane so that no two sides of any of the triangles are parallel. Compute the maximum number of points of intersection among all five triangles.,"['Any two of the triangles intersect in at most six points, because each side of one triangle can intersect the other triangle in at most two points. To count the total number of intersections among the five triangles, note that there are $\\left(\\begin{array}{l}5 \\\\ 2\\end{array}\\right)=10$ ways to select a pair of triangles, and each pair may result in 6 intersections. Thus $10 \\times 6=60$ is an upper bound.\n\nThis can be achieved, for example, by taking six equilateral triangles of equal size, centered at a single point, and rotating them different amounts so that no three sides intersect at a single point. Thus the answer is 60.']",['60'],False,,Numerical, 2735,Combinatorics,,"$\quad$ Let $S$ be the set of four-digit positive integers for which the sum of the squares of their digits is 17 . For example, $2023 \in S$ because $2^{2}+0^{2}+2^{2}+3^{2}=17$. Compute the median of $S$.","['In order for the sums of the squares of four digits to be 17 , the digits must be either $0,2,2$, and 3 , or $0,0,1$, and 4 , in some order. If the leading digit is 2 , there are $3 !=6$ possible four-digit numbers. If the leading digit is 1,3 , or 4 , there are $\\frac{3 !}{2 !}=3$ possible four-digit numbers. In total, there are $6+3 \\cdot 3=15$ four-digit integers in $S$, and the median will be the eighth least. The least eight integers in $S$, from least to greatest, are: 1004, 1040, 1400, 2023, 2032, 2203, 2230, 2302. Thus the median of $S$ is 2302.']",['2302'],False,,Numerical, 2736,Geometry,,"Let $E U C L I D$ be a hexagon inscribed in a circle of radius 5 . Given that $E U=U C=L I=I D=6$, and $C L=D E$, compute $C L$.","[""Let $C L=x$. Because the quadrilaterals $E U C L$ and $L I D E$ are congruent, $\\overline{E L}$ is a diameter of the circle in which the hexagon is inscribed, so $E L=10$. Furthermore, because $\\overline{E L}$ is a diameter of the circle, it follows that the inscribed $\\angle E U L$ is a right angle, hence $U L=8$.\n\n\n\n\n\nUsing Ptolemy's Theorem for cyclic quadrilaterals and the fact that $\\triangle E C L$ is also a right triangle,\n\n$$\n\\begin{aligned}\n& U C \\cdot E L+E U \\cdot C L=E C \\cdot U L \\\\\n\\Longrightarrow & 6(10+x)=8 \\sqrt{100-x^{2}} \\\\\n\\Longrightarrow & 36(10+x)^{2}=64(10+x)(10-x) \\\\\n\\Longrightarrow & 6 \\sqrt{10+x}=8 \\sqrt{10-x} \\\\\n\\Longrightarrow & 36(10+x)=64(10-x) \\\\\n\\Longrightarrow & 360+36 x=640-64 x \\\\\n\\Longrightarrow & 100 x=280 \\\\\n\\Longrightarrow & x=\\frac{\\mathbf{1 4}}{\\mathbf{5}} .\n\\end{aligned}\n$$""]",['$\\frac{14}{5}$'],False,,Numerical, 2737,Combinatorics,,"The ARMLLexicon consists of 10 letters: $\{A, R, M, L, e, x, i, c, o, n\}$. A palindrome is an ordered list of letters that read the same backwards and forwards; for example, MALAM, n, oncecno, and MoM are palindromes. Compute the number of 15-letter palindromes that can be spelled using letters in the ARMLLexicon, among which there are four consecutive letters that spell out $A R M L$.","['Any 15-letter palindrome is determined completely by its first 8 letters, because the last 7 letters must be the first 7 in reverse. Such a palindrome contains the string $A R M L$ if and only if its first 8 letters contain either $A R M L$ or $L M R A$. (The string $A R M L$ cannot cross the middle of the palindrome, because the 7th and 9th letters must be the same.) It therefore suffices to count the number of 8-letter strings consiting of letters in the ARMLLexicon that contain either ARML or LMRA.\n\nThere are 5 possible positions for $A R M L$, and likewise with $L M R A$. For each choice of position, there are four remaining letters, which can be any letter in the ARMLLexicon (here, $W, X, Y$, and $Z$ are used to denote arbitrary letters that need not be distinct). This leads to the following table:\n\n\n\n| Word | Num. Possibilities |\n| :---: | :---: |\n| ARMLWXYZ | $10^{4}$ |\n| WARMLXYZ | $10^{4}$ |\n| WXARMLYZ | $10^{4}$ |\n| WXYARMLZ | $10^{4}$ |\n| WXYZARML | $10^{4}$ |\n| LMRAWXYZ | $10^{4}$ |\n| WLMRAXYZ | $10^{4}$ |\n| WXLMRAYZ | $10^{4}$ |\n| WXYLMRAZ | $10^{4}$ |\n| WXYZLMRA | $10^{4}$ |\n\nThis gives $10 \\cdot 10^{4}$ possible words, but each word with two of ARML or LMRA (e.g., ARMLARML or $A A R M L M R A$ ) is counted twice. There are four words with two of $A R M L$ or $L M R A$ that use all 8 letters, and four possible types of words that use 7 of the 8 positions and leave one ""free space"". This leads to the following table:\n\n| Word | Num. Possibilities |\n| :---: | :---: |\n| ARMLARML | 1 |\n| LMRALMRA | 1 |\n| ARMLLMRA | 1 |\n| LMRAARML | 1 |\n| ARMLMRAW | 10 |\n| LMRARMLW | 10 |\n| WARMLMRA | 10 |\n| WLMRARML | 10 |\n\nThus the total number of desired words is $10 \\cdot 10^{4}-4 \\cdot 10-4 \\cdot 1=\\mathbf{9 9 9 5 6}$.']",['99956'],False,,Numerical, 2738,Algebra,,Let $10^{y}$ be the product of all real numbers $x$ such that $\log x=\frac{3+\left\lfloor(\log x)^{2}\right\rfloor}{4}$. Compute $y$.,"['First, note that\n\n$$\n\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq(\\log x)^{2} \\Longrightarrow \\frac{3+\\left\\lfloor(\\log x)^{2}\\right\\rfloor}{4} \\leq \\frac{3+(\\log x)^{2}}{4}\n$$\n\nTherefore\n\n$$\n\\log x \\leq \\frac{(\\log x)^{2}+3}{4} \\Longrightarrow 0 \\leq(\\log x)^{2}-4 \\log x+3=(\\log x-1)(\\log x-3)\n$$\n\nThis implies either $\\log x \\leq 1$ or $\\log x \\geq 3$, so $0 \\leq(\\log x)^{2} \\leq 1$ or $(\\log x)^{2} \\geq 9$.\n\nIn the first case, $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=0$ or $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=1$, so $\\log x=\\frac{3}{4}$ or $\\log x=1$, hence $x=10^{3 / 4}$ or $x=10$.\n\nTo solve the second case, note that $\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\geq(\\log x)^{2}-1$, so $0 \\geq(\\log x)^{2}-4 \\log x+2$. The solutions to $t^{2}-4 t+2=0$ are $t=\\frac{4 \\pm \\sqrt{16-8}}{2}=2 \\pm \\sqrt{2}$ by the Quadratic Formula, so $2-\\sqrt{2} \\leq \\log x \\leq 2+\\sqrt{2}$. This implies that $6-4 \\sqrt{2} \\leq(\\log x)^{2} \\leq 6+4 \\sqrt{2}$, so $0 \\leq\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq 11$. However, this case is for $(\\log x)^{2} \\geq 9$, so the only possibilities that need to be considered are $9 \\leq\\left\\lfloor(\\log x)^{2}\\right\\rfloor \\leq 11$.\n\n- If $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=9$, then $\\log x=3$, so $x=10^{3}$.\n- If $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=10$, then $\\log x=\\frac{13}{4}$, so $x=10^{13 / 4}$.\n- Finally, if $\\left\\lfloor(\\log x)^{2}\\right\\rfloor=11$, then $\\log x=\\frac{7}{2}$, which yields $(\\log x)^{2}=\\frac{49}{4}>12$, so there are no solutions.\n\nThus the product of all possible values of $x$ is $y=10^{3 / 4} \\cdot 10 \\cdot 10^{13 / 4} \\cdot 10^{3}=10^{8}$, so $y=\\mathbf{8}$.']",['8'],False,,Numerical, 2739,Algebra,,"The solutions to the equation $x^{2}-180 x+8=0$ are $r_{1}$ and $r_{2}$. Compute $$ \frac{r_{1}}{\sqrt[3]{r_{2}}}+\frac{r_{2}}{\sqrt[3]{r_{1}}} $$","[""First note that the solutions of the given equation are real because the equation's discriminant is positive. By Vieta's Formulas, $r_{1}+r_{2}=180(*)$ and $r_{1} r_{2}=8(* *)$. The expression to be computed can be written with a common denominator as\n\n$$\n\\frac{\\sqrt[3]{r_{1}^{4}}+\\sqrt[3]{r_{2}^{4}}}{\\sqrt[3]{r_{1} r_{2}}}\n$$\n\nBy $(* *)$, the denominator is equal to $\\sqrt[3]{8}=2$. To compute the numerator, first let $S_{k}=\\sqrt[3]{r_{1}^{k}}+\\sqrt[3]{r_{2}^{k}}$, so that the numerator is $S_{4}$. Then note that\n\n$$\n\\begin{aligned}\n\\left(S_{1}\\right)^{3} & =r_{1}+3 \\sqrt[3]{r_{1}^{2} r_{2}}+3 \\sqrt[3]{r_{2}^{2} r_{1}}+r_{2} \\\\\n& =\\left(r_{1}+r_{2}\\right)+3 \\sqrt[3]{r_{1} r_{2}}\\left(\\sqrt[3]{r_{1}}+\\sqrt[3]{r_{2}}\\right) \\\\\n& =180+3 \\cdot 2 \\cdot S_{1}\n\\end{aligned}\n$$\n\nwhere $(*)$ and $(* *)$ are used to substitute values into the second equality. Next note that $S_{1}^{3}-6 S_{1}-180$ can be factored as $\\left(S_{1}-6\\right)\\left(S_{1}^{2}+6 S_{1}+30\\right)$. Because the polynomial $t^{2}+6 t+30$ has no real roots, the unique real solution to $(\\dagger)$ is $S_{1}=6$, so $\\sqrt[3]{r_{1}}+\\sqrt[3]{r_{2}}=6$. Square each side of the previous equation to obtain $S_{2}+2 \\sqrt[3]{r_{1} r_{2}}=36$, hence $S_{2}=36-2 \\cdot 2$; that is, $\\sqrt[3]{r_{1}^{2}}+\\sqrt[3]{r_{2}^{2}}=32$. Again, square both sides of this equation to obtain $\\sqrt[3]{r_{1}^{4}}+2 \\sqrt[3]{r_{1}^{2} r_{2}^{2}}+\\sqrt[3]{r_{2}^{4}}=1024$, so $S_{4}+2 \\sqrt[3]{r_{1}^{2} r_{2}^{2}}=1024$, from which $S_{4}=1024-2 \\cdot 4=1016$. Thus the desired expression equals $\\frac{S_{4}}{2}=\\frac{1016}{2}=\\mathbf{5 0 8}$.""]",['508'],False,,Numerical, 2740,Geometry,,"Circle $\omega$ is tangent to parallel lines $\ell_{1}$ and $\ell_{2}$ at $A$ and $B$ respectively. Circle $\omega_{1}$ is tangent to $\ell_{1}$ at $C$ and to $\omega$ externally at $P$. Circle $\omega_{2}$ is tangent to $\ell_{2}$ at $D$ and to $\omega$ externally at $Q$. Circles $\omega_{1}$ and $\omega_{2}$ are also externally tangent to each other. Given that $A Q=12$ and $D Q=8$, compute $C D$.","['Let $O, O_{1}$ and $O_{2}$ be the centers, and let $r, r_{1}$ and $r_{2}$ be the radii of the circles $\\omega, \\omega_{1}$, and $\\omega_{2}$, respectively. Let $R$ be the point of tangency between $\\omega_{1}$ and $\\omega_{2}$.\n\nLet $H_{1}$ and $H_{2}$ be the projections of $O_{1}$ and $O_{2}$ onto $\\overline{A B}$. Also, let $H$ be the projection of $O_{1}$ onto $\\overline{O_{2} H_{2}}$. Note that $O H_{1}=r-r_{1}, O H_{2}=r-r_{2}, O O_{1}=r+r_{1}, O O_{2}=r+r_{2}$, and $O_{1} O_{2}=r_{1}+r_{2}$. From the Pythagorean Theorem, it follows that $O_{1} H_{1}=2 \\sqrt{r r_{1}}$ and $O_{2} H_{2}=2 \\sqrt{r r_{2}}$. Similarly, applying the Pythagorean Theorem to triangle $O_{1} H O_{2}$ yields $\\left(O_{1} H\\right)^{2}+\\left(O_{2} H\\right)^{2}=\\left(O_{1} O_{2}\\right)^{2}$, which is equivalent to\n\n$$\n\\left(2 \\sqrt{r r_{2}}-2 \\sqrt{r r_{1}}\\right)^{2}+\\left(2 r-r_{1}-r_{2}\\right)^{2}=\\left(r_{1}+r_{2}\\right)^{2}\n$$\n\nwhich yields $r^{2}=4 r_{1} r_{2}$ after simplifying.\n\n\n\n\nNote that $\\overline{A O} \\| \\overline{O_{2} D}$, hence $\\angle A O Q \\cong \\angle D O_{2} Q$, which implies that isosceles triangles $A O Q$ and $D O_{2} Q$ are similar. Thus $\\angle A Q O \\cong \\angle D Q O_{2}$ and therefore points $A, Q$, and $D$ are collinear. Analogously, it follows that the points $B, P$, and $C$ are collinear, as are the points $C, R$, and $D$.\n\nIn right triangle $A B D, \\overline{B Q}$ is the altitude to $\\overline{A D}$. By similarity of triangles, it follows that $D Q \\cdot D A=B D^{2}$ and $A Q \\cdot A D=A B^{2}$. Hence $B D=4 \\sqrt{10}, A B=4 \\sqrt{15}$, and $r=2 \\sqrt{15}$. Because $\\frac{D O_{2}}{A O}=\\frac{D Q}{A Q}=\\frac{2}{3}$, it follows that $r_{2}=\\frac{4}{3} \\sqrt{15}$ and $r_{1}=\\frac{3}{4} \\sqrt{15}$.\n\nNote that $A C=2 \\sqrt{r r_{1}}=3 \\sqrt{10}, B D=2 \\sqrt{r r_{2}}=4 \\sqrt{10}$, and\n\n$$\nC D^{2}=A B^{2}+(B D-A C)^{2}=(4 \\sqrt{15})^{2}+(4 \\sqrt{10}-3 \\sqrt{10})^{2}=240+10=250\n$$\n\nwhich implies that $C D=\\mathbf{5} \\sqrt{\\mathbf{1 0}}$.\n\nAlternate Solution: Conclude that $r^{2}=4 r_{1} r_{2}$, as explained above. Note that $\\angle C A Q \\cong \\angle Q D B \\cong \\angle Q R D$, using the fact that the two given lines are parallel and $\\omega_{2}$ is tangent one of them at $D$. Quadrilateral $C A Q R$ is cyclic, so apply Power of a Point to obtain $D Q \\cdot D A=D R \\cdot D C$. Because $\\frac{r_{2}}{r}=\\frac{Q D}{Q A}=\\frac{2}{3}$, conclude that $r_{2}=2 x, r=3 x$, and hence $r_{1}=\\frac{9}{8} x$. It follows that $\\frac{D R}{C R}=\\frac{r_{2}}{r_{1}}=\\frac{16}{9}$ and $D R=\\frac{16}{25} \\cdot C D$. Thus\n\n$$\nD R \\cdot D C=\\frac{16}{25} \\cdot C D^{2}=D Q \\cdot D A=8 \\cdot 20\n$$\n\nhence $C D=5 \\sqrt{10}$.']",['$5 \\sqrt{10}$'],False,,Numerical, 2741,Geometry,,"Given quadrilateral $A R M L$ with $A R=20, R M=23, M L=25$, and $A M=32$, compute the number of different integers that could be the perimeter of $A R M L$.","['Notice that $\\triangle A R M$ is fixed, so the number of integers that could be the perimeter of $A R M L$ is the same as the number of integers that could be the length $A L$ in $\\triangle A L M$. By the Triangle Inequality, $32-25",['Note that $G H=A B-A H-B G=22-5-5=12$. Thus\n\n$$\n\\begin{aligned}\n{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\\\\n& =22^{2}+12^{2}+\\frac{1}{2} \\cdot 5 \\cdot 12+\\frac{1}{2} \\cdot 5 \\cdot 12 \\\\\n& =484+144+30+30 \\\\\n& =\\mathbf{6 8 8} .\n\\end{aligned}\n$$'],['688'],False,,Numerical, 2743,Geometry,,"Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$. ![](https://cdn.mathpix.com/cropped/2023_12_21_605741f6c8eedebee745g-1.jpg?height=372&width=290&top_left_y=405&top_left_x=955)",['Note that $G H=A B-A H-B G=22-5-5=12$. Thus\n\n$$\n\\begin{aligned}\n{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\\\\n& =22^{2}+12^{2}+\\frac{1}{2} \\cdot 5 \\cdot 12+\\frac{1}{2} \\cdot 5 \\cdot 12 \\\\\n& =484+144+30+30 \\\\\n& =\\mathbf{6 8 8} .\n\\end{aligned}\n$$'],['688'],False,,Numerical, 2744,Number Theory,,"Let $T=688$. Let $a$ be the least nonzero digit in $T$, and let $b$ be the greatest digit in $T$. In square $N O R M, N O=b$, and points $P_{1}$ and $P_{2}$ lie on $\overline{N O}$ and $\overline{O R}$, respectively, so that $O P_{1}=O P_{2}=a$. A circle centered at $O$ has radius $a$, and quarter-circular arc $\widehat{P_{1} P_{2}}$ is drawn. There is a circle that is tangent to $\widehat{P_{1} P_{2}}$ and to sides $\overline{M N}$ and $\overline{M R}$. The radius of this circle can be written in the form $x-y \sqrt{2}$, where $x$ and $y$ are positive integers. Compute $x+y$.","['Let $r$ and $Q$ denote the respective radius and center of the circle whose radius is concerned. Let this circle be tangent to arc $\\widehat{P_{1} P_{2}}$ at point $P$, and let it be tangent to sides $\\overline{M N}$ and $\\overline{M R}$ at points $T_{1}$ and $T_{2}$, respectively.\n\n\n\nNote that $Q$ lies on diagonal $\\overline{M O}$ because it is equidistant to $\\overline{M N}$ and $\\overline{M R}$. Points $Q, P$, and $O$ must be collinear because the circles centered at $Q$ and $O$ are mutually tangent at point $P$. It therefore follows that $P$ also lies on diagonal $\\overline{M O}$. Because triangles $Q T_{1} M$ and $Q T_{2} M$ are isosceles right triangles, it follows that $M Q=r \\sqrt{2}$. Thus\n\n$$\nb \\sqrt{2}=M O=M Q+Q P+P O=r \\sqrt{2}+r+a\n$$\n\nSolving this equation yields $r=a+2 b-(a+b) \\sqrt{2}$. With $T=688, a=6$ and $b=8$, so $r=22-14 \\sqrt{2}$, hence $x+y=22+14=\\mathbf{3 6}$.']",['36'],False,,Numerical, 2745,Geometry,,"Let $T=36$. Square $A B C D$ has area $T$. Points $M, N, O$, and $P$ lie on $\overline{A B}$, $\overline{B C}, \overline{C D}$, and $\overline{D A}$, respectively, so that quadrilateral $M N O P$ is a rectangle with $M P=2$. Compute $M N$.","['Let $A M=a$ and $A P=b$, and let $s=\\sqrt{T}$ be the side length of square $A B C D$. Then $M B=s-a$ and $D P=s-b$. Using the right angles of $M N O P$ and complementary acute angles in triangles $A M P, B N M$, $C O N$, and $D P O$, note that\n\n$$\n\\angle A M P \\cong \\angle B N M \\cong \\angle C O N \\cong D P O\n$$\n\nAlso note that $\\mathrm{m} \\angle B M N=180^{\\circ}-\\left(90^{\\circ}+\\mathrm{m} \\angle A M P\\right)$, so it also follows that\n\n$$\n\\angle B M N \\cong \\angle C N O \\cong \\angle D O P \\cong A P M\n$$\n\n\n\nThus, by side-angle-side congruence, it follows that $\\triangle A M P \\cong \\triangle C O N$ and $\\triangle B N M \\cong \\triangle D P O$. Moreover, by side-angle-side similarity, it follows that $\\triangle A M P \\sim \\triangle B N M \\sim \\triangle C O N \\sim \\triangle D P O$. Thus $B N=s-b, N C=b$, $C O=a$, and $O D=s-a$. The similarity relation implies $\\frac{A M}{B N}=\\frac{A P}{B M}$, so $\\frac{a}{s-b}=\\frac{b}{s-a}$. Cross-multiplying, rearranging, and simplifying yields $s(a-b)=(a+b)(a-b)$. Thus either $a=b$ or $s=a+b$. In the case where $a=b, A M=A P=\\frac{2}{\\sqrt{2}}=\\sqrt{2}$, so $M N=(s-\\sqrt{2}) \\sqrt{2}=s \\sqrt{2}-2$. With $T=36, s=6$, and the answer is thus $6 \\sqrt{\\mathbf{2}}-\\mathbf{2}$. For completeness, it remains to verify that for this particular value of $s$, the case where $s=a+b$ is impossible. Applying the Pythagorean Theorem in $\\triangle M A P$ yields $a^{2}+b^{2}=4$. Now if $s=6=a+b$, then by squaring, it would follow that $a^{2}+b^{2}+2 a b=36 \\Longrightarrow 4+2 a b=36 \\Longrightarrow a b=16$. But the equation $a+b=a+\\frac{16}{a}=6$ has no real solutions, thus $a+b \\neq 6$. (Alternatively, note that by the Arithmetic Mean-Geometric Mean Inequality, $a+\\frac{16}{a} \\geq 2 \\sqrt{a \\cdot \\frac{16}{a}}=8>6$.)']",['$6 \\sqrt{2}-2$'],False,,Numerical, 2746,Combinatorics,,"In a game, a player chooses 2 of the 13 letters from the first half of the alphabet (i.e., A-M) and 2 of the 13 letters from the second half of the alphabet (i.e., N-Z). Aditya plays the game, and then Ayesha plays the game. Compute the probability that Aditya and Ayesha choose the same set of four letters.",['The number of ways to choose 2 distinct letters out of 13 is $\\frac{13 \\cdot 12}{2}=78$. The probability of matching on both halves is therefore $\\frac{1}{78^{2}}=\\frac{1}{6084}$.'],['$\\frac{1}{6084}$'],False,,Numerical, 2747,Combinatorics,,"Let $T=\frac{1}{6084}$. Compute the least positive integer $n$ such that when a fair coin is flipped $n$ times, the probability of it landing heads on all $n$ flips is less than $T$.","['The problem is equivalent to finding the least integer $n$ such that $\\frac{1}{2^{n}}\\frac{1}{T}=6084$. Because $2^{12}=4096$ and $2^{13}=8192$, the answer is $\\mathbf{1 3}$.']",['13'],False,,Numerical, 2748,Number Theory,,Let $T=13$. Compute the least integer $n>2023$ such that the equation $x^{2}-T x-n=0$ has integer solutions.,"['The discriminant of the quadratic, $T^{2}+4 n$, must be a perfect square. Because $T$ and the discriminant have the same parity, and the leading coefficient of the quadratic is 1 , by the quadratic formula, the discriminant being a perfect square is sufficient to guarantee integer solutions. Before knowing $T$, note that $\\sqrt{4 \\cdot 2024}=$ $\\sqrt{8096}$ is slightly less than 90 because $90^{2}=8100$, and the square root must have the same parity as $T$. Because\n\n\n\n$T=13$, the square root must be greater than $\\sqrt{13^{2}+4 \\cdot 2023}=\\sqrt{8261}$, which is between 90 and 91 , so the desired square root is 91 . Hence $13^{2}+4 n=91^{2}$, so $n=\\mathbf{2 0 2 8}$.']",['2028'],False,,Numerical, 2749,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that any two consecutive odd positive integers are relatively prime to each other.","['Two consecutive odd integers differ by 2 . Thus any common divisor of the two integers must also divide 2 . However, the only prime divisor of 2 is 2 , and neither of the consecutive odd integers is a multiple of 2 . Therefore any two consecutive odd integers are relatively prime to each other, as desired.']",,True,,, 2750,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that every sequence of three consecutive positive integers contains a cromulent element.","['Consider the sequence $a-1, a, a+1$. Note that $a$ is relatively prime to $a-1$ and $a+1$ because any common divisor must divide their difference, and the differences are both 1 . Thus $a$ is a cromulent element of the sequence.']",,True,,, 2751,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that every sequence of four consecutive positive integers contains a cromulent element.","['First we prove that any two consecutive odd positive integers are relatively prime to each other.\n\nProof: Two consecutive odd integers differ by 2 . Thus any common divisor of the two integers must also divide 2 . However, the only prime divisor of 2 is 2 , and neither of the consecutive odd integers is a multiple of 2 . Therefore any two consecutive odd integers are relatively prime to each other, as desired.\n\nConsider a sequence of four consecutive integers. Notice that two of these integers will be odd, and by the proof above, these are relatively prime to each other. At most one of these two odd integers can be a multiple of 3 . Let $a$ be an odd integer in the sequence that is not a multiple of 3 . Then $a$ is cromulent by the following reasoning. Note that the difference between $a$ and any other element of the sequence is at most 3. Thus if $a$ shared a common factor greater than 1 with some other element of the sequence, this factor would have to be 2 or 3 . But because $a$ is odd and not a multiple of 3 , it follows that $a$ is cromulent.']",,True,,, 2752,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Show that every sequence of five consecutive positive integers contains a cromulent element.","['Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at least one other number in the sequence. There are several cases to consider, namely whether the sequence starts with an even number or an odd number.\n\nIf the sequence starts with an even number, then the second and fourth numbers are both odd, and at least one of them is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence because it is neither a multiple of 2 nor 3 and hence is at least 5 away from the nearest integer with a common factor. Thus the sequence contains a cromulent element.\n\nIf the sequence starts with an odd number, then again, it contains an odd number that is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence, thus the sequence contains a cromulent element. In fact, it contains two such numbers if the first or last number is a multiple of 3 , and if the middle number is a multiple of 3 , then all three odd elements are cromulent.']",,True,,, 2753,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=6$;","['First we prove that every sequence of five consecutive positive integers contains a cromulent element.\n\nProof: Consider a sequence of five consecutive integers. Exactly one number in such a sequence will be a multiple of 5 , but that number could also be a multiple of 2 and hence share a common factor with at least one other number in the sequence. There are several cases to consider, namely whether the sequence starts with an even number or an odd number.\n\nIf the sequence starts with an even number, then the second and fourth numbers are both odd, and at least one of them is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence because it is neither a multiple of 2 nor 3 and hence is at least 5 away from the nearest integer with a common factor. Thus the sequence contains a cromulent element.\n\nIf the sequence starts with an odd number, then again, it contains an odd number that is not a multiple of 3 and hence is relatively prime to all other numbers in the sequence, thus the sequence contains a cromulent element. In fact, it contains two such numbers if the first or last number is a multiple of 3 , and if the middle number is a multiple of 3 , then all three odd elements are cromulent.\n\n\nThe minimum number is 1 and the maximum number is 2 . One example of a sequence of length 6 with one cromulent element is $5,6,7,8,9$, 10, where 7 is the cromulent element. To show that it is not possible for\n\n\na sequence of six consecutive elements to have zero cromulent elements, consider two cases. If the sequence begins with an even number, that number is not cromulent, and one of the other five elements must be cromulent by the argument in the proof above. A similar argument establishes that one element must be cromulent if the sequence of length 6 begins with an odd number (and thus ends in an even number).\n\nOne example of a sequence of length 6 with two cromulent elements is $1,2,3,4,5,6$, where 1 and 5 are both cromulent.\n\nTo prove that a sequence of length 6 cannot have three cromulent elements, consider that the cromulent elements would all have to be odd, and one of those three would be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is not possible for all three odd elements to be cromulent.']","['1,2']",True,,Numerical, 2754,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Find the maximum and minimum possible number of cromulent elements in a sequence of $n$ consecutive positive integers with $n=7$.","['The minimum number is 1 and the maximum number is 3 . One example of a sequence of length 7 with one cromulent element is $4,5,6,7,8,9,10$, where 7 is the cromulent element. To show that it is not possible for such a sequence to have zero cromulent elements, consider two cases. If the sequence begins with an even number, then it contains three odd numbers. At most one of these is divisible by 3 , and at most one is divisible by 5 , so one of the odd numbers must be divisible by neither 3 nor 5 . This odd number differs by at most 6 from each other element of the sequence, so the only prime factors it can share with another element of the sequence are 2, 3, and 5 . Because it is divisible by none of these primes, it follows that the odd number in question is cromulent. Similarly, if the sequence begins with an odd number, then it contains four odd numbers; at most two of these are divisible by 3 , and at most one is divisible by 5 , so again, one odd number in the sequence must be divisible by neither 3 nor 5 . By the same argument, this element is cromulent.\n\nOne example of a sequence of length 7 with three cromulent elements is $1,2,3,4,5,6$, 7 , where 1,5 , and 7 are all cromulent.\n\nTo prove that a sequence of length 7 cannot have four cromulent elements, consider that the cromulent elements would all have to be odd. At least one of these four odd elements must be a multiple of 3 . Because one of the even elements must also be a multiple of 3 , it is thus not possible for all four odd elements to be cromulent.']","['1,3']",True,,Numerical, 2755,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=8$;","['In any sequence of eight consecutive integers, four of them will be even and hence not cromulent. The primes 3,5 , and 7 are the only numbers that could possibly divide more than one number in the sequence, with 5 and 7 each dividing at most one of the four odd numbers. Note that for 7 to divide two numbers in the sequence, they must be the first and last numbers because only those differ by 7 . The prime 3 can divide two odd numbers in the sequence, but only if they differ by 6 , which means they must be either the 1st and 7th or the $2 \\mathrm{nd}$ and 8th numbers in the list. Therefore one of the two odd multiples of 3 would also be the only candidate to be an odd multiple of 7 when there are two multiples of 7 , and so at least one of the four odd numbers must be cromulent.']",,True,,, 2756,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. Prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=9$.","['First we prove that there is at least one cromulent element in every sequence of $n$ consecutive positive integers with $n=8$;\n\nProof: In any sequence of eight consecutive integers, four of them will be even and hence not cromulent. The primes 3,5 , and 7 are the only numbers that could possibly divide more than one number in the sequence, with 5 and 7 each dividing at most one of the four odd numbers. Note that for 7 to divide two numbers in the sequence, they must be the first and last numbers because only those differ by 7 . The prime 3 can divide two odd numbers in the sequence, but only if they differ by 6 , which means they must be either the 1st and 7th or the $2 \\mathrm{nd}$ and 8th numbers in the list. Therefore one of the two odd multiples of 3 would also be the only candidate to be an odd multiple of 7 when there are two multiples of 7 , and so at least one of the four odd numbers must be cromulent.\n\nNow we turn to prove that given assumption. In a sequence of nine consecutive integers, again the only primes to check are $2,3,5$, and 7 . If there are five even numbers in the sequence, then at least one of the odd numbers is cromulent by similar reasoning to the proof above. The difference is that in this case the multiples of 7 can be either the 1st and 8th or the 2nd and 9th numbers, while the odd multiples of 3 must be the 2 nd and 8th numbers (because the 1st number is even) and hence overlap with the multiples of 7 again. If instead there are four even numbers in the sequence, then the 1st, 3rd, 5th, 7th, and 9th numbers are odd. Now 7 divides at most one of these odd numbers, 5 divides at most one, and 3 divides at most two, leaving at least one odd cromulent element.']",,True,,, 2757,Algebra,,"In a sequence of $n$ consecutive positive integers, where $n>1$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. The goal of this problem is to prove the following claim: Claim: For $k \geq 300$, there is a sequence of $k$ consecutive integers with no cromulent elements. Let $\pi(n)$ denote the number of primes less than or equal to $n$. You may use the following lemma without proof in the problems that follow. Lemma: For $x \geq 75, \pi(2 x)-\pi(x) \geq 2\left\lfloor\log _{2}(x)\right\rfloor+2$. One argument for proving the above claim begins as follows. Let $m=\left\lfloor\frac{k}{4}\right\rfloor \geq 75$, let $p_{1}, p_{2}, \ldots, p_{r}$ denote the primes in the set $\{1,2, \ldots, m\}$ (note that $p_{1}=2$ ), and let $p_{r+1}, p_{r+2}, \ldots, p_{s}$ denote the primes in the set $\{m+1, m+2, \ldots, 2 m\}$. Prove that if each of the $k$ consecutive integers in the sequence is divisible by at least one of the primes $p_{1}, p_{2}, \ldots, p_{s}$, then the sequence has no cromulent elements.","['Let the sequence of $k$ integers be $a_{1}, a_{2}, \\ldots, a_{k}$, where $a_{n}=a_{1}+(n-1)$ for $1 \\leq n \\leq k$. For any $a_{j} \\in\\left\\{a_{1}, a_{2}, \\ldots, a_{2 m}\\right\\}$, suppose that $a_{j}$ is divisible by some $p \\in\\left\\{p_{1}, p_{2}, \\ldots, p_{s}\\right\\}$. Then $a_{j+p}=a_{j}+p$, so $a_{j+p}$ is also divisible by $p$. Hence neither $a_{j}$ nor $a_{j+p}$ is cromulent because both numbers are divisible by $p$ (also note that $j+p<4 m \\leq k$, so $a_{j+p}$ is one of the $k$ integers in the sequence). Thus none of $a_{1}, a_{2}, \\ldots, a_{2 m}$ are cromulent. Similarly, for any $a_{j^{\\prime}} \\in\\left\\{a_{2 m+1}, a_{2 m+2}, \\ldots, a_{k}\\right\\}$, suppose that $a_{j^{\\prime}}$ is divisible by some $p^{\\prime} \\in\\left\\{p_{1}, p_{2}, \\ldots, p_{s}\\right\\}$. Then $a_{j^{\\prime}-p^{\\prime}}=a_{j^{\\prime}}-p^{\\prime}$ is also divisible by $p^{\\prime}$ (also note that $11$, an element of the sequence is said to be cromulent if it is relatively prime to all other numbers in the sequence. Every element of a sequence with $n=2$ is cromulent because any two consecutive integers are relatively prime to each other. The goal of this problem is to prove the following claim: Claim: For $k \geq 300$, there is a sequence of $k$ consecutive integers with no cromulent elements. Let $\pi(n)$ denote the number of primes less than or equal to $n$. You may use the following lemma without proof in the problems that follow. Lemma: For $x \geq 75, \pi(2 x)-\pi(x) \geq 2\left\lfloor\log _{2}(x)\right\rfloor+2$. One argument for proving the above claim begins as follows. Let $m=\left\lfloor\frac{k}{4}\right\rfloor \geq 75$, let $p_{1}, p_{2}, \ldots, p_{r}$ denote the primes in the set $\{1,2, \ldots, m\}$ (note that $p_{1}=2$ ), and let $p_{r+1}, p_{r+2}, \ldots, p_{s}$ denote the primes in the set $\{m+1, m+2, \ldots, 2 m\}$. Let $x$ be a solution to the system of congruences $x \equiv 1(\bmod 2)$ and $x \equiv 0\left(\bmod p_{2} p_{3} \cdots p_{r}\right)$. Then the integers $$ x-2 m, x-2 m+2, \ldots, x-2, x, x+2, \ldots, x+2 m-2, x+2 m $$ form a sequence of $2 m+1$ consecutive odd integers of the form $x \pm 2 y$, where $y$ varies from 0 to $m$. Prove that every number in the above sequence, except those in which $y$ is a power of 2 , is divisible by one of the primes $p_{2}, \ldots, p_{r}$.","['Note that $x \\equiv 1(\\bmod 2)$ implies that $x$ is odd. Because $x \\equiv 0\\left(\\bmod p_{2} p_{3} \\cdots p_{r}\\right)$, write $x=q p_{2} p_{3} \\cdots p_{r}$ for some odd integer $q$. Then $x$ is divisible by all of the primes $p_{2}, \\ldots, p_{r}$. The desired result will first be established for numbers of the form $x+2 y$, with $1 \\leq y \\leq m$. Because $x$ is divisible by each of $p_{2}, \\ldots, p_{r}$, it suffices to prove that $y$ is divisible by one of $p_{2}, \\ldots, p_{r}$ unless $y$ is a power of 2 . Note that because $p_{2}, \\ldots, p_{r}$ are all odd, if $y$ is a power of 2 (including $2^{0}=1$ ), then $2 y$ cannot be divisible by any of $p_{2}, \\ldots, p_{r}$. On the other hand, if $y$ has an odd factor greater than 1, by the Fundamental Theorem of Arithmetic, $y$ must be divisible by at least one of $p_{2}, \\ldots, p_{r}$ (note that $p_{2}=3$ and $p_{r} \\leq m$ ). Finally, because the desired result holds for $x+2 y$, with $1 \\leq y \\leq m$, it also holds for $x-2 y$ because $x-2 y=(x+2 y)-4 y$, thus completing the proof.']",,True,,, 2759,Number Theory,,"For an integer $n \geq 4$, define $a_{n}$ to be the product of all real numbers that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. Compute $$ \frac{a_{4}}{a_{5}}+\frac{a_{5}}{a_{6}}+\frac{a_{6}}{a_{7}}+\cdots+\frac{a_{2022}}{a_{2023}} . $$","['For an integer $n \\geq 4$, let $S_{n}$ denote the set of real numbers $x$ that are roots to at least one quadratic polynomial whose coefficients are positive integers that sum to $n$. (Note that $S_{n}$ is nonempty, as the polynomial $x^{2}+(n-2) x+1$ has a discriminant of $(n-2)^{2}-4$, which is nonnegative for $n \\geq 4$.) Then $a_{n}=\\prod_{x \\in S_{n}} x$.\n\nSuppose that $a, b$, and $c$ are positive integers and $x$ is a real solution to $a x^{2}+b x+c=0$. Then $x$ must be nonzero. (In fact, $x$ must be negative.) Dividing the above equation by $x^{2}$ yields $a+\\frac{b}{x}+\\frac{c}{x^{2}}=0$, thus $r=\\frac{1}{x}$ is a solution to the quadratic equation $c r^{2}+b r+a=0$. This shows that $x \\in S_{n}$ if and only if $\\frac{1}{x} \\in S_{n}$.\n\nOne might then think that $a_{n}$ must equal 1, because one can presumably pair up all elements in a given $S_{n}$ into $\\left\\{x, \\frac{1}{x}\\right\\}$ pairs. But there is a (negative) value of $x$ for which $x=\\frac{1}{x}$, namely $x=-1$. Therefore the value of $a_{n}$ depends only on whether $-1 \\in S_{n}$. It is readily seen via a parity argument that $-1 \\in S_{n}$ if and only if $n$ is even. If $n=2 k$, then the polynomial $x^{2}+k x+(k-1)$ has -1 as a root. (In fact, any quadratic polynomial whose middle coefficient is $k$ and whose coefficients sum to $2 k$ will work.) But if $n=2 k+1$, then $a(-1)^{2}+b(-1)+c=a-b+c=(a+b+c)-2 b=(2 k+1)-2 b$ will be odd, and so $-1 \\notin S_{n}$.\n\nThus $a_{n}=-1$ when $n$ is even, $a_{n}=1$ when $n$ is odd, and finally,\n\n$$\n\\frac{a_{4}}{a_{5}}+\\frac{a_{5}}{a_{6}}+\\frac{a_{6}}{a_{7}}+\\cdots+\\frac{a_{2022}}{a_{2023}}=\\underbrace{(-1)+(-1)+(-1)+\\cdots+(-1)}_{2019(-1) \\mathrm{s}}=-\\mathbf{2 0 1 9} .\n$$']",['-2019'],False,,Numerical, 2760,Combinatorics,,"Suppose that $u$ and $v$ are distinct numbers chosen at random from the set $\{1,2,3, \ldots, 30\}$. Compute the probability that the roots of the polynomial $(x+u)(x+v)+4$ are integers.","['Assume without loss of generality that $u>v$. The condition that $(x+u)(x+v)+4$ has integer roots is equivalent to the discriminant $(u+v)^{2}-4(u v+4)=(u-v)^{2}-16$ being a perfect square. This is possible if and only if $u-v=4$ or $u-v=5$. There are $(30-4)+(30-5)=26+25=51$ such ordered pairs $(u, v)$, so the answer is\n\n$$\n\\frac{51}{\\left(\\begin{array}{c}\n30 \\\\\n2\n\\end{array}\\right)}=\\frac{\\mathbf{1 7}}{\\mathbf{1 4 5}}\n$$']",['$\\frac{17}{145}$'],False,,Numerical, 2761,Geometry,,The degree-measures of the interior angles of convex hexagon TIEBRK are all integers in arithmetic progression. Compute the least possible degree-measure of the smallest interior angle in hexagon TIEBRK.,"['The sum of the measures of the interior angles of a convex hexagon is $(6-2)\\left(180^{\\circ}\\right)=720^{\\circ}$. Let the measures of the angles be $a, a+d, \\ldots, a+5 d$. This implies that $6 a+15 d=720 \\rightarrow 2 a+5 d=240 \\rightarrow 5 d=240-2 a$. Note that $a+5 d<180 \\rightarrow 240-a<180 \\rightarrow a>60$. By inspection, note that the least $a$ greater than 60 that produces an integer $d$ is $a=65 \\rightarrow d=22$. Thus the least possible degree-measure of the smallest angle is $65^{\\circ}$, and the hexagon has angles with degree-measures $65^{\\circ}, 87^{\\circ}, 109^{\\circ}, 131^{\\circ}, 153^{\\circ}$, and $175^{\\circ}$.']",['65'],False,,Numerical, 2762,Combinatorics,,"A six-digit natural number is ""sort-of-decreasing"" if its first three digits are in strictly decreasing order and its last three digits are in strictly decreasing order. For example, 821950 and 631631 are sort-of-decreasing but 853791 and 911411 are not. Compute the number of sort-of-decreasing six-digit natural numbers.","['If three distinct digits are chosen from the set of digits $\\{0,1,2, \\ldots, 9\\}$, then there is exactly one way to arrange them in decreasing order. There are $\\left(\\begin{array}{c}10 \\\\ 3\\end{array}\\right)=120$ ways to choose the first three digits and 120 ways to choose the last three digits. Thus the answer is $120 \\cdot 120=\\mathbf{1 4 4 0 0}$.']",['14400'],False,,Numerical, 2763,Number Theory,,"For each positive integer $N$, let $P(N)$ denote the product of the digits of $N$. For example, $P(8)=8$, $P(451)=20$, and $P(2023)=0$. Compute the least positive integer $n$ such that $P(n+23)=P(n)+23$.","['One can verify that no single-digit positive integer $n$ satisfies the conditions of the problem.\n\nIf $n$ has two digits, then $n+23$ cannot be a three-digit number; this can be verified by checking the numbers $n \\geq 88$, because if $n<88$, then one of the digits of $n+23$ is 0 . Therefore both $n$ and $n+23$ must be two-digit numbers, so the only possible carry for $n+23$ will occur in the tens place. If there is a carry for $n+23$, then $n=\\underline{a} \\underline{8}$ or $n=\\underline{a} \\underline{9}$, while $n+23=(a+3) 1$ or $n+23=(a+3) 2$, respectively (the case $n=\\underline{a} \\underline{7}$ is omitted because then $P(n+23)=0)$. In either case, $P(n+23)-8$ for all $x$, so any integer value of $f(x)$ must be at least -7 .\n\nWhen $x=3$, the remainder term is less than 1 , so $f(3)$ is less than -7 . But $f(4)=-\\frac{34}{5}>-7$, so there must be some value of $x$ between 3 and 4 for which $f(x)=-7$, so the least integer value of $f(x)$ is $\\mathbf{- 7}$. The reader may note that $f(x)=-7$ when $x \\approx 2.097$ and $x \\approx 3.970$.']",['-7'],False,,Numerical, 2765,Geometry,,"Suppose that noncongruent triangles $A B C$ and $X Y Z$ are given such that $A B=X Y=10, B C=$ $Y Z=9$, and $\mathrm{m} \angle C A B=\mathrm{m} \angle Z X Y=30^{\circ}$. Compute $[A B C]+[X Y Z]$.","['Because triangles $A B C$ and $X Y Z$ are noncongruent yet have two adjacent sides and an angle in common, the two triangles are the two possibilities in the ambiguous case of the Law of Sines. Without loss of generality, let triangle $A B C$ have obtuse angle $C$ and triangle $X Y Z$ have acute angle $Z$ so that $\\mathrm{m} \\angle C+\\mathrm{m} \\angle Z=$ $180^{\\circ}$. Place triangle $A B C$ so that $B$ and $Y$ coincide, and $C$ and $Z$ coincide. Because $\\mathrm{m} \\angle C$ and $\\mathrm{m} \\angle Z$ add up to $180^{\\circ}$, it follows that points $X, Z$, and $A$ all lie on the same line. The two triangles together then form $\\triangle A B X$, where $\\mathrm{m} \\angle B A X=\\mathrm{m} \\angle B X A=30^{\\circ}$ and $B X=A B=10$. Therefore the sum of the areas of the two triangles is equal to the area of triangle $A B X$, which is $\\frac{1}{2} \\cdot 10 \\cdot 10 \\cdot \\sin \\left(120^{\\circ}\\right)=\\frac{5 \\cdot 10 \\cdot \\sqrt{3}}{2}=\\mathbf{2 5} \\sqrt{\\mathbf{3}}$.\n\n\n\nFigure not drawn to scale.', 'As explained above, let $\\triangle A B C$ have obtuse angle $C$ and $\\triangle X Y Z$ have acute angle $Z$. By the Law of Sines, $\\sin (\\angle C)=\\sin (\\angle Z)=\\frac{5}{9}$. This implies $\\mathrm{m} \\angle X Y Z=\\frac{5 \\pi}{6}-\\arcsin \\left(\\frac{5}{9}\\right)$ and $\\mathrm{m} \\angle A B C=$ $\\arcsin \\left(\\frac{5}{9}\\right)-\\frac{\\pi}{6}$. The areas of the triangles are $[X Y Z]=\\frac{1}{2} \\cdot 10 \\cdot 9 \\cdot \\sin \\left(\\frac{5 \\pi}{6}-\\arcsin \\left(\\frac{5}{9}\\right)\\right)$ and $[A B C]=\\frac{1}{2} \\cdot 10 \\cdot 9$. $\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)-\\frac{\\pi}{6}\\right)$. By the angle subtraction rule, it follows that\n\n$$\n\\begin{aligned}\n\\sin \\left(\\frac{5 \\pi}{6}-\\arcsin \\left(\\frac{5}{9}\\right)\\right) & =\\sin \\left(\\frac{5 \\pi}{6}\\right) \\cos \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right)-\\cos \\left(\\frac{5 \\pi}{6}\\right) \\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right) \\text { and } \\\\\n\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)-\\frac{\\pi}{6}\\right) & =\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right) \\cos \\left(\\frac{\\pi}{6}\\right)-\\cos \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right) \\sin \\left(\\frac{\\pi}{6}\\right) .\n\\end{aligned}\n$$\n\nThe sum of the two sines is $\\sin \\left(\\arcsin \\left(\\frac{5}{9}\\right)\\right)\\left(\\cos \\left(\\frac{\\pi}{6}\\right)-\\cos \\left(\\frac{5 \\pi}{6}\\right)\\right)=\\frac{5}{9} \\cdot \\sqrt{3}$ because $\\sin \\left(\\frac{\\pi}{6}\\right)=\\sin \\left(\\frac{5 \\pi}{6}\\right)$. Finally, the sum of the areas of the two triangles is $\\frac{1}{2} \\cdot 10 \\cdot 9 \\cdot \\frac{5}{9} \\cdot \\sqrt{3}=25 \\sqrt{3}$.']",['$25 \\sqrt{3}$'],False,,Numerical, 2766,Combinatorics,,"The mean, median, and unique mode of a list of positive integers are three consecutive integers in some order. Compute the least possible sum of the integers in the original list.","['One possible list is $1,1,3,7$, which has mode 1 , median 2 , and mean 3 . The sum is $1+1+3+7=12$. A list with fewer than four numbers cannot produce a median and unique mode that are distinct from each other. To see this, first note that a list with one number has the same median and mode. In a list with two numbers, the mode is not unique if the numbers are different, and if the numbers are the same, the median and mode are equal. In a list of three numbers with a unique mode, the mode must occur twice. Hence the\n\n\n\nmode is equal to the middle number of the three, which is the median. Thus a list with a median and unique mode that are different from each other must contain at least four numbers.\n\nNow suppose that a list satisfying the given conditions sums to less than 12 . The mean must be greater than 1, and because the list contains at least four numbers, the mean must be exactly 2 . The median must also be greater than 1 , and if the mode is 4 , then the sum must be greater than 12 . Thus it remains to determine if a mean of 2 with mode 1 and median 3 can be achieved with a list of four or five positive integers. However, having two 1s in the list and a median of 3 forces the remaining numbers in each case to have a sum too large for a mean of 2 . The least possible sum is therefore $\\mathbf{1 2}$.']",['12'],False,,Numerical, 2767,Combinatorics,,"David builds a circular table; he then carves one or more positive integers into the table at points equally spaced around its circumference. He considers two tables to be the same if one can be rotated so that it has the same numbers in the same positions as the other. For example, a table with the numbers $8,4,5$ (in clockwise order) is considered the same as a table with the numbers 4, 5,8 (in clockwise order), but both tables are different from a table with the numbers 8, 5, 4 (in clockwise order). Given that the numbers he carves sum to 17 , compute the number of different tables he can make.","['The problem calls for the number of ordered partitions of 17 , where two partitions are considered the same if they are cyclic permutations of each other. Because 17 is prime, each ordered partition of 17 into $n$ parts will be a cyclic permutation of exactly $n$ such partitions (including itself), unless $n=17$. (If $n=17$, then all the numbers are 1s, and there is exactly one table David can make.) By the sticks and stones method, the number of ordered partitions of 17 into $n$ nonzero parts is $\\left(\\begin{array}{c}16 \\\\ n-1\\end{array}\\right)$, and this overcounts the number of tables by a factor of $n$, except when $n=17$. Thus the number of possible tables is\n\n$$\n1+\\sum_{n=1}^{16}\\left(\\begin{array}{c}\n16 \\\\\nn-1\n\\end{array}\\right) \\cdot \\frac{1}{n}=1+\\sum_{n=1}^{16}\\left(\\begin{array}{c}\n17 \\\\\nn\n\\end{array}\\right) \\cdot \\frac{1}{17}=1+\\frac{2^{17}-2}{17}=\\mathbf{7 7 1 1}\n$$']",['7711'],False,,Numerical, 2768,Geometry,,"In quadrilateral $A B C D, \mathrm{~m} \angle B+\mathrm{m} \angle D=270^{\circ}$. The circumcircle of $\triangle A B D$ intersects $\overline{C D}$ at point $E$, distinct from $D$. Given that $B C=4, C E=5$, and $D E=7$, compute the diameter of the circumcircle of $\triangle A B D$.","['Note that $\\mathrm{m} \\angle A+\\mathrm{m} \\angle C=90^{\\circ}$ in quadrilateral $A B C D$. Because quadrilateral $A B E D$ is cyclic, it follows that $\\mathrm{m} \\angle A D E+\\mathrm{m} \\angle A B E=180^{\\circ}$. Moreover, because $\\mathrm{m} \\angle A B E+\\mathrm{m} \\angle E B C+\\mathrm{m} \\angle A D E=270^{\\circ}$, it follows that $\\angle E B C$ is a right angle. Thus $B E=\\sqrt{C E^{2}-B C^{2}}=\\sqrt{5^{2}-4^{2}}=3$. Let $\\mathrm{m} \\angle B E C=\\theta$; then $\\cos \\theta=\\frac{3}{5}$ and $\\sin \\theta=\\frac{4}{5}$.\n\n\n\nApplying the Law of Cosines to $\\triangle B E D$ yields\n\n$$\nB D^{2}=3^{2}+7^{2}-2 \\cdot 3 \\cdot 7 \\cos \\left(180^{\\circ}-\\theta\\right)=3^{2}+7^{2}+2 \\cdot 3 \\cdot 7 \\cos \\theta=\\frac{416}{5}\n$$\n\nThus $B D=\\frac{4 \\sqrt{26}}{\\sqrt{5}}$. Let $R$ be the circumradius of $\\triangle A B D$ and $\\triangle B E D$. Then the requested diameter is $2 R$, and\n\n\n\napplying the Law of Sines to $\\triangle B E D$ yields\n\n$$\n2 R=\\frac{B D}{\\sin \\left(180^{\\circ}-\\theta\\right)}=\\frac{B D}{\\sin \\theta}=\\frac{4 \\sqrt{26}}{\\sqrt{5}} \\cdot \\frac{5}{4}=\\sqrt{\\mathbf{1 3 0}}\n$$']",['$\\sqrt{130}$'],False,,Numerical, 2769,Geometry,,"Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace. ![](https://cdn.mathpix.com/cropped/2023_12_21_5b86f26b7ea18a50f6b5g-1.jpg?height=330&width=307&top_left_y=1727&top_left_x=947)","['Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:\n\n$$\n\\mathcal{I}=\\left\\{A_{1}, A_{2}, \\ldots, A_{6}\\right\\} \\text { (i.e., the inner ring), } \\quad \\mathcal{O}=\\left\\{B_{1}, B_{2}, \\ldots, B_{6}\\right\\} \\text { (i.e., the outer ring). }\n$$\n\nwhere $A_{1}=A$. The two rings are connected by the edges $E=\\left\\{A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{6} B_{6}\\right\\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.\n\nFurther note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.\n\nIf all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.\n\nIf not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \\leq j \\leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.\n\nThe forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.\n\nEach undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \\cdot 2=16$.']",['16'],False,,Numerical, 2769,Geometry,,"Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace. ","['Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:\n\n$$\n\\mathcal{I}=\\left\\{A_{1}, A_{2}, \\ldots, A_{6}\\right\\} \\text { (i.e., the inner ring), } \\quad \\mathcal{O}=\\left\\{B_{1}, B_{2}, \\ldots, B_{6}\\right\\} \\text { (i.e., the outer ring). }\n$$\n\nwhere $A_{1}=A$. The two rings are connected by the edges $E=\\left\\{A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{6} B_{6}\\right\\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.\n\nFurther note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.\n\nIf all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.\n\nIf not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \\leq j \\leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.\n\nThe forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.\n\nEach undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \\cdot 2=16$.']",['16'],False,,Numerical, 2770,Algebra,,"Let $i=\sqrt{-1}$. The complex number $z=-142+333 \sqrt{5} i$ can be expressed as a product of two complex numbers in multiple different ways, two of which are $(57-8 \sqrt{5} i)(-6+5 \sqrt{5} i)$ and $(24+\sqrt{5} i)(-3+14 \sqrt{5} i)$. Given that $z=-142+333 \sqrt{5} i$ can be written as $(a+b \sqrt{5} i)(c+d \sqrt{5} i)$, where $a, b, c$, and $d$ are positive integers, compute the lesser of $a+b$ and $c+d$.","['Multiply each of the given parenthesized expressions by its complex conjugate to obtain\n\n$$\n\\begin{aligned}\n142^{2}+5 \\cdot 333^{2} & =\\left(57^{2}+5 \\cdot 8^{2}\\right)\\left(6^{2}+5 \\cdot 5^{2}\\right) \\\\\n& =\\left(24^{2}+5 \\cdot 1^{2}\\right)\\left(3^{2}+5 \\cdot 14^{2}\\right) \\\\\n& =\\left(a^{2}+5 b^{2}\\right)\\left(c^{2}+5 d^{2}\\right) .\n\\end{aligned}\n$$\n\nThe expression on the second line is equal to $581 \\cdot 989=7 \\cdot 83 \\cdot 23 \\cdot 43$ (one can perhaps factor 989 a little faster by noting that 23 divides $6^{2}+5 \\cdot 5^{2}=7 \\cdot 23$ but not 581 , so it must divide 989 ). Thus $a^{2}+5 b^{2}$ and $c^{2}+5 d^{2}$ must be a factor pair of this number. It is not possible to express $1,7,23,43$, or 83 in the form $x^{2}+5 y^{2}$ for integers $x, y$.\n\nLet $N=a^{2}+5 b^{2}$, and without loss of generality, assume that 7 divides $N$. From the above analysis, $N$ must be $7 \\cdot 23,7 \\cdot 43$, or $7 \\cdot 83$. By direct computation of checking all positive integers $b$ less than $\\sqrt{\\frac{N}{5}}$, the only possibilities for $(a, b)$ are:\n\n- when $N=7 \\cdot 23$, either $(9,4)$ or $(6,5)$;\n- when $N=7 \\cdot 43$, either $(16,3)$ or $(11,6)$; and\n- when $N=7 \\cdot 83$, either $(24,1)$ or $(9,10)$.\n\nNext, observe that\n\n$$\n\\frac{-142+333 \\sqrt{5} i}{a+b \\sqrt{5} i}=\\frac{(-142 a+1665 b)+(333 a+142 b) \\sqrt{5} i}{N}\n$$\n\nmust equal $c+d \\sqrt{5} i$, so $N$ must divide $-142 a+1665 b$ and $333 a+142 b$. But\n\n- 7 does not divide $333 \\cdot 9+142 \\cdot 4$ or $333 \\cdot 6+142 \\cdot 5$;\n- 43 does not divide $333 \\cdot 16+142 \\cdot 3$; and\n- 83 does not divide $333 \\cdot 9+142 \\cdot 10$.\n\nThus the only candidates are $(a, b)=(11,6)$ and $(a, b)=(24,1)$. Note that $(24,1)$ yields the second factorization given in the problem statement, which has a negative real part in one of its factors. Thus the only remaining candidate for $(a, b)$ is $(11,6)$, which yields $(c, d)=(28,15)$, thus the answer is $11+6=\\mathbf{1 7}$.']",['17'],False,,Numerical, 2771,Geometry,,"Parallelogram $A B C D$ is rotated about $A$ in the plane, resulting in $A B^{\prime} C^{\prime} D^{\prime}$, with $D$ on $\overline{A B^{\prime}}$. Suppose that $\left[B^{\prime} C D\right]=\left[A B D^{\prime}\right]=\left[B C C^{\prime}\right]$. Compute $\tan \angle A B D$.","[""Editor's Note: It was pointed out that the conditions of the problem determine two possible values of $\\tan \\angle A B D$ : one based on $\\mathrm{m} \\angle A<90^{\\circ}$, and the other based on $\\mathrm{m} \\angle A>90^{\\circ}$. A complete solution is provided below. We thank Matthew Babbitt and Silas Johnson for their contributions to this solution.\n\n\n\nLet $A B=x, B C=y$, and $\\mathrm{m} \\angle A=\\alpha$.\n\n\n\nIt then follows that\n\n\n\n$$\n\\left[A B D^{\\prime}\\right]=\\left\\{\\begin{array}{ll}\n\\frac{x y \\sin 2 \\alpha}{2} & \\text { if } \\alpha<90^{\\circ} \\\\\n\\frac{-x y \\sin 2 \\alpha}{2} & \\text { if } \\alpha>90^{\\circ}\n\\end{array} \\quad \\text { and } \\quad\\left[B^{\\prime} C D\\right]=\\frac{x(x-y) \\sin \\alpha}{2}\\right.\n$$\n\nBecause $\\overline{B C}, \\overline{A B^{\\prime}}$, and $\\overline{D^{\\prime} C^{\\prime}}$ are all parallel, it follows that $\\triangle B C C^{\\prime}$ and $\\triangle B C D^{\\prime}$ have the same height with respect to base $\\overline{B C}$, and thus $\\left[B C C^{\\prime}\\right]=\\left[B C D^{\\prime}\\right]$. Therefore $\\left[B C D^{\\prime}\\right]=\\left[A B D^{\\prime}\\right]$, and it follows that triangles $B C D^{\\prime}$ and $A B D^{\\prime}$ have the same height with respect to base $\\overline{B D^{\\prime}}$. Thus $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$. Let $M$ be the midpoint of $\\overline{A C}$. Consider the following two cases.\n\nCase 1: Suppose that $\\alpha<90^{\\circ}$. Because $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $M$ lies on $\\overleftrightarrow{B D^{\\prime}}$. But $\\overleftrightarrow{B D}$ also passes through the midpoint of $\\overline{A C}$ by parallelogram properties, so it follows that $D$ must lie on $\\overline{B D^{\\prime}}$. This implies that $\\left[A B D^{\\prime}\\right]$ must also equal $\\frac{y^{2} \\sin \\alpha}{2}+\\frac{x y \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1$ and $\\sin \\alpha=\\cos \\alpha=\\frac{\\sqrt{2}}{2}$. Finally, from right triangle $D^{\\prime} A B$ with legs in the ratio $1: \\sqrt{2}+1$, it follows that $\\tan (\\angle A B D)=\\tan \\left(\\angle A B D^{\\prime}\\right)=$ $\\sqrt{2}-1$.\n\nCase 2: Suppose that $\\alpha>90^{\\circ}$. The points $D$ and $D^{\\prime}$ lie on opposite sides of $\\overleftrightarrow{A B}$. Because $B C=A D^{\\prime}$ and points $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $A C B D^{\\prime}$ is either a parallelogram or an isosceles trapezoid. It cannot be the former because that would imply that $\\overleftrightarrow{D^{\\prime} A}\\|\\overleftrightarrow{B C}\\| \\overleftrightarrow{A D}$. Thus $A C B D^{\\prime}$ is an isosceles trapezoid. Then $\\left[B A D^{\\prime}\\right]=\\left[B M D^{\\prime}\\right]$. Because $B, M$, and $D$ are collinear and $B D: B M=2$, it follows that $\\left[B D D^{\\prime}\\right]=2 \\cdot\\left[B M D^{\\prime}\\right]$. Moreover, $\\left[B D D^{\\prime}\\right]=\\left[B A D^{\\prime}\\right]+[B A D]+\\left[D A D^{\\prime}\\right]$, so $\\left[B A D^{\\prime}\\right]=[B A D]+\\left[D A D^{\\prime}\\right]$. Thus $\\left[B A D^{\\prime}\\right]=\\frac{x y \\sin \\alpha}{2}+\\frac{y^{2} \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=-x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1, \\sin \\alpha=\\frac{\\sqrt{2}}{2}$, and $\\cos \\alpha=-\\frac{\\sqrt{2}}{2}$, so $\\alpha=135^{\\circ}$. Let $H$ be the foot of the perpendicular from $D$ to $\\overleftrightarrow{A B}$. Then $A D H$ is a $45^{\\circ}-45^{\\circ}-90^{\\circ}$ triangle with $H A=H D=\\frac{y}{\\sqrt{2}}$. Thus\n\n$$\n\\begin{aligned}\n\\tan \\angle A B D & =\\frac{D H}{B H}=\\frac{D H}{B A+A H} \\\\\n& =\\frac{y / \\sqrt{2}}{x+y / \\sqrt{2}}=\\frac{y}{x \\sqrt{2}+y} \\\\\n& =\\frac{y}{y(\\sqrt{2}+1)(\\sqrt{2})+y} \\\\\n& =\\frac{1}{(\\sqrt{2}+1)(\\sqrt{2})+1} \\\\\n& =\\frac{\\mathbf{3}-\\sqrt{\\mathbf{2}}}{\\mathbf{7}}\n\\end{aligned}\n$$"", 'Let $x, y$, and $\\alpha$ be as defined in the first solution. Then $C D=x$ because $A B C D$ is a parallelogram. Also note that $A B^{\\prime}=x, B^{\\prime} C^{\\prime}=y$, and $A D^{\\prime}=y$ because $A B C D$ and $A B^{\\prime} C^{\\prime} D^{\\prime}$ are congruent. Thus $D B^{\\prime}=A B^{\\prime}-A D=x-y$. Let $E$ be the intersection of $\\overleftrightarrow{A B}$ and $\\overleftrightarrow{C^{\\prime} D^{\\prime}}$, as shown in both configurations below.\n\n\nBecause $E$ lies on $\\overleftrightarrow{A B}$, it follows that $\\angle B^{\\prime} A E=180^{\\circ}-\\angle B A D=180^{\\circ}-\\alpha$. Thus, in quadrilateral $A B^{\\prime} C^{\\prime} E$, $\\overline{A B^{\\prime}} \\| \\overline{C^{\\prime} E}$ and $\\angle A B^{\\prime} C^{\\prime}=\\angle B^{\\prime} A E=180^{\\circ}-\\alpha$. Therefore quadrilateral $A B^{\\prime} C^{\\prime} E$ is an isosceles trapezoid. Hence $A E=B^{\\prime} C^{\\prime}=y$. It follows that $B E=B A+A E=x+y$. Therefore, from the sine area formula with respect to $\\angle C B E$,\n\n$$\n[B C E]=\\frac{1}{2} x(x+y) \\sin \\left(180^{\\circ}-\\alpha\\right)=\\frac{1}{2} x(x+y) \\sin \\alpha .\n$$\n\nBecause $\\overline{E C^{\\prime}} \\| \\overline{B C}$, it follows that $\\left[B C C^{\\prime}\\right]=[B C E]=\\frac{1}{2} x(x+y) \\sin \\alpha$. From the sine area formula with respect to $\\angle B A D^{\\prime}$ and $\\angle B^{\\prime} D C$, respectively,\n\n$$\n\\left[B A D^{\\prime}\\right]=\\frac{1}{2} x y|\\sin (2 \\alpha)|, \\quad\\left[B^{\\prime} D C\\right]=\\frac{1}{2} x(x-y) \\sin \\alpha\n$$\n\nThus\n\n$$\n\\frac{1}{2} x(x+y) \\sin \\alpha=\\frac{1}{2} x y|\\sin (2 \\alpha)|=\\frac{1}{2} x(x-y) \\sin \\alpha .\n$$\n\n\nBecause $\\overline{B C}, \\overline{A B^{\\prime}}$, and $\\overline{D^{\\prime} C^{\\prime}}$ are all parallel, it follows that $\\triangle B C C^{\\prime}$ and $\\triangle B C D^{\\prime}$ have the same height with respect to base $\\overline{B C}$, and thus $\\left[B C C^{\\prime}\\right]=\\left[B C D^{\\prime}\\right]$. Therefore $\\left[B C D^{\\prime}\\right]=\\left[A B D^{\\prime}\\right]$, and it follows that triangles $B C D^{\\prime}$ and $A B D^{\\prime}$ have the same height with respect to base $\\overline{B D^{\\prime}}$. Thus $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$. Let $M$ be the midpoint of $\\overline{A C}$. Consider the following two cases.\n\nCase 1: Suppose that $\\alpha<90^{\\circ}$. Because $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $M$ lies on $\\overleftrightarrow{B D^{\\prime}}$. But $\\overleftrightarrow{B D}$ also passes through the midpoint of $\\overline{A C}$ by parallelogram properties, so it follows that $D$ must lie on $\\overline{B D^{\\prime}}$. This implies that $\\left[A B D^{\\prime}\\right]$ must also equal $\\frac{y^{2} \\sin \\alpha}{2}+\\frac{x y \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1$ and $\\sin \\alpha=\\cos \\alpha=\\frac{\\sqrt{2}}{2}$. Finally, from right triangle $D^{\\prime} A B$ with legs in the ratio $1: \\sqrt{2}+1$, it follows that $\\tan (\\angle A B D)=\\tan \\left(\\angle A B D^{\\prime}\\right)=$ $\\sqrt{2}-1$.\n\nCase 2: Suppose that $\\alpha>90^{\\circ}$. The points $D$ and $D^{\\prime}$ lie on opposite sides of $\\overleftrightarrow{A B}$. Because $B C=A D^{\\prime}$ and points $A$ and $C$ are equidistant from $\\overleftrightarrow{B D^{\\prime}}$, it follows that $A C B D^{\\prime}$ is either a parallelogram or an isosceles trapezoid. It cannot be the former because that would imply that $\\overleftrightarrow{D^{\\prime} A}\\|\\overleftrightarrow{B C}\\| \\overleftrightarrow{A D}$. Thus $A C B D^{\\prime}$ is an isosceles trapezoid. Then $\\left[B A D^{\\prime}\\right]=\\left[B M D^{\\prime}\\right]$. Because $B, M$, and $D$ are collinear and $B D: B M=2$, it follows that $\\left[B D D^{\\prime}\\right]=2 \\cdot\\left[B M D^{\\prime}\\right]$. Moreover, $\\left[B D D^{\\prime}\\right]=\\left[B A D^{\\prime}\\right]+[B A D]+\\left[D A D^{\\prime}\\right]$, so $\\left[B A D^{\\prime}\\right]=[B A D]+\\left[D A D^{\\prime}\\right]$. Thus $\\left[B A D^{\\prime}\\right]=\\frac{x y \\sin \\alpha}{2}+\\frac{y^{2} \\sin \\alpha}{2}=\\frac{\\left(x y+y^{2}\\right) \\sin \\alpha}{2}$.\n\nThus $x(x-y) \\sin \\alpha=-x y \\sin 2 \\alpha=\\left(x y+y^{2}\\right) \\sin \\alpha$, which implies $x: y=\\sqrt{2}+1, \\sin \\alpha=\\frac{\\sqrt{2}}{2}$, and $\\cos \\alpha=-\\frac{\\sqrt{2}}{2}$, so $\\alpha=135^{\\circ}$. Let $H$ be the foot of the perpendicular from $D$ to $\\overleftrightarrow{A B}$. Then $A D H$ is a $45^{\\circ}-45^{\\circ}-90^{\\circ}$ triangle with $H A=H D=\\frac{y}{\\sqrt{2}}$. Thus\n\n$$\n\\begin{aligned}\n\\tan \\angle A B D & =\\frac{D H}{B H}=\\frac{D H}{B A+A H} \\\\\n& =\\frac{y / \\sqrt{2}}{x+y / \\sqrt{2}}=\\frac{y}{x \\sqrt{2}+y} \\\\\n& =\\frac{y}{y(\\sqrt{2}+1)(\\sqrt{2})+y} \\\\\n& =\\frac{1}{(\\sqrt{2}+1)(\\sqrt{2})+1} \\\\\n& =\\frac{\\mathbf{3}-\\sqrt{\\mathbf{2}}}{\\mathbf{7}}\n\\end{aligned}\n$$']","['$\\sqrt{2}-1$,$\\frac{3-\\sqrt{2}}{7}$']",True,,Numerical, 2772,Number Theory,,"Compute the least integer greater than 2023 , the sum of whose digits is 17 .","['A candidate for desired number is $\\underline{2} \\underline{0} \\underline{X} \\underline{Y}$, where $X$ and $Y$ are digits and $X+Y=15$. To minimize this number, take $Y=9$. Then $X=6$, and the desired number is 2069 .']",['2069'],False,,Numerical, 2773,Algebra,,"Let $T$ = 2069, and let $K$ be the sum of the digits of $T$. Let $r$ and $s$ be the two roots of the polynomial $x^{2}-18 x+K$. Compute $|r-s|$.","[""Note that $|r-s|=\\sqrt{r^{2}-2 r s+s^{2}}=\\sqrt{(r+s)^{2}-4 r s}$. By Vieta's Formulas, $r+s=-(-18)$ and $r s=K$, so $|r-s|=\\sqrt{18^{2}-4 K}$. With $T=2069, K=17$, and the answer is $\\sqrt{324-68}=\\sqrt{256}=16$.""]",['16'],False,,Numerical, 2774,Geometry,,"Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$. ","['Let $2 x$ be the side length of the squares. Then the intersection of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ ).']",['7'],False,,Numerical, 2774,Geometry,,"Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$. ![](https://cdn.mathpix.com/cropped/2023_12_21_a3766b499ee5b1dfa2f0g-1.jpg?height=241&width=290&top_left_y=411&top_left_x=1487)","['Let $2 x$ be the side length of the squares. Then the intersection of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ ).']",['7'],False,,Numerical, 2775,Algebra,,"Let $T=$ 7, and let $K=9 T$. Let $A_{1}=2$, and for $n \geq 2$, let $$ A_{n}= \begin{cases}A_{n-1}+1 & \text { if } n \text { is not a perfect square } \\ \sqrt{n} & \text { if } n \text { is a perfect square. }\end{cases} $$ Compute $A_{K}$.","['Let $\\lfloor\\sqrt{n}\\rfloor=x$. Then $n$ can be written as $x^{2}+y$, where $y$ is an integer such that $0 \\leq y<2 x+1$. Let $m$ be the greatest perfect square less than or equal to $9 T$. Then the definition of the sequence and the previous observation imply that $A_{K}=A_{9 T}=\\sqrt{m}+(9 T-m)=\\lfloor\\sqrt{9 T}\\rfloor+\\left(9 T-\\lfloor\\sqrt{9 T}\\rfloor^{2}\\right)$. With $T=7, K=9 T=63$, $\\lfloor\\sqrt{9 T}\\rfloor=7$, and the answer is therefore $7+\\left(63-7^{2}\\right)=\\mathbf{2 1}$.']",['21'],False,,Numerical, 2776,Number Theory,,Let $T=$ 21. The number $20^{T} \cdot 23^{T}$ has $K$ positive divisors. Compute the greatest prime factor of $K$.,"['Write $20^{T} \\cdot 23^{T}$ as $2^{2 T} \\cdot 5^{T} \\cdot 23^{T}$. This number has $K=(2 T+1)(T+1)^{2}$ positive divisors. With $T=21, K=43 \\cdot 22^{2}$. The greatest prime factor of $K$ is $\\mathbf{4 3}$.']",['43'],False,,Numerical, 2777,Number Theory,,Let $T=43$. Compute the positive integer $n \neq 17$ for which $\left(\begin{array}{c}T-3 \\ 17\end{array}\right)=\left(\begin{array}{c}T-3 \\ n\end{array}\right)$.,"['Using the symmetry property of binomial coefficients, the desired value of $n$ is $T-3-17=T-20$. With $T=43$, the answer is $\\mathbf{2 3}$.']",['23'],False,,Numerical, 2778,Number Theory,,Let $T=23$ . Compute the units digit of $T^{2023}+T^{20}-T^{23}$.,"['Assuming that $T$ is a positive integer, because units digits of powers of $T$ cycle in groups of at most 4, the numbers $T^{2023}$ and $T^{23}$ have the same units digit, hence the number $T^{2023}-T^{23}$ has a units digit of 0 , and the answer is thus the units digit of $T^{20}$. With $T=23$, the units digit of $23^{20}$ is the same as the units digit of $3^{20}$, which is the same as the units digit of $3^{4}=81$, so the answer is $\\mathbf{1}$.']",['1'],False,,Numerical, 2779,Geometry,,"In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$. ","['Using the Pythagorean Theorem, $I G=\\sqrt{(I L)^{2}-(L G)^{2}}=\\sqrt{41-25}=4$, and $G K=\\sqrt{(I K)^{2}-(I G)^{2}}=$ $\\sqrt{25-16}=3$.']",['3'],False,,Numerical, 2779,Geometry,,"In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$. ![](https://cdn.mathpix.com/cropped/2023_12_21_a3766b499ee5b1dfa2f0g-1.jpg?height=225&width=330&top_left_y=1210&top_left_x=1277)","['Using the Pythagorean Theorem, $I G=\\sqrt{(I L)^{2}-(L G)^{2}}=\\sqrt{41-25}=4$, and $G K=\\sqrt{(I K)^{2}-(I G)^{2}}=$ $\\sqrt{25-16}=3$.']",['3'],False,,Numerical, 2780,Combinatorics,,Let $T=$ 3. Suppose that $T$ fair coins are flipped. Compute the probability that at least one tails is flipped.,"['The probability of flipping all heads is $\\left(\\frac{1}{2}\\right)^{T}$, so the probability of flipping at least one tails is $1-\\frac{1}{2^{T}}$. With $T=3$, the desired probability is $1-\\frac{1}{8}=\\frac{7}{8}$.']",['$\\frac{7}{8}$'],False,,Numerical, 2781,Algebra,,"Let $T=$ $\frac{7}{8}$. The number $T$ can be expressed as a reduced fraction $\frac{m}{n}$, where $m$ and $n$ are positive integers whose greatest common divisor is 1 . The equation $x^{2}+(m+n) x+m n=0$ has two distinct real solutions. Compute the lesser of these two solutions.","['The left-hand side of the given equation can be factored as $(x+m)(x+n)$. The two solutions are therefore $-m$ and $-n$, so the answer is $\\min \\{-m,-n\\}$. With $T=\\frac{7}{8}, m=7, n=8$, and $\\min \\{-7,-8\\}$ is $\\mathbf{- 8}$.']",['-8'],False,,Numerical, 2782,Algebra,,"Let $T=$ -8, and let $i=\sqrt{-1}$. Compute the positive integer $k$ for which $(-1+i)^{k}=\frac{1}{2^{T}}$.","['Note that $(-1+i)^{2}=1+2 i-1=2 i$. Thus $(-1+i)^{4}=(2 i)^{2}=-4$, and $(-1+i)^{8}=(-4)^{2}=16$. The expression $\\frac{1}{2^{T}}$ is a power of 16 if $T$ is a negative multiple of 4 . With $T=-8, \\frac{1}{2^{-8}}=2^{8}=16^{2}=\\left((-1+i)^{8}\\right)^{2}=$ $(-1+i)^{16}$, so the desired value of $k$ is $\\mathbf{1 6}$.']",['16'],False,,Numerical, 2783,Algebra,,Let $T=$ 16. Compute the value of $x$ that satisfies $\log _{4} T=\log _{2} x$.,"['By the change of base rule and a property of $\\operatorname{logs}, \\log _{4} T=\\frac{\\log _{2} T}{\\log _{2} 4}=\\frac{\\log _{2} T}{2}=\\log _{2} \\sqrt{T}$. Thus $x=\\sqrt{T}$, and with $T=16, x=4$.']",['4'],False,,Numerical, 2784,Geometry,,"Let $T=$ 4. Pyramid $L E O J S$ is a right square pyramid with base $E O J S$, whose area is $T$. Given that $L E=5 \sqrt{2}$, compute $[L E O]$.","['Let the side length of square base $E O J S$ be $2 x$, and let $M$ be the midpoint of $\\overline{E O}$. Then $\\overline{L M} \\perp \\overline{E O}$, and $L M=\\sqrt{(5 \\sqrt{2})^{2}-x^{2}}$ by the Pythagorean Theorem. Thus $[L E O]=\\frac{1}{2} \\cdot 2 x \\sqrt{(5 \\sqrt{2})^{2}-x^{2}}=$\n\n\n\n$x \\sqrt{(5 \\sqrt{2})^{2}-x^{2}}$. With $T=4, x=1$, and the answer is $1 \\cdot \\sqrt{50-1}=\\mathbf{7}$.']",['7'],False,,Numerical, 2785,Number Theory,,Let $T=$ 7. Compute the units digit of $T^{2023}+(T-2)^{20}-(T+10)^{23}$.,"['Note that $T$ and $T+10$ have the same units digit. Because units digits of powers of $T$ cycle in groups of at most 4 , the numbers $T^{2023}$ and $(T+10)^{23}$ have the same units digit, hence the number $T^{2023}-(T+10)^{23}$ has a units digit of 0 , and the answer is thus the units digit of $(T-2)^{20}$. With $T=7$, the units digit of $5^{20}$ is 5 .']",['5'],False,,Numerical, 2786,Geometry,,"Let $r=1$ and $R=5$. A circle with radius $r$ is centered at $A$, and a circle with radius $R$ is centered at $B$. The two circles are internally tangent. Point $P$ lies on the smaller circle so that $\overline{B P}$ is tangent to the smaller circle. Compute $B P$.","['Draw radius $A P$ and note that $A P B$ is a right triangle with $\\mathrm{m} \\angle A P B=90^{\\circ}$. Note that $A B=R-r$ and $A P=r$, so by the Pythagorean Theorem, $B P=\\sqrt{(R-r)^{2}-r^{2}}=\\sqrt{R^{2}-2 R r}$. With $r=1$ and $R=5$, it follows that $B P=\\sqrt{\\mathbf{1 5}}$.']",['$\\sqrt{15}$'],False,,Numerical, 2787,Number Theory,,Compute the largest prime divisor of $15 !-13$ !.,"['Factor 15 ! -13 ! to obtain $13 !(15 \\cdot 14-1)=13$ ! $\\cdot 209$. The largest prime divisor of 13 ! is 13 , so continue by factoring $209=11 \\cdot 19$. Thus the largest prime divisor of 15 ! - 13 ! is 19 .']",['19'],False,,Numerical, 2788,Geometry,,"Three non-overlapping squares of positive integer side lengths each have one vertex at the origin and sides parallel to the coordinate axes. Together, the three squares enclose a region whose area is 41 . Compute the largest possible perimeter of the region.","['Proceed in two steps: first, determine the possible sets of side lengths for the squares; then determine which arrangement of squares produces the largest perimeter. Let the side lengths of the squares be positive integers $m \\geq n \\geq p$. Then $m^{2}+n^{2}+p^{2}=41$, so $m \\leq 6$, and because $3^{2}+3^{2}+3^{2}<41$, it follows that $m>3$. If $m=6$, then $n^{2}+p^{2}=5$, so $n=2$ and $p=1$. If $m=5$, then $n^{2}+p^{2}=16$, which has no positive integral solutions. If $m=4$, then $n^{2}+p^{2}=25$, which is possible if $n=4$ and $p=3$. So the two possible sets of values are $m=6, n=2, p=1$ or $m=4, n=4, p=3$.\n\nFirst consider $m=6, n=2, p=1$. Moving counterclockwise around the origin, one square is between the other two; by symmetry, it suffices to consider only the three possibilities for this ""middle"" square. If the middle square is the 6-square, then each of the other two squares has a side that is a subset of a side of the 6 -square. To compute the total perimeter, add the perimeters of the three squares and subtract twice the lengths of the shared segments (because they contribute 0 to the perimeter). Thus the total perimeter is $4 \\cdot 6+4 \\cdot 2+4 \\cdot 1-2 \\cdot 2-2 \\cdot 1=30$. If the middle square is the 2 -square, then one of its sides is a subset of the 6 -square\'s side, and one of its sides is a superset of the 1 -square\'s side, for a total perimeter of $4 \\cdot 6+4 \\cdot 2+4 \\cdot 1-2 \\cdot 2-2 \\cdot 1=$ 30. But if the middle square is the 1-square, then two of its sides are subsets of the other squares\' sides, and the total perimeter is $4 \\cdot 6+4 \\cdot 2+4 \\cdot 1-2 \\cdot 1-2 \\cdot 1=32$.\n\nIf $m=4, n=4$, and $p=3$, similar logic to the foregoing suggests that the maximal perimeter is obtained when the smallest square is between the other two, yielding a total perimeter of $4 \\cdot 4+4 \\cdot 4+4 \\cdot 3-2 \\cdot 3-2 \\cdot 3=32$. Either of the other two arrangements yields a total perimeter of $4 \\cdot 4+4 \\cdot 4+4 \\cdot 3-2 \\cdot 3-2 \\cdot 4=30$. So the maximum perimeter is $\\mathbf{3 2}$.', 'Let the side lengths be $a, b$, and $c$, and let $P$ be the perimeter. If the $a \\times a$ square is placed in between the other two (going either clockwise or counterclockwise around the origin), then\n\n$$\nP=3 b+|b-a|+2 a+|c-a|+3 c \\text {. }\n$$\n\nTo obtain a more symmetric expression, note that for any real numbers $x$ and $y$,\n\n$$\n|x-y|=\\max \\{x, y\\}-\\min \\{x, y\\}=x+y-2 \\min \\{x, y\\}\n$$\n\nUsing this identity,\n\n$$\nP=4 a+4 b+4 c-2 \\min \\{a, b\\}-2 \\min \\{a, c\\} .\n$$\n\nThus $P$ is the sum of the perimeters of the three, less twice the overlaps. To maximize $P$, choose $a$ to be the smallest of the three, which leads to $P=4 b+4 c$.\n\n\n\nAs in the first solution, the two possible sets of values are $c=6, b=2, a=1$ and $c=b=4$, $a=3$.\n\nIn the first case, the maximum length of the boundary is $P=4 \\cdot 2+4 \\cdot 6=32$, and in the second case it is $P=4 \\cdot 4+4 \\cdot 4=32$. So the maximum perimeter is $\\mathbf{3 2}$.']",['32'],False,,Numerical, 2789,Geometry,,"A circle with center $O$ and radius 1 contains chord $\overline{A B}$ of length 1 , and point $M$ is the midpoint of $\overline{A B}$. If the perpendicular to $\overline{A O}$ through $M$ intersects $\overline{A O}$ at $P$, compute $[M A P]$.","['Draw auxiliary segment $\\overline{O B}$, as shown in the diagram below.\n\n\n\nTriangle $O A B$ is equilateral, so $\\mathrm{m} \\angle O A B=60^{\\circ}$. Then $\\triangle M A P$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle with hypotenuse $A M=1 / 2$. Thus $A P=1 / 4$ and $M P=\\sqrt{3} / 4$, so\n\n$$\n\\begin{aligned}\n{[M A P] } & =\\frac{1}{2}\\left(\\frac{1}{4}\\right)\\left(\\frac{\\sqrt{3}}{4}\\right) \\\\\n& =\\frac{\\sqrt{3}}{\\mathbf{3 2}} .\n\\end{aligned}\n$$']",['$\\frac{\\sqrt{3}}{32}$'],False,,Numerical, 2790,Number Theory,,$\quad$ Suppose that $p$ and $q$ are two-digit prime numbers such that $p^{2}-q^{2}=2 p+6 q+8$. Compute the largest possible value of $p+q$.,"['Subtract from both sides and regroup to obtain $p^{2}-2 p-\\left(q^{2}+6 q\\right)=8$. Completing both squares yields $(p-1)^{2}-(q+3)^{2}=0$. The left side is a difference of two squares; factor to obtain $((p-1)+(q+3))((p-1)-(q+3))=0$, whence $(p+q+2)(p-q-4)=0$. For positive primes $p$ and $q$, the first factor $p+q+2$ must also be positive. Therefore the second factor $p-q-4$ must be zero, hence $p-4=q$. Now look for primes starting with 97 and working downward. If $p=97$, then $q=93$, which is not prime; if $p=89$, then $q=85$, which is also not prime. But if $p=83$, then $q=79$, which is prime. Thus the largest possible value of $p+q$ is $83+79=\\mathbf{1 6 2}$.']",['162'],False,,Numerical, 2791,Algebra,,The four zeros of the polynomial $x^{4}+j x^{2}+k x+225$ are distinct real numbers in arithmetic progression. Compute the value of $j$.,"['Let the four zeros be $p \\leq q \\leq r \\leq s$. The coefficient of $x^{3}$ is 0 , so $p+q+r+s=0$. The mean of four numbers in arithmetic progression is the mean of the middle two numbers, so $q=-r$. Then the common difference is $r-q=r-(-r)=2 r$, so $s=r+2 r=3 r$ and $p=q-2 r=-3 r$. Therefore the four zeros are $-3 r,-r, r, 3 r$. The product of\n\n\n\nthe zeros is $9 r^{4}$; referring to the original polynomial and using the product of roots formula gives $9 r^{4}=225$. Thus $r=\\sqrt{5}$, the zeros are $-3 \\sqrt{5},-\\sqrt{5}, \\sqrt{5}, 3 \\sqrt{5}$, and the polynomial can be factored as $(x-\\sqrt{5})(x+\\sqrt{5})(x-3 \\sqrt{5})(x+3 \\sqrt{5})$. Expanding this product yields $\\left(x^{2}-5\\right)\\left(x^{2}-45\\right)=x^{4}-50 x^{2}+225$, so $j=-50$.', 'Proceed as in the original solution, finding the values $-3 \\sqrt{5},-\\sqrt{5}, \\sqrt{5}$, and $3 \\sqrt{5}$ for the zeros. By the sums and products of roots formulas, the coefficient of $x^{2}$ is the sum of all six possible products of pairs of roots:\n\n$$\n(-3 \\sqrt{5})(-\\sqrt{5})+(-3 \\sqrt{5})(\\sqrt{5})+(-3 \\sqrt{5})(3 \\sqrt{5})+(-\\sqrt{5})(\\sqrt{5})+(-\\sqrt{5})(3 \\sqrt{5})+(\\sqrt{5})(3 \\sqrt{5})\n$$\n\nObserving that some of these terms will cancel yields the simpler expression\n\n$$\n(-3 \\sqrt{5})(3 \\sqrt{5})+(-\\sqrt{5})(\\sqrt{5})=-45+-5=-50\n$$']",['-50'],False,,Numerical, 2792,Algebra,,"Compute the smallest positive integer $n$ such that $$ n,\lfloor\sqrt{n}\rfloor,\lfloor\sqrt[3]{n}\rfloor,\lfloor\sqrt[4]{n}\rfloor,\lfloor\sqrt[5]{n}\rfloor,\lfloor\sqrt[6]{n}\rfloor,\lfloor\sqrt[7]{n}\rfloor, \text { and }\lfloor\sqrt[8]{n}\rfloor $$ are distinct.","['Inverting the problem, the goal is to find seven positive integers $a2012,2012$ !! $0$. Then the angle between the sides $\\overline{0 z}$ and $\\overline{0 z^{-1}}$ is $2 \\theta$, and the side lengths are $r$ and $r^{-1}$, so the area of the parallelogram is\n\n$$\n\\frac{35}{37}=r \\cdot r^{-1} \\cdot \\sin (2 \\theta)=\\sin 2 \\theta\n$$\n\nNote that $0<\\theta<\\pi / 2$, so $0<2 \\theta<\\pi$, and there are two values of $\\theta$ that satisfy this equation. Adding the expressions for $z$ and $z^{-1}$ and calculating the absolute value yields\n\n$$\n\\begin{aligned}\n\\left|z+\\frac{1}{z}\\right|^{2} & =\\left(r+r^{-1}\\right)^{2} \\cos ^{2} \\theta+\\left(r-r^{-1}\\right)^{2} \\sin ^{2} \\theta \\\\\n& =\\left(r^{2}+r^{-2}\\right)\\left(\\cos ^{2} \\theta+\\sin ^{2} \\theta\\right)+2 r \\cdot r^{-1}\\left(\\cos ^{2} \\theta-\\sin ^{2} \\theta\\right) \\\\\n& =r^{2}+r^{-2}+2 \\cos 2 \\theta .\n\\end{aligned}\n$$\n\nMinimize the terms involving $r$ using the Arithmetic-Geometric Mean inequality:\n\n$$\nr^{2}+r^{-2} \\geq 2 \\sqrt{r^{2} \\cdot r^{-2}}=2\n$$\n\nwith equality when $r^{2}=r^{-2}$, that is, when $r=1$. For the term involving $\\theta$, recall that there are two possible values:\n\n$$\n\\cos 2 \\theta= \\pm \\sqrt{1-\\sin ^{2} 2 \\theta}= \\pm \\sqrt{\\frac{37^{2}-35^{2}}{37^{2}}}= \\pm \\frac{\\sqrt{(37+35)(37-35)}}{37}= \\pm \\frac{12}{37}\n$$\n\nTo minimize this term, take the negative value, yielding\n\n$$\nd^{2}=2-2 \\cdot \\frac{12}{37}=\\frac{\\mathbf{5 0}}{\\mathbf{3 7}}\n$$', 'If $z=x+y i$, then compute $1 / z$ by rationalizing the denominator:\n\n$$\n\\frac{1}{z}=\\frac{x-y i}{x^{2}+y^{2}}=\\frac{x}{x^{2}+y^{2}}+\\frac{-y}{x^{2}+y^{2}} i\n$$\n\nThe area of the parallelogram is given by the absolute value of the $2 \\times 2$ determinant\n\n$$\n\\left|\\begin{array}{cc}\nx & y \\\\\nx /\\left(x^{2}+y^{2}\\right) & -y /\\left(x^{2}+y^{2}\\right)\n\\end{array}\\right|=\\frac{1}{x^{2}+y^{2}}\\left|\\begin{array}{cc}\nx & y \\\\\nx & -y\n\\end{array}\\right|=\\frac{-2 x y}{x^{2}+y^{2}}\n$$\n\n\n\nThat is,\n\n$$\n\\frac{2 x y}{x^{2}+y^{2}}=\\frac{35}{37}\n$$\n\nCalculation shows that\n\n$$\n\\left|z+\\frac{1}{z}\\right|^{2}=\\left(x+\\frac{x}{x^{2}+y^{2}}\\right)^{2}+\\left(y-\\frac{y}{x^{2}+y^{2}}\\right)^{2}=\\left(x^{2}+y^{2}\\right)+\\frac{1}{x^{2}+y^{2}}+2\\left(\\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\\right) .\n$$\n\nAs in the previous solution, the sum of the first two terms is at least 2 , when $x^{2}+y^{2}=1$. The trick for relating the third term to the area is to express both the third term and the area in terms of the ratio\n\n$$\nt=\\frac{y}{x} .\n$$\n\nIndeed,\n\n$$\n\\frac{2 x y}{x^{2}+y^{2}}=\\frac{2 t}{1+t^{2}} \\quad \\text { and } \\quad \\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\\frac{1-t^{2}}{1+t^{2}}=\\frac{(1+t)(1-t)}{1+t^{2}}\n$$\n\nAs in the previous solution, assume without loss of generality that $z$ is in the first quadrant, so that $t>0$. As found above,\n\n$$\n\\frac{2 t}{1+t^{2}}=\\frac{35}{37}\n$$\n\nIt is not difficult to solve for $t$ using the quadratic formula, but the value of $t$ is not needed to solve the problem. Observe that\n\n$$\n\\frac{(1 \\pm t)^{2}}{1+t^{2}}=1 \\pm \\frac{2 t}{1+t^{2}}=1 \\pm \\frac{35}{37},\n$$\n\nso that\n\n$$\n\\left(\\frac{1-t^{2}}{1+t^{2}}\\right)^{2}=\\frac{(1+t)^{2}}{1+t^{2}} \\cdot \\frac{(1-t)^{2}}{1+t^{2}}=\\frac{72}{37} \\cdot \\frac{2}{37}=\\left(\\frac{12}{37}\\right)^{2}\n$$\n\nIn order to minimize $d$, take the negative square root, leading to\n\n$$\nd^{2}=2+2 \\cdot \\frac{1-t^{2}}{1+t^{2}}=2-\\frac{24}{37}=\\frac{\\mathbf{5 0}}{\\mathbf{3 7}}\n$$']",['$\\frac{50}{37}$'],False,,Numerical, 2795,Combinatorics,,"One face of a $2 \times 2 \times 2$ cube is painted (not the entire cube), and the cube is cut into eight $1 \times 1 \times 1$ cubes. The small cubes are reassembled randomly into a $2 \times 2 \times 2$ cube. Compute the probability that no paint is showing.","['Call each $1 \\times 1 \\times 1$ cube a cubelet. Then four cubelets are each painted on one face, and the other four cubelets are completely unpainted and can be ignored. For each painted cubelet, the painted face can occur in six positions, of which three are hidden from the outside, so the probability that a particular painted cubelet has no paint showing is $3 / 6=1 / 2$. Thus the probability that all four painted cubelets have no paint showing is $(1 / 2)^{4}=\\frac{1}{\\mathbf{1 6}}$.']",['$\\frac{1}{16}$'],False,,Numerical, 2796,Geometry,,"In triangle $A B C, A B=B C$. A trisector of $\angle B$ intersects $\overline{A C}$ at $D$. If $A B, A C$, and $B D$ are integers and $A B-B D=7$, compute $A C$.","['Let $E$ be the point where the other trisector of $\\angle B$ intersects side $\\overline{A C}$. Let $A B=B C=a$, and let $B D=B E=d$. Draw $X$ on $\\overline{B C}$ so that $B X=d$. Then $C X=7$.\n\n\n\nThe placement of point $X$ guarantees that $\\triangle B E X \\cong \\triangle B D E$ by Side-Angle-Side. Therefore $\\angle B X E \\cong \\angle B E X \\cong \\angle B D E$, and so $\\angle C X E \\cong \\angle A D B \\cong \\angle C E B$. By Angle-Angle, $\\triangle C E X \\sim \\triangle C B E$. Let $E X=c$ and $E C=x$. Then comparing ratios of corresponding sides yields\n\n$$\n\\frac{c}{d}=\\frac{7}{x}=\\frac{x}{d+7}\n$$\n\nUsing the right proportion, $x^{2}=7(d+7)$. Because $d$ is an integer, $x^{2}$ is an integer, so either $x$ is an integer or irrational. The following argument shows that $x$ cannot be irrational. Applying the Angle Bisector Theorem to $\\triangle B C D$ yields $D E=c=\\frac{d}{d+7} \\cdot x$. Then $A C=2 x+c=$ $x\\left(2+\\frac{d}{d+7}\\right)$. Because the expression $\\left(2+\\frac{d}{d+7}\\right)$ is rational, $A C$ will not be an integer if $x$ is irrational.\n\nHence $x$ is an integer, and because $x^{2}$ is divisible by $7, x$ must also be divisible by 7 . Let $x=7 k$ so that $d=c k$. Rewrite the original proportion using $7 k$ for $x$ and $c k$ for $d$ :\n\n$$\n\\begin{aligned}\n\\frac{c}{d} & =\\frac{x}{d+7} \\\\\n\\frac{c}{c k} & =\\frac{7 k}{c k+7} \\\\\n7 k^{2} & =c k+7 \\\\\n7 k & =c+\\frac{7}{k} .\n\\end{aligned}\n$$\n\n\n\nBecause the left side of this last equation represents an integer, $7 / k$ must be an integer, so either $k=1$ or $k=7$. The value $k=1$ gives the extraneous solution $c=0$. So $k=7$, from which $c=48$. Then $d=336$ and $A C=2 x+c=2 \\cdot 49+48=\\mathbf{1 4 6}$.']",['146'],False,,Numerical, 2797,Algebra,,"The rational number $r$ is the largest number less than 1 whose base-7 expansion consists of two distinct repeating digits, $r=0 . \underline{A} \underline{B} \underline{A} \underline{B} \underline{A} \underline{B} \ldots$ Written as a reduced fraction, $r=\frac{p}{q}$. Compute $p+q$ (in base 10).","['In base 7, the value of $r$ must be $0.656565 \\ldots=0 . \\overline{65}_{7}$. Then $100_{7} \\cdot r=65 . \\overline{65}_{7}$, and $\\left(100_{7}-1\\right) r=$ $65_{7}$. In base $10,65_{7}=6 \\cdot 7+5=47_{10}$ and $100_{7}-1=7^{2}-1=48_{10}$. Thus $r=47 / 48$, and $p+q=95$.']",['95'],False,,Numerical, 2798,Geometry,,"Let $T=95$. Triangle $A B C$ has $A B=A C$. Points $M$ and $N$ lie on $\overline{B C}$ such that $\overline{A M}$ and $\overline{A N}$ trisect $\angle B A C$, with $M$ closer to $C$. If $\mathrm{m} \angle A M C=T^{\circ}$, then $\mathrm{m} \angle A C B=U^{\circ}$. Compute $U$.","['Because $\\triangle A B C$ is isosceles with $A B=A C, \\mathrm{~m} \\angle A B C=U^{\\circ}$ and $\\mathrm{m} \\angle B A C=(180-2 U)^{\\circ}$. Therefore $\\mathrm{m} \\angle M A C=\\left(\\frac{180-2 U}{3}\\right)^{\\circ}=\\left(60-\\frac{2}{3} U\\right)^{\\circ}$. Then $\\left(60-\\frac{2}{3} U\\right)+U+T=180$, so $\\frac{1}{3} U=$ $120-T$ and $U=3(120-T)$. Substituting $T=95$ yields $U=\\mathbf{7 5}$.']",['75'],False,,Numerical, 2799,Combinatorics,,"Let $T=75$. At Wash College of Higher Education (Wash Ed.), the entering class has $n$ students. Each day, two of these students are selected to oil the slide rules. If the entering class had two more students, there would be $T$ more ways of selecting the two slide rule oilers. Compute $n$.","['With $n$ students, Wash Ed. can choose slide-rule oilers in $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)=\\frac{n(n-1)}{2}$ ways. With $n+2$ students, there would be $\\left(\\begin{array}{c}n+2 \\\\ 2\\end{array}\\right)=\\frac{(n+2)(n+1)}{2}$ ways of choosing the oilers. The difference is $\\frac{(n+2)(n+1)}{2}-\\frac{n(n-1)}{2}=T$. Simplifying yields $\\frac{\\left(n^{2}+3 n+2\\right)-\\left(n^{2}-n\\right)}{2}=2 n+1=T$, so $n=\\frac{T-1}{2}$. Because $T=75, n=37$.']",['37'],False,,Numerical, 2800,Number Theory,,"Compute the least positive integer $n$ such that the set of angles $$ \left\{123^{\circ}, 246^{\circ}, \ldots, n \cdot 123^{\circ}\right\} $$ contains at least one angle in each of the four quadrants.","['The first angle is $123^{\\circ}$, which is in Quadrant II, the second $\\left(246^{\\circ}\\right)$ is in Quadrant III, and the third is in Quadrant I, because $3 \\cdot 123^{\\circ}=369^{\\circ} \\equiv 9^{\\circ} \\bmod 360^{\\circ}$. The missing quadrant is IV, which is $270^{\\circ}-246^{\\circ}=24^{\\circ}$ away from the second angle in the sequence. Because $3 \\cdot 123^{\\circ} \\equiv 9^{\\circ} \\bmod 360^{\\circ}$, the terminal ray of the $(n+3)^{\\mathrm{rd}}$ angle is rotated $9^{\\circ}$ counterclockwise from the $n^{\\text {th }}$ angle. Thus three full cycles are needed to reach Quadrant IV starting from the second angle: the fifth angle is $255^{\\circ}$, the eighth angle is $264^{\\circ}$, and the eleventh angle is $273^{\\circ}$. So $n=11$.']",['11'],False,,Numerical, 2801,Number Theory,,"Let $T=11$. In ARMLvania, license plates use only the digits 1-9, and each license plate contains exactly $T-3$ digits. On each plate, all digits are distinct, and for all $k \leq T-3$, the $k^{\text {th }}$ digit is at least $k$. Compute the number of valid ARMLvanian license plates.","['There are 9 valid one-digit plates. For a two-digit plate to be valid, it has to be of the form $\\underline{A} \\underline{B}$, where $B \\in\\{2, \\ldots, 9\\}$, and either $A \\in\\{2, \\ldots, 9\\}$ with $A \\neq B$ or $A=1$. So there are 8 ways to choose $B$ and $8-1+1=8$ ways to choose $A$, for a total of $8 \\cdot 8=64$ plates. In general, moving from the last digit to the first, if there are $k$ ways to choose digit $n$, then there are $k-1$ ways to choose digit $n-1$ from the same set of possibilities as digit $n$ had, plus one additional way, for a total of $k-1+1=k$ choices for digit $n-1$. So if a license plate has $d$ digits, there are $10-d$ choices for the last digit and for each digit before it, yielding $(10-d)^{d}$ possible $d$-digit plates. Using $d=T-3=8$, there are $2^{8}=\\mathbf{2 5 6}$ plates.']",['256'],False,,Numerical, 2802,Algebra,,Let $T=256$. Let $\mathcal{R}$ be the region in the plane defined by the inequalities $x^{2}+y^{2} \geq T$ and $|x|+|y| \leq \sqrt{2 T}$. Compute the area of region $\mathcal{R}$.,"['The first inequality states that the point $(x, y)$ is outside the circle centered at the origin with radius $\\sqrt{T}$, while the second inequality states that $(x, y)$ is inside the tilted square centered at the origin with diagonal $2 \\sqrt{2 T}$. The area of the square is $4 \\cdot \\frac{1}{2}(\\sqrt{2 T})^{2}=4 T$, while the area of the circle is simply $\\pi T$, so the area of $\\mathcal{R}$ is $4 T-\\pi T=\\mathbf{1 0 2 4}-\\mathbf{2 5 6 \\pi}$.']",['$1024-256 \\pi$'],False,,Numerical, 2803,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Consider the hideout map $M$ below. Show that one Cop can always catch the Robber.","['Have the Cop stay at $A$ for 2 days. If the Robber is not at $A$ the first day, he must be at one of $B_{1}-B_{6}$, and because the Robber must move along an edge every night, he will be forced to go to $A$ on day 2 .']",,True,,, 2803,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Consider the hideout map $M$ below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=350&width=401&top_left_y=324&top_left_x=908) Show that one Cop can always catch the Robber.","['Have the Cop stay at $A$ for 2 days. If the Robber is not at $A$ the first day, he must be at one of $B_{1}-B_{6}$, and because the Robber must move along an edge every night, he will be forced to go to $A$ on day 2 .']",['证明题，略'],True,,Need_human_evaluate, 2804,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The map shown below is $\mathcal{C}_{6}$, the cyclic graph with six hideouts. Show that three Cops are sufficient to catch the Robber on $\mathcal{C}_{6}$, so that $C\left(\mathcal{C}_{6}\right) \leq 3$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912)","['The Cops should stay at $\\left\\{A_{1}, A_{3}, A_{5}\\right\\}$ for 2 days. If the Robber evades capture the first day, he must have been at an even-numbered hideout. Because he must move, he will be at an odd-numbered hideout the second day. Equivalently, the Cops could stay at $\\left\\{A_{2}, A_{4}, A_{6}\\right\\}$ for 2 days.']",['证明题，略'],True,,Need_human_evaluate, 2804,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The map shown below is $\mathcal{C}_{6}$, the cyclic graph with six hideouts. Show that three Cops are sufficient to catch the Robber on $\mathcal{C}_{6}$, so that $C\left(\mathcal{C}_{6}\right) \leq 3$. ","['The Cops should stay at $\\left\\{A_{1}, A_{3}, A_{5}\\right\\}$ for 2 days. If the Robber evades capture the first day, he must have been at an even-numbered hideout. Because he must move, he will be at an odd-numbered hideout the second day. Equivalently, the Cops could stay at $\\left\\{A_{2}, A_{4}, A_{6}\\right\\}$ for 2 days.']",,True,,, 2805,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Show that for all maps $M, C(M) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Find $C(M)$ for the map below. ","['First we prove that for all maps $M, C(M) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Show that $C(M) \leq 3$ for the map below. ","['The following strategy guarantees capture using three Cops for four consecutive days, so $C(M) \\leq 3$. Position three Cops at $\\{B, E, H\\}$ for two days, which will catch any Robber who starts out at $B, C, D, E, F, G$, or $H$, because a Robber at $C$ or $D$ would have to move to either $B$ or $E$, and a Robber at $F$ or $G$ would have to move to either $E$ or $H$. So if the Robber is not yet caught, he must have been at $I, J, K, L$, or $A$ for those two days. In this case, an analogous argument shows that placing Cops at $B, H, K$ for two consecutive days will guarantee a capture.']",,True,,, 2807,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. Show that $C(M) \leq 3$ for the map below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=353&width=353&top_left_y=2095&top_left_x=932)","['The following strategy guarantees capture using three Cops for four consecutive days, so $C(M) \\leq 3$. Position three Cops at $\\{B, E, H\\}$ for two days, which will catch any Robber who starts out at $B, C, D, E, F, G$, or $H$, because a Robber at $C$ or $D$ would have to move to either $B$ or $E$, and a Robber at $F$ or $G$ would have to move to either $E$ or $H$. So if the Robber is not yet caught, he must have been at $I, J, K, L$, or $A$ for those two days. In this case, an analogous argument shows that placing Cops at $B, H, K$ for two consecutive days will guarantee a capture.']",['证明题，略'],True,,Need_human_evaluate, 2808,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. If $\mathcal{K}_{n}$ is a map with $n$ hideouts in which every hideout is adjacent to every other hideout (also called the complete map on $n$ hideouts), determine $C\left(\mathcal{K}_{n}\right)$.","['First we prove that for all maps $M, C(M)n-2$.']",['$C\\left(\\mathcal{K}_{n}\\right)>n-2$'],False,,Need_human_evaluate, 2809,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) A path on $n$ hideouts is a map with $n$ hideouts, connected in one long string. (More formally, a map is a path if and only if two hideouts are adjacent to exactly one hideout each and all other hideouts are adjacent to exactly two hideouts each.) It is denoted by $\mathcal{P}_{n}$. The maps $\mathcal{P}_{3}$ through $\mathcal{P}_{6}$ are shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_a4ab1664c3b96ebcfbd1g-1.jpg?height=174&width=1548&top_left_y=951&top_left_x=337) Show that $D\left(\mathcal{P}_{n}, 1\right) \leq 2 n$ for $n \geq 3$.","[""The following argument shows that $C\\left(\\mathcal{P}_{n}\\right)=1$, and that capture occurs in at most $2 n-4$ days. It helps to draw the hideout map as in the following diagram, so that odd-numbered hideouts are all on one level and even-numbered hideouts are all on another level; the case where $n$ is odd is shown below. (A similar argument applies where $n$ is even.)\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_fbf8df629daaa704cef9g-1.jpg?height=203&width=631&top_left_y=2240&top_left_x=801)\n\n\n\nEach night, the Robber has the choice of moving diagonally left or diagonally right on this map, but he is required to move from top to bottom or vice-versa.\n\nFor the first $n-2$ days, the Cop should search hideouts $A_{2}, A_{3}, \\ldots, A_{n-1}$, in that order. For the next $n-2$ days, the Cop should search hideouts $A_{n-1}, A_{n-2}, \\ldots, A_{2}$, in that order.\n\nThe Cop always moves from top to bottom, or vice-versa, except that he searches $A_{n-1}$ two days in a row. If the Cop is lucky, the Robber chose an even-numbered hideout on the first day. In this case, the Cop and the Robber are always on the same level (top or bottom) for the first half ( $n-2$ days) of the search. The Cop is searching from left to right, and the Robber started out to the right of the Cop (or at $A_{2}$, the first hideout searched), so eventually the Robber is caught (at $A_{n-1}$, if not earlier).\n\nIf the Robber chose an odd-numbered hideout on the first day, then the Cop and Robber will be on different levels for the first half of the search, but on the same level for the second half. For the second half of the search, the Robber is to the left of the Cop (or possibly at $A_{n-1}$, which would happen if the Robber moved unwisely or if he spent day $n-2$ at hideout $A_{n}$ ). But now the Cop is searching from right to left, so again, the Cop will eventually catch the Robber.\n\nThe same strategy can be justified without using the zig-zag diagram above. Suppose that the Robber is at hideout $A_{R}$ on a given day, and the Cop searches hideout $A_{C}$. Whenever the Cop moves to the next hideout or the preceding hideout, $C$ changes by \\pm 1 , and the Robber's constraint forces $R$ to change by \\pm 1 . Thus if the Cop uses the strategy above, on each of the first $n-2$ days, either the difference $R-C$ stays the same or it decreases by 2 . If on the first day $R-C$ is even, either $R-C$ is 0 (the Robber was at $A_{2}$ and is caught immediately) or is positive (because $C=2$ and $R \\geq 4$ ). Because the difference $R-C$ is even and decreases (if at all) by 2 each day, it cannot go from positive to negative without being zero. If on the first day $R-C$ is odd, then the Robber avoids capture through day $n-2$ (because $R-C$ is still odd), but then on day $n-1, R$ changes (by \\pm 1 ) while $C$ does not. So $R-C$ is now even (and is either 0 or negative), and henceforth either remains the same or increases by 2 each day, so again, $R-C$ must be zero at some point between day $n-1$ and day $2 n-4$, inclusive.""]",['证明题，略'],True,,Need_human_evaluate, 2809,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. A path on $n$ hideouts is a map with $n$ hideouts, connected in one long string. (More formally, a map is a path if and only if two hideouts are adjacent to exactly one hideout each and all other hideouts are adjacent to exactly two hideouts each.) It is denoted by $\mathcal{P}_{n}$. The maps $\mathcal{P}_{3}$ through $\mathcal{P}_{6}$ are shown below. Show that $D\left(\mathcal{P}_{n}, 1\right) \leq 2 n$ for $n \geq 3$.","[""The following argument shows that $C\\left(\\mathcal{P}_{n}\\right)=1$, and that capture occurs in at most $2 n-4$ days. It helps to draw the hideout map as in the following diagram, so that odd-numbered hideouts are all on one level and even-numbered hideouts are all on another level; the case where $n$ is odd is shown below. (A similar argument applies where $n$ is even.)\n\n\n\n\n\nEach night, the Robber has the choice of moving diagonally left or diagonally right on this map, but he is required to move from top to bottom or vice-versa.\n\nFor the first $n-2$ days, the Cop should search hideouts $A_{2}, A_{3}, \\ldots, A_{n-1}$, in that order. For the next $n-2$ days, the Cop should search hideouts $A_{n-1}, A_{n-2}, \\ldots, A_{2}$, in that order.\n\nThe Cop always moves from top to bottom, or vice-versa, except that he searches $A_{n-1}$ two days in a row. If the Cop is lucky, the Robber chose an even-numbered hideout on the first day. In this case, the Cop and the Robber are always on the same level (top or bottom) for the first half ( $n-2$ days) of the search. The Cop is searching from left to right, and the Robber started out to the right of the Cop (or at $A_{2}$, the first hideout searched), so eventually the Robber is caught (at $A_{n-1}$, if not earlier).\n\nIf the Robber chose an odd-numbered hideout on the first day, then the Cop and Robber will be on different levels for the first half of the search, but on the same level for the second half. For the second half of the search, the Robber is to the left of the Cop (or possibly at $A_{n-1}$, which would happen if the Robber moved unwisely or if he spent day $n-2$ at hideout $A_{n}$ ). But now the Cop is searching from right to left, so again, the Cop will eventually catch the Robber.\n\nThe same strategy can be justified without using the zig-zag diagram above. Suppose that the Robber is at hideout $A_{R}$ on a given day, and the Cop searches hideout $A_{C}$. Whenever the Cop moves to the next hideout or the preceding hideout, $C$ changes by \\pm 1 , and the Robber's constraint forces $R$ to change by \\pm 1 . Thus if the Cop uses the strategy above, on each of the first $n-2$ days, either the difference $R-C$ stays the same or it decreases by 2 . If on the first day $R-C$ is even, either $R-C$ is 0 (the Robber was at $A_{2}$ and is caught immediately) or is positive (because $C=2$ and $R \\geq 4$ ). Because the difference $R-C$ is even and decreases (if at all) by 2 each day, it cannot go from positive to negative without being zero. If on the first day $R-C$ is odd, then the Robber avoids capture through day $n-2$ (because $R-C$ is still odd), but then on day $n-1, R$ changes (by \\pm 1 ) while $C$ does not. So $R-C$ is now even (and is either 0 or negative), and henceforth either remains the same or increases by 2 each day, so again, $R-C$ must be zero at some point between day $n-1$ and day $2 n-4$, inclusive.""]",,True,,, 2810,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Determine $W(M)$ for each of the maps in figure a and figure b. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=350&width=355&top_left_y=1603&top_left_x=926) figure a ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=353&width=353&top_left_y=2095&top_left_x=932) figure b","['or the map $M$ from figure a, $W(M)=7$. The most efficient strategy is to use 7 Cops to blanket all the hideouts on the first day. Any strategy using fewer than 7 Cops would require 6 Cops on each of two consecutive days: given that any hideout can be reached from any other hideout, leaving more than one hideout unsearched on one day makes it is impossible to eliminate any hideouts the following day. So any other strategy would require a minimum of 12 Cop workdays.\n\nFor the map $M$ from figure b, $W(M)=8$. The strategy outlined in $2 \\mathrm{~b}$ used three Cops for a maximum of four workdays, yielding 12 Cop workdays. The most efficient strategy is to use 4 Cops, positioned at $\\{B, E, H, K\\}$ for 2 days each. The following argument demonstrates that 8 Cop workdays is in fact minimal. First, notice that there is no advantage to searching one of the hideouts between vertices of the square (for example, $C$ ) without searching the other hideout between the same vertices (for example, $D$ ). The Robber can reach $C$ on day $n$ if and only if he is at either $B$ or $E$ on day $n-1$, and in either case he could just as well go to $D$ instead of $C$. So there is no situation in which the Robber is certain to be caught at $C$ rather than at $D$. Additionally, the Robber\'s possible locations on day $n+1$ are the same whether he is at $C$ or $D$ on day $n$, so searching one rather than the other fails to rule out any locations for future days. So any successful strategy that involves searching $C$ should also involve searching $D$ on the same day, and similarly for $F$ and $G, I$ and $J$, and $L$ and $A$. On the other hand, if the Robber must be at one of $C$ and $D$ on day $n$, then he must be at either $B$ or $E$ on day $n+1$, because those are the only adjacent hideouts. So any strategy that involves searching both hideouts of one of the off-the-square pairs on day $n$ is equivalent to a strategy that searches the adjacent on-the-square hideouts on day $n+1$; the two strategies use the same number of Cops for the same number of workdays. Thus the optimal number of Cop workdays can be achieved using strategies that only search the ""corner"" hideouts $B, E, H, K$.\n\nRestricting the search to only those strategies searching corner hideouts $B, E, H, K$, a total of 8 workdays can be achieved by searching all four hideouts on two consecutive days: if the Robber is at one of the other eight hideouts the first day, he must move to one of the two adjacent corner hideouts the second day. But each of these corner hideouts is adjacent to two other corner hideouts. So if only one hideout is searched, for no matter how many consecutive days, the following day, the Robber could either be back at the previously-searched hideout or be at any other hideout: no possibilities are ruled out. If two adjacent corners are searched, the Cops do no better, as the following argument shows. Suppose that $B$ and $E$ are both searched for two consecutive days. Then the Cops can rule out $B, E, C$, and $D$ as possible locations, but if the Cops then switch to searching either $H$ or $K$ instead of $B$ or $E$, the Robber can go back to $C$ or $D$ within two days. So searching two adjacent corner hideouts for two days is fruitless and costs four Cop workdays. Searching diagonally opposite corner hideouts is even less fruitful, because doing so rules out none of the other hideouts as possible Robber locations. Using three Cops each day, it is easy to imagine scenarios in which the Robber evades capture for three days before being caught: for example, if $B, E, H$ are searched for two consecutive days, the Robber goes from $I$ or $J$ to $K$ to $L$. Therefore if three Cops are used, four days are required for a total of 12 Cop workdays.\n\nIf there are more than four Cops, the preceding arguments show that the number of Cops must be even to produce optimal results (because there is no advantage to searching one hideout between vertices of the square without searching the other). Using six Cops with four at corner hideouts yields no improvement, because the following day, the Robber could get to any of the four corner hideouts, requiring at least four Cops the second day, for ten Cop workdays. If two or fewer Cops are at corner hideouts, the situation is even worse, because if the Robber is not caught that day, he has at least nine possible hideouts the following day (depending on whether the unsearched corners are adjacent or diagonally opposite to each other). Using eight Cops (with four at corner vertices) could eliminate one corner vertex as a possible location for the second day (if the non-corner hideouts searched are on adjacent sides of the square), but eight Cop workdays have already been used on the first day. So 8 Cop workdays is minimal.']","['12, 8']",True,,Numerical, 2810,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Determine $W(M)$ for each of the maps in figure a and figure b. figure a figure b","['or the map $M$ from figure a, $W(M)=7$. The most efficient strategy is to use 7 Cops to blanket all the hideouts on the first day. Any strategy using fewer than 7 Cops would require 6 Cops on each of two consecutive days: given that any hideout can be reached from any other hideout, leaving more than one hideout unsearched on one day makes it is impossible to eliminate any hideouts the following day. So any other strategy would require a minimum of 12 Cop workdays.\n\nFor the map $M$ from figure b, $W(M)=8$. The strategy outlined in $2 \\mathrm{~b}$ used three Cops for a maximum of four workdays, yielding 12 Cop workdays. The most efficient strategy is to use 4 Cops, positioned at $\\{B, E, H, K\\}$ for 2 days each. The following argument demonstrates that 8 Cop workdays is in fact minimal. First, notice that there is no advantage to searching one of the hideouts between vertices of the square (for example, $C$ ) without searching the other hideout between the same vertices (for example, $D$ ). The Robber can reach $C$ on day $n$ if and only if he is at either $B$ or $E$ on day $n-1$, and in either case he could just as well go to $D$ instead of $C$. So there is no situation in which the Robber is certain to be caught at $C$ rather than at $D$. Additionally, the Robber\'s possible locations on day $n+1$ are the same whether he is at $C$ or $D$ on day $n$, so searching one rather than the other fails to rule out any locations for future days. So any successful strategy that involves searching $C$ should also involve searching $D$ on the same day, and similarly for $F$ and $G, I$ and $J$, and $L$ and $A$. On the other hand, if the Robber must be at one of $C$ and $D$ on day $n$, then he must be at either $B$ or $E$ on day $n+1$, because those are the only adjacent hideouts. So any strategy that involves searching both hideouts of one of the off-the-square pairs on day $n$ is equivalent to a strategy that searches the adjacent on-the-square hideouts on day $n+1$; the two strategies use the same number of Cops for the same number of workdays. Thus the optimal number of Cop workdays can be achieved using strategies that only search the ""corner"" hideouts $B, E, H, K$.\n\nRestricting the search to only those strategies searching corner hideouts $B, E, H, K$, a total of 8 workdays can be achieved by searching all four hideouts on two consecutive days: if the Robber is at one of the other eight hideouts the first day, he must move to one of the two adjacent corner hideouts the second day. But each of these corner hideouts is adjacent to two other corner hideouts. So if only one hideout is searched, for no matter how many consecutive days, the following day, the Robber could either be back at the previously-searched hideout or be at any other hideout: no possibilities are ruled out. If two adjacent corners are searched, the Cops do no better, as the following argument shows. Suppose that $B$ and $E$ are both searched for two consecutive days. Then the Cops can rule out $B, E, C$, and $D$ as possible locations, but if the Cops then switch to searching either $H$ or $K$ instead of $B$ or $E$, the Robber can go back to $C$ or $D$ within two days. So searching two adjacent corner hideouts for two days is fruitless and costs four Cop workdays. Searching diagonally opposite corner hideouts is even less fruitful, because doing so rules out none of the other hideouts as possible Robber locations. Using three Cops each day, it is easy to imagine scenarios in which the Robber evades capture for three days before being caught: for example, if $B, E, H$ are searched for two consecutive days, the Robber goes from $I$ or $J$ to $K$ to $L$. Therefore if three Cops are used, four days are required for a total of 12 Cop workdays.\n\nIf there are more than four Cops, the preceding arguments show that the number of Cops must be even to produce optimal results (because there is no advantage to searching one hideout between vertices of the square without searching the other). Using six Cops with four at corner hideouts yields no improvement, because the following day, the Robber could get to any of the four corner hideouts, requiring at least four Cops the second day, for ten Cop workdays. If two or fewer Cops are at corner hideouts, the situation is even worse, because if the Robber is not caught that day, he has at least nine possible hideouts the following day (depending on whether the unsearched corners are adjacent or diagonally opposite to each other). Using eight Cops (with four at corner vertices) could eliminate one corner vertex as a possible location for the second day (if the non-corner hideouts searched are on adjacent sides of the square), but eight Cop workdays have already been used on the first day. So 8 Cop workdays is minimal.']","['12, 8']",True,,Numerical, 2811,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Let $M$ be a map with $n \geq 3$ hideouts. Prove that $2 \leq W(M) \leq n$, and that these bounds cannot be improved. In other words, prove that for each $n \geq 3$, there exist maps $M_{1}$ and $M_{2}$ such that $W\left(M_{1}\right)=2$ and $W\left(M_{2}\right)=n$.","['A single Cop can only search one hideout in a day, so as long as $M$ has two or more hideouts, there is no strategy that guarantees that a lone Cop captures the Robber the first day. Then either more than one Cop will have to search on the first day, or a lone Cop will have to search for at least 2 days; in either of these cases, $W(M) \\geq 2$. On the other hand, $n$ Cops can always guarantee a capture simply by searching all $n$ hideouts on the first day, so $W(M) \\leq n$.\n\nTo show that the lower bound cannot be improved, consider the star on $n$ hideouts; that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout. (The map in 1a is $\\mathcal{S}_{7}$.) Then $W\\left(\\mathcal{S}_{n}\\right)=2$, because a single Cop can search the central hideout for 2 days; if the Robber is at one of the outer hideouts on the first day, he must go to the central hideout on the second day. For the upper bound, the complete map $\\mathcal{K}_{n}$ is an example of a map with $W(M)=n$. As argued above, the minimum number of Cops needed to guarantee a catch is $n-1$, and using $n-1$ Cops requires two days, for a total of $2 n-2$ Cop workdays. So the optimal strategy for $\\mathcal{K}_{n}$ (see 2c) is to use $n$ Cops for a single day.']",['证明题，略'],True,,Need_human_evaluate, 2811,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Let $M$ be a map with $n \geq 3$ hideouts. Prove that $2 \leq W(M) \leq n$, and that these bounds cannot be improved. In other words, prove that for each $n \geq 3$, there exist maps $M_{1}$ and $M_{2}$ such that $W\left(M_{1}\right)=2$ and $W\left(M_{2}\right)=n$.","['A single Cop can only search one hideout in a day, so as long as $M$ has two or more hideouts, there is no strategy that guarantees that a lone Cop captures the Robber the first day. Then either more than one Cop will have to search on the first day, or a lone Cop will have to search for at least 2 days; in either of these cases, $W(M) \\geq 2$. On the other hand, $n$ Cops can always guarantee a capture simply by searching all $n$ hideouts on the first day, so $W(M) \\leq n$.\n\nTo show that the lower bound cannot be improved, consider the star on $n$ hideouts; that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout. (The map in 1a is $\\mathcal{S}_{7}$.) Then $W\\left(\\mathcal{S}_{n}\\right)=2$, because a single Cop can search the central hideout for 2 days; if the Robber is at one of the outer hideouts on the first day, he must go to the central hideout on the second day. For the upper bound, the complete map $\\mathcal{K}_{n}$ is an example of a map with $W(M)=n$. As argued above, the minimum number of Cops needed to guarantee a catch is $n-1$, and using $n-1$ Cops requires two days, for a total of $2 n-2$ Cop workdays. So the optimal strategy for $\\mathcal{K}_{n}$ (see 2c) is to use $n$ Cops for a single day.']",,True,,, 2812,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Prove that if $M$ is bipartite, then $C(M) \leq n / 2$.","['Suppose $M$ is bipartite, and let $\\mathcal{A}$ and $\\mathcal{B}$ be the sets of hideouts referenced in the definition. Because $\\mathcal{A}$ and $\\mathcal{B}$ are disjoint, either $|\\mathcal{A}| \\leq n / 2$ or $|\\mathcal{B}| \\leq n / 2$ or both. Without loss of generality, suppose that $|\\mathcal{A}| \\leq n / 2$. Then position Cops at each hideout in $\\mathcal{A}$ for two days. If the Robber was initially on a hideout in $\\mathcal{B}$, he must move the following day, and because no hideout in $\\mathcal{B}$ is connected to any other hideout in $\\mathcal{B}$, his new hideout must be a hideout in $\\mathcal{A}$.']",['证明题，略'],True,,Need_human_evaluate, 2812,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Prove that if $M$ is bipartite, then $C(M) \leq n / 2$.","['Suppose $M$ is bipartite, and let $\\mathcal{A}$ and $\\mathcal{B}$ be the sets of hideouts referenced in the definition. Because $\\mathcal{A}$ and $\\mathcal{B}$ are disjoint, either $|\\mathcal{A}| \\leq n / 2$ or $|\\mathcal{B}| \\leq n / 2$ or both. Without loss of generality, suppose that $|\\mathcal{A}| \\leq n / 2$. Then position Cops at each hideout in $\\mathcal{A}$ for two days. If the Robber was initially on a hideout in $\\mathcal{B}$, he must move the following day, and because no hideout in $\\mathcal{B}$ is connected to any other hideout in $\\mathcal{B}$, his new hideout must be a hideout in $\\mathcal{A}$.']",,True,,, 2813,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Prove that $C(M) \leq n / 2$ for any map $M$ with the property that, for all hideouts $H_{1}$ and $H_{2}$, either all paths from $H_{1}$ to $H_{2}$ contain an odd number of edges, or all paths from $H_{1}$ to $H_{2}$ contain an even number of edges.","['The given condition actually implies that the graph is bipartite. Let $A_{1}$ be a hideout in $M$, and let $\\mathcal{A}$ be the set of all hideouts $V$ such that all paths from $A_{1}$ to $V$ have an even number of edges, as well as $A_{1}$ itself; let $\\mathcal{B}$ be the set of all other hideouts in $M$. Notice that there are no edges from $A_{1}$ to any other element $A_{i}$ in $\\mathcal{A}$, because if there were, that edge would create a path from $A_{1}$ to $A_{i}$ with an odd number of edges (namely 1). So $A_{1}$ has edges only to hideouts in $\\mathcal{B}$. In fact, there can be no edge from any hideout $A_{i}$ in $\\mathcal{A}$ to any other hideout $A_{j}$ in $\\mathcal{A}$, because if there were, there would be a path with an odd number of edges from $A_{1}$ to $A_{j}$ via $A_{i}$. Similarly, there can be no edge from any hideout $B_{i}$ in $\\mathcal{B}$ to any other hideout $B_{j}$ in $\\mathcal{B}$, because if there were such an edge, there would be a path from $A_{1}$ to $B_{j}$ with an even number of edges via $B_{i}$. Thus hideouts in $\\mathcal{B}$ are connected only to hideouts in $\\mathcal{A}$ and vice versa; hence $M$ is bipartite.']",['证明题，略'],True,,Need_human_evaluate, 2813,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Prove that $C(M) \leq n / 2$ for any map $M$ with the property that, for all hideouts $H_{1}$ and $H_{2}$, either all paths from $H_{1}$ to $H_{2}$ contain an odd number of edges, or all paths from $H_{1}$ to $H_{2}$ contain an even number of edges.","['The given condition actually implies that the graph is bipartite. Let $A_{1}$ be a hideout in $M$, and let $\\mathcal{A}$ be the set of all hideouts $V$ such that all paths from $A_{1}$ to $V$ have an even number of edges, as well as $A_{1}$ itself; let $\\mathcal{B}$ be the set of all other hideouts in $M$. Notice that there are no edges from $A_{1}$ to any other element $A_{i}$ in $\\mathcal{A}$, because if there were, that edge would create a path from $A_{1}$ to $A_{i}$ with an odd number of edges (namely 1). So $A_{1}$ has edges only to hideouts in $\\mathcal{B}$. In fact, there can be no edge from any hideout $A_{i}$ in $\\mathcal{A}$ to any other hideout $A_{j}$ in $\\mathcal{A}$, because if there were, there would be a path with an odd number of edges from $A_{1}$ to $A_{j}$ via $A_{i}$. Similarly, there can be no edge from any hideout $B_{i}$ in $\\mathcal{B}$ to any other hideout $B_{j}$ in $\\mathcal{B}$, because if there were such an edge, there would be a path from $A_{1}$ to $B_{j}$ with an even number of edges via $B_{i}$. Thus hideouts in $\\mathcal{B}$ are connected only to hideouts in $\\mathcal{A}$ and vice versa; hence $M$ is bipartite.']",,True,,, 2814,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. A map $M$ is called $k$-perfect if its hideout set $H$ can be partitioned into equal-sized subsets $\mathcal{A}_{1}, \mathcal{A}_{2}, \ldots, \mathcal{A}_{k}$ such that for any $j$, the hideouts of $\mathcal{A}_{j}$ are only adjacent to hideouts in $\mathcal{A}_{j+1}$ or $\mathcal{A}_{j-1}$. (The indices are taken modulo $k$ : the hideouts of $\mathcal{A}_{1}$ may be adjacent to the hideouts of $\mathcal{A}_{k}$.) Show that if $M$ is $k$-perfect, then $C(M) \leq \frac{2 n}{k}$.","['Because a Robber in $\\mathcal{A}_{i}$ can only move to a hideout in $\\mathcal{A}_{i-1}$ or $\\mathcal{A}_{i+1}$, this map is essentially the same as the cyclic map $\\mathcal{C}_{k}$. So the Cops should apply a similar strategy. First, position $n / k$ Cops at the hideouts of set $\\mathcal{A}_{1}$ and $n / k$ Cops at the hideouts of set $\\mathcal{A}_{2}$. On day 2 , leave the first $n / k$ Cops at $\\mathcal{A}_{1}$, but move the second group of Cops to $\\mathcal{A}_{3}$. Continue until the second set of Cops is at $\\mathcal{A}_{k}$; then ""wait"" one turn (search $\\mathcal{A}_{k}$ again), and then search backward $\\mathcal{A}_{k-1}$, $\\mathcal{A}_{k-2}$, etc.']",['证明题，略'],True,,Need_human_evaluate, 2814,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. A map $M$ is called $k$-perfect if its hideout set $H$ can be partitioned into equal-sized subsets $\mathcal{A}_{1}, \mathcal{A}_{2}, \ldots, \mathcal{A}_{k}$ such that for any $j$, the hideouts of $\mathcal{A}_{j}$ are only adjacent to hideouts in $\mathcal{A}_{j+1}$ or $\mathcal{A}_{j-1}$. (The indices are taken modulo $k$ : the hideouts of $\mathcal{A}_{1}$ may be adjacent to the hideouts of $\mathcal{A}_{k}$.) Show that if $M$ is $k$-perfect, then $C(M) \leq \frac{2 n}{k}$.","['Because a Robber in $\\mathcal{A}_{i}$ can only move to a hideout in $\\mathcal{A}_{i-1}$ or $\\mathcal{A}_{i+1}$, this map is essentially the same as the cyclic map $\\mathcal{C}_{k}$. So the Cops should apply a similar strategy. First, position $n / k$ Cops at the hideouts of set $\\mathcal{A}_{1}$ and $n / k$ Cops at the hideouts of set $\\mathcal{A}_{2}$. On day 2 , leave the first $n / k$ Cops at $\\mathcal{A}_{1}$, but move the second group of Cops to $\\mathcal{A}_{3}$. Continue until the second set of Cops is at $\\mathcal{A}_{k}$; then ""wait"" one turn (search $\\mathcal{A}_{k}$ again), and then search backward $\\mathcal{A}_{k-1}$, $\\mathcal{A}_{k-2}$, etc.']",,True,,, 2815,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Find an example of a map $M$ with 2012 hideouts such that $C(M)=17$ and $W(M)=34$, or prove that no such map exists.","['There are many examples. Perhaps the simplest to describe is the complete bipartite map on the hideouts $\\mathcal{A}=\\left\\{A_{1}, A_{2}, \\ldots, A_{17}\\right\\}$ and $\\mathcal{B}=\\left\\{B_{1}, B_{2}, \\ldots, B_{1995}\\right\\}$, which is often denoted $M=\\mathcal{K}_{17,1995}$. That is, let $M$ be the map with hideouts $\\mathcal{A} \\cup \\mathcal{B}$ such that $A_{i}$ is adjacent to $B_{j}$ for all $i$ and $j$, and that $A_{i}$ is not adjacent to $A_{j}$, nor is $B_{i}$ adjacent to $B_{j}$, for any $i$ and $j$.\n\nIf 17 Cops search the 17 hideouts in $\\mathcal{A}$ for two consecutive days, then they are guaranteed to catch the Robber. This shows that $C(M) \\leq 17$ and that $W(M) \\leq 34$. If fewer than 17 Cops search on a given day, then the Robber has at least one safe hideout: he has exactly one choice if the Cops search 16 of the hideouts $A_{i}$ and the Robber was hiding at some $B_{j}$ the previous day, 1995 choices if the Robber was hiding at some $A_{i}$ the previous day. Thus $C(M)=17$.\n\nOn the other hand, fewer than 34 Cop workdays cannot guarantee catching the Robber. Unless the Cops search every $A_{i}$ (or every $B_{j}$ ) on a given day, they gain no information about where the Robber is (unless he is unlucky enough to be caught).\n\nThe complete bipartite map is not the simplest in terms of the number of edges. Try to find a map with 2012 hideouts and as few edges as possible that has a Cop number of 17 and a workday number of 34 .']",,True,,, 2815,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$. Find an example of a map $M$ with 2012 hideouts such that $C(M)=17$ and $W(M)=34$, or prove that no such map exists.","['There are many examples. Perhaps the simplest to describe is the complete bipartite map on the hideouts $\\mathcal{A}=\\left\\{A_{1}, A_{2}, \\ldots, A_{17}\\right\\}$ and $\\mathcal{B}=\\left\\{B_{1}, B_{2}, \\ldots, B_{1995}\\right\\}$, which is often denoted $M=\\mathcal{K}_{17,1995}$. That is, let $M$ be the map with hideouts $\\mathcal{A} \\cup \\mathcal{B}$ such that $A_{i}$ is adjacent to $B_{j}$ for all $i$ and $j$, and that $A_{i}$ is not adjacent to $A_{j}$, nor is $B_{i}$ adjacent to $B_{j}$, for any $i$ and $j$.\n\nIf 17 Cops search the 17 hideouts in $\\mathcal{A}$ for two consecutive days, then they are guaranteed to catch the Robber. This shows that $C(M) \\leq 17$ and that $W(M) \\leq 34$. If fewer than 17 Cops search on a given day, then the Robber has at least one safe hideout: he has exactly one choice if the Cops search 16 of the hideouts $A_{i}$ and the Robber was hiding at some $B_{j}$ the previous day, 1995 choices if the Robber was hiding at some $A_{i}$ the previous day. Thus $C(M)=17$.\n\nOn the other hand, fewer than 34 Cop workdays cannot guarantee catching the Robber. Unless the Cops search every $A_{i}$ (or every $B_{j}$ ) on a given day, they gain no information about where the Robber is (unless he is unlucky enough to be caught).\n\nThe complete bipartite map is not the simplest in terms of the number of edges. Try to find a map with 2012 hideouts and as few edges as possible that has a Cop number of 17 and a workday number of 34 .']",['证明题，略'],True,,Need_human_evaluate, 2816,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. ![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836) Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. ![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912) Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Find an example of a map $M$ with 4 or more hideouts such that $W(M)=3$, or prove that no such map exists.","['We denote a map as star, that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout.\n\nNo such map exists. Let $M$ be a map with at least four hideouts. The proof below shows that either $W(M)>3$ or else $M$ is a star, in which case $W(M)=2$.\n\nFirst, suppose that there are four distinct hideouts $A_{1}, A_{2}, B_{1}$, and $B_{2}$ such that $A_{1}$ and $A_{2}$ are adjacent, as are $B_{1}$ and $B_{2}$. If the Cops make fewer than four searches, then they cannot search $\\left\\{A_{1}, A_{2}\\right\\}$ twice and also search $\\left\\{B_{1}, B_{2}\\right\\}$ twice. Without loss of generality, assume the Cops search $\\left\\{A_{1}, A_{2}\\right\\}$ at most once. Then the Robber can evade capture by moving from $A_{1}$ to $A_{2}$ and back again, provided that he is lucky enough to start off at the right one. Thus\n\n\n\n$W(M)>3$ in this case. For the second case, assume that whenever $A_{1}$ is adjacent to $A_{2}$ and $B_{1}$ is adjacent to $B_{2}$, the four hideouts are not distinct. Start with two adjacent hideouts, and call them $A_{1}$ and $A_{2}$. Consider any hideout, say $B$, that is not one of these two. Then $B$ must be adjacent to some hideout, say $C$. By assumption, $A_{1}, A_{2}, B$, and $C$ are not distinct, so $C=A_{1}$ or $C=A_{2}$. That is, every hideout $B$ is adjacent to $A_{1}$ or to $A_{2}$.\n\nIf there are hideouts $B_{1}$ and $B_{2}$, distinct from $A_{1}$ and $A_{2}$ and from each other, such that $B_{1}$ is adjacent to $A_{1}$ and $B_{2}$ is adjacent to $A_{2}$, then it clearly violates the assumption of this case. That is, either every hideout $B$, distinct from $A_{1}$ and $A_{2}$, is adjacent to $A_{1}$, or every such hideout is adjacent to $A_{2}$. Without loss of generality, assume the former.\n\nBecause every hideout is adjacent to $A_{1}$ (except for $A_{1}$ itself), the foregoing proves that the map $M$ is a star. It remains to show that there are no ""extra"" edges in the map. For the sake of contradiction, suppose that $B_{1}$ and $B_{2}$ are adjacent hideouts, both distinct from $A_{1}$. Because the map has at least four hideouts, choose one distinct from these three and call it $A_{2}$. Then these four hideouts violate the assumption of this case.']",['证明题，略'],True,,Need_human_evaluate, 2816,Combinatorics,,"This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3. Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 . The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit. Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$. The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught. The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$. Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays. Find an example of a map $M$ with 4 or more hideouts such that $W(M)=3$, or prove that no such map exists.","['We denote a map as star, that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout.\n\nNo such map exists. Let $M$ be a map with at least four hideouts. The proof below shows that either $W(M)>3$ or else $M$ is a star, in which case $W(M)=2$.\n\nFirst, suppose that there are four distinct hideouts $A_{1}, A_{2}, B_{1}$, and $B_{2}$ such that $A_{1}$ and $A_{2}$ are adjacent, as are $B_{1}$ and $B_{2}$. If the Cops make fewer than four searches, then they cannot search $\\left\\{A_{1}, A_{2}\\right\\}$ twice and also search $\\left\\{B_{1}, B_{2}\\right\\}$ twice. Without loss of generality, assume the Cops search $\\left\\{A_{1}, A_{2}\\right\\}$ at most once. Then the Robber can evade capture by moving from $A_{1}$ to $A_{2}$ and back again, provided that he is lucky enough to start off at the right one. Thus\n\n\n\n$W(M)>3$ in this case. For the second case, assume that whenever $A_{1}$ is adjacent to $A_{2}$ and $B_{1}$ is adjacent to $B_{2}$, the four hideouts are not distinct. Start with two adjacent hideouts, and call them $A_{1}$ and $A_{2}$. Consider any hideout, say $B$, that is not one of these two. Then $B$ must be adjacent to some hideout, say $C$. By assumption, $A_{1}, A_{2}, B$, and $C$ are not distinct, so $C=A_{1}$ or $C=A_{2}$. That is, every hideout $B$ is adjacent to $A_{1}$ or to $A_{2}$.\n\nIf there are hideouts $B_{1}$ and $B_{2}$, distinct from $A_{1}$ and $A_{2}$ and from each other, such that $B_{1}$ is adjacent to $A_{1}$ and $B_{2}$ is adjacent to $A_{2}$, then it clearly violates the assumption of this case. That is, either every hideout $B$, distinct from $A_{1}$ and $A_{2}$, is adjacent to $A_{1}$, or every such hideout is adjacent to $A_{2}$. Without loss of generality, assume the former.\n\nBecause every hideout is adjacent to $A_{1}$ (except for $A_{1}$ itself), the foregoing proves that the map $M$ is a star. It remains to show that there are no ""extra"" edges in the map. For the sake of contradiction, suppose that $B_{1}$ and $B_{2}$ are adjacent hideouts, both distinct from $A_{1}$. Because the map has at least four hideouts, choose one distinct from these three and call it $A_{2}$. Then these four hideouts violate the assumption of this case.']",,True,,, 2817,Geometry,,Triangle $A B C$ has $\mathrm{m} \angle A>\mathrm{m} \angle B>\mathrm{m} \angle C$. The angle between the altitude and the angle bisector at vertex $A$ is $6^{\circ}$. The angle between the altitude and the angle bisector at vertex $B$ is $18^{\circ}$. Compute the degree measure of angle $C$.,"['Let the feet of the altitudes from $A$ and $B$ be $E$ and $D$, respectively, and let $F$ and $G$ be the intersection points of the angle bisectors with $\\overline{A C}$ and $\\overline{B C}$, respectively, as shown below.\n\n\n\nThen $\\mathrm{m} \\angle G A E=6^{\\circ}$ and $\\mathrm{m} \\angle D B F=18^{\\circ}$. Suppose $\\mathrm{m} \\angle F B C=x^{\\circ}$ and $\\mathrm{m} \\angle C A G=y^{\\circ}$. So $\\mathrm{m} \\angle C A E=(y+6)^{\\circ}$ and $\\mathrm{m} \\angle C B D=(x+18)^{\\circ}$. Considering right triangle $B D C$, $\\mathrm{m} \\angle C=90^{\\circ}-(x+18)^{\\circ}=(72-x)^{\\circ}$, while considering right triangle $A E C, \\mathrm{~m} \\angle C=$ $90^{\\circ}-(y+6)^{\\circ}=(84-y)^{\\circ}$. Thus $84-y=72-x$ and $y-x=12$. Considering $\\triangle A B E$, $\\mathrm{m} \\angle E A B=(y-6)^{\\circ}$ and $\\mathrm{m} \\angle E B A=2 x^{\\circ}$, so $(y-6)+2 x=90$, or $2 x+y=96$. Solving the system yields $x=28, y=40$. Therefore $\\mathrm{m} \\angle A=80^{\\circ}$ and $\\mathrm{m} \\angle B=56^{\\circ}$, so $\\mathrm{m} \\angle C=44^{\\circ}$.', 'From right triangle $A B E, 90^{\\circ}=\\left(\\frac{1}{2} A-6^{\\circ}\\right)+B$, and from right triangle $A B D, 90^{\\circ}=\\left(\\frac{1}{2} B-18^{\\circ}\\right)+A$. Adding the two equations gives $180^{\\circ}=\\frac{3}{2}(A+B)-24^{\\circ}$, so $A+B=\\frac{2}{3} \\cdot 204^{\\circ}=136^{\\circ}$ and $C=180^{\\circ}-(A+B)=44^{\\circ}$.']",['$44^{\\circ}$'],False,,Numerical, 2818,Algebra,,"Compute the number of ordered pairs of integers $(b, c)$, with $-20 \leq b \leq 20,-20 \leq c \leq 20$, such that the equations $x^{2}+b x+c=0$ and $x^{2}+c x+b=0$ share at least one root.","['Let $r$ be the common root. Then $r^{2}+b r+c=r^{2}+c r+b \\Rightarrow b r-c r=b-c$. So either $b=c$ or $r=1$. In the latter case, $1+b+c=0$, so $c=-1-b$.\n\nThere are 41 ordered pairs where $b=c$. If $c=-1-b$ and $-20 \\leq b \\leq 20$, then $-21 \\leq c \\leq 19$. Therefore there are 40 ordered pairs $(b,-1-b)$ where both terms are in the required intervals. Thus there are $41+40=\\mathbf{8 1}$ solutions.']",['81'],False,,Numerical, 2819,Combinatorics,,"A seventeen-sided die has faces numbered 1 through 17, but it is not fair: 17 comes up with probability $1 / 2$, and each of the numbers 1 through 16 comes up with probability $1 / 32$. Compute the probability that the sum of two rolls is either 20 or 12.","['The rolls that add up to 20 are $17+3,16+4,15+5,14+6,13+7,12+8,11+9$, and $10+10$. Accounting for order, the probability of $17+3$ is $\\frac{1}{2} \\cdot \\frac{1}{32}+\\frac{1}{32} \\cdot \\frac{1}{2}=2 \\cdot \\frac{1}{2} \\cdot \\frac{1}{32}=\\frac{32}{1024}$. The combination $10+10$ has probability $\\frac{1}{32} \\cdot \\frac{1}{32}=\\frac{1}{1024}$; the other six combinations have probability $2 \\cdot \\frac{1}{32} \\cdot \\frac{1}{32}=\\frac{2}{1024}$, for a total of $\\frac{32+1+6 \\cdot 2}{1024}=\\frac{45}{1024}$ (again, accounting for two possible orders per combination). The rolls that add up to 12 are $1+11,2+10,3+9,4+8,5+7,6+6$, all\n\n\n\nof which have probability $2 \\cdot \\frac{1}{32} \\cdot \\frac{1}{32}=\\frac{2}{1024}$ except the last, which has probability $\\left(\\frac{1}{32}\\right)^{2}$, for a total of $\\frac{11}{1024}$. Thus the probability of either sum appearing is $\\frac{45}{1024}+\\frac{11}{1024}=\\frac{56}{1024}=\\frac{\\mathbf{7}}{\\mathbf{1 2 8}}$.']",['$\\frac{7}{128}$'],False,,Numerical, 2820,Geometry,,"In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs. ![](https://cdn.mathpix.com/cropped/2023_12_21_78e54ff35f7d827e777fg-1.jpg?height=368&width=374&top_left_y=602&top_left_x=924)","['Because $A B=16, A C=B C=\\frac{16}{\\sqrt{2}}=8 \\sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \\sqrt{2}-8$. Each large arc has measure $45^{\\circ}$ and the small arc has measure $90^{\\circ}$. Therefore the area enclosed by each large arc is $\\frac{45}{360} \\cdot \\pi \\cdot 8^{2}=8 \\pi$, and the area enclosed by the small arc is $\\frac{90}{360} \\cdot \\pi \\cdot(8 \\sqrt{2}-8)^{2}=48 \\pi-32 \\pi \\sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \\pi-32 \\pi \\sqrt{2}$. On the other hand, the area of the triangle is $\\frac{1}{2}(8 \\sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \\pi+32 \\pi \\sqrt{2}$.']",['$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$'],False,,Numerical, 2820,Geometry,,"In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs. ","['Because $A B=16, A C=B C=\\frac{16}{\\sqrt{2}}=8 \\sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \\sqrt{2}-8$. Each large arc has measure $45^{\\circ}$ and the small arc has measure $90^{\\circ}$. Therefore the area enclosed by each large arc is $\\frac{45}{360} \\cdot \\pi \\cdot 8^{2}=8 \\pi$, and the area enclosed by the small arc is $\\frac{90}{360} \\cdot \\pi \\cdot(8 \\sqrt{2}-8)^{2}=48 \\pi-32 \\pi \\sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \\pi-32 \\pi \\sqrt{2}$. On the other hand, the area of the triangle is $\\frac{1}{2}(8 \\sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \\pi+32 \\pi \\sqrt{2}$.']",['$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$'],False,,Numerical, 2821,Number Theory,,"Compute the number of ordered pairs of integers $(a, b)$ such that $10$. Because both graphs are symmetric about the $y$-axis, the other two points of intersection are $A=(-a-h, b+h)$ and $B=(-a, b)$, and $a>0$.\n\nIn terms of these coordinates, the distances are $A B=C D=\\sqrt{2} h$ and $B C=2 a$. Thus the condition $A B=B C=C D$ holds if and only if $\\sqrt{2} h=2 a$, or $h=\\sqrt{2} a$.\n\nThe foregoing uses the condition that $C$ and $D$ lie on a line of slope 1 , so now use the remaining equation and subtract:\n\n$$\n\\begin{aligned}\nb & =a^{2}-a-12 \\\\\nb+h & =(a+h)^{2}-(a+h)-12 \\\\\nh & =2 a h+h^{2}-h\n\\end{aligned}\n$$\n\nBecause the points are distinct, $h \\neq 0$. Dividing by $h$ yields $2-2 a=h=\\sqrt{2} a$. Thus $a=\\frac{2}{2+\\sqrt{2}}=2-\\sqrt{2}$.\n\nFinally, because $C$ lies on the two graphs, $b=a^{2}-a-12=-8-3 \\sqrt{2}$ and $k=a-b=$ $10+2 \\sqrt{2}$.']",['$10+2 \\sqrt{2}$'],False,,Numerical, 2825,Combinatorics,,"The zeros of $f(x)=x^{6}+2 x^{5}+3 x^{4}+5 x^{3}+8 x^{2}+13 x+21$ are distinct complex numbers. Compute the average value of $A+B C+D E F$ over all possible permutations $(A, B, C, D, E, F)$ of these six numbers.","['There are $6 !=720$ permutations of the zeros, so the average value is the sum, $S$, divided by 720. Setting any particular zero as $A$ leaves $5 !=120$ ways to permute the other five zeros, so over the 720 permutations, each zero occupies the $A$ position 120 times. Similarly, fixing any ordered pair $(B, C)$ of zeros allows $4 !=24$ permutations of the other four zeros, and $B C=C B$ means that each value of $B C$ occurs 48 times. Finally, fixing any ordered triple $(D, E, F)$ allows $3 !=6$ permutations of the other variables, and there are $3 !=6$ equivalent arrangements within each product $D E F$, so that the product of any three zeros occurs 36 times within the sum. Let $S_{1}=A+B+C+D+E+F$ (i.e., the sum of the zeros taken singly), $S_{2}=A B+A C+\\cdots+A F+B C+\\cdots+E F$ (i.e., the sum of the zeros taken two at a time), and $S_{3}=A B C+A B D+\\cdots+D E F$ be the sum of the zeros three at a time. Then $S=120 S_{1}+48 S_{2}+36 S_{3}$. Using the sums and products of roots formulas, $S_{1}=-2 / 1=-2$, $S_{2}=3 / 1=3$, and $S_{3}=-5 / 1=-5$. Thus $S=120(-2)+48(3)+36(-5)=-276$. The average value is thus $-\\frac{276}{720}=-\\frac{\\mathbf{2 3}}{\\mathbf{6 0}}$.']",['$-\\frac{23}{60}$'],False,,Numerical, 2826,Geometry,,"Given noncollinear points $A, B, C$, segment $\overline{A B}$ is trisected by points $D$ and $E$, and $F$ is the midpoint of segment $\overline{A C} . \overline{D F}$ and $\overline{B F}$ intersect $\overline{C E}$ at $G$ and $H$, respectively. If $[D E G]=18$, compute $[F G H]$. ![](https://cdn.mathpix.com/cropped/2023_12_21_2af37857c7a17f2e18a0g-1.jpg?height=461&width=870&top_left_y=447&top_left_x=671)","[""Compute the desired area as $[E G F B]-[E H B]$. To compute the area of concave quadrilateral $E G F B$, draw segment $\\overline{B G}$, which divides the quadrilateral into three triangles, $\\triangle D E G, \\triangle B D G$, and $\\triangle B G F$. Then $[B D G]=[D E G]=18$ because the triangles have equal bases and heights. Because $D, G$, and $F$ are collinear, to compute $[B G F]$ it suffices to find the ratio $D G / G F$. Use Menelaus's Theorem on $\\triangle A D F$ with Menelaus Line $\\overline{E C}$ to obtain\n\n$$\n\\frac{A E}{E D} \\cdot \\frac{D G}{G F} \\cdot \\frac{F C}{C A}=1\n$$\n\n\n\nBecause $E$ and $F$ are the midpoints of $\\overline{A D}$ and $\\overline{C A}$ respectively, $A E / E D=1$ and $F C / C A=$ $1 / 2$. Therefore $D G / G F=2 / 1$, and $[B G F]=\\frac{1}{2}[B D G]=9$. Thus $[E G F B]=18+18+9=45$.\n\nTo compute $[E H B]$, consider that its base $\\overline{E B}$ is twice the base of $\\triangle D E G$. The ratio of their heights equals the ratio $E H / E G$ because the altitudes from $H$ and $G$ to $\\overleftrightarrow{B E}$ are parallel to each other. Use Menelaus's Theorem twice more on $\\triangle A E C$ to find these values:\n\n$$\n\\begin{gathered}\n\\frac{A D}{D E} \\cdot \\frac{E G}{G C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E G}{G C}=\\frac{1}{2} \\Rightarrow E G=\\frac{1}{3} E C, \\text { and } \\\\\n\\frac{A B}{B E} \\cdot \\frac{E H}{H C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E H}{H C}=\\frac{2}{3} \\Rightarrow E H=\\frac{2}{5} E C .\n\\end{gathered}\n$$\n\nTherefore $\\frac{E H}{E G}=\\frac{2 / 5}{1 / 3}=\\frac{6}{5}$. Thus $[E H B]=\\frac{6}{5} \\cdot 2 \\cdot[D E G]=\\frac{216}{5}$. Thus $[F G H]=45-\\frac{216}{5}=\\frac{9}{5}$."", ""The method of mass points leads to the same results as Menelaus's Theorem, but corresponds to the physical intuition that masses on opposite sides of a fulcrum balance if and only if the products of the masses and their distances from the fulcrum are equal (in physics-speak, the net torque is zero). If a mass of weight 1 is placed at vertex $B$ and masses of weight 2 are placed at vertices $A$ and $C$, then $\\triangle A B C$ balances on the line $\\overleftrightarrow{B F}$ and also on the line $\\overleftrightarrow{C E}$. Thus it balances on the point $H$ where these two lines intersect. Replacing the masses at $A$ and $C$ with a single mass of weight 4 at their center of mass $F$, the triangle still balances at $H$. Thus $B H / H F=4$.\n\nNext, consider $\\triangle B E F$. Placing masses of weight 1 at the vertices $B$ and $E$ and a mass of weight 4 at $F$, the triangle balances at $G$. A similar argument shows that $D G / G F=2$ and that $E G / G H=5$. Because $\\triangle D E G$ and $\\triangle F H G$ have congruent (vertical) angles at $G$, it follows that $[D E G] /[F H G]=(D G / F G) \\cdot(E G / H G)=2 \\cdot 5=10$. Thus $[F G H]=[D E G] / 10=$ $\\frac{18}{10}=\\frac{9}{5}$.""]",['$\\frac{9}{5}$'],False,,Numerical, 2826,Geometry,,"Given noncollinear points $A, B, C$, segment $\overline{A B}$ is trisected by points $D$ and $E$, and $F$ is the midpoint of segment $\overline{A C} . \overline{D F}$ and $\overline{B F}$ intersect $\overline{C E}$ at $G$ and $H$, respectively. If $[D E G]=18$, compute $[F G H]$. ","[""Compute the desired area as $[E G F B]-[E H B]$. To compute the area of concave quadrilateral $E G F B$, draw segment $\\overline{B G}$, which divides the quadrilateral into three triangles, $\\triangle D E G, \\triangle B D G$, and $\\triangle B G F$. Then $[B D G]=[D E G]=18$ because the triangles have equal bases and heights. Because $D, G$, and $F$ are collinear, to compute $[B G F]$ it suffices to find the ratio $D G / G F$. Use Menelaus's Theorem on $\\triangle A D F$ with Menelaus Line $\\overline{E C}$ to obtain\n\n$$\n\\frac{A E}{E D} \\cdot \\frac{D G}{G F} \\cdot \\frac{F C}{C A}=1\n$$\n\n\n\nBecause $E$ and $F$ are the midpoints of $\\overline{A D}$ and $\\overline{C A}$ respectively, $A E / E D=1$ and $F C / C A=$ $1 / 2$. Therefore $D G / G F=2 / 1$, and $[B G F]=\\frac{1}{2}[B D G]=9$. Thus $[E G F B]=18+18+9=45$.\n\nTo compute $[E H B]$, consider that its base $\\overline{E B}$ is twice the base of $\\triangle D E G$. The ratio of their heights equals the ratio $E H / E G$ because the altitudes from $H$ and $G$ to $\\overleftrightarrow{B E}$ are parallel to each other. Use Menelaus's Theorem twice more on $\\triangle A E C$ to find these values:\n\n$$\n\\begin{gathered}\n\\frac{A D}{D E} \\cdot \\frac{E G}{G C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E G}{G C}=\\frac{1}{2} \\Rightarrow E G=\\frac{1}{3} E C, \\text { and } \\\\\n\\frac{A B}{B E} \\cdot \\frac{E H}{H C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E H}{H C}=\\frac{2}{3} \\Rightarrow E H=\\frac{2}{5} E C .\n\\end{gathered}\n$$\n\nTherefore $\\frac{E H}{E G}=\\frac{2 / 5}{1 / 3}=\\frac{6}{5}$. Thus $[E H B]=\\frac{6}{5} \\cdot 2 \\cdot[D E G]=\\frac{216}{5}$. Thus $[F G H]=45-\\frac{216}{5}=\\frac{9}{5}$."", ""The method of mass points leads to the same results as Menelaus's Theorem, but corresponds to the physical intuition that masses on opposite sides of a fulcrum balance if and only if the products of the masses and their distances from the fulcrum are equal (in physics-speak, the net torque is zero). If a mass of weight 1 is placed at vertex $B$ and masses of weight 2 are placed at vertices $A$ and $C$, then $\\triangle A B C$ balances on the line $\\overleftrightarrow{B F}$ and also on the line $\\overleftrightarrow{C E}$. Thus it balances on the point $H$ where these two lines intersect. Replacing the masses at $A$ and $C$ with a single mass of weight 4 at their center of mass $F$, the triangle still balances at $H$. Thus $B H / H F=4$.\n\nNext, consider $\\triangle B E F$. Placing masses of weight 1 at the vertices $B$ and $E$ and a mass of weight 4 at $F$, the triangle balances at $G$. A similar argument shows that $D G / G F=2$ and that $E G / G H=5$. Because $\\triangle D E G$ and $\\triangle F H G$ have congruent (vertical) angles at $G$, it follows that $[D E G] /[F H G]=(D G / F G) \\cdot(E G / H G)=2 \\cdot 5=10$. Thus $[F G H]=[D E G] / 10=$ $\\frac{18}{10}=\\frac{9}{5}$.""]",['$\\frac{9}{5}$'],False,,Numerical, 2827,Algebra,,Let $N=\left\lfloor(3+\sqrt{5})^{34}\right\rfloor$. Compute the remainder when $N$ is divided by 100 .,"['Let $\\alpha=3+\\sqrt{5}$ and $\\beta=3-\\sqrt{5}$, so that $N=\\left\\lfloor\\alpha^{34}\\right\\rfloor$, and let $M=\\alpha^{34}+\\beta^{34}$. When the binomials in $M$ are expanded, terms in which $\\sqrt{5}$ is raised to an odd power have opposite signs, and so cancel each other out. Therefore $M$ is an integer. Because $0<\\beta<1,0<\\beta^{34}<1$, and so $M-1<\\alpha^{34}\\mathrm{m} \\angle A=20^{\\circ}$, so $A B>B C$. The third possibility for $\\triangle P A B$ is that $P A=P B$, implying that the perpendicular bisector of $\\overline{A B}$ intersects $\\odot B$, which only occurs if $B C / A B \\geq 1 / 2$ (although if $B C / A B=1 / 2$, the triangle is degenerate). But $B C / A B=2 \\cos 80^{\\circ}$, and the given approximation $\\cos 80^{\\circ} \\approx 0.17$ implies that $B C / A B \\approx 0.34$. Hence the perpendicular bisector of $\\overline{A B}$ does not intersect $\\odot B$. Thus the assumption $P B=B C$ yields only one additional location for $P, P_{5}$. Similarly, $P C=B C$ yields exactly one more location, $P_{6}$, for a total of $\\mathbf{6}$ points. All six points, and their associated triangles, are pictured below.\n\n\n\n']",['6'],False,,Numerical, 2829,Algebra,,"If $\lceil u\rceil$ denotes the least integer greater than or equal to $u$, and $\lfloor u\rfloor$ denotes the greatest integer less than or equal to $u$, compute the largest solution $x$ to the equation $$ \left\lfloor\frac{x}{3}\right\rfloor+\lceil 3 x\rceil=\sqrt{11} \cdot x $$","['Let $f(x)=\\left\\lfloor\\frac{x}{3}\\right\\rfloor+\\lceil 3 x\\rceil$. Observe that $f(x+3)=f(x)+1+9=f(x)+10$. Let $g(x)=f(x)-\\frac{10}{3} x$. Then $g$ is periodic, because $g(x+3)=f(x)+10-\\frac{10 x}{3}-\\frac{10 \\cdot 3}{3}=g(x)$. The graph of $g$ is shown below:\n\n\n\nBecause $g(x)$ is the (vertical) distance between the graph of $y=f(x)$ and the line $y=\\frac{10}{3} x$, the fact that $g$ is periodic implies that $f$ always stays within some fixed distance $D$ of the line $y=\\frac{10}{3} x$. On the other hand, because $\\frac{10}{3}>\\sqrt{11}$, the graph of $y=\\frac{10}{3} x$ gets further and further away from the graph of $y=\\sqrt{11} x$ as $x$ increases. Because the graph of $y=f(x)$ remains near $y=\\frac{10}{3} x$, the graph of $y=f(x)$ drifts upward from the line $y=\\sqrt{11} x$.\n\nFor each integer $n$, define the open interval $I_{n}=\\left(\\frac{n-1}{3}, \\frac{n}{3}\\right)$. In fact, $f$ is constant on $I_{n}$, as the following argument shows. For $x \\in I_{n}, \\frac{n}{9}-\\frac{1}{9}<\\frac{x}{3}<\\frac{n}{9}$. Because $n$ is an integer, there are no integers between $\\frac{n}{9}-\\frac{1}{9}$ and $\\frac{n}{9}$, so $\\left\\lfloor\\frac{x}{3}\\right\\rfloor$ is constant; similarly, $\\lceil 3 x\\rceil$ is constant on the same intervals. Let $l_{n}$ be the value of $f$ on the interval $I_{n}$, and let $L_{n}=f\\left(\\frac{n}{3}\\right)$, the value at the right end of the interval $I_{n}$. If $n$ is not a multiple of 9 , then $l_{n}=L_{n}$, because as $x$ increases from $n-\\varepsilon$ to $n$, the floor function does not increase. This means that $f$ is actually constant on the half-closed interval $\\left(\\frac{n-1}{3}, \\frac{n}{3}\\right]$. If neither $n$ nor $n+1$ are multiples of 9 , then $l_{n+1}=l_{n}+1$. However if $n$ is a multiple of 9 , then $L_{n}=l_{n}+1$ and $l_{n+1}=L_{n}+1$. (The value of $f(x)$ increases when $x$ increases from $n-\\varepsilon$ to $n$, as well as going from $n$ to $n+\\varepsilon$.)\n\nHence on each interval of the form $(3 n-3,3 n)$, the graph of $f$ looks like 9 steps of height 1 and width $\\frac{1}{3}$, all open on the left and closed on the right except for the last step, which is open on both ends. Between the intervals $(3 n-3,3 n)$ and $(3 n, 3 n+3), f(x)$ increases by 2 , with $f(3 n)$ halfway between steps. This graph is shown below:\n\n\n\n\n\nOn each interval $(3 n-3,3 n)$, the average rate of change is $3<\\sqrt{11}$, so the steps move down relative $y=\\sqrt{11} x$ within each interval. At the end of each interval, the graph of $f$ rises relative to $y=\\sqrt{11} x$. Thus the last intersection point between $f(x)$ and $\\sqrt{11} x$ will be on the ninth step of one of these intervals. Suppose this intersection point lies in the interval $(3 k-3,3 k)$. The ninth step is of height $10 k-1$. Set $x=3 k-r$, where $r<\\frac{1}{3}$. Then the solution is the largest $k$ for which\n\n$$\n\\begin{aligned}\n10 k-1 & =\\sqrt{11}(3 k-r) \\quad\\left(01$, so the equation cannot hold for large values of $n$. To make this explicit, write\n\n$$\n\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor=\\frac{n}{3 \\sqrt{11}}-r \\quad \\text { and } \\quad\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil=\\frac{3 n}{\\sqrt{11}}+s\n$$\n\nwhere $r$ and $s$ are real numbers between 0 and 1. (If $n \\neq 0$, then $r$ and $s$ are strictly between 0 and 1.) Then\n\n$$\n\\begin{aligned}\n1>r-s & =\\left(\\frac{n}{3 \\sqrt{11}}-\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor\\right)-\\left(\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil-\\frac{3 n}{\\sqrt{11}}\\right) \\\\\n& =\\left(\\frac{n}{3 \\sqrt{11}}+\\frac{3 n}{\\sqrt{11}}\\right)-\\left(\\left\\lfloor\\frac{n}{3 \\sqrt{11}}\\right\\rfloor+\\left\\lceil\\frac{3 n}{\\sqrt{11}}\\right\\rceil\\right) \\\\\n& =n\\left(\\frac{1}{3 \\sqrt{11}}+\\frac{3}{\\sqrt{11}}-1\\right),\n\\end{aligned}\n$$\n\nso $n<1 /\\left(\\frac{1}{3 \\sqrt{11}}+\\frac{3}{\\sqrt{11}}-1\\right)=99+30 \\sqrt{11}=198.45 \\ldots$\n\nUse trial and error with $n=198,197,196, \\ldots$, to find the value of $n$ that works. Computing the first row of the following table to three decimal digits, and computing both $\\frac{1}{3 \\sqrt{11}}$ and $\\frac{3}{\\sqrt{11}}$ to the same degree of accuracy, allows one to calculate the remaining rows with acceptable round-off errors.\n\n| $n$ | $n /(3 \\sqrt{11})$ | $3 n / \\sqrt{11}$ |\n| :---: | :---: | :---: |\n| | | |\n| 198 | 19.900 | 179.098 |\n| 197 | 19.799 | 178.193 |\n| 196 | 19.699 | 177.289 |\n| 195 | 19.598 | 176.384 |\n| 194 | 19.498 | 175.480 |\n| 193 | 19.397 | 174.575 |\n| 192 | 19.297 | 173.671 |\n| 191 | 19.196 | 172.766 |\n| 190 | 19.096 | 171.861 |\n| 189 | 18.995 | 170.957 |\n\nBecause $n=189=18+171$, the final answer is $x=\\frac{\\mathbf{1 8 9} \\sqrt{\\mathbf{1 1}}}{\\mathbf{1 1}}$.']",['$\\frac{189 \\sqrt{11}}{11}$'],False,,Numerical, 2830,Algebra,,"If $x, y$, and $z$ are positive integers such that $x y=20$ and $y z=12$, compute the smallest possible value of $x+z$.","['Note that $x$ and $z$ can each be minimized by making $y$ as large as possible, so set $y=$ $\\operatorname{lcm}(12,20)=4$. Then $x=5, z=3$, and $x+z=\\mathbf{8}$.']",['8'],False,,Numerical, 2831,Geometry,,"Let $T=8$. Let $A=(1,5)$ and $B=(T-1,17)$. Compute the value of $x$ such that $(x, 3)$ lies on the perpendicular bisector of $\overline{A B}$.","['The midpoint of $\\overline{A B}$ is $\\left(\\frac{T}{2}, 11\\right)$, and the slope of $\\overleftrightarrow{A B}$ is $\\frac{12}{T-2}$. Thus the perpendicular bisector of $\\overline{A B}$ has slope $\\frac{2-T}{12}$ and passes through the point $\\left(\\frac{T}{2}, 11\\right)$. Thus the equation of the perpendicular bisector of $\\overline{A B}$ is $y=\\left(\\frac{2-T}{12}\\right) x+\\left(11-\\frac{2 T-T^{2}}{24}\\right)$. Plugging $y=3$ into this equation and solving for $x$ yields $x=\\frac{96}{T-2}+\\frac{T}{2}$. With $T=8$, it follows that $x=\\frac{96}{6}+\\frac{8}{2}=16+4=\\mathbf{2 0}$.']",['20'],False,,Numerical, 2832,Number Theory,,Let T be a rational number. Let $N$ be the smallest positive $T$-digit number that is divisible by 33 . Compute the product of the last two digits of $N$.,"['The sum of the digits of $N$ must be a multiple of 3 , and the alternating sum of the digits must be a multiple of 11 . Because the number of digits of $N$ is fixed, the minimum $N$ will have the alternating sum of its digits equal to 0 , and therefore the sum of the digits of $N$ will be even, so it must be 6 . Thus if $T$ is even, then $N=1 \\underbrace{0 \\ldots .02}_{T-30^{\\prime} \\mathrm{s}}$, and if $T$ is odd, then $N=1 \\underbrace{0 \\ldots 0}_{T-30^{\\prime} \\mathrm{s}} 32$. Either way, the product of the last two digits of $N$ is 6 (independent of $T$ ).']",['6'],False,,Numerical, 2833,Geometry,,"Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$. ","['Note that $D E S H$ is a trapezoid with height $\\frac{T}{2}$. Because $\\overline{A S}$ and $\\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\\frac{T}{6}$. Thus $S H=T-2 \\cdot \\frac{T}{6}=\\frac{2 T}{3}$. Thus $[D E S H]=\\frac{1}{2}\\left(T+\\frac{2 T}{3}\\right) \\cdot \\frac{T}{2}=\\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 .']",['15'],False,,Numerical, 2833,Geometry,,"Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$. ![](https://cdn.mathpix.com/cropped/2023_12_21_da8b9ad9706d1f3f1497g-1.jpg?height=372&width=390&top_left_y=909&top_left_x=1540)","['Note that $D E S H$ is a trapezoid with height $\\frac{T}{2}$. Because $\\overline{A S}$ and $\\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\\frac{T}{6}$. Thus $S H=T-2 \\cdot \\frac{T}{6}=\\frac{2 T}{3}$. Thus $[D E S H]=\\frac{1}{2}\\left(T+\\frac{2 T}{3}\\right) \\cdot \\frac{T}{2}=\\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 .']",['15'],False,,Numerical, 2834,Algebra,,"Let $T=15$. For complex $z$, define the function $f_{1}(z)=z$, and for $n>1, f_{n}(z)=$ $f_{n-1}(\bar{z})$. If $f_{1}(z)+2 f_{2}(z)+3 f_{3}(z)+4 f_{4}(z)+5 f_{5}(z)=T+T i$, compute $|z|$.","['Because $\\overline{\\bar{z}}=z$, it follows that $f_{n}(z)=z$ when $n$ is odd, and $f_{n}(z)=\\bar{z}$ when $n$ is even. Taking $z=a+b i$, where $a$ and $b$ are real, it follows that $\\sum_{k=1}^{5} k f_{k}(z)=15 a+3 b i$. Thus $a=\\frac{T}{15}, b=\\frac{T}{3}$, and $|z|=\\sqrt{a^{2}+b^{2}}=\\frac{|T| \\sqrt{26}}{15}$. With $T=15$, the answer is $\\sqrt{\\mathbf{2 6}}$.']",['$\\sqrt{26}$'],False,,Numerical, 2835,Number Theory,,"Let $T=\sqrt{26}$. Compute the number of ordered pairs of positive integers $(a, b)$ with the property that $a b=T^{20} \cdot 210^{12}$, and the greatest common divisor of $a$ and $b$ is 1 .","[""If the prime factorization of $a b$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \\ldots p_{k}^{e_{k}}$, where the $p_{i}$ 's are distinct primes and the $e_{i}$ 's are positive integers, then in order for $\\operatorname{gcd}(a, b)$ to equal 1 , each $p_{i}$ must be a divisor of exactly one of $a$ or $b$. Thus the desired number of ordered pairs is $2^{k}$ because there are 2 choices for each prime divisor (i.e., $p_{i} \\mid a$ or $p_{i} \\mid b$ ). With $T=\\sqrt{26}$, it follows that $(\\sqrt{26})^{20} \\cdot 210^{12}=\\left(2^{10} \\cdot 13^{10}\\right) \\cdot 210^{12}=2^{22} \\cdot 3^{12} \\cdot 5^{12} \\cdot 7^{12} \\cdot 13^{10}$. Thus there are five distinct prime divisors, and the answer is $2^{5}=\\mathbf{3 2}$.""]",['32'],False,,Numerical, 2836,Algebra,,"Let $T=32$. Given that $\sin \theta=\frac{\sqrt{T^{2}-64}}{T}$, compute the largest possible value of the infinite series $\cos \theta+\cos ^{2} \theta+\cos ^{3} \theta+\ldots$.","['Using $\\sin ^{2} \\theta+\\cos ^{2} \\theta=1$ gives $\\cos ^{2} \\theta=\\frac{64}{T^{2}}$, so to maximize the sum, take $\\cos \\theta=\\frac{8}{|T|}$. Using the formula for the sum of an infinite geometric series gives $\\frac{8 /|T|}{1-8 /|T|}=\\frac{8}{|T|-8}$. With $T=32$, the answer is $\\frac{8}{24}=\\frac{1}{3}$.']",['$\\frac{1}{3}$'],False,,Numerical, 2837,Geometry,,"Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$. ![](https://cdn.mathpix.com/cropped/2023_12_21_da8b9ad9706d1f3f1497g-1.jpg?height=304&width=531&top_left_y=1943&top_left_x=1472)","['Note that $\\overline{D Y}$ and $\\overline{K P}$ are both perpendicular to line $\\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\\sqrt{D K^{2}-D J^{2}}=$ $\\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \\sqrt{R r}$. With $R=450$ and $r=\\frac{1}{3}$, the answer is $2 \\sqrt{150}=\\mathbf{1 0} \\sqrt{\\mathbf{6}}$.']",['$10 \\sqrt{6}$'],False,,Numerical, 2837,Geometry,,"Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$. ","['Note that $\\overline{D Y}$ and $\\overline{K P}$ are both perpendicular to line $\\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\\sqrt{D K^{2}-D J^{2}}=$ $\\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \\sqrt{R r}$. With $R=450$ and $r=\\frac{1}{3}$, the answer is $2 \\sqrt{150}=\\mathbf{1 0} \\sqrt{\\mathbf{6}}$.']",['$10 \\sqrt{6}$'],False,,Numerical, 2838,Geometry,,"Let $T=\frac{9}{17}$. When $T$ is expressed as a reduced fraction, let $m$ and $n$ be the numerator and denominator, respectively. A square pyramid has base $A B C D$, the distance from vertex $P$ to the base is $n-m$, and $P A=P B=P C=P D=n$. Compute the area of square $A B C D$.","[""By the Pythagorean Theorem, half the diagonal of the square is $\\sqrt{n^{2}-(n-m)^{2}}=\\sqrt{2 m n-m^{2}}$. Thus the diagonal of the square is $2 \\sqrt{2 m n-m^{2}}$, and the square's area is $4 m n-2 m^{2}$. With $T=\\frac{9}{17}, m=9, n=17$, and the answer is 450 .""]",['450'],False,,Numerical, 2839,Combinatorics,,"Let $T=-14$, and let $d=|T|$. A person whose birthday falls between July 23 and August 22 inclusive is called a Leo. A person born in July is randomly selected, and it is given that her birthday is before the $d^{\text {th }}$ day of July. Another person who was also born in July is randomly selected, and it is given that his birthday is after the $d^{\text {th }}$ day of July. Compute the probability that exactly one of these people is a Leo.","['Note that there are 9 days in July in which a person could be a Leo (July 23-31). Let the woman (born before the $d^{\\text {th }}$ day of July) be called Carol, and let the man (born after the $d^{\\text {th }}$ day of July) be called John, and consider the possible values of $d$. If $d \\leq 21$, then Carol will not be a Leo, and the probability that John is a Leo is $\\frac{9}{31-d}$. If $d=22$ or 23 , then the probability is 1 . If $d \\geq 24$, then John will be a Leo, and Carol will not be a Leo with probability $1-\\frac{d-23}{d-1}$. With $T=-14$, the first case applies, and the desired probability is $\\frac{\\mathbf{9}}{\\mathbf{1 7}}$.']",['$\\frac{9}{17}$'],False,,Numerical, 2840,Algebra,,"Let $T=-10$. Given that $\log _{2} 4^{8 !}+\log _{4} 2^{8 !}=6 ! \cdot T \cdot x$, compute $x$.","['Note that $4^{8 !}=2^{2 \\cdot 8 !}$, thus $\\log _{2} 4^{8 !}=2 \\cdot 8$ !. Similarly, $\\log _{4} 2^{8 !}=\\frac{8 !}{2}$. Thus $2 \\cdot 8 !+\\frac{8 !}{2}=$ $6 !\\left(2 \\cdot 7 \\cdot 8+7 \\cdot \\frac{8}{2}\\right)=6 ! \\cdot 140$. Thus $140=T x$, and with $T=-10, x=\\mathbf{- 1 4}$.']",['-14'],False,,Numerical, 2841,Algebra,,"Let $T=20$. For some real constants $a$ and $b$, the solution sets of the equations $x^{2}+(5 b-T-a) x=T+1$ and $2 x^{2}+(T+8 a-2) x=-10 b$ are the same. Compute $a$.","['Divide each side of the second equation by 2 and equate coefficients to obtain $5 b-T-a=$ $\\frac{T}{2}+4 a-1$ and $T+1=-5 b$. Thus $b=\\frac{T+1}{-5}$, and plugging this value into the first equation yields $a=-\\frac{T}{2}$. With $T=20$, the answer is $\\mathbf{- 1 0}$.']",['-10'],False,,Numerical, 2842,Algebra,,"Let T be a rational number, and let $K=T-2$. If $K$ workers can produce 9 widgets in 1 hour, compute the number of workers needed to produce $\frac{720}{K}$ widgets in 4 hours.","['Because $T$ workers produce 9 widgets in 1 hour, 1 worker will produce $\\frac{9}{T}$ widgets in 1 hour. Thus 1 worker will produce $\\frac{36}{T}$ widgets in 4 hours. In order to produce $\\frac{720}{T}$ widgets in 4 hours, it will require $\\frac{720 / T}{36 / T}=\\mathbf{2 0}$ workers (independent of $T$ ).']",['20'],False,,Numerical, 2843,Algebra,,"Let $T=2018$, and append the digits of $T$ to $\underline{A} \underline{A} \underline{B}$ (for example, if $T=17$, then the result would be $\underline{1} \underline{\underline{A}} \underline{A} \underline{B}$ ). If the resulting number is divisible by 11 , compute the largest possible value of $A+B$.","[""Let $R$ be the remainder when $T$ is divided by 11 . Note that the alternating sum of the digits of the number must be divisible by 11 . This sum will be congruent $\\bmod 11$ to $B-A+A-R=$ $B-R$, thus $B=R$. Because $A$ 's value is irrelevant, to maximize $A+B$, set $A=9$ to yield $A+B=9+R$. For $T=2018, R=5$, and the answer is $9+5=\\mathbf{1 4}$.""]",['14'],False,,Numerical, 2844,Algebra,,"Given that April $1^{\text {st }}, 2012$ fell on a Sunday, what is the next year in which April $1^{\text {st }}$ will fall on a Sunday?","['Note that $365=7 \\cdot 52+1$. Thus over the next few years after 2012 , the day of the week for April $1^{\\text {st }}$ will advance by one day in a non-leap year, and it will advance by two days in a leap year. Thus in six years, the day of the week will have rotated a complete cycle, and the answer is 2018 .']",['2018'],False,,Numerical, 2845,Number Theory,,"Let $p$ be a prime number. If $p$ years ago, the ages of three children formed a geometric sequence with a sum of $p$ and a common ratio of 2 , compute the sum of the children's current ages.","[""Let $x, 2 x$, and $4 x$ be the ages of the children $p$ years ago. Then $x+2 x+4 x=p$, so $7 x=p$. Since $p$ is prime, $x=1$. Thus the sum of the children's current ages is $(1+7)+(2+7)+(4+7)=\\mathbf{2 8}$.""]",['28'],False,,Numerical, 2846,Number Theory,,"Define a reverse prime to be a positive integer $N$ such that when the digits of $N$ are read in reverse order, the resulting number is a prime. For example, the numbers 5, 16, and 110 are all reverse primes. Compute the largest two-digit integer $N$ such that the numbers $N, 4 \cdot N$, and $5 \cdot N$ are all reverse primes.","['Because $N<100,5 \\cdot N<500$. Since no primes end in 4, it follows that $5 \\cdot N<400$, hence $N \\leq 79$. The reverses of $5 \\cdot 79=395,4 \\cdot 79=316$, and 79 are 593,613 , and 97 , respectively. All three of these numbers are prime, thus 79 is the largest two-digit integer $N$ for which $N$, $4 \\cdot N$, and $5 \\cdot N$ are all reverse primes.']",['79'],False,,Numerical, 2847,Combinatorics,,"Some students in a gym class are wearing blue jerseys, and the rest are wearing red jerseys. There are exactly 25 ways to pick a team of three players that includes at least one player wearing each color. Compute the number of students in the class.","['Let $r$ and $b$ be the number of students wearing red and blue jerseys, respectively. Then either we choose two blues and one red or one blue and two reds. Thus\n\n$$\n\\begin{aligned}\n& \\left(\\begin{array}{l}\nb \\\\\n2\n\\end{array}\\right)\\left(\\begin{array}{l}\nr \\\\\n1\n\\end{array}\\right)+\\left(\\begin{array}{l}\nb \\\\\n1\n\\end{array}\\right)\\left(\\begin{array}{l}\nr \\\\\n2\n\\end{array}\\right)=25 \\\\\n\\Rightarrow & \\frac{r b(b-1)}{2}+\\frac{b r(r-1)}{2}=25 \\\\\n\\Rightarrow & r b((r-1)+(b-1))=50 \\\\\n\\Rightarrow & r b(r+b-2)=50 .\n\\end{aligned}\n$$\n\nNow because $r, b$, and $r+b-2$ are positive integer divisors of 50 , and $r, b \\geq 2$, we have only a few possibilities to check. If $r=2$, then $b^{2}=25$, so $b=5$; the case $r=5$ is symmetric. If $r=10$, then $b(b+8)=5$, which is impossible. If $r=25$, then $b(b+23)=2$, which is also impossible. So $\\{r, b\\}=\\{2,5\\}$, and $r+b=7$.']",['7'],False,,Numerical, 2848,Geometry,,"Point $P$ is on the hypotenuse $\overline{E N}$ of right triangle $B E N$ such that $\overline{B P}$ bisects $\angle E B N$. Perpendiculars $\overline{P R}$ and $\overline{P S}$ are drawn to sides $\overline{B E}$ and $\overline{B N}$, respectively. If $E N=221$ and $P R=60$, compute $\frac{1}{B E}+\frac{1}{B N}$.","['We observe that $\\frac{1}{B E}+\\frac{1}{B N}=\\frac{B E+B N}{B E \\cdot B N}$. The product in the denominator suggests that we compare areas. Let $[B E N]$ denote the area of $\\triangle B E N$. Then $[B E N]=\\frac{1}{2} B E \\cdot B N$, but because $P R=P S=60$, we can also write $[B E N]=[B E P]+[B N P]=\\frac{1}{2} \\cdot 60 \\cdot B E+\\frac{1}{2} \\cdot 60 \\cdot B N$. Therefore $B E \\cdot B N=60(B E+B N)$, so $\\frac{1}{B E}+\\frac{1}{B N}=\\frac{B E+B N}{B E \\cdot B N}=\\frac{1}{\\mathbf{6 0}}$. Note that this value does not depend on the length of the hypotenuse $\\overline{E N}$; for a given location of point $P, \\frac{1}{B E}+\\frac{1}{B N}$ is invariant.', 'Using similar triangles, we have $\\frac{E R}{P R}=\\frac{P S}{S N}=\\frac{B E}{B N}$, so $\\frac{B E-60}{60}=$ $\\frac{60}{B N-60}=\\frac{B E}{B N}$ and $B E^{2}+B N^{2}=221^{2}$. Using algebra, we find that $B E=204, B N=85$, and $\\frac{1}{204}+\\frac{1}{85}=\\frac{1}{60}$.']",['$\\frac{1}{60}$'],False,,Numerical, 2849,Algebra,,$\quad$ Compute all real values of $x$ such that $\log _{2}\left(\log _{2} x\right)=\log _{4}\left(\log _{4} x\right)$.,"['If $y=\\log _{a}\\left(\\log _{a} x\\right)$, then $a^{a^{y}}=x$. Let $y=\\log _{2}\\left(\\log _{2} x\\right)=\\log _{4}\\left(\\log _{4} x\\right)$. Then $2^{2^{y}}=4^{4^{y}}=$ $\\left(2^{2}\\right)^{\\left(2^{2}\\right)^{y}}=2^{2^{2 y+1}}$, so $2 y+1=y, y=-1$, and $x=\\sqrt{\\mathbf{2}}$. (This problem is based on one submitted by ARML alum James Albrecht, 1986-2007.)', 'Raise 4 (or $2^{2}$ ) to the power of both sides to get $\\left(\\log _{2} x\\right)^{2}=\\log _{4} x$. By the change of base formula, $\\frac{(\\log x)^{2}}{(\\log 2)^{2}}=\\frac{\\log x}{2 \\log 2}$, so $\\log x=\\frac{\\log 2}{2}$, thus $x=2^{1 / 2}=\\sqrt{\\mathbf{2}}$.', 'Let $x=4^{a}$. The equation then becomes $\\log _{2}(2 a)=\\log _{4} a$. Raising 4 to the power of each side, we get $4 a^{2}=a$. Since $a \\neq 0$, we get $4 a=1$, thus $a=\\frac{1}{4}$ and $x=\\sqrt{2}$.']",['$\\sqrt{2}$'],False,,Numerical, 2850,Number Theory,,"Let $k$ be the least common multiple of the numbers in the set $\mathcal{S}=\{1,2, \ldots, 30\}$. Determine the number of positive integer divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\mathcal{S}$.","['We know that $k=2^{4} \\cdot 3^{3} \\cdot 5^{2} \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\cdot 23 \\cdot 29$. It is not difficult to see that the set $\\mathcal{T}_{1}=\\left\\{\\frac{k}{2}, \\frac{k}{3}, \\frac{k}{5}, \\frac{k}{17}, \\frac{k}{19}, \\frac{k}{23}, \\frac{k}{29}\\right\\}$ comprises all divisors of $k$ that are divisible by exactly 29 of the numbers in the set $\\mathcal{S}$. Let $\\mathcal{P}=\\{2,3,5,17,19,23,29\\}$. Then\n\n$$\n\\mathcal{T}_{2}=\\left\\{\\frac{k}{p_{1} p_{2}}, \\text { where } p_{1} \\text { and } p_{2} \\text { are distinct elements of } \\mathcal{P}\\right\\}\n$$\n\nconsists of divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\\mathcal{S}$. There are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)=21$ elements in $\\mathcal{T}_{2}$.\n\nFurthermore, note that $\\frac{k}{7}$ is only divisible by 26 of the numbers in $\\mathcal{S}$ (since it is not divisible by $7,14,21$, or 28 ) while $\\frac{k}{11}$ and $\\frac{k}{13}$ are each divisible by 28 of the numbers in $\\mathcal{S}$. We can also rule out $\\frac{k}{4}$ (27 divisors: all but 8,16 , and 24 ), $\\frac{k}{9}$ (27 divisors), $\\frac{k}{25}$ (24 divisors), and all other numbers, thus the answer is $21+2=\\mathbf{2 3}$.']",['23'],False,,Numerical, 2851,Number Theory,,"Let $A$ and $B$ be digits from the set $\{0,1,2, \ldots, 9\}$. Let $r$ be the two-digit integer $\underline{A} \underline{B}$ and let $s$ be the two-digit integer $\underline{B} \underline{A}$, so that $r$ and $s$ are members of the set $\{00,01, \ldots, 99\}$. Compute the number of ordered pairs $(A, B)$ such that $|r-s|=k^{2}$ for some integer $k$.","['Because $|(10 A+B)-(10 B+A)|=9|A-B|=k^{2}$, it follows that $|A-B|$ is a perfect square. $|A-B|=0$ yields 10 pairs of integers: $(A, B)=(0,0),(1,1), \\ldots,(9,9)$.\n\n$|A-B|=1$ yields 18 pairs: the nine $(A, B)=(0,1),(1,2), \\ldots,(8,9)$, and their reverses.\n\n$|A-B|=4$ yields 12 pairs: the six $(A, B)=(0,4),(1,5), \\ldots,(5,9)$, and their reverses.\n\n$|A-B|=9$ yields 2 pairs: $(A, B)=(0,9)$ and its reverse.\n\nThus the total number of possible ordered pairs $(A, B)$ is $10+18+12+2=\\mathbf{4 2}$.']",['42'],False,,Numerical, 2852,Combinatorics,,"For $k \geq 3$, we define an ordered $k$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{k}\right)$ to be special if, for every $i$ such that $1 \leq i \leq k$, the product $x_{1} \cdot x_{2} \cdot \ldots \cdot x_{k}=x_{i}^{2}$. Compute the smallest value of $k$ such that there are at least 2009 distinct special $k$-tuples.","[""The given conditions imply $k$ equations. By taking the product of these $k$ equations, we have $\\left(x_{1} x_{2} \\ldots x_{k}\\right)^{k-1}=x_{1} x_{2} \\ldots x_{k}$. Thus it follows that either $x_{1} x_{2} \\ldots x_{k}=0$ or $x_{1} x_{2} \\ldots x_{k}= \\pm 1$. If $x_{1} x_{2} \\ldots x_{k}=0$, then some $x_{j}=0$, and by plugging this into each of the equations, it follows that all of the $x_{i}$ 's are equal to 0 . Note that we cannot have $x_{1} x_{2} \\ldots x_{k}=-1$, because the left hand side equals $x_{1}\\left(x_{2} \\ldots x_{k}\\right)=x_{1}^{2}$, which can't be negative, because the $x_{i}$ 's are all given as real. Thus $x_{1} x_{2} \\ldots x_{k}=1$, and it follows that each $x_{i}$ is equal to either 1 or -1 . Because the product of the $x_{i}$ 's is 1 , there must be an even number of -1 's. Furthermore, by picking any even number of the $x_{i}$ 's to be -1 , it can be readily verified that the ordered $k$-tuple $\\left(x_{1}, x_{2}, \\ldots, x_{k}\\right)$ is special. Thus there are\n\n$$\n\\left(\\begin{array}{c}\nk \\\\\n0\n\\end{array}\\right)+\\left(\\begin{array}{l}\nk \\\\\n2\n\\end{array}\\right)+\\left(\\begin{array}{l}\nk \\\\\n4\n\\end{array}\\right)+\\ldots+\\left(\\begin{array}{c}\nk \\\\\n2\\lfloor k / 2\\rfloor\n\\end{array}\\right)\n$$\n\nspecial non-zero $k$-tuples. By considering the binomial expansion of $(1+1)^{k}+(1-1)^{k}$, it is clear that the above sum of binomial coefficients equals $2^{k-1}$. Thus there are a total of\n\n\n\n$2^{k-1}+1$ special $k$-tuples. Because $2^{10}=1024$ and $2^{11}=2048$, the inequality $2^{k-1}+1 \\geq 2009$ is first satisfied when $k=\\mathbf{1 2}$."", 'Use a recursive approach. Let $S_{k}$ denote the number of special non-zero $k$-tuples. From the analysis in the above solution, each $x_{i}$ must be either 1 or -1 . It can easily be verified that $S_{3}=4$. For $k>3$, suppose that $x_{k}=1$ for a given special $k$-tuple. Then the $k$ equations that follow are precisely the equation $x_{1} x_{2} \\ldots x_{k-1}=1$ and the $k-1$ equations that follow for the special $(k-1)$-tuple $\\left(x_{1}, x_{2}, \\ldots, x_{k-1}\\right)$. Because $x_{1} x_{2} \\ldots x_{k-1}=1$ is consistent for a special $(k-1)$-tuple, and because this equation imposes no further restrictions, we conclude that there are $S_{k-1}$ special $k$-tuples in which $x_{k}=1$.\n\nIf, on the other hand, $x_{k}=-1$ for a given special $k$-tuple, then consider the $k$ equations that result, and make the substitution $x_{1}=-y_{1}$. Then the $k$ resulting equations are precisely the same as the $k$ equations obtained in the case where $x_{k}=1$, except that $x_{1}$ is replaced by $y_{1}$. Thus $\\left(x_{1}, x_{2}, \\ldots, x_{k-1},-1\\right)$ is special if and only if $\\left(y_{1}, x_{2}, \\ldots, x_{k-1}\\right)$ is special, and thus there are $S_{k-1}$ special $k$-tuples in which $x_{k}=-1$.\n\nThus the recursion becomes $S_{k}=2 S_{k-1}$, and because $S_{3}=4$, it follows that $S_{k}=2^{k-1}$.']",['12'],False,,Numerical, 2853,Geometry,,A cylinder with radius $r$ and height $h$ has volume 1 and total surface area 12. Compute $\frac{1}{r}+\frac{1}{h}$.,"['Since $\\pi r^{2} h=1$, we have $h=\\frac{1}{\\pi r^{2}}$ and $\\pi r^{2}=\\frac{1}{h}$. Consequently,\n\n$$\n2 \\pi r h+2 \\pi r^{2}=12 \\Rightarrow(2 \\pi r)\\left(\\frac{1}{\\pi r^{2}}\\right)+2\\left(\\frac{1}{h}\\right)=12 \\Rightarrow \\frac{2}{r}+\\frac{2}{h}=12 \\Rightarrow \\frac{1}{r}+\\frac{1}{h}=\\mathbf{6}\n$$', 'The total surface area is $2 \\pi r h+2 \\pi r^{2}=12$ and the volume is $\\pi r^{2} h=1$. Dividing, we obtain $\\frac{12}{1}=\\frac{2 \\pi r h+2 \\pi r^{2}}{\\pi r^{2} h}=\\frac{2}{r}+\\frac{2}{h}$, thus $\\frac{1}{r}+\\frac{1}{h}=\\frac{12}{2}=\\mathbf{6}$.']",['6'],False,,Numerical, 2854,Algebra,,"If $6 \tan ^{-1} x+4 \tan ^{-1}(3 x)=\pi$, compute $x^{2}$.","[""$\\quad$ Let $z=1+x i$ and $w=1+3 x i$, where $i=\\sqrt{-1}$. Then $\\tan ^{-1} x=\\arg z$ and $\\tan ^{-1}(3 x)=\\arg w$, where $\\arg z$ gives the measure of the angle in standard position whose terminal side passes through $z$. By DeMoivre's theorem, $6 \\tan ^{-1} x=\\arg \\left(z^{6}\\right)$ and $4 \\tan ^{-1}(3 x)=\\arg \\left(w^{6}\\right)$. Therefore the equation $6 \\tan ^{-1} x+4 \\tan ^{-1}(3 x)=\\pi$ is equivalent to $z^{6} \\cdot w^{4}=a$, where $a$ is a real number (and, in fact, $a<0$ ). To simplify somewhat, we can take the square root of both sides, and get $z^{3} \\cdot w^{2}=0+b i$, where $b$ is a real number. Then $(1+x i)^{3}(1+3 x i)^{2}=$ $0+b i$. Expanding each binomial and collecting real and imaginary terms in each factor yields $\\left(\\left(1-3 x^{2}\\right)+\\left(3 x-x^{3}\\right) i\\right)\\left(\\left(1-9 x^{2}\\right)+6 x i\\right)=0+b i$. In order that the real part of the product be 0 , we have $\\left(1-3 x^{2}\\right)\\left(1-9 x^{2}\\right)-\\left(3 x-x^{3}\\right)(6 x)=0$. This equation simplifies to $1-30 x^{2}+33 x^{4}=0$, yielding $x^{2}=\\frac{15 \\pm 8 \\sqrt{3}}{33}$. Notice that $\\frac{15 \\pm 8 \\sqrt{3}}{33} \\approx 1$, which would mean that $x \\approx 1$, and $\\operatorname{so} \\tan ^{-1}(x) \\approx \\frac{\\pi}{4}$, which is too large, since $6 \\cdot \\frac{\\pi}{4}>\\pi$. (It can be verified that this value for $x$ yields a value of $3 \\pi$ for the left side of the equation.) Therefore we are left with $x^{2}=\\frac{15-8 \\sqrt{3}}{\\mathbf{3 3}}$. To verify that this answer is reasonable, consider that $\\sqrt{3} \\approx 1.73$, so that $15-8 \\sqrt{3} \\approx 1.16$, and so $x^{2} \\approx \\frac{7}{200}=0.035$. Then $x$ itself is a little less than 0.2 , and so\n\n\n\n$\\tan ^{-1} x \\approx \\frac{\\pi}{15}$. Similarly, $3 x$ is about 0.6 , so $\\tan ^{-1}(3 x)$ is about $\\frac{\\pi}{6} \\cdot 6 \\cdot \\frac{\\pi}{15}+4 \\cdot \\frac{\\pi}{6}$ is reasonably close to $\\pi$."", 'Recall that $\\tan (a+b)=\\frac{\\tan a+\\tan b}{1-\\tan a \\tan b}$, thus $\\tan (2 a)=\\frac{2 \\tan a}{1-\\tan ^{2} a}$ and\n\n$$\n\\tan (3 a)=\\tan (2 a+a)=\\frac{\\frac{2 \\tan a}{1-\\tan ^{2} a}+\\tan a}{1-\\frac{2 \\tan a}{1-\\tan ^{2} a} \\cdot \\tan a}=\\frac{2 \\tan a+\\tan a-\\tan ^{3} a}{1-\\tan ^{2} a-2 \\tan ^{2} a}=\\frac{3 \\tan a-\\tan ^{3} a}{1-3 \\tan ^{2} a}\n$$\n\nBack to the problem at hand, divide both sides by 2 to obtain $3 \\tan ^{-1} x+2 \\tan ^{-1}(3 x)=\\frac{\\pi}{2}$. Taking the tangent of the left side yields $\\frac{\\tan \\left(3 \\tan ^{-1} x\\right)+\\tan \\left(2 \\tan ^{-1}(3 x)\\right)}{1-\\tan \\left(3 \\tan ^{-1} x\\right) \\tan \\left(2 \\tan ^{-1}(3 x)\\right)}$. We know that the denominator must be 0 since $\\tan \\frac{\\pi}{2}$ is undefined, thus $1=\\tan \\left(3 \\tan ^{-1} x\\right) \\tan \\left(2 \\tan ^{-1}(3 x)\\right)=$ $\\frac{3 x-x^{3}}{1-3 x^{2}} \\cdot \\frac{2 \\cdot 3 x}{1-(3 x)^{2}}$ and hence $\\left(1-3 x^{2}\\right)\\left(1-9 x^{2}\\right)=\\left(3 x-x^{3}\\right)(6 x)$. Simplifying yields $33 x^{4}-$ $30 x^{2}+1=0$, and applying the quadratic formula gives $x^{2}=\\frac{15 \\pm 8 \\sqrt{3}}{33}$. The "" + "" solution is extraneous: as noted in the previous solution, $x=\\frac{15+8 \\sqrt{3}}{33}$ yields a value of $3 \\pi$ for the left side of the equation), so we are left with $x^{2}=\\frac{\\mathbf{1 5}-\\mathbf{8} \\sqrt{\\mathbf{3}}}{\\mathbf{3 3}}$.']",['$\\frac{15-8 \\sqrt{3}}{33}$'],False,,Numerical, 2855,Geometry,,A rectangular box has dimensions $8 \times 10 \times 12$. Compute the fraction of the box's volume that is not within 1 unit of any of the box's faces.,"['Let the box be defined by the product of the intervals on the $x, y$, and $z$ axes as $[0,8] \\times$ $[0,10] \\times[0,12]$ with volume $8 \\times 10 \\times 12$. The set of points inside the box that are not within 1 unit of any face is defined by the product of the intervals $[1,7] \\times[1,9] \\times[1,11]$ with volume $6 \\times 8 \\times 10$. This volume is $\\frac{6 \\times 8 \\times 10}{8 \\times 10 \\times 12}=\\frac{1}{2}$ of the whole box.']",['$\\frac{1}{2}$'],False,,Numerical, 2856,Algebra,,Let $T=T N Y W R$. Compute the largest real solution $x$ to $(\log x)^{2}-\log \sqrt{x}=T$.,"['Let $u=\\log x$. Then the given equation can be rewritten as $u^{2}-\\frac{1}{2} u-T=0 \\rightarrow 2 u^{2}-u-2 T=0$. This quadratic has solutions $u=\\frac{1 \\pm \\sqrt{1+16 T}}{4}$. As we are looking for the largest real solution for $x$ (and therefore, for $u$ ), we want $u=\\frac{1+\\sqrt{1+16 T}}{4}=1$ when $T=\\frac{1}{2}$. Therefore, $x=10^{1}=\\mathbf{1 0}$.']",['10'],False,,Numerical, 2857,Combinatorics,,Let $T=T N Y W R$. Kay has $T+1$ different colors of fingernail polish. Compute the number of ways that Kay can paint the five fingernails on her left hand by using at least three colors and such that no two consecutive fingernails have the same color.,"['There are $T+1$ possible colors for the first nail. Each remaining nail may be any color except that of the preceding nail, that is, there are $T$ possible colors. Thus, using at least two colors, there are $(T+1) T^{4}$ possible colorings. The problem requires that at least three colors be used, so we must subtract the number of colorings that use only two colors. As before, there are $T+1$ possible colors for the first nail and $T$ colors for the second. With only two colors, there are no remaining choices; the colors simply alternate. The answer is therefore $(T+1) T^{4}-(T+1) T$, and with $T=10$, this expression is equal to $110000-110=\\mathbf{1 0 9 8 9 0}$.']",['109890'],False,,Numerical, 2858,Number Theory,,"Compute the number of ordered pairs $(x, y)$ of positive integers satisfying $x^{2}-8 x+y^{2}+4 y=5$.","['Completing the square twice in $x$ and $y$, we obtain the equivalent equation $(x-4)^{2}+(y+2)^{2}=$ 25 , which describes a circle centered at $(4,-2)$ with radius 5 . The lattice points on this circle are points 5 units up, down, left, or right of the center, or points 3 units away on one axis and 4 units away on the other. Because the center is below the $x$-axis, we know that $y$ must increase by at least 2 units; $x$ cannot decrease by 4 or more units if it is to remain positive. Thus, we have:\n\n$$\n\\begin{aligned}\n& (x, y)=(4,-2)+(-3,4)=(1,2) \\\\\n& (x, y)=(4,-2)+(0,5)=(4,3) \\\\\n& (x, y)=(4,-2)+(3,4)=(7,2) \\\\\n& (x, y)=(4,-2)+(4,3)=(8,1) .\n\\end{aligned}\n$$\n\nThere are $\\mathbf{4}$ such ordered pairs.']",['4'],False,,Numerical, 2859,Number Theory,,Let $T=T N Y W R$ and let $k=21+2 T$. Compute the largest integer $n$ such that $2 n^{2}-k n+77$ is a positive prime number.,"['If $k$ is positive, there are only four possible factorizations of $2 n^{2}-k n+77$ over the integers, namely\n\n$$\n\\begin{aligned}\n& (2 n-77)(n-1)=2 n^{2}-79 n+77 \\\\\n& (2 n-1)(n-77)=2 n^{2}-145 n+77 \\\\\n& (2 n-11)(n-7)=2 n^{2}-25 n+77 \\\\\n& (2 n-7)(n-11)=2 n^{2}-29 n+77\n\\end{aligned}\n$$\n\n\n\nBecause $T=4, k=29$, and so the last factorization is the correct one. Because $2 n-7$ and $n-11$ are both integers, in order for their product to be prime, one factor must equal 1 or -1 , so $n=3,4,10$, or 12 . Checking these possibilities from the greatest downward, $n=12$ produces $17 \\cdot 1=17$, which is prime. So the answer is $\\mathbf{1 2}$.']",['12'],False,,Numerical, 2860,Geometry,,"Let $T=T N Y W R$. In triangle $A B C, B C=T$ and $\mathrm{m} \angle B=30^{\circ}$. Compute the number of integer values of $A C$ for which there are two possible values for side length $A B$.","['By the Law of Cosines, $(A C)^{2}=T^{2}+(A B)^{2}-2 T(A B) \\cos 30^{\\circ} \\rightarrow(A B)^{2}-2 T \\cos 30^{\\circ}(A B)+$ $\\left(T^{2}-(A C)^{2}\\right)=0$. This quadratic in $A B$ has two positive solutions when the discriminant and product of the roots are both positive. Thus $\\left(2 T \\cos 30^{\\circ}\\right)^{2}-4\\left(T^{2}-(A C)^{2}\\right)>0$, and $\\left(T^{2}-(A C)^{2}\\right)>0$. The second inequality implies that $A C0$, so $T / 2a_{3}$. This is impossible, so the 3 -signature $(123,321)$ is impossible.']",,True,,, 2867,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. List three other impossible 3-signatures that have exactly two windows.","['There are 18 impossible 3-signatures with two windows. In nine of these, the first window indicates that $a_{2}a_{3}$ (a decrease). In the other nine, the end of the first window indicates a decrease (that is, $\\left.a_{2}>a_{3}\\right)$, but the beginning of the second window indicates an increase $\\left(a_{2}a_{3}$ (a decrease). In the other nine, the end of the first window indicates a decrease (that is, $\\left.a_{2}>a_{3}\\right)$, but the beginning of the second window indicates an increase $\\left(a_{2} A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Consider the shape below: Find the 2-signature that corresponds to this shape.","['The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$.']","['$(12,21,21,21,12)$']",False,,Tuple, 2868,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Consider the shape below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=212&width=350&top_left_y=878&top_left_x=931) Find the 2-signature that corresponds to this shape.","['The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$.']","['$(12,21,21,21,12)$']",False,,Tuple, 2869,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) List all 5-labels with 2-signature $(12,12,21,21)$.","['12543,13542,14532,23541,24531,34521']","['12543,13542,14532,23541,24531,34521']",True,,Numerical, 2869,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: List all 5-labels with 2-signature $(12,12,21,21)$.","['12543,13542,14532,23541,24531,34521']","['12543,13542,14532,23541,24531,34521']",True,,Numerical, 2870,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Find a formula for the number of $(2 n+1)$-labels with the 2 -signature $$ (\underbrace{12,12, \ldots, 12}_{n}, \underbrace{21,21, \ldots, 21}_{n}) $$","['The answer is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. The shape of this signature is a wedge: $n$ up steps followed by $n$ down steps. The wedge for $n=3$ is illustrated below:\n\n\n\nThe largest number in the label, $2 n+1$, must be placed at the peak in the center. If we choose the numbers to put in the first $n$ spaces, then they must be placed in increasing order. Likewise, the remaining $n$ numbers must be placed in decreasing order on the downward sloping piece of the shape. Thus there are exactly $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ such labels.\n\n']",['$\\binom{2n}{n}$'],False,,Expression, 2870,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Find a formula for the number of $(2 n+1)$-labels with the 2 -signature $$ (\underbrace{12,12, \ldots, 12}_{n}, \underbrace{21,21, \ldots, 21}_{n}) $$","['The answer is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. The shape of this signature is a wedge: $n$ up steps followed by $n$ down steps. The wedge for $n=3$ is illustrated below:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_844f8caecf3f9cf70b8eg-1.jpg?height=455&width=862&top_left_y=781&top_left_x=726)\n\nThe largest number in the label, $2 n+1$, must be placed at the peak in the center. If we choose the numbers to put in the first $n$ spaces, then they must be placed in increasing order. Likewise, the remaining $n$ numbers must be placed in decreasing order on the downward sloping piece of the shape. Thus there are exactly $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ such labels.']",['$\\binom{2n}{n}$'],False,,Need_human_evaluate, 2871,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Compute the number of 5-labels with 2 -signature $(12,21,12,21)$.","[""The answer is 16 . We have a shape with two peaks and a valley in the middle. The 5 must go on one of the two peaks, so we place it on the first peak. By the shape's symmetry, we will double our answer at the end to account for the 5 -labels where the 5 is on the other peak.\n\n\n\nThe 4 can go to the left of the 5 or at the other peak. In the first case, shown below left, the 3 must go at the other peak and the 1 and 2 can go in either order. In the latter case, shown below right, the 1,2 , and 3 can go in any of 3 ! arrangements.\n\n\n\n\nSo there are $2 !+3 !=8$ possibilities. In all, there are 165 -labels (including the ones where the 5 is at the other peak).""]",['16'],False,,Numerical, 2871,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Compute the number of 5-labels with 2 -signature $(12,21,12,21)$.","[""The answer is 16 . We have a shape with two peaks and a valley in the middle. The 5 must go on one of the two peaks, so we place it on the first peak. By the shape's symmetry, we will double our answer at the end to account for the 5 -labels where the 5 is on the other peak.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_844f8caecf3f9cf70b8eg-1.jpg?height=215&width=591&top_left_y=1741&top_left_x=859)\n\nThe 4 can go to the left of the 5 or at the other peak. In the first case, shown below left, the 3 must go at the other peak and the 1 and 2 can go in either order. In the latter case, shown below right, the 1,2 , and 3 can go in any of 3 ! arrangements.\n![](https://cdn.mathpix.com/cropped/2023_12_21_844f8caecf3f9cf70b8eg-1.jpg?height=216&width=1298&top_left_y=2166&top_left_x=518)\n\n\n\nSo there are $2 !+3 !=8$ possibilities. In all, there are 165 -labels (including the ones where the 5 is at the other peak).""]",['16'],False,,Numerical, 2872,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Determine the number of 9-labels with 2-signature $$ (12,21,12,21,12,21,12,21) \text {. } $$ Justify your answer.","['The answer is 7936. The shape of this 2-signature has four peaks and three intermediate valleys:\n\n\n\nWe will solve this problem by building up from smaller examples. Let $f_{n}$ equal the number of $(2 n+1)$-labels whose 2 -signature consists of $n$ peaks and $n-1$ intermediate valleys. In part (b) we showed that $f_{2}=16$. In the case where we have one peak, $f_{1}=2$. For the trivial case (no peaks), we get $f_{0}=1$. These cases are shown below.\n\n1\n\n\n\n\n\nSuppose we know the peak on which the largest number, $2 n+1$, is placed. Then that splits our picture into two shapes with fewer peaks. Once we choose which numbers from $1,2, \\ldots, 2 n$ to place each shape, we can compute the number of arrangements of the numbers on each shape, and then take the product. For example, if we place the 9 at the second peak, as shown below, we get a 1-peak shape on the left and a 2-peak shape on the right.\n\n\n\nFor the above shape, there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways to pick the three numbers to place on the left-hand side, $f_{1}=2$ ways to place them, and $f_{2}=16$ ways to place the remaining five numbers on the right.\n\nThis argument works for any $n>1$, so we have shown the following:\n\n$$\nf_{n}=\\sum_{k=1}^{n}\\left(\\begin{array}{c}\n2 n \\\\\n2 k-1\n\\end{array}\\right) f_{k-1} f_{n-k}\n$$\n\n\n\nSo we have:\n\n$$\n\\begin{aligned}\n& f_{1}=\\left(\\begin{array}{l}\n2 \\\\\n1\n\\end{array}\\right) f_{0}^{2}=2 \\\\\n& f_{2}=\\left(\\begin{array}{l}\n4 \\\\\n1\n\\end{array}\\right) f_{0} f_{1}+\\left(\\begin{array}{l}\n4 \\\\\n3\n\\end{array}\\right) f_{1} f_{0}=16 \\\\\n& f_{3}=\\left(\\begin{array}{l}\n6 \\\\\n1\n\\end{array}\\right) f_{0} f_{2}+\\left(\\begin{array}{l}\n6 \\\\\n3\n\\end{array}\\right) f_{1}^{2}+\\left(\\begin{array}{l}\n6 \\\\\n5\n\\end{array}\\right) f_{2} f_{0}=272 \\\\\n& f_{4}=\\left(\\begin{array}{l}\n8 \\\\\n1\n\\end{array}\\right) f_{0} f_{3}+\\left(\\begin{array}{l}\n8 \\\\\n3\n\\end{array}\\right) f_{1} f_{2}+\\left(\\begin{array}{l}\n8 \\\\\n5\n\\end{array}\\right) f_{2} f_{1}+\\left(\\begin{array}{l}\n8 \\\\\n7\n\\end{array}\\right) f_{3} f_{0}=7936 .\n\\end{aligned}\n$$']",['7936'],False,,Numerical, 2872,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Determine the number of 9-labels with 2-signature $$ (12,21,12,21,12,21,12,21) \text {. } $$ Justify your answer.","['The answer is 7936. The shape of this 2-signature has four peaks and three intermediate valleys:\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b4f25e2667a6f4e35f44g-1.jpg?height=176&width=1152&top_left_y=535&top_left_x=579)\n\nWe will solve this problem by building up from smaller examples. Let $f_{n}$ equal the number of $(2 n+1)$-labels whose 2 -signature consists of $n$ peaks and $n-1$ intermediate valleys. In part (b) we showed that $f_{2}=16$. In the case where we have one peak, $f_{1}=2$. For the trivial case (no peaks), we get $f_{0}=1$. These cases are shown below.\n\n1\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b4f25e2667a6f4e35f44g-1.jpg?height=219&width=357&top_left_y=1037&top_left_x=835)\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b4f25e2667a6f4e35f44g-1.jpg?height=214&width=350&top_left_y=1037&top_left_x=1251)\n\nSuppose we know the peak on which the largest number, $2 n+1$, is placed. Then that splits our picture into two shapes with fewer peaks. Once we choose which numbers from $1,2, \\ldots, 2 n$ to place each shape, we can compute the number of arrangements of the numbers on each shape, and then take the product. For example, if we place the 9 at the second peak, as shown below, we get a 1-peak shape on the left and a 2-peak shape on the right.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_b4f25e2667a6f4e35f44g-1.jpg?height=220&width=1155&top_left_y=1755&top_left_x=577)\n\nFor the above shape, there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways to pick the three numbers to place on the left-hand side, $f_{1}=2$ ways to place them, and $f_{2}=16$ ways to place the remaining five numbers on the right.\n\nThis argument works for any $n>1$, so we have shown the following:\n\n$$\nf_{n}=\\sum_{k=1}^{n}\\left(\\begin{array}{c}\n2 n \\\\\n2 k-1\n\\end{array}\\right) f_{k-1} f_{n-k}\n$$\n\n\n\nSo we have:\n\n$$\n\\begin{aligned}\n& f_{1}=\\left(\\begin{array}{l}\n2 \\\\\n1\n\\end{array}\\right) f_{0}^{2}=2 \\\\\n& f_{2}=\\left(\\begin{array}{l}\n4 \\\\\n1\n\\end{array}\\right) f_{0} f_{1}+\\left(\\begin{array}{l}\n4 \\\\\n3\n\\end{array}\\right) f_{1} f_{0}=16 \\\\\n& f_{3}=\\left(\\begin{array}{l}\n6 \\\\\n1\n\\end{array}\\right) f_{0} f_{2}+\\left(\\begin{array}{l}\n6 \\\\\n3\n\\end{array}\\right) f_{1}^{2}+\\left(\\begin{array}{l}\n6 \\\\\n5\n\\end{array}\\right) f_{2} f_{0}=272 \\\\\n& f_{4}=\\left(\\begin{array}{l}\n8 \\\\\n1\n\\end{array}\\right) f_{0} f_{3}+\\left(\\begin{array}{l}\n8 \\\\\n3\n\\end{array}\\right) f_{1} f_{2}+\\left(\\begin{array}{l}\n8 \\\\\n5\n\\end{array}\\right) f_{2} f_{1}+\\left(\\begin{array}{l}\n8 \\\\\n7\n\\end{array}\\right) f_{3} f_{0}=7936 .\n\\end{aligned}\n$$']",['7936'],False,,Numerical, 2873,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Prove that the following signature is possible. $(123,132,213)$,","['The signature is possible, because it is the 3 -signature of 12435']",['证明题，略'],True,,Need_human_evaluate, 2873,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Prove that the following signature is possible. $(123,132,213)$,","['The signature is possible, because it is the 3 -signature of 12435']",,True,,, 2874,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Prove that the following signature is impossible: $(321,312,213)$.","['The signature is impossible. Let a 5 -label be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. The second window of (ii) implies $a_{3}a_{4}$, a contradiction.']",,True,,, 2874,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Prove that the following signature is impossible: $(321,312,213)$.","['The signature is impossible. Let a 5 -label be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. The second window of (ii) implies $a_{3}a_{4}$, a contradiction.']",['证明题，略'],True,,Need_human_evaluate, 2875,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) For a general $n$, determine the number of distinct possible $p$-signatures.","['The answer is $p ! \\cdot p^{n-p}$.\n\nCall two consecutive windows in a $p$-signature compatible if the last $p-1$ numbers in the first label and the first $p-1$ numbers in the second label (their ""overlap"") describe the same ordering. For example, in the $p$-signature $(. ., 2143,2431, \\ldots), 2143$ and 2431 are compatible. Notice that the last three digits of 2143 and the first three digits of 2431 can be described by the same 3-label, 132 .\n\nTheorem: A signature $\\sigma$ is possible if and only if every pair of consecutive windows is compatible.\n\nProof: $(\\Rightarrow)$ Consider a signature $\\sigma$ describing a $p$-label $L$. If some pair in $\\sigma$ is not compatible, then there is some string of $p-1$ numbers in our label $L$ that has two different $(p-1)$-signatures. This is impossible, since the $p$-signature is well-defined.\n\n$(\\Leftarrow)$ Now suppose $\\sigma$ is a $p$-signature such that that every pair of consecutive windows is compatible. We need to show that there is at least one label $L$ with $S_{p}[L]=\\sigma$. We do so by induction on the number of windows in $\\sigma$, using the results from $5(\\mathrm{~b})$.\n\nLet $\\sigma=\\left\\{\\omega_{1}, \\omega_{2}, \\ldots, \\omega_{k+1}\\right\\}$, and suppose $\\omega_{1}=a_{1}, a_{2}, \\ldots, a_{p}$. Set $L_{1}=\\omega_{1}$.\n\nSuppose that $L_{k}$ is a $(p+k-1)$-label such that $S_{p}\\left[L_{k}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k}\\right\\}$. We will construct $L_{k+1}$ for which $S_{p}\\left[L_{k+1}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$.\n\nAs in $5(\\mathrm{~b})$, denote by $L_{k}^{(j)}$ the label $L_{k}$ with a $j+0.5$ appended; we will eventually renumber the elements in the label to make them all integers. Appending $j+0.5$ does not affect any of the non-terminal windows of $S_{p}\\left[L_{k}\\right]$, and as $j$ varies from 0 to $p-k+1$ the final window of $S_{p}\\left[L_{k}^{(j)}\\right]$ varies over each of the $p$ windows compatible with $\\omega_{k}$. Since $\\omega_{k+1}$ is compatible with $\\omega_{k}$, there exists some $j$ for which $S_{p}\\left[L_{k}^{(j)}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$. Now we renumber as follows: set $L_{k+1}=S_{k+p}\\left[L_{k}^{(j)}\\right]$, which replaces $L_{k}^{(j)}$ with the integers 1 through $k+p$ and preserves the relative order of all integers in the label.\n\nBy continuing this process, we conclude that the $n$-label $L_{n-p+1}$ has $p$-signature $\\sigma$, so $\\sigma$ is possible.\n\nTo count the number of possible $p$-signatures, we choose the first window ( $p$ ! choices), then choose each of the remaining $n-p$ compatible windows ( $p$ choices each). In all, there are $p ! \\cdot p^{n-p}$ possible $p$-signatures.']",['$p ! \\cdot p^{n-p}$'],False,,Expression, 2875,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: For a general $n$, determine the number of distinct possible $p$-signatures.","['The answer is $p ! \\cdot p^{n-p}$.\n\nCall two consecutive windows in a $p$-signature compatible if the last $p-1$ numbers in the first label and the first $p-1$ numbers in the second label (their ""overlap"") describe the same ordering. For example, in the $p$-signature $(. ., 2143,2431, \\ldots), 2143$ and 2431 are compatible. Notice that the last three digits of 2143 and the first three digits of 2431 can be described by the same 3-label, 132 .\n\nTheorem: A signature $\\sigma$ is possible if and only if every pair of consecutive windows is compatible.\n\nProof: $(\\Rightarrow)$ Consider a signature $\\sigma$ describing a $p$-label $L$. If some pair in $\\sigma$ is not compatible, then there is some string of $p-1$ numbers in our label $L$ that has two different $(p-1)$-signatures. This is impossible, since the $p$-signature is well-defined.\n\n$(\\Leftarrow)$ Now suppose $\\sigma$ is a $p$-signature such that that every pair of consecutive windows is compatible. We need to show that there is at least one label $L$ with $S_{p}[L]=\\sigma$. We do so by induction on the number of windows in $\\sigma$, using the results from $5(\\mathrm{~b})$.\n\nLet $\\sigma=\\left\\{\\omega_{1}, \\omega_{2}, \\ldots, \\omega_{k+1}\\right\\}$, and suppose $\\omega_{1}=a_{1}, a_{2}, \\ldots, a_{p}$. Set $L_{1}=\\omega_{1}$.\n\nSuppose that $L_{k}$ is a $(p+k-1)$-label such that $S_{p}\\left[L_{k}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k}\\right\\}$. We will construct $L_{k+1}$ for which $S_{p}\\left[L_{k+1}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$.\n\nAs in $5(\\mathrm{~b})$, denote by $L_{k}^{(j)}$ the label $L_{k}$ with a $j+0.5$ appended; we will eventually renumber the elements in the label to make them all integers. Appending $j+0.5$ does not affect any of the non-terminal windows of $S_{p}\\left[L_{k}\\right]$, and as $j$ varies from 0 to $p-k+1$ the final window of $S_{p}\\left[L_{k}^{(j)}\\right]$ varies over each of the $p$ windows compatible with $\\omega_{k}$. Since $\\omega_{k+1}$ is compatible with $\\omega_{k}$, there exists some $j$ for which $S_{p}\\left[L_{k}^{(j)}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$. Now we renumber as follows: set $L_{k+1}=S_{k+p}\\left[L_{k}^{(j)}\\right]$, which replaces $L_{k}^{(j)}$ with the integers 1 through $k+p$ and preserves the relative order of all integers in the label.\n\nBy continuing this process, we conclude that the $n$-label $L_{n-p+1}$ has $p$-signature $\\sigma$, so $\\sigma$ is possible.\n\nTo count the number of possible $p$-signatures, we choose the first window ( $p$ ! choices), then choose each of the remaining $n-p$ compatible windows ( $p$ choices each). In all, there are $p ! \\cdot p^{n-p}$ possible $p$-signatures.']",['$p ! \\cdot p^{n-p}$'],False,,Expression, 2876,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) If a randomly chosen $p$-signature is 575 times more likely of being impossible than possible, determine $p$ and $n$.","['The answer is $n=7, p=5$.\n\nLet $P$ denote the probability that a randomly chosen $p$-signature is possible. We are\n\n\n\ngiven that $1-P=575$, so $P=\\frac{1}{576}$. We want to find $p$ and $n$ for which\n\n$$\n\\begin{aligned}\n\\frac{p ! \\cdot p^{n-p}}{(p !)^{n-p+1}} & =\\frac{1}{576} \\\\\n\\frac{p^{n-p}}{(p !)^{n-p}} & =\\frac{1}{576} \\\\\n((p-1) !)^{n-p} & =576\n\\end{aligned}\n$$\n\nThe only factorial that has 576 as an integer power is $4 !=\\sqrt{576}$. Thus $p=5$ and $n-p=2 \\Rightarrow n=7$.']","['7,5']",True,,Numerical, 2876,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: If a randomly chosen $p$-signature is 575 times more likely of being impossible than possible, determine $p$ and $n$.","['The answer is $n=7, p=5$.\n\nLet $P$ denote the probability that a randomly chosen $p$-signature is possible. We are\n\n\n\ngiven that $1-P=575$, so $P=\\frac{1}{576}$. We want to find $p$ and $n$ for which\n\n$$\n\\begin{aligned}\n\\frac{p ! \\cdot p^{n-p}}{(p !)^{n-p+1}} & =\\frac{1}{576} \\\\\n\\frac{p^{n-p}}{(p !)^{n-p}} & =\\frac{1}{576} \\\\\n((p-1) !)^{n-p} & =576\n\\end{aligned}\n$$\n\nThe only factorial that has 576 as an integer power is $4 !=\\sqrt{576}$. Thus $p=5$ and $n-p=2 \\Rightarrow n=7$.']","['7,5']",True,,Numerical, 2877,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Show that $(312,231,312,132)$ is not a unique 3 -signature.",['The $p$-signature is not unique because it equals both $S_{3}[625143]$ and $S_{3}[635142]$.'],['证明题，略'],True,,Need_human_evaluate, 2877,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Show that $(312,231,312,132)$ is not a unique 3 -signature.",['The $p$-signature is not unique because it equals both $S_{3}[625143]$ and $S_{3}[635142]$.'],,True,,, 2878,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) Show that $(231,213,123,132)$ is a unique 3 -signature.","['Let $L=a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We have $a_{4} A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Show that $(231,213,123,132)$ is a unique 3 -signature.","['Let $L=a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We have $a_{4} A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Find two 5-labels with unique 2-signatures.",['12345 and 54321 are the only ones.'],"['12345, 54321']",True,,Numerical, 2880,Combinatorics,,"An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following: $$ \underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213 $$ Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write $$ S_{3}[263415]=(132,312,231,213) $$ More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label. In this power question, you will be asked to analyze some of the properties of labels and signatures. We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ : ![](https://cdn.mathpix.com/cropped/2023_12_21_793ce3b512229d118445g-1.jpg?height=236&width=442&top_left_y=2213&top_left_x=836) A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: ![](https://cdn.mathpix.com/cropped/2023_12_21_54786ae6b210123098f4g-1.jpg?height=257&width=464&top_left_y=419&top_left_x=825) For a general $n \geq 2$, compute all $n$-labels that have unique 2 -signatures.","['The $n$-labels with unique 2 -signatures are $1,2, \\ldots, n$ and $n, n-1, \\ldots, 1$, and their respective 2-signatures are $(12,12, \\ldots, 12)$ and $(21,21, \\ldots, 21)$.\n\nProof: (Not required for credit.) Let $L=a_{1}, a_{2}, \\ldots, a_{n}$. The first signature above implies that $a_{1} A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Prove that $S_{5}[495138627]$ is unique.","['Let $L=a_{1}, \\ldots, a_{9}$ and suppose $S_{5}[L]=S_{5}[495138627]=\\left(\\omega_{1}, \\ldots, \\omega_{5}\\right)$. Then we get the following inequalities:\n\n| $a_{4} A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Determine the smallest $p$ for which the 20-label $$ L=3,11,8,4,17,7,15,19,6,2,14,1,10,16,5,12,20,9,13,18 $$ has a unique $p$-signature.","['The answer is $p=16$. To show this fact we will need to extend the idea from part 8(b) about ""linking"" inequalities forced by the various windows:\n\nTheorem: A $p$-signature for an $n$-label $L$ is unique if and only if for every $k A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below: Show that for each $k \geq 2$, the number of unique $2^{k-1}$-signatures on the set of $2^{k}$-labels is at least $2^{2^{k}-3}$.","[""Let $s_{k}$ denote the number of such unique signatures. We proceed by induction with base case $k=2$. From $8(\\mathrm{c})$, a 2-signature for a label $L$ is unique if and only if consecutive numbers in $L$ appear together in some window. Because $k=2$, the consecutive numbers must be adjacent\n\n\nin the label. The $2^{2}$-labels 1234 and 4321 satisfy this condition, ${ }^{1}$ so their $2^{1}$-signatures are unique. Thus we have shown that $s_{2} \\geq 2=2^{2^{2}-3}$, and the base case is established.\n\nNow suppose $s_{k} \\geq 2^{2^{k}-3}$ for some $k \\geq 2$. Let $L_{k}$ be a $2^{k}$-label with a unique $2^{k-1}$-signature. Write $L_{k}=\\left(a_{1}, a_{2}, \\ldots, a_{2^{k}}\\right)$. We will expand $L_{k}$ to form a $2^{k+1}$ label by replacing each $a_{i}$ above with the numbers $2 a_{i}-1$ and $2 a_{i}$ (in some order). This process produces a valid $2^{k+1}$-label, because the numbers produced are all the integers from 1 to $2^{k+1}$. Furthermore, different $L_{k}$ 's will produce different labels: if the starting labels differ at place $i$, then the new labels will differ at places $2 i-1$ and $2 i$. Therefore, each starting label produces $2^{2^{k}}$ distinct $2^{k+1}$ labels through this process. Summarizing, each valid $2^{k}$-label can be expanded to produce $2^{2^{k}}$ distinct $2^{k+1}$-labels, none of which could be obtained by expanding any other $2^{k}$-label.\n\nIt remains to be shown that the new label has a unique $2^{k-1}$-signature. Because $L_{k}$ has a unique $2^{k-1}$-signature, for all $i \\leq 2^{k}-1$, both $i$ and $i+1$ appeared in some $2^{k-1}$-window. Therefore, there were fewer than $2^{k-1}-1$ numbers between $i$ and $i+1$. When the label is expanded, $2 i$ and $2 i-1$ are adjacent, $2 i+1$ and $2 i+2$ are adjacent, and $2 i$ and $2 i+1$ are fewer than $2 \\cdot\\left(2^{k-1}-1\\right)+2=2^{k}$ places apart. Thus, every pair of adjacent integers is within some $2^{k}$-window.\n\nSince each pair of consecutive integers in our new $2^{k+1}$-label coexists in some $2^{k}$-window for every possible such expansion of $L_{k}$, that means all $2^{2^{k}}$ ways of expanding $L_{k}$ to a $2^{k+1}$-label result in labels with unique $2^{k}$-signatures. We then get\n\n$$\n\\begin{aligned}\ns_{k+1} & \\geq 2^{2^{k}} \\cdot s_{k} \\\\\n& \\geq 2^{2^{k}} \\cdot 2^{2^{k}-3} \\\\\n& =2^{2^{k+1}-3}\n\\end{aligned}\n$$\n\nwhich completes the induction.""]",,True,,, 2884,Geometry,,"In $\triangle A B C, D$ is on $\overline{A C}$ so that $\overline{B D}$ is the angle bisector of $\angle B$. Point $E$ is on $\overline{A B}$ and $\overline{C E}$ intersects $\overline{B D}$ at $P$. Quadrilateral $B C D E$ is cyclic, $B P=12$ and $P E=4$. Compute the ratio $\frac{A C}{A E}$.","['Let $\\omega$ denote the circle that circumscribes quadrilateral $B C D E$. Draw in line segment $\\overline{D E}$. Note that $\\angle D P E$ and $\\angle C P B$ are congruent, and $\\angle D E C$ and $\\angle D B C$ are congruent, since they cut off the same arc of $\\omega$. Therefore, $\\triangle B C P$ and $\\triangle E D P$ are similar. Thus $\\frac{B C}{D E}=\\frac{B P}{E P}=$ $\\frac{12}{4}=3$.\n\nBecause $\\angle B C E$ and $\\angle B D E$ cut off the same arc of $\\omega$, these angles are congruent. Let $\\alpha$ be the measure of these angles. Similarly, $\\angle D C E$ and $\\angle D B E$ cut off the same arc of $\\omega$. Let $\\beta$ be the measure of these angles. Since $B D$ is an angle bisector, $\\mathrm{m} \\angle C B D=\\beta$.\n\nNote that $\\mathrm{m} \\angle A D E=180^{\\circ}-\\mathrm{m} \\angle B D E-\\mathrm{m} \\angle B D C$. It follows that\n\n$$\n\\begin{aligned}\n\\mathrm{m} \\angle A D E & =180^{\\circ}-\\mathrm{m} \\angle B D E-\\left(180^{\\circ}-\\mathrm{m} \\angle C B D-\\mathrm{m} \\angle B C D\\right) \\\\\n\\Rightarrow \\mathrm{m} \\angle A D E & =180^{\\circ}-\\mathrm{m} \\angle B D E-\\left(180^{\\circ}-\\mathrm{m} \\angle C B D-\\mathrm{m} \\angle B C E-\\mathrm{m} \\angle D C E\\right) \\\\\n\\Rightarrow \\mathrm{m} \\angle A D E & =180^{\\circ}-\\alpha-\\left(180^{\\circ}-\\beta-\\alpha-\\beta\\right) \\\\\n\\Rightarrow \\mathrm{m} \\angle A D E & =2 \\beta=\\mathrm{m} \\angle C B D .\n\\end{aligned}\n$$\n\nThus $\\angle A D E$ is congruent to $\\angle C B D$, and it follows that $\\triangle A D E$ is similar to $\\triangle A B C$. Hence $\\frac{B C}{D E}=\\frac{A C}{A E}$, and by substituting in given values, we have $\\frac{A C}{A E}=\\mathbf{3}$.']",['3'],False,,Numerical, 2885,Combinatorics,,"Complete the following ""cross-number puzzle"", where each ""Across"" answer represents a fourdigit number, and each ""Down"" answer represents a three-digit number. No answer begins with the digit 0. ## Across: 1. $\underline{A} \underline{B} \underline{C} \underline{D}$ is the cube of the sum of the digits in the answer to 1 Down. 2. From left to right, the digits in $E \underline{F} \underline{G} \underline{H}$ are strictly decreasing. 3. From left to right, the digits in $\underline{I} \underline{J} K \underline{L}$ are strictly decreasing. ## Down: 1. $\underline{A} \underline{E}$ I is a perfect fourth power. 2. $\underline{B} \underline{F} \underline{J}$ is a perfect square. 3. The digits in $\underline{C} G \underline{K}$ form a geometric progression. 4. $\underline{D} \underline{H} \underline{L}$ has a two-digit prime factor. ![](https://cdn.mathpix.com/cropped/2023_12_21_94af3fb67559aed23a01g-1.jpg?height=480&width=545&top_left_y=1454&top_left_x=839)","['From 1 Down, $\\underline{A} \\underline{E} \\underline{I}=256$ or 625 , either of which make $\\underline{A} \\underline{B} \\underline{C} \\underline{D}=2197$, so $\\underline{A} \\underline{E} \\underline{I}=256$.\n\nFrom $\\underline{B}=1$ together with 2 Down, $\\underline{B} \\underline{F} \\underline{J}=121$ or 144 . But $\\underline{J}=1$ does not work because then 6 Across could not be satisfied. Therefore $\\underline{B} \\underline{F} \\underline{J}=144$.\n\nFrom $\\underline{C}=9$ together with 5 Across and 3 Down, we have $\\underline{C} \\underline{G} \\underline{K}=931$.\n\nFrom $\\underline{D}=7$ together with 5 and 6 Across, we get $\\underline{D} \\underline{H} \\underline{L}=720$ or 710 , but only 710 has a two-digit prime factor.']",['| ${ }^{1} 2$ | ${ }^{2} 1$ | 9 | 7 |\n| :--- | :--- | :--- | :--- |\n| ${ }^{5} 5$ | 4 | 3 | 1 |\n| ${ }^{6} 6$ | 4 | 1 | 0 |'],False,,Need_human_evaluate, 2886,Geometry,,"In rectangle $M N P Q$, point $A$ lies on $\overline{Q N}$. Segments parallel to the rectangle's sides are drawn through point $A$, dividing the rectangle into four regions. The areas of regions I, II, and III are integers in geometric progression. If the area of $M N P Q$ is 2009 , compute the maximum possible area of region I. ","['Because $A$ is on diagonal $\\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\\frac{A B}{A X}=\\frac{Q Y}{Q C}=$ $\\frac{A C}{A Y} \\Rightarrow A B \\cdot A Y=A C \\cdot A X$. Therefore, we have $2009=[\\mathrm{I}]+2[\\mathrm{II}]+[\\mathrm{III}]$.\n\nLet the common ratio of the geometric progression be $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get\n\n$$\n2009=a q^{2}+2 a p q+a p^{2} \\Rightarrow 7^{2} \\cdot 41=a(p+q)^{2}\n$$\n\nThus we must have $p+q=7$ and $a=41$. Since $[\\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\\frac{p}{q}=\\frac{1}{6}$, giving $[\\mathrm{I}]=41 \\cdot 36=\\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 .']",['1476'],False,,Numerical, 2886,Geometry,,"In rectangle $M N P Q$, point $A$ lies on $\overline{Q N}$. Segments parallel to the rectangle's sides are drawn through point $A$, dividing the rectangle into four regions. The areas of regions I, II, and III are integers in geometric progression. If the area of $M N P Q$ is 2009 , compute the maximum possible area of region I. ![](https://cdn.mathpix.com/cropped/2023_12_21_5bb283f04ad556d7c2dcg-1.jpg?height=398&width=624&top_left_y=492&top_left_x=794)","['Because $A$ is on diagonal $\\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\\frac{A B}{A X}=\\frac{Q Y}{Q C}=$ $\\frac{A C}{A Y} \\Rightarrow A B \\cdot A Y=A C \\cdot A X$. Therefore, we have $2009=[\\mathrm{I}]+2[\\mathrm{II}]+[\\mathrm{III}]$.\n\nLet the common ratio of the geometric progression be $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get\n\n$$\n2009=a q^{2}+2 a p q+a p^{2} \\Rightarrow 7^{2} \\cdot 41=a(p+q)^{2}\n$$\n\nThus we must have $p+q=7$ and $a=41$. Since $[\\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\\frac{p}{q}=\\frac{1}{6}$, giving $[\\mathrm{I}]=41 \\cdot 36=\\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 .']",['1476'],False,,Numerical, 2887,Combinatorics,,"Let $N$ be a six-digit number formed by an arrangement of the digits $1,2,3,3,4,5$. Compute the smallest value of $N$ that is divisible by 264 .","['Note that $264=3 \\cdot 8 \\cdot 11$, so we will need to address all these factors. Because the sum of the digits is 18 , it follows that 3 divides $N$, regardless of how we order the digits of $N$. In order for 8 to divide $N$, we need $N$ to end in $\\underline{O} 12, \\underline{O} 52, \\underline{E} 32$, or $\\underline{E} 24$, where $O$ and $E$ denote odd and even digits. Now write $N=\\underline{U} \\underline{V} \\underline{W} \\underline{X} \\underline{Y} \\underline{Z}$. Note that $N$ is divisible by 11 if and only if $(U+W+Y)-(V+X+Z)$ is divisible by 11. Because the sum of the three largest digits is only 12 , we must have $U+W+Y=V+X+Z=9$.\n\nBecause $Z$ must be even, this implies that $V, X, Z$ are $2,3,4$ (in some order). This means $Y \\neq 2$, and so we must have $Z \\neq 4 \\Rightarrow Z=2$. Of the three remaining possibilities, $\\underline{E} 32$ gives the smallest solution, 135432.']",['135432'],False,,Numerical, 2888,Geometry,,"In triangle $A B C, A B=4, B C=6$, and $A C=8$. Squares $A B Q R$ and $B C S T$ are drawn external to and lie in the same plane as $\triangle A B C$. Compute $Q T$.","['Set $\\mathrm{m} \\angle A B C=x$ and $\\mathrm{m} \\angle T B Q=y$. Then $x+y=180^{\\circ}$ and so $\\cos x+\\cos y=0$. Applying the Law of Cosines to triangles $A B C$ and $T B Q$ gives $A C^{2}=A B^{2}+B C^{2}-2 A B \\cdot B C \\cos x$ and $Q T^{2}=B T^{2}+B Q^{2}-2 B T \\cdot B Q \\cos y$, which, after substituting values, become $8^{2}=$ $4^{2}+6^{2}-48 \\cos x$ and $Q T^{2}=4^{2}+6^{2}-48 \\cos y$.\n\nAdding the last two equations yields $Q T^{2}+8^{2}=2\\left(4^{2}+6^{2}\\right)$ or $Q T=\\mathbf{2} \\sqrt{\\mathbf{1 0}}$.']",['$2 \\sqrt{10}$'],False,,Numerical, 2889,Combinatorics,,"The numbers $1,2, \ldots, 8$ are placed in the $3 \times 3$ grid below, leaving exactly one blank square. Such a placement is called okay if in every pair of adjacent squares, either one square is blank or the difference between the two numbers is at most 2 (two squares are considered adjacent if they share a common side). If reflections, rotations, etc. of placements are considered distinct, compute the number of distinct okay placements. ","[""We say that two numbers are neighbors if they occupy adjacent squares, and that $a$ is a friend of $b$ if $0<|a-b| \\leq 2$. Using this vocabulary, the problem's condition is that every pair of neighbors must be friends of each other. Each of the numbers 1 and 8 has two friends, and each number has at most four friends.\n\nIf there is no number written in the center square, then we must have one of the cycles in the figures below. For each cycle, there are 8 rotations. Thus there are 16 possible configurations with no number written in the center square.\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 4 | - | 5 |\n| 6 | 8 | 7 |\n\n\n| 3 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | - | 4 |\n| 7 | 8 | 6 |\n\nNow assume that the center square contains the number $n$. Because $n$ has at least three neighbors, $n \\neq 1$ and $n \\neq 8$. First we show that 1 must be in a corner. If 1 is a neighbor of $n$, then one of the corners neighboring 1 must be empty, because 1 has only two friends ( 2 and $3)$. If $c$ is in the other corner neighboring 1 , then $\\{n, c\\}=\\{2,3\\}$. But then $n$ must have three\n\n\n\nmore friends $\\left(n_{1}, n_{2}, n_{3}\\right)$ other than 1 and $c$, for a total of five friends, which is impossible, as illustrated below. Therefore 1 must be in a corner.\n\n| - | 1 | $c$ |\n| :--- | :--- | :--- |\n| $n_{1}$ | $n$ | $n_{2}$ |\n| | $n_{3}$ | |\n\nNow we show that 1 can only have one neighbor, i.e., one of the squares adjacent to 1 is empty. If 1 has two neighbors, then we have, up to a reflection and a rotation, the configuration shown below. Because 2 has only one more friend, the corner next to 2 is empty and $n=4$. Consequently, $m_{1}=5$ (refer to the figure below). Then 4 has one friend (namely 6) left with two neighbors $m_{2}$ and $m_{3}$, which is impossible. Thus 1 must have exactly one neighbor. An analogous argument shows that 8 must also be at a corner with exactly one neighbor.\n\n| 1 | 2 | - |\n| :--- | :--- | :--- |\n| 3 | $n$ | $m_{3}$ |\n| $m_{1}$ | $m_{2}$ | |\n\nTherefore, 8 and 1 must be in non-opposite corners, with the blank square between them. Thus, up to reflections and rotations, the only possible configuration is the one shown at left below.\n\n| 1 | - | 8 |\n| :--- | :--- | :--- |\n| $m$ | | |\n| | | |\n\n\n| 1 | - | 8 |\n| :---: | :---: | :---: |\n| $2 / 3$ | $4 / 5$ | $6 / 7$ |\n| $3 / 2$ | $5 / 4$ | $7 / 6$ |\n\nThere are two possible values for $m$, namely 2 and 3 . For each of the cases $m=2$ and $m=3$, the rest of the configuration is uniquely determined, as illustrated in the figure above right. We summarize our process: there are four corner positions for 1; two (non-opposite) corner positions for 8 (after 1 is placed); and two choices for the number in the square neighboring 1 but not neighboring 8 . This leads to $4 \\cdot 2 \\cdot 2=16$ distinct configurations with a number written in the center square.\n\nTherefore, there are 16 configurations in which the center square is blank and 16 configurations with a number in the center square, for a total of $\\mathbf{3 2}$ distinct configurations.""]",['32'],False,,Numerical, 2889,Combinatorics,,"The numbers $1,2, \ldots, 8$ are placed in the $3 \times 3$ grid below, leaving exactly one blank square. Such a placement is called okay if in every pair of adjacent squares, either one square is blank or the difference between the two numbers is at most 2 (two squares are considered adjacent if they share a common side). If reflections, rotations, etc. of placements are considered distinct, compute the number of distinct okay placements. ![](https://cdn.mathpix.com/cropped/2023_12_21_2fb354bef09ddaf774e2g-1.jpg?height=252&width=263&top_left_y=1072&top_left_x=974)","[""We say that two numbers are neighbors if they occupy adjacent squares, and that $a$ is a friend of $b$ if $0<|a-b| \\leq 2$. Using this vocabulary, the problem's condition is that every pair of neighbors must be friends of each other. Each of the numbers 1 and 8 has two friends, and each number has at most four friends.\n\nIf there is no number written in the center square, then we must have one of the cycles in the figures below. For each cycle, there are 8 rotations. Thus there are 16 possible configurations with no number written in the center square.\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 4 | - | 5 |\n| 6 | 8 | 7 |\n\n\n| 3 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | - | 4 |\n| 7 | 8 | 6 |\n\nNow assume that the center square contains the number $n$. Because $n$ has at least three neighbors, $n \\neq 1$ and $n \\neq 8$. First we show that 1 must be in a corner. If 1 is a neighbor of $n$, then one of the corners neighboring 1 must be empty, because 1 has only two friends ( 2 and $3)$. If $c$ is in the other corner neighboring 1 , then $\\{n, c\\}=\\{2,3\\}$. But then $n$ must have three\n\n\n\nmore friends $\\left(n_{1}, n_{2}, n_{3}\\right)$ other than 1 and $c$, for a total of five friends, which is impossible, as illustrated below. Therefore 1 must be in a corner.\n\n| - | 1 | $c$ |\n| :--- | :--- | :--- |\n| $n_{1}$ | $n$ | $n_{2}$ |\n| | $n_{3}$ | |\n\nNow we show that 1 can only have one neighbor, i.e., one of the squares adjacent to 1 is empty. If 1 has two neighbors, then we have, up to a reflection and a rotation, the configuration shown below. Because 2 has only one more friend, the corner next to 2 is empty and $n=4$. Consequently, $m_{1}=5$ (refer to the figure below). Then 4 has one friend (namely 6) left with two neighbors $m_{2}$ and $m_{3}$, which is impossible. Thus 1 must have exactly one neighbor. An analogous argument shows that 8 must also be at a corner with exactly one neighbor.\n\n| 1 | 2 | - |\n| :--- | :--- | :--- |\n| 3 | $n$ | $m_{3}$ |\n| $m_{1}$ | $m_{2}$ | |\n\nTherefore, 8 and 1 must be in non-opposite corners, with the blank square between them. Thus, up to reflections and rotations, the only possible configuration is the one shown at left below.\n\n| 1 | - | 8 |\n| :--- | :--- | :--- |\n| $m$ | | |\n| | | |\n\n\n| 1 | - | 8 |\n| :---: | :---: | :---: |\n| $2 / 3$ | $4 / 5$ | $6 / 7$ |\n| $3 / 2$ | $5 / 4$ | $7 / 6$ |\n\nThere are two possible values for $m$, namely 2 and 3 . For each of the cases $m=2$ and $m=3$, the rest of the configuration is uniquely determined, as illustrated in the figure above right. We summarize our process: there are four corner positions for 1; two (non-opposite) corner positions for 8 (after 1 is placed); and two choices for the number in the square neighboring 1 but not neighboring 8 . This leads to $4 \\cdot 2 \\cdot 2=16$ distinct configurations with a number written in the center square.\n\nTherefore, there are 16 configurations in which the center square is blank and 16 configurations with a number in the center square, for a total of $\\mathbf{3 2}$ distinct configurations.""]",['32'],False,,Numerical, 2890,Geometry,,"An ellipse in the first quadrant is tangent to both the $x$-axis and $y$-axis. One focus is at $(3,7)$, and the other focus is at $(d, 7)$. Compute $d$.","['See the diagram below. The center of the ellipse is $C=\\left(\\frac{d+3}{2}, 7\\right)$. The major axis of the ellipse is the line $y=7$, and the minor axis is the line $x=\\frac{d+3}{2}$. The ellipse is tangent to the coordinate axes at $T_{x}=\\left(\\frac{d+3}{2}, 0\\right)$ and $T_{y}=(0,7)$. Let $F_{1}=(3,7)$ and $F_{2}=(d, 7)$. Using the locus definition of an ellipse, we have $F_{1} T_{x}+F_{2} T_{x}=F_{1} T_{y}+F_{2} T_{y}$; that is,\n\n$$\n2 \\sqrt{\\left(\\frac{d-3}{2}\\right)^{2}+7^{2}}=d+3 \\quad \\text { or } \\quad \\sqrt{(d-3)^{2}+14^{2}}=d+3\n$$\n\nSquaring both sides of the last equation gives $d^{2}-6 d+205=d^{2}+6 d+9$ or $196=12 d$, so $d=\\frac{49}{3}$.\n\n']",['$\\frac{49}{3}$'],False,,Numerical, 2891,Geometry,,"Let $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} A_{8}$ be a regular octagon. Let $\mathbf{u}$ be the vector from $A_{1}$ to $A_{2}$ and let $\mathbf{v}$ be the vector from $A_{1}$ to $A_{8}$. The vector from $A_{1}$ to $A_{4}$ can be written as $a \mathbf{u}+b \mathbf{v}$ for a unique ordered pair of real numbers $(a, b)$. Compute $(a, b)$.","['We can scale the octagon so that $A_{1} A_{2}=\\sqrt{2}$. Because the exterior angle of the octagon is $45^{\\circ}$, we can place the octagon in the coordinate plane with $A_{1}$ being the origin, $A_{2}=(\\sqrt{2}, 0)$, and $A_{8}=(1,1)$.\n\n\n\nThen $A_{3}=(1+\\sqrt{2}, 1)$ and $A_{4}=(1+\\sqrt{2}, 1+\\sqrt{2})$. It follows that $\\mathbf{u}=\\langle\\sqrt{2}, 0\\rangle, \\mathbf{v}=\\langle-1,1\\rangle$, and\n\n$$\n\\overrightarrow{A_{1} A_{4}}=\\langle 1+\\sqrt{2}, 1+\\sqrt{2}\\rangle=a\\langle\\sqrt{2}, 0\\rangle+b\\langle-1,1\\rangle=\\langle a \\sqrt{2}-b, b\\rangle .\n$$\n\nThus $b=\\sqrt{2}+1$ and $a \\sqrt{2}-b=\\sqrt{2}+1$, or $a=2+\\sqrt{2}$, so $(a, b)=(2+\\sqrt{2}, \\sqrt{2}+1)$.', 'Extend $\\overline{A_{1} A_{2}}$ and $\\overline{A_{5} A_{4}}$ to meet at point $Q$; let $P$ be the intersection of $\\widehat{A_{1} Q}$ and $\\overleftrightarrow{A_{6} A_{3}}$. Then $A_{1} A_{2}=\\|\\mathbf{u}\\|, A_{2} P=\\|\\mathbf{u}\\| \\sqrt{2}$, and $P Q=\\|\\mathbf{u}\\|$, so $A_{1} Q=(2+\\sqrt{2})\\|\\mathbf{u}\\|$.\n\n\n\nBecause $A_{1} Q A_{4}$ is a $45^{\\circ}-45^{\\circ}-90^{\\circ}$ right triangle, $A_{4} Q=\\frac{A_{1} Q}{\\sqrt{2}}=(\\sqrt{2}+1)\\|\\mathbf{u}\\|$. Thus $\\overrightarrow{A_{1} A_{4}}=\\overrightarrow{A_{1} Q}+\\overrightarrow{Q A_{4}}$, and because $\\|\\mathbf{u}\\|=\\|\\mathbf{v}\\|$, we have $(a, b)=(2+\\sqrt{2}, \\sqrt{2}+\\mathbf{1})$.']","['$\\quad(2+\\sqrt{2}, 1+\\sqrt{2})$']",False,,Numerical, 2892,Number Theory,,Compute the integer $n$ such that $20091024$, if $p_{i} \\geq 11$, then $a_{i}=1$ and $1+p_{i}$ must be a power of 2 that is no greater than 1024. The possible values of $p_{i}$, with $p_{i} \\geq 11$, are 31 and 127 (as 5 divides 255, 7 divides 511, and 3 divides 1023).\n\nIf $p_{1}<11$, then $p_{i}$ can be $3,5,7$. It is routine to check that $a_{i}=1$ and $p_{i}=3$ or 7 .\n\nThus $a_{i}=1$ for all $i$, and the possible values of $p_{i}$ are $3,7,31,127$. The only combinations of these primes that yield 1024 are $(1+3) \\cdot(1+7) \\cdot(1+31)\\left(\\right.$ with $\\left.n=2^{k} \\cdot 3 \\cdot 7 \\cdot 31=651 \\cdot 2^{k}\\right)$ and $(1+7) \\cdot(1+127)$ (with $n=7 \\cdot 127=889 \\cdot 2^{k}$ ). Thus $n=651 \\cdot 2^{2}=\\mathbf{2 6 0 4}$ is the unique value of $n$ satisfying the conditions of the problem.']",['2604'],False,,Numerical, 2893,Geometry,,"Points $A, R, M$, and $L$ are consecutively the midpoints of the sides of a square whose area is 650. The coordinates of point $A$ are $(11,5)$. If points $R, M$, and $L$ are all lattice points, and $R$ is in Quadrant I, compute the number of possible ordered pairs $(x, y)$ of coordinates for point $R$.","['Write $x=11+c$ and $y=5+d$. Then $A R^{2}=c^{2}+d^{2}=\\frac{1}{2} \\cdot 650=325$. Note that $325=18^{2}+1^{2}=17^{2}+6^{2}=15^{2}+10^{2}$. Temporarily restricting ourselves to the case where $c$ and $d$ are both positive, there are three classes of solutions: $\\{c, d\\}=\\{18,1\\},\\{c, d\\}=\\{17,6\\}$, or $\\{c, d\\}=\\{15,10\\}$. In fact, $c$ and $d$ can be negative, so long as those values do not cause $x$ or $y$ to be negative. So there are 10 solutions:\n\n| $(c, d)$ | $(x, y)$ |\n| :---: | :---: |\n| $(18,1)$ | $(29,6)$ |\n| $(18,-1)$ | $(29,4)$ |\n| $(1,18)$ | $(12,23)$ |\n| $(-1,18)$ | $(10,23)$ |\n| $(17,6)$ | $(28,11)$ |\n| $(6,17)$ | $(17,22)$ |\n| $(-6,17)$ | $(5,22)$ |\n| $(15,10)$ | $(26,15)$ |\n| $(10,15)$ | $(21,20)$ |\n| $(-10,15)$ | $(1,20)$ |']",['10'],False,,Numerical, 2894,Geometry,,"The taxicab distance between points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $$ d\left(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)\right)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|+\left|z_{1}-z_{2}\right| . $$ The region $\mathcal{R}$ is obtained by taking the cube $\{(x, y, z): 0 \leq x, y, z \leq 1\}$ and removing every point whose taxicab distance to any vertex of the cube is less than $\frac{3}{5}$. Compute the volume of $\mathcal{R}$.","['For a fixed vertex $V$ on the cube, the locus of points on or inside the cube that are at most $\\frac{3}{5}$ away from $V$ form a corner at $V$ (that is, the right pyramid $V W_{1} W_{2} W_{3}$ in the figure shown at left below, with equilateral triangular base $W_{1} W_{2} W_{3}$ and three isosceles right triangular lateral faces $V W_{1} W_{2}, V W_{2} W_{3}, V W_{3} W_{1}$ ). Thus $\\mathcal{R}$ is formed by removing eight such congruent corners from the cube. However, each two neighboring corners share a common region along their shared edge. This common region is the union of two smaller right pyramids, each similar to the original corners. (See the figure shown at right below.)\n\n\nWe compute the volume of $\\mathcal{R}$ as\n\n$$\n1-8 \\cdot \\frac{1}{6}\\left(\\frac{3}{5}\\right)^{3}+12 \\cdot 2 \\cdot \\frac{1}{6}\\left(\\frac{1}{10}\\right)^{3}=\\frac{\\mathbf{1 7 9}}{\\mathbf{2 5 0}}\n$$']",['$\\frac{179}{250}$'],False,,Numerical, 2895,Algebra,,"$\quad$ Let $a$ and $b$ be real numbers such that $$ a^{3}-15 a^{2}+20 a-50=0 \quad \text { and } \quad 8 b^{3}-60 b^{2}-290 b+2575=0 $$ Compute $a+b$.","['Each cubic expression can be depressed - that is, the quadratic term can be eliminated-by substituting as follows. Because $(a-p)^{3}=a^{3}-3 a^{2} p+3 a p^{2}-p^{3}$, setting $p=-\\frac{(-15)}{3}=5$ and substituting $c+p=a$ transforms the expression $a^{3}-15 a^{2}+20 a-50$ into the equivalent expression $(c+5)^{3}-15(c+5)^{2}+20(c+5)-50$, which simplifies to $c^{3}-55 c-200$. Similarly, the substitution $d=b-\\frac{5}{2}$ yields the equation $d^{3}-55 d=-200$. [This procedure, which is analogous to completing the square, is an essential step in the algebraic solution to the general cubic equation.]\n\nConsider the function $f(x)=x^{3}-55 x$. It has three zeros, namely, 0 and $\\pm \\sqrt{55}$. Therefore, it has a relative maximum and a relative minimum in the interval $[-\\sqrt{55}, \\sqrt{55}]$. Note that for $0 \\leq x \\leq 5.5,|f(x)|<\\left|x^{3}\\right|<5.5^{3}=166.375$, and for $5.5198$, there is a unique real number $x_{0}$ such that $f\\left(x_{0}\\right)=m$.\n\nIn particular, since $200>198$, the values of $c$ and $d$ are uniquely determined. Because $f(x)$ is odd, we conclude that $c=-d$, or $a+b=\\frac{\\mathbf{1 5}}{\\mathbf{2}}$.', 'Set $a=x-b$ and substitute into the first equation. We get\n\n$$\n\\begin{aligned}\n(x-b)^{3}-15(x-b)^{2}+20(x-b)-50 & =0 \\\\\n-b^{3}+b^{2}(3 x-15)+b\\left(-3 x^{2}+30 x-20\\right)+\\left(x^{3}-15 x^{2}+20 x-50\\right) & =0 \\\\\n8 b^{3}+b^{2}(-24 x+120)+b\\left(24 x^{2}-240 x+160\\right)-8\\left(x^{3}-15 x^{2}+20 x-50\\right) & =0 .\n\\end{aligned}\n$$\n\nIf we equate coefficients, we see that\n\n$$\n\\begin{aligned}\n-24 x+120 & =-60 \\\\\n24 x^{2}-240 x+160 & =-290 \\\\\n-8\\left(x^{3}-15 x^{2}+20 x-50\\right) & =2575\n\\end{aligned}\n$$\n\nare all satisfied by $x=\\frac{15}{2}$. This means that any real solution $b$ to the second equation yields a real solution of $\\frac{15}{2}-b$ to the first equation. We can follow the reasoning of the previous solution to establish the existence of exactly one real solution to the second cubic equation. Thus $a$ and $b$ are unique, and their sum is $\\left(\\frac{15}{2}-b\\right)+b=\\frac{\\mathbf{1 5}}{\\mathbf{2}}$.']",['$\\frac{15}{2}$'],False,,Numerical, 2896,Number Theory,,"For a positive integer $n$, define $s(n)$ to be the sum of $n$ and its digits. For example, $s(2009)=2009+2+0+0+9=2020$. Compute the number of elements in the set $\{s(0), s(1), s(2), \ldots, s(9999)\}$.","['If $s(10 x)=a$, then the values of $s$ over $\\{10 x+0,10 x+1, \\ldots, 10 x+9\\}$ are $a, a+2, a+4, \\ldots, a+18$. Furthermore, if $x$ is not a multiple of 10 , then $s(10(x+1))=a+11$. This indicates that the values of $s$ ""interweave"" somewhat from one group of 10 to the next: the sets alternate between even and odd. Because the $s$-values for starting blocks of ten differ by 11, consecutive blocks of the same parity differ by 22 , so the values of $s$ do not overlap. That is, $s$ takes on 100 distinct values over any range of the form $\\{100 y+0,100 y+1, \\ldots, 100 y+99\\}$.\n\nFirst determine how many values are repeated between consecutive hundreds. Let $y$ be an integer that is not a multiple of 10 . Then the largest value for $s(100 y+k)(0 \\leq k \\leq 99)$ is $100 y+(s(y)-y)+99+s(99)=100 y+s(y)-y+117$, whereas the smallest value in the next group of 100 is for\n\n$$\n\\begin{aligned}\ns(100(y+1)) & =100(y+1)+(s(y+1)-(y+1))=100 y+(s(y)+2)-(y+1)+100 \\\\\n& =100 y+s(y)-y+101\n\\end{aligned}\n$$\n\nThis result implies that the values for $s(100 y+91)$ through $s(100 y+99)$ match the values of $s(100 y+100)$ through $s(100 y+108)$. So there are 9 repeated values.\n\nNow determine how many values are repeated between consecutive thousands. Let $z$ be a digit, and consider $s(1000 z+999)$ versus $s(1000(z+1))$. The first value equals\n\n$$\n1000 z+(s(z)-z)+999+s(999)=1000 z+z+1026=1001 z+1026\n$$\n\nThe latter value equals $1000(z+1)+(s(z+1)-(z+1))=1001(z+1)=1001 z+1001$. These values differ by an odd number. We have overlap between the $982,983, \\ldots, 989$ terms and the $000,001, \\ldots, 007$ terms. We also have overlap between the $992,993, \\ldots, 999$ terms and the $010,011, \\ldots, 017$ terms, for a total of 16 repeated values in all.\n\nThere are 90 instances in which we have 9 repeated terms, and 9 instances in which we have 16 repeated terms, so there are a total of $10000-90 \\cdot 9-9 \\cdot 16=\\mathbf{9 0 4 6}$ unique values.']",['9046'],False,,Numerical, 2897,Geometry,,"Quadrilateral $A R M L$ is a kite with $A R=R M=5, A M=8$, and $R L=11$. Compute $A L$.","['Let $K$ be the midpoint of $\\overline{A M}$. Then $A K=K M=8 / 2=4, R K=\\sqrt{5^{2}-4^{2}}=3$, and $K L=11-3=8$. Thus $A L=\\sqrt{A K^{2}+K L^{2}}=\\sqrt{4^{2}+8^{2}}=4 \\sqrt{5}$.']",['$4 \\sqrt{5}$'],False,,Numerical, 2898,Algebra,,"Let $T=4 \sqrt{5}$. If $x y=\sqrt{5}, y z=5$, and $x z=T$, compute the positive value of $x$.","['Multiply the three given equations to obtain $x^{2} y^{2} z^{2}=5 T \\sqrt{5}$. Thus $x y z= \\pm \\sqrt[4]{125 T^{2}}$, and the positive value of $x$ is $x=x y z / y z=\\sqrt[4]{125 T^{2}} / 5=\\sqrt[4]{T^{2} / 5}$. With $T=4 \\sqrt{5}$, we have $x=\\mathbf{2}$.']",['2'],False,,Numerical, 2899,Combinatorics,,$\quad$ Let $T=2$. In how many ways can $T$ boys and $T+1$ girls be arranged in a row if all the girls must be standing next to each other?,"['First choose the position of the first girl, starting from the left. There are $T+1$ possible positions, and then the positions for the girls are all determined. There are $(T+1)$ ! ways to arrange the girls, and there are $T$ ! ways to arrange the boys, for a total of $(T+1) \\cdot(T+1) ! \\cdot T !=$ $((T+1) !)^{2}$ arrangements. With $T=2$, the answer is $\\mathbf{3 6}$.']",['36'],False,,Numerical, 2900,Geometry,,"$\triangle A B C$ is on a coordinate plane such that $A=(3,6)$, $B=(T, 0)$, and $C=(2 T-1,1-T)$. Let $\ell$ be the line containing the altitude to $\overline{B C}$. Compute the $y$-intercept of $\ell$.","['The slope of $\\overleftrightarrow{B C}$ is $\\frac{(1-T)-0}{(2 T-1)-T}=-1$, and since $\\ell$ is perpendicular to $\\overleftrightarrow{B C}$, the slope of $\\ell$ is 1. Because $\\ell$ passes through $A=(3,6)$, the equation of $\\ell$ is $y=x+3$, and its $y$-intercept is 3 (independent of $T$ ).']",['3'],False,,Numerical, 2901,Geometry,,"Let $T=3$. In triangle $A B C, A B=A C-2=T$, and $\mathrm{m} \angle A=60^{\circ}$. Compute $B C^{2}$.","['By the Law of Cosines, $B C^{2}=A B^{2}+A C^{2}-2 \\cdot A B \\cdot A C \\cdot \\cos A=T^{2}+(T+2)^{2}-2 \\cdot T \\cdot(T+2) \\cdot \\frac{1}{2}=$ $T^{2}+2 T+4$. With $T=3$, the answer is 19 .']",['19'],False,,Numerical, 2902,Algebra,,"Let $T=19$. Let $\mathcal{S}_{1}$ denote the arithmetic sequence $0, \frac{1}{4}, \frac{1}{2}, \ldots$, and let $\mathcal{S}_{2}$ denote the arithmetic sequence $0, \frac{1}{6}, \frac{1}{3}, \ldots$ Compute the $T^{\text {th }}$ smallest number that occurs in both sequences $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$.","['$\\mathcal{S}_{1}$ consists of all numbers of the form $\\frac{n}{4}$, and $\\mathcal{S}_{2}$ consists of all numbers of the form $\\frac{n}{6}$, where $n$ is a nonnegative integer. Since $\\operatorname{gcd}(4,6)=2$, the numbers that are in both sequences are of the form $\\frac{n}{2}$, and the $T^{\\text {th }}$ smallest such number is $\\frac{T-1}{2}$. With $T=19$, the answer is 9 .']",['9'],False,,Numerical, 2903,Combinatorics,,"$\quad$ Let $T=9$. An integer $n$ is randomly selected from the set $\{1,2,3, \ldots, 2 T\}$. Compute the probability that the integer $\left|n^{3}-7 n^{2}+13 n-6\right|$ is a prime number.","['Let $P(n)=n^{3}-7 n^{2}+13 n-6$, and note that $P(n)=(n-2)\\left(n^{2}-5 n+3\\right)$. Thus $|P(n)|$ is prime if either $|n-2|=1$ and $\\left|n^{2}-5 n+3\\right|$ is prime or if $\\left|n^{2}-5 n+3\\right|=1$ and $|n-2|$ is prime. Solving $|n-2|=1$ gives $n=1$ or 3 , and solving $\\left|n^{2}-5 n+3\\right|=1$ gives $n=1$ or 4 or $\\frac{5 \\pm \\sqrt{17}}{2}$. Note that $P(1)=1, P(3)=-3$, and $P(4)=-2$. Thus $|P(n)|$ is prime only when $n$ is 3 or 4 , and if $T \\geq 2$, then the desired probability is $\\frac{2}{2 T}=\\frac{1}{T}$. With $T=9$, the answer is $\\frac{\\mathbf{1}}{\\mathbf{9}}$.']",['$\\frac{1}{9}$'],False,,Numerical, 2904,Algebra,,"Let $A=\frac{1}{9}$, and let $B=\frac{1}{25}$. In $\frac{1}{A}$ minutes, 20 frogs can eat 1800 flies. At this rate, in $\frac{1}{B}$ minutes, how many flies will 15 frogs be able to eat?","['In $\\frac{1}{A}$ minutes, 1 frog can eat $1800 / 20=90$ flies; thus in $\\frac{1}{B}$ minutes, 1 frog can eat $\\frac{A}{B} \\cdot 90$ flies. Thus in $\\frac{1}{B}$ minutes, 15 frogs can eat $15 \\cdot 90 \\cdot \\frac{A}{B}$ flies. With $A=\\frac{1}{9}$ and $B=\\frac{1}{25}$, this simplifies to $15 \\cdot 250=\\mathbf{3 7 5 0}$.']",['3750'],False,,Numerical, 2905,Algebra,,"Let $T=5$. If $|T|-1+3 i=\frac{1}{z}$, compute the sum of the real and imaginary parts of $z$.","['Let $t=|T|$. Note that $z=\\frac{1}{t-1+3 i}=\\frac{1}{t-1+3 i} \\cdot \\frac{t-1-3 i}{t-1-3 i}=\\frac{t-1-3 i}{t^{2}-2 t+10}$. Thus the sum of the real and imaginary parts of $z$ is $\\frac{t-1}{t^{2}-2 t+10}+\\frac{-3}{t^{2}-2 t+10}=\\frac{|T|-4}{|T|^{2}-2|T|+10}$. With $T=5$, the answer is $\\frac{1}{25}$.']",['$\\frac{1}{25}$'],False,,Numerical, 2906,Geometry,,"Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units. ![](https://cdn.mathpix.com/cropped/2023_12_21_8fe750e0fde7a21c6855g-1.jpg?height=293&width=477&top_left_y=2111&top_left_x=1472)","['Note that $\\mathrm{m} \\angle A E B=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-m \\widehat{C D})=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-\\mathrm{m} \\angle C O D)$. Also note that $\\mathrm{m} \\angle C O D=$ $360^{\\circ}-(\\mathrm{m} \\angle A O C+\\mathrm{m} \\angle B O D+\\mathrm{m} \\angle A O B)=360^{\\circ}-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O A C\\right)-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O B D\\right)-$ $\\mathrm{m} \\widehat{A B}=2(\\mathrm{~m} \\angle O A C+\\mathrm{m} \\angle O B D)-\\mathrm{m} \\widehat{A B}$. Thus $\\mathrm{m} \\angle A E B=\\mathrm{m} \\widehat{A B}-\\mathrm{m} \\angle O A C-\\mathrm{m} \\angle O B D=$ $\\frac{T}{4}-10^{\\circ}-5^{\\circ}$, and with $T=80$, the answer is 5 .']",['5'],False,,Numerical, 2906,Geometry,,"Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units. ","['Note that $\\mathrm{m} \\angle A E B=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-m \\widehat{C D})=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-\\mathrm{m} \\angle C O D)$. Also note that $\\mathrm{m} \\angle C O D=$ $360^{\\circ}-(\\mathrm{m} \\angle A O C+\\mathrm{m} \\angle B O D+\\mathrm{m} \\angle A O B)=360^{\\circ}-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O A C\\right)-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O B D\\right)-$ $\\mathrm{m} \\widehat{A B}=2(\\mathrm{~m} \\angle O A C+\\mathrm{m} \\angle O B D)-\\mathrm{m} \\widehat{A B}$. Thus $\\mathrm{m} \\angle A E B=\\mathrm{m} \\widehat{A B}-\\mathrm{m} \\angle O A C-\\mathrm{m} \\angle O B D=$ $\\frac{T}{4}-10^{\\circ}-5^{\\circ}$, and with $T=80$, the answer is 5 .']",['5'],False,,Numerical, 2907,Algebra,,"Let $T=10$. Ann spends 80 seconds climbing up a $T$ meter rope at a constant speed, and she spends 70 seconds climbing down the same rope at a constant speed (different from her upward speed). Ann begins climbing up and down the rope repeatedly, and she does not pause after climbing the length of the rope. After $T$ minutes, how many meters will Ann have climbed in either direction?","['In 150 seconds (or 2.5 minutes), Ann climbs up and down the entire rope. Thus in $T$ minutes, she makes $\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor$ round trips, and therefore climbs $2 T\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor$ meters. After making all her round trips, there are $t=60\\left(T-2.5\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor\\right)$ seconds remaining. If $t \\leq 80$, then the remaining distance climbed is $T \\cdot \\frac{t}{80}$ meters, and if $t>80$, then the distance climbed is $T+T \\cdot\\left(\\frac{t-80}{70}\\right)$ meters. In general, the total distance in meters that Ann climbs is\n\n$$\n2 T\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor+T \\cdot \\min \\left(1, \\frac{60\\left(T-2.5\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor\\right)}{80}\\right)+T \\cdot \\max \\left(0, \\frac{60\\left(T-2.5\\left\\lfloor\\frac{T}{2.5}\\right\\rfloor\\right)-80}{70}\\right) .\n$$\n\nWith $T=10$, Ann makes exactly 4 round trips, and therefore climbs a total of $4 \\cdot 2 \\cdot 10=\\mathbf{8 0}$ meters.']",['80'],False,,Numerical, 2908,Algebra,,Let $T=800$. Simplify $2^{\log _{4} T} / 2^{\log _{16} 64}$.,"['Note that $2^{\\log _{4} T}=4^{\\left(\\frac{1}{2} \\log _{4} T\\right)}=4^{\\log _{4} T^{\\frac{1}{2}}}=\\sqrt{T}$. Letting $\\log _{16} 64=x$, we see that $2^{4 x}=2^{6}$, thus $x=\\frac{3}{2}$, and $2^{x}=\\sqrt{8}$. Thus the given expression equals $\\sqrt{\\frac{T}{8}}$, and with $T=800$, this is equal to 10 .']",['10'],False,,Numerical, 2909,Algebra,,"Let $P(x)=x^{2}+T x+800$, and let $r_{1}$ and $r_{2}$ be the roots of $P(x)$. The polynomial $Q(x)$ is quadratic, it has leading coefficient 1, and it has roots $r_{1}+1$ and $r_{2}+1$. Find the sum of the coefficients of $Q(x)$.","['Let $Q(x)=x^{2}+A x+B$. Then $A=-\\left(r_{1}+1+r_{2}+1\\right)$ and $B=\\left(r_{1}+1\\right)\\left(r_{2}+1\\right)$. Thus the sum of the coefficients of $Q(x)$ is $1+\\left(-r_{1}-r_{2}-2\\right)+\\left(r_{1} r_{2}+r_{1}+r_{2}+1\\right)=r_{1} r_{2}$. Note that $T=-\\left(r_{1}+r_{2}\\right)$ and $800=r_{1} r_{2}$, so the answer is $\\mathbf{8 0 0}$ (independent of $T$ ). [Note: With $T=108,\\left\\{r_{1}, r_{2}\\right\\}=\\{-8,-100\\}$.']",['800'],False,,Numerical, 2910,Geometry,,"Let $T=12$. Equilateral triangle $A B C$ is given with side length $T$. Points $D$ and $E$ are the midpoints of $\overline{A B}$ and $\overline{A C}$, respectively. Point $F$ lies in space such that $\triangle D E F$ is equilateral and $\triangle D E F$ lies in a plane perpendicular to the plane containing $\triangle A B C$. Compute the volume of tetrahedron $A B C F$.","['The volume of tetrahedron $A B C F$ is one-third the area of $\\triangle A B C$ times the distance from $F$ to $\\triangle A B C$. Since $D$ and $E$ are midpoints, $D E=\\frac{B C}{2}=\\frac{T}{2}$, and the distance from $F$ to $\\triangle A B C$ is $\\frac{T \\sqrt{3}}{4}$. Thus the volume of $A B C F$ is $\\frac{1}{3} \\cdot \\frac{T^{2} \\sqrt{3}}{4} \\cdot \\frac{T \\sqrt{3}}{4}=\\frac{T^{3}}{16}$. With $T=12$, the answer is $\\mathbf{1 0 8}$.']",['108'],False,,Numerical, 2911,Geometry,,"In triangle $A B C, A B=5, A C=6$, and $\tan \angle B A C=-\frac{4}{3}$. Compute the area of $\triangle A B C$.","['Let $s=\\sin \\angle B A C$. Then $s>0$ and $\\frac{s}{-\\sqrt{1-s^{2}}}=-\\frac{4}{3}$, which gives $s=\\frac{4}{5}$. The area of triangle $A B C$ is therefore $\\frac{1}{2} \\cdot A B \\cdot A C \\cdot \\sin \\angle B A C=\\frac{1}{2} \\cdot 5 \\cdot 6 \\cdot \\frac{4}{5}=\\mathbf{1 2}$.']",['12'],False,,Numerical, 2912,Number Theory,,Compute the number of positive integers less than 25 that cannot be written as the difference of two squares of integers.,"['Suppose $n=a^{2}-b^{2}=(a+b)(a-b)$, where $a$ and $b$ are integers. Because $a+b$ and $a-b$ differ by an even number, they have the same parity. Thus $n$ must be expressible as the product of two even integers or two odd integers. This condition is sufficient for $n$ to be a difference of squares, because if $n$ is odd, then $n=(k+1)^{2}-k^{2}=(2 k+1) \\cdot 1$ for some integer $k$, and if $n$ is a multiple of 4 , then $n=(k+1)^{2}-(k-1)^{2}=2 k \\cdot 2$ for some integer $k$. Therefore any integer of the form $4 k+2$ for integral $k$ cannot be expressed as the difference of two squares of integers, hence the desired integers in the given range are $2,6,10,14,18$, and 22 , for a total of 6 values.', 'Suppose that an integer $n$ can be expressed as the difference of squares of two integers, and let the squares be $a^{2}$ and $(a+b)^{2}$, with $a, b \\geq 0$. Then\n\n$$\n\\begin{aligned}\n& n=(a+b)^{2}-a^{2}=2 a b+b^{2} \\\\\n& =2 a+1 \\quad(b=1) \\\\\n& =4 a+4 \\quad(b=2) \\\\\n& =6 a+9 \\quad(b=3) \\\\\n& =8 a+16 \\quad(b=4) \\\\\n& =10 a+25 \\quad(b=5) .\n\\end{aligned}\n$$\n\nSetting $b=1$ generates all odd integers. If $b=3$ or $b=5$, then the values of $n$ are still odd, hence are already accounted for. If $b=2$, then the values of $4 a+4=4(a+1)$ yield all multiples of $4 ; b=8$ yields multiples of 8 (hence are already accounted for). The remaining integers are even numbers that are not multiples of $4: 2,6,10,14,18,22$, for a total of 6 such numbers.']",['6'],False,,Numerical, 2913,Algebra,,"For digits $A, B$, and $C,(\underline{A} \underline{B})^{2}+(\underline{A} \underline{C})^{2}=1313$. Compute $A+B+C$.","['Because $10 A \\leq \\underline{A} \\underline{B}<10(A+1), 200 A^{2}<(\\underline{A} \\underline{B})^{2}+(\\underline{A} \\underline{C})^{2}<200(A+1)^{2}$. So $200 A^{2}<$ $1313<200(A+1)^{2}$, and $A=2$. Note that $B$ and $C$ must have opposite parity, so without loss of generality, assume that $B$ is even. Consider the numbers modulo 10: for any integer $n, n^{2} \\equiv 0,1,4,5,6$, or $9 \\bmod 10$. The only combination whose sum is congruent to $3 \\bmod 10$ is $4+9$. So $B=2$ or 8 and $C=3$ or 7 . Checking cases shows that $28^{2}+23^{2}=1313$, so $B=8, C=3$, and $A+B+C=\\mathbf{1 3}$.', 'Rewrite $1313=13 \\cdot 101=\\left(3^{2}+2^{2}\\right)\\left(10^{2}+1^{2}\\right)$. The two-square identity states:\n\n$$\n\\begin{aligned}\n\\left(a^{2}+b^{2}\\right)\\left(x^{2}+y^{2}\\right) & =(a x+b y)^{2}+(a y-b x)^{2} \\\\\n& =(a y+b x)^{2}+(a x-b y)^{2}\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n1313=(30+2)^{2}+(3-20)^{2} & =32^{2}+17^{2} \\\\\n& =(3+20)^{2}+(30-2)^{2}=23^{2}+28^{2}\n\\end{aligned}\n$$\n\n\n\nHence $A=2, B=3, C=8$, and $A+B+C=\\mathbf{1 3}$.']",['13'],False,,Numerical, 2914,Geometry,,"Points $P, Q, R$, and $S$ lie in the interior of square $A B C D$ such that triangles $A B P, B C Q$, $C D R$, and $D A S$ are equilateral. If $A B=1$, compute the area of quadrilateral $P Q R S$.","['$P Q R S$ is a square with diagonal $\\overline{R P}$. Extend $\\overline{R P}$ to intersect $\\overline{A B}$ and $\\overline{C D}$ at $M$ and $N$ respectively, as shown in the diagram below.\n\n\n\nThen $\\overline{M P}$ is an altitude of $\\triangle A B P$ and $\\overline{R N}$ is an altitude of $\\triangle C D R$. Adding lengths, $M P+R N=M R+2 R P+P N=1+R P$, so $R P=\\sqrt{3}-1$. Therefore $[P Q R S]=\\frac{1}{2}(R P)^{2}=$ $2-\\sqrt{3}$.']",['$2-\\sqrt{3}$'],False,,Numerical, 2915,Algebra,,"For real numbers $\alpha, B$, and $C$, the zeros of $T(x)=x^{3}+x^{2}+B x+C \operatorname{are~}^{2} \alpha$, $\cos ^{2} \alpha$, and $-\csc ^{2} \alpha$. Compute $T(5)$.","['Use the sum of the roots formula to obtain $\\sin ^{2} \\alpha+\\cos ^{2} \\alpha+-\\csc ^{2} \\alpha=-1$, so $\\csc ^{2} \\alpha=2$, and $\\sin ^{2} \\alpha=\\frac{1}{2}$. Therefore $\\cos ^{2} \\alpha=\\frac{1}{2}$. T(x) has leading coefficient 1 , so by the factor theorem, $T(x)=\\left(x-\\frac{1}{2}\\right)\\left(x-\\frac{1}{2}\\right)(x+2)$. Then $T(5)=\\left(5-\\frac{1}{2}\\right)\\left(5-\\frac{1}{2}\\right)(5+2)=\\frac{567}{4}$.']",['$\\frac{567}{4}$'],False,,Numerical, 2916,Geometry,,"Let $\mathcal{R}$ denote the circular region bounded by $x^{2}+y^{2}=36$. The lines $x=4$ and $y=3$ partition $\mathcal{R}$ into four regions $\mathcal{R}_{1}, \mathcal{R}_{2}, \mathcal{R}_{3}$, and $\mathcal{R}_{4}$. $\left[\mathcal{R}_{i}\right]$ denotes the area of region $\mathcal{R}_{i}$. If $\left[\mathcal{R}_{1}\right]>\left[\mathcal{R}_{2}\right]>\left[\mathcal{R}_{3}\right]>\left[\mathcal{R}_{4}\right]$, compute $\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]-\left[\mathcal{R}_{3}\right]+\left[\mathcal{R}_{4}\right]$.","['Draw the lines $x=-4$ and $y=-3$, creating regions $\\mathcal{R}_{21}, \\mathcal{R}_{22}, \\mathcal{R}_{11}, \\mathcal{R}_{12}, \\mathcal{R}_{13}, \\mathcal{R}_{14}$ as shown below.\n\n\n\n\n\nThen $\\left[\\mathcal{R}_{21}\\right]=\\left[\\mathcal{R}_{4}\\right]=\\left[\\mathcal{R}_{13}\\right],\\left[\\mathcal{R}_{22}\\right]=\\left[\\mathcal{R}_{14}\\right]$, and $\\left[\\mathcal{R}_{3}\\right]=\\left[\\mathcal{R}_{12}\\right]+\\left[\\mathcal{R}_{13}\\right]$. Therefore\n\n$$\n\\begin{aligned}\n{\\left[\\mathcal{R}_{1}\\right]-\\left[\\mathcal{R}_{2}\\right]-\\left[\\mathcal{R}_{3}\\right]+\\left[\\mathcal{R}_{4}\\right] } & =\\left(\\left[\\mathcal{R}_{1}\\right]-\\left[\\mathcal{R}_{2}\\right]\\right)-\\left(\\left[\\mathcal{R}_{3}\\right]-\\left[\\mathcal{R}_{4}\\right]\\right) \\\\\n& =\\left(\\left[\\mathcal{R}_{1}\\right]-\\left[\\mathcal{R}_{13}\\right]-\\left[\\mathcal{R}_{14}\\right]\\right)-\\left(\\left[\\mathcal{R}_{12}\\right]+\\left[\\mathcal{R}_{13}\\right]-\\left[\\mathcal{R}_{21}\\right]\\right) \\\\\n& =\\left(\\left[\\mathcal{R}_{11}\\right]+\\left[\\mathcal{R}_{12}\\right]\\right)-\\left[\\mathcal{R}_{12}\\right] \\\\\n& =\\left[\\mathcal{R}_{11}\\right] .\n\\end{aligned}\n$$\n\nThis last region is simply a rectangle of height 6 and width 8 , so its area is 48 .']",['48'],False,,Numerical, 2917,Algebra,,"Let $x$ be a real number in the interval $[0,360]$ such that the four expressions $\sin x^{\circ}, \cos x^{\circ}$, $\tan x^{\circ}, \cot x^{\circ}$ take on exactly three distinct (finite) real values. Compute the sum of all possible values of $x$.","['If the four expressions take on three different values, exactly two of the expressions must have equal values. There are $\\left(\\begin{array}{l}4 \\\\ 2\\end{array}\\right)=6$ cases to consider:\n\nCase 1: $\\sin x^{\\circ}=\\cos x^{\\circ}$ : Then $\\tan x^{\\circ}=\\cot x^{\\circ}=1$, violating the condition that there be three distinct values.\n\nCase 2: $\\sin x^{\\circ}=\\tan x^{\\circ}$ : Because $\\tan x^{\\circ}=\\frac{\\sin x^{\\circ}}{\\cos x^{\\circ}}$, either $\\cos x^{\\circ}=1$ or $\\sin x^{\\circ}=0$. However, in both of these cases, $\\cot x^{\\circ}$ is undefined, so it does not have a real value.\n\nCase 3: $\\sin x^{\\circ}=\\cot x^{\\circ}$ : Then $\\sin x^{\\circ}=\\frac{\\cos x^{\\circ}}{\\sin x^{\\circ}}$, and so $\\sin ^{2} x^{\\circ}=\\cos x^{\\circ}$. Rewrite using the Pythagorean identity to obtain $\\cos ^{2} x^{\\circ}+\\cos x^{\\circ}-1=0$, so $\\cos x^{\\circ}=\\frac{-1+\\sqrt{5}}{2}$ (the other root is outside the range of $\\cos )$. Because $\\cos x^{\\circ}>0$, this equation has two solutions in $[0,360]$ : an angle $x_{0}^{\\circ}$ in the first quadrant and the angle $\\left(360-x_{0}\\right)^{\\circ}$ in the fourth quadrant. The sum of these two values is 360 .\n\nCase 4: $\\cos x^{\\circ}=\\tan x^{\\circ}$ : Use similar logic as in the previous case to obtain the equation $\\sin ^{2} x^{\\circ}+$ $\\sin x^{\\circ}-1=0$, so now $\\sin x^{\\circ}=\\frac{-1+\\sqrt{5}}{2}$. Because $\\sin x^{\\circ}>0$, this equation has two solutions, one an angle $x_{0}^{\\circ}$ in the first quadrant, and the other its supplement $\\left(180-x_{0}\\right)^{\\circ}$ in the second quadrant. The sum of these two values is 180 .\n\nCase 5: $\\cos x^{\\circ}=\\cot x^{\\circ}$ : In this case, $\\tan x^{\\circ}$ is undefined for reasons analogous to those in Case 2.\n\nCase 6: $\\tan x^{\\circ}=\\cot x^{\\circ}$ : Thus $\\tan ^{2} x^{\\circ}=1$, hence $\\tan x^{\\circ}= \\pm 1$. If $\\tan x^{\\circ}=1$, then $\\sin x^{\\circ}=\\cos x^{\\circ}$, which yields only two distinct values. So $\\tan x^{\\circ}=-1$, which occurs at $x=135$ and $x=315$. The sum of these values is 450 .\n\nThe answer is $360+180+450=\\mathbf{9 9 0}$.', 'Consider the graphs of all four functions; notice first that 0, 90, 180, 270 are not solutions because either $\\tan x^{\\circ}$ or $\\cot x^{\\circ}$ is undefined at each value.\n\n\n\n\n\nStart in the first quadrant. Let $x_{1}$ and $x_{2}$ be the values of $x$ such that $\\cos x^{\\circ}=\\tan x^{\\circ}$ and $\\sin x^{\\circ}=\\cot ^{\\circ}$, respectively, labeled $A$ and $B$ in the diagram. Because $\\cos x^{\\circ}=\\sin (90-x)^{\\circ}$ and $\\cot x^{\\circ}=\\tan (90-x)^{\\circ}, x_{1}+x_{2}=90$. One can also see that the graphs of $y=\\cot x^{\\circ}$ and $y=\\tan x^{\\circ} \\operatorname{cross}$ at $x=45$, but so do the graphs of $y=\\sin x^{\\circ}$ and $y=\\cos x^{\\circ}$. So at $x=45$, there are only two distinct values, not three.\n\n\n\nIn the second quadrant, $\\tan x^{\\circ}=\\cot x^{\\circ}$ when $x=135$. Also, because $\\tan x^{\\circ}$ increases from $-\\infty$ to 0 while $\\cos x^{\\circ}$ decreases from 0 to -1 , there exists a number $x_{3}$ such that $\\tan x_{3}^{\\circ}=\\cos x_{3}^{\\circ}$ (marked point $C$ in the diagram above).\n\n\n\n\n\nIn the third quadrant, $\\tan x^{\\circ}$ and $\\cot x^{\\circ}$ are positive, while $\\sin x^{\\circ}$ and $\\cos x^{\\circ}$ are negative; the only place where graphs cross is at $x=225$, but this value is not a solution because the four trigonometric functions have only two distinct values.\n\n\n\nIn the fourth quadrant, $\\tan x^{\\circ}=\\cot x^{\\circ}=-1$ when $x=315$. Because $\\sin x^{\\circ}$ is increasing from -1 to 0 while $\\cot x^{\\circ}$ is decreasing from 0 to $-\\infty$, there exists a number $x_{4}$ such that $\\sin x_{4}^{\\circ}=\\cot x_{4}^{\\circ}$ (marked $D$ in the diagram above). Because $\\cos x^{\\circ}=\\sin (90-x)^{\\circ}=\\sin (450-x)^{\\circ}$ and $\\cot x^{\\circ}=\\tan (90-x)^{\\circ}=\\tan (450-x)^{\\circ}$, the values $x_{3}$ and $x_{4}$ are symmetrical around $x=225$, that is, $x_{3}+x_{4}=450$.\n\nThe sum is $\\left(x_{1}+x_{2}\\right)+(135+315)+\\left(x_{3}+x_{4}\\right)=90+450+450=\\mathbf{9 9 0}$.']",['990'],False,,Numerical, 2918,Algebra,,"Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic sequence, and let $b_{1}, b_{2}, b_{3}, \ldots$ be a geometric sequence. The sequence $c_{1}, c_{2}, c_{3}, \ldots$ has $c_{n}=a_{n}+b_{n}$ for each positive integer $n$. If $c_{1}=1, c_{2}=4, c_{3}=15$, and $c_{4}=2$, compute $c_{5}$.","['Let $a_{2}-a_{1}=d$ and $\\frac{b_{2}}{b_{1}}=r$. Using $a=a_{1}$ and $b=b_{1}$, write the system of equations:\n\n$$\n\\begin{aligned}\na+b & =1 \\\\\n(a+d)+b r & =4 \\\\\n(a+2 d)+b r^{2} & =15 \\\\\n(a+3 d)+b r^{3} & =2 .\n\\end{aligned}\n$$\n\nSubtract the first equation from the second, the second from the third, and the third from the fourth to obtain three equations:\n\n$$\n\\begin{aligned}\nd+b(r-1) & =3 \\\\\nd+b\\left(r^{2}-r\\right) & =11 \\\\\nd+b\\left(r^{3}-r^{2}\\right) & =-13\n\\end{aligned}\n$$\n\nNotice that the $a$ terms have canceled. Repeat to find the second differences:\n\n$$\n\\begin{aligned}\nb\\left(r^{2}-2 r+1\\right) & =8 \\\\\nb\\left(r^{3}-2 r^{2}+r\\right) & =-24\n\\end{aligned}\n$$\n\nNow divide the second equation by the first to obtain $r=-3$. Substituting back into either of these two last equations yields $b=\\frac{1}{2}$. Continuing in the same vein yields $d=5$ and $a=\\frac{1}{2}$. Then $a_{5}=\\frac{41}{2}$ and $b_{5}=\\frac{81}{2}$, so $c_{5}=\\mathbf{6 1}$.']",['61'],False,,Numerical, 2919,Geometry,,"In square $A B C D$ with diagonal $1, E$ is on $\overline{A B}$ and $F$ is on $\overline{B C}$ with $\mathrm{m} \angle B C E=\mathrm{m} \angle B A F=$ $30^{\circ}$. If $\overline{C E}$ and $\overline{A F}$ intersect at $G$, compute the distance between the incenters of triangles $A G E$ and $C G F$.","['Let $M$ be the midpoint of $\\overline{A G}$, and $I$ the incenter of $\\triangle A G E$ as shown below.\n\n\n\nBecause $\\frac{A B}{A C}=\\sin 45^{\\circ}$ and $\\frac{E B}{A B}=\\frac{E B}{B C}=\\tan 30^{\\circ}$,\n\n$$\n\\begin{aligned}\nA E & =A B-E B=A B\\left(1-\\tan 30^{\\circ}\\right) \\\\\n& =\\sin 45^{\\circ}\\left(1-\\tan 30^{\\circ}\\right) \\\\\n& =\\frac{\\sin 45^{\\circ} \\cos 30^{\\circ}-\\cos 45^{\\circ} \\sin 30^{\\circ}}{\\cos 30^{\\circ}} \\\\\n& =\\frac{\\sin \\left(45^{\\circ}-30^{\\circ}\\right)}{\\cos 30^{\\circ}} \\\\\n& =\\frac{\\sin 15^{\\circ}}{\\cos 30^{\\circ}} .\n\\end{aligned}\n$$\n\n\n\nNote that $\\frac{A M}{A E}=\\cos 30^{\\circ}$ and $\\frac{A M}{A I}=\\cos 15^{\\circ}$. Therefore\n\n$$\n\\begin{aligned}\n\\frac{A I}{A E} & =\\frac{\\cos 30^{\\circ}}{\\cos 15^{\\circ}} \\\\\n& =\\frac{\\sin 60^{\\circ}}{\\cos 15^{\\circ}} \\\\\n& =\\frac{2 \\sin 30^{\\circ} \\cos 30^{\\circ}}{\\cos 15^{\\circ}} \\\\\n& =\\frac{2\\left(2 \\sin 15^{\\circ} \\cos 15^{\\circ}\\right) \\cos 30^{\\circ}}{\\cos 15^{\\circ}} \\\\\n& =4 \\sin 15^{\\circ} \\cos 30^{\\circ} .\n\\end{aligned}\n$$\n\nThus $A I=\\left(4 \\sin 15^{\\circ} \\cos 30^{\\circ}\\right)\\left(\\frac{\\sin 15^{\\circ}}{\\cos 30^{\\circ}}\\right)=4 \\sin ^{2} 15^{\\circ}=4 \\cdot\\left(\\frac{1-\\cos 30^{\\circ}}{2}\\right)=2-\\sqrt{3}$. Finally, the desired distance is $2 I G=2 A I=4-2 \\sqrt{3}$.']",['$4-2 \\sqrt{3}$'],False,,Numerical, 2920,Geometry,,"Let $a, b, m, n$ be positive integers with $a m=b n=120$ and $a \neq b$. In the coordinate plane, let $A=(a, m), B=(b, n)$, and $O=(0,0)$. If $X$ is a point in the plane such that $A O B X$ is a parallelogram, compute the minimum area of $A O B X$.","['The area of parallelogram $A O B X$ is given by the absolute value of the cross product $|\\langle a, m\\rangle \\times\\langle b, n\\rangle|=|a n-m b|$. Because $m=\\frac{120}{a}$ and $n=\\frac{120}{b}$, the desired area of $A O B X$ equals $120\\left|\\frac{a}{b}-\\frac{b}{a}\\right|$. Note that the function $f(x)=x-\\frac{1}{x}$ is monotone increasing for $x>1$. (Proof: if $x_{1}>x_{2}>0$, then $f\\left(x_{1}\\right)-f\\left(x_{2}\\right)=\\left(x_{1}-x_{2}\\right)+\\frac{x_{1}-x_{2}}{x_{1} x_{2}}$, where both terms are positive because $x_{1} x_{2}>0$.) So the minimum value of $[A O B X]$ is attained when $\\frac{a}{b}$ is as close as possible to 1 , that is, when $a$ and $b$ are consecutive divisors of 120. By symmetry, consider only $a\n\nBecause the hyperbola is concave up, $[O A C]+[O C B]<[O A B]$, so in particular, $[O A C]<$ $[O A B]$. Thus, if $[O A B]$ is minimal, there can be no point $C$ with integer coordinates between $A$ and $B$ on the hyperbola.']",['44'],False,,Numerical, 2921,Combinatorics,,"Let $\mathcal{S}$ be the set of integers from 0 to 9999 inclusive whose base- 2 and base- 5 representations end in the same four digits. (Leading zeros are allowed, so $1=0001_{2}=0001_{5}$ is one such number.) Compute the remainder when the sum of the elements of $\mathcal{S}$ is divided by 10,000.","[""The remainders of an integer $N$ modulo $2^{4}=16$ and $5^{4}=625$ uniquely determine its remainder modulo 10000. There are only 16 strings of four 0's and 1's. In addition, because 16 and 625 are relatively prime, it will be shown below that for each such string $s$, there exists exactly one integer $x_{s}$ in the range $0 \\leq x_{s}<10000$ such that the base- 2 and base- 5 representations of $x_{s}$ end in the digits of $s$ (e.g., $x_{1001}$ is the unique positive integer less than 10000 such that $x$ 's base- 5 representation and base- 2 representation both end in 1001).\n\nHere is a proof of the preceding claim: Let $p(s)$ be the number whose digits in base 5 are the string $s$, and $b(s)$ be the number whose digits in base 2 are the string $s$. Then the system $x \\equiv$ $p(s) \\bmod 625$ and $x \\equiv b(s) \\bmod 16$ can be rewritten as $x=p(s)+625 m$ and $x=b(s)+16 n$ for integers $m$ and $n$. These reduce to the Diophantine equation $16 n-625 m=p(s)-b(s)$, which has solutions $m, n$ in $\\mathbb{Z}$, with at least one of $m, n \\geq 0$. Assuming without loss of generality that $m>0$ yields $x=p(s)+625 m \\geq 0$. To show that there exists an $x_{s}<10000$ and that it is unique, observe that the general form of the solution is $m^{\\prime}=m-16 t, n^{\\prime}=n+625 t$. Thus if $p(s)+625 m>10000$, an appropriate $t$ can be found by writing $0 \\leq p(s)+625(m-16 t)<10000$, which yields $p(s)+625 m-10000<10000 t \\leq p(s)+625 m$. Because there are exactly 10000 integers in that interval, exactly one of them is divisible by 10000 , so there is exactly one value of $t$ satisfying $0 \\leq p(s)+625(m-16 t)<10000$, and set $x_{s}=625(m-16 t)$.\n\nTherefore there will be 16 integers whose base- 2 and base- 5 representations end in the same four digits, possibly with leading 0 's as in the example. Let $X=x_{0000}+\\cdots+x_{1111}$. Then $X$ is congruent modulo 16 to $0000_{2}+\\cdots+1111_{2}=8 \\cdot\\left(1111_{2}\\right)=8 \\cdot 15 \\equiv 8$. Similarly, $X$ is congruent modulo 625 to $0000_{5}+\\cdots+1111_{5}=8 \\cdot 1111_{5}=2 \\cdot 4444_{5} \\equiv 2 \\cdot(-1)=-2$.\n\nSo $X$ must be $8(\\bmod 16)$ and $-2(\\bmod 625)$. Noticing that $625 \\equiv 1(\\bmod 16)$, conclude that the answer is $-2+10 \\cdot 625=\\mathbf{6 2 4 8}$.""]",['6248'],False,,Numerical, 2922,Number Theory,,"If $A, R, M$, and $L$ are positive integers such that $A^{2}+R^{2}=20$ and $M^{2}+L^{2}=10$, compute the product $A \cdot R \cdot M \cdot L$.","['The only positive integers whose squares sum to 20 are 2 and 4 . The only positive integers whose squares sum to 10 are 1 and 3 . Thus $A \\cdot R=8$ and $M \\cdot L=3$, so $A \\cdot R \\cdot M \\cdot L=\\mathbf{2 4}$.']",['24'],False,,Numerical, 2923,Geometry,,"Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$. ![](https://cdn.mathpix.com/cropped/2023_12_21_daa62eb0f647b311d7dag-1.jpg?height=352&width=290&top_left_y=732&top_left_x=1357)","['Let $\\mathrm{m} \\angle A=x$. Then $\\mathrm{m} \\angle P=\\mathrm{m} \\angle Q=T x$, and $(2 T+1) x=180^{\\circ}$, so $x=\\frac{180^{\\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_44186b4d618c78311efdg-1.jpg?height=439&width=393&top_left_y=710&top_left_x=909)\n\nThen $\\mathrm{m} \\angle P O Q=2 \\mathrm{~m} \\angle P A Q=2\\left(\\frac{180^{\\circ}}{2 T+1}\\right)=\\frac{360^{\\circ}}{2 T+1}$. Because $\\mathrm{m} \\angle P O Q=\\frac{360^{\\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\\mathbf{4 9}$.']",['49'],False,,Numerical, 2923,Geometry,,"Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$. ","['Let $\\mathrm{m} \\angle A=x$. Then $\\mathrm{m} \\angle P=\\mathrm{m} \\angle Q=T x$, and $(2 T+1) x=180^{\\circ}$, so $x=\\frac{180^{\\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.\n\n\n\nThen $\\mathrm{m} \\angle P O Q=2 \\mathrm{~m} \\angle P A Q=2\\left(\\frac{180^{\\circ}}{2 T+1}\\right)=\\frac{360^{\\circ}}{2 T+1}$. Because $\\mathrm{m} \\angle P O Q=\\frac{360^{\\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\\mathbf{4 9}$.']",['49'],False,,Numerical, 2924,Number Theory,,"Let $T=49$. Compute the last digit, in base 10, of the sum $$ T^{2}+(2 T)^{2}+(3 T)^{2}+\ldots+\left(T^{2}\right)^{2} $$","['Let $S$ be the required sum. Factoring $T^{2}$ from the sum yields\n\n$$\n\\begin{aligned}\nS & =T^{2}\\left(1+4+9+\\ldots+T^{2}\\right) \\\\\n& =T^{2}\\left(\\frac{T(T+1)(2 T+1)}{6}\\right) \\\\\n& =\\frac{T^{3}(T+1)(2 T+1)}{6} .\n\\end{aligned}\n$$\n\nFurther analysis makes the final computation simpler. If $T \\equiv 0,2$, or $3 \\bmod 4$, then $S$ is even. Otherwise, $S$ is odd. And if $T \\equiv 0,2$, or $4 \\bmod 5$, then $S \\equiv 0 \\bmod 5$; otherwise, $S \\equiv 1 \\bmod 5$. These observations yield the following table:\n\n| $T \\bmod 4$ | $T \\bmod 5$ | $S \\bmod 10$ |\n| :---: | :---: | :---: |\n| $0,2,3$ | $0,2,4$ | 0 |\n| $0,2,3$ | 1,3 | 6 |\n| 1 | $0,2,4$ | 5 |\n| 1 | 1,3 | 1 |\n\nBecause $T=49$, the value corresponds to the third case above; the last digit is $\\mathbf{5}$.']",['5'],False,,Numerical, 2925,Combinatorics,,A fair coin is flipped $n$ times. Compute the smallest positive integer $n$ for which the probability that the coin has the same result every time is less than $10 \%$.,"['After the first throw, the probability that the succeeding $n-1$ throws have the same result is $\\frac{1}{2^{n-1}}$. Thus $\\frac{1}{2^{n-1}}<\\frac{1}{10} \\Rightarrow 2^{n-1}>10 \\Rightarrow n-1 \\geq 4$, so $n=5$ is the smallest possible value.']",['5'],False,,Numerical, 2926,Algebra,,Let $T=5$. Compute the smallest positive integer $n$ such that there are at least $T$ positive integers in the domain of $f(x)=\sqrt{-x^{2}-2 x+n}$.,"['Completing the square under the radical yields $\\sqrt{n+1-(x+1)^{2}}$. The larger zero of the radicand is $-1+\\sqrt{n+1}$, and the smaller zero is negative because $-1-\\sqrt{n+1}<0$, so the $T$ positive integers in the domain of $f$ must be $1,2,3, \\ldots, T$. Therefore $-1+\\sqrt{n+1} \\geq T$. Hence $\\sqrt{n+1} \\geq T+1$, and $n+1 \\geq(T+1)^{2}$. Therefore $n \\geq T^{2}+2 T$, and substituting $T=5$ yields $n \\geq 35$. So $n=\\mathbf{3 5}$ is the smallest such value.']",['35'],False,,Numerical, 2927,Algebra,,Let $T=35$. Compute the smallest positive real number $x$ such that $\frac{\lfloor x\rfloor}{x-\lfloor x\rfloor}=T$.,"['If $\\frac{\\lfloor x\\rfloor}{x-\\lfloor x\\rfloor}=T$, the equation can be rewritten as follows:\n\n$$\n\\begin{aligned}\n\\frac{x-\\lfloor x\\rfloor}{\\lfloor x\\rfloor} & =\\frac{1}{T} \\\\\n\\frac{x}{\\lfloor x\\rfloor}-1 & =\\frac{1}{T} \\\\\n\\frac{x}{\\lfloor x\\rfloor} & =\\frac{T+1}{T} .\n\\end{aligned}\n$$\n\nNow $00$. Let these plots have radius $r$. Because $P_{1}$ is tangent to $U, y+r=1$. Because $P_{1}$ is tangent to $C$, the distance from $(0,1-r)$ to $\\left(\\frac{1}{2}, 0\\right)$ is $r+\\frac{1}{2}$. Therefore\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{2}\\right)^{2}+(1-r)^{2} & =\\left(r+\\frac{1}{2}\\right)^{2} \\\\\n1-2 r & =r \\\\\nr & =\\frac{1}{3} \\\\\ny=1-r & =\\frac{2}{3} .\n\\end{aligned}\n$$\n\nThus the plots are centered at $\\left(0, \\pm \\frac{2}{3}\\right)$ and have radius $\\frac{1}{3}$.']",['证明题，略'],True,,Need_human_evaluate, 2928,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Without using Descartes' Circle Formula, Show that the circles marked off and sold on day 1 are centered at $\left(0, \pm \frac{2}{3}\right)$ with radius $\frac{1}{3}$.","['By symmetry, $P_{1}, P_{2}$, the two plots sold on day 1 , are centered on the $y$-axis, say at $(0, \\pm y)$ with $y>0$. Let these plots have radius $r$. Because $P_{1}$ is tangent to $U, y+r=1$. Because $P_{1}$ is tangent to $C$, the distance from $(0,1-r)$ to $\\left(\\frac{1}{2}, 0\\right)$ is $r+\\frac{1}{2}$. Therefore\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{2}\\right)^{2}+(1-r)^{2} & =\\left(r+\\frac{1}{2}\\right)^{2} \\\\\n1-2 r & =r \\\\\nr & =\\frac{1}{3} \\\\\ny=1-r & =\\frac{2}{3} .\n\\end{aligned}\n$$\n\nThus the plots are centered at $\\left(0, \\pm \\frac{2}{3}\\right)$ and have radius $\\frac{1}{3}$.']",,True,,, 2929,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Without using Descartes' Circle Formula, Find the combined area of the six remaining curvilinear territories after day 1.","['The four ""removed"" circles have radii $\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{3}$ so the combined area of the six remaining curvilinear territories is:\n\n$$\n\\pi\\left(1^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}\\right)=\\frac{5 \\pi}{18}\n$$']",['$\\frac{5 \\pi}{18}$'],False,,Numerical, 2929,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Without using Descartes' Circle Formula, Find the combined area of the six remaining curvilinear territories after day 1.","['The four ""removed"" circles have radii $\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{3}$ so the combined area of the six remaining curvilinear territories is:\n\n$$\n\\pi\\left(1^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}\\right)=\\frac{5 \\pi}{18}\n$$']",['$\\frac{5 \\pi}{18}$'],False,,Numerical, 2930,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Determine the number of curvilinear territories remaining at the end of day 3.","[""At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3.""]",['54'],False,,Numerical, 2930,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Determine the number of curvilinear territories remaining at the end of day 3.","[""At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3.""]",['54'],False,,Numerical, 2931,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $X_{n}$ be the number of plots sold on day $n$. Find a formula for $X_{n}$ in terms of $n$.","['Each day, every curvilinear territory is divided into three smaller curvilinear territories. Let $R_{n}$ be the number of regions at the end of day $n$. Then $R_{0}=2$, and $R_{n+1}=3 \\cdot R_{n}$. Thus $R_{n}$ is a geometric sequence, so $R_{n}=2 \\cdot 3^{n}$. For $n>0$, the number of plots sold on day $n$ equals the number of territories existing at the end of day $n-1$, i.e., $X_{n}=R_{n-1}$, so $X_{0}=X_{1}=2$, and for $n>1, X_{n}=2 \\cdot 3^{n-1}$.']","['$X_{0}=X_{1}=2$, and for $n>1, X_{n}=2 \\cdot 3^{n-1}$']",True,,Need_human_evaluate, 2932,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Determine the total number of plots sold up to and including day $n$.","['The total number of plots sold up to and including day $n$ is\n\n$$\n\\begin{aligned}\n2+\\sum_{k=1}^{n} X_{k} & =2+2 \\sum_{k=1}^{n} 3^{k-1} \\\\\n& =2+2 \\cdot\\left(1+3+3^{2}+\\ldots+3^{n-1}\\right) \\\\\n& =3^{n}+1\n\\end{aligned}\n$$\n\nAlternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \\geq 0$,\n\n$$\n\\begin{aligned}\n\\left(3^{n}+1\\right)+X_{n+1} & =\\left(3^{n}+1\\right)+2 \\cdot 3^{n} \\\\\n& =3 \\cdot 3^{n}+1 \\\\\n& =3^{n+1}+1 .\n\\end{aligned}\n$$']",['$3^{n}+1$'],False,,Expression, 2932,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Determine the total number of plots sold up to and including day $n$.","['The total number of plots sold up to and including day $n$ is\n\n$$\n\\begin{aligned}\n2+\\sum_{k=1}^{n} X_{k} & =2+2 \\sum_{k=1}^{n} 3^{k-1} \\\\\n& =2+2 \\cdot\\left(1+3+3^{2}+\\ldots+3^{n-1}\\right) \\\\\n& =3^{n}+1\n\\end{aligned}\n$$\n\nAlternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \\geq 0$,\n\n$$\n\\begin{aligned}\n\\left(3^{n}+1\\right)+X_{n+1} & =\\left(3^{n}+1\\right)+2 \\cdot 3^{n} \\\\\n& =3 \\cdot 3^{n}+1 \\\\\n& =3^{n+1}+1 .\n\\end{aligned}\n$$']",['$3^{n}+1$'],False,,Expression, 2933,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Two unit circles and a circle of radius $\frac{2}{3}$ are mutually externally tangent. Compute all possible values of $r$ such that a circle of radius $r$ is tangent to all three circles.","[""Use Descartes' Circle Formula with $a=b=1$ and $c=\\frac{3}{2}$ to solve for $d$ :\n\n$$\n\\begin{aligned}\n2 \\cdot\\left(1^{2}+1^{2}+\\left(\\frac{3}{2}\\right)^{2}+d^{2}\\right) & =\\left(1+1+\\frac{3}{2}+d\\right)^{2} \\\\\n\\frac{17}{2}+2 d^{2} & =\\frac{49}{4}+7 d+d^{2} \\\\\nd^{2}-7 d-\\frac{15}{4} & =0\n\\end{aligned}\n$$\n\nfrom which $d=\\frac{15}{2}$ or $d=-\\frac{1}{2}$. These values correspond to radii of $\\frac{2}{15}$, a small circle nestled between the other three, or 2 , a large circle enclosing the other three.\n\nAlternatively, start by scaling the kingdom with the first four circles removed to match the situation given. Thus the three given circles are internally tangent to a circle of radius $r=2$ and curvature $d=-\\frac{1}{2}$. Descartes' Circle Formula gives a quadratic equation for $d$, and the sum of the roots is $2 \\cdot\\left(1+1+\\frac{3}{2}\\right)=7$, so the second root is $7+\\frac{1}{2}=\\frac{15}{2}$, corresponding to a circle of radius $r=\\frac{2}{15}$.""]","['$2$, $\\frac{2}{15}$']",True,,Numerical, 2933,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Two unit circles and a circle of radius $\frac{2}{3}$ are mutually externally tangent. Compute all possible values of $r$ such that a circle of radius $r$ is tangent to all three circles.","[""Use Descartes' Circle Formula with $a=b=1$ and $c=\\frac{3}{2}$ to solve for $d$ :\n\n$$\n\\begin{aligned}\n2 \\cdot\\left(1^{2}+1^{2}+\\left(\\frac{3}{2}\\right)^{2}+d^{2}\\right) & =\\left(1+1+\\frac{3}{2}+d\\right)^{2} \\\\\n\\frac{17}{2}+2 d^{2} & =\\frac{49}{4}+7 d+d^{2} \\\\\nd^{2}-7 d-\\frac{15}{4} & =0\n\\end{aligned}\n$$\n\nfrom which $d=\\frac{15}{2}$ or $d=-\\frac{1}{2}$. These values correspond to radii of $\\frac{2}{15}$, a small circle nestled between the other three, or 2 , a large circle enclosing the other three.\n\nAlternatively, start by scaling the kingdom with the first four circles removed to match the situation given. Thus the three given circles are internally tangent to a circle of radius $r=2$ and curvature $d=-\\frac{1}{2}$. Descartes' Circle Formula gives a quadratic equation for $d$, and the sum of the roots is $2 \\cdot\\left(1+1+\\frac{3}{2}\\right)=7$, so the second root is $7+\\frac{1}{2}=\\frac{15}{2}$, corresponding to a circle of radius $r=\\frac{2}{15}$.""]","['$2$, $\\frac{2}{15}$']",True,,Numerical, 2934,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Given three mutually tangent circles with curvatures $a, b, c>0$, suppose that $(a, b, c, 0)$ does not satisfy Descartes' Circle Formula. Show that there are two distinct values of $r$ such that there is a circle of radius $r$ tangent to the given circles.","[""Apply Descartes' Circle Formula to yield\n\n$$\n(a+b+c+x)^{2}=2 \\cdot\\left(a^{2}+b^{2}+c^{2}+x^{2}\\right),\n$$\n\na quadratic equation in $x$. Expanding and rewriting in standard form yields the equation\n\n$$\nx^{2}-p x+q=0\n$$\n\nwhere $p=2(a+b+c)$ and $q=2\\left(a^{2}+b^{2}+c^{2}\\right)-(a+b+c)^{2}$.\n\nThe discriminant of this quadratic is\n\n$$\n\\begin{aligned}\np^{2}-4 q & =8(a+b+c)^{2}-8\\left(a^{2}+b^{2}+c^{2}\\right) \\\\\n& =16(a b+a c+b c)\n\\end{aligned}\n$$\n\nThis last expression is positive because it is given that $a, b, c>0$. Therefore the quadratic has two distinct real roots, say $d_{1}$ and $d_{2}$. These usually correspond to two distinct radii, $r_{1}=\\frac{1}{\\left|d_{1}\\right|}$ and $r_{2}=\\frac{1}{\\left|d_{2}\\right|}$.\n\nOne possible exception case to consider is if $r_{2}=r_{1}$, which would occur if $d_{2}=-d_{1}$. This case can be ruled out because $d_{1}+d_{2}=p=2(a+b+c)$, which must be positive if $a, b, c>0$. (Notice too that this inequality rules out the possibility that both circles have negative curvature, so that there cannot be two distinct circles to which the given circles are internally tangent.)\n\nWhen both roots $d_{1}$ and $d_{2}$ are positive, the three given circles are externally tangent to both fourth circles. When one is positive and one is negative, the three given circles are internally tangent to one circle and externally tangent to the other.\n\n\n\nWhile the foregoing answers the question posed, it is interesting to examine the result from a geometric perspective: why are there normally two possible fourth circles? Consider the case when one of $a, b$, and $c$ is negative (i.e., two circles are internally tangent to a third). Let $A$ and $B$ be circles internally tangent to $C$. Then $A$ and $B$ partition the remaining area of $C$ into two c-triangles, each of which has an incircle, providing the two solutions.\n\nIf, as in the given problem, $a, b, c>0$, then all three circles $A, B$, and $C$, are mutually externally tangent. In this case, the given circles bound a c-triangle, which has an incircle, corresponding to one of the two roots. The complementary arcs of the given circles bound an infinite region, and this region normally contains a second circle tangent to the given circles. To demonstrate this fact geometrically, consider shrink-wrapping the circles: the shrink-wrap is the border of the smallest convex region containing all three circles. (This region is called the convex hull of the circles). There are two cases to address. If only two circles are touched by the shrink-wrap, then one circle is wedged between two larger ones and completely enclosed by their common tangents. In such a case, a circle can be drawn so that it is tangent to all three circles as shown in the diagram below (shrink-wrap in bold; locations of fourth circle marked at $D_{1}$ and $D_{2}$ ).\n\n\n\nOn the other hand, if the shrink-wrap touches all three circles, then it can be expanded to make a circle tangent to and containing $A, B$, and $C$, as shown below.\n\n\n\n""]",,True,,, 2934,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Given three mutually tangent circles with curvatures $a, b, c>0$, suppose that $(a, b, c, 0)$ does not satisfy Descartes' Circle Formula. Show that there are two distinct values of $r$ such that there is a circle of radius $r$ tangent to the given circles.","[""Apply Descartes' Circle Formula to yield\n\n$$\n(a+b+c+x)^{2}=2 \\cdot\\left(a^{2}+b^{2}+c^{2}+x^{2}\\right),\n$$\n\na quadratic equation in $x$. Expanding and rewriting in standard form yields the equation\n\n$$\nx^{2}-p x+q=0\n$$\n\nwhere $p=2(a+b+c)$ and $q=2\\left(a^{2}+b^{2}+c^{2}\\right)-(a+b+c)^{2}$.\n\nThe discriminant of this quadratic is\n\n$$\n\\begin{aligned}\np^{2}-4 q & =8(a+b+c)^{2}-8\\left(a^{2}+b^{2}+c^{2}\\right) \\\\\n& =16(a b+a c+b c)\n\\end{aligned}\n$$\n\nThis last expression is positive because it is given that $a, b, c>0$. Therefore the quadratic has two distinct real roots, say $d_{1}$ and $d_{2}$. These usually correspond to two distinct radii, $r_{1}=\\frac{1}{\\left|d_{1}\\right|}$ and $r_{2}=\\frac{1}{\\left|d_{2}\\right|}$.\n\nOne possible exception case to consider is if $r_{2}=r_{1}$, which would occur if $d_{2}=-d_{1}$. This case can be ruled out because $d_{1}+d_{2}=p=2(a+b+c)$, which must be positive if $a, b, c>0$. (Notice too that this inequality rules out the possibility that both circles have negative curvature, so that there cannot be two distinct circles to which the given circles are internally tangent.)\n\nWhen both roots $d_{1}$ and $d_{2}$ are positive, the three given circles are externally tangent to both fourth circles. When one is positive and one is negative, the three given circles are internally tangent to one circle and externally tangent to the other.\n\n\n\nWhile the foregoing answers the question posed, it is interesting to examine the result from a geometric perspective: why are there normally two possible fourth circles? Consider the case when one of $a, b$, and $c$ is negative (i.e., two circles are internally tangent to a third). Let $A$ and $B$ be circles internally tangent to $C$. Then $A$ and $B$ partition the remaining area of $C$ into two c-triangles, each of which has an incircle, providing the two solutions.\n\nIf, as in the given problem, $a, b, c>0$, then all three circles $A, B$, and $C$, are mutually externally tangent. In this case, the given circles bound a c-triangle, which has an incircle, corresponding to one of the two roots. The complementary arcs of the given circles bound an infinite region, and this region normally contains a second circle tangent to the given circles. To demonstrate this fact geometrically, consider shrink-wrapping the circles: the shrink-wrap is the border of the smallest convex region containing all three circles. (This region is called the convex hull of the circles). There are two cases to address. If only two circles are touched by the shrink-wrap, then one circle is wedged between two larger ones and completely enclosed by their common tangents. In such a case, a circle can be drawn so that it is tangent to all three circles as shown in the diagram below (shrink-wrap in bold; locations of fourth circle marked at $D_{1}$ and $D_{2}$ ).\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_6e3e0c1749fe30e12b6bg-1.jpg?height=889&width=1181&top_left_y=1277&top_left_x=518)\n\nOn the other hand, if the shrink-wrap touches all three circles, then it can be expanded to make a circle tangent to and containing $A, B$, and $C$, as shown below.\n\n\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_a1e4d46b9ec7c4ff99e7g-1.jpg?height=986&width=992&top_left_y=236&top_left_x=621)""]",['证明题，略'],True,,Need_human_evaluate, 2935,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation.","['In this case, the fourth ""circle"" is actually a line tangent to all three circles, as shown in the diagram below.\n\n']",,True,,, 2935,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation.","['In this case, the fourth ""circle"" is actually a line tangent to all three circles, as shown in the diagram below.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_a1e4d46b9ec7c4ff99e7g-1.jpg?height=417&width=952&top_left_y=1667&top_left_x=627)']",['证明题，略'],True,,Need_human_evaluate, 2936,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$. Prove that $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$.","['Note that\n\n$$\n\\begin{aligned}\n\\frac{1}{\\rho} & =\\frac{\\phi^{2}-\\phi}{\\phi+\\sqrt{\\phi}} \\\\\n& =\\phi-\\sqrt{\\phi}\n\\end{aligned}\n$$\n\n\n\nTherefore\n\n$$\n\\begin{aligned}\n\\left(\\rho-\\frac{1}{\\rho}\\right)^{2} & =(2 \\sqrt{\\phi})^{2}=4 \\phi \\\\\n& =2\\left(\\rho+\\frac{1}{\\rho}\\right) .\n\\end{aligned}\n$$\n\nMultiplying both sides of the equation by $\\rho^{2}$ gives $\\left(\\rho^{2}-1\\right)^{2}=2\\left(\\rho^{3}+\\rho\\right)$. Expand and isolate $\\rho^{4}$ to obtain $\\rho^{4}=2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1$.\n\nAlternate Proof: Because $\\phi^{2}=\\phi+1$, any power of $\\rho$ can be expressed as an integer plus integer multiples of $\\sqrt{\\phi}, \\phi$, and $\\phi \\sqrt{\\phi}$. In particular,\n\n$$\n\\begin{aligned}\n\\rho^{2} & =\\phi^{2}+2 \\phi \\sqrt{\\phi}+\\phi \\\\\n& =2 \\phi \\sqrt{\\phi}+2 \\phi+1, \\\\\n\\rho^{3} & =4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4, \\text { and } \\\\\n\\rho^{4} & =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 .\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1 & =2(4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4)+2(2 \\phi \\sqrt{\\phi}+2 \\phi+1)+2 \\rho-1 \\\\\n& =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 \\\\\n& =\\rho^{4}\n\\end{aligned}\n$$']",['证明题，略'],True,,Need_human_evaluate, 2936,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$. Prove that $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$.","['Note that\n\n$$\n\\begin{aligned}\n\\frac{1}{\\rho} & =\\frac{\\phi^{2}-\\phi}{\\phi+\\sqrt{\\phi}} \\\\\n& =\\phi-\\sqrt{\\phi}\n\\end{aligned}\n$$\n\n\n\nTherefore\n\n$$\n\\begin{aligned}\n\\left(\\rho-\\frac{1}{\\rho}\\right)^{2} & =(2 \\sqrt{\\phi})^{2}=4 \\phi \\\\\n& =2\\left(\\rho+\\frac{1}{\\rho}\\right) .\n\\end{aligned}\n$$\n\nMultiplying both sides of the equation by $\\rho^{2}$ gives $\\left(\\rho^{2}-1\\right)^{2}=2\\left(\\rho^{3}+\\rho\\right)$. Expand and isolate $\\rho^{4}$ to obtain $\\rho^{4}=2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1$.\n\nAlternate Proof: Because $\\phi^{2}=\\phi+1$, any power of $\\rho$ can be expressed as an integer plus integer multiples of $\\sqrt{\\phi}, \\phi$, and $\\phi \\sqrt{\\phi}$. In particular,\n\n$$\n\\begin{aligned}\n\\rho^{2} & =\\phi^{2}+2 \\phi \\sqrt{\\phi}+\\phi \\\\\n& =2 \\phi \\sqrt{\\phi}+2 \\phi+1, \\\\\n\\rho^{3} & =4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4, \\text { and } \\\\\n\\rho^{4} & =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 .\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1 & =2(4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4)+2(2 \\phi \\sqrt{\\phi}+2 \\phi+1)+2 \\rho-1 \\\\\n& =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 \\\\\n& =\\rho^{4}\n\\end{aligned}\n$$']",,True,,, 2937,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$. Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have common ratio $\rho$.","[""If the radii are in geometric progression, then so are their reciprocals (i.e., curvatures). Without loss of generality, let $(a, b, c, d)=\\left(a, a r, a r^{2}, a r^{3}\\right)$ for $r>1$. By Descartes' Circle Formula,\n\n$$\n\\left(a+a r+a r^{2}+a r^{3}\\right)^{2}=2\\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\\right)\n$$\n\nCancel $a^{2}$ from both sides of the equation to obtain\n\n$$\n\\left(1+r+r^{2}+r^{3}\\right)^{2}=2\\left(1+r^{2}+r^{4}+r^{6}\\right) .\n$$\n\nBecause $1+r+r^{2}+r^{3}=(1+r)\\left(1+r^{2}\\right)$ and $1+r^{2}+r^{4}+r^{6}=\\left(1+r^{2}\\right)\\left(1+r^{4}\\right)$, the equation can be rewritten as follows:\n\n$$\n\\begin{aligned}\n(1+r)^{2}\\left(1+r^{2}\\right)^{2} & =2\\left(1+r^{2}\\right)\\left(1+r^{4}\\right) \\\\\n(1+r)^{2}\\left(1+r^{2}\\right) & =2\\left(1+r^{4}\\right) \\\\\nr^{4}-2 r^{3}-2 r^{2}-2 r+1 & =0 .\n\\end{aligned}\n$$\n\nUsing the identity from 4a, $r=\\rho$ is one solution; because the polynomial is palindromic, another real solution is $r=\\rho^{-1}=\\phi-\\sqrt{\\phi}$, but this value is less than 1 . The product of the corresponding linear factors is $r^{2}-2 \\phi r+\\phi^{2}-\\phi=r^{2}-2 \\phi r+1$. Division verifies that the other quadratic factor of the polynomial is $r^{2}+(2 \\phi-2) r+1$, which has no real roots because $(\\sqrt{5}-1)^{2}<1$.""]",['证明题，略'],True,,Need_human_evaluate, 2937,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$. Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have common ratio $\rho$.","[""If the radii are in geometric progression, then so are their reciprocals (i.e., curvatures). Without loss of generality, let $(a, b, c, d)=\\left(a, a r, a r^{2}, a r^{3}\\right)$ for $r>1$. By Descartes' Circle Formula,\n\n$$\n\\left(a+a r+a r^{2}+a r^{3}\\right)^{2}=2\\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\\right)\n$$\n\nCancel $a^{2}$ from both sides of the equation to obtain\n\n$$\n\\left(1+r+r^{2}+r^{3}\\right)^{2}=2\\left(1+r^{2}+r^{4}+r^{6}\\right) .\n$$\n\nBecause $1+r+r^{2}+r^{3}=(1+r)\\left(1+r^{2}\\right)$ and $1+r^{2}+r^{4}+r^{6}=\\left(1+r^{2}\\right)\\left(1+r^{4}\\right)$, the equation can be rewritten as follows:\n\n$$\n\\begin{aligned}\n(1+r)^{2}\\left(1+r^{2}\\right)^{2} & =2\\left(1+r^{2}\\right)\\left(1+r^{4}\\right) \\\\\n(1+r)^{2}\\left(1+r^{2}\\right) & =2\\left(1+r^{4}\\right) \\\\\nr^{4}-2 r^{3}-2 r^{2}-2 r+1 & =0 .\n\\end{aligned}\n$$\n\nUsing the identity from 4a, $r=\\rho$ is one solution; because the polynomial is palindromic, another real solution is $r=\\rho^{-1}=\\phi-\\sqrt{\\phi}$, but this value is less than 1 . The product of the corresponding linear factors is $r^{2}-2 \\phi r+\\phi^{2}-\\phi=r^{2}-2 \\phi r+1$. Division verifies that the other quadratic factor of the polynomial is $r^{2}+(2 \\phi-2) r+1$, which has no real roots because $(\\sqrt{5}-1)^{2}<1$.""]",,True,,, 2938,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$.","['The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$']",['证明题，略'],True,,Need_human_evaluate, 2938,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$.","['The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$']",,True,,, 2939,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Show that by area, $12 \%$ of the kingdom is sold on day 2 .","['On day 2, six plots are sold: two with curvature 15 from the configuration $(2,2,3,15)$, and four with curvature 6 from the configuration $(-1,2,3,6)$. The total area sold on day 2 is therefore\n\n$$\n2 \\cdot \\frac{\\pi}{15^{2}}+4 \\cdot \\frac{\\pi}{6^{2}}=\\frac{3}{25} \\pi\n$$\n\nwhich is exactly $12 \\%$ of the unit circle.']",['证明题，略'],True,,Need_human_evaluate, 2939,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Show that by area, $12 \%$ of the kingdom is sold on day 2 .","['On day 2, six plots are sold: two with curvature 15 from the configuration $(2,2,3,15)$, and four with curvature 6 from the configuration $(-1,2,3,6)$. The total area sold on day 2 is therefore\n\n$$\n2 \\cdot \\frac{\\pi}{15^{2}}+4 \\cdot \\frac{\\pi}{6^{2}}=\\frac{3}{25} \\pi\n$$\n\nwhich is exactly $12 \\%$ of the unit circle.']",,True,,, 2940,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Find the areas of the circles removed on day 3.","['Day 3 begins with two circles of curvature 15 from the configuration $(2,2,3,15)$, and four circles of curvature 6 from the configuration $(-1,2,3,6)$. Consider the following two cases:\n\nCase 1: $(a, b, c, d)=(2,2,3,15), s=22$\n\n- $a=2: a^{\\prime}=2 s-3 a=\\mathbf{3 8}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{3 8}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{3 5}$\n- $d=15: d^{\\prime}=2 s-3 d=-1$, which is the configuration from day 1 .\n\nCase 2: $(a, b, c, d)=(-1,2,3,6), s=10$\n\n- $a=-1: a^{\\prime}=2 s-3 a=\\mathbf{2 3}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{1 4}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{1 1}$\n- $d=6: d^{\\prime}=2 s-3 d=2$, which is the configuration from day 1 .\n\n\n\nSo the areas of the plots removed on day 3 are:\n\n$$\n\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\text { and } \\frac{\\pi}{11^{2}}\n$$\n\nThere are two circles with area $\\frac{\\pi}{35^{2}}$, and four circles with each of the other areas, for a total of 18 plots.']","['$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$']",True,,Numerical, 2940,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Find the areas of the circles removed on day 3.","['Day 3 begins with two circles of curvature 15 from the configuration $(2,2,3,15)$, and four circles of curvature 6 from the configuration $(-1,2,3,6)$. Consider the following two cases:\n\nCase 1: $(a, b, c, d)=(2,2,3,15), s=22$\n\n- $a=2: a^{\\prime}=2 s-3 a=\\mathbf{3 8}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{3 8}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{3 5}$\n- $d=15: d^{\\prime}=2 s-3 d=-1$, which is the configuration from day 1 .\n\nCase 2: $(a, b, c, d)=(-1,2,3,6), s=10$\n\n- $a=-1: a^{\\prime}=2 s-3 a=\\mathbf{2 3}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{1 4}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{1 1}$\n- $d=6: d^{\\prime}=2 s-3 d=2$, which is the configuration from day 1 .\n\n\n\nSo the areas of the plots removed on day 3 are:\n\n$$\n\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\text { and } \\frac{\\pi}{11^{2}}\n$$\n\nThere are two circles with area $\\frac{\\pi}{35^{2}}$, and four circles with each of the other areas, for a total of 18 plots.']","['$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$']",True,,Numerical, 2941,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Show that the plots sold on day 3 have mean curvature of 23.","['Because 18 plots were sold on day 3 , the mean curvature is\n\n$$\n\\frac{2(38+38+35)+4(23+14+11)}{18}=23 .\n$$']",,True,,, 2941,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Show that the plots sold on day 3 have mean curvature of 23.","['Because 18 plots were sold on day 3 , the mean curvature is\n\n$$\n\\frac{2(38+38+35)+4(23+14+11)}{18}=23 .\n$$']",['证明题，略'],True,,Need_human_evaluate, 2942,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Prove that the curvature of each circular plot is an integer.","['Proceed by induction. The base case, that all curvatures prior to day 2 are integers. Using the formula $a^{\\prime}=2 s-3 a$, if $a, b, c, d$, and $s$ are integers on day $n$, then $a^{\\prime}, b^{\\prime}, c^{\\prime}$, and $d^{\\prime}$ are integer curvatures on day $n+1$, proving inductively that all curvatures are integers.']",,True,,, 2942,Geometry,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Prove that the curvature of each circular plot is an integer.","['Proceed by induction. The base case, that all curvatures prior to day 2 are integers. Using the formula $a^{\\prime}=2 s-3 a$, if $a, b, c, d$, and $s$ are integers on day $n$, then $a^{\\prime}, b^{\\prime}, c^{\\prime}$, and $d^{\\prime}$ are integer curvatures on day $n+1$, proving inductively that all curvatures are integers.']",['证明题，略'],True,,Need_human_evaluate, 2943,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Prove that the center of each circular plot has coordinates $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature of the plot.","['It suffices to show that for each circle $C$ with curvature $c$ and center $z_{C}$ (in the complex plane), $c z_{C}$ is of the form $u+i v$ where $u$ and $v$ are integers. If this is the case, then each center is of the form $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$.\n\nProceed by induction. To check the base case, check the original kingdom and the first four plots: $U, C, C^{\\prime}, P_{1}$, and $P_{2}$. Circle $U$ is centered at $(0,0)$, yielding $-1 \\cdot z_{U}=0+0 i$. Circles $C$ and $C^{\\prime}$ are symmetric about the $y$-axis, so it suffices to check just one of them. Circle $C$ has radius $\\frac{1}{2}$ and therefore curvature 2 . It is centered at $\\left(\\frac{1}{2}, \\frac{0}{2}\\right)$, yielding $2 z_{C}=2\\left(\\frac{1}{2}+0 i\\right)=1$. Circles $P_{1}$ and $P_{2}$ are symmetric about the $x$-axis, so it suffices to check just one of them. Circle $P_{1}$ has radius $\\frac{1}{3}$ and therefore curvature 3 . It is centered at $\\left(\\frac{0}{3}, \\frac{2}{3}\\right)$, yielding $3 z_{P_{3}}=0+2 i$.\n\nFor the inductive step, suppose that $a z_{A}, b z_{B}, c z_{C}, d z_{D}$ have integer real and imaginary parts. Then by closure of addition and multiplication in the integers, $a^{\\prime} z_{A^{\\prime}}=2 \\hat{s}-3 a z_{A}$ also has integer real and imaginary parts, and similarly for $b^{\\prime} z_{B^{\\prime}}, c^{\\prime} z_{C^{\\prime}}, d^{\\prime} z_{D^{\\prime}}$.\n\nSo for all plots $A$ sold, $a z_{A}$ has integer real and imaginary parts, so each is centered at $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$ where $u$ and $v$ are integers, and $c$ is the curvature.']",,True,,, 2943,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Prove that the center of each circular plot has coordinates $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature of the plot.","['It suffices to show that for each circle $C$ with curvature $c$ and center $z_{C}$ (in the complex plane), $c z_{C}$ is of the form $u+i v$ where $u$ and $v$ are integers. If this is the case, then each center is of the form $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$.\n\nProceed by induction. To check the base case, check the original kingdom and the first four plots: $U, C, C^{\\prime}, P_{1}$, and $P_{2}$. Circle $U$ is centered at $(0,0)$, yielding $-1 \\cdot z_{U}=0+0 i$. Circles $C$ and $C^{\\prime}$ are symmetric about the $y$-axis, so it suffices to check just one of them. Circle $C$ has radius $\\frac{1}{2}$ and therefore curvature 2 . It is centered at $\\left(\\frac{1}{2}, \\frac{0}{2}\\right)$, yielding $2 z_{C}=2\\left(\\frac{1}{2}+0 i\\right)=1$. Circles $P_{1}$ and $P_{2}$ are symmetric about the $x$-axis, so it suffices to check just one of them. Circle $P_{1}$ has radius $\\frac{1}{3}$ and therefore curvature 3 . It is centered at $\\left(\\frac{0}{3}, \\frac{2}{3}\\right)$, yielding $3 z_{P_{3}}=0+2 i$.\n\nFor the inductive step, suppose that $a z_{A}, b z_{B}, c z_{C}, d z_{D}$ have integer real and imaginary parts. Then by closure of addition and multiplication in the integers, $a^{\\prime} z_{A^{\\prime}}=2 \\hat{s}-3 a z_{A}$ also has integer real and imaginary parts, and similarly for $b^{\\prime} z_{B^{\\prime}}, c^{\\prime} z_{C^{\\prime}}, d^{\\prime} z_{D^{\\prime}}$.\n\nSo for all plots $A$ sold, $a z_{A}$ has integer real and imaginary parts, so each is centered at $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$ where $u$ and $v$ are integers, and $c$ is the curvature.']",['证明题，略'],True,,Need_human_evaluate, 2944,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Show that $P^{(3)}$ dominates $P^{(2)}$ and that $P^{(2)}$ dominates $P^{(1)}$.","['The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$\n\n\nLet $s=a+b+c+d$. From above induction, it follows that\n\n$$\n\\begin{aligned}\n\\mathcal{C}\\left(P^{(1)}\\right) & =\\left(b, c, d, a^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(2)}\\right) & =\\left(a, c, d, b^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(3)}\\right) & =\\left(a, b, d, c^{\\prime}\\right)\n\\end{aligned}\n$$\n\nwhere $a^{\\prime}=2 s-3 a, b^{\\prime}=2 s-3 b$, and $c^{\\prime}=2 s-3 c$. Because $a \\leq b \\leq c$, it follows that $c^{\\prime} \\leq b^{\\prime} \\leq a^{\\prime}$. Therefore $P^{(3)}$ dominates $P^{(2)}$ and $P^{(2)}$ dominates $P^{(1)}$.']",,True,,, 2944,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Show that $P^{(3)}$ dominates $P^{(2)}$ and that $P^{(2)}$ dominates $P^{(1)}$.","['The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$\n\n\nLet $s=a+b+c+d$. From above induction, it follows that\n\n$$\n\\begin{aligned}\n\\mathcal{C}\\left(P^{(1)}\\right) & =\\left(b, c, d, a^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(2)}\\right) & =\\left(a, c, d, b^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(3)}\\right) & =\\left(a, b, d, c^{\\prime}\\right)\n\\end{aligned}\n$$\n\nwhere $a^{\\prime}=2 s-3 a, b^{\\prime}=2 s-3 b$, and $c^{\\prime}=2 s-3 c$. Because $a \\leq b \\leq c$, it follows that $c^{\\prime} \\leq b^{\\prime} \\leq a^{\\prime}$. Therefore $P^{(3)}$ dominates $P^{(2)}$ and $P^{(2)}$ dominates $P^{(1)}$.']",['证明题，略'],True,,Need_human_evaluate, 2945,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Prove that $P^{(1)}$ dominates $Q^{(1)}$.","[""Because $\\mathcal{C}\\left(P^{(1)}\\right)=\\left(b, c, d, a^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(1)}\\right)=\\left(x, y, z, w^{\\prime}\\right)$, it is enough to show that $a^{\\prime} \\leq w^{\\prime}$. $a$ and $a^{\\prime}$ are the two roots of the quadratic given by Descartes' Circle Formula:\n\n$$\n(X+b+c+d)^{2}=2\\left(X^{2}+b^{2}+c^{2}+d^{2}\\right)\n$$\n\nSolve by completing the square:\n\n$$\n\\begin{aligned}\nX^{2}-2(b+c+d) X+2\\left(b^{2}+c^{2}+d^{2}\\right) & =(b+c+d)^{2} \\\\\n(X-(b+c+d))^{2} & =2(b+c+d)^{2}-2\\left(b^{2}+c^{2}+d^{2}\\right) \\\\\n& =4(b c+b d+c d) .\n\\end{aligned}\n$$\n\nThus $a, a^{\\prime}=b+c+d \\pm 2 \\sqrt{b c+b d+c d}$.\n\nBecause $a \\leq b \\leq c \\leq d$, and only $a$ can be less than zero, $a$ must get the minus sign, and $a^{\\prime}$ gets the plus sign:\n\n$$\na^{\\prime}=b+c+d+2 \\sqrt{b c+b d+c d} \\text {. }\n$$\n\nSimilarly,\n\n$$\nw^{\\prime}=x+y+z+2 \\sqrt{x y+x z+y z} .\n$$\n\nBecause $P$ dominates $Q$, each term in the expression for $a^{\\prime}$ is less than or equal to the corresponding term in the expression for $w^{\\prime}$, thus $a^{\\prime} \\leq w^{\\prime}$.""]",['证明题，略'],True,,Need_human_evaluate, 2945,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Prove that $P^{(1)}$ dominates $Q^{(1)}$.","[""Because $\\mathcal{C}\\left(P^{(1)}\\right)=\\left(b, c, d, a^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(1)}\\right)=\\left(x, y, z, w^{\\prime}\\right)$, it is enough to show that $a^{\\prime} \\leq w^{\\prime}$. $a$ and $a^{\\prime}$ are the two roots of the quadratic given by Descartes' Circle Formula:\n\n$$\n(X+b+c+d)^{2}=2\\left(X^{2}+b^{2}+c^{2}+d^{2}\\right)\n$$\n\nSolve by completing the square:\n\n$$\n\\begin{aligned}\nX^{2}-2(b+c+d) X+2\\left(b^{2}+c^{2}+d^{2}\\right) & =(b+c+d)^{2} \\\\\n(X-(b+c+d))^{2} & =2(b+c+d)^{2}-2\\left(b^{2}+c^{2}+d^{2}\\right) \\\\\n& =4(b c+b d+c d) .\n\\end{aligned}\n$$\n\nThus $a, a^{\\prime}=b+c+d \\pm 2 \\sqrt{b c+b d+c d}$.\n\nBecause $a \\leq b \\leq c \\leq d$, and only $a$ can be less than zero, $a$ must get the minus sign, and $a^{\\prime}$ gets the plus sign:\n\n$$\na^{\\prime}=b+c+d+2 \\sqrt{b c+b d+c d} \\text {. }\n$$\n\nSimilarly,\n\n$$\nw^{\\prime}=x+y+z+2 \\sqrt{x y+x z+y z} .\n$$\n\nBecause $P$ dominates $Q$, each term in the expression for $a^{\\prime}$ is less than or equal to the corresponding term in the expression for $w^{\\prime}$, thus $a^{\\prime} \\leq w^{\\prime}$.""]",,True,,, 2946,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347) Circle $A$ has curvature 2; circle $B$ has curvature 1 . ![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386) Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Prove that $P^{(3)}$ dominates $Q^{(3)}$.","['Because $\\mathcal{C}\\left(P^{(3)}\\right)=\\left(a, b, d, c^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(3)}\\right)=\\left(w, x, z, y^{\\prime}\\right)$, it suffices to show that $c^{\\prime} \\leq y^{\\prime}$. If $a \\geq 0$, but if $a<0$, then there is more to be done.\n\nArguing as in 9b, $c, c^{\\prime}=a+b+d \\pm 2 \\sqrt{a b+a d+b d}$. If $a<0$, then the other three circles are internally tangent to the circle of curvature $a$, so this circle has the largest radius. In particular, $\\frac{1}{|a|}>\\frac{1}{b}$. Thus $b>|a|=-a$, which shows that $a+b>0$. Therefore $c$ must get the minus sign, and $c^{\\prime}$ gets the plus sign. The same argument applies to $y$ and $y^{\\prime}$.\n\nWhen $a<0$, it is also worth considering whether the square roots are defined (and real). In fact, they are. Consider the diameters of the circles with curvatures $b$ and $d$ along the line through the centers of these circles. These two diameters form a single segment inside the circle with curvature $a$, so the sum of the diameters is at most the diameter of that circle: $\\frac{2}{b}+\\frac{2}{d} \\leq \\frac{2}{|a|}$. It follows that $-a d-a b=|a| d+|a| b \\leq b d$, or $a b+a d+b d \\geq 0$. This is the argument of the square root in the expressions for $c$ and $c^{\\prime}$. An analogous argument shows that the radicands are nonnegative in the expressions for $b$ and $b^{\\prime}$.\n\nThe foregoing shows that\n\n$$\nc^{\\prime}=a+b+d+2 \\sqrt{a b+a d+b d}\n$$\n\nand, by an analogous argument for $w<0$,\n\n$$\ny^{\\prime}=w+x+z+2 \\sqrt{w x+w z+x z} .\n$$\n\nIt remains to prove that $c^{\\prime} \\leq y^{\\prime}$. Note that only $a$ and $w$ may be negative; $b, c, d, x, y$, and $z$ are all positive. There are three cases.\n\n\n\n(i) If $0 \\leq a \\leq w$, then $a b \\leq w x, a d \\leq w z$, and $b d \\leq x z$, so $c^{\\prime} \\leq y^{\\prime}$.\n\n(ii) If $a<0 \\leq w$, then $a b+a d+b d \\leq b d$, and $b d \\leq x z \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$. (As noted above, both radicands are nonnegative.)\n\n(iii) If $a \\leq w<0$, then it has already been established that $a+b$ is positive. Analogously, $a+d, w+x$, and $w+z$ are positive. Furthermore, $a^{2} \\geq w^{2}$. Thus $(a+b)(a+d)-a^{2} \\leq$ $(w+x)(w+z)-w^{2}$, which establishes that $a b+a d+b d \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$.']",['证明题，略'],True,,Need_human_evaluate, 2946,Combinatorics,,"A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short. This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner. Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius. Circle $A$ has curvature 2; circle $B$ has curvature 1 . Circle $A$ has curvature 2; circle $B$ has curvature -1 . Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then $$ (a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right), $$ where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$. Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states $$ \left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) . $$ Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper. Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term ""dominates"" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.) Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations $$ \begin{aligned} \mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\ \mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\ \mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right), \end{aligned} $$ Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$. Prove that $P^{(3)}$ dominates $Q^{(3)}$.","['Because $\\mathcal{C}\\left(P^{(3)}\\right)=\\left(a, b, d, c^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(3)}\\right)=\\left(w, x, z, y^{\\prime}\\right)$, it suffices to show that $c^{\\prime} \\leq y^{\\prime}$. If $a \\geq 0$, but if $a<0$, then there is more to be done.\n\nArguing as in 9b, $c, c^{\\prime}=a+b+d \\pm 2 \\sqrt{a b+a d+b d}$. If $a<0$, then the other three circles are internally tangent to the circle of curvature $a$, so this circle has the largest radius. In particular, $\\frac{1}{|a|}>\\frac{1}{b}$. Thus $b>|a|=-a$, which shows that $a+b>0$. Therefore $c$ must get the minus sign, and $c^{\\prime}$ gets the plus sign. The same argument applies to $y$ and $y^{\\prime}$.\n\nWhen $a<0$, it is also worth considering whether the square roots are defined (and real). In fact, they are. Consider the diameters of the circles with curvatures $b$ and $d$ along the line through the centers of these circles. These two diameters form a single segment inside the circle with curvature $a$, so the sum of the diameters is at most the diameter of that circle: $\\frac{2}{b}+\\frac{2}{d} \\leq \\frac{2}{|a|}$. It follows that $-a d-a b=|a| d+|a| b \\leq b d$, or $a b+a d+b d \\geq 0$. This is the argument of the square root in the expressions for $c$ and $c^{\\prime}$. An analogous argument shows that the radicands are nonnegative in the expressions for $b$ and $b^{\\prime}$.\n\nThe foregoing shows that\n\n$$\nc^{\\prime}=a+b+d+2 \\sqrt{a b+a d+b d}\n$$\n\nand, by an analogous argument for $w<0$,\n\n$$\ny^{\\prime}=w+x+z+2 \\sqrt{w x+w z+x z} .\n$$\n\nIt remains to prove that $c^{\\prime} \\leq y^{\\prime}$. Note that only $a$ and $w$ may be negative; $b, c, d, x, y$, and $z$ are all positive. There are three cases.\n\n\n\n(i) If $0 \\leq a \\leq w$, then $a b \\leq w x, a d \\leq w z$, and $b d \\leq x z$, so $c^{\\prime} \\leq y^{\\prime}$.\n\n(ii) If $a<0 \\leq w$, then $a b+a d+b d \\leq b d$, and $b d \\leq x z \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$. (As noted above, both radicands are nonnegative.)\n\n(iii) If $a \\leq w<0$, then it has already been established that $a+b$ is positive. Analogously, $a+d, w+x$, and $w+z$ are positive. Furthermore, $a^{2} \\geq w^{2}$. Thus $(a+b)(a+d)-a^{2} \\leq$ $(w+x)(w+z)-w^{2}$, which establishes that $a b+a d+b d \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$.']",,True,,, 2947,Combinatorics,,"Let set $S=\{1,2,3,4,5,6\}$, and let set $T$ be the set of all subsets of $S$ (including the empty set and $S$ itself). Let $t_{1}, t_{2}, t_{3}$ be elements of $T$, not necessarily distinct. The ordered triple $\left(t_{1}, t_{2}, t_{3}\right)$ is called satisfactory if either (a) both $t_{1} \subseteq t_{3}$ and $t_{2} \subseteq t_{3}$, or (b) $t_{3} \subseteq t_{1}$ and $t_{3} \subseteq t_{2}$. Compute the number of satisfactory ordered triples $\left(t_{1}, t_{2}, t_{3}\right)$.","['Let $T_{1}=\\left\\{\\left(t_{1}, t_{2}, t_{3}\\right) \\mid t_{1} \\subseteq t_{3}\\right.$ and $\\left.t_{2} \\subseteq t_{3}\\right\\}$ and let $T_{2}=\\left\\{\\left(t_{1}, t_{2}, t_{3}\\right) \\mid t_{3} \\subseteq t_{1}\\right.$ and $\\left.t_{3} \\subseteq t_{2}\\right\\}$. Notice that if $\\left(t_{1}, t_{2}, t_{3}\\right) \\in T_{1}$, then $\\left(S \\backslash t_{1}, S \\backslash t_{2}, S \\backslash t_{3}\\right) \\in T_{2}$, so that $\\left|T_{1}\\right|=\\left|T_{2}\\right|$. To count $T_{1}$, note that if $t_{1} \\subseteq t_{3}$ and $t_{2} \\subseteq t_{3}$, then $t_{1} \\cup t_{2} \\subseteq t_{3}$. Now each set $t_{3}$ has $2^{\\left|t_{3}\\right|}$ subsets; $t_{1}$ and $t_{2}$ could be any of these, for a total of $\\left(2^{\\left|t_{3}\\right|}\\right)^{2}=4^{\\left|t_{3}\\right|}$ possibilities given a particular subset $t_{3}$. For $n=0,1, \\ldots, 6$, if $\\left|t_{3}\\right|=n$, there are $\\left(\\begin{array}{l}6 \\\\ n\\end{array}\\right)$ choices for the elements of $t_{3}$. So the total number of elements in $T_{1}$ is\n\n$$\n\\begin{aligned}\n\\left|T_{1}\\right| & =\\sum_{k=0}^{6}\\left(\\begin{array}{l}\n6 \\\\\nk\n\\end{array}\\right) 4^{k} \\\\\n& =(4+1)^{6}=15625\n\\end{aligned}\n$$\n\nby the Binomial Theorem. However, $T_{1} \\cap T_{2} \\neq \\emptyset$, because if $t_{1}=t_{2}=t_{3}$, the triple $\\left(t_{1}, t_{2}, t_{3}\\right)$ satisfies both conditions and is in both sets. Therefore there are 64 triples that are counted in both sets. So $\\left|T_{1} \\cup T_{2}\\right|=2 \\cdot 15625-64=\\mathbf{3 1 1 8 6}$.', 'Let $T_{1}$ and $T_{2}$ be defined as above. Then count $\\left|T_{1}\\right|$ based on the number $n$ of elements in $t_{1} \\cup t_{2}$. There are $\\left(\\begin{array}{l}6 \\\\ n\\end{array}\\right)$ ways to choose those $n$ elements. For each element $a$ in $t_{1} \\cup t_{2}$, there are three possibilities: $a \\in t_{1}$ but not $t_{2}$, or $a \\in t_{2}$ but not $t_{1}$, or $a \\in t_{1} \\cap t_{2}$. Then for each element $b$ in $S \\backslash\\left(t_{1} \\cup t_{2}\\right)$, there are two possibilities: either $b \\in t_{3}$, or $b \\notin t_{3}$. Combine these observations in the table below:\n\n| $\\left\\|t_{1} \\cup t_{2}\\right\\|$ | Choices for
$t_{1} \\cup t_{2}$ | Ways of dividing
between $t_{1}$ and $t_{2}$ | $\\left\\|S \\backslash\\left(t_{1} \\cup t_{2}\\right)\\right\\|$ | Choices for $t_{3}$ | Total |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 0 | 1 | 1 | 6 | $2^{6}$ | 64 |\n| 1 | 6 | 3 | 5 | $2^{5}$ | 576 |\n| 2 | 15 | $3^{2}$ | 4 | $2^{4}$ | 2160 |\n| 3 | 20 | $3^{3}$ | 3 | $2^{3}$ | 4320 |\n| 4 | 15 | $3^{4}$ | 2 | $2^{2}$ | 4860 |\n| 5 | 6 | $3^{5}$ | 1 | $2^{1}$ | 2916 |\n| 6 | 1 | $3^{6}$ | 0 | $2^{0}$ | 729 |\n\nThe total is 15625 , so $\\left|T_{1}\\right|=\\left|T_{2}\\right|=15625$. As noted in the first solution, there are 64 triples that are counted in both $T_{1}$ and $T_{2}$, so $\\left|T_{1} \\cup T_{2}\\right|=2 \\cdot 15625-64=\\mathbf{3 1 1 8 6}$.']",['31186'],False,,Numerical, 2948,Geometry,,"Let $A B C D$ be a parallelogram with $\angle A B C$ obtuse. Let $\overline{B E}$ be the altitude to side $\overline{A D}$ of $\triangle A B D$. Let $X$ be the point of intersection of $\overline{A C}$ and $\overline{B E}$, and let $F$ be the point of intersection of $\overline{A B}$ and $\overleftrightarrow{D X}$. If $B C=30, C D=13$, and $B E=12$, compute the ratio $\frac{A C}{A F}$.","[""Extend $\\overline{A D}$ to a point $M$ such that $\\overline{C M} \\| \\overline{B E}$ as shown below.\n\n\n\nBecause $C D=A B=13$ and $B E=12=C M, A E=D M=5$. Then $A C=\\sqrt{35^{2}+12^{2}}=$ $\\sqrt{1369}=37$. Because $\\overline{E X} \\| \\overline{C M}, X E / C M=A E / A M=\\frac{1}{7}$. Thus $E X=\\frac{12}{7}$ and $X B=\\frac{72}{7}$, from which $E X / X B=\\frac{1}{6}$. Apply Menelaus's Theorem to $\\triangle A E B$ and Menelaus line $\\overline{F D}$ :\n\n$$\n\\begin{aligned}\n\\frac{A D}{E D} \\cdot \\frac{E X}{X B} \\cdot \\frac{B F}{F A} & =1 \\\\\n\\frac{30}{25} \\cdot \\frac{1}{6} \\cdot \\frac{13-F A}{F A} & =1 \\\\\n\\frac{13-F A}{F A} & =5 .\n\\end{aligned}\n$$\n\nThus $F A=\\frac{13}{6}$. The desired ratio is:\n\n$$\n\\frac{37}{13 / 6}=\\frac{\\mathbf{2 2 2}}{\\mathbf{1 3}}\n$$"", ""After calculating $A C$ as above, draw $\\overline{B D}$, intersecting $\\overline{A C}$ at $Y$. Because the diagonals of a parallelogram bisect each other, $D Y=Y B$. Then apply Ceva's Theorem to $\\triangle A B D$ and concurrent cevians $\\overline{A Y}, \\overline{B E}, \\overline{D F}$ :\n\n$$\n\\begin{aligned}\n& \\frac{A E}{E D} \\cdot \\frac{D Y}{Y B} \\cdot \\frac{B F}{F A}=1 \\\\\n& \\frac{5}{25} \\cdot 1 \\cdot \\frac{13-F A}{F A}=1\n\\end{aligned}\n$$\n\nThus $F A=\\frac{13}{6}$, and the desired ratio is $\\frac{\\mathbf{2 2 2}}{\\mathbf{1 3}}$."", 'By AA similarity, note that $\\triangle A F X \\sim \\triangle C D X$ and $\\triangle A E X \\sim \\triangle C B X$. Thus $\\frac{A F}{C D}=\\frac{A X}{X C}=\\frac{A E}{C B}$. Thus $\\frac{A F}{13}=\\frac{A E}{C B}=\\frac{5}{30}$, so $A F=\\frac{13}{6}$, and the answer follows after calculating $A C$, as in the first solution.']",['$\\frac{222}{13}$'],False,,Numerical, 2949,Number Theory,,Compute the sum of all positive two-digit factors of $2^{32}-1$.,"['Using the difference of squares, $2^{32}-1=\\left(2^{16}-1\\right)\\left(2^{16}+1\\right)$. The second factor, $2^{16}+1$, is the Fermat prime 65537 , so continue with the first factor:\n\n$$\n\\begin{aligned}\n2^{16}-1 & =\\left(2^{8}+1\\right)\\left(2^{8}-1\\right) \\\\\n2^{8}-1 & =\\left(2^{4}+1\\right)\\left(2^{4}-1\\right) \\\\\n2^{4}-1 & =15=3 \\cdot 5\n\\end{aligned}\n$$\n\n\n\nBecause the problem does not specify that the two-digit factors must be prime, the possible two-digit factors are $17,3 \\cdot 17=51,5 \\cdot 17=85$ and $3 \\cdot 5=15$, for a sum of $17+51+85+15=\\mathbf{1 6 8}$.']",['168'],False,,Numerical, 2950,Algebra,,"Compute all ordered pairs of real numbers $(x, y)$ that satisfy both of the equations: $$ x^{2}+y^{2}=6 y-4 x+12 \quad \text { and } \quad 4 y=x^{2}+4 x+12 $$","['Rearrange the terms in the first equation to yield $x^{2}+4 x+12=6 y-y^{2}+24$, so that the two equations together yield $4 y=6 y-y^{2}+24$, or $y^{2}-2 y-24=0$, from which $y=6$ or $y=-4$. If $y=6$, then $x^{2}+4 x+12=24$, from which $x=-6$ or $x=2$. If $y=-4$, then $x^{2}+4 x+12=-16$, which has no real solutions because $x^{2}+4 x+12=(x+2)^{2}+8 \\geq 8$ for all real $x$. So there are two ordered pairs satisfying the system, namely $(-6,6)$ and $(2,6)$.']","['$(-6,6)$, $(2,6)$']",True,,Tuple, 2951,Algebra,,"Define $\log ^{*}(n)$ to be the smallest number of times the log function must be iteratively applied to $n$ to get a result less than or equal to 1 . For example, $\log ^{*}(1000)=2$ since $\log 1000=3$ and $\log (\log 1000)=\log 3=0.477 \ldots \leq 1$. Let $a$ be the smallest integer such that $\log ^{*}(a)=3$. Compute the number of zeros in the base 10 representation of $a$.","['If $\\log ^{*}(a)=3$, then $\\log (\\log (\\log (a))) \\leq 1$ and $\\log (\\log (a))>1$. If $\\log (\\log (a))>1$, then $\\log (a)>10$ and $a>10^{10}$. Because the problem asks for the smallest such $a$ that is an integer, choose $a=10^{10}+1=10,000,000,001$, which has 9 zeros.']",['9'],False,,Numerical, 2952,Combinatorics,,"An integer $N$ is worth 1 point for each pair of digits it contains that forms a prime in its original order. For example, 6733 is worth 3 points (for 67,73 , and 73 again), and 20304 is worth 2 points (for 23 and 03). Compute the smallest positive integer that is worth exactly 11 points. [Note: Leading zeros are not allowed in the original integer.]","['If a number $N$ has $k$ base 10 digits, then its maximum point value is $(k-1)+(k-2)+\\cdots+1=$ $\\frac{1}{2}(k-1)(k)$. So if $k \\leq 5$, the number $N$ is worth at most 10 points. Therefore the desired number has at least six digits. If $100,0000$. Then if $d(x)$ denotes the number of positive divisors of $x$,\n\n$$\nd\\left(n^{n}\\right)=\\left(a_{1} n+1\\right)\\left(a_{2} n+1\\right) \\cdots\\left(a_{k} n+1\\right) \\geq(n+1)^{k}\n$$\n\nNote that if $n \\geq 99$ and $k \\geq 3$, then $d\\left(n^{n}\\right) \\geq 100^{3}=10^{6}$, so $102=2 \\cdot 3 \\cdot 17$ is an upper bound for the solution. Look for values less than 99, using two observations: (1) all $a_{i} \\leq 6$\n\n\n\n(because $p^{7}>99$ for all primes); and (2) $k \\leq 3$ (because $2 \\cdot 3 \\cdot 5 \\cdot 7>99$ ). These two facts rule out the cases $k=1$ (because $(*)$ yields $\\left.d \\leq(6 n+1)^{1}<601\\right)$ and $k=2$ (because $\\left.d\\left(n^{n}\\right) \\leq(6 n+1)^{2}<601^{2}\\right)$.\n\nSo $k=3$. Note that if $a_{1}=a_{2}=a_{3}=1$, then from $(*), d\\left(n^{n}\\right)=(n+1)^{3}<10^{6}$. So consider only $n<99$ with exactly three prime divisors, and for which not all exponents are 1 . The only candidates are 60,84 , and 90 ; of these, $n=84$ is the smallest one that works:\n\n$$\n\\begin{aligned}\n& d\\left(60^{60}\\right)=d\\left(2^{120} \\cdot 3^{60} \\cdot 5^{60}\\right)=121 \\cdot 61 \\cdot 61<125 \\cdot 80 \\cdot 80=800,000 \\\\\n& d\\left(84^{84}\\right)=d\\left(2^{168} \\cdot 3^{84} \\cdot 7^{84}\\right)=169 \\cdot 85 \\cdot 85>160 \\cdot 80 \\cdot 80=1,024,000\n\\end{aligned}\n$$\n\nTherefore $n=\\mathbf{8 4}$ is the least positive integer $n$ such that $d\\left(n^{n}\\right)>1,000,000$.']",['84'],False,,Numerical, 2955,Combinatorics,,"Given an arbitrary finite sequence of letters (represented as a word), a subsequence is a sequence of one or more letters that appear in the same order as in the original sequence. For example, $N, C T, O T T$, and CONTEST are subsequences of the word CONTEST, but NOT, ONSET, and TESS are not. Assuming the standard English alphabet $\{A, B, \ldots, Z\}$, compute the number of distinct four-letter ""words"" for which $E E$ is a subsequence.","[""Divide into cases according to the number of $E$ 's in the word. If there are only two $E$ 's, then the word must have two non- $E$ letters, represented by ?'s. There are $\\left(\\begin{array}{l}4 \\\\ 2\\end{array}\\right)=6$ arrangements of two $E$ 's and two ?'s, and each of the ?'s can be any of 25 letters, so there are $6 \\cdot 25^{2}=3750$ possible words. If there are three $E$ 's, then the word has exactly one non- $E$ letter, and so there are 4 arrangements times 25 choices for the letter, or 100 possible words. There is one word with four $E$ 's, hence a total of 3851 words.""]",['3851'],False,,Numerical, 2956,Geometry,,"Six solid regular tetrahedra are placed on a flat surface so that their bases form a regular hexagon $\mathcal{H}$ with side length 1 , and so that the vertices not lying in the plane of $\mathcal{H}$ (the ""top"" vertices) are themselves coplanar. A spherical ball of radius $r$ is placed so that its center is directly above the center of the hexagon. The sphere rests on the tetrahedra so that it is tangent to one edge from each tetrahedron. If the ball's center is coplanar with the top vertices of the tetrahedra, compute $r$.","['Let $O$ be the center of the sphere, $A$ be the top vertex of one tetrahedron, and $B$ be the center of the hexagon.\n\n\n\nThen $B O$ equals the height of the tetrahedron, which is $\\frac{\\sqrt{6}}{3}$. Because $A$ is directly above the centroid of the bottom face, $A O$ is two-thirds the length of the median of one triangular face, so $A O=\\frac{2}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\sqrt{3}}{3}$. The radius of the sphere is the altitude to hypotenuse $\\overline{A B}$ of $\\triangle A B O$, so the area of $\\triangle A B O$ can be represented in two ways: $[A B O]=\\frac{1}{2} A O \\cdot B O=\\frac{1}{2} A B \\cdot r$. Substitute given and computed values to obtain $\\frac{1}{2}\\left(\\frac{\\sqrt{3}}{3}\\right)\\left(\\frac{\\sqrt{6}}{3}\\right)=\\frac{1}{2}(1)(r)$, from which $r=\\frac{\\sqrt{18}}{9}=\\frac{\\sqrt{2}}{3}$.']",['$\\frac{\\sqrt{2}}{3}$'],False,,Numerical, 2957,Combinatorics,,"Derek starts at the point $(0,0)$, facing the point $(0,1)$, and he wants to get to the point $(1,1)$. He takes unit steps parallel to the coordinate axes. A move consists of either a step forward, or a $90^{\circ}$ right (clockwise) turn followed by a step forward, so that his path does not contain any left turns. His path is restricted to the square region defined by $0 \leq x \leq 17$ and $0 \leq y \leq 17$. Compute the number of ways he can get to $(1,1)$ without returning to any previously visited point.","['Divide into cases according to the number of right turns Derek makes.\n\n- There is one route involving only one turn: move first to $(0,1)$ and then to $(1,1)$.\n- If he makes two turns, he could move up to $(0, a)$ then to $(1, a)$ and then down to $(1,1)$. In order to do this, $a$ must satisfy $1","['Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nCompute $[A K M]$ as $\\frac{1}{2}$ base $\\cdot$ height, using base $\\overline{A M}$.\n\n\n\nBecause of (*), $\\triangle A Y O \\sim \\triangle A K L$. To compute $A M$, notice that in $\\triangle A Y O, A O=A M-r$, while in $\\triangle A K L$, the corresponding side $A L=A M+M L=A M+2$. Therefore:\n\n$$\n\\begin{aligned}\n\\frac{A O}{A L} & =\\frac{Y O}{K L} \\\\\n\\frac{A M-\\frac{5}{4}}{A M+2} & =\\frac{5 / 4}{3}\n\\end{aligned}\n$$\n\nfrom which $A M=\\frac{25}{7}$. Draw the altitude of $\\triangle A K M$ from vertex $K$, and let $h$ be its length. In right triangle $O K L, h$ is the altitude to the hypotenuse, so $\\frac{h}{3}=\\sin (\\angle K L O)=\\frac{r}{r+2}$. Hence $h=\\frac{15}{13}$. Therefore $[A K M]=\\frac{1}{2} \\cdot \\frac{25}{7} \\cdot \\frac{15}{13}=\\frac{375}{182}$.', 'Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nBy the Power of the Point Theorem, $L K^{2}=L M \\cdot L R$, so\n\n$$\n\\begin{aligned}\nL R & =\\frac{9}{2} \\\\\nR M & =L R-L M=\\frac{5}{2} \\\\\nO L & =r+M L=\\frac{13}{4}\n\\end{aligned}\n$$\n\nFrom (*), we know that $\\triangle A Y O \\sim \\triangle A K L$. Hence by $(\\dagger)$,\n\n$$\n\\frac{A L}{A O}=\\frac{A L}{A L-O L}=\\frac{K L}{Y O}=\\frac{3}{5 / 4}=\\frac{12}{5}, \\quad \\text { thus } \\quad A L=\\frac{12}{7} \\cdot O L=\\frac{12}{7} \\cdot \\frac{13}{4}=\\frac{39}{7}\n$$\n\nHence $A M=A L-2=\\frac{25}{7}$. The ratio between the areas of triangles $A K M$ and $R K M$ is equal to\n\n$$\n\\frac{[A K M]}{[R K M]}=\\frac{A M}{R M}=\\frac{25 / 7}{5 / 2}=\\frac{10}{7}\n$$\n\nThus $[A K M]=\\frac{10}{7} \\cdot[R K M]$.\n\nBecause $\\angle K R L$ and $\\angle M K L$ both subtend $\\widehat{K M}, \\triangle K R L \\sim \\triangle M K L$. Therefore $\\frac{K R}{M K}=\\frac{L K}{L M}=$ $\\frac{3}{2}$. Thus let $K R=3 x$ and $M K=2 x$ for some positive real number $x$. Because $R M$ is a diameter of $\\omega$ (see left diagram below), $\\mathrm{m} \\angle R K M=90^{\\circ}$. Thus triangle $R K M$ is a right triangle with hypotenuse $\\overline{R M}$. In particular, $13 x^{2}=K R^{2}+M K^{2}=R M^{2}=\\frac{25}{4}$, so $x^{2}=\\frac{25}{52}$ and $[R K M]=\\frac{R K \\cdot K M}{2}=3 x^{2}$. Therefore\n\n$$\n[A K M]=\\frac{10}{7} \\cdot[R K M]=\\frac{10}{7} \\cdot 3 \\cdot \\frac{25}{52}=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$\n\n', 'Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\n\nLet $U$ and $V$ be the respective feet of the perpendiculars dropped from $A$ and $M$ to $\\overleftrightarrow{K L}$. From (*), $\\triangle A K L$ can be dissected into two infinite progressions of triangles: one progression of triangles similar to $\\triangle O K L$ and the other similar to $\\triangle Y O K$, as shown in the right diagram above. In both progressions, the corresponding sides of the triangles have common ratio equal to\n\n$$\n\\frac{Y O}{K L}=\\frac{5 / 4}{3}=\\frac{5}{12}\n$$\n\n\n\nThus\n\n$$\nA U=\\frac{5}{4}\\left(1+\\frac{5}{12}+\\left(\\frac{5}{12}\\right)^{2}+\\cdots\\right)=\\frac{5}{4} \\cdot \\frac{12}{7}=\\frac{15}{7}\n$$\n\nBecause $\\triangle L M V \\sim \\triangle L O K$, and because $L O=\\frac{13}{4}$ by $(\\dagger)$,\n\n$$\n\\frac{M V}{O K}=\\frac{L M}{L O}, \\quad \\text { thus } \\quad M V=\\frac{O K \\cdot L M}{L O}=\\frac{\\frac{5}{4} \\cdot 2}{\\frac{13}{4}}=\\frac{10}{13}\n$$\n\nFinally, note that $[A K M]=[A K L]-[K L M]$. Because $\\triangle A K L$ and $\\triangle K L M$ share base $\\overline{K L}$,\n\n$$\n[A K M]=\\frac{1}{2} \\cdot 3 \\cdot\\left(\\frac{15}{7}-\\frac{10}{13}\\right)=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$']",['$\\frac{375}{182}$'],False,,Numerical, 2959,Geometry,,"Points $A$ and $L$ lie outside circle $\omega$, whose center is $O$, and $\overline{A L}$ contains diameter $\overline{R M}$, as shown below. Circle $\omega$ is tangent to $\overline{L K}$ at $K$. Also, $\overline{A K}$ intersects $\omega$ at $Y$, which is between $A$ and $K$. If $K L=3, M L=2$, and $\mathrm{m} \angle A K L-\mathrm{m} \angle Y M K=90^{\circ}$, compute $[A K M]$ (i.e., the area of $\triangle A K M$ ). ![](https://cdn.mathpix.com/cropped/2023_12_21_32b17155940ff395abf0g-1.jpg?height=396&width=740&top_left_y=1057&top_left_x=736)","['Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=404&width=742&top_left_y=351&top_left_x=735)\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nCompute $[A K M]$ as $\\frac{1}{2}$ base $\\cdot$ height, using base $\\overline{A M}$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=407&width=746&top_left_y=1428&top_left_x=733)\n\nBecause of (*), $\\triangle A Y O \\sim \\triangle A K L$. To compute $A M$, notice that in $\\triangle A Y O, A O=A M-r$, while in $\\triangle A K L$, the corresponding side $A L=A M+M L=A M+2$. Therefore:\n\n$$\n\\begin{aligned}\n\\frac{A O}{A L} & =\\frac{Y O}{K L} \\\\\n\\frac{A M-\\frac{5}{4}}{A M+2} & =\\frac{5 / 4}{3}\n\\end{aligned}\n$$\n\nfrom which $A M=\\frac{25}{7}$. Draw the altitude of $\\triangle A K M$ from vertex $K$, and let $h$ be its length. In right triangle $O K L, h$ is the altitude to the hypotenuse, so $\\frac{h}{3}=\\sin (\\angle K L O)=\\frac{r}{r+2}$. Hence $h=\\frac{15}{13}$. Therefore $[A K M]=\\frac{1}{2} \\cdot \\frac{25}{7} \\cdot \\frac{15}{13}=\\frac{375}{182}$.', 'Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=404&width=742&top_left_y=351&top_left_x=735)\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nBy the Power of the Point Theorem, $L K^{2}=L M \\cdot L R$, so\n\n$$\n\\begin{aligned}\nL R & =\\frac{9}{2} \\\\\nR M & =L R-L M=\\frac{5}{2} \\\\\nO L & =r+M L=\\frac{13}{4}\n\\end{aligned}\n$$\n\nFrom (*), we know that $\\triangle A Y O \\sim \\triangle A K L$. Hence by $(\\dagger)$,\n\n$$\n\\frac{A L}{A O}=\\frac{A L}{A L-O L}=\\frac{K L}{Y O}=\\frac{3}{5 / 4}=\\frac{12}{5}, \\quad \\text { thus } \\quad A L=\\frac{12}{7} \\cdot O L=\\frac{12}{7} \\cdot \\frac{13}{4}=\\frac{39}{7}\n$$\n\nHence $A M=A L-2=\\frac{25}{7}$. The ratio between the areas of triangles $A K M$ and $R K M$ is equal to\n\n$$\n\\frac{[A K M]}{[R K M]}=\\frac{A M}{R M}=\\frac{25 / 7}{5 / 2}=\\frac{10}{7}\n$$\n\nThus $[A K M]=\\frac{10}{7} \\cdot[R K M]$.\n\nBecause $\\angle K R L$ and $\\angle M K L$ both subtend $\\widehat{K M}, \\triangle K R L \\sim \\triangle M K L$. Therefore $\\frac{K R}{M K}=\\frac{L K}{L M}=$ $\\frac{3}{2}$. Thus let $K R=3 x$ and $M K=2 x$ for some positive real number $x$. Because $R M$ is a diameter of $\\omega$ (see left diagram below), $\\mathrm{m} \\angle R K M=90^{\\circ}$. Thus triangle $R K M$ is a right triangle with hypotenuse $\\overline{R M}$. In particular, $13 x^{2}=K R^{2}+M K^{2}=R M^{2}=\\frac{25}{4}$, so $x^{2}=\\frac{25}{52}$ and $[R K M]=\\frac{R K \\cdot K M}{2}=3 x^{2}$. Therefore\n\n$$\n[A K M]=\\frac{10}{7} \\cdot[R K M]=\\frac{10}{7} \\cdot 3 \\cdot \\frac{25}{52}=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_352f1bc0cb3b408f3284g-1.jpg?height=364&width=1484&top_left_y=1650&top_left_x=367)', 'Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=404&width=742&top_left_y=351&top_left_x=735)\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\n\nLet $U$ and $V$ be the respective feet of the perpendiculars dropped from $A$ and $M$ to $\\overleftrightarrow{K L}$. From (*), $\\triangle A K L$ can be dissected into two infinite progressions of triangles: one progression of triangles similar to $\\triangle O K L$ and the other similar to $\\triangle Y O K$, as shown in the right diagram above. In both progressions, the corresponding sides of the triangles have common ratio equal to\n\n$$\n\\frac{Y O}{K L}=\\frac{5 / 4}{3}=\\frac{5}{12}\n$$\n\n\n\nThus\n\n$$\nA U=\\frac{5}{4}\\left(1+\\frac{5}{12}+\\left(\\frac{5}{12}\\right)^{2}+\\cdots\\right)=\\frac{5}{4} \\cdot \\frac{12}{7}=\\frac{15}{7}\n$$\n\nBecause $\\triangle L M V \\sim \\triangle L O K$, and because $L O=\\frac{13}{4}$ by $(\\dagger)$,\n\n$$\n\\frac{M V}{O K}=\\frac{L M}{L O}, \\quad \\text { thus } \\quad M V=\\frac{O K \\cdot L M}{L O}=\\frac{\\frac{5}{4} \\cdot 2}{\\frac{13}{4}}=\\frac{10}{13}\n$$\n\nFinally, note that $[A K M]=[A K L]-[K L M]$. Because $\\triangle A K L$ and $\\triangle K L M$ share base $\\overline{K L}$,\n\n$$\n[A K M]=\\frac{1}{2} \\cdot 3 \\cdot\\left(\\frac{15}{7}-\\frac{10}{13}\\right)=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$']",['$\\frac{375}{182}$'],False,,Numerical, 2960,Number Theory,,"Let $N$ be a perfect square between 100 and 400 , inclusive. What is the only digit that cannot appear in $N$ ?","['When the perfect squares between 100 and 400 inclusive are listed out, every digit except 7 is used. Note that the perfect squares 100, 256, 289, 324 use each of the other digits.']",['7'],False,,Numerical, 2961,Number Theory,,"Let $T=7$. Let $A$ and $B$ be distinct digits in base $T$, and let $N$ be the largest number of the form $\underline{A} \underline{B} \underline{A}_{T}$. Compute the value of $N$ in base 10 .","['To maximize $\\underline{A} \\underline{B} \\underline{A}_{T}$ with $A \\neq B$, let $A=T-1$ and $B=T-2$. Then $\\underline{A} \\underline{B}^{A} \\underline{A}_{T}=$ $(T-1) \\cdot T^{2}+(T-2) \\cdot T^{1}+(T-1) \\cdot T^{0}=T^{3}-T-1$. With $T=7$, the answer is 335 .']",['335'],False,,Numerical, 2962,Number Theory,,"Let T be an integer. Given a nonzero integer $n$, let $f(n)$ denote the sum of all numbers of the form $i^{d}$, where $i=\sqrt{-1}$, and $d$ is a divisor (positive or negative) of $n$. Compute $f(2 T+1)$.","['Let $n=2^{m} r$, where $r$ is odd. If $m=0$, then $n$ is odd, and for each $d$ that divides $n$, $i^{d}+i^{-d}=i^{d}+\\frac{i^{d}}{\\left(i^{2}\\right)^{d}}=0$, hence $f(n)=0$ when $n$ is odd. If $m=1$, then for each $d$ that divides $n, i^{d}+i^{-d}$ equals 0 if $d$ is odd, and -2 if $d$ is even. Thus when $n$ is a multiple of 2 but not 4 , $f(n)=-2 P$, where $P$ is the number of positive odd divisors of $n$. Similarly, if $m=2$, then $f(n)=0$, and in general, $f(n)=2(m-2) P$ for $m \\geq 1$. Because $T$ is an integer, $2 T+1$ is odd, hence the answer is $\\mathbf{0}$. [Note: If $r=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdot \\ldots \\cdot p_{k}^{a_{k}}$, where the $p_{i}$ are distinct odd primes, it is well known that $P=\\left(a_{1}+1\\right)\\left(a_{2}+1\\right) \\ldots\\left(a_{k}+1\\right)$.]']",['0'],False,,Numerical, 2963,Algebra,,"Let $T=0$. Compute the real value of $x$ for which there exists a solution to the system of equations $$ \begin{aligned} x+y & =0 \\ x^{3}-y^{3} & =54+T . \end{aligned} $$","['$\\quad$ Plug $y=-x$ into the second equation to obtain $x=\\sqrt[3]{\\frac{54+T}{2}}$. With $T=0, x=\\sqrt[3]{27}=3$.']",['3'],False,,Numerical, 2964,Geometry,,"Let $T=3$. In $\triangle A B C, A C=T^{2}, \mathrm{~m} \angle A B C=45^{\circ}$, and $\sin \angle A C B=\frac{8}{9}$. Compute $A B$.","['From the Law of Sines, $\\frac{A B}{\\sin \\angle A C B}=\\frac{A C}{\\sin \\angle A B C}$. Thus $A B=\\frac{8}{9} \\cdot \\frac{T^{2}}{1 / \\sqrt{2}}=\\frac{8 \\sqrt{2}}{9} \\cdot T^{2}$. With $T=3, A B=\\mathbf{8} \\sqrt{\\mathbf{2}}$.']",['$8 \\sqrt{2}$'],False,,Numerical, 2965,Geometry,,"Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle. ","['Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \\sqrt{2}$, note that $r>2 \\sqrt{2} \\rightarrow 3 r>6 \\sqrt{2}=\\sqrt{72}$. The least integer greater than $\\sqrt{72}$ is 9 .']",['9'],False,,Numerical, 2965,Geometry,,"Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle. ![](https://cdn.mathpix.com/cropped/2023_12_21_431bf83118cb8d582b13g-1.jpg?height=347&width=434&top_left_y=1569&top_left_x=1580)","['Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \\sqrt{2}$, note that $r>2 \\sqrt{2} \\rightarrow 3 r>6 \\sqrt{2}=\\sqrt{72}$. The least integer greater than $\\sqrt{72}$ is 9 .']",['9'],False,,Numerical, 2966,Algebra,,"Let $T=9$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is an arithmetic progression, $d$ is the common difference, $a_{T}=10$, and $a_{K}=2010$, where $K>T$. If $d$ is an integer, compute the value of $K$ such that $|K-d|$ is minimal.","['Note that $a_{T}=a_{1}+(T-1) d$ and $a_{K}=a_{1}+(K-1) d$, hence $a_{K}-a_{T}=(K-T) d=2010-10=$ 2000. Thus $K=\\frac{2000}{d}+T$, and to minimize $\\left|T+\\frac{2000}{d}-d\\right|$, choose a positive integer $d$ such that $\\frac{2000}{d}$ is also an integer and $\\frac{2000}{d}-d$ is as close as possible to $-T$. Note that $T>0$, so $\\frac{2000}{d}-d$ should be negative, i.e., $d^{2}>2000$ or $d>44$. The value of $T$ determines how far apart $\\frac{2000}{d}$ and $d$ need to be. For example, if $T$ is close to zero, then choose $d$ such that $\\frac{2000}{d}$ and $d$ are close to each other. With $T=9$, take $d=50$ so that $\\frac{2000}{d}=40$ and $|K-d|=|49-50|=1$. Thus $K=49$.']",['49'],False,,Numerical, 2967,Algebra,,"Let $A$ be the number you will receive from position 7 , and let $B$ be the number you will receive from position 9 . There are exactly two ordered pairs of real numbers $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ that satisfy both $|x+y|=6(\sqrt{A}-5)$ and $x^{2}+y^{2}=B^{2}$. Compute $\left|x_{1}\right|+\left|y_{1}\right|+\left|x_{2}\right|+\left|y_{2}\right|$.","['Note that the graph of $x^{2}+y^{2}=B^{2}$ is a circle of radius $|B|$ centered at $(0,0)$ (as long as $\\left.B^{2}>0\\right)$. Also note that the graph of $|x+y|=6(\\sqrt{A}-5)$ is either the line $y=-x$ if $A=25$, or the graph consists of two parallel lines with slope -1 if $A>25$. In the former case, the\n\n\nline $y=-x$ intersects the circle at the points $\\left( \\pm \\frac{|B|}{\\sqrt{2}}, \\mp \\frac{|B|}{\\sqrt{2}}\\right)$. In the latter case, the graph is symmetric about the origin, and in order to have exactly two intersection points, each line must be tangent to the circle, and the tangency points are $\\left(\\frac{|B|}{\\sqrt{2}}, \\frac{|B|}{\\sqrt{2}}\\right)$ and $\\left(-\\frac{|B|}{\\sqrt{2}},-\\frac{|B|}{\\sqrt{2}}\\right)$. In either case, $\\left|x_{1}\\right|+\\left|y_{1}\\right|+\\left|x_{2}\\right|+\\left|y_{2}\\right|=2 \\sqrt{2} \\cdot|B|$, and in the case where the graph is two lines, this is also equal to $12(\\sqrt{A}-5)$. Thus if $A \\neq 25$, then only one of $A$ or $B$ is needed to determine the answer. With $A=49$ and $B=6 \\sqrt{2}$, the answer is $2 \\sqrt{2} \\cdot 6 \\sqrt{2}=12(\\sqrt{49}-5)=\\mathbf{2 4}$.']",['24'],False,,Numerical, 2968,Geometry,,"Let $T=23$. In triangle $A B C$, the altitude from $A$ to $\overline{B C}$ has length $\sqrt{T}, A B=A C$, and $B C=T-K$, where $K$ is the real root of the equation $x^{3}-8 x^{2}-8 x-9=0$. Compute the length $A B$.","['Rewrite the equation as $x^{3}-1=8\\left(x^{2}+x+1\\right)$, so that $(x-1)\\left(x^{2}+x+1\\right)=8\\left(x^{2}+x+1\\right)$. Because $x^{2}+x+1$ has no real zeros, it can be canceled from both sides of the equation to obtain $x-1=8$ or $x=9$. Hence $B C=T-9$, and $A B^{2}=(\\sqrt{T})^{2}+\\left(\\frac{T-9}{2}\\right)^{2}=T+\\left(\\frac{T-9}{2}\\right)^{2}$. Substitute $T=23$ to obtain $A B=\\sqrt{72}=\\mathbf{6} \\sqrt{\\mathbf{2}}$.']",['$6 \\sqrt{2}$'],False,,Numerical, 2969,Geometry,,Let $T=8$. A cube has volume $T-2$. The cube's surface area equals one-eighth the surface area of a $2 \times 2 \times n$ rectangular prism. Compute $n$.,"[""The cube's side length is $\\sqrt[3]{T}$, so its surface area is $6 \\sqrt[3]{T^{2}}$. The rectangular prism has surface area $2(2 \\cdot 2+2 \\cdot n+2 \\cdot n)=8+8 n$, thus $6 \\sqrt[3]{T^{2}}=1+n$. With $T=8, n=6 \\sqrt[3]{64}-1=\\mathbf{2 3}$.""]",['23'],False,,Numerical, 2970,Number Theory,,"Let $T=98721$, and let $K$ be the sum of the digits of $T$. Let $A_{n}$ be the number of ways to tile a $1 \times n$ rectangle using $1 \times 3$ and $1 \times 1$ tiles that do not overlap. Tiles of both types need not be used; for example, $A_{3}=2$ because a $1 \times 3$ rectangle can be tiled with three $1 \times 1$ tiles or one $1 \times 3$ tile. Compute the smallest value of $n$ such that $A_{n} \geq K$.","[""Consider the rightmost tile of the rectangle. If it's a $1 \\times 1$ tile, then there are $A_{n-1}$ ways to tile the remaining $1 \\times(n-1)$ rectangle, and if it's a $1 \\times 3$ tile, then there are $A_{n-3}$ ways to tile the remaining $1 \\times(n-3)$ rectangle. Hence $A_{n}=A_{n-1}+A_{n-3}$ for $n>3$, and $A_{1}=A_{2}=1, A_{3}=2$. Continuing the sequence gives the following values:\n\n| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $A_{n}$ | 1 | 1 | 2 | 3 | 4 | 6 | 9 | 13 | 19 | 28 |\n\nWith $T=98721, K=27$, hence the answer is 10 .""]",['10'],False,,Numerical, 2971,Combinatorics,,"Let $T=3$, and let $K=T+2$. Compute the largest $K$-digit number which has distinct digits and is a multiple of 63.","['Let $N_{K}$ be the largest $K$-digit number which has distinct digits and is a multiple of 63 . It can readily be verified that $N_{1}=0, N_{2}=63$, and $N_{3}=945$. For $K>3$, compute $N_{K}$ using the following strategy: start with the number $M_{0}=\\underline{9} \\underline{8} \\underline{7} \\ldots(10-K)$; let $M_{1}$ be the largest multiple of 63 not exceeding $M_{0}$. That is, to compute $M_{1}$, divide $M_{0}$ by 63 and discard the remainder: $M_{0}=1587 \\cdot 63+44$, so $M_{1}=M_{0}-44=1587 \\cdot 63$. If $M_{1}$ has distinct digits, then $N_{K}=M_{1}$. Otherwise, let $M_{2}=M_{1}-63, M_{3}=M_{2}-63$, and so on; then $N_{K}$ is the first term of the sequence $M_{1}, M_{2}, M_{3}, \\ldots$ that has distinct digits. Applying this strategy gives $N_{4}=9765, N_{5}=98721, N_{6}=987651$, and $N_{7}=9876510$. With $T=3, K=5$, and the answer is $\\mathbf{9 8 7 2 1}$.']",['98721'],False,,Numerical, 2972,Algebra,,"Let $T\neq 0$. Suppose that $a, b, c$, and $d$ are real numbers so that $\log _{a} c=\log _{b} d=T$. Compute $$ \frac{\log _{\sqrt{a b}}(c d)^{3}}{\log _{a} c+\log _{b} d} $$","['Note that $a^{T}=c$ and $b^{T}=d$, thus $(a b)^{T}=c d$. Further note that $(a b)^{3 T}=(\\sqrt{a b})^{6 T}=(c d)^{3}$, thus $\\log _{\\sqrt{a b}}(c d)^{3}=6 T$. Thus the given expression simplifies to $\\frac{6 T}{2 T}=\\mathbf{3}$ (as long as $T \\neq 0$ ).']",['3'],False,,Numerical, 2973,Algebra,,"Let $T=2030$. Given that $\mathrm{A}, \mathrm{D}, \mathrm{E}, \mathrm{H}, \mathrm{S}$, and $\mathrm{W}$ are distinct digits, and that $\underline{\mathrm{W}} \underline{\mathrm{A}} \underline{\mathrm{D}} \underline{\mathrm{E}}+\underline{\mathrm{A}} \underline{\mathrm{S}} \underline{\mathrm{H}}=T$, what is the largest possible value of $\mathrm{D}+\mathrm{E}$ ?","['First note that if $T \\geq 10000$, then $\\mathrm{W}=9$ and $\\mathrm{A} \\geq 5$. If $T<10000$ and $x$ is the leading digit of $T$, then either $\\mathrm{W}=x$ and $\\mathrm{A} \\leq 4$ or $\\mathrm{W}=x-1$ and $\\mathrm{A} \\geq 5$. With $T=2030$, either $\\underline{\\mathrm{W}} \\underline{\\mathrm{A}}=20$\n\n\nor $\\underline{W} \\underline{A}=15$. In either case, $\\underline{D} \\underline{E}+\\underline{S} \\underline{H}=30$. Considering values of $D+E$, there are three possibilities to consider:\n\n$\\mathrm{D}+\\mathrm{E}=11: \\underline{\\mathrm{D}} \\underline{\\mathrm{E}}=29, \\underline{\\mathrm{S}} \\underline{\\mathrm{H}}=01$, which duplicates digits;\n\n$\\mathrm{D}+\\mathrm{E}=10: \\underline{\\mathrm{D}} \\underline{\\underline{E}}=28, \\underline{\\mathrm{S}} \\underline{\\underline{H}}=02$ or $\\underline{\\mathrm{D}} \\underline{E}=19, \\underline{\\mathrm{S}} \\underline{\\mathrm{H}}=11$, both of which duplicate digits;\n\n$\\mathrm{D}+\\mathrm{E}=9: \\quad \\underline{\\mathrm{D}} \\underline{\\mathrm{E}}=27, \\underline{\\mathrm{S}} \\underline{\\mathrm{H}}=03$, in which no digits are duplicated if $\\underline{\\mathrm{W}} \\underline{\\mathrm{A}}=15$.\n\nTherefore the answer is $\\mathbf{9}$.']",['9'],False,,Numerical, 2974,Number Theory,,Let $f(x)=2^{x}+x^{2}$. Compute the smallest integer $n>10$ such that $f(n)$ and $f(10)$ have the same units digit.,"['The units digit of $f(10)$ is the same as the units digit of $2^{10}$. Because the units digits of powers of 2 cycle in groups of four, the units digit of $2^{10}$ is 4 , so the units digit of $f(10)$ is 4 . Note that $n$ must be even, otherwise, the units digit of $f(n)$ is odd. If $n$ is a multiple of 4 , then $2^{n}$ has 6 as its units digit, which means that $n^{2}$ would need to have a units digit of 8 , which is impossible. Thus $n$ is even, but is not a multiple of 4 . This implies that the units digit of $2^{n}$ is 4 , and so $n^{2}$ must have a units digit of 0 . The smallest possible value of $n$ is therefore 30 .']",['30'],False,,Numerical, 2975,Geometry,,"In rectangle $P A U L$, point $D$ is the midpoint of $\overline{U L}$ and points $E$ and $F$ lie on $\overline{P L}$ and $\overline{P A}$, respectively such that $\frac{P E}{E L}=\frac{3}{2}$ and $\frac{P F}{F A}=2$. Given that $P A=36$ and $P L=25$, compute the area of pentagon $A U D E F$.","['For convenience, let $P A=3 x$ and let $P L=5 y$. Then the given equations involving ratios of segment lengths imply that $P E=3 y, E L=2 y, P F=2 x$, and $F A=x$. Then $[P A U L]=(3 x)(5 y)=15 x y$ and\n\n$$\n\\begin{aligned}\n{[A U D E F] } & =[P A U L]-[P E F]-[E L D] \\\\\n& =15 x y-\\frac{1}{2}(3 y)(2 x)-\\frac{1}{2}(2 y)\\left(\\frac{3 x}{2}\\right) \\\\\n& =15 x y-3 x y-\\frac{3 x y}{2} \\\\\n& =\\frac{21 x y}{2} .\n\\end{aligned}\n$$\n\nBecause $15 x y=36 \\cdot 25$, it follows that $3 x y=36 \\cdot 5=180$ and that $\\frac{21 x y}{2}=\\frac{7}{2}(3 x y)=\\frac{7}{2} \\cdot 180=\\mathbf{6 3 0}$.']",['630'],False,,Numerical, 2976,Geometry,,Rectangle $A R M L$ has length 125 and width 8. The rectangle is divided into 1000 squares of area 1 by drawing in gridlines parallel to the sides of $A R M L$. Diagonal $\overline{A M}$ passes through the interior of exactly $n$ of the 1000 unit squares. Compute $n$.,"['Notice that 125 and 8 are relatively prime. Examining rectangles of size $a \\times b$ where $a$ and $b$ are small and relatively prime suggests an answer of $a+b-1$. To see that this is the case, note that other than the endpoints, the diagonal does not pass through any vertex of any unit square. After the first square, it must enter each subsequent square via a vertical or horizontal side. By continuity, the total number of these sides is the sum of the $a-1$ interior vertical lines and $b-1$ interior horizontal lines. The diagonal passes through $(a-1)+(b-1)=a+b-2$ additional squares, so the total is $a+b-1$. Because 125 and 8 are relatively prime, it follows that $N=125+8-1=\\mathbf{1 3 2}$.\n\nRemark: As an exercise, the reader is encouraged to show that the answer for general $a$ and $b$ is $a+b-\\operatorname{gcd}(a, b)$.']",['132'],False,,Numerical, 2977,Number Theory,,Compute the least integer $n>1$ such that the product of all positive divisors of $n$ equals $n^{4}$.,"['Note that every factor pair $d$ and $\\frac{n}{d}$ have product $n$. For the product of all such divisor pairs to equal $n^{4}$, there must be exactly 4 divisor pairs, or 8 positive integer divisors. A number has 8 positive integer divisors if it is of the form $a^{3} b^{1}$ or $a^{7}$ where $a$ and $b$ are distinct primes. The prime factorization $a^{3} b^{1}(a \\neq b)$ provides a set of divisors each of which has 4 options for using $a\\left(a^{0}, a^{1}, a^{2}, a^{3}\\right)$ and an independent 2 options for using $b\\left(b^{0}, b^{1}\\right)$. Using the least values $(a, b)=(2,3), a^{3} b^{1}=24$. If instead the prime factorization is $a^{7}$ (having divisors $a^{0}, a^{1}, a^{2}, \\ldots, a^{7}$ ), the least answer would be $2^{7}=128$. Thus the answer is 24 .']",['24'],False,,Numerical, 2978,Combinatorics,,Each of the six faces of a cube is randomly colored red or blue with equal probability. Compute the probability that no three faces of the same color share a common vertex.,"['There are $2^{6}=64$ colorings of the cube. Let $r$ be the number of faces that are colored red. Define a monochromatic vertex to be a vertex of the cube for which the three faces meeting there have the same color. It is clear that a coloring without a monochromatic vertex is only possible in the cases $2 \\leq r \\leq 4$. If $r=2$ or $r=4$, the only colorings that do not have a monochromatic vertex occur when two opposing faces are colored with the minority color (red in the $r=2$ case, blue in the $r=4$ case). Because there are 3 pairs of opposite\n\n\n\nfaces of a cube, there are 3 colorings without a monochromatic vertex if $r=2$ and another 3 such colorings if $r=4$. For the $r=3$ colorings, of which there are 20, the only cases in which there are monochromatic vertices occur when opposing faces are monochromatic, but in different colors. There are $2^{3}=8$ such colorings, leaving $20-8=12$ colorings that do not have a monochromatic vertex. Therefore $3+3+12=18$ of the 64 colorings have no monochromatic vertex, and the answer is $\\frac{\\mathbf{9}}{\\mathbf{3 2}}$.']",['$\\frac{9}{32}$'],False,,Numerical, 2979,Geometry,,"Scalene triangle $A B C$ has perimeter 2019 and integer side lengths. The angle bisector from $C$ meets $\overline{A B}$ at $D$ such that $A D=229$. Given that $A C$ and $A D$ are relatively prime, compute $B C$.","['Let $B C=a, A C=b, A B=c$. Also, let $A D=e$ and $B D=f$. Then $a+b+e+f=2019$, the values $a, b$, and $e+f$ are integers, and by the Angle Bisector Theorem, $\\frac{e}{f}=\\frac{b}{a}$. So $b=\\frac{a e}{f}=\\frac{229 a}{f}$. Because 229 is prime and $\\operatorname{gcd}(b, e)=1$, conclude that $f$ must be an integer multiple of 229 . So let $f=229 x$ for some integer $x$. Then $a=b \\cdot x$ and $a+b+c=2019$ implies $2019=b x+b+229+229 x=(b+229)(1+x)$. Because $2019=673 \\cdot 3$, it follows that $b=444$ and $x=2$, from which $B C=a=\\mathbf{8 8 8}$.']",['888'],False,,Numerical, 2980,Algebra,,"Given that $a$ and $b$ are positive and $$ \lfloor 20-a\rfloor=\lfloor 19-b\rfloor=\lfloor a b\rfloor, $$ compute the least upper bound of the set of possible values of $a+b$.","['Let the common value of the three expressions in the given equation be $N$. Maximizing $a+b$ involves making at least one of $a$ and $b$ somewhat large, which makes the first two expressions for $N$ small. So, to maximize $a+b$, look for the least possible value of $N$. One can show that $N=14$ is not possible because that would require $a>5$ and $b>4$, which implies $a b>20$. But $N=15$ is possible by setting $a=4+x, b=3+y$, where $0A>R$ and $R']",['$\\frac{4}{3}$'],False,,Numerical, 2983,Number Theory,,"Given that $a, b, c$, and $d$ are positive integers such that $$ a ! \cdot b ! \cdot c !=d ! \quad \text { and } \quad a+b+c+d=37 $$ compute the product $a b c d$.","['Without loss of generality, assume $a \\leq b \\leq c\\sqrt{64}=8$. Because $8.1^{2}=65.61$ and $8.15^{2}=66.4225>66$, conclude that $81<10 \\sqrt{66}<81.5$, hence $10 a$ rounded to the nearest integer is 81 , and the answer is $81-14=\\mathbf{6 7}$.']",['67'],False,,Numerical, 2989,Algebra,,"Let $T=67$. A group of children and adults go to a rodeo. A child's admission ticket costs $\$ 5$, and an adult's admission ticket costs more than $\$ 5$. The total admission cost for the group is $\$ 10 \cdot T$. If the number of adults in the group were to increase by $20 \%$, then the total cost would increase by $10 \%$. Compute the number of children in the group.","[""Suppose there are $x$ children and $y$ adults in the group and each adult's admission ticket costs $\\$ a$. The given information implies that $5 x+a y=10 T$ and $5 x+1.2 a y=11 T$. Subtracting the first equation from the second yields $0.2 a y=T \\rightarrow a y=5 T$, so from the first equation, $5 x=5 T \\rightarrow x=T$. With $T=67$, the answer is 67 .""]",['67'],False,,Numerical, 2990,Geometry,,"Let $T=67$. Rectangles $F A K E$ and $F U N K$ lie in the same plane. Given that $E F=T$, $A F=\frac{4 T}{3}$, and $U F=\frac{12}{5}$, compute the area of the intersection of the two rectangles.","['Without loss of generality, let $A, U$, and $N$ lie on the same side of $\\overline{F K}$. Applying the Pythagorean Theorem to triangle $A F K$, conclude that $F K=\\frac{5 T}{3}$. Comparing the altitude to $\\overline{F K}$ in triangle $A F K$ to $\\overline{U F}$, note that the intersection of the two rectangles will be a triangle with area $\\frac{2 T^{2}}{3}$ if $\\frac{4 T}{5} \\leq \\frac{12}{5}$, or $T \\leq 3$. Otherwise, the intersection will be a trapezoid. In this case, using similarity, the triangular regions of $F U N K$ that lie outside of FAKE each have one leg of length $\\frac{12}{5}$ and the others of lengths $\\frac{16}{5}$ and $\\frac{9}{5}$, respectively. Thus their combined areas $\\frac{1}{2} \\cdot \\frac{12}{5}\\left(\\frac{16}{5}+\\frac{9}{5}\\right)=6$, hence the area of the intersection is $\\frac{5 T}{3} \\cdot \\frac{12}{5}-6=4 T-6$. With $T=67$, the answer is therefore $\\mathbf{2 6 2}$.']",['262'],False,,Numerical, 2991,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Compute the $E(6,1)$","['$E(6,1)=6$. Note that at least six minutes are required because exactly one switch is flipped each minute. By flipping all six switches (in any order) in the first six minutes, the door will open in six minutes.']",['6'],False,,Numerical, 2992,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Compute the $E(6,2)$","['$E(6,2)=3$. The sequence $\\{1,2\\},\\{3,4\\},\\{5,6\\}$ will allow Elizabeth to escape the room in three minutes. It is not possible to escape the room in fewer than three minutes because every switch must be flipped, and that requires at least $\\frac{6}{2}=3$ minutes.']",['3'],False,,Numerical, 2993,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Compute the $E(7,3)$","['$E(7,3)=3$. First, note that $E(7,3) \\geq 3$, because after only two minutes, it is impossible to flip each switch at least once. It is possible to escape in three minutes with the sequence $\\{1,2,3\\},\\{1,4,5\\}$, and $\\{1,6,7\\}$.']",['3'],False,,Numerical, 2994,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Compute the $E(9,5)$","['$E(9,5)=3$. Notice that $E(9,5) \\neq 1$ because each switch must be flipped at least once, and only five switches can be flipped in one minute. Notice also that $E(9,5) \\neq 2$ because after two minutes, there have been 10 flips, but in order to escape the room, each switch must be flipped at least once, and this requires 9 of the 10 flips. However, the tenth flip of a switch returns one of the nine switches to the off position, so it is not possible for Elizabeth to escape in two minutes. In three minutes, however, Elizabeth can escape with the sequence $\\{1,2,3,4,5\\},\\{1,2,3,6,7\\},\\{1,2,3,8,9\\}$.']",['3'],False,,Numerical, 2995,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Find the following in terms of $n$. $E(n, 2)$ for positive even integers $n$","['If $n$ is even, then $E(n, 2)=\\frac{n}{2}$. This is the minimum number of minutes required to flip each switch at least once, and Elizabeth can clearly escape in $\\frac{n}{2}$ minutes by flipping each switch exactly once.']",['$\\frac{n}{2}$'],False,,Expression, 2996,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Find the following in terms of $n$. $E(n, n-2)$ for $n \geq 5$","['If $n \\geq 5$, then $E(n, n-2)=3$. Note that Elizabeth cannot flip every switch in one minute, and after two minutes, some switch (in fact, many switches) must be flipped exactly twice. However, Elizabeth can escape in three minutes using the sequence $\\{1,4,5, \\ldots, n\\},\\{2,4,5, \\ldots, n\\},\\{3,4,5, \\ldots, n\\}$.']",['3'],False,,Numerical, 2997,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Find an integer value of $k$ with $1\\frac{n}{2}$, some switches must be flipped twice in the first two minutes, and so $E(n, k) \\neq 2$.\n\nHere is a strategy by which Elizabeth can escape in three minutes. Note that because $n+k$ is even, it follows that $3 k-n=(k+n)+2(k-n)$ is also even. Then let $3 k-n=2 b$. Note that $b>0$ (because $k>\\frac{n}{2}$ ), and $b74$, it will require at least 75 minutes to flip each switch once. Furthermore, $E(2020,27) \\geq 76$ because the prove above implies that $E(2020,27)$ is even.\n\nTo solve the puzzle in exactly 76 minutes, use the following strategy. For the first 33 minutes, flip switch 1, along with the first 26 switches that have not yet been flipped. The end result is that lights 1 through $26 \\cdot 33+1=859$ are on, and the remaining 1161 lights are off. Note that $1161=27 \\cdot 43$, so it takes 43 minutes to flip each remaining switch exactly once, for a total of 76 minutes, as desired.']",['76'],False,,Numerical, 3001,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Find the $E(2001,501)$","['$E(2001,501)=5$. First, note that three minutes is not enough time to flip each switch once. In four minutes, Elizabeth can flip each switch once, but has three flips left over. Because there are an odd number of leftover flips to distribute among the 2001 switches, some switch must get an odd number of leftover flips, and thus an even number of total flips. Thus $E(2001,501)>4$.\n\nTo solve the puzzle in five minutes, Elizabeth can flip the following sets of switches:\n\n- in the first minute, $\\{1,2,3, \\ldots, 501\\}$;\n- in the second minute, $\\{1,2,3, \\ldots, 102\\}$ and $\\{502,503,504, \\ldots, 900\\}$;\n- in the third minute, $\\{1,2,3, \\ldots, 102\\}$ and $\\{901,902,903, \\ldots, 1299\\}$;\n- in the fourth minute, $\\{1,2,3, \\ldots, 100\\}$ and $\\{1300,1301,1302, \\ldots, 1700\\}$;\n- in the fifth minute, $\\{1,2,3, \\ldots, 100\\}$ and $\\{1701,1702,1703, \\ldots, 2001\\}$.\n\nThis results in switches $1,2,3, \\ldots, 100$ being flipped five times, switches 101 and 102 being flipped three times, and the remaining switches being flipped exactly once, so that all the lights are on at the end of the fifth minute.']",['5'],False,,Numerical, 3002,Combinatorics,,"Elizabeth is in an ""escape room"" puzzle. She is in a room with one door which is locked at the start of the puzzle. The room contains $n$ light switches, each of which is initially off. Each minute, she must flip exactly $k$ different light switches (to ""flip"" a switch means to turn it on if it is currently off, and off if it is currently on). At the end of each minute, if all of the switches are on, then the door unlocks and Elizabeth escapes from the room. Let $E(n, k)$ be the minimum number of minutes required for Elizabeth to escape, for positive integers $n, k$ with $k \leq n$. For example, $E(2,1)=2$ because Elizabeth cannot escape in one minute (there are two switches and one must be flipped every minute) but she can escape in two minutes (by flipping Switch 1 in the first minute and Switch 2 in the second minute). Define $E(n, k)=\infty$ if the puzzle is impossible to solve (that is, if it is impossible to have all switches on at the end of any minute). For convenience, assume the $n$ light switches are numbered 1 through $n$. Show that if $n$ and $k$ are both even and $k \leq \frac{n}{2}$, then $E(n, k)=\left\lceil\frac{n}{k}\right\rceil$.","['First, if $k$ divides $n$, then $E(n, k)=\\frac{n}{k}=\\left\\lceil\\frac{n}{k}\\right\\rceil$. Assume then that $k$ does not divide $n$. Then let $r=\\left\\lceil\\frac{n}{k}\\right\\rceil$, which implies $(r-1) k2$, there exists an ordered pair $(n, k)$ for which $I(n, k)>x$.","['First, we prove that if $n$ is even and $k$ is odd, then $E(n, k)=E(n,n-k)$.\n\n\nBecause $n$ is even, and because each switch must be flipped an odd number of times in order to escape, the total number of flips is even. Because $k$ must be odd, $E(n, k)$ must be even. To show this, consider the case where $E(n, k)$ is odd. If $E(n, k)$ is odd, then an odd number of flips happen an odd number of times, resulting in an odd number of total flips. This is a contradiction because $n$ is even.\n\nCall a switch ""non-flipped"" in any given minute if it is not among the switches flipped in that minute. Because $E(n, k)$ (i.e., the total number of minutes) is even, and each switch is flipped an odd number of times, each switch must also be non-flipped an odd number of times. Therefore any sequence of flips that solves the "" $(n, k)$ puzzle"" can be made into a sequence of flips that solves the "" $(n, n-k)$ "" puzzle by interchanging flips and non-flips. These sequences last for the same number of minutes, and therefore $E(n, k)=E(n, n-k)$.\n\n\nLet $n=2 x$ and $k=2 x-1$ for some positive integer $x$. Then $n$ is even and $k$ is odd, so the above prove applies, and $E(n, k)=E(n, n-k)=E(n, 1)=2 x$. Therefore $I(n, k)=2 x-\\frac{2 x}{2 x-1}$. Because $x>2$, it follows that $\\frac{2 x}{2 x-1}<2x$, as desired.']",,True,,, 3006,Geometry,,"Regular tetrahedra $J A N E, J O H N$, and $J O A N$ have non-overlapping interiors. Compute $\tan \angle H A E$.","['First note that $\\overline{J N}$ is a shared edge of all three pyramids, and that the viewpoint for the figure below is from along the line that is the extension of edge $\\overline{J N}$.\n\n\n\nLet $h$ denote the height of each pyramid. Let $X$ be the center of pyramid JOAN, and consider the plane passing through $H, A$, and $E$. By symmetry, the altitude in pyramid $J O H N$ through $H$ and the altitude in pyramid $J A N E$ through $E$ pass through $X$. Thus points $H, X$, and $A$ are collinear, as are points $E, X$, and $O$. Hence $A H=O E=2 h$. Using the result that the four medians in a tetrahedron are concurrent and divide each other in a $3: 1$ ratio, it follows that $A X=O X=\\frac{3 h}{4}$ and $X E=O E-O X=\\frac{5 h}{4}$. Applying the Law of Cosines to triangle $A X E$ yields $\\cos \\angle X A E=\\cos \\angle H A E=\\frac{2-2 h^{2}}{3 h}$. Suppose, without loss of generality, that the common side length of the pyramids is 1 . Then $h=\\sqrt{\\frac{2}{3}}$ and $\\cos \\angle H A E=\\frac{\\sqrt{6}}{9}$. Hence $\\sin \\angle H A E=\\frac{\\sqrt{75}}{9}$ and therefore $\\tan \\angle H A E=\\frac{5 \\sqrt{2}}{\\mathbf{2}}$.']",['$\\frac{5 \\sqrt{2}}{2}$'],False,,Numerical, 3007,Number Theory,,"Each positive integer less than or equal to 2019 is written on a blank sheet of paper, and each of the digits 0 and 5 is erased. Compute the remainder when the product of the remaining digits on the sheet of paper is divided by 1000 .","[""Count the digits separately by position, noting that 1 is irrelevant to the product. There are a total of 20 instances of the digit 2 in the thousands place. The digit 0 only occurs in the hundreds place if the thousands digit is 2 , so look at the numbers 1 through 1999. Each non-zero digit contributes an equal number of times, so there are 200 each of $1,2,3,4,6,7,8,9$. The same applies to the tens digit, except there can be the stray digit of 1 among the numbers 2010 through 2019, but again, these do not affect the product. In the units place, there are 202 of each of the digits. Altogether, there are 602 each of $2,3,4,6,7,8$, 9, along with 20 extra instances of the digit 2 . Note that $9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 4 \\cdot 3 \\cdot 2=3024 \\cdot 24=72,576$ leaves a remainder of 576 when divided by 1000 . Also $2^{20}=1024^{2} \\equiv 24^{2}(\\bmod 1000)$, so $2^{20}$ contributes another factor of 576 . The answer is therefore the remainder when $576^{603}$ is divided by 1000 . This computation can be simplified by using the Chinese Remainder Theorem with moduli 8 and 125 , whose product is 1000 . Note $576^{603} \\equiv 0(\\bmod 8)$ because 576 is divisible by 8 . Also $576 \\equiv 76(\\bmod 125)$. By Euler's totient theorem, $576^{100} \\equiv 1(\\bmod 125)$, so $576^{603} \\equiv 76^{3}(\\bmod 125)$. This can quickly be computed by noting that $76^{3}=(75+1)^{3}=75^{3}+3 \\cdot 75^{2}+3 \\cdot 75+1 \\equiv 3 \\cdot 75+1 \\equiv-24(\\bmod 125)$. Observing that $-24 \\equiv 0(\\bmod 8)$, it follows that $576^{603} \\equiv-24(\\bmod 1000)$, hence the desired remainder is 976 .""]",['976'],False,,Numerical, 3008,Number Theory,,"Compute the third least positive integer $n$ such that each of $n, n+1$, and $n+2$ is a product of exactly two (not necessarily distinct) primes.","['Define a positive integer $n$ to be a semiprime if it is a product of exactly two (not necessarily distinct) primes. Define a lucky trio to be a sequence of three consecutive integers, $n, n+1, n+2$, each of which is a semiprime. Note that a lucky trio must contain exactly one multiple of 3. Also note that the middle number in a lucky trio must be even. To see this, note that if the first and last numbers in a lucky trio were both even, then exactly one of these numbers would be a multiple of 4 . But neither $2,3,4$ nor 4,5,6 is a lucky trio, and if a list of three consecutive integers contains a multiple of 4 that is greater than 4 , this number cannot be a semiprime. Using this conclusion and because $3,4,5$ is not a lucky trio, it follows that the middle number of a lucky trio cannot be a multiple of 4 . Hence it is necessary that a lucky trio has the form $4 k+1,4 k+2,4 k+3$, for some positive integer $k$, with $2 k+1$ being a prime. Note that $k \\neq 1(\\bmod 3)$ because when $k=1$, the sequence $5,6,7$ is not a lucky trio, and when $k>1,4 k+2$ would be a multiple of 6 greater than 6 , hence it cannot be a semiprime. Trying $k=2,3,5,6,8,9, \\ldots$ allows one to eliminate sequences of three consecutive integers that are not lucky trios, and if lucky trios are ordered by their least elements, one finds that the first three lucky trios are 33,34,35; 85,86,87; and 93,94,95. Hence the answer is 93.']",['93'],False,,Numerical, 3009,Geometry,,"The points $(1,2,3)$ and $(3,3,2)$ are vertices of a cube. Compute the product of all possible distinct volumes of the cube.","['The distance between points $A(1,2,3)$ and $B(3,3,2)$ is $A B=\\sqrt{(3-1)^{2}+(3-2)^{2}+(2-3)^{2}}=\\sqrt{6}$. Denote by $s$ the side length of the cube. Consider three possibilities.\n\n- If $\\overline{A B}$ is an edge of the cube, then $A B=s$, so one possibility is $s_{1}=\\sqrt{6}$.\n- If $\\overline{A B}$ is a face diagonal of the cube, then $A B=s \\sqrt{2}$, so another possibility is $s_{2}=\\sqrt{3}$.\n- If $\\overline{A B}$ is a space diagonal of the cube, then $A B=s \\sqrt{3}$, so the last possibility is $s_{3}=\\sqrt{2}$.\n\nThe answer is then $s_{1}^{3} s_{2}^{3} s_{3}^{3}=\\left(s_{1} s_{2} s_{3}\\right)^{3}=6^{3}=\\mathbf{2 1 6}$.']",['216'],False,,Numerical, 3010,Combinatorics,,"Eight students attend a Harper Valley ARML practice. At the end of the practice, they decide to take selfies to celebrate the event. Each selfie will have either two or three students in the picture. Compute the minimum number of selfies so that each pair of the eight students appears in exactly one selfie.","['The answer is 12 . To give an example in which 12 selfies is possible, consider regular octagon $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6} P_{7} P_{8}$. Each vertex of the octagon represents a student and each of the diagonals and sides of the octagon represents a pair of students. Construct eight triangles $P_{1} P_{2} P_{4}, P_{2} P_{3} P_{5}, P_{3} P_{4} P_{6}, \\ldots, P_{8} P_{1} P_{3}$. Each of the segments in the forms of $\\overline{P_{i} P_{i+1}}, \\overline{P_{i} P_{i+2}}, \\overline{P_{i} P_{i+3}}$ appears exactly once in these eight triangles. Taking 8 three-person selfies (namely $\\left.\\left\\{P_{1}, P_{2}, P_{4}\\right\\},\\left\\{P_{2}, P_{3}, P_{5}\\right\\}, \\ldots,\\left\\{P_{8}, P_{1}, P_{3}\\right\\}\\right)$ and 4 two-person selfies (namely $\\left.\\left\\{P_{1}, P_{5}\\right\\},\\left\\{P_{2}, P_{6}\\right\\},\\left\\{P_{3}, P_{7}\\right\\},\\left\\{P_{4}, P_{8}\\right\\}\\right)$ gives a total of 12 selfies, completing the desired task.\n\nA diagram of this construction is shown below. Each of the eight triangles is a different color, and each of the two-person selfies is represented by a dotted diameter.\n\n\n\nIt remains to show fewer than 12 selfies is impossible. Assume that the students took $x$ three-person selfies and $y$ two-person selfies. Each three-person selfie counts 3 pairs of student appearances (in a selfie), and each two-person selfie counts 1 pair of student appearances (in a selfie). Together, these selfies count $3 x+y$ pairs of student appearances. There are $\\left(\\begin{array}{l}8 \\\\ 2\\end{array}\\right)=28$ pairs of student appearances. Hence $3 x+y=28$. The number of\n\n\n\nselfies is $x+y=28-2 x$, so it is enough to show that $x \\leq 8$.\n\nAssume for contradiction there are $x \\geq 9$ three-person selfies; then there are at least $3 \\cdot 9=27$ (individual) student appearances on these selfies. Because there are 8 students, some student $s_{1}$ had at least $\\lceil 27 / 8\\rceil$ appearances; that is, $s_{1}$ appeared in at least 4 of these three-person selfies. There are $2 \\cdot 4=8$ (individual) student appearances other than $s_{1}$ on these 4 selfies. Because there are only 7 students besides $s_{1}$, some other student $s_{2}$ had at least $[8 / 7\\rceil$ (individual) appearances on these 4 selfies; that is, $s_{2}$ appeared (with $s_{1}$ ) in at least 2 of these 4 three-person selfies, violating the condition that each pair of the students appears in exactly one selfie. Thus the answer is $\\mathbf{1 2}$.']",['12'],False,,Numerical, 3011,Algebra,,"$\quad$ Compute the least positive value of $t$ such that $$ \operatorname{Arcsin}(\sin (t)), \operatorname{Arccos}(\cos (t)), \operatorname{Arctan}(\tan (t)) $$ form (in some order) a three-term arithmetic progression with a nonzero common difference.","['For $0 \\leq t<\\pi / 2$, all three values are $t$, so the desired $t$ does not lie in this interval.\n\nFor $\\pi / 2\n\nThus if the three numbers are to form an arithmetic progression, they should satisfy\n\n$$\nt-\\pi<\\pi-t\n\nIf instead $\\mathrm{m} \\angle B=\\theta$, then it follows that $\\mathrm{m} \\angle A P Q=\\mathrm{m} \\angle B A P+\\mathrm{m} \\angle A B P=2 \\theta$, and hence $\\mathrm{m} \\angle C=2 \\theta$. So $\\triangle A B C$ has angles of measures $5 \\theta, 2 \\theta, \\theta$, and thus $\\theta=22.5^{\\circ}$. Hence $\\mathrm{m} \\angle B=\\theta=\\mathbf{2 2 . 5}$.\n\n']",['$\\frac{45}{2}$'],False,,Numerical, 3013,Algebra,,"Consider the system of equations $$ \begin{aligned} & \log _{4} x+\log _{8}(y z)=2 \\ & \log _{4} y+\log _{8}(x z)=4 \\ & \log _{4} z+\log _{8}(x y)=5 . \end{aligned} $$ Given that $x y z$ can be expressed in the form $2^{k}$, compute $k$.","['Note that for $n>0, \\log _{4} n=\\log _{64} n^{3}$ and $\\log _{8} n=\\log _{64} n^{2}$. Adding together the three given equations and using both the preceding facts and properties of logarithms yields\n\n$$\n\\begin{aligned}\n& \\log _{4}(x y z)+\\log _{8}\\left(x^{2} y^{2} z^{2}\\right)=11 \\\\\n\\Longrightarrow & \\log _{64}(x y z)^{3}+\\log _{64}(x y z)^{4}=11 \\\\\n\\Longrightarrow & \\log _{64}(x y z)^{7}=11 \\\\\n\\Longrightarrow & 7 \\log _{64}(x y z)=11 .\n\\end{aligned}\n$$\n\nThe last equation is equivalent to $x y z=64^{11 / 7}=2^{66 / 7}$, hence the desired value of $k$ is $\\frac{\\mathbf{6 6}}{\\mathbf{7}}$.']",['$\\frac{66}{7}$'],False,,Numerical, 3014,Combinatorics,,A complex number $z$ is selected uniformly at random such that $|z|=1$. Compute the probability that $z$ and $z^{2019}$ both lie in Quadrant II in the complex plane.,"['For convenience, let $\\alpha=\\pi / 4038$. Denote by\n\n$$\n0 \\leq \\theta<2 \\pi=8076 \\alpha\n$$\n\nthe complex argument of $z$, selected uniformly at random from the interval $[0,2 \\pi)$. Then $z$ itself lies in Quadrant II if and only if\n\n$$\n2019 \\alpha=\\frac{\\pi}{2}<\\theta<\\pi=4038 \\alpha\n$$\n\nOn the other hand, $z^{2019}$ has argument 2019日, and hence it lies in Quadrant II if and only if there is some integer $k$ with\n\n$$\n\\begin{gathered}\n\\frac{\\pi}{2}+2 k \\pi<2019 \\theta<\\pi+2 k \\pi \\\\\n\\Longleftrightarrow(4 k+1) \\cdot \\frac{\\pi}{2}<2019 \\theta<(4 k+2) \\cdot \\frac{\\pi}{2} \\\\\n\\Longleftrightarrow(4 k+1) \\alpha<\\theta<(4 k+2) \\alpha .\n\\end{gathered}\n$$\n\nBecause it is also true that $2019 \\alpha<\\theta<4038 \\alpha$, the set of $\\theta$ that satisfies the conditions of the problem is the union of intervals:\n\n$$\n(2021 \\alpha, 2022 \\alpha) \\cup(2025 \\alpha, 2026 \\alpha) \\cup \\cdots \\cup(4037 \\alpha, 4038 \\alpha)\n$$\n\nThere are 505 such intervals, the $j^{\\text {th }}$ interval consisting of $(4 j+2017) \\alpha<\\theta<(4 j+2018) \\alpha$. Each interval has length $\\alpha$, so the sum of the intervals has length $505 \\alpha$. Thus the final answer is\n\n$$\n\\frac{505 \\alpha}{2 \\pi}=\\frac{505}{2 \\cdot 4038}=\\frac{\\mathbf{5 0 5}}{\\mathbf{8 0 7 6}} .\n$$']",['$\\frac{505}{8076}$'],False,,Numerical, 3015,Number Theory,,Compute the least positive integer $n$ such that the sum of the digits of $n$ is five times the sum of the digits of $(n+2019)$.,"['Let $S(n)$ denote the sum of the digits of $n$, so that solving the problem is equivalent to solving $S(n)=5 S(n+2019)$. Using the fact that $S(n) \\equiv n(\\bmod 9)$ for all $n$, it follows that\n\n$$\n\\begin{aligned}\nn & \\equiv 5(n+2019) \\equiv 5(n+3)(\\bmod 9) \\\\\n4 n & \\equiv-15(\\bmod 9) \\\\\nn & \\equiv 3(\\bmod 9)\n\\end{aligned}\n$$\n\nThen $S(n+2019) \\equiv 6(\\bmod 9)$. In particular, $S(n+2019) \\geq 6$ and $S(n) \\geq 5 \\cdot 6=30$. The latter inequality implies $n \\geq 3999$, which then gives $n+2019 \\geq 6018$. Thus if $n+2019$ were a four-digit number, then $S(n+2019) \\geq 7$. Moreover, $S(n+2019)$ can only be 7, because otherwise, $S(n)=5 S(n+2019) \\geq 40$, which is impossible (if $n$ has four digits, then $S(n)$ can be no greater than 36). So if $n+2019$ were a four-digit number, then $S(n+2019)=7$ and $S(n)=35$. But this would imply that the digits of $n$ are $8,9,9,9$ in some order, contradicting the assumption that $n+2019$ is a four-digit number. On the other hand, if $n+2019$ were a five-digit number such that $S(n+2019) \\geq 6$, then the least such value of $n+2019$ is 10005 , and indeed, this works because it corresponds to $n=\\mathbf{7 9 8 6}$, the least possible value of $n$.']",['7986'],False,,Numerical, 3016,Geometry,,"$\quad$ Compute the greatest real number $K$ for which the graphs of $$ (|x|-5)^{2}+(|y|-5)^{2}=K \quad \text { and } \quad(x-1)^{2}+(y+1)^{2}=37 $$ have exactly two intersection points.","['The graph of the second equation is simply the circle of radius $\\sqrt{37}$ centered at $(1,-1)$. The first graph is more interesting, and its behavior depends on $K$.\n\n- For small values of $K$, the first equation determines a set of four circles of radius $\\sqrt{K}$ with centers at $(5,5),(5,-5),(-5,5)$, and $(-5,-5)$. Shown below are versions with $K=1, K=4$, and $K=16$.\n\n- However, when $K>25$, the graph no longer consists of four circles! As an example, for $K=36$, the value $x=5$ gives $(|y|-5)^{2}=36$; hence $|y|=-1$ or $|y|=6$. The first option is impossible; the graph ends up ""losing"" the portions of the upper-right circle that would cross the $x$ - or $y$-axes compared to the graph for $(x-5)^{2}+(y-5)^{2}=36$. The graph for $K=36$ is shown below.\n\n\n\n- As $K$ continues to increase, the ""interior"" part of the curve continues to shrink, until at $K=50$, it simply comprises the origin, and for $K>50$, it does not exist. As examples, the graphs with $K=50$ and $K=64$ are shown below.\n\n\n\n\nOverlay the graph of the circle of radius $\\sqrt{37}$ centered at $(1,-1)$ with the given graphs. When $K=25$, this looks like the following graph.\n\n\n\nNote that the two graphs intersect at $(0,5)$ and $(-5,0)$, as well as four more points (two points near the positive $x$-axis and two points near the negative $y$-axis). When $K$ is slightly greater than 25 , this drops to four intersection points. The graph for $K=27$ is shown below.\n\n\n\nThus for the greatest $K$ for which there are exactly two intersection points, those two intersection points should be along the positive $x$ - and negative $y$-axes. If the intersection point on the positive $x$-axis is at $(h, 0)$, then $(h-1)^{2}+(0+1)^{2}=37$ and $(h-5)^{2}+(0-5)^{2}=K$. Thus $h=7$ and $K=\\mathbf{2 9}$']",['29'],False,,Numerical, 3017,Algebra,,"To morph a sequence means to replace two terms $a$ and $b$ with $a+1$ and $b-1$ if and only if $a+10$. Therefore the only possible values of $d$ are $7,5,3,1$; thus there are at most four possibilities for the morphed sequence, shown in the table below. Denote these four sequences by $A, B, C, D$.\n\n| | $T$ | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $d=7:$ | $A$ | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 |\n| $d=5:$ | $B$ | 16 | 21 | 26 | 31 | 36 | 41 | 46 | 51 | 56 | 61 |\n| $d=3:$ | $C$ | 25 | 28 | 31 | 34 | 37 | 40 | 43 | 46 | 49 | 52 |\n| $d=1:$ | $D$ | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 |\n\nStep 2. Given any two sequences $X=\\left(x_{1}, \\ldots, x_{10}\\right)$ and $Y=\\left(y_{1}, \\ldots, y_{10}\\right)$ with $\\sum_{i=1}^{10} x_{i}=\\sum_{i=1}^{10} y_{i}=385$, define the taxicab distance\n\n$$\n\\rho(X, Y)=\\sum_{i=1}^{10}\\left|x_{i}-y_{i}\\right|\n$$\n\nObserve that if $X^{\\prime}$ is a morph of $X$, then $\\rho\\left(X^{\\prime}, Y\\right) \\geq \\rho(X, Y)-2$. Therefore the number of morphs required to transform $T$ into some sequence $Z$ is at least $\\frac{1}{2} \\rho(T, Z)$. Now\n\n$$\n\\frac{1}{2} \\rho(T, A)=\\frac{1}{2} \\sum_{i=1}^{10}\\left|i^{2}-7 i\\right|=56\n$$\n\nand also $\\rho(T, A)<\\min (\\rho(T, B), \\rho(T, C), \\rho(T, D))$. Thus at least 56 morphs are needed to obtain sequence $A$ (and more morphs would be required to obtain any of sequences $B, C$, or $D$ ).\n\nStep 3. To conclude, it remains to verify that one can make 56 morphs and arrive from $T$ to $A$. One of many possible constructions is given below.\n\n| $T$ | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |\n| ---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 6 morphs | 1 | 4 | 9 | 16 | 25 | 42 | 49 | 58 | 81 | 100 |\n| 2 morphs | 1 | 4 | 9 | 16 | 27 | 42 | 49 | 56 | 81 | 100 |\n| 8 morphs | 1 | 4 | 9 | 16 | 35 | 42 | 49 | 56 | 73 | 100 |\n| 10 morphs | 1 | 4 | 9 | 26 | 35 | 42 | 49 | 56 | 63 | 100 |\n| 2 morphs | 1 | 4 | 9 | 28 | 35 | 42 | 49 | 56 | 63 | 98 |\n| 12 morphs | 1 | 4 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 86 |\n| 10 morphs | 1 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 76 |\n| 6 morphs | 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 |\n\nTherefore the least number of morphs needed to transform $T$ into an arithmetic progression is $\\mathbf{5 6}$.']",['56'],False,,Numerical, 3018,Geometry,,"Triangle $A B C$ is inscribed in circle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at point $T$. The tangent to $\omega$ at $A$ intersects the perpendicular bisector of $\overline{A T}$ at point $P$. Given that $A B=14, A C=30$, and $B C=40$, compute $[P B C]$.","[""To begin, denote by $R$ the radius of $\\omega$. The semiperimeter of triangle $A B C$ is 42 , and then applying Heron's formula yields\n\n$$\n[A B C]=\\frac{14 \\cdot 30 \\cdot 40}{4 R}=\\sqrt{42 \\cdot 28 \\cdot 12 \\cdot 2}=168\n$$\n\nfrom which it follows that $R=\\frac{14 \\cdot 30 \\cdot 40}{4 \\cdot 168}=25$.\n\nNow consider the point circle with radius zero centered at $T$ in tandem with the circle $\\omega$. Because $P A=P T$, it follows that $P$ lies on the radical axis of these circles. Moreover, the midpoints of $\\overline{T B}$ and $\\overline{T C}$ lie on this radical axis as well. Thus $P$ lies on the midline of $\\triangle T B C$ that is parallel to $\\overline{B C}$.\n\n\n\nTo finish, let $O$ denote the center of $\\omega$ and $M$ the midpoint of $\\overline{B C}$. By considering right triangle $T B O$ with altitude $\\overline{B M}$, it follows that $M T \\cdot M O=M B^{2}$, but also $M O=\\sqrt{O B^{2}-M B^{2}}=\\sqrt{25^{2}-20^{2}}=15$, so\n\n$$\nM T=\\frac{M B^{2}}{M O}=\\frac{400}{15}=\\frac{80}{3}\n$$\n\nThus the distance from $P$ to $\\overline{B C}$ is $\\frac{1}{2} M T=\\frac{40}{3}$. Finally,\n\n$$\n[P B C]=\\frac{1}{2} \\cdot \\frac{40}{3} \\cdot B C=\\frac{\\mathbf{8 0 0}}{\\mathbf{3}}\n$$""]",['$\\frac{800}{3}$'],False,,Numerical, 3019,Algebra,,"Given that $a, b, c$, and $d$ are integers such that $a+b c=20$ and $-a+c d=19$, compute the greatest possible value of $c$.",['Adding the two given equations yields $b c+c d=c(b+d)=39$. The greatest possible value of $c$ therefore occurs when $c=\\mathbf{3 9}$ and $b+d=1$.'],['39'],False,,Numerical, 3020,Combinatorics,,"Let $T$ = 39. Emile randomly chooses a set of $T$ cards from a standard deck of 52 cards. Given that Emile's set contains no clubs, compute the probability that his set contains three aces.","[""Knowing that 13 of the cards are not in Emile's set, there are $\\left(\\begin{array}{c}39 \\\\ T\\end{array}\\right)$ ways for him to have chosen a set of $T$ cards. Given that Emile's set contains no clubs, the suits of the three aces are fixed (i.e., diamonds, hearts, and spades). The number of possible sets of cards in which these three aces appear is therefore $\\left(\\begin{array}{c}36 \\\\ T-3\\end{array}\\right)$. The desired probability is therefore $\\left(\\begin{array}{c}36 \\\\ T-3\\end{array}\\right) /\\left(\\begin{array}{c}39 \\\\ T\\end{array}\\right)$. With $T=39$, this probability is $1 / 1=\\mathbf{1}$, which is consistent with the fact that Emile's set contains all cards in the deck that are not clubs, hence he is guaranteed to have all three of the remaining aces.""]",['1'],False,,Numerical, 3021,Geometry,,"Let $T=1$. In parallelogram $A B C D, \frac{A B}{B C}=T$. Given that $M$ is the midpoint of $\overline{A B}$ and $P$ and $Q$ are the trisection points of $\overline{C D}$, compute $\frac{[A B C D]}{[M P Q]}$.",['Let $C D=3 x$ and let $h$ be the length of the altitude between bases $\\overline{A B}$ and $\\overline{C D}$. Then $[A B C D]=3 x h$ and $[M P Q]=\\frac{1}{2} x h$. Hence $\\frac{[A B C D]}{[M P Q]}=\\mathbf{6}$. Both the position of $M$ and the ratio $\\frac{A B}{B C}=T$ are irrelevant.'],['6'],False,,Numerical, 3022,Algebra,,Let $T=6$. Compute the value of $x$ such that $\log _{T} \sqrt{x-7}+\log _{T^{2}}(x-2)=1$.,"['It can readily be shown that $\\log _{a} b=\\log _{a^{2}} b^{2}$. Thus it follows that $\\log _{T} \\sqrt{x-7}=\\log _{T^{2}}(x-7)$. Hence the left-hand side of the given equation is $\\log _{T^{2}}(x-7)(x-2)$ and the equation is equivalent to $(x-7)(x-2)=T^{2}$, which is equivalent to $x^{2}-9 x+14-T^{2}=0$. With $T=6$, this equation is $x^{2}-9 x-22=0 \\Longrightarrow(x-11)(x+2)=0$. Plugging $x=-2$ into the given equation leads to the first term of the left-hand side having a negative radicand and the second term having an argument of 0 . However, one can easily check that $x=\\mathbf{1 1}$ indeed satisfies the given equation.']",['11'],False,,Numerical, 3023,Number Theory,,"Let $T=11$. Let $p$ be an odd prime and let $x, y$, and $z$ be positive integers less than $p$. When the trinomial $(p x+y+z)^{T-1}$ is expanded and simplified, there are $N$ terms, of which $M$ are always multiples of $p$. Compute $M$.","['A general term in the expansion of $(p x+y+z)^{T-1}$ has the form $K(p x)^{a} y^{b} z^{c}$, where $a, b$, and $c$ are nonnegative integers such that $a+b+c=T-1$. Using the ""stars and bars"" approach, the number of nonnegative integral solutions to $a+b+c=T-1$ is the number of arrangements of $T-1$ stars and 2 bars in a row (the bars act has separators and the "" 2 "" arises because it is one less than the number of variables in the equation). Thus there are $\\left(\\begin{array}{c}T+1 \\\\ 2\\end{array}\\right)$ solutions. Each term will be a multiple of $p$ unless $a=0$. In this case, the number of terms that are not multiples of $p$ is the number of nonnegative integral solutions to the equation $b+c=T-1$, which is $T$ ( $b$ can range from 0 to $T-1$ inclusive, and then $c$ is fixed). Hence $M=\\left(\\begin{array}{c}T+1 \\\\ 2\\end{array}\\right)-T=\\frac{T^{2}-T}{2}$. With $T=11$, the answer is $\\mathbf{5 5}$']",['55'],False,,Numerical, 3024,Algebra,,"Let $T=55$. Compute the value of $K$ such that $20, T-5, K$ is an increasing geometric sequence and $19, K, 4 T+11$ is an increasing arithmetic sequence.","['The condition that $20, T-5, K$ is an increasing geometric sequence implies that $\\frac{T-5}{20}=\\frac{K}{T-5}$, hence $K=\\frac{(T-5)^{2}}{20}$. The condition that $19, K, 4 T+11$ is an increasing arithmetic sequence implies that $K-19=4 T+11-K$, hence $K=2 T+15$. With $T=55$, each of these equations implies that $K=\\mathbf{1 2 5}$. Note that the two equations can be combined and solved without being passed a value of $T$. A quadratic equation results, and its roots are $T=55$ or $T=-5$. However, with $T=-5$, neither of the given sequences is increasing.']",['125'],False,,Numerical, 3025,Geometry,,"Let $T=125$. Cube $\mathcal{C}_{1}$ has volume $T$ and sphere $\mathcal{S}_{1}$ is circumscribed about $\mathcal{C}_{1}$. For $n \geq 1$, the sphere $\mathcal{S}_{n}$ is circumscribed about the cube $\mathcal{C}_{n}$ and is inscribed in the cube $\mathcal{C}_{n+1}$. Let $k$ be the least integer such that the volume of $\mathcal{C}_{k}$ is at least 2019. Compute the edge length of $\mathcal{C}_{k}$.","['In general, let cube $\\mathcal{C}_{n}$ have edge length $x$. Then the diameter of sphere $\\mathcal{S}_{n}$ is the space diagonal of $\\mathcal{C}_{n}$, which has length $x \\sqrt{3}$. This in turn is the edge length of cube $\\mathcal{C}_{n+1}$. Hence the edge lengths of $\\mathcal{C}_{1}, \\mathcal{C}_{2}, \\ldots$ form an increasing geometric sequence with common ratio $\\sqrt{3}$ and volumes of $\\mathcal{C}_{1}, \\mathcal{C}_{2}, \\ldots$ form an increasing geometric sequence with common ratio $3 \\sqrt{3}$. With $T=125$, the edge length of $\\mathcal{C}_{1}$ is 5 , so the sequence of edge lengths of the cubes is $5,5 \\sqrt{3}, 15, \\ldots$, and the respective sequence of the volumes of the cubes is $125,375 \\sqrt{3}, 3375, \\ldots$. Hence $k=3$, and the edge length of $\\mathcal{C}_{3}$ is $\\mathbf{1 5}$.']",['15'],False,,Numerical, 3026,Geometry,,"Square $K E N T$ has side length 20 . Point $M$ lies in the interior of $K E N T$ such that $\triangle M E N$ is equilateral. Given that $K M^{2}=a-b \sqrt{3}$, where $a$ and $b$ are integers, compute $b$.","['Let $s$ be the side length of square $K E N T$; then $M E=s$. Let $J$ be the foot of the altitude from $M$ to $\\overline{K E}$. Then $\\mathrm{m} \\angle J E M=30^{\\circ}$ and $\\mathrm{m} \\angle E M J=60^{\\circ}$. Hence $M J=\\frac{s}{2}, J E=\\frac{s \\sqrt{3}}{2}$, and $K J=K E-J E=s-\\frac{s \\sqrt{3}}{2}$. Applying the Pythagorean Theorem to $\\triangle K J M$ implies that $K M^{2}=\\left(s-\\frac{s \\sqrt{3}}{2}\\right)^{2}+\\left(\\frac{s}{2}\\right)^{2}=2 s^{2}-s^{2} \\sqrt{3}$. With $s=20$, the value of $b$ is therefore $s^{2}=\\mathbf{4 0 0}$.']",['400'],False,,Numerical, 3027,Algebra,,"Let $T$ be a rational number. Let $a, b$, and $c$ be the three solutions of the equation $x^{3}-20 x^{2}+19 x+T=0$. Compute $a^{2}+b^{2}+c^{2}$.","[""According to Vieta's formulas, $a+b+c=-(-20)=20$ and $a b+b c+c a=19$. Noting that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+c a)$, it follows that $a^{2}+b^{2}+c^{2}=20^{2}-2 \\cdot 19=\\mathbf{3 6 2}$. The value of $T$ is irrelevant.""]",['362'],False,,Numerical, 3028,Algebra,,Let $T=362$ and let $K=\sqrt{T-1}$. Compute $\left|(K-20)(K+1)+19 K-K^{2}\right|$.,['The expression inside the absolute value bars simplifies to $K^{2}-19 K-20+19 K-K^{2}=-20$. Hence the answer is $\\mathbf{2 0}$ and the value of $K(=\\sqrt{361}=19)$ is not needed.'],['20'],False,,Numerical, 3029,Geometry,,"Let $T=20$. In $\triangle L E O, \sin \angle L E O=\frac{1}{T}$. If $L E=\frac{1}{n}$ for some positive real number $n$, then $E O=$ $n^{3}-4 n^{2}+5 n$. As $n$ ranges over the positive reals, compute the least possible value of $[L E O]$.","['Note that $[L E O]=\\frac{1}{2}(\\sin \\angle L E O) \\cdot L E \\cdot E O=\\frac{1}{2} \\cdot \\frac{1}{T} \\cdot \\frac{1}{n} \\cdot\\left(n^{3}-4 n^{2}+5 n\\right)=\\frac{n^{2}-4 n+5}{2 T}$. Because $T$ is a constant, the least possible value of $[L E O]$ is achieved when the function $f(n)=n^{2}-4 n+5$ is minimized.\n\n\n\nThis occurs when $n=-(-4) /(2 \\cdot 1)=2$, and the minimum value is $f(2)=1$. Hence the desired least possible value of $[L E O]$ is $\\frac{1}{2 T}$, and with $T=20$, this is $\\frac{1}{40}$.']",['$\\frac{1}{40}$'],False,,Numerical, 3030,Algebra,,"Let $T=\frac{1}{40}$. Given that $x, y$, and $z$ are real numbers such that $x+y=5, x^{2}-y^{2}=\frac{1}{T}$, and $x-z=-7$, compute $x+z$","['Note that $x^{2}-y^{2}=(x+y)(x-y)=5(x-y)$, hence $x-y=\\frac{1}{5 T}$. Then $x+z=(x+y)+(x-y)+(z-x)=$ $5+\\frac{1}{5 T}+7=12+\\frac{1}{5 T}$. With $T=\\frac{1}{40}$, the answer is thus $12+8=\\mathbf{2 0}$.']",['20'],False,,Numerical, 3031,Number Theory,,Let $T=20$. The product of all positive divisors of $2^{T}$ can be written in the form $2^{K}$. Compute $K$.,"['When $n$ is a nonnegative integer, the product of the positive divisors of $2^{n}$ is $2^{0} \\cdot 2^{1} \\cdot \\ldots \\cdot 2^{n-1} \\cdot 2^{n}=$ $2^{0+1+\\cdots+(n-1)+n}=2^{n(n+1) / 2}$. Because $T=20$ is an integer, it follows that $K=\\frac{T(T+1)}{2}=\\mathbf{2 1 0}$.']",['210'],False,,Numerical, 3032,Combinatorics,,"Let $T=210$. At the Westward House of Supper (""WHS""), a dinner special consists of an appetizer, an entrée, and dessert. There are 7 different appetizers and $K$ different entrées that a guest could order. There are 2 dessert choices, but ordering dessert is optional. Given that there are $T$ possible different orders that could be placed at the WHS, compute $K$.","['Because dessert is optional, there are effectively $2+1=3$ dessert choices. Hence, by the Multiplication Principle, it follows that $T=7 \\cdot K \\cdot 3$, thus $K=\\frac{T}{21}$. With $T=210$, the answer is 10 .']",['10'],False,,Numerical, 3033,Algebra,,"Let $S=15$ and let $M=10$ . Sam and Marty each ride a bicycle at a constant speed. Sam's speed is $S \mathrm{~km} / \mathrm{hr}$ and Marty's speed is $M \mathrm{~km} / \mathrm{hr}$. Given that Sam and Marty are initially $100 \mathrm{~km}$ apart and they begin riding towards one another at the same time, along a straight path, compute the number of kilometers that Sam will have traveled when Sam and Marty meet.","[""In km/hr, the combined speed of Sam and Marty is $S+M$. Thus one can determine the total time they traveled and use this to determine the number of kilometers that Sam traveled. However, this is not needed, and there is a simpler approach. Suppose that Marty traveled a distance of $d$. Then because Sam's speed is $\\frac{S}{M}$ of Marty's speed, Sam will have traveled a distance of $\\frac{S}{M} \\cdot d$. Thus, together, they traveled $d+\\frac{S}{M} \\cdot d$. Setting this equal to 100 and solving yields $d=\\frac{100 M}{M+S}$. Thus Sam traveled $\\frac{S}{M} \\cdot d=\\frac{100 S}{M+S}$. With $S=15$ and $M=10$, this is equal to $60 \\mathrm{~km}$.""]",['60'],False,,Numerical, 3034,Number Theory,,Compute the $2011^{\text {th }}$ smallest positive integer $N$ that gains an extra digit when doubled.,"['Let $S$ be the set of numbers that gain an extra digit when doubled. First notice that the numbers in $S$ are precisely those whose first digit is at least 5 . Thus there are five one-digit numbers in $S, 50$ two-digit numbers in $S$, and 500 three-digit numbers in $S$. Therefore 5000 is the $556^{\\text {th }}$ smallest number in $S$, and because all four-digit numbers greater than 5000 are in $S$, the $2011^{\\text {th }}$ smallest number in $S$ is $5000+(2011-556)=\\mathbf{6 4 5 5}$.']",['6455'],False,,Numerical, 3035,Geometry,,"In triangle $A B C, C$ is a right angle and $M$ is on $\overline{A C}$. A circle with radius $r$ is centered at $M$, is tangent to $\overline{A B}$, and is tangent to $\overline{B C}$ at $C$. If $A C=5$ and $B C=12$, compute $r$.","['Let $N$ be the point of tangency of the circle with $\\overline{A B}$ and draw $\\overline{M B}$, as shown below.\n\n\n\nBecause $\\triangle B M C$ and $\\triangle B M N$ are right triangles sharing a hypotenuse, and $\\overline{M N}$ and $\\overline{M C}$ are radii, $\\triangle B M C \\cong \\triangle B M N$. Thus $B N=12$ and $A N=1$. Also $\\triangle A N M \\sim \\triangle A C B$ because the right triangles share $\\angle A$, so $\\frac{N M}{A N}=\\frac{C B}{A C}$. Therefore $\\frac{r}{1}=\\frac{12}{5}$, so $r=\\frac{\\mathbf{1 2}}{\\mathbf{5}}$.', 'Let $r$ denote the radius of the circle, and let $D$ be the foot of the perpendicular from $O$ to $\\overline{A B}$. Note that $\\triangle A B C \\sim \\triangle A O D$. Thus $\\frac{A B}{A O}=\\frac{B C}{D O} \\Longrightarrow \\frac{13}{5-r}=\\frac{12}{r}$, and $r=\\frac{\\mathbf{1 2}}{\\mathbf{5}}$.']",['$\\frac{12}{5}$'],False,,Numerical, 3036,Number Theory,,"The product of the first five terms of a geometric progression is 32 . If the fourth term is 17 , compute the second term.","['Let $a$ be the third term of the geometric progression, and let $r$ be the common ratio. Then the product of the first five terms is\n\n$$\n\\left(a r^{-2}\\right)\\left(a r^{-1}\\right)(a)(a r)\\left(a r^{2}\\right)=a^{5}=32\n$$\n\nso $a=2$. Because the fourth term is $17, r=\\frac{17}{a}=\\frac{17}{2}$. The second term is $a r^{-1}=\\frac{2}{17 / 2}=\\frac{4}{17}$.']",['$\\frac{4}{17}$'],False,,Numerical, 3037,Geometry,,"Polygon $A_{1} A_{2} \ldots A_{n}$ is a regular $n$-gon. For some integer $k\n\nThen $\\left[A_{1} A_{2} O\\right]=\\frac{1}{4}\\left[A_{1} A_{2} A_{k} A_{k+1}\\right]=\\frac{1}{n}\\left[A_{1} A_{2} \\ldots A_{n}\\right]=60$. So $\\frac{1}{4}(6)=\\frac{1}{n}(60)$, and $n=40$.""]",['40'],False,,Numerical, 3038,Combinatorics,,"A bag contains 20 lavender marbles, 12 emerald marbles, and some number of orange marbles. If the probability of drawing an orange marble in one try is $\frac{1}{y}$, compute the sum of all possible integer values of $y$.","['Let $x$ be the number of orange marbles. Then the probability of drawing an orange marble is $\\frac{x}{x+20+12}=\\frac{x}{x+32}$. If this probability equals $\\frac{1}{y}$, then $y=\\frac{x+32}{x}=1+\\frac{32}{x}$. This expression represents an integer only when $x$ is a factor of 32 , thus $x \\in\\{1,2,4,8,16,32\\}$. The corresponding $y$-values are $33,17,9,5,3$, and 2 , and their sum is $\\mathbf{6 9}$.']",['69'],False,,Numerical, 3039,Algebra,,"Compute the number of ordered quadruples of integers $(a, b, c, d)$ satisfying the following system of equations: $$ \left\{\begin{array}{l} a b c=12,000 \\ b c d=24,000 \\ c d a=36,000 \end{array}\right. $$","['From the first two equations, conclude that $d=2 a$. From the last two, $3 b=2 a$. Thus all solutions to the system will be of the form $(3 K, 2 K, c, 6 K)$ for some integer $K$. Substituting these expressions into the system, each equation now becomes $c K^{2}=2000=2^{4} \\cdot 5^{3}$. So $K^{2}$ is of the form $2^{2 m} 5^{2 n}$. There are 3 choices for $m$ and 2 for $n$, so there are 6 values for $K^{2}$, which means there are 12 solutions overall, including negative values for $K$.\n\nAlthough the problem does not require finding them, the twelve values of $K$ are $\\pm 1, \\pm 2, \\pm 4$, $\\pm 5, \\pm 10, \\pm 20$. These values yield the following quadruples $(a, b, c, d)$ :\n\n$$\n\\begin{aligned}\n& (3,2,2000,6),(-3,-2,2000,-6), \\\\\n& (6,4,500,12),(-6,-4,500,-12), \\\\\n& (12,8,125,24),(-12,-8,125,-24), \\\\\n& (15,10,80,30),(-15,-10,80,-30), \\\\\n& (30,20,20,60),(-30,-20,20,-60), \\\\\n& (60,40,5,120),(-60,-40,5,-120) .\n\\end{aligned}\n$$']",['12'],False,,Numerical, 3040,Algebra,,Let $n$ be a positive integer such that $\frac{3+4+\cdots+3 n}{5+6+\cdots+5 n}=\frac{4}{11}$. Compute $\frac{2+3+\cdots+2 n}{4+5+\cdots+4 n}$.,"['In simplifying the numerator and denominator of the left side of the equation, notice that\n\n$$\n\\begin{aligned}\nk+(k+1)+\\cdots+k n & =\\frac{1}{2}(k n(k n+1)-k(k-1)) \\\\\n& =\\frac{1}{2}(k(n+1)(k n-k+1))\n\\end{aligned}\n$$\n\nThis identity allows the given equation to be transformed:\n\n$$\n\\begin{aligned}\n\\frac{3(n+1)(3 n-3+1)}{5(n+1)(5 n-5+1)} & =\\frac{4}{11} \\\\\n\\frac{3(n+1)(3 n-2)}{5(n+1)(5 n-4)} & =\\frac{4}{11} \\\\\n\\frac{3 n-2}{5 n-4} & =\\frac{20}{33}\n\\end{aligned}\n$$\n\nSolving this last equation yields $n=14$. Using the same identity twice more, for $n=14$ and $k=2$ and $k=4$, the desired quantity is $\\frac{2(2 n-1)}{4(4 n-3)}=\\frac{\\mathbf{2 7}}{\\mathbf{1 0 6}}$.']",['$\\frac{27}{106}$'],False,,Numerical, 3041,Algebra,,"The quadratic polynomial $f(x)$ has a zero at $x=2$. The polynomial $f(f(x))$ has only one real zero, at $x=5$. Compute $f(0)$.","['Let $f(x)=a(x-b)^{2}+c$. The graph of $f$ is symmetric about $x=b$, so the graph of $y=f(f(x))$ is also symmetric about $x=b$. If $b \\neq 5$, then $2 b-5$, the reflection of 5 across $b$, must be a zero of $f(f(x))$. Because $f(f(x))$ has exactly one zero, $b=5$.\n\nBecause $f(2)=0$ and $f$ is symmetric about $x=5$, the other zero of $f$ is $x=8$. Because the zeros of $f$ are at 2 and 8 and $f(5)$ is a zero of $f$, either $f(5)=2$ or $f(5)=8$. The following argument shows that $f(5)=8$ is impossible. Because $f$ is continuous, if $f(5)=8$, then $f\\left(x_{0}\\right)=2$ for some $x_{0}$ in the interval $2\n\nThe following argument shows that the first case is impossible. By the Triangle Inequality on $\\triangle A B O$, the radius $r_{1}$ of circle $\\omega_{1}$ must be at least 20 . But because $B$ is outside $\\omega_{1}, B O>r_{1}$, which is impossible, because $B O=17$. So $B$ must be inside the circle.\n\nConstruct point $D$ on minor arc $A O$ of circle $\\omega_{2}$, so that $A D=O B$ (and therefore $\\left.D O=B C\\right)$.\n\n\n\nBecause $A, D, O, B$ all lie on $\\omega_{2}$, Ptolemy's Theorem applies to quadrilateral $A D O B$.\n\n\n\n\n\nTherefore $A D \\cdot O B+O D \\cdot A B=A O \\cdot D B=r_{1}^{2}$. Substituting $A D=O B=17, D O=B C=7$, and $A B=37$ yields $r_{1}^{2}=37 \\cdot 7+17^{2}=548$. Thus the area of $\\omega_{1}$ is $\\mathbf{5 4 8 \\pi}$.""]",['$548 \\pi$'],False,,Numerical, 3044,Algebra,,Compute the number of integers $n$ for which $2^{4}<8^{n}<16^{32}$.,"['$8^{n}=2^{3 n}$ and $16^{32}=2^{128}$. Therefore $4<3 n<128$, and $2 \\leq n \\leq 42$. Thus there are 41 such integers $n$.']",['41'],False,,Numerical, 3045,Number Theory,,Let $T=41$. Compute the number of positive integers $b$ such that the number $T$ has exactly two digits when written in base $b$.,"['If $T$ has more than one digit when written in base $b$, then $b \\leq T$. If $T$ has fewer than three digits when written in base $b$, then $b^{2}>T$, or $b>\\sqrt{T}$. So the desired set of bases $b$ is $\\{b \\mid \\sqrt{T}1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$.","['$$\n\\begin{aligned}\n& \\mathrm{Pa}(1,1)+\\mathrm{Pa}(2,1)+\\mathrm{Pa}(3,1)+\\mathrm{Pa}(4,1)+\\mathrm{Pa}(5,1)=1+2+3+4+5=\\mathbf{1 5} \\\\\n& \\mathrm{Pa}(2,2)+\\mathrm{Pa}(3,2)+\\mathrm{Pa}(4,2)+\\mathrm{Pa}(5,2)+\\mathrm{Pa}(6,2)=1+3+6+10+15=\\mathbf{3 5} \\\\\n& \\mathrm{Pa}(3,3)+\\mathrm{Pa}(4,3)+\\mathrm{Pa}(5,3)+\\mathrm{Pa}(6,3)+\\mathrm{Pa}(7,3)=1+4+10+20+35=\\mathbf{7 0} \\\\\n& \\mathrm{Pa}(4,4)+\\mathrm{Pa}(5,4)+\\mathrm{Pa}(6,4)+\\mathrm{Pa}(7,4)+\\mathrm{Pa}(8,4)=1+5+15+35+70=\\mathbf{1 2 6}\n\\end{aligned}\n$$']","['15, 35, 70, 126']",True,,Numerical, 3050,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$.","['$$\n\\begin{aligned}\n& \\mathrm{Pa}(1,1)+\\mathrm{Pa}(2,1)+\\mathrm{Pa}(3,1)+\\mathrm{Pa}(4,1)+\\mathrm{Pa}(5,1)=1+2+3+4+5=\\mathbf{1 5} \\\\\n& \\mathrm{Pa}(2,2)+\\mathrm{Pa}(3,2)+\\mathrm{Pa}(4,2)+\\mathrm{Pa}(5,2)+\\mathrm{Pa}(6,2)=1+3+6+10+15=\\mathbf{3 5} \\\\\n& \\mathrm{Pa}(3,3)+\\mathrm{Pa}(4,3)+\\mathrm{Pa}(5,3)+\\mathrm{Pa}(6,3)+\\mathrm{Pa}(7,3)=1+4+10+20+35=\\mathbf{7 0} \\\\\n& \\mathrm{Pa}(4,4)+\\mathrm{Pa}(5,4)+\\mathrm{Pa}(6,4)+\\mathrm{Pa}(7,4)+\\mathrm{Pa}(8,4)=1+5+15+35+70=\\mathbf{1 2 6}\n\\end{aligned}\n$$']","['15, 35, 70, 126']",True,,Numerical, 3051,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$.","['Notice that $\\mathrm{Pa}(n, n)+\\operatorname{Pa}(n+1, n)+\\cdots+\\operatorname{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\\mathrm{Pa}(n, n)$ and the right side is $\\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\\mathrm{Pa}(n, n)+\\mathrm{Pa}(n+1, n)+\\cdots+\\mathrm{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, then adding $\\mathrm{Pa}(n+k+1, n)$ to both sides yields $\\mathrm{Pa}(n+k+1, n)+\\mathrm{Pa}(n+k+1, n+1)=$ $\\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\\mathrm{Pa}$.']","['$m=n+k+1$, $j=n+1$']",True,,Expression, 3051,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$.","['Notice that $\\mathrm{Pa}(n, n)+\\operatorname{Pa}(n+1, n)+\\cdots+\\operatorname{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\\mathrm{Pa}(n, n)$ and the right side is $\\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\\mathrm{Pa}(n, n)+\\mathrm{Pa}(n+1, n)+\\cdots+\\mathrm{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, then adding $\\mathrm{Pa}(n+k+1, n)$ to both sides yields $\\mathrm{Pa}(n+k+1, n)+\\mathrm{Pa}(n+k+1, n+1)=$ $\\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\\mathrm{Pa}$.']","['$m=n+k+1$, $j=n+1$']",True,,Expression, 3052,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Prove that $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ for all nonnegative integers $n$.","['By definition of $\\mathrm{Pa}, \\operatorname{Pa}(n, 0)=\\operatorname{Pa}(n, n)=1$ for all nonnegative integers $n$, and this value is odd, so $\\operatorname{PaP}(n, 0)=\\operatorname{PaP}(n, n)=1$ by definition.']",,True,,, 3052,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Prove that $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ for all nonnegative integers $n$.","['By definition of $\\mathrm{Pa}, \\operatorname{Pa}(n, 0)=\\operatorname{Pa}(n, n)=1$ for all nonnegative integers $n$, and this value is odd, so $\\operatorname{PaP}(n, 0)=\\operatorname{PaP}(n, n)=1$ by definition.']",['证明题，略'],True,,Need_human_evaluate, 3053,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Compute rows $n=0$ to $n=8$ of $\mathrm{PaP}$.",['![](https://cdn.mathpix.com/cropped/2023_12_21_86f320d0441923b1aeb4g-1.jpg?height=463&width=938&top_left_y=1243&top_left_x=645)'],['![](https://cdn.mathpix.com/cropped/2023_12_21_86f320d0441923b1aeb4g-1.jpg?height=463&width=938&top_left_y=1243&top_left_x=645)'],False,,Need_human_evaluate, 3054,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ If $n=2^{j}$ for some nonnegative integer $j$, and $01$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ If $n=2^{j}$ for some nonnegative integer $j$, and $01$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Let $j \geq 0$, and suppose $n \geq 2^{j}$. Prove that $\mathrm{Pa}(n, k)$ has the same parity as the sum $\operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right)$, i.e., either both $\operatorname{Pa}(n, k)$ and the given sum are even, or both are odd.","['Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $01$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ Let $j \geq 0$, and suppose $n \geq 2^{j}$. Prove that $\mathrm{Pa}(n, k)$ has the same parity as the sum $\operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right)$, i.e., either both $\operatorname{Pa}(n, k)$ and the given sum are even, or both are odd.","['Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $01$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ If $j$ is an integer such that $2^{j} \leq n<2^{j+1}$, and $k<2^{j}$, prove that $$ \operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right) . $$","['Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $01$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Consider the parity of each entry: define $$ \operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases} $$ If $j$ is an integer such that $2^{j} \leq n<2^{j+1}$, and $k<2^{j}$, prove that $$ \operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right) . $$","['Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $01$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) Compute the next three rows of Clark's Triangle.",['![](https://cdn.mathpix.com/cropped/2023_12_21_60cd80f97141be4c26ddg-1.jpg?height=159&width=979&top_left_y=1666&top_left_x=622)'],['![](https://cdn.mathpix.com/cropped/2023_12_21_60cd80f97141be4c26ddg-1.jpg?height=159&width=979&top_left_y=1666&top_left_x=622)'],False,,Need_human_evaluate, 3058,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. If $\mathrm{Cl}(n, 1)=a n^{2}+b n+c$, determine the values of $a, b$, and $c$.","['Using the given values yields the system of equations below.\n\n$$\n\\left\\{\\begin{array}{l}\n\\mathrm{Cl}(1,1)=1=a(1)^{2}+b(1)+c \\\\\n\\mathrm{Cl}(2,1)=7=a(2)^{2}+b(2)+c \\\\\n\\mathrm{Cl}(3,1)=19=a(3)^{2}+b(3)+c\n\\end{array}\\right.\n$$\n\nSolving this system, $a=3, b=-3, c=1$.']","['$3,-3,1$']",True,,Numerical, 3058,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) If $\mathrm{Cl}(n, 1)=a n^{2}+b n+c$, determine the values of $a, b$, and $c$.","['Using the given values yields the system of equations below.\n\n$$\n\\left\\{\\begin{array}{l}\n\\mathrm{Cl}(1,1)=1=a(1)^{2}+b(1)+c \\\\\n\\mathrm{Cl}(2,1)=7=a(2)^{2}+b(2)+c \\\\\n\\mathrm{Cl}(3,1)=19=a(3)^{2}+b(3)+c\n\\end{array}\\right.\n$$\n\nSolving this system, $a=3, b=-3, c=1$.']","['$3,-3,1$']",True,,Numerical, 3059,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. Prove the formula $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1$.","['Use induction on $n$. For $n=1,2,3$, the values above demonstrate the theorem. If $\\mathrm{Cl}(n, 1)=$ $3 n^{2}-3 n+1$, then $\\mathrm{Cl}(n+1,1)=\\mathrm{Cl}(n, 0)+\\mathrm{Cl}(n, 1)=6 n+\\left(3 n^{2}-3 n+1\\right)=\\left(3 n^{2}+6 n+3\\right)-$ $(3 n+3)+1=3(n+1)^{2}-3(n+1)+1$.']",,True,,, 3059,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) Prove the formula $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1$.","['Use induction on $n$. For $n=1,2,3$, the values above demonstrate the theorem. If $\\mathrm{Cl}(n, 1)=$ $3 n^{2}-3 n+1$, then $\\mathrm{Cl}(n+1,1)=\\mathrm{Cl}(n, 0)+\\mathrm{Cl}(n, 1)=6 n+\\left(3 n^{2}-3 n+1\\right)=\\left(3 n^{2}+6 n+3\\right)-$ $(3 n+3)+1=3(n+1)^{2}-3(n+1)+1$.']",['证明题，略'],True,,Need_human_evaluate, 3060,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. Compute $\mathrm{Cl}(11,2)$.","['$\\mathrm{Cl}(11,2)=1000$.']",['1000'],False,,Numerical, 3060,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) Compute $\mathrm{Cl}(11,2)$.","['$\\mathrm{Cl}(11,2)=1000$.']",['1000'],False,,Numerical, 3061,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) Find and justify a formula for $\mathrm{Cl}(n, 2)$ in terms of $n$.","['$\\mathrm{Cl}(n, 2)=(n-1)^{3}$. Use induction on $n$. First, rewrite $\\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1=n^{3}-(n-1)^{3}$, and notice that $\\mathrm{Cl}(2,2)=1=(2-1)^{3}$. Then if $\\mathrm{Cl}(n, 2)=(n-1)^{3}$, using the recursive definition, $\\mathrm{Cl}(n+1,2)=\\mathrm{Cl}(n, 2)+\\mathrm{Cl}(n, 1)=(n-1)^{3}+\\left(n^{3}-(n-1)^{3}\\right)=n^{3}$.']","['$\\mathrm{Cl}(n, 2)=(n-1)^{3}$. Use induction on $n$. First, rewrite $\\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1=n^{3}-(n-1)^{3}$, and notice that $\\mathrm{Cl}(2,2)=1=(2-1)^{3}$. Then if $\\mathrm{Cl}(n, 2)=(n-1)^{3}$, using the recursive definition, $\\mathrm{Cl}(n+1,2)=\\mathrm{Cl}(n, 2)+\\mathrm{Cl}(n, 1)=(n-1)^{3}+\\left(n^{3}-(n-1)^{3}\\right)=n^{3}$.']",True,,Need_human_evaluate, 3062,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) Compute $\mathrm{Cl}(11,3)$.","['$\\mathrm{Cl}(11,3)=2025$.']",['2025'],False,,Numerical, 3062,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. Compute $\mathrm{Cl}(11,3)$.","['$\\mathrm{Cl}(11,3)=2025$.']",['2025'],False,,Numerical, 3063,Combinatorics,,"The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or ""Mountain of Gems"". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle (""PT"") and its variants. Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below. PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is: $$ \begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases} $$ The first four rows of Clark's Triangle are given below. ![](https://cdn.mathpix.com/cropped/2023_12_21_c41919b703ea0b966244g-1.jpg?height=412&width=1152&top_left_y=1602&top_left_x=476) Find and justify a formula for $\mathrm{Cl}(n, 3)$ in terms of $n$.","['Notice that $\\mathrm{Cl}(3,3)=1=1^{3}$, and then for $n>3, \\mathrm{Cl}(n, 3)=\\mathrm{Cl}(n-1,2)+\\mathrm{Cl}(n-1,3)$; replacing $\\mathrm{Cl}(n-1,3)$ analogously on the right side yields the summation\n\n$$\n\\mathrm{Cl}(n, 3)=\\mathrm{Cl}(n-1,2)+\\mathrm{Cl}(n-2,2)+\\ldots+\\mathrm{Cl}(3,2)+1\n$$\n\nNow we prove that $\\mathrm{Cl}(n, 2)=(n-1)^{3}$. Use induction on $n$. First, rewrite $\\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1=n^{3}-(n-1)^{3}$, and notice that $\\mathrm{Cl}(2,2)=1=(2-1)^{3}$. Then if $\\mathrm{Cl}(n, 2)=(n-1)^{3}$, using the recursive definition, $\\mathrm{Cl}(n+1,2)=\\mathrm{Cl}(n, 2)+\\mathrm{Cl}(n, 1)=(n-1)^{3}+\\left(n^{3}-(n-1)^{3}\\right)=n^{3}$.\n\nBy the formula proved above, \\mathrm{Cl}(n-1,2)=(n-2)^{3}$, so this formula is equivalent to\n\n$$\n\\mathrm{Cl}(n, 3)=(n-2)^{3}+(n-3)^{3}+\\ldots+2^{3}+1\n$$\n\nUse the identity $1^{3}+2^{3}+\\cdots+m^{3}=\\frac{m^{2}(m+1)^{2}}{4}$ and substitute $n-2$ for $m$ to obtain $\\mathrm{Cl}(n, 3)=$ $\\frac{(n-2)^{2}(n-1)^{2}}{4}$.']","['Notice that $\\mathrm{Cl}(3,3)=1=1^{3}$, and then for $n>3, \\mathrm{Cl}(n, 3)=\\mathrm{Cl}(n-1,2)+\\mathrm{Cl}(n-1,3)$; replacing $\\mathrm{Cl}(n-1,3)$ analogously on the right side yields the summation\n\n$$\n\\mathrm{Cl}(n, 3)=\\mathrm{Cl}(n-1,2)+\\mathrm{Cl}(n-2,2)+\\ldots+\\mathrm{Cl}(3,2)+1\n$$\n\nNow we prove that $\\mathrm{Cl}(n, 2)=(n-1)^{3}$. Use induction on $n$. First, rewrite $\\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1=n^{3}-(n-1)^{3}$, and notice that $\\mathrm{Cl}(2,2)=1=(2-1)^{3}$. Then if $\\mathrm{Cl}(n, 2)=(n-1)^{3}$, using the recursive definition, $\\mathrm{Cl}(n+1,2)=\\mathrm{Cl}(n, 2)+\\mathrm{Cl}(n, 1)=(n-1)^{3}+\\left(n^{3}-(n-1)^{3}\\right)=n^{3}$.\n\nBy the formula proved above, \\mathrm{Cl}(n-1,2)=(n-2)^{3}$, so this formula is equivalent to\n\n$$\n\\mathrm{Cl}(n, 3)=(n-2)^{3}+(n-3)^{3}+\\ldots+2^{3}+1\n$$\n\nUse the identity $1^{3}+2^{3}+\\cdots+m^{3}=\\frac{m^{2}(m+1)^{2}}{4}$ and substitute $n-2$ for $m$ to obtain $\\mathrm{Cl}(n, 3)=$ $\\frac{(n-2)^{2}(n-1)^{2}}{4}$.']",True,,Need_human_evaluate, 3064,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute the entries in the next two rows of Leibniz's Triangle.",['![](https://cdn.mathpix.com/cropped/2023_12_21_d6baba47070a814ce4efg-1.jpg?height=134&width=730&top_left_y=280&top_left_x=746)'],['![](https://cdn.mathpix.com/cropped/2023_12_21_d6baba47070a814ce4efg-1.jpg?height=134&width=730&top_left_y=280&top_left_x=746)'],False,,Need_human_evaluate, 3065,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute Le(17,1).","['$\\operatorname{Le}(17,1)=\\operatorname{Le}(16,0)-\\operatorname{Le}(17,0)=\\frac{1}{17}-\\frac{1}{18}=\\frac{1}{306}$.']",['$\\frac{1}{306}$'],False,,Numerical, 3065,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute Le(17,1).","['$\\operatorname{Le}(17,1)=\\operatorname{Le}(16,0)-\\operatorname{Le}(17,0)=\\frac{1}{17}-\\frac{1}{18}=\\frac{1}{306}$.']",['$\\frac{1}{306}$'],False,,Numerical, 3066,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute $\operatorname{Le}(17,2)$.","['$\\operatorname{Le}(17,2)=\\operatorname{Le}(16,1)-\\operatorname{Le}(17,1)=\\operatorname{Le}(15,0)-\\operatorname{Le}(16,0)-\\operatorname{Le}(17,1)=\\frac{1}{2448}$.']",['$\\frac{1}{2448}$'],False,,Numerical, 3066,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute $\operatorname{Le}(17,2)$.","['$\\operatorname{Le}(17,2)=\\operatorname{Le}(16,1)-\\operatorname{Le}(17,1)=\\operatorname{Le}(15,0)-\\operatorname{Le}(16,0)-\\operatorname{Le}(17,1)=\\frac{1}{2448}$.']",['$\\frac{1}{2448}$'],False,,Numerical, 3067,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Find and justify a formula for $\operatorname{Le}(n, 1)$ in terms of $n$.","['$\\operatorname{Le}(n, 1)=\\operatorname{Le}(n-1,0)-\\operatorname{Le}(n, 0)=\\frac{1}{n}-\\frac{1}{n+1}=\\frac{1}{n(n+1)}$.']","['$\\operatorname{Le}(n, 1)=\\operatorname{Le}(n-1,0)-\\operatorname{Le}(n, 0)=\\frac{1}{n}-\\frac{1}{n+1}=\\frac{1}{n(n+1)}$.']",False,,Need_human_evaluate, 3068,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute $\sum_{n=1}^{2011} \operatorname{Le}(n, 1)$.","['Because $\\operatorname{Le}(n, 1)=\\frac{1}{n}-\\frac{1}{n+1}$,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2011} \\operatorname{Le}(i, 1) & =\\sum_{i=1}^{2011}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\\\\n& =\\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots+\\left(\\frac{1}{2010}-\\frac{1}{2011}\\right)+\\left(\\frac{1}{2011}-\\frac{1}{2012}\\right) \\\\\n& =1-\\frac{1}{2012} \\\\\n& =\\frac{2011}{2012} .\n\\end{aligned}\n$$']",['$\\frac{2011}{2012}$'],False,,Numerical, 3068,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Compute $\sum_{n=1}^{2011} \operatorname{Le}(n, 1)$.","['Because $\\operatorname{Le}(n, 1)=\\frac{1}{n}-\\frac{1}{n+1}$,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2011} \\operatorname{Le}(i, 1) & =\\sum_{i=1}^{2011}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\\\\n& =\\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots+\\left(\\frac{1}{2010}-\\frac{1}{2011}\\right)+\\left(\\frac{1}{2011}-\\frac{1}{2012}\\right) \\\\\n& =1-\\frac{1}{2012} \\\\\n& =\\frac{2011}{2012} .\n\\end{aligned}\n$$']",['$\\frac{2011}{2012}$'],False,,Numerical, 3069,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Find and justify a formula for $\operatorname{Le}(n, 2)$ in terms of $n$.","['$\\operatorname{Le}(n, 2)=\\operatorname{Le}(n-1,1)-\\operatorname{Le}(n, 1)=\\frac{1}{n(n-1)}-\\frac{1}{n(n+1)}=\\frac{2}{(n-1)(n)(n+1)}$. Note that this result appears in the table as a unit fraction because at least one of the integers $n-1, n, n+1$ is even.']","['$\\operatorname{Le}(n, 2)=\\operatorname{Le}(n-1,1)-\\operatorname{Le}(n, 1)=\\frac{1}{n(n-1)}-\\frac{1}{n(n+1)}=\\frac{2}{(n-1)(n)(n+1)}$. Note that this result appears in the table as a unit fraction because at least one of the integers $n-1, n, n+1$ is even.']",True,,Need_human_evaluate, 3070,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=1}^{\infty} \operatorname{Le}(i, 1)=\operatorname{Le}(n, k)$, determine the values of $n$ and $k$.","['Extending the result of $8 \\mathrm{~b}$ gives\n\n$$\n\\sum_{i=1}^{n} \\operatorname{Le}(i, 1)=\\frac{1}{1}-\\frac{1}{n}\n$$\n\nso as $n \\rightarrow \\infty, \\sum_{i=1}^{n} \\operatorname{Le}(i, 1) \\rightarrow 1$. This value appears as $\\operatorname{Le}(0,0)$, so $n=k=0$.']","['$0,0$']",True,,Numerical, 3070,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=1}^{\infty} \operatorname{Le}(i, 1)=\operatorname{Le}(n, k)$, determine the values of $n$ and $k$.","['Extending the result of $8 \\mathrm{~b}$ gives\n\n$$\n\\sum_{i=1}^{n} \\operatorname{Le}(i, 1)=\\frac{1}{1}-\\frac{1}{n}\n$$\n\nso as $n \\rightarrow \\infty, \\sum_{i=1}^{n} \\operatorname{Le}(i, 1) \\rightarrow 1$. This value appears as $\\operatorname{Le}(0,0)$, so $n=k=0$.']","['$0,0$']",True,,Numerical, 3071,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, compute expressions for $n$ and $k$ in terms of $m$.",['$n=k=m-1$.'],"['$m-1,m-1$']",True,,Expression, 3071,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, compute expressions for $n$ and $k$ in terms of $m$.",['$n=k=m-1$.'],"['$m-1,m-1$']",True,,Expression, 3072,Combinatorics,,"Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$.","[""Because in general $\\operatorname{Le}(i, m)=\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)$, a partial sum can be rewritten as follows:\n\n$$\n\\begin{aligned}\n\\sum_{i=m}^{n} \\operatorname{Le}(i, m)= & \\sum_{i=m}^{n}(\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)) \\\\\n= & (\\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(m, m-1))+(\\operatorname{Le}(m, m-1)-\\operatorname{Le}(m+1, m-1))+ \\\\\n& \\cdots+(\\operatorname{Le}(n-1, m-1)-\\operatorname{Le}(n, m-1)) \\\\\n= & \\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(n, m-1) .\n\\end{aligned}\n$$\n\nBecause the values of $\\operatorname{Le}(n, m-1)$ get arbitrarily small as $n$ increases (proof: $\\operatorname{Le}(i, j)<$ $\\operatorname{Le}(i-1, j-1)$ by construction, so $\\left.\\operatorname{Le}(n, m-1)<\\operatorname{Le}(n-m+1,0)=\\frac{1}{n-m+1}\\right)$, the limit of these partial sums is $\\operatorname{Le}(m-1, m-1)$. So $n=k=m-1$.\n\n\n\nNote: This result can be extended even further. In fact, for every value of $k For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$.","[""Because in general $\\operatorname{Le}(i, m)=\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)$, a partial sum can be rewritten as follows:\n\n$$\n\\begin{aligned}\n\\sum_{i=m}^{n} \\operatorname{Le}(i, m)= & \\sum_{i=m}^{n}(\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)) \\\\\n= & (\\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(m, m-1))+(\\operatorname{Le}(m, m-1)-\\operatorname{Le}(m+1, m-1))+ \\\\\n& \\cdots+(\\operatorname{Le}(n-1, m-1)-\\operatorname{Le}(n, m-1)) \\\\\n= & \\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(n, m-1) .\n\\end{aligned}\n$$\n\nBecause the values of $\\operatorname{Le}(n, m-1)$ get arbitrarily small as $n$ increases (proof: $\\operatorname{Le}(i, j)<$ $\\operatorname{Le}(i-1, j-1)$ by construction, so $\\left.\\operatorname{Le}(n, m-1)<\\operatorname{Le}(n-m+1,0)=\\frac{1}{n-m+1}\\right)$, the limit of these partial sums is $\\operatorname{Le}(m-1, m-1)$. So $n=k=m-1$.\n\n\n\nNote: This result can be extended even further. In fact, for every value of $k0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Find three distinct sets of positive integers $\{a, b, c, d\}$ with $a1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_b0415e12b85f62aa86deg-1.jpg?height=420&width=1201&top_left_y=1015&top_left_x=473) As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question. Leibniz's Harmonic Triangle: Consider the triangle formed by the rule $$ \begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases} $$ This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below. ![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470) For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$. Find and prove a closed formula (that is, a formula with a fixed number of terms and no ""..."") for $\operatorname{Le}(n, k)$ in terms of $n, k$, and the $\mathrm{Pa}$ function.","['The formula is $\\operatorname{Le}(n, k)=\\frac{1}{(n+1) \\cdot \\operatorname{Pa}(n, k)}$, or equivalently, $\\frac{1}{(k+1) \\cdot \\operatorname{Pa}(n+1, k+1)}$. Because $\\operatorname{Pa}(n, 0)=$ $\\operatorname{Pa}(n, n)=1$, when $k=0$ or $k=n$, the formula is equivalent to the definition of $\\operatorname{Le}(n, 0)=$ $\\operatorname{Le}(n, n)=\\frac{1}{n+1}$.\n\nTo prove the formula for $1 \\leq k \\leq n-1$, use induction on $k$. The base case $k=0$ was proved above. If the formula holds for a particular value of $k\n\nThen $P P^{\\prime}=M Q^{\\prime}=p$ and $Q M=q-p$, while $P Q=p+q$ and $P M=P^{\\prime} Q^{\\prime}$. By the Pythagorean Theorem, $(q-p)^{2}+P^{\\prime} Q^{\\prime 2}=(p+q)^{2}$, so $q=\\frac{\\left(P^{\\prime} Q^{\\prime}\\right)^{2}}{4 p}$. Thus $4 p q=P^{\\prime} Q^{\\prime 2}=12^{2}$. Similarly, $4 p r=P^{\\prime} R^{\\prime 2}=6^{2}$ and $4 q r=Q^{\\prime} R^{\\prime 2}=12^{2}$. Dividing the first equation by the third shows that $p=r$ (which can also be inferred from the symmetry of $\\triangle P^{\\prime} Q^{\\prime} R^{\\prime}$ ) and the equation $p r=9$ yields 3 as their common value; substitute in either of the other two equations to obtain $q=12$. Therefore the sides of $\\triangle P Q R$ are $P Q=Q R=12+3=15$ and $P R=6$. The altitude to $\\overline{P R}$ has length $\\sqrt{15^{2}-3^{2}}=6 \\sqrt{6}$, so the triangle's area is $\\frac{1}{2}(6)(6 \\sqrt{6})=\\mathbf{1 8} \\sqrt{\\mathbf{6}}$.""]",['$18 \\sqrt{6}$'],False,,Numerical, 3076,Algebra,,Let $f(x)=x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots$. Compute the coefficient of $x^{10}$ in $f(f(x))$.,"['By the definition of $f$,\n\n$$\nf(f(x))=f(x)+(f(x))^{2}+(f(x))^{4}+(f(x))^{8}+\\cdots\n$$\n\nConsider this series term by term. The first term, $f(x)$, contains no $x^{10}$ terms, so its contribution is 0 . The second term, $(f(x))^{2}$, can produce terms of $x^{10}$ in two ways: as $x^{2} \\cdot x^{8}$ or as $x^{8} \\cdot x^{2}$. So its contribution is 2 .\n\nNow consider the third term:\n\n$$\n\\begin{aligned}\n(f(x))^{4}= & f(x) \\cdot f(x) \\cdot f(x) \\cdot f(x) \\\\\n= & \\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) \\cdot\\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) \\cdot \\\\\n& \\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) \\cdot\\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\\cdots\\right) .\n\\end{aligned}\n$$\n\nEach $x^{10}$ term in the product is the result of multiplying four terms whose exponents sum to 10 , one from each factor of $f(x)$. Thus this product contains a term of $x^{10}$ for each quadruple\n\n\n\nof nonnegative integers $(i, j, k, l)$ such that $2^{i}+2^{j}+2^{k}+2^{l}=10$; the order of the quadruple is relevant because rearrangements of the integers correspond to choosing terms from different factors. Note that none of the exponents can exceed 2 because $2^{3}+2^{0}+2^{0}+2^{0}>10$. Therefore $i, j, k, l \\leq 2$. Considering cases from largest values to smallest yields two basic cases. First, $10=4+4+1+1=2^{2}+2^{2}+2^{0}+2^{0}$, which yields $\\frac{4 !}{2 ! \\cdot 2 !}=6$ ordered quadruples. Second, $10=4+2+2+2=2^{2}+2^{1}+2^{1}+2^{1}$, which yields 4 ordered quadruples. Thus the contribution of the $(f(x))^{4}$ term is $6+4=10$.\n\nThe last term to consider is $f(x)^{8}$, because $(f(x))^{n}$ contains no terms of degree less than $n$. An analogous analysis to the case of $(f(x))^{4}$ suggests that the expansion of $(f(x))^{8}$ has an $x^{10}$ term for every ordered partition of 10 into a sum of eight powers of two. Up to order, there is only one such partition: $2^{1}+2^{1}+2^{0}+2^{0}+2^{0}+2^{0}+2^{0}+2^{0}$, which yields $\\frac{8 !}{6 ! \\cdot 2 !}=28$ ordered quadruples.\n\nTherefore the coefficient of $x^{10}$ is $2+10+28=\\mathbf{4 0}$.']",['40'],False,,Numerical, 3077,Algebra,,Compute $\left\lfloor 100000(1.002)^{10}\right\rfloor$.,"['Consider the expansion of $(1.002)^{10}$ as $(1+0.002)^{10}$. Using the Binomial Theorem yields the following:\n\n$$\n(1+0.002)^{10}=1+\\left(\\begin{array}{c}\n10 \\\\\n1\n\\end{array}\\right)(0.002)+\\left(\\begin{array}{c}\n10 \\\\\n2\n\\end{array}\\right)(0.002)^{2}+\\left(\\begin{array}{c}\n10 \\\\\n3\n\\end{array}\\right)(0.002)^{3}+\\cdots+(0.002)^{10} .\n$$\n\nHowever, when $k>3$, the terms $\\left(\\begin{array}{c}10 \\\\ k\\end{array}\\right)(0.002)^{k}$ do not affect the final answer, because $0.002^{4}=$ $0.000000000016=\\frac{16}{10^{12}}$, and the maximum binomial coefficient is $\\left(\\begin{array}{c}10 \\\\ 5\\end{array}\\right)=252$, so\n\n$$\n\\left(\\begin{array}{c}\n10 \\\\\n4\n\\end{array}\\right)(0.002)^{4}+\\left(\\begin{array}{c}\n10 \\\\\n5\n\\end{array}\\right)(0.002)^{5}+\\cdots+(0.002)^{10}<\\frac{252 \\cdot 16}{10^{12}}+\\frac{252 \\cdot 16}{10^{12}}+\\cdots+\\frac{252 \\cdot 16}{10^{12}},\n$$\n\nwhere the right side of the inequality contains seven terms, giving an upper bound of $\\frac{7 \\cdot 252 \\cdot 16}{10^{12}}$. The numerator is approximately 28000 , but $\\frac{28000}{10^{12}}=2.8 \\times 10^{-8}$. So even when multiplied by $100000=10^{5}$, these terms contribute at most $3 \\times 10^{-3}$ to the value of the expression before rounding.\n\nThe result of adding the first four terms $(k=0$ through $k=3)$ and multiplying by 100,000 is given by the following sum:\n\n$$\n100000+10(200)+45(0.4)+120(0.0008)=100000+2000+18+0.096=102018.096 .\n$$\n\nThen the desired quantity is $\\lfloor 102018.096\\rfloor=\\mathbf{1 0 2 , 0 1 8}$.']",['102018'],False,,Numerical, 3078,Algebra,,"If $1, x, y$ is a geometric sequence and $x, y, 3$ is an arithmetic sequence, compute the maximum value of $x+y$.","['The common ratio in the geometric sequence $1, x, y$ is $\\frac{x}{1}=x$, so $y=x^{2}$. The arithmetic sequence $x, y, 3$ has a common difference, so $y-x=3-y$. Substituting $y=x^{2}$ in the equation yields\n\n$$\n\\begin{aligned}\nx^{2}-x & =3-x^{2} \\\\\n2 x^{2}-x-3 & =0\n\\end{aligned}\n$$\n\nfrom which $x=\\frac{3}{2}$ or -1 . The respective values of $y$ are $y=x^{2}=\\frac{9}{4}$ or 1 . Thus the possible values of $x+y$ are $\\frac{15}{4}$ and 0 , so the answer is $\\frac{\\mathbf{1 5}}{\\mathbf{4}}$.']",['$\\frac{15}{4}$'],False,,Numerical, 3079,Algebra,,"Define the sequence of positive integers $\left\{a_{n}\right\}$ as follows: $$ \left\{\begin{array}{l} a_{1}=1 \\ \text { for } n \geq 2, a_{n} \text { is the smallest possible positive value of } n-a_{k}^{2}, \text { for } 1 \leq k0$, this equation can be simplified to\n\n$$\nb^{3}+X \\cdot b+Y=19 b^{2}+31 b+17\n$$\n\nThus $Y=17$ and $b^{2}+X=19 b+31$, from which $b(b-19)=31-X$. The expression on the left side is positive (because $b>19$ ) and the expression on the right side is at most 31 (because $X>0$ ), so the only possible solution is $b=20, X=11$. The answer is 20 .']",['20'],False,,Numerical, 3081,Geometry,,"Some portions of the line $y=4 x$ lie below the curve $y=10 \pi \sin ^{2} x$, and other portions lie above the curve. Compute the sum of the lengths of all the segments of the graph of $y=4 x$ that lie in the first quadrant, below the graph of $y=10 \pi \sin ^{2} x$.","['Notice first that all intersections of the two graphs occur in the interval $0 \\leq x \\leq \\frac{5 \\pi}{2}$, because the maximum value of $10 \\pi \\sin ^{2} x$ is $10 \\pi$ (at odd multiples of $\\frac{\\pi}{2}$ ), and $4 x>10 \\pi$ when $x>\\frac{5 \\pi}{2}$. The graphs are shown below.\n\n\n\nWithin that interval, both graphs are symmetric about the point $A=\\left(\\frac{5 \\pi}{4}, 5 \\pi\\right)$. For the case of $y=10 \\pi \\sin ^{2} x$, this symmetry can be seen by using the power-reducing identity $\\sin ^{2} x=$ $\\frac{1-\\cos 2 x}{2}$. Then the equation becomes $y=5 \\pi-5 \\pi \\cos 2 x$, which has amplitude $5 \\pi$ about the line $y=5 \\pi$, and which crosses the line $y=5 \\pi$ for $x=\\frac{\\pi}{4}, \\frac{3 \\pi}{4}, \\frac{5 \\pi}{4}, \\ldots$ Label the points of intersection $A, B, C, D, E, F$, and $O$ as shown. Then $\\overline{A B} \\cong \\overline{A C}, \\overline{B D} \\cong \\overline{C E}$, and $\\overline{O D} \\cong \\overline{E F}$. Thus\n\n$$\n\\begin{aligned}\nB D+A C+E F & =O D+D B+B A \\\\\n& =O A .\n\\end{aligned}\n$$\n\nBy the Pythagorean Theorem,\n\n$$\n\\begin{aligned}\nO A & =\\sqrt{\\left(\\frac{5 \\pi}{4}\\right)^{2}+(5 \\pi)^{2}} \\\\\n& =\\frac{5 \\pi}{4} \\sqrt{1^{2}+4^{2}} \\\\\n& =\\frac{5 \\pi}{\\mathbf{4}} \\sqrt{\\mathbf{1 7}}\n\\end{aligned}\n$$']",['$\\frac{5 \\pi}{4} \\sqrt{17}$'],False,,Numerical, 3082,Geometry,,"In equilateral hexagon $A B C D E F, \mathrm{~m} \angle A=2 \mathrm{~m} \angle C=2 \mathrm{~m} \angle E=5 \mathrm{~m} \angle D=10 \mathrm{~m} \angle B=10 \mathrm{~m} \angle F$, and diagonal $B E=3$. Compute $[A B C D E F]$, that is, the area of $A B C D E F$.","['Let $\\mathrm{m} \\angle B=\\alpha$. Then the sum of the measures of the angles in the hexagon is:\n\n$$\n\\begin{aligned}\n720^{\\circ} & =\\mathrm{m} \\angle A+\\mathrm{m} \\angle C+\\mathrm{m} \\angle E+\\mathrm{m} \\angle D+\\mathrm{m} \\angle B+\\mathrm{m} \\angle F \\\\\n& =10 \\alpha+5 \\alpha+5 \\alpha+2 \\alpha+\\alpha+\\alpha=24 \\alpha .\n\\end{aligned}\n$$\n\n\n\nThus $30^{\\circ}=\\alpha$ and $\\mathrm{m} \\angle A=300^{\\circ}$, so the exterior angle at $A$ has measure $60^{\\circ}=\\mathrm{m} \\angle D$. Further, because $A B=C D$ and $D E=A F$, it follows that $\\triangle C D E \\cong \\triangle B A F$. Thus\n\n$$\n[A B C D E F]=[A B C E F]+[C D E]=[A B C E F]+[A B F]=[B C E F] .\n$$\n\n\n\nTo compute $[B C E F]$, notice that because $\\mathrm{m} \\angle D=60^{\\circ}, \\triangle C D E$ is equilateral. In addition,\n\n$$\n\\begin{aligned}\n150^{\\circ} & =\\mathrm{m} \\angle B C D \\\\\n& =\\mathrm{m} \\angle B C E+\\mathrm{m} \\angle D C E=\\mathrm{m} \\angle B C E+60^{\\circ} .\n\\end{aligned}\n$$\n\nTherefore $\\mathrm{m} \\angle B C E=90^{\\circ}$. Similarly, because the hexagon is symmetric, $\\mathrm{m} \\angle C E F=90^{\\circ}$, so quadrilateral $B C E F$ is actually a square with side length 3 . Thus $C E=\\frac{B E}{\\sqrt{2}}=\\frac{3}{\\sqrt{2}}$, and $[A B C D E F]=[B C E F]=\\frac{9}{2}$.\n\nAlternate Solution: Calculate the angles of the hexagon as in the first solution. Then proceed as follows.\n\nFirst, $A B C D E F$ can be partitioned into four congruent triangles. Because the hexagon is equilateral and $\\mathrm{m} \\angle A B C=\\mathrm{m} \\angle A F E=30^{\\circ}$, it follows that $\\triangle A B C$ and $\\triangle A F E$ are congruent isosceles triangles whose base angles measure $75^{\\circ}$. Next, $\\mathrm{m} \\angle A B C+\\mathrm{m} \\angle B C D=30^{\\circ}+150^{\\circ}=$ $180^{\\circ}$, so $\\overline{A B} \\| \\overline{C D}$. Because these two segments are also congruent, quadrilateral $A B C D$ is a parallelogram. In particular, $\\triangle C D A \\cong \\triangle A B C$. Similarly, $\\triangle E D A \\cong \\triangle A F E$.\n\nNow let $a=A C=A E$ be the length of the base of these isosceles triangles, and let $b=A B$ be the length of the other sides (or of the equilateral hexagon). Because the four triangles are congruent, $[A B C D E F]=[A B C]+[A C D]+[A D E]+[A E F]=4[A B C]=4 \\cdot \\frac{1}{2} b^{2} \\sin 30^{\\circ}=b^{2}$. Applying the Law of Cosines to $\\triangle A B C$ gives $a^{2}=b^{2}+b^{2}-2 b^{2} \\cos 30^{\\circ}=(2-\\sqrt{3}) b^{2}$. Because $4-2 \\sqrt{3}=(\\sqrt{3}-1)^{2}$, this gives $a=\\left(\\frac{\\sqrt{3}-1}{\\sqrt{2}}\\right) b$. Using the given length $B E=3$ and applying the Law of Cosines to $\\triangle A B E$ gives\n\n$$\n\\begin{aligned}\n9 & =a^{2}+b^{2}-2 a b \\cos 135^{\\circ} \\\\\n& =a^{2}+b^{2}+\\sqrt{2} a b \\\\\n& =(2-\\sqrt{3}) b^{2}+b^{2}+(\\sqrt{3}-1) b^{2} \\\\\n& =2 b^{2} .\n\\end{aligned}\n$$\n\nThus $[A B C D E F]=b^{2}=\\frac{9}{2}$.']",['$\\frac{9}{2}$'],False,,Numerical, 3083,Geometry,,"The taxicab distance between points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ is defined as $d(A, B)=$ $\left|x_{A}-x_{B}\right|+\left|y_{A}-y_{B}\right|$. Given some $s>0$ and points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$, define the taxicab ellipse with foci $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ to be the set of points $\{Q \mid d(A, Q)+d(B, Q)=s\}$. Compute the area enclosed by the taxicab ellipse with foci $(0,5)$ and $(12,0)$, passing through $(1,-1)$.","['Let $A=(0,5)$ and $B=(12,0)$, and let $C=(1,-1)$. First compute the distance sum: $d(A, C)+d(B, C)=19$. Notice that if $P=(x, y)$ is on the segment from $(0,-1)$ to $(12,-1)$, then $d(A, P)+d(B, P)$ is constant. This is because if $0\n\n\n\nThe simplest way to compute the polygon\'s area is to subtract the areas of the four corner triangles from that of the enclosing rectangle. The enclosing rectangle\'s area is $14 \\cdot 7=98$, while each triangle has area $\\frac{1}{2} \\cdot 1 \\cdot 1=\\frac{1}{2}$. Thus the area is $98-4 \\cdot \\frac{1}{2}=\\mathbf{9 6}$.']",['96'],False,,Numerical, 3084,Algebra,,"The function $f$ satisfies the relation $f(n)=f(n-1) f(n-2)$ for all integers $n$, and $f(n)>0$ for all positive integers $n$. If $f(1)=\frac{f(2)}{512}$ and $\frac{1}{f(1)}=2 f(2)$, compute $f(f(4))$.","['Substituting yields $\\frac{512}{f(2)}=2 f(2) \\Rightarrow(f(2))^{2}=256 \\Rightarrow f(2)=16$. Therefore $f(1)=\\frac{1}{32}$. Using the recursion, $f(3)=\\frac{1}{2}$ and $f(4)=8$. So $f(f(4))=f(8)$. Continue to apply the recursion:\n\n$$\nf(5)=4, \\quad f(6)=32, \\quad f(7)=128, \\quad f(8)=\\mathbf{4 0 9 6} .\n$$\n\nAlternate Solution: Let $g(n)=\\log _{2} f(n)$. Then $g(n)=g(n-1)+g(n-2)$, with initial conditions $g(1)=g(2)-9$ and $-g(1)=1+g(2)$. From this, $g(1)=-5$ and $g(2)=4$, and from the recursion,\n\n$$\ng(3)=-1, \\quad g(4)=3\n$$\n\nso $f(4)=2^{g(4)}=8$. Continue to apply the recursion:\n\n$$\ng(5)=2, \\quad g(6)=5, \\quad g(7)=7, \\quad g(8)=12\n$$\n\nBecause $g(f(4))=12$, it follows that $f(f(4))=2^{12}=\\mathbf{4 0 9 6}$.']",['4096'],False,,Numerical, 3085,Algebra,,"Frank Narf accidentally read a degree $n$ polynomial with integer coefficients backwards. That is, he read $a_{n} x^{n}+\ldots+a_{1} x+a_{0}$ as $a_{0} x^{n}+\ldots+a_{n-1} x+a_{n}$. Luckily, the reversed polynomial had the same zeros as the original polynomial. All the reversed polynomial's zeros were real, and also integers. If $1 \leq n \leq 7$, compute the number of such polynomials such that $\operatorname{GCD}\left(a_{0}, a_{1}, \ldots, a_{n}\right)=1$.","['When the coefficients of a polynomial $f$ are reversed to form a new polynomial $g$, the zeros of $g$ are the reciprocals of the zeros of $f: r$ is a zero of $f$ if and only if $r^{-1}$ is a zero of $g$. In this case, the two polynomials have the same zeros; that is, whenever $r$ is a zero of either, so must be $r^{-1}$. Furthermore, both $r$ and $r^{-1}$ must be real as well as integers, so $r= \\pm 1$. As the only zeros are \\pm 1 , and the greatest common divisor of all the coefficients is 1 , the polynomial must have leading coefficient 1 or -1 . Thus\n\n$$\n\\begin{aligned}\nf(x) & = \\pm(x \\pm 1)(x \\pm 1) \\cdots(x \\pm 1) \\\\\n& = \\pm(x+1)^{k}(x-1)^{n-k}\n\\end{aligned}\n$$\n\nIf $A_{n}$ is the number of such degree $n$ polynomials, then there are $n+1$ choices for $k, 0 \\leq k \\leq n$. Thus $A_{n}=2(n+1)$. The number of such degree $n$ polynomials for $1 \\leq n \\leq 7$ is the sum:\n\n$$\nA_{1}+A_{2}+\\ldots+A_{7}=2(2+3+\\ldots+8)=2 \\cdot 35=\\mathbf{7 0}\n$$']",['70'],False,,Numerical, 3086,Geometry,,"Given a regular 16-gon, extend three of its sides to form a triangle none of whose vertices lie on the 16-gon itself. Compute the number of noncongruent triangles that can be formed in this manner.","['Label the sides of the polygon, in order, $s_{0}, s_{1}, \\ldots, s_{15}$. First note that two sides of the polygon intersect at a vertex if and only if the sides are adjacent. So the sides chosen must be nonconsecutive. Second, if nonparallel sides $s_{i}$ and $s_{j}$ are extended, the angle of intersection is determined by $|i-j|$, as are the lengths of the extended portions of the segments. In other words, the spacing of the extended sides completely determines the shape of the triangle. So the problem reduces to selecting appropriate spacings, that is, finding integers $a, b, c \\geq 2$ whose sum is 16 . However, diametrically opposite sides are parallel, so (for example) the sides $s_{3}$ and $s_{11}$ cannot both be used. Thus none of $a, b, c$ may equal 8 . Taking $s_{0}$ as the first side, the second side would be $s_{0+a}=s_{a}$, and the third side would be $s_{a+b}$, with $c$ sides between $s_{a+b}$ and $s_{0}$. To eliminate reflections and rotations, specify additionally that $a \\geq b \\geq c$. The allowable partitions are in the table below.\n\n| $a$ | $b$ | $c$ | triangle |\n| :---: | :---: | :---: | :---: |\n| 12 | 2 | 2 | $s_{0} s_{12} s_{14}$ |\n| 11 | 3 | 2 | $s_{0} s_{11} s_{14}$ |\n| 10 | 4 | 2 | $s_{0} s_{10} s_{14}$ |\n| 10 | 3 | 3 | $s_{0} s_{10} s_{13}$ |\n| 9 | 5 | 2 | $s_{0} s_{9} s_{14}$ |\n| 9 | 4 | 3 | $s_{0} s_{9} s_{13}$ |\n| 7 | 7 | 2 | $s_{0} s_{7} s_{14}$ |\n| 7 | 6 | 3 | $s_{0} s_{7} s_{13}$ |\n| 7 | 5 | 4 | $s_{0} s_{7} s_{12}$ |\n| 6 | 6 | 4 | $s_{0} s_{6} s_{12}$ |\n| 6 | 5 | 5 | $s_{0} s_{6} s_{11}$ |\n\nThus there are $\\mathbf{1 1}$ distinct such triangles.']",['11'],False,,Numerical, 3087,Geometry,,"Two square tiles of area 9 are placed with one directly on top of the other. The top tile is then rotated about its center by an acute angle $\theta$. If the area of the overlapping region is 8 , compute $\sin \theta+\cos \theta$.","['In the diagram below, $O$ is the center of both squares $A_{1} A_{2} A_{3} A_{4}$ and $B_{1} B_{2} B_{3} B_{4}$. Let $P_{1}, P_{2}, P_{3}, P_{4}$ and $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ be the intersections of the sides of the squares as shown. Let $H_{A}$ be on $\\overline{A_{3} A_{4}}$ so that $\\angle A_{3} H_{A} O$ is right. Similarly, let $H_{B}$ be on $\\overline{B_{3} B_{4}}$ such that $\\angle B_{3} H_{B} O$ is right. Then the angle by which $B_{1} B_{2} B_{3} B_{4}$ was rotated is $\\angle H_{A} O H_{B}$. Extend $\\overline{O H_{B}}$ to meet $\\overline{A_{3} A_{4}}$ at $M$.\n\n\n\nBoth $\\triangle H_{A} O M$ and $\\triangle H_{B} P_{3} M$ are right triangles sharing acute $\\angle M$, so $\\triangle H_{A} O M \\sim \\triangle H_{B} P_{3} M$. By an analogous argument, both triangles are similar to $\\triangle B_{3} P_{3} Q_{3}$. Thus $\\mathrm{m} \\angle Q_{3} P_{3} B_{3}=\\theta$. Now let $B_{3} P_{3}=x, B_{3} Q_{3}=y$, and $P_{3} Q_{3}=z$. By symmetry, notice that $B_{3} P_{3}=B_{2} P_{2}$ and that $P_{3} Q_{3}=P_{2} Q_{3}$. Thus\n\n$$\nx+y+z=B_{3} Q_{3}+Q_{3} P_{2}+P_{2} B_{2}=B_{2} B_{3}=3 .\n$$\n\nBy the Pythagorean Theorem, $x^{2}+y^{2}=z^{2}$. Therefore\n\n$$\n\\begin{aligned}\nx+y & =3-z \\\\\nx^{2}+y^{2}+2 x y & =9-6 z+z^{2} \\\\\n2 x y & =9-6 z .\n\\end{aligned}\n$$\n\nThe value of $x y$ can be determined from the areas of the four triangles $\\triangle B_{i} P_{i} Q_{i}$. By symmetry, these four triangles are congruent to each other. Their total area is the area not in both squares, i.e., $9-8=1$. Thus $\\frac{x y}{2}=\\frac{1}{4}$, so $2 x y=1$. Applying this result to the above equation,\n\n$$\n\\begin{aligned}\n1 & =9-6 z \\\\\nz & =\\frac{4}{3}\n\\end{aligned}\n$$\n\n\n\nThe desired quantity is $\\sin \\theta+\\cos \\theta=\\frac{x}{z}+\\frac{y}{z}$, and\n\n$$\n\\begin{aligned}\n\\frac{x}{z}+\\frac{y}{z} & =\\frac{x+y+z}{z}-\\frac{z}{z} \\\\\n& =\\frac{3}{z}-1 \\\\\n& =\\frac{\\mathbf{5}}{\\mathbf{4}}\n\\end{aligned}\n$$']",['$\\frac{5}{4}$'],False,,Numerical, 3088,Number Theory,,"Suppose that neither of the three-digit numbers $M=\underline{4} \underline{A} \underline{6}$ and $N=\underline{1} \underline{B} \underline{7}$ is divisible by 9 , but the product $M \cdot N$ is divisible by 9 . Compute the largest possible value of $A+B$.","['In order for the conditions of the problem to be satisfied, $M$ and $N$ must both be divisible by 3 , but not by 9 . Thus the largest possible value of $A$ is 5 , and the largest possible value of $B$ is 7 , so $A+B=\\mathbf{1 2}$.']",['12'],False,,Numerical, 3089,Geometry,,Let $T=12$. Each interior angle of a regular $T$-gon has measure $d^{\circ}$. Compute $d$.,"['From the angle sum formula, $d^{\\circ}=\\frac{180^{\\circ} \\cdot(T-2)}{T}$. With $T=12, d=\\mathbf{1 5 0}$.']",['150'],False,,Numerical, 3090,Algebra,,"Suppose that $r$ and $s$ are the two roots of the equation $F_{k} x^{2}+F_{k+1} x+F_{k+2}=0$, where $F_{n}$ denotes the $n^{\text {th }}$ Fibonacci number. Compute the value of $(r+1)(s+1)$.","['$\\quad$ Distributing, $(r+1)(s+1)=r s+(r+s)+1=\\frac{F_{k+2}}{F_{k}}+\\left(-\\frac{F_{k+1}}{F_{k}}\\right)+1=\\frac{F_{k+2}-F_{k+1}}{F_{k}}+1=\\frac{F_{k}}{F_{k}}+1=\\mathbf{2}$.']",['2'],False,,Numerical, 3091,Algebra,,"Let $T=2$. Compute the product of $-T-i$ and $i-T$, where $i=\sqrt{-1}$.","['Multiplying, $(-T-i)(i-T)=-(i+T)(i-T)=-\\left(i^{2}-T^{2}\\right)=1+T^{2}$. With $T=2,1+T^{2}=\\mathbf{5}$.']",['5'],False,,Numerical, 3092,Combinatorics,,Let $T=5$. Compute the number of positive divisors of the number $20^{4} \cdot 11^{T}$ that are perfect cubes.,"['Let $N=20^{4} \\cdot 11^{T}=2^{8} \\cdot 5^{4} \\cdot 11^{T}$. If $m \\mid N$, then $m=2^{a} \\cdot 5^{b} \\cdot 11^{c}$ where $a, b$, and $c$ are nonnegative integers such that $a \\leq 8, b \\leq 4$, and $c \\leq T$. If $m$ is a perfect cube, then $a, b$, and $c$ must be divisible by 3 . So $a=0,3$, or $6 ; b=0$ or 3 , and $c \\in\\{0,3, \\ldots, 3 \\cdot\\lfloor T / 3\\rfloor\\}$. There are a total of $3 \\cdot 2 \\cdot(\\lfloor T / 3\\rfloor+1)$ possible values of $m$. For $T=5,\\lfloor T / 3\\rfloor+1=2$, so the number of possible values of $m$ is $\\mathbf{1 2}$.']",['12'],False,,Numerical, 3093,Geometry,,"$\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles. ","['The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\\frac{T}{2}$. Thus the altitude to the base has length $\\sqrt{\\left(\\frac{3 T}{2}\\right)^{2}-\\left(\\frac{T}{2}\\right)^{2}}=T \\sqrt{2}$. Thus the area of the triangle is $\\frac{1}{2} \\cdot\\left(\\frac{T}{2}+\\frac{T}{2}\\right) \\cdot T \\sqrt{2}=\\frac{T^{2} \\sqrt{2}}{2}$. With $T=12$, the area is $\\mathbf{7 2} \\sqrt{\\mathbf{2}}$.']",['$72 \\sqrt{2}$'],False,,Numerical, 3093,Geometry,,"$\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles. ![](https://cdn.mathpix.com/cropped/2023_12_21_9a82243db0768b0d2f1cg-1.jpg?height=436&width=632&top_left_y=1273&top_left_x=1343)","['The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\\frac{T}{2}$. Thus the altitude to the base has length $\\sqrt{\\left(\\frac{3 T}{2}\\right)^{2}-\\left(\\frac{T}{2}\\right)^{2}}=T \\sqrt{2}$. Thus the area of the triangle is $\\frac{1}{2} \\cdot\\left(\\frac{T}{2}+\\frac{T}{2}\\right) \\cdot T \\sqrt{2}=\\frac{T^{2} \\sqrt{2}}{2}$. With $T=12$, the area is $\\mathbf{7 2} \\sqrt{\\mathbf{2}}$.']",['$72 \\sqrt{2}$'],False,,Numerical, 3094,Algebra,,"Let $T=72 \sqrt{2}$, and let $K=\left(\frac{T}{12}\right)^{2}$. In the sequence $0.5,1,-1.5,2,2.5,-3, \ldots$, every third term is negative, and the absolute values of the terms form an arithmetic sequence. Compute the sum of the first $K$ terms of this sequence.","['The general sequence looks like $x, x+d,-(x+2 d), x+3 d, x+4 d,-(x+5 d), \\ldots$ The sum of the first three terms is $x-d$; the sum of the second three terms is $x+2 d$; the sum of the third three terms is $x+5 d$, and so on. Thus the sequence of sums of terms $3 k-2,3 k-1$, and $3 k$ is an arithmetic sequence. Notice that $x=d=0.5$ and so $x-d=0$. If there are $n$ triads of terms of the original sequence, then their common difference is 1.5 and their sum is $n \\cdot\\left(\\frac{0+0+(n-1) \\cdot 1.5}{2}\\right) \\cdot T=72 \\sqrt{2}$, so $K=72$, and $n=24$. Thus the desired sum is 414.']",['414'],False,,Numerical, 3095,Algebra,,"Let $A$ be the sum of the digits of the number you will receive from position 7 , and let $B$ be the sum of the digits of the number you will receive from position 9 . Let $(x, y)$ be a point randomly selected from the interior of the triangle whose consecutive vertices are $(1,1),(B, 7)$ and $(17,1)$. Compute the probability that $x>A-1$.","['Let $P=(1,1), Q=(17,1)$, and $R=(B, 7)$ be the vertices of the triangle, and let $X=(B, 1)$ be the foot of the perpendicular from $R$ to $\\overleftrightarrow{P Q}$. Let $M=(A-1,1)$ and let $\\ell$ be the vertical line through $M$; then the problem is to determine the fraction of the area of $\\triangle P Q R$ that lies to the right of $\\ell$.\n\nNote that $B \\geq 0$ and $A \\geq 0$ because they are digit sums of integers. Depending on their values, the line $\\ell$ might intersect any two sides of the triangle or none at all. Each case\n\n\n\nrequires a separate computation. There are two cases where the computation is trivial. First, when $\\ell$ passes to the left of or through the leftmost vertex of $\\triangle P Q R$, which occurs when $A-1 \\leq \\min (B, 1)$, the probability is 1 . Second, when $\\ell$ passes to the right of or through the rightmost vertex of $\\triangle P Q R$, which occurs when $A-1 \\geq \\max (B, 17)$, the probability is 0 . The remaining cases are as follows.\n\nCase 1: The line $\\ell$ intersects $\\overline{P Q}$ and $\\overline{P R}$ when $1 \\leq A-1 \\leq 17$ and $A-1 \\leq B$.\n\nCase 2: The line $\\ell$ intersects $\\overline{P Q}$ and $\\overline{Q R}$ when $1 \\leq A-1 \\leq 17$ and $A-1 \\geq B$.\n\nCase 3: The line $\\ell$ intersects $\\overline{P R}$ and $\\overline{Q R}$ when $17 \\leq A-1 \\leq B$.\n\nNow proceed case by case.\n\nCase 1: Let $T$ be the point of intersection of $\\ell$ and $\\overline{P R}$. Then the desired probability is $[M Q R T] /[P Q R]=1-[P M T] /[P Q R]$. Since $\\triangle P M T \\sim \\triangle P X R$ and the areas of similar triangles are proportional to the squares of corresponding sides, $[P M T] /[P X R]=(P M / P X)^{2}$. Since $\\triangle P X R$ and $\\triangle P Q R$ both have height $X R$, their areas are proportional to their bases: $[P X R] /[P Q R]=P X / P Q$. Taking the product, $[P M T] /[P Q R]=(P M / P X)^{2}(P X / P Q)=$ $\\frac{P M^{2}}{P X \\cdot P Q}=\\frac{(A-2)^{2}}{(B-1)(17-1)}$, and the final answer is\n\n$$\n\\frac{[M Q R T]}{[P Q R]}=1-\\frac{[P M T]}{[P Q R]}=1-\\frac{(A-2)^{2}}{16(B-1)}\n$$\n\nCase 2: Let $U$ be the point of intersection of $\\ell$ and $\\overline{Q R}$. A similar analysis to the one in the previous case yields\n\n$$\n\\frac{[M Q U]}{[P Q R]}=\\frac{[M Q U]}{[X Q R]} \\cdot \\frac{[X Q R]}{[P Q R]}=\\left(\\frac{M Q}{X Q}\\right)^{2} \\frac{X Q}{P Q}=\\frac{(18-A)^{2}}{16(17-B)}\n$$\n\nCase 3: Let $T$ be the point of intersection of $\\ell$ and $\\overline{P R}$ and let $U$ be the point of intersection of $\\ell$ and $\\overline{Q R}$ as in the previous cases. Let $S$ be the point on $\\overline{P R}$ such that $\\overline{Q S} \\perp \\overline{P Q}$. Then $\\triangle T U R \\sim \\triangle S Q R$, so the areas of these two triangles are proportional to the squares of the corresponding altitudes $M X$ and $Q X$. Thinking of $\\overleftrightarrow{P R}$ as the common base, $\\triangle S Q R$ and $\\triangle P Q R$ have a common altitude, so the ratio of their areas is $S R / P R$. Since $\\triangle P Q S \\sim$ $\\triangle P X R, P S / P R=P Q / P X$ and so $\\frac{S R}{P R}=1-\\frac{P S}{P R}=1-\\frac{P Q}{P X}=\\frac{Q X}{P X}$. Therefore the desired probability is\n\n$$\n\\frac{[T U R]}{[P Q R]}=\\frac{[T U R]}{[S Q R]} \\cdot \\frac{[S Q R]}{[P Q R]}=\\left(\\frac{M X}{Q X}\\right)^{2} \\frac{Q X}{P X}=\\frac{(B-A+1)^{2}}{(B-17)(B-1)}\n$$\n\nUsing the answers from positions 7 and $9, A=4+1+4=9$ and $B=2+7=9$. The first case applies, so the probability is\n\n$$\n1-\\frac{(9-2)^{2}}{16(9-1)}=1-\\frac{49}{128}=\\frac{\\mathbf{7 9}}{\\mathbf{1 2 8}}\n$$']",['$\\frac{79}{128}$'],False,,Numerical, 3096,Algebra,,"Let $T=9.5$. If $\log _{2} x^{T}-\log _{4} x=\log _{8} x^{k}$ is an identity for all $x>0$, compute the value of $k$.","['Note that in general, $\\log _{b} c=\\log _{b^{n}} c^{n}$. Using this identity yields $\\log _{2} x^{T}=\\log _{2^{2}}\\left(x^{T}\\right)^{2}=$ $\\log _{4} x^{2 T}$. Thus the left hand side of the given equation simplifies to $\\log _{4} x^{2 T-1}$. Express each side in base 64: $\\log _{4} x^{2 T-1}=\\log _{64} x^{6 T-3}=\\log _{64} x^{2 k}=\\log _{8} x^{k}$. Thus $k=3 T-\\frac{3}{2}$. With $T=9.5, k=\\mathbf{2 7}$.']",['27'],False,,Numerical, 3097,Geometry,,"Let $T=16$. An isosceles trapezoid has an area of $T+1$, a height of 2 , and the shorter base is 3 units shorter than the longer base. Compute the sum of the length of the shorter base and the length of one of the congruent sides.","['Let $x$ be the length of the shorter base of the trapezoid. The area of the trapezoid is $\\frac{1}{2} \\cdot 2$. $(x+x+3)=T+1$, so $x=\\frac{T}{2}-1$. Drop perpendiculars from each vertex of the shorter base to the longer base, and note that by symmetry, the feet of these perpendiculars lie $\\frac{3}{2}=1.5$ units away from their nearest vertices of the trapezoid. Hence the congruent sides have length $\\sqrt{1.5^{2}+2^{2}}=2.5$. With $T=16, x=7$, and the desired sum of the lengths is $\\mathbf{9 . 5}$.']",['9.5'],False,,Numerical, 3098,Combinatorics,,Let $T=10$. Susan flips a fair coin $T$ times. Leo has an unfair coin such that the probability of flipping heads is $\frac{1}{3}$. Leo gets to flip his coin the least number of times so that Leo's expected number of heads will exceed Susan's expected number of heads. Compute the number of times Leo gets to flip his coin.,"['The expected number of heads for Susan is $\\frac{T}{2}$. If Leo flips his coin $N$ times, the expected number of heads for Leo is $\\frac{N}{3}$. Thus $\\frac{N}{3}>\\frac{T}{2}$, so $N>\\frac{3 T}{2}$. With $T=10$, the smallest possible value of $N$ is $\\mathbf{1 6}$.']",['16'],False,,Numerical, 3099,Algebra,,"Let $T=1$. Dennis and Edward each take 48 minutes to mow a lawn, and Shawn takes 24 minutes to mow a lawn. Working together, how many lawns can Dennis, Edward, and Shawn mow in $2 \cdot T$ hours? (For the purposes of this problem, you may assume that after they complete mowing a lawn, they immediately start mowing the next lawn.)","['Working together, Dennis and Edward take $\\frac{48}{2}=24$ minutes to mow a lawn. When the three of them work together, it takes them $\\frac{24}{2}=12$ minutes to mow a lawn. Thus they can mow 5 lawns per hour. With $T=1$, they can mow $5 \\cdot 2=\\mathbf{1 0}$ lawns in 2 hours.']",['10'],False,,Numerical, 3100,Algebra,,Let T be a rational number. Compute $\sin ^{2} \frac{T \pi}{2}+\sin ^{2} \frac{(5-T) \pi}{2}$.,['Note that $\\sin \\frac{(5-T) \\pi}{2}=\\cos \\left(\\frac{\\pi}{2}-\\frac{(5-T) \\pi}{2}\\right)=\\cos \\left(\\frac{T \\pi}{2}-2 \\pi\\right)=\\cos \\frac{T \\pi}{2}$. Thus the desired quantity is $\\sin ^{2} \\frac{T \\pi}{2}+\\cos ^{2} \\frac{T \\pi}{2}=\\mathbf{1}$ (independent of $T$ ).'],['1'],False,,Numerical, 3101,Algebra,,Let $T=11$. Compute the value of $x$ that satisfies $\sqrt{20+\sqrt{T+x}}=5$.,"['Squaring each side gives $20+\\sqrt{T+x}=25$, thus $\\sqrt{T+x}=5$, and $x=25-T$. With $T=11$, $x=14$.']",['14'],False,,Numerical, 3102,Geometry,,The sum of the interior angles of an $n$-gon equals the sum of the interior angles of a pentagon plus the sum of the interior angles of an octagon. Compute $n$.,"['Using the angle sum formula, $180^{\\circ} \\cdot(n-2)=180^{\\circ} \\cdot 3+180^{\\circ} \\cdot 6=180^{\\circ} \\cdot 9$. Thus $n-2=9$, and $n=11$.']",['11'],False,,Numerical,